Question: As shown in the figure, $\triangle ABC$ is an isosceles right triangle with $\angle ABC=90^\circ$. $\triangle ABC$ is folded along a straight line such that the corresponding point of vertex $A$, denoted as ${A}^{\prime}$, lands exactly on side $BC$. This crease intersects $AB$ and $AC$ at points $D$ and $E$, respectively. The bisector of $\angle ABC$ intersects ${A}^{\prime}D$ at point $F$. Line $EF$ is drawn. If $CE={A}^{\prime}B$, what is the degree measure of $\angle {A}^{\prime}EF$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is an isosceles right-angled triangle, we fold $\triangle ABC$ along a straight line such that vertex $A'$ falls exactly on side $BC$, where this fold line crosses $AB$ and $AC$ at points D and E, respectively,\\
$\therefore$ $\angle A = \angle C = \angle DA'E = 45^\circ$,\\
$\angle FA'B + \angle EA'C = 180^\circ - 45^\circ = 135^\circ$,\\
Since BF bisects $\angle ABC$,\\
$\therefore$ $\angle A'BF = \frac{1}{2} \angle ABC = 45^\circ$,\\
$\therefore$ $\angle BFA' + \angle BA'F = 180^\circ - 45^\circ = 135^\circ$,\\
$\therefore$ $\angle BFA' = \angle EA'C$,\\
In $\triangle A'FB$ and $\triangle EA'C$,\\
$\left\{\begin{array}{l}
\angle BFA' = \angle EA'C\\
\angle A'BF = \angle C\\
A'B = CE
\end{array}\right.$\\
$\therefore$ $\triangle A'FB \cong \triangle EA'C$ (AAS),\\
$\therefore$ $A'F = A'E$,\\
$\therefore$ $\angle A'EF = \angle A'FE = \frac{1}{2}(180^\circ - \angle DA'E) = \frac{1}{2}(180^\circ - 45^\circ) = \boxed{67.5^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $D$ is a point on $BC$, $CA=CD$, $CE$ bisects $\angle ACB$, intersects $AB$ at point $E$, and $DE$ is connected. If $\angle A=100^\circ$ and $\angle B=45^\circ$, what is the measure of $\angle BED$ in degrees?

Solution:
\textbf{Solution:} Given that $CE$ bisects $\angle ACB$, \\
thus $\angle ACE = \angle DCE$, \\
in $\triangle ACE$ and $\triangle DCE$, \\
\[
\left\{
\begin{array}{l}
CA = CD \\
\angle ACE = \angle DCE \\
CE = CE
\end{array}
\right.,
\]
therefore $\triangle ACE \cong \triangle DCE$ (SAS), \\
thus $\angle CDE = \angle A = 100^\circ$, \\
since $\angle B = 45^\circ$, \\
therefore $\angle BED = \angle CDE - \angle B = 100^\circ - 45^\circ = \boxed{55^\circ}$,


Question: As shown in the figure, in the right triangle $ABC$, $\angle ACB=90^\circ$, $\angle A>\angle B$. $CD$ is the altitude on the hypotenuse, and $CE$ is the angle bisector of $\triangle ABC$. $FG$ is the perpendicular bisector of side $AB$, and $FG$ intersects sides $BC$ and $AB$ at points $F$ and $G$, respectively. If $\angle DCE=\angle B$, what is the value of $\frac{BF}{FC}$?

Solution:
\textbf{Solution:} As shown in the figure, connect $AF$,

$\because$ $CD$ is the altitude on the hypotenuse of right $\triangle ABC$,

$\therefore$ $\angle ACD + \angle CAB = 90^\circ$.

$\because$ $\angle ACB = 90^\circ$,

$\therefore$ $\angle CAB + \angle B = 90^\circ$,

$\therefore$ $\angle ACD = \angle B$,

$\therefore$ $\angle ACD = \angle DCE = \angle B$.

$\because$ $CE$ is the bisector of $\angle ACB$,

$\therefore$ $\angle ACE = \frac{1}{2}\angle ACB = 45^\circ$,

$\therefore$ $\angle ACD + \angle DCE = 45^\circ$,

$\therefore$ $\angle B = \angle ACD = \angle DCE = \frac{1}{2} \times 45^\circ = 22.5^\circ$.

$\because$ $FG$ is the perpendicular bisector of side $AB$,

$\therefore$ $AF = BF$,

$\therefore$ $\angle FAB = \angle B = 22.5^\circ$,

$\therefore$ $\angle AFC = \angle B + \angle FAB = 45^\circ$,

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

$\therefore$ $\frac{AF}{FC} = \sqrt{2}$, that is, $\frac{BF}{FC} = \boxed{\sqrt{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACD=125^\circ$, $\angle B=40^\circ$, then what is the measure of $\angle A$?

Solution:
\textbf{Solution:} Since $\angle ACD$ is the exterior angle of $\triangle ABC$, and $\angle ACD=125^\circ$, $\angle B=40^\circ$,
\[
\therefore \angle A=\angle ACD-\angle B=\boxed{85^\circ}
\]


Question: As shown in the figure, in $\triangle ABC$, $CD$ is the altitude from $C$ to side $AB$, $BE$ bisects $\angle ABC$ and intersects $CD$ at point $E$, $BC=5$. If the area of $\triangle BCE$ is 5, what is the length of $ED$?

Solution:
\textbf{Solution:} Construct EF $\perp$ BC at F, \\
since CD is the altitude on side AB, and BE bisects $\angle$ABC, intersecting CD at point E, \\
thus $DE=EF$, \\
since $S_{\triangle BCE}=\frac{1}{2}\times BC\times EF=5$, \\
it follows that $\frac{1}{2}\times 5\times EF=5$, \\
hence, $EF=DE=\boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, CE is the angle bisector of $\triangle ABC$, and $\angle AEC=105^\circ$. What is the measure of $\angle B$ in degrees?

Solution:
\textbf{Solution:} Since CE bisects $\angle ACB$ and $\angle ACB=90^\circ$, \\
it follows that $\angle BCE=\frac{1}{2}\angle ACB=45^\circ$, \\
Given that $\angle AEC=105^\circ$, \\
it implies that $\angle B=\angle AEC-\angle BCE=\boxed{60^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $\angle A=90^\circ$, $\angle C=30^\circ$, and $AD \perp BC$ at $D$. $BE$ is the bisector of $\angle ABC$, and it intersects $AD$ at point $P$. If $AP=3$, what is the length of $AC$?

Solution:
\textbf{Solution:} Given $\because \angle A=90^\circ$, $\angle C=30^\circ$, \\
$\therefore \angle ABC=60^\circ$, \\
$\because BE$ is the bisector of $\angle ABC$, \\
$\therefore \angle ABP=\angle DBP=30^\circ$, \\
$\because AD\perp BC$, \\
$\therefore \angle BAD=30^\circ$, \\
$\therefore \angle PAB=\angle PBA$, \\
$\therefore BP=AP=3$, \\
In $Rt\triangle PBD$, $PD=\frac{1}{2}PB=\frac{3}{2}$, \\
$\therefore AD=AP+PD=\frac{9}{2}$, \\
In $Rt\triangle ADC$, $AC=2AD=\boxed{9}$.


Question: As shown in the figure, in $ \triangle ABC$, it is known that $ \angle A=58^\circ, \angle B=60^\circ$. What is the degree measure of $ \angle ADC$?

Solution:
\textbf{Solution:} Given: $\because$ $\angle 1=58^\circ$, $\angle B=60^\circ$\\
$\therefore$ $\angle ADC=\angle 1+\angle B=58^\circ+60^\circ=\boxed{118^\circ}$


Question: Given: As shown in the diagram, in $\triangle ABC$, $\angle ACB = 90^\circ$, and $BC = BD$, $AC = AE$, then what is the degree measure of $\angle DCE$?

Solution:
\textbf{Solution:} Given that AC=AE and BC=BD,\\
let $\angle$AEC=$\angle$ACE=x and $\angle$BDC=$\angle$BCD=y,\\
then $\angle$A=180$^\circ$-2x and $\angle$B=180$^\circ$-2y,\\
since $\angle$ACB+$\angle$A+$\angle$B=$180^\circ$,\\
thus 90$^\circ$+(180$^\circ$-2x)+(180$^\circ$-2y)=180$^\circ$,\\
hence x+y=135$^\circ$,\\
therefore, $\angle$DCE=180$^\circ$-($\angle$AEC+$\angle$BDC)=180$^\circ$-(x+y)=\boxed{45^\circ}.


Question: As shown in the diagram, point $D$ is on side $AB$ of $\triangle ABC$, and point $E$ is the midpoint of $CD$. Connect $AE$ and $BE$. If the area of $\triangle ABC$ is 8, what is the area of the shaded region?

Solution:
\textbf{Solution:} Since point E is the midpoint of CD, \\
it follows that the area of $\triangle ACE$ equals the area of $\triangle ADE$, and the area of $\triangle BCE$ equals the area of $\triangle BDE$, \\
hence, the area of the shaded region equals $\frac{1}{2}$ the area of $\triangle ABC$ = $\frac{1}{2} \times 8 = \boxed{4}$.


Question: As shown in the figure, the two medians $AD$ and $BE$ of $\triangle ABC$ intersect at point $F$. Connecting $CF$, if the area of $\triangle ABF$ is $8$, what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Given that the two medians $AD$ and $BE$ of $\triangle ABC$ intersect at point $F$,\\
$\therefore 2EF=BF$,\\
$\because$ the area of $\triangle ABF$ is $8$,\\
$\therefore$ the area of $\triangle AEF$ is $4$,\\
$\therefore$ the area of $\triangle ABE$ is $12$.\\
$\because AE$ is the median of $\triangle ABC$,\\
$\therefore$ the area of $\triangle ABC$ is $2$ times the area of $\triangle ABE$ which equals $\boxed{24}$.


Question: As shown in the figure, $ \angle AOB=30^\circ$, point P lies on the bisector of $ \angle AOB$, the perpendicular bisector of $ OP$ intersects $ OA$ and $ OB$ at points M and N, respectively. Point E is a point on $ OA$ different from point M, and $ PE=ON=2$. What is the area of $ \triangle POE$?

Solution:
\textbf{Solution:} As shown in the figure, connect PM, draw PC $\perp$ OA at point C through point P, suppose MN intersects OP at point D,\\
$\because \angle AOB=30^\circ$, OP is the angle bisector of $\angle AOB$,\\
$\therefore \angle MOP=15^\circ$\\
$\because$ MN is the perpendicular bisector of OP\\
$\therefore MN\perp OD$, $MP=MO$\\
$\therefore \angle ODM=\angle ODN=90^\circ$, $\angle MOP=\angle MPO=15^\circ$\\
Also, $\because OD=OD$\\
$\therefore \triangle OMD\cong \triangle OND$\\
$\therefore OM=ON$\\
$\because PE=ON=2$\\
$\therefore PE=OM$\\
$\because MP=MO$\\
$\therefore MP=PE=2$\\
$\because \angle PMA=\angle MOP+\angle MPO=30^\circ$\\
$\therefore \angle PEC=\angle PME=30^\circ$\\
$\because PC\perp ME$\\
$\therefore CE=CM$, $PC=\frac{1}{2}PE=1$\\
In $\triangle CPE$, $CE=\sqrt{3}$\\
$\therefore OE=OM+ME=OM+2CE=2+2\sqrt{3}$\\
$\therefore {S}_{\triangle POE}=\frac{1}{2}\cdot OE\cdot PC = \boxed{1+\sqrt{3}}$


Question: As shown in the figure, two bamboo poles $AB$ and $DB$ are leaning against the wall $CE$ at angles such that $\angle CAB = 33^\circ$ and $\angle CDB = 21^\circ$. What is the measure of $\angle ABD$?

Solution:
\textbf{Solution:} Since $\angle CAB$ is an exterior angle of $\triangle ABD$ and $\angle CAB=33^\circ$, $\angle CDB=21^\circ$,\\
then $\angle CAB=\angle ABD+\angle CDB$,\\
thus, $\angle ABD=\angle CAB-\angle CDB=12^\circ$,\\
therefore, the answer is: $\boxed{12^\circ}$.


Question: As shown in the diagram, in the right triangle $\triangle ABC$ the two perpendicular sides satisfy $BC=6$ and $AC=8$. Three squares are constructed on the three sides of right triangle $\triangle ABC$. If the areas of the four shaded regions are denoted as $S_{1}$, $S_{2}$, $S_{3}$, and $S_{4}$ respectively, what is the value of $S_{4}$? What is the value of $S_{2}+S_{3}-S_{1}$?

Solution:
\textbf{Solution:} Given that quadrilateral ACFE and quadrilateral ABGD are squares, and the right-angle sides of $\triangle ACB$ are BC and AC,

$\therefore$AE=AC, $\angle EAD+\angle DAC=\angle CAB+\angle DAC=90^\circ$, $\angle E=\angle ACB=90^\circ$, AD=AB, AB=BG, $\angle ABK=\angle BGH=90^\circ$, $\angle KAB+\angle ABC=\angle HBG+\angle ABC=90^\circ$,

$\therefore$$\angle EAD=\angle CAB$, $\angle KAB=\angle HBG$,

Given that in $\triangle EAD$ and $\triangle CAB$,

$\left\{\begin{array}{l}
\angle E=\angle ACB \\
\angle EAD=\angle CAB \\
AE=AC
\end{array}\right.$

$\therefore$$\triangle EAD\cong \triangle CAB$,

Given BC=6, AC=8,

$\therefore$S$_{4}$=S$_{\triangle ABC}=\frac{1}{2}AC\cdot BC=\frac{1}{2}\times 6\times 8=\boxed{24}$,

Given that in $\triangle ABK$ and $\triangle BGH$,

$\left\{\begin{array}{l}
\angle ABK=\angle BGH \\
AB=BG \\
\angle KAB=\angle HBG
\end{array}\right.$

$\therefore$$\triangle ABK\cong \triangle BGH$,

$\therefore$S$_{\triangle ABC}=S_{1}$,

Let the area of quadrilateral ADHC be $m$, and the area of $\triangle BCK$ be $n$,

Given quadrilateral ACFE, quadrilateral ABGD, and quadrilateral BJIC are all squares, and the right-angle sides of $\triangle ACB$ are BC and AC,

$\therefore$${S}_{3}+{S}_{4}+m=AC^{2}$, ${S}_{2}+n=BC^{2}$, ${S}_{1}+m+n+{S}_{\triangle ABC}=AB^{2}$, $AC^{2}+BC^{2}=AB^{2}$

$\therefore$S$_{2}$+S$_{3}$−S$_{1}$

=$\left({S}_{3}+{S}_{4}+m\right)+\left({S}_{2}+n\right)−\left({S}_{\triangle ABC}+m+n+{S}_{1}\right)$

=$AC^{2}+BC^{2}−AB^{2}$

=$\boxed{0}$


Question: As shown in the figure, in $\triangle ABC$, points D and E are located on sides AC and BC respectively, and $AD=DE$, $AB=BE$, $\angle A=70^\circ$, what is the measure of $\angle CED$ in degrees?

Solution:
\textbf{Solution:} Since $AD=DE$, and $AB=BE$\\
Also $BD= BD$\\
Therefore, $\triangle ABD \cong \triangle EBD$ (SSS)\\
Thus, $\angle BED = \angle A = 70^\circ$\\
Therefore, $\angle CED = 180^\circ - \angle BED = 180^\circ - 70^\circ = \boxed{110^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle B=15^\circ$. Point D is the midpoint of AB, and $DE \perp AB$ intersects BC at point E. Given $BE=8\text{cm}$, what is the length of $AC$ in cm?

Solution:
\textbf{Solution:} Since point D is the midpoint of AB, and DE is perpendicular to AB, intersecting BC at point E, \\
therefore DE bisects AB perpendicularly, \\
thus AE=BE=8, \\
therefore $\angle B=\angle BAE=15^\circ$, \\
since $\angle AEC=\angle B+\angle BAE$, \\
therefore $\angle AEC=30^\circ$, \\
since $\angle ACB=90^\circ$, and AE=8, \\
therefore AC=\boxed{4}.


Question: As shown in the figure, point D is a point on side AC of $\triangle ABC$. Points B and C are symmetric with respect to DE. If $AC=6$ and $AD=2$, what is the length of segment BD?

Solution:
\textbf{Solution:} \quad Since $AC=6$ and $AD=2$, \\
$\therefore CD=AC-AD=6-2=4$, \\
Since $B$ and $C$ are symmetrical about $DE$, \\
$\therefore DB=DC=\boxed{4}$.


