Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $\angle A=36^\circ$, $BD$ bisects $\angle ABC$ and intersects $AC$ at point $D$, $CE$ bisects $\angle ACB$ and intersects $BD$ at point $E$. If $AD= 5\sqrt{5}-5$, what is the length of $BE$?

Solution:
\textbf{Solution:} Given $\because$ $\angle A=36^\circ$ and $AB=AC$,\\
$\therefore$ $\angle ABC=\angle ACB=72^\circ$,\\
$\because$ BD bisects $\angle ABC$ and CE bisects $\angle ACB$,\\
$\therefore$ $\angle ABD=\angle DBC=36^\circ$ and $\angle ACE=\angle BCE=36^\circ$,\\
$\therefore$ $BD=AD=5\sqrt{5}-5$ and $BE=CE=CD$,\\
$\because$ $\angle BDC=\angle CDE$\\
and $\angle DBC=\angle DCE$,\\
$\therefore$ $\triangle BDC \sim \triangle CDE$,\\
$\therefore$ $\frac{DE}{CD}=\frac{CD}{BD}$,\\
$\therefore$ $\frac{BD-BE}{BE}=\frac{BE}{BD}$,\\
Solving this gives: $BE^2+(5\sqrt{5}-5)BE-(5\sqrt{5}-5)^2=0$,\\
Solutions are: $BE=15-5\sqrt{5}$ or $BE=-10$ (discard the negative value),\\
$\therefore$ $BE=\boxed{15-5\sqrt{5}}$.


Question: As shown in the figure, in right triangle $ \triangle ABC$, $\angle ABC=90^\circ$, $AB=BC=2$, $AE$ is the median on the side $BC$, and $BD$ is drawn perpendicular to $AE$ at point $H$ and intersects $AC$ at point $D$. What is the length of $AD$?

Solution:
\textbf{Solution:} As shown in the figure, draw $DF\perp BC$ at point $F$ through point $D$,\\
$\because$ $AE$ is the median of $BC$, $BC=2$,\\
$\therefore BE=\frac{1}{2}BC=1$,\\
$\because$ in $Rt\triangle ABC$, $\angle ABC=90^\circ$, $AB=BC=2$,\\
$\therefore AC=\sqrt{AB^{2}+BC^{2}}=2\sqrt{2}$, $\angle C=45^\circ$,\\
$\therefore Rt\triangle CDF$ is an isosceles right triangle,\\
$\therefore CF=DF$, $CD=\sqrt{CF^{2}+DF^{2}}=\sqrt{2}CF$,\\
Let $CF=DF=x(x>0)$, then $BF=2-x$, $CD=\sqrt{2}x$,\\
$\because \angle ABC=90^\circ$, $BD\perp AE$,\\
$\therefore \angle ABD+\angle DBF=\angle ABD+\angle EAB=90^\circ$,\\
$\therefore \angle DBF=\angle EAB$,\\
In $\triangle BDF$ and $\triangle AEB$, $\left\{\begin{array}{l}\angle DBF=\angle EAB \\ \angle BFD=\angle ABE=90^\circ \end{array}\right.$,\\
$\therefore \triangle BDF\sim \triangle AEB$,\\
$\therefore \frac{BF}{AB}=\frac{DF}{BE}$, i.e., $\frac{2-x}{2}=\frac{x}{1}$,\\
Solving gives $x=\frac{2}{3}$,\\
$\therefore CD=\sqrt{2}x=\frac{2}{3}\sqrt{2}$,\\
$\therefore AD=AC-CD=2\sqrt{2}-\frac{2}{3}\sqrt{2}=\boxed{\frac{4}{3}\sqrt{2}}$,


Question: In the figure, within $\triangle ABC$ and $\triangle ADE$, $\angle ACB=\angle AED=90^\circ$, $\angle ABC=\angle ADE$. Connecting BD and CE, if $AC:BC=3:4$, what is $BD:CE$?

Solution:
\textbf{Solution:} Given that $\angle ACB=90^\circ$ and $AC\colon BC=3\colon 4$, we have $\frac{AB}{AC}=\frac{5}{3}$.\\
Since $\angle ACB=\angle AED=90^\circ$ and $\angle ABC=\angle ADE$, we conclude that $\triangle ABC \sim \triangle ADE$, leading to $\frac{AB}{AD}=\frac{AC}{AE}$ and $\angle DAE=\angle BAC$. Therefore, $\angle DAB=\angle EAC$, implying that $\triangle DAB\sim \triangle EAC$, and thus $\frac{BD}{CE}=\frac{AB}{AC}=\boxed{\frac{5}{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle A=30^\circ$, and $BC=2$. Let $D$ be any point on $AC$, and let $F$ be the midpoint of $AB$. Draw line $BD$. Let $E$ be a point on $BD$ such that $CD^{2}=DE\cdot BD$. Draw line $EF$. What is the minimum value of $EF$?

Solution:
\textbf{Solution:} In $\triangle CED$ and $\triangle BDC$,\\
$\because CD^2=DE\cdot BD$,\\
$\therefore \frac{CD}{DB}=\frac{DE}{DC}$,\\
$\because \angle EDC=\angle CDB$,\\
$\therefore \triangle CDE\sim \triangle BDC$,\\
$\therefore \angle DEC=\angle DCB=90^\circ$,\\
$\therefore \angle BEC=180^\circ-\angle DEC=90^\circ$,\\
As shown in the diagram, let Q be the midpoint of BC, then EQ= $\frac{1}{2}$ BC=1,\\
$\because \angle ACB=90^\circ$ and $\angle A=30^\circ$, BC=2,\\
$\therefore AB=2BC=4$,\\
$\therefore AC= \sqrt{AB^2-BC^2}=\sqrt{4^2-2^2}=2\sqrt{3}$,\\
$\because F$ is the midpoint of AB, and Q is the midpoint of BC,\\
$\therefore FQ= \frac{1}{2} AC= \sqrt{3}$,\\
$\because EF\geq FQ-EQ$, when and only when points E, F, and Q are collinear, EF can reach $\sqrt{3}-1$,\\
$\therefore The minimum value of EF is \boxed{\sqrt{3}-1}$.


Question: As shown in the figure, in $\triangle ABC$, $EG \parallel BC$, if $\frac{EG}{BC} = \frac{1}{3}$, then what is the value of $\frac{AF}{AD}$?

Solution:
\textbf{Solution:} Given $EG \parallel BC$,\\
thus $\triangle AEG \sim \triangle ABC$,\\
so $\frac{AE}{AB}=\frac{EG}{BC}=\frac{1}{3}$,\\
Given $EG \parallel BC$,\\
thus $\triangle AEF \sim \triangle ABD$,\\
so $\frac{AF}{AD}=\frac{AE}{AB}=\boxed{\frac{1}{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $DE\parallel BC$, $S_{\triangle ADE} = S_{\text{trapezoid } DBCE}$, then what is the ratio $DE: BC$?

Solution:
\textbf{Solution:} Given that, $S_{\triangle ADE}=S_{\text{trapezoid} DBCE}$,\\
then $S_{\triangle ADE} : S_{\triangle ABC} = 1 : 2$,\\
since $DE \parallel BC$,\\
it follows that $\triangle ADE \sim \triangle ABC$,\\
let the similarity ratio be $k$,\\
then the ratio of the areas is $k^2 = \frac{1}{2}$,\\
thus the similarity ratio is $\frac{\sqrt{2}}{2}$,\\
therefore $DE : BC = \boxed{\frac{\sqrt{2}}{2}}$.


Question: As shown in the figure, it is known that $AB \parallel CD \parallel EF$ and $BD: DF = 2: 5$. What is the value of $\frac{AC}{AE}$?

Solution:
\textbf{Solution:} Given $\because$ AB$\parallel$CD$\parallel$EF,\\
$\therefore$ AC:CE=BD:DF,\\
$\because$ BD:DF$=2:5$,\\
$\therefore$ AC:CE= BD:DF$=2:5$, i.e., CE$=\frac{5}{2}$AC,\\
$\therefore$ AE$=\frac{7}{2}$AC,\\
$\therefore$ AC:AE$=2:7=\boxed{\frac{2}{7}}$.


Question: As shown in the figure, it is known that quadrilateral $ABCD$ is a rectangle, point $E$ is on the extension line of $BA$, and $AE=AD$, $EC$ intersects $AD$, $BD$ at points $F$, $G$ respectively. If $AF=AB$, what is the value of $AD\colon AB$?

Solution:
\textbf{Solution:} Let quadrilateral ABCD be a rectangle.\\
Therefore, DC $\parallel$ AB,\\
thus $\angle$DCE = $\angle$AEC, $\angle$CDA = $\angle$EAD\\
therefore, $\triangle EAF \sim \triangle CDF$\\
hence, $\frac{AE}{AF} = \frac{CD}{DF}$\\
Let AB = AF = CD = x, AE = AD = y,\\
then we have $x^2 - y^2 + xy = 0$\\
Divide both sides of the equation by $x^2$, we have $1 - \left(\frac{y}{x}\right)^2 + \frac{y}{x} = 0$\\
Let $\frac{y}{x}$ be t, then we have $t^2 - t - 1 = 0$\\
Solving this, we get $t_1 = \frac{1 + \sqrt{5}}{2}$, $t_2 = \frac{1 - \sqrt{5}}{2}$ (discard this root)\\
thus, $t = \frac{y}{x} = \frac{1 + \sqrt{5}}{2}$\\
Therefore, $\frac{AD}{AB} = \boxed{\frac{1 + \sqrt{5}}{2}}$.


Question: As shown in the figure, straight lines AB and CD intersect at point O. If $\angle 1 + \angle 2 = 100^\circ$, what is the measure of $\angle BOC$?

Solution:
\textbf{Solution:} Since $\angle 1+\angle 2=100^\circ$ and $\angle 1=\angle 2$, \\
it follows that $\angle 1=\angle 2=50^\circ$. \\
Since $\angle 1+\angle BOC=180^\circ$, \\
it follows that $\angle BOC=\boxed{130^\circ}$.


Question: As shown in the figure, the lines $AB$ and $CD$ intersect at point $O$, with $OE\perp CD$. If $\angle AOE=50^\circ$, what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Since $OE \perp CD$,\\
it follows that $\angle EOC=90^\circ$,\\
which means $\angle EOA+\angle AOC=90^\circ$,\\
Since $\angle AOE=50^\circ$,\\
it follows that $\angle AOC=90^\circ-50^\circ=40^\circ$,\\
Therefore, $\angle BOD=\angle AOC=\boxed{40^\circ}$.


Question: As shown in the diagram, lines AB and CD intersect at point O, and OA bisects $\angle EOC$ where $\angle EOC=110^\circ$. What is the measure of $\angle BOC$?

Solution:
\textbf{Solution:} Since OA bisects $\angle EOC$ and $\angle EOC=110^\circ$,\\
it follows that $\angle AOC= \frac{1}{2}\angle COE=55^\circ$\\
Given $\angle AOC+\angle BOC$ form a straight angle,\\
it implies $\angle BOC=180^\circ-55^\circ=\boxed{125^\circ}$


Question: As shown in the figure, point $O$ is on the line $AB$, $OD$ bisects $\angle COB$, and $\angle AOC=118^\circ$. What is the degree measure of $\angle COD$?

Solution:
\textbf{Solution:} Since $\angle AOC$ and $\angle BOC$ are adjacent supplementary angles,\\
we have $\angle AOC+\angle BOC=180^\circ$,\\
Given that $\angle AOC=118^\circ$,\\
thus $\angle BOC=180^\circ-118^\circ=62^\circ$,\\
Since $OD$ bisects $\angle BOC$,\\
therefore $\angle COD=\frac{1}{2}\angle BOC=\boxed{31^\circ}$.


Question: As shown in the figure, it is known that the length of line segment $AB=20\,\text{cm}$, point C is on line $AB$, and $AC=4\,\text{cm}$. M and N are the midpoints of $AC$ and $BC$, respectively. What is the length of $MN$? in cm.

Solution:
\textbf{Solution:} Given that $AB=20\,cm$ and $AC=4\,cm$,\\
thus, $BC=AB-AC=20-4=16\,cm$,\\
since M and N are the midpoints of $AC$ and $BC$, respectively,\\
it follows that $MC=\frac{1}{2}AC$ and $CN=\frac{1}{2}BC$,\\
thus, $MN=\frac{1}{2}(AC+BC)=\frac{1}{2}(4+16)=\boxed{10}\,cm$.


Question: As shown in the figure, in $\triangle ABC$, $AB=3$, $AC=7$, and $D$, $E$ are the midpoints of $AB$, $AC$, respectively. What could be the length of $DE$?

Solution:
\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively, \\
it follows that $DE$ is the midsegment of $\triangle ABC$, \\
hence $DE=\frac{1}{2}BC$. \\
In $\triangle ABC$, with $AB=3$ and $AC=7$, the triangle inequality theorem implies that $7-3<BC<7+3$, i.e., $4<BC<10$, \\
thus $2<DE<5$. \\
Therefore, the answer is: $\boxed{2<DE<5}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle CAB=90^\circ$. Three squares are constructed on the sides of $\triangle ABC$. A line is drawn from point H perpendicular to $ED$ at point J, and $HE$ is connected. Extend $HE$ to intersect $FC$ at point K. If $K$ is the midpoint of $FC$, and the area of $\triangle ABC$ minus the area of $\triangle EHJ$ equals 16, then what is the length of $AB$?

Solution:
\textbf{Solution:} Extend $HJ$ and $FC$ to intersect at point N.\\
$\because$ Quadrilateral $EABD$ is a square and $HJ\perp ED$,\\
$\therefore$ Quadrilateral $ECNJ$ is a rectangle.\\
$\therefore$ $CE=NJ$, $EJ=CN$.\\
$CE\parallel HN$, $ED\parallel CN$.\\
$\therefore$ $\angle ECH=\angle MHN$.\\
$\because$ $\angle ECH+\angle BCA=90^\circ$ and $\angle CBA+\angle BCA=90^\circ$,\\
$\therefore$ $\angle ECH=\angle CBA$.\\
$\therefore$ $\angle CBA=\angle MHN$.\\
$\therefore$ $\triangle HCN\cong \triangle BCA$.\\
$\therefore$ $CN=CA$, $HN=AB$.\\
Let $CN=CA=a$, $HN=AB=b$.\\
$\therefore$ $CE=b-a$, $HJ=HN-JN=b-(b-a)=a$.\\
$\therefore$ $HJ=EJ=a$.\\
$\therefore$ $\angle HEJ=45^\circ$.\\
Also, $\because$ $ED\parallel KN$,\\
$\therefore$ $\angle EKC=45^\circ$.\\
$\therefore$ $\triangle EKC$ is an isosceles right triangle.\\
$\therefore$ $CK=CE=b-a$.\\
$\because$ K is the midpoint of $CF$, we have $CK=\frac{1}{2}a$.\\
$\therefore$ $\frac{1}{2}a=b-a$.\\
$\therefore$ $b=\frac{3}{2}a$.\\
$\because$ ${S}_{\triangle ABC}-{S}_{\triangle EHJ}=16$,\\
$\therefore$ $\frac{1}{2}\times a\times b-\frac{1}{2}a^2=16$.\\
$\therefore$ Solving gives $a=8$.\\
$\therefore$ $b=\frac{3}{2}\times 8=\boxed{12}$.


Question: As shown in the figure, the graph of the inverse proportion function $y=\frac{1}{x}$ intersects the sides AB and BC of the rectangle ABCO at points E and F, respectively, where points A and C are on the coordinate axes. If $AB=(n+1)AE$, what is the area of the quadrilateral OEBF?

