Question: As illustrated, four small squares with side length $b$ are placed around a square with side length $a$ named $EFGH$, forming a larger square $ABCD$. The shaded parts of the figure form an icon resembling a "red badge". Let the area of the shaded part be denoted as $S_1$, and the area of each small square with side length $b$ be denoted as $S_2$. If $S_1 = 6S_2$, what is the value of $\frac{a}{b}$?

Solution:
\textbf{Solution:} As shown in the diagram, we label the required intersections with letters:\\
$S_{\triangle DGI}=\frac{1}{2}(a+b)\times b=\frac{1}{2}ab+\frac{1}{2}b^{2}$,\\
$S_{\triangle KMD}=S_{ABCD}-S_{\triangle DMC}-S_{\triangle DKA}-S_{\triangle KBM}$\\
$=(a+2b)^{2}-\frac{1}{2}(a+b)(a+2b)-\frac{1}{2}(a+2b)(a+b)-\frac{1}{2}b^{2}$\\
$=ab+\frac{3}{2}b^{2}$,\\
$S_{\triangle MNC}=\frac{1}{2}(a+b)\times b=\frac{1}{2}ab+\frac{1}{2}b^{2}$,\\
Therefore, $S_{1}=S_{\triangle DGI}+S_{\triangle KMD}+S_{\triangle MNC}=2ab+\frac{5}{2}b^{2}$,\\
$S_{2}=b^{2}$,\\
Since $S_{1}=6S_{2}$,\\
therefore $2ab+\frac{5}{2}b^{2}=6b^{2}$,\\
Simplifying yields: $2a=\frac{7}{2}b$,\\
Therefore, $\frac{a}{b}=\boxed{\frac{7}{4}}$.


Question: As shown in the figure, all vertices of the shape are on the grid points of a $3 \times 3$ square grid. Then, what is the sum of $\angle 1+\angle 2$?

Solution:
\textbf{Solution:} As shown in the diagram, \\
from the given conditions, in right triangles $\triangle ABC$ and $\triangle EFC$, \\
$\because \left\{\begin{array}{l}AB=EF\\ \angle B=\angle EFC=90^\circ\\ BC=FC\end{array}\right.$  \\
$\therefore$ right triangles $\triangle ABC\cong \triangle EFC$ (SAS) \\
$\therefore \angle 3=\angle 1$ \\
$\because \angle 2+\angle 3=45^\circ$ \\
$\therefore \angle 1+\angle 2=\angle 3+\angle 2=\boxed{45^\circ}$


Question: As shown in the figure, in rhombus $ABCD$, $DE \perp AB$ with the foot of the perpendicular being point $E$, and $CE$ is drawn. If $AE=2$ and $\angle DCE=30^\circ$, what is the side length of the rhombus?

Solution: \textbf{Solution}: Let $DE = x$. Since quadrilateral $ABCD$ is a rhombus, it follows that $AB \parallel CD$, and therefore $\angle CDE = \angle AED = 90^\circ$. Given that $\angle DCE = 30^\circ$, it follows that $CD = \sqrt{3}DE = \sqrt{3}x$. Consequently, $AD = CD = \sqrt{3}x$. In right triangle $\triangle AED$, $AD^{2} = DE^{2} + AE^{2}$, which implies $(\sqrt{3}x)^{2} = x^{2} + 2^{2}$. Solving for $x$ gives $x = \sqrt{2}$, therefore $CD = \sqrt{3}x = \sqrt{6}$. Hence, the answer is: $\boxed{\sqrt{6}}$.


Question: As shown in the diagram, in rectangle $ABCD$, $AB=4$, $BC=6$. The perpendicular bisector of diagonal $AC$ intersects $AC$, $AD$, and $BC$ at points $O$, $E$, and $F$, respectively. Connecting $AF$ and $CE$, then what is the value of $\frac{AE}{BF}$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rectangle, it follows that AD$\parallel$BC and $\angle B=90^\circ$. Since EF is the perpendicular bisector of AC, we have AF=FC. Moreover, since AD$\parallel$BC, it follows that $\angle$EAO=$\angle$FCO. In $\triangle$AOE and $\triangle$COF, we have
\[
\left\{
\begin{array}{l}
\angle EAO=\angle FCO\\
OA=OC\\
\angle AOE=\angle COF
\end{array}
\right.
\]
Therefore, $\triangle$AOE$\cong$$\triangle$COF by ASA, which implies AE=FC. Consequently, AF=AE=FC. Letting AE=$x$, we find BF=BC−FC=6−$x$. In right triangle ABF, we have AB$^{2}$+BF$^{2}$=AF$^{2}$. Thus, $4^{2}$+$x^{2}$=(6−$x$)$^{2}$, which solves to $x=\frac{13}{3}$. Therefore, BF=BC−FC=6−$\frac{13}{3}$=$\frac{5}{3}$. Hence, $\frac{AE}{BF}=\boxed{\frac{13}{5}}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the rhombus OABC has its side OA on the positive half of the $x$-axis, and the graph of the inverse proportional function $y=\frac{k}{x}(x>0)$ passes through the midpoint $D$ of the diagonal OB and the vertex $C$. If the area of the rhombus OABC is 12, what is the value of $k$?

Solution:
\textbf{Solution:} Let the coordinates of point A be $(a, 0)$, and the coordinates of point C be $\left(c, \frac{k}{c}\right)$, \\
$\because$ the area of the rhombus is 12, \\
$\therefore a \cdot \frac{k}{c}=12$, \\
$\because$ D is the midpoint of the diagonal OB, \\
$\therefore$ the coordinates of point D are: $\left(\frac{a+c}{2}, \frac{k}{2c}\right)$, \\
$\therefore \frac{a+c}{2} \cdot \frac{k}{2c}=k$, \\
thus $a=3c$, \\
$\therefore 3c \cdot \frac{k}{c}=12$, \\
solving this gives $k=\boxed{4}$.


Question: As shown in the figure, given the line segment AB, arcs are drawn with A and B as centers, respectively, and with radii greater than half the length of segment AB. The arcs intersect at points C and D. Connect AC, BC, BD, CD, and AD. Given that $AB=3$ and $CD=4$, what is the perimeter of the quadrilateral $ACBD$?

Solution:
\textbf{Solution:} As shown in the figure, AC and CD intersect at point O.\\
From the construction steps, it is known that CD is the perpendicular bisector of line segment AB,\\
$\because$AC=AD=BC=BD,\\
$\therefore$The quadrilateral ACBD is a rhombus,\\
$\because$AB=3, CD=4,\\
$\therefore$AO=$\frac{1}{2}$AB=$\frac{3}{2}$, OD=$\frac{1}{2}$CD=2,\\
$\therefore$AD=$\sqrt{AO^{2}+OB^{2}}$=$\sqrt{\left(\frac{3}{2}\right)^{2}+2^{2}}$=$\frac{5}{2}$,\\
$\therefore$The perimeter of rhombus ACBD=4AD=4$\times\frac{5}{2}$=$\boxed{10}$.


Question: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $H$ is the midpoint of side $AD$. If the perimeter of rhombus $ABCD$ is $40$, what is the length of $OH$?

Solution:
Solution: Since the perimeter of the rhombus ABCD is 40,\\
therefore $CD=10$,\\
and since H is the midpoint of side AD and point O is the midpoint of AC,\\
therefore $OH$ is the median of $\triangle ACD$,\\
therefore $OH=\frac{1}{2}CD=\frac{1}{2} \times 10=\boxed{5}$.


Question: As shown in the figure, the area of square $ABCD$ is $5$ and the area of square $BEFG$ is $4$. What is the area of $\triangle GCE$?

Solution:
\textbf{Solution:} Given that the area of square ABCD is 5, and the area of square BEFG is 4,\\
it follows that the side length of square ABCD is $\sqrt{5}$, and the side length of square BEFG is 2,\\
therefore, CE＝$\sqrt{5}-2$,\\
the area of $\triangle GCE$＝$\frac{1}{2}$CE·BG＝$\frac{1}{2}\times(\sqrt{5}-2)\times2=\boxed{\sqrt{5}-2}$


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, a square ABDE is constructed externally on the hypotenuse AB of the triangle, and the diagonals of the square meet at point $O$. Line OC is connected. Given that $AC=3$ and $OC=4\sqrt{2}$, what is the length of BC?

Solution:
\textbf{Solution:} As shown in the figure, draw line OF $\perp$ BC at F, and draw line AM $\perp$ OF at M,\\
$\because$ $\angle ACB=90^\circ$,\\
$\therefore$ $\angle AMO=\angle OFB=90^\circ$, $\angle ACB=\angle CFM=\angle AMF=90^\circ$,\\
$\therefore$ Quadrilateral ACFM is a rectangle,\\
$\therefore$ AM=CF, AC=MF=3,\\
$\because$ Quadrilateral ABDE is a square,\\
$\therefore$ $\angle AOB=90^\circ$, OA=OB,\\
$\therefore$ $\angle AOM+\angle BOF=90^\circ$,\\
Also $\because$ $\angle AMO=90^\circ$,\\
$\therefore$ $\angle AOM+\angle OAM=90^\circ$,\\
$\therefore$ $\angle BOF=\angle OAM$,\\
$\therefore$ $\triangle AOM \cong \triangle OBF$ (AAS),\\
$\therefore$ AM=OF, OM=FB,\\
$\therefore$ OF=CF,\\
$\because$ $\angle CFO=90^\circ$,\\
$\therefore$ $\triangle CFO$ is an isosceles right-angle triangle,\\
$\because$ OC=$4\sqrt{2}$,\\
By the Pythagorean theorem: CF=OF=4,\\
$\therefore$ BF=OF-MF=4-3=1,\\
$\therefore$ BC=CF+BF=4+1=\boxed{5}.


Question: As shown in the figure, in $\triangle ABC$, D and E are the midpoints of AB and AC, respectively. Extend DE to the point F such that EF=DE. Given AB=10 and BC=8, what is the perimeter of the quadrilateral BCFD?

Solution:
\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively,\\
it follows that $DE \parallel BC$ and $DE = \frac{1}{2} BC$, $BD = \frac{1}{2} AB = 5$,\\
since $DE = EF$,\\
it follows that $DE + EF = DF = BC$,\\
therefore, quadrilateral $DBCF$ is a parallelogram,\\
hence, the perimeter of quadrilateral $BCFD$ is $2(BD + DF) = 2(5 + 8) = \boxed{26}$.


Question: As shown in the figure, in $\triangle ABC$, the perimeter of $\triangle DEF$ formed by the three medians is $15\text{cm}$. What is the perimeter of $\triangle ABC$ in $\text{cm}$?

Solution:
\textbf{Solution:} Since DE is the median of $\triangle ABC$,\\
it follows that DE=$\frac{1}{2}$BC, similarly EF=$\frac{1}{2}$AB, and DF=$\frac{1}{2}$AC,\\
thus $\frac{DE}{BC}=\frac{EF}{AB}=\frac{DF}{AC}=\frac{1}{2}$,\\
therefore $\triangle ABC \sim \triangle DEF$,\\
hence the perimeter of $\triangle ABC$: the perimeter of $\triangle DEF$=2:1,\\
thus the perimeter of $\triangle ABC$=2×15=\boxed{30}\text{cm}.


Question: As shown in the figure, a rectangular piece of paper ABCD is folded along AE, where the width AB of the paper is 8 cm and the length BC is 10 cm. When folded along AE, the vertex D falls on point F on side BC. What is the length of CE in cm?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle, \\
hence AD = BC = 10cm, DC = AB = 8cm, $\angle$B = $\angle$D = $\angle$C = $90^\circ$, \\
since when folded along AE, vertex D falls on point F on side BC, \\
thus AF = AD = 10cm, DE = EF, \\
in $\triangle ABF$, BF = $\sqrt{AF^{2}-AB^{2}}$ = $\sqrt{10^{2}-8^{2}}$ = 6 (cm), \\
hence CF = BC - BF = 10 - 6 = 4 (cm), \\
let CE = x cm, then DE = EF = (8-x) cm, \\
in $\triangle CEF$, since $CF^{2}$ + $CE^{2}$ = $EF^{2}$, \\
thus $4^{2}$ + $x^{2}$ = $(8-x)^{2}$, \\
solving gives x = $\boxed{3}$, \\
that is, the length of CE is 3cm.


Question: As shown in the figure, in the Cartesian coordinate system, the area of the rhombus OABC is 12. Point $B$ is on the $y$-axis, and point $C$ is on the graph of the inverse proportion function $y=\frac{k}{x}$. What is the value of $k$?

Solution:
\textbf{Solution:} As shown in the figure, we connect AC to intersect the y-axis at point D,\\
$\because$ rhombus OABC,\\
$\therefore$ AC$\perp$OB, and CD=AD=$\frac{1}{2}$AC, BD=OD=$\frac{1}{2}$BO,\\
$\because$ the area of rhombus OABC is 12,\\
$\therefore$ AC$\cdot$BO=24,\\
$\therefore$ CD$\cdot$OD=6,\\
$\therefore$ $\left|k\right|=6$,\\
Furthermore, $\because$ the graph of the inverse proportional function is located in the second quadrant,\\
$\therefore$ k$<$0, that is, $k=\boxed{-6}$.


Question: As shown in the figure, the vertices $C$ and $A$ of the square $ABCO$ are on the $x$-axis and $y$-axis, respectively. $BC$ is the diagonal of the rhombus $BDCE$. If $\angle D = 60^\circ$ and $BC = 2$, what are the coordinates of point $D$?

Solution:
\textbf{Solution:} As illustrated, draw line DF perpendicular to BC at point F from point D,\\
$\because$ rhombus BDCE,\\
$\therefore$ BD = CD,\\
Also $\because$ $\angle$D = 60$^\circ$,\\
$\therefore$ $\triangle$BDC is an equilateral triangle,\\
$\therefore$ $\angle$DFC = 90$^\circ$, BF = CF = 1, so the y-coordinate of point D is: 1,\\
$\therefore$ in $\triangle$DFC, DF = $\sqrt{CD^{2}-CF^{2}} = \sqrt{2^{2}-1^{2}} = \sqrt{3}$,\\
$\because$ square ABCD, BC = 2,\\
$\therefore$ OC = 2,\\
$\therefore$ the x-coordinate of point D is: OC + DF = 2 + $\sqrt{3}$,\\
Thus, the answer is: $(2+\sqrt{3}, 1) = \boxed{(2+\sqrt{3}, 1)}$.


Question: As shown in the figure, in rectangle $ABCD$, $CE \perp BD$ with point $E$ being the foot of the perpendicular. Connect $AE$. If $\angle DCE : \angle ECB = 3 : 1$, then what is the measure of $\angle ACE$?

Solution:
\textbf{Solution:} Given that rectangle $ABCD$, with $CE \perp BD$, \\
hence, $\angle DCB = \angle CEB = \angle CED = 90^\circ$, and $OC = OA = OD = OB$, \\
therefore, $\angle OCB = \angle OBC$, \\
since $\angle DCE : \angle ECB = 3 : 1$ and $\angle ECB + \angle DCE = 90^\circ$, \\
thus, $4\angle ECB = 90^\circ$, \\
therefore, $\angle ECB = 22.5^\circ$, \\
hence, $\angle OCB = \angle OBC = 90^\circ - \angle ECB = 90^\circ - 22.5^\circ = 67.5^\circ$, \\
hence, $\angle OCE = \angle OCB - \angle ECB = 67.5^\circ - 22.5^\circ = 45^\circ$, \\
therefore, $\angle ACE = \boxed{45^\circ}$.


Question: As shown in the figure, within rectangle $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. If $\angle COD=52^\circ$, then what is the degree measure of $\angle CAD$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\
it follows that $OA=OB=OC=OD$,\\
hence $\angle CAD=\angle ADO$,\\
thus $\angle COD=2\angle CAD$,\\
therefore $\angle CAD=\boxed{26^\circ}$.


Question: As shown in the figure, $C$, $D$, $E$ are three points on the line segment $AB$, where $E$ is the midpoint of $BD$, $AC = \frac{1}{2} CD$, $AB = 5$, and $EB = 1$. What is the length of $CD$?

Solution:
\textbf{Solution:} Since $E$ is the midpoint of $BD$, $EB=1$\\
Therefore, $BD=2BE=2$\\
Since $AB=5$\\
Therefore, $AD=AB-BD=5-2=3$\\
Since $AC=\frac{1}{2}CD$,\\
Therefore, $AD=AC+CD=\frac{1}{2}CD+CD=3$, solving gives: $CD=\boxed{2}$.


