Question: As shown in the figure, $ \angle DAC = 2x$, $ \angle ACB = 4x$, $ \angle ABC = 3x$, and $AD = BC$, then what is the value of $ \angle BAD$?

Solution:
\textbf{Solution:} Extend BC to point E such that CE=BD,\\
$\therefore$BD+CD=CE+CD,\\
$\therefore$BC=DE,\\
$\because$AD=BC,\\
$\therefore$AD=DE,\\
$\therefore \angle E=\angle DAE$,\\
$\because \angle ACD=4x$,\\
$\therefore \angle CAE+\angle E=4x$,\\
$\therefore \angle E=4x-\angle CAE=\angle DAE$,\\
$\therefore 4x-\angle CAE =2x+\angle CAE$,\\
$\therefore \angle CAE =x$, $\angle DAE =\angle E =3x$,\\
$\because \angle ABC =3x$,\\
$\therefore \angle ABC =\angle E$,\\
$\therefore AB=AE$,\\
$\therefore \triangle ABD \cong \triangle AEC$,\\
$\therefore \angle ADB=\angle ACE$,\\
$\therefore \angle ADC=\angle ACD=4x$,\\
$\because \angle ADC+\angle ACD+\angle DAC =180^\circ$,\\
$\therefore 4x+4x+2x=180^\circ$,\\
Solving for: $x=18^\circ$,\\
$\therefore \angle BAD= \boxed{18^\circ}$.


Question: As shown in the figure, $\angle 2 : \angle 3 : \angle 4 = 3 : 9 : 7$, then what is the degree measure of $\angle 1$?

Solution:
\textbf{Solution:} Given that $\angle 2 : \angle 3 : \angle 4 = 3 : 9 : 7$,\\
let $\angle 2 = 3\beta$, $\angle 3 = 9\beta$, and $\angle 4 = 7\beta$,\\
since $\angle 2 + \angle 3 = 180^\circ$,\\
thus $3\beta + 9\beta = 180^\circ$,\\
therefore $\beta = 15^\circ$,\\
therefore $\angle 2 = 45^\circ$ and $\angle 4 = 105^\circ$,\\
therefore $\angle 1 = 180^\circ - \angle 2 - \angle 4 = \boxed{30^\circ}$.


Question: As shown in the figure, $ \triangle ABC \cong \triangle ADE$, where $AB=8$, $AC=5$, $BC=6$. What is the length of $CD$?

Solution:
\textbf{Solution:} Since $\triangle ABC \cong \triangle ADE$, AB=8, AC=5, BC=6,\\
thus AD=AB=8,\\
therefore CD=AD-AC=8-5=\boxed{3}.


Question: As shown in the figure, in $\triangle ABC$, $\angle C = 90^\circ$, $AD$ bisects $\angle BAC$ and intersects $BC$ at point $D$. If $BD : CD = 3 : 2$ and the distance from point $D$ to $AB$ is 8, what is the length of $BC$?

Solution:
\textbf{Solution}: Draw $DE\perp AB$ at point $E$ through point $D$,\\
since $AD$ bisects $\angle BAC$, $\angle C=90^\circ$, and $DE\perp AB$,\\
it follows that $CD=DE=8$,\\
also, since $BD:CD=3:2$,\\
it follows that $BD=12$,\\
hence, $BC=BD+DC=\boxed{20}$.


Question: In the book "Gougu Juyu," the Qing Dynasty mathematician Mei Wending proved the Pythagorean theorem by using four congruent right triangles to piece together square ABCD, as shown in the figure. By connecting CE, if $CE=5$ and $BE=4$, what is the side length of square ABCD?

Solution:
As shown in the figure:

From the four congruent right triangles, it is known that $BE=CF=4$,  and $AE=BF$,

By Pythagorean theorem, $EF=\sqrt{CE^{2}-CF^{2}}=\sqrt{5^{2}-4^{2}}=3$,

$\therefore BF=BE-EF=4-3=1$,

By Pythagorean theorem, $AB=\sqrt{AE^{2}+BE^{2}}=\sqrt{1^{2}+4^{2}}=\boxed{\sqrt{17}}$.


Question: As shown in the figure, it is known that $\angle A=\angle B=90^\circ$, $AB=6$, and E, F are moving points on segment $AB$ and ray $BD$ respectively, with $BF=2BE$. Point G is on ray $AC$, and $EG$ is connected. If $\triangle AEG$ is congruent to $\triangle BEF$, what is the length of segment $AG$?

Solution:
\textbf{Solution:}
(1) As shown in the figure:\\
When $\triangle GAE \cong \triangle EBF$, we have: $AG=BE$, $AE=BF$\\
Since $BF=2BE$,\\
it follows that $AE=2BE$,\\
Since $AB=AE+BE=3BE=6$,\\
it follows that $BE=2$,\\
Hence $AG=BE=\boxed{2}$;\\
(2) When $\triangle GAE \cong \triangle FBE$, we have: $AE=BE$, $AG=BF$\\
Since $AB=AE+BE=2BE=6$,\\
it follows that $BE=3$,\\
Since $BF=2BE$,\\
it follows that $AG=2BE=\boxed{6}$;\\


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle A > \angle B$, the $\triangle ABC$ is first folded along the crease $CE$ so that point $A$ can fall on $BC$. After flattening, $\angle B$ is folded along the crease $GF$ making point $B$ coincide with point $A$, where $FG$ intersects sides $BC$ and $AB$ at points $F$ and $G$, respectively. $CD$ is the height from the hypotenuse. If $\angle DCE = \angle B$, then what is $\frac{BF}{CE}$?

Solution:
\textbf{Solution:} Connect $AF$, \\
$\because$ when $\triangle ABC$ is folded for the first time along crease $CE$, making point $A$ fall onto $BC$, \\
$\therefore$ $\angle ACE=\angle BCE=45^\circ$, \\
$\because$ by folding $\angle B$ along crease $GF$, making point $B$ coincide with point $A$, \\
$\therefore$ $\angle B=\angle FAB$, $FA=FB$, \\
$\because$ $\angle ACD+\angle DCB=\angle B+\angle DCB=90^\circ$, \\
$\therefore$ $\angle ACD=\angle B$, \\
$\because$ $\angle DCE=\angle B$, \\
$\therefore$ $\angle ACD=\angle DCE=\angle B=\frac{1}{2}\angle ACE=22.5^\circ$, \\
$\therefore$ $\angle AFC=\angle B+\angle FAB=2\angle B=45^\circ$, \\
$\therefore$ $\triangle$ AFC is an isosceles right triangle, \\
Let $AC=CF=a$, then $AF=\sqrt{a^2+a^2}=\sqrt{2}a$, \\
$\because$ $\angle CAB=90^\circ-\angle B=67.5^\circ$, $\angle CEA=\angle B+\angle BCE=67.5^\circ$, \\
that is $\angle CAE=\angle CEA$, \\
$\therefore$ CA=CE, \\
$\therefore$ $\frac{BF}{CE}=\frac{AF}{CA}=\frac{\sqrt{2}a}{a}=\boxed{\sqrt{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=20\,cm$, $DE$ is perpendicular bisector of $AB$ with the foot of the perpendicular as $E$, and intersects $AC$ at $D$. If $BC=15\,cm$, then what is the perimeter of $\triangle DBC$?

Solution:
\textbf{Solution:} Since $DE$ bisects $AB$ perpendicularly,\\
it follows that $AD=BD$,\\
hence the perimeter of $\triangle DBC = BD+DC+BC = AD+DC+BC = AC+BC$,\\
since $AB=AC=20\,cm$, $BC=15\,cm$,\\
therefore, the perimeter of $\triangle DBC =AC+BC=20+15=\boxed{35}\,cm$.


Question: In $\triangle ABC$, $\angle B=40^\circ$, $D$ is a point on side $BC$. The triangle is folded along $AD$ such that $AC$ falls on $AB$ and point $C$ coincides with point $E$. If $\triangle BDE$ is a right-angled triangle, what is the measure of $\angle C$?

Solution:
\textbf{Solution:} When $\angle BED=90^\circ$, $\angle C=\angle AED=90^\circ$.\\
When $\angle BDE=90^\circ$, $\angle C=\angle AED=\angle B+\angle BDE=130^\circ$.\\
In summary, the measure of $\angle C$ is either $90^\circ$ or $130^\circ$.\\
Therefore, the answer is: $\boxed{90^\circ}$ or $130^\circ$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $AB=AC$, $AD$ bisects $\angle BAC$, and point $E$ is the midpoint of $AB$. If $AC=6$, what is the length of $DE$?

Solution:
\textbf{Solution:} \\
Since $AB=AC$ and $AD$ bisects $\angle BAC$, \\
it follows that $D$ is the midpoint of $BC$, \\
Since $E$ is the midpoint of $AB$, \\
it follows that $DE$ is the median of the triangle, \\
Given $AC=6$, \\
it follows that $DE=\boxed{3}$.


Question: In right triangle $ABC$, $\angle B=90^\circ$, $AB=4$, $BC=8$. Points $D$ and $E$ are located on sides $AC$ and $BC$, respectively. If $\triangle ABC$ is folded along $DE$ such that points $A$ and $C$ coincide, what is the length of $EC$?

Solution:
\textbf{Solution:} Let $EC = x$, then $BE = 8 - x$,\\
since folding $\triangle ABC$ along $DE$ causes point $A$ to coincide with point $C$,\\
it follows that $AE = EC = x$,\\
In $\text{Rt}\triangle ABE$, we have $AB^2 + BE^2 = AE^2$,\\
$4^2 + (8 - x)^2 = x^2$,\\
solving for $x$ yields $x = \boxed{5}$,\\
therefore, $EC = 5$.


Question: In quadrilateral $ABCD$, $AD\parallel BC$, $DE\perp BC$ with the foot of the perpendicular being point $E$, connect $AC$ intersecting $DE$ at point $F$, point $G$ is the midpoint of $AF$, and $\angle ACD=2\angle ACB$. If $DG=4$ and $EC=1$, what is the length of $DE$?

Solution:
\textbf{Solution:} Given that $AD \parallel BC$ and $DE \perp BC$,\\
it follows that $AD \perp DE$, and $\angle CAD = \angle ACB$,\\
Given that $\angle ACD = 2\angle ACB$,\\
it follows that $\angle ACD = 2\angle CAD$,\\
Given that $AD \perp DE$ and $G$ is the midpoint of $AF$,\\
it follows that $GA = GD$,\\
therefore, $\angle GAD = \angle GDA$,\\
thus, $\angle DGC = \angle DCG$,\\
therefore, $DC = DG = 4$,\\
thus, $DE= \sqrt{CD^{2}-EC^{2}} = \boxed{\sqrt{15}}$.


Question: In the trapezoid $ABCD$ shown in the figure, $\angle ABC = \angle BCD$, $AD \parallel BC$, and $BD$ bisects $\angle ABC$. If $AD = 3$ and $BC = 7$, what is the length of $BD$?

Solution:
\textbf{Solution:} As shown in the figure, draw AE$\perp$BC at point D, and draw DF$\perp$BC at point F, then the quadrilateral AEFD is a rectangle.\\
In trapezoid ABCD, $\angle ABC=\angle BCD$, AD$\parallel$BC,\\
$\therefore$AB=CD, $\angle ADB=\angle DBC$,\\
$\because$BD bisects $\angle ABC$,\\
$\therefore$$\angle ABD=\angle DBC=\angle ADB$,\\
$\therefore$AB=AD=3,\\
$\because$AD$\parallel$CB, AE$\perp$CB, DF$\perp$BC,\\
$\therefore$AD=DF,\\
$\therefore$right $\triangle AEB\cong$ right $\triangle DFC$ (by HL),\\
$\therefore$BE=CF,\\
$\because$quadrilateral AEFD is a rectangle,\\
$\therefore$AD=EF=3,\\
$\therefore$BE=CF= $\frac{1}{2}$(7−3)=2,\\
$\therefore$AE=DF= $\sqrt{AB^{2}-BE^{2}}=\sqrt{3^{2}-2^{2}}=\sqrt{5}$,\\
$\therefore$BD= $\sqrt{DF^{2}+BF^{2}}=\sqrt{(\sqrt{5})^{2}+5^{2}}=\boxed{\sqrt{30}}$,


Question: As shown in the figure, it is known that $AB \parallel CF$, and E is the midpoint of $DF$. If $AB = 8\,cm$ and $CF = 5\,cm$, then how many centimeters is $BD$?

Solution:
\textbf{Solution:} Given that $AB\parallel CF$,\\
thus $\angle A=\angle ACF$, $\angle F=\angle ADE$,\\
since E is the midpoint of $DF$,\\
thus $DE=EF$,\\
in $\triangle ADE$ and $\triangle CFE$, $\left\{\begin{array}{l}
\angle A=\angle ECF\\
\angle ADE=\angle F\\
DE=EF
\end{array}\right.$,\\
therefore $\triangle ADE\cong \triangle CFE$ (by AAS),\\
thus $CF=AD=5cm$,\\
thus $BD=AB-AD=8-5=\boxed{3cm}$,


Question: As shown in the figure, in the rectangle $ABCD$, $AB=3$, $BC=3\sqrt{3}$, and $\angle CBD=30^\circ$. Point $M$ is a point on the ray $BD$ (not coinciding with points $B$, $D$). Line $AM$ is drawn, and through point $M$, $MN \perp AM$ intersects line $BC$ at point $N$. If $\triangle BMN$ is an isosceles triangle, what is the length of $BN$?

Solution:
\textbf{Solution:} Connect AN,\\
since quadrilateral ABCD is a rectangle, it follows that $\angle ABN=90^\circ$,\\
since $MN\perp AM$, it follows that $\angle AMN=90^\circ$,\\
hence $\angle BNM+\angle BAM=180^\circ$,\\
since $\angle BAM<90^\circ$, it follows that $\angle BNM>90^\circ$,\\
since $\triangle BMN$ is an isosceles triangle, there only exists the case where $BN=MN$,\\
in $Rt\triangle ABN$ and $Rt\triangle AMN$, $\left\{\begin{array}{l}AN=AN\\ BN=MN\end{array}\right.$,\\
hence $Rt\triangle ABN\cong Rt\triangle AMN\left(HL\right)$,\\
thus $AB=AM$,\\
since $\angle ABN=90^\circ$, $\angle CBD=30^\circ$, hence $\angle ABM=60^\circ$,\\
thus $\triangle ABM$ is an equilateral triangle, hence $\angle BAM=60^\circ$,\\
thus $\angle BAN=\angle MAN=\frac{1}{2}\angle BAM=30^\circ$, hence $AN=2BN$,\\
since $AB^2+BN^2=AN^2$, and $AB=3$,\\
hence $3^2+BN^2=(2BN)^2$, hence $BN=\boxed{\sqrt{3}}$.


Question: As shown in the diagram, points A, D, and C are on the same straight line, and $\triangle ABC \cong \triangle DBE$. If $\angle A = 60^\circ, \angle C = 30^\circ$, then what is the measure of $\angle DBC$?

Solution:
\textbf{Solution:} Given $\because$ $\triangle ABC\cong \triangle DBE$ and $\angle A=60^\circ$,\\
$\therefore$ $AB=BD$,\\
$\therefore$ $\angle ADB=\angle A=60^\circ$,\\
$\therefore$ $\angle ABD=60^\circ$,\\
$\because$ $\angle A=60^\circ$ and $\angle C=30^\circ$,\\
$\therefore$ $\angle ABC=90^\circ$,\\
$\therefore$ $\angle DBC=\angle ABC-\angle ABD=\boxed{30^\circ}$.


Question: As shown in the figure, line segment $AB=8\,\text{cm}$, ray $AN \perp AB$ with the foot of the perpendicular at point A. Point C is a moving point on the ray, and isosceles right triangles are formed with $AC$ and $BC$ as the legs, resulting in $\triangle ACD$ and $\triangle BCE$ respectively. $DE$ is connected and intersects ray $AN$ at point M. What is the length of $CM$?

