Question:
As shown in the figure, in the known $\triangle ABC$, proceed with the following steps:
(1) With centers at B and C, respectively, draw arcs with radii longer than $\frac{1}{2}BC$, intersecting at points M and N;
(2) Draw line MN intersecting AB at D, and connect CD. If $AC = 5$ and $AB = 12$, what is the perimeter of $\triangle ACD$?

Solution:
\textbf{Solution:} From the construction, we have that $MN$ bisects $BC$ perpendicularly,\\
$\therefore DB=DC$,\\
$\therefore$ the perimeter of $\triangle ACD = AC+AD+CD = AC+AD+BD = AC+AB = 5+12 = \boxed{17}$.


Question: As shown in the figure, $AD$ is the median of $ \triangle ABC$, $CE$ is the median of $ \triangle ACD$. If the area of $ \triangle ABC$ is $12\text{cm}^2$, what is the area of $ \triangle CDE$?

Solution:
\textbf{Solution:} Since $AD$ is the median of side $BC$ of $\triangle ABC$, and the area of $\triangle ABD$ is $12\text{cm}^2$,\\
therefore, the area of $\triangle ADC$ is: $\frac{1}{2}\times 12=6\text{cm}^2$,\\
Since $CE$ is the median of side $AD$ of $\triangle ACD$,\\
therefore, the area of $\triangle CDE$ is: $\frac{1}{2}\times 6=\boxed{3}\text{cm}^2$.


Question: As shown in the figure, $OP$ bisects $\angle AOB$, where point $E$ is on $OA$ with $OE = 4$, and the distance from point $P$ to $OB$ is $2$. What is the area of $\triangle POE$?

Solution:
\textbf{Solution:} Since OP bisects $\angle$AOB, and the distance from point P to OB is 2, \\
$\therefore$ the distance from point P to OA is also 2, \\
$\therefore$ $ {S}_{\triangle OEP}=\frac{1}{2}\times 4\times 2=\boxed{4}$


Question: As shown in the figure, $AB \parallel DE$, and points $B$, $C$, $D$ are on the same line. If $\angle BCE = 55^\circ$ and $\angle E = 25^\circ$, then what is the measure of $\angle B$?

Solution:
\textbf{Solution:} Since $\angle BCE=55^\circ$ and $\angle E=25^\circ$, we have $\angle BCE=\angle E+\angle D$. \\
Therefore, $\angle D=\angle BCE-\angle E=55^\circ-25^\circ=30^\circ$. \\
Thus, $AB\parallel DE$. \\
Consequently, $\angle B=\angle D$. \\
Hence, $\angle B=\boxed{30^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, D is a point on $BC$, $\angle 1=\angle 2$, $\angle 3=\angle 4$, $\angle BAC=108^\circ$. What is the measure of $\angle DAC$?

Solution:
\textbf{Solution:} Let $ \angle 1=\angle 2=x$,

$ \therefore \angle 3=\angle 4=\angle 1+\angle 2=2x$,

$ \because \angle DAC+\angle 3+\angle 4=180^\circ$,

$ \therefore \angle DAC=180^\circ−4x$,

$ \because \angle BAC=108^\circ$,

$ \therefore x+180^\circ−4x=108^\circ$,

$ \therefore x=24^\circ$,

$ \therefore \angle DAC=180^\circ−4\times 24^\circ=\boxed{84^\circ}$


Question: As illustrated, let $\triangle ABC$ be an equilateral triangle. Draw a perpendicular from point $D$ on side $AB$ to $BC$ at point $E$. Construct $EF \perp DE$, intersecting $AC$ at point $F$. Draw $FG \perp AC$, meeting $BC$ at point $G$, where $FG$ and $DE$ intersect at point $M$. If $DM = ME$ and $GE = 3$, what is the length of $AB$?

Solution:
\textbf{Solution:} As shown in the figure, draw MN $\perp$ BC at point N, \\
$\because \triangle ABC$ is an equilateral triangle, \\
$\therefore \angle B=\angle C=60^\circ$, \\
$\because DE\perp AB$, $EF\perp DE$, $FG\perp AC$, \\
$\therefore \angle BDE=90^\circ$, $\angle DEF=90^\circ$, $\angle GFC=90^\circ$, \\
$\therefore \angle DEB=180^\circ-90^\circ-60^\circ=30^\circ$, $\angle FGC=180^\circ-90^\circ-60^\circ=30^\circ$, \\
$\therefore \angle FEC=180^\circ-90^\circ-30^\circ=60^\circ$, $\angle DEB=\angle FGC$, \\
$\therefore GN=EN=\frac{1}{2}GE=\frac{3}{2}$, \\
In $Rt\triangle MNE$, $\angle MEN=30^\circ$, $EN=\frac{3}{2}$, \\
$\therefore \frac{NE}{ME}=\cos30^\circ=\frac{\sqrt{3}}{2}$, \\
$\therefore ME=\frac{\frac{3}{2}}{\frac{\sqrt{3}}{2}}=\sqrt{3}$, \\
$\because DM=ME$, \\
$\therefore DE=2ME=2\sqrt{3}$, \\
In $Rt\triangle BDE$, $\angle DEB=30^\circ$, \\
$\therefore BE=\frac{DE}{\frac{\sqrt{3}}{2}}=\frac{2\sqrt{3}}{\frac{\sqrt{3}}{2}}=4$, \\
$\because \angle C=60^\circ$, $\angle FEC=60^\circ$, \\
$\therefore \triangle EFC$ is an equilateral triangle, \\
$\therefore \angle EFC=60^\circ$, $EF=CE$, \\
$\therefore \angle GFE=90^\circ-60^\circ=30^\circ=\angle FGC$, \\
$\therefore EF=GE=3$, \\
$\therefore CE=3$, \\
$\therefore BC=BE+CE=4+3=7$, \\
$\therefore AB=BC=\boxed{7}$, \\


Question: As shown in the figure, in the isosceles right triangle $ABC$, $\angle BAC=90^\circ$, $D$ is the midpoint of $BC$, $E$ is a point on the side $AC$ (not coinciding with the endpoints), a line is drawn through point $E$ such that $EG \perp BC$ at point $G$, and $EH \perp AD$ at point $H$, a line is drawn through point $B$ such that $BF \parallel AC$ intersecting the extension of $EG$ at point $F$. If $AG=3$, what is the area of the shaded region?

Solution:
\textbf{Solution:} Let $DG=a$, $CG=b$, then $CD=a+b$, \\
since $\triangle ABC$ is an isosceles right triangle, $\angle BAC=90^\circ$, \\
thus $\angle ABC=\angle ACB=45^\circ$, $AB=AC$, \\
also since $D$ is the midpoint of $BC$, \\
thus $BD=AD=CD=a+b$, $BC=2BD=2(a+b)$, \\
since $EG\perp BC$, $EH\perp AD$, \\
thus quadrilateral DGEH is a rectangle, $\angle GEC=45^\circ$, \\
thus $DH=EG=CG=b$, \\
since $BF\parallel AC$, \\
thus $\angle FBG=\angle ACB=45^\circ$, \\
since $EF\perp BC$, \\
thus $\angle F=45^\circ$, \\
thus $GF=BG=BD+DG=a+b+a=2a+b$, \\
by the Pythagorean theorem, $AD^2+DG^2=AG^2$, \\
thus $(a+b)^2+a^2=9$, \\
which simplifies to $2a^2+2ab+b^2=9$, \\
According to the problem, ${S}_{\text{shaded}}={S}_{\triangle ABC}+{S}_{\triangle BGF}-{S}_{\text{rectangle} DGEH}$\\
$=\frac{1}{2}BC\cdot AD+\frac{1}{2}BG\cdot GF-DG\cdot DH$\\
$=BD\cdot AD+\frac{1}{2}BG^2-DG\cdot DH$\\
$=(a+b)^2+\frac{1}{2}(2a+b)^2-ab$\\
$=a^2+2ab+b^2+2a^2+ab+\frac{1}{2}b^2-ab$\\
$=\frac{3}{2}(2a^2+2ab+b^2)$\\
$=\frac{3}{2}\times 9$\\
$=\boxed{13.5}$


Question: As shown in the figure, it is known that $\triangle OCA \cong \triangle OBD$ and $\angle A=30^\circ$, $\angle AOC=80^\circ$. What is the measure of $\angle B$?

Solution:
\textbf{Solution:} Since $\triangle OCA \cong \triangle OBD$, $\angle A=30^\circ$, $\angle AOC=80^\circ$,
$\therefore \angle B=\angle C=180^\circ-\angle A-\angle AOC=180^\circ-30^\circ-80^\circ=\boxed{70^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=60^\circ$, $BC=16$, point D lies on the extension of BA, $CA=CD$, and $BD=10$. What is the length of $AD$?

Solution:
\textbf{Solution:} As shown in the figure, draw CE $\perp$ AD at point E,\\
$\therefore$ $\angle$BEC = $90^\circ$,\\
$\because$ $\angle$ABC = $60^\circ$,\\
$\therefore$ $\angle$BEC = $30^\circ$,\\
$\therefore$ BE = $\frac{1}{2}$BC = 8,\\
$\therefore$ DE = BD - BE = 10 - 8 = 2,\\
$\because$ CA = CD,\\
$\therefore$ AE = DE = 2,\\
$\therefore$ AD = \boxed{4}.


Question: As shown in the figure, the straight line $CD$ is the perpendicular bisector of the line segment $AB$, and $P$ is a point on the line $CD$. Given that the length of segment $PA$ is $3\,\text{cm}$, what is the length of segment $PB$?

Solution:
\textbf{Solution}: Since line CD is the perpendicular bisector of segment AB, \\
it follows that PA=PB, \\
Since PA=3cm, \\
it follows that PB=\boxed{3cm}.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=90^\circ$, $\angle A=60^\circ$. Point $D$ is on side $AC$ with $\angle DBC=30^\circ$ and $AC=12\,\text{cm}$. What is the perimeter of $\triangle ABD$?

Solution:
\textbf{Solution:} Since $\triangle ABC\cong \triangle ADE$, and $\angle D=30^\circ$,\\
it follows that $\angle B=\angle D=30^\circ$,\\
Since $\angle C=70^\circ$,\\
then $\angle BAC=180^\circ-\angle B-\angle C=180^\circ-70^\circ-30^\circ=80^\circ$,\\
Since $\triangle ABC\cong \triangle ADE$,\\
it follows that $\angle EAD=\angle BAC=80^\circ$,\\
Also, given that $\angle CAD=35^\circ$,\\
then $\angle EAC=\angle EAD-\angle DAC=80^\circ-35^\circ=45^\circ$,\\
Thus, the answer is: $\boxed{B}$.


Question: As shown in the figure, in $\triangle ABC$, it is known that points $D$, $E$, and $F$ are respectively the midpoints of $BC$, $AD$, and $EC$, and the area of $\triangle ABC$, ${S}_{\triangle ABC}=8$. What is the area $S$ of the shaded part?

Solution:
\textbf{Solution:}\\
Since point D is the midpoint of BC,\\
therefore, the area of $\triangle ABD$ equals the area of $\triangle ACD$ equals $\frac{1}{2}$ the area of $\triangle ABC$,\\
Since point E is the midpoint of AD,\\
therefore, the area of $\triangle BDE$ equals $\frac{1}{2}$ the area of $\triangle ABD$ equals $\frac{1}{4}$ the area of $\triangle ABC$, and the area of $\triangle CDE$ equals $\frac{1}{2}$ the area of $\triangle ACD$ equals $\frac{1}{4}$ the area of $\triangle ABC$,\\
therefore, the area of $\triangle BCE$ equals the area of $\triangle BDE$ plus the area of $\triangle CDE$, which equals $\frac{1}{2}$ the area of $\triangle ABC$,\\
Since point F is the midpoint of EC,\\
therefore, the area of $\triangle BEF$ equals $\frac{1}{2}$ the area of $\triangle BCE$ equals $\frac{1}{2} \times \frac{1}{2} \times 8 = \boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, it is known that $BA=BC$, $\angle B=120^\circ$, and the perpendicular bisector $DE$ of $AB$ intersects $AC$ at point D. If $AC=6\,cm$, what is the length of $AD$?

Solution:
\textbf{Solution:} Connect $BD$,\\
since the perpendicular bisector $DE$ of $AB$ intersects $AC$ at point D,\\
it follows that $AD=BD$,\\
hence $\angle A=\angle ABD$,\\
since $BA=BC$, $\angle B=120^\circ$,\\
it follows $\angle A=\angle C=\frac{1}{2}\left(180^\circ-120^\circ\right)=30^\circ$,\\
thus $\angle ABD=30^\circ$,\\
hence $\angle CBD=90^\circ$,\\
thus $CD=2BD$,\\
hence $CD=2AD$,\\
thus $AC=AD+CD=AD+2AD=3AD$,\\
further, since $AC=6\text{cm}$,\\
it follows $AD=\boxed{2\text{cm}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AD$ bisects $\angle CAB$, if $AB=10$, $CD=3$, what is the area of $\triangle ABD$?

Solution:
\textbf{Solution:} As shown in the figure, draw $DE \perp AB$ at point E, \\
since $\angle C=90^\circ$, $AD$ bisects $\angle BAC$, \\
thus $DE=CD=3$, \\
therefore, the area of $\triangle ABD = \frac{1}{2}AB\cdot DE=\frac{1}{2}\times 10\times 3=\boxed{15}$.


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle with side length 2, and the area of $\triangle ABC$ is $\sqrt{3}$. $D$ and $E$ are the midpoints of $BC$ and $AC$, respectively; $P$ is a moving point on $AD$. What is the minimum value of $PE+PC$?

Solution:
\textbf{Solution:} As shown in the figure, draw $BE$ intersecting $AD$ at point $P'$,\\
since $\triangle ABC$ is an equilateral triangle, with $AB=2$, and $AD$ being the altitude on side $BC$, and E being the midpoint of $AC$,\\
therefore, $AD$ and $BE$ are the perpendicular bisectors of sides $BC$ and $AC$ of the equilateral triangle $ABC$ respectively,\\
therefore, $P'B=P'C$,\\
therefore, $P'E+P'C=P'E+P'B=BE$,\\
based on the principle that the line segment between two points is the shortest,\\
when point P is at $P'$, $PE+PC$ has its minimum value, which is exactly the length of $BE$.\\
since the area of $\triangle ABC$ is equal to $\sqrt{3}$, and the side length is 2,\\
therefore, $\frac{1}{2}\times 2\times BE=\sqrt{3}$,\\
therefore, $BE=\sqrt{3}$,\\
thus, the minimum value of $PE+PC$ is $\boxed{\sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, DE is the perpendicular bisector of the line segment AB. If the perimeter of $\triangle ABC$ is 18, and the length of the line segment AE is 4, then what is the perimeter of $\triangle BCD$?

Solution:
\textbf{Solution:} Since the perimeter of $\triangle ABC$ is $18$, \\
it follows that $AC+BC+AB=18$, \\
Since $DE$ is the perpendicular bisector of segment $AB$, and $AE=4$, \\
it follows that $AB=2AE=8$, and $DA=DB$, \\
thus, $AC+BC=10$, \\
therefore, the perimeter of $\triangle BCD$ = $BD+CD+BC=AD+CD+BC=AC+BC=\boxed{10}$


Question: As shown in the figure, fold $\triangle ABC$ along the segment $DE$ so that point $B$ falls on point $F$; if $AC \parallel DE$, $\angle A = 70^\circ$, $AB = AC$, then what is the degree measure of $\angle CEF$?

Solution:
\textbf{Solution}: We fold triangle $ABC$ along the line segment $DE$, such that point $B$ falls on point $F$,\\
thus, triangle $BDE \cong FDE$,\\
hence, $\angle DEB = \angle DEF$,\\
since $\angle A=70^\circ$, and $AB=AC$,\\
thus, $\angle B = \angle C = \frac{1}{2} \times (180^\circ - 70^\circ) = 55^\circ$,\\
since $AC \parallel DE$,\\
hence, $\angle DEB = \angle C = 55^\circ = \angle DEF$,\\
thus, $\angle FEC = 180^\circ - \angle DEB - \angle DEF = \boxed{70^\circ}$.