Question: As shown in the figure, point O lies on line AB, and ray OD bisects $\angle AOC$. If $\angle AOD = 20^\circ$, what is the measure of $\angle COB$ in degrees?

Solution:
\textbf{Solution:} Since OD bisects $\angle$AOC,\\
it follows that $\angle$AOC = 2$\angle$AOD = 40$^\circ$,\\
therefore, $\angle$COB = 180$^\circ$ - $\angle$COA = \boxed{140^\circ}


Question: As shown in the figure, points C, M, and N are on line segment AB, with $AC=12$, $BC=6$, $AM=\frac{1}{3}AC$, and $BN=\frac{1}{3}BC$. What is the length of line segment MN?

Solution:
\textbf{Solution:}\\
Since $AC=12$,\\
it follows that $AM=\frac{1}{3}AC=4$,\\
Since $BC=6$,\\
it follows that $BN=\frac{1}{3}BC=2$,\\
Therefore, $MN=AB-AM-BN=AC+BC-AM-BN=12+6-4-2=\boxed{12}$.


Question: As shown in the figure, $\angle AOB$ is a right angle, $\angle AOC = 40^\circ$, and $OD$ bisects $\angle BOC$. What is the measure of $\angle AOD$ in degrees?

Solution:
\textbf{Solution:} Because $\angle BOC = \angle AOB - \angle AOC$, \\
therefore $\angle BOC = 90^\circ - 40^\circ = 50^\circ$, \\
Because OD bisects $\angle BOC$, \\
therefore $\angle COD = \frac{1}{2}\angle BOC = \frac{1}{2} \times 50^\circ = 25^\circ$, \\
therefore $\angle AOD = \angle AOC + \angle COD = 40^\circ + 25^\circ = \boxed{65^\circ}$.


Question: As shown in the figure, $O$ is a point on line $AC$, $OB$ is a ray, $OD$ bisects $\angle AOB$, $OE$ is within $\angle BOC$, $\angle BOE = \frac{1}{3}\angle EOC$, $\angle DOE = 60^\circ$, then what is the degree measure of $\angle EOC$?

Solution:
\textbf{Solution:} Let $\angle BOE$ be $x^\circ$, then $\angle DOB = 60^\circ - x^\circ$,\\
since $\angle BOE = \frac{1}{3}\angle EOC$,\\
it follows that $\angle EOC = 3\angle BOE = 3x^\circ$,\\
given that OD bisects $\angle AOB$,\\
we get $\angle AOB = 2\angle DOB$,\\
thus, we have $3x + x + 2(60 - x) = 180$,\\
solving the equation, we find $x = 30^\circ$,\\
therefore, $\angle EOC = \boxed{90^\circ}$.


Question: As shown in the figure, it is known that points $A$, $B$, and $C$ are on the same line segment, $M$ is the midpoint of segment $AC$, $N$ is the midpoint of segment $BC$, and $AM=5\, \text{cm}$, $CN=3\, \text{cm}$. What is the length of line segment $AB$ in centimeters?

Solution:
\textbf{Solution:} Since $M$ is the midpoint of line segment $AC$ and $N$ is the midpoint of line segment $BC$,\\
it follows that $MC = AM = 5\,\text{cm}$, $BN = CN = 3\,\text{cm}$,\\
therefore $AB = AM + MC + CN + NB = \boxed{16}\,\text{cm}$.


Question: As shown in the figure, if $AC=48$ and M is the midpoint of $AC$, with $AB=\frac{1}{3}AC$, what is the length of $BM$?

Solution:
\textbf{Solution:} Given that $AC=48$ and M is the midpoint of $AC$, $AB=\frac{1}{3}AC$\\
Therefore, $AM=\frac{1}{2}AC=\frac{1}{2}\times 48=24$, $AB=\frac{1}{3}\times 48=16$\\
Hence, $BM=AM-AB=24-16=\boxed{8}$


Question: As shown in the figure, $\angle AOB$ is a right angle, $\angle BOC = 36^\circ$, and $OD$ bisects $\angle AOC$. What is the measure of $\angle BOD$ in degrees?

Solution:
\textbf{Solution:} Since $\angle AOB$ is a right angle, and $\angle BOC=36^\circ$, \\
then $\angle AOC=90^\circ+36^\circ=126^\circ$ \\
Since $OD$ bisects $\angle AOC$, \\
then $\angle COD=\frac{1}{2}\angle AOC=63^\circ$, \\
therefore, $\angle BOD=\angle COD-\angle BOC=63^\circ-36^\circ=\boxed{27^\circ}$


Question: As shown in the figure, it is known that $\angle AOC : \angle BOC = 1 : 5$. OD is the angle bisector of $\angle AOB$, and $\angle COD = 36^\circ$. What is the degree measure of $\angle AOC$?

Solution:
\textbf{Solution:} Given that $\angle AOC : \angle BOC = 1 : 5$,\\
let $\angle AOC = x$, then $\angle BOC = 5x$,\\
thus $\angle AOB = \angle AOC + \angle BOC = x + 5x = 6x$,\\
since OD bisects $\angle AOB$,\\
thus $\angle AOD = \frac{1}{2} \angle AOB = 3x$,\\
thus $\angle COD = \angle AOD - \angle AOC$,\\
thus $3x - x = 36^\circ$,\\
solving this gives $x = 18^\circ$,\\
thus $\angle AOC = \boxed{18^\circ}$.


Question: As shown in the figure, $O$ is a point on the straight line $AB$, $\angle AOC = 53^\circ 17'$, then what is the degree measure of $\angle BOC$?

Solution:
\textbf{Solution:} Since $\angle AOC=53^\circ 17'$, \\
thus $\angle BOC=180^\circ-53^\circ 17'$ \\
$=179^\circ 60'-53^\circ 17'$ $=\boxed{126^\circ 43'}$.


Question: As shown in the figure, $AB=12\,cm$, $C$ is the midpoint of $AB$, and point $D$ is on segment $AC$, with $AD\colon CB=1\colon 3$, what is the length of $DB$ in $cm$?

Solution:
\textbf{Solution:} Given $\because$AB=12cm, and C is the midpoint of AB,\\
$\therefore$AC=BC=$\frac{1}{2}$AB=6(cm),\\
$\because$AD$\colon$ CB=1$\colon$ 3,\\
$\therefore$AD=2cm,\\
$\therefore$DC=AC−AD=4(cm),\\
$\therefore$DB=DC+BC=\boxed{10}(cm).


Question: As shown in the diagram, B is a point on line segment AC, and $AB=24\,cm$, $BC=\frac{1}{3}AB$. If O is the midpoint of line segment AC, then what is the length of line segment $OB$ in $cm$?

Solution:
\textbf{Solution:} Given that $AB=24cm$, and $BC=\frac{1}{3}AB$,\\
therefore, $BC=\frac{1}{3} \times 24=8cm$,\\
thus, $AC=AB+BC=24+8=32cm$,\\
since $O$ is the midpoint of the line segment $AC$,\\
thus, $AO=\frac{1}{2}AB=\frac{1}{2} \times 32=16cm$,\\
therefore, $OB=AB-AO=24-16=\boxed{8cm}$.


Question: As shown in the figure, it is known that the line segment $AB=12\, \text{cm}$, point $C$ is on the line segment $AB$, $AC=2BC$, then how many centimeters is $BC$?

Solution:
\textbf{Solution:} Let $BC=x\, \text{cm}$, then $AC=2x\, \text{cm}$,\\
$x+2x=12$,\\
solving for $x$, we get: $x=4$,\\
$\therefore BC=\boxed{4}\, \text{cm}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, and $\angle A=100^\circ$. BD bisects $\angle ABC$. What is the measure of $\angle ADB$?

Solution:
\textbf{Solution:} Given that $AB=AC$ and $\angle A=100^\circ$,

$\therefore \angle ABC=\angle ACB=(180^\circ-100^\circ)\times \frac{1}{2}=40^\circ$,

$\because$ BD bisects $\angle ABC$,

$\therefore \angle ABD=\angle CBD=20^\circ$,

$\therefore \angle ADB=180^\circ-20^\circ-100^\circ=\boxed{60^\circ}$.


Question: As shown in the figure, \(C\) is the midpoint of segment \(AB\), \(D\) is a point on \(CB\), \(E\) is the midpoint of segment \(DB\), \(EB = 1\,cm\), and \(EB = \frac{1}{5}AB\), then what is the length of segment \(CE\) in centimeters?

Solution:
Solution: Since $EB=1\text{cm}$, and $EB=\frac{1}{5}AB$,\\
it follows that $AB=5EB=5\text{cm}$, \\
Since C is the midpoint of the segment $AB$, \\
we have $CB=\frac{1}{2}AB=\frac{5}{2}\text{cm}$, \\
thus $CE=CB-EB=\frac{5}{2}-1=\boxed{\frac{3}{2}}\text{cm}$.


Question: A right-angle triangle is placed as shown in the figure, with point C located on the extension line of $FD$, $AB \parallel CF$, $\angle F = \angle ACB = 90^\circ$. What is the degree measure of $\angle DBC$?

Solution:
\textbf{Solution:} Given the problem, we can deduce: $\angle EDF = 45^\circ$, $\angle ABC = 30^\circ$,\\
since AB $\parallel$ CF,\\
thus $\angle ABD = \angle EDF = 45^\circ$,\\
therefore $\angle DBC = 45^\circ - 30^\circ = \boxed{15^\circ}$.


Question: As shown in the figure, point $C$ is on line $AB$, and $CD$ bisects $\angle ACE$. If $\angle 1 = 63^\circ$, then what is the measure of $\angle BCE$ in degrees?

Solution:
\textbf{Solution:} Since $CD$ bisects $\angle ACE$, and $\angle 1=63^\circ$,

it follows that $\angle ACE=2\angle 1=126^\circ$,

Since $\angle ACE+\angle BCE=180^\circ$,

it follows that $\angle BCE=180^\circ−126^\circ=\boxed{54^\circ}$.


Question: As shown in the figure, line $AB$ intersects with line $CD$ at point $O$, with $OE \perp AB$. Given that $\angle BOD = 30^\circ$, what is the value of $\angle COE$?

Solution:
\textbf{Solution:} Given that $OE\perp AB$\\
Hence, $\angle AOE=90^\circ$\\
Given that $\angle BOD=30^\circ$\\
Therefore, $\angle AOC=\angle BOD=30^\circ$\\
Thus, $\angle COE=\angle AOC+\angle AOE=\boxed{120^\circ}$


Question: As shown in the figure, the right-angled vertices O of a pair of set squares are made to coincide. If the ratio of $\angle COB$ to $\angle DOA$ is $2\colon 7$ and $OP$ bisects $\angle DOA$, then what is the degree measure of $\angle POC$?

Solution:
\[
\begin{aligned}
&\because \angle COB + \angle DOA = \angle COB + \angle COA + \angle COB + \angle DOB \\
&= \angle AOB + \angle COD = 180^\circ \\
&\therefore \angle DOA = 180^\circ \times \frac{7}{2+7} = 140^\circ \\
&\because OP \text{ bisects } \angle DOA \\
&\therefore \angle DOP = 70^\circ \\
&\therefore \angle POC = \boxed{20^\circ}.
\end{aligned}
\]


Question: As shown in the figure, point $C$ is a point on line segment $AB$, with $AC = 10\, \text{cm}$, $CB = 5\, \text{cm}$. $M$, $N$ are the midpoints of $AC$ and $BC$, respectively. What is the length of line segment $MN$ in centimeters?

Solution:
\textbf{Solution:} Given that $AC=10\text{cm}$, $CB=5\text{cm}$, and $M$, $N$ are the midpoints of $AC$, $BC$ respectively,\\
thus, $CM=\frac{1}{2}AC=5\text{cm}$, $CN=\frac{1}{2}CB=2.5\text{cm}$,\\
hence $MN=CM+CN=\boxed{7.5}\text{cm}$.


Question: As shown in the figure, $D$ is the midpoint of $AC$, $CB = 4\,cm$, $DB = 7\,cm$, what is the length of $AB$ in $cm$?

Solution:
Solution: Since $DC=DB-CB=7-4=3\,cm$,\\
and since point D is the midpoint of AC,\\
it follows that $AD=DC=3\,cm$\\
Therefore, $AB=AD+DB = 3+7=\boxed{10}\,cm$


Question: As shown in the figure, $AB$ is the diameter of semicircle $O$, and chord $AC$ makes an angle of $30^\circ$ with $AB$. Given that $AC=CD$ and $OA=2$, what is the length of $AC$?

Solution:
\textbf{Solution:} As shown in the figure, connect $BC$, \\
since $AB$ is the diameter of $\odot O$, \\
therefore, $\angle ACB=90^\circ$, \\
since $\angle BAC=30^\circ$ and $OA=OB=2$, \\
therefore $AC=AB\cdot\cos30^\circ=4\times \frac{\sqrt{3}}{2}=\boxed{2\sqrt{3}}.$


Question: As shown in the figure, A, B, and C are three points on the circle $\odot O$, and $\angle OAB=64^\circ$, what is the degree measure of $\angle ACB$?

Solution:
\textbf{Solution:} Since $OA = OB$,\\
it follows that $\angle OAB = \angle OBA = 64^\circ$,\\
thus $\angle AOB = 180^\circ - 64^\circ - 64^\circ = 52^\circ$,\\
therefore, $\angle ACB = \frac{1}{2}\angle AOB = \boxed{26^\circ}$.


Question: As shown in the figure, the radius of the circle $\odot O$ is 2, and the chord $AB=2\sqrt{3}$. Point C is a moving point on chord AB, and the length of OC is an integer. What is the length of OC?

Solution:
\textbf{Solution:} Draw $OH \perp AB$ at $H$, as shown in the figure, \\
since $OH \perp AB$, \\
therefore $AH=BH$, \\
thus $AH=BH=\frac{1}{2}AB=\frac{1}{2} \times 2\sqrt{3} = \sqrt{3}$, \\
in $\triangle BOH$, where $OB=2$, and $BH=\sqrt{3}$, \\
therefore $OH=\sqrt{OB^{2}-BH^{2}}=\sqrt{2^{2}-\left(\sqrt{3}\right)^{2}}=1$, \\
since point $C$ is a moving point on chord $AB$, \\
therefore $1 \le OC \le 2$, \\
since the length of $OC$ is an integer, \\
thus $OC=\boxed{1}$ or $\boxed{2}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in $\odot O$. If the quadrilateral $AOCD$ is a rhombus, what is the degree measure of $\angle B$? .

Solution:
\textbf{Solution:} Since quadrilateral ABCD is inscribed in circle O,\\
it follows that $\angle B+\angle D=180^\circ$,\\
Since quadrilateral OACD is a rhombus,\\
it follows that $\angle AOC=\angle D$,\\
By the inscribed angle theorem, we have $\angle B= \frac{1}{2}\angle AOC$,\\
Thus $\angle B+2\angle B=180^\circ$,\\
Solving, we find $\angle B=\boxed{60^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, points C and D lie on $\odot O$, $\angle BCD=15^\circ$, what is the degree measure of $\angle AOD$?

Solution:
\textbf{Solution:} Since $\angle BOD=2\angle BCD$ and $\angle BCD=15^\circ$, \\
it follows that $\angle BOD=30^\circ$, \\
thus $\angle AOD=180^\circ-\angle BOD=150^\circ$, \\
therefore, $\angle AOD=\boxed{150^\circ}$.


Question: As shown in the figure, in the right triangle $\triangle ABC$, $\angle ABC=90^\circ$, $AB=6$, $BC=5$. $P$ is a moving point inside $\triangle ABC$, and it satisfies $\angle PAB=\angle PBC$. What is the minimum value of the length of the segment $CP$?

Solution:
\textbf{Solution:} Given that $\because$ $\angle ABC=90^\circ$ and $\angle PAB=\angle PBC$, \\
$\therefore$ $\angle PBA+\angle PBC=90^\circ$ and $\angle PBA+\angle PAB=90^\circ$, \\
$\therefore$ $\angle BPA=90^\circ$, \\
$\therefore$ point P lies on the circle with diameter AB, as shown in the figure, with O as the center of the circle. Connect OC, with P being the intersection point of OC and circle O, thus CP represents the minimum value. \\
$\because$ AB=6, \\
$\therefore$ OB=OP=3, \\
$\because$ BC=5, \\
$\therefore$ OC=$\sqrt{OB^2+BC^2}=\sqrt{3^2+5^2}=\sqrt{34}$, \\
$\therefore$ CP=$\sqrt{34}-3$, \\
Therefore, the answer is: $CP=\boxed{\sqrt{34}-3}$.