Solution:
\textbf{Solution:} Construct $EF \perp AN$ at $F$, as shown in the figure.\\
Since $AN \perp AB$, and $\triangle BCE$ and $\triangle ACD$ are isosceles right triangles,\\
thus $\angle BAC = \angle BCE = \angle ACD =\angle CFE = 90^\circ$, $BC = CE$, $AC = CD$.\\
Therefore, $\angle ABC + \angle ACB = 90^\circ$, $\angle FCE + \angle ACB = 90^\circ$,\\
thus $\angle ABC = \angle FCE$,\\
In $\triangle ABC$ and $\triangle FCE$\\
$\left\{\begin{array}{l} \angle BAC = \angle CFE \\ \angle ABC = \angle FCE \\ BC = CE \end{array}\right.$\\
Therefore, $\triangle ABC \cong \triangle FCE$\\
Therefore, $AB = FC = 8\,cm$, $AC = FE$\\
Therefore, $CD = FE$\\
In $\triangle DCM$ and $\triangle EFM$\\
$\left\{\begin{array}{l} \angle DMC = \angle EMF \\ \angle DCM = \angle EFM = 90^\circ \\ CD =
FE \end{array}\right.$\\
Therefore, $\triangle DCM \cong \triangle EFM$\\
Therefore, $CM = FM = \frac{1}{2}FC = \boxed{4cm}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ bisects $\angle BAC$, $AE \perp BC$ at point E, $\angle B = 40^\circ$, $\angle DAE = 15^\circ$, what is the measure of $\angle C$?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $AE\perp BC$, \\
it follows that $\angle AEB=90^\circ$, \\
since $\angle B=40^\circ$, \\
it follows that $\angle BAE = 90^\circ - 40^\circ = 50^\circ$, \\
since $\angle DAE = 15^\circ$, \\
it follows that $\angle BAD = 50^\circ - 15^\circ = 35^\circ$, \\
since $AD$ bisects $\angle BAC$, \\
it follows that $\angle BAC = 2\angle BAD = 70^\circ$, \\
thus, $\angle C = 180^\circ - \angle B - \angle BAC = \boxed{70^\circ}$;


Question: As shown in the figure is an axisymmetric figure. If $\angle A=36^\circ, \angle C'=24^\circ$, then what is the degree measure of $\angle B$?

Solution:
\textbf{Solution:} Since the figure is an axisymmetric figure, \\
therefore, $\angle C = \angle C' = 24^\circ$,\\
since $\angle A = 36^\circ$,\\
therefore, $\angle B = 180^\circ - \angle A - \angle C = \boxed{120^\circ}$.


Question: As shown in the figure, point C is on $BE$, $\angle B = \angle E = \angle ACF$, $AC = CF$, $AB = 4$, and $EF = 6$. What is the length of $BE$?

Solution:
\textbf{Solution:} Since $\angle B+\angle BAC=\angle ACE=\angle FCE+\angle ACF$,\\
and since $\angle B=\angle ACF$,\\
it follows that $\angle BAC=\angle FCE$,\\
Since $\angle B=\angle E$, $AC=CF$,\\
it follows that $\triangle ABC\cong \triangle CEF$,\\
therefore $AB=CE$, $BC=EF$,\\
Since $AB=4$, $EF=6$,\\
it follows that $BE=BC+CE=6+4=\boxed{10}$.


Question: As shown in the figure, in $ \triangle MPN $, H is the intersection of the altitudes $ MQ $ and $ NR $, and $ MQ = NQ $. Given $ PQ = 3 $ and $ NQ = 7 $, what is the length of $ MH $?

Solution:
\textbf{Solution:} Let $H$ be the intersection of altitudes $MQ$ and $NR$,\\
therefore $NR\perp PM$, and $MQ\perp PN$,\\
therefore $\angle PMQ + \angle MHR = 90^\circ$, and $\angle HNQ + \angle QHN = 90^\circ$.\\
Since $\angle MHR = \angle QHN$,\\
therefore $\angle PMQ = \angle HNQ$.\\
Also, since $MQ = NQ = 7$ and $\angle PQM = \angle HQN = 90^\circ$,\\
therefore $\triangle PMQ \cong \triangle HNQ$ (by ASA),\\
thus $PQ = HQ = 3$,\\
hence $MH = MQ - HQ = 7 - 3 = \boxed{4}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ is the bisector of $\angle BAC$, and $E$ is a point on $AD$, with $EF \perp BC$ at point $F$. If $\angle C = 35^\circ$ and $\angle DEF = 15^\circ$, what is the measure of $\angle B$?

Solution:
\textbf{Solution:} Since $EF\perp BC$ and $\angle DEF=15^\circ$, we have $\angle ADB=90^\circ-15^\circ=75^\circ$. \\
Since $\angle C=35^\circ$, it follows that $\angle CAD=75^\circ-35^\circ=40^\circ$. \\
Since $AD$ is the bisector of $\angle BAC$, we have $\angle BAC=2\angle CAD=80^\circ$, hence $\angle B=180^\circ-\angle BAC-\angle C=180^\circ-80^\circ-35^\circ=\boxed{65^\circ}$.


Question: As shown in the figure, in $ \triangle ABC$, D is a point on $BC$, $AD=BD$, $\angle B=40^\circ$, and $\triangle ADE$ is symmetrical to $\triangle ADB$ with respect to $AD$. What is the measure of $\angle CDE$?

Solution:
\textbf{Solution:} Since $AD=BD$ and $\angle B=40^\circ$, \\
it follows that $\angle B=40^\circ=\angle BAD$, \\
therefore, $\angle ADB=180^\circ-\left(\angle BAD+\angle B\right)=100^\circ$ and $\angle ADC=\angle B+\angle BAD=80^\circ$, \\
since $\triangle ADE$ is symmetric to $\triangle ADB$ with respect to $AD$, \\
it follows that $\angle ADE=\angle ADB=100^\circ$, \\
thus, $\angle CDE=\angle ADE-\angle ADC=100^\circ-80^\circ=\boxed{20^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B = \angle C$. Point $D$ is on side $AB$. A line is drawn through $D$, with $DE \perp BC$ intersecting $BC$ at point $E$ and extending to intersect the extended line of $CA$ at point $F$. Given $AC = 5$ and $BD = 3$, what is the length of $AF$?

Solution:
\textbf{Solution:} Given that $DE\perp BC$,\\
it follows that $\angle DEB=\angle FEC=90^\circ$,\\
thus $\angle B+\angle BDE=90^\circ$, $\angle C+\angle F=90^\circ$,\\
since $\angle B=\angle C$,\\
it results in $\angle F=\angle BDE$,\\
and since $\angle FDA=\angle BDE$,\\
we have $\angle FDA=\angle F$,\\
hence $AF=AD$,\\
as $\angle B=\angle C$ and $AC=5$,\\
it implies $AB=AC=5$,\\
given $BD=3$,\\
it follows $AD=AB−BD=5−3=2$,\\
thus $AF=AD=\boxed{2}$.


Question: As shown in the figure, in the square $ABCD$ with a side length of 2, points $E$ and $F$ are the midpoints of sides $AB$ and $BC$, respectively. Connect $EC$ and $FD$. Points $G$ and $H$ are the midpoints of $EC$ and $FD$, respectively. Connect $GH$. What is the length of $GH$?

Solution:
\textbf{Solution:} As illustrated, let $DF$ and $CE$ intersect at $O$,\\
$\because$ Quadrilateral $ABCD$ is a square,\\
$\therefore \angle B=\angle DCF=90^\circ$, $BC=CD=AB=2$,\\
$\because$ Points $E$ and $F$ are the midpoints of sides $AB$ and $BC$ respectively,\\
$\therefore$ $BE=CF=1$,\\
$\therefore \triangle CBE \cong \triangle DCF$ (SAS),\\
$\therefore CE=DF$, $\angle BCE=\angle CDF$,\\
$\because \angle CDF+\angle CFD=90^\circ$,\\
$\therefore \angle BCE+\angle CFD=90^\circ$,\\
$\therefore \angle COF=90^\circ$,\\
$\therefore$ $DF \perp CE$,\\
$\therefore CE=DF=\sqrt{2^2+1^2}=\sqrt{5}$,\\
$\because$ Points $G$ and $H$ are the midpoints of $EC$ and $FD$ respectively,\\
$\therefore$ $CG=FH=\frac{\sqrt{5}}{2}$,\\
$\because \angle DCF=90^\circ$, $CO \perp DF$,\\
$\therefore CF^2=OF\cdot DF$,\\
$\therefore OF=\frac{CF^2}{DF}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$,\\
$\therefore$ $OH=\frac{3\sqrt{5}}{10}$, $OD=\frac{4\sqrt{5}}{5}$,\\
$\because OC^2=OF\cdot OD$,\\
$\therefore OC=\sqrt{\frac{\sqrt{5}}{5}\times \frac{4\sqrt{5}}{5}}=\frac{2\sqrt{5}}{5}$,\\
$\therefore$ $OG=CG-OC=\frac{\sqrt{5}}{2}-\frac{2\sqrt{5}}{5}=\frac{\sqrt{5}}{10}$,\\
$\therefore$ $HG=\sqrt{OG^2+OH^2}=\sqrt{\left(\frac{\sqrt{5}}{10}\right)^2+\left(\frac{3\sqrt{5}}{10}\right)^2}=\boxed{\frac{\sqrt{2}}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $D$ is the midpoint of the hypotenuse $AB$, and $DE \perp AB$ intersects $AC$ at point $E$. If $AC=8$ and $BC=6$, then what is the length of $DE$?

Solution:
\textbf{Solution:} In right triangle $\triangle ABC$, we have
\[AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{8^{2}+6^{2}}=10,\]
since point D is the midpoint of AB,
\[AD=\frac{1}{2}AB=5;\]
since DE$\perp$AB,
\[\angle ADE=\angle C=90^\circ,\]
and since $\angle A=\angle A$,
\[\triangle ADE\sim \triangle ACB,\]
hence
\[\frac{DE}{BC}=\frac{AD}{AC},\]
which gives
\[\frac{DE}{6}=\frac{5}{8}.\]
Solving this, we find $DE=\boxed{\frac{15}{4}}$.


Question: As shown in the figure, in $\triangle ABC$, D and E are points on AB and AC, respectively. AF bisects $\angle BAC$ and intersects DE at point G and intersects BC at point F. If $\angle AED=\angle B$, and $AG: GF=3:2$, what is the ratio of $DE: BC$?

Solution:
\textbf{Solution:} Given that $\angle DAE = \angle CAB$ and $\angle AED = \angle B$,\\
it follows that $\triangle ADE \sim \triangle ACB$,\\
since $GA$ and $FA$ are the angle bisectors of $\triangle ADE$ and $\triangle ABC$ respectively,\\
therefore $\frac{DE}{BC} = \frac{AG}{AF}$ (the ratio of the corresponding angle bisectors is equal to the similarity ratio),\\
$AG:FG = 3:2$,\\
thus, $AG:AF = 3:5$,\\
hence, $DE:BC = \boxed{3:5}$.


Question: As shown in the figure, in right-angled triangle $ABC$ with $\angle C=90^\circ$, $D$ is the midpoint of $AB$, $AD=5$, $BC=8$, and $E$ is a moving point on line $BC$. After folding $\triangle BDE$ along line $ED$, point $B$ falls at point $F$. When $FD \perp BC$, what is the length of segment $BE$?

Solution:
\textbf{Solution:} Suppose when point F is below BC, DF intersects BC at point P, as shown in Figure 1:\\
$\because$D is the midpoint of AB,\\
$\therefore$AB $=$ 2AD $=$ 10,\\
$\because$$\angle$C $= 90^\circ$,\\
$\therefore$AC $=$ $\sqrt{AB^{2}-BC^{2}}$ $=$ $\sqrt{10^{2}-8^{2}}$ $=$ 6,\\
$\because$FD$\perp$BC, $\angle$C $= 90^\circ$,\\
$\therefore$FD$\parallel$AC, and D is the midpoint of AB,\\
$\therefore$$\triangle PBD\sim \triangle CBA$\\
$\therefore$$\frac{BD}{BA}$ $=$ $\frac{BP}{BC}$ $=$ $\frac{DP}{AC}$ $=$ $\frac{1}{2}$,\\
$\therefore$BP $=$ PC $=$ $\frac{1}{2}$BC $=$ 4, DP $=$ $\frac{1}{2}$AC $=$ 3,\\
$\because$$\triangle BDE$ is folded along line ED,\\
$\therefore$FD $=$ BD $=$ 5, FE $=$ BE,\\
$\therefore$FP $=$ FD - DP $=$ 5 - 3 $=$ 2,\\
In $\triangle FPE$, EF$^{2}$ $=$ FP$^{2}$ + PE$^{2}$,\\
$\therefore$BE$^{2}$ $=$ $2^{2}$ + $(4-BE)^{2}$,\\
Solving: BE $=$ $\boxed{\frac{5}{2}}$;\\
Suppose when point F is above BC, the extension of FD intersects BC at point P, as shown in Figure 2,\\
$\because$FD = BD = 5, DP $=$ $\frac{1}{2}$AC $=$ 3,\\
$\therefore$FP $=$ DP + FD $=$ 3 + 5 $=$ 8,\\
In $\triangle EFP$, EF$^{2}$ $=$ FP$^{2}$ + EP$^{2}$,\\
$\therefore$BE$^{2}$ $=$ 64 + $(BE-4)^{2}$,\\
Solving: BE $=$ 10;


Question: As shown in the figure, in $ \triangle ABC$, M and N are points on the sides AB and AC, respectively, where AM = $ \frac{1}{4}$AB and AN = $ \frac{1}{4}$AC. What is the ratio of the area of $ \triangle AMN$ to the area of the quadrilateral MBCN?

Solution:
\textbf{Solution:} Since $AM = \frac{1}{4}AB$ and $AN = \frac{1}{4}AC$, and $\angle MAN = \angle BAC$,

$\therefore \triangle AMN \sim \triangle ABC$,

$\therefore \frac{S_{\triangle AMN}}{S_{\triangle ABC}} = \left(\frac{AM}{AB}\right)^2 = \frac{1}{16}$,

$\therefore S_{\triangle AMN} = \frac{1}{16}S_{\triangle ABC}$,

$\therefore \frac{S_{\triangle AMN}}{S_{\text{quadrilateral }MBCN}} = \frac{S_{\triangle AMN}}{S_{\triangle ABC} - S_{\triangle AMN}} = \boxed{\frac{1}{15}}$.


Question: As shown in the figure, parallelogram $ABCD$ has diagonals $AC$ and $BD$ intersecting at point $O$. Point $F$ is the midpoint of $BC$, and connecting $DF$ intersects $AC$ at point $E$. What is the ratio of $DE:EF$?

Solution:
\textbf{Solution:} As shown in the figure, connect OF, \\
$\because$ Quadrilateral ABCD is a parallelogram,\\
$\therefore$ O is the midpoint of BD,\\
$\because$ F is the midpoint of BC,\\
$\therefore$ OF is the median of $\triangle$BCD,\\
$\therefore$ CD $=$ 2OF, and OF $\parallel$ CD,\\
$\therefore$ $\triangle$DEC $\sim$ $\triangle$FEO,\\
$\therefore$ DE$\colon$EF $=$ CD$\colon$OF $=$ \boxed{2\colon1}.


Question: As illustrated, the graph of the inverse proportion function $y=\frac{k}{x}$ (where $k>0$) intercepts in the first quadrant at points A(1, 6) and B(3, b). The line $AB$ intersects the x-axis at point C, and D is a point on the line segment $OA$. If $CA\cdot AD=AB\cdot AO$, and by connecting $CD$, let the areas of $\triangle ADC$ and $\triangle DOC$ be $S_{1}$ and $S_{2}$ respectively, then what is the value of $S_{1}-S_{2}$?

Solution:
\textbf{Solution:} Let $\because$ A(1, 6) is on the graph of the inverse proportion function $y=\frac{6}{x}$,\\
$\therefore$ k=6, the analytical expression of the inverse proportion function is: $y=\frac{6}{x}$,\\
$\because$ B(3, b) is on the graph of the inverse proportion function $y=\frac{6}{x}$,\\
$\therefore$ b=2, that is, B(3, 2).\\
Let the line AB be: $y=mx+n(m\ne 0)$,\\
$\therefore$
$\begin{cases}
m+n=6\\
3m+n=2
\end{cases}$, solving we get:
$\begin{cases}
m=-2\\
n=8
\end{cases}$,\\
$\therefore$ the analytical expression for line AB is: $y=-2x+8$.\\
$\therefore$ For $y=-2x+8$, when $y=0$, i.e., $-2x+8=0$,\\
solving we get: $x=4$,\\
$\therefore$ C(4, 0),\\
$\therefore$ ${S}_{\triangle AOC}=\frac{1}{2}OC\cdot y_{A}=\frac{1}{2}\times 4\times 6=12$,\\
$\because$ $CA\cdot AD=AB\cdot AO$,\\
$\therefore$ $\frac{CA}{AB}=\frac{AO}{AD}$,\\
Furthermore, $\because$ $\angle DAB=\angle OAC$,\\
$\therefore$ $\triangle DAB\sim \triangle OAC$,\\
$\therefore$ $\angle ADB=\angle AOC$,\\
$\therefore$ $DB\parallel OC$,\\
$\therefore$ $y_{D}=y_{B}=2$,\\
$\therefore$ ${S}_{\triangle DOC}=\frac{1}{2}OC\cdot y_{D}=\frac{1}{2}\times 4\times 2=4$,\\
$\therefore$ ${S}_{\triangle ADC}={S}_{\triangle AOC}-{S}_{\triangle DOC}=12-4=8$,\\
$\therefore$ ${S}_{1}-{S}_{2}={S}_{\triangle ADC}-{S}_{\triangle DOC}=8-4=\boxed{4}$.