Solution:
\textbf{Solution:} As shown in the diagram, connect OB.\\
$\because$ AB=$(n+1)$AE\\
$\because$ BE＝nAE,\\
$\therefore$ S$_{\triangle OBE}$＝n$\cdot$S$_{\triangle OAE}$,\\
$\because$ E, F lie on y＝$\frac{1}{x}$ and quadrilateral AOCB is a rectangle,\\
$\therefore$ S$_{\triangle AEO}$＝S$_{\triangle OCF}$＝$\frac{1}{2}$, S$_{\triangle OBC}$＝S$_{\triangle OBA}$,\\
$\therefore$ S$_{\triangle OBE}$＝S$_{\triangle OBF}$＝$\frac{1}{2}$n,\\
$\therefore$ S$_{\text{quadrilateral}OEBF}$＝\boxed{n}.


Question: As shown in the diagram of rhombus ABCD, where $\angle BAD=120^\circ$ and $AC=4$, what is the perimeter of the rhombus?

Solution:
\textbf{Solution:} In rhombus ABCD, we have AB=AC\\
$\because$ $\angle$BAD = 120$^\circ$\\
$\therefore$ $\angle$ABC = 60$^\circ$\\
$\therefore$ $\triangle ABC$ is an equilateral triangle\\
That is, AB=AC=BC=4\\
Thus, the perimeter of the rhombus is \boxed{16}


Question: As shown in the figure, AC is the diagonal of square ABCD, E is a point on BC, with $BE=1$. By folding $\triangle ABE$ along AE so that point B falls on point F on AC, what is the length of AB?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a square, \\
$\therefore$ AB = BC, $\angle$BCD = $\angle$B = $90^\circ$, $\angle$ECF = $\frac{1}{2}$ $\angle$BCD = $45^\circ$,\\
By the property of folding: $\angle$AFE = $\angle$B = $90^\circ$, FE = BE = 1,\\
$\therefore$ $\angle$CFE = $90^\circ$,\\
$\therefore$ $\triangle$ CEF is an isosceles right-angled triangle,\\
$\therefore$ CE = $\sqrt{2}$ FE = $\sqrt{2}$,\\
$\therefore$ BC = BE + CE = 1 + $\sqrt{2}$,\\
$\therefore$ AB = BC = \boxed{1 + \sqrt{2}};


Question: As shown in the figure, $P$ is any point on the diagonal $AC$ of the square $ABCD$, with $PE \perp AB$ at $E$, and $PF \perp BC$ at $F$. If $AC= \sqrt{2}$, then what is the value of $PE+PF$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a square,\\
thus $\angle A=90^\circ$, and $AB=BC$,\\
thus $AB^2+BC^2=AC^2$,\\
since $AC=\sqrt{2}$,\\
thus $AB=BC=1$,\\
since quadrilateral $ABCD$ is a square,\\
thus $\angle BAC=\angle BCA=45^\circ$,\\
since $PE\perp AB$ at $E$, and $PF\perp BC$ at $F$,\\
thus quadrilateral $PEBF$ is a rectangle, and $\triangle AEP$ and $\triangle PFC$ are isosceles right triangles,\\
thus $PF=BE$, $PE=AE$,\\
thus $PE+PF=AB=\boxed{1}$.


Question: As shown in the figure, in the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Given that $AC = 8$ and $BD = 12$, what is the area of the rhombus $ABCD$?

Solution:
\textbf{Solution:} In the rhombus $ABCD$, $AC=8$, $BD=12$,\\
since $AC\perp BD$,\\
therefore the area of the rhombus $ABCD$ $=\frac{1}{2}AC\cdot BD=\frac{1}{2}\times 8\times 12=\boxed{48}$.


Question: In the diagram, within $\triangle ABC$, $\angle A=90^\circ$, $AB=8$, $AC=6$, point $D$ is a moving point on $BC$. Lines are drawn from $D$ such that $DE\perp AB$ and $DF\perp AC$, with $E$ and $F$ being the feet of the perpendiculars, respectively. What is the minimum value of $EF$?

Solution:
\textbf{Solution:} As shown in the figure, connect CD,
\[
\because \angle BAC = 90^\circ, AB = 8, AC = 6,
\]
\[
\therefore BC = \sqrt{AB^2 + AC^2} = 10,
\]
\[
\because DE \perp AB, DF \perp CA,
\]
\[
\therefore \angle DEA = \angle DFA = 90^\circ,
\]
\[
\because \angle BAC = 90^\circ,
\]
\[
\therefore Quadrilateral FDEA is a rectangle,
\]
\[
\therefore EF = DA,
\]
Given that a perpendicular segment is the shortest, when DA $\perp$ BC, the length of segment DA is the minimum, and hence the length of segment EF is also the minimum,
At this point, $S_{\triangle ABC} = \frac{1}{2}AB \cdot AC = \frac{1}{2}BC \cdot AD$,
\[
\text{indicating } \frac{1}{2} \times 8 \times 6 = \frac{1}{2} \times 10 \cdot AD
\]
Solving gives: DA $= \frac{24}{5} = 4.8$
\[
\therefore \text{The minimum value of } EF \text{ is } \boxed{4.8}
\]


Question: As shown in the figure, in the parallelogram \(ABCD\), the diagonals \(AC\) and \(BD\) intersect at point \(O\). Point \(E\) is the midpoint of \(AD\). If \(OE = 10\), what is the length of \(AB\)?

Solution:
\textbf{Solution:} Since the quadrilateral $ABCD$ is a parallelogram, \\
$\therefore OB=OD$, which means point $O$ is the midpoint of $BD$, \\
Given that point $E$ is the midpoint of $AD$, \\
$\therefore OE=\frac{1}{2}AB$, \\
Given $OE=10$, \\
$\therefore AB=2OE=\boxed{20}$.


Question: As shown in the figure, quadrilateral $ABCD$ is a rectangle, point $E$ is on the extension of segment $CB$, connecting $DE$ intersects $AB$ at point $F$, $\angle AED = 2\angle DEC$, $G$ is the midpoint of $DF$, if $BE=1, DF=8$, then what is the length of $AB$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\
it follows that $\angle ABC=\angle BAD=90^\circ$, $AD\parallel BC$,\\
therefore, $\angle ADG=\angle DEC$,\\
since G is the midpoint of $DF$ and $DF=8$,\\
it follows that AG=FG=DG=4,\\
thus, $\angle GAD=\angle GDA=\angle DEC$,\\
since $\angle AGE=\angle GAD+\angle GDA=2\angle DEC$ and $\angle AED=2\angle DEC$,\\
it follows that $\angle AEG=\angle AGE$,\\
therefore, AE=AG=4,\\
in right triangle $ABE$,\\
$AB=\sqrt{AE^{2}-EB^{2}}=\sqrt{4^{2}-1^{2}}=\sqrt{15}$.\\
Therefore, $AB=\boxed{\sqrt{15}}$.


Question: As shown in the figure, in the square $ABCD$, the diagonal $AC=4$. What is the perimeter of the square $ABCD$?

Solution:
\textbf{Solution:} Given that the diagonal $AC=4$,\\
hence $\sqrt{AB^{2}+BC^{2}}=4$,\\
since in square ABCD $AB=BC$,\\
thus $AB=BC=2\sqrt{2}$,\\
therefore, the perimeter of square ABCD is: $2\sqrt{2}\times 4=\boxed{8\sqrt{2}}$.


Question: As shown in the figure of rhombus $ABCD$, if $\angle ABC=80^\circ$, then what is the measure of $\angle BDA$?

Solution:
\textbf{Solution:} Since ABCD is a rhombus,\\
$\therefore$ $\angle ADC = \angle ABC = 80^\circ$, and DB bisects $\angle ADC$ and $\angle ABC$,\\
$\therefore$ $\angle BDA = \angle BDC = 40^\circ$,\\
Therefore, $\angle BDA = \boxed{40^\circ}$


Question: As shown in the figure, in $ \triangle ABC$, $\angle BAC=120^\circ$, point D is the midpoint of BC, and point E is a point on AC, such that $ AB+AE=EC$. If $ DE=2$, what is the length of AB?

Solution:
\textbf{Solution:} Extend $CA$ to point $F$ such that $AF=AB$, and connect $BF$.\\
Since $\angle BAC=120^\circ$,\\
it follows that $\angle BAF=180^\circ-\angle BAC=180^\circ-120^\circ=60^\circ$,\\
and since $AF=AB$,\\
triangle $ABF$ is equilateral,\\
thus $AB=AF=BF$,\\
Given $AB+AE=EC$,\\
it follows that $AF+AE=EC$, namely $EF=EC$,\\
hence $E$ is the midpoint of $FC$,\\
Given $D$ is the midpoint of $BC$,\\
it follows that $DE$ is the midline of triangle $BCF$,\\
thus $BF=2DE$,\\
Given $DE=2$,\\
thus $AB=BF=\boxed{4}$.


Question: As shown in the figure, quadrilateral $OABC$ is a rectangular piece of paper placed in a Cartesian coordinate system, where $O$ is the origin, point $A$ is on the negative half-axis of the $x$-axis, and point $C$ is on the positive half-axis of the $y$-axis, with $OA=10$ and $OC=8$. Point $D$ is taken on side $OC$, and the paper is folded along $AD$, with point $O$ landing on point $E$ on side $BC$. What are the coordinates of point $D$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\
thus AB=OC=8 and BC=OA=10,\\
since the paper is folded along AD, bringing point O to point E on side BC,\\
thus AE=AO=10 and DE=DO,\\
in right triangle $\triangle ABE$, where AB=8 and AE=10,\\
thus BE=$\sqrt{AE^{2}-AB^{2}}=6$,\\
thus CE=BC-BE=4,\\
let OD=x, then DE=x, and DC=8-x,\\
in right triangle $\triangle CDE$, since $DE^{2}=CD^{2}+CE^{2}$,\\
thus $x^{2}=(8-x)^{2}+4^{2}$,\\
thus x=5,\\
thus the coordinates of point D are \boxed{(0, 5)}.


Question: As shown in the figure, in $ \triangle ABC$, $ \angle ACB=90^\circ$, $BC=4$, and $CD$ is the median. Extend $CB$ to point $F$ so that $BF=BC$, and connect $DF$. Let $E$ be the midpoint of $DF$, and connect $BE$. If $BE= 2$, what is the length of $AC$?

Solution:
\textbf{Solution:} Given $\because BF=BC$, and E is the midpoint of DF,\\
$\therefore BE=\frac{1}{2}CD$,\\
$\because BE=2$,\\
$\therefore CD=4$,\\
$\because$ In $\triangle ABC$ with $\angle ACB=90^\circ$, BC=4, and CD being the median,\\
$\therefore AB=2CD=8$, $AC=\sqrt{AB^{2}-BC^{2}}=\sqrt{8^{2}-4^{2}}=\boxed{4\sqrt{3}}$.


Question: As shown in the figure, the quadrilateral ABCD is a rhombus, with the diagonal AC = 8 and BD = 6. If DH $\perp$ AB at point H, what is the length of DH?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, AC=8, and BD=6, \\
therefore AC$\perp$BD, OA=OC=4, and OB=OD=3, \\
thus AB=5cm, \\
therefore the area of rhombus ABCD, $S_{\text{rhombus} ABCD} = \frac{1}{2} AC \cdot BD = AB \cdot DH$, \\
thus $DH = \frac{AC \cdot BD}{2AB} = \frac{6 \times 8}{2 \times 5} = \boxed{4.8}$.


Question: As shown in the figure, in the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, and $H$ is the midpoint of side $AD$. If the perimeter of the rhombus $ABCD$ is $24$, what is the length of $OH$?

Solution:
\textbf{Solution:} Since in the rhombus $ABCD$, the perimeter is 24, \\
it follows that $OA=OC$, $CD=24\div 4=6$, \\
since $H$ is the midpoint of side $AD$, \\
it follows that $OH=\frac{1}{2}CD=\boxed{3}$.


Question: As shown in the figure, in the right-angled triangle $ABC$, where $\angle ACB = 90^\circ$, $AC = 3$, and $BC = 4$, point $M$ is a point on side $AB$ (not coinciding with points $A$ or $B$). Construct $ME \perp AC$ at point $E$ and $MF \perp BC$ at point $F$. What is the minimum value of $EF$?

Solution:
\textbf{Solution:} Connect CM, \\
$\because$ ME $\perp$ AC, and MF $\perp$ BC, \\
$\therefore \angle$ MEC = $\angle$ MFC = 90$^\circ$, \\
$\because \angle$ C = 90$^\circ$, \\
$\therefore$ Quadrilateral ECFM is a rectangle, \\
$\therefore$ EF = CM, \\
When CM $\perp$ AB, CM is at its shortest, as shown in the figure: \\
When CM $\perp$ AB, \\
$S_{\triangle ABC} = \frac{1}{2}AC \cdot BC = \frac{1}{2}AB \cdot CM$, \\
$\therefore \frac{1}{2} \times 3 \times 4 = \frac{1}{2}CM \cdot AB$, \\
$\because$ In Rt $\triangle$ ABC, \\
$AB = \sqrt{AC^2 + BC^2} = \sqrt{3^2 + 4^2} = 5$, \\
$\therefore \frac{1}{2} \times 3 \times 4 = \frac{1}{2}CM \times 5$, \\
$\therefore$ CM = 2.4, \\
$\therefore$ The minimum value of CM is 2.4, \\
$\therefore$ EF = CM = 2.4, \\
$\therefore$ The minimum value of EF is $\boxed{2.4}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=4$, and $BC=3$. A square $ABDE$ is constructed with $AB$ as its side. What is the area of square $ABDE$?

Solution:
\textbf{Solution:} In $\triangle ABC$, $\angle C=90^\circ$, $AC=4$, and $BC=3$,\\
$\therefore$ $AB=\sqrt{AC^2+BC^2}=\sqrt{4^2+3^2}=5$,\\
$\therefore$ the area of the square $ABDE$ is $AB^2=5^2=\boxed{25}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is a rhombus with the coordinates of points $A$ and $B$ being $(\sqrt{3}, 0)$ and $(0, 1)$, respectively. If points $C$ and $D$ are on the coordinate axes, what is the perimeter of the rhombus $ABCD$?

Solution:
Solution: Since the coordinates of points $A$ and $B$ are $(\sqrt{3}, 0)$ and $(0,1)$ respectively,\\
$\therefore OB=1$, $OA=\sqrt{3}$,\\
$\because \angle BOA=90^\circ$,\\
$\therefore AB=\sqrt{OB^2+OA^2}=\sqrt{1+3}=2$,\\
$\because$ quadrilateral $ABCD$ is a rhombus,\\
$\therefore AB=BC=CD=DA=2$,\\
$\therefore$ the perimeter of rhombus $ABCD$ is equal to $4\times 2=\boxed{8}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=6$, $AD=10$. If point $E$ is on side $BC$ and $AE$ bisects $\angle DEB$, what is the length of $BE$?

Solution:
\textbf{Solution}: In rectangle $ABCD$, \\
$\because$ $AD\parallel BC$, \\
$\therefore$ $\angle AEB=\angle EAD$, \\
Also, $\because$ $AE$ bisects $\angle DEB$, \\
$\therefore$ $\angle AEB=\angle AED$, \\
$\therefore$ $\angle EAD=\angle AED$, \\
$\therefore$ $AD=DE=10$, \\
In right triangle $\triangle CDE$, \\
$\because$ $DE=10$, $DC=AB=6$, \\
According to the Pythagorean theorem, we have: $CE=\sqrt{DE^{2}-DC^{2}}=\sqrt{10^{2}-6^{2}}=8$, \\
Also, $\because$ $BC=AD=10$, \\
$\therefore$ $BE=BC-CE=10-8=\boxed{2}$.


Question: As shown in the diagram, in the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, and $E$ is the midpoint of $AB$, with $AC=8$ and $BD=6$. What is the length of $OE$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\
$\therefore$AC$\perp$BD, AO=CO, DO=BO, \\
Since AC$=8$, BD$=6$, \\
$\therefore$AO$= \frac{1}{2}AC=\frac{1}{2}\times 8=4$, OD$= \frac{1}{2}BD=\frac{1}{2}\times 6=3$, \\
In $\triangle AOD$, AD$= \sqrt{OD^{2}+OA^{2}}=\sqrt{3^{2}+4^{2}}=5$, \\
Also, since point E is the midpoint of side AB, and O is the midpoint of BD, \\
$\therefore$OE$= \frac{1}{2}AB=\frac{1}{2}\times 5=\frac{5}{2}=\boxed{2.5}$.