Question: As shown in the figure, BD is the diagonal of rhombus ABCD with side length \(a\), where \( \angle BAD=60^\circ\). Points M and N start simultaneously from points A and B, respectively, moving towards points B and D at the same speed along AB and BD. Lines DM and AN are drawn and intersect at point P. Line BP is then drawn. What is the minimum value of BP?

Solution:
\textbf{Solution:} Extend $CD$ such that $CD=DG$,\\
$\because$ Quadrilateral $ABCD$ is a rhombus, $\angle BAD=60^\circ$\\
$\therefore$ $\angle ADG=60^\circ$, $AD=GD$\\
$\therefore$ $\triangle ADG$ is an equilateral triangle\\
In $\triangle ADM$ and $\triangle BAN$\\
$\because$ $\left\{\begin{array}{l}AD=BA\\ \angle DAM=\angle ABN=60^\circ\\ AM=BN\end{array}\right.$\\
$\therefore$ $\triangle ADM\cong \triangle BAN\left(SAS\right)$\\
$\therefore$ $\angle ADM=\angle BAN$\\
$\because$ $\angle DAP+\angle BAN=60^\circ$\\
$\therefore$ $\angle GAD+\angle DAP+\angle GDA+\angle ADP=180^\circ$\\
$\therefore$ Points A, P, D, G are concyclic, the radius of the circle is $\frac{\frac{AD}{2}}{\cos30^\circ}=\frac{\sqrt{3}a}{3}$\\
$GB=2AB\cos30^\circ=\sqrt{3}a$\\
$\therefore$ When points G, P, B are collinear, namely when GP is the diameter of the circle, $BP$ is the shortest,\\
$\therefore$ $BP=GB−GP=\sqrt{3}a−2\times \frac{\sqrt{3}a}{3}=\boxed{\frac{\sqrt{3}a}{3}}$


Question: As shown in the figure, the vertices $A$ and $C$ of the rectangle $ABCO$ are located on the $x$-axis and $y$-axis, respectively. The coordinates of point $B$ are $\left(4, 3\right)$. Circle $M$ is the inscribed circle of $\triangle AOC$. Points $N$ and $P$ are moving points on circle $M$ and the $x$-axis, respectively. What is the minimum value of $BP+PN$?

Solution:
\textbf{Solution:} Let point $B^{\prime}$ be the symmetrical point of B with respect to the x-axis. Draw $MB^{\prime}$, which intersects $\odot M$ at point N and the x-axis at point P. Draw $MQ \perp$ x-axis through point M, intersecting the x-axis at point E. Draw $B^{\prime}Q \perp MQ$ through point $B^{\prime}$,

$\because$ Point B and point $B^{\prime}$ are symmetrical with respect to the x-axis,

$\therefore$ $PB + PN = PB^{\prime} + PN$,

The minimum value is obtained when N, P, and $B^{\prime}$ are on the same line and pass through point M.

In the right triangle $\triangle ABC$, $AC= \sqrt{OA^{2}+OC^{2}}=5$,

Since $\odot M$ is the inscribed circle of $\triangle AOC$, let the radius of $\odot M$ be r,

$\therefore$ $S_{\triangle AOC} = \frac{1}{2}(3r + 4r + 5r) = \frac{1}{2} \times 3 \times 4$,

Solving for r gives $r=1$,

$\therefore$ $ME=MN=1$,

$\therefore$ $QB^{\prime}=4-1=3$, $QM=3+1=4$,

$\therefore$ $MB^{\prime}=5$,

$\therefore$ $PB^{\prime}+PN=5-1=4$,

Hence, the minimum value of $PB + PN$ is $\boxed{4}$.


Question: As shown in the figure, in the Cartesian coordinate system \(xOy\), it is known that the chord of \(\odot D\) intersecting the \(y\)-axis has a length of 6, and the center of the circle is \(D(2, 4)\). What is the length of the shortest chord passing through point \(B(2, 3)\)?

Solution:
\textbf{Solution:} Let the intersection points of circle $D$ with the y-axis be $E$ and $A$. Connect $DE$. Draw $DC \perp y$-axis at $C$,

$\because$ The chord length of circle $D$ intersecting with the y-axis is 6,

$\therefore$ $AE = 6$,

$\therefore$ $CE = 3$,

$\because$ $D(2, 4)$,

$\therefore$ $CD = 2$,

$\therefore$ $DE= \sqrt{CD^{2}+CE^{2}}=\sqrt{2^{2}+3^{2}}=\sqrt{13}$,

$\because$ $D(2, 4)$, $B(2, 3)$,

$\therefore$ $DB \parallel y$-axis,

$\therefore$ Among all the chords through point $B(2, 3)$, the shortest chord is perpendicular to $DB$,

Draw $MN \perp DB$intersecting circle $D$ at $M$ and $N$. Connect $DN$,

In $\triangle DBN$, $\angle DBN=90^\circ$, $DB=1$, $DN= \sqrt{13}$,

$\therefore$ $BN= \sqrt{DN^{2}-BD^{2}}=2\sqrt{3}$,

$\therefore$ $MN=2BN= 4\sqrt{3}$,

Hence, among all the chords through point $B(2, 3)$, the shortest chord length is $\boxed{4\sqrt{3}}$.


Question: As shown in the figure, a circle $\odot O$ with a radius of 2 is tangent to the regular hexagon ABCDEF at points C and F. What is the area of the shaded region in the figure?

Solution:
\textbf{Solution:} Connect $OF$, $OC$, $CF$, from point $O$ draw $OH\perp ED$ at point H, it intersects $FC$ at point P,\\
in the quadrilateral $OCDH$, $OC\perp CD$, $OH\perp HD$, $\angle CDH=120^\circ$,\\
$\therefore$ $\angle OCD=90^\circ$, $\angle DHO=90^\circ$,\\
$\therefore$ $\angle COH=360^\circ-\angle OCD-\angle CDH-\angle DHO=360^\circ-90^\circ-120^\circ-90^\circ=60^\circ$,\\
Similarly, $\angle FOH=60^\circ$,\\
Since $OC=OF$,\\
$\therefore$ $OP$ is the perpendicular bisector of $FC$,\\
In $\triangle OPC$, $\angle OPC=90^\circ$, $\angle COP=60^\circ$, $OC=2$,\\
$\therefore$ $\angle OCP=180^\circ-\angle OPC-\angle COP=180^\circ-90^\circ-60^\circ=30^\circ$,\\
$\therefore$ $OP=\frac{1}{2}\times OC=\frac{1}{2}\times 2=1$,\\
$PC=\sqrt{OC^2-OP^2}=\sqrt{2^2-1^2}=\sqrt{3}$,\\
$\therefore$ $FC=2PC=2\times \sqrt{3}=2\sqrt{3}$,\\
Draw $DM\perp FC$ from point D, draw $EN\perp FC$ from point E,\\
$\therefore$ $\angle DMC=\angle ENF=90^\circ$,\\
Because $\angle OCD=90^\circ$, $\angle OCP=30^\circ$,\\
$\therefore$ $\angle PCD=\angle OCD-\angle OCP=90^\circ-30^\circ=60^\circ$,\\
Similarly, $\angle PFE=60^\circ$,\\
In $\triangle DMC$, $\angle MDC=180^\circ-\angle DMC-\angle MCD=180^\circ-90^\circ-60^\circ=30^\circ$,\\
$\therefore$ $MC=\frac{1}{2}DC$,\\
In $\triangle EFN$, $\angle FEN=180^\circ-\angle EFN-\angle ENF=180^\circ-90^\circ-60^\circ=30^\circ$,\\
$\therefore$ $NF=\frac{1}{2}EF$,\\
$\therefore$ $FC=FN+NM+MC=2ED=2\sqrt{3}$,\\
Since $EF=DE=CD=NM$,\\
$\therefore$ $ED=CD=EF=NM=\sqrt{3}$,\\
$MC=NF=\frac{1}{2}(FC-NM)=\frac{1}{2}\times (2\sqrt{3}-\sqrt{3})=\frac{\sqrt{3}}{2}$,\\
$\therefore$ $M=\sqrt{CD^2-MC^2}=\sqrt{\sqrt{3}^2-\left(\frac{\sqrt{3}}{2}\right)^2}=\frac{3}{2}$,\\
Then $PH=DM=\frac{3}{2}$,\\
$\therefore$ ${S}_{OFEDC}={S}_{\triangle OFC}+{S}_{FEDC}=\frac{1}{2}\times 2\sqrt{3}\times 1+\frac{1}{2}\times (\sqrt{3}+2\sqrt{3})\times \frac{3}{2}=\frac{13\sqrt{3}}{4}$,\\
$S_{\stackrel{\frown}{OFC}}=\frac{\frac{3\pi}{2}}{2\pi}\cdot\pi\cdot 2^2=\frac{3\pi}{2}$,\\
$\therefore$ The area of the shaded part = ${S}_{OFEDC}-{S}_{\stackrel{\frown}{OFC}}=\frac{13\sqrt{3}}{4}-\frac{3\pi}{2}=\boxed{\frac{13\sqrt{3}}{4}-\frac{3\pi}{2}}$,


Question: As shown in the diagram, within circle $\odot O$ of radius 5, chord $AB=6$, $OC\perp AB$ at point $D$, and intersects $\odot O$ at point $C$, what is the length of $CD$?

Solution:
\textbf{Solution:} Draw a line from point O to point A,\\
since AB=6, and OC $\perp$ AB at point D,\\
therefore, AD=$\frac{1}{2}$AB=$\frac{1}{2}\times 6=3$,\\
since the radius of circle O is 5,\\
therefore, OD=$\sqrt{OA^{2}-AD^{2}}=\sqrt{5^{2}-3^{2}}=4$,\\
therefore, CD=OC-OD=5-4=\boxed{1}.


Question: As shown in the figure, PA and PB are tangent to the circle $O$ at points A and B, respectively. C is a moving point on the major arc AB. If $\angle P = 50^\circ$, then what is the degree measure of $\angle ACB$?

Solution:
\textbf{Solution:} Draw $OA, OB$ as shown in the diagram,\\
$\because$ PA and PB are tangent to $\odot O$ respectively\\
$\therefore \angle OAP=\angle OBP=90^\circ$\\
$\therefore \angle AOB=360^\circ-\angle OAP-\angle OBP-\angle P=130^\circ$\\
$\because \overset{\frown}{AB}=\overset{\frown}{AB}$\\
$\therefore \angle ACB=\frac{1}{2}\angle AOB=\boxed{65^\circ}$


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the points A, B, and C have integer coordinates for both their x- and y-values. If an arc is drawn through these three points, what are the coordinates of the center of the arc?

Solution:
\textbf{Solution:} According to the corollary of the Perpendicular Diameter Theorem: the perpendicular bisector of a chord always passes through the center of the circle,\\
the perpendicular bisectors of chords AB and BC can be drawn, and their intersection point is the center of the circle.\\
As shown in the figure, the center of the circle is $\boxed{(2,1)}$.


Question: As shown in the diagram, AB is the diameter of the circle with center O, and the chord CD is perpendicular to AB at point H. If AB = 10 and CD = 8, what is the length of OH?

Solution:
\textbf{Solution:} Given that AB is the diameter of $\odot$O, and chord CD is perpendicular to AB at point H, if AB=10, and CD=8, \\
$\therefore CH=\frac{1}{2}CD=4$, $OC=\frac{1}{2}AB=5$\\
In $\triangle OHC$, $ OH=\sqrt{OC^{2}-CH^{2}}=\sqrt{5^{2}-4^{2}}=\boxed{3}$


Question: As shown in the diagram, within $\odot O$, $A$, $B$, and $C$ are three points on $\odot O$. If $\angle AOB=70^\circ$, what is the degree measure of $\angle C$?

Solution:
\textbf{Solution:} Given: $\angle AOB$ and $\angle ACB$ both subtend $\overset{\frown}{AB}$, and $\angle AOB=70^\circ$, \\
$\therefore \angle C=\frac{1}{2}\angle AOB=\boxed{35^\circ}$.


Question: As shown in the figure, AB is the diameter of semicircle O with AB = 4. Points C and D are on the semicircle, and $OC \perp AB$. It is given that $\overset{\frown}{BD} = 2\overset{\frown}{CD}$. Point P is a moving point on OC. What is the minimum value of $BP + DP$?

Solution:
\textbf{Solution:} As shown in the figure, connect AD, PA, PD, OD. \\
$\because$ OC $\perp$ AB, OA = OB, \\
$\therefore$ PA = PB, $\angle$ COB = $90^\circ$, \\
$\because$ $\overset{\frown}{BD} = 2\overset{\frown}{CD}$, \\
$\therefore$ $\angle$ DOB = $\frac{2}{3} \times 90^\circ = 60^\circ$, \\
$\because$ OD = OB, \\
$\therefore$ $\triangle$ OBD is an equilateral triangle, \\
$\therefore$ $\angle$ ABD = $60^\circ$, \\
$\because$ AB is the diameter, \\
$\therefore$ $\angle$ ADB = $90^\circ$, \\
$\therefore$ AD = AB $\cdot \sin\angle$ ABD = $2\sqrt{3}$, \\
$\because$ PB + PD = PA + PD $\geq$ AD, \\
$\therefore$ PD + PB $\geq 2\sqrt{3}$, \\
$\therefore$ The minimum value of PD + PB is $\boxed{2\sqrt{3}}$.


Question: As shown in the figure, points A, B, and C are three points on a circle $\odot$O with a radius of 4. If $\angle BAC=45^\circ$, what is the length of chord BC?

Solution:
\textbf{Solution:} Connect OB and OC.\\
Since $\angle$BOC = 2$\angle$BAC and $\angle$BAC = 45$^\circ$,\\
it follows that $\angle$BOC = 90$^\circ$,\\
Since OB = OC = 4,\\
it follows that BC = $\sqrt{4^2 + 4^2} = 4\sqrt{2}$,\\
Therefore, the answer is: $BC = \boxed{4\sqrt{2}}$.


Question: As shown in the figure, a circular pipe in a residential area has ruptured, and the repair workers are preparing to replace a section of the pipe. They have measured the width of the sewage water surface to be $80\, \text{cm}$, and the distance from the water surface to the top of the pipe to be $20\, \text{cm}$. What should be the inner diameter of the new pipe that the repair workers need to prepare, in $\text{cm}$?

Solution:
\textbf{Solution:} As shown in the diagram, let the center of the pipe be point O and the top of the pipe be point A, with BD representing the surface of the sewage. Connect AO, where AO intersects BD perpendicularly at point C.\\
Let AO=OB=r, then OC=r−20, and BC=$\frac{1}{2}BD=40$.\\
We have $OB^{2}=OC^{2}+BC^{2}$,\\
$r^{2}=(r−20)^{2}+40^{2}$,\\
Simplifying, we get $r=\boxed{50}$.\\
Therefore, the internal diameter of the new pipe is $100\text{cm}$.


Question: In $\triangle ABC$, $\angle BAC=60^\circ$, $\angle ABC=45^\circ$, $AB=2$, and $D$ is a moving point on line segment $BC$. With $AD$ as the diameter, draw $\odot O$ intersecting $AB$, $AC$ at $E$, $F$ respectively. Connect $EF$. What is the minimum length of line segment $EF$?

Solution:
\textbf{Solution:} As shown in the figure, connect OE, OF, and draw OH$\perp$EF through point O with H as the foot,\\
$\therefore$ $EH=\frac{1}{2}EF$,\\
$\because$ OE=OF, OH$\perp$EF, $\angle$BAC$=60^\circ$,\\
$\therefore$ $\angle EOH=\angle FOH=\frac{1}{2}\angle EOF=\angle BAC=60^\circ$,\\
$\therefore$ $\angle$OEH$=30^\circ$,\\
$\therefore$ OH$=\frac{1}{2}OE$,\\
$\therefore$ $EH=\sqrt{OE^2-OH^2}=\frac{\sqrt{3}}{2}OE$,\\
$\therefore$ $EF=\sqrt{3}OE$,\\
$\therefore$ To minimize EF, that is to minimize the radius OE, which means to minimize the diameter AD,\\
$\therefore$ According to the property of the perpendicular segment, when AD is the height from side BC of $\triangle ABC$, the diameter AD is the shortest,\\
$\because$ In $\triangle ADB$, $\angle ABC=45^\circ$, AB$=2$,\\
$\therefore$ AD$=$BD, $BD^2+AD^2=AB^2$,\\
$\therefore$ $2AD^2=4$,\\
$\therefore$ AD$=$BD$=\sqrt{2}$,\\
$\therefore$ $EF=\frac{\sqrt{3}}{2}AD=\boxed{\frac{\sqrt{6}}{2}}$


Question: As shown in the figure, $AB$ is the tangent to circle $O$ at point $A$, $OB$ intersects circle $O$ at point $C$, point $D$ is on circle $O$, and $AD$, $CD$, $OA$ are connected. If $\angle ADC = 25^\circ$, what is the degree measure of $\angle B$?

