Question: As shown in the figure, in $\triangle ABC$, $CD$ is the altitude to the hypotenuse $AB$, and $\angle ACD=40^\circ$. What is the measure of $\angle B$?

Solution: \textbf{Solution:} Given that in the right-angled triangle $ABC$, $\angle ACB=90^\circ$, $\angle ACD=40^\circ$, it follows that $\angle DCB=50^\circ$. Since $CD$ is the altitude from the hypotenuse $AB$, we have $\angle CDB=90^\circ$. In the right-angled triangle $CDB$: $\angle B=180^\circ-\angle DCB-\angle CDB=180^\circ-50^\circ-90^\circ=\boxed{40^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, point D is on $BC$, and $DE\perp AB$ at point E, $AE=AC$. If $BC=4$ and $DE=1.5$, what is the value of $BD$?

Solution:
\textbf{Solution:} Since $DE\perp AB$,\\
it follows that $\angle AED=\angle C=90^\circ$,\\
since $AE=AC$ and $AD=AD$,\\
it follows that $\triangle ACD\cong \triangle AED\left(HL\right)$,\\
hence $DE=CD=1.5$,\\
thus $BD=BC-CD=4-1.5=\boxed{2.5}$.


Question: As shown in the figure, it is known that there is an equilateral triangle $\triangle ABC$, point $D$ is on the outside of $AC$, $BD$ is rotated clockwise by $60^\circ$ around point $B$ to $BF$, point $F$ corresponds to point $D$, connect $AF$, $AD$, with $AD=2$, $\angle CBD=15^\circ$, and $\angle AFB=30^\circ$, then what is the length of $AF$?

Solution:
\[ \textbf{Solution:} \quad \text{As shown in the figure, connect } DF, \text{ extend } FA \text{ to intersect } BD \text{ at } E, \]
\[ \because BD \text{ rotates clockwise around point } B \text{ by } 60^\circ \text{ to } BF, \]
\[ \therefore BD = BF, \angle DBF = 60^\circ, \]
\[ \therefore \triangle BDF \text{ is an equilateral triangle, } \angle BFD = 60^\circ. \]
\[ \because \angle BFA = 30^\circ, \]
\[ \therefore \angle DFA = 30^\circ, \]
\[ \therefore FE \text{ perpendicularly bisects } BD, \]
\[ \therefore AD = AB, \]
\[ \because \angle ABC = 60^\circ, \angle CBD = 15^\circ, \]
\[ \therefore \angle ABD = 45^\circ \]
\[ \therefore \triangle DAB \text{ is an isosceles right triangle, } \]
\[ \therefore DE = AE = \frac{\sqrt{2}}{2}AD = \sqrt{2}, \]
\[ \therefore BD = DF = 2\sqrt{2} \]
\[ \therefore EF = \sqrt{DF^{2} - DE^{2}} = \sqrt{6} \]
\[ AF = EF - AE = \boxed{\sqrt{6} - \sqrt{2}}. \]


Question: As shown in the figure, $AD$ is the angle bisector of $\triangle ABC$, $\angle C=90^\circ$, $CD=3\text{cm}$, and point P is a moving point on $AB$. Connect $DP$. What is the minimum value of $DP$ in cm?

Solution:
\textbf{Solution:} Construct $DP'\perp AB$ at $P'$,\\
since $AD$ is the angle bisector of $\triangle ABC$, $\angle C=90^\circ$, and $DP'\perp AB$,\\
therefore $DP'=DC=3$ cm,\\
hence, the minimum value of $DP$ is $\boxed{3}$ cm.


Question: As shown in the figure, in $\triangle ABC$, $CD$ is the altitude on the hypotenuse $AB$, and $\angle ACD=40^\circ$, then what is the degree measure of $\angle B$?

Solution:
Solution: Since in the right-angled triangle $ABC$, $CD$ is the altitude on the hypotenuse $AB$\\
Therefore, $\angle ACB=\angle CDB=90^\circ$\\
Since $\angle ACD=40^\circ$\\
Therefore, $\angle BCD=90^\circ-40^\circ=50^\circ$\\
In $\triangle BCD$, $\angle B=180^\circ-90^\circ-50^\circ=\boxed{40^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $\angle A=32^\circ$, and E is a point on side AC. First, fold $\triangle ABE$ along BE, after the fold, side AB of $\triangle ABE$ intersects AC at point D. Then fold $\triangle BCD$ along BD, such that point C precisely lands on BE. If at this moment $\angle CDB=82^\circ$, what is the degree measure of $\angle B$ in the original triangle?

Solution:
\textbf{Solution}: As shown in the figure,\\
due to the properties of folding, we have: $\angle GDB = \angle CDB = 82^\circ$, $\angle FBE = \angle ABE = \angle ABG$,\\
since $\angle GDB$ is an external angle of $\triangle BDF$,\\
therefore, $\angle DBF = \angle GDB - \angle F = 82^\circ - 32^\circ = 50^\circ$,\\
therefore, $\angle FBE = \angle ABE = 25^\circ$,\\
therefore, $\angle FBG = 3 \times 25^\circ = 75^\circ$,\\
thus, the angle $B$ of the original triangle is $\boxed{75^\circ}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=9$ and $AD=14$. Points $E$ and $F$ are located on $AD$ and $BC$ respectively, with $AE=CF=1$. Point $G$ is a moving point on side $DC$, and point $H$ is a moving point on side $AB$. What is the minimum value of $EG+HG+HF$?