Question: As shown in the figure, the side length of square $ABCD$ is 8, $M$ is on $DC$, and $DM=2$, $N$ is a moving point on $AC$, then what is the minimum value of $DN+MN$?

Solution:
\textbf{Solution:} Let's connect BN, BD, and BM, where BM intersects AC at point E, \\
given that ABCD is a square, then AC and BD are perpendicular bisectors of each other, \\
$\therefore ND=NB$, \\
when point N and E are not coincident, in $\triangle NBM$ we have $NB+NM>BM$, \\
when point N and E coincide, $NB+NM=BM$, \\
$\therefore NB+NM\geq BM$, hence the minimum value of $DN+MN$ is $BM$, \\
since ABCD is a square, we have $BC=CD=8$, $\angle BCD=90^\circ$, \\
$\therefore CM=CD-DM=8-2=6$, \\
$\therefore BM= \sqrt{BC^{2}+CM^{2}}=\boxed{10}$, \\
thus, the minimum value of $DN+MN$ is $10$.


Question: As shown in the diagram, two square pieces of paper with areas of 8 and 16 are placed without overlapping inside the rectangle $ABCD$. What is the area of the blank part in the diagram?

Solution:
Solution: Given the information, we find that:
the length of rectangle ABCD is $\sqrt{16}+\sqrt{8}=4+2\sqrt{2}$, and its width is 4,
$\therefore$ the area of rectangle ABCD is $4\left(4+2\sqrt{2}\right)=16+8\sqrt{2}$,
$\therefore$ the area of the blank part is: $16+8\sqrt{2}-16-8=\boxed{8\sqrt{2}-8}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=90^\circ$, BF is the median to the side AC, and DE is the median of $\triangle ABC$. If DE$=10$, then what is the length of BF?

Solution:
\textbf{Solution:} Since DE is the median of $\triangle ABC$, if DE=10,\\
then AC=2DE=20,\\
In the right triangle $\triangle ABC$, where $\angle ABC=90^\circ$, BF is the median to side AC,\\
thus BF=$\frac{1}{2}$AC=\boxed{10}.


Question: As shown in the diagram, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle A=30^\circ$. Points D and E are the midpoints of sides AB and AC, respectively. Extend BC to point F so that CF = $\frac{1}{2}$BC. If EF = 4, what is the length of DE?

Solution:
\textbf{Solution:}
Draw $DC$,\\
since points D and E are respectively the midpoints of sides AB and AC,\\
it follows that $DE\parallel BC$ and $DE=\frac{1}{2}BC$\\
Extending BC to point F such that $CF=\frac{1}{2}BC$,\\
it follows that $CF=DE$ and $CF\parallel DE$\\
Therefore, quadrilateral $CDEF$ is a parallelogram\\
Thus $CD=EF=4$\\
Therefore $AB=2CD=8$\\
Since $\angle ACB=90^\circ$ and $\angle A=30^\circ$,\\
it follows that $BC=\frac{1}{2}AB=4$\\
Therefore $DE=\frac{1}{2}BC=\boxed{2}$


Question: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, and $OE \perp AB$ at point $E$. If $\angle ADC = 130^\circ$, what is the measure of $\angle AOE$?

Solution:
\textbf{Solution:} In rhombus $ABCD$, $\angle ADC=130^\circ$, \\
$\therefore \angle BAD=180^\circ-130^\circ=50^\circ$, \\
$\therefore \angle BAO= \frac{1}{2}\angle BAD= \frac{1}{2}\times50^\circ=25^\circ$, \\
$\because OE\perp AB$, \\
$\therefore \angle AOE=90^\circ-\angle BAO=90^\circ-25^\circ=\boxed{65^\circ}$.


Question: As shown in the diagram, in $\triangle ABC$, $\angle BAC=90^\circ$, $AB=3$, and $AC=4$. Squares $ABED$, $ACHI$, and $BCGF$ are constructed externally on sides $AB$, $AC$, and $BC$ of $\triangle ABC$, respectively. Lines $ED$ and $HI$ intersect at point $J$. Line $KF$ is drawn through point $F$ parallel to $HI$ and intersects $DE$ at point $K$. Line $GM$ is drawn through point $G$ parallel to $DE$ and intersects $HI$ and $KF$ at points $M$ and $L$, respectively. What is the area of quadrilateral $KLMJ$?

Solution:
\textbf{Solution:} In the right triangle $ABC$, $\angle BAC=90^\circ$, $AB=3$, $AC=4$. By the Pythagorean theorem, we have $BC=5$.\\
Since quadrilaterals $ABED$, $ACHI$, $BCGF$ are all squares,\\
it follows that each quadrilateral has four $90^\circ$ angles, with four sides being parallel and equal.\\
Since $KF\parallel HI$ and $GM\parallel DE$,\\
it follows that $\angle JKL=\angle KJM=90^\circ$.\\
Therefore, quadrilaterals $KLMJ$ and $DGIA$ are rectangles.\\
Hence, $ED=DA=AB=3$, $AI=AC=IH=4$, $BC=BF=CG=5$.\\
It follows that $DJ=AI=4$, $JI=DA=3$.\\
Extending $AC$ to $O$ gives $CO\perp ML$.\\
Since $\angle ACB+\angle GCO=90^\circ=\angle ACB+\angle ABC$,\\
it follows that $\angle ABC=\angle GCO$.\\
Since $\angle BAC=\angle COG$ and $BC=CG$,\\
it follows that $\triangle ABC\cong \triangle OCG$ (AAS).\\
Hence, $AB=OC=3$.\\
Thus, $\angle COG=90^\circ$.\\
Since $\angle JML=90^\circ$,\\
it follows that quadrilateral $COMH$ is a rectangle.\\
Therefore, $CO=HM=3$. Similarly, it can be shown that quadrilateral $EKQB$ is a rectangle.\\
It follows that $KE=QB=4$.\\
Hence, the area of quadrilateral $KLMJ$ is $HJ\cdot JM=(KE+ED+DJ)\cdot (JI+IH+HM)=(4+3+4)\times (3+4+3)=\boxed{110}$.


Question: As shown in the figure, E is a point on the diagonal BD of a square ABCD with a side length of 1, and $BE=BC$. Let P be any point on CE, and let PQ be perpendicular to BC at point Q, and PR be perpendicular to BE at point R. What is the value of $PQ+PR$?

Solution:
\textbf{Solution:} Draw BP and construct CM $\perp$ BD through C, \\
$\because$BC=BE, \\
$\therefore$Area of $\triangle BCE$ = Area of $\triangle BPE$ + Area of $\triangle BPC$ \\
= BC $\times$ PQ $\times \frac{1}{2}$ + BE $\times$ PR $\times \frac{1}{2}$ \\
= BE $\times$ (PQ+PR) $\times \frac{1}{2}$ \\
= BE $\times$ CM $\times \frac{1}{2}$, \\
$\therefore$ PQ+PR=CM, \\
$\because$ BE=BC=1, and BD is the diagonal of the square ABCD, \\
$\therefore$ BD=$\sqrt{2}$BC=$\sqrt{2}$, \\
$\because$ BC=CD, and CM$\perp$ BD, \\
$\therefore$ M is the midpoint of BD, \\
$\because$ $\triangle$BDC is a right triangle, \\
$\therefore$ CM=$\frac{1}{2}$BD=$\frac{\sqrt{2}}{2}$, \\
thus, the value of PQ+PR is $\boxed{\frac{\sqrt{2}}{2}}$.


Question: As shown in the figure, in parallelogram $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, and point $E$ is the midpoint of $BC$. If $OE = 6\,cm$, what is the length of $AB$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\
it follows that $AO = CO$, \\
since point E is the midpoint of CB, \\
thus OE is the median of $\triangle ABC$, \\
therefore $AB = 2OE$, \\
since $OE = 6\,cm$, \\
it follows that $AB = \boxed{12cm}$.


Question: As shown in the figure, the diagonals \(AC\) and \(BD\) of the rhombus \(ABCD\) intersect at point \(O\), where \(AC=8\) and \(BD=6\). If point \(E\) is the midpoint of \(BC\), what is the length of \(OE\)?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\
then OC=OA=$\frac{1}{2}AC=4$ and OB=OD=$\frac{1}{2}BD=3$, $\angle$BOC=$90^\circ$.\\
Thus, AB=$\sqrt{OA^{2}+OB^{2}}=5$.\\
Since point E is the midpoint of AB, and O is the midpoint of AC,\\
then OE is the median of $\triangle ABC$,\\
thus OE=$\frac{1}{2}AB=\boxed{\frac{5}{2}}$.


Question: As shown in the figure, in the Cartesian coordinate system, the point $A(3, 3)$ forms a rhombus $AOBC$ with $OA$ as one of its sides. What are the coordinates of point $C$?

Solution:
\textbf{Solution:} Let's draw AE$\perp$x-axis at E, and CF$\perp$x-axis at F,\\
then quadrilateral AEFC is a rectangle,\\
$\therefore$AE=CF, and AC=EF,\\
$\because$ point A is (3, 3),\\
$\therefore$AE=OE=3,\\
$\therefore$OA=$3\sqrt{2}$,\\
$\because$ quadrilateral AOBC is a rhombus,\\
$\therefore$AC=OA=$3\sqrt{2}$,\\
$\therefore$CF=AE=3, and OF=3+$3\sqrt{2}$,\\
$\therefore$ the coordinates of point C are (3+$3\sqrt{2}$, 3),\\
so, the answer is: The coordinates of point C are $\boxed{(3+3\sqrt{2}, 3)}$.


Question: As shown in the figure, the graph of a linear function intersects the y-axis at the point $\left(0, 2\right)$ and the x-axis at the point $\left(-1, 0\right)$. What is the range of the independent variable $x$ when $y \le 0$?

Solution:
\textbf{Solution:} From the graph, it is observed that when $y\le 0$, the graph of the linear function that meets the condition is below the x-axis,\\
as shown by the bold part in the figure,\\
$\because$ this part of the graph is to the left of the line $x=-1$,\\
$\therefore$ when $y\le 0$, the range of the independent variable x is $x\le -1$.\\
Hence, the answer is: \boxed{x\le -1}.


Question: As shown in the figure, the graphs of the linear function $y_{1}=mx\ (m\ne 0)$ and $y_{2}=x+3$ intersect at point $A(-1, 2)$. What is the solution set of the inequality $mx>x+3$ with respect to $x$?

Solution:
\textbf{Solution:} \\
Since $mx > x + 3$, \\
it follows that $y_1 > y_2$, \\
thus $x < -1$. \\
Hence, the answer is $\boxed{x < -1}$.


Question: As shown in the figure, the line $y=-\frac{3}{4}x+3$ intersects the x-axis and y-axis at points A and B, respectively. Point C is on the line segment OA, and the line segment OB is folded along BC, with point O falling on point D on the side AB. What is the equation of line BC?

Solution:
\textbf{Solution:} Since the line $y=-\frac{3}{4}x+3$ intersects the $x$ and $y$ axes at points $A$ and $B$ respectively,

let $x=0$, then $y=3$, let $y=0$, then $x=4$,

therefore, point $A(4,0)$, point $B(0,3)$,

therefore, OA=4, OB=3,

therefore, AB= $\sqrt{OB^{2}+OA^{2}}=5$,

since segment $OB$ is folded along $BC$, point $O$ falls on the point $D$ on side $AB$,

therefore, OB=BD=3, OC=CD, $\angle BOC=\angle BDC=90^\circ$,

therefore, AD=AB−BD=2,

since $AC^{2}=AD^{2}+CD^{2}$,

therefore, $(4−OC)^{2}=2^{2}+OC^{2}$,

therefore, OC=1.5,

therefore, point $C(1.5,0)$,

Assume the equation of line $BC$ is: $y=kx+3$,

therefore, $0=1.5k+3$,

therefore, $k=−2$,

therefore, the equation of line $BC$ is: $y=-2x+3$.

Hence, the answer is: $y=-2x+3=\boxed{-2x+3}$.


Question: The graph of the line $y_1=k_1x+b$ and the line $y_2=k_2x$ in the same Cartesian coordinate system is shown in the figure. What is the solution set of the inequality $k_1x+b\le k_2x$ with respect to $x$?

Solution: \textbf{Solution:} Given that ${k}_{1}x+b\le {k}_{2}x$ graphically represents that the line ${y}_{1}={k}_{1}x+b$ is below the line ${y}_{2}={k}_{2}x$. From the graph, it is known that the solution set is $\boxed{x\le -3}$.


Question: As shown in the figure, it is known that the line $y_1=x+a$ and $y_2=kx+b$ intersect at point $P(-1,2)$. What is the correct solution set of the inequality $x+a>kx+b$ with respect to $x$?

Solution:
\textbf{Solution:} Given the lines $y_1=x+a$ and $y_2=kx+b$ intersect at point P ($-1$, $2$), \\
for the inequality $x+a>kx+b$ to hold, it is necessary that $y_1=x+a$ is above $y_2=kx+b$,\\
According to the graph, when $x>-1$, $y_1=x+a$ is above $y_2=kx+b$,\\
Therefore, the solution set for the inequality $x+a>kx+b$ is $x>-1$,\\
Thus, the answer is: $\boxed{x>-1}$.


Question: As shown in the figure, the line $y=kx+3$ passes through the point $(2,0)$. What is the solution set of the inequality $kx+3>0$ with respect to $x$?

Solution:
\textbf{Solution:} Since the line $y=kx+3$ passes through the point $(2,0)$,\\
it follows that the solution set of the inequality $kx+3>0$ with respect to $x$ is $\boxed{x<2}$.


Question: As shown in the figure, the equation of line $l_1$ is $y_1=k_1x+b_1$, and the equation of line $l_2$ is $y_2=k_2x+b_2$. What is the solution set of the inequality $k_1x+b_1<k_2x+b_2$?

Solution:
\textbf{Solution:} Solving the inequality $k_1x + b_1 < k_2x + b_2$ implies that the line $l_1$ is below the line $l_2$.\\
From the graph of the function, we know that the range of $x$ is $x > -2$,\\
meaning the solution set for the inequality $k_1x + b_1 < k_2x + b_2$ is $x > -2$,\\
thus the answer is: $\boxed{x > -2}$.


Question: As shown in the figure, the lines $y=-x+b$ and $y=kx$ are represented in the same Cartesian coordinate system. What is the solution set of the inequality $-x+b<kx$ with respect to $x$?

Solution:
\textbf{Solution:} From the graph, it is known that the graphs of the two lines intersect at ($-1$, $-3$), \\
$\because$ $-x+b<kx$, \\
$\therefore$ select the part to the right of the intersection point, \\
thus, for $x>-1$, $-x+b<kx$, \\
so the answer is: $\boxed{x>-1}$


Question: As shown in the figure, the straight line $y=x+b$ intersects with the coordinate axes at two points. What is the range of $x$ when $y>0$?

Solution:
\textbf{Solution:} From the graph, it is known that the intersection point of the line with the x-axis is ($-2, 0$),\\
when $y>0$, the graph is above the x-axis, on the right side of the intersection point with the x-axis,\\
therefore, when $x>-2$, $y>0$\\
Hence, the answer is: $\boxed{x>-2}$


Question: As shown in the figure, a linear function, $y=-x-2$, intersects with the graph of $y=kx+b$ at point $P\left(2, n\right)$. What then is the solution set of the inequality $kx+b<-x-2$ in terms of $x$?