Solution:
\textbf{Solution:} Since $\angle ADC=25^\circ$,\\
thus, $\angle AOC=50^\circ$,\\
since $AB$ is a tangent of $\odot O$ at point A,\\
thus, $\angle OAB=90^\circ$,\\
therefore, $\angle B=90^\circ-\angle AOC=90^\circ-50^\circ=\boxed{40^\circ}$.


Question: As shown in the figure, the circumcircle $O$ of $\triangle ABC$ has a radius of $2\text{cm}$. If $BC=2\text{cm}$, what is the measure of $\angle A$ in degrees?

Solution:
\textbf{Solution:} Connect OB and OC,\\
since the radius of circle O is 2cm, and BC=2cm,\\
therefore OB=OC=BC,\\
thus, $\triangle$OBC is an equilateral triangle,\\
therefore, $\angle$BOC=60$^\circ$,\\
therefore, $\angle$A=$\frac{1}{2}\angle$BOC=30$^\circ$,\\
hence, the answer is: $\boxed{30^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $BC=4$, a circle $\odot A$ with center A and radius 2 is tangent to $BC$ at point D, intersects $AB$ at point E and $AC$ at point F. Point P is a point on $\odot A$, and $\angle EPF = 40^\circ$. What is the area of the shaded region in the figure?

Solution:
\[ \text{\bf Solution:} \text{ As shown in the figure, connect AD.} \]
\[ \because \angle EPF=40^\circ, \]
\[ \therefore \angle EAF=2\angle EPF=80^\circ, \]
\[ \therefore S_{\text{sector} EAF}=\frac{80\pi \times AE^{2}}{360}=\frac{80\pi \times 2^{2}}{360}=\frac{8\pi }{9}. \]
\[ \because \text{BC is the tangent of } \odot O \text{ with the tangent point at D}, \]
\[ \therefore AD\perp BC, AD=2, \]
\[ \therefore S_{\triangle ABC}=\frac{1}{2}AD\cdot BC=\frac{1}{2}\times 2\times 4=4. \]
\[ \because S_{\text{shaded}}=S_{\triangle ABC}-S_{\text{sector} EAF}, \]
\[ \therefore S_{\text{shaded}}=4-\frac{8\pi }{9}. \]
\[ \text{Therefore, the answer is: } \boxed{4-\frac{8\pi }{9}}. \]


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, the chord $CD\perp AB$ at point $H$, $CD=8$, $OA=5$, what is the length of $AH$?

Solution:
\textbf{Solution:} As shown in the figure, connect OC. \\
$\because$ AB is the diameter of $\odot$O, and chord CD $\perp$ AB at point H with CD=8, \\
$\therefore$ $CH=DH=\frac{1}{2}CD=4$, $\angle$OHC$=90^\circ$, \\
$\because$ OC=OA=5, \\
$\therefore$ $OH=\sqrt{OC^{2}-CH^{2}}=3$, \\
$\therefore$ AH=OA+OH=$\boxed{8}$.


Question: As shown in the diagram, for the inscribed quadrilateral $ABCD$ in the circle $\odot O$, if $\angle BCD=130^\circ$, then what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in circle $O$, \\
thus $\angle A = 180^\circ - \angle C = 50^\circ$, \\
therefore $\angle BOD = 2\angle A = \boxed{100^\circ}$


Question: As shown in the figure, there are two points A and B on the outer edge of the protractor, and their readings are $70^\circ$ and $40^\circ$ respectively. What is the degree measure of $\angle 1$?

Solution:
\textbf{Solution:} Since $\angle$AOB $= 70^\circ - 40^\circ = 30^\circ$, \\
therefore $\angle 1 = \frac{1}{2} \angle$AOB $= \boxed{15^\circ}$


Question: As shown in the figure, PA and PB are tangent to circle \(O\) at points A and B, respectively. Point E is a point on circle \(O\), and \(\angle AEB = 60^\circ\). What is the measure of \(\angle P\)?

Solution:
\textbf{Solution:} Draw OA, BO;\\
Since $\angle AOB = 2\angle E = 120^{\circ}$,\\
thus $\angle OAP = \angle OBP = 90^{\circ}$,\\
therefore $\angle P = 180^{\circ} - \angle AOB = \boxed{60^{\circ}}$.


Question: As shown in the figure, $O$ is the circumcenter of $\triangle ABC$, and $\angle ABC=40^\circ$, $\angle ACB=70^\circ$, what is the measure of $\angle BOC$?

Solution:
\textbf{Solution:} Given: $\because$ $\angle ABC=40^\circ$ and $\angle ACB=70^\circ$,\\
$\therefore$ $\angle BAC=180^\circ-40^\circ-70^\circ=70^\circ$,\\
$\because$ O is the circumcenter of $\triangle ABC$,\\
$\therefore$ the circle with center O and radius OB is the circumcircle of $\triangle ABC$,\\
$\therefore$ $\angle BOC=2\angle BAC=\boxed{140^\circ}$.


Question: As shown in the figure, in circle $O$, $AB$ is the diameter, and the length of chord $AC$ is 5 cm. Point $D$ is on the circle, and $\angle ADC=30^\circ$. What is the radius of circle $O$ in cm?

Solution:
\textbf{Solution:} Connect \textit{BC} as shown in the figure:\\
$\because$ $\angle ADC=30^\circ$,\\
$\therefore$ $\angle ABC=\angle ADC=30^\circ$,\\
$\because$ \textit{AB} is the diameter,\\
$\therefore$ $\angle ACB=90^\circ$,\\
$\because$ \textit{AC} $=5\text{cm}$,\\
$\therefore$ \textit{AB} $=2AC=10\text{cm}$,\\
$\therefore$ the radius of $\odot O$ is $\boxed{5}\text{cm}$;


Question: As shown in the figure, the circumcircle of quadrilateral $ABCD$ is $\odot O$, $BC=CD$, $\angle DAC=35^\circ$. What is the measure of $\angle BDC$?

Solution:
\textbf{Solution:} Since $\angle DAC=35^\circ$, \\
then $\angle DBC=\angle DAC=35^\circ$, \\
because $BC=CD$, \\
thus $\angle BDC=\angle DBC=35^\circ$, \\
Therefore, the answer is: $\boxed{35^\circ}$.


Question: As shown in the figure, the central angle of sector OAB is $110^\circ$, and C is a point on $\overset{\frown}{AB}$. What is the degree measure of $\angle C$?

Solution:
\textbf{Solution:} Construct $\overset{\frown}{AB}$'s corresponding central angle $\angle ADB$, as shown in the diagram,\\
$\therefore \angle ADB = \frac{1}{2}\angle AOB = \frac{1}{2} \times 110^\circ = 55^\circ$,\\
$\because \angle ADB + \angle C = 180^\circ$,\\
$\therefore \angle C = 180^\circ - 55^\circ = \boxed{125^\circ}$.


Question: As shown in the figure, $AE=AB$, $\angle E=\angle B$, $EF=BC$. If $\angle EAB=52^\circ$, then what is the measure of $\angle EFA$?

Solution: \textbf{Solution:} In $\triangle AEF$ and $\triangle ABC$, since $AE=AB$, $\angle E=\angle B$, and $EF=BC$, then $\triangle AEF \cong \triangle ABC$ by SAS, thus $\angle EAF=\angle BAC$, $AF=AC$, and $\angle C=\angle EFA$. Therefore, $\angle EAF-\angle BAF=\angle BAC-\angle BAF$, that is, $\angle FAC=\angle EAB=52^\circ$. Consequently, $\angle C=\angle EFA= \frac{1}{2}(180^\circ-\angle FAC)=\boxed{64^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $\angle B=22.5^\circ$, $AC=2$. Taking points $A$ and $B$ as centers, arcs are drawn with radii longer than $\frac{1}{2}AB$, intersecting at points $M$ and $N$. The line $MN$ intersects $BC$ at point $D$. Connecting $AD$, what is the length of $BD$?

Solution:
\textbf{Solution:} Given the construction, $MN$ bisects $AB$ perpendicularly, hence $DA=DB$, \\
$\therefore \angle DAB=\angle B=22.5^\circ$, \\
$\therefore \angle CDA=22.5^\circ+22.5^\circ=45^\circ$, \\
$\because \angle C=90^\circ$, \\
$\therefore \triangle ACD$ is an isosceles right triangle, \\
$\therefore AD=\sqrt{2}AC=2\sqrt{2}$, \\
$\therefore BD=\boxed{2\sqrt{2}}$.


Question: As shown in the figure, point $D$ lies on side $BC$ of $\triangle ABC$, with $AC=AB=BD$, $AD=CD$. What is the measure of $\angle BAC$ in degrees?

Solution: \textbf{Solution:} \boxed{108}


Question: As shown in the figure, $OP$ bisects $\angle MON$, and point $C$ is any point on $OP$ with $CA\perp ON$ at foot $A$. The perpendicular bisector of line segment $OA$, denoted as $BG$, intersects $OM$ at point $B$ and intersects $OA$ at point $G$. Given $AB=6$ and $AC=3$, what is the area of $\triangle OBC$?

Solution: \textbf{Solution:} As shown in the figure, draw line $CE\perp OM$ at point E. Since $OP$ bisects $\angle MON$, $CA\perp ON$, and $CE\perp OM$, it follows that $CE=CA=3$. Since $BG$ is the perpendicular bisector of $OA$, we have $BA=BO=6$. The area of $\triangle OBC$ is given by $S_{\triangle OBC}=\frac{1}{2}\times OB\times CE=\frac{1}{2}\times 6\times 3=\boxed{9}$.


Question: As shown in the figure, rays $BA$ and $CA$ intersect at point $A$, with $BC$ connected. Given that $AB=AC$ and $\angle B=40^\circ$, what is the value of $x$?

Solution: \textbf{Solution:} Since $AB = AC$ and $\angle B = 40^\circ$, it follows that $\angle C = \angle B = 40^\circ$, hence $x^\circ = 40^\circ + 40^\circ = \boxed{80^\circ}$.


Question: As shown in the figure, within $\triangle ABC$, $AB=AC=10\text{cm}$, $DE$ is a perpendicular bisector of $AB$ with the foot of the perpendicular being at $E$, and it intersects $AC$ at $D$. If the perimeter of $\triangle DBC$ is $18\text{cm}$, what is the length of $BC$ in $\text{cm}$?

Solution:
\textbf{Solution:} Since the perimeter of $\triangle DBC$ equals BC + BD + CD $= 18\,\text{cm}$,\\
and since DE perpendicularly bisects AB,\\
thus AD $=$ BD,\\
hence BC + AD + CD $= 18\,\text{cm}$,\\
since AC $=$ AD + DC $= 10\,\text{cm}$,\\
therefore BC $= 18 - 10 = \boxed{8}\,\text{cm}$.


Question: As shown in the figure, $\triangle ABC$ is inscribed in a semicircle $O$ with radius $\sqrt{5}$, where $AB$ is the diameter. Point $M$ is the midpoint of the arc $\overset{\frown}{AC}$. Connecting $BM$ intersects $AC$ at point $E$. $AD$ bisects $\angle CAB$ and intersects $BM$ at point $D$, and $D$ is the midpoint of $BM$. What is the length of $BC$?

Solution:
\textbf{Solution:} As shown in the diagram, construct $MH\perp AB$ at $H$, connect $AM$, $OM$, where $OM$ intersects $AC$ at $F$. \\
$\because AB$ is the diameter,\\
$\therefore \angle AMB=90^\circ$, $\angle ACB=90^\circ$,\\
$\therefore \angle CAB+\angle CBA=90^\circ$,\\
$\because$ point $M$ is the midpoint of $\overset{\frown}{AC}$,\\
$\therefore \overset{\frown}{CM}=\overset{\frown}{AM}$\\
$\therefore \angle CBM=\angle ABM$,\\
$\because \angle CAD=\angle BAD$,\\
$\therefore \angle DAB+\angle DBA=\frac{1}{2}(\angle CAB+\angle CBA)=45^\circ$,\\
$\therefore \angle ADB=180^\circ-(\angle DAB+\angle DBA)=135^\circ$,\\
$\because \angle ADM=180^\circ-\angle ADB=45^\circ$,\\
$\therefore MA=MD$,\\
$\because DM=DB$,\\
$\therefore BM=2AM$, let $AM=x$, then $BM=2x$,\\
$\because AB=2\sqrt{5}$,\\
$\therefore x^2+4x^2=20$,\\
$\therefore x=2$ (the negative root has been discarded),\\
$\therefore AM=2$, $BM=4$,\\
$\because \frac{1}{2}\cdot AM\cdot BM=\frac{1}{2}\cdot AB\cdot MH$,\\
$\therefore MH=\frac{2\times 4}{2\sqrt{5}}=\frac{4\sqrt{5}}{5}$,\\
$\therefore OH=\sqrt{OM^2-MH^2}=\sqrt{(\sqrt{5})^2-\left(\frac{4\sqrt{5}}{5}\right)^2}=\frac{3\sqrt{5}}{5}$,\\
$\because \overset{\frown}{CM}=\overset{\frown}{AM}$,\\
$\therefore OM\perp AC$,\\
$\therefore AF=FC$,\\
$\because OA=OB$,\\
$\therefore BC=2OF$,\\
$\because \angle OHM=\angle OFA=90^\circ$, $\angle AOF=\angle MOH$, $OA=OM$,\\
$\therefore \triangle OAF\cong \triangle OMH\left(AAS\right)$,\\
$\therefore OF=OH=\frac{3\sqrt{5}}{5}$,\\
$\therefore BC=2OF=\boxed{\frac{6\sqrt{5}}{5}}$.


Question: As shown in the figure, within circle $O$ with a radius of $10\,\text{cm}$, $AB=16\,\text{cm}$, and $OC\perp$ chord $AB$ at point $C$, what is the length of $OC$ in $\text{cm}$?

Solution:
\textbf{Solution:} Join OA, as shown in the figure,\\
$\because OC\perp AB$,\\
$\therefore AC=BC= \frac{1}{2}AB=8$,\\
In Rt$\triangle$ OAC, $OC=\sqrt{OA^{2}-AC^{2}}=\sqrt{10^{2}-8^{2}}=\boxed{6}\ (\text{cm})$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, with $AB=2$. $OC$ is the radius of $\odot O$, and $OC\perp AB$. Point $D$ lies on $\overset{\frown}{AC}$, and $\overset{\frown}{AD}=2\overset{\frown}{CD}$. Point $P$ is a moving point on the radius $OC$. What is the minimum value of $AP+PD$?

Solution:
\textbf{Solution:} Construct line BD, intersecting OC at point P. Draw lines OD, AP, AD.\\
Since $OC\perp AB$ and point $P$ is a movable point on the radius $OC$,\\
it follows that $AP=PB$\\
Since $DP+AP=DP+PB\ge DB$\\
the minimum value of $AP+PD$ is the length of $DB$,\\
hence, $\overset{\frown}{AC}=\overset{\frown}{CB}$, $\angle AOC=90^\circ$\\
Since $\overset{\frown}{AD}=2\overset{\frown}{CD}$\\
it follows that $\overset{\frown}{AD}=\frac{2}{3}\overset{\frown}{AC}$\\
Therefore, $\angle AOD=\frac{2}{3}\angle AOC=60^\circ$\\
Since $\overset{\frown}{AD}=\overset{\frown}{AD}$\\
it follows that $\angle ABD=30^\circ$\\
In right triangle $\triangle ADC$\\
$AB=2$, $AD=\frac{1}{2}AB=1$\\
Therefore, $BD=\sqrt{3}$\\
Hence, the minimum value of $AP+PD$ is $\boxed{\sqrt{3}}$


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in the circle $\odot O$. If $\angle ADC=85^\circ$, then what is the measure of $\angle B$ in degrees?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in circle $O$,\\
it follows that $\angle ADC + \angle B = 180^\circ$,\\
Since $\angle ADC = 85^\circ$,\\
then $\angle B = 180^\circ - \angle ADC$\\
$= 180^\circ - 85^\circ$\\
$= \boxed{95^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, $M$ is a point on $AB$, and $OA=6$, $AM=2$. The chord $PQ$ passes through point $M$, and the minor arc $PQ$ is folded along the chord $PQ$ to just pass through the center $O$. What is the length of $PM$?