Solution:
\textbf{Solution:} Let point $E'$ be the symmetric point of $E$ with respect to line $CD$, and point $F'$ be the symmetric point of $F$ with respect to line $AB$,\\
thus, $EG= E'G$ and $HF=HF'$, \\
$\therefore EG+HG+HF=E'G+HG+HF'$, \\
Connect points $E'$ and $F'$, intersecting lines $AB$ and $CD$ at points $H$ and $G$ respectively, the minimum value of $EG+HG+HF$ is exactly the length of $E'F'$, \\
Draw a line from $E'$ such that $E'H\perp BC$ and intersects the extension line of $BC$ at point $H$, \\
thus, $F'H=14+2\times 13=40$, $E'H =9$, \\
In $\triangle E'HF'$, by the Pythagorean theorem, we obtain $E'F'=\sqrt{9^2+40^2}=\boxed{41}$, \\


Question: As shown in the figure, the line $a\parallel b$. Place a right-angled triangle, which includes a $30^\circ$ angle, in the position as shown in the figure. If $\angle 2 = 2\angle 1$, then what is the degree measure of $\angle 3$?

Solution:
\textbf{Solution}: As shown in the diagram,\\
from the given condition we have $\angle 1+\angle 2=90^\circ$,\\
since $\angle 2=2\angle 1$,\\
thus $3\angle 1=90^\circ$,\\
therefore $\angle 1=30^\circ$,\\
hence $\angle 2=2\angle 1=60^\circ$,\\
since $a\parallel b$,\\
therefore $\angle 2=\angle ABD=60^\circ$,\\
since $\angle ABD=\angle C+\angle BDC$ and $\angle C=30^\circ$,\\
thus $\angle BDC=30^\circ$,\\
therefore $\angle 3=180^\circ−30^\circ=\boxed{150^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AD \perp BC$ at point $D$, and $AE$ bisects $\angle CAD$ and intersects $BC$ at $E$. If $\angle C= 60^\circ$, then what is the measure of $\angle DEA$ in degrees?

Solution:
\textbf{Solution:} Given that $AD$ is the height from $D$ to side $BC$, and $\angle C=60^\circ$,\\
$\therefore \angle CAD=90^\circ-\angle C=90^\circ-60^\circ=30^\circ$;\\
Given that $AE$ is the bisector of $\angle BAC$,\\
$\therefore \angle CAE=\angle DAE= \frac{1}{2}\angle DAC=15^\circ$,\\
$\therefore \angle DEA=\angle CAE+\angle C=15^\circ+60^\circ=\boxed{75^\circ}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $\angle BAD=\angle ABD=\angle BCD=45^\circ$, and $BC=\sqrt{2}$. What is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Draw $BE\perp DC$ at point $E$, and draw $CF\perp AB$, intersecting the extension line of $AB$ at point $F$,\\
$\therefore$ $\angle$DFB=$\angle$BFC=90$^\circ$, $\because$ $\angle$BAD=$\angle$ABD=$\angle$BCD=45$^\circ$, $\therefore$ $\angle$ADB=90$^\circ$, $AD=BD$, let $BD=x$, then $AB=\sqrt{2}x$, in $\triangle BEC$, $BE=CE$, $\therefore$ 2$BE^{2}$=$BC^{2}$=$(\sqrt{2})^{2}$, solving gives: $BE=1$, $\because$ $\angle$CDB+$\angle$CBD=180$^\circ$-45$^\circ$=135$^\circ$, $\angle$CBF+$\angle$DBC=180$^\circ$-45$^\circ$=135$^\circ$, $\therefore$ $\angle$CBF=$\angle$CDB, $\therefore$ $\triangle BDE\sim \triangle CBF$, $\therefore$ $\frac{BE}{CF}=\frac{BD}{CB}$, $\therefore$ $\frac{1}{CF}=\frac{x}{\sqrt{2}}$, solving gives: $CF=\frac{\sqrt{2}}{x}$, $\therefore$ the area of $\triangle ABC$ is $\frac{1}{2}AB\cdot CF=\frac{1}{2}\times \sqrt{2}x\times \frac{\sqrt{2}}{x}=\boxed{1}$.


Question: As shown in the diagram, in $\triangle ABC$, $\angle B=80^\circ$, $\angle C=70^\circ$. If $\triangle ABC$ is folded along $EF$, with point $A$ falling on $A'$ inside the shape, what is the measure of $\angle 1+\angle 2$?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $\angle B=80^\circ$ and $\angle C=70^\circ$, \\
therefore $\angle A=180^\circ−\angle B−\angle C=180^\circ−80^\circ−70^\circ=30^\circ$, \\
hence $\angle A'=30^\circ$, \\
thus $\angle A'EF+\angle A'FE=180^\circ−\angle A'=180^\circ−30^\circ=150^\circ$, \\
since $\triangle AFE$ is created by folding $\triangle A'FE$, \\
thus $\angle AEF+\angle AFE=\angle A'EF+\angle A'FE=180^\circ−\angle A'=150^\circ$, \\
therefore, $\angle 1+\angle 2=360^\circ−\angle B−\angle C−\left(\angle AEF+\angle AFE\right)=360^\circ−80^\circ−70^\circ−150^\circ=\boxed{60^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $\angle ABC=70^\circ$. Point D is the midpoint of AB, and point E is a moving point on side AC. $\triangle ADE$ is folded along DE, placing point A at point A'. When $A'E \parallel BC$, then what is the degree measure of $\angle ADE$?

Solution:
\textbf{Solution:} As shown in the figure, when $A'E\parallel BC$,\\
$\therefore \angle A'EA=\angle C=90^\circ$,\\
$\because \angle ABC=70^\circ$,\\
$\therefore \angle A=90^\circ-70^\circ=20^\circ$,\\
By folding, we know: $\angle A'ED=\angle AED=\frac{1}{2}\angle A'EA=45^\circ$,\\
$\therefore \angle ADE=180^\circ-\angle A-\angle AED=180^\circ-20^\circ-45^\circ=\boxed{115^\circ}$.


Question: As shown in the figure, $AD$ is the median of $\triangle ABC$, with points $E$ and $F$ located on $AD$, and $AE=EF=DF$. Connect $BE$ and $CF$. If the area of $\triangle ABC$ is 18, what is the area of the shaded region?

