Question: As shown in the figure, it is known that in $\triangle ABC$, with side $BC=6$ and height $AD=3$, the vertices $E$ and $F$ of the square $EFGH$ lie on side $BC$, while vertices $H$ and $G$ lie on sides $AB$ and $AC$, respectively. What is the length of the side of this square?

Solution:
\textbf{Solution:} As shown in the diagram:\\
$ \because $ Quadrilateral $EFMN$ is a square,\\
$ \therefore HG\parallel BC$, $HG=EF$,\\
$ \therefore \angle AHG=\angle B$,\\
$ \because \angle BAC=\angle BAC$,\\
$ \therefore \triangle AHG\sim \triangle ABC$,\\
$ \therefore \frac{AH}{AB}=\frac{HG}{BC}$,\\
Also $ \because AD\perp BC$,\\
$ \therefore AD\perp HG$, $HG=EF=MD$,\\
$ \because HG\parallel BC$,\\
$ \therefore \frac{AM}{AD}=\frac{AH}{AB}$, that is $\frac{AM}{AD}=\frac{HG}{BC}$,\\
Let $HG=x$, then $AM=AD-MD=3-x$,\\
$ \therefore \frac{3-x}{3}=\frac{x}{6}$, solving this gives: $x=\boxed{2}$,\\
$ \therefore $ The side length of this square is $2$.


Question: As shown in the figure, it is known that $AD \parallel BC$, $S_{\triangle AOD} = 4$, $S_{\triangle BOC} = 9$, then what is the area of $\triangle ABO$?

Solution:
\textbf{Solution:} Since $AD\parallel BC$,\\
it follows that $\triangle AOD\sim \triangle COB$,\\
hence $\frac{{S}_{\triangle AOD}}{{S}_{\triangle COB}}=\left(\frac{OA}{OC}\right)^{2}$, that is, $\left(\frac{OA}{OC}\right)^{2}=\frac{4}{9}$,\\
from which we find $\frac{OA}{OC}=\frac{2}{3}$,\\
thus $\frac{{S}_{\triangle AOB}}{{S}_{\triangle BOC}}=\frac{OA}{OC}=\frac{2}{3}$, that is, $\frac{{S}_{\triangle ABO}}{9}=\frac{2}{3}$,\\
solving this, we get ${S}_{\triangle ABO}=\boxed{6}$.


Question: As shown in the figure, quadrilaterals $ABCD$, $CDEF$, and $EFGH$ are all squares. What is the sum of $\angle ACB+\angle AFB+\angle AGB$ in degrees?

Solution:
\textbf{Solution:} Let the side length of the square be $a$, \\
$AC=\sqrt{a^2+a^2}=\sqrt{2}a$, \\
$\because \frac{CA}{CF}=\frac{\sqrt{2}a}{a}=\sqrt{2}$, $\frac{CG}{AC}=\frac{2a}{\sqrt{2}a}=\sqrt{2}$, \\
$\therefore \frac{AC}{CF}=\frac{CG}{AC}$, \\
$\because \angle ACF=\angle ACF$, \\
$\therefore \triangle ACF\sim \triangle GCA$; \\
$\therefore \angle AGB=\angle CAF$, \\
$\because \angle AGB+\angle AFB=\angle CAF+\angle AFB=\angle ACB=45^\circ$, \\
$\therefore \angle ACB + \angle AFB + \angle AGB = 45^\circ + 45^\circ = \boxed{90^\circ}$.


Question: As shown in the diagram, in parallelogram $ABCD$, we have $CE:BE=1:3$, and the area of triangle $EFC$ is denoted as $S_{\triangle EFC}=1$. What is the area of triangle $ABC$, denoted as $S_{\triangle ABC}$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram,\\
therefore $AD=BC$, $AD\parallel BC$, ${S}_{\triangle ACD}={S}_{\triangle ABC}=\frac{1}{2}{S}_{\text{parallelogram} ABCD}$,\\
therefore $\triangle ADF\sim \triangle CEF$,\\
therefore $\frac{CE}{AD}=\frac{EF}{DF}$,\\
since $CE\colon BE=1\colon 3$,\\
therefore $CE\colon CB=1\colon 4$,\\
therefore $CE\colon AD=1\colon 4$,\\
therefore $\frac{CE}{AD}=\frac{EF}{DF}=\frac{1}{4}$,\\
therefore ${S}_{\triangle CDF}=4{S}_{\triangle EFC}=4$, $\frac{{S}_{\triangle ADF}}{{S}_{\triangle EFC}}={\left(\frac{AD}{CE}\right)}^{2}=16$,\\
therefore ${S}_{\triangle ADC}={S}_{\triangle ADF}+{S}_{\triangle CDF}=20$,\\
therefore ${S}_{\triangle ABC}={S}_{\triangle ACD}=\boxed{20}$


Question: As shown in the figure, it is known that the similarity ratio between $\triangle ADE$ and $\triangle ABC$ is $1:2$, and the area of $\triangle ADE$ is 3. What is the area of the quadrilateral $DBCE$?

Solution:
Since the similarity ratio between $\triangle ADE$ and $\triangle ABC$ is $1\colon 2$,\\
the ratio of their areas is $1\colon 4$.\\
Therefore, the area ratio between $\triangle ADE$ and quadrilateral $DBCE$ is $1\colon 3$.\\
Since the area of $\triangle ADE$ is 3,\\
the area of quadrilateral $DBCE$ is $\boxed{9}$.


Question: As shown in the figure, in trapezoid $ABCD$, where $AD \parallel BC$ and $BD^2 = AD \cdot BC$, if $AB : CD = 1 : 2$, what is the ratio of $AD : BC$?

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

it follows that $\angle ADB=\angle DBC$.

Since $BD^{2}=AD\cdot BC$,

it follows that $\frac{BD}{BC}=\frac{AD}{BD}$,

therefore $\triangle ADB\sim \triangle DBC$,

hence $\frac{BD}{BC}=\frac{AD}{BD}=\frac{AB}{CD}=\frac{1}{2}$,

thus $BD=2AD$, $BC=2BD$,

hence $BC=4AD$,

thus $AD:BC=\boxed{1:4}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AD\parallel BC\parallel EF$. If $AE=3$, $AB=8$, $CD=10$, then what is the length of $CF$?

Solution:
\[
\textbf{Solution:} \quad \text{Since } AD\parallel BC\parallel EF,
\]
\[
\therefore \frac{AE}{AB}=\frac{DF}{CD},
\]
\[
\text{Since } AE=3, AB=8, CD=10,
\]
\[
\therefore \frac{3}{8}=\frac{DF}{10},
\]
\[
\text{Solving for } DF, \text{ we get } DF=\frac{15}{4},
\]
\[
\therefore CF=CD-DF=10-\frac{15}{4}=\boxed{\frac{25}{4}}.
\]


Question: As shown in the figure, given $\triangle ABC\sim \triangle ADE$, $AD=6$, $BD=3$, $DE=4$, what is the length of $BC$?

Solution:
\textbf{Solution}: Since $AD=6$ and $BD=3$,\\
then $AB=9$,\\
Since $\triangle ABC\sim \triangle ADE$ and $DE=4$,\\
then $AD: AB = DE: BC$,\\
thus $BC=\frac{AB\cdot DE}{AD}$,\\
Since $DE=4$,\\
thus $BC=\frac{9\times 4}{6}=\boxed{6}$.


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle with $AB=3$. Point $D$ is a moving point on side $BC$, $\angle ADE=60^\circ$, and $DE$ intersects $AC$ at point $E$. What is the maximum value of the length of segment $CE$?

Solution:
\textbf{Solution:}

Since $\triangle ABC$ is an equilateral triangle, \\
therefore, AB=BC=3, $\angle B=\angle C=60^\circ$, \\
since $\angle ADC=\angle ADE+\angle EDC=\angle B+\angle BAD$, $\angle ADE=60^\circ$, \\
therefore, $\angle BAD=\angle CDE$, \\
thus, $\triangle ABD\sim \triangle DCE$, \\
so, AB$\colon$CD=BD$\colon$CE, \\
let BD=x, thus CD=3−x, \\
so, 3$\colon$(3−x)=x$\colon$CE, \\
thus, CE$=-\frac{1}{3}x^2+x=-\frac{1}{3}\left(x-\frac{3}{2}\right)^2+\frac{3}{4}$ \\
Since $a=-\frac{1}{3}<0$, the graph opens downward, \\
thus, the maximum value of CE is $\boxed{\frac{3}{4}}$.


Question: As shown in the figure, in $ \triangle ABC$, D and F are the trisection points on side $AB$, and E and G are the trisection points on side $AC$. If $DE=2$, then how much is $BC$?

Solution:
\textbf{Solution:} Since $D,F$ are the trisection points on side $AB$, and $E,G$ are the trisection points on side $AC$,\\
it follows that $DE\parallel FG\parallel BC$,\\
which means $\triangle ADE\sim \triangle ABC$,\\
thus, $\frac{AD}{AB}=\frac{DE}{BC}$,\\
leading to $\frac{1}{3}=\frac{DE}{BC}$,\\
since $DE=2$,\\
it follows that $BC=\boxed{6}$.


Question: As shown in the diagram, $DE \parallel BC$, and $AD:DB = 2:1$, what is the ratio of similarity between $\triangle ADE$ and $\triangle ABC$?

Solution:
\textbf{Solution:} Given that $DE\parallel BC$ and $AD:DB=2:1$, \\
it follows that $\triangle ADE\sim \triangle ABC$ and $AD:AB=2:3$, \\
hence, the ratio of similarity between $\triangle ADE$ and $\triangle ABC$ is $\boxed{2:3}$.


Question: As shown in the diagram, in $\triangle ABC$, $\angle ACB=90^\circ$, $BC=4$, $CA=6$. Circle $C$ has a radius of 2, and $P$ is a moving point on the circle. $AP$ and $BP$ are connected. What is the minimum value of $AP+\frac{1}{2}BP$?

Solution:
\textbf{Solution:} As shown in the figure,\\
take a point $D$ on $CB$, such that $CD=1$, then we have $\frac{CD}{CP}=\frac{CP}{CB}=\frac{1}{2}$,\\
and since $\angle PCD=\angle BCP$,\\
therefore, $\triangle PCD\sim \triangle BCP$,\\
thus, $\frac{PD}{BP}=\frac{1}{2}$,\\
hence, $PD=\frac{1}{2}BP$,\\
therefore, $AP+\frac{1}{2}BP=AP+PD$.\\
To minimize $AP+\frac{1}{2}BP$, $AP+PD$ must be minimized,\\
when points $A$, $P$, $D$ are on the same line, $AP+PD$ is minimized,\\
that is, the minimum value of $AP+\frac{1}{2}BP$ is $AD$,\\
in $\triangle ACD$, where $CD=1$, $AC=6$,\\
therefore, $AD=\sqrt{AC^2+CD^2}=\boxed{\sqrt{37}}$,\\
the minimum value of $AP+\frac{1}{2}BP$ is $\sqrt{37}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=12$, $BC=5$. A square $CDEF$ is cut out from the triangle. What is the length of the side of square $CDEF$?

Solution:
\textbf{Solution:} Since quadrilateral CDEF is a square, \\
it follows that EF $\parallel$ CD, and EF = FC = CD = DE. Let EF = x, then AF = 12 - x, \\
therefore, $\triangle$ AFE $\sim$ $\triangle$ ACB, \\
thus, $\frac{EF}{BC} = \frac{AF}{AC}$, \\
which leads to $\frac{x}{5} = \frac{12 - x}{12}$, \\
solving this gives x = $\boxed{\frac{60}{17}}$, \\
meaning the side length of square CDEF is $\frac{60}{17}$.


Question: As shown in the figure, point E is the midpoint on side CD of square $ABCD$. The diagonals intersect at point O. If BE intersects AC at point F, what is the ratio of $CF:OF$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a square,\\
it follows that $AB=CD$ and $AB\parallel CD$,\\
since $E$ is the midpoint of $CD$,\\
it follows that $CE=\frac{1}{2}CD=\frac{1}{2}AB$,\\
since the diagonals of square $ABCD$ intersect at $O$,\\
it follows that $AO=CO=OF+CF$,\\
since $AB\parallel CD$,\\
it follows that $\triangle ABF\sim \triangle CEF$,\\
hence, $\frac{CE}{AB}=\frac{CF}{AF}=\frac{1}{2}$,\\
since $AF=AO+OF$ and $AO=OF+CF$,\\
it follows that $\frac{CF}{AF}=\frac{CF}{AO+OF}=\frac{CF}{CF+2OF}=\frac{1}{2}$,\\
thus, $2CF=CF+2OF$,\\
therefore, $CF=2OF$,\\
hence, $\frac{CF}{OF}=\boxed{\frac{2}{1}}$.


Question: As shown in the diagram, P is a moving point on side AD of a square $ABCD$ with side length 6 (P does not coincide with A or D). Connect $CP$, and draw $BH\perp CP$ from point B. Fold $\triangle DPC$ along the straight line containing $CP$ to obtain $\triangle CPD^{\prime}$. Extend $PD^{\prime}$ to meet the extension of $CB$ at point G. When $BH = \frac{18\sqrt{13}}{13}$, what is the length of $PG$?

Solution:
\textbf{Solution:}

In right triangle $BCH$, $\angle BHC=90^\circ$, $BH=\frac{18\sqrt{13}}{13}$, $BC=6$,

according to the Pythagorean theorem, $CH=\sqrt{BC^2-BH^2}=\sqrt{6^2-\left(\frac{18\sqrt{13}}{13}\right)^2}=\frac{12\sqrt{13}}{13}$,

since square $ABCD$, $DC=6$,

thus $AD\parallel BC$,

therefore $\angle DPC=\angle PCB$,

therefore $\triangle PDC\sim \triangle CHB$,

therefore $\frac{PD}{CH}=\frac{DC}{HB}$,

therefore $\frac{PD}{\frac{12\sqrt{13}}{13}}=\frac{6}{\frac{18\sqrt{13}}{13}}$,

solving yields $PD=4$,

since $\triangle DPC$ is reflected along line $CP$ to get $\triangle CPD'$,

therefore $PD'=PD=4$, $D'C=DC=6$, $\angle D'PC=\angle DPC=\angle PCB$,

thus $GP=GC$,

let $GP=GC=x$, then $GD'=GP-PD'=x-4$,

in right triangle $CD'G$, $\angle CD'G=90^\circ$,

according to the Pythagorean theorem, $GC^2=GD'^2+CD'^2$,

thus $x^2=(x-4)^2+6^2$,

which gives $8x=52$,

solving yields $x=\boxed{\frac{13}{2}}$,

thus, $PG=\frac{13}{2}$.


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

Solution:
\textbf{Solution:} Given $\because AB\parallel CD\parallel EF$,\\
$\therefore \frac{AC}{CE}=\frac{BD}{DF}$,\\
$\because AC=6, CE=3, DF=2$,\\
$\therefore \frac{6}{3}=\frac{BD}{2}$,\\
Solving this, we get: $BD=\boxed{4}$.


Question: As shown in the figure, in triangle $ABC$, $DE \parallel BC$, $AD = 2$, $AB = 6$, $AE = 3$, then what is the length of $AC$?

Solution:
\textbf{Solution:} \\
\[
\because DE\parallel BC, \\
\therefore \triangle ADE\sim \triangle ABC, \\
\therefore \frac{AD}{AB}=\frac{AE}{AC}, \\
\therefore \frac{2}{6}=\frac{3}{AC}, \\
\therefore AC=\boxed{9}
\]


Question: As shown in the figure, in rectangle $ABCD$, $AD=2$, $AB=4$, and $EF\perp AC$, intersecting $AB$ and $CD$ at $E$ and $F$ respectively, what is the minimum value of $AF+CE$?