Question: As shown in the figure, in $\triangle ABC$, $AD \perp BC$ at point $D$, $AD$ and $BE$ intersect at point $F$, and $AC=BF$, $DF=DC$. If $\angle ABE=15^\circ$, what is the degree measure of $\angle DBF$?

Solution:
\textbf{Solution:} Since $AD \perp BC$, \\
therefore $\angle BDF = \angle ADC = 90^\circ$, \\
In right triangles $\triangle BDF$ and $\triangle ADC$, \\
$\left\{\begin{array}{l}AC=BF\\ DC=DF\end{array}\right.$, \\
therefore right $\triangle BDF \cong \triangle ADC$ (HL), \\
therefore $AD = BD$, \\
therefore $\angle ABD = \angle DAB = 45^\circ$, \\
since $\angle ABE = 15^\circ$, \\
therefore $\angle DBF = \angle ABD - \angle ABE = 45^\circ - 15^\circ = \boxed{30^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ is the altitude from $A$ to side $BC$, and $BE$ is the altitude from $B$ to side $AC$, with $AD$ and $BE$ intersecting at point $F$. If $BF=AC$, $BD=8$, and $CD=3$, what is the length of segment $AF$?

Solution:
\textbf{Solution:}

Since AD is the height from vertex A to side BC, and BE is the height from vertex B to side AC,

$\therefore \angle ADC=\angle FDB=90^\circ$, $\angle AEB=90^\circ$,

$\therefore \angle 1+\angle C=90^\circ$, $\angle 1+\angle 2=90^\circ$,

$\therefore \angle 2=\angle C$,

Since $\angle 2=\angle 3$,

$\therefore \angle 3=\angle C$,

In $\triangle ADC$ and $\triangle BDF$,

$\left\{\begin{array}{l}
\angle 3=\angle C \\
\angle FDB=\angle CDA \\
BF=AC
\end{array}\right.$,

$\therefore \triangle BDF \cong \triangle ADC$ (AAS),

$\therefore FD=CD$, $AD=BD$,

Since CD=3, BD=8,

$\therefore AD=8$, $DF=3$,

$\therefore AF=8-3=\boxed{5}$.


Question: \textit{As shown in the figure, in an equilateral $\triangle ABC$, $BD$ is the bisector of $\angle ABC$, point $E$ is the midpoint of $BC$, and point $P$ is a movable point on $BD$. Lines $PE$ and $PC$ are drawn. When the value of $PE + PC$ is at its minimum, what is the degree measure of $\angle EPC$?}

Solution:
\textbf{Solution:} Since $\triangle ABC$ is an equilateral triangle, and BD is the bisector of $\angle ABC$, \\
therefore, point D is the midpoint of AC, and BD$\perp$AC, \\
thus, points A and C are symmetric with respect to BD, \\
as shown in the figure, draw AE, intersecting BD at P, the length of line segment AE is the minimum value of PE+PC, \\
since point E is the midpoint of side BC, \\
thus AE$\perp$BC, \\
since $\angle ABC=60^\circ$, and BD is the bisector of $\angle ABC$, \\
thus $\angle PBE=30^\circ$, \\
therefore, $\angle BPE=60^\circ$, \\
since in $\triangle BPE$ and $\triangle CPE$, \\
$ \left\{\begin{array}{l}PE=PE \\ \angle PEB=\angle PEC=90^\circ \\ BE=CE \end{array}\right.$, \\
therefore, $\triangle BPE\cong \triangle CPE$ (SAS), \\
therefore, $\angle EPC=\angle BPE=60^\circ$. \\
Hence, the answer is: $\boxed{60^\circ}$.


Question: As shown in the figure, in $ \triangle ABC$, $ \angle C=90^\circ$, AD bisects $ \angle BAC$, $ DE\perp AB$ with foot E, $ AB=12$, $ AC=8$, what is the length of BE?

Solution:
\textbf{Solution:} Given $\because$ AD bisects $\angle$BAC,\\
$\therefore$$\angle$DAE=$\angle$DAC\\
$\because$$\angle$C=90$^\circ$, DE$\perp$AB,\\
$\therefore$$\angle$AED=$\angle$ACD=90$^\circ$,\\
Also, AD=AD,\\
$\therefore$ $\triangle$ADE$\cong$ $\triangle$ADC,\\
$\therefore$AE=AC,\\
$\because$AB=12, AC=8,\\
$\therefore$BE=AB−AE=AB−AC=12−8=\boxed{4}.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $\angle B=15^\circ$, the perpendicular bisector of AB intersects AB at E and intersects BC at D. Given that BD=8, what is the length of AC?

Solution:
\textbf{Solution:} As shown in the figure: \\
Since in $\triangle ABC$, $\angle C=90^\circ$, $\angle B=15^\circ$, \\
therefore $\angle BAC=180^\circ-\angle C-\angle B=180^\circ-90^\circ-15^\circ=75^\circ$. \\
Draw AD. \\
Since ED is the perpendicular bisector of AB, therefore AD=BD=8, $\angle B=\angle 1=15^\circ$, \\
therefore $\angle 2=\angle BAC-\angle 1=75^\circ-15^\circ=60^\circ$. \\
In $\triangle ACD$, $\angle 2=60^\circ$, $\angle C=90^\circ$, \\
therefore $\angle 3=180^\circ-\angle C-\angle 2=180^\circ-90^\circ-60^\circ=30^\circ$. \\
Therefore, $AC=\frac{1}{2}AD=\frac{1}{2}BD=\frac{1}{2}\times 8=\boxed{4}$.


Question: In the diagram, within $\triangle ABC$, where $AB=10$, $AC=8$, the angle bisectors of $\angle ABC$ and $\angle ACB$ intersect at point D. A line through D, $EF\parallel BC$, intersects $AB$ at point E, and intersects $AC$ at point F. What is the perimeter of $\triangle AEF$?

Solution:
\textbf{Solution:} Since $EF\parallel BC$\\
$\therefore \angle BDE=\angle DBC$\\
Since $BD$ bisects $\angle ABC$\\
$\therefore \angle DBC=\angle DBE$\\
$\therefore \angle BDE=\angle DBE$\\
$\therefore BE=DE$\\
Similarly, it can be proved that $CF=DF$\\
Since $AB=10$, $AC=8$\\
$\therefore$ the perimeter of $\triangle AEF = AE+AF+DE+DF=AE+AF+BE+CF=AB+AC=\boxed{18}$


Question: As shown in the figure, in $\triangle ABC$, it is known that $\angle ABC=90^\circ$. With $AC$ as the right angle side, $\triangle ACD$ is constructed outside ($\angle CAD=90^\circ$). Semicircles are drawn outside with diameters $AB$, $BC$, $CD$, $DA$ respectively, and their areas are denoted as $S_1$, $S_2$, $S_3$, and $S_4$. It is known that $S_1=3$, $S_2=1$, and $S_3=7$. What is the value of $S_4$?

Solution:
\textbf{Solution:} From the given problem, we have
\begin{align*}
S_1 &= \frac{1}{8}AB^2\pi = 3, \\
S_2 &= \frac{1}{8}BC^2\pi = 1, \\
S_3 &= \frac{1}{8}CD^2\pi = 7, \\
S_4 &= \frac{1}{8}AD^2\pi,
\end{align*}
Using the Pythagorean theorem, we can derive:
\begin{align*}
AC^2 &= BC^2 + AB^2, \\
AD^2 &= CD^2 - AC^2,
\end{align*}
Hence,
\begin{align*}
AD^2 &= CD^2 - BC^2 - AB^2, \\
\end{align*}
From the above equations, we reach:
\begin{align*}
AB^2 &= \frac{24}{\pi}, \\
BC^2 &= \frac{8}{\pi}, \\
CD^2 &= \frac{56}{\pi},
\end{align*}
Therefore,
\begin{align*}
AD^2 &= CD^2 - BC^2 - AB^2 = \frac{24}{\pi},
\end{align*}
Thus,
\begin{align*}
S_4 &= \frac{1}{8}AD^2\pi = \frac{1}{8} \times \frac{24}{\pi} \times \pi = \boxed{3}.
\end{align*}


Question: As shown in the diagram, in $\triangle ABC$, $DE$ is the perpendicular bisector of $AC$, $AB=4$, and $BC=7$. What is the perimeter of $\triangle ABD$?

Solution:
\textbf{Solution:} Given that $DE$ is the perpendicular bisector of $AC$,\\
$\therefore AD=CD$,\\
$\therefore BC=BD+CD=BD+AD$,\\
$\therefore$ the perimeter of $\triangle ABD$ $=AB+BD+AD=AB+BC=4+7=\boxed{11}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, D is the midpoint of AB, $AC=5$, and $BC=12$. What is the length of $CD$?

Solution:
\textbf{Solution:} In right triangle $\triangle ACB$, $\angle ACB=90^\circ$, $AC=5$, $BC=12$,\\
$\therefore AB= \sqrt{AC^{2}+BC^{2}}=\sqrt{5^{2}+12^{2}}=13$,\\
$\because D$ is the midpoint of $AB$,\\
$\therefore AD=BD$,\\
$\therefore CD= \frac{1}{2} AB=\boxed{6.5}$.


Question: As shown in the figure, in the isosceles right triangle $\triangle ABC$ with leg length of $4$, $D$ and $E$ are two moving points on sides $AB$ and $AC$ respectively, and $AE=BD$. What is the minimum value of $CD+BE$?

Solution:
\textbf{Solution:} Construct CF $\perp$ AC such that CF=AC=4. Connect EF and BF, intersecting AC at point O.

$\because$ AB=AC, AE=BD,

$\therefore$ AD=CE,

In $\triangle ACD$ and $\triangle CEF$,

$\left\{\begin{array}{l}
AD=CE\\
\angle DAC=\angle ECF\\
AC=CF
\end{array}\right.$

$\therefore$ $\triangle ACD \cong \triangle CEF$ (SAS),

$\therefore$ CD=FE,

$\therefore$ BE+CD=BE+EF$\geq$BF,

When points B, E, and F are collinear, the value of BE+CD=BF is minimal,

$\because$ $\angle$BAC=$\angle$ACF=90$^\circ$, AB=AC, AC=CF,

$\therefore$ AB$\parallel$CF,

$\therefore$ quadrilateral ABCF is a parallelogram,

$\therefore$ BF=2OB, AO=OC=2,

In Rt $\triangle ABO$, $BO=\sqrt{AB^{2}+AO^{2}}=\sqrt{4^{2}+2^{2}}=2\sqrt{5}$

$\therefore$ BF=$\boxed{4\sqrt{5}}$.


Question: As shown in the figure, given a rectangular paper $ABCD$, where point $E$ is on side $AB$ and $BE=2, BC=3$, if $\triangle CBE$ is folded along the line $CE$ such that point $B$ falls on point $G$, and extending $EG$ intersects $CD$ at point $F$, then what is the length of segment $FG$?

Solution:
\textbf{Solution:} Given that by folding the triangle $\triangle CBE$ along the line $CE$, with $BE=2$ and $BC=3$, \\
it follows that $\angle BEC=\angle FEC$ and $\angle CGE=\angle B=90^\circ$. Hence, $BE=GE=2$ and $BC=CG=3$. \\
Since we are dealing with the rectangle $ABCD$, \\
it implies that $AB\parallel CD$, \\
which leads to the conclusion that $\angle BEC=\angle FCE$, \\
therefore, $\angle FEC=\angle FCE$, \\
indicating that $FE=FC$. \\
Let $FG=x$, then $FC=x+2$, \\
thus, in the right triangle $\triangle FGC$, we have $(x+2)^2=x^2+3^2$, \\
solving this, we get $x=\frac{5}{4}$, which means $FG=\boxed{\frac{5}{4}}$.


Question: As shown in the figure, two right-angled triangles with angles $45^\circ$ and $30^\circ$ are placed on the same plane. If $AB=2\sqrt{2}$, what is the length of $BD$?

Solution:
\textbf{Solution:} Given that $\triangle ABC$ is a right-angled triangle with a $45^\circ$ angle, and $AB=2\sqrt{2}$, \\
therefore $AC^2+BC^2=AB^2$, \\
thus $AC=BC=2$, \\
since $\triangle DBC$ is a right-angled triangle with a $30^\circ$ angle, and $\angle D=30^\circ$, $BC=2$, \\
therefore $BD=2BC=\boxed{4}$.


Question: As shown in the figure, fold a rectangular strip of paper successively along the line segments $EF$ and $HI$, with $EF\parallel HI$, to form a "Z" shaped pattern. Given that $\angle DFE = 60^\circ$ and $AE = 2\text{cm}$, in order for point $H$ and point $K$ to be on the extensions of $AD$ and $EF$ respectively (not coincident with $D$ or $F$), what is the length of $AB$ in centimeters?

Solution:
\textbf{Solution:} As shown in the figure, connect $DH$ and $FK$, where points H and K are on the extensions of $AD$ and $EF$ (not coinciding with D and F), respectively, and point M is a point on the extension of $EH$.\\
In the rectangle strip $ABCD$, $\angle A=\angle ADC=90^\circ$, $AB\parallel CD$,\\
$\therefore$ $\angle DFE=\angle BEF=60^\circ$, $EH\parallel FI$,\\
From the folding, it is known that: $\angle GEF=\angle BEF=60^\circ$,\\
$\therefore$ $\angle AEG=60^\circ$,\\
$\therefore$ $\angle AHE=30^\circ$,\\
$\therefore$ $AE=\frac{1}{2}EH$,\\
$\because$ $EH\parallel FI$,\\
$\therefore$ $\angle EHI+\angle GEF=180^\circ$,\\
$\therefore$ $\angle EHI=120^\circ$,\\
$\therefore$ $\angle MHI=60^\circ$,\\
From the folding, it is known that: $\angle JHI=\angle MHI=60^\circ$,\\
$\therefore$ $\angle EHK=60^\circ$,\\
$\because$ $\angle GEF=60^\circ$,\\
$\therefore$ $\angle EKH=60^\circ$,\\
$\therefore$ $\triangle EHK$ is an equilateral triangle,\\
$\therefore$ $EH=HK=EK$,\\
$\because$ $AE=\frac{1}{2}EH$, $AE=2\text{cm}$\\
$\therefore$ $EH=HK=EK=4\text{cm}$,\\
From the folding, it is known: $AE+EH+HK=AB$,\\
$\therefore$ $AB=2+4+4=\boxed{10}\text{cm}$.


Question: As shown in the figure, in triangle $ABC$, $AD \perp BC$, $CE \perp AB$, with the perpendicular feet being D and E respectively. $AD$ and $CE$ intersect at point H. Given that $EH = EB = 5$ and $CH = 2$, then what is the length of $AE$?

Solution:
\textbf{Solution:} Given: $EH=EB=5$, $CH=2$,\\
Therefore $EC=EH+CH=7$,\\
Since $AD\perp BC$, $CE\perp AB$,\\
Thus $\angle ADB=\angle AEH=90^\circ$,\\
Since $\angle AHE=\angle CHD$,\\
Thus $\angle BAD=\angle BCE$,\\
In $\triangle HEA$ and $\triangle BEC$, $\left\{\begin{array}{l}\angle EAH=\angle BCE \\ \angle AEH=\angle BEC=90^\circ \\ EH=EB\end{array}\right.$,\\
Therefore $\triangle HEA\cong \triangle BEC\left(AAS\right)$,\\
Hence $AE=EC=\boxed{7}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, point D is on the right angle side $BC$, $AD$ bisects $\angle BAC$, $DE$ is the perpendicular bisector of $AB$, $CD=8\,cm$, then what is the length of $BD$ in $cm$?

Solution:
Solution: Since $DE$ is the perpendicular bisector of side $AB$ of $\triangle ABC$,\\
it follows that $AD=BD$,\\
therefore, $\angle B=\angle DAB$,\\
since $AD$ bisects $\angle BAC$ and $\angle C=90^\circ$,\\
it follows that $\angle DAB=\angle DAC$ and $DC=DE=8$,\\
therefore, $\angle B=\angle DAB=\angle DAC$,\\
hence, $\angle B=30^\circ$,\\
in right triangle $Rt\triangle BDE$ with $\angle B=30^\circ$ and $DE=8$,\\
it follows that $DB=\boxed{16}$.


Question: As shown in the figure, $AD$ perpendicularly bisects $BC$ at $D$, $\angle BAC=45^\circ$, $CE \perp AB$ at $E$, intersecting $AD$ at $F$, and $BD=2$. What is the length of $AF$?