Solution:
\textbf{Solution:} Since $AD$ is the median of $\triangle ABC$, and the area of $\triangle ABC$, $S_{\triangle ABC}=18$,\\
then $S_{\triangle ABD}=S_{\triangle ADC}=\frac{1}{2}S_{\triangle ABC}$,\\
since $S_{\triangle ABC}=18$,\\
then $S_{\triangle ABD}=S_{\triangle ADC}=9$,\\
since $AE=EF=DF$,\\
then $S_{\triangle ABE}=\frac{1}{3}S_{\triangle ABD}=3$, $S_{\triangle DCF}=\frac{1}{3}S_{\triangle ADC}=3$,\\
therefore, the area of the shaded region $S_{\text{shaded}}=S_{\triangle ABE}+S_{\triangle DCF}=3+3=\boxed{6}$.


Question: Place a right-angle triangle set as shown in the diagram, with the short right-angle side of the triangle containing the $30^\circ$ angle coinciding with one of the right-angle sides of the triangle containing the $45^\circ$ angle. What is the measure of $\angle \alpha$?

Solution:
\textbf{Solution:} As shown in the figure, $ \angle ABC=45^\circ$, $ \angle D=30^\circ$, \\
$ \because \angle ABC+\angle DBC=180^\circ$, \\
$ \therefore \angle DBC=135^\circ$, \\
$ \because \angle DBC+\angle D+\angle BFD=180^\circ$, \\
$ \therefore \angle BFD=15^\circ$, \\
$ \therefore \angle \alpha =180^\circ−15^\circ=\boxed{165^\circ}$.


Question: As shown in the figure, $\angle BAC=30^\circ$, $AM$ is the bisector of $\angle BAC$, through $M$ draw $ME\parallel BA$ intersecting $AC$ at $E$, draw $MD\perp BA$ with $D$ as the foot of the perpendicular, and $ME=10\,\text{cm}$, then how long is $MD$ in $cm$?

Solution:
\textbf{Solution:} Draw $MF \perp AC$ at $F$,\\
since $AM$ is the angle bisector of $\angle BAC$,\\
thus $MD=MF$, $\angle BAM = \angle CAM$,\\
since $ME \parallel BA$,\\
thus $\angle AME = \angle BAM$,\\
thus $\angle CAM = \angle AME = \frac{1}{2} \angle BAC = \frac{1}{2} \times 30^\circ = 15^\circ$,\\
since $\angle CEM$ is the exterior angle of $\triangle AME$,\\
thus $\angle CEM = \angle CAM + \angle AME = 15^\circ + 15^\circ = 30^\circ$,\\
in right-angled $\triangle MEF$, $\angle FEM = 30^\circ$,\\
thus $MF = \frac{1}{2} ME = \frac{1}{2} \times 10 = \boxed{5} \, cm$,\\
thus $MD = MF = 5cm$.


Question: As shown in the figure, in $\triangle ABF$, $\angle AFB=90^\circ$. The vertex $E$ of $\triangle CDE$ is on the side $BF$ of $\triangle ABF$, point $C$ is on the extension of $BF$, $\angle C=\angle B$, and $CD=CE$. If $\angle A=36^\circ$, what is the degree measure of $\angle D$?

Solution:
\textbf{Solution:} In right triangle $ABF$, $\angle B=90^\circ-\angle A=90^\circ-36^\circ=54^\circ$,\\
$\therefore \angle C=\angle B=54^\circ$.\\
Because $CD=CE$,\\
$\therefore \angle D=\angle DEC$,\\
In triangle $CDE$,\\
$\angle C+\angle D+\angle DEC=180^\circ$,\\
$\therefore \angle D=\boxed{63^\circ}$.


Question: In $\triangle ABC$, where $BC=2$ and $AC=5$, a perpendicular line is drawn from side $AB$ to meet $AC$ at point $F$ and the extended line of $BC$ at point $E$. Connect $AE$. If it happens that $EF=AB$, what is the length of $AE$?

Solution:
\textbf{Solution:}

Given that $AB\perp DE$,

It follows that $\angle BDE=\angle ACB=90^\circ$,

Thus $\angle B+\angle BAC=\angle B+\angle BED=90^\circ$,

Therefore, $\angle BAC=\angle CEF$,

Since $\angle ACB=\angle ECF=90^\circ$ and $AB=EF$,

It means that $\triangle ABC\cong \triangle EFC$ (AAS),

Hence, $CE=AC=5$,

In $\triangle ACE$,

It results in $AE=\sqrt{AC^2+CE^2}=\sqrt{5^2+5^2}=\boxed{5\sqrt{2}}$.


Question: As shown in the figure, point $P$ is a point inside $\angle AOB$. The symmetric point of $P$ with respect to $OA$ is $C$, and the symmetric point of $P$ with respect to $OB$ is $D$. Connect $CD$ intersecting $OA$, $OB$ at points $M$ and $N$ respectively, and then connect $PM$, $PN$. If $\angle AOB=50^\circ$, what is the measure of $\angle MPN$ in degrees?

Solution:
\textbf{Solution:} Connect PC, PM, PD, PN. \\
$\because$ Point P is symmetrical to point C with respect to line OA, and to point D with respect to line OB, \\
$\therefore OC=OP=OD$, $\angle AOP=\angle AOC$, $\angle POB=\angle BOD$, \\
$\because \angle AOB=50^\circ$, \\
$\therefore \angle COD=2\angle AOB=100^\circ$. $\because OC=OP$, $\angle AOP=\angle AOC$, $OM=OM$ $\therefore \triangle OCM\cong \triangle OPM$, $\therefore \angle OCM=\angle OPM$, similarly $\angle ODN=\angle OPN$ \\
$\therefore \angle MPN=\angle OPM+\angle OPN=\angle OCM+\angle ODN=180^\circ−2\angle AOB=\boxed{80^\circ}$.