Solution:
\textbf{Solution:} Let $DF=x$, then $FC=4-x$,\\
Draw $CG\parallel EF$ through point C, and let $CG=EF$, connect $FG$,\\
When points A, F, G are collinear, find the minimum value of $AF+FG$;\\
$\because$ $CG\parallel EF$, and $CG=EF$,\\
$\therefore$ quadrilateral $CEFG$ is a parallelogram;\\
$\therefore$ $EC\parallel FG$, $EC=FG$,\\
Furthermore, since points A, F, G are collinear,\\
$\therefore$ $AF\parallel EC$,\\
Also, since quadrilateral $ABCD$ is a rectangle,\\
$\therefore$ $AE\parallel DC$, $\angle D=90^\circ$,\\
$\therefore$ quadrilateral $AECF$ is a parallelogram,\\
$\therefore$ $OA=OC$, $OE=OF$,\\
Moreover, since $EF\perp AC$,\\
$AF=CF=4-x$,\\
In $\triangle ADF$, by the Pythagorean theorem, we have:\\
$AD^2+DF^2=AF^2$,\\
Also, since $AD=2$, $DF=x$, then $FC=4-x$,\\
$\therefore$ $2^2+x^2=(4-x)^2$,\\
Solving this, we get: $x=\frac{3}{2}$,\\
$\therefore$ $AF=\frac{5}{2}$,\\
In $\triangle ADC$, by the Pythagorean theorem, we have:\\
$AD^2+DC^2=AC^2$,\\
$\therefore$ $AC=2\sqrt{5}$,\\
$\therefore$ $AO=\frac{1}{2}AC=\sqrt{5}$,\\
Moreover, since $OF\parallel CG$,\\
$\therefore$ $\triangle AOF\sim \triangle ACG$,\\
$\therefore$ $\frac{AO}{AC}=\frac{AF}{AG}$,\\
$\therefore$ $AG=5$,\\
Furthermore, since $AG=AF+FG$, $FG=EC$,\\
$\therefore$ $AF+EC=\boxed{5}$.


Question: As shown in the figure, the length of segment $BC$ is $8$. With $B$ as the center and $4$ as the radius, a circle $\odot B$ is drawn. Point $A$ is a moving point on $\odot B$. Segment $AC$ is connected, and then segment $AC$ is rotated $90^\circ$ counterclockwise about point $A$ to obtain segment $AD$. Connect $DB$. What is the minimum value of $DB$?

Solution:
\textbf{Solution:} Connect DC, AB, and rotate segment BC clockwise around point B by $90^\circ$ to get segment BF. Connect CF, DF,\\
$\therefore$ $\triangle BCF$ is an isosceles right triangle,\\
$\therefore$ $CF=\sqrt{BC^2+BF^2}=\sqrt{8^2+8^2}=8\sqrt{2}$;\\
$\because$ rotating segment AC around point A clockwise by $90^\circ$ gets segment AD,\\
$\therefore$ $\triangle ADC$ is an isosceles right triangle,\\
$\therefore$ $\triangle ADC \sim \triangle CBF$,\\
$\therefore$ $\frac{DC}{AC}=\frac{CF}{AB}$,\\
$\because$ $\angle BCF=\angle ACD$,\\
$\therefore$ $\angle DCF=\angle ACB$,\\
$\therefore$ $\triangle DCF \sim \triangle ACB$,\\
$\therefore$ $\frac{AB}{DF}=\frac{BC}{CF}$ thus $\frac{4}{DF}=\frac{8}{8\sqrt{2}}$,\\
Solving this, $DF=4\sqrt{2}$;\\
As shown in the diagram, point D moves on circle F. If circle F intersects BF at point E, when point D moves to point E, the length of BD is the smallest,\\
$\therefore$ The minimal value is $BE=BF−EF=BF−DF=8−4\sqrt{2}$,\\
$\therefore$ The minimal value of $DB$ is $\boxed{8−4\sqrt{2}}$.


Question: As shown in the figure, a rectangle is divided into three congruent smaller rectangles. For each smaller rectangle to be similar to the original rectangle, what should the ratio $AD\colon CD$ be?

Solution:
\textbf{Solution:} As shown in the figure,\\
$\because$ a rectangle is divided into three congruent smaller rectangles,\\
$\therefore$ AB=3AE,\\
Let AD=y, AE=x, then AB=CD=3x,\\
$\because$ each smaller rectangle is similar to the original rectangle,\\
$\therefore$ rectangle ABCD$\sim$rectangle ADFE,\\
$\therefore$ $\frac{AD}{AE}=\frac{CD}{AD}$ i.e., $\frac{y}{x}=\frac{3x}{y}$,\\
Solving this gives: $y=\sqrt{3}x$\\
$\therefore$ $\frac{AD}{CD}=\frac{y}{3x}=\frac{\sqrt{3}x}{3x}=\frac{1}{\sqrt{3}}$, i.e., AD:CD=$\boxed{1:\sqrt{3}}$


Question: As shown in the figure, there is a right-angled triangular scrap ABC, where $\angle BAC=90^\circ$. Now, a rectangular strip DEFG is cut from it, where E and F are on BC, G is on AB, and D is on AC. If $BF=4.5\text{cm}$, and $CE=2\text{cm}$, then what is the length of the strip GF in centimeters?

Solution:
\textbf{Solution:} Given $\because$ $\angle BAC=90^\circ$,\\
$\therefore$ $\angle AGD+\angle ADG=90^\circ$,\\
$\because$ rectangle GFED,\\
$\therefore$ $\angle BFG=\angle DEC=\angle GFE=\angle DEF=\angle GDE=90^\circ$, DG$\parallel$BC, FG=DE,\\
$\therefore$ $\angle ADG+\angle CDE=90^\circ$, $\angle AGD=\angle B$,\\
$\therefore$ $\angle AGD=\angle CDE$,\\
$\therefore$ $\angle B=\angle CDE$,\\
$\therefore$ $\triangle BFG\sim \triangle DEC$,\\
$\therefore$ $\frac{FG}{CE}=\frac{BF}{DE}$\\
$\therefore$ FG$^{2}=BF\cdot CE=4.5\times 2=9$\\
Solving for FG yields $\boxed{3}$ (taking the positive value)


Question: As shown in the figure, in the circle $\odot O$, the chord $AC=\sqrt{15}$. When the minor arc $\stackrel{\frown}{AC}$ is folded along $AC$, it intersects the diameter $AB$ at $D$, where $DB=\frac{1}{2}$. What is the length of the diameter $AB$?

Solution:
\textbf{Solution:} Draw $CE\perp AB$ at point $E$, and connect $CB$, $CD$,
\[
\because \overset{\frown}{AC}=\overset{\frown}{ADC},
\]
\[
\therefore \angle B=\angle A+\angle ACD=\angle CDB,
\]
\[
\therefore CD=CB,
\]
\[
\therefore DE=BE=\frac{1}{2}BD=\frac{1}{4};
\]
\[
\because AB \text{ is the diameter},
\]
\[
\therefore \angle ACB=\angle BEC=90^\circ,
\]
\[
\because \angle B=\angle B,
\]
\[
\therefore \triangle ACE\sim \triangle CBE,
\]
\[
\therefore \frac{CE}{AE}=\frac{BE}{CE},
\]
\[
\therefore CE^2=AE\cdot BE=\frac{1}{4}AE;
\]
\[
\because AE^2=AC^2−AE^2=15−\frac{1}{4}AE
\]
Solving this, we find $AE=\frac{15}{4},$
\[
\therefore AB=AE+BE=\frac{15}{4}+\frac{1}{4}=\boxed{4}.
\]


Question: As shown in the figure, point $C$ is the midpoint of the semicircle, $AB$ is the diameter, and point $D$ is a point on the semicircle. $AC$ and $BD$ intersect at point $E$. If $AD=2$ and $BD=6$, what is the length of $AC$? What is the length of $CD$?

Solution: \textbf{Solution:} Since $AB$ is the diameter, it follows that $\angle ADB = \angle ACB = 90^\circ$. Within $\triangle ADB$, given AD = 2 and BD = 6, it follows $AB = \sqrt{2^2 + 6^2} = 2\sqrt{10}$. Since point $C$ is the midpoint of the semicircle, it follows that $\angle CAB = 45^\circ$. Therefore, $AB = \sqrt{2}AC = \sqrt{2}BC$, implying $AC = BC = 2\sqrt{5}$. Also, since $\angle DEA = \angle CEB$ and $\angle DAE = \angle CBE$, it follows that $\triangle BEC \sim \triangle AED$. Hence, $CE:DE = BE:AE = BC:AD = 2\sqrt{5}:2$, which leads to $CE = \sqrt{5}DE$ and $BE = \sqrt{5}AE$. Since $BE = BD - DE$ and $AE = AC - CE$, we have $6 - DE = \sqrt{5}(2\sqrt{5} - CE) = \sqrt{5}(2\sqrt{5} - \sqrt{5}DE)$. Solving this, we find $DE = 1$, thus $CE = \sqrt{5}$ and $AE = 2\sqrt{5} - \sqrt{5} = \sqrt{5}$. Since $\angle DCA = \angle DBA$ and $\angle CED = \angle BEA$, it follows that $\triangle DEC \sim \triangle AEB$. Consequently, $DE:EA = CD:AB$, leading to $CD = \frac{DE \cdot AB}{EA} = \frac{1 \times 2\sqrt{10}}{\sqrt{5}} = 2\sqrt{2}$. Therefore, the solutions are: $AC = \boxed{2\sqrt{5}}$ and $CD = \boxed{2\sqrt{2}}$.


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

Solution:
\textbf{Solution:} As shown in the figure, \\
$\because$AC=2, BC=3,\\
$\therefore$AB=AC+BC=2+3=5,\\
$\therefore$AB=EF,\\
In $\triangle ABG$ and $\triangle EFG$,\\
$\left\{\begin{array}{l}
\angle F=\angle ABG\\
\angle FGE=\angle AGB\\
EF=AB
\end{array}\right.$\\
$\therefore$$\triangle ABG\cong \triangle EFG$(AAS),\\
$\therefore$FG=BG=$\frac{1}{2}$BF=2.5;\\
$\because$CH$\parallel$BG,\\
$\therefore$$\triangle ACH\sim \triangle ABG$,\\
$\frac{AC}{AB}=\frac{CH}{BG}$, that is $\frac{2}{5}=\frac{CH}{2.5}$\\
Solving this: CH=1,\\
$\therefore$Area of trapezoid CBGH$=\frac{(CH+BG)\cdot BC}{2}=\frac{3\times (1+2.5)}{2}=\frac{21}{4}$,\\
$\therefore$Area of the shaded region$=3^{2}-\frac{21}{4}=\boxed{\frac{15}{4}}$.


Question: In $\triangle ABC$, points D and E are on AB and AC respectively, and $DE \parallel BC$. If $AD=2$, $AB=3$, and $DE=4$, what is the length of $BC$?

Solution:
\textbf{Solution:} Since $DE \parallel BC$, \\
it follows that $\angle ADE = \angle ABC$, $\angle AED = \angle ACB$, \\
therefore, $\triangle ADE \sim \triangle ABC$, \\
thus, $\frac{BC}{DE} = \frac{AB}{AD}$, that is $\frac{BC}{4} = \frac{3}{2}$, \\
hence, $BC = \boxed{6}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, $AB=2\sqrt{10}$. Connect $AB$ and extend it to $C$, also connect $OC$. If it satisfies $OC^2=BC\cdot AC$ and $\tan\alpha =3$, what are the coordinates of point $C$?

Solution:
\textbf{Solution:}

Given that
\[ OC^2 = BC \cdot AC, \]
it follows that
\[ \frac{OC}{BC} = \frac{AC}{OC}, \]
and given that
\[ \angle C = \angle C, \]
it implies
\[ \triangle OBC \sim \triangle AOC, \]
therefore
\[ \angle A = \angle COB, \]
Given that
\[ \alpha + \angle COB = 90^\circ, \quad \angle A + \angle ABO = 90^\circ, \]
it follows that
\[ \angle ABO = \alpha, \]
Given that
\[ \tan \alpha = 3, \]
it follows that
\[ \tan \angle ABO = \frac{OA}{OB} = 3, \]
therefore
\[ OA = 3OB, \]
Given that
\[ AB = 2\sqrt{10}, \]
by the Pythagorean theorem, we have
\[ OA^2 + OB^2 = AB^2, \]
which implies
\[ (3OB)^2 + OB^2 = (2\sqrt{10})^2, \]
Solving this, we get
\[ OB = 2, \]
therefore
\[ OA = 6, \]
therefore
\[ \tan \angle A = \frac{OB}{OA} = \frac{1}{3}, \]
As shown in the figure, draw $CD \perp x$-axis at point $D$.

Given that
\[ \tan \alpha = 3, \]
let $C(-m, 3m)$, with $m > 0$.

Therefore
\[ AD = 6 + m, \]
Given that
\[ \tan \angle A = \frac{1}{3}, \]
it follows that
\[ \frac{CD}{AD} = \frac{1}{3}, \]
leading to
\[ \frac{3m}{6+m} = \frac{1}{3}, \]
Solving this, we get
\[ m = \frac{3}{4}, \]
Upon checking, $m = \frac{3}{4}$ is a solution of the original equation.

Therefore
\[ -m = -\frac{3}{4}, \quad 3m = \frac{9}{4}, \]
Thus, the coordinates of point $C$ are
\[ \boxed{\left(-\frac{3}{4}, \frac{9}{4}\right)}. \]


Question: As shown in the diagram, it is known that line $l_1 \parallel l_2 \parallel l_3$, and lines $m$ and $n$ intersect lines $l_1$, $l_2$, and $l_3$ at points A, B, C, D, E, and F, respectively. If $DE = 3$ and $DF = 8$, what is the value of $\frac{BC}{AC}$?

Solution:
\textbf{Solution:} Since $l_1 \parallel l_2 \parallel l_3$,\\
it follows that $\frac{EF}{DF} = \frac{BC}{AC}$,\\
given that $DE=3$ and $DF=8$,\\
thus $\frac{BC}{AC} = \frac{8-3}{8} = \boxed{\frac{5}{8}}$.


Question: As shown in the figure, to measure the height of a flagpole, a comprehensive practice group designed the following plan: Use a 2.5m long bamboo pole as a measuring tool, move the bamboo pole to keep it parallel to the flagpole, so that the top ends of the bamboo pole and the flagpole cast shadows that exactly meet at the same point on the ground. At this moment, if the distance from the bamboo pole to this point is 5m, and the distance from the bamboo pole to the flagpole is 20m, what is the height of the flagpole in meters?

Solution:
\textbf{Solution:} From the figure, we can see that,\\
let the height of the flagpole be $x$ meters,\\
$\because DE\parallel BC$\\
$\therefore \triangle ADE\sim \triangle ABC$\\
$\therefore \frac{2.5}{5}=\frac{x}{5+20}$\\
$\therefore x=\boxed{12.5}$


Question: As shown in the figure, $AB \parallel CE$, $\angle ABC = 30^\circ$, $\angle BDE = 45^\circ$, what is the measure of $\angle DBC$ in degrees?

Solution:
\textbf{Solution:} Since $AB\parallel CE$, and $\angle ABC=30^\circ$,\\
it follows that $\angle ABC=\angle BCE=30^\circ$,\\
since $\angle BDE=45^\circ$,\\
it implies that $\angle DBC=\angle BDE-\angle BCE=45^\circ-30^\circ=\boxed{15^\circ}$.


Question: As shown in the figure, it is known that the line $l_{1} \parallel l_{2}$, $\angle A=125^\circ$, $\angle B=85^\circ$, and $\angle 1$ is $4^\circ$ larger than $\angle 2$. What is the degree measure of $\angle 1$?

Solution:
\textbf{Solution:} Extend AB to intersect the two parallel lines at C and D, \\
since $l_1 \parallel l_2$ and $\angle A=125^\circ$, $\angle B=85^\circ$, \\
it follows that $\angle 4+\angle 2=85^\circ$ and $\angle 1+\angle 3=125^\circ$, with $\angle 3+\angle 4=180^\circ$, \\
thus, $85^\circ-\angle 2+125^\circ-\angle 1=180^\circ$, \\
therefore $\angle 1+\angle 2=30^\circ$, \\
since $\angle 1$ is $4^\circ$ larger than $\angle 2$, \\
it follows that $\angle 2=\angle 1-4^\circ$, \\
hence, $2\angle 1=34^\circ$, \\
thus, $\angle 1=\boxed{17^\circ}$;


Question: As shown in the figure, $OC$ bisects $\angle AOB$, and $\angle BOC=22^\circ$. What is the measure of $\angle AOB$?