Solution:
\textbf{Solution:} Given that $AD$ vertically bisects $BC$, and $BD=2$, \\
it follows that $BC=2BD=4$, \\
since $CE\perp AB$, \\
it follows that $\angle CEB=90^\circ$, \\
thus, $\angle AEC=90^\circ$, \\
given that $\angle BAC=45^\circ$, \\
it follows that $\angle ACE=90^\circ-45^\circ=45^\circ$, \\
hence, $\angle EAC=\angle ACE$, \\
thus, $AE=CE$, \\
since $AD$ vertically bisects $BC$, \\
it follows that $\angle ADB=90^\circ$, \\
thus, $\angle B+\angle BAD=90^\circ$, \\
given that $\angle B+\angle BCE=90^\circ$, \\
it follows that $\angle BAD=\angle BCE$, \\
in $\triangle AEF$ and $\triangle CEB$, we have $\left\{\begin{array}{l}\angle AEF=\angle CEB \\ AE=CE \\ \angle EAF=\angle BCE\end{array}\right.$, \\
therefore, $\triangle AEF\cong \triangle CEB\left(ASA\right)$, \\
hence, $AF=BC= \boxed{4}$.


Question: As shown in the figure, points B and D are on the ray AM, points C and E are on the ray AN, and AB=BC=CD=DE. Given that $\angle EDM=84^\circ$, what is the measure of $\angle A$ in degrees?

Solution:
Solution: Since $AB=BC=CD=DE$, \\
it follows that $\angle A=\angle BCA$, $\angle CBD=\angle BDC$, $\angle ECD=\angle CED$,\\
according to the exterior angle theorem, $\angle A+\angle BCA=\angle CBD$, $\angle A+\angle CDB=\angle ECD$, $\angle A+\angle CED=\angle EDM$,\\
additionally, since $\angle EDM=84^\circ$,\\
thus, $\angle A+3\angle A=84^\circ$,\\
solving this gives $\angle A=\boxed{21^\circ}$.


Question: As shown in the figure, $AB\perp BC$, $CD\perp BC$, with the perpendicular feet being $B$, $C$ respectively. Let $P$ be a point on the line segment $BC$. Connect $PA$ and $PD$. Given that $AB=5$, $DC=4$, and $BC=12$, what is the minimum value of $AP+DP$?

Solution:
\textbf{Solution:} As shown in the figure, let's construct point $A'$ symmetrical to point $A$ with respect to $BC$. Draw $A'D$ intersecting $BC$ at point $P$. Through point $A'$, draw $A'M \perp DC$ intersecting at point $M$,
\[
\therefore AP = A'P,
\]
\[
\therefore AP + PD = A'P + PD,
\]
When points $A'$, $P$, and $D$ are collinear, $A'P + PD = A'D$. At this moment, the value of $A'P + PD$ is minimized,
\[
\because AB = 5, \; DC = 4, \; BC = 12,
\]
\[
\therefore AM = 12, \; DM = 5 + 4 = 9,
\]
In $\triangle A'DM$, $A'D = \sqrt{A'M^{2} + DM^{2}} = \sqrt{12^{2} + 9^{2}} = 15$,
\[
\therefore \text{The minimum value of } AP + PD \text{ is } \boxed{15}.
\]


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$. Point $D$ is on $AC$, and $\angle DBC=\angle A$. If $AC=4$ and $AB=5$, what is the length of $BD$?

Solution:
\textbf{Solution:} In the right triangle $ABC$, by the Pythagorean theorem, we have $BC=\sqrt{AB^{2}-AC^{2}}=3$,\\
since $\angle DBC = \angle A$, and $\angle C = \angle C$,\\
thus $\triangle ABC \sim \triangle BDC$,\\
therefore $\frac{AC}{BC}=\frac{AB}{BD}$, that is, $\frac{4}{3}=\frac{5}{BD}$,\\
hence $BD=\boxed{\frac{15}{4}}$.


Question: As shown in the figure, it is known that quadrilateral $ABCD$ is inscribed in circle $\odot O$, with radius $AO=6$. The diagonals $AC$ and $BD$ intersect at point $E$, and $AB=BD$, $EC=2$. Then what is the length of $AD$?

Solution:
\textbf{Solution:} Connect BO and extend it to meet AD at point F, draw OD, and CD, \\
$\because$BA=BD, OA=OD, \\
$\therefore$BF is the perpendicular bisector of line segment AD, \\
$\therefore$BF$\perp$AD, \\
$\because$AC is the diameter of $\odot$O, \\
$\therefore$$\angle$ADC$=90^\circ$, \\
that is, AD$\perp$DC, \\
$\therefore$BF$\parallel$CD, \\
$\therefore$$\triangle $BOE$\sim $$\triangle $DCE, \\
$\therefore$$\frac{OB}{CD}=\frac{EO}{EC}$, \\
$\because$AO=6, EC=2, \\
$\therefore$OB=6, OC=6, \\
$\therefore$OE=4, \\
$\therefore$$\frac{6}{CD}=\frac{4}{2}$, \\
Solving, we get CD=3, \\
In the right triangle ADC, $\angle$ADC$=90^\circ$, AC=12, CD=3, \\
$\therefore$AD=$\sqrt{AC^2−CD^2}=\sqrt{12^2−3^2}=\boxed{3\sqrt{15}}$, \\


Question: As shown in the figure, the line $y=-\frac{1}{2}x$ intersects the x-axis and y-axis at points B and A, respectively. Point C is a moving point on the x-axis. A circle $\odot C$ with center C and radius $\sqrt{5}$ is drawn. When $\odot C$ is tangent to line AB, what are the coordinates of point C?

Solution:
\textbf{Solution:}
(1) When point C is on the negative half of the x-axis, let C be $(t,0)$. Construct CM $\perp$ AB at M, as shown,\\
For\\
$y=-\frac{1}{2}x+1$, when $x=0$, $y=1$; when $y=0$, $x=2$\\
$\therefore$ A$(0,1)$, B$(2,0)$\\
$\therefore$ OA$=1$, OB$=2$,\\
$\therefore$ BC$=2-t$\\
By the Pythagorean theorem,\\
$AB=\sqrt{OA^2+OB^2}=\sqrt{1^2+2^2}=\sqrt{5}$\\
Since line AB is tangent to circle C,\\
$\therefore$ $\angle$CMB$=90^\circ$\\
Also $\angle$ABO$=\angle$CBM,\\
$\triangle$BMC$\sim$$\triangle$BOA,\\
$\therefore$ $\frac{AB}{BC}=\frac{AO}{CM}$, i.e.,\\
$\frac{\sqrt{5}}{2-t}=\frac{1}{\sqrt{5}}$\\
Solving this,\\
$t=-3$\\
$\therefore$ The coordinates of point C are ($-3$,0)\\
(2) When point C is on the positive half of the x-axis, let C be $(t,0)$. Construct CN $\perp$ AB at N, as shown,\\
$\therefore$ BC$=t-2$\\
Since $\angle$ABO$=\angle$CBN,\\
$\angle$CNB$=\angle$AOB\\
$\therefore$ $\triangle$BNC$\sim$$\triangle$BOA,\\
$\therefore$ $\frac{AB}{BC}=\frac{AO}{CN}$, i.e.,\\
$\frac{\sqrt{5}}{t-2}=\frac{1}{\sqrt{5}}$\\
Solving this,\\
$t=7$\\
$\therefore$ The coordinates of point C are $(7,0)$\\
In summary, the coordinates of point C are $(-3,0)$ or $(7,0)$\\
Therefore, the answer is: $\boxed{(-3,0) \text{ or } (7,0)}$.


Question: As shown in the figure, AB is the diameter of the semicircle O, D is a point on the semicircle O, C is the midpoint of the arc $\overset{\frown}{BD}$, connecting AC intersects BD at point E, connecting AD, if $BE = 4DE$ and $CE = 6$, what is the length of AB?

Solution:
\textbf{Solution:} As shown in the figure, extend OC to intersect BD at K, and connect BC.\\
Since $\overset{\frown}{CD} = \overset{\frown}{BC}$,\\
it follows that OC$\perp$BD.\\
Given BE=4DE,\\
we can assume DE=k, BE=4k, then DK=BK=2.5k, EK=1.5k.\\
Since AB is the diameter,\\
it implies $\angle$ADK=$\angle$DKC=$\angle$ACB=$90^\circ$,\\
thus AD$\parallel$CK,\\
which leads to AE: EC=DE: EK,\\
therefore, AE: 6=k: 1.5k,\\
hence, AE=4.\\
Since $\triangle$ECK$\sim$$\triangle$EBC,\\
it follows that EC$^{2}$=EK$\cdot$EB,\\
thus, 36=1.5k$\cdot$4k.\\
Given k>0,\\
we find k=$\sqrt{6}$,\\
therefore, BC=$\sqrt{{BE}^{2}-{EC}^{2}}=\sqrt{96-36}=2\sqrt{15}$,\\
hence, AB=$\sqrt{{AC}^{2}+{BC}^{2}}=\sqrt{{10}^{2}+{(2\sqrt{15})}^{2}}=\boxed{4\sqrt{10}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $E$ is a point on $CD$. Connect $BE$ and extend it to meet the extended line of $AD$ at point $F$. If $DE:EC=2:3$, then what is the value of $\frac{S_{\triangle DEF}}{S_{\triangle ABF}}$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram,\\
it follows that AB$\parallel$CD and AB＝CD,\\
thus $\triangle DEF\sim \triangle ABF$,\\
hence $\frac{S_{\triangle DEF}}{S_{\triangle ABF}} = \left(\frac{DE}{AB}\right)^2$,\\
since DE:EC＝2:3,\\
it follows that DE:CD＝DE:AB＝2:5,\\
therefore, S$_{\triangle DEF}$:S$_{\triangle ABF}$＝4:25.\\
Thus the answer is: $\boxed{4:25}$.


Question: As shown in the figure, in $\triangle ABC$, $BC=120$, the height $AD=60$, and the square $EFGH$ with one side on $BC$, points $E$, $F$ respectively on $AB$, $AC$, $AD$ intersects $EF$ at point $N$, what is the length of $AN$?

Solution:
\textbf{Solution:} Let the side lengths of square EFGH be EF = EH = x,\\
since EFGH is a square,\\
thus $\angle$HEF = $\angle$EHG = $90^\circ$, and EF$\parallel$BC,\\
thus $\triangle$AEF $\sim$ $\triangle$ABC,\\
since AD is the altitude of $\triangle$ABC,\\
thus $\angle$HDN = $90^\circ$,\\
thus quadrilateral EHDN is a rectangle,\\
thus DN = EH = x,\\
since $\triangle$AEF $\sim$ $\triangle$ABC,\\
thus  $\frac{AN}{AD}=\frac{EF}{BC}$ (the ratio of altitudes is equal to the scale factor),\\
since BC = 120, AD = 60,\\
thus AN = 60−x,\\
thus $\frac{60−x}{60}$ = $\frac{x}{120}$,\\
solving for x gives: x = 40,\\
thus AN=60−x=60−40=\boxed{20}.


Question: As shown in the figure, $\triangle A'B'C'$ and $\triangle ABC$ are similar figures with $O$ as the center of similarity. If $OA'=A'A$, then what is the ratio of the areas of $\triangle A'B'C'$ to $\triangle ABC$?

Solution:
\textbf{Solution:} Given that, $\triangle ABC$ and $\triangle A'B'C'$ are similar figures,\\
$\therefore \triangle ABC \sim \triangle A'B'C'$, and the ratio AB$\colon$ A'B'=OA$\colon$ AA'=1$\colon$ 2,\\
$\therefore$ the ratio of the areas of $\triangle A'B'C'$ to $\triangle ABC$ is: $1\colon 4$.\\
Therefore, the ratio of the areas is $\boxed{1\colon 4}$.


Question: As shown in the figure, in right-angled triangle $\triangle ABC$, with $\angle ACB=90^\circ$, and $CD\perp AB$ at $D$, if $AD=4$ and $BD=9$, what is the length of $CD$?

Solution:
\textbf{Solution:} Given: $\because$ $\angle ACD + \angle BCD = \angle BCD + \angle B$,\\
$\therefore$ $\angle ACD = \angle B$,\\
Also, $\angle ADC = \angle BDC = 90^\circ$,\\
$\therefore$ $\triangle ADC \sim \triangle BDC$,\\
$\therefore$ $\frac{AD}{CD} = \frac{CD}{BD}$,\\
$\therefore$ $CD^2 = AD \cdot BD = 36$,\\
$\therefore$ $CD = \boxed{6}$ or $-6$ (discard).


Question: As shown in the figure, D and E are points on sides AB and BC of $\triangle ABC$, respectively, and $DE\parallel AC$. AE and CD intersect at point O. If $\frac{S_{\triangle DOE}}{S_{\triangle COA}}=\frac{1}{25}$, what is the value of $\frac{S_{\triangle BDE}}{S_{\triangle CDE}}$?

Solution:
\textbf{Solution:} Given that $DE\parallel AC$,\\
$\therefore \triangle DEO\sim \triangle CAO$,\\
$\therefore \frac{{S}_{\triangle DEO}}{{S}_{\triangle CAO}}=\left(\frac{DE}{AC}\right)^{2}=\frac{1}{25}$,\\
$\therefore \frac{DE}{AC}=\frac{BE}{BC}=\frac{1}{5}$,\\
$\therefore \frac{BE}{EC}=\frac{1}{4}$,\\
$\therefore \frac{{S}_{\triangle BED}}{{S}_{\triangle DEC}}=\boxed{\frac{1}{4}}$.


Question: As shown in the figure, $ \triangle ABC$ is an equilateral triangle with a side length of 6. $CE$ bisects the exterior angle $\angle ACF$. Point $D$ is on $AC$, line $BD$ is extended to intersect $CE$ at point $E$. If $CD= \frac{1}{2} AD$, then what is the value of $ \frac{CE}{BC}$? What is the value of $BE$?

Solution:
\textbf{Solution:} Given that $ \triangle ABC$ is an equilateral triangle with a side length of 6,\\
$ \therefore AB=BC=6$, $\angle A=\angle ACB=60^\circ$,\\
since $CE$ bisects the exterior angle $\angle ACF$,\\
$ \therefore \angle FCE=\angle DCE=\frac{1}{2}\angle ACF=\frac{1}{2}(180^\circ-\angle ACB)=60^\circ$,\\
in $\triangle ABD$ and $\triangle CED$, $\left\{\begin{array}{l}\angle ADB=\angle CDE \\ \angle A=\angle DCE=60^\circ \end{array}\right.$,\\
$ \therefore \triangle ABD\sim \triangle CED$,\\
$ \therefore \frac{CE}{AB}=\frac{CD}{AD}=\frac{1}{2}$, that is $\frac{CE}{6}=\frac{1}{2}$,\\
solving this gives $CE=3$,\\
$ \therefore \frac{CE}{BC}=\frac{3}{6}=\boxed{\frac{1}{2}}$,\\
as shown, draw $EG\perp BF$ at point $G$,\\
in $Rt\triangle CEG$, $\angle CEG=90^\circ-\angle FCE=30^\circ$,\\
$ \therefore CG=\frac{1}{2}CE=\frac{3}{2}$, $EG=\sqrt{CE^2-CG^2}=\frac{3}{2}\sqrt{3}$,\\
$ \therefore BG=BC+CG=\frac{15}{2}$,\\
in $Rt\triangle BEG$, $BE=\sqrt{BG^2+EG^2}=\sqrt{\left(\frac{15}{2}\right)^2+\left(\frac{3}{2}\sqrt{3}\right)^2}=\boxed{3\sqrt{7}}$.


Question: As shown in the trapezoid ABCD, AD $\parallel$ BC, $\angle BAD = 90^\circ$, and the diagonal BD $\perp$ DC. If AD = 4 and BC = 9, what is the length of BD?

Solution:
\textbf{Solution:} Given: $\because$ AD$\parallel$BC $\therefore$ $\angle$ADB=$\angle$CBD (When two lines are parallel, the corresponding alternate interior angles are equal)\\
$\because$ $\angle$BAD=$90^\circ$, BD$\perp$DC, $\therefore$ $\angle$BAD=$\angle$BDC=$90^\circ$,\\
$\because$ $\angle$ADB=$\angle$CBD, $\angle$BAD=$\angle$BDC, $\therefore$ $\triangle$ABD$\sim$$\triangle$DCB (Two triangles are similar if two angles are equal respectively)\\
$\therefore$ $\frac{BD}{BC}=\frac{AD}{BD}$ (Corresponding sides of similar triangles are proportional)\\
$\therefore$ $BD^2=AD\times BC$, $\because$AD=4, BC=9\\
$\therefore$BD=\boxed{6}


Question: As shown in the figure, $AB \parallel CD \parallel EF$, $AF$ intersects $BE$ at point $G$, and $AG=2$, $GD=1$, $DF=5$. What is the value of $\frac{BC}{CE}$?