Question: As shown in the figure, fold a rectangular strip $ABCD$ along $EF$. If $\angle 1=65^\circ$, what is the measure of $\angle AEG$ in degrees?

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

$\therefore \angle DEF = \angle 1 = 65^\circ$,

By the property of folding, $\angle DEF = \angle DEF = 65^\circ$,

$\therefore \angle AEG = 180^\circ - 2\times65^\circ = \boxed{50^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $O$ is a point inside $\triangle ABC$ such that the distances from point $O$ to the three sides of $\triangle ABC$ are equal. Given that $\angle BOC = 134^\circ$, what is the measure of $\angle A$?

Solution:
\textbf{Solution:} Since the point O is equidistant from the sides of $\triangle ABC$,\\
it follows that BO bisects $\angle ABC$ and CO bisects $\angle ACB$,\\
therefore $\angle A = 180^\circ - (\angle ABC + \angle ACB) = 180^\circ - 2(\angle OBC + \angle OCB) = 180^\circ - 2 \times (180^\circ - \angle BOC) = 180^\circ - 2 \times (180^\circ - 134^\circ) = \boxed{88^\circ}$


Question: In the Cartesian coordinate system, $Q$ is a moving point on the line $y=-\frac{1}{2}x+2$. When $Q$ is rotated clockwise by $90^\circ$ around point $P(1, 0)$ to get point $Q'$, and $OQ'$ is drawn, what is the minimum value of $OQ'$?

Solution:
\textbf{Solution:} Draw $QM \perp x$-axis at point M through point Q and draw $Q^{\prime}N \perp x$-axis at point N through point $Q^{\prime}$, \\
$\therefore$ $\angle PMQ = \angle Q^{\prime}NP = 90^\circ$, \\
$\because$ $\angle QPM + \angle MPQ^{\prime} = 90^\circ$, $\angle QPM + \angle PQM = 90^\circ$, \\
$\therefore$ $\angle MPQ^{\prime} = \angle PQM$, \\
In $\triangle PQM$ and $\triangle Q^{\prime}NP$, \\
$\left\{ \begin{array}{l}
\angle PMQ = \angle Q^{\prime}NP, \\
\angle MPQ^{\prime} = \angle PQM, \\
PQ = PQ^{\prime}
\end{array} \right.$ \\
$\therefore$ $\triangle PQM \cong \triangle Q^{\prime}NP$ (AAS), \\
$\therefore$ PN = QM, $Q^{\prime}N$ = PM, \\
Let point Q be $\left(m, -\frac{1}{2}m + 2\right)$, \\
$\therefore$ PM = $|m - 1|$, QM = $\left|-\frac{1}{2}m + 2\right|$, \\
$\therefore$ ON = $\left|3 - \frac{1}{2}m\right|$, \\
$\therefore$ point $Q^{\prime} \left(3 - \frac{1}{2}m, 1 - m\right)$, \\
$\therefore$ $OQ^{\prime}{}^{2} = \left(3 - \frac{1}{2}m\right)^{2} + \left(1 - m\right)^{2} = \frac{5}{4}(m - 2)^{2} + 5$, \\
$\because$ $\frac{5}{4} > 0$, \\
$\therefore$ the parabola opens upwards, \\
$\therefore$ when $m = 2$, $OQ^{\prime}{}^{2}$ has its minimum value, which is 5, \\
$\therefore$ the minimum value of $OQ^{\prime}$ is $\boxed{\sqrt{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AD$ bisects $\angle CAB$, $BD=2CD$, and the distance from point D to $AB$ is 5.6. What is the length of $BC$ in $cm$?

Solution:
\textbf{Solution:} As shown in the figure, construct DE $\perp$ AB at E, \\
$\because$ $\angle C=90^\circ$, \\
$\therefore$ CD $\perp$ AC, \\
$\because$ AD bisects $\angle$BAC, \\
$\therefore$ CD = DE, \\
$\because$ the distance from D to AB is equal to 5.6 cm, \\
$\therefore$ CD = DE = 5.6 cm, \\
Furthermore, $\because$ BD = 2CD, \\
$\therefore$ BD = 11.2 cm, \\
$\therefore$ BC = 5.6 + 11.2 = $\boxed{16.8}$ cm,


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $AD$ is the angle bisector of $\triangle ABC$, and $E$ is the midpoint of $AC$. Connect $DE$. If $DE = 3$, then what is the value of $AB$?

Solution:
\textbf{Solution:} Since $AB=AC$ and $AD$ bisects $\angle BAC$, \\
it follows that $D$ is the midpoint of $BC$, \\
and since $E$ is the midpoint of $AC$, \\
it follows that $AB=2DE=\boxed{6}$.


Question: In the Cartesian coordinate system, \( \triangle AOB\) is an equilateral triangle, and the coordinates of point \( B\) are \((2,0)\). If \( \triangle AOB\) is rotated \(90^\circ\) counterclockwise around the origin, what are the coordinates of point \( A'\)?

Solution:
\textbf{Solution:}

Given by rotation, $\triangle A'B'O \cong \triangle ABO$,\\
Through point $A'$ draw $A'D \perp$ y-axis at $D$,\\
Since $\triangle ABC$ is an equilateral triangle,\\
Therefore, $OD=DB'= \frac{1}{2}OB' =1$,\\
Hence, $A'{D}=\sqrt{A'{O}^{2}-OD^{2}}=\sqrt{2^{2}-1^{2}}=\sqrt{3}$,\\
Therefore, the coordinates of point $A'$ are $(-\sqrt{3}, 1)$,\\
So, the answer is $\boxed{(-\sqrt{3}, 1)}$.


Question: As shown in the figure, within $\triangle ABC$, we have $AB=AC$ and $\angle A=36^\circ$. Point $D$ is located on $AC$, and $BD=BC$. What is the measure of $\angle BDC$?