Solution: \textbf{Solution:} From the graph, it is known that when $x<2$, the graph of the line $y=kx+b$ is below the graph of the line $y=-x-2$. Therefore, the solution set of the inequality $kx+b<-x-2$ with respect to $x$ is $\boxed{x<2}$.


Question: As shown in the figure, in the Cartesian coordinate system, the two diagonals $AC$ and $BD$ of the quadrilateral $\square ABCD$ intersect at the origin $O$ of the Cartesian coordinate system. The coordinates of point $D$ are $\left(2, 1\right)$. What are the coordinates of point $B$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a parallelogram, and $O$ is the intersection point of diagonals $AC$ and $BD$,\\
it follows that points $B$ and $D$ are symmetrical with respect to the origin $O$,\\
since the coordinates of point $D$ are $\left(2, 1\right)$,\\
thus, the coordinates of point $B$ are $\boxed{\left(-2, -1\right)}$.


Question: Given the figure, the graph of the linear function $y_{1}=x+b$ intersects with the graph of the linear function $y_{2}=kx+4$ at point $P(1, 3)$. What is the solution set of the inequality $x+b\le kx+4$ with respect to $x$?

Solution:
\textbf{Solution:} From the graph, we have, when $x \leq 1$, $x + b \leq kx + 4$,

which means the solution set of the inequality $x + b \leq kx + 4$ with respect to $x$ is $\boxed{x \leq 1}$.


Question: As shown in the figure, the line $y=kx+b$ intersects the line $y=mx+n$ at point $C(1, 5)$. What is the solution set for the inequality $mx+n<kx+b$?

Solution:
\textbf{Solution:} From the graph, we can see that to the left of point C, the line $y=mx+n$ is above the line $y=kx+b$, and to the right of point C, the line $y=mx+n$ is below the line $y=kx+b$,\\
$\because$ The line $y=kx+b$ intersects the line $y=mx+n$ at point C(1, 5),\\
$\therefore$ The solution set of the inequality $mx+n<kx+b$ is $\boxed{x>1}$.


Question: As shown in the figure, the vertices of rectangle $OABC$ are $O(0, 0)$ and $B(-2, 2\sqrt{3})$. If the rectangle rotates counterclockwise about point $O$ at a rate of $60^\circ$ per second, what are the coordinates of the diagonal intersection point $D$ at the 2021st second?

Solution:
\textbf{Solution:} As shown in the figure, according to the problem, we know that $D(-1,\sqrt{3})$, and the trajectory of point $D$ is a circle centered at $O$ with $OD$ as the radius. Construct $DM\perp x$-axis, where $M$ is the foot of the perpendicular.\\
$\therefore$ $\tan\angle DOM=\frac{DM}{OM}=\sqrt{3}$\\
$\therefore$ $\angle DOM=60^\circ$\\
$\because$ $2021\times 60^\circ=336\times 360^\circ+300^\circ$\\
$\therefore$ it is known that at the 2021st second, the intersection point of the diagonals of the rectangle, $D$, is at point $E$, with $\angle DOE=60^\circ$\\
$\therefore$ $D$ and $E$ are symmetric with respect to the $y$-axis\\
$\therefore$ $E(1,\sqrt{3})$\\
Therefore, the answer is: $E(1,\sqrt{3})$, which means the coordinates of $D$ are $\boxed{E(1,\sqrt{3})}$


Question: Given the figure, the graphs of the linear functions $y_{1}=kx+4$ and $y_{2}=x+m$ intersect at point $P\left(1, 3\right)$. What is the solution set of the inequality $kx+4<x+m$ with respect to $x$?

Solution:
\textbf{Solution:} Given that $x>1$, the graph of the function $y_1=kx+4$ lies entirely below the graph of the function $y_2=x+m$, \\
$\therefore$ the solution set of the inequality $kx+4<x+m$ with respect to $x$ is $x>1$, \\
thus, the answer is: $\boxed{x>1}$.


Question: If the graph of the linear function \(y=kx+b(k\ne 0)\) is shown as in the figure, what is the solution set of the inequality \(kx+b<0\)?

Solution:
\textbf{Solution}: Given $y=kx+b\ (k \neq 0)$ intersects the $x$-axis at $(4,0)$,\\
$\therefore kx+b<0$ has the solution set: $x>4$,\\
thus, the answer is: $\boxed{x>4}$.


Question: As shown in the figure, the intersection point of the lines $y_1=k_1x+a$ and $y_2=k_2x+b$ is $(4, 8)$. What is the range of values of $x$ for which $y_1<y_2$?

Solution:
\textbf{Solution:} From the graph, it can be observed that when $x<4$, the line $y_1$ is below the line $y_2$. \\
Therefore, the range of $x$ for which $y_1<y_2$ is: $\boxed{x<4}$.


Question: Given that $y$ is a function of $x$, and the graph of the function is as shown, what is the range of the independent variable $x$ when $y>0$?

Solution:
\textbf{Solution:} When $y>0$, which means the part above the x-axis,\\
$\therefore$ the range of the independent variable $x$ is divided into two parts: $\boxed{x<-1$ and $1<x<2}$.\\
Therefore, the answer is: $x<-1$ and $1<x<2$.


Question: As shown in the figure, the horizontal distance between teaching buildings $AB$ and $CD$ is $BD=a\,m$. At point $C$ on top of teaching building $CD$, the angle of elevation to point $A$ on top of teaching building $AB$ is measured to be $\alpha$, and the angle of depression to point $B$ at the bottom of teaching building $AB$ is measured to be $\beta$. What is the height of teaching building $AB$?

Solution:
\textbf{Solution:} Construct $CE\perp AB$,\\
since $BD=a\,m$,\\
thus $CE=BD=a\,m$,\\
in $\triangle BCE$, $BE=CE\tan \beta = a\tan\beta$,\\
in $\triangle ACE$, $AE=CE\tan \alpha = a\tan\alpha$,\\
therefore $AB=a\tan\beta + a\tan\alpha$,\\
hence the height of $AB$ is $\boxed{a\tan\beta + a\tan\alpha}$.


Question: As shown in the figure, $A_1B_1$ is the orthogonal projection of the line segment $AB$ on the projection plane $P$, $AB=20\,\text{cm}$, $\angle ABB_1=70^\circ$. What is the length of the projection $A_1B_1$?

Solution:
\textbf{Solution:} Construct $AC \perp BB_1$ at point C, \\
$\therefore$ Quadrilateral $AA_1B_1C$ is a rectangle, \\
$\therefore AC=A_1B_1$, \\
In $\triangle ABC$, $AC=AB\cdot \sin\angle ABB_1=20\sin70^\circ$, \\
$\therefore A_1B_1=20\sin70^\circ$, \\
Therefore, the answer is: $A_1B_1=\boxed{20\sin70^\circ}$.


Question: As shown in the figure, quadrilateral $ABCD$ is a square with a side length of 4. A semicircle $\overset{\frown}{AB}$ is drawn with $B$ as the center and $BC$ as the radius. Point $E$ is located inside the quadrilateral, and $BE\perp CE$, $\angle BCE=30^\circ$. Connect $AE$. Calculate the area of the shaded region.

Solution:
\textbf{Solution:} Construct $EM\perp BC$ at point $M$, and $EN\perp AB$ at point $N$, as shown in the diagram.

Since $BE\perp CE$,

$\therefore \angle BEC=90^\circ$,

Since $\angle BCE=30^\circ$,

$\therefore \angle EBC=60^\circ$,

Since $EM\perp BC$,

$\therefore$ in $\triangle EMC$,

$\tan \angle ECM=\frac{EM}{MC}=\tan30^\circ=\frac{\sqrt{3}}{3}$,

$\therefore MC=\sqrt{3}EM$,

$\therefore$ in $\triangle EBM$,

$\tan \angle EBM=\frac{EM}{BM}=\tan60^\circ=\sqrt{3}$,

$\therefore BM=\frac{\sqrt{3}}{3}EM$,

Since $BM+MC=BC=4$,

$\therefore \frac{\sqrt{3}}{3}EM+\sqrt{3}EM=4$,

$\therefore EM=\sqrt{3}$,

$\therefore BM=\frac{\sqrt{3}}{3}EM=\frac{\sqrt{3}}{3}\times\sqrt{3}=1$,

Since $NE\perp AB$, $EM\perp BC$, and $\angle ABC=90^\circ$,

$\therefore$ quadrilateral $BMEN$ is a rectangle,

$\therefore NE=BM=1$,

Since $AB=BC=4$, $\angle ABC=90^\circ$,

$\therefore Area_{\triangle ABE}=\frac{1}{2}\times AB\times NE=\frac{1}{2}\times 4\times 1=2$, $Area_{\triangle BEC}=\frac{1}{2}\times BC\times EM=\frac{1}{2}\times 4\times\sqrt{3}=2\sqrt{3}$,

$Area_{\text{sector} ABC}=\pi\times AB^{2}\times\frac{90^\circ}{360^\circ}=\pi\times4^{2}\times\frac{1}{4}=4\pi$

$\therefore Area_{\text{shaded}}=4\pi-2-2\sqrt{3}=\boxed{4\pi-2-2\sqrt{3}}$


Question: As shown in the figure, in rhombus ABCD, where AB=2, and $\angle A=120^\circ$, points P, Q, and K are arbitrarily located on segments BC, CD, and BD, respectively. What is the minimum value of $PK+QK$?

Solution:
\textbf{Solution}: As shown in the figure, construct point $P^{\prime}$ symmetric to point P with respect to BD, and draw the intersection K between line $P^{\prime}Q$ and BD,\\
$\because$ Quadrilateral ABCD is a rhombus, AB=2, $\angle A=120^\circ$,\\
$\therefore$ AD=2, $\angle ADC=60^\circ$,\\
Draw AE$\perp$CD at E, then AE=$P^{\prime}Q$,\\
$\because$ AE=AD$\cdot\sin60^\circ=2\times\frac{\sqrt{3}}{2}=\sqrt{3}$,\\
$\therefore$ the distance from point $P^{\prime}$ to CD is $\sqrt{3}$,\\
$\therefore$ the minimum value of PK+QK is $\boxed{\sqrt{3}}$.


Question: As shown in the figure, in rectangle $ABCD$, point $E$ lies on $AD$, and $EC$ bisects $\angle BED$, $AB=\sqrt{3}$, $\angle ABE=30^\circ$, what is the length of $DE$?

Solution:
Solution: Since quadrilateral $ABCD$ is a rectangle,\\
it follows that $AB=CD=\sqrt{3}$, and $\angle A=\angle D=90^\circ$,\\
Since $\angle ABE=30^\circ$,\\
it follows that $\angle AEB=60^\circ$,\\
hence $\angle BED=120^\circ$,\\
Since $EC$ bisects $\angle BED$,\\
it follows that $\angle DEC=60^\circ$,\\
hence $\angle DCE=30^\circ$,\\
thus $CD=\sqrt{3}DE=\sqrt{3}$,\\
therefore, $DE=\boxed{1}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C = 90^\circ$, $\angle B = 30^\circ$, and $AB = 8$, what is the length of $BC$?

Solution:
\textbf{Solution:} In right triangle $\triangle ABC$, we have $\cos B = \frac{BC}{AB}$. \\
Thus, $BC = AB \cdot \cos B = 8 \times \cos 30^\circ = 8 \times \frac{\sqrt{3}}{2} = \boxed{4\sqrt{3}}$.


Question: As shown in the figure, there is a slope AB with the height of the top B from the ground BC being 30m. The gradient of slope AB, \(i = 1 \colon 2.5\), what is the horizontal distance AC of this slope?

Solution:
\textbf{Solution:} Since the slope $AB$ has a gradient $i = 1:2.5$, \\
hence $BC:AC = 1:2.5$, \\
since $BC = 30m$, \\
therefore $AC = 30 \times 2.5 = \boxed{75}m$


Question: As shown in the figure, the vertices of $\triangle ABC$ are all located on the grid points of a square grid. What is the value of $\tan C$?

Solution: \textbf{Solution:} As shown in the figure, in the lattice $\triangle ADC$, we have $AD=2$, $DC=4$, and $\tan C= \frac{AD}{DC} = \frac{2}{4}=\boxed{\frac{1}{2}}$.


Question: As shown in the figure, it is known that $AB$ is the diameter of $\odot O$, $BC$ is tangent to $\odot O$ at point $B$. Connect $AC$ and $OC$. If $\sin\angle BAC=\frac{1}{3}$, what is the value of $\tan\angle BOC$?

Solution:
\textbf{Solution:} Given that $AB$ is the diameter of circle $O$, and $BC$ is tangent to circle $O$ at point $B$, \\
$\therefore AB \perp BC$, \\
$\therefore \angle ABC = 90^\circ$, \\
$\because \sin\angle BAC = \frac{1}{3}$, \\
$\therefore$ let $BC = x$, $AC = 3x$, \\
$\therefore AB = \sqrt{AC^{2} - BC^{2}} = \sqrt{(3x)^{2} - x^{2}} = 2\sqrt{2}x$, \\
$\therefore OB = \frac{1}{2}AB = \sqrt{2}x$, \\
$\therefore \tan\angle BOC = \frac{BC}{OB} = \frac{x}{\sqrt{2}x} = \boxed{\frac{\sqrt{2}}{2}}$.


Question: As shown in the figure, $A$, $B$, $C$ are vertices of a small square, and the side length of each small square is 1. What is the value of $\sin\angle ACB$?

Solution:
\textbf{Solution:} As shown in the figure, draw AD $\perp$ BC at point D.\\
In $\triangle BCE$, $BC = \sqrt{BE^2 + CE^2} = \sqrt{3^2 + 1^2} = \sqrt{10}$.\\
$\therefore AC = BC = \sqrt{10}$.\\
$\therefore CH \perp AB$, and $CH = 3$.\\
$\because \frac{1}{2}BC \times AD = \frac{1}{2}AB \times CH$,\\
$\therefore AD = \frac{AB \times CH}{BC} = \frac{2 \times 3}{\sqrt{10}} = \frac{3\sqrt{10}}{5}$.\\
In $\triangle ADC$, $\sin\angle ACB = \frac{AD}{AC} = \frac{3\sqrt{10}}{5} \div \sqrt{10} = \boxed{\frac{3}{5}}$.


Question: As illustrated, one day, a large fishing vessel operating in the South China Sea suddenly encountered a dangerous situation. Upon receiving the distress signal, the relevant departments immediately dispatched a helicopter and a fishery administration ship that was patrolling in the South China Sea for rescue. When the helicopter reached a height of $3000\,m$ above the sea surface at point $C$, the depression angle to the fishery administration ship at point $A$ was measured to be $45^\circ$, and the depression angle to the distressed fishing vessel at point $B$ was measured to be $30^\circ$. What is the distance $AB$ between the fishery administration ship and the fishing vessel at this moment?

Solution:
\textbf{Solution:} As shown in the figure, according to the given information, CE$\parallel$BD,\\
$\therefore \angle CBA=30^\circ$, $\angle CAD=45^\circ$, and CD=3000m,\\
In $\triangle ACD$, AD=CD=3000m,\\
In $\triangle BCD$,\\
BD=$ \frac{CD}{\tan \angle CBA}$=$ \frac{3000}{\frac{\sqrt{3}}{3}}$=$ 3000\sqrt{3}$m,\\
$\therefore$AB=BD−AD=$ 3000\sqrt{3}$−3000=$ 3000\left(\sqrt{3}−1\right)$ (m),\\
Therefore, the answer is: $AB=\boxed{3000\left(\sqrt{3}-1\right)}$ (m).


Question: The photo wall made up of four squares is shown in Figure 1, and its planar geometry is shown in Figure 2. In it, the areas of the squares $ABCD$, $DEFG$, and $BIJK$ are $4\,\text{dm}^2$, $4\,\text{dm}^2$, and $16\,\text{dm}^2$ respectively. What is the area of square $AGHI$?