Solution:
\textbf{Solution}: We have $AD=AG+GD=3$, \\
since $AB\parallel CD\parallel EF$, \\
thus $\frac{BC}{CE}=\frac{AD}{DF}=\boxed{\frac{3}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $AD \perp BC$ with the foot of the perpendicular being D, $AD=5$, $BC=10$. Quadrilaterals $EFGH$ and $HGNM$ are both squares, and points E, F, G, N, M are all on the sides of $\triangle ABC$. What is the ratio of the area of $\triangle AEM$ to the area of quadrilateral $BCME$?

Solution:
\textbf{Solution:} Let quadrilaterals EFGH and HGNM both be squares, and $AD\perp BC$, \\
therefore quadrilateral EFDP is a rectangle, and $AP\perp EM$,\\
let us set $PD=EF=EH=HM=x$,\\
therefore $EM=2x$, $AP=5-x$,\\
since $EM\parallel BC$,\\
therefore $\angle AEM=\angle B$, $\angle AME=\angle C$,\\
therefore $\triangle AEM\sim \triangle ABC$,\\
since AP and AD are altitudes of $\triangle AEM$ and $\triangle ABC$ respectively,\\
therefore $\frac{AP}{AD}=\frac{EM}{BC}$, substituting we get: $\frac{5-x}{5}=\frac{2x}{10}$,\\
solving yields: $x=\frac{5}{2}$, which means $PD=\frac{5}{2}$,\\
therefore $AP=5-\frac{5}{2}=\frac{5}{2}$,\\
therefore $\frac{AP}{AD}=\frac{\frac{5}{2}}{5}=\frac{1}{2}$,\\
therefore $\frac{S_{\triangle AEM}}{S_{\triangle ABC}}=\left(\frac{AP}{AD}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}$,\\
since $S_{\triangle ABC}=S_{\triangle AEM}+S_{\text{quadrilateral} BCME}$,\\
therefore $S_{\triangle AEM} \colon S_{\text{quadrilateral} BCME}=1\colon 3$.\\
Hence, the answer is: $\boxed{1\colon 3}$.


Question: As shown in the figure, in square $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $E$ is the midpoint of $OB$. Connect $AE$ and extend it to intersect $BC$ at point $F$. If the area of $\triangle BEF$ is $1$, what is the area of square $ABCD$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a square,\\
it follows that OB=OD and AD$\parallel$BC,\\
thus, $\triangle BEF\sim \triangle DEA$,\\
which means $\frac{BE}{ED}=\frac{EF}{AE}$,\\
since E is the midpoint of OB,\\
it implies $\frac{BE}{ED}=\frac{EF}{AE}=\frac{1}{3}$,\\
hence, the area ratio $S_{\triangle BEF}: S_{\triangle AEB}=EF:AE=\frac{1}{3}$,\\
given that the area of $\triangle BEF$ is 1,\\
thus, the area of $\triangle AEB$ is 3,\\
considering $\frac{BE}{ED}=\frac{1}{3}$,\\
the area ratio $S_{\triangle AEB}: S_{\triangle AED}=\frac{1}{3}$,\\
as a result, the area of $\triangle AED$ is 9,\\
thus, the area of $\triangle ABD$=9+3=12,\\
therefore, the area of square ABCD=$12\times2=\boxed{24}$.


Question: As shown in the figure, in the Cartesian coordinate system, the coordinates of points $P$ and $A$ are (1, 0) and (2, 4), respectively. Point $B$ is a moving point on the $y$-axis. A line passing through point $A$ is perpendicular to $AB$ and intersects the $x$-axis at point $C$. Point $M$ is the midpoint of segment $BC$. What is the minimum value of $PM$?

Solution:
\textbf{Solution:} Draw $AF\perp y$-axis at point $F$, connect $AM$, $OM$\\
Since $\angle BAC=\angle BOC=90^\circ$, and $M$ is the midpoint of $BC$,\\
thus $AM=OM$\\
Therefore, point $M$ is on the perpendicular bisector of segment $AO$\\
Construct the perpendicular bisector of segment $AO$ to intersect the $y$-axis and $x$-axis at points $D$, $E$. When $PM\perp DE$, $PM$ is minimized\\
Connect $AD$, then $AD=OD$\\
Given $A(2,4)$\\
Therefore $AF=2$, $OF=4$\\
Let $OD=AD=t$, then $FD=4-t$,\\
Since $FD^2+AF^2=AD^2$\\
Thus $(4-t)^2+2^2=t^2$\\
Therefore $t=\frac{5}{2}$\\
Therefore $OD=\frac{5}{2}$\\
Since $\angle FOA+\angle AOE=90^\circ$, $\angle AOE+\angle OED=90^\circ$\\
Therefore $\angle FOA=\angle OED$\\
Since $\angle AFO=\angle DOE=90^\circ$\\
Therefore $\triangle FAO\sim \triangle ODE$\\
Therefore $\frac{AF}{OD}=\frac{OF}{OE}$, i.e., $AF\cdot OE=OD\cdot OF$,\\
Therefore $OE=5$\\
Given $P(1,0)$\\
Therefore $PE=4$\\
Since in $\triangle AFO$, $OA=\sqrt{OF^2+AF^2}=\sqrt{4^2+2^2}=2\sqrt{5}$\\
When $PM\perp DE$, $PM$ is minimized\\
Therefore $\angle PME=\angle AFO=90^\circ$\\
$\triangle PME\sim \triangle AFO$\\
Therefore $\frac{PM}{AF}=\frac{PE}{OA}$\\
Therefore $\frac{PM}{2}=\frac{4}{2\sqrt{5}}$\\
Therefore $PM=\boxed{\frac{4}{5}\sqrt{5}}$


Question: As shown, it is known that $\triangle ABC$ and $\triangle A'B'C'$ are similar figures with point $C$ as the center of similitude, and the ratio of the perimeters of $\triangle ABC$ and $\triangle A'B'C'$ is $1\colon 2$. The coordinate of point $C$ is $(-1,0)$. If the x-coordinate of the corresponding point $B'$ of point $B$ is $5$, then what is the x-coordinate of point $B$?

Solution:
\textbf{Solution:} As shown in the diagram, draw $\overline{BD} \perp$ x-axis at point D, and draw $\overline{B^{\prime}H} \perp$ x-axis at point H, then $\overline{BD} \parallel \overline{B^{\prime}H}$,\\
$\therefore \angle DBC = \angle HB^{\prime}C$, $\angle BDC = \angle B^{\prime}HC$,\\
$\therefore \triangle BCD \sim \triangle B^{\prime}CH$,\\
$\therefore \frac{CD}{CH} = \frac{BC}{B^{\prime}C}$,\\
$\because$ the perimeter ratio of $\triangle ABC$ to $\triangle A^{\prime}B^{\prime}C$ is $1:2$,\\
$\therefore \frac{BC}{B^{\prime}C} = \frac{1}{2}$,\\
$\therefore \frac{CD}{CH} = \frac{1}{2}$,\\
$\because$ the coordinates of point C are ($-1$, $0$), and the x-coordinate of the corresponding point $B^{\prime}$ of point B is 5,\\
$\therefore$ OC = 1, OH = 5,\\
$\therefore$ CH = 6,\\
$\therefore CD = \frac{1}{2}CH$ = 3,\\
$\therefore$ OD = OC + CD = 1 + 3 = 4,\\
$\therefore$ the x-coordinate of point B is $\boxed{-4}$.


Question: As shown in the figure, a rectangle I, a small square, and an L-shaped figure composed of four squares of the same size are placed inside rectangle ABCD. One vertex of rectangle I is located at the vertex E of a square within the L-shaped figure, with its other vertices on the sides of rectangle ABCD. The squares within the L-shape have three vertices precisely on the sides of rectangle ABCD, with another vertex coinciding with a vertex of the small square. If rectangle I is similar to rectangle ABCD, what is the value of \(AB:BC\)?

Solution:
\textbf{Solution:} Let the side length of the four identical squares be $a$, and $DF=x$, that is, $GH=IJ=KJ=a$, $IH=EG=2a$,

In $\triangle IJH$, we have $JH= \sqrt{IJ^2+IH^2}=\sqrt{5}a$,

Since quadrilateral ABCD and quadrilateral MEFD are rectangles, and quadrilateral NBLK is a rectangle,

The 4 identical small square templates are placed as shown in the figure,

Therefore, $\angle B=\angle C=\angle KLJ=\angle EFG=\angle JIH=90^\circ$,

Therefore, $\angle LKJ+\angle LJK=\angle IJH+\angle IHJ=\angle GHC+\angle HGC=\angle EGF+\angle GEF=90^\circ$,

$\angle KJL+\angle IJH=\angle IHJ+\angle GHC=\angle EGF+\angle HGC=90^\circ$,

Therefore, $\angle KJL=\angle IHJ=\angle HGC=\angle GEF$,

Therefore, $\triangle KJL\cong \triangle HGC$, $\triangle JHI\sim \triangle HGC\sim \triangle GEF$,

Therefore, $BL=KL=HC$, $LJ=GC$,

$\frac{IJ}{FG}=\frac{IH}{EF}=\frac{JH}{EG}$,

$\frac{IJ}{HC}=\frac{IH}{GC}=\frac{JH}{GH}$, that is,

$\frac{a}{FG}=\frac{2a}{EF}=\frac{\sqrt{5}a}{2a}$,

$\frac{a}{HC}=\frac{2a}{GC}=\frac{\sqrt{5}a}{a}$,

Therefore, $FG= \frac{2\sqrt{5}a}{5}$, $EF= \frac{4\sqrt{5}a}{5}$, $BL=KL=HC= \frac{\sqrt{5}a}{5}$, $LJ=GC= \frac{2\sqrt{5}a}{5}$,

Therefore, $CD=DF+FG+GC=x+ \frac{2\sqrt{5}a}{5}+\frac{2\sqrt{5}a}{5} = \frac{4\sqrt{5}a}{5}+x$,

$BC=BL+LJ+JH+HC= \frac{\sqrt{5}a}{5} + \frac{2\sqrt{5}a}{5}+\sqrt{5}a+\frac{\sqrt{5}a}{5} = \frac{9\sqrt{5}a}{5}$,

When rectangle MEFD $\sim$ rectangle ABCD,

$\frac{CD}{DF}=\frac{BC}{EF}$, that is,

$\frac{\frac{4\sqrt{5}a}{5}+x}{x}=\frac{\frac{9\sqrt{5}a}{5}}{\frac{4\sqrt{5}a}{5}}$,

Solving gives: $x= \frac{16\sqrt{5}a}{25}$,

The ratio of AB:BC is

$\frac{\frac{16\sqrt{5}a}{25}+\frac{4\sqrt{5}a}{5}}{\frac{9\sqrt{5}a}{5}}=\boxed{\frac{4}{5}}$;

When rectangle MDFE $\sim$ rectangle ABCD,

$\frac{CD}{EF}=\frac{BC}{DF}$, that is,

$\frac{\frac{4\sqrt{5}a}{5}+x}{\frac{4\sqrt{5}a}{5}}=\frac{\frac{9\sqrt{5}a}{5}}{x}$,

Solving gives: $x= \frac{10\sqrt{2}-2\sqrt{5}}{5}a$ (negative values discarded),

The ratio of AB:BC is

$\frac{\frac{10\sqrt{2}-2\sqrt{5}}{5}a+\frac{4\sqrt{5}a}{5}}{\frac{9\sqrt{5}a}{5}}$;


Question: As shown in the figure, point A is a moving point on the parabola \(y=\frac{1}{8}x^{2}\) that does not coincide with the origin O. \(AB\perp x\) axis at point B, draw a perpendicular line from point B to \(OA\) and extend it to intersect the y-axis at point C. Connect \(AC\), then what is the length of segment \(OC\)? What is the minimum value of \(AC\)?

Solution:
\textbf{Solution}: Let point A be $(a, \frac{1}{8}a^{2})$, then the coordinates of point B are $(a, 0)$,

$\therefore$ OB = $|a|$, AB = $\frac{1}{8}a^{2}$,

$\because$ $\angle$ABO = $\angle$BOC = $90^\circ$,

$\therefore$ $\angle$AOB + $\angle$OBC = $90^\circ$, $\angle$OBC + $\angle$BCO = $90^\circ$,

$\therefore$ $\angle$AOB = $\angle$BCO,

$\therefore$ $\triangle$AOB $\sim$ $\triangle$BCO,

$\therefore$ $\frac{OB}{CO}=\frac{AB}{BO}$,

$\therefore$ OB$^{2}$ = CO$\cdot$AB, i.e., a$^{2}$ = $\frac{1}{8}a^{2}\cdot$CO,

Solving for CO gives CO = 8,

$\therefore$ C(0, 8),

$\because$ AC$^{2}$ = (xC - xA)$^{2}$ + (yC - yA)$^{2}$ = a$^{2}$ + $\frac{1}{64}a^{4}$ - 2a$^{2}$ + 64 = $\frac{1}{64}(a^{4} - 64a^{2}) + 64 = \frac{1}{64}(a^{2} - 32)^{2} + 48$,

$\therefore$ when a$^{2}$ = 32, AC$^{2}$ = 48 is the minimum value, hence AC = $4\sqrt{3}$.

Thus, the answer is: CO = \boxed{8}, and the minimum value of AC is \boxed{4\sqrt{3}}.


Question: As shown in the figure, the rectangle $ABCD$ is divided into 5 congruent rectangles. If all these 5 rectangles are similar to rectangle $ABCD$, what is the value of $AD\colon AB$?

Solution:
\textbf{Solution:} Let AE = a, \\
$\because$ the five small rectangles are congruent, \\
$\therefore$ AD = 5AE = 5a, \\
$\because$ each small rectangle is similar to rectangle ABCD, \\
$\therefore$  $\frac{AD}{AB}$ =  $\frac{AB}{AE}$, \\
$\therefore$AB$^{2}$=AD$\cdot$AE=5AE$^{2}$=5a$^{2}$, \\
AB =  $\sqrt{5}$ a, \\
$\therefore$AD : AB = 5a : $\sqrt{5}$ a = $\boxed{\sqrt{5}}$.


Question: As shown in the figure, in $ \square ABCD$, E is the midpoint of $AB$, F is on $AD$, and $AF:AD=1:3$, $EF$ intersects $AC$ at G. If $AC=40$, then what is the value of $AG$?

Solution:
\textbf{Solution:} Let the midpoint of AC be O, and connect EO, \\
since E is the midpoint of AB, \\
$\therefore$EO$\parallel$BC and EO= $\frac{1}{2}$ BC, \\
also AD$\parallel$BC, \\
$\therefore$AF$\parallel$EO, \\
$\therefore$$\triangle$AFG$\sim$$\triangle$OEG, \\
$\therefore$ $\frac{AG}{GO}=\frac{AF}{OE}$.\\
Since AC=40 and O is the midpoint, \\
$\therefore$OA=20, then GO=20−AG\\
Since AF$\colon$ AD=1$\colon$ 3 and AD=BC, \\
$\therefore$ $\frac{AG}{GO}=\frac{AF}{EO}=\frac{2}{3}$, \\
$\therefore$ $\frac{AG}{OG}=\frac{AG}{AO−AG}=\frac{2}{3}$.\\
$\therefore$ $\frac{AG}{20−AG}=\frac{2}{3}$, \\
Solving gives $AG=\boxed{8}$.


Question: As shown in the figure, the quadrilateral ADEF is a rhombus, and $AB=6$, $AC=4$. Then, what is the value of $DE$?

Solution:
\[ \textbf{Solution:} \]
\[ \because \text{ Quadrilateral ADEF is a rhombus,} \]
\[ \therefore AD=DE, DE\parallel AC, \]
\[ \therefore \triangle BDE \sim \triangle BAC, \]
\[ \therefore \frac{DE}{AC}=\frac{BD}{AB}, \]
\[ \therefore \frac{DE}{4}=\frac{AB-AD}{AB}, \]
\[ \therefore \frac{DE}{4}=\frac{6-DE}{6}, \]
\[ \text{Solving gives } DE=\boxed{2.4}. \]


Question: As shown in the figure, three squares with side lengths of 2, 3, and 5, respectively, are arranged as shown in the figure. What is the area of the shaded region in the figure?