Solution:
\textbf{Solution:} Since $AB=AC$ and $\angle A=36^\circ$,

it follows that $ \angle ABC=\angle ACB=\frac{1}{2}(180^\circ−\angle A)=72^\circ$.

Since $BD=BC$,

it follows that $\angle BDC=\angle ACB=\boxed{72^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=90^\circ$, $\angle C=30^\circ$, $DE$ is perpendicular to and bisects $AC$, intersecting $BC$ at point $E$, with $CE=2$, then how much is $BC$?

Solution:
\textbf{Solution:} Since DE is the perpendicular bisector of AC, and CE=2,\\
it follows that AE=CE,\\
which means $\angle C=\angle$EAC$=30^\circ$,\\
therefore $\angle$BAE$=30^\circ$,\\
hence $EB=\frac{1}{2}EA=1$,\\
thus BC=BE+EC$=\boxed{3}$.


Question: As shown in the figure, the angle bisectors of $ \angle ABC$ and $\angle ACB$ intersect at point F. A line through F is drawn parallel to $BC$, intersecting $AB$ at point D and $AC$ at point E. Given that $BD = 3\,cm$ and $EC = 2\,cm$, what is the length of $DE$ in centimeters?

Solution:
\textbf{Solution:}\\
Since $BF$ is the bisector of $\angle ABC$,\\
it follows that $\angle DBF=\angle CBF$,\\
Since $DE \parallel BC$,\\
it then implies $\angle DFB=\angle CBF$,\\
Thus, $\angle DBF=\angle DFB$,\\
Hence, $DF=BD=3\, cm$,\\
Similarly, it can be concluded that $EF=EC=2\, cm$,\\
Therefore, $DE=DF+EF=\boxed{5\, cm}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C = 90^\circ$, $\angle B = 35^\circ$. The perpendicular bisector MN of segment AB intersects AB at point E and intersects BC at point D. If AD is connected, what is the measure of $\angle DAC$ in degrees?

Solution:
\textbf{Solution:} Since $\angle B=35^\circ$ and $\angle C=90^\circ$, \\
it follows that $\angle CAB=55^\circ$, \\
Since MN is the perpendicular bisector of line segment AB, \\
it follows that AD=BD, \\
Thus, $\angle B=\angle BAD=35^\circ$, \\
Hence, $\angle DAC=\boxed{20^\circ}$.


Question: As shown in the figure, it is known that $CB\perp AD$, $AE\perp CD$, with the feet of the perpendiculars being B and E respectively, $AE$ and $BC$ intersect at point F. If $AB=BC=8$ and $CF=2$, by connecting DF, what is the area of the shaded region in the diagram?

Solution:
\textbf{Solution:} Given that $CB \perp AD$ and $AE \perp CD$, \\
it follows that $\angle ABF = \angle CEF = 90^\circ$, \\
and since $\angle AFB = \angle CFE$, \\
we have $\angle A = \angle C$. \\
In $\triangle ABF$ and $\triangle CBD$, we have \\
$\left\{\begin{array}{l}
\angle A = \angle C \\
AB = BC \\
\angle ABF = \angle CBD = 90^\circ
\end{array}\right.$, \\
hence $\triangle ABF \cong \triangle CBD$ (ASA), \\
therefore $BD = BF$, \\
given $AB = BC = 8$, $CF = 2$, \\
it results that $BD = BF = 8 - 2 = 6$, \\
thus, the area of the shaded region ${S}_{\text{shaded}} = {S}_{\triangle CBD} - {S}_{\triangle FBD} = \frac{1}{2} \times 6 \times 8 - \frac{1}{2} \times 6 \times 6 = \boxed{6}$.


Question: As shown in the figure, $ \angle MON=35^\circ$, point P is on the side $ ON$ of $ \angle MON$. With point P as the center and $ PO$ as the radius, an arc is drawn, intersecting $ OM$ at point A. When $ AP$ is connected, what is the degree measure of $ \angle APN$?

Solution:
\textbf{Solution:} From the diagram, we know that $PO=PA$,\\
$\therefore \angle PAO = \angle O = 35^\circ$,\\
$\therefore \angle APN = \angle O + \angle PAO = \boxed{70^\circ}$.


Question: As shown in the figure, in the triangular paper $ABC$, $\angle C=90^\circ$, $\angle A=30^\circ$, and $AC=9$. Fold the paper so that point $C$ falls on point $D$ on side $AB$, and the fold line $BE$ intersects $AC$ at point $E$. What is the length of the fold line $BE$?

Solution:
\textbf{Solution:} Given that $\angle C=90^\circ$ and $\angle A=30^\circ$,\\
it follows that $\angle ABC=60^\circ$.\\
Since the paper is folded such that point C lies on point D on side AB,\\
it follows that $\angle CBE=\angle ABE=30^\circ$, CE=DE, and $\angle ADE=90^\circ$,\\
which implies that $\angle A=\angle ABE$ and AE=2CE\\
Therefore, AE=BE=\boxed{6}.


Question: As shown in the figure, $BD$ bisects $\angle ABC$, and $DE\parallel BC$ intersects $BA$ at point $E$. If $DE=\frac{5}{2}$, then what is the value of $EB$?

Solution:
\textbf{Solution:} Since BD bisects $\angle$ABC,\\
it follows that $\angle$CBD=$\angle$ABD,\\
Since DE$\parallel$BC,\\
it follows that $\angle$BDE=$\angle$CBD,\\
thus $\angle$BDE=$\angle$DBE,\\
which means BE=DE,\\
Since DE= $\frac{5}{2}$,\\
it follows that EB= $\boxed{\frac{5}{2}}$.


Question: As shown in the figure, $\triangle ABC \cong \triangle DEF$, $BE=5$, $BF=1$, what is the length of $CF$?