Question: As shown in the figure, if $\angle A = 31^\circ$, then what is the value of $\angle A + \angle B + \angle C + \angle D + \angle E$?

Solution:
\textbf{Solution:} According to the given conditions,\\
\[ \because \angle AFG+\angle AGF=180^\circ-\angle A, \angle CFG+\angle AFG=180^\circ, \angle EGF+\angle AGF=180^\circ\]
\[ \therefore \angle CFG+\angle EGF=360^\circ-\left(\angle AFG+\angle AGF\right)=360^\circ-\left(180^\circ-\angle A\right)=180^\circ+\angle A\]
Also \[ \because \angle CFG=\angle B+\angle C, \angle EGF=\angle D+\angle E,\]
\[ \therefore \angle A+\angle B+\angle C+\angle D+\angle E\]
\[ =\angle A+\angle CFG+\angle EGF\]
\[ =180^\circ+2\angle A\]
\[ =180^\circ+2\times 31^\circ\]
\[ =\boxed{242^\circ}\]


Question: As shown in the figure, in \( \triangle ABC\), \(DE\) is the perpendicular bisector of side \(AC\). If \(BC=8\, \text{cm}\) and \(AB=10\, \text{cm}\), what is the perimeter of \( \triangle EBC\)?

Solution:
\textbf{Solution:} Given that $DE$ is the perpendicular bisector of side $AC$ in $\triangle ABC$,\\
thus $EA=EC$,\\
therefore $BE+EC=BE+EA=AB$,\\
given $BC=8$ cm and $AB=10$ cm,\\
thus the perimeter of $\triangle EBC$ is $BC+BE+EC=8+10=\boxed{18}$ cm.


Question: As shown in the figure, $D$ is a point on $AB$, $DF$ intersects $AC$ at point $E$, $DE=FE$, $FC \parallel AB$, $AB=5$, $BD=1$. What is the length of $CF$?

Solution:
\textbf{Solution:} Given $FC\parallel AB$,\\
it follows that $\angle F = \angle ADE$, $\angle FCE = \angle A$,\\
In $\triangle CFE$ and $\triangle ADE$,\\
$\left\{ \begin{array}{l} \angle F = \angle ADE \\ \angle FCE = \angle A \\ FE = DE \end{array} \right.$,\\
thus $\triangle CFE \cong \triangle ADE$ (AAS),\\
thus $CF = AD$,\\
since $AB = 5$, $BD = 1$,\\
it follows that $AD = AB - BD = 5 - 1 = 4$,\\
thus $CF = \boxed{4}$,\\
hence the length of $CF$ is $4$.


Question: As shown in the diagram, given rectangle ABCD, point E is on the side AB, and $BE=2$, $BC=3$. Fold $\triangle CBE$ along the line CE so that point B falls on point G, and extend EG to intersect CD at point F. What is the length of the segment FG?

Solution:
\textbf{Solution:} Given that when $\triangle CBE$ is folded along line CE such that point B falls on point G, \\
it follows that $\angle BEC = \angle GEC$, $GE = BE = 2$, and $CG = BC = 3$, \\
since quadrilateral ABCD is a rectangle, \\
it implies $CD \parallel AB$, \\
thus $\angle BEC = \angle FCE$, \\
leading to $\angle GEC = \angle FCE$, \\
therefore, $CF = EF$, \\
let $FG = x$, then $CF = EF = x + 2$, \\
in $\triangle CFG$, we have $FG^2 + CG^2 = CF^2$, \\
which gives $x^2 + 3^2 = (x + 2)^2$, \\
solving for $x$ yields $x = \boxed{\frac{5}{4}}$.


Question: As shown in the figure, point E is on $AC$, $\triangle ABC\cong \triangle DAE$, $BC=3$, $DE=7$, then what is the length of $CE$?

Solution:
\textbf{Solution:} Since it follows from the problem statement that $\triangle ABC\cong \triangle DAE$,

we have $AE=BC=3$, $AC=DE=7$,

thus $CE=AC-AE=7-3=\boxed{4}$


Question: As shown in the figure, $AD$ is the median of $\triangle ABC$. It is known that the perimeter of $\triangle ABD$ is $25\,\text{cm}$, and $AB$ is $7\,\text{cm}$ longer than $AC$. What is the perimeter of $\triangle ACD$?

Solution:
\textbf{Solution:} Since $AD$ is the median of $\triangle ABC$,\\
it follows that $BD=CD$,\\
therefore, the difference in the perimeters of $\triangle ABD$ and $\triangle ACD$ is $(AB+BD+AD)-(AC+CD+AD)=AB-AC$.\\
Since the perimeter of $\triangle ABD$ is $25\,cm$ and $AB$ is $7\,cm$ longer than $AC$,\\
the perimeter of $\triangle ACD$ is: $25-7=\boxed{18}\,(cm)$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ bisects $\angle BAC$, $AB=4$, $AC=2$. If the area of $\triangle ACD$ is equal to $3$, what is the area of $\triangle ABD$?

Solution:
\textbf{Solution:} Draw $DE \perp AB$ at $E$ and $DF \perp AC$ at $F$, as shown in the figure,\\
since $AD$ bisects $\angle BAC$,\\
thus $DE = DF$,\\
since the area of $\triangle ACD$ = $\frac{1}{2} \cdot DF \cdot AC = 3$,\\
thus $DF = \frac{2 \times 3}{AC} = 3$,\\
thus $DE = 3$.\\
Therefore, the area of $\triangle ABD$ = $\frac{1}{2} \cdot DE \cdot AB = \frac{1}{2} \times 3 \times 4 = \boxed{6}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=90^\circ$, $AB=3$, and $AC=4$. Now, fold $\triangle ABC$ along $BD$ such that point A lies exactly on $BC$. What is the length of $CD$?

Solution:
\textbf{Solution:} Let $CD=x$, then $AD=A'D=4-x$,\\
In $\triangle ABC$ right-angled at $B$,\\
$BC=\sqrt{AB^{2}+AC^{2}}=\sqrt{3^{2}+4^{2}}=5$,\\
$\therefore A'C=BC-A'B=BC-AB=5-3=2$,\\
In $\triangle A'DC$ right-angled at $D$,\\
$A'D^{2}+A'C^{2}=CD^{2}$  \\
i.e., ${(4-x)}^{2}+2^{2}=x^{2}$\\
Solving gives: $x=\boxed{\frac{5}{2}}$


Question: As shown in the figure, there are two slides of equal length. The height \(AC\) of the left slide is equal to the horizontal length \(DF\) of the right slide. What is the sum of the degrees of the incline angles \(\angle ABC\) and \(\angle DFE\)?

Solution:
\textbf{Solution:} Given $\angle ABC + \angle DFE = 90^\circ$, the reasoning is as follows:\\
From the problem, we know that both $\triangle ABC$ and $\triangle DEF$ are right-angled triangles, with $BC = EF$ and $AC = DF$.\\
In $\triangle ABC$ and $\triangle DEF$,\\
$\left\{\begin{array}{l}
BC = EF\\
AC = DF
\end{array}\right.$,\\
$\therefore \triangle ABC \cong \triangle DEF$ (by HL criterion),\\
$\therefore \angle ABC = \angle DEF$,\\
$\because \angle DEF + \angle DFE = 90^\circ$,\\
$\therefore \angle ABC + \angle DFE = \boxed{90^\circ}$.


Question: As shown in the diagram, in $\triangle ABC$, if $\triangle ABD$ and $\triangle ACE$ are constructed with $AB$ and $AC$ as their respective sides, and $\angle DAB = \angle CAE = \alpha$, $AD = AB$, $AC = AE$, and $DC$ and $BE$ intersect at point $P$, with $AP$ connected, then what is the measure of $\angle APC$?

Solution:
\textbf{Solution:} As illustrated, construct $AF\perp CD$ at point F, and $AG\perp BE$ at point G, then $\angle AFC=\angle AGE=90^\circ$,\\
$\because \angle DAB=\angle CAE=\alpha$,\\
$\therefore \angle DAC=\angle BAE=\alpha +\angle BAC$,\\
In $\triangle DAC$ and $\triangle BAE$,\\
$ \left\{\begin{array}{l}AD=AB\\ \angle DAC=\angle BAE\\ AC=AE\end{array}\right.$,\\
$\therefore \triangle DAC\cong \triangle BAE$ (SAS),\\
$\therefore \angle ACF=\angle AEG$,\\
In $\triangle ACF$ and $\triangle AEG$\\
$ \left\{\begin{array}{l}\angle AFC=\angle AGE\\ \angle ACF=\angle AEG\\ AC=AE\end{array}\right.$,\\
$\therefore \triangle ACF\cong \triangle AEG$ (AAS),\\
$\therefore AF=AG$,\\
$\therefore$ point A is on the bisector of $\angle DPE$,\\
$\therefore \angle APE=\angle APD=\frac{1}{2}\angle DPE$,\\
$\because \angle CPE+\angle ACF=\angle CAE+\angle AEG=\angle AHP$,\\
$\therefore \angle CPE=\angle CAE=\alpha$,\\
$\therefore \angle APE=\frac{1}{2}\angle DPE=\frac{1}{2}(180^\circ-\angle CPE)=90^\circ-\frac{1}{2}\alpha$,\\
$\therefore \angle APC=\angle APE+\angle CPE=90^\circ-\frac{1}{2}\alpha+\alpha=90^\circ+\frac{1}{2}\alpha$,\\
$\therefore$ the measure of $\angle APC$ is $90^\circ+\frac{1}{2}\alpha$.\\
Therefore, the measure of $\angle APC$ is $\boxed{90^\circ+\frac{1}{2}\alpha}$.


Question: As shown in the figure, what is the sum of $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7$?

Solution:
\textbf{Solution:} As shown in the figure, \\
by the internal angle sum of a quadrilateral, we have $\angle 2+\angle 3+\angle 5+\angle 8=360^\circ$, $\angle 6+\angle 7+\angle 9+\angle 10=360^\circ$, \\
$\therefore \angle 2+\angle 3+\angle 5+\angle 8+\angle 6+\angle 7+\angle 9+\angle 10=720^\circ$, \\
$\because \angle 8+\angle 9=180^\circ$, $\angle 10=\angle 1+\angle 4$, \\
$\therefore \angle 1+\angle 2+\angle 3+\angle 5+\angle 8+\angle 6+\angle 7=720^\circ-180^\circ=\boxed{540^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=46^\circ$, $\angle C=52^\circ$, $AD$ bisects $\angle BAC$, intersects $BC$ at point $D$, $DE \parallel AB$, and intersects $AC$ at point $E$, then what is the measure of $\angle ADE$?

Solution:
\textbf{Solution:} Given that $\angle B=46^\circ$ and $\angle C=52^\circ$,\\
thus $\angle BAC=180^\circ-\angle B-\angle C=180^\circ-46^\circ-52^\circ=82^\circ$,\\
and since AD bisects $\angle BAC$,\\
thus $\angle BAD=\frac{1}{2}\times82^\circ=41^\circ$,\\
since DE$\parallel$AB,\\
thus $\angle ADE=\boxed{41^\circ}$.


Question: As shown in the figure, $OP$ bisects $\angle AOB$, $E$ is a point on $OA$, $OE=4$, and the distance from $P$ to $OB$ is 2. What is the area of $\triangle OPE$?

Solution:
\textbf{Solution:} As shown in the figure, draw PD$\perp$OB at point D and PC$\perp$OA at C, \\
since $OP$ is the bisector of $\angle AOB$, and the distance from P to OB is 2, \\
therefore $PC=PD=2$, \\
since $OE=4$, \\
therefore the area of $\triangle OPE$ is $\frac{1}{2}OE\cdot PC=\frac{1}{2}\times 4\times 2=\boxed{4}$.


Question: As shown in the figure, $AD$ is the median of $\triangle ABC$, $CE \parallel AB$ intersects the extension of $AD$ at point $E$, $AB=5$, $AC=7$, then what are the possible values of $AD$?

Solution:
\textbf{Solution:} Since $AD$ is the median of $\triangle ABC$, \\
thus $CD=BD$, \\
since $CE \parallel AB$, \\
thus $\angle DCE=\angle DBA$, \\
in $\triangle CDE$ and $\triangle BDA$, \\
$\left\{\begin{array}{l}
\angle CDE=\angle BDA \\
CD=BD \\
\angle DCE=\angle DBA
\end{array}\right.$, \\
therefore $\triangle CDE\cong \triangle BDA$ (ASA), \\
thus $EC=AB=5$, \\
since $7-5<AE<7+5$, \\
thus $2<2AD<12$, \\
thus $\boxed{1<AD<6}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=\angle C$, point $D$ is on side $BC$, and point $E$ is on $AC$, $\angle ADE=\angle AED$. If $\angle BAD=40^\circ$, what is the measure of $\angle CDE$?

Solution:
\textbf{Solution:} Since $\angle ADC$ is an exterior angle of $\triangle ABD$, \\
we have $\angle ADC=\angle BAD+\angle B=40^\circ+\angle B$, \\
Since $\angle AED$ is an exterior angle of $\triangle DEC$, \\
we have $\angle AED=\angle C+\angle EDC$, \\
Since $\angle ADE=\angle AED$, \\
we have $\angle ADE=\angle C+\angle EDC$, \\
Since $\angle ADC=\angle ADE+\angle EDC=\angle C+2\angle EDC$, \\
and since $\angle B=\angle C$, \\
we have $\angle B+40^\circ=\angle B+2\angle EDC$, \\
Therefore, $\angle EDC=\boxed{20^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, point $D$ is on $AB$. When folded along $CD$, point $A$ falls on point $E$ on side $BC$. If $\angle B=26^\circ$, what is the measure of $\angle CDE$?

Solution:
\textbf{Solution:} Given that $\angle ACB=90^\circ$ and $\angle B=26^\circ$, \\
thus $\angle A=90^\circ-26^\circ=64^\circ$. \\
By the fold, $\angle CDE=\angle ADC$ and $\angle ACD=\angle BCD=45^\circ$, \\
therefore $\angle ADC=180^\circ-45^\circ-64^\circ=71^\circ$, \\
thus $\angle CDE=\angle ADC=\boxed{71^\circ}$.


Question: As illustrated in the diagram, line $m\parallel n$, $\triangle ABC$ is an equilateral triangle with vertex $B$ on line $n$. Line $m$ intersects $AB$ at point $E$ and $AC$ at point $F$. Given that $\angle 1=140^\circ$, what is the measure of $\angle 2$?

Solution:
\textbf{Solution:} As shown in the figure: \\
Since $\triangle ABC$ is an equilateral triangle,\\
it follows that $\angle A=\angle ABC=\angle C=60^\circ$.\\
In $\triangle AFE$, since $\angle 1=\angle A+\angle AEF=140^\circ$,\\
it follows that $\angle AEF=140^\circ-60^\circ=80^\circ$,\\
hence, $\angle DEB=\angle AEF=80^\circ$,\\
Since $m\parallel n$,\\
it follows that $\angle 2+\angle DEB=180^\circ$,\\
thus, $\angle 2=180^\circ-80^\circ=\boxed{100^\circ}$.


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle, $\angle CBD = 90^\circ$, $BD = BC$, and $AD$ is connected and intersects $BC$ at point $E$. What is the measure of $\angle CAD$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is an equilateral triangle, \\
therefore AB＝BC, and $\angle ABC = \angle CAB = 60^\circ$, \\
since BD＝BC, \\
therefore AB＝BD, \\
therefore $\angle BAD = \angle BDA$, \\
since $\angle CBD = 90^\circ$, \\
therefore $\angle ABD = 90^\circ + 60^\circ = 150^\circ$, \\
therefore $\angle BDA = 15^\circ$, \\
therefore $\angle CAD = \angle CAB - \angle BDA = 60^\circ - 15^\circ = \boxed{45^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $\angle BAC=120^\circ$, $BC=2\sqrt{3}$, with $D$ being the midpoint of $BC$, and $DE\perp AB$, what is the area of $\triangle EBD$?