Solution:
\textbf{Solution:} Draw $DM \perp AG$ at point $M$, and draw $IN \perp BA$ intersecting the extension of $BA$ at point $N$,\\
since the areas of squares $ABCD$, $DEFG$, and $BIJK$ are $4 \, \text{square meters}$, $4 \, \text{square meters}$, and $16 \, \text{square meters}$ respectively,\\
therefore, $AD = 2$, $DG = 2$, $BI = 4$, $\angle IAG = \angle BAD = 90^\circ$,\\
therefore, $\angle IAB + \angle MAD = 180^\circ$,\\
also, since $\angle IAB + \angle IAN = 180^\circ$,\\
therefore, $\angle IAN = \angle MAD$,\\
let $AG = 2x$,\\
since $DA = DG = 2$, $DM \perp AG$,\\
therefore, $\cos \angle MAD = \frac{AM}{AD} = \frac{x}{2}$,\\
since $\cos \angle IAN = \frac{AN}{AI} = \frac{AN}{2x}$,\\
therefore, $\frac{x}{2} = \frac{AN}{2x}$,\\
therefore, $AN = x^2$,\\
therefore, $NI^2 = AI^2 - AN^2 = (2x)^2 - (x^2)^2 = 4x^2 - x^4$,\\
since $BI = 4$, $BN = BA + AN = 2 + x^2$, $\angle BNI = 90^\circ$,\\
therefore, $4^2 = (2+x^2)^2 + 4x^2 - x^4$,\\
solving this, we get $x^2 = \frac{3}{2}$,\\
therefore, the area of square $AGHI$ is: $(2x)^2 = 4x^2 = 4 \times \frac{3}{2} = \boxed{6}$.


Question: As shown in the figure, there is a clock face. If we connect points $B$ and $D$ that represent 2 o'clock and 10 o'clock respectively, and extend this line to intersect the tangent line through the 8 o'clock point at point $P$, and if the length of the tangent line $PC=2$, what is the radius of the clock face?

Solution:
\textbf{Solution:} Let the center of the clock be point $O$, and connect $BC$ and $OD$,\\
From the given information,\\
Point $O$ is on $BC$, $\angle DOC=2\times 30^\circ=60^\circ$,\\
$\therefore \angle DBC=\frac{1}{2}\angle DOC=30^\circ$,\\
$\because$ Line $PC$ is tangent to circle $O$ at point $C$,\\
$\therefore \angle BCP=90^\circ$,\\
$\therefore BC=\frac{CP}{\tan 30^\circ}=\frac{2}{\frac{\sqrt{3}}{3}}=2\sqrt{3}$,\\
$\therefore$ The radius of the clock face is $\boxed{\sqrt{3}}$.


Question: As shown in the figure, in $ \triangle ABC$, $\angle ACB=90^\circ$, $AC=BC$, point D is a moving point on side $AC$, connecting $BD$, a circle with $CD$ as its diameter intersects $BD$ at point E. If the length of $AB$ is 4, what is the minimum length of segment $AE$?

Solution:
\textbf{Solution:} Join CE. Since CD is the diameter,\\
$\therefore \angle CED=90^\circ=\angle BEC$,\\
$\therefore$ point E moves on the circle $\odot O$ with the midpoint O of BC as its center and BC as its diameter,\\
Join AO, intersecting $\odot O$ at point E, then at this moment AE=AO−OE is the minimum,\\
$\because \angle ACB=90^\circ$, $AC=BC$, $AB=4$,\\
$\therefore \angle ABC=\angle BAC=45^\circ$,\\
$\therefore AC=BC=AB\cdot\sin45^\circ=2\sqrt{2}$, $OB=OC=OE=\sqrt{2}$,\\
$\therefore AO=\sqrt{(2\sqrt{2})^2+(\sqrt{2})^2}=\sqrt{10}$,\\
$\therefore AE=\sqrt{10}-\sqrt{2}=\boxed{\sqrt{10}-\sqrt{2}}$.


Question: As shown in Figure 1, in $ \triangle DEF$, $\angle DEF=90^\circ$, $\angle D=30^\circ$. Let B be a moving point on the hypotenuse DF. Construct AB $\perp$ DF at B, intersecting side DE (or side EF) at point A. Let BD $= x$, and the area of $\triangle ABD = y$. Figure 2 is the graph of the function of y with respect to x. What is the maximum area of $\triangle ABD$?

Solution:
\textbf{Solution:} It can be observed from the diagram that point A reaches point E, the area of $\triangle ABD$ is maximized. At this juncture, $DB=12$, \\
$AB=DB\cdot \tan 30^\circ=12\times \frac{\sqrt{3}}{3}=4\sqrt{3}$, \\
$\therefore S_{\triangle ABD}=\frac{1}{2}\times 12\times 4\sqrt{3}=\boxed{24\sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $\cos A=\frac{1}{3}$, then what is the value of $\sin B$?

Solution:
\textbf{Solution:} Since $\angle C=90^\circ$,\\
it follows that $\angle A+\angle B=90^\circ$,\\
hence $\sin B=\cos A=\boxed{\frac{1}{3}}$.


Question: As shown in the figure, the top of a building is equipped with a billboard CD. Xiao Ming measures the angle of elevation to the bottom of the billboard D to be $60^\circ$ at point A at the foot of the hill. While walking up the hill to point B, he measures the angle of elevation to the top of the billboard C to be $45^\circ$. Given that the slope angle of AB is $30^\circ$, $AB=10$ meters, and $AE=15$ meters, what is the height of the billboard CD? (in meters)

Solution:
\textbf{Solution:} Draw perpendicular lines from point B to AE and DE, with the perpendicular feet being G and F, respectively, as shown in the diagram. \\
In right triangle $\triangle ABG$, $\angle BAG=30^\circ$ \\
$\therefore BG=\frac{1}{2}AB=5$ meters, $AG=AB\cdot \cos 30^\circ=10\times \frac{\sqrt{3}}{2}=5\sqrt{3}$ (meters) \\
$\therefore EG=AG+AE=(5\sqrt{3}+15)$ meters \\
$\because BG\perp AE$, $BF\perp ED$, $AE\perp ED$ \\
$\therefore$ Quadrilateral BGEF is a rectangle. \\
$\therefore EF=BG=5$ meters, $BF=EG=(5\sqrt{3}+15)$ meters \\
$\because \angle CBF=45^\circ$, $BF\perp ED$ \\
$\therefore \angle BCF=\angle CBF=45^\circ$ \\
$\therefore CF=BF=(5\sqrt{3}+15)$ meters \\
In right triangle $\triangle DAE$, $\angle DAE=60^\circ$, AE=15 meters \\
$\therefore DE=AE\cdot \tan\angle DAE=15\sqrt{3}$(meters) \\
$\therefore CD=CF+EF−DE=5\sqrt{3}+15+5−15\sqrt{3}=\boxed{20−10\sqrt{3}}$ meters


Question: As shown in the figure, in $\triangle ABC$, $BC=\sqrt{6}+\sqrt{2}$, $\angle C=45^\circ$, $AB=\sqrt{2}AC$, what is the length of $AC$?

Solution:
\textbf{Solution:} Draw line AD perpendicular to BC through point A, with D being the foot of the perpendicular, as shown in the diagram.\\
Let AC = x, then AB = $\sqrt{2}x$.\\
In $\triangle ACD$, AD = AC$\cdot\sin C = \frac{\sqrt{2}}{2}x$,\\
CD = AC$\cdot\cos C = \frac{\sqrt{2}}{2}x$;\\
In $\triangle ABD$, AB = $\sqrt{2}x$, AD = $\frac{\sqrt{2}}{2}x$,\\
$\therefore$BD = $\sqrt{AB^{2}-AD^{2}} = \frac{\sqrt{6}}{2}x$,\\
$\therefore$BC = BD + CD = $\frac{\sqrt{6}}{2}x + \frac{\sqrt{2}}{2}x = \sqrt{6} + \sqrt{2}$,\\
$\therefore x = \boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$. If $AB=5$, what is the length of $BC$?

Solution:
\textbf{Solution:} In right-angled triangle $\triangle ABC$, where $\angle C=90^\circ$ and $AB=5$, \\
it follows that $\sin A= \frac{BC}{AB}$, \\
thus $BC=AB\sin A=5\sin A$. \\
Therefore, $BC=\boxed{5\sin A}$.


Question: As shown in the figure, the waterfront slope $AB$ of the dam cross-section has a slope ratio of $1\colon \sqrt{3}$. If the dam height $BC$ is 4m, what is the length of $AC$?

Solution:
\textbf{Solution:} According to the given information, we have $\frac{BC}{AC}=\frac{1}{\sqrt{3}}$,\\
$\therefore$ $\frac{4}{AC}=\frac{1}{\sqrt{3}}$,\\
Solving this, we get: $AC=\boxed{4\sqrt{3}}$m.


Question: As shown in the diagram, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=BC=8$. Fold $\triangle ABC$ so that point A coincides with point D on side BC, with EF as the crease. If $AE=5$, what is the value of $\sin\angle BFD$?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $\angle ACB=90^\circ$ and AC=BC=8, \\
it follows that $\angle A=\angle B$, \\
By the properties of folding: $\triangle AEF\cong \triangle DEF$, \\
Hence, $\angle EDF=\angle A$, \\
Thus, $\angle EDF=\angle B$, \\
Therefore, $\angle CDE+\angle BDF+\angle EDF=\angle BFD+\angle BDF+\angle B=180^\circ$, \\
Hence, $\angle CDE=\angle BFD$. \\
Moreover, since AE=DE=5, \\
it follows that CE=8−5=3, \\
Therefore, in right-angled $\triangle ECD$, $\sin\angle CDE= \frac{CE}{ED}=\frac{3}{5}$, \\
Therefore, $\sin\angle BFD=\boxed{\frac{3}{5}}$.


Question: As shown in the figure, Xiaoming measured the elevation angle of the flagpole, $\angle DAC$, to be $30^\circ$ from point A on the school playground. He then moved 10 meters in the direction of AC to point B, where he measured the elevation angle, $\angle DBC$, to be $45^\circ$. What is the height of the flagpole, DC?

Solution:
\textbf{Solution:} Let the height of the flagpole $DC=x$ meters,\\
according to the given problem, we have $\angle ACD=90^\circ$, $AB=10$ meters,\\
since $\angle DBC=45^\circ$,\\
therefore $\angle BDC=90^\circ-\angle DBC=45^\circ$,\\
thus $\angle BDC=\angle DBC=45^\circ$,\\
hence $BC=CD=x$ meters,\\
thus $AC=AB+BC=(10+x)$ meters,\\
since $\angle DAC=30^\circ$,\\
therefore $\tan\angle DAC=\frac{CD}{AB}=\frac{\sqrt{3}}{3}$,\\
hence $\frac{x}{x+10}=\frac{\sqrt{3}}{3}$,\\
thus $3x=\sqrt{3}x+10\sqrt{3}$,\\
hence $(3-\sqrt{3})x=10\sqrt{3}$,\\
therefore $x=5(\sqrt{3}+1)$,\\
upon verification, when $x=5(\sqrt{3}+1)$, $x+10=5\sqrt{3}+15\ne 0$,\\
thus the height of the flagpole DC is $\boxed{5(\sqrt{3}+1)}$ meters.


Question: As shown in the figure, the cross-section of a certain dam of a reservoir is a trapezoid $ABCD$, with the height of the dam $DE = 5m$, and the slope ratio of slope $BC$ is $5\colon 12$. What is the length of slope $BC$?

Solution:
\textbf{Solution:} As shown in the diagram, draw line CF$\perp$AB through point C, with F being the foot of the perpendicular. Thus,\\
$\because$ Dam height $DE=5m$, and CF$\perp$AB,\\
$\therefore$DE=CF=5cm\\
Moreover, the slope ratio of slope $BC$ is $5:12$\\
$\therefore$BF=12cm,\\
In $\triangle BCF$\\
$BC= \sqrt{CF^{2}+BF^{2}}$\\
$= \sqrt{5^{2}+12^{2}}$\\
$=\boxed{13}$cm


Question: In right-angled triangle $ABC$, where $\angle C=90^\circ$, if $AC=6$ and $BC=8$, then what is the value of $\cos A$?

Solution:
\textbf{Solution:} In a right triangle $ABC$, with $\angle C=90^\circ$, $AC=6$, and $BC=8$,\\
it follows that $AB=\sqrt{AC^2+BC^2}=\sqrt{6^2+8^2}=10$,\\
therefore, $\cos A= \frac{AC}{AB}=\frac{6}{10}=\boxed{\frac{3}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AB=5$, and $AC=4$, what is the value of $\tan B$?

Solution:
\textbf{Solution:} Since $\angle C=90^\circ$ and $AB=5$, $AC=4$, \\
thus $BC=\sqrt{5^2-4^2}=3$,\\
therefore, $\tan B=\frac{AC}{BC}=\frac{4}{3}=\boxed{\frac{4}{3}}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, $C$ and $D$ are points on $\odot O$, and $\angle CDB=30^\circ$. A tangent to $\odot O$ passing through point $C$ intersects the extension of $AB$ at point $E$. What is the value of $\sin E$?

Solution:
\textbf{Solution:} Draw the line OC, \\
since $CE$ is a tangent to the circle $O$, \\
it follows that $OC \perp CE$, \\
that is, $\angle OCE = 90^\circ$, \\
since $\angle CDB = 30^\circ$, and angles $\angle COB$ and $\angle CDB$ are the central and inscribed angles that intercept the same arc $BC$, respectively, \\
it follows that $\angle COB = 2\angle CDB = 60^\circ$, \\
therefore $\angle E = 90^\circ - \angle COB = 30^\circ$, \\
therefore $\sin\angle E = \boxed{\frac{1}{2}}$.


Question: As shown in the figure, port A is located to the true east of observation point O, with OA = 4km. A ship sets sail from port A and arrives at point B after traveling in a direction of 15° north of east for a certain distance. At this moment, it is measured from observation point O that the ship is in a direction of 60° north of east. The question is, what is the distance the ship has traveled (i.e., the length of AB)?

Solution: \textbf{Solution:} Construct $AD \perp OB$ through point A, so that $AD= \frac{1}{2}OA=2\,\text{km}$. According to the problem statement, $\triangle ABD$ is an isosceles right triangle, hence $AB=2\sqrt{2}\,\text{km}$. Therefore, the distance traveled by the ship (i.e., the length of $AB$) is $AB=\boxed{2\sqrt{2}}\,\text{km}$.


Question: As shown in the figure, within a grid of small squares with side length $1$, points $A, B, C, D$ are all at the vertices of these small squares. Lines $AB$ and $CD$ intersect at point $O$. What is the value of $\cos\angle BOD$?

Solution:
\textbf{Solution:} Draw $CE$ and $DE$ as shown in the figure:\\
Since it is known from the figure that $\angle 1=\angle 2=\angle 3=\angle 4=\angle ABE=45^\circ$\\
Thus, $\angle CED=\angle 2+\angle 3=90^\circ$, $AB\parallel CE$\\
Therefore, $\angle BOD=\angle DCE$\\
Since the side length of the small square is $1$\\
Therefore, in $Rt\triangle CDE$, $CE=\sqrt{1^2+1^2}=\sqrt{2}$, $CD=\sqrt{1^2+3^2}=\sqrt{10}$\\
Therefore, $\cos\angle DCE=\frac{CE}{CD}=\frac{\sqrt{2}}{\sqrt{10}}=\frac{\sqrt{5}}{5}$\\
Therefore, $\cos\angle BOD=\cos\angle DCE=\boxed{\frac{\sqrt{5}}{5}}$.


Question: As shown in the figure, the line $y=\frac{1}{2}x-1$ intersects the x-axis at point B, and the hyperbola $y=\frac{k}{x}$ (where $x>0$) intersects at point A. A perpendicular line from point B to the x-axis intersects the hyperbola $y=\frac{k}{x}$ at point C, and $AB=AC$. What is the value of $k$?