Solution:
\textbf{Solution:} Since $BE=5$, and $BF=1$,\\
it follows that $EF=BE-BF=4$,\\
Since $\triangle ABC \cong \triangle DEF$,\\
it follows that $BC=EF=4$,\\
Therefore, $CF=BC-BF=4-1=\boxed{3}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, and $\angle A=30^\circ$, it then follows that $AB=2BC$. Based on this conclusion, further consider: if $AC=2$ and point $D$ is a moving point on side $AB$, what is the minimum value of $CD+ \frac{1}{2} AD$?

Solution:
\textbf{Solution:} Construct ray AG such that $\angle BAG=30^\circ$,\\
Draw $DE \perp AG$ at E from D, and draw $CF \perp AG$ at F from C,\\
$\therefore DE= \frac{1}{2} AD$,\\
$\therefore CD+ \frac{1}{2} AD=CD+DE \geq CF$,\\
$\because \angle CAG=\angle CAB+\angle BAG=60^\circ$, $AC=2$,\\
$\therefore \angle ACF=30^\circ$,\\
$\therefore AF=1$,\\
$\therefore CF= \sqrt{AC^2-AF^2}=\sqrt{3}$,\\
$\therefore$ The minimum value of $CD+ \frac{1}{2} AD$ is $\boxed{\sqrt{3}}$.


Question: As shown in the figure, it is known that the perimeter of $\triangle ABC$ is 22, $PB$ and $PC$ bisect $\angle ABC$ and $\angle ACB$ respectively, $PD \perp BC$ at D, and $PD = 3$, what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} As shown in the diagram, connect AP, and through point P, draw PE$\perp$AB at point E, and PF$\perp$AC at point F,\\
$\because$ PB and PC bisect $\angle ABC$ and $\angle ACB$ respectively,\\
$PD\perp BC$ at D,\\
$\therefore$ PD=PE, PD=PF,\\
$\therefore$ PD=PE=PF=3,\\
$\because$ the perimeter of $\triangle ABC$ is 22,\\
$\therefore$ the area of $\triangle ABC$ is\\
$\frac{1}{2}AB\times PE+\frac{1}{2}BC\times PD+\frac{1}{2}AC\times PF=\frac{1}{2}\times PD\times (AB+BC+AC)=\frac{1}{2}\times 3\times 22=\boxed{33}$.


Question: As shown in the figure, in $\triangle ABC$, $BE$ bisects $\angle ABC$, and $AE\perp BE$ at point $E$. The area of $\triangle BCE$ is $2$. What is the area of $\triangle ABC$?

Solution:
\textbf{Solution}: Extend AE to intersect BC at D, \\
$\because$ BE bisects $\angle ABC$, \\
$\therefore$ $\angle ABE=\angle DBE$, \\
$\because$ $AE\perp BE$, \\
$\therefore$ $\angle AEB=\angle DEB=90^\circ$, \\
In $\triangle ABE$ and $\triangle DBE$, \\
$\begin{array}{c}
\angle ABE=\angle DBE\\
EB=EB\\
\angle AEB=\angle DEB
\end{array}$, \\
$\therefore$ $\triangle ABE\cong \triangle DBE$, \\
$\therefore$ $AE=ED$, \\
$\therefore$ ${S}_{\triangle ABE}={S}_{\triangle DBE}$, \\
${S}_{\triangle ACE}={S}_{\triangle DCE}$, \\
$\therefore$ ${S}_{\triangle ABC}=2+{S}_{\triangle CBE}=\boxed{4}$


Question: In the figure, in the equilateral triangle $ABC$, the bisector of $\angle ABC$ intersects with the bisector of $\angle ACB$ at point $D$. Through point $D$, line segment $EF$ is drawn parallel to $BC$, intersecting $AB$ at $E$ and $AC$ at $F$, with $EF=4$. What is the length of $BC$?

Solution:
\textbf{Solution:} Given that in equilateral triangle ABC, EF$\parallel$BC,\\
thus $\angle$AEF=$\angle$ABC=$\angle$AFE=$\angle$ACB=$60^\circ$, and AB=AC=BC,\\
which implies that $\triangle$AEF is also an equilateral triangle,\\
therefore AE=EF=FA=4,\\
resulting in EB=FC,\\
since the bisector of $\angle$ABC intersects with the bisector of $\angle$ACB at point D, and EF$\parallel$BC,\\
it follows that $\angle$EBD=$\angle$DBC=$\angle$EDB=$\angle$FCD=$\angle$FDC=$\angle$DCB=$30^\circ$,\\
hence EB=ED=DF=FC,\\
given EF=4,\\
it leads to EB=ED=DF=FC=2,\\
therefore AB=AC=BC=AE+EB=\boxed{6},


Question: As shown in the figure, in the Cartesian coordinate system, the edges $OB$ and $OA$ of the rectangle $AOBC$ are on the $x$-axis and $y$-axis, respectively, and point $D$ is on edge $BC$. When the rectangle is folded along $AD$, point $C$ falls exactly on point $E$ on edge $OB$. If point $A$ is at $(0, 8)$ and point $B$ is at $(10, 0)$, what are the coordinates of point $D$?

Solution: \textbf{Solution:} Given $A(0,8)$ and point $B(10,0)$, $\therefore$ $OA=BC=8$, $OB=AC=10$. Let $BD=a$, then $CD=8-a$. According to the problem, $CD=DE=8-a$. Knowing from folding, $AE=AC=10$, $\therefore$ $OE=\sqrt{AE^{2}-AO^{2}}=\sqrt{10^{2}-8^{2}}=6$, $\therefore$ $BE=OB-OE=10-6=4$. Since $\angle DBE=90^\circ$, $\therefore$ $a^{2}+4^{2}=(8-a)^{2}$. Solving this, we find $a=3$. Therefore, the coordinates of point D are $(10,3)$. Hence, the answer is: $\boxed{(10,3)}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $\angle C=60^\circ$, BD bisects $\angle ABC$, and AD bisects the external angle $\angle CAE$. What is the degree measure of $\angle D$?