Solution:
\textbf{Solution:} Since AB=AC and $\angle$BAC=120$^\circ$, \\
it follows that $\angle$B=$\angle$C=30$^\circ$, \\
Since D is the midpoint of BC and BC=$2\sqrt{3}$, \\
then in right $\triangle$BED, BD=$\frac{1}{2}$BC=$\sqrt{3}$, thus DE=$\frac{\sqrt{3}}{2}$, BE=$\frac{3}{2}$. \\
Therefore, the area of $\triangle$EBD is $\frac{1}{2}\times \frac{\sqrt{3}}{2}\times \frac{3}{2}=\boxed{\frac{3\sqrt{3}}{8}}$.


Question: As shown in the figure, CD is the median of the base of the isosceles triangle $\triangle ABC$, BE bisects $\angle ABC$ and intersects CD at point E. Given that AC = 8 and DE = 2, what is the area of $\triangle BCE$?

Solution:
\textbf{Solution:} Construct EF $\perp$ BC at F, \\
since AC = BC = 8, and CD is the median of the isosceles triangle $\triangle ABC$ on the base, \\
therefore CD $\perp$ AB, \\
since BE bisects $\angle ABC$, ED $\perp$ AB, and EF $\perp$ BC, \\
therefore EF = DE = 2, \\
thus, the area of $\triangle BCE$ = $\frac{1}{2} \times BC \times EF = \frac{1}{2} \times 8 \times 2 = \boxed{8}$.


Question: As shown in the figure, \( \triangle ADE \cong \triangle BCF \), \( AD = 10 \) cm, \( CD = 6 \) cm, then what is the length of \( BD \)?

Solution:
\textbf{Solution:} Since $\triangle ADE \cong \triangle BCF$,\\
it follows that $AD = BC = 10\text{cm}$,\\
Given that $BD = BC - CD$, and $CD = 6\text{cm}$,\\
we have $BD = 10 - 6 = \boxed{4}\text{cm}$.


Question: As shown in the figure, isosceles right triangles are constructed externally on each side of $\triangle ABC$ using the sides of $\triangle ABC$ as the right angles. If $AB = \sqrt{3}$, what is the area of the shaded region in the figure?

Solution:
\textbf{Solution:} Since $ABC$ is a right-angled triangle,\\
we have $AC^{2}+BC^{2}=AB^{2}=3$,\\
thus, the area of the shaded region $S_{\text{shaded}}= \frac{1}{2} AC^{2}+ \frac{1}{2} BC^{2}+ \frac{1}{2} AB^{2}= \frac{1}{2} (AC^{2}+BC^{2})+ \frac{1}{2} AB^{2}= \frac{1}{2} AB^{2}+ \frac{1}{2} AB^{2}=AB^{2}=\boxed{3}.$


Question: As shown in the figure, $\triangle ABC \cong \triangle DEF$, with points B, E, C, F lying on the same straight line. If $BC=7$ and $EC=4$, then what is the length of $CF$?

Solution:
\textbf{Solution:} Given that $\triangle ABC \cong \triangle DEF$,\\
hence $BC = EF$.\\
Since points B, E, C, and F are on the same line, and $BC = 7$, $EC = 4$,\\
therefore $CF = EF - EC = BC - EC = 7 - 4 = \boxed{3}$.


Question: As shown in the figure, in the right-angled triangle paper ABC, where $\angle ACB = 90^\circ$ and $\angle A = 50^\circ$, it is folded along the midpoint CE of side AB, making point A fall on point $A'$, then what is the degree measure of $\angle A'EB$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is a right-angled triangle, and CE is the median, \\
thus $AE=CE=BE$, \\
by the property of folding, we obtain \\
$AE=A'E$, $\angle AEC=\angle A'EC$, \\
therefore, $AE=CE=BE=A'E$, \\
since $\angle A=50^\circ$, \\
hence $\angle ACE=50^\circ$, \\
therefore, $\angle AEC=\angle A'EC=180^\circ-50^\circ-50^\circ=80^\circ$, \\
since $\angle BEC=50^\circ+50^\circ=100^\circ$, \\
hence $\angle A'EB=100^\circ-80^\circ=\boxed{20^\circ}$;


Question: As shown in the figure, B is on AC, D is on CE, $AD=BD=BC$, $\angle ACE=25^\circ$, what is the degree measure of $\angle ADE$?

Solution:
\textbf{Solution}: Since $BD=BC$, $\angle ACE=25^\circ$,\\
hence $\angle BDC=\angle C=25^\circ$,\\
thus $\angle ABD=50^\circ$,\\
since $AD=BD$,\\
thus $\angle A=\angle ABD=50^\circ$,\\
therefore $\angle ADE=\angle A+\angle C=\boxed{75^\circ}$.


Question: In the figure, in triangle $ABC$, $\angle C=90^\circ$, the angle bisector of $\angle CAB$, $AD$, intersects $BC$ at $D$, $DE\perp AB$ at $E$, $DE=2\,\text{cm}$, and $DB=4\,\text{cm}$. What is the length of $BC$?

Solution:
\textbf{Solution:} Given that $\because$AD bisects $\angle$CAB, and DE$\perp$AB at E, with $\angle$C=$90^\circ$,\\
$\therefore$CD=DE=2,\\
Also $\because$ $DB=4\text{cm}$,\\
$\therefore$BC=BD+CD=4+2=\boxed{6}\text{(cm)};


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=130^\circ$, the perpendicular bisectors of $AB$ and $AC$ intersect $BC$ at points $E$ and $F$, and intersect $AB$ and $AC$ at points $D$ and $G$, respectively. What is the measure of $\angle EAF$?

Solution:
\textbf{Solution:} \\
Since DE bisects AB perpendicularly, and FG bisects AC perpendicularly, \\
it follows that EB=EA, and FA=FC, \\
thus, $\angle$BAE=$\angle$B, $\angle$FAC=$\angle$C, \\
Since in $\triangle ABC$, $\angle$BAC=$130^\circ$, \\
it follows that $\angle$B+$\angle$C=$50^\circ$, \\
thus, $\angle$BAE+$\angle$FAC=$50^\circ$, \\
therefore, $\angle$EAF=$\angle$BAC−($\angle$BAE+$\angle$FAC)=$\boxed{80^\circ}$.


Question: As shown in the figure, $AD$ is the angle bisector of $\triangle ABC$, $DE \perp AB$ at point E, the area of $\triangle ABC$ is 9, $DE=2$, and $AB=5$. What is the length of $AC$?

Solution:
Solution: As shown in the figure, construct $DF \perp AC$,\\
since $DE \perp AB$ and $AD$ is the angle bisector of $\triangle ABC$,\\
therefore $DE = DF = 2$.\\
Since $S_{\triangle ABC} = S_{\triangle ABD} + S_{\triangle ACD}$, $AB = 5$, $S_{\triangle ABC} = 9$,\\
therefore $S_{\triangle ABC} = \frac{1}{2}AB \times DE + \frac{1}{2}AB \times DF$\\
that is $9 = \frac{1}{2} \times 2 \times (5 + AC)$\\
Solving gives $AC = \boxed{4}$.


Question: As shown in the figure, it is known that the length of the fishing rod $AC$ is $10\mathrm{m}$, and the length of the fishing line $BC$ exposed above the water surface is $6\mathrm{m}$. A fisher wants to check the situation on the hook, so he rotates the fishing rod $AC$ to the position $A{C}^{\prime}$, at this time, the length of the fishing line ${B}^{\prime}{C}^{\prime}$ exposed above the water surface is $8\mathrm{m}$. What is the length of $B{B}^{\prime}$?

Solution:
\textbf{Solution:} Given $\because$AC=10m, BC=6m, $\angle$ABC$=90^\circ$,\\
$\therefore$AB= $\sqrt{AC^{2}-BC^{2}}=\sqrt{10^{2}-6^{2}}=8$ m,\\
$\because$AC$^{\prime }$=10m, B$^{\prime }$C$^{\prime }$=8m, $\angle$AB$^{\prime }$C$^{\prime }$=$90^\circ$,\\
$\therefore$AB$^{\prime }$= $\sqrt{AC{'}^{2}-B'C{'}^{2}}=\sqrt{10^{2}-8^{2}}=6$ m,\\
$\therefore$BB$^{\prime }$=AB−AB$^{\prime }$=$\boxed{2}$m;


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, $AC = BC$. A straight line DE is drawn outside $\triangle ABC$ through point C, and perpendicular lines are drawn from points A and B to line DE, with the feet of the perpendiculars being E and D, respectively. If $AE=4$ and $BD=3$, then what is the length of $DE$?

Solution:
\textbf{Solution:} Given that $\angle ACB=90^\circ$, \\
it follows that $\angle ACE + \angle BCD = 90^\circ$.\\
Since $\angle ACE + \angle CAE = 90^\circ$,\\
it implies that $\angle BCD = \angle CAE$.\\
In $\triangle ACE$ and $\triangle CBD$, \\
$\begin{array}{c}
\angle CAE = \angle BCD,\\
\angle E = \angle D = 90^\circ,\\
AC = BC,
\end{array}$\\
hence, $\triangle ACE \cong \triangle CBD$ by AAS,\\
thus, CE = BD = 3, CD = AE = 4,\\
therefore, DE = CE + CD = \boxed{7}.


Question: Given a right-angled $\triangle ABC$, where $\angle ACB=90^\circ$, $AC=BC$, and $AB=4\sqrt{2}$, let $D$ be the midpoint of $BC$, and $E$ be a point on line segment $AB$. Connect $CE$ and $DE$. What is the minimum value of $CE+DE$?

Solution:
\textbf{Solution:} As shown in the figure, construct the symmetric point \(C'\) of point \(C\) with respect to \(AB\), and connect \(C'D\), which intersects \(AB\) at point \(E\). Construct \(DF \perp CC'\) at point \(F\),\\
then \(CE = C'E\), \(CE + DE = C'E + DE\), and the line segment \(C'D\) is the minimum value of \(CE + DE\).\\
Since \(\angle ACB = 90^\circ\) and \(AC = BC\), \(AB = 4 \sqrt{2}\),\\
thus \(AC = BC = 4\).\\
Since \(D\) is the midpoint of \(BC\),\\
thus \(CD = BD = \frac{1}{2} BC = \frac{1}{2} \times 4 = 2\).\\
thus \(CF = DF = \sqrt{2}\), \(CC' = 2CG = 2 \times 2\sqrt{2} = 4\sqrt{2}\)\\
thus \(C'F = CC' - CF = 4\sqrt{2} - \sqrt{2} = 3\sqrt{2}\)\\
thus \(C'D = \sqrt{C'F^2 + FD^2} = \sqrt{(3\sqrt{2})^2 + (\sqrt{2})^2} = \boxed{2\sqrt{5}}\)


Question: As shown in the figure, what is the sum of $ \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6$ in degrees?

Solution:
Solution: As shown in the diagram,\\
based on the given information, we have:\\
$\angle 9=\angle 4+\angle 6$, $\angle 10=\angle 1+\angle 5$,\\
$\because$ $\angle 2+\angle 3+\angle 9+\angle 10=360^\circ$,\\
$\therefore$ $\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6=\boxed{360^\circ}$.


Question: As shown in the figure, in $ \triangle MNP$, $ \angle P=60^\circ$, $ MN=NP$, $ MQ\perp PN$ with the foot being Q. Extend MN to G, taking $ NG=NQ$. If the perimeter of $ \triangle MNP$ is 12 and $ MQ=m$, what is the perimeter of $ \triangle MGQ$?

Solution:
\textbf{Solution:} Given that $\angle P=60^\circ$,\\
$MN=NP$,\\
thus, $\triangle PMN$ is an equilateral triangle,\\
given that $MQ\perp PN$,\\
thus, $QN=PQ= \frac{1}{2}MN$, $\angle QMN=30^\circ$, $\angle QNM=60^\circ$,\\
given that $NG=NQ$, therefore $\angle GQN=\angle G=30^\circ$, $QN=NG= \frac{1}{2}MN$,\\
thus, $\angle QMN=\angle G=30^\circ$,\\
thus, $QM=QG$,\\
given that the perimeter of $\triangle MNP$ is 12,\\
$MQ=m$,\\
thus, MN=4, QN=NC=2, QM=QG=m,\\
thus, the perimeter of $\triangle MGQ$ is QM+QG+MN+NG=$\boxed{6+2m}$.


Question: As shown in the figure, in $\triangle ABC$, point E is on the extension line of BC. The angle bisectors of $\angle ABC$ and $\angle ACE$ intersect at point D, $\angle D=15^\circ$. What is the degree measure of $\angle A$?

Solution:
\textbf{Solution:} \\
Since the bisector of $\angle ABC$ intersects the bisector of $\angle ACE$ at point D, \\
Therefore, $\angle CBD=\frac{1}{2}\angle ABC$, $\angle ECD=\frac{1}{2}\angle ACE$, \\
Since $\angle ACE=\angle A +\angle ABC$, $\angle ECD=\angle CBD+\angle D$, \\
Therefore, $\angle CBD + \angle D=\frac{1}{2}(\angle A + \angle ABC)=\frac{1}{2}(\angle A+2\angle CBD)$, \\
Therefore, $\angle D=\frac{1}{2}\angle A$, \\
Since $\angle D=15^\circ$, \\
Therefore, $\angle A=\boxed{30^\circ}$.


Question: As shown in the figure, the lines $l_{1}$ and $l_{2}$ intersect two sides $AB$ and $BC$ of $\triangle ABC$ respectively, and $l_{1} \parallel l_{2}$. If $\angle B=35^\circ$ and $\angle 1=105^\circ$, what is the measure of $\angle 2$?

Solution:
\textbf{Solution:} Since $\angle B = 35^\circ$ and $\angle 1 = 105^\circ$,

it follows that $\angle 3 = 180 - \angle 1 - \angle B = 40^\circ$,

Given $l_{1} \parallel l_{2}$,

therefore $\angle 2 = \angle 3 = \boxed{40^\circ}$.


Question: As shown in the figure, $Rt\triangle AOB \cong Rt\triangle CDA$, and the coordinates of points A and B are $(-1, 0)$ and $B(0, 2)$, respectively. What is the length of $OD$?

Solution:
\textbf{Solution:} Since $A(-1,0)$ and $B(0,2)$, \\
hence $OA=1$, $OB=2$, \\
since $\triangle AOB \cong \triangle CDA$, \\
hence $OB=AD=2$, \\
therefore $OD=AD+AO=2+1=\boxed{3}$.


Question: As shown in the figure, points $D$ and $E$ are the midpoints of sides $BC$ and $AB$ of the equilateral triangle $\triangle ABC$, respectively, with $AD=6$. If $F$ is a moving point on $AD$, what is the minimum value of $BF+EF$?

Solution:
\textbf{Solution:} Connect CE, intersecting AD at F, and connect BF, then BF+EF is the minimum (according to the fact that the shortest distance between two points is a straight line; the perpendicular distance from a point to a line is the shortest),\\
$\because$E is the midpoint of AB, $\triangle ABC$ is an equilateral triangle,\\
$\therefore CE\perp AB$\\
Since C and B are symmetric about AD, then BF+EF=CF,\\
$\because$In the equilateral $\triangle ABC$, BD=CD,\\
$\therefore$AD$\perp$BC,\\
$\therefore$AD is the perpendicular bisector of BC (three lines in one),\\
$\therefore$C and B are symmetric about the line AD,\\
$\therefore$CF=BF,\\
That is, BF+EF=CF+EF=CE,\\
$\because$AD$\perp$BC, CE$\perp$AB,\\
$\therefore$$\angle$ADB=$\angle$CEB=$90^\circ$,\\
In $\triangle ADB$ and $\triangle CEB$,\\
$\left\{\begin{array}{l}
\angle ADB=\angle CEB\\
\angle ABD=\angle CBE\\
AB=CB
\end{array}\right.$ ,\\
$\therefore$$\triangle ADB\cong \triangle CEB$ (AAS),\\
$\therefore$CE=AD=6,\\
That is, BF+EF=\boxed{6}.


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle. Point E is the midpoint of AC. A line through point E is perpendicular to AB at point F. The line BC is extended to intersect the extension of EF at point D. If EF = 1, what is the length of DF?