Question: As shown in the diagram, it is known that $M$ and $N$ are two points on the side $BC$ of $\triangle ABC$, and satisfy $BM=MN=NC$. A line parallel to $AC$ intersects the extensions of $AB$, $AM$, and $AN$ at points $D$, $E$, and $F$, respectively. What is the value of $\frac{EF}{DE}$?

Solution:
\textbf{Solution:} Draw line $MG \parallel DF$ with point $G$ on $AB$, and draw line $NH \parallel DF$ with point $H$ on $AB$, where $NH$ intersects $AM$ at $I$,

then we have $MG \parallel DF \parallel NH \parallel AC$

$\because$ $GM \parallel NH$,

$\therefore$ $\triangle BMG \sim \triangle BNH$

$\therefore$ $\frac{MG}{NH} = \frac{BM}{BN}$

Furthermore, $\because$ $BM = \frac{1}{2}BN$,

$\therefore$ $\frac{MG}{NH} = \frac{BM}{BN} = \frac{1}{2}$

$\because$ $MG \parallel NH \parallel AC$, $MN = NC$

$\therefore$ $\frac{GH}{AH} = \frac{MN}{NC} = 1$

$\therefore$ $\frac{AH}{AG} = \frac{1}{2}$

$\because$ $MG \parallel NH$

$\therefore$ $\triangle AHI \sim \triangle AGM$

$\therefore$ $\frac{IH}{MG} = \frac{AH}{AG} = \frac{1}{2}$

Additionally, $\because$ $\frac{MG}{NH} = \frac{1}{2}$

$\therefore$ $\frac{IH}{NH} = \frac{IH}{MG} \cdot \frac{MG}{NH} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

$\therefore$ $\frac{IH}{IN} = \frac{1}{3}$

Also, $\because$ $DF \parallel NH$

$\therefore$ $\triangle AHI \sim \triangle ADE$, $\triangle ANI \sim \triangle AFE$,

$\therefore$ $\frac{IH}{ED} = \frac{AI}{AE} = \frac{IN}{EF}$

$\therefore$ $\frac{ED}{EF} = \frac{IH}{IN} = \frac{1}{3}$

$\therefore$ $\frac{EF}{DE} = \boxed{3}$


Question: As shown, both $\triangle ABC$ and $\triangle ADE$ are isosceles right-angled triangles, with $\angle BAC = \angle DAE = 90^\circ$, and $AB = 4$. Point $F$ is the midpoint of $DE$. If point $E$ is a moving point on line $BC$, and $BF$ is connected, what is the minimum value of $BF$?

Solution:
\textbf{Solution:} Since both $\triangle ABC$ and $\triangle ADE$ are isosceles right-angled triangles,\\
it follows that $\triangle ABC\sim \triangle ADE$,\\
hence $\angle ADE = \angle ABE$,\\
therefore, points A, D, B, E are concyclic,\\
since $\angle DAE = 90^\circ$,\\
it follows that $\angle DBE = 90^\circ$,\\
since F is the midpoint of DE,\\
it follows that BF = $\frac{1}{2}$DE,\\
therefore, when DE is at its minimum, the value of BF is also minimal,\\
since if point E is a moving point on line BC,\\
it follows that when AE$\perp$BC, AE is at its minimum, at this time, DE is also minimal,\\
since $\angle BAC = 90^\circ$, AB = 4 = AC,\\
it follows that BC = $4\sqrt{2}$,\\
therefore, AE = $\frac{AB \cdot AC}{BC} = \frac{4 \times 4}{4\sqrt{2}} = 2\sqrt{2}$,\\
therefore, DE = 4,\\
therefore, BF = $\boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, $DE \parallel BC$, $G$ is a point on $BC$, and line $AG$ intersects $DE$ at point $F$. It is known that $AF = 2$, $AG = 6$, and $EC = 5$. What is the length of $AC$?

Solution:
\textbf{Solution:} Since $DE \parallel BC$,\\
it follows that $\frac{AF}{FG} = \frac{AE}{EC}$, that is, $\frac{2}{4} = \frac{AE}{5}$,\\
therefore, $AE = \frac{5}{2}$,\\
hence, $AC = AE + EC = \frac{5}{2} + 5 = \boxed{\frac{15}{2}}$.


Question: As shown in the figure, there is a point $A(3, 0)$ on the x-axis. Point $D$ is the symmetric point of point $A$ with respect to the y-axis. Point $B$ is on the graph of the inverse proportional function $y=\frac{k}{x}$ $(x<0)$. When connecting $BD$ and intersecting the graph of the inverse proportional function at point $C$, if $OC\parallel AB$, and the area of $\triangle ABD$ is $24$, what is the value of $k$?

Solution:
\textbf{Solution:} Let $BE \perp x$-axis at point E, $CF \perp x$-axis at point F, \\
since point D is the symmetric point of point A with respect to the y-axis, \\
thus the coordinates of D are $(-3, 0)$, \\
thus $AD=6$, \\
since the area of $\triangle ABD = \frac{1}{2}BE \cdot AD = 24$, \\
thus $BE=8$, \\
since $OC \parallel AB$, \\
thus $\triangle DCO \sim \triangle DBA$, \\
thus $\frac{OC}{AB} = \frac{DO}{AD} = \frac{1}{2}$, \\
since $\triangle OCF \sim \triangle ABE$, \\
thus $\frac{CF}{BE} = \frac{OC}{AB} = \frac{OF}{AE} = \frac{1}{2}$, \\
thus $CF = \frac{1}{2}BE = 4$, \\
since B and C are on the graph, \\
thus $B\left(\frac{k}{8},8\right)$, $C\left(\frac{k}{4},4\right)$, \\
since $AE = 3 - \frac{k}{8}$, $OF = -\frac{k}{4}$, \\
thus $\frac{-\frac{k}{4}}{3-\frac{k}{8}} = \frac{1}{2}$, \\
solving gives $k = \boxed{-8}$.


Question: As shown in the figure, within rectangle $ABCD$, $E$ is the midpoint of $BC$. Connect $AE$ and draw $EF \perp AE$ intersecting $DC$ at point $F$. If $AB=4$ and $BC=6$, what is the length of $DF$?

Solution:
\textbf{Solution:} As shown in the figure, since quadrilateral ABCD is a rectangle,\\
hence AB = CD = 4, and $\angle$B = $\angle$C = $90^\circ$,\\
since BC = 6, and E is the midpoint of BC,\\
hence BE = EC = 3,\\
since EF $\perp$ AE,\\
hence $\angle$AEF = $90^\circ$,\\
hence $\angle$BAE = $90^\circ$ - $\angle$AEB = $\angle$CEF,\\
since $\triangle$ABE $\sim$ $\triangle$ECF,\\
hence $\frac{AB}{EC} = \frac{BE}{CF}$,\\
hence CF = $\frac{BE \cdot EC}{AB}$ = $\frac{3 \times 3}{4}$ = $\frac{9}{4}$,\\
hence DF = 4 - $\frac{9}{4}$ = $\boxed{\frac{7}{4}}$,


Question: The large square $ABCD$, as shown in the figure, is composed of four congruent right-angled triangles and a small square. The diagonal $EF$ of the small square is extended bidirectionally to intersect the side $AB$ and the extension of side $BC$ at points $G$ and $H$, respectively. If the ratio of the area of the large square to that of the small square is $5$, and $GH = 2\sqrt{5}$, then what is the side length of the large square?

Solution:
\textbf{Solution:} As shown in the diagram:\\
$\because$ The ratio of the area of the large square to the small square is 5,\\
$\therefore$ $\frac{AD}{EM} = \sqrt{5}$,\\
$\therefore$ AD = $\sqrt{5}$EM,\\
Let EM = a, AE = b, then AD = $\sqrt{5}$a,\\
By Pythagoras' theorem: AE$^2$ + DE$^2$ = AD$^2$,\\
$\therefore$ b$^2$ + (a + b)$^2$ = ($\sqrt{5}$a)$^2$,\\
$\therefore$ 2b$^2$ + 2ab - 4a$^2$ = 0,\\
(b - a)(b + 2a) = 0,\\
$\because$ b + 2a $\neq$ 0,\\
$\therefore$ b - a = 0,\\
$\therefore$ b = a,\\
$\therefore$ AE = DM = a,\\
As shown in the diagram, extend BF to meet CD at N,\\
$\because$ BN $\parallel$ DE, CF = FM,\\
$\therefore$ DN = CN,\\
$\therefore$ EN = $\frac{1}{2}$DM = $\frac{1}{2}$a,\\
$\because$ PN $\parallel$ BG,\\
$\therefore$ $\frac{FN}{BF} = \frac{PN}{BG} = \frac{FP}{GF} = \frac{\frac{1}{2}a}{2a} = \frac{1}{4}$,\\
Let PN = x, then BG = 4x,\\
$\because$ DE = BF, $\angle$BFG = $\angle$DEF, $\angle$BGF = $\angle$DPE,\\
$\therefore$ $\triangle$BFG $\cong$ $\triangle$DEP (AAS),\\
$\therefore$ PD = BG = 4x,\\
Similarly, EG = FP,\\
$\therefore$ DN = 3x = CN,\\
$\therefore$ PC = 2x,\\
$\because$ CP $\parallel$ BG,\\
$\therefore$ $\frac{CP}{BG} = \frac{PH}{GH}$, that is $\frac{2x}{4x} = \frac{PH}{2\sqrt{5}}$,\\
$\therefore$ PH = PG = $\sqrt{5}$,\\
$\because$ $\frac{FP}{FG} = \frac{1}{4}$,\\
$\therefore$ EF = $\sqrt{2}$a = $\frac{3}{5}$GP = $\frac{3}{5}\sqrt{5}$,\\
$\therefore$ a = $\frac{3\sqrt{10}}{10}$,\\
$\therefore$ AD = $\sqrt{5}$a = $\boxed{\frac{3\sqrt{2}}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, with $DE \parallel BC$, $AD=EC$, $BD=4$, and $AE=3$, what is the length of $AB$?

Solution:
\textbf{Solution:} Since $DE\parallel BC$,\\
it follows that $\frac{AD}{BD}=\frac{AE}{EC}$,\\
since $AD=EC$,\\
then $AD^2=AE\cdot BD=3\times 4=12$,\\
thus $AD=2\sqrt{3}$,\\
therefore $AB=AD+BD=2\sqrt{3}+4$,\\
hence the answer is: $AB=\boxed{2\sqrt{3}+4}$.


Question: As shown in Figure 1, a rectangular piece of paper $ABCD$, with points $E$ and $F$ located on $AB$ and $CD$ respectively, and points $G$, $H$ located on $AF$ and $EC$ respectively. Now, cut the paper along $AF$, $GH$, $EC$, and reassemble it into the rectangle shown in Figure 2. Given that the ratio $DF:AD=5:12$ and $GH=6$, what is the length of $AD$?

Solution:
\[ \textbf{Solution:} \text{ As shown in the figure, let } DF = 5x, \text{ according to the problem, we have } AD = 12x, \]
\[ AF = \sqrt{AD^{2} + DF^{2}} = 13x, \]
\[ \text{In figure 2, since } \angle CHA = \angle FDA = 90^\circ \text{ and } \angle CAH = \angle FAD, \]
\[ \therefore \triangle ADF \sim \triangle AHC, \]
\[ \therefore \frac{AD}{AH} = \frac{DF}{HC} = \frac{AF}{AC}, \]
\[ \therefore \frac{12x}{6+12x} = \frac{5x}{HC} = \frac{13x}{FC+13x}, \]
\[ \therefore HC = 5x + \frac{5}{2}, FC = \frac{13}{2}, \]
\[ \therefore \text{ The area of the rectangle formed as shown in figure 2 } = AH \times HC = \left(12x + 6\right) \left(5x + \frac{5}{2} \right) = 60\left(x + \frac{1}{2}\right)^{2}, \]
\[ \text{In figure 1,} \]
\[ CD = DF + FC = 5x + \frac{13}{2}, \]
\[ \text{The area of the original rectangle } = AD \times DC = 12x\left(5x + \frac{13}{2}\right), \]
\[ \therefore 60\left(x + \frac{1}{2}\right)^{2} = 12x\left(5x + \frac{13}{2}\right), \]
\[ \text{Solving yields } x = \frac{5}{6}, \]
\[ \therefore AD = 12x = 12 \times \frac{5}{6} = \boxed{10}. \]


Question: As shown in the figure, square $ABCD$ has a side length of 10, and contains 6 identical smaller squares within. The vertices $E$, $F$, $G$, and $H$ of a small square are located on sides $AD$, $AB$, $BC$, and $CD$ respectively. What is the length of $DH$?

Solution:
\textbf{Solution:} As illustrated in the figure:\\
$\because$ The side length of square ABCD is 10,\\
$\therefore \angle A=\angle B=90^\circ$, $AB=10$,\\
Draw GP$\perp$AD at point G, with P being the foot of the perpendicular, thus $\angle 4=\angle 5=90^\circ$,\\
$\therefore$ Quadrilateral APGB is a rectangle,\\
$\therefore \angle 2+\angle 3=90^\circ$, $PG=AB=10$,\\
$\because$ Six squares of identical size are placed within the large square as shown in the figure,\\
$\therefore \angle 1+\angle 2=90^\circ$,\\
$\therefore \angle 1=\angle 3$,\\
$\therefore \triangle BGF \sim \triangle PGE$,\\
$\therefore \frac{BG}{PG}=\frac{FG}{EG}$,\\
$\therefore \frac{BG}{10}=\frac{1}{5}$,\\
$\therefore GB=2$,\\
$\therefore AP=2$,\\
Similarly, $DE=2$,\\
$\therefore PE=AD-AP-DE=6$,\\
$\therefore EG=\sqrt{10^2+6^2}=2\sqrt{34}$,\\
$\therefore$ The side length of the small square is $\frac{2\sqrt{34}}{5}$,\\
$\therefore DH=\sqrt{EH^2-DE^2}=\sqrt{\left(\frac{2\sqrt{34}}{5}\right)^2-2^2}=\boxed{\frac{6}{5}}$.


Question: As shown in the figure, $AB\parallel CD\parallel EF$. If $AC: CE = 2: 3$ and $BF = 10$, what is the length of the line segment $DF$?

Solution:
Solution: Since $AB\parallel CD\parallel EF$,\\
hence $\frac{BD}{DF}=\frac{AC}{CE} = \frac{2}{3}$,\\
Since $BF=10$,\\
thus $DF=10 \times \frac{3}{5} = \boxed{6}$.


Question: As shown in the figure, it is known that $AB\parallel CD\parallel EF$. If $AC=6$, $CE=3$, and $DF=2$, what is the length of $BD$?

Solution:
\textbf{Solution:} Since $AB\parallel CD\parallel EF$, \\
therefore $\frac{AC}{CE}=\frac{BD}{DF}$, \\
since $AC=6$, $CE=3$, $DF=2$, \\
therefore $\frac{6}{3}=\frac{BD}{2}$, \\
solving, we get: $BD=\boxed{4}$.


Question: As shown in the figure, it is known that point $A(0, 2)$, $B(4, 0)$, point $C$ is on the $x$ axis, $CD\perp x$ axis, intersecting line segment $AB$ at point $D$, and point $D$ does not coincide with points $A$, $B$. Fold $\triangle ABO$ along $CD$, making point $B$ fall at point $E$ on the $x$ axis. Let the abscissa of point $C$ be $x$. Then, when $\triangle ADE$ is a right-angled triangle, what is the value of $x$?

Solution:
\textbf{Solution:} Given points A(0, 2), and B(4, 0),\\
$\therefore OA=2$, $OB=4$. Consider two cases:\\
(1) When $\angle DAE=90^\circ$,\\
then $\angle EAO+\angle BAO=90^\circ$,\\
$\because \angle AOB=90^\circ$,\\
$\therefore \angle ABO+\angle BAO=90^\circ$,\\
$\therefore \angle EAO=\angle ABO$,\\
also, $\because \angle AOE=\angle BOA=90^\circ$,\\
$\therefore \triangle AOE \sim \triangle BOA$,\\
$\therefore \frac{OE}{OA}=\frac{OA}{OB}=\frac{1}{2}$,\\
$\therefore OE=\frac{1}{2}OA=1$,\\
$\therefore BE=OE+OB=5$,\\
By the property of folding: $CE=CB=\frac{1}{2}BE=2.5$,\\
$\therefore OC=CE-OE=1.5$,\\
$\therefore x=1.5$;\\
(2) When $\angle AED=90^\circ$,\\
Similar to (1), $\triangle AOE \sim \triangle BOA$,\\
$\therefore \frac{OE}{OA}=\frac{OA}{OB}=\frac{1}{2}$,\\
$\therefore OE=\frac{1}{2}OA=1$,\\
$\therefore BE=OB-OE=3$,\\
By the property of folding: $CE=CB=\frac{1}{2}BE=1.5$,\\
$\therefore OC=CE+OE=2.5$,\\
$\therefore x=2.5$;\\
In summary, when $\triangle ADE$ is a right triangle, the value of $x$ can be either $\boxed{2.5}$ or $\boxed{1.5}$.