Solution:
\textbf{Solution:} The line $y=\frac{1}{2}x-1$ intersects the x-axis at point B, thus: B(2,0),\\
Since AB=AC and BC is perpendicular to the x-axis, point A lies on the perpendicular bisector of BC, hence: C(2, $\frac{k}{2}$), A(4, $\frac{k}{4}$),\\
Substituting point A into the line equation $y=x-1$ yields: $k=\boxed{4}$.


Question: As shown in the figure, the axis of symmetry of the parabola is the line $x=1$, and it intersects the $x$-axis at the point $A(-1,0)$. What are the coordinates of the other intersection point?

Solution:
\textbf{Solution:} The axis of symmetry of the parabola is the line $x=1$. Given that the coordinates of point $A$ are $(-1,0)$,\\
the symmetry of the parabola allows us to find the coordinates of the other intersection point with the $x$-axis as $\boxed{(3,0)}$.


Question: As shown in the figure, in pentagon $ABCDE$, $AB \parallel CD$, and $\angle 1$, $\angle 2$, $\angle 3$ are the exterior angles of $\angle BAE$, $\angle AED$, and $\angle EDC$ respectively. If $\angle 1=32^\circ$ and $\angle 3=60^\circ$, what is the value of $\angle 2$?

Solution:
\textbf{Solution:} As shown in the figure,\\
since AB $\parallel$ CD,\\
therefore, $\angle$4 + $\angle$5 = 180$^\circ$,\\
also, since $\angle$1 + $\angle$3 = 92$^\circ$,\\
according to the exterior angle sum theorem of polygons, $\angle$1 + $\angle$2 + $\angle$3 + $\angle$4 + $\angle$5 = 360$^\circ$,\\
thus, $\angle$2 = 360$^\circ$ - 180$^\circ$ - 92$^\circ$ = \boxed{88^\circ}.


Question: As shown in the figure, the side AB of rectangle ABCD is parallel to the $y$-axis, and the coordinates of vertex $A$ are $(1, 2)$. Points $B$ and $D$ lie on the graph of the inverse proportion function $y=\frac{6}{x} \, (x>0)$. What is the area of rectangle ABCD?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle, with vertex A's coordinates being (1, 2),\\
we assume the coordinates of points B and D to be (1, y) and (x, 2) respectively,\\
since points B and D lie on the graph of the inverse proportion function $y=\frac{6}{x}$ (x>0),\\
we get y=6, x=3,\\
thus, point B is (1, 6), and point D is (3, 2),\\
hence, AB=6-2=4, AD=3-1=2,\\
therefore, the area of rectangle ABCD is AB$\cdot$AD=4$\times$2=$\boxed{8}$.


Question: As shown in the figure, the rhombus $ABCD$ has a side length of $10$, and $\angle A=60^\circ$. By successively connecting the midpoints of each side of the rhombus $ABCD$, quadrilateral ${A}_{1}{B}_{1}{C}_{1}{D}_{1}$ is obtained; by successively connecting the midpoints of each side of quadrilateral ${A}_{1}{B}_{1}{C}_{1}{D}_{1}$, quadrilateral ${A}_{2}{B}_{2}{C}_{2}{D}_{2}$ is obtained; by successively connecting the midpoints of each side of quadrilateral ${A}_{2}{B}_{2}{C}_{2}{D}_{2}$, quadrilateral ${A}_{3}{B}_{3}{C}_{3}{D}_{3}$ is obtained; and so on. Following this pattern, what is the perimeter of the quadrilateral ${A}_{2n}{B}_{2n}{C}_{2n}{D}_{2n}$?

Solution:
\textbf{Solution:} By utilizing the properties of mid-quadrilaterals, we know quadrilaterals $A_{1}B_{1}C_{1}D_{1}$ and $A_{3}B_{3}C_{3}D_{3}$ are rectangles, while $A_{2}B_{2}C_{2}D_{2}$ and $A_{4}B_{4}C_{4}D_{4}$ are rhombuses.\\
Since the side length of rhombus ABCD is 10, and $\angle A=60^\circ$,\\
therefore, $\triangle ADB$ and $\triangle AA_{1}D_{1}$ are equilateral triangles, and the perimeter of rhombus ABCD is $10\times 4=40$;\\
therefore, the perimeter of rhombus $A_{2}B_{2}C_{2}D_{2}$ is $\frac{1}{2}\times 40=20$;\\
The perimeter of rhombus $A_{4}B_{4}C_{4}D_{4}$ is $\frac{1}{{2}^{2}}\times 40$;\\
$ \dots $\\
therefore, the perimeter of quadrilateral ${A}_{2n}{B}_{2n}{C}_{2n}{D}_{2n}$ is $40\times \frac{1}{{2}^{n}}=5\times 8\times \frac{1}{{2}^{n}}=\boxed{\frac{5}{{2}^{n-3}}}$.


Question: As shown in the figure, fold a rectangular paper ABCD in the method illustrated twice, to form an isosceles right-angled triangle BEF. If \(BC = 1\), what is the length of \(AB\)?

Solution:
\textbf{Solution:} As illustrated,\\
since quadrilateral ABCD is a rectangle,\\
thus $\angle A = \angle ADC = \angle C = \angle B = 90^\circ$,\\
thus $BC = AD = 1$, $CD = AB$\\
After the first fold,\\
$\angle A = \angle DA^{\prime}E = 90^\circ$,\\
quadrilateral DAEA$^{\prime}$ is a square,\\
thus $\angle ADE = 45^\circ = \angle AED$\\
thus $AE = AD = 1$,\\
thus $DE = \sqrt{{1}^{2}+{1}^{2}} = \sqrt{2}$;\\
After the second fold,\\
$CD = DE = \sqrt{2}$\\
thus $AB = \boxed{\sqrt{2}}$.


Question: As shown in the figure, in square $ABCD$, $E$ is a point on side $CD$, and $F$ is a point on the extension of side $BC$, with $CE=CF$. If $\angle BEC=80^\circ$, what is the degree measure of $\angle EFD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a square,\\
it follows that BC=CD, and $\angle$BCE=$\angle$DCF=90$^\circ$,\\
since CE=CF,\\
it follows that $\triangle$BCE$\cong$$\triangle$DCF, $\angle$CFE=$\angle$CEF=45$^\circ$,\\
therefore $\angle$CFD=$\angle$BEC=80$^\circ$,\\
therefore $\angle$EFD=$\angle$CFD−$\angle$CFE=80$^\circ$−45$^\circ=\boxed{35^\circ}$.


Question: As shown in the figure, the side length of the square $ABCD$ is $2\, \text{cm}$. What is the area of the shaded region?

Solution: \textbf{Solution:} It is known from symmetry that $S_{\text{shaded area}} = \frac{1}{2}S_{\text{square} ABCD} = \frac{1}{2} \times 2 \times 2 = \boxed{2}\text{cm}^2$.


Question: As shown in the figure, the diagonals of square $ABCD$ intersect at point $O$. $E$ is any point on $BC$, $EG \perp BD$ at point $G$, and $EF \perp AC$ at point $F$. If $AC=10$, what is the value of $EG+EF$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a square,\\
$\therefore \angle OBC = \angle OCB = 45^\circ$, AC$\perp$BD, and OC=$\frac{1}{2}$AC=5,\\
$\because EG \perp$BD and EF$\perp$AC,\\
$\therefore quadrilateral EFOG is a rectangle, and  \triangle BEG$, $\triangle CEF$ are isosceles right triangles,\\
$\therefore EG=FO, CF=EF,\\
\therefore EG+EF=FO+CF=CO=\boxed{5}$


Question: As shown in the figure, the side length of the square cardboard ABCD is 4, with E and F being the midpoints of AB and BC, respectively. If it is cut along the dashed line as shown in the lower left figure and assembled into a "small villa" as shown below, what is the area of the shaded part?

Solution:
\textbf{Solution:} According to the diagram, the top of the "small villa" is an isosceles triangle, whose area is half of the square ABCD's area, and the area of the "small villa's" bottom is half of the square ABCD's area as well, and the bottom consists of two equal rectangles.\\
Therefore, the area of the shaded part in the diagram is $\frac{1}{4}$ of the square ABCD's area,\\
i.e., the area of the shaded part = $\frac{1}{4}S_{\text{square ABCD}}=\frac{1}{4}\times 4^{2}=\boxed{4}$.


Question: As shown in the figure, in rhombus $ABCD$, we have $AE=1$, and $AF=BE=2$. If $P$ is a moving point on diagonal $BD$, what is the minimum value of $EP+FP$?

Solution:
\textbf{Solution:} As shown in the diagram, draw $EE'\perp BC$, intersecting $BD$ at $E'$, then draw line $E'F$ intersecting $BD$ at $P'$, and connect $EP'$,\\
since quadrilateral $ABCD$ is a rhombus,\\
therefore $\angle ABD = \angle CBD$,\\
therefore $BD$ is the perpendicular bisector of $EE'$,\\
therefore $P'E = P'E'$,\\
therefore $PE + PF > P'E' + P'F' = E'F$,\\
therefore the minimum value of $EP + FP$ is $E'F$,\\
since $B'E = BE = AF$, $BE \parallel AF$,\\
therefore quadrilateral $ABE'F$ is a parallelogram,\\
therefore $E'F = AB = AE + BE = \boxed{3}$.


Question: As shown in the figure, EF passes through the intersection point O of the diagonals of rectangle ABCD, and intersects AD and BC at points E and F, respectively. Given that AB = 3 and BC = 4, what is the area of the shaded region in the figure?

Solution:
\textbf{Solution:} Given rectangle ABCD, \\
it follows that BO=DO=AO=CO, and AD$\parallel$BC, \\
hence, $\angle$EDO=$\angle$FBO, \\
since also $\angle$EOD=$\angle$FOB, \\
it follows that $\triangle$EDO$\cong$$\triangle$FBO (by ASA), \\
thus, the area of $\triangle$FBO = the area of $\triangle$EDO, \\
hence, the area of the shaded region = S$_{\triangle BOC}$= $\frac{1}{4}$S$_{\text{rectangle }ABCD}$= $\frac{1}{4}\times4\times3=\boxed{3}$.


Question: As shown in Figure 1, in rectangle $ABCD$, point $P$ starts from point $C$ and moves in the direction of $C \rightarrow D \rightarrow A \rightarrow B$ and stops at point $B$. Let the distance traveled by point $P$ be $x$, and the area of $\triangle PBC$ be $y$. Given the functional relationship between $y$ and $x$ as shown in Figure 2, what is the area of rectangle $ABCD$?

Solution:
Solution: From the problem statement, we know that: when point P is on side CD, \(y\) increases as \(x\) increases;\\
when point P is on side AD, \(y\) does not change as \(x\) changes;\\
when point P is on side AB, \(y\) decreases as \(x\) increases.\\
Considering the graph of a linear function, we find that CD=5, AD=6,\\
\(\therefore\) the area of rectangle ABCD is: AD\(\times\)CD=5\(\times\)6=\boxed{30}.


Question: As shown in the figure, in quadrilateral $ABCD$, the diagonals $AC \perp BD$ intersect at point $O$, and points $E$, $F$, $G$, $H$ are the midpoints of sides $AD$, $AB$, $BC$, $CD$, respectively. If $AC=8$ and $BD=6$, what is the area of quadrilateral $EFGH$?

Solution:
\textbf{Solution}: Since points $E$, $F$ are the midpoints of the sides $AD$, $AB$ of the quadrilateral $ABCD$, respectively,\\
it follows that $EF \parallel BD$ and $EF = \frac{1}{2}BD = 3.$\\
Similarly, it is found that $EH \parallel AC \parallel GF$, and $EH = GF = \frac{1}{2}AC = 4$,\\
Moreover, since $AC \perp BD$,\\
it follows that $EF \parallel GH$, $FG \parallel HE$ and $EF \perp FG$.\\
The quadrilateral $EFGH$ is a rectangle.\\
Therefore, the area of quadrilateral $EFGH$ is $EF \cdot EH = 3 \times 4 = \boxed{12}$, which means the area of quadrilateral $EFGH$ is 12.


Question: As shown in the figure, in quadrilateral $ABCD$, $AD \parallel BC$, $\angle ABC + \angle DCB = 90^\circ$, and $BC = 2AD$. Squares are constructed externally on the sides $AB$, $BC$, and $DC$, with areas denoted as $S_{1}$, $S_{2}$, and $S_{3}$ respectively. Given that $S_{1} = 4$ and $S_{3} = 12$, what is the value of $S_{2}$?

Solution:
\textbf{Solution:} Given $S_{1}=4$ and $S_{3}=12$, \\
it follows that $AB=2$ and $CD=2\sqrt{3}$. \\
Draw AE $\parallel$ CD intersecting BC at E, \\
hence $\angle AEB=\angle DCB$, \\
since $AD\parallel BC$, \\
the quadrilateral AECD is a parallelogram, \\
thus $CE=AD$ and $AE=CD=2\sqrt{3}$, \\
given $\angle ABC+\angle DCB=90^\circ$, \\
it implies $\angle AEB+\angle ABC=90^\circ$, \\
thus $\angle BAE=90^\circ$, \\
therefore $BE=\sqrt{AB^{2}+AE^{2}}=\sqrt{2^{2}+12}=4$, \\
since $BC=2AD$, \\
it follows that $BC=2BE=8$, \\
therefore $S_{2}=(8)^{2}=\boxed{64}$.


Question: As shown in the diagram, in rectangle \(ABCD\), \(AB=6\), \(BC=8\), \(M\) is any point on \(AD\), and \(ME \perp AC\) at point \(E\), \(MF \perp BD\) at point \(F\), then what is the value of \(ME+MF\)?

Solution:
\textbf{Solution:} As shown in the figure, take point O as the intersection of AC and BD, and connect OM,\\
$S_{\triangle AOD}=S_{\triangle AOM}+S_{\triangle MOD}= \frac{1}{2}$OA$\cdot$ME+ $\frac{1}{2}$OD$\cdot$MF= $\frac{1}{2}+$OA(ME+MF),\\
$\because$ Quadrilateral ABCD is a rectangle,\\
$\therefore$ $\angle$ABC$=90^\circ$,\\
$\therefore$ AC= $\sqrt{AB^{2}+BC^{2}}=10$,\\
$\therefore$ OA= $\frac{1}{2}$AC=5,\\
$S_{\triangle AOD}= \frac{1}{4}S_{\text{rectangle}ABCD}=12$,\\
$\therefore$ ME+MF=12÷ $\frac{1}{2}$OA= $\boxed{\frac{24}{5}}$.


Question: As shown in the figure, the side length of square $ABCD$ is 4. Point $E$ is a moving point on line segment $BC$. Connect $AE$, and rotate $AE$ clockwise by $90^\circ$ around point $E$ to $EF$. Connect $BF$, and take the midpoint $M$ of $BF$. If point $E$ moves from point $B$ to point $C$, what is the length of the path traveled by point $M$?

Solution:
\textbf{Solution:} Because rotating line AE clockwise by $90^\circ$ around point E brings it to line EF,\\
therefore EF$\perp$AE.\\
When point E is located at point B, point M is at the midpoint G of BC,\\
when point E is at point C, point M is at the midpoint H of CD, the path traveled by point M is the length of GH\\
Because the side length of square ABCD is 4,\\
therefore $GH=2\sqrt{2}$,\\
thus, the length of the path traveled by point M is $\boxed{2\sqrt{2}}$.


Question: As shown in the figure, by changing the interior angles of the square $ABCD$, the square is transformed into a rhombus $AB{C}^{\prime }{D}^{\prime }$. If $\angle {D}^{\prime }AB=30^\circ$, what is the ratio of the area of the rhombus $AB{C}^{\prime }{D}^{\prime }$ to the area of the square $ABCD$?