Solution:
\textbf{Solution:} As shown in the diagram, \\
$\because$VB$\parallel$ED, and the side lengths of the three squares are 2, 3, and 5 respectively, \\
$\therefore$VB$\colon$ DE=AB$\colon$ AD, that is, VB$\colon$ 5=2$\colon$ (2+3+5)=1$\colon$ 5, \\
$\therefore$VB=1, \\
$\because$CF$\parallel$ED, \\
$\therefore$CF$\colon$ DE=AC$\colon$ AD, that is, CF$\colon$ 5=5$\colon$ 10, \\
$\therefore$CF=2.5, \\
$\because$Area of trapezoid $VBFC$= $\frac{1}{2}$(BV+CF)$\cdot$BC= $\frac{21}{4}$, \\
$\therefore$ the area of the shaded part=Area of square $BCQW$−Area of trapezoid $VBCF$= $\boxed{\frac{15}{4}}$.


Question: As shown in the figure, $E$ is a point on the extension of side $BA$ of $\square ABCD$, $CE$ intersects $AD$ at point $F$, $AE=1$, $AB=2$, $BC=3$. What is the value of $AF$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\
then AD $\parallel$ BC, \\
thus $\triangle EAF \sim \triangle EBC$, \\
hence $\frac{AE}{BE} = \frac{AF}{BC}$, \\
since $AE=1$, $AB=2$, $BC=3$, \\
thus $BE=3$, \\
hence $\frac{1}{3} = \frac{AF}{3}$, \\
therefore $AF = \boxed{1}$.


Question: As shown in the figure, it is known that $\triangle ABC \sim \triangle AMN$, where point M is the midpoint of AC, $AB=6$, and $AC=8$. What is the length of $AN$?

Solution:
\textbf{Solution:} Since $\triangle ABC \sim \triangle AMN$, \\
then $\frac{AB}{AM} = \frac{AC}{AN}$, \\
Since M is the midpoint of AC, and AB$=6$, AC$=8$, \\
then AM$=$MC$=4$, \\
thus $\frac{6}{4} = \frac{8}{AN}$, \\
solving gives AN$=\boxed{\frac{16}{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, the side $DE$ of rectangle $DEFG$ is on side $AB$, and the vertices $F$ and $G$ are on sides $BC$ and $AC$, respectively. If the areas of $\triangle BEF$, $\triangle ADG$, and $\triangle CFG$ are respectively 1, 2, and 3, then what is the area of rectangle $DEFG$?

Solution:
\textbf{Solution:} As shown in the diagram, draw $CH\perp AB$, intersecting AB and GF at points H and M, respectively.\\
Since $CH\perp AB$,\\
it follows that $CM\perp GF$,\\
Since quadrilateral DEFG is a rectangle,\\
it follows that $GF\parallel AB$,\\
hence $\triangle CGF\sim \triangle CAB$,\\
therefore $\frac{GF}{AB}=\frac{CM}{CH}$,\\
Given that the areas of $\triangle BEF$ and $\triangle ADG$ are 1 and 2, respectively, with equal heights,\\
it follows that $\frac{BE}{AD}=\frac{1}{2}$,\\
Let $DE=x$, $EF=y$, $BE=a$, then $AD=2a$, the height of $\triangle CGF$ is $CM=\frac{6}{x}$, $CH=y+\frac{6}{x}$,\\
$\frac{x}{x+3a}=\frac{\frac{6}{x}}{y+\frac{6}{x}}$,\\
which simplifies to: $18a=x^2y$ (1),\\
Since $\frac{1}{2}ay=1$,\\
it follows that $ay=2$ (2),\\
Substituting (2) into (1) yields: $x^2y^2=36$,\\
therefore $xy=6$ (disregarding $xy=-6$),\\
thus, the area of the rectangle is $\boxed{6}$.


Question: As shown in the figure, Xiaoming placed a plane mirror on the ground and selected a suitable position, just to see the top of the flagpole in the plane mirror. At this time, the horizontal distance between Xiaoming and the plane mirror is $2\,m$, and the horizontal distance between the bottom of the flagpole and the plane mirror is $12\,m$. If the distance between Xiaoming's eyes and the ground is $1.5\,m$, what is the height of the flagpole in meters?

Solution:
\textbf{Solution:} As shown in the figure,\\
$BC=2m$, $CE=16m$, $AB=1.5m$,\\
By the given information, $\angle ACB=\angle DCE$,\\
Since $\angle ABC=\angle DEC$,\\
Therefore, $\triangle ACB\sim \triangle DCE$,\\
Thus, $\frac{AB}{DE}=\frac{BC}{CE}$, i.e., $\frac{1.5}{DE}=\frac{2}{16}$,\\
Therefore, $DE=\boxed{9}$m.


Question: As shown in the figure, it is known that in $\triangle ABC$, $AC=6\,cm$, and points D and E are the midpoints of AC and BC, respectively. DE is connected, and there is a point F on DE such that $EF=1\,cm$. AF and CF are connected. If $AF \perp CF$, then what is the length of $AB$ in $cm$?

Solution:
Solution: Since $AF\perp CF$ and point D is the midpoint of AC,\\
$\therefore DF=\frac{1}{2}AC$\\
Given that $AC=6cm$\\
$\therefore DF=3cm$\\
Since points D and E are the midpoints of AC and BC, respectively,\\
$\therefore DE=\frac{1}{2}AB$\\
Given that $EF=1cm$\\
$\therefore DE=EF+FD=4cm$\\
$\therefore AB=2DE=\boxed{8cm}$


Question: As shown in the figure, in the parallelogram $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Point $E$ is taken on the extension line of $DC$ such that $CE=\frac{1}{2}CD$. Line $OE$ intersects $BC$ at point $F$. If $BC=12$, what is the length of $CF$?

Solution:
\textbf{Solution:} Construct $OM\parallel BC$ intersecting $CD$ at $M$,\\
since in parallelogram $ABCD$, $BC=12$,\\
therefore, $BO=DO$,\\
therefore, $CM=DM=\frac{1}{2}CD$, $OM=\frac{1}{2}BC=6$\\
since $CE=\frac{1}{2}CD$,\\
therefore, $CE=CM$,\\
since $OM\parallel BC$,\\
therefore, $CF$ is the midline of $\triangle EMO$, that is $CF=\frac{1}{2}OM=\boxed{3}$;


Question: As shown in the diagram, in rhombus $ABCD$, $\angle B=60^\circ$, and the diagonal $AC=10$. What is the perimeter and the area of the rhombus?

Solution:
\textbf{Solution:} Draw line BD intersecting AC at O, as shown in the figure:\\
$\because$ Quadrilateral $ABCD$ is a rhombus,\\
$\therefore$ AB=AC, $AC\perp BD$,\\
$\because$ $\angle ABC=60^\circ$,\\
$\therefore$ $\triangle ABC$ is an equilateral triangle,\\
$\therefore$ AC=AB=10, AO=5,\\
By Pythagoras' theorem, BO=$5\sqrt{3}$,\\
$\therefore$ The perimeter of the rhombus is $4\times 10=\boxed{40}$, the area of the rhombus is $2AO\times BO=10\times 5\sqrt{3}=\boxed{50\sqrt{3}}$.


Question: As shown in the figure, in the rhombus $ABCD$, $E$ is a point on $CD$. The line $AE$ intersects the diagonal $BD$ at point $F$, and $CF$ is connected. If $\angle AED = 40^\circ$, what is the degree measure of $\angle BCF$?

Solution:
\textbf{Solution:} Given: Quadrilateral ABCD is a rhombus,\\
$\therefore$ AB = CB, AB $\parallel$ DC, $\angle$ABF = $\angle$CBF,\\
Since AB = CB, $\angle$ABF = $\angle$CBF, BF = BF,\\
$\therefore$ $\triangle$ABF $\cong$ $\triangle$CBF (by SAS),\\
$\therefore$ $\angle$BAF = $\angle$BCF,\\
Since $\angle$AED = $40^\circ$, AD $\parallel$ BC,\\
$\therefore$ $\angle$AED = $\angle$BAF,\\
$\therefore$ $\angle$BCF = \boxed{40^\circ},


Question: As shown in the diagram, in rhombus $ABCD$, the diagonals $AC=6$ and $BD=8$, $M$ and $N$ are respectively the midpoints of $BC$ and $CD$, $P$ is a moving point on line segment $BD$. What is the minimum value of $PM+PN$?

Solution:
\textbf{Solution:}

Let point $Q$ be the symmetric point of $M$ with respect to $BD$. Connect $NQ$ which intersects $BD$ at $P$, and connect $MP$. At this time, the value of $MP+NP$ is minimized. Connect $AC$,\\
$\because$ Quadrilateral $ABCD$ is a rhombus,\\
$\therefore$ $AC \perp BD$, $\angle QBP = \angle MBP$,\\
which means $Q$ is on $AB$,\\
$\because$ $MQ \perp BD$,\\
$\therefore$ $AC \parallel MQ$,\\
$\because$ $M$ is the midpoint of $BC$,\\
$\therefore$ $Q$ is the midpoint of $AB$,\\
$\because$ $N$ is the midpoint of $CD$ and quadrilateral $ABCD$ is a rhombus,\\
$\therefore$ $BQ \parallel CD$, $BQ = CN$,\\
$\therefore$ Quadrilateral $BQNC$ is a parallelogram,\\
$\therefore$ $NQ = BC$,\\
$\because$ $AQ = CN$, $\angle QAP = \angle PCN$, $\angle APQ = \angle CPN$,\\
$\therefore$ $\triangle APQ \cong \triangle CPN$ (AAS),\\
$\therefore$ $AP = PC$,\\
$\because$ Quadrilateral $ABCD$ is a rhombus,\\
$\therefore$ $CP = \frac{1}{2}AC = 3$, $BP = \frac{1}{2}BD = 4$,\\
In $\triangle BPC$, by the Pythagorean theorem: $BC=5$,\\
that is, $NQ=5$,\\
$\therefore$ $MP+NP=QP+NP=QN=\boxed{5}$,


Question: Given: As shown in the figure, segment $AB=6\text{cm}$, point $P$ is a moving point on segment $AB$, with sides $AP$, $BP$ to construct equilateral $\triangle APC$ and equilateral $\triangle BPD$ respectively on $AB$, connect $CD$, point $M$ is the midpoint of $CD$. When point $P$ moves from point $A$ to point $B$, what is the length of the path traced by point $M$ in centimeters?

Solution:
\textbf{Solution:} As shown in the diagram, extend AC and BD to meet at H. Through point M, draw GN $\parallel$ AB intersecting AH at G and BH at N,\\
$\because$ $\triangle APC$ and $\triangle BPD$ are both equilateral triangles,\\
$\therefore$ $\angle A=\angle B=\angle DPB=\angle CPA=60^\circ$,\\
$\therefore$ AH $\parallel$ PD and BH $\parallel$ CP,\\
$\therefore$ quadrilateral CPDH is a parallelogram,\\
$\therefore$ CD and HP bisect each other,\\
$\therefore$ M is the midpoint of PH,\\
Thus, as P moves, M always remains at the midpoint of HP, so the trajectory of M is indeed the median of $\triangle HAB$, that is, the segment GN,\\
$\therefore$ $GN=\frac{1}{2}AB=\boxed{3}$ cm.


Question: As shown in the figure, points $E$ and $F$ are the midpoints of sides $AB$ and $AC$ of $\triangle ABC$, respectively. $BD$ bisects $\angle ABC$ and intersects $EF$ at point $D$. Connect $AD$. Given $AB=6$ and $BC=8$, what is the length of $DF$? What is the measure of $\angle ADB$ in degrees?

Solution:
\textbf{Solution:} Given that points $E$ and $F$ are the midpoints of sides $AB$ and $AC$ of $\triangle ABC$ respectively, and $BC=8$, $AB=6$, \\
it follows that $EF\parallel BC$ and $EF=\frac{1}{2}BC=4$, and $AE=BE=3$, \\
therefore, $\angle EDB=\angle DBC$, \\
since $BD$ bisects $\angle ABC$, \\
it follows that $\angle EBD=\angle DBC$, \\
hence, $\angle EBD=\angle EDB$, \\
thus, $DE=BE=3$, \\
therefore, $DF=EF-DE=1$, \\
since $AE=DE=BE=3$, \\
it follows that $\angle EAD=\angle EDA$, and since $\angle EBD=\angle EDB$, \\
therefore, $\angle BAD+\angle ABD+\angle ADB=2(\angle ADE+\angle BDE)=180^\circ$, \\
thus, $\angle ADE+\angle BDE=90^\circ=\angle ADB.$\\
Consequently, the answer is: $DF=\boxed{1}$, $\angle ADB=\boxed{90^\circ}$.


Question: As shown in the figure, the diagonals AC and BD of rectangle ABCD intersect at point O. A line is drawn through point O such that $OE \perp AC$, intersecting AD at point E. Another line is drawn through point E such that $EF \perp BD$, with the foot of the perpendicular at F, where $EF=1$, $OE=2$, and $BD=4\sqrt{3}$. What is the area of rectangle ABCD?

Solution:
\textbf{Solution}: Since quadrilateral ABCD is a rectangle,\\
it follows that $OD = OB = OA = OC = \frac{1}{2}BD = 2\sqrt{3}$,\\
since $EF \perp BD$, and $OE \perp AC$,\\
thus, the area of $\triangle AOE = \frac{1}{2}AO \times OE = \frac{1}{2} \times 2\sqrt{3} \times 2 = 2\sqrt{3}$,\\
the area of $\triangle DOE = \frac{1}{2}OD \times EF = \frac{1}{2} \times 2\sqrt{3} \times 1 = \sqrt{3}$,\\
hence, the area of $\triangle AOD = {S}_{\triangle AOE} + {S}_{\triangle DOE} = 2\sqrt{3} + \sqrt{3} = 3\sqrt{3}$,\\
since quadrilateral ABCD is a rectangle,\\
therefore, the area of rectangle ABCD = $4 \times {S}_{\triangle AOD} = 4 \times 3\sqrt{3} = \boxed{12\sqrt{3}}$.


Question: As shown in the figure, point A is on the line segment BG. The quadrilateral ABCD and the quadrilateral DEFG are both squares with areas of 10 and 19 respectively. What is the area of $\triangle CDE$?

Solution:
\textbf{Solution:} Draw EH$\perp$CD at point H.\\
$\because \angle$ADG+$\angle$GDH=$\angle$EDH+$\angle$GDH,\\
$\therefore \angle$ADG=$\angle$EDH.\\
Also, $\because$DG=DE, $\angle$DAG=$\angle$DHE.\\
$\therefore \triangle ADG\cong \triangle HDE$.\\
$\therefore$HE=AG.\\
$\because$Both quadrilaterals ABCD and DEFG are squares, with areas 5 and 9, respectively. That is, AD$^{2}$=5, DG$^{2}$=9.\\
$\therefore$In the right-angled $\triangle ADG$,\\
AG=$\sqrt{DG^{2}-AD^{2}}=\sqrt{9-5}=2$,\\
$\therefore$EH=AG=2.\\
$\therefore$The area of $\triangle CDE$ is $\frac{1}{2}$CD$\cdot$EH=$\frac{1}{2}\times\sqrt{10}\times2=\boxed{\frac{\sqrt{10}}{2}}$.


Question: Consider placing the squares $A_1B_1C_1O$, $A_2B_2C_2C_1$, $A_3B_3C_3C_2$ as shown in the figure, where the points $A_1$, $A_2$, $A_3,\dots$ and the points $C_1$, $C_2$, $C_3,\dots$ lie on the line $y=x+1$ and the x-axis, respectively. What are the coordinates of point $B_3$? What are the coordinates of point $B_{2022}$?

Solution:
\textbf{Solution:} When $x=0$, $y=x+1=1$, \\
$\therefore$ the coordinates of point $A_1$ are $(0,1)$, \\
$\therefore$ $OA_1=1$. \\
$\because$ Quadrilateral $A_1B_1C_1O$ is a square, \\
$\therefore$ $OC_1=B_1C_1=OA_1=1$, \\
$\therefore$ the coordinates of point $B_1$ are $(1,1)$, and the coordinates of point $C_1$ are $(1,0)$. \\
When $x=1$, $y=x+1=2$, \\
$\therefore$ the coordinates of point $A_2$ are $(1,2)$, \\
$\therefore$ $A_2C_1=2$. \\
$\because$ $A_2B_2C_2C_1$ is a square, \\
$\therefore$ $C_1C_2=B_2C_2=A_2C_1=2$, \\
$\therefore$ the coordinates of point $B_2$ are $(3,2)$, and the coordinates of point $C_2$ are $(3,0)$. \\
Similarly, we can obtain that the coordinates of point $B_3$ are $\boxed{(7,4)}$, \\
$\therefore$ the coordinates of point $B_n$ are $\left(2^n-1,2^{n-1}\right)$. \\
$\therefore$ the coordinates of point $B_{2022}$ are $\boxed{\left(2^{2022}-1,2^{2021}\right)}$.