Solution:
\textbf{Solution:} Connect BE, \\
since $\triangle ABC$ is an equilateral triangle and point E is the midpoint of AC, \\
it follows that $\angle ABC=60^\circ$, $\angle ABE=\angle CBE=30^\circ$, \\
since EF$\perp$AB, \\
it follows that $\angle D=90^\circ-\angle ABC=30^\circ$, that is, $\angle D=\angle CBE=30^\circ$, \\
thus BE=DE, \\
in $\triangle BEF$, EF=1, \\
hence BE=2EF=2, \\
thus BE=DE=2, \\
therefore DF=EF+DE=\boxed{3},


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, AE is the bisector of the external angle $\angle BAD$ of $\triangle ABC$, BF bisects $\angle ABC$ and intersects the extended line of AE at point F. What is the measure of $\angle BFE$?

Solution:
\textbf{Solution:} Let $\angle ABF = x$, \\
since $BF$ bisects $\angle ABC$, \\
then $\angle ABC = 2\angle ABF = 2x$, \\
since $\angle C = 90^\circ$, \\
then $\angle DAB = \angle C + \angle ABC = 90^\circ+2x$, \\
since $AE$ is the bisector of $\angle BAD$, \\
then $\angle EAB = 45^\circ+x$, \\
since $\angle EAB = \angle ABF + \angle BFE$, \\
then $\angle BFE = \boxed{45^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $AC=BC$, $\angle A=40^\circ$. Observing the traces of the construction with compass and straightedge in the figure, what is the measure of $\angle BCG$?

Solution:
\textbf{Solution:} Since $AC=BC$ \\
it follows that $\angle B=\angle A=40^\circ$ \\
Thus, $\angle ACB=180^\circ-2\angle A=100^\circ$ \\
From compass and straightedge construction, we know that $CF$ is perpendicular to $AB$, and according to the congruence of medians, altitudes, and angle bisectors in an isosceles triangle, we get $\angle BCG= \frac{1}{2}\angle ACB=50^\circ$ \\
Therefore, $\angle BCG=\boxed{50^\circ}$


Question: As shown in the figure, in $ \triangle ABC$, EF is the perpendicular bisector of AB, intersecting AB at point D. If BF = 6 and CF = 2, then what is the length of AC?

Solution:
\textbf{Solution:} Since EF is the perpendicular bisector of AB, and BF = 6, \\
we have AF=BF=6, \\
and since CF = 2, \\
it follows that AC=AF+CF=\boxed{8}.


Question: As shown in the figure, in $\triangle ABC$, $AB=BC$, $AD \perp BC$ at point $D$, $CE$ bisects $\angle ACB$, intersects $AB$ at point $E$, and intersects $AD$ at point $P$. If $\angle B=x^\circ$, what is the measure of $\angle APE$?

Solution:
\textbf{Solution:} Since $AB=BC$ and $\angle B=x^\circ$,\\
it follows that $\angle BAC=\angle ACB$,\\
thus $\angle ACB= \frac{1}{2}(180^\circ-\angle B)= \frac{1}{2}(180^\circ-x^\circ)=90^\circ- \frac{1}{2}x^\circ$;\\
Since $CE$ bisects $\angle ACB$,\\
then $\angle DCP= \frac{1}{2}\angle ACB= \frac{1}{2}(90^\circ- \frac{1}{2}x^\circ)=45^\circ- \frac{1}{4}x^\circ$;\\
Since $AD \perp BC$,\\
then $\angle ADC=90^\circ$,\\
thus $\angle APE=\angle DPC=90^\circ-\angle DCP=90^\circ-(45^\circ- \frac{1}{4}x^\circ)=45^\circ+ \frac{1}{4}x^\circ.$\\
Therefore, the answer is: $\boxed{45^\circ+\frac{1}{4}x^\circ}$.


Question: As shown in the figure, $\triangle ABC \cong \triangle EDC$, $BC \perp CD$, and points A, D, E lie on the same straight line. Given that $\angle ACB = 25^\circ$, what is the measure of $\angle ADC$?

Solution:
\textbf{Solution:} Since BC$\perp$CD and $\angle$ACB$=25^\circ$,\\
it follows that $\angle$ACD$=65^\circ$,\\
Since $\triangle ABC \cong \triangle EDC$,\\
it implies AC=CE and $\angle$DCE= $\angle$ACB$=25^\circ$,\\
thus $\angle$ACE$=90^\circ$,\\
therefore $\angle$E=$\angle$CAE$= \frac{1}{2}\times(180^\circ-90^\circ)=45^\circ$,\\
consequently, $\angle$ADC=$\angle$DCE+$\angle$E$=\boxed{70^\circ}$.


Question: As shown in the figure, the four vertices of the square piece of paper \(ABCD\) lie on four parallel lines \(l_1\), \(l_2\), \(l_3\), and \(l_4\), respectively. The distances between each pair of adjacent lines among these four lines are \(h_1\), \(h_2\), and \(h_3\) sequentially (\(h_1>0\), \(h_2>0\), \(h_3>0\)). Given \(h_1=5\), \(h_2=2\), what is the area \(S\) of the square \(ABCD\)?

Solution:
\textbf{Solution:} As shown in the figure, draw AF $\perp l_{3}$ through point A, intersecting $l_{2}$, $l_{3}$ at points E, F, respectively, and draw CG $\perp l_{3}$ through point C, intersecting $l_{3}$, $l_{2}$ at points G, H, respectively,

$\because$ $l_{1}\parallel l_{2}\parallel l_{3}\parallel l_{4}$,

$\therefore$ $\angle AEB = \angle AFD = \angle DGC = \angle BHC = 90^\circ$,

$\because$ Quadrilateral ABCD is a square,

$\therefore$ AB=BC=CD=DA, $\angle BAD = \angle ADC = \angle DCB = \angle CBA = 90^\circ$,

$\therefore$ $\angle ABE = \angle DAF = \angle CDG = \angle BCH$,

$\therefore$ $\Delta ABE \cong \Delta DAF \cong \Delta CDG \cong \Delta BCH$,

$\therefore$ BE=AF=$h_{1}+h_{2}$,

$\therefore$ $S_{\Delta ABE} = S_{\Delta DAF} = S_{\Delta CDG} = S_{\Delta BCH} = \frac{1}{2}h_{1}(h_{1}+h_{2})$,

$\therefore$ $S = 4S_{\Delta ABE} + S_{\text{square }EFGH} = 4 \times \frac{1}{2}h_{1}(h_{1}+h_{2}) + h_{2}^2 = \boxed{74}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=5$, $BC=12$. Rotating $\triangle ABC$ $60^\circ$ clockwise around point $B$ results in $\triangle BDE$. When $DC$ is connected and intersects $AB$ at point $F$, what is the sum of the perimeters of $\triangle ACF$ and $\triangle BDF$?

Solution:
\textbf{Solution:} Given that $\triangle BDE$ is obtained by rotating $\triangle BCA$, \\
thus BD = BC = 12. \\
Because $\angle CBD = 60^\circ$, \\
thus $\triangle BCD$ is an equilateral triangle, \\
therefore CD = BC = 12. \\
In $\text{Rt}\triangle ABC$, with $\angle ACB = 90^\circ$, AC = 5, and BC = 12, \\
thus AB = $\sqrt{AC^{2}+BC^{2}}$ = 13, \\
therefore the perimeter of $\triangle ACF$ + the perimeter of $\triangle BDF$ = AC + CF + AF + BF + DF + BD = AC + AB + CD + BD = 5 + 13 + 12 + 12 = \boxed{42}.


Question: As shown in the figure, $\triangle ABC \cong \triangle ADE$, $\angle C = 40^\circ$, then what is the degree measure of $\angle E$?

Solution:
\textbf{Solution:}
Since $\triangle ABC \cong \triangle ADE$,\\
therefore $\angle E = \angle C = 40^\circ$.\\
Hence, $\angle E = \boxed{40^\circ}$.


Question: As shown in the figure, there is a triangular paper ABC, with point D lying on side BC. Connect AD and fold $\triangle ABD$ along AD to obtain $\triangle AED$, where DE intersects AC at point F. If point F is the midpoint of DE, $AD=9$, $EF=2.5$, and the area of $\triangle AEF$ is 9, then what is the distance from point F to BC?

Solution:
\textbf{Solution:} As shown in the figure, connect BE, intersecting AD at point O. Through point E, draw $EH\perp BC$ at point H, and from point F, draw $FG\perp BC$ at point G,\\
from the folding, it is known that AB=AE, $\angle BAO=\angle EAO$, BD=DE,\\
furthermore $\because$AO=AO,\\
$\therefore$ $\triangle BAO\cong \triangle EAO(SAS)$,\\
$\therefore$BO=EO, $\angle BOA=\angle EOA$,\\
$\therefore$ $AD\perp BE$.\\
$\because$ point F is the midpoint of DE, EF=2.5,\\
$\therefore$DF=EF=2.5, BD=DE=5,\\
$\therefore$ $\triangle ADF$ and $\triangle AEF$ have the same base and height,\\
$\therefore$ ${S}_{\triangle ADE}={S}_{\triangle AEF}+{S}_{\triangle ADF}=18$.\\
$\because$ ${S}_{\triangle ADE}=\frac{1}{2}AD\cdot OE$,\\
$\therefore$ $\frac{1}{2}\times 9\times OE=18$,\\
solving gives: $EO=4$.\\
$\therefore$ in $Rt\triangle ODE$, $OD=\sqrt{DE^{2}-OE^{2}}=\sqrt{5^{2}-4^{2}}=3$,\\
$\because$ $BE=OB+OE=8$.\\
$\therefore$ ${S}_{\triangle BDE}=\frac{1}{2}BE\cdot OD=\frac{1}{2}\times 8\times 3=12$.\\
Moreover, since ${S}_{\triangle BDE}=\frac{1}{2}BD\cdot EH=\frac{1}{2}\times 5\times EH$,\\
$\therefore$ $\frac{1}{2}\times 5\times EH=12$,\\
solving gives: $EH=4.8$.\\
$\because$ point F is the midpoint of DE, $EH\perp BC$, $FG\perp BC$,\\
$\therefore$FG is the midline of $\triangle DEH$,\\
$\therefore$ $FG=\frac{1}{2}EH=\boxed{2.4}$.


Question: As shown in the figure, \(CM\) is the median of \(\triangle ABC\), and \(AM=4\,cm\), what is the length of \(BM\)?

Solution:
Solution: Since $CM$ is the median of $\triangle ABC$ and $AM=4\text{cm}$, \\
therefore $BM=AM=4\text{cm}$,\\
so $BM=\boxed{4\text{cm}}$


Question: As shown in the figure, it is known that $AB \perp AC$, $AB=AC$, $DE$ passes through point $A$, and $CD \perp DE$, $BE \perp DE$, with the feet of the perpendiculars being points $D$ and $E$, respectively, $CD=5$, $BE=3$. What is the length of $DE$?

Solution:
\textbf{Solution:} Given $\because \angle CAB=90^\circ$,\\
$\therefore \angle DAC+\angle BAE=90^\circ$,\\
$\because CD\perp DE, BE\perp DE$,\\
$\therefore \angle D=\angle E=90^\circ$,\\
$\therefore \angle DAC+\angle DCA=90^\circ$,\\
$\therefore \angle BAE=\angle DCA$,\\
$\because AB=AC$,\\
$\therefore \triangle ADC\cong \triangle BEA(AAS)$,\\
$\therefore AD=BE, CD=AE$,\\
$\because CD=5, BE=3$,\\
$\therefore AD=3, AE=5$,\\
$\therefore DE=\boxed{8}$


Question: As shown in the figure, it is known that $AB \perp AC$, $AB = AC$, and $DE$ passes through point A, with $CD \perp DE$, $BE \perp DE$, where the feet of the perpendiculars are points D and E, respectively. Given that $CD = 5$ and $BE = 3$, what is the length of $DE$?

Solution:
\textbf{Solution:} Since $AB\perp AC$, $CD\perp DE$, and $BE\perp DE$, \\
it follows that $\angle BAC=\angle D=\angle E=90^\circ$, \\
therefore, $\angle CAD+\angle BAE=90^\circ$ and $\angle DCA+\angle CAD=90^\circ$, \\
hence, $\angle DCA=\angle EAB$, \\
In $\triangle ADC$ and $\triangle BEA$, we have \\
$\left\{\begin{array}{l}
\angle D=\angle E \\
\angle DCA=\angle EAB \\
AC=AB
\end{array}\right.$, \\
thus, $\triangle ADC\cong \triangle BEA$ by AAS, \\
which implies $DC=AE$, $AD=BE$, \\
since $DE=AD+AE$, \\
it follows that $DE=DC+BE=5+3=\boxed{8}$.


Question: As shown in the figure, the number represented by point $A$ is $x$, then what is the value of $x$?

Solution:
\textbf{Solution:} According to the question, as illustrated,

since $Rt\triangle BCD$ is an isosceles right triangle, and $BC=BD=1$,

therefore, $CD=\sqrt{BC^2+BD^2}=\sqrt{1^2+1^2}=\sqrt{2}$,

also, since arc $\overset{\frown}{AD}$ is part of a circle with $CD$ as its radius,

therefore, $CA=CD=\sqrt{2}$,

since it is at the origin of the coordinate axis,

therefore, the number represented by point $A$ is $-\sqrt{2}$, that is, $x=\boxed{-\sqrt{2}}$.


Question: In the figure, within $\triangle ABC$, $\angle C=90^\circ$. If $BC=20$, $AD$ bisects $\angle BAC$ and intersects $BC$ at point D, and $BD:CD=3:2$, then what is the length of distance $DE$ from point D to line segment $AB$?

Solution:
\textbf{Solution:} Given that $BC=20$ and $BD:CD=3:2$, \\
thus $CD=\frac{2}{5}CB=8$, \\
since $AD$ bisects $\angle BAC$ and $\angle C=90^\circ$, with $DE$ being the distance from point D to line segment $AB$, \\
hence $DE=CD=\boxed{8}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $\angle A=120^\circ$, $BC=12\,cm$. The perpendicular bisector of $AB$ intersects $BC$ at point M and intersects $AB$ at point E. The perpendicular bisector of $AC$ intersects $BC$ at point N and intersects $AC$ at point F. What is the length of $MN$?

Solution:
\textbf{Solution:} Draw $AM$ and $AN$, \\
since the perpendicular bisector of $AB$ intersects $BC$ at point $M$, and $AB$ at point $E$, and the perpendicular bisector of $AC$ intersects $BC$ at point $N$, and $AC$ at point $F$, \\
thus $AM=BM$, $AN=CN$, \\
therefore, $\angle BAM=\angle B$, $\angle CAN=\angle C$, \\
since $AB=AC$ and $\angle A=120^\circ$, \\
therefore, $\angle B=\angle C=30^\circ$, \\
thus, $\angle AMN=\angle ANM=60^\circ$, \\
therefore, $\triangle AMN$ is an equilateral triangle, \\
thus, $AM=MN=AN$, \\
thus, $BM=MN=BN$, \\
since $BC=12\text{cm}$, \\
therefore, $MN=\frac{1}{3}BC=\boxed{4\text{cm}}$.


Question: As shown in the figure, a set of triangular boards are stacked together as depicted. What is the degree measure of $\angle \alpha$ in the figure?

Solution:
\textbf{Solution:} As shown in the figure,\\
$\because$ $\angle DBC = 90^\circ$ and $\angle C = 45^\circ$ and $\angle A = 30^\circ$,\\
$\therefore$ $\angle BDC = 90^\circ - \angle C = 45^\circ$,\\
$\therefore$ $\angle ADE = \angle BDC = 45^\circ$,\\
$\therefore$ $\angle \alpha = \angle A + \angle ADE = 30^\circ + 45^\circ = \boxed{75^\circ}$.


Question: As shown in the figure, $\triangle ABD$ and $\triangle CBD$, $\angle ADB=90^\circ$, $\angle ABD=\angle DBC$, $AD=DC=1$. If $AB=4$, what is the length of $BC$?