Solution:
\textbf{Solution:} As shown in the figure, draw $D'M\perp AB$ at point M through point $D'$,\\
$\because \angle D'AB=30^\circ$,\\
$\therefore D'M=\frac{1}{2}AD'$,\\
$\because$ Quadrilateral $ABC'D'$ is a rhombus,\\
$\therefore AD'=AB$,\\
$\therefore$ The area of rhombus $ABC'D'$ is $AB\times D'M= AB\times \frac{1}{2}AD'=AB\times \frac{1}{2}AB=\frac{1}{2}AB^2$,\\
$\because$ The area of square $ABCD=AB^2$,\\
$\therefore$ The ratio of the area of rhombus $ABC'D'$ to the area of square $ABCD$ is $\boxed{\frac{1}{2}}$.


Question: As shown in the figure, in diamond $ABCD$, $\angle BAD=80^\circ$, the perpendicular bisector of $AB$ intersects diagonal $AC$ at point $F$ and $AB$ at point $E$. Connecting $DF$, what is the value of $\angle CDF$?

Solution:
\textbf{Solution:} As shown in the figure, connect BF, \\
$\because$ quadrilateral ABCD is a rhombus, \\
$\therefore$ $\angle BAC=\frac{1}{2}\angle BAD=\frac{1}{2}\times 80^\circ=40^\circ$, \\
$BC=CD$, $\angle DCF=\angle BCF$, AD$\parallel$BC, \\
$\therefore$ $\angle ABC=180^\circ-\angle BAD=100^\circ$, \\
$\because$EF is the perpendicular bisector of AB, \\
$\therefore$AF=BF, \\
$\therefore$ $\angle ABF=\angle BAC=40^\circ$, \\
$\therefore$ $\angle CBF=\angle ABC-\angle ABF=60^\circ$, \\
In $\triangle DCF$ and $\triangle BCF$, \\
$\left\{\begin{array}{c}DC=BC\\ \angle DCF=\angle BCF\\ CF=CF\end{array}\right.$, \\
$\therefore$ $\triangle DCF\cong \triangle BCF$ (SAS), \\
$\therefore$ $\angle CDF=\angle CBF= \boxed{60^\circ}$


Question: As shown in the figure, the areas of the two smaller squares are 9 and 16, respectively. What is the area of the square represented by the letter A?

Solution: \textbf{Solution:} The area of the square represented by the letter A is: $16+9= \boxed{25}$.


Question: As shown in the figure, $D$ is a point inside $\triangle ABC$ with $BD \perp CD$, $AD=6$, $BD=4$, $CD=3$, and $E$, $F$, $G$, $H$ are the midpoints of $AB$, $AC$, $CD$, and $BD$, respectively. What is the perimeter of the quadrilateral $EFGH$?

Solution:
\textbf{Solution:} Given $BD\perp DC$, with $BD=4$ and $CD=3$, by the Pythagorean theorem, we have $BC=\sqrt{BD^2+CD^2}=5$.\\
Since $E$, $F$, $G$, $H$ are the midpoints of $AB$, $AC$, $CD$, and $BD$, respectively,\\
it follows that $HG=\frac{1}{2}BC=EF$ and $EH=FG=\frac{1}{2}AD$.\\
Given $AD=6$,\\
therefore $EF=HG=2.5$ and $EH=GF=3$,\\
hence the perimeter of quadrilateral $EFGH$ is $EF+FG+HG+EH=2\times(2.5+3)=\boxed{11}$.


Question: As shown in the figure, in $ \square ABCD$, $AD=8$, points $E$ and $F$ are the midpoints of $BD$ and $CD$ respectively, what is the length of $EF$?

Solution:
\textbf{Solution}: Since quadrilateral ABCD is a parallelogram, and $AD=8$, \\
then $BC=AD=8$.\\
Since points E and F are midpoints of BD and CD, respectively, \\
then $EF=\frac{1}{2}BC=\frac{1}{2}\times 8=\boxed{4}$.


Question: As shown in the figure, the graph of the inverse proportion function $y=\frac{2}{x}$ passes through the midpoint $D$ of side AB of the rectangle OABC. What is the area of the rectangle OABC?

Solution:
\textbf{Solution:} As shown in the figure, draw $DE \perp OC$ at point $E$,\\
since the graph of the inverse proportion function $y=\frac{2}{x}$ passes through the midpoint $D$ of side $AB$ of rectangle $OABC$,\\
therefore, $S_{\text{rectangle }OABC}=2S_{\text{rectangle }OADE}$,\\
also, since $k=2$,\\
therefore, $S_{\text{rectangle }OADE}=2$,\\
therefore, $S_{\text{rectangle }OABC}=\boxed{4}$.


Question: As shown in the figure, the diagonals AC and BD of the square ABCD intersect at point O, and OA = 3. What is the area of this square?

Solution:
\textbf{Solution:} Given the square ABCD, with OA=3,\\
it follows that AC=BD=6, and AO$\perp$BO,\\
hence the area of the square is: $\frac{1}{2}$AC$\times$BD=$\boxed{18}$.


Question: As shown in the figure, in $ABCD$, $E$ is a point on side $BC$. Taking $AE$ as a side, square $AEFG$ is constructed. If $\angle BAE = 40^\circ$ and $\angle CEF = 15^\circ$, then what is the measure of $\angle D$?

Solution:
\textbf{Solution:} Given that square $AEFG$,\\
it follows that $\angle AEF=90^\circ$,\\
given $\angle BAE=40^\circ$, $\angle CEF=15^\circ$,\\
thus, $\angle AEC=\angle AEF+\angle CEF=\angle BAE+\angle B$, which means $90^\circ+15^\circ=40^\circ+\angle B$,\\
so, $\angle B=65^\circ$,\\
given parallelogram $ABCD$,\\
it follows that $\angle D=\angle B=\boxed{65^\circ}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is a square, $BE \perp EF$, $DF \perp EF$, with $BE = 2.5 \, \text{dm}$, and $DF = 4 \, \text{dm}$. What is the length of $EF$?

Solution:
\textbf{Solution:} Since ABCD is a square,\\
it follows that BC=CD, and BC$\perp$CD,\\
therefore, $\angle$BCE+$\angle$FCD=$90^\circ$,\\
also, since BE$\perp$EF and DF$\perp$EF,\\
it follows that $\angle$BEC=$\angle$CFD=$90^\circ$,\\
therefore, $\angle$BCE+$\angle$EBC=$90^\circ$,\\
thus, $\angle$EBC=$\angle$FCD,\\
hence, $\triangle$BEC$\cong$$\triangle$CFD,\\
therefore, BE=CF=2.5dm, and DF=EC=4dm,\\
thus, EF=EC+CF=$\boxed{6.5}$dm.


Question: As shown in the figure, in diamond $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, and $E$ is the midpoint of $BC$. If we connect $OE$, what must the length of $OE$ be?

Solution:
\textbf{Solution}: Since ABCD is a rhombus, \\
therefore, AC $\perp$ BD, i.e., $\angle$BOC = $90^\circ$, \\
Also, since E is the midpoint of BC, \\
therefore OE = BE = EC = $\frac{1}{2}$BC. \\
Hence, $OE = \boxed{\frac{1}{2}BC}$.


Question: As shown in the figure, in diamond ABCD, AB=4, $\angle B=60^\circ$, AE$\perp$BC, and AF$\perp$CD, with the perpendicular feet being points E and F respectively. If we connect EF, what is the area of $\triangle AEF$?

Solution:
\textbf{Solution:} As shown in the figure, draw line AH $\perp$ EF at point H,\\
$\because$ rhombus ABCD,\\
$\therefore$ BC=CD, $\angle$B=$\angle$D=$60^\circ$,\\
$\therefore$ $\angle$BAD=$180^\circ-60^\circ=120^\circ$,\\
$\because$ AE$\perp$BC, AF$\perp$CD,\\
$\therefore$ $\angle$AEB=$\angle$AFD=$90^\circ$,\\
$\therefore$ $\angle$BAE=$\angle$DAF=$30^\circ$,\\
$\therefore$ $\angle$EAF=$\angle$BAD-$\angle$BAE-$\angle$DAF=$120^\circ-30^\circ-30^\circ=60^\circ$,\\
Also, $\because$ BC$\cdot$AE=CD$\cdot$AF,\\
$\therefore$ AE=AF,\\
$\therefore$ $\triangle$AEF is an equilateral triangle,\\
$\therefore$ AE=EF=AF, EH=HF=$\frac{1}{2}$EF, AH=$\sqrt{3}$EH, $\because$ AB=4,\\
$\therefore$ in $\triangle$RtAEB, BE=$\frac{1}{2}$AB=2,\\
$\therefore$ EF=AF=AE=$\sqrt{3}$BE=2$\sqrt{3}$,\\
$\therefore$ AH=$\sqrt{3}$EH=3,\\
$\therefore$ Area of $\triangle$AEF=$\frac{1}{2}$EF$\cdot$AH=$\frac{1}{2} \times 2\sqrt{3} \times 3 = \boxed{3\sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $BC=20$, $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively, $F$ is a point on $DE$, $DF=4$, connecting $AF$ and $CF$. If $\angle AFC=90^\circ$, then what is the length of $AC$?

Solution:
\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively,

it follows that $DE$ is the midline of $\triangle ABC$,

therefore $DE= \frac{1}{2}BC=10$,

thus $EF=DE−DF=10−4=6$,

in the right triangle $\triangle AFC$, $AE=EC$,

therefore $AC=2EF=\boxed{12}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $\angle A=30^\circ$. The perpendicular bisector of AC intersects AC and AB at points D and F, respectively. A perpendicular line is drawn from point B to DF, and the foot of the perpendicular is E. If BC=2, what is the area of the quadrilateral BCDE?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $\angle C=90^\circ$ and $\angle A=30^\circ$, $BC=2$,\\
hence $AC= \sqrt{3}BC=2\sqrt{3}$,\\
since DE is perpendicular bisector of AC, with $\angle CDE=90^\circ$,\\
then $AD=DC= \frac{1}{2}AC=\sqrt{3}$,\\
since BE$\perp$ED,\\
thus $\angle C=\angle CDE=\angle E=90^\circ$,\\
therefore, quadrilateral BCDE is a rectangle,\\
thus the area of quadrilateral BCDE=BC$\cdot$DC=$2\times\sqrt{3}=\boxed{2\sqrt{3}}$.


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

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\
it follows that AD=CD, and $\angle CAD=\angle BAC=52^\circ$,\\
therefore, $\angle ADC=180^\circ - \angle CAD - \angle ACD=180^\circ - 52^\circ - 52^\circ=76^\circ$,\\
hence, $\angle CDO= \frac{1}{2}\angle ADC=\boxed{38^\circ}$.


Question: As shown in the figure, the line $AB$ intersects with the line $MN$ at point $O$, $OC \perp AB$, $OA$ bisects $\angle MOD$, if $\angle BON = 20^\circ$, what is the measure of $\angle COD$?

Solution:
\textbf{Solution:} Since $\angle BON=20^\circ$\\
Thus, $\angle AOM=\angle BON=20^\circ$\\
Since $OA$ bisects $\angle MOD$,\\
Thus, $\angle AOD=\angle AOM=20^\circ$\\
Since $OC\perp AB$\\
Thus, $\angle COD=90^\circ-\angle AOD=90^\circ-20^\circ=\boxed{70^\circ}$


Question: As shown in the figure, point C is the midpoint of segment $AB$, and point D is on segment $CB$. If $AB=10$ and $DB=2$, then what is the length of segment $CD$?

Solution:
\textbf{Solution:} Since point $C$ is the midpoint of line segment $AB$ and $AB=10$, \\
it follows that $BC=\frac{1}{2}AB=5$, \\
Furthermore, since $DB=2$, \\
it follows that $CD=BC-DB=\boxed{3}$.


Question: As shown in the figure, there are three points $D$, $B$, and $E$ sequentially on the line segment $AC$, among which point $B$ is the midpoint of line segment $AC$, $AD=BE$, if $AC=8$, then what is the length of $DE$?

Solution:
\textbf{Solution:}
Given that points D, B, and E lie sequentially on segment $AC$,

it follows that $DE = DB + BE$,

since $AD = BE$,

thus $DE = DB + AD = AB$,

given $AC = 8$, and B is the midpoint of segment $AC$,

it follows that $AB = \frac{1}{2}AC = \frac{1}{2} \times 8 = 4$,

therefore, $DE = AB = \boxed{4}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=90^\circ$, $\angle A=60^\circ$, D is a point on side $AC$, $\angle DBC=30^\circ$, $AC=12\,cm$, then what is the perimeter of $\triangle ABD$?

Solution:
\textbf{Solution:} The net of this solid consists entirely of triangles,\\
$\therefore$ this solid is a triangular pyramid.\\
Hence, the answer is: \boxed{A}


Question: As shown in the figure, a right-angled triangle is placed with point C located on the extension of $FD$, $AB \parallel CF$, $\angle F = \angle ACB = 90^\circ$, $\angle E = 30^\circ$, and $\angle A = 45^\circ$. What is the measure of $\angle CBD$?

Solution:
\textbf{Solution:} Given: $\because$ $\angle A=45^\circ$,\\
$\therefore$ $\angle ABC=45^\circ$,\\
$\because$ $AB\parallel CF$,\\
$\therefore$ $\angle BCD=45^\circ$,\\
$\because$ $\angle E=30^\circ$,\\
$\therefore$ $\angle EDF=60^\circ$,\\
$\therefore$ $\angle BDC=180^\circ-\angle EDF=120^\circ$\\
In $\triangle BCD$, $\angle BCD=45^\circ$, $\angle BDC=120^\circ$,\\
$\therefore$ $\angle CBD=180^\circ-\angle BDC-\angle BCD=15^\circ$,\\
Therefore, the answer is: $\boxed{15^\circ}$.


Question: As shown in the figure, segment $AB=16\, \text{cm}$. A point $C$ is taken on $AB$, where $M$ is the midpoint of $AB$ and $N$ is the midpoint of $AC$. If $MN=3\, \text{cm}$, what is the length of segment $AC$?

Solution:
\textbf{Solution:} Let $CM = a$, and $CN = CM + MN = a + 3$, \\
Since $M$ is the midpoint of $AB$, and $N$ is the midpoint of $AC$, \\
Therefore, $AM = \frac{1}{2}AB = \frac{1}{2} \times 16 = 8$, and $AN = CN = a + 3$, \\
Since $AM = AN + MN = 8$, this implies $a + 3 + 3 = 8$, \\
Therefore, $a = 2$, \\
Therefore, $AC = AM + CM = 8 + 2 = \boxed{10}$.


Question: As shown in the diagram, it is known that $ON$ and $OM$ respectively bisect $\angle AOC$ and $\angle BON$. If $\angle MON=20^\circ$ and $\angle AOM=35^\circ$, what is the measure of $\angle AOB$?

Solution:
\textbf{Solution:} Since $ON$ bisects $\angle AOC$ and $OM$ bisects $\angle BON$,\\
it follows that $\angle AON=\angle NOC=\frac{1}{2}\angle AOC$ and $\angle BOM=\angle NOM=\frac{1}{2}\angle BON$.\\
Given that $\angle MON=20^\circ$ and $\angle AOM=35^\circ$,\\
we have $\angle AON=\angle AOM-\angle MON=35^\circ-20^\circ=15^\circ$.\\
Therefore, $\angle AOC=2\times 15^\circ=30^\circ$ and $\angle BOM=\angle NOM=20^\circ$.\\
Hence, $\angle AOB=\angle BOM+\angle MON+\angle AON=20^\circ+20^\circ+15^\circ=\boxed{55^\circ}$.


Question: As shown in the figure, $AB=24\,cm$, $C$ is the midpoint of $AB$, point $D$ is on line segment $AC$, and $AD:CB=1:3$, what is the length of $CD$?