Solution:
\textbf{Solution:} Draw OF $\perp$ PQ at point N, intersecting circle O at point F,\\
$\therefore$ PN=QN, $\because$ the minor arc PQ folds along the chord PQ to exactly pass through the center O, $\therefore$ ON=$\frac{1}{2}$OF=$\frac{1}{2}$OQ=3,\\
$\because$ AM=2, $\therefore$ OM=OA−AM=6−2=4,\\
$\therefore$ QN=PN=$\sqrt{OQ^{2}−ON^{2}}=\sqrt{6^{2}−3^{2}}=3\sqrt{3}$,\\
$\therefore$ MN=$\sqrt{OM^{2}−ON^{2}}=\sqrt{4^{2}−3^{2}}=\sqrt{7}$,\\
$\therefore$ PM=PN−MN=$3\sqrt{3}−\sqrt{7}$. Thus, the answer is: $PM=\boxed{3\sqrt{3}−\sqrt{7}}$.


Question: In rectangle $ABCD$, $AB = 3$, $AD = 5$, the point $E$ is on side $AD$ with $AE = 4$. Point $P$ is inside the rectangle and $\angle APE = 90^\circ$. Point $M$ is on side $BC$. Connecting $PA$, $DM$, what is the minimum value of $PM + DM$?

Solution:
\textbf{Solution:} As shown in the diagram, \\
$\because$ $\angle APE=90^\circ$, \\
$\therefore$ point P moves on circle O with diameter $AE$, \\
Construct point G as the mirror image of D with respect to $BC$, connect $OG$ to intersect circle $O$ at point P and $BC$ at M, the minimum value of $PM+DM$ is the length of $PG$, \\
$\because$ quadrilateral $ABCD$ is a rectangle, \\
$\therefore$ $CG=CD=AB=3$, $\angle ADC=90^\circ$, \\
In $\triangle ODG$, $DG=CD+CG=6$, $OD=AD−OA=5−2=3$, \\
$\therefore$ $OG=\sqrt{OD^{2}+DG^{2}}=3\sqrt{5}$, \\
$\therefore$ $PG=OG−OP=3\sqrt{5}−2$, \\
$\therefore$ the minimum value of $PM+DM$ is: $\boxed{3\sqrt{5}−2}$.


Question: As shown in the figure, in right-angled triangle $ABC$ with $\angle ABC=90^\circ$, $AB=4$, and $BC=3$, point $D$ is a moving point on circle $\odot A$ with radius $2$, and point $M$ is the midpoint of $CD$. What is the maximum value of $BM$?

Solution:
\textbf{Solution:} As shown in the figure, take N as the midpoint of AC, and connect MN, BN,\\
$\because$ in $\triangle ABC$, $\angle ABC=90^\circ$, AB=4, BC=3,\\
$\therefore AC=\sqrt{3^2+4^2}=5$\\
$\because$AN=NC=$\frac{1}{2}AC=\frac{5}{2}$,\\
$\therefore$BN=$\frac{1}{2}AC=\frac{5}{2}$\\
$\because$ point M is the midpoint of CD,\\
$\therefore$DM=MC,\\
$\therefore$MN=$\frac{1}{2}AD=1$\\
$\therefore$BM$\leq$BN+NM,\\
$\therefore$BM$\leq\frac{5}{2}+1=\boxed{\frac{7}{2}}$,\\
i.e., the maximum value of BM is $\frac{7}{2}$.


Question: As shown in the figure, points $A, B, C$ are on the circle $\odot O$, $\angle B=55^\circ$, and $\angle BOC=70^\circ$. What is the measure of $\angle C$ in degrees?

Solution:
\textbf{Solution:} Since $\angle BOC=70^\circ$ and it subtends the same arc $\overset{\frown}{BC}$,\\
it follows that $\angle A=\frac{1}{2}\angle BOC=35^\circ$,\\
Since $\angle B+\angle A+\angle BDA=180^\circ$,\\
it follows that $\angle ADB=90^\circ$,\\
and hence $\angle ODC=90^\circ$,\\
therefore $\angle C=180^\circ-\angle ODC-\angle BOC=20^\circ$.\\
Thus, the answer is: $\boxed{20}$.


Question: As shown in the figure, let $\odot O$ have a diameter $AB=6$, with $CD$ being a chord of $\odot O$ and $CD\perp AB$ at the perpendicular foot $P$. Connect $AC$ and $AD$. If $\triangle ACD$ is an equilateral triangle, what is the length of $CD$?

Solution:
\textbf{Solution:} Connect $OC$, as shown in the diagram:\\
$\because$ $CD\perp AB$,\\
$\therefore$ $PC=PD=\frac{1}{2}CD$, $\angle OPC=90^\circ$,\\
$\because$ the diameter $AB$ of $\odot O$ is $6$,\\
$\therefore$ $OC=3$,\\
$\because$ $\triangle ACD$ is an equilateral triangle,\\
$\therefore$ $\angle COP=60^\circ$, $\angle OCP=30^\circ$,\\
$\therefore$ $OP=\frac{3}{2}$,\\
By the Pythagorean theorem we get:\\
$PC=\sqrt{OC^{2}-OP^{2}}=\sqrt{3^{2}-\left(\frac{3}{2}\right)^{2}}=\sqrt{\frac{27}{4}}=\frac{3\sqrt{3}}{2}$,\\
$\therefore$ $CD=2PC=\boxed{3\sqrt{3}}$.


Question: The cross-section of a drainpipe is shown in the figure below. Given that the radius of the drainpipe is $OB=10$, and the width of the water surface $AB=12$. If more water is added, such that the width of the water surface $AB$ changes to $16$, then by how much does the height of the water surface $AB$ increase?

Solution:
\textbf{Solution:} Given $EF \parallel AB$, $EF=16$. Draw $OC \perp AB$ at point $C$, intersecting $EF$ at point $G$. Connect $OE$.

$\therefore OC \perp EF$,

$\therefore \angle OGE=\angle OCB=90^\circ$, $BC=\frac{1}{2}AB=6$, $EG=\frac{1}{2}EF=8$,

In $\triangle OCB$, $OC=\sqrt{OB^{2}-CB^{2}}=\sqrt{10^{2}-6^{2}}=8$;

In $\triangle EOG$, $OG=\sqrt{OE^{2}-EG^{2}}=\sqrt{10^{2}-8^{2}}=6$

When $EF$ and $AB$ are on the same side of the center $O$, $CG=OC-OG=8-6=2$;

When $EF$ and $AB$ are on opposite sides of the center $O$, $CG=OC+OG=8+6=14$;

$\therefore$ When the width of the water surface $AB$ changes to 16, the height by which the water surface $AB$ rises is $\boxed{2}$ or 14.


Question: As shown in the figure, points A, B, and C are all on $\odot O$. If $\angle AOC = \angle ABC$, then what is the degree measure of $\angle A + \angle C$?

Solution:
\paragraph{Solution:} Let's look at the diagram: Select a point $D$ on the major arc $AC$, and connect $AD$ and $CD$, \\
$\therefore$ $\angle ADC=\frac{1}{2}\angle AOC$, $\angle ADC+\angle ABC=180^\circ$\\
$\because$ $\angle AOC=\angle ABC$\\
$\therefore$ $\frac{1}{2}\angle ABC+\angle ABC=180^\circ$, which yields: $\angle ABC=120^\circ$\\
$\because$ in quadrilateral $ABCD$\\
$\therefore$ $\angle A+\angle C+\angle ABC+\angle AOC=360^\circ$\\
$\therefore$ $\angle A+\angle C=360^\circ-2\angle ABC=\boxed{120^\circ}$.


Question: As shown in the figure, $PM$ and $PN$ are tangent to $\odot O$ at points $A$ and $B$ respectively, and $C$ is a point on $\odot O$ that is different from $A$ and $B$. Connect $AC$ and $BC$. If $\angle P=58^\circ$, what is the measure of $\angle ACB$ in degrees?

Solution:
\textbf{Solution:} Connect $OA$ and $OB$ as shown in the figure:\\
Since $PM$ and $PN$ are tangent to circle $O$ at points A and B, respectively,\\
it follows that $OA\perp PA$ and $OB\perp PB$,\\
hence $\angle OAP=\angle OBP=90^\circ$,\\
thus $\angle AOB=360^\circ-90^\circ-90^\circ-58^\circ=122^\circ$,\\
When point C is on the major arc $AB$, $\angle ACB=\frac{1}{2}\angle AOB=\frac{1}{2}\times 122^\circ=\boxed{61^\circ}$,\\
When point C is on the minor arc $AB$, $\angle ACB=180^\circ-61^\circ=\boxed{119^\circ}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in the circle $\odot O$. If one of its exterior angles $\angle DCE=68^\circ$, what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is inscribed in circle O,\\
it follows that $\angle A = \angle DCE = 68^\circ$,\\
by the Inscribed Angle Theorem, we find $\angle BOD = 2\angle A = \boxed{136^\circ}$


Question: As shown in the figure, the radius of the semicircle is 5. The vertex of the $30^\circ$ angle of the triangle plate is placed on the semicircle, and the two sides of this angle intersect with the semicircle at points A and B, respectively. What is the length of $AB$?

Solution:
\textbf{Solution:} As shown in the figure,\\
let point O be the center of the circle, connect $AO$, $BO$,\\
$\therefore$ $AO=BO=5$,\\
$\because$ $\angle ACB=30^\circ$,\\
$\therefore$ $\angle AOB=60^\circ$,\\
$\therefore$ $\triangle AOB$ is an equilateral triangle,\\
$\therefore$ $AB=AO=BO=\boxed{5}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\odot O$, with $BE$ being the diameter of circle $\odot O$. Connect $AE$ and $BD$. If $\angle BCD=115^\circ$, what is the measure of $\angle EBD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is inscribed in circle $O$, and $\angle BCD=115^\circ$, \\
it follows that $\angle BAD=180^\circ-115^\circ=65^\circ$\\
Draw line DE, \\
Since arc $\overset{\frown}{BD}=\overset{\frown}{BD}$, \\
it follows that $\angle BED=65^\circ$\\
Since BE is the diameter of circle $O$, \\
it follows that $\angle BDE=90^\circ$\\
Therefore, $\angle EBD=90^\circ-65^\circ=\boxed{25^\circ}$


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and $C$, $D$ are two points on $\odot O$. If $\angle BAC=40^\circ$, then what is the degree measure of $\angle ADC$?

Solution:
\textbf{Solution:} Connect $CB$, \\
$\because$ $AB$ is the diameter of $\odot O$, $\angle BAC = 40^\circ$\\
$\therefore$ $\angle ACB = 90^\circ$, \\
$\because$ $\angle BAC = 40^\circ$, \\
$\therefore$ $\angle CBA = 50^\circ$, \\
$\because$ quadrilateral $ABCD$ is inscribed in $\odot O$, \\
$\therefore$ $\angle ADC + \angle CBA = 180^\circ$, \\
$\therefore$ $\angle ADC = \boxed{130^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=6$, $BC=8$. $\triangle ABC$ is rotated counterclockwise about point $A$ to obtain $\triangle AB'C'$, such that point $C'$ falls on side $AB$. When connecting $BB'$, what is the value of $\sin\angle BB'C'$?

Solution:
\textbf{Solution:} Because $\angle C=90^\circ$, $AC=6$, and $BC=8$, \\
therefore $AB=\sqrt{AC^{2}+BC^{2}}=10$, \\
Also, because $AC'=AC=6$, \\
therefore $BC'=AB-AC'=4$, \\
Because $B'C'=BC=8$, $\angle B'C'B=90^\circ$, \\
therefore $BB'=\sqrt{B'C'^{2}+BC'^{2}}=4\sqrt{5}$, \\
therefore $\sin\angle BB'C'=\frac{BC'}{BB'}=\boxed{\frac{\sqrt{5}}{5}}$.


Question: There is a rectangular piece of paper ABCD, with the known dimensions $AB=2\sqrt{2}$ and $AD=4$. On it, there is a semicircle with $AD$ as the diameter (as shown in Figure 1). Let $E$ be a point on side $AB$. When the paper is folded along $DE$ such that point $A$ exactly falls on $BC$, what is the area of the part of the semicircle that is still exposed (as shown in Figure 2, the shaded area)?

Solution:
\textbf{Solution:} Let the center of the circle containing the shaded region be $O$, and let $AD$ intersect the semicircle at point $F$, as shown in the figure. Connect $OF$,

By the property of folds, we know: $AD=BC=4$,

By the property of rectangles, we know: $CD=AB=2\sqrt{2}$, $\angle DCA=90^\circ$,

$\therefore AC=\sqrt{AD^{2}-CD^{2}}=\sqrt{4^{2}-(2\sqrt{2})^{2}}=2\sqrt{2}=CD$,

$\therefore \triangle ADC$ is an isosceles right triangle,

$\therefore \angle DAC=45^\circ$,

Because $OD\parallel BC$,

$\therefore \angle ODF=\angle DAC=45^\circ$,

Because $OD=OF=2$,

$\therefore \angle OFD=\angle ODF=45^\circ$,

$\therefore \angle DOF=180^\circ-45^\circ-45^\circ=90^\circ$,

$\therefore S_{\text{shaded}}=S_{\text{sector} DOF}-S_{\triangle DOF}=\frac{90^\circ \pi \times 2^{2}}{360}-\frac{1}{2}\times 2\times 2=\boxed{\pi -2}$.


Question: As illustrated, $\triangle OAB$ and $\triangle ODC$ are similar figures with point O as the center of similarity and a similarity ratio of $1\colon2$. $\angle OCD=90^\circ$, and $CO=CD$. If $A(1, 0)$, what are the coordinates of point C?

Solution:
\textbf{Solution:} Draw line $AC$, \\
since $CO=CD$ and $\angle OCD=90^\circ$, \\
therefore $\triangle OCD$ is an isosceles right triangle, \\
thus $\angle COD=45^\circ$, \\
since $\triangle OAB$ and $\triangle ODC$ are similar figures with center of similarity at $O$ and a similarity ratio of $1:2$, \\
thus $2OA=OD$, and $\triangle OAB$ is an isosceles right triangle, \\
hence $A$ is the midpoint of $OD$, \\
therefore $AC \perp OD$ (the three lines of an isosceles triangle coincide), \\
thus $\triangle OAC$ is an isosceles right triangle, \\
since $A(1,0)$, \\
thus $OA=AC=1$, \\
hence the coordinates of point $C$ are $\boxed{(1,1)}$.


Question: In the figure, in $\triangle ABC$, $\angle B=40^\circ$. If $\triangle ABC$ is rotated counterclockwise around point A to $\triangle ADE$, such that point B falls on point D, which is on the extension line of BC, then what is the measure of $\angle BDE$?

Solution:
\textbf{Solution:} Since it is known from the property of rotation that $AB=AD$ and $\angle ADE=\angle B=40^\circ$, in $\triangle ABD$, as $AB=AD$, according to the property of isosceles triangles, we have $\angle ADB=\angle B=40^\circ$. Therefore, $\angle BDE=\angle ADE+\angle ADB=80^\circ$.\\
Hence, the answer is: $\boxed{80^\circ}$.


Question: In $\triangle ABC$, $\angle C = 90^\circ$, $AC = 3$, and points $D$ and $E$ are located on sides $AC$ and $AB$ respectively. $\triangle ADE$ is folded along line $DE$ such that point $A$ lands on side $BC$, denoted as point $F$. If $CF=1$ and $DF \parallel AB$, what is the length of $AB$?