Solution:
\textbf{Solution:} Draw DE$\perp$AB at point E, and DF$\perp$BC, intersecting the extension of BC at point F, \\
$\therefore$ $\angle$CFD$=90^\circ$, \\
$\because$ $\angle$ABD$=\angle$DBC, \\
$\therefore$ DE=DF; \\
In right triangles $\triangle$DEB and $\triangle$DFB, \\
$\left\{\begin{array}{l}BD=BD\\ DE=DF\end{array}\right.$\\
$\therefore$ right $\triangle$DEB$\cong$ right $\triangle$DFB (HL)\\
$\therefore$ BE=BF; \\
In right $\triangle$ABD, \\
$BD=\sqrt{AB^{2}-AD^{2}}=\sqrt{4^{2}-1^{2}}=\sqrt{15}$; \\
$\because$ $S_{\triangle ABC}=\frac{1}{2}AD\cdot BD=\frac{1}{2}AB\cdot DE$ \\
$\therefore$ DE=DF$=\frac{1\times \sqrt{15}}{4}=\frac{\sqrt{15}}{4}$, \\
In right $\triangle$BDE, \\
$BE=BF=\sqrt{BD^{2}-DE^{2}}=\sqrt{\left(\sqrt{15}\right)^{2}-\left(\frac{\sqrt{15}}{4}\right)^{2}}=\frac{15}{4}$, \\
$CF=\sqrt{DC^{2}-DF^{2}}=\sqrt{1^{2}-\left(\frac{\sqrt{15}}{4}\right)^{2}}=\frac{1}{4}$ \\
$\therefore$ $BC=BF-CF=\frac{15}{4}-\frac{1}{4}=\boxed{\frac{7}{2}}$.


Question: As shown in the figure, $\triangle ABC \cong \triangle ADE$, the extension of $BC$ intersects $DE$ at point $F$, $\angle B=30^\circ$, $\angle AED=110^\circ$, $\angle DAC=10^\circ$, what is the measure of $\angle DFB$?

Solution:
\textbf{Solution:} Since $\triangle ABC \cong \triangle ADE$,\\
it follows that $\angle AED = \angle ACB = 110^\circ$, and $\angle D = \angle B = 30^\circ$,\\
therefore $\angle ACF = 180^\circ - 110^\circ = 70^\circ$,\\
Given the sum of angles in a triangle is $180^\circ$, we have: $\angle DAC + \angle ACF = \angle D + \angle DFC$,\\
thus $\angle DFC = 70^\circ + 10^\circ - 30^\circ = \boxed{50^\circ}$


Question: As shown in the diagram, the perimeter of $\triangle ABC$ is 19. Points $D$ and $E$ are on side $BC$. The bisector of $\angle ABC$ is perpendicular to $AE$ at point $N$. The bisector of $\angle ACB$ is perpendicular to $AD$ at point $M$. If $BC = 7$, what is the length of $DE$?

Solution:
\textbf{Solution:} Given that the angle bisector of $\angle ABC$ is perpendicular to $AE$, with the foot of the perpendicular being point $N$, \\
it follows that $\angle ABN=\angle EBN$, and $\angle BNA=\angle BNE=90^\circ$, \\
In $\triangle BNA$ and $\triangle BNE$, \\
since $\left\{\begin{array}{l}\angle ABN=\angle EBN \\ BN=BN \\ \angle BNA=\angle BNE\end{array}\right.$, \\
it implies $\triangle BNA\cong \triangle BNE\left(ASA\right)$, \\
thus $BA=BE$, \\
Similarly, it can be proved that $\triangle CAM\cong \triangle CDM\left(ASA\right)$, \\
hence, $AC=DC$.\\
Given that the perimeter of $\triangle ABC$ is 19, \\
it follows that $AB+AC+BC=19$, \\
Since $BA=BE$ and $AC=DC$, \\
it follows that $BE+DC+BC=19$, \\
thus $BD+DE+DE+EC+BC=19$, \\
Given $BC=7$, \\
it follows that $BD+DE+DE+EC=19-BC=12$, \\
thus $BD+DE+EC+DE=BC+DE=12$, \\
which means $DE=12-BC=12-7=\boxed{5}$.


Question: As shown in the figure, $AB \parallel CD$, $\angle 1 = 30^\circ$, $\angle 2 = 40^\circ$, then what is the degree measure of $\angle 3$?

Solution:
\textbf{Solution:} Given $\because$ AB$\parallel$CD, $\angle 1=30^\circ$,\\
$\therefore$ $\angle A=\angle 1=30^\circ$,\\
Also, $\because$ $\angle 2=40^\circ$,\\
$\therefore$ $\angle 3=\angle A+\angle 2=70^\circ$,\\
Therefore, $\angle 3=\boxed{70^\circ}$


Question: As shown in the diagram, in the equilateral $\triangle ABC$, $D$ is the midpoint of side $BC$. An arc is drawn with $A$ as the center and $AD$ as the radius, intersecting side $AC$ at point $E$. What is the degree measure of $\angle ADE$?

Solution:
\textbf{Solution:} Let's consider in equilateral $\triangle ABC$, with D being the midpoint of side BC,\\
$\therefore$ $\angle DAC=\frac{1}{2}\angle BAC=\frac{1}{2}\times 60^\circ=30^\circ$,\\
According to the problem statement, we know $AD=AE$,\\
$\therefore$ $\angle ADE=\angle AED$,\\
$\therefore$ $\angle ADE=\frac{1}{2}\times (180^\circ-\angle DAC)=\frac{1}{2}\times (180^\circ-30^\circ)=\boxed{75^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $BC=4\text{cm}$. A point $E$ is chosen on $AC$ such that $EC=BC$. A line $EF$ is drawn perpendicular to $AC$, and $CF$ is connected so that $CF=AB$. If $EF=10\text{cm}$, what is the length of $AE$?

Solution:
\textbf{Solution:} Given that $EF\perp AC$,\\
it follows that $\angle CEF = \angle ACB = 90^\circ$,\\
In right $\triangle ACB$ and right $\triangle FEC$,\\
$\left\{\begin{array}{l}
AB = FC \\
BC = CE
\end{array}\right.$,\\
hence, right $\triangle ACB \cong$ right $\triangle FEC$,\\
thus, AC = $EF = 10\text{cm}$, EC = $BC = 4\text{cm}$,\\
hence, AE = AC - EC = $\boxed{6\text{cm}}$,


Question: As shown in the figure, in $\triangle ABC$, $CD$ is the median to the hypotenuse $AB$. If $\angle A=20^\circ$, what is the measure of $\angle BDC$?

Solution:
\textbf{Solution:} Given that $\angle ACB=90^\circ$ and $CD$ is the median to the hypotenuse $AB$,\\
it follows that $BD=CD=AD$,\\
therefore, $\angle A=\angle DCA=20^\circ$,\\
thus, $\angle BDC=\angle A+\angle DCA=20^\circ+20^\circ=\boxed{40^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=85^\circ$, $\angle B=30^\circ$. Arcs are drawn with radii greater than half of the length of $AB$, centered at points A and B, respectively. The arcs intersect at points M and N, intersect $BC$ at point D. When $AD$ is connected, what is the measure of $\angle CAD$?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $\angle B=30^\circ$, $\angle C=85^\circ$,\\
hence $\angle BAC=180^\circ-\angle B-\angle C=65^\circ$,\\
From the construction, it is known that $MN$ is the perpendicular bisector of $AB$,\\
hence $DA=DB$,\\
thus $\angle DAB=\angle B=30^\circ$,\\
therefore, $\angle CAD=\angle BAC-\angle DAB=\boxed{35^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, it is known that points D, E, and F are the midpoints of sides BC, AD, and CE, respectively, and the area of $\triangle ABC$ is 4. What is the area of $\triangle BEF$?

Solution:
\textbf{Solution:} Given that point D is the midpoint of BC, and the area of $\triangle ABC$ is 4,\\
it follows that the area of $\triangle ABD = \triangle ADC = \frac{1}{2}\times$ the area of $\triangle ABC = \frac{1}{2}\times4=2$,\\
Given that point E is the midpoint of AD,\\
it follows that the area of $\triangle BDE = \frac{1}{2}\times$ the area of $\triangle ABD = \frac{1}{2}\times2=1$, and the area of $\triangle CDE = \frac{1}{2}\times$ the area of $\triangle ADC = \frac{1}{2}\times2=1$,\\
therefore, the area of $\triangle BEC = $ the area of $\triangle BED + \triangle CDE = 2$,\\
Given that point F is the midpoint of CE,\\
it follows that the area of $\triangle BEF = \frac{1}{2}\times$the area of $\triangle BEC = \frac{1}{2}\times2 = \boxed{1}$.


Question: As shown in the figure, the angle bisectors of $\angle ABC$ and $\angle ACB$ intersect at point $F$. Through point $F$, draw $EG \parallel BC$ intersecting $AB$ and $AC$ at points $E$ and $G$, respectively. If $BE=6$ and $CG=10$, what is the length of segment $EG$?

Solution:
\textbf{Solution:} Since $EG\parallel BC$,

it follows that $\angle EFB=\angle FBC$,

since $BF$ bisects $\angle ABC$,

it follows that $\angle EBF=\angle FBC$,

therefore, $\angle EBF=\angle EFB$,

hence $EF=BE$,

similarly, $FG=GC$,

therefore, $EG=EF+FG=BE+GC=\boxed{16}$.


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle with a side length of 4. $AD$ is the median on side $BC$, and $F$ is a moving point on segment $AD$. $E$ is a point on side $AC$. If $AE = 2$, when the sum of $EF+CF$ attains its minimum value, what is the measure of $\angle ECF$?

Solution:
\textbf{Solution:} Draw $EM \parallel BC$ passing through $E$ and intersecting $AD$ at $N$,\\
since $AC=4$ and $AE=2$,\\
thus $EC=2=AE$,\\
thus $AM=BM=2$,\\
thus $AM=AE$,\\
since $AD$ is the median of side $BC$, and $\triangle ABC$ is an equilateral triangle,\\
thus $AD \perp BC$,\\
since $EM \parallel BC$,\\
thus $AD \perp EM$,\\
since $AM=AE$,\\
thus $E$ and $M$ are symmetric with respect to $AD$,\\
draw $CM$ intersecting $AD$ at $F$, and connect $EF$,\\
at this moment, the sum of $EF+CF$ is minimized,\\
since $\triangle ABC$ is an equilateral triangle,\\
thus $\angle ACB=60^\circ$, and $AC=BC$,\\
since $AM=BM$,\\
thus $\angle ECF=\frac{1}{2}\angle ACB=30^\circ$,\\
therefore, the answer is: $\boxed{30^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AD \perp BC$ at point $D$. Point $E$ is on $AC$, and $BE$ intersects $AD$ at point $F$. If $BF=AC$ and $DF=DC$, what is the sum of $\angle 1$ and $\angle 2$?

Solution:
\textbf{Solution:} Since $AD\perp BC$ at point $D$,\\
it follows that $\angle BDF=\angle ADC=90^\circ$,\\
In $Rt\triangle BDF$ and $Rt\triangle ADC$,\\
$\left\{\begin{array}{l}BF=AC\\ DF=DC\end{array}\right.$,\\
therefore $Rt\triangle BDF \cong Rt\triangle ADC\left(HL\right)$,\\
thus $\angle DBF=\angle 2$, $BD=AD$,\\
hence $\angle DBA=\angle DAB=45^\circ$,\\
thus $\angle 1+\angle 2=\angle 1+\angle DBF=\angle DBA=45^\circ$,\\
therefore the sum of $\angle 1$ and $\angle 2$ is $\boxed{45^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $AD$ is the altitude, $E$ is the midpoint of $AB$. Given that the area of $\triangle ABC$ is $8$, what is the area of $\triangle ADE$?

Solution:
\textbf{Solution:} Given that $AB=AC$ and $AD$ is the altitude, \\
thus $BD=CD$, $AD\perp BC$, \\
thus $S_{\triangle ABD}=S_{\triangle ACD}=\frac{1}{2} S_{\triangle ABC}=4$, \\
given that $E$ is the midpoint of $AB$, \\
thus $AE=BE$, \\
thus $S_{\triangle ADE}=S_{\triangle BDE}=\frac{1}{2} S_{\triangle ABD}=\boxed{2}$.


Question: As shown in the diagram, there is a right-angled triangle paper piece, where $\angle C=90^\circ$, $BC=6$, and $AC=8$. Now, fold $\triangle ABC$ as shown in the diagram, making points A and B coincide. The folding line is DE. What is the length of $CE$?

Solution:
\textbf{Solution:} In the right-angled triangle $ACB$, we have $AC=8$ and $BC=6$,
\[
\therefore AB= \sqrt{AC^{2}+BC^{2}}=\sqrt{6^{2}+8^{2}}=10.
\]
According to the principle of folding invariance, $\triangle EDA\cong \triangle EDB$,
\[
\therefore EA=EB.
\]
Therefore, in the right-angled triangle $BCE$, let $CE=x$, then $BE=AE=8-x$,
\[
\therefore BE^{2}=BC^{2}+CE^{2},
\]
\[
\therefore (8-x)^{2}=6^{2}+x^{2},
\]
solving this, we get $x=\boxed{\frac{7}{4}}$.


Question: As shown in the figure, E is the midpoint of the hypotenuse AB of $\triangle ABC$ and $\triangle ABD$, with $AC = BC$ and $\angle DBA = 25^\circ$. What is the measure of $\angle DCE$?

Solution:
\textbf{Solution:} Given that point $E$ is the midpoint of the hypotenuse $AB$ of the right triangle $ABD$,\\
hence, $ED=EB=\frac{1}{2}AB$,\\
therefore, $\angle EDB=\angle DBA=25^\circ$,\\
thus, $\angle DEA=\angle EDB+\angle DBA=50^\circ$,\\
since point $E$ is also the midpoint of the hypotenuse $AB$ of the right triangle $ABC$, and $AC=BC$,\\
hence, $EC=\frac{1}{2}AB$, and $CE\perp AB$,\\
thus, $\angle AEC=90^\circ$,\\
therefore, $\angle DEC=\angle DEA+\angle AEC=140^\circ$, and since $ED=EC$,\\
hence, $\angle DCE=\frac{1}{2}\times \left(180^\circ-140^\circ\right)=\boxed{20^\circ}$.


Question: As shown in the figure, in quadrilateral ABCD, connect AC and BD. If $\triangle ABC$ is an equilateral triangle, with $AB=BD$ and $\angle ABD=20^\circ$, then what is the measure of $\angle BDC$?

Solution:
\textbf{Solution:} Given an equilateral $\triangle ABC$ where $AB=BC$, $\angle ABC=60^\circ$, \\
since $AB=BD$, $\angle ABD=20^\circ$, \\
it follows that $BD=BC$, $\angle CBD=60^\circ-20^\circ=40^\circ$, \\
thus, $\angle BDC=(180^\circ-40^\circ)\div 2=\boxed{70^\circ}$.


Question: As shown in the figure, all the shaded quadrilaterals are squares, and all triangles are right-angled triangles. If the areas of squares $A$, $B$, and $C$ are $4$, $8$, and $6$ respectively, what is the area of square $D$?

Solution:
Solution: As shown in the diagram,\\
since all the shaded quadrilaterals are squares and all the triangles are right-angled triangles, the areas of squares A, B, C are 4, 8, 6, respectively,\\
therefore, EF$^{2}$ = 4 + 8 = 12, GH$^{2}$ = area of square D - 6, with EF$^{2}$ = GH$^{2}$,\\
therefore, GH$^{2}$ = area of square D - 6 = 12\\
Upon solving: the area of square D = $\boxed{18}$.


Question: As illustrated, in triangle $ABC$, $\angle ACB=90^\circ$, $\angle A=20^\circ$, $\triangle ABC \cong \triangle A'B'C'$. If $A'B'$ precisely passes through point B, and $A'C'$ intersects $AB$ at $D$, what is the measure of $\angle BDC$?

Solution:
\textbf{Solution:} Given $\triangle ABC \cong \triangle A'B'C'$,

it follows that $\angle A = \angle A' = 20^\circ$, $\angle ACB = \angle A'CB' = 90^\circ$, CB = CB',

since $\angle B' = \angle CBA = 70^\circ$,

and because CB = CB',

it follows that $\angle B' = \angle B'BC = 70^\circ$,

thus $\angle BCB' = 180^\circ - 70^\circ - 70^\circ = 40^\circ$,

hence $\angle BCD = 90^\circ - 40^\circ = 50^\circ$,

consequently $\angle BDC = 180^\circ - \angle CBD - \angle BCD = 180^\circ - 50^\circ - 70^\circ = \boxed{60^\circ}$.