Solution:
\textbf{Solution:} Since BD bisects $\angle ABC$ and AD bisects the exterior angle $\angle CAE$,\\
it follows that $\angle DBA= \frac{1}{2}\angle ABC$ and $\angle DAE= \frac{1}{2}\angle CAE$,\\
Given that $\angle CAE=\angle ABC+\angle C$,\\
and $\angle DAE=\angle ABD+\angle D$,\\
it can be derived that $\angle D= \frac{1}{2}\angle CAE- \frac{1}{2} \angle ABD= \frac{1}{2} (\angle CAE-\angle ABD)= \frac{1}{2} \angle C=\frac{1}{2}\times 60^\circ=\boxed{30^\circ}$;


Question: As shown in the figure, it is known that $\triangle ABC$ is an equilateral triangle, $BD$ is the median of $\triangle ABC$. Extend $BC$ to $E$ so that $CE=CD$, and connect $DE$. What is the measure of $\angle BDE$?

Solution:
\textbf{Solution:} Given that $\triangle ABC$ is an equilateral triangle and BD is the median,\\
therefore, $\angle BDC=90^\circ$, $\angle ACB=60^\circ$\\
thus, $\angle ACE=180^\circ-\angle ACB=180^\circ-60^\circ=120^\circ$,\\
since CE=CD,\\
therefore, $\angle CDE=\angle CED=30^\circ$,\\
thus, $\angle BDE=\angle BDC+\angle CDE=90^\circ+30^\circ=\boxed{120^\circ}$


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

Solution:
\textbf{Solution:} Connect PC,

$\because$ $\angle ABC = \angle ACB$,

$\therefore$ AB = AC,

$\because$ D is the midpoint of BC,

$\therefore$ AD is the perpendicular bisector of BC, BD = $\frac{1}{2}$BC = 6

$\therefore$ BP = CP,

$AD = \sqrt{AB^{2} - BD^{2}} = \sqrt{10^{2} - 6^{2}} = 8$

$\therefore$ EP + BP = EP + CP

To minimize the value of EP + BP, utilizing the fact that the shortest distance between two points is a straight line and the shortest segment is the perpendicular, it is known that when points E, P, C are collinear and CE$\perp$AB, the value of EP + BP is minimized, with the minimum value being the length of EC;

$\because$ $S_{\triangle ABC} = \frac{1}{2}AB \cdot CE = \frac{1}{2}CB \cdot AD$,

$\therefore$ 10CE = 12 $\times$ 8

Solving for it: CE = \boxed{9.6}.


Question: As shown in the graph, in the Cartesian coordinate system, $\triangle OAB$ is translated rightward along the x-axis to obtain $\triangle O'A'B'$. The coordinates of point A are $(0, 4)$, and the corresponding point A' lies on the line $y =\frac{5}{4}x-1$. Point B is on the angle bisector of $\angle A'AO$. If the area of quadrilateral $AA'B'B$ is 4, what are the coordinates of point B?

Solution:
\textbf{Solution:} According to the properties of translation, we have: $AA'=BB'$, $AA'\parallel BB'$, and the ordinate of $A'$ is 4,\\
$\because$ point $A'$ lies on the line $y=\frac{5}{4}x-1$,\\
$\therefore$ 4=$\frac{5}{4}x-1$,\\
$\therefore$ x=4,\\
$\therefore$ AA'=BB'=4,\\
$\because$ the area of the quadrilateral $AA'B'B$ is 4,\\
$\therefore$ the distance from point B to $AA'$ is 1,\\
$\because$ point B is on the angle bisector of $\angle A'AO$,\\
$\therefore$ the distance from point B to OA is 1,\\
$\therefore$ the coordinates of point B' are $\boxed{(5,3)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the equation of line $MN$ is $y=-x+3$. Point $A$ is located on line segment $MN$ and satisfies $AN=2AM$. Point $B$ is on the $x$-axis. When $\triangle AOB$ is an isosceles triangle with $OA$ as one of its equal sides, what are the coordinates of point $B$?

Solution:
\textbf{Solution:} As illustrated in the diagram, draw AC $\perp$ OM and AD $\perp$ ON,\\
let x=0, then y=3,\\
let y=0, then x=3,\\
$\therefore$ M(0, 3), N(3, 0),\\
$\therefore$ OM=ON=3,\\
$\therefore$ MN=$3\sqrt{2}$, $\angle M=45^\circ$,\\
$\because$ AN = 2AM,\\
$\therefore$ AM=$\sqrt{2}$,\\
$\therefore$ AC=CM=1,\\
$\therefore$ OC=2,\\
$\therefore$ OA=$\sqrt{1^2+2^2}=\sqrt{5}$,\\
When point B is on the positive x-axis, OB=OA=$\sqrt{5}$, hence point B$_1$ ($\sqrt{5}$, 0),\\
When point B is on the negative x-axis, OB=OA=$\sqrt{5}$, hence point B$_2$ ($-\sqrt{5}$, 0),\\
When AB = OA, OD=$\sqrt{(\sqrt{5})^2-2^2}=1$,\\
$\therefore$ OB=2OD=2,\\
$\therefore$ point B$_3$ (2, 0),\\
$\therefore$ the coordinates of point B are $\boxed{(2,0)}$ or ($\sqrt{5}$, 0) or ($-\sqrt{5}$, 0).


Question: As shown in the diagram, in the right-angled triangle $ABC$ with $\angle A=90^\circ$, $AB=3$, and $AC=4$, triangle $ABC$ is folded along $BD$ such that point $A$ precisely falls on $BC$. What is the length of $CD$?