Solution:
\textbf{Solution:} Given that $\triangle ADE$ is folded along the line de, point A falls on side BC, denoted as point F, with CF=1,\\
let $DC=x$, then $DF=3-x$,\\
In $Rt\triangle DCF$,\\
$DC^2+FC^2=DF^2$,\\
we get $x^2+1^2=(3-x)^2$,\\
Solving this gives: $x=\frac{4}{3}$, so $DF=3-\frac{4}{3}=\frac{5}{3}$,\\
Because $DF\parallel AB$,\\
$\triangle DCF \sim \triangle ACB$,\\
thus $\frac{DC}{AC}=\frac{DF}{AB}$,\\
which gives $\frac{\frac{4}{3}}{3}=\frac{\frac{5}{3}}{AB}$,\\
Solved to get: $AB=\boxed{\frac{15}{4}}$.


Question: As shown in the figure, in the Cartesian coordinate system, $\triangle OAB$ is an isosceles right-angled triangle with $\angle OAB=90^\circ$. The coordinate of point B is $(0, 2\sqrt{2})$, and point A is in the first quadrant. When the triangle is translated to the right along the $x$ axis to get $\triangle O'B'C$, the coordinate of point B' is $(2\sqrt{2}, 2\sqrt{2})$. What is the area of the region swept by $\triangle OAB$ during the translation process?

Solution:
\textbf{Solution:} Given that the coordinates of point B' are $(2\sqrt{2}, 2\sqrt{2})$, and after shifting the triangle along the x-axis to the right, we obtain the right-angled triangle $\triangle O'B'C$, \\
$\therefore$AC=BB'=$2\sqrt{2}$,\\
$\because$ $\triangle OAB$ is an isosceles right-angled triangle,\\
$\therefore$ the coordinates of A are $(\sqrt{2}, \sqrt{2})$, and OA=AB=O'C=CB'=2,\\
$\therefore$ the area of the shape swept by $\triangle OAB$ during the shifting process is: $2\sqrt{2}\times 2\sqrt{2}+\frac{1}{2}\times 2\times 2=\boxed{10}$.


Question: In real life, we often see many "standard" rectangles, such as the covers of our textbooks, A4 printing paper, etc. In fact, the ratio of the length to the width of these rectangles is $\sqrt{2}: 1$. We might as well call such rectangles "standard rectangles". In the "standard rectangle" $ABCD$, as shown in the figure, point $Q$ is on $DC$, and $DQ=AD$. If $G$ is a moving point on side $BC$, when the perimeter of $\triangle AGQ$ is minimized, what is the value of $\frac{CG}{GB}$?

Solution:
\textbf{Solution:} Let DC=$\sqrt{2}x$, DQ=AD=$x$,\\
$\therefore$ $CQ=(\sqrt{2}-1)x$\\
$\because$ In rectangle ABCD,\\
$\therefore$ $\angle D=\angle DCB=\angle B=90^\circ$, $AB=DC=\sqrt{2}x$, $BC=AD=x$,\\
$\therefore$ $AQ=\sqrt{AD^2+DQ^2}=\sqrt{2}x$,\\
As shown in the figure, construct the symmetric point E of point Q with respect to BC, and connect AE to intersect BC at point M,\\
$\therefore$ GQ=GE, CQ=CE=$(\sqrt{2}-1)x$\\
$\therefore$ AQ+QG+AG=$\sqrt{2}x+AG+EG\ge \sqrt{2}x+AE$,\\
$\therefore$ When points A, G, and E are collinear, the perimeter of $\triangle AGQ$ is minimized,\\
At this time, point G should be located at point M in the figure;\\
$\because$ In rectangle ABCD, $\angle QCG=90^\circ$,\\
$\therefore$ Point E lies on the extension line of QC,\\
$\therefore$ CE$\parallel$AB,\\
$\therefore$ $\frac{CM}{MB}=\frac{CE}{AB}=\frac{(\sqrt{2}-1)x}{\sqrt{2}x}=\frac{2-\sqrt{2}}{2}$,\\
which means $\frac{CG}{GB}=\boxed{\frac{2-\sqrt{2}}{2}}$.


Question: As shown in the rectangle $ABCD$, where $AB = 6$, $AD = 10$, point $E$ is moving on side $AD$. If triangle $DEC$ is folded along $EC$ such that point $D$ is placed at point $D'$, and if there exists a relationship between two sides of triangle $DEC$ where one side is twice the length of the other, what is the distance from point $D'$ to side $AD$?

Solution:
Solution:Let's solve the problem step by step.
(1) When $DE=\frac{1}{2}CE$, construct $D'H\perp AD$ at H, as shown in the diagram: \\
Since rectangle ABCD, AB=6, \\
$\therefore$$\angle$D=$90^\circ$, CD=6, \\
In $\triangle DCE$, $DE^{2}+CD^{2}=CE^{2}$, CD=6, $DE=\frac{1}{2}CE$, \\
$\therefore$$DE^{2}+{6}^{2}=(2DE)^{2}$, \\
$\therefore$$DE=2\sqrt{3}$, \\
Since $DE=\frac{1}{2}CE$, $\angle$D=$90^\circ$, \\
$\therefore$$\angle$DCE=$30^\circ$, $\angle$DEC=$60^\circ$, \\
Since $\triangle DEC$ is folded along EC causing D to fall on D', \\
$\therefore$$\angle$D'EC=$\angle$DEC=$60^\circ$, $D'E=2\sqrt{3}$, \\
$\therefore$$\angle$D'EH=$180^\circ$-$\angle D'EC-\angle$DEC=$60^\circ$, in right $\triangle D'HE$, $D'H=D'E\cdot\sin\angle D'EH=2\sqrt{3}\times \sin60^\circ=3$; \\
(2) When $DE=\frac{1}{2}DC$, construct D'H$\perp$AD at H, extend HD' to intersect BC at F, as shown in the diagram: \\
In right $\triangle DCE$, $DE^{2}+CD^{2}=CE^{2}$, CD=6, $DE=\frac{1}{2}CD$, \\
$\therefore$$DE=3$, $CE=3\sqrt{5}$, \\
Since $\triangle DEC$ is folded along EC causing D to fall on D', \\
$\therefore$$\angle$ED'C=$\angle$D=$90^\circ$, DE=D'E, CD=CD', \\
$\therefore$$\angle$HD'E=$90^\circ$-$\angle$FD'C=$\angle$FCD', \\
and $\angle$D'HE=$\angle$CFD'=$90^\circ$, \\
$\therefore$$\triangle D'HE\sim \triangle CFD'$, \\
$\therefore$$\frac{HE}{D'F}=\frac{D'H}{CF}=\frac{D'E}{CD'}$, \\
Since $DE=\frac{1}{2}DC$, \\
$\therefore$$\frac{DE}{DC}=\frac{1}{2}$, \\
$\therefore$$\frac{HE}{D'F}=\frac{D'H}{CF}=\frac{D'E}{CD'}=\frac{1}{2}$, \\
Let HE=x, D'H=y, then D'F=2x, CF=2y, \\
Since D'H$\perp$AD, and ABCD is a rectangle, \\
$\therefore$Quadrilateral HFCD is a rectangle, \\
$\therefore$$\begin{cases} 2x+y=6 \\ 2y=3+x \end{cases}$ \\
Solving gives $y=\frac{12}{5}$, \\
$\therefore$D'H=$\boxed{\frac{12}{5}}$; \\
(3) When $DC=\frac{1}{2}CE$, construct $D'H\perp AD$ at H, as shown in the diagram: \\
In right $\triangle DCE$, $DE^{2}+CD^{2}=CE^{2}$


Question: As illustrated, a square OABC with a side length of 3 rotates counterclockwise by $n^\circ$ ($0<n<90$) around point O to form square ODEF. DE intersects BC at point P, and the extension of ED intersects AB at point Q and the extension of OA at point M. If $AQ=1$, what is the value of $PD$?

Solution:
\textbf{Solution:} As shown in the figure, connect OQ and OP,

In $\triangle OPD$ and $\triangle OPC$,

$\because \angle PDO=\angle C=90^\circ$,

then $\left\{\begin{array}{l}
OP=OP\\
OD=OC
\end{array}\right.$,

$\therefore \triangle OPD\cong \triangle OPC$,

$\therefore PD=PC$,

Similarly, $\triangle OQA\cong \triangle OQD$, $QA=QD$,

Let $PD=x$, then $PC=x$, $BP=3-x$, $PQ=1+x$,

In $Rt\triangle BPQ$, $BP^2+BQ^2=PQ^2$,

i.e., $(3-x)^2+(3-1)^2=(1+x)^2$,

yielding $x=\boxed{\frac{3}{2}}$,

thus $PD=\frac{3}{2}$.


Question: After folding a circular paper (with center at point O) along diameter $MN$, and then folding it into six equal parts as shown in figure 1 to obtain figure 2, if we cut along the dashed line $AB$ and then unfold $\triangle AOB$ to get a hexagon star as shown in figure 3. If $\angle CDE = 75^\circ$, what is the degree measure of $\angle OBA$?

Solution:
\textbf{Solution:} Connect OC, EO\\
Due to the folding property, we have: $\angle$EOC=$ \frac{360^\circ}{12}=30^\circ$, EC=DC, OC bisects $\angle$ECD\\
$\therefore$ $\angle$ECO=$ \frac{1}{2}\angle ECD=\frac{1}{2}(180^\circ−2\times 75^\circ)=15^\circ$\\
$\therefore$ $\angle$OEC=$180^\circ$−$\angle$ECO−$\angle$EOC=$135^\circ$\\
Thus, the measure of $\angle OBA$ is $\boxed{135^\circ}$


Question: As shown in the figure, it is known that the side length of square $ABCD$ is $4$, and point $E$ is a point inside the square. Connect $BE$ and $CE$, and $\angle ABE = \angle BCE$. Point $F$ is a moving point on side $AB$, connecting $FD$ and $FE$. What is the minimum value of the length of $FD+FE$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a square, \\
$\therefore \angle ABC = 90^\circ$, \\
$\therefore \angle ABE + \angle CBE = 90^\circ$, \\
Since $\angle ABE = \angle BCE$, \\
$\therefore \angle BCE + \angle CBE = 90^\circ$, \\
$\therefore \angle BEC = 90^\circ$, \\
$\therefore$ point E moves on the semicircle with BC as the diameter, \\
As shown in the diagram, let the midpoint of BC be O, and construct the square APGB symmetric to square ABCD with respect to line AB, then the corresponding point of D is P, \\
Connect PO, intersecting AB at F and the semicircle O at E, then the length of segment EP is the minimum value of $FD+FE$, OE $= 4$, \\
Since $\angle G = 90^\circ$, PG $=$ BG $=$ AB $= 4$, \\
$\therefore$ OG $= 6$, \\
$\therefore$ OP $= \sqrt{PG^2 + OG^2} = \sqrt{4^2 + 6^2} = 2\sqrt{13}$, \\
$\therefore$ EP $= 2\sqrt{13} - 2$, \\
$\therefore$ The minimum value of the length of FD+FE is $\boxed{2\sqrt{13} - 2}$.


Question: In the rhombus $ABCD$ as shown in the figure, $AB=2$ and $\angle BAD=60^\circ$. The rhombus $ABCD$ is rotated counterclockwise around point $A$ to get the rhombus $AEFG$, where point $E$ is on $AC$, and $EF$ intersects $CD$ at point $P$. What is the length of $DP$?

Solution:
\textbf{Solution:} As shown in the figure, connect BD intersecting AC at point O, \\
$\because$ Quadrilateral ABCD is a rhombus, \\
$\therefore$ CD=AB=2, $\angle$BCD=$\angle$BAD=$60^\circ$, $\angle$ACD=$\angle$BAC=$\frac{1}{2}\angle$BAD=$30^\circ$, OA=OC, AC$\perp$BD, \\
$\therefore$ OB=$\frac{1}{2}$AB=1, \\
$\therefore$ OA=$\sqrt{3}$OB=$\sqrt{3}$, \\
$\therefore$ AC=2OA=2$\sqrt{3}$, \\
By the property of rotation: AE=AB=2, $\angle$EAG=$\angle$BAD=$60^\circ$, \\
$\therefore$ CE=AC−AE=2$\sqrt{3}$−2, \\
$\because$ Quadrilateral AEFG is a rhombus, \\
$\therefore$ EF$\parallel$AG, \\
$\therefore$ $\angle$CEP=$\angle$EAG=$60^\circ$, \\
$\therefore$ $\angle$CEP+$\angle$ACD=$90^\circ$, \\
$\therefore$ $\angle$CPE=$90^\circ$, \\
$\therefore$ PE=$\frac{1}{2}$CE=$\sqrt{3}$−1, \\
$\therefore$ PC=$\sqrt{3}$PE=3−$\sqrt{3}$, \\
$\therefore$ DP=CD−PC=2−(3−$\sqrt{3}$)=$\boxed{\sqrt{3}-1}$.


Question: In the equilateral $\triangle ABC$, let D be a point on side AC. Connect BD and rotate $\triangle BCD$ $60^\circ$ counterclockwise around point B to obtain $\triangle BAE$. Connect ED. What is the measure of $\angle DEB$?; If $BC=5$ and $BD=4$, what is the perimeter of $\triangle ADE$?

Solution:
\textbf{Solution:} Given that $\triangle BCD$ is rotated $60^\circ$ counterclockwise about point B to obtain $\triangle BAE$,

it follows that $BD=BE$, $\angle DBE=60^\circ$, and AE=CD.

Therefore, $\triangle BDE$ is an equilateral triangle.

Thus, $DE=BD=4$, and $\angle DEB=\boxed{60^\circ}$.

Given also that $\triangle ABC$ is an equilateral triangle,

we have $AC=BC=5$.

Therefore, the perimeter of $\triangle AED$ is $AE+AD+DE=CD+AD+DE=AC+DE=5+4=\boxed{9}$.


Question: As shown in the figure, given a rectangular piece of paper ABCD, with \(AB=6\), \(BC=20\), and \(M\) being the midpoint of \(BC\), point \(P\) moves along the polyline \(B-A-D\). If the paper is folded to the right along the crease \(MP\), making point \(B\) fall on the edge of the rectangle, what is the length of the crease \(MP\)?

Solution:
Solution:
(1) As shown in the figure, when point P is on side AB, draw ME $\perp$ AD at E, so quadrilateral ABME is a rectangle,\\
$\therefore$ EM = AB = 6, AE = BM,\\
Also, $\because$ BC = 20, and M is the midpoint of BC,\\
$\therefore$ by folding: B$^{\prime }$M = BM = AE = 10,\\
In Rt$\triangle$EMB$^{\prime}$, according to the Pythagorean theorem: B$^{\prime}$E = $\sqrt{10^{2}-6^{2}} = 8$,\\
$\therefore$ AB$^{\prime} = $ AE - B$^{\prime }$E = 2,\\
Let BP = x, then AP = 6 - x, PB$^{\prime }$ = x,\\
In Rt$\triangle$PAB$^{\prime}$, according to the Pythagorean theorem: PB$^{\prime }$$^{2} =$ AP$^{2}+$AB$^{\prime }$$^{2}$,\\
Thus x$^{2} = $($6 - x)^{2} + 2^{2}$,\\
Solving for x, we find x = $\frac{10}{3}$,\\
$\therefore$ PB = $\frac{10}{3}$,\\
In Rt$\triangle$BMP, according to the Pythagorean theorem: PM = $\sqrt{\left(\frac{10}{3}\right)^{2} + 10^{2}} = \frac{10}{3}\sqrt{10}$;\\
(2) As shown in the figure, when point P is on side AD, draw ME $\perp$ AD at E, so quadrilateral ABME is a rectangle,\\
$\therefore$ EM = AB = 6,\\
Also, $\because$ BC = 20, and M is the midpoint of BC,\\
$\therefore$ by folding: B$^{\prime}$M = BM = 10,\\
In Rt$\triangle$EMB$^{\prime}$, according to the Pythagorean theorem: B$^{\prime}$E = $\sqrt{10^{2} - 6^{2}} = 8$,\\
Since AD $\parallel$ BC, $\angle$DPM = $\angle$BMP,\\
By folding, $\angle$PMB$^{\prime}$ = $\angle$BMP,\\
$\therefore$$\angle$DPM = $\angle$PMB$^{\prime}$,\\
$\therefore$ B$^{\prime}$M = B$^{\prime}$P = 10,\\
$\therefore$ PE = 10 - 8 = 2,\\
In Rt$\triangle$PEM, according to the Pythagorean theorem: PM = $\sqrt{2^{2} + 6^{2}} = 2\sqrt{10}$;\\
(3) As shown in the figure, when point B$^{\prime}$ coincides with point C, because $\angle$A = $\angle$B = $\angle$BMP = $90^\circ$, quadrilateral ABMP is a rectangle,\\
At this point, PM = AB = 6.\\
In summary, the length of the crease MP is: $\frac{5}{3}\sqrt{13}$ or $2\sqrt{10}$ or \boxed{6}.