Question: As shown in the figure, in $ \triangle ABC$, $\angle ACB=90^\circ$. $BD=AB$ is taken on the extension line of $BC$, and $AD$ is connected. A perpendicular $BE\perp AD$ is drawn from point $B$ to point $E$, intersecting $AC$ at point $F$. $DF$ is connected. Point $P$ is a moving point on the ray $BE$. If $AC=9$ and $BC=12$, what is the length of $BP$ when $\triangle APB$ is similar to $\triangle AFD$?

Solution:
\textbf{Solution:} Connect AP. \\
In $\triangle ACB$, $\angle ACB=90^\circ$, $AC=9$, $BC=12$, \\
$\therefore AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{9^{2}+12^{2}}=15$, \\
$\therefore BD=BA=15$, $CD=BD-BC=15-12=3$, \\
$\therefore AD=\sqrt{CD^{2}+AC^{2}}=\sqrt{3^{2}+9^{2}}=3\sqrt{10}$, \\
$\therefore BE\perp AD$, \\
$\therefore AE=ED$, \\
$\therefore FA=FD$, \\
Let $FA=FD=x$, then we have $x^{2}=(9-x)^{2}+3^{2}$, \\
$\therefore x=5$, \\
$\therefore FA=FD=5$, \\
Because $\angle AEF=\angle FCB=90^\circ$, $\angle AFE=\angle CFB$, \\
$\therefore \angle EAF=\angle ABP$, \\
When $\triangle APB \sim \triangle AFD$, $\frac{BP}{DF}=\frac{AB}{AD}$, \\
$\therefore \frac{BP}{5}=\frac{15}{3\sqrt{10}}$, \\
$\therefore BP=\frac{5\sqrt{10}}{2}$. \\
When $\triangle AFD \sim \triangle PAB$, $\frac{AD}{BP}=\frac{DF}{AB}$, \\
$\therefore \frac{3\sqrt{10}}{BP}=\frac{5}{15}$, \\
$\therefore BP=9\sqrt{10}$. \\
In conclusion, the length of $BP$ is $\boxed{\frac{5\sqrt{10}}{2}}$ or $\boxed{9\sqrt{10}}$.


Question: As shown in the figure, in the right-angled $\triangle ABC$ with $\angle C=90^\circ$, $AC=8$, and $BC=6$, point $M$ moves from point $C$ along the segment $CA$ towards point $A$, and $BM$ is connected, with $MN\perp BM$ intersecting side $AB$ at point $N$. If $CM=2$, then how long is the segment $AN$? When point $M$ moves from $C$ to the midpoint of $AC$, what is the length of the path traversed by point $N$?

Solution:
\textbf{Solution:}

(1) As shown in the diagram, construct $ND\perp AC$, it is easy to determine $AB=10$, and it is found that $\triangle BCM \sim \triangle MDN$, \\
$\therefore \frac{MD}{ND}=\frac{BC}{MC}=\frac{6}{2}=3$, let $ND=x$, then $MD=3x$, \\
thus $AD=AC-CM-MD=6-3x$, it is easy to determine $\triangle DNA \sim \triangle CBA$, \\
$\therefore \frac{AD}{AC}=\frac{ND}{BC}=\frac{AN}{AB}$, \\
$\therefore \frac{6-3x}{8}=\frac{x}{6}=\frac{AN}{10}$, \\
$\therefore AN=\boxed{\frac{30}{13}}$. \\
(2) Following the same reasoning as in (1), we obtain $AN=\boxed{\frac{40}{17}}$.


Question: As shown in the figure, in $\triangle ABC$, $BC=20$, points $B_1$, $B_2$, $B_3$, $B_4$ and points $C_1$, $C_2$, $C_3$, $C_4$ are the quintisection points of $AB$, $AC$ respectively. What is the value of $B_1C_1+B_2C_2+B_3C_3+B_4C_4$?

Solution:
\[ \text{Solution:} \quad \because \frac{AB_4}{AB} = \frac{AC_4}{AC} = \frac{4}{5}, \]
\[ \angle A = \angle A, \]
\[ \therefore \triangle AB_4C_4 \sim \triangle ABC, \]
\[ \therefore \frac{B_4C_4}{BC} = \frac{AB_4}{AB} = \frac{4}{5}, \]
\[ \therefore B_4C_4 = \frac{4}{5}BC = \frac{4}{5} \times 20 = 16, \]
Similarly, $B_1C_1 = 4$,
\[ B_2C_2 = 8, \]
\[ B_3C_3 = 12, \]
\[ \therefore B_1C_1 + B_2C_2 + B_3C_3 + B_4C_4 = \boxed{40}. \]


Question: As shown in the figure, the hyperbola $y=\frac{k}{x}$ passes through the midpoint A of the hypotenuse of the right triangle $\triangle BOC$, and intersects with BC at point D. Given that the area of $\triangle BOD$ is 21, what is the value of $k$?

Solution:
\textbf{Solution:} As shown in the figure, construct $AE\perp$ to the $x$-axis, then the area of $\triangle AOE$ and $\triangle DOC$ is $\frac{1}{2}|k|$, \\
$\therefore$ the area of the quadrilateral $BAEC$ and $\triangle BOD$ is $21$, \\
$\because$ $AE\perp x$ axis, \\
$\angle BCO=90^\circ$, and point A is the midpoint of OB, \\
$\therefore$ $\angle AOE = \angle BOC$, \\
$\angle AEO = \angle BCO = 90^\circ$ \\
$\therefore$ $\triangle AOE \sim \triangle BOC$, \\
$\therefore$ $\frac{{S}_{\triangle AOE}}{{S}_{\triangle BOC}} = \left(\frac{OA}{OB}\right)^{2} = \frac{1}{4}$, \\
$\because$ the area of quadrilateral $BAEC$ times the area of $\triangle AOE$ equals the area of $\triangle BOC$, \\
$\therefore$ $\frac{{S}_{\triangle AOE}}{{S}_{\text{quadrilateral }BAEC}} = \frac{{S}_{\triangle AOE}}{21} = \frac{1}{3}$, \\
$\therefore$ ${S}_{\triangle AOE} = 7$, \\
$\therefore$ $\frac{1}{2}|k| = 7$, \\
Solving for $k$, we get: $k = \pm 14$, \\
$\because$ the inverse proportion function passes through the first quadrant, \\
$\therefore$ $k = \boxed{14}$.


Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is on $AD$. Connect $CE$ and extend it to intersect the extension of $BA$ at point $F$. If the ratio of $AE$ to $AD$ is $2:3$, and $CD=2$, what is the length of $AF$?

Solution:
Solution: Since quadrilateral ABCD is a parallelogram, \\
it follows that AB $\parallel$ CD, \\
therefore, $\triangle AFE\sim \triangle DCE$,\\
thus $\frac{AE}{DE}=\frac{AF}{CD}$, \\
since AE:AD=2:3 and CD=2, \\
it follows that $\frac{2}{1}=\frac{AF}{2}$, \\
therefore, AF=\boxed{4}.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=2\angle B$, and $CD$ bisects $\angle ACB$. If $AD=2$, $BD=3$, what is the length of $AC$?

Solution:
\textbf{Solution}:\\
Since $CD$ bisects $\angle ACB$,\\
it follows that $\angle ACD=\angle BCD=\frac{1}{2}\angle ACB$.\\
Given $\angle ACB=2\angle B$,\\
it follows that $\angle ACD=\angle B$.\\
Considering $\triangle ADC$ and $\triangle ACB$, we have\\
$\left\{\begin{array}{l}
\angle ACD=\angle B \\
\angle A=\angle A
\end{array}\right.$\\
Hence, $\triangle ADC\sim \triangle ACB$\\
which implies $\frac{AD}{AC}=\frac{AC}{AB}$, i.e.,\\
$AC^{2}=AD\cdot AB$.\\
Given $AD=2, BD=3$,\\
it follows that $AB=AD+BD=5$\\
and thus $AC^{2}=AD\cdot AB=10$.\\
Since the length of a segment is a positive number, $AC=\sqrt{10}$. Therefore, the solution is $AC=\boxed{\sqrt{10}}$.


Question: As shown in the figure, $l_{1}\parallel l_{2}\parallel l_{3}$. If $AB=2$, $BC=3$, $AD=1$, and $CF=4$, what is the length of $BE$?

Solution:
\textbf{Solution:} Construct line $DN \parallel AC$ through point D to intersect $BE$ at point $M$, and $CF$ at point $N$;\\
$\therefore$ Quadrilaterals $ABMD$ and $BCNM$ are both parallelograms\\
$\therefore$ $AB=DM=2$,\\
$BC=MN=3$,\\
$AD=BM=CN=1$\\
From the problem, we know\\
$\frac{AB}{BC}=\frac{DE}{EF}=\frac{2}{3}$\\
$\because BE\parallel CF$\\
$\therefore$ $\triangle DME\sim \triangle DNF$\\
$\therefore$ $\frac{ME}{NF}=\frac{DE}{DF}=\frac{DM}{DN}=\frac{DM}{DM+MN}=\frac{2}{5}$\\
$\because NF=CF-CN=4-1=3$\\
$\therefore$ $ME=\frac{6}{5}$\\
$\therefore$ $BE=BM+ME=1+\frac{6}{5}=\boxed{\frac{11}{5}}$


Question: As shown in the figure, within the circle $\odot O$, the arcs $\overset{\frown}{AB} = \overset{\frown}{AC}$, $AB=10$, $BC=12$, and $D$ is a point on the arc $\overset{\frown}{BC}$ with $CD=5$. What is the length of $AD$?

Solution:
\textbf{Solution:} Draw $AE\perp BC$ at $E$, and draw $CF\perp AD$ at $F$, then $\angle AEB=\angle CFD=90^\circ$,

$\because$ $\stackrel{\frown}{AB} = \stackrel{\frown}{AC}$, and $AB=10$,

$\therefore$ $\angle ACB=\angle B=\angle D$, and $AB=AC=10$,

$\because$ $AE\perp BC$ and $BC=12$,

$\therefore$ $BE=CE=6$,

$\therefore$ $AE=\sqrt{AB^{2}-BE^{2}}=\sqrt{10^{2}-6^{2}}=8$,

$\because$ $\angle B=\angle D$ and $\angle AEB=\angle CFD=90^\circ$,

$\therefore$ $\triangle ABE\sim \triangle CDF$,

$\therefore$ $\frac{AB}{CD}=\frac{BE}{DF}=\frac{AE}{CF}$,

$\because$ $AB=10$, $CD=5$, $BE=6$, $AE=8$,

$\therefore$ $\frac{10}{5}=\frac{6}{DF}=\frac{8}{CF}$,

Solving gives: $DF=3$, $CF=4$,

In $\triangle AFC$, $\angle AFC=90^\circ$, $AC=10$, $CF=4$,

thus $AF=\sqrt{AC^{2}-CF^{2}}=\sqrt{10^{2}-4^{2}}=2\sqrt{21}$,

$\therefore$ $AD=DF+AF=3+2\sqrt{21}$.

Therefore, the answer is: $AD=\boxed{3+2\sqrt{21}}$.


Question: As shown in the figure, in trapezoid $ABCD$, where $AD\parallel BC$ and $AC$ intersects $BD$ at point $O$, if $BC = 2AD$, what is the value of the ratio $S_{\triangle AOD} : S_{\triangle BOC}$?

Solution:
\textbf{Solution:} As shown in the figure below:

$\because$AD $\parallel$ BC,

$\therefore\triangle AOD\sim \triangle COB$,

$\therefore \frac{S_{\triangle AOD}}{S_{\triangle COB}}=\left(\frac{AD}{BC}\right)^2$

$\because$BC＝2AD,

$\therefore \frac{S_{\triangle AOD}}{S_{\triangle COB}}=\left(\frac{AD}{2AD}\right)^2=\left(\frac{1}{2}\right)^2=\boxed{\frac{1}{4}}$


Question: As shown in the figure, in rectangle $ABCD$, $E$ and $F$ are the midpoints of sides $AB$ and $AD$, respectively, and $BF$ intersects $EC$ and $ED$ at points $M$ and $N$, respectively. Given that $AB=4$ and $BC=6$, what is the length of $MN$?

Solution:
\textbf{Solution:} As shown in Figure 1, extend CE and DA to intersect at point Q, \\
$\because$ Quadrilateral ABCD is a rectangle, and BC=6, \\
$\therefore$ $\angle$BAD=$90^\circ$, AD=BC=6, and AD$\parallel$ BC, \\
$\because$ F is the midpoint of AD, \\
$\therefore$ AF=DF=3, \\
In $\triangle$BAF, by the Pythagorean theorem, we have: $BF=\sqrt{AB^{2}+AF^{2}}=\sqrt{4^{2}+3^{2}}=5$, \\
$\because$ AD $\parallel$ BC, \\
$\therefore$ $\angle$Q=$\angle$ECB, \\
$\because$ E is the midpoint of AB, and AB=4, \\
$\therefore$ AE=BE=2, \\
In $\triangle$QAE and $\triangle$CBE, \\
$\begin{array}{c}
\angle QEA=\angle BEC\\
\angle Q=\angle ECB\\
AE=BE
\end{array}$\\
$\therefore$ $\triangle$QAE$\cong$$\triangle$CBE (AAS), \\
$\therefore$ AQ=BC=6, thus QF=6+3=9, \\
$\because$ AD $\parallel$ BC, \\
$\therefore$ $\triangle$QMF$\sim$$\triangle$CMB, \\
$\therefore$ $\frac{FM}{BM}=\frac{QF}{BC}=\frac{9}{6}$, \\
$\because$ BF=5, \\
$\therefore$ BM=2, FM=3, \\
As shown in Figure 2, extend BF and CD to intersect at W, \\
Similarly, AB=DW=4, CW=8, and BF=FW=5, \\
$\because$ AB $\parallel$ CD, \\
$\therefore$ $\triangle$BNE$\sim$$\triangle$WND, \\
$\therefore$ $\frac{BN}{NW}=\frac{BE}{DW}$, \\
$\therefore$ $\frac{BN}{5-BN+5}=\frac{2}{4}$, \\
Solving this, we get: BN= $\frac{10}{3}$, \\
$\therefore$ MN=BN−BM= $\frac{10}{3}-2= \boxed{\frac{4}{3}}$.


Question: As shown in the figure, $\triangle ABC$ is folded along the line $AD$ so that point $B$ coincides with point $E$ on side $AC$. If $AB=AD=4$ and $AC=6$, what is the length of $DC$?

Solution:
\textbf{Solution:} Due to the property of folding, we have $\angle BAD=\angle DAE$, $\angle ADB-\angle ADE$, $AB=AE=4$,\\
since $AB=AD=4$,\\
thus $\angle B=\angle ADB$,\\
let $\angle B=\angle ADB=x$,\\
in $\triangle ABD$, we have $\angle BAD=180^\circ-2x$,\\
since $\angle EDC=180^\circ-\angle ADB-\angle ADE=180^\circ-2x$,\\
thus $\angle BAD=\angle EDC$,\\
since $\angle BAD=\angle DAE$,\\
thus $\angle DAE=\angle EDC$,\\
since $\angle C=\angle C$,\\
thus $\triangle DCE\sim \triangle ACD$,\\
therefore, $\frac{EC}{DC}=\frac{DC}{AC}=\frac{6-4}{DC}=\frac{DC}{6}$,\\
hence, $DC=\boxed{2\sqrt{3}}$.


Question: As shown in the figure, $AD$ is the angle bisector of $\triangle ABC$, $\angle ADE = \angle B$, $AE=3$, and $CE=1$. What is the length of $CD$?

Solution:
\textbf{Solution:} \\
Since $AD$ is the angle bisector of $\triangle ABC$, \\
then $\angle BAD=\angle DAC$, \\
and since $\angle ADE=\angle B$, \\
thus $\angle AED=\angle ADB$, \\
thus $\angle CED=\angle CDA$, \\
and since $\angle C=\angle C$, \\
thus $\triangle ADC\sim \triangle DEC$, \\
therefore $\frac{CE}{CD}=\frac{CD}{AC}$, \\
Given that $AE=3$ and $CE=1$, \\
thus $AC=AE+EC=3+1=4$, \\
therefore $\frac{1}{CD}=\frac{CD}{4}$, i.e., $CD^2=4$, \\
Solving gives $CD=2$ or $CD=-2$ (discard this), \\
thus $CD=\boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=90^\circ$, $AB=8$, $BC=6$, and D, E, F are points on sides AB, BC, and AC, respectively. Given that $\angle BED + \angle C = 90^\circ$ and $\triangle BED$ is symmetrical to $\triangle FED$ with respect to DE, what is the length of DE?