Solution:
\textbf{Solution:} Since OC bisects $\angle AOB$ and $\angle BOC=22^\circ$,\\
it follows that $\angle AOB =2\angle BOC=44^\circ$.\\
Therefore, $\angle AOB = \boxed{44^\circ}$.


Question: As shown in the figure, $\angle AOB = 120^\circ$, and $OC$ bisects $\angle AOB$. If $\angle COD = 20^\circ$, what is the measure of $\angle BOD$ in degrees?

Solution:
\textbf{Solution:} Given that $\angle AOB=120^\circ$ and OC bisects $\angle AOB$.\\
Therefore, $\angle BOC= \frac{1}{2}\angle AOB=60^\circ$.\\
Given that $\angle COD=20^\circ$,\\
Therefore, $\angle BOD=\angle BOC -\angle COD$ \\
$=60^\circ-20^\circ$ \\
$=\boxed{40^\circ}$


Question: As shown in the figure, $AD = \frac{1}{2}DB$, and $E$ is the midpoint of $BC$, with $BE = \frac{1}{5}AC = 2\, \text{cm}$. How long is segment $DE$ in centimeters?

Solution:
\textbf{Solution:} By the problem statement, we know $DB=\frac{2}{3}AB$, $BE=EC=2\text{cm}$, and $AC=10\text{cm}$. \\
$\therefore$ $AB=AC-BC=6\text{cm}$. \\
$\therefore$ $DB=\frac{2}{3}AB=4\text{cm}$. \\
$\therefore$ $DE=DB+BE=\boxed{6}\text{cm}$.


Question: As shown in the diagram, $P$ is a point on line segment $MN$, and $Q$ is the midpoint of line segment $PN$. If $MN = 10$ and $MP = 6$, what is the length of $MQ$?

Solution:
\textbf{Solution:} Since $MN = 10$ and $MP = 6$, \\
it follows that $NP = MN - MP = 4$, \\
because point Q is the midpoint of NP, \\
it follows that $PQ = QN = \frac{1}{2} NP = 2$, \\
thus $MQ = MP + PQ = 6 + 2 = \boxed{8}$.


Question: As shown in the figure, point O is on line AB, and ray OC bisects $\angle AOD$. If $\angle AOC = 35^\circ$, then what is the measure of $\angle BOD$ in degrees?

Solution:
\textbf{Solution:} Since line OC bisects $\angle$DOA, \\
$\therefore$ $\angle$AOD = 2$\angle$AOC, \\
Since $\angle$COA = $35^\circ$, \\
$\therefore$ $\angle$DOA = $70^\circ$, \\
$\therefore$ $\angle$BOD = $180^\circ - 70^\circ = \boxed{110^\circ}$,


Question: As shown in the figure, $O$ is a point on $AB$, $OD$ bisects $\angle BOC$, $\angle 1=20^\circ$. What is the degree measure of $\angle 2$?

Solution:
\textbf{Solution:} Given that $\because \angle 1=20^\circ$,\\
$\therefore \angle BOC=180^\circ-\angle 1=180^\circ-20^\circ=160^\circ$,\\
$\because$ OD bisects $\angle BOC$,\\
$\therefore \angle 2=\boxed{80^\circ}$.


Question: As shown in the figure, point C is on line segment AB, point D is the midpoint of line segment AC, and point C is the one-fourth division point of line segment BD. If $CB = 2$, what is the length of line segment AB?

Solution:
\textbf{Solution:} Since point D is the midpoint of segment AC, and point C is the one-fourth point of segment BD\\
$\therefore AD=BD=\frac{1}{2}AB$, $BC=\frac{1}{4}BD$,\\
$\therefore BC=\frac{1}{8}AB$,\\
Since $CB=2$,\\
$\therefore AB=\boxed{16}$


Question: As shown in the figure, within $\angle AOB$, $OD$ is the bisector of $\angle BOC$, and $OE$ is the bisector of $\angle AOC$. If $\angle AOB=140^\circ$, then what is the measure of $\angle EOD$ in degrees?

Solution:
\textbf{Solution:} Since OD is the bisector of $\angle$BOC, and OE is the bisector of $\angle$AOC, \\
therefore, $\angle$DOC=$\frac{1}{2}$$\angle$BOC, $\angle$EOC=$\frac{1}{2}$$\angle$AOC, \\
thus, $\angle$DOE=$\angle$DOC+$\angle$EOC=$\frac{1}{2}$$\angle$AOB, \\
since $\angle$AOB=$140^\circ$, \\
therefore, $\angle$EOD=\boxed{70^\circ}.


Question: As shown in the figure, it is known that $OD$ bisects $\angle AOC$, and $OE$ bisects $\angle COB$. If $\angle AOD = 20^\circ$ and $\angle EOB = 40^\circ$, what is the measure of $\angle AOB$ in degrees?

Solution:
\textbf{Solution:} Since OD bisects $\angle AOC$ and OE bisects $\angle COB$, \\
$\therefore \angle AOC = 2\angle AOD$ and $\angle COB = 2\angle EOB$, \\
Since $\angle AOD = 20^\circ$ and $\angle EOB = 40^\circ$. \\
$\therefore \angle AOC = 2 \times 20^\circ = 40^\circ$ and $\angle BOC = 2 \times 40^\circ = 80^\circ$, \\
$\therefore \angle AOB = \angle AOC + \angle BOC = 40^\circ + 80^\circ = \boxed{120^\circ}$


Question: As shown in the figure, $O$ is a point on the line $AB$, $\angle COD$ is a right angle, $OE$ bisects $\angle BOC$, $\angle AOC=40^\circ$, then what is the degree measure of $\angle DOE$?

Solution:
\textbf{Solution:} Given $\because$ $\angle AOC = 40^\circ$,\\
$\therefore$ $\angle BOC = 180^\circ - \angle AOC = 140^\circ$,\\
$\because$ OE bisects $\angle BOC$,\\
$\therefore$ $\angle COE = \angle BOE = 70^\circ$,\\
$\because$ $\angle COD$ is a right angle, that is $\angle COD = 90^\circ$,\\
$\therefore$ $\angle DOE = \angle COD - \angle COE = \boxed{20^\circ}$.


Question: As shown in the figure, extend the line segment AB to C such that $BC = \frac{1}{2}AB$. Let D be the midpoint of the line segment AC. If $DC = 3$, then what is the length of AB?

Solution:
\textbf{Solution:}

Since $D$ is the midpoint of $AC$, we have

\[AC = 2DC\]
\[= 2 \times 3\]
\[= 6\]

Also, since $BC = \frac{1}{2}AB$ and $AC = AB + BC$,

\[\therefore BC = \frac{1}{3}AC\]
\[= \frac{1}{3} \times 6\]
\[= 2\]

From the relationship between the sums and differences of line segments, we have

\[AB = AC - BC\]
\[= 6 - 2\]
\[= AB = \boxed{4}\]


Question: As shown in the figure, it is known that $\angle AOB$ is a right angle, $OC$ is the bisector of $\angle AOB$, and $\angle AOD = 2\angle BOD$. What is the measure of $\angle COD$?

Solution:
\textbf{Solution:} Since $\angle AOB$ is a right angle, and OC bisects $\angle AOB$,

$\therefore$ $\angle AOB=90^\circ$, $\angle AOC=\angle BOC=45^\circ$,

Since $\angle AOD=2\angle BOD$, and $\angle AOD+\angle BOD=90^\circ$,

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

$\therefore$ $\angle COD=\angle BOC−\angle DOB=45^\circ−30^\circ=\boxed{15^\circ}$.


Question: As shown in the diagram, a pair of set squares is overlaid. Given that $\angle OAB=\angle OCD=90^\circ$, $\angle AOB=45^\circ$, and $\angle COD=60^\circ$, with $OB$ bisecting $\angle COD$, what is the measurement of $\angle AOC$ in degrees?

Solution:
\textbf{Solution:} Given that OB bisects $\angle$COD and $\angle$COD = 60°,\\
it follows that $\angle$BOC = 30°,\\
and since $\angle$AOB = 45°,\\
it implies $\angle$AOC = $\angle$AOB - $\angle$BOC = 45° - 30° = \boxed{15^\circ}.


Question: As shown in the figure, points $B$ and $C$ divide the line segment $AD$ into three parts in the ratio $2:5:3$, with $M$ being the midpoint of $AD$. Given that $BM=6\,\text{cm}$, what is the length of $CM$ in centimeters?

Solution:
Solution: Let AB=2x cm, BC=5x cm, CD=3x cm,\\
$\therefore$AD=AB+BC+CD=10x cm,\\
$\because$ M is the midpoint of AD,\\
$\therefore$AM=MD= $\frac{1}{2}$ AD=5x cm,\\
$\therefore$BM=AM−AB=5x−2x=3x cm,\\
$\because$BM=6 cm,\\
$\therefore$x=2 cm,\\
hence CM=BC−BM=5×2−6=\boxed{4}cm.


Question: As shown in the figure, $a\parallel b$, $\angle 1=\angle 2$, $\angle 3=40^\circ$, then what is the degree measure of $\angle 4$?

Solution:
\textbf{Solution:} Given that $a \parallel b$,\\
it follows that $\angle 1 + \angle 2 + \angle 3 = 180^\circ$,\\
since $\angle 3 = 40^\circ$,\\
then $\angle 1 + \angle 2 = 180^\circ - 40^\circ = 140^\circ$,\\
given that $\angle 1 = \angle 2$,\\
thus $\angle 2 = 70^\circ$,\\
since $a \parallel b$,\\
it concludes that $\angle 4 = \angle 2 = \boxed{70^\circ}$.


Question: As shown in the figure, there is a triangular piece $ABC$, folded along the median $AD$ on the side $BC$, such that point $C$ falls on point $E$, and $AE$ intersects $BC$ at point $F$. If $AD$ is perpendicular to $AB$, and $AB=2$, $AD=3$, what is the length of $AF$?

Solution: \textbf{Solution:} As shown in the figure, draw $CG \perp AD$ at point $G$, and draw $DM \perp AD$, $DN \perp AD$ with the perpendicular foot at $D$. Since $AD$ is the median of $\triangle ABC$, it is easy to prove $\triangle ABD \cong \triangle GCD$. Therefore, $CG=AB=2$, $GD=AD=3$, hence $AC=\sqrt{AG^{2}+CG^{2}}=\sqrt{6^{2}+2^{2}}=2\sqrt{10}$. Since $CG \perp AD$ and $DM \perp AD$, it follows that $DM \parallel CG$, therefore $\triangle AMD \sim \triangle ACG$. Since $GD=AD$, it follows that $AM=MC=\frac{1}{2}AC=\sqrt{10}$, $DM=\frac{1}{2}CG=1$. Folding makes it easy to prove $\triangle AMD \cong \triangle AND$, therefore $AN=AM=\sqrt{10}$, $DN=DM=1$. Since $AD \perp AB$, $DN \perp AD$, it follows that $AB \parallel DN$, therefore $\frac{AF}{FN}=\frac{AB}{DN}=\frac{2}{1}$, hence $AF=\frac{2}{3}AN=\frac{2\sqrt{10}}{3}$. Therefore, the answer is: $\boxed{\frac{2\sqrt{10}}{3}}$.


Question: As shown in the figure, in $\triangle ABC$, point D is on side $AB$, and $\angle ACD = \angle ABC$. If $AC = \sqrt{3}$ and $AD = 1$, what is the length of $DB$?

Solution:
\textbf{Solution:} Given that $\angle ACD = \angle ABC$ and $\angle A = \angle A$,\\
it follows that $\triangle ABC \sim \triangle ACD$,\\
hence, $\frac{AD}{AC} = \frac{AC}{AB}$,\\
given that AC = $\sqrt{3}$ and AD = 1,\\
it implies that $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{AB}$,\\
therefore, AB = 3,\\
thus, BD = AB - AD = 3 - 1 = \boxed{2}.


Question: As shown in the figure, in rectangle $ABCD$, $BD$ is the diagonal. The rectangle $ABCD$ is folded along the lines containing $BE$ and $BF$, making point $A$ coincide with point $M$ on $BD$, and point $C$ with point $N$ on $BD$, then $EF$ is drawn. Given that $AB=3$ and $BC=4$, what is the length of $EF$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle, \\
$\therefore$AB=CD=3, AD=BC=4, $\angle$A=$\angle$C=$\angle$EDF=90$^\circ$,\\
$\therefore$BD=$\sqrt{AB^2+AD^2}=\sqrt{3^2+4^2}=5$,\\
Since rectangle ABCD is folded along the straight line containing BE, making point A fall on point M on BD,\\
$\therefore$AE=EM, $\angle$A=$\angle$BME=90$^\circ$,\\
$\therefore$$\angle$EMD=90$^\circ$,\\
Since $\angle$EDM=$\angle$ADB,\\
$\therefore$$\triangle EDM\sim \triangle BDA$,\\
$\therefore$$\frac{ED}{BD}=\frac{EM}{AB}$,\\
Let DE=x, then AE=EM=4−x,\\
$\therefore$$\frac{x}{5}=\frac{4−x}{3}$,\\
Solving for x, we get x=$\frac{5}{2}$,\\
$\therefore$DE=$\frac{5}{2}$,\\
Similarly, $\triangle DNF\sim \triangle DCB$,\\
$\therefore$$\frac{DF}{BD}=\frac{NF}{BC}$,\\
Let DF=y, then CF=NF=3−y,\\
$\therefore$$\frac{y}{5}=\frac{3−y}{4}$,\\
Solving for y, we get y=$\frac{5}{3}$,\\
$\therefore$DF=$\frac{5}{3}$,\\
$\therefore$EF=$\sqrt{DE^2+DF^2}=\sqrt{(\frac{5}{2})^2+(\frac{5}{3})^2}=\boxed{\frac{5\sqrt{13}}{6}}$.


Question: In parallelogram $ABCD$, $E$ is the trisection point of $AD$ close to $D$. Connect $BE$ and intersect $AC$ at point $F$. Given $AC = 12$, what is the length of $AF$?

Solution:
\textbf{Solution:} Given in $\square ABCD$, where AD = BC and AD$\parallel$BC,\\
since E is the trisection point of AD,\\
it follows that AE = $\frac{2}{3}$ AD = $\frac{2}{3}$ BC,\\
since AD$\parallel$BC,\\
it implies $\triangle AEF\sim \triangle CBF$,\\
therefore, $\frac{AF}{FC}=\frac{AE}{BC}=\frac{2}{3}$,\\
given AC = 12,\\
thus, AF = $\frac{2}{5}\times 12=\boxed{\frac{24}{5}}$;


Question: As shown in the figure, in rectangle $ABCD$, $E$ and $F$ are the midpoints of the sides $CD$ and $AD$ respectively, $CF$ intersects $EA$ and $EB$ at points $M$ and $N$ respectively. Given $AB=8$ and $BC=12$, what is the length of $MN$?

Solution:
\textbf{Solution:} As shown in the figure, extend BE and AD to meet at point Q. Since quadrilateral ABCD is a rectangle, $BC=12$, thus $\angle BAD=90^\circ$, $AD=BC=12$, $AD\parallel BC$. Since F is the midpoint of AD, it follows that $AF=DF=6$. In $\triangle DCF$, by the Pythagorean theorem, we get: $CF=\sqrt{DF^2+DC^2}=\sqrt{6^2+8^2}=10$. Since $AD\parallel BC$, it follows that $\angle Q=\angle EBC$. Since E is the midpoint of CD and $CD=8$, it follows that $CE=DE=4$. In $\triangle DQE$ and $\triangle CBE$, we have $\begin{cases}
\angle DEQ=\angle BEC\\
\angle Q=\angle EBC\\
DE=CE
\end{cases}$, thus $\triangle DQE\cong \triangle CBE$ (AAS), hence $DQ=BC=12$, that is $QF=12+6=18$. Since $AD\parallel BC$, it follows that $\triangle QNF\sim \triangle BNC$, hence $\frac{FN}{CN}=\frac{QF}{BC}=\frac{18}{12}=\frac{3}{2}$. Since $CF=10$, we get $CN=4$, $FN=6$. As illustrated, extend CF and BA to meet at W. Since $WB\parallel CD$, it follows that $\angle W=\angle DCF$. In $\triangle WAF$ and $\triangle CDF$, we have $\begin{cases}
\angle W=\angle DCF\\
AF=DF\\
\angle AFW=\angle CFD
\end{cases}$, thus $\triangle WAF\cong \triangle CDF$ (AAS), hence $CD=AW=8$, $BW=16$, $CF=FW=10$. Because $AB\parallel CD$, it follows that $\triangle CME\sim \triangle WMA$, hence $\frac{CM}{MW}=\frac{CE}{AW}$, solving this we get: $CM=\frac{20}{3}$, hence $MN=CM-CN=\frac{20}{3}-4=\boxed{\frac{8}{3}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AB \parallel CD$, $\angle B = 90^\circ$, $AB = 1$, $CD = 2$, $BC = m$. Point $P$ is a moving point on side $BC$. If $\triangle PAB$ is similar to $\triangle PCD$, and there are exactly two points $P$ that satisfy this condition, what is the value of $m$?