Question: As shown in the figure, $\angle MON=40^\circ$. With O as the center and 4 as the radius, an arc is drawn intersecting $OM$ at point A and $ON$ at point B. Arcs with centers at point A and point B, respectively, and with a radius greater than half the length of $AB$, intersect inside $\angle MON$ at point C. A ray $OC$ is drawn intersecting $\overset{\frown}{AB}$ at point D. Let E be a moving point on $OA$, and connect $BE$ and $DE$. What is the minimum perimeter of the shaded area?

Solution:
\textbf{Solution:} Given the problem statement: OC bisects $\angle MON$,\\
$\therefore$ $\angle BOD = \frac{1}{2}\angle MON = 20^\circ$,\\
$\therefore$ the length of $\overset{\frown}{BD}$ $= \frac{20\pi \times 4}{180} = \frac{4}{9}\pi$,\\
Construct the symmetric point D' of D about OM, draw BD' intersecting OM at point E', and connect OD', then BE'+ DE'= BE'+ D'E'=BD'. At this time, the minimum value of BE+DE= BD',\\
$\because$ $\angle AOD' = \angle AOD = \angle BOD = 20^\circ$,\\
$\therefore$ $\angle BOD' = 60^\circ$,\\
$\because$ OD'=OD=OB,\\
$\therefore$ $\triangle BOD'$ is an equilateral triangle,\\
$\therefore$ BD'=OB=4,\\
$\therefore$ the minimum perimeter of the shaded area $= \frac{4}{9}\pi + 4 = \boxed{\frac{4}{9}\pi + 4}$.


Question: As shown in the figure, in $\triangle ABC$, points $D$ and $E$ are on $BC$ and $AC$ respectively. $\triangle CDE$ is folded along $DE$ to result in $\triangle FDE$. Lines $BF$ and $CF$ are drawn, with $\angle BFC=90^\circ$. If $EF\parallel AB$, $AB=4\sqrt{3}$, and $EF=10$, what is the length of $AE$?

Solution:
\textbf{Solution:} As shown in the figure, extend $ED$ to meet $CF$ at point $G$,\\
let $\angle BFE=\alpha$\\
By folding, it is known that $DG\perp CF$,\\
since $BF\perp CF$,\\
thus $ED\parallel BF$,\\
therefore $\angle FED=\angle BFE=\alpha$,\\
extend $EA$, $FB$ to meet at point $M$,\\
since $AB\parallel EF$,\\
therefore $\angle BAC=\angle FEC=2\alpha$, $\angle ABM=\angle BFE=\alpha$,\\
therefore $\angle M=\angle BAC-\angle ABM=\alpha$,\\
since $\angle M=\angle BFE=\alpha$, $\angle M=\angle ABM=\alpha$,\\
thus $EM=EF=10$, $AM=AB=4\sqrt{3}$,\\
therefore $AE=EM-AM=10-4\sqrt{3}=\boxed{10-4\sqrt{3}}$.


Question: As shown in the figure, quadrilateral AEFH and quadrilateral ABCD are similar figures, with a similarity ratio of $\frac{2}{3}$. If the area of quadrilateral ABCD is 900 cm$^2$, then what is the area of quadrilateral AEFH in cm$^2$?

Solution:
\textbf{Solution:} Because quadrilateral AEFH and quadrilateral ABCD are similar figures with a similarity ratio of $\frac{2}{3}$,\\
therefore, the ratio of the area of quadrilateral AEFH to the area of quadrilateral ABCD is $4\colon 9$,\\
furthermore, since the area of quadrilateral ABCD is $900\text{cm}^{2}$,\\
therefore, the area of quadrilateral AEFH is $\boxed{400\text{cm}^{2}}$.


Question: In square $ABCD$, where $AB = 2$, let $E$ be the midpoint of $AB$. If $\triangle ADE$ is folded along $DE$ to form $\triangle FDE$, and $FH \perp BC$ with the foot of the perpendicular being $H$, then what is the length of $FH$?

Solution:
\textbf{Solution:} As shown in the figure, draw MN$\parallel$BC through point F, intersecting AB at M and CD at N,\\
$\because$ Quadrilateral ABCD is a square\\
$\therefore \angle B = \angle C = 90^\circ$\\
$\because$MN$\parallel$BC\\
$\therefore \angle AMF = \angle B = 90^\circ$, $\angle DNF=\angle C= 90^\circ$\\
$\therefore \angle EMF = \angle DNF = 90^\circ$\\
By folding: AD= DF = 2, AE = EF, $\angle A = \angle EFD= 90^\circ$,\\
$\therefore \angle EFM + \angle DFN= \angle DFN+ \angle NDF= 90^\circ$\\
$\therefore \angle EFM=\angle NDF$\\
$\therefore \triangle EMF\sim \triangle FND$\\
$\therefore \frac{EF}{DF}=\frac{FM}{DN}=\frac{EM}{FN}$\\
$\because$In square ABCD, AB = 2, E is the midpoint of AB\\
$\therefore$AE= BE = EF = 1\\
Let FH = BM=x, then EM= 1 – x, FN = 2EM = 2 (1 − x)= 2 – 2x\\
$\therefore$FM = 2 − FN = 2 −(2 −2 x)= 2 x\\
In $\triangle EMF$, by Pythagorean theorem:\\
EF$^2$= EM$^2$+ FM$^2$\\
$\therefore 1^2=(1−x)^2 +(2x)^2$\\
Solving yields: $x_1=0$, $x_2 = \frac{2}{5}$\\
$\therefore FH= \boxed{\frac{2}{5}}$


Question: As shown in the figure, in the Cartesian coordinate system, the coordinates of points $A$ and $B$ are $(2,0)$ and $(0,-4)$, respectively. If segment $AB$ is rotated $90^\circ$ clockwise around point $A$ to obtain segment $AC$, then what are the coordinates of point $C$?

Solution:
\textbf{Solution:} As shown in the figure, draw CD $\perp$ x-axis at point D,\\
$\because$ The coordinates of points A and B are (2, 0) and (0, -4) respectively,\\
$\therefore$ OA=2, OB=4,\\
From the property of rotation, we get: $\angle BAC=90^\circ$, AC=AB,\\
$\because$ $\angle ADC=\angle AOB=90^\circ$,\\
$\therefore$ $\angle CAD=\angle B$,\\
$\therefore$ $\triangle ACD \cong \triangle BAO$,\\
$\therefore$ CD=OA=2, AD=BO=4,\\
$\therefore$ OD=AD−OA=2,\\
$\therefore$ The coordinates of point C are $\boxed{(-2, 2)}$.


Question: As shown in the figure, in the isosceles right-angled triangle $ABC$ with $\angle C=90^\circ$ and $AC=BC=4\sqrt{2}$, points $D$ and $E$ are on sides $BC$ and $AB$ respectively, and segment $DE$ is drawn. Triangle $BDE$ is folded along $DE$ to obtain triangle $B^{\prime}DE$, with point $B$ precisely landing on the midpoint of $AC$. Let $DE$ intersect $BB^{\prime}$ at point $F$, what is the length of $DE$?

Solution:
\textbf{Solution: }

Given by the fold, we have
\[BD=B'D,\] \[BF=B'F,\]
\[DF \perp BF,\]
\[\because BC=AC=4\sqrt{2},\]
\[B'\] is the midpoint of AC,
\[\therefore CB'=2\sqrt{2},\]
In \(\triangle BCB'\),
\[BB'=\sqrt{BC^2+B'C^2}=\sqrt{(4\sqrt{2})^2+(2\sqrt{2})^2}=2\sqrt{10},\]
\[\therefore BF=\sqrt{10},\]
Let BD=\(x\), then \(CD=4\sqrt{2}-x\),
\[B'D=x,\]
In \(\triangle CDB'\),
\[B'D^2=CD^2+B'C^2,\]
\[\therefore x^2=(4\sqrt{2}-x)^2+(2\sqrt{2})^2,\]
Solving this, we get \(x=\frac{5\sqrt{2}}{2}\),
\[\therefore BD=\frac{5\sqrt{2}}{2},\]
In \(\triangle BDF\), \(DF=\sqrt{BD^2-BF^2}=\sqrt{\left(\frac{5\sqrt{2}}{2}\right)^2-(\sqrt{10})^2}=\frac{\sqrt{10}}{2},\)

Draw \(B'G \perp AB\) at point G, as shown in the diagram: \(\because \angle C=90^\circ\), AC=BC,
\[\therefore \angle A=\angle ABC=45^\circ,\]
\[\therefore \angle AB'G=90^\circ-45^\circ=45^\circ,\]
\[\therefore \angle A=\angle AB'G,\]
\[\therefore AG=B'G,\]
\[\because AB'=2\sqrt{2},\]
\[\therefore AG=B'G=AB'\times \sin 45^\circ=2\sqrt{2}\times \frac{\sqrt{2}}{2}=2,\]
Let \(BE=y\), then \(GE=6-y\),
\(B'E=y,\)
In \(\triangle B'GE\),
\[GE^2+B'G^2=B'E^2,\]
\[\therefore (6-y)^2+2^2=y^2,\]
Solving this, we get \(y=\frac{10}{3},\)
\[\therefore BE=\frac{10}{3},\]
In \(\triangle BEF\), \(EF^2=BE^2-BF^2\),
\[\therefore EF^2=\left(\frac{10}{3}\right)^2-(\sqrt{10})^2=\frac{10}{9},\]
\[\therefore EF=\frac{\sqrt{10}}{3},\]
\[\therefore ED=DF+EF=\boxed{\frac{5\sqrt{10}}{6}}.\]


Question: In the Cartesian coordinate system, given point $A(2, 3)$ and $B(5, -2)$, with the origin $O$ as the center of dilation, if $\triangle ABO$ is enlarged to twice its original size, what are the coordinates of point $B'$?

Solution:
\textbf{Solution:} Given that the origin O is the center of similitude, and $\triangle ABO$ is magnified by a factor of 2 with $B(5, -2)$, \\
\textbf{therefore,} the coordinates of the corresponding point $B'$ of point $B$ are\\
$(5\times 2, -2\times 2)$ or $(5\times (-2), -2\times (-2))$, that is, $(10, -4)$ or $(-10, 4)$,\\
Hence, the answer is: $\boxed{(10, -4)$ or $(-10, 4)}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=6$, $AC=BC=5$. $\triangle ABC$ is rotated clockwise around point $A$ to obtain $\triangle ADE$, with the angle of rotation being $\alpha$ ($0^\circ<\alpha<180^\circ$), where point $B$ corresponds to point $D$, and point $C$ corresponds to point $E$. When the angle between the lines $AB$ and $AD$ equals $\angle ACB$, what is the length of $BE$?

Solution:
\textbf{Solution:} As shown in the diagram, connect BD,

$\because$ $AC=BC$

$\therefore$ $\angle CAB=\angle CBA$

Based on the property of rotation,

$\angle CBA=\angle ADE$,

$AD=AB$

$\therefore$ $\angle ADB=\angle ABD$

$\because$ $\angle ACB=\angle DAB$

$\therefore$ $\angle CAB=\angle CBA=\angle ADE=\angle ADB=\angle ABD$

$\therefore$ $D$, $E$, $B$ are collinear

$\because$ $\frac{AC}{AD}=\frac{BC}{AB}$,

$\angle ACB=\angle DAB$

$\therefore$ $\triangle ACB\sim \triangle DAB$

$\therefore$ $\frac{AC}{AD}=\frac{AB}{DB}$, thus

$\frac{5}{6}=\frac{6}{DB}$

Solving gives

$DB=\frac{36}{5}$

$\therefore$ $BE=DB-DE=\frac{36}{5}-5=\boxed{\frac{11}{5}}$


Question: As shown in the figure, the regular hexagon $ABCDEF$ is obtained through a similarity transformation from the regular hexagon $A^{\prime}B^{\prime}C^{\prime}D^{\prime}E^{\prime}F^{\prime}$. Given that $AB=3$ and $B^{\prime}C^{\prime}=1$, what is the ratio of the area of the regular hexagon $A^{\prime}B^{\prime}C^{\prime}D^{\prime}E^{\prime}F^{\prime}$ to that of the regular hexagon $ABCDEF$?

Solution:
\textbf{Solution:} Given that the hexagon $ABCDEF$ is obtained from the hexagon $A'B'C'D'E'F'$ through a similarity transformation,\\
thus hexagon $ABCDEF \sim$ hexagon $A'B'C'D'E'F'$,\\
hence, the ratio of the areas of the hexagons $A'B'C'D'E'F'$ to $ABCDEF$ is equal to $(1\colon 3)^{2}=\boxed{1\colon 9}$.


Question: As shown in the figure, $\triangle ABC$ is rotated $40^\circ$ clockwise around point $A$ to get $\triangle ADE$. If $AC \perp DE$, what is the measure of $\angle C$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is rotated $40^\circ$ clockwise around point $A$ to get $\triangle ADE$\\
then $\angle CAE=40^\circ$, and $\angle C = \angle E$\\
Since $AC \perp DE$\\
then $\angle E=50^\circ$\\
thus $\angle C=\boxed{50^\circ}$


Question: As shown in the figure, $\triangle ABC$ is rotated $60^\circ$ clockwise around point $A$ to obtain $\triangle AED$. If the length of segment $AB=4$, what is the length of $BE$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is rotated $60^\circ$ clockwise around point A to get $\triangle AED$,\\
it follows that $AB=AE$ and $\angle BAE=60^\circ$.\\
Thus, $\triangle AEB$ is an equilateral triangle,\\
which means $BE=AB$.\\
Given $AB=4$,\\
it follows that $BE=\boxed{4}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=4$, $BC=3$. $\triangle ABC$ is rotated $90^\circ$ counterclockwise around point $C$ to obtain $\triangle DEC$. $P$ is a moving point on line segment $DE$. Taking point $P$ as the center and the length of $PD$ as the radius, a circle $\odot P$ is drawn. When $\odot P$ is tangent to the sides of $\triangle ABC$, what is the radius of $\odot P$?

Solution:
\textbf{Solution:}

(1) When $\odot P$ is tangent to AC at point F, as shown in the figure, connect PF, then PF $\perp$ AC,\\
Let DP=PF=x,\\
$\because$ $\angle ACB=90^\circ$, AC=4, BC=3,\\
$\therefore$ AB=$\sqrt{AC^2+BC^2}=5$,\\
$\because$ $\triangle ABC$ is rotated 90$^\circ$ counterclockwise around point C to get $\triangle DEC$,\\
$\therefore$ $\triangle DCE \cong \triangle ACB$,\\
$\therefore$ DE=AB=5, PE=DE−DP=5−x, CE=CB=3, DC=AC=4,\\
$\because$ $\angle ACB=90^\circ$, PF$\perp$AC,\\
$\therefore$ PF$\parallel$DC,\\
$\therefore$ $\triangle EPF \sim \triangle EDC$,\\
$\therefore$ $\frac{EP}{ED}=\frac{PF}{DC}$, that is, $\frac{5−x}{5}=\frac{x}{4}$, solving for x yields $x=\boxed{\frac{20}{9}}$, i.e., the radius is $\frac{20}{9}$;\\
(2) When $\odot P$ is tangent to AB at point F, as shown in the figure, BF$\perp$AB,\\
$\because$ $\angle A=\angle BDF$, $\angle ABC=\angle DBF$,\\
$\therefore$ $\triangle ABC \sim \triangle DBF$,\\
$\therefore$ $\frac{AB}{DB}=\frac{AC}{DF}$, that is, $\frac{5}{7}=\frac{4}{DF}$, solving for DF yields DF=$\frac{28}{5}$,\\
$\therefore$ PD=PF=$\frac{1}{2}$DF=$\boxed{\frac{14}{5}}$, i.e., the radius is $\frac{14}{5}$.


Question: As shown in the figure, $\triangle ABC$ and $\triangle ADE$ are both equilateral triangles, where point $E$ is the incenter of $\triangle ABC$. Taking $E$ as the center and the length of $DE$ as the radius, an arc is drawn to intersect $BC$ at point $B$. Then, the arc $DB$ is rotated $60^\circ$ counterclockwise around point $A$ to the position of arc $EC$. Given $AB=1$, what is the area of the shaded region in the figure?