Solution:
\textbf{Solution:} Draw $FN\perp AB$, with the foot at N, through point E draw $EM\perp NF$, intersecting the extended line of NF at point M,\\
$\therefore$ $\angle FND=\angle FME=90^\circ$,\\
$\because$ $\angle B=90^\circ$,\\
$\therefore$ Quadrilateral NBEM is a rectangle,\\
$\therefore$ $NB=ME$,\\
$\because$ $\angle BED+\angle C=90^\circ$, $\angle C+\angle A=90^\circ$,\\
$\therefore$ $\angle BED=\angle A$,\\
$\because$ $\angle B=\angle B$,\\
$\therefore$ $\triangle BED\sim \triangle BAC$,\\
$\therefore$ $\frac{BD}{BE}=\frac{BC}{BA}=\frac{6}{8}=\frac{3}{4}$,\\
$\because$ Triangle $BED$ and $\triangle FED$ are symmetric with respect to DE,\\
$\therefore$ $\triangle BED\cong \triangle FED$,\\
$\therefore$ $\angle DFE=\angle B=90^\circ$, $DF=BD$, $EF=BE$,\\
$\therefore$ $\frac{DF}{EF}=\frac{BD}{BE}=\frac{3}{4}$,\\
$\because$ $\angle FME=90^\circ$,\\
$\therefore$ $\angle MEF+\angle MFE=90^\circ$,\\
$\because$ $\angle MFE+\angle NFD=90^\circ$,\\
$\therefore$ $\angle MEF=\angle NFD$,\\
$\therefore$ $\triangle NDF\sim \triangle MFE$,\\
$\therefore$ $\frac{NF}{ME}=\frac{DF}{FE}=\frac{3}{4}$,\\
$\therefore$ Let $NF=3x$, $ME=4x$,\\
$\because$ $\angle ANF=\angle B=90^\circ$, $\angle A=\angle A$,\\
$\therefore$ $\triangle ANF\sim \triangle ABC$,\\
$\therefore$ $\frac{NF}{BC}=\frac{AN}{AB}$,\\
$\therefore$ $\frac{3x}{6}=\frac{8-4x}{8}$,\\
$\therefore$ $x=1$,\\
$\therefore$ $NF=3$, $ME=NB=4$,\\
Let $BD=DF=y$,\\
then $ND=NB-BD=4-y$,\\
In $\triangle NDF$, $NF^{2}+ND^{2}=DF^{2}$,\\
$\therefore$ $3^{2}+(4-y)^{2}=y^{2}$,\\
$\therefore$ $y=\frac{25}{8}$,\\
$\therefore$ $BD=\frac{25}{8}$,\\
$\because$ $\angle B=90^\circ$, $AB=8$, $BC=6$,\\
$\therefore$ $AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{8^{2}+6^{2}}=10$,\\
$\because$ $\triangle BED\sim \triangle BAC$,\\
$\therefore$ $\frac{DE}{AC}=\frac{BD}{BC}$,\\
$\therefore$ $DE=\boxed{\frac{125}{24}}$.


Question: As shown in the figure, in quadrilateral \(ABCD\), \(AD \parallel BC\), \(\angle BAD=90^\circ\), and diagonals \(BD \perp DC\), \(AD=4\), \(BC=9\), what is the length of \(BD\)?

Solution:
\textbf{Solution:} Given that $AD \parallel BC$ and $BD \perp DC$, \\
$\therefore \angle ADB = \angle DBC$, and $\angle BDC = 90^\circ$, \\
$\therefore \angle BAD = \angle BDC = 90^\circ$, \\
$\therefore \triangle ADB \sim \triangle DBC$, \\
$\therefore \frac{AD}{BD} = \frac{BD}{BC}$, \\
$\therefore BD^{2} = AD \cdot BC = 4 \times 9 = 36$, \\
$\therefore BD = \boxed{6}$ (negative values are discarded).


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $CD\perp AB$ at point $D$, it is known that $AD=5$, $BD=4$, what is the length of $BC$?

Solution:
\textbf{Solution:} Since $CD \perp AB$,\\
then $\angle BDC = 90^\circ$,\\
thus $\angle BDC = \angle BCA$,\\
Since $\angle B = \angle B$,\\
thus $\triangle BDC \sim \triangle BCA$,\\
therefore $\frac{BD}{BC}=\frac{BC}{BA}$, that is $\frac{4}{BC}=\frac{BC}{4+5}$,\\
Solving gives: $BC=\boxed{6}$ (negative values have been discarded).


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in $\circ O$, and $\angle CBE$ is one of its exterior angles. If $\angle CBE=58^\circ$, what is the degree measure of $\angle AOC$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is inscribed in circle $O$,\\
it follows that $\angle ADC+\angle ABC=180^\circ$,\\
since $\angle CBE+\angle ABC=180^\circ$ and $\angle CBE=58^\circ$,\\
then $\angle ADC=\angle CBE=58^\circ$,\\
By the Inscribed Angle Theorem, $\angle AOC=2\angle ADC=\boxed{116^\circ}$.


Question: As shown in the figure, observe the compass and straightedge constructions in the diagram. If $\angle FMO = 50^\circ$, what is the measure of $\angle FOE$?

Solution:
\textbf{Solution:} From the drawing, we know that PQ perpendicularly bisects FM, \\
$\therefore$ point E is the midpoint of the arc $\overset{\frown}{FM}$, \\
$\therefore$ $\overset{\frown}{FE} = \overset{\frown}{EM}$, \\
$\therefore$ $\angle MOE = \angle BOE = \frac{1}{2}\angle AOB$, \\
Also, $\because \angle FMO = 50^\circ$, $\angle OFM = 90^\circ$, \\
$\therefore$ $\angle AOB = 40^\circ$, \\
$\therefore$ $\angle FOE = \boxed{20^\circ}$,


Question: As shown in the figure, given rectangle $ABCD$, with $AB=4\sqrt{3}$ and $BC=3$, point $P$ is a moving point on the right side of line $AB$, and $\angle APB=60^\circ$, $CE=2\sqrt{3}-3$. What is the maximum distance of point $P$ to line $AB$? What is the minimum value of $PE$?

Solution: \textbf{Solution:} Given the problem, point $P$ is on the major arc $AB$ of the circumcircle of the equilateral triangle with side $AB$. If we draw $OF \perp AB$ at point $F$, then $OF$ is the perpendicular bisector of $AB$. Therefore, when point $P$ is on $OF$, the distance from point $P$ to line $AB$ is maximized, with the maximum value being the length of $PF$. At this time, $\triangle PAB$ is an equilateral triangle, therefore $PB=PA=AB=4\sqrt{3}$, $\angle PBA=60^\circ$, therefore $PF=\frac{\sqrt{3}}{2}PB=6$. When points $P$, $E$, and $O$ are collinear, the value of $PE$ is minimized. As shown in the figure, connect $OB$. In $\triangle BFO$, $\angle BOF=60^\circ$, $\angle FBO=30^\circ$, $BF=CG=\frac{1}{2}AB=2\sqrt{3}$, therefore $OF=\frac{\sqrt{3}}{3}BF=2$, therefore $OG=3-2=1$, $GE=CG-CE=3$, therefore $OE=\sqrt{1^2+3^2}=\sqrt{10}$, therefore $PE_{\min}=OP-OE=4-\sqrt{10}$. Thus, the answers are: $PF=\boxed{6}$, $PE_{\min}=\boxed{4-\sqrt{10}}$.


Question: As shown in the figure, for the quadrilateral $ABCD$ inscribed in a circle, if $\angle A=60^\circ$, then what is the measure of $\angle C$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in a circle, \\
it follows that $\angle A + \angle C = 180^\circ$, \\
since $\angle A = 60^\circ$, \\
it follows that $\angle C = 180^\circ - 60^\circ = \boxed{120^\circ}$.


Question: As shown in the figure, the square $ABCD$ and the equilateral triangle $\triangle AEF$ are both inscribed in the circle $O$, with $EF$ intersecting $BC$ and $CD$ at points $G$ and $H$ respectively. If $AE=6$, what is the radius of circle $O$? And what is the length of $EG$?

Solution:
\textbf{Solution:} As shown in Figure 1, draw lines OA, OE, and through point O draw line OP $\perp$ AE at P, \\
thus AP = PE = $\frac{1}{2}$AE = 3, \\
$\because$ $\triangle AEF$ is an equilateral triangle, \\
$\therefore$ $\angle AOE = 120^\circ$, \\
$\because$ OA = OE, \\
$\therefore$ $\angle OAP = 30^\circ$, \\
$\therefore$ OA = $\frac{AP}{\cos 30^\circ}$ = $2\sqrt{3}$; \\
Connect BD and AC, with AC intersecting EF at Q, and connect OF, \\
then AC $\perp$ EF, \\
$\therefore$ EQ = $\frac{1}{2}$EF = 3, \\
In $\triangle OQF$, $\angle OFQ = 30^\circ$, \\
$\therefore$ OQ = $\frac{1}{2}$OF = $\sqrt{3}$, \\
$\therefore$ CQ = OC - OQ = $\sqrt{3}$, \\
$\because$ Quadrilateral ABCD is a square, \\
$\therefore$ $\angle GCQ = 45^\circ$, \\
$\therefore$ GQ = CQ = $\sqrt{3}$, \\
$\therefore$ EG = EQ - QG = $3 - \sqrt{3}$, \\
Hence, the answer is: The radius of circle O is $\boxed{2\sqrt{3}}$; The length of $EG$ is $\boxed{3 - \sqrt{3}}$.


Question: As shown in the figure, AB is the diameter of the semicircle O, and $\angle BAC=40^\circ$. What is the measure of $\angle D$ in degrees?

Solution:
\textbf{Solution:} Let $AB$ be the diameter of the semicircle O,\\
$\therefore$ $\angle ACB = 90^\circ$,\\
$\therefore$ $\angle BAC + \angle B = 90^\circ$,\\
Since $\angle BAC = 40^\circ$,\\
$\therefore$ $\angle B = 90^\circ - \angle BAC = 90^\circ - 40^\circ = 50^\circ$,\\
Since $\angle B + \angle D = 180^\circ$,\\
$\therefore$ $\angle D = 180^\circ - 50^\circ = \boxed{130^\circ}$.


Question: As shown in the diagram, in circle $\odot O$, the diameter $AB\perp$ chord $CD$. An arc is drawn with $C$ as the center and $CD$ as the radius, intersecting the diameter $AB$ at point $E$. Connect $CE$ and extend it to meet $\odot O$ at point $F$. Connect $DF$. If $AB=8$, what is the length of $DF$?

Solution:
\textbf{Solution:} Construct diameter DG passing through point D, and connect DE and GF, \\
since a arc is drawn with center C and radius CD intersecting diameter AB at point E, \\
it follows that CD=CE, \\
since diameter AB is perpendicular to chord CD, \\
it follows that AB perpendicularly bisects CD, \\
thus, we have CE=DE=CD, \\
therefore, $\triangle CDE$ is an equilateral triangle, \\
which implies $\angle C=60^\circ$, \\
since $\overset{\frown}{DF}=\overset{\frown}{DF}$, \\
it follows that $\angle G=\angle C=60^\circ$, \\
since DG is a diameter, \\
it follows that $\angle DFG=90^\circ$, \\
therefore, $\angle GDF=90^\circ-60^\circ=30^\circ$, \\
hence, FG=$\frac{1}{2}$DG=$\frac{1}{2}$×8=4, \\
therefore, DF=$\sqrt{DG^{2}-FG^{2}}=\sqrt{8^{2}-4^{2}}=\boxed{4\sqrt{3}}$.


Question: As shown in the figure, $\odot P$ intersects the x-axis at points $A(-5,0)$ and $B(1,0)$, and intersects the positive y-axis at point $C$. If $\angle ACB=60^\circ$, what is the ordinate of point $C$?

Solution:
\textbf{Solution:} As shown in the figure, draw $PH\perp AB$ at point $H$, and $PD\perp OC$ at point $D$. Connect $PA$, $PB$, and $PC$.\\
Since $A(-5,0)$ and $B(1,0)$,\\
thus $OA=5$ and $OB=1$,\\
therefore, $AB=6$,\\
since $PH\perp AB$,\\
therefore, $AH=BH=\frac{1}{2}AB=3$,\\
thus $OH=2$,\\
because $\angle ACB=60^\circ$,\\
thus $\angle APB=2\angle ACB=2\times60^\circ=120^\circ$,\\
thus $\angle APH=60^\circ$, $\angle PAH=30^\circ$,\\
since in $\triangle PAH$ right-angled, $PH=\frac{\sqrt{3}}{3}AH=\sqrt{3}$,\\
thus $PA=2PH=2\sqrt{3}$,\\
because $\angle PHO=\angle PDO=\angle HOD=90^\circ$,\\
thus quadrilateral $PHOD$ is a rectangle,\\
thus $OD=PH=\sqrt{3}$, $PD=OH=2$,\\
since in right-angled $\triangle PCD$, $PC=PA=2\sqrt{3}$, $PD=2$,\\
thus $CD=\sqrt{PC^{2}-PD^{2}}=\sqrt{(2\sqrt{3})^{2}-2^{2}}=2\sqrt{2}$,\\
thus $OC=OD+CD=\sqrt{3}+2\sqrt{2}$,\\
since point $C$ is on the positive semi-axis of $y$,\\
thus the $y$-coordinate of point $C$ is $\boxed{\sqrt{3}+2\sqrt{2}}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, D and C are the trisection points of arc BE, $\angle COD=32^\circ$. What is the degree measure of $\angle E$?

Solution:
\textbf{Solution:} Since D and C are the trisecting points of the arc BE,\\
thus, arc DE = arc CD = arc BC,\\
thus, $\angle$BOC = $\angle$COD = $\angle$DOE = $32^\circ$,\\
thus, $\angle$BOE = $3 \times 32^\circ = 96^\circ$,\\
thus, $\angle$A = $\frac{1}{2}\angle$BOE = $48^\circ$.\\
Therefore, the answer is: $\boxed{48^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, where $AC=3$, $BC=4\sqrt{2}$, and $\angle ACB=45^\circ$, let $D$ be a moving point within $\triangle ABC$. Let $\odot O$ be the circumcircle of $\triangle ACD$. The line $BD$ intersects $\odot O$ at point $P$ and intersects $BC$ at point $E$. If the arc $AE=CP$, what is the minimum value of $AD$?

Solution:
\textbf{Solution:} As shown in the figure,\\
$\because$ arc AE = arc CP,\\
$\therefore$ $\angle$CDP=$\angle$ACB=$45^\circ$,\\
$\therefore$ $\angle$BDC=$135^\circ$,\\
$\therefore$ when AD passes through $O_{1}$, AD has its minimum value,\\
$\because$ $\angle$BDC=$135^\circ$,\\
$\therefore$ $\angle$BO$_{1}$C=$90^\circ$,\\
$\because$ BC=$4\sqrt{2}$,\\
$\therefore$ O$_{1}$C=O$_{1}$B=4=O$_{1}$D,\\
$\therefore$ $\angle$O$_{1}$CB=$45^\circ$,\\
$\therefore$ $\angle$AO$_{1}$C=$45^\circ+45^\circ=90^\circ$,\\
$\because$ AC=3,\\
$\therefore$ AO$_{1}$=5,\\
$\therefore$ AD=AO$_{1}$−O$_{1}$D=5−4=\boxed{1}.


Question: As shown in the figure, the regular pentagon ABCDE is inscribed in circle $O$. What is the measure of $\angle CAD$ in degrees?

Solution:
\textbf{Solution}: Since pentagon ABCDE is a regular pentagon,\\
therefore, \(\angle BAE=108^\circ\), and \(BC=CD=DE\),\\
therefore, \(\angle BAC=\angle CAD=\angle DAE=\frac{1}{3}\angle BAE=36^\circ\),\\
thus, the answer is: \(\boxed{36^\circ}\).


Question: As shown in the figure, within circle $\odot O$, diameter $AB \perp CD$, and $\angle A=26^\circ$, what is the degree measure of the inscribed angle subtended by chord $AC$?

Solution:
\textbf{Solution:} As shown in the diagram: Connect OC, BC, take a point E on arc AC, and connect AE, CE,\\
$\because OA=OC$,\\
$\therefore \angle AOC=180^\circ-2\times 26^\circ=128^\circ$,\\
$\therefore \angle B=\frac{1}{2}\angle AOC=\frac{1}{2}\times 128^\circ=64^\circ$,\\
$\therefore \angle E=180^\circ-\angle B=116^\circ$,\\
$\therefore$ the measure of the inscribed angle facing chord AC is $\boxed{64^\circ}$ or $116^\circ$.