Question: As shown in the figure, a large square ABCD is formed by four congruent right-angled triangles and a small square. A perpendicular line DF is drawn from point D to intersect the extension of the diagonal EF of the small square at point G. Connect BG. If the area of the large square is 5 times the area of the small square, what is the value of $ \frac{BG}{BE}$?

Solution:
Solution:Let the extension lines of $BE$ and $GD$ intersect at point $M$, and $BM$ intersect $CF$ at point $N$, as shown in the diagram.\\
Suppose the side length of the small square is $x$, then the side length of the large square is $\sqrt{5}x$,\\
Let $CN = t$, then $BE = t$,\\
In $\triangle BCN$, since $CN = t$, $BN = t + x$, $BC = \sqrt{5}x$,\\
Therefore, $x^2 + (t + x)^2 = (\sqrt{5}x)^2$,\\
Solving gives $t = x$,\\
Therefore, $BE = CN = x$,\\
Since $ED$ is the diagonal of the small square,\\
Therefore, $\angle FEN = \angle EFN = 45^\circ$,\\
Therefore, $\angle GFD = 45^\circ$,\\
Since $GD \perp DF$,\\
Therefore, $\triangle GDF$ is an isosceles right triangle,\\
Therefore, $\angle FGD = 45^\circ$, $DG = DF = x$,\\
Since $\angle GEM = \angle EGM = 45^\circ$,\\
Therefore, $\triangle MGE$ is an isosceles right triangle,\\
Therefore, $ME = MG = 2x$,\\
In $\triangle BMG$, since $BM = 3x, GM = 2x$,\\
Therefore, $BG= \sqrt{(3x)^2 + (2x)^2} = \sqrt{13}x$,\\
Therefore, $\frac{BG}{BE} = \frac{\sqrt{13}x}{x} = \boxed{\sqrt{13}}$.


Question: As shown in the figure, point $D$ is inside $\triangle ABC$, where $CD$ bisects $\angle ACB$, $BD\perp CD$, and $\angle A = \angle ABD$. If $AC = 7$ and $BC = 4$, what is the length of $BD$?

Solution:
\textbf{Solution:} Extend BD to meet AC at point E,\\
$\because$ $\angle A = \angle ABD$,\\
$\therefore$ BE = AE,\\
$\because$ BD $\perp$ CD,\\
$\therefore$ BE $\perp$ CD,\\
$\because$ CD bisects $\angle ACB$,\\
$\therefore$ $\angle BCD = \angle ECD$,\\
$\therefore$ $\angle EBC = \angle BEC$,\\
$\therefore$ $\triangle BEC$ is an isosceles triangle,\\
$\therefore$ BC = CE,\\
$\because$ BE $\perp$ CD,\\
$\therefore$ 2BD = BE,\\
$\because$ AC = 7, BC = 4,\\
$\therefore$ CE = 4,\\
$\therefore$ AE = AC - EC = 7 - 4 = 3,\\
$\therefore$ BE = 3,\\
$\therefore$ BD = $\boxed{\frac{3}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, $AC = BC$, $BE \perp CE$ at point $E$, and $AD \perp CE$ at point $D$. Given $DE = 5$ and $AD = 9$, what is the length of $BE$?

Solution:
\textbf{Solution}: We have $\angle ACB = 90^\circ$ and $BE \perp CE$,\\
therefore, $\angle BCE + \angle ACD = 90^\circ$ and $\angle BCE + \angle CBE = 90^\circ$,\\
thus, $\angle ACD = \angle CBE$,\\
in $\triangle BEC$ and $\triangle CDA$,\\
\[
\left\{ \begin{array}{l}
\angle BEC = \angle CDA \\
\angle CBE = \angle ACD \\
BC = AC
\end{array} \right.,
\]
therefore, $\triangle ACD \cong \triangle CBE$ (by AAS),\\
thus, EC = AD = 9, and BE = DC,\\
since DE = 5,\\
thus, CD = EC - DE = 4,\\
therefore, BE = \boxed{4}.


Question: As shown in the diagram, given that the area of the square $ABCD$ is 64 square centimeters and $DE = 10$ cm, what is the length of $CE$?

Solution: \textbf{Solution:} Since the area of square ABCD is 64 square centimeters, $\therefore$ $\angle ADC=90^\circ$ and $DC=8$ cm. Since $DE=10$ cm, $\therefore$ $CE=\sqrt{DE^{2}+DC^{2}}=\sqrt{8^{2}+10^{2}}=\boxed{2\sqrt{41}}$ (cm).


Question: As shown in the figure, $ \angle ACD=90^\circ$, $ \angle D=15^\circ$, point B is on the perpendicular bisector of $ AD$, if $ AC=4$, then what is the length of $ AB$?

Solution:
\textbf{Solution:} Since point B is on the perpendicular bisector of $AD$,\\
it follows that $AB=BD$,\\
thus, $\angle BAD=\angle D=15^\circ$,\\
therefore, $\angle ABC=\angle BAD+\angle D=30^\circ$,\\
since $\angle ACD=90^\circ$ and $AC=4$,\\
it follows that $AB=2AC=\boxed{8}$;


Question: As shown in the diagram, in $\triangle ABC$, $\angle B=\angle C$, $AB=3$, what is the length of $AC$?

Solution:
\textbf{Solution:} Since $\angle B = \angle C$ and AB=3, \\
it follows that AB=AC=\boxed{3}.


Question: As shown in the diagram, $AB \perp BC$ at point $B$, $AE \perp DE$ at point $E$, $AB = AE$, $\angle ACB = \angle ADE$, $\angle ACD = 65^\circ$, $\angle BAD = 75^\circ$. What is the measure of $\angle BAE$?

Solution:
\textbf{Solution:} Given that $AB\perp BC$ and $AE\perp DE$,\\
thus $\angle B=\angle E=90^\circ$,\\
in $\triangle ABC$ and $\triangle AED$,\\
\[
\left\{\begin{array}{l}
\angle B=\angle E\\
\angle ACB=\angle ADE\\
AB=AE
\end{array}\right.,
\]
therefore, $\triangle ABC \cong \triangle AED$ (by AAS),\\
hence $\angle BAC=\angle EAD$, $AC=AD$\\
Given $\angle ACD=65^\circ$,\\
it follows that $\angle ACD=\angle ADC=65^\circ$\\
Hence, $\angle CAD=180^\circ-\angle ACD-\angle ADC=50^\circ$\\
Given $\angle BAD=75^\circ$,\\
it implies that $\angle BAC=\angle EAD=\angle BAD-\angle CAD=75^\circ-50^\circ=25^\circ$,\\
Therefore, $\angle BAE=\angle BAD+\angle DAE=\angle BAD+\angle BAC=75^\circ+25^\circ=\boxed{100^\circ}$.


Question: As shown in the figure, $ \angle A=15^\circ$, $ AB=BC=CD=DE=EF$, what is the value of $ \angle DEF$?

Solution:
\textbf{Solution:} Given that $AB=BC=CD=DE=EF$, \\
$\therefore \angle BCA=\angle A=15^\circ$, \\
$\therefore \angle CBD=\angle BDC=\angle BCA+\angle A=15^\circ+15^\circ=30^\circ$, \\
$\therefore \angle ECD=\angle CED=\angle A+\angle CDB=45^\circ$, \\
$\therefore \angle EDF=\angle EFD=\angle A+\angle CED=60^\circ$, \\
$\therefore \angle DEF=180^\circ-60^\circ-60^\circ=\boxed{60^\circ}$.


Question: As shown in the figure, $\triangle ABC$ is similar to $\triangle DEF$ with point O as the center of similarity. If $OD=3OA$ and the area of $\triangle ABC$ is $4$, what is the area of $\triangle DEF$?

Solution:
\textbf{Solution:}\\
Since $\triangle ABC$ and $\triangle DEF$ are similar,\\
thus $AC\parallel DF$,\\
thus $\triangle OAC\sim \triangle ODF$,\\
thus $\frac{AC}{DF}=\frac{OA}{OD}=\frac{1}{3}$,\\
thus $\frac{{S}_{\triangle ABC}}{{S}_{\triangle DEF}}={(\frac{AC}{DF})}^{2}=\frac{1}{9}$,\\
since ${S}_{\triangle ABC}=4$,\\
therefore ${S}_{\triangle DEF}=\boxed{36}$.


Question: As shown in the diagram, in the Cartesian coordinate system, $\triangle OAB$ is scaled up with the origin $O$ as the center of similarity to obtain $\triangle OCD$. If $B(0, 1)$ and $D(0, 3)$, what is the ratio of the areas of $\triangle OAB$ to $\triangle OCD$?

Solution:
\textbf{Solution:} Let $\triangle OAB$ be enlarged from the origin $O$ to form $\triangle OCD$, where B(0,1), D(0,3),\\
$\therefore OB\colon OD=1\colon 3$\\
$\therefore$ The ratio of the areas of $\triangle OAB$ and $\triangle OCD$ is $\boxed{1\colon 9}$


Question: As shown in the figure, $a\parallel b\parallel c$, $\frac{AC}{CE}=\frac{1}{2}$, and $DF=12$, what is the length of $BD$?

Solution:
\textbf{Solution:} Since $a\parallel b\parallel c$,\\
it follows that $\frac{AC}{CE}=\frac{BD}{DF}=\frac{1}{2}$,\\
since $DF=12$,\\
it follows that $\frac{BD}{12}=\frac{1}{2}$,\\
therefore, $BD=\boxed{6}$.


Question: As shown in the figure, in $ \triangle ABC$, $ DE\parallel BC$. If $ \frac{AD}{DB}=\frac{2}{3}$, then what is the value of $ \frac{DE}{BC}$?

Solution:
\textbf{Solution:} Given that $\because \frac{AD}{DB}=\frac{2}{3}$,\\
$\therefore \frac{AD}{AB}=\frac{2}{5}$,\\
since $DE\parallel BC$,\\
$\therefore \frac{DE}{BC}=\frac{AD}{AB}=\boxed{\frac{2}{5}}$


Question: As shown in the figure, $\triangle ABC$ is translated downward along the line containing the median $AD$ from side $BC$ to the position of $\triangle ABC$. It is known that the area of $\triangle ABC$ is $9$, and the area of the shaded triangle is $4$. If $AA'=1$, then what is the length of $A'D$?

Solution:
\textbf{Solution:} Let $AB$, $AC$ intersect with $BE$ at points $E$, $F$, \\
since $\triangle ABC$ is translated downwards along the line containing the median $AD$ from the side $BC$ to the position of $\triangle ABC$, \\
therefore $\triangle ABC \sim \triangle AEF$, \\
therefore $\frac{S_{\triangle ABC}}{S_{\triangle AEF}}=\left(\frac{AD}{AD}\right)^2$, \\
since the area of $\triangle ABC$ is $9$, and the area of the shaded part of the triangle is $4$, $AA=1$, \\
therefore $\frac{9}{4}=\left(\frac{AD+1}{AD}\right)^2$, \\
solving this gives $AD=\boxed{2}$.


Question: As shown in the diagram, the straight lines $l_{1} \parallel l_{2} \parallel l_{3}$ intersect the straight lines $m$ and $n$ at points A, B, C, D, E, F respectively. Given that $AB=4$, $BC=6$, $DE=2$, what is the length of $EF$?

Solution:
\textbf{Solution:} Since $l_1\parallel l_2\parallel l_3$,\\
it follows that $\frac{AB}{BC}=\frac{DE}{EF}$,\\
given $AB=4, BC=6, DE=2$,\\
thus $\frac{4}{6}=\frac{2}{EF}$,\\
solving gives $EF=\boxed{3}$.


Question: As shown in the figure, it is known that $AD$ is the angle bisector of $\triangle ABC$, and $DE \parallel AB$ intersects $AC$ at $E$. If $\frac{AE}{EC}=\frac{3}{5}$, then what is the value of $\frac{AC}{AB}$?

Solution:
\textbf{Solution:} Since $DE\parallel AB$,\\
$\therefore$ $\angle ADE=\angle BAD$,\\
Since $AD$ is the bisector of $\angle A$ in $\triangle ABC$,\\
$\therefore$ $\angle BAD=\angle EAD$,\\
$\therefore$ $\angle EAD=\angle ADE$,\\
$\therefore$ $AE=DE$,\\
Since $\frac{AE}{EC}=\frac{3}{5}$,\\
$\therefore$ $\frac{EC}{DE}=\frac{5}{3}$,\\
Since $DE\parallel AB$,\\
$\therefore$ $\triangle CDE\sim \triangle CBA$,\\
$\therefore$ $\frac{DE}{AB}=\frac{EC}{AC}$,\\
$\therefore$ $\frac{AC}{AB}=\frac{EC}{DE}=\boxed{\frac{5}{3}}$.


Question: As shown in the figure, the line of symmetry of rectangle ABCD intersects AB at point E and CD at point F respectively. If rectangle AEFD is similar to rectangle ABCD, what is the value of AB$\colon$ BC?

Solution:
\textbf{Solution:} Since $ABCD$ is a rectangle,\\
it follows that $AD=BC$,\\
since the axes of symmetry of rectangle $ABCD$ intersect $AB$ at point $E$, and $CD$ at point $F$,\\
it follows that $AE=\frac{1}{2}AB$,\\
since rectangle $AEFD$ is similar to rectangle $ABCD$,\\
it follows that $\frac{AB}{BC}=\frac{AD}{AE}$,\\
thus $\frac{AB}{BC}=\frac{BC}{\frac{1}{2}AB}$,\\
hence $\frac{1}{2}AB^2=BC^2$,\\
therefore $AB^2=2BC^2$,\\
thus $AB=\sqrt{2}BC$,\\
hence $AB\colon BC=\boxed{\sqrt{2}}$.


Question: According to legend, the ancient Greek mathematician and astronomer Thales once used the principle of similar triangles by erecting a wooden pole at the top of the shadow of a pyramid. With the help of sunlight, he constructed two similar triangles to measure the height of the pyramid. Given that the wooden pole $EF$ is $2m$ long and its shadow $FD$ is $3m$ long, and $OA$ is measured to be $201m$, what is the height of the pyramid $BO$?

Solution:
\textbf{Solution:} Let the height of the pyramid be $x$ meters, \\
According to the problem, we have: $\frac{OB}{OA}=\frac{EF}{FD}$, \\
$\frac{x}{201}=\frac{2}{3}$, \\
Solving this equation, we get: $x=\boxed{134}$.


Question: As shown in the diagram, a point A is selected on the opposite side of the river. Then, on this side of the river, points B and C are chosen such that $AB \perp BC$. Another point E is selected such that $EC \perp BC$, and the line of sight determines that BC and AE intersect at point D. At this moment, it is measured that $BD=120\,m$, $DC=60\,m$, and $EC=50\,m$. What is the distance between the two sides of the river, that is, the length of AB?

Solution:
\textbf{Solution:} Given that $AB\perp BC$, $EC\perp BC$,\\
$\therefore AB\parallel CE$\\
$\therefore \triangle ABD\sim \triangle ECD$\\
$\therefore \frac{BD}{DC}=\frac{AB}{EC}$\\
Given $BD=120m$, $DC=60m$, $EC=50m$,\\
$\therefore AB=\frac{BD\cdot EC}{DC}=\frac{120\times 50}{60}=\boxed{100}m$


Question: As shown in the figure, $AD \parallel BE \parallel CF$, $AB=3$, $AC=9$, $DE=2$, then what is the value of $EF$?

Solution:
\textbf{Solution:} Given that $AD\parallel BE\parallel CF$,\\
it follows that $\frac{AB}{BC}=\frac{DE}{EF}$,\\
Given that $AB=3$, $AC=9$, $DE=2$,\\
we have $BC=6$,\\
thus, $\frac{3}{6}=\frac{2}{EF}$,\\
therefore, $EF=\boxed{4}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $E$ is the midpoint of $AB$, and $EC$ intersects $BD$ at point $F$. What is the ratio of the area of $\triangle BEF$ to the area of $\triangle BCF$?