Question: As shown in the figure, in rhombus $ABCD$, $AC$ and $BD$ intersect at point $O$. If $AC=8$ and $BD=6$, then what is the area of the rhombus $ABCD$?

Solution:
\textbf{Solution:} Since the quadrilateral $ABCD$ is a rhombus, \\
$\therefore AC\perp BD$, \\
Since $AC=8, BD=6$, \\
$\therefore$ the area of the rhombus $ABCD$ is $\frac{1}{2}AC\times BD=\frac{1}{2}\times 8\times 6=\boxed{24}$.


Question: As shown in the figure, point $E$ is on the side $BC$ of the rhombus $ABCD$, and $\angle DAE=\angle B=70^\circ$, then what is the degree measure of $\angle CDE$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rhombus,\\
it follows that $AD\parallel BC$,\\
thus $\angle AEB=\angle DAE=\angle B=70^\circ$,\\
which means $AE=AB=AD$,\\
In $\triangle AED$, $AE=AD$, and $\angle DAE=70^\circ$,\\
thus $\angle ADE=55^\circ$,\\
Also, since $\angle B=70^\circ$,\\
it follows that $\angle ADC=70^\circ$,\\
thus $\angle CDE=\angle ADC-\angle ADE=15^\circ$.\\
Therefore, the answer is: $\boxed{15^\circ}$


Question: As shown in the figure, $AC$ and $BD$ are the diagonals of the quadrilateral $ABCD$, points $E$ and $F$ are the midpoints of $AD$ and $BC$ respectively, and points $M$ and $N$ are the midpoints of $AC$ and $BD$ respectively. Connecting $EM$, $MF$, $FN$, $NE$ in sequence, if $AB=CD=2$, then what is the perimeter of the quadrilateral $ENFM$?

Solution:
\textbf{Solution:} Given that points E and F are the midpoints of $AD$ and $BC$ respectively, and points M and N are the midpoints of $AC$ and $BD$ respectively, \\
$\therefore EN$, $MF$, $NF$, $ME$ are the medians of $\triangle ABD$, $\triangle ABC$, $\triangle BCD$, $\triangle ADC$ respectively, \\
$\because AD=CD=2$, \\
$\therefore EN=MF=\frac{1}{2}AB=1$, $NF=ME=\frac{1}{2}CD=1$, \\
The perimeter of quadrilateral $ENFM$ equals $EN+MF+NF+ME=1+1+1+1=\boxed{4}$.


Question: As shown in the figure, in $\triangle ABC$, $BD$ bisects $\angle ABC$, $CD \perp BD$ with the foot of the perpendicular being D, and E is the midpoint of $AC$. If $AB=30$, $BC=18$, what is the length of $DE$?

Solution:
\textbf{Solution:} As shown in the figure:\\
Extend CD to intersect AB at F.\\
In $\triangle BDC$ and $\triangle BDF$, we have\\
$\left\{\begin{array}{l}
\angle DBC=\angle DBF\\
BD=BD\\
\angle BDC=\angle BDF=90^\circ
\end{array}\right.$\\
$\therefore \triangle BDC \cong \triangle BDF$ (ASA)\\
$\therefore BF=BC=18$, CD=DF\\
$\therefore AF=AB-BF=12$,\\
$\because E$ is the midpoint of AC,\\
$\therefore CE=EA$\\
$\because CD=DF$,\\
$\therefore DE=\frac{1}{2}AF=\boxed{6}.$


Question: As shown in the figure, square $ABCD$ has a side length of $4$, point $M$ is on $AB$ with $AM=1$, and $N$ is a moving point on $BD$. What is the minimum value of $AN+MN$?

Solution:
\textbf{Solution:} As shown in the figure, we connect MC and CN,

since the quadrilateral ABCD is a square,

hence A and C are symmetric about BD,

therefore AN = CN,

therefore AN + MN = CN + NM $\geq$ CM,

in triangle BMC,

since $\angle$BCM $= 90^\circ$, BC = 4, BM = AB - AM = 4 - 1 = 3,

therefore CM = $\sqrt{MB^{2}+BC^{2}}$ = 5.

therefore AN + MN $\geq$ 5,

thus, the minimum value of $AN+MN$ is $\boxed{5}$.


Question: As shown in the figure, the diagonals AC and BD of the quadrilateral $ABCD$ intersect at point O. P is the midpoint of side AB, and $OP=2$. What is the length of BC?

Solution:
\textbf{Solution:} \\
Since $ABCD$ is a square, \\
it follows that $OA = OC$, \\
hence, $O$ is the midpoint of $AC$, \\
Since $P$ is the midpoint on side $AB$, \\
it follows that $OP$ is the median of $\triangle ABC$, \\
hence, $BC = 2OP = 2\times2 = \boxed{4}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, $AB=5$, $AC=12$. $P$ is a moving point on side $BC$, $PE\perp AB$ at $E$, $PF\perp AC$ at $F$, and $M$ is the midpoint of $EF$. What is the range of values for $AM$?

Solution:
\textbf{Solution:} Let's consider that PE$\perp$AB and PF$\perp$AC, and $\angle$BAC$=90^\circ$, \\
therefore, quadrilateral AEPF is a rectangle, \\
since M is the midpoint of EF, \\
thus AM$= \frac{1}{2}$AP, and when AP$\perp$BC, the value of AP is minimized, and when point P coincides with point C, the value of AC is maximized, \\
thus AP$= \frac{AB\times AC}{BC}=\frac{60}{13}$, and the maximum value of AP is 12, \\
therefore, $ \frac{30}{13}\le AM<6$. \\
Hence, the answer is: $ \boxed{\frac{30}{13}\le AM<6}$.


Question: As shown in the figure, a hobby group intends to determine the distance between two points, $A$ and $B$, which are separated by a pond. They select a point $C$ outside the line $AB$, connect $AC$ and $BC$, and identify their midpoints $D$ and $E$, respectively, then connect $DE$. Now, it's measured that $AC=30m$, $BC=40m$, and $DE=24m$. What is the distance between points $A$ and $B$ in meters?

Solution:
\textbf{Solution:} It is given that since $D$ and $E$ are the midpoints of $AC$ and $BC$ respectively,\\
therefore $DE$ is the midline of $\triangle ABC$,\\
therefore $AB=2DE$\\
since $DE=24m$\\
therefore $AB=\boxed{48m}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=8$, $BC=6$. Points D and E are respectively the midpoints of $AC$ and $AB$. When connecting $DE$, what is the length of $DE$?

Solution:
\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AC$ and $AB$ respectively, \\
it follows that $DE$ is the midline of $\triangle ABC$, \\
therefore $DE = \frac{1}{2}BC = \frac{1}{2} \times 6 = \boxed{3}$.


Question: As shown in the figure, in $ \triangle ABC$, $ \angle BAC=60^\circ$, $ \angle ABC=45^\circ$. AD bisects $ \angle BAC$ and intersects BC at point D. P is a moving point on line AB. Connect DP and construct the parallelogram DPQB with DP and DB as adjacent sides. Connect CQ. If $ AC=6$, what is the minimum value of CQ?

Solution:
Solution:As shown in Figure 1, draw $CO \perp AB$ at O and $DH \perp AB$ at H.

In $\triangle ACO$, $\angle CAB=60^\circ$,

$\therefore \angle ACO=30^\circ$,

$\therefore AO=\frac{1}{2}AC=3$,

$\therefore OC=\sqrt{AC^2-AO^2}=3\sqrt{3}$,

In $\triangle BCO$, $\angle CBA=45^\circ$,

$\therefore OB=CO=3\sqrt{3}$,

$\therefore AB=AO+OB=3\sqrt{3}+3$,

Since $AD$ bisects $\angle CAB$,

$\therefore \angle DAB=\frac{1}{2}\angle CAB=30^\circ$,

In $\triangle DHB$, $\angle CBA=45^\circ$,

Let $DH=HB=a$,

$\therefore AD=2DH=2a$,

$\therefore AH=\sqrt{AD^2-DH^2}=\sqrt{3}a$,

$\therefore AB=AH+BH=\sqrt{3}a+a$,

$\therefore \sqrt{3}a+a=3\sqrt{3}+3$,

$\therefore a=3$,

$\therefore DH=3$,

As shown in Figure 2, draw $QG \perp AB$ at G and connect DQ intersecting AB at M,

Since quadrilateral DPQB is a parallelogram,

$\therefore DM=QM$,

In $\triangle QGM$ and $\triangle DHM$,

$\left\{\begin{array}{l}
\angle QGM=\angle DHM=90^\circ \\
\angle QMG=\angle DMH \\
QM=DM
\end{array}\right.$,

$\therefore \triangle QGM \cong \triangle DHM$ (AAS),

$\therefore QG=DH=3$,

Hence, the distance from Q to line AB is always 3,

Therefore, point Q moves on a line parallel to AB, and the distance between the two lines is 3,

Based on the shortest distance along the perpendicular,

When points C, O, and Q are on the same line, at that moment, CQ is minimal, as shown in Figure 3,

The minimum value is: $CO+3=3\sqrt{3}+3$,

Therefore, the answer is: $\boxed{3\sqrt{3}+3}$.


Question: As shown in the quadrilateral ABCD, $\angle A=90^\circ$, AB = 12, and AD = 5. Points M and N are the moving points (including endpoints, but point M does not coincide with point B) on the line segments BC and AB, respectively. Points E and F are the midpoints of DM and MN, respectively. What is the maximum length of EF?

Solution:
\textbf{Solution:} Connect DN and DB\\
$\because$ Points E and F are the midpoints of DM and MN respectively,\\
$\therefore$ EF is the midline of $\triangle MDN$\\
$\therefore$ EF$=\frac{1}{2}DN$\\
According to the problem, when N coincides with point B, DN is at its maximum, at this time, the value of EF is maximal\\
By the Pythagorean theorem, DB$= \sqrt{AD^{2}+AB^{2}}=13$\\
$\therefore$ The maximum value of EF is $\boxed{6.5}$


Question: A rectangular paper $MNPQ$ is folded in the manner shown such that vertex $Q$ coincides with $N$, with the fold line being $AB$, $MN=3$, and $MQ=9$. What is the length of the fold line $AB$?

Solution:
\textbf{Solution:} Connect NQ as shown in the diagram:\\
Since quadrilateral MNPQ is a rectangle,\\
thus, $\angle M=90^\circ$, $PN\parallel MQ$,\\
thus, $NQ=\sqrt{MN^2+MQ^2}=\sqrt{3^2+9^2}=3\sqrt{10}$,\\
According to the folding, AB is perpendicular bisector of NQ,\\
thus, $\angle QOA=\angle M=90^\circ$,\\
$OQ=ON=\frac{1}{2}NQ=\frac{3\sqrt{10}}{2}$,\\
Since $\angle AQO=\angle NOM$,\\
thus, $\triangle AQO\sim \triangle NQM$,\\
thus, $\frac{AO}{MN}=\frac{OQ}{MQ}$,\\
which means $\frac{AO}{3}=\frac{\frac{3\sqrt{10}}{2}}{9}$,\\
thus, $AO=\frac{\sqrt{10}}{2}$,\\
Since $PN\parallel MQ$,\\
thus, $\angle AQO=\angle BNO$,\\
Since in $\triangle AQO$ and $\triangle BNO$
$\left\{\begin{array}{l}
\angle AQO=\angle BNO \\
OQ=ON \\
\angle AOQ=\angle BON
\end{array}\right.$,\\
thus, $\triangle AQO\cong \triangle BNO$,\\
thus, AO=BO,\\
thus, $AB=2AO=2\times \frac{\sqrt{10}}{2}=\boxed{\sqrt{10}}$.


Question: As shown in the figure, in rectangle $ABCD$, $AD=8$, $AB=4$, and $E$, $F$ are points on sides $AD$ and $BC$, respectively (points $E$, $F$ do not coincide with the vertices). If the rectangle is folded along line $EF$, such that point $B$ coincides with point $D$, and the corresponding point of point $A$ is point $G$, then what is the length of line segment $EF$?

Solution:
\textbf{Solution:} Construct line EM $\perp$ BC at point M,\\
since quadrilateral ABCD is a rectangle,\\
therefore $\angle$A=$\angle$B=$\angle$C=$\angle$ADC=$90^\circ$, CD=AB=4, AD$\parallel$BC, BC=AD=8,\\
since EM$\perp$BC,\\
therefore $\angle$BME=$\angle$A=$\angle$B=$\angle$C=$\angle$ADC=$90^\circ$,\\
therefore quadrilaterals ABME and CDEM are both rectangles,\\
therefore EM=CD=4, CM=DE,\\
since the rectangle is folded along line $EF$, point $B$ coincides exactly with point $D$, and the corresponding point of $A$ is point $G$,\\
therefore DF=BF, $\angle$BFE=$\angle$DFE,\\
therefore CF=BC−BF=8−BF=8−DF,\\
In right triangle $\triangle$CDF, $CF^{2}+CD^{2}=DF^{2}$,\\
therefore $(8−DF)^{2}$+$4^{2}$=$DF^{2}$, solving gives: DF=5,\\
therefore CF=3,\\
since AD$\parallel$BC,\\
therefore $\angle$DEF=$\angle$BFE,\\
therefore $\angle$DEF=$\angle$DFE,\\
therefore DE=DF=5,\\
therefore CM=DE=5,\\
therefore MF=CM−CF=5−3=2,\\
In right triangle $\triangle$EFM, by the Pythagorean theorem,\\
$EF=\sqrt{EM^{2}+MF^{2}}=\sqrt{4^{2}+2^{2}}=\boxed{2\sqrt{5}}$.


Question: As shown in the figure, $O$ is the midpoint of the diagonal $AC$ of rectangle $ABCD$, and $E$ is a point on $AB$. After folding $\triangle BCE$ along $CE$, point $B$ coincides exactly with point $O$. If $BC = 3$, what is the length of the crease $CE$?

Solution:
\textbf{Solution:} As shown in the diagram, we connect $OB$, \\
since quadrilateral $ABCD$ is a rectangle, and O is the midpoint of the diagonal AC of rectangle ABCD, \\
it follows that $BO=CO$, and $\angle ABC=90^\circ$, \\
since folding $\triangle BCE$ along CE exactly overlaps point B with point O, \\
it follows that $BC=OC$, and $\angle BCE=\angle OCE$, \\
thus, $OB=OC=BC$, \\
which means $\triangle OBC$ is an equilateral triangle, \\
therefore, $\angle OCB=60^\circ$, \\
therefore, $\angle BCE=\angle OCE=30^\circ$, \\
In the right triangle $\triangle BCE$, $CE=2EB$, and $CB=3$, \\
$CE^2=BC^2+EB^2$, \\
which implies $CE^2=\left(\frac{1}{2}CE\right)^2+3^2$, \\
solving this gives us $CE=\boxed{2\sqrt{3}}$.


Question: As shown in the figure, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, $\angle ABD = 60^\circ$, $BD = 4$. What is the area of rectangle $ABCD$?

Solution:
\textbf{Solution:} In rectangle $ABCD$, $\angle BAD=90^\circ$ and $AC=BD=4$,\\
$OA=\frac{1}{2}AC$, $OB=\frac{1}{2}BD=2$,\\
thus $OA=OB$,\\
and since $\angle ABD=60^\circ$,\\
therefore $\triangle AOB$ is an equilateral triangle,\\
thus $AB=OB=2$,\\
since in $\triangle ABD$ with $BD=4$,\\
therefore $AD=\sqrt{4^2-2^2}=2\sqrt{3}$,\\
hence, the area of rectangle $ABCD$ is $2\times 2\sqrt{3}=\boxed{4\sqrt{3}}$.


Question: As shown in the rectangle $ABCD$, where $AB=2$ and $AD=4$, point $P$ is distinct from points $A$ and $D$. Perpendiculars to $AC$ and $BD$ are drawn from point $P$, meeting them at points $E$ and $F$, respectively. What is the value of $PE+PF$?

Solution:
\textbf{Solution:} As shown in the diagram, connect OP, \\
$\because$ AB = 2, AD = 4, \\
by the Pythagorean theorem, we have BD = $\sqrt{2^2+4^2}=2\sqrt{5}$, \\
$S_{\triangle ABD}=\frac{1}{2}\times AB \times AD=\frac{1}{2}\times 2\times 4=4$, \\
in rectangle ABCD, OA = OD = OB = $\frac{1}{2}$BD = $\sqrt{5}$, \\
$\because S_{\triangle AOD}=S_{\triangle AOP}+S_{\triangle DOP}=\frac{1}{2}S_{\triangle ABD}$, \\
$\therefore \frac{\sqrt{5}}{2}PE+\frac{\sqrt{5}}{2}PF=2$, \\
$\therefore PE+PF=\boxed{\frac{4\sqrt{5}}{5}}$.