Solution:
\textbf{Solution}: When $\angle BAP = \angle CPD$, then $\triangle PAB \sim \triangle PDC$,\\
$\therefore \frac{AB}{DC} = \frac{BP}{PC}$, that is, $\frac{1}{2} = \frac{x}{m-x}$,\\
$\therefore x = \frac{m}{3}$;\\
When $\angle BAP = \angle CPD$, then $\triangle PAB \sim \triangle DPC$,\\
$\therefore \frac{AB}{CP} = \frac{BP}{CD}$, that is, $\frac{1}{m-x} = \frac{x}{2}$,\\
$\therefore x^2 - mx + 2 = 0$,\\
$\because$ there are exactly two points P that satisfy the condition,\\
$\therefore$ the value of $x$ from (1) is a solution in (2), $\left(\frac{m}{3}\right)^2 - m \times \frac{m}{3} + 2 = 0$, yielding $m = 3$ or $-3$ (discard);\\
Equation in (2) has two equal real roots,\\
$\therefore \Delta = m^2 - 4 \times 2 = 0$, yielding $m = 2\sqrt{2}$ or $-2\sqrt{2}$ (discard).\\
Hence, the answer is: $m = \boxed{3}$ or $\boxed{2\sqrt{2}}$.


Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is on side $BC$, and $AE$ intersects diagonal $BD$ at point $F$. If $BE:BC=2:3$, then what is the value of $\frac{EF}{AE}$?

Solution:
\textbf{Solution}: Since quadrilateral ABCD is a parallelogram, \\
therefore BC$\parallel$AD and BC=AD, \\
therefore $\frac{BE}{BC} = \frac{BE}{AD}$, $\triangle BEF\sim \triangle DAF$, \\
therefore $\frac{BE}{AD}=\frac{BF}{FD}=\frac{EF}{AF}=\frac{2}{3}$, \\
therefore $AF=\frac{3}{2}EF$, \\
therefore $AE=AF+EF=\frac{5}{2}EF$, \\
therefore $\frac{EF}{AE}=\boxed{\frac{2}{5}}$;


Question: As shown in the figure, $A$, $B$, $C$, and $D$ are four points on the circle $\odot O$. $AB=AC$, $AD$ intersects $BC$ at point $E$, $AE=4$, and $ED=6$. What is the length of $AC$?

Solution:
\[ \textbf{Solution:} \text{Since } AB=AC, \]
\[ \therefore \angle ABE=\angle ACE, \]
\[ \text{Since } \angle ACE=\angle ADB, \]
\[ \therefore \angle ABE=\angle ADB, \]
\[ \text{Since } \angle BAE=\angle DAB, \]
\[ \therefore \triangle ABE\sim \triangle ADB, \]
\[ \therefore \frac{AB}{AD}=\frac{AE}{AB}, \]
\[ \text{Thus, } AB^2=AD\cdot AE, \]
\[ \text{Since } AE=4, ED=6, \]
\[ \therefore AD=10, \]
\[ \therefore AB=\sqrt{AD\cdot AE}=\sqrt{10\times 4}=2\sqrt{10}, \]
\[ \therefore AC=AB=2\sqrt{10}, \]
\[ \text{Hence, the answer is: } AC=\boxed{2\sqrt{10}}. \]


Question: As shown in the figure, two lines are intersected by three parallel lines, with $DE = 4$, $EF = 6$, and $AB = 2$. What is the value of $AC$?

Solution:
\textbf{Solution:} Given: $l_1 \parallel l_2$,\\
Therefore, $\frac{DE}{EF}=\frac{AB}{BC}$,\\
Therefore, $\frac{4}{6}=\frac{2}{BC}$,\\
Therefore, $BC=3$,\\
Therefore, $AC=AB+BC=2+3=\boxed{5}$.


Question: As shown in the figure, the side length of square $ABCD$ is $12$, and the radius of $\odot B$ is $6$. Point $P$ is a moving point on $\odot B$. What is the minimum value of $PD + \frac{1}{2}PC$?

Solution:
\textbf{Solution:} As shown in the diagram, take $BE=3$ on $BC$ and connect $BP$, $PE$,

$\because$ the side length of square $ABCD$ is 12 and the radius of $\odot B$ is 6,

$\therefore$ $BC=12=CD$, $BP=6$, $EC=9$,

$\because$ $\frac{BP}{BC}=\frac{1}{2}=\frac{BE}{BP}$, and $\angle PBE=\angle PBC$,

$\therefore$ $\triangle PBE \sim \triangle CBP$,

$\therefore$ $\frac{BE}{BP}=\frac{PE}{PC}=\frac{1}{2}$,

$\therefore$ $PE=\frac{1}{2}PC$,

$\therefore$ $PD+\frac{1}{2}PC=PD+PE$,

$\therefore$ when points $D$, $P$, and $E$ are collinear, $PD+PE$ has the minimum value, that is, $PD+\frac{1}{2}PC$ has the minimum value,

$\therefore$ the minimum value of $PD+\frac{1}{2}PC$ is $DE=\sqrt{DC^2+CE^2}=\boxed{15}$,


Question: As shown in the figure, point $P$ is a point on the bisector $OC$ of $\angle MON$, with $P$ as the vertex, the sides of $\angle APB$ intersect rays $OM$ and $ON$ at points $A$ and $B$, respectively. If $\angle APB$ always satisfies $OA \cdot OB = OP^2$ while rotating around point $P$, we call $\angle APB$ the associated angle of $\angle MON$. If $\angle MON = 50^\circ$ and $\angle APB$ is the associated angle of $\angle MON$, what is the degree measure of $\angle APB$?

Solution:
\textbf{Solution:} \\
Since $OA \cdot OB = OP^{2}$, \\
it follows that $\frac{OB}{OP} = \frac{OP}{OA}$, \\
Since $OP$ bisects $\angle MON$, \\
it follows that $\angle AOP = \angle BOP$, \\
thus, $\triangle AOP \sim \triangle POB$, \\
therefore, $\angle OPA = \angle OBP$, \\
In $\triangle OBP$, $\angle BOP = \frac{1}{2}\angle MON = 25^\circ$, \\
thus, $\angle OBP + \angle OPB = 180^\circ - 25^\circ = 155^\circ$, \\
hence, $\angle OPA + \angle OPB = 155^\circ$, \\
which means $\angle APB = \boxed{155^\circ}$.


Question: As the figure shows, $AD$ is the median of $\triangle ABC$, and $E$ is a point on $AD$ with $DE=2AE$. The extension of $CE$ intersects $AB$ at point $F$. If $AF=1.6\text{cm}$, then what is the length of $AB$ in $\text{cm}$?

Solution:
\textbf{Solution:} Let DM be drawn parallel to CF and intersect AB at M,\\
since $DM\parallel CF$,\\
then $\frac{BD}{DC}=\frac{BM}{MF}$,\\
since AD is the median of $\triangle ABC$,\\
then D is the midpoint of BC,\\
thus BD=DC,\\
thus $\frac{BD}{DC}=\frac{BM}{MF}=1$,\\
thus BM=MF,\\
since EF$\parallel$MD,\\
then $\triangle AFE\sim \triangle AMD$,\\
also since DE=2AE,\\
then $\frac{AF}{AM}=\frac{AE}{AD}=\frac{AE}{DE+AE}=\frac{AE}{2AE+AE}=\frac{1}{3}$,\\
thus AM=3AF,\\
since AM=AF+MF=3AF,\\
then MF=2AF,\\
since BM=MF,\\
then BM=MF=2AF,\\
thus AB=AF+MF+BF=5AF,\\
since AF=1.6,\\
thus AB=$5\times 1.6$=\boxed{8},


Question: As shown in the figure, the line $DE$ parallel to $BC$ divides $\triangle ABC$ into two parts with an area ratio of 1$\colon$4. What is the value of $\frac{AD}{AB}$?

Solution:
\textbf{Solution:} Given that line DE divides $\triangle ABC$ into two parts with an area ratio of 1:4,\\
it follows that $S_{\triangle ADE} = \frac{1}{5}S_{\triangle ABC}$,\\
Since DE$\parallel$BC,\\
it implies that $\triangle ADE \sim \triangle ABC$,\\
thus $\frac{S_{\triangle ADE}}{S_{\triangle ABC}}=\left(\frac{AD}{AB}\right)^2=\frac{1}{5}$,\\
therefore $\frac{AD}{AB}=\boxed{\frac{\sqrt{5}}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, D and E are points on sides BC and AC, respectively, and AD intersects BE at point F. If E is the midpoint of AC, and the ratio of BD to DC is 2:3, what is the value of AF: FD?

Solution:
\textbf{Solution:} Let line DH be drawn through point D such that DH $\parallel$ AC, intersecting BE at H.\\
$\therefore \triangle DHF\sim \triangle AEF$, $\triangle BDH\sim \triangle BCE$,\\
$\therefore \frac{DH}{AE}=\frac{DF}{AF}$, $\frac{DH}{CE}=\frac{BD}{BC}$,\\
$\because$ If E is the midpoint of AC,\\
$\therefore$ CE＝AE,\\
$\therefore \frac{DF}{AF}=\frac{BD}{BC}$,\\
$\because$ The ratio BD: DC＝2: 3,\\
$\therefore$ The ratio BD: BC＝2: 5,\\
$\therefore$ The ratio DF: AF＝2: 5,\\
$\therefore$ The ratio AF: FD＝$\boxed{\frac{5}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=10$ and $AD\perp BC$ at point $D$, $AD=8$. If point $E$ is the centroid of $\triangle ABC$, and point $F$ is the centroid of $\triangle ACD$, then what is the area of $\triangle AEF$?

Solution:
Solution:As shown in the figure, extend AF to meet BC at point M, \\
$\because AB=AC=10, AD\perp BC, AD=8$, \\
$\therefore CD=BD=\sqrt{AB^2-AD^2}=6$, \\
$\because$ point E is the centroid of $\triangle ABC$, point F is the centroid of $\triangle ACD$, \\
$\therefore \frac{AE}{DE}=2, \frac{AF}{FM}=2, DM=\frac{1}{2}CD=3$, \\
$\therefore \frac{AE}{AD}=\frac{AF}{AM}=\frac{2}{3}, AE=\frac{2}{3}AD=\frac{16}{3}$, \\
Furthermore, $\because \angle EAF=\angle DAM$, \\
$\therefore \triangle AEF\sim \triangle ADM$, \\
$\therefore \frac{EF}{DM}=\frac{AE}{AD}=\frac{2}{3}, \angle AEF=\angle ADM=90^\circ$, \\
$\therefore \frac{EF}{3}=\frac{2}{3}$, \\
Solving gives $EF=2$, \\
Hence, the area of $\triangle AEF$ is $\frac{1}{2}AE\cdot EF=\frac{1}{2}\times \frac{16}{3}\times 2=\boxed{\frac{16}{3}}$.


Question: As shown in the figure, the sidelength of square $ABCD$ is $4$. Let the point $E$ be a moving point on side $AD$, and the point $F$ be on side $CD$ with the constraint that the length of segment $EF$ is $4$. Let the point $G$ be the midpoint of segment $EF$. Connecting $BG$ and $CG$, what is the minimum value of $BG+\frac{1}{2}CG$?

Solution:
\textbf{Solution:} As shown in the diagram, \\
in right triangle $\triangle DEF$, where $G$ is the midpoint of $EF$, \\
$\therefore DG = \frac{1}{2}EF = 2$, \\
$\therefore$ point $G$ moves on a circle with $D$ as the center and 2 as the radius, \\
Cut $DI = 1$ on $CD$, and connect $GI$, \\
$\therefore \frac{DI}{DG} = \frac{DG}{CD} = \frac{1}{2}$, \\
$\therefore \angle GDI = \angle CDG$, \\
$\therefore \triangle GDI \sim \triangle CDG$, \\
$\therefore \frac{IG}{CG} = \frac{DI}{DG} = \frac{1}{2}$, \\
$\therefore IG = \frac{1}{2}CG$, \\
$\therefore BG + \frac{1}{2}CG = BG + IG \geq BI$, \\
$\therefore$ when $B$, $G$, and $I$ are collinear, $BG + \frac{1}{2}CG$ is minimized $=$ $BI$, \\
In right $\triangle BCI$, where $CI = 3$ and $BC = 4$, \\
$\therefore BI = \boxed{5}$.


Question: Arrange 2020 squares with side length 1 as shown in the figure. The points A, $A_{1}$, $A_{2}$, $A_{3}$, \ldots, $A_{2020}$ and points M, $M_{1}$, $M_{2}$, \ldots, $M_{2019}$ are the vertices of the squares. The lines connecting AM$_{1}$, AM$_{2}$, AM$_{3}$, \ldots, AM$_{2019}$ intersect the sides of the squares $A_{1}M$, $A_{2}M_{1}$, $A_{3}M_{2}$, \ldots, $A_{2019}M_{2018}$ at the points $N_{1}$, $N_{2}$, $N_{3}$, \ldots, $N_{2019}$, respectively. The area of the quadrilateral $M_{1}N_{1}A_{1}A_{2}$ is denoted as $S_{1}$, the area of the quadrilateral $M_{2}N_{2}A_{2}A_{3}$ is denoted as $S_{2}$, \ldots, then what is the area of $S_{2019}$?

Solution:
\textbf{Solution:} As shown in the figure, assume the upper left vertex of the leftmost square is denoted as O.\\
$\because$ The 2019 squares, each with a side length of 1, are arranged as illustrated.\\
$\therefore$ OA$\parallel$MA$_{1}$$\parallel$M$_{1}$A$_{2}$$\parallel$M$_{2}$A$_{3}$$\parallel$…$\parallel$M$_{2018}$A$_{2019}$.\\
$\therefore$ $\triangle$M$_{1}$MN$_{1}$$\sim$$\triangle$M$_{1}$OA.\\
$\therefore$ $\frac{MM_{1}}{OM_{1}}=\frac{MN_{1}}{OA}=\frac{1}{2}$.\\
$\therefore$MN$_{1}$=$\frac{1}{2}$,\\
$\therefore$The area of quadrilateral M$_{1}$N$_{1}$A$_{1}$A$_{2}$, S$_{1}$=$1-\frac{1}{2}\times1\times\frac{1}{2}=\frac{3}{4}$;\\
Similarly, it follows that: $ \frac{M_{1}M_{2}}{OM_{2}}=\frac{M_{1}N_{2}}{OA}=\frac{1}{3}$.\\
$\therefore$The area of quadrilateral M$_{2}$N$_{2}$A$_{2}$A$_{3}$, S$_{2}$=$1-\frac{1}{2}\times1\times\frac{1}{3}=\frac{5}{6}$;\\
…\\
$\therefore$The area of quadrilateral MnNnAnAn$_{+1}$, Sn=$1-\frac{1}{2(n+1)}=\frac{2n+1}{2n+2}$.\\
$\therefore$S$_{2019}$=$\boxed{\frac{4039}{4040}}$;


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

Solution:
\textbf{Solution:} Let the side length EF=EH=x of the square EFGH, \\
$\because$ EFGH is a square, \\
$\therefore$ $\angle$HEF=$\angle$EHG=90$^\circ$, EF$\parallel$BC, \\
$\therefore$ $\triangle$AEF$\sim$ $\triangle$ABC, \\
$\because$ AD is the altitude of $\triangle$ABC, \\
$\therefore$ $\angle$HDN=90$^\circ$, \\
$\therefore$ the quadrilateral EHDN is a rectangle, \\
$\therefore$ DN=EH=x, \\
$\because$ $\triangle$AEF$\sim$ $\triangle$ABC, \\
$\therefore$ $\frac{AN}{AD}=\frac{EF}{BC}$, \\
$\because$ BC=120, AD=60, \\
$\therefore$ AN=60−x, \\
$\therefore$ $\frac{60−x}{60}=\frac{x}{120}$, \\
Solving for x, we get: x=40, \\
$\therefore$ AN=60−x=60−40=\boxed{20}.