Solution:
\textbf{Solution:} Given that $ABCD$ is a parallelogram,\\
it follows that $BE\parallel CD$, \\
thus $\angle BEC=\angle DCF$, $\angle EBF=\angle CDF$, \\
which means $\triangle EBF \sim \triangle CDF$,\\
hence $\frac{EF}{CF}=\frac{BE}{CD}=\frac{1}{2}$,\\
For triangles $BEF$ and $BCF$ taking $EF$ and $CF$ as the base respectively, the heights are the same, \\
therefore $S_{\triangle BEF} : S_{\triangle BCF} = EF : CF = 1:2$, \\
Thus, the answer is: $\boxed{1:2}$.


Question: As shown in the figure, $\triangle ABC\sim \triangle EDC$, given that $\angle ABC=90^\circ$, $\frac{AB}{BC}=\frac{1}{2}$, what is the value of $\frac{AE}{BD}$?

Solution:
\textbf{Solution:} Given that $\triangle ABC \sim \triangle EDC$,\\
it follows that $\frac{AC}{EC}=\frac{BC}{DC}$ and $\angle ACB=\angle ECD$,\\
hence $\frac{AC}{EC}=\frac{BC}{DC}$ and $\angle ACE=\angle BCD$,\\
implying $\triangle AEC \sim \triangle BDC$,\\
which leads to $\frac{AE}{BD}=\frac{AC}{BC}$,\\
since $\angle ABC=90^\circ$ and $\frac{AB}{BC}=\frac{1}{2}$,\\
it results in $\frac{AB^2}{BC^2}=\frac{1}{4}$,\\
thus $\frac{AB^2+BC^2}{BC^2}=\frac{5}{4}$, which means $\frac{AC^2}{BC^2}=\frac{5}{4}$,\\
therefore $\frac{AC}{BC}=\frac{\sqrt{5}}{2}$,\\
concluding that $\frac{AE}{BD}=\boxed{\frac{\sqrt{5}}{2}}$.


Question: As shown in the figure, in parallelogram $ABCD$, $DE: EC = 4: 1$, and $AE$ intersects $BD$ at point $F$, then what is the ratio of $S_{\triangle DEF}$ to $S_{\triangle BAF}$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\
$\therefore$DC$\parallel$AB, DC=AB, \\
Since DE: EC=4: 1, \\
$\therefore$ $\frac{DE}{DC}=\frac{4}{5}$, \\
$\therefore$ $\frac{DE}{AB}=\frac{4}{5}$, \\
Since DE$\parallel$AB, \\
$\therefore$ $\triangle DEF\sim \triangle BAF$, \\
$\therefore$ $\frac{S_{\triangle DEF}}{S_{\triangle BAF}}=\left(\frac{DE}{AB}\right)^2=\boxed{\frac{16}{25}}$


Question: As shown in the figure, in $\triangle ABC$, points $D$ and $E$ are on sides $AB$ and $AC$, respectively, with $DE \parallel BC$ and $BC = 6$. Given that the area of quadrilateral $BCED$ is $8$ times the area of $\triangle ADE$, what is the length of $DE$?

Solution:
\textbf{Solution:} Since ${S}_{\text{quadrilateral} BCED}=8{S}_{\triangle ADE}$,
it follows that ${S}_{\triangle ABC}={S}_{\triangle ADE}+{S}_{\text{quadrilateral} BCED}={S}_{\triangle ADE}+8{S}_{\triangle ADE}=9{S}_{\triangle ADE}$,
thus $\frac{{S}_{\triangle ADE}}{{S}_{\triangle ABC}}=\frac{1}{9}$,
Since DE$\parallel$BC, it follows that $\angle$ADE=$\angle$B, $\angle$AED=$\angle$C,
which implies $\triangle ADE\sim \triangle ABC$,
therefore $\frac{{S}_{\triangle ADE}}{{S}_{\triangle ABC}}=\left(\frac{DE}{BC}\right)^{2}=\frac{1}{9}$,
hence $\frac{DE}{BC}=\frac{1}{3}$,
Since BC＝6, it follows that $DE=\frac{1}{3}BC=\frac{1}{3}\times 6=\boxed{2}$.


Question: As shown in the figure, Master Zhao uses a piece of triangular tinplate to cut out a piece of square tinplate for reserve. In $\triangle ABC$, $BC=120$, the height $AD=80$, the square $EFGH$ has its side $GH$ on side $BC$, and $E, F$ are respectively on sides $AB, AC$. Then, what is the length of the side of the square $EFGH$?

Solution:
\textbf{Solution:} Set the side length of the square part as $x$.\\
In square $EFGH$, $EF\parallel BC$, $EH\parallel AD$\\
$\therefore \angle AEF=\angle ABC$, $\angle EAF=\angle BAC$; $\angle BHE=\angle BDA$, $\angle B=\angle B$\\
$\therefore \triangle AEF\sim \triangle ABC$, $\triangle BEH\sim \triangle BAD$\\
$\therefore \frac{EF}{BC}=\frac{AE}{AB}$, $\frac{EH}{AD}=\frac{BE}{BA}$\\
$\therefore \frac{EF}{BC}+\frac{EH}{AD}=\frac{AE}{AB}+\frac{BE}{BA}=1$\\
$\therefore \frac{x}{120}+\frac{x}{80}=1$ \\
Solving for $x$ gives: $x=\boxed{48}$


Question: As shown in the figure, in $ \square ABCD $, with $ AD=6 $, E is a moving point on AD, and M, N are the midpoints of BE and CE, respectively. What is the length of MN?

Solution:
\textbf{Solution:} As shown in the figure, in parallelogram $ABCD$, we have $BC=AD=6$.\\
Since $M$, $N$ are the midpoints of $BE$, $CE$, respectively,\\
therefore $MN$ is the midline of $\triangle EBC$,\\
therefore $MN=\frac{1}{2}BC=\boxed{3}$.


Question: As shown in the figure, fold $\triangle ABC$ along its median $DE$, with point $A$ landing on point $F$. If $\angle C = 120^\circ$ and $\angle A = 20^\circ$, what is the degree measure of $\angle FEB$?

Solution:
\textbf{Solution:} Given that $\angle C=120^\circ$ and $\angle A=20^\circ$,\\
it follows that $\angle B=40^\circ$,\\
since DE is the median line of $\triangle ABC$,\\
it implies that DE$\parallel$BC,\\
thus, $\angle B+\angle DEB=180^\circ$, and $\angle B=\angle AED=\angle DEF=40^\circ$\\
which leads to $\angle DEB=140^\circ$,\\
hence, $\angle FEB=\angle DEB-\angle DEF=\boxed{100^\circ}$.


Question: As shown in the figure, in rhombus $ABCD$, the lengths of the two diagonals are $AC=6$ and $BD=8$. What is the area of this rhombus?

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


Question: As shown in the figure, in the rhombus $ABCD$, $P$ is a point on the diagonal $AC$. If $AB = 4$, $PA = 5$, and $PC = 2$, what is the length of $PB$?

Solution:
\textbf{Solution:} As shown in the diagram, connect BD, intersecting AC at point O,\\
it is easy to obtain that BD perpendicularly bisects AC\\
$\because AC=7$\\
$\therefore AO=\frac{7}{2}$, $PO=\frac{3}{2}$,\\
in $\triangle ABO$ with a right angle at B, $BO=\sqrt{4^2-\left(\frac{7}{2}\right)^2}=\frac{\sqrt{15}}{2}$\\
$\therefore$ in $\triangle PBO$ with a right angle at B, $PB=\sqrt{BO^2+PO^2}=\boxed{\sqrt{6}}$


Question: As shown in the diagram, rectangle $ABCD$ is folded along the diagonal $BD$, with point $C$ falling on point $E$. $BE$ intersects $AD$ at point $F$. Given that $\angle BDC=62^\circ$, what is the measure of $\angle DFE$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rectangle, \\
$\therefore$ $AD\parallel BC$ and $\angle ADC=90^\circ$, \\
Given $\angle BDC=62^\circ$, \\
$\therefore$ $\angle FDB=90^\circ-\angle BDC=90^\circ-62^\circ=28^\circ$, \\
Since $AD\parallel BC$, \\
$\therefore$ $\angle CBD=\angle FDB=28^\circ$, \\
Since rectangle ABCD is folded along the diagonal BD, \\
$\therefore$ $\angle FBD=\angle CBD=28^\circ$, \\
$\therefore$ $\angle DFE=\angle FBD+\angle FDB=28^\circ+28^\circ=\boxed{56^\circ}$.


Question: As shown in the figure, in rectangle $ABCD$, where $AB=24$ and $BC=25$, an arc is drawn with point $B$ as the center and the length of $BC$ as the radius, intersecting side $AD$ at point $E$. What is the length of $AE$?

Solution:
\textbf{Solution:}
Connect BE as shown in the figure,\\
as given, BE=BC=25,\\
since quadrilateral ABCD is a rectangle,\\
therefore, $\angle A=\angle D=90^\circ$, AB=DC=24, AD=BC=25,\\
in $\triangle ABE$,\\
$AE=\sqrt{BE^{2}-AB^{2}}=\sqrt{25^{2}-24^{2}}=\boxed{7}$.


Question: As shown in the figure, the distance from the midpoint \(M\) of one side of the rhombus \(ABCD\) to the intersection point \(O\) of the diagonals is \(5 \, \text{cm}\). What is the perimeter of the rhombus \(ABCD\)?

Solution:
Solution: Since quadrilateral ABCD is a rhombus,\\
it follows that $\angle AOB=90^\circ$,\\
since M is the midpoint of side AB,\\
it follows that AB=2OM=10,\\
therefore, the perimeter of rhombus ABCD is $10\times 4=\boxed{40}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=2$, point $E$ is on side $AD$, $EB$ bisects $\angle AEC$, and $\angle DEC=30^\circ$. What is the length of $AE$?

Solution:
\textbf{Solution:} In rectangle ABCD, we have AD=BC, CD=AB=2, AD$\parallel$BC,\\
$\because$ $\angle$DEC＝$30^\circ$,\\
$\therefore$ $\angle$BCE=$\angle$DEC=$30^\circ$, $\angle$AEC=$150^\circ$, CE=2CD=4,\\
$\therefore$ $DE=\sqrt{CE^{2}−CD^{2}}=2\sqrt{3}$,\\
$\because$ EB bisects $\angle$AEC,\\
$\therefore$ $\angle BEC=\frac{1}{2}\angle AEC=75^\circ$,\\
$\therefore$ $\angle$CBE=$75^\circ$,\\
$\therefore$ $\angle$CBE=$\angle$BEC,\\
$\therefore$ CE=BC=AD=4,\\
$\therefore$ $AE=AD−DE=4−2\sqrt{3}$.\\
Thus, the answer is: $AE=\boxed{4−2\sqrt{3}}$.


Question: As shown in the figure, in parallelogram \(ABCD\), diagonal \(AC \perp BC\), \(M\) is on the bisector of \(\angle CAD\), and \(AM \perp DM\), point \(N\) is the midpoint of \(CD\), connect \(MN\), if \(AD = 12\), \(MN = 2\). What is the length of \(AB\)?

Solution:
\textbf{Solution:} Extend $DM$ to intersect $AC$ at $E$, \\
$\because AM$ bisects $\angle CAD$ and $AM \perp DM$, \\
$\angle DAM = \angle EAM$, $\angle AMD = \angle AME = 90^\circ$, \\
In $\triangle ADM$ and $\triangle AEM$, \\
$\left\{\begin{array}{l} \angle DAM = \angle EAM \hfill \\ AM = AM \hfill \\ \angle AMD = \angle AME \hfill \end{array}\right.$, \\
$\therefore \triangle ADM \cong \triangle AEM$ (ASA),\\
$\therefore DM = EM$, $AE = AD = 12$,\\
$\therefore$ Point $M$ is the midpoint of $DE$,\\
$\because N$ is the midpoint of $CD$,\\
$\therefore MN$ is the midline of $\triangle CDE$,\\
$\because MN = 2$,\\
$\therefore CE = 2MN = 4$,\\
$\therefore AC = AE + CE = 12 + 4 = 16$,\\
In the parallelogram $ABCD$, $AB = CD$, $AD \parallel BC$, $AC \perp BC$,\\
$\therefore AC \perp AD$,\\
$\therefore \angle CAD = 90^\circ$,\\
$\therefore AB = CD = \sqrt{AD^2 + AC^2} = \sqrt{12^2 + 16^2} = \boxed{20}$.


Question: In the figure, in $\triangle ABC$, $AB=CB=6$, $BD\perp AC$ at point D, $F$ is on $BC$ and $BF=2$. Connect AF, with E being the midpoint of AF. Connect DE. What is the length of DE?

Solution:
\textbf{Solution:} Given that $CB=6, BF=2$, \\
thus $CF=CB-BF=4$, \\
since $AB=CB=6, BD\perp AC$, \\
it follows that $AD=CD$ (the three sides of an isosceles triangle are equal), meaning point $D$ is the midpoint of $AC$,\\
since $E$ is the midpoint of $AF$,\\
thus $DE$ is the midsegment of $\triangle ACF$,\\
therefore $DE=\frac{1}{2}CF=\boxed{2}$.


Question: As shown in the figure, in the square $ABCD$ with side length 8, $E$ and $F$ are moving points on sides $AB$ and $BC$, respectively, and $EF=6$. Let $M$ be the midpoint of $EF$, and $P$ be a moving point on side $AD$. What is the minimum value of $CP+PM$?

Solution:
\textbf{Solution:} Extend CD to $C'$, making $C'D = CD$,\\
$CP + PM = C'P + PM$,\\
When $C'$, P, and M are collinear, the value of $C'P + PM$ is minimized,\\
According to the problem, the trajectory of point M is on the circular arc with B as the center and radius 3,\\
The minimum distance from a point outside a circle $C'$ to a point M on the circle, $C'M = C'B - 3$,\\
since $BC = CD = 8$,\\
thus $CC' = 16$,\\
thus $C'B = \sqrt{CC'^2 + BC^2} = \sqrt{16^2 + 8^2} = 8\sqrt{5}$,\\
therefore, the minimum value of $CP + PM$ is $8\sqrt{5} - 3 = \boxed{8\sqrt{5} - 3}$.


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

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\
it follows that $OA=OC$, \\
since point E is the midpoint of side AB, \\
it follows that $OE$ is the median of $\triangle ABC$, \\
hence $BC=2OE=2\times 6=\boxed{12}$


Question: As shown in the figure, in rectangle $ABCD$, where $AB=2$ and $AD=1$, point $M$ is on the side $CD$. If $AM$ bisects $\angle DMB$, what is the length of $DM$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\
hence $CD=AB=2$, and $AB\parallel CD$, $BC=AD=1$, $\angle C=90^\circ$,\\
thus $\angle BAM=\angle AMD$,\\
since $AM$ bisects $\angle DMB$,\\
thus $\angle AMD=\angle AMB$,\\
hence $\angle BAM=\angle AMB$,\\
thus $BM=AB=2$,\\
hence $CM= \sqrt{MB^{2}-BC^{2}}=\sqrt{2^{2}-1^{2}}=\sqrt{3}$,\\
thus $DM=CD−CM=2−\sqrt{3}=\boxed{2-\sqrt{3}}$.


Question: As shown in the figure, within square $ABCD$, point $P$ is on side $AB$, $AE\perp DP$ at point $E$, and $CF\perp DP$ at point $F$. If $AE=4$ and $CF=7$, what is the length of $EF$?

Solution:
\textbf{Solution:} Since $ABCD$ is a square,\\
therefore, $\angle ADC=90^\circ$, and $CD=DA$,\\
therefore, $\angle CDF+\angle ADE=90^\circ$,\\
since $AE\perp DP$, and $CF\perp DP$,\\
therefore, $\angle AED=\angle DFC=90^\circ$,\\
therefore, $\angle DAE+\angle ADE=90^\circ$,\\
therefore, $\angle CDF=\angle DAE$,\\
in $\triangle CDF$ and $\triangle DAE$,\\
since $\angle DFC=\angle AED=90^\circ$, $\angle CDF=\angle DAE$, $CD=DA$,\\
therefore, $\triangle CDF \cong \triangle DAE$ (AAS),\\
therefore, $CF=DE=7$, $DF=AE=4$,\\
therefore, $EF=DE-DF=7-4=\boxed{3}$.