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in the circle $\odot O$. Given that $\angle ADC = 150^\circ$, what is the measure of $\angle AOC$ in degrees?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is inscribed in circle O, \\
it follows that $\angle B + \angle ADC = 180^\circ$, \\
Given that $\angle ADC = 150^\circ$, \\
it follows that $\angle B = 180^\circ - 150^\circ = 30^\circ$. \\
Therefore, $\angle AOC = 2\angle B = \boxed{60^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and chord $CD \perp AB$ at $P$. If $CD = AP = 8$, what is the radius of $\odot O$?

Solution:
\textbf{Solution:} As shown in the diagram, draw line $OC$.\\
Since $AP=8$,\\
it follows that $AP=OA+OP=OC+OP=8$.\\
Let $OP=x$, then $OC=8-x$.\\
Since $AB$ is the diameter of the circle $\odot O$, and chord $CD$ is perpendicular to $AB$ at point $P$,\\
it follows that $CP=\frac{1}{2}CD=4$.\\
In $\triangle OCP$, $\angle OPC=90^\circ$,\\
hence $OP^2+CP^2=OC^2$.\\
Therefore, $(8-x)^2=x^2+4^2$.\\
Therefore, $x=3$.\\
Therefore, $OC=8-x=5$.\\
Therefore, the radius of the circle $\odot O$ is $\boxed{5}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, a circle with its center at the origin $O$ passes through point $A(2, 0)$. The line $y=\frac{\sqrt{3}}{3}x+\sqrt{3}$ intersects the circle $O$ at points $B$ and $C$. What is the length of chord $BC$?

Solution:
\textbf{Solution:} Let us consider the line $y=\frac{\sqrt{3}}{3}x+\sqrt{3}$ intersecting the two coordinate axes at points D and E. Draw OM$\perp$BC at point M and connect OB,\\
When $x=0$, $y=\sqrt{3}$,\\
$\therefore$D$\left(0, \sqrt{3}\right)$,\\
When $y=0$, $x=-3$,\\
$\therefore$ point E$(-3,0)$,\\
$\therefore$ $DE=\sqrt{\left(\sqrt{3}\right)^{2}+3^{2}}=2\sqrt{3}$,\\
$\therefore$DE=2OD,\\
$\therefore$$\angle$DEO$=30^\circ$,\\
$\therefore$OM$=\frac{1}{2}$OE$=\frac{3}{2}$,\\
$\because$ point A$(2,0)$,\\
$\therefore$OA=OB=2,\\
$\therefore$ $BM=\sqrt{OB^{2}-OM^{2}}=\sqrt{4-\frac{9}{4}}=\frac{\sqrt{7}}{2}$,\\
$\because$OM$\perp$BC,\\
$\therefore$BC=2BM$=\boxed{\sqrt{7}}$.


Question: As shown in the figure, the inscribed angle $\angle ACB = 130^\circ$, then what is the degree measure of the central angle $\angle AOB$?

Solution:
\textbf{Solution:} As shown in the figure, take point D on the major arc AB, and connect AD and BD,\\
$\because$ Quadrilateral ACBD is a cyclic quadrilateral,\\
$\therefore$ $\angle$D=180$^\circ$−$\angle$ACB=180$^\circ$−130$^\circ$=50$^\circ$,\\
$\because$ $ \overset{\frown}{AB}=\overset{\frown}{AB}$  , \\
$\therefore$ $\angle$AOB=2$\angle$D=2×50$^\circ$=\boxed{100^\circ}.


Question: As shown in the figure, it is known that $\triangle ABC$ is an inscribed triangle of $\odot O$, and $AD$ is the diameter of $\odot O$. Connect $BD$, and $BC$ bisects $\angle ABD$. If $AD=6$, then what is the length of $\overset{\frown}{CD}$?

Solution:
\textbf{Solution:} Connect OC, \\
$\because$AD is the diameter, and AD=6 \\
$\therefore$$\angle$ABD=90$^\circ$, OD=3 \\
$\because$BC bisects $\angle$ABD, \\
$\therefore$$\angle$DBC= $ \frac{1}{2}$$\angle$ABD=45$^\circ$, \\
$\because$$ \overset{\frown}{CD}=\overset{\frown}{CD}$, \\
$\therefore$$\angle$DOC=2$\angle$DBC=90$^\circ$, \\
$\therefore$ the length of $ \overset{\frown}{CD}$ is $ \frac{90^\circ\pi \times 3}{180}=\boxed{1.5\pi} $.


Question: As shown in the diagram, a quadrilateral is inscribed in $\odot O$. If one of its exterior angles $\angle DCE=46^\circ$, then what is the measure of $\angle A$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in circle $\odot O$,\\
it follows that $\angle A+\angle BCD=180^\circ$,\\
Since $\angle DCE+\angle BCD=180^\circ$,\\
it follows that $\angle A=\angle DCE=46^\circ$,\\
Therefore, the answer is: $\boxed{46^\circ}$.


Question: As shown in the figure, points $A$, $B$, $C$, $D$, $E$ are all on the circle $\odot O$, with $AC=AE$, and $\angle D = 128^\circ$. Then, what is the degree measure of $\angle B$?

Solution:
\textbf{Solution:} Connect AC and CE, \\
$\because$ Quadrilateral ACDE is inscribed in circle O, \\
$\therefore$ $\angle$CAE + $\angle$C = 180$^\circ$, \\
$\therefore$ $\angle$CAE = 180$^\circ$ − 128$^\circ$ = 52$^\circ$, \\
$\because$ AC = CE, \\
$\therefore$ $\angle$ACE = $\angle$AEC = $ \frac{1}{2}$(180$^\circ$ − 52$^\circ$) = 64$^\circ$, \\
$\because$ Quadrilateral ABCE is inscribed in circle O, \\
$\therefore$ $\angle$B + $\angle$AEC = 180$^\circ$, \\
$\therefore$ $\angle$B = 180$^\circ$ − 64$^\circ$ = \boxed{116^\circ}.


Question: As shown in the figure, quadrilateral ABCD is inscribed in $\odot O$, and point M is on the extension of AD. If $\angle AOC = 142^\circ$, what is the degree measure of $\angle CDM$?

Solution:
\textbf{Solution:} Given that $\angle AOC=142^\circ$,\\
it follows that $\angle B=71^\circ$,\\
since quadrilateral ABCD is inscribed in $\odot O$,\\
it implies that $\angle B+\angle ADC=180^\circ$,\\
thus $\angle ADC=142^\circ$,\\
which means $\angle CDM=180^\circ-142^\circ=71^\circ$,\\
hence, the answer is: $\boxed{71^\circ}$.


Question: As shown in the figure, point $A$ is a point on the circle $\odot O$, $OD\perp$ chord $BC$ at point $D$. If $\angle BAC=60^\circ$ and $OD=1$, then what is the length of $BC$?

Solution:
\textbf{Solution:} As shown in the figure, connect OC.\\
Since $\angle BAC = 60^\circ$,\\
thus $\angle BOC = 2\angle BAC = 120^\circ$,\\
Since OB = OC,\\
thus $\angle OBC = \angle OCB = 30^\circ$,\\
Since OD$\perp$BC,\\
thus BD = CD,\\
In $\triangle BOD$, with $\angle OBC = 30^\circ$, and OD = 1,\\
thus BD = $\sqrt{3}$OD = $\sqrt{3}$,\\
therefore BC = 2BD = $\boxed{2\sqrt{3}}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in a semicircle, where $AB$ is the diameter, and the arc $DC$ is equal to the arc $CB$. If $\angle C = 110^\circ$, then what is the degree measure of $\angle ABC$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is inscribed in a semicircle, and $\angle C=110^\circ$,
\[
\therefore \angle DAB=180^\circ-\angle C=70^\circ,
\]
since arc DC equals arc CB,
\[
\therefore CD=CB,
\]
\[
\therefore \angle CDB=\angle CBD=35^\circ,
\]
since AB is the diameter,
\[
\therefore \angle ADB=90^\circ,
\]
\[
\therefore \angle ABD=90^\circ-\angle ADB=90^\circ-70^\circ=20^\circ,
\]
\[
\therefore \angle ABC=\angle ABD+\angle CBD=20^\circ+35^\circ=\boxed{55^\circ}.
\]


Question: As shown in the figure, $AB$ is the diameter of circle $O$. Points $C$ and $D$ are on circle $O$, and segments $AC$, $BC$, and $CD$ are connected respectively. Given that $\angle DOB=150^\circ$, what is the measure of $\angle ACD$?

Solution:
\textbf{Solution:} Since $\angle DOB=150^\circ$ and $\angle AOD+\angle DOB=180^\circ$,\\
it follows that $\angle AOD=30^\circ$,\\
thus $\angle ACD=\frac{1}{2}\angle AOD=15^\circ$.\\
Therefore, the answer is: $\boxed{15^\circ}$.


Question: As shown in the figure, $AB$ is a chord of the circle $\odot O$, and point $C$ is a point on the circle $\odot O$. Given that $\angle ACB = 55^\circ$, what is the measure of $\angle AOB$?

Solution:
\textbf{Solution:} Given that $\angle ACB = 55^\circ$, \\
$\therefore \angle AOB = 2\angle ACB = 110^\circ$, \\
$\therefore \angle AOB = \boxed{110^\circ}$.


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=ax^2-4ax+2(a>0)$ intersects the $y$-axis at point $A$. Point $B$ is the symmetric point of $A$ with respect to the axis of symmetry, and point $C$ is the vertex of the parabola. If the circumcircle of $\triangle ABC$ passes through the origin $O$, what are the coordinates of point $C$?

Solution: \textbf{Solution:} Connect OB, which intersects the axis of symmetry of the parabola at point $O^{\prime}$. The axis of symmetry of the parabola is the line $x=2$. When $x=0$, we have $y=2$. Therefore, point A is $(0,2)$. Since point A and B are symmetrical about the axis of symmetry, point B is $(4,2)$. AB is parallel to the x-axis, therefore $\angle OAB = 90^\circ$. Since the circumcircle of $\triangle ABC$ passes through the origin $O$, OB is the diameter of circle $O^{\prime}$. The coordinate of point $O^{\prime}$ is $(2,1)$. Therefore, $OB = \sqrt{2^2+4^2} = 2\sqrt{5}$, $O^{\prime}B=O^{\prime}C=\sqrt{5}$. Therefore, the coordinate of point C is $\boxed{(2, 1-\sqrt{5})}$.


Question: As shown in the figure, $\triangle ABC$ is inscribed in $\odot O$, where $CD$ is the diameter of $\odot O$. Connect $AD$. If $CD = 2AD$ and $AB = BC = 6$, then what is the radius of $\odot O$?

Solution:
Solution: Since CD is a diameter, \\
it follows that $\angle DAC=90^\circ$, \\
since CD=2AD, \\
it follows that $\angle ACD=30^\circ$, $\angle D=60^\circ$, \\
since $\overset{\frown}{AC} = \overset{\frown}{AC}$, \\
it follows that $\angle D=\angle B=60^\circ$, \\
since AB=BC, \\
it follows that $\triangle ABC$ is an equilateral triangle, \\
therefore BC=AC=6; \\
thus $AD^{2}+AC^{2}=CD^{2}$, that is, $AD^{2}+36=4AD^{2}$ \\
Solving this, we find AD=$2\sqrt{3}$. \\
Therefore, the radius of the circle is $\boxed{2\sqrt{3}}$.


Question: As shown in the diagram, in square $ABCD$, circle $O$ with diameter $AB$ intersects the midpoint of side $BC$ at point $E$ and intersects diagonal $AC$ at point $F$. Draw $EG \perp AC$ with the foot of the perpendicular at $M$. If $FA=3CF$, what is the value of $\frac{CG}{GD}$?

Solution:
\textbf{Solution:} As shown in the figure, connect BF and extend GE, where AB intersects at point H, let AC intersect GE at point M,\\
since AB is the diameter of circle O,\\
therefore $\angle AFB=90^\circ$,\\
since quadrilateral ABCD is a parallelogram,\\
therefore $DC\parallel AB$, $DC=AB$,\\
therefore $\angle DCE=\angle CBH$, $\angle CGE=\angle EHB$,\\
since E is the midpoint of BC,\\
therefore $CE=BE=\frac{1}{2}CB$,\\
therefore $\triangle GEC\cong \triangle HEB$ (AAS),\\
therefore $GC=BH$,\\
since $EG\perp AC$, $BF\perp AC$,\\
therefore $ME\parallel BF$,\\
therefore $CM:CF=CE:CB=\frac{1}{2}$,\\
therefore let $CM=MF=a$, $CF=2a$,\\
therefore $FA=3CF=6a$,\\
therefore $CM:MA=a:7a=\frac{1}{7}$,\\
since $CD\parallel AB$\\
therefore $\triangle GMC\sim \triangle HMA$,\\
therefore $GC:AH=MC:MA=\frac{1}{7}$,\\
therefore $GC:AB=\frac{1}{6}$,\\
therefore $GC:CD=\frac{1}{6}$,\\
therefore $GC:DG=\boxed{\frac{1}{5}}$.


Question: As shown in the figure, point A is on the semicircle O, and BC is the diameter. If $\angle ABC=40^\circ$ and BC = 2, then what is the length of $\overset{\frown}{AC}$?

Solution:
\textbf{Solution:} Given $\angle AOC = 2\angle ABC = 80^\circ$, by the arc length formula, we have\\
the length of $\overset{\frown}{AC}$ is $\frac{80\pi \times 1}{180} = \boxed{\frac{4}{9}\pi}$.


Question: In the circle $\odot O$, points C and D lie on $\odot O$, and are distributed on opposite sides of diameter AB. Extending CO intersects chord BD at point E. If $\angle DEC=120^\circ$, and point A is the midpoint of $\overset{\frown}{DC}$, then what is the degree measure of $\overset{\frown}{DC}$?

Solution:
\textbf{Solution:} As shown in the figure, join OD, \\
$\because$ point A is the midpoint of $ \overset{\frown}{DC}$, and AB is the diameter, \\
$\therefore$ AB$\perp$CD, and $\angle$AOC = $\angle$AOD, \\
$\therefore$ $\angle$AOD = 2$\angle$B, \\
Let $\angle$B = x, then $\angle$AOC = $\angle$AOD = 2x, and $\angle$ODB = $\angle$B = x, \\
$\because$ $\angle$DEC = $120^\circ$, and $\angle$COD = $\angle$DEC + $\angle$ODB, \\
$\therefore$ 4x = x + $120^\circ$, \\
Solving for x, we get x = $40^\circ$, \\
$\therefore$ 4x = $160^\circ$, \\
That is, $\angle$COD = $160^\circ$, \\
$\therefore$ the degree of $ \overset{\frown}{CD}$ is $\boxed{160^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC<BC$, and $CD$ bisects $\angle ACB$ and intersects $AB$ at point $D$. A circle $\odot O$ is drawn with $DB$ as the diameter, which intersects $CD$ and $BC$ at points $E$ and $F$ respectively. Connect $BE$ and $EF$. What is the degree measure of $\angle EBF$? If $DE=DC$ and $BC=8$, what is the length of $EF$?

Solution:
\textbf{Solution:} As shown in the figure, connect DF and construct EG$\perp$BF at point G,\\
$\because$ $\angle ACB=90^\circ$ and CD bisects $\angle ACB$,\\
$\therefore$ $\angle DCB= \frac{1}{2} \angle ACB=45^\circ$,\\
$\because$ BD is the diameter of circle $O$,\\
$\therefore$ $\angle DFB=\angle DEB=90^\circ$,\\
$\because$ $\angle CFD=180^\circ-\angle DFB=90^\circ$,\\
$\therefore$ $\angle FDC=\angle FCD=45^\circ$,\\
$\therefore$ $\angle EBF=180^\circ-\angle EDF=\angle FDC=45^\circ$;\\
$\because$ $\angle CEB=90^\circ$ and $\angle EBC=\angle ECB=45^\circ$, BC=8, DE=DC,\\
$\therefore$ CE=BE,\\
$\therefore$ CG=BG= $\frac{1}{2}$ BC=4,\\
$\therefore$ EG= $\frac{1}{2}$ BC=4,\\
$\because$ $\angle CFD=\angle CGE=90^\circ$,\\
$\therefore$ DF$\parallel$EG,\\
$\therefore$ $\frac{FG}{FC}=\frac{DE}{DC} =1$,\\
$\therefore$ FG=FC= $\frac{1}{2}$ CG=2,\\
$\therefore$ EF= $\sqrt{EG^2+FG^2}=\sqrt{4^2+2^2}=\boxed{2\sqrt{5}}$.


Question: As shown in the figure, within the circle $\odot O$, $OA=2$, and $\angle ACB=30^\circ$, what is the length of the chord $AB$?