Question: As shown in the figure, the right triangle $\triangle ABC$ has $\angle ACB=90^\circ$, with $AC=BC=3$. A square $CDEF$ with $C$ as the vertex (the four vertices $C$, $D$, $E$, and $F$ are arranged counterclockwise) can freely rotate around point $C$, and $CD=2$. Connect $AF$ and $BD$. During the rotation of square $CDEF$, what is the minimum value of $BD+\frac{2}{3}AD$?

Solution:
\textbf{Solution}: As shown in the figure, take a point \(M\) on \(AC\) such that \(CM=\frac{4}{3}\). Connect \(DM\) and \(AD\).\\
\(\because\) \(CD=2\), \(CM=\frac{4}{3}\), \(CA=3\),\\
\(\therefore\) \(CD^{2}=CM\cdot CA\),\\
\(\therefore\) \(\frac{CD}{CA}=\frac{CM}{CD}\),\\
\(\because\) \(\angle DCM=\angle ACD\),\\
\(\therefore\) \(\triangle DCM \sim \triangle ACD\),\\
\(\therefore\) \(\frac{DM}{AD}=\frac{CD}{AC}=\frac{2}{3}\),\\
\(\therefore\) \(DM=\frac{2}{3}AD\),\\
\(\therefore\) \(BD+\frac{2}{3}AD=BD+DM\),\\
\(\therefore\) when \(B\), \(D\), \(M\) are collinear, the value of \(BD+\frac{2}{3}AD\) is minimized,\\
\(\therefore\) The minimum value = \(\sqrt{CB^{2}+CM^{2}}=\sqrt{3^{2}+(\frac{4}{3})^{2}}=\boxed{\frac{\sqrt{97}}{3}}\).


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle with a side length of $1$. By folding $\triangle ABC$ along the line $AC$, we obtain $\triangle AB^{\prime}C$. Then, by translating $\triangle AB^{\prime}C$ along the line $AC$, we obtain $\triangle A^{\prime}B^{\prime\prime}C^{\prime}$. What is the minimum perimeter of $\triangle BB^{\prime\prime}C^{\prime}$?

Solution:
\textbf{Solution:} Let $E$ be the symmetric point of $B'$ with respect to point $B$, and let the extension of $B'B$ intersect $AC$ at point $O$. Connect $EC$,\\
$\because$ translating $\triangle ABC'$ along line $AC$, we obtain $\triangle AB''C'$,\\
$\therefore$ we can keep $\triangle ABC'$ stationary and translate point $B$ parallel to line $AC$,\\
$\therefore$ the problem of finding the minimum perimeter of $\triangle BB''C'$ is transformed into finding the minimum perimeter of $\triangle BB'C$,\\
$\therefore$ when points $E$, $B$, and $C$ are collinear, the sum of $BB'+BC$ is minimized, being equal to the length of $CE$,\\
$\because$ both $\triangle ABC$ and $\triangle AB'C$ are equilateral triangles,\\
$\therefore$ $AB=BC=CB'=AB'$,\\
$\therefore$ the quadrilateral $ABCB'$ is a rhombus,\\
$\therefore$ $BB'\perp AC$, and $OC=\frac{1}{2}AC=\frac{1}{2}$,\\
$\therefore$ $BO=B'O=\frac{\sqrt{3}}{2}$,\\
$\therefore$ $OE=BE+OB=\sqrt{3} +\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$,\\
In $\triangle CEO$, by the Pythagorean theorem:\\
$CE=\sqrt{OE^2+OC^2}=\sqrt{\left(\frac{3\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2}=\sqrt{7}$,\\
$\therefore$ the minimum perimeter of $\triangle BB'C$ is: $\sqrt{7}+1$,\\
which means the minimum perimeter of $\triangle BB''C'$ is: $\boxed{\sqrt{7}+1}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AD \parallel BC$, $\angle B = 70^\circ$, $\angle C = 40^\circ$, $DE \parallel AB$ intersects $BC$ at point $E$. If $AD = 5\,cm$, $BC = 17\,cm$, then how many $cm$ is $CD$?

Solution:
\textbf{Solution:} Given that $AD\parallel BC$, $DE\parallel AB$,\\
therefore, quadrilateral ABED is a parallelogram,\\
thus, BE=AD=5,\\
therefore, EC=BC-BE=17cm-5cm=12cm,\\
since $DE\parallel AB$,\\
we have $\angle DEC=\angle B=70^\circ$,\\
given that $\angle C=40^\circ$,\\
thus, $\angle EDC=180^\circ-\angle DEC-\angle C=70^\circ$,\\
therefore, $\angle EDC=\angle DEC$,\\
thus, CD=CE=\boxed{12cm}.


Question: Given: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, $OE\parallel DC$ intersects $BC$ at point $E$, $AC=6$, and $BD=8$. What is the length of $OE$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, with AC=6, and BD=8,\\
it follows that OA=OC=$\frac{1}{2}$AC=3, and OB=$\frac{1}{2}$BD=4, also AC$\perp$BD,\\
therefore AB=$\sqrt{OA^{2}+OB^{2}}$=5,\\
since OE$\parallel$DC$\parallel$AB,\\
it follows that OE is the midline of $\triangle ABC$,\\
thus OE=$\frac{1}{2}$AB=\boxed{2.5}.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $BC=3\sqrt{2}$, and $AC=4\sqrt{2}$. The bisectors of $\angle CAB$ and $\angle CBA$ intersect at point $P$. Points $D$ and $E$ are located on sides $AC$ and $BC$, respectively (neither coincides with point $C$), and it is given that $\angle DPE=45^\circ$. What is the distance from point $P$ to side $AB$? What is the perimeter of $\triangle CDE$?

Solution:
\textbf{Solution:} As shown in the figure, draw $PG \perp AB$ at $G$, $PH \perp AC$ at $H$, and $PM \perp BC$ at $M$. Take a point $M'$ on $AH$ such that $M'H=EM$, and connect $PM'$,\\
$\because$ the angle bisectors of $\angle CAB$ and $\angle CBA$ intersect at point $P$,\\
$\therefore$ $PG=PM=PH$,\\
$\because$ $\angle PMC=\angle C=\angle PHC=90^\circ$,\\
$\therefore$ quadrilateral $PMCH$ is a square,\\
$\therefore$ $CM=CH=PH=PM$, $\angle MPH=90^\circ$,\\
$\because$ $\angle DPE=45^\circ$,\\
$\therefore$ $\angle MPE+\angle DPH=45^\circ$,\\
In $\triangle PME$ and $\triangle PHM'$,\\
$\begin{cases} PM=PH \\ \angle PME=\angle PHM'=90^\circ \\ ME=HM' \end{cases}$,\\
$\therefore$ $\triangle PME \cong \triangle PHM'$ (SAS),\\
$\therefore$ $PE=PM'$, $\angle MPE=\angle HPM'$,\\
$\therefore$ $\angle HPM'+\angle DPH=\angle DPH+\angle EPM=45^\circ=\angle DPE$,\\
In $\triangle DPM'$ and $\triangle DPE$,\\
$\begin{cases} PD=PD \\ \angle DPM'=\angle DPE \\ PE=PM' \end{cases}$,\\
$\therefore$ $\triangle DPM' \cong \triangle DPE$ (SAS),\\
$\therefore$ $DE=DM'=DH+HM'=DH+ME$,\\
$\therefore$ the perimeter of $\triangle CDE=CD+CE+DE=CD+CE+DH+EM=CH+CM$,\\
$\because$ $\angle PGB=\angle PMB=90^\circ$, $\angle PBG=\angle PBM$,\\
$\therefore$ $\angle BPG=\angle BPM$,\\
$\therefore$ $BG=BM$,\\
Similarly, we get: $AG=AH$,\\
In $\triangle ACB$, $\angle C=90^\circ$, $AC=4\sqrt{2}$, $BC=3\sqrt{2}$,\\
By Pythagorean theorem: $AB=\sqrt{(4\sqrt{2})^2+(3\sqrt{2})^2}=5\sqrt{2}$,\\
$\therefore$ $AG+BG=AH+BM=5\sqrt{2}$,\\
$\because$ $AC+BC=3\sqrt{2}+4\sqrt{2}=7\sqrt{2}$,\\
$\therefore$ $CH+CM=AC+BC−AH−BM=7\sqrt{2}−5\sqrt{2}=2\sqrt{2}$,\\
$\therefore$ the distance from point $P$ to side $AB$ is $PG=\boxed{\sqrt{2}}$, the perimeter of $\triangle CDE$ is $\boxed{2\sqrt{2}}$.


Question: As shown in the figure, $\angle 1$ and $\angle 2$ are vertically opposite angles, $\angle 1 = 180^\circ - a$, $\angle 2 = 35^\circ$, then $a=$?

Solution:
\textbf{Solution:} Since $\angle 1$ and $\angle 2$ are vertical angles,\\
it follows that $\angle 1 = \angle 2$,\\
Since $\angle 1=180^\circ-a$  and  $\angle 2=35^\circ$,\\
it follows that $180^\circ-a=35^\circ$,\\
hence, $a=\boxed{145^\circ}$


Question: As illustrated in the diagram, Island C is located at $50^\circ$ east of north from Island A, and Island C is at $40^\circ$ west of north from Island B. What is the angle $\angle ACB$ between Islands A and B as seen from Island C?

Solution:
\textbf{Solution:} Draw line CD through point C such that CD$\parallel$AE, \\
since AE$\parallel$BF, \\
therefore CD$\parallel$AE$\parallel$BF, \\
therefore $\angle$ACD＝$\angle$EAC＝50°, $\angle$BCD＝$\angle$CBF＝40°, \\
therefore $\angle$ACB＝$\angle$ACD＋$\angle$BCD＝$50^\circ+40^\circ$＝\boxed{90^\circ}.


Question: In the regular hexagon $ABCDEF$, the diagonals $AC$, $BD$ intersect at point $M$. What is the value of $\frac{AM}{CM}$?

Solution:
\textbf{Solution:} Since hexagon ABCDEF is a regular hexagon, \\
$\therefore \angle BCD=\angle ABC=120^\circ$, AB=BC=CD, \\
$\therefore \angle BAC=\angle ACB=\angle CBD=\angle CDB=30^\circ$, \\
$\therefore CM=BM$, $\angle ABM=90^\circ$, \\
Let $BM=a$, then $CM=a$, \\
$\therefore AM=2BM=2a$, \\
$\therefore \frac{AM}{CM}=\frac{2a}{a}=\boxed{2}$.


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

Solution:
\textbf{Solution:} Since the regular pentagon $ABCDE$ is inscribed in circle $\odot O$, \\
therefore, $\angle AED=\frac{(5-2)\times 180^\circ}{5}=108^\circ$, and $AE=DE$,\\
hence, $\angle DAE=\angle ADE=\frac{1}{2}\times (180^\circ-108^\circ)=\boxed{36^\circ}$.


Question: As shown in the figure, it is known that the radius of $\odot O$ is 1, and the point $P$ is a point outside $\odot O$, with $OP=2$. If $PT$ is a tangent to $\odot O$ with $T$ as the point of tangency, and $OT$ is connected, then what is the length of $PT$?

Solution:
\textbf{Solution:} Since $PT$ is a tangent to the circle $\odot O$ with $T$ as the point of tangency, \\
it follows that $\angle OTP=90^\circ$. \\
Thus, $PT=\sqrt{OP^{2}-OT^{2}}$. \\
Given that the radius of $\odot O$ is 1, \\
it follows that $OT=1$. \\
Therefore, $PT=\sqrt{OP^{2}-OT^{2}}=\sqrt{2^{2}-1}=\boxed{\sqrt{3}}$.


Question: As shown in the figure, $AB$ and $AC$ are the sides of an inscribed square and an inscribed equilateral triangle in the circle $\odot O$, respectively, and $BC$ is one side of an inscribed regular $n$-sided polygon in the circle. What is the value of $n$?

Solution:
\textbf{Solution:} As shown in the figure, draw lines AO, BO, and CO.\\
Since AB and AC are respectively one side of a square and an equilateral triangle inscribed in $\odot$O,\\
it follows that $ \angle AOB=\frac{360^\circ}{4}=90^\circ$ and $ \angle AOC=\frac{360^\circ}{3}=120^\circ$,\\
therefore $ \angle BOC=30^\circ$,\\
hence $ n=\frac{360^\circ}{30^\circ}=\boxed{12}$.


Question: As shown in the figure, the vertices A and B of the square $ABCD$ both lie on the circle $\odot O$, and side $CD$ is tangent to the circle $\odot O$ at point E. If the radius of the circle $\odot O$ is 1, what is the side length of the square $ABCD$?

Solution:
\textbf{Solution:} Let's extend line $EO$ and have it intersect $AB$ at point $F$, and also connect $OA$,\\
$\because$ side $CD$ of square $ABCD$ is tangent to circle $O$ at point E,\\
$\therefore$ $OE\perp CD$, $AB\parallel CD$, $\angle BAD=\angle ADC=90^\circ$,\\
$\therefore$ $OF\perp AB$,\\
$\therefore$ quadrilateral $EFAD$ forms a rectangle,\\
$\therefore$ $EF=AD$,\\
$\because$ vertices A and B of square $ABCD$ are located on circle $O$,\\
$\therefore$ $AF=\frac{1}{2}AB$,\\
$\because$ the radius of circle $O$ is 1,\\
$\therefore$ $OE=OA=1$,\\
Let the side length of the square be $a$, then: $OF=EF-OE=a-1$, $AF=\frac{a}{2}$,\\
In right triangle $\triangle OFA$: $OA^{2}=OF^{2}+AF^{2}$,\\
which gives: $1^{2}=(a-1)^{2}+\left(\frac{a}{2}\right)^{2}$,\\
Solving this equation results in: $a=0$ (discard) or $a=\frac{8}{5}$;\\
$\therefore$ the side length of the square is: $\boxed{\frac{8}{5}}$.


Question: As shown in the figure, $A$, $B$, $C$, and $D$ are four consecutive vertices of a regular polygon, and $O$ is the center of the regular polygon. If $\angle ADB = 12^\circ$, how many sides does this regular polygon have?

Solution:
Solution: As shown in the figure, let the circumcircle of the polygon be $\odot O$, and connect OA and OB,\\
$\because$ $\angle ADB = 12^\circ$,\\
$\therefore$ $\angle AOB = 2\angle ADB = 24^\circ$,\\
and $360^\circ \div 24^\circ = 15$,\\
$\therefore$ the polygon is a regular pentadecagon,\\
thus, the answer is: $\boxed{15}$.


Question: As shown in the figure, line segment AB passes through the center O and intersects the circle $\odot O$ at points A and C, $\angle B=30^\circ$. The line BD is tangent to the circle $\odot O$ at point D. What is the degree measure of $\angle ADB$?

Solution:
\textbf{Solution:} Connect OD, as shown in the diagram,\\
BD is tangent to $\odot$O at D,\\
$\therefore$ $\angle$ODB$=90^\circ$,\\
$\because$ $\angle$B$=30^\circ$\\
$\therefore$ $\angle$DOB$=60^\circ$\\
$\therefore$ $\angle$A$=30^\circ$\\
$\therefore$ $\angle$ADB$=180^\circ-\angle$B$-\angle$A=$180^\circ-30^\circ-30^\circ$=\boxed{120^\circ}


Question: As shown in the figure, the incircle $\odot O$ of $\triangle ABC$ is tangent to side $BC$ at point $D$, $\angle A=70^\circ$, $\angle C=80^\circ$. Connect $OB$ and $OD$. What is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Given $\because$ $\angle A=70^\circ$ and $\angle C=80^\circ$, \\
$\therefore$ $\angle ABC=180^\circ-70^\circ-80^\circ=30^\circ$, \\
$\because$ the incircle $\odot$O of $\triangle ABC$ touches side $BC$ at point D, \\
$\therefore$ $OB$ bisects $\angle ABC$, and $OD\perp BC$, \\
$\therefore$ $\angle OBD=\frac{1}{2}\angle ABC=\frac{1}{2}\times 30^\circ=15^\circ$, \\
$\therefore$ $\angle BOD=90^\circ-\angle OBD=\boxed{75^\circ}$.