Solution:
\textbf{Solution:} Given $AB=24cm$ and $C$ is the midpoint of $AB$,\\
it follows that $AC=BC=\frac{1}{2}AB=12cm$,\\
given $AD\colon CB=1\colon 3$,\\
it follows that $AD=\frac{1}{3}CB=4cm$,\\
therefore, $CD=AC-AD=12-4=\boxed{8cm}$


Question: As shown in the figure, line segment $AB=24\,cm$, C is a point on $AB$, and $AC=15\,cm$, O is the midpoint of $AB$. What is the length of line segment $OC$?

Solution:
\textbf{Solution:} Since AB=24cm, and point O is the midpoint of AB,\\
then OB=$\frac{1}{2}$AB=$\frac{1}{2}\times 24=12cm$,\\
Since BC=AB−AC,\\
then BC=24−15=9;\\
Since OC=OB−BC,\\
then OC=12−9=$\boxed{3}$cm.


Question: As shown in the figure, $AB \parallel CD$, $\angle FGB = 155^\circ$, $FG$ bisects $\angle EFD$, what is the size of $\angle BEF$?

Solution:
\textbf{Solution:} Since $AB\parallel CD$ and $\angle FGB=155^\circ$,\\
it follows that $\angle BEF+\angle EFD=180^\circ$,\\
hence $\angle GFD=180^\circ-\angle FGB=180^\circ-155^\circ=25^\circ$,\\
Given that $FG$ bisects $\angle EFD$,\\
it implies $\angle EFD=2\angle GFD=2\times 25^\circ=50^\circ$,\\
thus $\angle BEF=180^\circ-\angle EFD=180^\circ-50^\circ=\boxed{130^\circ}$.


Question: As shown in the figure, points $C$ and $D$ are two points on line segment $AB$, with $AC:CD:BD=3:6:4$. If $AB=13$, what is the length of $CD$?

Solution:
\textbf{Solution:} Given $\because$ $AC: CD: BD = 3: 6: 4$, and $AB = 13$,

$\therefore$ $CD = \frac{6}{3+6+4}AB = \boxed{6}$


Question: As shown in the diagram, point $O$ is on line $AB$, and ray $OD$ is the bisector of $\angle AOC$. If $\angle COB = 40^\circ$, what is the degree measure of $\angle DOC$?

Solution:
\textbf{Solution:} Since $\angle COB=40^\circ$, \\
it follows that $\angle AOC=180^\circ-\angle COB=140^\circ$, \\
because OD is the angle bisector of $\angle AOC$, \\
therefore $\angle DOC= \frac{1}{2}\angle AOC= \frac{1}{2}\times140^\circ=\boxed{70^\circ}$.


Question: As shown in the diagram, in $ \triangle ABC$, $AD$ bisects $ \angle BAC$ and $AE$ is the altitude. If $ \angle B=40^\circ$ and $ \angle C=60^\circ$, what is the measure of $ \angle EAD$?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $\angle B=40^\circ$ and $\angle C=60^\circ$, \\
therefore $\angle BAC=180^\circ-\angle B-\angle C=180^\circ-40^\circ-60^\circ=80^\circ$ \\
Since $AD$ bisects $\angle BAC$, \\
thus $\angle BAD=\angle CAD=40^\circ$ \\
Also, since $AE$ is the altitude, \\
thus $\angle BAE=90^\circ-\angle B=90^\circ-40^\circ=50^\circ$ \\
Therefore, $\angle EAD=\angle BAE-\angle BAD=50^\circ-40^\circ=\boxed{10^\circ}$


Question: As shown in the figure, point C is on segment $AB$, and point D is the midpoint of $BC$, with $AB=30\,cm$ and $AC=4CD$. What is the length of $AC$ in $cm$?

Solution:
\textbf{Solution:} Given that point $D$ is the midpoint of $BC$, we have \\
$BC = 2CD = 2BD$, \\
By the segment addition and subtraction property, we have \\
$AB = AC + BC$, which means $4CD + 2CD = 30$, \\
Solving for $CD$ gives $CD = 5$, \\
Therefore, $AC = 4CD = 4 \times 5 = \boxed{20}\,\text{cm}$.


Question: As shown in the figure, point $C$ is on the line segment $AB$, and point $D$ is the midpoint of the line segment $BC$. Given that $AB=10$ and $AC=6$, what is the length of the line segment $BD$?

Solution:
\textbf{Solution.} Since point D is the midpoint of line segment $BC$, and $AB=10$, $AC=6$,
\[\therefore BD = \frac{1}{2}BC =\frac{1}{2}\left(AB-AC\right) =\frac{1}{2}\left(10-6\right)=\boxed{2}.\]


Question: As shown in the figure, the straight line AO is perpendicular to OB with the foot of the perpendicular at O. The length of segment AO is 3, and the length of BO is 4. Taking point A as the center and the length of AB as the radius, an arc is drawn to intersect the straight line AO at point C. What is the length of OC?

Solution:
\textbf{Solution:} Since $AO \perp OB$, and the length of segment AO = 3, BO = 4,\\
therefore, in $\triangle AOB$, we have $AB = \sqrt{AO^{2} + OB^{2}} = 5$,\\
according to the problem, we know: AC = AB = 5,\\
therefore, $OC = AC - AO = \boxed{2}$.


Question: As shown in the figure, it is known that points B and C divide the line segment AD from left to right into three parts in the ratio $2:4:3$. If M is the midpoint of AD and BM $= 5\,\text{cm}$, then what is the length of the line segment MC?

Solution:
\textbf{Solution:} Given, let $AB=2x$, $BC=4x$, $CD=3x$\\
Since M is the midpoint of AD,\\
it follows that $AM=MD=\frac{1}{2}AD=\frac{1}{2}(2x+4x+3x)=\frac{9}{2}x$\\
Since $BM=5cm$,\\
it follows that $AD-AB-MC-CD=5$\\
Thus, $9x-2x-MC-3x=5$\\
Therefore, $MC=4x-5$\\
Since $MC=MD-CD=\frac{9}{2}x-3x=\frac{3}{2}x$,\\
it follows that $4x-5=\frac{3}{2}x$\\
Thus, $\frac{3}{2}x=4x-5$\\
Thus, $x=2$\\
Therefore, $MC=\frac{3}{2}\times 2=\boxed{3}$


Question: As shown in the figure, $\angle AOE = 100^\circ$, $\angle BOF = 80^\circ$, OE bisects $\angle BOC$, and OF bisects $\angle AOC$. What is the measure of $\angle EOF$?

Solution:
\textbf{Solution:}

Since $OE$ bisects $\angle BOC$, and $OF$ bisects $\angle AOC$\\
Therefore, $\angle AOB=2\angle EOF$\\
Also, since $\angle AOB=\angle BOF+\angle AOE-\angle EOF=80^\circ+100^\circ-\angle EOF=180^\circ-\angle EOF$\\
Therefore, $2\angle EOF=180^\circ-\angle EOF$\\
Solving this gives $\angle EOF=\boxed{60^\circ}$


Question: As shown in the figure, the line segment $AD=21\,cm$, point $B$ is on line segment $AD$, $C$ is the midpoint of $BD$, and $AB=\frac{1}{3}CD$, what is the length of $BC$?

Solution:
\textbf{Solution:} Let $AB=x$, then $CD=3x$,\\
since C is the midpoint of BD,\\
thus $BC=CD=3x$,\\
therefore $3x+3x+x=21$,\\
solving this equation yields $x=3$,\\
$BC=3\times 3=\boxed{9}$ cm.


Question: As shown in the figure, $ \angle AOB : \angle BOC : \angle COD = 2 : 3 : 4$, the rays $OM$ and $ON$ bisect $ \angle AOB$ and $ \angle COD$ respectively. If $ \angle MON$ is a right angle, what is the measure of $ \angle COD$ in degrees?

Solution:
\textbf{Solution:} Let $\angle AOB=2x^\circ$, then $\angle BOC=3x^\circ$, $\angle COD=4x^\circ$,\\
since ray OM and ON bisect $\angle AOB$ and $\angle COD$ respectively,\\
therefore $\angle BOM= \frac{1}{2} \angle AOB=x^\circ$\\
$\angle CON= \frac{1}{2} \angle COD=2x^\circ$\\
since $\angle MON=90^\circ$\\
therefore $\angle CON+\angle BOC+\angle BOM=90^\circ$\\
therefore $2x+3x+x=90$\\
Solving for x, we get: $x=15$\\
therefore $\angle COD=4x =15^\circ \times 4=\boxed{60^\circ}$.


Question: As shown in the figure, point \(C\) divides the line segment \(AB\) into two parts in the left-to-right order, in the ratio \(2:3\). Point \(D\) is the midpoint of \(AB\). If \(CD=2\), what is the length of the line segment \(AB\)?

Solution:
\textbf{Solution}: Let $AC = 2x$, then $BC = 3x$, \\
$\therefore AB = AC + BC = 5x$, \\
$\because$ point D is the midpoint of AB, \\
$\therefore AD = \frac{1}{2}AB = 2.5x$, \\
$\therefore CD = AD - AC = 2.5x - 2x = 0.5x$, \\
$\because CD = 2$, \\
$\therefore 0.5x = 2$, \\
$\therefore x = 4$, \\
$\therefore AB = 5x = \boxed{20}$,


Question: As shown in the figure, $AB$ is tangent to $\odot O$ at point $B$, extending $AO$ intersects $\odot O$ at point $C$, connecting $BC$. If $\angle A=40^\circ$, what is the value of $\angle C$?

Solution:
\textbf{Solution:} Given that AB is a tangent to circle O at point B,\\
it follows that OB$\perp$AB, meaning $\angle$ABO=$90^\circ$,\\
hence $\angle$AOB=$50^\circ$ (the two acute angles of a right triangle are complementary),\\
Also, given that point C is on the extension of AO and lies on circle O,\\
thus $\angle$C=$\frac{1}{2}\angle$AOB=$\boxed{25^\circ}$ (the angle subtended by an arc at the circumference is half that at the center).


Question: As shown in the figure, AB is the diameter of $\odot O$, C and D are two points on $\odot O$, $\angle CDB=30^\circ$, and BC=4.5, then what is the length of AB?

Solution:
\textbf{Solution:} As shown in the figure, connect $AC$.\\
Since $AB$ is the diameter of circle $O$,\\
it follows that $\angle ACB=90^\circ$,\\
Since $\angle CAB=\angle CDB=30^\circ$, and $BC=4.5$,\\
it follows that $AB=2BC=\boxed{9}$.


Question: As shown in the figure, in circle $\odot O$, C and D are two points on circle $\odot O$, and AB is the diameter of $\odot O$. Given that $\angle AOC = 130^\circ$, what is the measure of $\angle BDC$?

Solution:
\textbf{Solution:} Since $\angle AOC=130^\circ$ and AB is the diameter of $\odot O$,\\
it follows that $\angle BOC=180^\circ-\angle AOC=50^\circ$,\\
thus $\angle BDC= \frac{1}{2}\angle BOC=\boxed{25^\circ}$.


Question: As shown in the figure, A, B, and C are three points on the circle $\odot O$. Given that $\angle ABC=20^\circ$, what is the measure of $\angle AOC$?

Solution:
\textbf{Solution:} Since $\angle ABC=20^\circ$, \\
then $\angle AOC= 2\angle ABC = \boxed{40^\circ}$;


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=90^\circ$, $\angle A=60^\circ$, D is a point on side $AC$, $\angle DBC=30^\circ$, $AC=12\,cm$, then what is the perimeter of $\triangle ABD$?

Solution:
\textbf{Solution:} Connect $CO$, $AO$, passing through point $O$, construct $OM\perp AB$ at $M$, $ON\perp CD$ at $N$,\\
then quadrilateral $OMEN$ is a rectangle,\\
since $OM\perp AB$, $ON\perp CD$,\\
and $AB=CD$,\\
it follows that $CN=\frac{1}{2}CD$, $AM=\frac{1}{2}AB$,\\
since $CO=AO$,\\
it follows that right triangle $CNO\cong$ right triangle $AMO$ (HL),\\
thus $NO=MO$,\\
hence $NO=MO=EM$,\\
seeing that $OM\perp AB$,\\
it follows that $AM=MB=\frac{1}{2}AB=\frac{1}{2}(AE+BE)=5$,\\
therefore, $EM=AM-AE=5-3=2$,\\
thus $ON=EM=\boxed{2}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, $C$ and $D$ are points on $\odot O$, $\angle CDB=25^\circ$. A tangent to $\odot O$ passing through point $C$ intersects the extension of $AB$ at point $E$. What is the measure of $\angle E$?

Solution:
\textbf{Solution:} Draw line OC,\\
$\because$ CE is the tangent to $\odot$O,\\
$\therefore$ OC$\perp$CE,\\
which means $\angle$OCE$=90^\circ$,\\
$\because$ $\angle$COB$=2\angle$CDB$=50^\circ$,\\
$\therefore$ $\angle$E$=90^\circ-\angle$COB$=\boxed{40^\circ}$.


Question: As shown in the figure, the incircle $\odot O$ of $\triangle ABC$ is tangent to $AB$, $BC$, $CA$ at points $D$, $E$, and $F$, respectively. If $\angle DEF = 52^\circ$, then what is the degree measure of $\angle A$?

Solution:
\textbf{Solution:} As shown in the figure, connect OD, and OF, \\
$\because$ The incircle $\odot O$ of $\triangle ABC$ is tangent to the sides AB, BC, CA at points D, E, F, respectively,\\
$\therefore$ $\angle ADO=\angle AFO=90^\circ$；\\
$\therefore$ $\angle A+\angle DOF=180^\circ$，\\
$\because$ $\angle DEF=52^\circ$, and $\angle DOF$ and $\angle DEF$ are the central angle and the inscribed angle subtended by the same arc $\overset{\frown}{DF}$, respectively,\\
$\therefore$ $\angle DOF=2\angle DEF=104^\circ$；\\
$\therefore$ $\angle A=180^\circ-\angle DOF=\boxed{76^\circ}$.


Question: As shown in the figure, $E$ is a moving point on the side $CD$ of square $ABCD$, with $AB=2$. Connect $BE$, and draw $AF \perp BE$, which intersects $BC$ at $F$ and $BE$ at $G$. Connect $CG$. When $CG$ is at its minimum value, what is the length of $CF$?

Solution:
\textbf{Solution:} As shown in Figure 1, take the midpoint O of AB, and connect OG and OC.\\
$\because$ Quadrilateral ABCD is a square,\\
$\therefore \angle ABC=90^\circ$,\\
$\because$AB=2,\\
$\therefore$OB=OA=1,\\
$\therefore$OC=$\sqrt{1^2+2^2}=\sqrt{5}$,\\
$\because$AF$\perp$BE,\\
$\therefore \angle AGB=90^\circ$,\\
$\therefore$ point G lies on the circle $\odot$O with AB as the diameter,\\
$\because$AO=OB,\\
$\therefore$OG=$\frac{1}{2}$AB=1,\\
$\therefore$ when O, G, and C are collinear, the value of CG is minimum, and the minimum value=$\sqrt{5}-1$ (as shown in Figure 2),\\
$\because$OB=OG=1,\\
$\therefore \angle OBG=\angle OGB$,\\
$\because$AB$\parallel$CD,\\
$\therefore \angle OBG=\angle CEG$,\\
$\because \angle OGB=\angle CGE$,\\
$\therefore \angle CGE=\angle CEG$,\\
$\therefore$CE=CG=$\sqrt{5}-1$,\\
$\because \angle ABF=\angle BCE=\angle AGB=90^\circ$,\\
$\therefore \angle BAF+\angle ABG=90^\circ$, $\angle ABG+\angle CBE=90^\circ$,\\
$\therefore \angle BAF=\angle CBE$,\\
$\because$AB=BC,\\
$\therefore \triangle ABF \cong \triangle BCE$ (ASA),\\
$\therefore$BF=CE=$\sqrt{5}-1$,\\
$\therefore$CF=BC−BF=2−($\sqrt{5}-1$)=3−$\sqrt{5}$,\\
Therefore, $CF = \boxed{3-\sqrt{5}}$