Solution:
\textbf{Solution:} Since $\angle AOB=2\angle ACB$ and $\angle ACB=30^\circ$,\\
it follows that $\angle AOB=60^\circ$,\\
Since $OA=OB$,\\
it follows that $\triangle OAB$ is an equilateral triangle,\\
Therefore, $AB=OA= \boxed{2}$.


Question: As shown in the figure, points $A, B, C$ are on the circle $\odot O$ with $\angle A=45^\circ$. What is the degree measure of $\angle BOC$?

Solution:
\textbf{Solution:}

Since, as shown in the figure: $\angle A = 45^\circ$, \\
therefore, $\angle BOC = 2\angle A = 90^\circ$, \\
thus, the answer is: $\boxed{90^\circ}$.


Question: As shown in the figure, in circle $\odot O$, $OA=2$, and $\angle ACB=30^\circ$, what is the length of chord $AB$?

Solution:
\textbf{Solution:} Since $\overset{\frown}{AB}=\overset{\frown}{AB}$,\\
it follows that $\angle AOB=2\angle ACB=2\times 30^\circ=60^\circ$,\\
since $OA=OB$,\\
thus $\triangle AOB$ is an equilateral triangle,\\
therefore $OA=AB=\boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=120^\circ$, $AB=AC=6$, $D$ is a moving point on side $AB$ (not coinciding with point $B$). Connect $CD$, and rotate line segment $CD$ $90^\circ$ counterclockwise around point $D$ to obtain $DE$. Connect $BE$. What is the maximum value of $S_{\triangle BDE}$?

Solution:
Solution: Draw a line segment CM$\perp$AB at M through point C, and draw EN$\perp$AB at N through point E, \\
therefore, $\angle END=\angle DMC=90^\circ$, \\
therefore, $\angle EDN+\angle DEN=90^\circ$, \\
since $\angle EDC=90^\circ$, \\
therefore, $\angle EDN+\angle CDM=90^\circ$, \\
therefore, $\angle DEN=\angle CDM$, \\
in $\triangle EDN$ and $\triangle DCM$, \\
$\left\{\begin{array}{l}
\angle DEN=\angle CDM\\
\angle END=\angle DMC\\
ED=DC
\end{array}\right.$, \\
therefore, $\triangle EDN \cong \triangle DCM$ (AAS), \\
therefore, $EN=DM$, \\
since $\angle BAC=120^\circ$, \\
therefore, $\angle MAC=60^\circ$, \\
therefore, $\angle ACM=30^\circ$, \\
therefore, $AM=\frac{1}{2}AC=\frac{1}{2}\times 6=3$, \\
therefore, $BM=AB+AM=6+3=9$, \\
let $BD=x$, then $EN=DM=9-x$, \\
therefore, $S_{\triangle BDE}=\frac{1}{2}BD\cdot EN=\frac{1}{2}x(9-x)=-\frac{1}{2}(x-4.5)^2+\frac{81}{8}$, \\
therefore, when $BD=4.5$, $S_{\triangle BDE}$ reaches its maximum value of $\boxed{\frac{81}{8}}$.


Question: As shown in the figure, the graph of the inverse proportion function $y=\frac{k}{x}$ is depicted. Point A is on the graph of this function, and AB is perpendicular to the x-axis with B being the foot of the perpendicular. If the area of $\triangle AOB$ is 2, what is the value of $k$?

Solution:
\textbf{Solution:} Let $A\left(x_{0},y_{0}\right)$. As $S_{\triangle AOB}=2$ we have\\
$\frac{1}{2}\left|x_{0}\right|\left|y_{0}\right|=2$,\\
thus $\left|x_{0}\right|\left|y_{0}\right|=4$.\\
Since point A is in the second quadrant,\\
we have $x_{0}+y_{0}=-4$,\\
$\because$ point A lies on the graph of the function,\\
thus $y_{0}=\frac{k}{x_{0}}$,\\
which leads to $k=x_{0}y_{0}=\boxed{-4}$.


Question: As shown in the figure, the parabola $y=-\frac{1}{4}x^{2}+x+3$ intersects the $x$-axis at points $A$ and $B$, and intersects the $y$-axis at point $C$. Point $F$ is the vertex of the parabola. There is a point $G$ on the axis of symmetry of the parabola. When is the coordinate of point $G$ such that $\triangle AFG$ is an isosceles triangle?

Solution:
\textbf{Solution:} When $y=0$, we have $-\frac{1}{4}x^{2}+x+3=0$,\\
which gives us $x_{1}=-2$ and $x_{2}=6$,\\
$\therefore A(-2,0)$ and $B(6,0)$,\\
$\because y=-\frac{1}{4}x^{2}+x+3=-\frac{1}{4}(x-2)^{2}+4$,\\
$\therefore$ the axis of symmetry is the line $x=2$, and the coordinates of vertex $F$ are $(2,4)$,\\
$\therefore AF=\sqrt{(2+2)^{2}+(4-0)^{2}}=4\sqrt{2}$,\\
Let point $G(2,t)$,\\
If $\triangle AFG$ is an isosceles triangle,\\
(1) For $GA=GF$,\\
$(2+2)^{2}+t^{2}=(4-t)^{2}$,\\
solving this gives $t=0$,\\
$\therefore G(2,0)$;\\
(2) For $FG=FA=4\sqrt{2}$,\\
$\therefore G(2,4+4\sqrt{2})$ or $(2,4-4\sqrt{2})$;\\
(3) For $AG=AF$, point $G$ is symmetric to point $F$ with respect to the $x$-axis,\\
$\therefore$ the coordinates of point $G$ are $(2,-4)$,\\
In summary, the coordinates of point $G$ are $\boxed{(2,0)}$ or $(2,-4)$ or $(2,4+4\sqrt{2})$ or $(2,4-4\sqrt{2})$.


Question: Consider point $P$ on the parabola $y=-x^2+x+2$ in the first quadrant. Perpendicular lines are drawn from point $P$ to the $x$-axis and the $y$-axis, with the feet of the perpendiculars being points $A$ and $B$, respectively. What is the maximum perimeter of the quadrilateral $OAPB$?

Solution:
\textbf{Solution:} Solve when $y=0$, we have $-x^{2}+x+2=0$\\
Solving this, we get $x_{1}=-1$ and $x_{2}=2$,\\
Therefore, when $0<x<2$ then $y>0$\\
Let the point $P(x, -x^{2}+x+2)$,\\
Therefore, the perimeter of the rectangle $OAPB$ is $2(x-x^{2}+x+2)=-2(x-1)^{2}+6$,\\
Since the parabola opens downwards,\\
Therefore, when $x=1$ the maximum perimeter of the rectangle $OAPB$ is $\boxed{6}$.


Question: As shown in the figure, an athlete is shot putting. The shot is released at point A and lands at point B. It is known that the trajectory followed by the shot is a parabola described by the equation \(y =-\frac{1}{12}(x-4)^{2}+3\). What is the horizontal distance from the athlete to the landing point B of the shot in meters?

Solution:
\textbf{Solution:} Let $y=0$, \\
thus $-\frac{1}{12}(x-4)^2+3=0$, \\
we obtain: $x=-2$ or $x=10$, \\
$\therefore B(10,0)$, \\
$\therefore$ the horizontal distance from the shot put landing point B to the athlete is $\boxed{10}$ meters.


Question: As shown in the figure, quadrilateral $OABC$ is a rectangle, with point $A$ in the third quadrant. The point symmetrical to point $A$ with respect to $OB$ is point $D$. Points $B$ and $D$ are both on the graph of the function $y=-\frac{3\sqrt{2}}{x}$ $(x<0)$. $BE\perp y$-axis at point $E$. If the extension line of $DC$ intersects the $y$-axis at point $F$, and when the area of rectangle $OABC$ is $6$, what is the value of $\frac{OF}{OE}$?

Solution:
\textbf{Solution:} As shown in the figure, connect BF and DO,\\
due to symmetry and the properties of rectangles, it's easy to see that $\triangle OAB \cong \triangle BCO \cong \triangle BDO$,\\
$\therefore$BO$\parallel$DF,\\
$\therefore S_{\triangle BFE} = S_{\triangle BDO} = \frac{1}{2}S_{\text{rectangle} OABC} = 3$,\\
Moreover, since point B lies on the graph of the inverse proportional function $y = -\frac{3\sqrt{2}}{x}$,\\
$\therefore S_{\triangle BEO} = \frac{3\sqrt{2}}{2}$,\\
$\therefore \frac{OF}{OE} = \frac{S_{\triangle BFE}}{S_{\triangle BEO}} = \frac{3}{\frac{3\sqrt{2}}{2}} = \boxed{\sqrt{2}}$.


Question: As shown in the graph, the quadratic function $y_{1}=ax^{2}+bx+c$ ($a>0$) intersects with the linear function $y_{2}=kx+m$ ($k\neq 0$) at points $A(-2,4)$ and $B(8,2)$. What is the range of values of $x$ for which $y_{1}>y_{2}$?

Solution:
\textbf{Solution:} From the given conditions, we have: when $x<-2$ or $x>8$, $y_{1}>y_{2}$,\\
$\therefore$ the range of $x$ for which $y_{1}>y_{2}$ holds true is $x<-2$ or $x>8$.\\
Therefore, the solution is: $\boxed{x<-2$ or $x>8}$.


Question: As shown in the figure, points $A$ and $B$ are on the hyperbola $y=\frac{3}{x}$. Perpendiculars are drawn from points $A$ and $B$ to the $x$-axis and the $y$-axis, respectively. If the area of the shaded region is $S_{\text{shaded}}=2$, what is the value of $S_{1}+S_{2}$?

Solution:
\textbf{Solution:} Given $y= \frac{3}{x}$,\\
the area of the shaded region, $S_{\text{shaded}}=2$,\\
therefore $S_{1}+S_{\text{shaded}}=3=S_{2}+S_{\text{shaded}}$,\\
thus $S_{1}=1$, $S_{2}=1$,\\
hence $S_{1}+S_{2}=\boxed{2}$.


Question: As shown in the figure, the parabola $y = -x^2 - 4x + 5$ intersects the $x$-axis at points $A$ and $B$, and intersects the $y$-axis at point $C$. Point $P$ is any point on the axis of symmetry of the parabola. If points $D$, $E$, and $F$ are the midpoints of $BC$, $BP$, and $PC$, respectively, and $DE$ and $DF$ are connected, what is the minimum value of $DE + DF$?

Solution:
\textbf{Solution:} Let points D, E, F be the midpoints of BC, BP, and PC, respectively.\\
Therefore, DE and DF are the midlines of $\triangle PBC$,\\
Hence, $DE=\frac{1}{2}PC$, $DF=\frac{1}{2}PB$,\\
Thus, $DE+DF=\frac{1}{2}(PB+PC)$,\\
When the value of PB+PC is minimum, the value of DE+DF is also at its minimum,\\
When $y=0$, solving $-x^2-4x+5=0$, we find $x_1=-5$ and $x_2=1$, thus $A(-5,0)$, $B(1,0)$,\\
When $x=0$, $y=-x^2-4x+5=5$, hence $C(0,5)$,\\
Connect AC and the symmetry axis of the parabola at point P',\\
Since $P'A=P'B$,\\
It follows that $P'B+P'C=P'A+P'C=AC$,\\
Therefore, at this point, the value of $P'B+P'C$ is at its minimum, which is $\sqrt{5^2+5^2}=5\sqrt{2}$,\\
Hence, the minimum value of DE+DF is $\boxed{\frac{5\sqrt{2}}{2}}$.


Question: As shown in the figure, the parabola $y=\frac{1}{3}x^{2}+\frac{8}{3}x-3$ intersects the x-axis at points $A$ and $B$, and intersects the y-axis at point $C$. Point $D$ is a moving point on the parabola within the third quadrant. When $\angle ACD+2\angle ABC=180^\circ$, what are the coordinates of point $D$?

Solution:
\textbf{Solution:} See figure below. \\
When $y=0$, $\frac{1}{3}x^{2}+\frac{8}{3}x-3=0$, \\
Solving this, we get: $x=-9$ or $1$, \\
$\therefore$ A$(-9,0)$, B$(1,0)$, \\
$\therefore$ AB$=10$, \\
When $x=0$, $y=3$, \\
$\therefore$ C$(0,-3)$, \\
$\because$ AC$^{2}=9^{2}+3^{2}=90$, $BC^{2}=1^{2}+3^{2}=10$, $AB^{2}=100$, \\
$\therefore$ AC$^{2}+BC^{2}=AB^{2}$, \\
$\therefore$ $\angle$ACB$=90^\circ$, \\
$\therefore$ $\angle$BAC$+\angle$ABC$=90^\circ$, \\
$\therefore$ $2\angle$BAC$+2\angle$ABC$=180^\circ$, \\
$\because$ $\angle$ACD$+2\angle$ABC$=180^\circ$ \\
$\therefore$ $2\angle$BAC$=\angle$ACD, \\
Construct AE$\perp$x-axis, intersecting the extension of CD at E, and construct the bisector of $\angle$ACD, intersecting AE at F, then $\angle$ACF$=\angle$BAC, \\
$\therefore$ CF$\parallel$AB, \\
$\therefore$ CF$\perp$AE, \\
$\therefore$ AF$=EF=BC=3$, \\
$\therefore$ E$(-9,-6)$, \\
Assume the equation of the line CD is $y=kx-3$, \\
Substituting the coordinates of E, we get $-6=-9k-3$, \\
$\therefore$ $k=\frac{1}{3}$, \\
$\therefore$ The equation of line CD is $y=\frac{1}{3}x-3$, \\
$\therefore$ $\left\{\begin{array}{l}y=\frac{1}{3}x-3\\ y=\frac{1}{3}x^{2}+\frac{8}{3}x-3\end{array}\right.$ \\
Solving these, we get: $\left\{\begin{array}{l}x_{1}=0\\ y_{1}=-3\end{array}\right.$, $\left\{\begin{array}{l}x_{2}=-7\\ y_{2}=-\frac{16}{3}\end{array}\right.$ \\
$\therefore$ The point D is $\boxed{\left(-7,-\frac{16}{3}\right)}$.


Question: As illustrated, it is known that the parabola $y=-\frac{3}{16}(x+4)(x-4)$ intersects the $x$ axis at points $A$ and $B$, and intersects the $y$ axis at point $C$. The radius of $\odot C$ is 2. Let $G$ be a moving point on $\odot C$, and $P$ be the midpoint of $AG$. What is the minimum value of $AG$? What is the maximum value of $OP$?

Solution: \textbf{Solution:} As shown in the figure, when $y=0$, we have $0=-\frac{3}{16}(x+4)(x-4)$. Solving this yields: $x_{1}=-4$, $x_{2}=4$. Therefore, we have point A ($-4$, $0$) and point B ($4$, $0$). Hence, $OB=OA=4$. When $x=0$, we have $y=3$. Therefore, point C ($0$, $3$) and hence $OC=3$; Since the radius of circle C is 2, therefore $CG=2$. Since point P is the midpoint of AG and point O is the midpoint of AB, we have $OP=\frac{1}{2}BG$. Hence, when BG is at its maximum, the value of OP is maximum. Therefore, when points B, C, G lie on the same straight line, BG is at its maximum, consequently, OP reaches its maximum value. In $\triangle$BOC, we have $BC=\sqrt{3^2+4^2}=5$. Therefore, $BG=CG+BC=2+5=7$. Thus, the maximum value of OP is $\boxed{\frac{7}{2}}$; As illustrated, connecting AC and intersecting circle C at point $G_{1}$, then point G coincides with point $G_{1}$. At this moment, the length of AG is the shortest. Since $OC$ perpendicularly bisects $AB$, we have $AC=BC=5$. Thus, the minimum value of AG is $5-2=\boxed{3}$.


Question: As shown in the figure, points $A\left(5a-1, 2\right)$ and $B\left(8, 1\right)$ both lie on the graph of the inverse proportion function $y=\frac{k}{x}\left(k\ne 0\right)$. Point $P$ is a moving point on the line $y=x$. What is the minimum value of $PA+PB$?

Solution:
\textbf{Solution:} Since points A ($5a-1$, 2) and B (8, 1) are both on the graph of the inverse proportion function $y=\frac{k}{x}$,\\
we have $(5a-1) \times 2 = 8$,\\
thus, $a = 1$,\\
therefore, A (4, 2) and B (8, 1),\\
therefore, the symmetrical point of A about the line $y=x$ is A' (2, 4),\\
thus, $AB = \sqrt{(8-2)^{2} + (1-4)^{2}} = 3\sqrt{5}$,\\
therefore, the minimum value of $PA + PB$ is $\boxed{3\sqrt{5}}$.