Question: As shown in the figure, PA and PB are tangents to the circle $\odot O$ with points A and B as the points of tangency. AC is a diameter of $\odot O$. If $\angle BAC=25^\circ$, what is the degree measure of $\angle P$?

Solution:
\textbf{Solution:} Connect OB, \\
$\because$ PA, PB are the tangents to circle $O$ with A and B as the points of tangency, \\
$\therefore$ $\angle OAP = \angle OBP = 90^\circ$, \\
$\because$ $\angle BAC = 25^\circ$, \\
$\therefore$ $\angle BOC = 50^\circ$, \\
$\therefore$ $\angle AOB = 180^\circ - 50^\circ = 130^\circ$, \\
$\therefore$ $\angle P = 360^\circ - 90^\circ - 90^\circ - 130^\circ = \boxed{50^\circ}$.


Question: As shown in the figure, if $AB$ and $AC$ are respectively one side of an inscribed equilateral triangle and one side of an inscribed square of circle $O$, and if $BC$ is one side of an inscribed regular $n$-sided polygon of circle $O$, then what is the value of $n$?

Solution:
\textbf{Solution:} Draw lines OA, OC, and OB. \\
Since AB and AC are respectively one side of a regular triangle and a regular square inscribed in circle O, \\
therefore, $\angle$AOC = $360^\circ \div 4 = 90^\circ$, $\angle$AOB = $360^\circ \div 3 = 120^\circ$, \\
therefore, $\angle$BOC = $\angle$AOB - $\angle$AOC = $120^\circ - 90^\circ = 30^\circ$, \\
therefore, n = $360^\circ \div 30^\circ = \boxed{12}$, \\
therefore, BC is definitely one side of a regular dodecagon inscribed in circle O.


Question: As shown in the figure, the vertices A, B, C of the rhombus OABC are on the circle $\odot O$. A tangent to $\odot O$ passing through point B intersects the extension of OA at point D. If the radius of $\odot O$ is 1, what is the length of BD?

Solution: \textbf{Solution:} Draw line OB, then OB=OA=1, since BD is the tangent of circle O, thus $\angle$OBD=90$^\circ$, since quadrilateral OABC is a rhombus, thus OA=OB, therefore $\triangle$OAB is an equilateral triangle, hence $\angle$AOB=60$^\circ$, therefore $\angle$ADB=30$^\circ$, hence OD=2OB=2, therefore BD=$ \sqrt{OA^{2}-OB^{2}}$=$ \boxed{\sqrt{3}}$.


Question: As shown in the figure, the radius $OA$ of $\odot O$ is 2, and $B$ is a moving point on $\odot O$ (not coinciding with point $A$). A tangent line $BC$ to $\odot O$ passing through point $B$ is drawn, where $BC=OA$. The lines $OC$ and $AC$ are connected. When $\triangle OAC$ is a right triangle, what is the length of its hypotenuse?

Solution:
\textbf{Solution:} Since BC is the tangent of $\odot$O, \\
$\therefore \angle$OBC$=90^\circ$, \\
Since BC=OA, \\
$\therefore$ OB=BC=2, \\
$\therefore \triangle$OBC is an isosceles right triangle, \\
$\therefore \angle$BCO=$45^\circ$, \\
$\therefore \angle$ACO$\leq 45^\circ$, \\
Since when $\triangle$OAC is a right triangle, (1) $\angle$AOC=$90^\circ$, connect OB, \\
$\therefore$ OC=$ \sqrt{2}$ OB=$2 \sqrt{2}$, \\
$\therefore$ AC=$ \sqrt{OA^{2}+OC^{2}}=\sqrt{2^{2}+(2\sqrt{2})^{2}}$=$2 \sqrt{3}$; \\
(2) When $\triangle$OAC is a right triangle, $\angle$OAC=$90^\circ$, connect OB, \\
Since BC is the tangent of $\odot$O, \\
$\therefore \angle$CBO=$\angle$OAC=$90^\circ$, \\
Since BC=OA=OB, \\
$\therefore \triangle$OBC is an isosceles right triangle, \\
$\therefore OC=2\sqrt{2}$, \\
Hence, the answer is: $AC=\boxed{2 \sqrt{3}}$ or $2 \sqrt{2}$.


Question: As shown in the figure, the ratio of the radii of two concentric circles is $3\colon 5$. $AB$ is the diameter of the larger circle, and the chord $BC$ of the larger circle is tangent to the smaller circle. If $AC=12$, what is the length of $BC$?

Solution:
\textbf{Solution:} As shown in the diagram, let chord BC be tangent to the smaller circle at point D, and connect OD,\\
$\therefore$OD$\perp$BC,\\
$\therefore$BD=CD,\\
$\because$O is the midpoint of AB,\\
$\therefore$OD is the median of $\triangle ABC$,\\
$\therefore$OD=$\frac{1}{2}$AC=$\frac{1}{2}\times 12=6$,\\
$\because$the ratio of the radii of the two concentric circles is 3:5,\\
$\therefore$the radius of the larger circle is 10,\\
$\therefore$AB=20,\\
$\because$AB is the diameter of the larger circle,\\
$\therefore$$\angle$ACB=$90^\circ$,\\
$\therefore$BC=$\sqrt{AB^{2}-AC^{2}}=\sqrt{20^{2}-12^{2}}=\boxed{16}$.


Question: As shown in the figure, P is a point outside the circle $\odot O$. The tangents PA and PB are drawn from P to $\odot O$. If $\angle APB=50^\circ$, what is the degree measure of $\overset{\frown}{AB}$?

Solution:
\textbf{Solution:} Since PA and PB are tangents to the circle O,\\
it follows that $\angle PAO=\angle PBO=90^\circ$,\\
and since $\angle APB=50^\circ$,\\
then $\angle AOB=360^\circ-\angle APB-\angle PAO-\angle PBO=130^\circ$,\\
thus, the degree measure of $\overset{\frown}{AB}$ is $\boxed{130^\circ}$.


Question: As shown in the figure, it is known that the radius of $ \odot P$ is 1, and the center P moves along the parabola $ y=-\frac{1}{2}x^{2}+1$. When $ \odot P$ is tangent to the x-axis, what is the x-coordinate of the center P?

Solution:
\textbf{Solution:} When $y=1$, we have $1=-\frac{1}{2}x^2+1$, giving $x=0$.\\
When $y=-1$, we have $-1=-\frac{1}{2}x^2+1$, giving $x= \pm 2$.\\
Therefore, the answer is: $x=\boxed{2}$ or $x=\boxed{-2}$ or $x=\boxed{0}$.


Question: As shown in the figure, from a point $P$ outside the circle $\odot O$, rays $PA$ and $PB$ are drawn to tangentially touch the circle $\odot O$ at points $A$ and $B$, respectively, with $\angle P = 50^\circ$. Point $C$ is on the minor arc $AB$. A tangent to circle $\odot O$ at point $C$ intersects $PA$ and $PB$ at points $D$ and $E$, respectively. What is the measure of $\angle DOE$ in degrees?

Solution:
\textbf{Solution:} As shown in the diagram, connect OA, OC, OB,

since PA, PB, DE are tangential to the circle at points A, B, and E, respectively,

therefore, $OA \perp PA$,

$OB \perp PB$,

$OC \perp DE$,

since $\angle P = 50^\circ$,

therefore, $\angle AOB = 180^\circ - \angle P = 130^\circ$,

since $OA = OB = OC$,

therefore, DO bisects $\angle ADC$, EO bisects $\angle BEC$,

therefore, $\angle ADO = \angle CDO$,

$\angle CEO = \angle BEO$,

therefore, $\angle AOD = \angle COD$,

$\angle COE = \angle BOE$,

therefore, $\angle DOE = \angle COD + \angle COE = \frac{1}{2}\angle AOC + \frac{1}{2}\angle BOC = \frac{1}{2}\angle AOB = \boxed{65^\circ}$,


Question: As shown in the figure, within circle $O$, $AB$ is a side of the inscribed regular hexagon in circle $O$, and $BC$ is a side of the inscribed regular decagon in circle $O$. What is the measure of $\angle ABC$ in degrees?

Solution:
\textbf{Solution:} As shown in the diagram, connect AO, BO, CO,

$\because$ AB is a side of the inscribed hexagon in $\odot O$,

$\therefore$ $\angle AOB = \frac{360^\circ}{6} = 60^\circ$,

$AO = BO$,

$\therefore$ $\angle ABO = \frac{1}{2}(180^\circ-60^\circ) = 60^\circ$,

$\because$ BC is a side of the inscribed decagon in $\odot O$,

$\therefore$ $\angle BOC = \frac{360^\circ}{10} = 36^\circ$, BO=CO,

$\therefore$ $\angle CBO = \frac{1}{2}(180^\circ-36^\circ) = 72^\circ$,

$\therefore$ $\angle ABC = \angle ABO + \angle CBO = 60^\circ + 72^\circ = \boxed{132^\circ}$.


Question: As shown in the figure, for a regular hexagon $ABCDEF$ inscribed in a circle, with the radius $OA=4$, what is the length of the side of the hexagon?

Solution:
\textbf{Solution:} Draw line OB, as shown in the diagram,\\
$\because$ Hexagon ABCDEF is inscribed in circle $O$,\\
$\therefore \angle$AOB=$ \frac{360^\circ}{6}=60^\circ$, OA=OB,\\
$\therefore \triangle$OAB is an equilateral triangle,\\
$\therefore$AB=OA= \boxed{4},\\
i.e., the side length of this regular hexagon is 4.


Question: As shown in the figure, \(PA\) and \(PB\) are tangents to circle \(\odot O\) at points \(A\) and \(B\), respectively, and \(\angle P=70^\circ\). If point \(C\) is on circle \(\odot O\) and does not coincide with \(A\) or \(B\), what is the measure of \(\angle ACB\)?

Solution:
\textbf{Solution:} Draw OA, OB, \\
$ \because $ PA, PB are tangents to circle $O$ at points A, B, and $ \angle P=70^\circ$, \\
$ \therefore \angle AOB=360^\circ-90^\circ-90^\circ-\angle P=110^\circ$\\
When C is on the major arc $ \overset{\frown}{ACB}$,\\
$ \angle ACB=\frac{1}{2}\angle AOB=55^\circ$\\
When point C is on the minor arc $ \overset{\frown}{AB}$, quadrilateral ACBC$_{1}$ is a circumscribed quadrilateral\\
Then $ \angle A{C}_{1}B+\angle C=180^\circ$\\
$ \therefore \angle A{C}_{1}B=125^\circ$\\
Hence, the answer is: $\boxed{55^\circ}$ or $\boxed{125^\circ}$.


Question: As shown in the figure, AB is a chord of the semicircle O, DE is the diameter. The tangent BC passing through point B is tangent to $\odot$O at point B, and intersects the extension line of DE at point C. Connect BD. If the quadrilateral OABC is a parallelogram, then what is the degree measure of $\angle BDC$?

Solution:
Solution: Since $BC$ is the tangent of circle $O$,\\
therefore $\angle OBC=90^\circ$,\\
Since quadrilateral $ABCO$ is a parallelogram,\\
therefore $AO=BC$,\\
Also since $AO=BO$,\\
therefore $BO=BC$,\\
therefore $\angle BOC=\angle BCO=45^\circ$,\\
therefore $\angle BDC=\boxed{22.5^\circ}$.


Question: As shown in the figure, divide the circle $ \odot O$ into six equal arcs and connect the division points sequentially to obtain the regular hexagon ABCDEF. If the circumference of $ \odot O$ is $ 12\pi $, what is the side length of this regular hexagon?

Solution:
\textbf{Solution:} As shown in the figure, connect OA, OB, OC, OD, OE, OF. \\
$\because$ the hexagon ABCDEF is regular, \\
$\therefore$ AB = BC = CD = DE = EF = FA, $\angle$AOB = $\angle$BOC = $\angle$COD = $\angle$DOE = $\angle$EOF = $\angle$FOA = $60^\circ$, \\
$\therefore$ $\triangle$AOB, $\triangle$BOC, $\triangle$DOC, $\triangle$EOD, $\triangle$EOF, $\triangle$AOF are all equilateral triangles, \\
$\because$ the circumference of $\odot O$ is $12\pi$, \\
$\therefore$ the radius of $\odot O$ is $\frac{12\pi}{2\pi}=6$, \\
thus, the side length of the regular hexagon is $\boxed{6}$.


Question: As shown in the figure, regular pentagon ABCDE is inscribed in circle $O$, and point P is on the minor arc $\overset{\frown}{BC}$ (point P does not coincide with point C). What is the measure of $\angle CPD$?

Solution:
\textbf{Solution:} Draw lines OC and OD,\\
$ \because $ the regular pentagon ABCDE is inscribed in $\odot$O,\\
$ \therefore \angle COD=\frac{360^\circ}{5}=72^\circ$,\\
$ \therefore \angle CPD=\frac{1}{2}\angle COD=\boxed{36^\circ}$.


Question: As shown in the figure, $AB$ is a tangent to the circle $\odot O$ with $A$ as the point of tangency. The line $OB$ intersects $\odot O$ at point $C$. Point $D$ is on the circle $\odot O$. Connect $AD$, $CD$, and $OA$. If $\angle ADC=25^\circ$, what is the degree measure of $\angle ABO$?

Solution:
\textbf{Solution:} Given that $AB$ is the tangent of circle $O$,

it follows that $AB \perp OA$,

which implies $\angle OAB = 90^\circ$.

Given that $\angle ADC = 25^\circ$,

it follows that $\angle AOB = 2\angle ADC = 50^\circ$,

therefore, $\angle ABO = 90^\circ - 50^\circ = \boxed{40^\circ}$.


Question: As shown in the figure, the radius of the circle $ \odot O$ is 2 cm, and a regular hexagon is inscribed in $ \odot O$. What is the area of the shaded region in the figure?

Solution:
\textbf{Solution:} As shown in the diagram, connect BO, CO, and OA.\\
From the given information, both $\triangle OBC$ and $\triangle AOB$ are equilateral triangles,\\
$\therefore \angle AOB = \angle OBC = 60^\circ$,\\
$\therefore OA \parallel BC$,\\
$\therefore$ The area of $\triangle OBC$ is equal to the area of $\triangle ABC$,\\
$\therefore$ The area of the shaded region is equal to the area of the sector OBC $= \frac{60^\circ \pi \times 2^2}{360^\circ} = \boxed{\frac{2\pi}{3}}$.


Question: As shown in the figure, the famous scenic spot Sheng'an Pavilion in the Western Hills of Kunming is founded on a regular hexagon with a radius of 3m. What is the perimeter of the regular hexagon in meters?

Solution:
\textbf{Solution:} As shown in the figure, connect $OA$ and $OB$,\\
since the radius of the regular hexagon is $3m$,\\
therefore $OA=OB=3m$,\\
since the figure is a regular hexagon,\\
therefore $\angle AOB=60^\circ$,\\
therefore $\triangle AOB$ is an equilateral triangle,\\
therefore $AB=3m$,\\
therefore the perimeter of the regular hexagon is: $3\times 6=\boxed{18m}$.


Question: As shown in the figure, hexagon ABCDEF is a regular hexagon inscribed in a circle with a radius of 6. What is the length of $\overset{\frown}{CD}$?

Solution:
\textbf{Solution:} Connect OC and OD, \\
$\because$ hexagon ABCDEF is a regular hexagon, \\
$\therefore$ $\angle$COD $=$ 360$^\circ$ $\times$ $\frac{1}{6}$ $=$ 60$^\circ$, \\
$\because$ OD $=$ 2, \\
$\therefore$ the length of arc $\overset{\frown}{CD}$ is $ \frac{60\pi \times 2}{180}=\boxed{2\pi} $.