Question: As shown in the figure, in rectangle $ABCD$, we have $AD=1$, $AB>1$, and $AG$ bisects $\angle BAD$. Perpendiculars are drawn from points B and C to $AG$ at points E and F, respectively. What is the value of $\left(AE-GF\right)$?

Solution:
\textbf{Solution:} Let AE = x,

since quadrilateral ABCD is a rectangle,

therefore, $\angle BAD = \angle D = 90^\circ$, CD = AB,

since AG bisects $\angle BAD$,

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

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

therefore, DG = AD = 1,

therefore, AG = $\sqrt{2}$AD = $\sqrt{2}$,

Similarly: BE = AE = x, CD = AB = $\sqrt{2}x$,

therefore, CG = CD - DG = $\sqrt{2}x - 1$,

Similarly: CG = $\sqrt{2}$FG,

therefore, FG = $\frac{\sqrt{2}}{2}$CG = x - $\frac{\sqrt{2}}{2}$,

therefore, AE - GF = x - (x - $\frac{\sqrt{2}}{2}$) = $\boxed{\frac{\sqrt{2}}{2}}$.


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

Solution:
\textbf{Solution:} In quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $O$,\\
$\therefore$ $BO=DO$,\\
$\because$ point $E$ is the midpoint of side $BC$,\\
$\therefore$ $OE$ is the median of $\triangle ABC$,\\
$\therefore$ $OE = \frac{1}{2}AB = \boxed{2}$.


Question: As shown in the figure, a rectangular sheet of paper is folded along EF, with points D and C corresponding to positions D' and C', respectively. If $\angle EFC'=115^\circ$, what is the measure of $\angle AED'$?

Solution:
\textbf{Solution:} Given $AD \parallel BC$ and $\angle EFC' = 115^\circ$,\\
$\therefore \angle DEF = 180^\circ - 115^\circ = 65^\circ$.\\
Since $\angle D'EF$ is formed by folding $\angle DEF$,\\
$\therefore \angle DED' = 2\angle DEF = 130^\circ$,\\
$\therefore \angle AED' = 180^\circ - 130^\circ = \boxed{50^\circ}$.


Question: As shown in the figure, $\triangle ABC$ and $\triangle A^{\prime}B^{\prime}C^{\prime}$ are similar, with the center of similarity at point O, and $\frac{A^{\prime}C^{\prime}}{AC}=\frac{2}{3}$. If the area of $\triangle ABC$ is 9, what is the area of $\triangle A^{\prime}B^{\prime}C^{\prime}$?

Solution:
\textbf{Solution}: Since $\triangle ABC$ and $\triangle A^{\prime}B^{\prime}C^{\prime}$ are similar, with the center of similarity being point O, and $\frac{A'C'}{AC}=\frac{2}{3}$,\\
it follows that $\frac{S_{\triangle A^{\prime}B^{\prime}C^{\prime}}}{S_{\triangle ABC}}=\left(\frac{A^{\prime}C^{\prime}}{AC}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}$,\\
Since the area of $\triangle ABC$ is 9,\\
therefore, the area of $\triangle A^{\prime}B^{\prime}C^{\prime}$ is $\boxed{4}$.


Question: As shown in the figure, the line MN is an axis of symmetry of the rectangle ABCD, with point E located on side AD. When $\triangle BAE$ is folded along BE, the corresponding point of A, denoted as F, falls on the line MN. If $AB=4$, what is the length of BE?

Solution:
\textbf{Solution:} Given that MN is an axis of symmetry of rectangle ABCD, \\
it follows that $AM=BM=\frac{1}{2}AB=2$, and $MN\parallel AD$, \\
hence, $\angle AEF+\angle MFE=180^\circ$, $\angle BMF=\angle BAE=90^\circ$, \\
by the properties of folding, we have $BF=AB=4$, $\angle AEB=\angle FEB$, $\angle BFE=\angle A=90^\circ$, \\
therefore, $\sin\angle BFM=\frac{AB}{BF}=\frac{1}{2}$, \\
hence, $\angle BFM=30^\circ$, \\
thus, $\angle MFE=60^\circ$, \\
and so, $\angle AEF=120^\circ$, \\
therefore, $\angle AEB=60^\circ$, \\
thus, $BE=\frac{AB}{\sin\angle AEB}=\boxed{\frac{8\sqrt{3}}{3}}$.


Question: In the figure, within $\triangle ABC$ and $\triangle ADE$, we have $\angle BAC=\angle DAE=90^\circ$, $AC=AD=3$, and $AB=AE=5$. Connect $BD$ and $CE$, and rotate $\triangle ADE$ around point $A$ for one complete circle. During the rotation, when $\angle DBA$ is at its maximum, what is the area of $\triangle ACE$?

Solution:
\textbf{Solution:} From the given information, we know that the trajectory of point D is a circle with center at point A and radius equal to AD. \\
When the line BD is tangent to the trajectory circle of point D, $\angle DBA$ reaches its maximum value, at which point $\angle BDA=90^\circ$, as shown in the figure. \\
Draw line CF $\perp$ AE at F. \\
Since $\angle DAE=90^\circ$ and $\angle BAC=90^\circ$, \\
it follows that $\angle CAF=\angle BAD$. \\
In $\triangle ABD$, by the Pythagorean theorem, BD=$\sqrt{5^2-3^2}=4$. \\
Therefore, from $\sin \angle CAF=\sin \angle BAD$, we get: \\
$\frac{CF}{AC}=\frac{BD}{AB}$, \\
that is, $\frac{CF}{3}=\frac{4}{5}$, \\
solving this yields: CF=$\frac{12}{5}$, \\
thus, the area of triangle ACE = $\frac{1}{2}\times \frac{12}{5}\times 5=\boxed{6}$.


Question: As shown in the figure, the four corners of the rectangular paper $ABCD$ are folded inward to precisely form a quadrilateral $EFGH$ with no gaps and no overlaps. If $EH=3$ and $EF=4$, what is the length of side $BC$?

Solution:
\textbf{Solution:} We have $\because \angle HEM=\angle AEH$ and $\angle BEF=\angle FEM$, \\
$\therefore \angle HEF=\angle HEM+\angle FEM=\frac{1}{2}\times 180^\circ=90^\circ$, \\
Similarly, we find: $\angle EHG=\angle HGF=\angle EFG=90^\circ$, \\
$\therefore$ Quadrilateral $EFGH$ is a rectangle, \\
$\therefore GH=EF$ and $GH\parallel EF$, \\
$\therefore \angle GHN=\angle EFM$, \\
In $\triangle GHN$ and $\triangle EFM$ we have\\
$\left\lbrace\begin{array}{l}
\angle GNH=\angle EMF\\
\angle NHG=\angle MFE\\
HG=EF
\end{array}\right.$, \\
$\therefore \triangle GHN\cong \triangle EFM(AAS)$, \\
$\therefore HN=MF=HD$, \\
$\therefore AD=AH+HD=HM+MF=HF$, \\
$HF=\sqrt{EH^2+EF^2}=\sqrt{3^2+4^2}=5$, \\
$\therefore BC=AD=\boxed{5}$.


Question: As shown in the diagram, in square $ABCD$, side $BC$ is rotated counterclockwise around point $B$ to $BC'$, and lines $CC'$ and $DC'$ are drawn. If $\angle CC'D = 90^\circ$ and $BC' = 2\sqrt{5}$, what is the length of segment $C'D$?

Solution:
\textbf{Solution:} As shown in the figure, draw $BE\perp CC'$ at point E,\\
since quadrilateral ABCD is a square,\\
therefore $BC=CD$, $\angle BCD=90^\circ$,\\
therefore $\angle BCE + \angle C'CD = 90^\circ$,\\
since $\angle BCE + \angle CBE = 90^\circ$,\\
therefore $\angle C'CD = \angle CBE$,\\
also since $\angle BEC = \angle CC'D = 90^\circ$,\\
therefore $\triangle BCE \cong \triangle CDC'$ (AAS),\\
therefore $CE=C'D$, $BE=CC'$,\\
since side BC is rotated counterclockwise around point B to BC',\\
therefore $BC=BC'$,\\
also since $BE\perp CC'$,\\
therefore $CE=C'E=C'D= \frac{1}{2}CC'= \frac{1}{2}BE$,\\
since $BC'= 2\sqrt{5}$,\\
therefore $CE^2+BE^2=CE^2+(2CE)^2=BC^2=(2\sqrt{5})^2$,\\
solving for: CE=2,\\
therefore, the length of segment C'D is $\boxed{2}$.


Question: As shown, in $\triangle ABC$, $AB=6$. After rotating $\triangle ABC$ $40^\circ$ counterclockwise around point $A$, we get $\triangle ADE$. The path traced by point $B$ is $\overset{\frown}{BD}$. What is the area of the shaded region in the figure?

Solution:
\textbf{Solution:} From the property of rotation, we know that $S_{\triangle ADE} = S_{\triangle ABC}$,\\
thus, the area of the shaded part $= S_{\triangle ADE} + S_{\text{sector} DAB} - S_{\triangle ABC}$\\
$= S_{\text{sector} DAB}$\\
$= \frac{40\pi \times 6^2}{360}$\\
$= \boxed{4\pi}$,


Question: As shown in the figure, in quadrilateral $ABCD$, $\angle BAD=110^\circ$, $\angle B=\angle D=90^\circ$. Find points $M$ and $N$ on $BC$ and $CD$ respectively, such that the perimeter of $\triangle AMN$ is minimal. What is the measure of $\angle AMN + \angle ANM$?

Solution:
\textbf{Solution:} As shown in the figure, construct the symmetric point of point A about line $BC$ as $A^{\prime}$, and the symmetric point about line $CD$ as $A^{\prime \prime}$,\\
join $A^{\prime}A^{\prime \prime}$, and let the intersection points with $BC$ and $CD$ be the required points M and N, respectively,\\
$\because \angle BAD=110^\circ$, $\angle B=\angle D=90^\circ$,\\
$\therefore \angle A^{\prime} +\angle A^{\prime \prime} =180^\circ-110^\circ=70^\circ$,\\
from the property of axial symmetry, we get: $\angle A^{\prime} =\angle A^{\prime}AM$, $\angle A^{\prime \prime} =\angle A^{\prime \prime}AN$,\\
$\therefore \angle AMN+\angle ANM=2(\angle A^{\prime} +\angle A^{\prime \prime})=2\times 70^\circ=\boxed{140^\circ}$.


Question: As shown in the figure, $\triangle ABC$ is similar to $\triangle DEF$, with point O being the center of similitude. Given that $OA=1$ and $OD=3$, if the perimeter of $\triangle ABC$ is 3, what is the perimeter of $\triangle DEF$?

Solution:
\textbf{Solution}: Since $\triangle ABC$ is similar to $\triangle DEF$ and $OA=1$, $OD=3$, \\
thus $OA: OD=1: 3$ and $AC\parallel DF$, \\
therefore $\triangle OAC\sim \triangle ODF$, \\
hence $OA: OD=AC: DF=1: 3$, \\
since $\triangle ABC$ is similar to $\triangle DEF$, \\
therefore $\triangle ABC\sim \triangle DEF$, \\
thus the perimeter ratio of $\triangle ABC$ to $\triangle DEF$ is $1: 3$, \\
therefore the perimeter of $\triangle DEF$ is $3\times 3=\boxed{9}$.


Question: As shown in the figure, in square ABCD, the side BC is rotated counterclockwise around point B to $BC'$, and $CC'$, $DC'$ are connected. If $\angle CC'D=90^\circ$ and $C'D=2$, what is the length of the line segment BC?

Solution:
\textbf{Solution:} As shown in the figure, take point O as the midpoint of segment CC$^{\prime}$ and connect BO. \\
According to the condition, BC = BC$^{\prime}$ \\
$\therefore$BO$\perp$CC$^{\prime}$ \\
$\therefore$$\angle$BOC = $90^\circ$ \\
In square ABCD, \\
BC = CD, $\angle$BCD = $90^\circ$ \\
$\therefore$$\angle$OCB + $\angle$C$^{\prime}$CD = $90^\circ$ \\
Also, since $\angle$CC$^{\prime}$D = $90^\circ$ \\
$\therefore$$\angle$C$^{\prime}$DC + $\angle$C$^{\prime}$CD = $90^\circ$ \\
$\therefore$$\angle$OCB = $\angle$C$^{\prime}$DC \\
In Rt$\triangle$OBC and Rt$\triangle$C$^{\prime}$CD, \\
$\left\{\begin{array}{l}\angle OCB = \angle C^{\prime}DC \\ \angle BOC = \angle CC^{\prime}D \\ BC = CD \end{array}\right.$ \\
$\therefore$Rt$\triangle$OBC$\cong$ Rt$\triangle$C$^{\prime}$CD (AAS) \\
$\therefore$OC = C$^{\prime}$D = 2 \\
$\therefore$CC$^{\prime}$ = 2 OC = $2 \times 2$ = 4 \\
$\therefore$BO = CC$^{\prime}$ = 4 \\
In Rt$\triangle$BOC, \\
BC = $\sqrt{BO^{2} + OC^{2}}$ = $\sqrt{4^{2} + 2^{2}}$ = $\boxed{2\sqrt{5}}$


Question: As shown in the figure, equilateral $\triangle ABC$ has a side length of $4\sqrt{3}$. The angle bisectors of $\angle ABC$ and $\angle ACB$ intersect at point O. $\triangle OBC$ is rotated $30^\circ$ counterclockwise around point O to obtain $\triangle OB_{1}C_{1}$. $B_{1}C_{1}$ intersects BC at point D, and $B_{1}C_{1}$ intersects AC at point E. What is the length of DE?

Solution:
\textbf{Solution:} Draw OH$\perp$BC at H and OB$_{1}$ intersecting BC at M. Through M, draw MF$\perp$BO at F, as shown in the diagram below:\\
Since $\triangle ABC$ is an equilateral triangle, and OB, OC are the angle bisectors of $\angle ABC$, $\angle ACB$ respectively,\\
therefore, $\angle 1= \frac{1}{2}\angle ABC=30^\circ$, $\angle 3= \frac{1}{2}\angle ACB=30^\circ$,\\
thus, $\triangle OBC$ is an isosceles triangle, and by "the three lines coincide", \\
BH=CH=$ \frac{1}{2}$BC=$ 2\sqrt{3}$,\\
therefore, BO=$ \frac{2\sqrt{3}}{3}$BH=4,\\
since $\triangle OBC$ is rotated $30^\circ$ counterclockwise around point O to get $\triangle OB_{1}C_{1}$,\\
therefore, $\angle 2=30^\circ=\angle 1$,\\
thus, $\triangle OBM$ is an isosceles triangle, and by "the three lines coincide", \\
BF=$ \frac{1}{2}$BO=2,\\
thus, MO=BM=$ \frac{2\sqrt{3}}{3}$BF=$ \frac{4\sqrt{3}}{3}$,\\
thus, MB$_{1}$=OB$_{1}$−OM=OB−OM=$ 4−\frac{4\sqrt{3}}{3}$,\\
moreover, by the rotation $\angle B=\angle B_{1}=30^\circ$, and the vertically opposite angle $\angle BMO=\angle DMB_{1}=120^\circ$,\\
therefore, $\angle MDB_{1}=180^\circ-\angle B_{1}-\angle DMB_{1}=180^\circ-30^\circ-120^\circ=30^\circ$,\\
therefore, $\triangle MB_{1}D$ is an isosceles triangle,\\
therefore, MD=MB$_{1}=$ $4-\frac{4\sqrt{3}}{3}$,\\
therefore, CD=BC−MD−BM=$ 4\sqrt{3}-(4-\frac{4\sqrt{3}}{3})-\frac{4\sqrt{3}}{3}=4\sqrt{3}-4$,\\
since the vertically opposite angle $\angle EDC=\angle MDB_{1}=30^\circ$, and $\angle ACB=60^\circ$,\\
therefore, $\angle DEC=180^\circ-\angle EDC-\angle ACB=90^\circ$,\\
therefore, $\triangle CDE$ is a right-angled triangle with angles $30^\circ$, $60^\circ$, $90^\circ$,\\
therefore, DE=$ \frac{\sqrt{3}}{2}$CD=$ \frac{\sqrt{3}}{2}(4\sqrt{3}-4)=\boxed{6-2\sqrt{3}}$.