Question: As shown in the figure, in rectangle $OABC$, where $OA=4$ and $AB=3$, point $D$ is on side $BC$, and $CD=3DB$. Point $E$ is a point on side $OA$, and $DE$ is connected. When quadrilateral $ABDE$ is folded along $DE$, and the symmetric point $A'$ of point $A$ exactly falls on side $OC$, then what is the length of $OE$?

Solution:
\textbf{Solution:} Connect $A^{\prime}D$, $AD$,\\
since quadrilateral $OABC$ is a rectangle,\\
thus BC=OA=4, OC=AB=3, $\angle C=\angle B=\angle O=90^\circ$,\\
since $CD=3DB$,\\
thus $CD=3$, $BD=1$,\\
thus $CD=AB$,\\
since quadrilateral $ABDE$ is folded along $DE$, if the symmetric point of $A^{\prime}$ exactly falls on side $OC$,\\
thus $A^{\prime}D=AD$, $A^{\prime}E=AE$,\\
in $\triangle A^{\prime}CD$ and $\triangle DBA$, $\left\{\begin{array}{l}CD=AB \\ A^{\prime}D=AD\end{array}\right.$,\\
thus $\triangle A^{\prime}CD\cong \triangle DBA$ (HL),\\
thus $A^{\prime}C=BD=1$,\\
thus $A^{\prime}O=2$,\\
since $A^{\prime}O^{2}+OE^{2}=A^{\prime}E^{2}$,\\
thus $2^{2}+OE^{2}=(4-OE)^{2}$,\\
thus $OE=\boxed{\frac{3}{2}}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=4$, $BC=8$. The rectangle $ABCD$ is folded along $EF$ so that point $C$ coincides with point $A$. What is the length of the crease $EF$?

Solution:
\textbf{Solution:}\\
Let $BE = x$, then $CE = BC - BE = 8 - x$, \\
$\because$ After folding along $EF$, point $C$ coincides with point $A$, \\
$\therefore$ $AE = CE = 8 - x$, \\
In $\triangle ABE$, $AB^{2} + BE^{2} = AE^{2}$, that is $4^{2} + x^{2} = (8 - x)^{2}$, \\
Solving it, we get: $x = 3$, \\
$\therefore$ $AE = 8 - 3 = 5$, \\
Due to the property of folding, $\angle AEF = \angle CEF$, \\
$\because$ $AD \parallel BC$, \\
$\therefore$ $\angle AFE = \angle CEF$, \\
$\therefore$ $\angle AEF = \angle AFE$, \\
$\therefore$ $AE = AF = 5$, \\
Draw a line from point $E$ to $H$ perpendicular to $AD$ at $H$, then quadrilateral $ABEH$ is a rectangle, \\
$\therefore$ $EH = AB = 4$, $AH = BE = 3$, \\
$\therefore$ $FH = AF - AH = 5 - 3 = 2$, \\
In $\triangle EFH$, $EF = \sqrt{4^{2} + 2^{2}} = \boxed{2\sqrt{5}}$.


Question: As shown in the figure, line $L$ passes through vertex $A$ of the square $ABCD$. Lines are drawn from points $B$ and $D$ such that $DE \perp l$ at point $E$ and $BF \perp l$ at point $F$. If $DE=4$ and $BF=5$, what is the length of $EF$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a square,\\
it follows that $AD=AB$ and $\angle DAB=90^\circ$,\\
therefore $\angle BAF+\angle DAE=90^\circ$\\
Given that $DE\perp l$,\\
it follows that $\angle DAE+\angle ADE=90^\circ$\\
therefore, $\angle BAF=\angle ADE$\\
Given that $DE\perp l$ and $BF\perp l$,\\
it follows that $\angle AFB=\angle AED=90^\circ$\\
In triangles $DAE$ and $ABF$,\\
given that $\left\{\begin{array}{l}\angle AED=\angle AFB \\ \angle BAF=\angle ADE \\ AD=AB \end{array}\right.$,\\
it follows that $\triangle DAE\cong \triangle ABF$\\
therefore, $AF=DE=4$, $AE=BF=5$\\
therefore, $EF=AF+AE=4+5=\boxed{9}$


Question: As shown in the figure, within square $ABCD$, an equilateral triangle $\triangle ADE$ is constructed. By connecting $BE$ and $CE$, what is the measure of $\angle CBE$ in degrees?

Solution:
\textbf{Solution:} Since $\triangle ADE$ is an equilateral triangle,\\
it follows that $AE=DE=AD$, and $\angle ADE=\angle EAD=\angle AED=60^\circ$,\\
Since quadrilateral ABCD is a square,\\
it follows that $AB=AD=DC$, and $\angle BAD=\angle ADC=90^\circ$,\\
thus $\angle BAE=\angle CDE=90^\circ-60^\circ=30^\circ$, and $AB=AE=DE=CD$,\\
hence $\angle ABE=\frac{1}{2}\left(180^\circ-30^\circ\right)=75^\circ=\angle DCE$,\\
thus $\angle CBE=90^\circ-75^\circ=\boxed{15^\circ}$.


Question: As shown in the diagram, within the rhombus $ABCD$, the diagonals $AC=6$ and $BD=8$. A line is drawn from point A such that $AE\perp BC$. What is the length of $AE$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rhombus,\\
$\therefore$ $CO=\frac{1}{2}AC=3$, $BO=\frac{1}{2}BD=4$, and $AO\perp BO$,\\
$\therefore$ $BC=\sqrt{CO^{2}+BO^{2}}=\sqrt{9+16}=5$,\\
Since the area of rhombus $ABCD$, $S_{\text{rhombus} ABCD}=\frac{1}{2}AC\cdot BD=BC\cdot AE$,\\
$\therefore$ $AE=\frac{\frac{1}{2}AC\cdot BD}{BC}=\boxed{\frac{24}{5}}$.


Question: As shown in the figure, the diagonals of rhombus $ABCD$ intersect at point $O$, with $AC=12$ and $BD=16$. Point $P$ lies on side $BC$ and does not coincide with points $B$ or $C$. Through $P$, construct $PE \perp AC$ at $E$ and $PF \perp BD$ at $F$, and connect $EF$. What is the minimum value of $EF$?

Solution:
\textbf{Solution:} Connect $OP$,\\
\textit{Since} quadrilateral $ABCD$ is a rhombus, $AC=12$, $BD=16$\\
\textit{Therefore} $AC\perp BD$,\\
$BO=\frac{1}{2}BD=8$, $OC=\frac{1}{2}AC=6$\\
\textit{Therefore} $BC=\sqrt{OB^2+OC^2}=\sqrt{64+36}=10$\\
\textit{Since} $PE\perp AC$, $PF\perp BD$, $AC\perp BD$\\
\textit{Therefore} quadrilateral $OEPF$ is a rectangle,\\
\textit{Therefore} $FE=OP$\\
When $OP\perp BC$, $OP$ is at its minimum,\\
Then $S_{\triangle OBC}=\frac{1}{2}OB\cdot OC=\frac{1}{2}BC\cdot OP$\\
\textit{Therefore} $OP=\frac{6\times 8}{10}=4.8$\\
\textit{Therefore} the minimum value of $EF$ is $\boxed{4.8}$


Question: As shown in the figure, it is known that in $\triangle ABC$, $D$ and $E$ are the midpoints of $AB$ and $AC$, respectively. Connect $DE$ and extend it to $F$ such that $EF=DE$. Connect $CF$. If $\angle B=45^\circ$, what is the measure of $\angle F$?

Solution:
\textbf{Solution:} Since point E is the midpoint of AC, and point D is the midpoint of AB, \\
therefore $DE \parallel BC$ and $BC=2DE$. \\
Since EF=DE, \\
therefore $DF=2DE$, \\
therefore $DF=BC$, \\
therefore quadrilateral BCFD is a parallelogram, \\
therefore $\angle F=\angle B=45^\circ$. \\
Hence, the answer is: $\boxed{45^\circ}$.


Question: As shown in the figure, in rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Point $E$ is the midpoint of side $AB$. If $OE=4$, what is the length of $BC$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\
it follows that AB = BC = CD = AD, and BD$\perp$AC,\\
Furthermore, since point E is the midpoint of side AB,\\
it follows that OE = AE = EB $ =\frac{1}{2}$ AB,\\
Therefore, BC = AB = 2OE = $4\times 2$ = \boxed{8}


Question: As shown in the figure, $\triangle ABH$, $\triangle CBG$, $\triangle CDF$, and $\triangle DAE$ are four congruent right-angled triangles. If $BH=8$ and $EG=2\sqrt{2}$, then what is the length of $AB$?

Solution:
\textbf{Solution:} Since $\triangle ABH$, $\triangle CBG$, $\triangle CDF$, and $\triangle DAE$ are four congruent right-angled triangles, with $BH=8$,\\
it follows that $BH=AE=8$, $BG=AH$, and $\angle AHB=90^\circ$,\\
therefore $BH-BG=AE-AH$, which means $HG=HE$, and $\angle EHG=90^\circ$,\\
in right-angled triangle $HEG$, by Pythagoras' theorem, we have $HG^2+HE^2=GE^2$,\\
since $EG=2\sqrt{2}$,\\
it follows that $HG=2=HE$,\\
therefore $AH=8-2=6$,\\
in right-angled triangle $ABH$, by Pythagoras' theorem, we get $AB=\sqrt{AH^2+BH^2}=\sqrt{6^2+8^2}=\boxed{10}$.


Question: As shown in the figure, in rhombus $ABCD$, $\angle A=60^\circ$, $E$ is the midpoint of $AD$, and $F$ is the midpoint of $AB$. If $EF=\sqrt{5}$, what is the perimeter of rhombus $ABCD$?

Solution:
\textbf{Solution:} Since $E$ and $F$ are the midpoints of $AD$ and $AB$ respectively, \\
$\therefore EF=\frac{1}{2}BD$, \\
Also, since $EF=\sqrt{5}$, \\
$\therefore BD=2\sqrt{5}$, \\
Since quadrilateral $ABCD$ is a rhombus with $\angle A=60^\circ$, \\
$\therefore \triangle ABD$ and $\triangle BDC$ are both equilateral triangles, \\
$\therefore AB=2\sqrt{5}$, \\
$\therefore$ the side length of the rhombus $ABCD$ is $2\sqrt{5}\times 4=\boxed{8\sqrt{5}}$.


Question: As shown in the figure, $O$ is the midpoint of the diagonal $AC$ of rectangle $ABCD$, and $M$ is the midpoint of $AD$. If $AB=5$ and $AD=12$, what is the perimeter of quadrilateral $ABOM$?

Solution:
\textbf{Solution:} Since $O$ is the midpoint of the diagonal $AC$ of rectangle $ABCD$, and $M$ is the midpoint of $AD$,\\
therefore $OM = \frac{1}{2}CD = \frac{1}{2}AB = 2.5$,\\
and since $AB=5$, $AD=12$,\\
therefore $AC = \sqrt{5^{2}+12^{2}}=13$,\\
Since $O$ is the midpoint of the diagonal $AC$ of rectangle $ABCD$,\\
therefore $BO = \frac{1}{2}AC = 6.5$,\\
therefore, the perimeter of quadrilateral $ABOM$ is $AB+AM+BO+OM = 5+6+6.5+2.5 = \boxed{20}$


Question: In $\triangle ACB$, $\angle ABC=90^\circ$, $AB=4$, and $BC=3$. Squares are constructed externally on each of the three sides. Connecting the vertices of these squares forms a hexagon DEFGHI. What is the area of hexagon DEFGHI?

Solution:
\textbf{Solution:} As shown in the figure, \\
in $\triangle ABC$, since $AB=4$ and $BC=3$, \\
therefore $AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{4^{2}+3^{2}}=5$. \\
Since quadrilaterals ABDE, BCGF, and ACHM are all squares, \\
therefore $\angle ABD=\angle CBF=\angle BAE=\angle CAM=\angle ACH=\angle GCH=90^\circ$, BD=BA, AM=AC, CBN=CG, $S_{\text{square ABDE}}=4^{2}=16$, $S_{\text{square ACHM}}=5^{2}=25$, $S_{\text{square BCGF}}=3^{2}=9$, \\
therefore, $\angle DBF+\angle ABC=180^\circ$, $\angle MAE+\angle BAC=180^\circ$, $\angle ACB+\angle HCG=180^\circ$, \\
draw IM$\perp$DA to intersect the extension line of DA at M, \\
therefore $\angle M=\angle ABC=90^\circ$, \\
since $\angle DAI+\angle MAI=\angle DAI+\angle BAC=180^\circ$, \\
therefore $\angle IAM=\angle BAC$, \\
in $\triangle AMI$ and $\triangle BAC$, \\
$\left\{\begin{array}{l}\angle M=\angle ABC \\ \angle BAC=\angle MAI \\ AC=AI \end{array}\right.$ \\
therefore $\triangle AMI \cong \triangle ABC$, \\
hence AB=AM, \\
thus AD=AM, \\
thus $S_{\triangle AMI}=S_{\triangle ABC}=S_{\triangle ADI}$, \\
Similarly, $S_{\triangle BEF}=S_{\triangle ABC}$, $S_{\triangle CHG}=S_{\triangle ABC}$, \\
thus $S_{\triangle DBF}=S_{\triangle MAE}=S_{\triangle HCG}=S_{\triangle ABC}=\frac{1}{2}\times 3\times 4=6$, \\
therefore the area of the hexagon DEMHGF=$25+16+9+4\times6=\boxed{74}$.


Question: As shown in the figure, in $\triangle ABC$, $AE$ is the median to side $BC$. A line is drawn through point $C$ such that $CD \perp AE$, and it intersects the extension line of $AE$ at point $D$. Line $BD$ is drawn. If $AB = BD$ and the area of $\triangle BCD$ is 10, what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} As shown in the figure, construct $BF\perp AE$ at point $F$,

since $CD\perp AE$,

$\therefore \angle BFE=\angle CDE=90^\circ$,

since $AE$ is the median to side $BC$,

$\therefore BE=CE$,

In $\triangle BEF$ and $\triangle CED$,

$\left\{\begin{array}{l}
\angle BFE=\angle CDE \\
\angle BEF=\angle CED \\
BE=CE
\end{array}\right.$

$\therefore \triangle BEF \cong \triangle CED$ (AAS),

$\therefore FE=DE$, $BF=CD$,

since $AB = BD$,

$\therefore BF= DF$,

Let $FE=DE=m$, $BF=CD=n$, then $AF= DF= 2FE= 2m$,

$\therefore AE=AF+FE= 3m$,

since $S_{\triangle BCD}=S_{\triangle BDE}+S_{\triangle CDE}=\frac{1}{2}DE\cdot BF+\frac{1}{2}DE\cdot CD=\frac{1}{2}mn+\frac{1}{2}mn=mn$

and $S_{\triangle BCD}=10$,

$\therefore mn=10$,

$\therefore S_{\triangle ABC}=S_{\triangle ABE}+S_{\triangle ACE}=\frac{1}{2}AE\cdot BF+\frac{1}{2}AE\cdot CD=\frac{1}{2}\times 3mn+\frac{1}{2}\times 3mn=3mn=\boxed{30}$


Question: As shown in the figure, in $\triangle ABC$, $AD \perp BC$, $CE \perp AB$, with the feet of the perpendiculars being $D$ and $E$ respectively, $AD$ and $CE$ intersect at point $H$. Given that $AE=CE=5$ and $CH=2$, what is the length of $BE$?

Solution:
\textbf{Solution:} Since $AD \perp BC$, $CE \perp AB$,\\
therefore $\angle AEH = \angle BEC = \angle HDC = 90^\circ$,\\
thus $\angle EAH + \angle EHA = \angle ECB + \angle DHC = 90^\circ$\\
Since $\angle EHA = \angle DHC$,\\
thus $\angle EAH = \angle ECB$,\\
In $\triangle AEH$ and $\triangle CEB$,\\
$\left\{ \begin{array}{l} \angle EAH = \angle ECB \\ CE = AE \\ \angle AEH = \angle BEC \end{array} \right.$,\\
therefore, $\triangle AEH \cong \triangle CEB$ (ASA),\\
thus $BE = EH = CE - CH = 5 - 2 = \boxed{3}$,\\


Question: As shown in the figure, in $\triangle ABC$, point $P$ is the intersection of the perpendicular bisectors of $AB$ and $BC$, and point $Q$ is symmetric to point $P$ with respect to $AC$. Connect $PC$, $PQ$, and $CQ$. If in $\triangle PCQ$ there is an angle of $50^\circ$, then what is the measure of $\angle B$?