Question: As shown in the diagram, $\angle MON=30^\circ$, points $A_1, A_2, A_3, \ldots$ are on ray ON, points $B_1, B_2, B_3, \ldots$ are on ray OM, $\triangle A_1B_1A_2$, $\triangle A_2B_2A_3$, $\triangle A_3B_3A_4,\ldots$ are all equilateral triangles. The side length of the first equilateral triangle from the left is denoted as $a_1$, the side length of the second equilateral triangle is denoted as $a_2$, and so on. If $OA_1=1$, then what is the value of $a_{2021}$?

Solution:
\textbf{Solution:} As shown in the figure,\\
since $\triangle A_1B_1A_2$ is an equilateral triangle,\\
thus $A_1B_1=A_2B_1$, $\angle 3=\angle 4=\angle 12=60^\circ$,\\
therefore $\angle 2=120^\circ$,\\
since $\angle MON=30^\circ$,\\
thus $\angle 1=180^\circ-120^\circ-30^\circ=30^\circ$,\\
also since $\angle 3=60^\circ$,\\
thus $\angle 5=180^\circ-60^\circ-30^\circ=90^\circ$,\\
since $\angle MON=\angle 1=30^\circ$,\\
thus $OA_1=A_1B_1=1$, that is, the side length of $\triangle A_1B_1A_2$ is $a_1=1=2^0$;\\
therefore $A_2B_1=1$,\\
since $\triangle A_2B_2A_3$, $\triangle A_3B_3A_4$ are equilateral triangles,\\
thus $\angle 11=\angle 10=60^\circ$, $\angle 13=60^\circ$,\\
since $\angle 4=\angle 10=\angle 11=60^\circ$, $\angle 12=\angle 13=60^\circ$,\\
thus $A_1B_1\parallel A_2B_2\parallel A_3B_3$, $B_1A_2\parallel B_2A_3$,\\
thus $\angle 1=\angle 6=\angle 7=30^\circ$, $\angle 5=\angle 8=90^\circ$,\\
thus $A_2B_2=2B_1A_2=2=2^1$, that is, the side length of $\triangle A_2B_2A_3$ is $a_2=2^1$\\
Similarly, $B_3A_3=2B_2A_3=4=2^2$, that is, the side length of $\triangle A_3B_3A_4$ is $a_3=2^2$,\\
... ,\\
thus, the side length of $\triangle A_nB_nA_{n+1}$ is $a_n=2^{n-1}$,\\
thus, the side length of $\triangle A_{2021}B_{2021}A_{2022}$ is $a_{2021}=\boxed{2^{2020}}$.


Question: As shown in the figure, in $ \triangle ABC$, $AD$ is the median to side $BC$, $E$ is a point on side $AB$, and a line through point $C$ is drawn parallel to $AB$ intersecting the extension of $ED$ at point $F$. When $AD \perp BC$, $AE=1$, and $CF=2$, what is the length of $AC$?

Solution:
\textbf{Solution:}

Since $AD$ is the median to side $BC$ and $AD\perp BC$,

it follows that $AB=AC$.

Since $CF\parallel AB$,

it follows that $\angle B=\angle DCF$.

In $\triangle BED$ and $\triangle CFD$,

\[
\left\{\begin{array}{l}
\angle B=\angle DCF \\
BD=CD \\
\angle BDE=\angle CDF
\end{array}\right.
\]

Therefore, $\triangle BED \cong \triangle CFD$,

which implies that $CF=BE$.

Given $AE=1$ and $CF=2$,

thus $AB=AE+BE=AE+CF=1+2=3$,

which leads to $AC=AB=\boxed{3}$.


Question: As shown in the figure, in a triangular piece of paper $ABC$, $\angle B = 90^\circ$, $AB = 3$. Point $D$ is on side $BC$. When $\triangle ABD$ is folded along the straight line $AD$, point $B$ exactly falls on point $E$ on the side $AC$. If $AD = CD$, what is the length of $AC$?

Solution:
\textbf{Solution:} It follows from the property of folding that $AE=AB=3$,\\
$\angle AED=90^\circ$ \\
$\because AD=CD$ \\
$\therefore \triangle ADC$ is an isosceles triangle \\
$\because DE\perp AC$ \\
$\therefore EC=AE$ \\
$\therefore AC=\boxed{6}$


Question: As shown in the figure, it is known that $AE=BE$, $DE$ is the perpendicular bisector of $AB$, and $F$ is a point on $DE$. If $BF=11\text{cm}$ and $CF=3\text{cm}$, what is the length of $AC$ in cm?

Solution:
\textbf{Solution:} Since $AE=BE$, and $DE$ is the perpendicular bisector of $AB$,\\
hence, $DE$ is the median of $AB$,\\
therefore, $DE$ is the perpendicular bisector of $AB$,\\
since $F$ is a point on $DE$,\\
therefore, $AF=BF$,\\
thus, $AC = AF + CF = BF + CF$,\\
since $BF=11\,cm$, and $CF=3\,cm$,\\
therefore, $AC = \boxed{14}\,cm$


Question: As shown in the figure, in $\triangle ABC$, $\angle A=90^\circ$, $AB=AC$, $BD$ bisects $\angle ABC$, $CE \perp BD$ at $E$. If $BD=8$, what is the length of $CE$?

Solution:
\textbf{Solution:} Extend BA and CE to intersect at point F, \\
since $\angle BAC = 90^\circ$ and $CE \perp BD$, \\
therefore, $\angle BAC = \angle BEC = \angle FAC$, \\
since $\angle ABD + \angle ADB = 90^\circ$ and $\angle CDE + \angle ACF = 90^\circ$, \\
since $\angle ADB = \angle CDE$, \\
therefore, $\angle ABD = \angle ACF$, \\
in $\triangle ABD$ and $\triangle ACF$, \\
$\left\{\begin{array}{l}
\angle BAD = \angle CAF \\
AB = AC \\
\angle ABD = \angle ACF
\end{array}\right.$ \\
therefore, $\triangle ABD \cong \triangle ACF$, \\
therefore, BD = CF = 8, \\
since BD bisects $\angle ABC$, \\
therefore, $\angle ABE = \angle CBE$, \\
since CE$\perp$BD, \\
therefore, $\angle BEF = \angle BEC = 90^\circ$ \\
in $\triangle BEF$ and $\triangle BEC$, \\
$\left\{\begin{array}{l}
\angle FBE = \angle CBE \\
BE = BE \\
\angle BEF = \angle BEC
\end{array}\right.$ \\
therefore, $\triangle BEF \cong \triangle BEC$, \\
therefore, EF = EC, \\
therefore, EC = $\frac{1}{2}$CF = \boxed{4}.


Question: As shown in the figure, fold the rectangular paper ABCD along AE so that point D falls on point F on side BC. If $\angle AEF = 75^\circ$, what is the degree measure of $\angle BAF$?

Solution:
\textbf{Solution:} Given the problem, we have: $\angle AFE = \angle D = 90^\circ$, $\angle EAF = \angle DAE$, $\angle BAD = 90^\circ$,\\
$\because \angle AEF = 75^\circ$,\\
$\therefore \angle EAF = 90^\circ - \angle AEF = 15^\circ$,\\
$\therefore \angle DAE = 15^\circ$,\\
$\therefore \angle DAF = \angle DAE + \angle EAF = 30^\circ$,\\
$\therefore \angle BAF = \angle BAD - \angle DAF = \boxed{60^\circ}$.


Question: As shown in the figure, in $ \triangle ABC$, $ \angle ACB=90^\circ$, CD is the median to side AB, and $ CD+AB=12$, what is the length of AB?

Solution:
\textbf{Solution:} Since $\angle ACB=90^\circ$ and D is the midpoint of side AB, \\
it follows that $CD=\frac{1}{2}AB$, \\
and since $CD+AB=12$, \\
we have $AB=\boxed{8}$.


Question: An equilateral triangle, a right-angled triangle, and an isosceles triangle are placed as shown in the figure. The base angle of the isosceles triangle $\angle 3=80^\circ$, then what is the sum of $\angle 1 + \angle 2$?

Solution:
\textbf{Solution}: As shown in the figure,\\
from the equilateral triangle and the right triangle we can obtain $\angle 1+\alpha =120^\circ$, $\angle 2+\beta =90^\circ$,\\
$\therefore \angle 1+\angle 2+\alpha +\beta =90^\circ+120^\circ=210^\circ$,\\
and $\angle 3=\alpha +\beta$,\\
$\therefore \alpha +\beta =80^\circ$,\\
$\therefore \angle 1+\angle 2=210^\circ-80^\circ=\boxed{130^\circ}$.


Question: As shown in the figure, fold the equilateral triangle $\triangle ABC$ such that point $B$ falls exactly on point $D$ on side $AC$. The crease is $EF$, and $O$ is a moving point on crease $EF$. If $AD=1$ and $AC=3$, then what is the minimum perimeter of $\triangle OCD$?

Solution:
\textbf{Solution:} As shown in the figure, connect BD, OB, \\
$\because$ by folding the equilateral $\triangle ABC$ so that point B falls exactly on point D on side AC, \\
$\therefore$ EF is the axis of symmetry of BD, \\
$\therefore$ OB=OD, \\
$\because$ AD=1, AC=3, \\
$\therefore$ CD=2, \\
$\because$ the perimeter of $\triangle OCD$ = CD+OD+OC = 2+BO+OC, \\
$\therefore$ when points B, O, and C are collinear, the minimum perimeter of $\triangle OCD$ = 2+BC = \boxed{5},


Question: As shown, it is known that $ \triangle ABD \cong \triangle ACE$, $ \angle A=53^\circ$, $ \angle B=22^\circ$. Then, how many degrees is $ \angle AEC$?

Solution:
\textbf{Solution}: Since $\triangle ABD \cong \triangle ACE$,\\
it follows that $\angle C = \angle B = 22^\circ$,\\
therefore, in $\triangle AEC$, $\angle AEC = 180^\circ - \angle A - \angle C = 180^\circ - 53^\circ - 22^\circ = \boxed{105^\circ}$.


Question: As shown in the figure, $D$ is a point on side $AC$ of $\triangle ABC$. An arc is drawn with $A$ as the center and $AD$ as the radius, intersecting the extension line of $BA$ at point $E$, and $ED$ is connected. If $\angle B=60^\circ$ and $\angle C=70^\circ$, what is the degree measure of $\angle ADE$?

Solution:
\textbf{Solution:} Given that $\angle B=60^\circ$ and $\angle C=70^\circ$,\\
it follows that $\angle CAB=180^\circ-\angle B-\angle C=50^\circ$,\\
since an arc is drawn with A as the center and AD as the radius, intersecting the extension of BA at point E,\\
it implies AE=AD,\\
thus $\angle AED=\angle ADE$,\\
since $\angle AED+\angle ADE=\angle CAB$,\\
therefore $\angle ADE+\angle ADE=50^\circ$,\\
which gives $\angle ADE=\boxed{25^\circ}$.


Question: As shown in the figure, if $\triangle ABC \cong \triangle ADF$ and $\angle B=60^\circ$, $\angle C=20^\circ$, with point D on side BC, then what is the degree measure of $\angle DAC$?

Solution:
\textbf{Solution:} Since $\triangle ABC \cong \triangle ADF$,\\
it follows that $AB = AD$,\\
therefore, $\angle ADB = \angle B = 60^\circ$,\\
thus, $\angle DAC = \angle ADB - \angle C = \boxed{40^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $BD$ bisects $\angle ABC$. If $AB=8$, $BC=12$, and the area of $\triangle ABD$ is $16$, then what is the area of $\triangle CBD$?

Solution:
\textbf{Solution:} As shown in the diagram, draw $DE\perp AB$ at E, and $DF\perp BC$ at F, \\
$\because BD$ bisects $\angle ABC$, $DE\perp AB$, $DF\perp BC$, \\
$\therefore DE=DF$, \\
$\because {S}_{\triangle ABD}=\frac{1}{2}AB\times DE=16$, \\
$\therefore DE=4$, \\
$\therefore DF=4$, \\
$\because BC=12$, \\
$\therefore {S}_{\triangle BCD}=\frac{1}{2}\times BC\times DF=\frac{1}{2}\times 12\times 4=\boxed{24}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, point $D$ is the midpoint of $BC$, $\angle BAD=24^\circ$, $AD=AE$, what is the measure of $\angle EDC$ in degrees?

Solution:
\textbf{Solution:} Given $AB=AC$ and point $D$ is the midpoint of $BC$, with $\angle BAD=24^\circ$, \\
it follows that $\angle CAD=\angle BAD=24^\circ$, and $AD\perp BC$,\\
Given $AD=AE$,\\
it implies $\angle ADE=\angle AED=\frac{1}{2}\times (180^\circ-24^\circ)=78^\circ$,\\
thus $\angle EDC=90^\circ-\angle ADE=\boxed{12^\circ}$.


Question: As shown in the figure, \(ABCD\) is a rectangular ground, with length \(AB = 20m\) and width \(AD = 10m\). In the middle, there stands a brick wall with height \(MN = 2m\). A grasshopper needs to crawl from point \(A\) to point \(C\), and it must climb over the wall in the middle. What is the minimum distance it has to travel?

Solution:
Solution: As shown in the figure, unfolding the figure increases the length by 2MN,\\
if the original length increases by 4 meters, then AB=20+4=24(m),\\
connect AC,\\
$\because$ quadrilateral ABCD is a rectangle, with AB=24m and width AD=10m,\\
$\therefore$AC=$\sqrt{AB^{2}+BC^{2}}=\sqrt{24^{2}+10^{2}}=\boxed{26}$(m),\\
$\therefore$, for the grasshopper to crawl from point A to point C, it needs to travel at least 26m.


Question: As shown in the figure, D, E, and F are the midpoints of BC, AD, and CE, respectively. If $S_{\triangle ABC}=8\text{cm}^2$, what is $S_{\triangle DEF}$?

Solution:
\textbf{Solution:} Given that $S_{\triangle ABC}=8\,\text{cm}^2$ and D is the midpoint of BC,\\
it follows that $S_{\triangle ABD}=S_{\triangle ADC}=\frac{1}{2}S_{\triangle ABC}=\frac{1}{2}\times 8=4\,\text{cm}^2$,\\
since E is the midpoint of AD,\\
then $S_{\triangle DEC}=\frac{1}{2}S_{\triangle ADC}=\frac{1}{2}\times 4=2\,\text{cm}^2$,\\
since F is the midpoint of EC,\\
then $S_{\triangle DEF}=\frac{1}{2}S_{\triangle DEC}=\frac{1}{2}\times 2=\boxed{1}\,\text{cm}^2$,


Question: As shown in the figure, $ABC$ is an equilateral triangle with side length $4$, $AD$ is the median of side $BC$, and $F$ is a moving point on side $AD$, with $E$ being the midpoint of side $AC$. What is the measure of $\angle EFC$ when the perimeter of $\triangle ECF$ is minimized?

Solution:
\textbf{Solution:} Draw $EM\parallel BC$ through $E$, intersecting $AD$ at $N$,\\
$\therefore \angle AMN=\angle ABC$,\\
$\because$ the sides of equilateral triangle $ABC$ are of length 4, and $E$ is the midpoint of side $AC$.\\
$\therefore AC=AB=BC=4$, AE=2, $\angle ABC=60^\circ=\angle AMN$, $\angle AEM=\angle ACB=60^\circ$,\\
$\therefore EC=2=AE$, $\triangle AME$ is an equilateral triangle,\\
$\therefore AM=BM=2$,\\
$\therefore AM=AE$,\\
$\because AD$ is the median to side $BC$, and $\triangle ABC$ is an equilateral triangle,\\
$\therefore AD\perp BC$,\\
$\because EM\parallel BC$,\\
$\therefore AD\perp EM$,\\
$\because AM=AE$,\\
$\therefore $E and M are symmetrical about $AD$,\\
Connect $CM$ intersecting $AD$ at $F$, then connect $EF$,\\
under these conditions, the sum of $EF+CF$ is minimized, and the perimeter of $\triangle CEF$ is minimized,\\
$\because \triangle ABC$ is an equilateral triangle,\\
$\therefore \angle ACB=60^\circ$, AC=BC=AB,\\
$\because AM=BM$, $CM\perp AB$,\\
$\therefore \angle ECF= \frac{1}{2}\angle ACB=30^\circ$, $\angle AMC=90^\circ$,\\
$\therefore \angle CME=90^\circ-60^\circ=30^\circ$,\\
By the property of axial symmetry, we have: $\angle MEF=\angle FME=30^\circ$,\\
$\therefore \angle CFE=30^\circ+30^\circ=60^\circ$.\\
Thus, the answer is: $\boxed{60^\circ}$


Question: As shown, student Xiao Zhang holds the teacher's isosceles right-angled triangle ruler, placing it between two stacks of rectangular solid teaching aids, with $\angle ACB=90^\circ$ and $AC=BC$. If the height of each rectangular solid teaching aid is $6\text{cm}$, what is the length of the distance $DE$ in $\text{cm}$ between the two stacks of rectangular solid teaching aids?