Question: As shown in the figure, it is known that a parabola passing through point $B(4, 1)$, given by $y=\frac{1}{2}x^{2}-\frac{5}{2}x+c$, intersects with the coordinate axes at points A and C as shown. Lines AC, BC, and AB are connected. Within the first quadrant, a moving point M is on the parabola. A perpendicular line $AM$ is drawn from point M to meet the y-axis at point P. When point P is above point A, and $\triangle AMP$ is similar to $\triangle ABC$, what are the coordinates of point M?

Solution:
\textbf{Solution:} Substituting point B $(4,1)$ into $y=\frac{1}{2}x^{2}-\frac{5}{2}x+c$, we get:\\
$\frac{1}{2}\times 1^{2}-\frac{5}{2}\times 1+c=4$\\
$\therefore c=3$\\
Hence, the equation of the parabola is $y=\frac{1}{2}x^{2}-\frac{5}{2}x+3$\\
Letting x=0, we find y=3,\\
$\therefore$A$(0,3)$\\
Letting y=0, we get $\frac{1}{2}x^{2}-\frac{5}{2}x+3=0$\\
Solving this yields $x_{1}=2$, $x_{2}=3$\\
$\therefore$C$(3,0)$\\
$\therefore$AC$=\sqrt{(3-0)^{2}+(0-3)^{2}}=3\sqrt{2}$\\
Because B$(4,1)$\\
$\therefore$BC$=\sqrt{(4-3)^{2}+(1-0)^{2}}=\sqrt{2}$, AB$=\sqrt{(0-4)^{2}+(3-1)^{2}}=2\sqrt{5}$\\
$\therefore AC^{2}+BC^{2}=AB^{2}$\\
$\therefore \triangle ABC$ is a right triangle, and $\frac{BC}{AC}=\frac{1}{3}$,\\
Draw MG perpendicular to the y-axis at G from the point M, then $\angle$MGA=90$^\circ$,\\
Let the x-coordinate of point M be x, and since M is to the right of the y-axis, we have x>0, thus MG=x,\\
Because PM$\perp$MA, $\angle$ACB=90$^\circ$,\\
$\therefore \angle AMP=\angle ACB=90^\circ$,\\
(1) As shown in the figure, when $\angle MAP=\angle CBA$, then $\triangle MAP\sim \triangle CBA$,\\
$\therefore \frac{AM}{MP}=\frac{BC}{AC}=\frac{1}{3}$\\
Similarly, $\triangle AGM\sim \triangle AMP$\\
$\therefore \frac{AG}{MG}=\frac{AM}{MP}=\frac{1}{3}$\\
$\therefore$AG$=\frac{1}{3}$MG$=\frac{1}{3}$x, therefore M$(x,3+\frac{1}{3}x)$,\\
Substituting M$(x,3+\frac{1}{3}x)$ into $y=\frac{1}{2}x^{2}-\frac{5}{2}x+3$, we obtain\\
$\frac{1}{2}x^{2}-\frac{5}{2}x+3=3+\frac{1}{3}x$,\\
Solving this, $x_{1}=0$ (discard), $x_{2}=\frac{17}{3}$,\\
$\therefore$3+$\frac{1}{3}$x=3+$\frac{17}{9}=\frac{44}{9}$\\
$\therefore$M$\left(\frac{17}{3},\frac{44}{9}\right)$;\\
In summary, the coordinates of point M are $\boxed{\left(\frac{17}{3},\frac{44}{9}\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the perpendiculars to the x-axis and y-axis through the point $M(-3, 2)$ intersect the graph of the inverse proportion function $y=\frac{4}{x}$ at points A and B, respectively. What is the area of the quadrilateral MAOB?

Solution:
\textbf{Solution}: As shown in the figure\\
Substitute $y=2$ into $y=\frac{4}{x}$ to find $x=2$\\
$\therefore B\left(2,2\right)$, $C\left(0,2\right)$\\
Substitute $x=-3$ into $y=\frac{4}{x}$ to find $y=-\frac{4}{3}$\\
$\therefore A\left(-3,-\frac{4}{3}\right)$, $D\left(-3,0\right)$\\
$\therefore {S}_{\text{quadrilateral }MAOB}={S}_{\triangle BOC}+{S}_{\triangle AOD}+{S}_{\text{rectangle }MDOC}$\\
$=\frac{1}{2}OC\times BC+\frac{1}{2}OD\times AD+OD\times OC$\\
$=\frac{1}{2}\times 2\times 2+\frac{1}{2}\times 3\times \frac{4}{3}+3\times 2$\\
$=\boxed{10}$


Question: As shown in the figure, in the Cartesian coordinate system, the side \(OB\) of the quadrilateral \(AOBD\) coincides with the positive half of the x-axis, \(AD \parallel OB\), \(DB \perp x\)-axis, and the diagonals \(AB\), \(OD\) intersect at point \(M\). It is given that \(AD:OB = 2:3\), and the area of \(\triangle AMD\) is 4. If the graph of the inverse proportion function \(y=\frac{k}{x}\) exactly passes through point \(M\), what is the value of \(k\)?

Solution:
\textbf{Solution:} Construct MH $\perp$ OB at H.\\
Since AD $\parallel$ OB,\\
$\therefore$ $\triangle ADM \sim \triangle BOM$,\\
$\therefore$ $\frac{S_{\triangle ADM}}{S_{\triangle BOM}} = \left(\frac{AD}{OB}\right)^{2} = \frac{4}{9}$,\\
Since $S_{\triangle ADM} = 4$,\\
$\therefore$ $S_{\triangle BOM} = 9$,\\
Since DB $\perp$ OB, MH $\perp$ OB,\\
$\therefore$ MH $\parallel$ DB,\\
$\therefore$ $\frac{OH}{HB} = \frac{OM}{DM} = \frac{OB}{AD} = \frac{3}{2}$,\\
$\therefore$ OH = $\frac{3}{5}$OB,\\
$\therefore$ $S_{\triangle MOH} = \frac{3}{5} \times S_{\triangle OBM} = \frac{27}{5}$,\\
Since $\frac{k}{2} = \frac{27}{5}$,\\
$\therefore$ k = $\boxed{\frac{54}{5}}$.


Question: As shown, the area of the paper triangle $\triangle ABC$ is $12\,\text{cm}^2$, where one side $BC$ has a length of $6\,\text{cm}$. After two cuts, it is assembled into a seamless, non-overlapping rectangle $BCDE$. What is the perimeter of the rectangle in $\,\text{cm}$?

Solution:
\textbf{Solution:} Extend AT to intersect BC at point P,\\
$\because$AP$\perp$BC,\\
$\therefore$ $\frac{1}{2}$•BC•AP=12,\\
$\therefore$ $\frac{1}{2}$×6×AP=12,\\
$\therefore$AP=4 (cm),\\
According to the problem, AT=PT=2 (cm),\\
$\therefore$BE=CD=PT=2 (cm),\\
$\because$DE=BC=6cm,\\
$\therefore$The perimeter of rectangle BCDE is 6+6+2+2=\boxed{16} (cm).


Question: As shown in the figure, point $C$ is the midpoint of segment $BD$, $AD=3\,\text{cm}$, $AC=7\,\text{cm}$. What is the length of $AB$ in centimeters?

Solution:
\textbf{Solution:} Given that $AD=3\,cm$ and $AC=7\,cm$,

thus, $CD=AC-AD=4\,cm$,

since point $C$ is the midpoint of line segment $BD$,

it follows that $BC=CD=4\,cm$,

therefore, $AB=AC+BC=7\,cm+4\,cm=\boxed{11\,cm}$.


Question: As shown in the figure, point $C$ is on line segment $AB$, it is known that $AB=8$, $CB=3$. If $D$ is the midpoint of line segment $AB$, what is the length of line segment $CD$?

Solution:
\textbf{Solution:} Given that D is the midpoint of line segment AB, and AB=8,\\
it follows that $BD=\frac{1}{2}AB=4$,\\
since $CB=3$,\\
thus $CD=BD-CB=\boxed{1}$.


Question: As shown in the figure, within the quadrilateral $ABCD$, points E and F are the midpoints of sides AD and BC, respectively. Extend CD to point G such that DG=CD. Construct quadrilateral $DGME$ outside of quadrilateral $ABCD$ using DG and DE as sides. Connect BM to intersect AD at point N, and connect FN. If DG=DE=1 and $\angle ADC=60^\circ$, what is the length of FN?

Solution:
\textbf{Solution:} As shown in the figure: connect EF, AF\\
$\because$ quadrilateral $DGME$, $DG=CD=1$,\\
$\therefore$ $DC\parallel EF$,\\
$\because$ $\angle ADC=60^\circ$\\
$\therefore$ $\angle AEF=60^\circ$\\
$\because$ points E, F are the midpoints of sides AD, BC respectively\\
$\therefore$ $AB\parallel EF$, $AE=DE$\\
$\therefore$ Quadrilateral ABFE is a parallelogram, $\angle BAN=\angle MEN$\\
$\because$ $DG=DC$, $DG=DE$\\
$\therefore$ $AE=EF=AB=ME=1$\\
$\because$ $\angle AEF=60^\circ$\\
$\therefore$ $\triangle AEF$ is an equilateral triangle\\
In $\triangle ABN$ and $\triangle EMN$\\
$\left\{\begin{array}{l}
ME=AB \\
\angle MEN=\angle BAN \\
\angle MNE=\angle ANB
\end{array}\right.$\\
$\therefore$ $\triangle ABN \cong \triangle EMN$ (by SAA)\\
$\therefore$ $AN=NE$\\
$\therefore$ $NE=\frac{1}{2}AE=\frac{1}{2}$\\
$\therefore$ $FN\perp AE$\\
$\therefore$ $FN=\sqrt{1^2-\left(\frac{1}{2}\right)^2}=\boxed{\frac{\sqrt{3}}{2}}$


Question: As shown in the figure, in the rhombus $ABCD$, it is known that $\angle A : \angle B = 1 : 2$, and the perimeter of the rhombus is $16$. What is the area of the rhombus?

Solution:
\textbf{Solution:} As shown in the figure: \\
Draw DE$\perp$BC at point E through point D,\\
Since in rhombus ABCD, the perimeter is 16,\\
Therefore, AD=AB=4, $AB\parallel CD$,\\
Therefore, $\angle A + \angle B = 180^\circ$,\\
Since the ratio $\angle A\colon\angle B=1\colon2$,\\
Therefore, $\angle A = 60^\circ$,\\
Therefore, $\angle ADE = 30^\circ$,\\
Therefore, AE=$\frac{1}{2}AD=2$,\\
Therefore, DE=$\sqrt{AD^{2}-AE^{2}}=2\sqrt{3}$,\\
Therefore, the area $S$ of rhombus ABCD is $AB\cdot DE=\boxed{8\sqrt{3}}$.


Question: As shown in the figure, in the Cartesian coordinate system, the quadrilateral $AOBC$ is a rectangle with point $O(0,0)$, point $A(5,0)$, and point $B(0,3)$. The rectangle $AOBC$ is rotated clockwise around the point $A$ to get the rectangle $ADEF$, where $O$, $B$, $C$ correspond to $D$, $E$, $F$, respectively. Let $K$ be the intersection point of the diagonals of rectangle $AOBC$, what is the maximum area of $\triangle KDE$?

Solution:
\textbf{Solution:} Let $A(5, 0)$, and $B(0, 3)$,\\
thus $OA=5$, $OB=3$,\\
since quadrilateral $AOBC$ is a rectangle,\\
thus $AC=OB=3$, $OA=BC=5$, $\angle OBC=\angle C=90^\circ$,\\
thus $AB=\sqrt{OA^{2}+OB^{2}}=\sqrt{25+9}=\sqrt{34}$\\
thus $BK=AK=\frac{1}{2}AB=\frac{\sqrt{34}}{2}$\\
Since rectangle $ADEF$ is obtained by rotating rectangle $AOBC$,\\
thus $AD=AO=5$,\\
As shown in the figure, when point $D$ is on segment $BK$, the area of $\triangle DEK$ is the smallest,\\
When point $D$ is on the extension of $BA$, the area of $\triangle D'E'K$ is the largest,\\
The maximum area $=\frac{1}{2}\times D'E' \times KD'=\frac{1}{2}\times 3\times \left(5+\frac{\sqrt{34}}{2}\right)=\boxed{\frac{30+3\sqrt{34}}{4}}$.


Question: As shown in the diagram, in rectangle ABCD, \(AB = 4\), \(BC = 6\). Fold \(\triangle ABC\) along \(AC\), making point \(B\) coincide with point \(E\), and the intersection of \(CE\) and \(AD\) is point \(F\). What is the length of \(DF\)?

Solution:
\textbf{Solution:}
Let rectangle ABCD be folded along the diagonal AC so that $\triangle ABC$ falls into the position of $\triangle ACE$,\\
therefore, AE=AB and $\angle E=\angle B=90^\circ$,\\
Furthermore, since ABCD is a rectangle,\\
it follows that AB=CD,\\
thus, AE=DC,\\
and $\angle AFE=\angle DFC$,\\
in triangles $\triangle AEF$ and $\triangle CDF$,\\
\[
\left\{\begin{array}{l}
\angle AFE=\angle CFD \\
\angle E=\angle D \\
AE=CD
\end{array}\right.
\]
thus, $\triangle AEF\cong \triangle CDF$ (AAS),\\
hence, EF=DF;\\
Since ABCD is a rectangle,\\
it follows that AD=BC=6, and CD=AB=4,\\
Since $\triangle AEF\cong \triangle CDF$,\\
it follows that FC=FA,\\
Let FA=x, then FC=x, FD=6−x,\\
in $\triangle CDF$, CF$^{2}$=CD$^{2}$+DF$^{2}$,\\
that is, $x^{2}=4^{2}+(6−x)^{2}$,\\
solving leads to $x=\frac{13}{3}$,\\
therefore, $FD=6−x=\boxed{\frac{5}{3}}$.


Question: As shown in the figure, a rectangular piece of paper ABCD is folded along EF and MN, resulting in a pentagon GEFNM. Herein, the vertices A and D coincide at point G, and the overlapping section GHIJ forms a square, with vertex I located on EM. If FN equals $4\sqrt{5}$ cm and EM equals 10 cm, what is the length of side BC in centimeters?

Solution:
\textbf{Solution:} Construct IR$\perp$BC at R, as shown in the figure:\\
$\because$ A rectangle paper ABCD is folded along EF, MN, resulting in a pentagon GEFNM, where, vertex A coincides with D at point G, FN=$4\sqrt{5}$cm, EM=10cm,\\
$\therefore$ $\angle$IEF=$\angle$BFE=$\angle$EFJ, $\angle$IMN=$\angle$MNC=$\angle$MNH, AB=GJ=CD=HG,\\
$\therefore$ EI=FI, MI=NI,\\
Let EI=FI=xcm, then MI=NI=EM-EI=(10-x)cm,\\
$\because$ Quadrilateral GHIJ is a square,\\
$\therefore$ GH=HI=IJ=GJ, $\angle$HIJ=$90^\circ$=$\angle$FIN,\\
In $\triangle$FIN, $EI^{2}+NI^{2}=FN^{2}$,\\
$\therefore$ $x^{2}+(10-x)^{2}=(4\sqrt{5})^{2}$,\\
Solving this, we get $x=5-\sqrt{15}$ or $x=5+\sqrt{15}$,\\
Assume $x=5-\sqrt{15}$ (The result is the same for $x=5+\sqrt{15}$),\\
Then EI=FI=($5-\sqrt{15}$)cm, MI=NI=($5+\sqrt{15}$)cm,\\
$\because$2S$\triangle$FIN=FN$\cdot$IR=FI$\cdot$NI,\\
$\therefore$ $IR=\frac{FI\cdot NI}{FN}=\frac{(5-\sqrt{15})(5+\sqrt{15})}{4\sqrt{5}}=\frac{\sqrt{5}}{2}$cm,\\
$\therefore$ AB=CD=$\frac{\sqrt{5}}{2}$cm,\\
$\therefore$ GJ=HG=$\frac{\sqrt{5}}{2}$cm,\\
$\because$ Quadrilateral GHIJ is a square,\\
$\therefore$ HI=JI=$\frac{\sqrt{5}}{2}$cm,\\
$\therefore$ HN=HI+NI=$\left(\frac{\sqrt{5}}{2}+5+\sqrt{15}\right)$cm, $FJ=JI+FI=\left(\frac{\sqrt{5}}{2}+5-\sqrt{15}\right)$cm,\\
$\therefore$ CN=HN=$\left(\frac{\sqrt{5}}{2}+5+\sqrt{15}\right)$cm, BF=FJ=$\left(\frac{\sqrt{5}}{2}+5-\sqrt{15}\right)$cm,\\
$\therefore$ BC=CN+FN+BF=$\left(\frac{\sqrt{5}}{2}+5+\sqrt{15}+4\sqrt{5}+\frac{\sqrt{5}}{2}+5-\sqrt{15}\right)=\boxed{10+5\sqrt{5}}$cm.