Solution:
\textbf{Solution:} Connect AP and BP as shown in the diagram: \\
$\because$ Point P is the intersection of the perpendicular bisectors of AB and BC,\\
$\therefore$ PA = PB = PC,\\
$\therefore$ $\angle$PAB = $\angle$PBA, $\angle$PBC = $\angle$PCB, $\angle$PAC = $\angle$PCA,\\
$\because$ Point Q is symmetric to point P with respect to AC,\\
$\therefore$ PC = QC, $\angle$PCA = $\angle$QCA,\\
$\therefore$ $\angle$CPQ = $\angle$CQP,\\
(1) When $\angle$CPQ = $\angle$CQP = 50$^\circ$, $\angle$PCQ = 80$^\circ$,\\
$\therefore$ $\angle$PCA = 40$^\circ$,\\
$\therefore$ $\angle$PAC = 40$^\circ$,\\
$\therefore$ $\angle$PAB + $\angle$PBA + $\angle$PBC + $\angle$PCB = 180$^\circ$ - $\angle$PAC - $\angle$PCA = 100$^\circ$,\\
$\therefore$ 2$\angle$ABP + 2$\angle$PBC = 100$^\circ$,\\
$\therefore$ $\angle$ABP + $\angle$PBC = 50$^\circ$, i.e., $\angle$ABC = 50$^\circ$,\\
(2) When $\angle$PCQ = 50$^\circ$, $\angle$PCA = 25$^\circ$,\\
$\therefore$ $\angle$PAC = 25$^\circ$,\\
$\therefore$ $\angle$PAB + $\angle$PBA + $\angle$PBC + $\angle$PCB = 180$^\circ$ - $\angle$PAC - $\angle$PCA = 130$^\circ$,\\
$\therefore$ 2$\angle$ABP + 2$\angle$PBC = 130$^\circ$,\\
$\therefore$ $\angle$ABP + $\angle$PBC = 65$^\circ$, i.e., $\angle$ABC = 65$^\circ$,\\
In conclusion, $\angle$ABC is \boxed{50^\circ} or \boxed{65^\circ}.


Question: As shown in the figure, in $ \triangle ABC$, $ \angle ABC=\angle ACB$, D is the midpoint of BC, AD is connected, E is a point on AB, P is a point on AD, EP and BP are connected, $ AC=10$, $ BC=12$, what is the minimum value of $ EP+BP$?

Solution:
\textbf{Solution:} Given $\because$ $\triangle ABC$ is an isosceles triangle, and AD is the median to side BC,\\
$\therefore$ AD bisects BC at a right angle,\\
$\therefore$ points B and C are symmetric about AD,\\
$\therefore$ BP=CP,\\
A perpendicular from point C to AB is drawn, and the perpendicular foot is point E, with CE intersecting AD at point P, (the perpendicular distance is the shortest distance between a point and a line),\\
As shown in the figure, at this time, the sum of $BP+EP$ is minimal, and it is equal to the length of CE,\\
$\because$ D is the midpoint of BC, and BC=12,\\
$\therefore$ CD=$\frac{1}{2} \times 12=6$,\\
$\therefore$ AD=$\sqrt{AC^{2}-CD^{2}}=8$,\\
$\because$ $\angle ABC=\angle ACB$,\\
$\therefore$ AB=AC=10,\\
$\because$ $S_{\triangle ABC}=\frac{1}{2}BC \cdot AD=\frac{1}{2}AB \cdot CE$,\\
$\therefore$ CE=$\frac{BC \cdot AD}{AB}=\frac{12 \times 8}{10}=\frac{48}{5}$,\\
$\therefore$ the minimum value of $BP+EP$ is $\boxed{\frac{48}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 60^\circ$, $DE$ is the perpendicular bisector of the hypotenuse $AC$, intersecting $AB$ and $AC$ at points $D$ and $E$, respectively. If $BD = 2$, what is the length of $AD$?

Solution:
\textbf{Solution:} Given $\triangle ABC$, $\angle ACB=60^\circ$, \\
$\therefore \angle A=30^\circ$. \\
Since DE is the perpendicular bisector of AC, \\
$\therefore$ DA=DC, \\
hence $\angle DCA=\angle A=30^\circ$, \\
$\therefore \angle DCB=\angle ACB-\angle DCA$, \\
$=60^\circ-30^\circ$ \\
$=30^\circ$ \\
In $\triangle BCD$, $\angle DCB=30^\circ$, \\
$\therefore BD= \frac{1}{2}CD$, \\
$\therefore CD=2BD=4$, \\
$\therefore AD=\boxed{4}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $\angle A=30^\circ$, and the perpendicular bisector of $AB$ intersects $AB$ at $D$ and $AC$ at $E$. If $AC=9$ cm, what is the length of $AE$ in centimeters?

Solution:
\textbf{Solution:} As shown in the figure, connect $BE$,\\
$\because$ $\angle C=90^\circ$ and $\angle A=30^\circ$, the perpendicular bisector of $AB$ intersects $AB$ at $D$,\\
$\therefore \angle ABC=60^\circ$, $BE=AE$, $\angle A=\angle EBA=30^\circ$,\\
$\therefore \angle CBE=30^\circ$,\\
$\therefore BE=2CE=AE$,\\
$\therefore AC=CE+AE=9$,\\
$\therefore 3CE=9$, hence $CE=3$,\\
$\therefore AE=\boxed{6}$.


Question: As shown in the figure, in $\triangle ABC$, $CD$ is the bisector of $\angle ACB$. Extend $CD$ to point $E$ such that $DE = \frac{1}{2}CD$. Connect $BE$. If $AC = 2BC$ and the area of $\triangle BDE$ is $1$, what is the area of $\triangle ABC$?

Solution:
\[
\textbf{Solution:} \text{As shown in the figure, draw } DG \perp AC \text{ at } G, \text{ and } DH \perp CB \text{ at } H, \\
\because DE = \frac{1}{2}CD, \text{ and the area of } \triangle BDE \text{ is 1,} \\
\therefore S\triangle BCD = 2S\triangle BDE = 2, \\
\because CD \text{ is the angle bisector of } \angle ACB, DH \perp CB, \text{ and } DG \perp AC, \\
\therefore DG = DH, \\
\because AC = 2BC, S_{\triangle ACD} = \frac{1}{2}AC \cdot GD, S_{\triangle BCD} = \frac{1}{2}BC \cdot DH, \\
\therefore S\triangle ACD = 2S\triangle BCD, \\
\therefore S\triangle ACD = 4, \\
\therefore S\triangle ABC = S\triangle ACD + S\triangle BCD = 4 + 2 = \boxed{6}.
\]


Question: As shown in the figure, $OA_1=A_1A_2=A_2A_3=A_3A_4=A_4A_5=1$, $\angle OA_1A_2=\angle OA_2A_3=\angle OA_3A_4=\angle OA_4A_5=90^\circ$, what is the length of $OA_5$?

Solution:
\textbf{Solution:} Given that $\angle O{A}_{1}{A}_{2}=\angle O{A}_{2}{A}_{3}=\angle O{A}_{3}{A}_{4}=\angle O{A}_{4}{A}_{5}=90^\circ$ and $O{A}_{1}={A}_{1}{A}_{2}={A}_{2}{A}_{3}={A}_{3}{A}_{4}={A}_{4}{A}_{5}=1$, \\
$\therefore OA{}_{2}=\sqrt{O{{A}_{1}}^{2}+{A}_{1}{{A}_{2}}^{2}}=\sqrt{{1}^{2}+{1}^{2}}=\sqrt{2}$, \\
$OA{}_{3}=\sqrt{O{{A}_{2}}^{2}+{A}_{2}{{A}_{3}}^{2}}=\sqrt{{\left(\sqrt{2}\right)}^{2}+{1}^{2}}=\sqrt{3}$, \\
$OA{}_{4}=\sqrt{O{{A}_{3}}^{2}+{A}_{3}{{A}_{4}}^{2}}=\sqrt{{\left(\sqrt{3}\right)}^{2}+{1}^{2}}=2$, \\
$OA{}_{5}=\sqrt{O{{A}_{4}}^{2}+{A}_{4}{{A}_{5}}^{2}}=\sqrt{{2}^{2}+{1}^{2}}=\boxed{\sqrt{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, the perpendicular bisector of AB intersects AB and AC at points D and E, respectively. If $AC=8$ and $BD=5$, what is the area of $\triangle ADE$?

Solution:
\textbf{Solution:} Connect BE,\\
since DE is the perpendicular bisector of AB,\\
it follows that EA=EB, AD=DB=5,\\
since $\angle C=90^\circ$, AC=8, BD=5,\\
it follows that AB=2BD=10,\\
by the Pythagorean theorem, we have BC=$\sqrt{AB^{2}-AC^{2}}=\sqrt{10^{2}-8^{2}}=6$,\\
thus CE=8−AE=8−EB,\\
in $\triangle CBE$, we have $BE^{2}=CE^{2}+BC^{2}$, that is $BE^{2}=(8-BE)^{2}+36$,\\
solving this, we get $BE=\frac{25}{4}$, therefore, AE=$\frac{25}{4}$,\\
thus the area of $\triangle ABE$ is $\frac{1}{2}AE\times BC=\frac{1}{2}\times\frac{25}{4}\times6=\frac{75}{4}$,\\
thus the area of $\triangle ADE$ is $\frac{1}{2}$ area of $\triangle ABE$=$\boxed{\frac{75}{8}}$.


Question: In right triangle $ABC$, $\angle ACB=90^\circ$, $BC=2\,\text{cm}$, $CD\perp AB$. Point $E$ is taken on $AC$ such that $EC=2\,\text{cm}$. A line through point $E$ and perpendicular to $AC$ intersects the extension of $CD$ at point $F$. If $AE=3\,\text{cm}$, what is the length of $EF$ in centimeters?

Solution:
\textbf{Solution:}\\
$\because \angle ACB=90^\circ$\\
$\therefore \angle ECF+\angle BCD=90^\circ$\\
$\because CD\perp AB$\\
$\therefore \angle BCD+\angle B=90^\circ$\\
$\therefore \angle ECF=\angle B$\\
In $\triangle ABC$ and $\triangle FEC$\\
$\because \angle ECF=\angle B$, $EC=BC$, $\angle ACB=\angle FEC=90^\circ$\\
$\therefore \triangle ABC \cong \triangle FCE$ (ASA)\\
$\therefore AC=EF$\\
$\because AC=AE+CE=3+2=5\text{cm}$,\\
$\therefore EF=\boxed{5\text{cm}}$


Question: As shown in the figure, in rectangle $ABCD$, we have $AB=3$ and $BC=4$. Point $E$ is a point on side $BC$. Connect $AE$. Fold $\angle B$ along $AE$, making point $B$ fall on point $B^{\prime}$. When $\triangle CEB^{\prime}$ is a right-angled triangle, what is the length of $BE$?

Solution:
\textbf{Solution:} As shown in Figure 1, when A, B', and C are collinear, $\angle EB'C=90^\circ$.

Quadrilateral ABCD is a rectangle,

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

$\therefore AC=\sqrt{AB^{2}+CB^{2}}=\sqrt{3^{2}+4^{2}}=5$,

$\because AB=AB' =3$,

$\therefore CB' =5-3=2$, let $BE=EB' =x$, then $EC=4-x$,

In $\triangle CEB'$, $CE^{2}=B'E^{2}+B'C^{2}$,

$\therefore (4-x)^{2}=2^{2}+x^{2}$,

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

As shown in Figure 2, when point B' lies on AD, $\angle CEB' =90^\circ$,

At this time, quadrilateral ABEB' is a square,

$\therefore BE=AB=3$,

In conclusion, the values of $BE$ that meet the conditions are $\frac{3}{2}$ or 3.


Question: As shown in the figure, in the Cartesian coordinate system, points B and C are on the y-axis, and $\triangle ABC$ is an equilateral triangle with $AB=4$. The intersection point D of $AC$ and the x-axis has coordinates $(\sqrt{3}, 0)$. What are the coordinates of point A?

Solution:
\textbf{Solution:} As shown in the figure, draw AE$\perp$BC through point A, thus AE$\parallel$OD, \\
because $\triangle ABC$ is an equilateral triangle, \\
therefore $BC=AB=4$, $BE=CE=\frac{1}{2}BC=2$, \\
therefore $AE=\sqrt{AB^{2}-BE^{2}}=\sqrt{4^{2}-2^{2}}=2\sqrt{3}$, \\
because $D(\sqrt{3},0)$, \\
therefore $OD=\sqrt{3}$, \\
therefore point O is the midpoint of CE, \\
therefore $OE=\frac{1}{2}CE=1$, \\
therefore the coordinates of point A are $\boxed{(2\sqrt{3},1)}$.


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle. Points $P$ and $Q$ are taken on sides $AC$ and $BC$, respectively, such that $AP=CQ$, and $\angle ABP=20^\circ$. Lines $AQ$ and $BP$ intersect at point $O$. What is the measure of $\angle AQB$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is an equilateral triangle,\\
it follows that AB=AC, $\angle BAP=\angle ACQ=60^\circ$,\\
In $\triangle BAP$ and $\triangle ACQ$,\\
$\left\{\begin{array}{l}
AB=AC\\
\angle BAP=\angle ACQ\\
AP=CQ
\end{array}\right.$,\\
thus $\triangle BAP \cong \triangle ACQ$ (SAS),\\
therefore $\angle CAQ=\angle ABP=20^\circ$,\\
thus $\angle AQB=\angle C+\angle CAQ=60^\circ+20^\circ=\boxed{80^\circ}$.


Question: As shown in the figure, $ \triangle ABC \cong \triangle DEF$. If $ \angle A = 65^\circ$, what is the measure of $ \angle EDC$?

Solution:
\textbf{Solution:} Since $\triangle ABC \cong \triangle DEF$, \\
it follows that $\angle EDC = \angle A = 65^\circ$, \\
therefore, $\angle EDC = \boxed{65^\circ}$.


Question: As shown in the figure, in the equilateral triangle $ABC$, $BD$ is the median to the side $AC$, and through point $D$, $DE \perp BC$ at point $E$. If $AB=2$, then what is the length of $CE$?

Solution:
\textbf{Solution:} Given: $\because$ $\triangle ABC$ is an equilateral triangle, $AB=2$,\\
$\therefore$ $\angle ABC = \angle C = 60^\circ$, $AB = BC = AC = 2$,\\
Also $\because$ $BD$ is the median to side $AC$,\\
$\therefore$ $\angle DBC = 30^\circ$, $\angle BDC = 90^\circ$, $DC = 1$,\\
Also $\because$ $DE \perp BC$,\\
$\therefore$ $\angle EDC = 30^\circ$,\\
$\therefore$ $EC = \frac{1}{2} DC = \boxed{\frac{1}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, point $D$ is on side $BC$, $DB=AB=\frac{3}{5}BC$, $BE \perp AD$ at point $E$. If the area of $\triangle ABE$ is 6, what is the area of $\triangle EBC$?

Solution:
\textbf{Solution:} Since $DB=AB$ and $BE \perp AD$, \\
it follows that $AE=DE$, \\
therefore $S_{\triangle BDE}=S_{\triangle ABE}=6$, \\
since $DB=\frac{3}{5}BC$, \\
it therefore follows that $S_{\triangle BDE} : S_{\triangle EDC}=3:2$, \\
hence $S_{\triangle EDC}=4$, \\
thus $S_{\triangle BCE}=S_{\triangle BDE}+S_{\triangle EDC}=\boxed{10}$.


Question: As shown in the figure, in $\triangle ABC$, point $D$ is on side $AB$, $E$ is the midpoint of side $AC$, $CF\parallel AB$, and the extension lines of $CF$ and $DE$ intersect at point $F$. If $AB=4$ and $CF=3$, what is the length of $BD$?

Solution:
\textbf{Solution:} Since $CF\parallel AB$,\\
it follows that $\angle ADF=\angle CFD$ and $\angle DAC=\angle ACF$,\\
since $E$ is the midpoint of side $AC$,\\
it follows that $AE=EC$,\\
in $\triangle ADE$ and $\triangle CFE$,\\
we have $\left\{\begin{array}{l}
\angle ADF=\angle CFD \\
\angle DAC=\angle ACF \\
AE=EC
\end{array}\right.$,\\
therefore, $\triangle ADE\cong \triangle CFE$,\\
therefore, $AD=CF=3$,\\
therefore, $BD=AB−AD= \boxed{1}$.