Solution:
\textbf{Solution:} From the given, we have: $AC=BC$, $\angle ACB=90^\circ$, $AD\perp DE$, $BE\perp DE$,\\
$\therefore \angle ADC=\angle CEB=90^\circ$\\
$\therefore \angle ACD+\angle BCE=90^\circ$, $\angle ACD+\angle DAC=90^\circ$,\\
$\therefore \angle BCE=\angle DAC$,\\
In $\triangle ADC$ and $\triangle CEB$,\\
$\left\{\begin{array}{l}
\angle ADC=\angle CEB \\
\angle DAC=\angle BCE \\
AC=BC
\end{array}\right.$,\\
$\therefore \triangle ADC\cong \triangle CEB$ (AAS),\\
$\therefore CD=BE$, $AD=CE$,\\
$\because DE=CD+CE$,\\
$\therefore DE=BE+AD$,\\
$\because$ the thickness of a rectangular solid teaching aid is $6\text{cm}$,\\
$\therefore AD=24\text{cm}$, $BE=18\text{cm}$,\\
$\therefore$ the distance $DE$ between the two stacks of rectangular solid teaching aids is $24+18=\boxed{42}\text{cm}$.


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

Solution:
\textbf{Solution:} Draw EF $\perp$ BC at F through E, \\
since CD is the altitude from AB, and BE bisects $\angle ABC$, intersecting CD at point E, \\
thus DE=EF, \\
since the area of $\triangle BCE= \frac{1}{2}\times$BC$\times$EF=5, \\
thus $\frac{1}{2}\times$5$\times$EF=5, \\
therefore, EF=DE=\boxed{2}


Question: As shown in the figure, in an isosceles $\triangle ABC$ with $AB=AC=13$ and $BC=10$, $D$ is the midpoint of side $BC$, and $M$, $N$ are moving points on $AD$ and $AB$, respectively. What is the minimum value of $BM+MN$?

Solution:
\textbf{Solution:} As shown in the figure, draw $BH\perp AC$ at H, intersecting AD at point $M'$, and draw $M'N'\perp AB$ at $N'$, then $BM'+M'N'$ is the minimum value sought.\\
Since $AB=AC$, and D is the midpoint on side BC,\\
thus AD is the bisector of $\angle BAC$,\\
therefore $M'H=M'N'$,\\
thus BH is the shortest distance from point B to line AC (perpendicular segment is the shortest),\\
since $AB=AC=13$, $BC=10$, and D is the midpoint on side BC,\\
thus $AD\perp BC$,\\
therefore $AD=12$,\\
since the area of $\triangle ABC=\frac{1}{2}AC\times BH=\frac{1}{2}BC\times AD$,\\
thus $13\times BH=10\times 12$,\\
solving yields: $BH=\boxed{\frac{120}{13}}$.


Question: As shown in the figure, point $A$ is on the line segment $DE$, $AB\perp AC$ with the foot of the perpendicular at $A$, and $AB = AC$, $BD\perp DE$, $CE\perp DE$ with the perpendicular feet at $D$ and $E$ respectively. If $ED = 12$ and $BD = 8$, what is the length of $CE$?

Solution:
\textbf{Solution:} Given that $BD\perp DE$ and $CE\perp DE$,\\
it follows that $\angle D=\angle E=90^\circ$ and $\angle ABD+\angle BAD=90^\circ$,\\
since $AB\perp AC$,\\
it also follows that $\angle BAD+\angle EAC=90^\circ$,\\
therefore, $\angle ABD=\angle EAC$,\\
in $\triangle ABD$ and $\triangle CAE$, we have $\left\{\begin{array}{l}
\angle D=\angle E \\
AB=CA \\
\angle ABD=\angle EAC
\end{array}\right.$,\\
thus, $\triangle ABD \cong \triangle CAE$ by ASA,\\
resulting in $BD=AE=8$, and $AD=CE$,\\
therefore, $AD=ED-AE=12-8=4$,\\
hence, $CE=\boxed{4}$.


Question: As shown in the figure, $BD$ is the bisector of $\angle ABC$, $DE \perp AB$ at point $E$, $DF \perp BC$ at point $F$, $AB=12$, $BC=15$, and the area of $\triangle ABC$ is $36$. What is the length of $DE$?

Solution:
\textbf{Solution:} Given that $BD$ is the angle bisector of $\angle ABC$, and that $DE\perp AB$, $DF\perp BC$\\
Therefore, $DE=DF$\\
Since the area of $\triangle ABC = S_{\triangle ABD} + S_{\triangle BCD} = \frac{1}{2}BC\cdot DF + \frac{1}{2}AB\cdot DE = 36$, and $AB=12$, $BC=15$\\
Therefore, $\frac{1}{2} \times 12 \cdot DE + \frac{1}{2} \times 15 \cdot DF = 36$\\
Therefore, $6DE + \frac{15}{2}DF = 36$\\
Furthermore, since $DE=DF$\\
Therefore, $6DE + \frac{15}{2}DE = 36$\\
Therefore, $DE = \boxed{\frac{8}{3}}$


Question: As shown in the figure, place a right-angled triangle plate as indicated, keeping \(AB \parallel DE\). What is the degree measure of \(\angle ACD\)?

Solution:
\textbf{Solution:} From the given, we have: $\angle A= 30^\circ$, $\angle D= 45^\circ$,\\
As shown in the figure, let AB intersect CD at point F,\\
Since AB$\parallel$DE,\\
Therefore, $\angle AFD=\angle D= 45^\circ$,\\
Since $\angle AFD$ is an exterior angle of $\triangle ACF$,\\
Therefore, $\angle ACD=\angle AFD-\angle A = \boxed{15^\circ}$.


Question: As shown in the figure, in $ \triangle ABC$, the perpendicular bisector of side $AB$ intersects $AB$ at point D and intersects $AC$ at point E. Line $BE$ is drawn. If $AC=5$ and $BC=3$, what is the perimeter of $ \triangle BCE$?

Solution:
\textbf{Solution:} Given $DE$ is the perpendicular bisector of $AB$,\\
it follows that $AE=BE$,\\
Given $AC=5$,\\
it then follows that $BE+CE=AE+CE=AC=5$,\\
Given $BC=3$,\\
the perimeter of $\triangle BCE$ is $BE+CE+BC=5+3=\boxed{8}$.


Question: As shown in the figure, the edges AO and CO of the rectangle ABCO lie exactly on the coordinate axes, with AB = 4 and OA = 2. Point D is a point on line segment OC, and point E is a point on line segment AB. When folded along DE, such that point B coincides with point O and point C moves to point $C'$, what are the coordinates of point D at that moment?

Solution:
\textbf{Solution:} Since AB$\parallel$OC,\\
it follows that $\angle$BED=$\angle$EDO,\\
by the property of folding we get BE=OE and $\angle$BED=$\angle$OED,\\
thus $\angle$EDO=$\angle$OED,\\
which implies OE=OD,\\
Let BE=OE=x, then AE=4-x,\\
hence in the Rt$\triangle$AEO, by Pythagorean theorem, we have $2^2+(4-x)^2=x^2$,\\
solving for $x$ gives $x=2.5$,\\
therefore, OE=OD=2.5,\\
thus the coordinates of point D are $\boxed{(2.5, 0)}$;


Question: As shown in the figure, it is known that $\triangle ABC$ is an equilateral triangle, and points $B$, $C$, $D$, $E$ are on the same line, with $CG=CD$, $DF=DE$. What is the measure of $\angle E$ in degrees?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is an equilateral triangle,\\
therefore, $\angle ACB=60^\circ$,\\
since CG=CD,\\
therefore, $\angle CDG=\angle CGD=30^\circ$,\\
since DF=DE,\\
therefore, $\angle E=\angle DFE=\boxed{15^\circ}$.


Question: As shown in the diagram, paper triangle $ABC$ is folded along $DE$ such that point $A$ falls on point $A'$, and $A'B$ bisects $\angle ABC$, $A'C$ bisects $\angle ACB$. If $\angle BA'C=120^\circ$, what is the degree measure of $\angle 1+\angle 2$?

Solution:
\textbf{Solution:} Let's connect $AA'$,\\
since $A'B$ bisects $\angle ABC$, $A'C$ bisects $\angle ACB$,\\
thus $\angle A'BC = \frac{1}{2}\angle ABC$, $\angle A'CB = \frac{1}{2}\angle ACB$,\\
since $\angle BA'C = 120^\circ$,\\
thus $\angle A'BC + \angle A'CB = 180^\circ - 120^\circ = 60^\circ$,\\
thus $\angle ABC + \angle ACB = 120^\circ$,\\
thus $\angle BAC = 180^\circ - 120^\circ = 60^\circ$,\\
since it's folded along DE,\\
thus $\angle DAA' = \angle DA'A$, $\angle EAA' = \angle EA'A$,\\
since $\angle 1 = \angle DAA' + \angle DA'A = 2\angle DAA'$, $\angle 2 = \angle EAA' + \angle EA'A = 2\angle EAA'$,\\
thus $\angle 1 + \angle 2 = 2\angle DAA' + 2\angle EAA' = 2\angle BAC = 2\times 60^\circ = \boxed{120^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, the perpendicular bisector of AC intersects AC at E and BC at D. The perimeter of $\triangle ABD$ is $13\text{cm}$, and $AE=4.5\text{cm}$. What is the perimeter of $\triangle ABC$?

Solution:
\textbf{Solution:} Since DE is the perpendicular bisector of AC,\\
it follows that AD = CD,\\
therefore, the perimeter of $\triangle ABD$ equals to AB + BD + AD = AB + BD + CD = AB + BC,\\
also, since AE = 4.5cm,\\
it follows that AC = 2AE = 2×4.5 = 9cm,\\
therefore, the perimeter of $\triangle ABC$ = 13 + 9 = \boxed{22}cm.


Question: As shown in the figure, the perpendicular bisectors $m$ and $n$ of sides $AB$ and $AC$ of $\triangle ABC$ intersect at point $D$. Connect $CD$. If $\angle 1=39^\circ$, then what is the measure of $\angle BCD$ in degrees?

Solution:
\textbf{Solution:} As shown in the figure, connect AD, BD,\\
since the perpendicular bisectors m and n of sides AB and AC of $\triangle ABC$ intersect at point D,\\
therefore, AD=BD=CD,\\
since $\angle 1=39^\circ$,\\
therefore, $\angle EDF=180^\circ-\angle 1=141^\circ$,\\
thus, $\angle EDA+\angle ADF=141^\circ$,\\
since DA=DC, and DE$\perp$AC,\\
therefore, $\angle ADE=\angle CDE$,\\
similarly, it can be derived that $\angle ADF=\angle BDF$,\\
therefore, $\angle CDE+\angle BDF=141^\circ$,\\
therefore, $\angle CDB=360^\circ-\angle EDA-\angle ADF-\angle CDE-\angle BDF=78^\circ$,\\
since DC=DB,\\
therefore, $\angle DCB=\frac{180^\circ-\angle CDB}{2}=\boxed{51^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $\angle BAC=120^\circ$, $AD\perp BC$ at $D$, $DE\perp AC$ at $E$. If $AE=3$, what is the length of $CE$?

Solution:
\textbf{Solution:} Given: In $\triangle ABC$, $AB=AC$ and $\angle BAC=120^\circ$, \\
$\therefore$ $\angle B=\angle C=30^\circ$, \\
Since $AD\perp BC$ at $D$, \\
$\therefore$ $\angle ADC=90^\circ$, \\
$\therefore$ $\angle DAC=60^\circ$, \\
Since $DE\perp AC$, \\
$\therefore$ $\angle AED=90^\circ$, \\
$\therefore$ $\angle ADE=30^\circ$, \\
Given $AE=3$, \\
$\therefore$ $AD=2AE=6$, \\
$\therefore$ $AC=2AD=12$, \\
$\therefore$ $CE=AC-AE=\boxed{9}$.


Question: As shown in the figure, both $\triangle ABC$ and $\triangle CDE$ are equilateral triangles, with point E inside $\triangle ABC$. Line segments AE, BE, and BD are connected. If $\angle EBD=50^\circ$, what is the measure of $\angle AEB$?

Solution:
\textbf{Solution:}

Since $\triangle ABC$ is an equilateral triangle,

it follows that AC = BC, $\angle ACB=\angle ABC=\angle BAC=60^\circ$,

Since $\triangle CDE$ is an equilateral triangle,

it follows that CD = CE, $\angle DCE=60^\circ$

Therefore, $\angle ACB=\angle DCE$,

Therefore, $\angle ACB-\angle BCE=\angle DCE-\angle BCE$,

that is $\angle BCD=\angle ACE$,

In $\triangle ACE$ and $\triangle BCD$,

$\left\{\begin{array}{l}AC=BC\\ \angle ACE=\angle BCD\\ EC=CD\end{array}\right.$,

Therefore, $\triangle ACE \cong \triangle BCD$ (SAS),

Therefore, $\angle CAE=\angle CBD$,

Since $\angle EBD=\angle EBC+\angle CBD=50^\circ$,

Therefore, $\angle CAE+\angle EBC=50^\circ$,

Since $\angle CAB+\angle ABC=60^\circ+60^\circ=120^\circ$,

Therefore, $\angle ABE+\angle BAE=120^\circ-(\angle CAE+\angle EBC)=120^\circ-50^\circ=70^\circ$,

Therefore, $\angle AEB=180^\circ-(\angle ABE+\angle BAE)=180^\circ-70^\circ=\boxed{110^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=10$, $BC=12$, and $AD=8$, where $AD$ is the angle bisector of $\angle BAC$. If $P$ and $Q$ are moving points on $AD$ and $AC$ respectively, what is the minimum value of $PC+PQ$?

Solution:
\textbf{Solution:}

Since $AB=AC=10$ and $AD$ is the angle bisector of $\angle BAC$,\\
it follows that $AD\perp BC$ (as the three lines of an isosceles triangle coincide),\\
Let the point symmetrical to $Q$ with respect to line $AD$ be $Q'$, and connect $PQ'$,\\
Since $AD$ is the angle bisector of $\angle BAC$,\\
it follows that point $Q'$ lies on $AB$ (based on the properties of axial symmetry and angle bisectors),\\
therefore $PC+PQ=PC+PQ'$,\\
therefore, when $CQ'\perp AB$ and points C, P, and $Q'$ are collinear,\\
$PC+PQ$ has its minimum value, i.e., $PC+PQ=CQ'$,\\
Since $\frac{1}{2}\times BC\times AD=\frac{1}{2}\times AB\times CQ'$,\\
with $AB=10$, $BC=12$, $AD=8$,\\
it follows that $\frac{1}{2}\times 12\times 8=\frac{1}{2}\times 10\times CQ'$,\\
Solving this, $CQ'=9.6$,\\
therefore, the minimum value of $PC+PQ$ is $\boxed{9.6}$.


Question: As shown in the figure, in $\triangle ABC$, the line $DE$ is the perpendicular bisector of the side $AC$, and $AE$ is connected. If $AB=3$ and $BC=5$, what is the perimeter of $\triangle ABE$?

Solution:
\textbf{Solution:} Since $DE$ is the perpendicular bisector of side $AC$,\\
$\therefore AE=EC$,\\
$\therefore AB+BE+AE=AB+BE+EC=AB+BC=2+5=\boxed{7}$.