Solution:
\textbf{Solution:}

Connect BD and CE as shown in the diagram.\\
Due to the properties of rotation: BD=CE, $\overset{\frown}{BD}=\overset{\frown}{CE}$\\
$\therefore$ The area of the arc with BD as its chord is equal to the area of the arc with CE as its chord.\\
$\because$ $\triangle ABC$ is an equilateral triangle, and point $E$ is the incenter of $\triangle ABC$\\
$\therefore$ $\angle$BAE=$\angle$CAE= $\frac{1}{2}$ $\angle$BAC=$30^\circ$, AE=CE\\
$\therefore$ $\angle$ECA=$\angle$CAE=$30^\circ$\\
$\therefore$ $\angle$AEC=$180^\circ$−($\angle$ECA+$\angle$CAE)=$120^\circ$\\
$\because$ $\triangle ADE$ is an equilateral triangle\\
$\therefore$ AE=DE, $\angle$AED=$60^\circ$\\
$\therefore$ $\angle$AED+$\angle$AEC=$180^\circ$, $\angle$BAE+$\angle$AED=$90^\circ$\\
That is, DE and CE are collinear, and DE=CE=AE, AB$\perp$DE, let the foot of the perpendicular be F\\
$\therefore$ Point F is the midpoint of AB\\
$\therefore$ $AF=BF=\frac{1}{2}AB=\frac{1}{2}$\\
$\therefore$ $AE=AF÷\cos30^\circ=\frac{\sqrt{3}}{3}$\\
$\therefore$ $CD=DE+CD=\frac{2\sqrt{3}}{3}$\\
$\therefore$ ${S}_{\triangle DBC}=\frac{1}{2}CD·BF=\frac{1}{2}\times \frac{2\sqrt{3}}{3}\times \frac{1}{2}=\frac{\sqrt{3}}{6}$\\
Therefore, the area of the shaded part is $\boxed{\frac{\sqrt{3}}{6}}$


Question: As shown in the diagram, point B lies on the hyperbolic function $y=\frac{k}{x}$ (where $k<0$, $x<0$), and point A lies on the negative half of the x-axis. Connect points A and B, and rotate the line segment AB $90^\circ$ counterclockwise around point A. The corresponding point to B after the rotation is point C. Suppose point C also lies on the graph of the hyperbolic function $y=\frac{k}{x}$. Given that the y-coordinates of points B and C are 4 and 1 respectively, what is the value of $k$?

Solution:
\textbf{Solution}: As illustrated, draw BM perpendicular to the x-axis at point M through point B; draw CN perpendicular to the x-axis at point N through point C. \\
Hence, $BM=4$, $CN=1$, $\angle AMB=\angle CNA=90^\circ$, \\
$\therefore \angle BAM+\angle ABM=90^\circ$, \\
Due to the property of rotation: $\angle BAC=90^\circ$, $AB=CA$, \\
$\therefore \angle BAM+\angle CAN=90^\circ$,\\
$\therefore \angle ABM=\angle CAN$,\\
In $\triangle ABM$ and $\triangle CAN$, we have $\left\{\begin{array}{l}\angle AMB=\angle CNA=90^\circ \\ \angle ABM=\angle CAN \\ AB=CA \end{array}\right.$,\\
$\therefore \triangle ABM\cong \triangle CAN(AAS)$, \\
$\therefore BM=AN=4$, $AM=CN=1$, \\
Let $OA=a(a>0)$, hence $OM=OA-AM=a-1$, $ON=AN+OA=4+a$, \\
$\therefore B(1-a, 4)$, $C(-4-a, 1)$, \\
Substituting points $B(1-a, 4)$, $C(-4-a, 1)$ into the inverse proportion function $y=\frac{k}{x}$ yields: $4(1-a)=k=-4-a$, \\
Solving, we get $a=\frac{8}{3}$, \\
Thus, $k=-4-a=-4-\frac{8}{3}=\boxed{-\frac{20}{3}}$.


Question: As shown in the figure, $A(-1, 6)$ is a point on the hyperbola $y=\frac{k}{x}$ $(x<0)$. Let $P$ be a point on the positive half of the $y$-axis. Rotating point $A$ counterclockwise by $90^\circ$ around point $P$ precisely lands it on another point $B$ on the hyperbola. What are the coordinates of point $B$?

Solution:
\textbf{Solution:} As shown in the figure, draw $AM \perp OM$ at point M through point A, and draw $BN \perp OM$ at point N through point B, \\
$\therefore \angle PMA = \angle BNP = 90^\circ$, \\
$\because$ Rotating point A $90^\circ$ counterclockwise around point P precisely lands on another point B on the hyperbola, \\
$\therefore PA=PB$, $\angle APB=90^\circ$, \\
$\therefore \angle APM + \angle BPN = \angle BPN + \angle PBN = 90^\circ$, \\
$\therefore \angle APM = \angle PBN$, \\
$\therefore \triangle APM \cong \triangle BNP$ (AAS), \\
$\therefore AM=PN$, $PM=BN$, \\
$\because$ Point A $(-1,6)$ lies on the hyperbola $y=\frac{k}{x}$ ($x>0$), \\
$\therefore k = -1 \times 6 = -6$, $AM=PN=1$, \\
Let point P be $(m,0)$, then $OP=m$, \\
$\therefore PM=BN=6-m$, $ON=OP-PN=m-1$, \\
$\therefore$ The coordinates of point B are $(m-6, m-1)$, \\
$\therefore k = (m-6) \cdot (m-1) = -6$, solving for $m$ gives $m=3$ or $m=4$, \\
$\therefore$ The coordinates of point B are $\boxed{(-3, 2)}$ or $\boxed{(-2, 3)}$.


Question: As shown, the $\triangle ODC$ is obtained by rotating $\triangle OAB$ clockwise by $30^\circ$ around point $O$. If point $D$ falls exactly on $AB$, and the measure of $\angle AOC$ is $100^\circ$, then what is the measure of $\angle B$?

Solution:
\textbf{Solution:} Since $\triangle ODC$ is rotated clockwise by $30^\circ$ around point O to get $\triangle OAB$,\\
it follows that AO=DO, and $\angle AOD=\angle BOC=30^\circ$,\\
hence $\angle A=\angle ADO=(180^\circ-30^\circ)\div 2=75^\circ$,\\
Given that $\angle AOC=100^\circ$,\\
it then follows that $\angle DOB=\angle AOC-\angle AOD-\angle BOC=100^\circ-30^\circ-30^\circ=40^\circ$,\\
Since $\angle ADO=\angle B+\angle DOB$,\\
it follows that $\angle B=\angle ADO-\angle DOB=75^\circ-40^\circ=\boxed{35^\circ}$.


Question: As shown in the figures, two triangles of different sizes are each chosen and overlaid as in Figure 1 (with the right-angled vertices coinciding). It is given that $ \angle A=45^\circ$, $ \angle CDE=30^\circ$, and $ CE:BC=1:\sqrt{6}$. After rotating $ \triangle DCE$ $45^\circ$ clockwise around point $ C$, line $ ED$ intersects line $ AC$ at point $ O$. What is the ratio of $ OA:OC$?

Solution:
\textbf{Solution:} As shown in the diagram, draw OF $\perp$ CE through O,\\
by the property of rotation, we have $\angle$OCE$=45^\circ$,\\
in $\triangle OEF$,\\
since $\angle$CDE$=30^\circ$,\\
thus $\angle$OEF$=60^\circ$,\\
therefore OF$=EF\tan\angle$OEF$= \sqrt{3}EF$,\\
thus OF$=CF= \sqrt{3}EF$, and $OC=\sqrt{OF^{2}+CF^{2}}=\sqrt{6}EF$,\\
therefore CE$=EF+CF= (\sqrt{3}+1)EF$,\\
therefore $\frac{CE}{OC}=\frac{(\sqrt{3}+1)EF}{\sqrt{6}EF}=\frac{\sqrt{3}+1}{\sqrt{6}}$,\\
since CE$\colon BC=1\colon \sqrt{6}$,\\
thus $\frac{BC}{OC}=\frac{BC}{CE} \cdot \frac{CE}{OC}=\sqrt{6} \times \frac{\sqrt{3}+1}{\sqrt{6}}=\sqrt{3}+1$,\\
thus $\frac{OA}{OC}=\frac{AC-OC}{OC}=\frac{AC}{OC}-1=\frac{BC}{OC}-1=\boxed{\sqrt{3}}$,\\
thus $OA\colon OC=\sqrt{3}$.


Question: As shown in the figure, $O$ is the center of symmetry of $\square ABCD$, point $E$ is on the side $BC$, $AD=7$, and $BE=3$. Rotate $\triangle ABE$ around point $O$ by $180^\circ$. Let the corresponding point of $E$ be $E'$. What is the value of $\frac{S_{\triangle AEE'}}{S_{\square ABCD}}$?

Solution:
\textbf{Solution:} Construct $\triangle CD{E}^{\prime }$ to be symmetrical to $\triangle ABE$ with respect to point O, and connect $EE^{\prime}$.

Since $\triangle CD{E}^{\prime }$ is symmetrical to $\triangle ABE$ with respect to point O,

it follows that $BE=DE^{\prime }=3$,

Given that $AD=7$,

it follows that $AE^{\prime }=4$,

Let the height of quadrilateral $ABCD$ be $h$,

then the height of $\triangle AE{E}^{\prime }$ is also equal to $h$,

Thus, $\frac{S_{\triangle AE{E}^{\prime }}}{S_{\square ABCD}}=\frac{\frac{1}{2}\times 4h}{7h}=\boxed{\frac{2}{7}}.$


Question: As shown in the figure, in the square $ABCD$, $\triangle ABC$ is rotated counterclockwise around point $A$ to obtain $\triangle AB'C'$, with $AB'$ and $AC'$ intersecting the diagonal $BD$ at points $E$ and $F$, respectively. If $AE=6$, what is the value of $EF\cdot ED$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a square,\\
it follows that $\angle BAC=\angle ADB=45^\circ$, \\
since triangle ABC is rotated counterclockwise around point A to obtain triangle $AB'C'$, \\
it follows that $\angle EAF=\angle BAC=45^\circ$, \\
since $\angle AEF=\angle DEA$, \\
it follows that $\triangle AEF\sim \triangle DEA$\\
therefore $\frac{AE}{DE}=\frac{EF}{AE}$, \\
therefore $EF\cdot ED=AE^{2}$, \\
since $AE=6$, \\
therefore, the value of $EF\cdot ED$ is $\boxed{36}$;


Question: As shown in the figure, rectangle $ABCD$ is folded along diagonal $BD$, and point $C$ lands on point $E$. Connect $CE$ and intersect side $AD$ at point $F$. If $DF=1$ and $BC=4$, what is the length of $AE$?

Solution:
\textbf{Solution:} Based on the property of folding, we have $CE\perp BD$,\\
$\therefore \angle ADB + \angle DFC = 90^\circ$\\
$\because$ Quadrilateral $ABCD$ is a rectangle,\\
$\therefore \angle ADC = \angle BCD = 90^\circ$,\\
$\therefore \angle ADB + \angle CDB = 90^\circ$,\\
$\therefore \angle CDB = \angle DFC$,\\
$\because \angle CDF = \angle BCD = 90^\circ$,\\
$\therefore \triangle CDF \sim \triangle BCD$,\\
$\therefore \frac{FD}{CD} = \frac{CD}{BC}$, i.e., $\frac{1}{CD} = \frac{CD}{4}$\\
$\therefore CD = 2$ (discarding the negative value)\\
$\because$ Quadrilateral $ABCD$ is a rectangle, $\triangle BED$ is obtained by folding $\triangle BCD$,\\
$\therefore BC = BE = AD$, $\angle CBD = \angle DBE$, $AD \parallel BC$,\\
$\therefore \angle ADB = \angle DBC$,\\
$\therefore \angle ADB = \angle DBE$,\\
$\therefore \angle EAD = \angle ADB = \angle DBC$,\\
$\therefore AE \parallel BD$,\\
$\therefore CE \perp AE$,\\
In $\triangle DCB$, $\tan \angle DBC = \frac{CD}{BC} = \frac{2}{4} = \frac{1}{2}$,\\
$\because \angle EAD = \angle DBC$\\
$\therefore \tan \angle EAD = \frac{1}{2}$\\
In $\triangle AEF$, $EF = \frac{1}{2}AE$\\
By the Pythagorean theorem, $AE^2 = AF^2 - EF^2$\\
i.e., $AE^2 = (AD - FD)^2 - \left(\frac{1}{2}AE\right)^2$\\
$\therefore AE^2 = (4 - 1)^2 - \frac{1}{4}AE^2$\\
Solving yields: $AE = \boxed{\frac{6\sqrt{5}}{5}}$.


Question: As shown in the figure, quadrilateral $ABCD$ is a rectangular piece of paper, with $AB=4$. Point $G$ is the midpoint of $CD$. First, fold the rectangular paper $ABCD$ along a straight line passing through point $B$ such that point $A$ lands on point $F$ on $BC$, with the fold line being $BE$. Then, unfold the paper. Next, fold the rectangular paper $ABCD$ along $BG$ so that point $C$ precisely lands on point $H$ on $BE$, with the fold line being $BG$. After unfolding the paper, connect $EF$, $HG$, and $BG$. What is the length of $BC$?

Solution:
\textbf{Solution:} As shown in the figure, connect $EG$. Due to the properties of folding, it is known that quadrilateral $ABFE$ is a square and quadrilateral $CDEF$ is a rectangle.\\
$\therefore$AB=AE=4, $BH=BC$, $\angle BHG=\angle BCD=90^\circ$, $CG=HG$.\\
$\because$ $AB=4$,\\
$\therefore$ $BE=\sqrt{AB^2+AE^2}=\sqrt{4^2+4^2}=4\sqrt{2}$.\\
Also, $\because$ point $G$ is the midpoint of $CD$,\\
$\therefore$ $CG=HG=DG=2$.\\
In $Rt\ \triangle EHG$ and $Rt\ \triangle EDG$,\\
$\left\{\begin{array}{l}HG=DG,\\ EG=EG,\end{array}\right.$\\
$\therefore$ $Rt\ \triangle EHG\cong Rt\ \triangle EDG$ (HL).\\
$\therefore$ $EH=ED=FC$.\\
Let $FC=x$, then $BH=BC=4x$.\\
$\therefore$ $BE=4+2x$. By $4+2x=4\sqrt{2}$,\\
we solve to get $x=2\sqrt{2}-2$.\\
$\therefore$ $BC=\boxed{2\sqrt{2}+2}$.


Question: As shown in the diagram, a right-angled triangle paper has its two right-angled sides measuring $6$ and $8$, respectively. Now, it is folded such that point $A$ coincides with point $B$, with the crease being $DE$. What is the value of $DE$?

Solution:
Solution: According to the properties of the fold transformation, we know that $\triangle DBE \cong \triangle DAE$, $\therefore$AD=BD,\\
$\because$ the lengths of the two right-angled sides of the rectangular paper are 6 and 8, respectively, $\therefore$AB=10, $\tan A = \frac{3}{4}$,\\
In $\triangle ADE$, $DE = \tan A \times AD = \boxed{\frac{15}{4}}$.


Question: As shown in the figure, in the isosceles $\triangle ABC$, with $AB=AC=5$ and $BC=8$, point B is on the y-axis, $BC\parallel x$-axis. The graph of the inverse proportional function $y=\frac{k}{x}$ ($k>0, x>0$) passes through point A and intersects $BC$ at point D. If $AB=BD$, what is the value of $k$?

Solution:
\textbf{Solution:} Given that $AB=AC=5$, $BC=8$, \\
$\therefore BE=\frac{1}{2}BC=4$, \\
$\therefore AE=\sqrt{AB^{2}-BE^{2}}=\sqrt{5^{2}-4^{2}}=3$, \\
Let $OB=m$, \\
As $AB=AC=5$, $AB=BD$, \\
$\therefore$ Point A's coordinates are $(4, m+3)$, and Point D's coordinates are $(5, m)$, \\
Since the inverse proportion function $y=\frac{k}{x}$ $(k>0, x>0)$ passes through point A and intersects BC at point D, \\
$\therefore \left\{\begin{array}{l}m+3=\frac{k}{4}\\ m=\frac{k}{5}\end{array}\right.$, solving these equations yields: $m=12$, \\
$\therefore k=\boxed{60}$.