Solution: \textbf{Solution:} Let $CD=x$, then $AD=A^{\prime}D=4-x$. In the right-angled triangle $ABC$, $BC= \sqrt{AB^{2}+AC^{2}} =5$. Hence $A^{\prime}C=BC-AB=BC-A^{\prime}B=5-3=2$. In the right-angled triangle $A^{\prime}DC$: $AD^{2}+AC^{2}=CD^{2}$. That is, $(4-x)^{2}+2^{2}=x^{2}$. Solving this, we find: $x=\boxed{\frac{5}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=13$, and $BC=24$. Point $D$ is on $BC$ ($BD > CD$). $\triangle AED$ and $\triangle ACD$ are symmetric about line $AD$, with point $C$ being symmetric to point $E$. $AE$ intersects $BC$ at point $F$. $BE$ and $CE$ are connected. When $DE \perp BC$, what is the degree measure of $\angle ADE$? What is the length of $CE$?

Solution:
\textbf{Solution:} Construct $AH \perp BC$ at $H$, \\
since $AB=AC=13$, $BC=24$, \\
thus $BH=CH=12$, \\
thus $AH= \sqrt{AB^{2}-BH^{2}} =5$, \\
since $\triangle AED$ and $\triangle ACD$ are symmetrical about the line $AD$, \\
thus $\angle ADC=\angle ADE$, $CD=DE$, \\
since $DE\perp BC$, \\
thus $\angle BDE=90^\circ$, \\
thus $\angle ADE=90^\circ+\angle ADB=\angle ADC$, \\
thus $90^\circ+\angle ADB=180^\circ-\angle ADB$, \\
thus $\angle ADB=45^\circ$, \\
since $\angle AHC=90^\circ$, \\
thus $\angle ADB=\angle HAD=45^\circ$, \\
thus $AH=HD=5$, $\angle ADE=\angle ADB+\angle BDE=\boxed{135^\circ}$, \\
thus $BD=12+5=17$, \\
thus $CD=DE=24-17=7$, \\
thus $CE= \sqrt{CD^{2}+DE^{2}} =7\sqrt{2}$, \\
Hence, the answer is: $\boxed{135^\circ}$, $\boxed{7\sqrt{2}}$.


Question: As shown in the figure, the line $l_{1}\colon y=x+1$ intersects the x-axis at point A and intersects the line $l_{2}\colon y=\frac{1}{2}x+2$ at point B. Point C is a point on the x-axis. If $\triangle ABC$ is a right triangle, what is the x-coordinate of point C?

Solution:
\textbf{Solution:} Given the line $l_{1}\colon y=x+1$ intersects the x-axis at point A,\\
thus, A is at (-1, 0),\\
From $\left\{\begin{array}{l}y=x+1 \\ y=\frac{1}{2}x+2\end{array}\right.$, we get $\left\{\begin{array}{l}x=2 \\ y=3\end{array}\right.$,\\
therefore, B is at (2, 3),\\
When $\angle ACB=90^\circ$, the x-coordinate of point C is the same as that of B,\\
thus, C is at (2, 0);\\
When $\angle ABC=90^\circ$, then $AC^{2}=AB^{2}+BC^{2}$,\\
Let C be at (x, 0), then $AC^{2}=(x+1)^{2}$, $AB^{2}=(2+1)^{2}+3^{2}$, $BC^{2}=(2-x)^{2}+3^{2}$,\\
thus, $(x+1)^{2}=(2+1)^{2}+3^{2}+(2-x)^{2}+3^{2}$,\\
Solving it, we get $x=\boxed{5}$,\\
therefore, C is \boxed{(5, 0)},\\
In summary, the coordinates of point C are either (2, 0) or (5, 0).


Question: As shown in the diagram, $AB\parallel CD$, $AC$ bisects $\angle BAD$, $BD$ bisects $\angle ADC$, $AC$ and $BD$ intersect at point $E$, and $F$, $G$ are moving points on the line segments $AB$ and $AC$, respectively, with $AF=CG$. If $DE=1$ and $AB=2$, what is the minimum value of $DF+DG$?

Solution:
\textbf{Solution:} Connect BC.\\
Since AC bisects $\angle$BAD and BD bisects $\angle$ADC, and AB$\parallel$CD,\\
$\therefore$$\angle$DAC=$\angle$BAC, $\angle$ADB=$\angle$CDB, $\angle$AED=$180^\circ-180^\circ\div2=90^\circ$,\\
Since AB$\parallel$CD,\\
$\therefore$$\angle$DCA=$\angle$BAC,\\
$\therefore$$\angle$DCA=$\angle$DAC,\\
$\therefore$DA=DC,\\
Similarly: DA=BA,\\
$\therefore$DC=AB,\\
Since AB$\parallel$CD,\\
$\therefore$Quadrilateral ABCD is a parallelogram,\\
Since DA=DC,\\
$\therefore$Quadrilateral ABCD is a rhombus.\\
As shown in the figure. Take point B' on AC, such that AB'=AB, connect FB', construct the symmetric point D' of D about AB, connect D'F, DD'.\\
Construct B'H$\perp$CD at point H, construct B'M$\perp$DD' at point M.\\
$\therefore$DF=D'F,\\
Since AF=CG, $\angle$B'AF=$\angle$DCG, AB'=AB=CD,\\
$\therefore$$\triangle$B'AF$\cong$DCG (SAS),\\
$\therefore$B'F=DG,\\
$\therefore$DF+DG=D'F+B'F,\\
$\therefore$When points B', F, D' are on the same line, DF+DG=D'F+B'F takes the minimum value of B'D'.\\
Since DE=1, AD=AB=2,\\
$\therefore$$\angle$DAE=$30^\circ$, $\angle$ADE=$60^\circ$,\\
$\therefore$AC= $\sqrt{3}$ AD=2$\sqrt{3}$, CB'=2$\sqrt{3}$−2,\\
$\therefore$B'H= $\frac{1}{2}$ B'C=$\sqrt{3}$−1, CH=$\sqrt{3}$ B'H=3−$\sqrt{3}$,\\
$\therefore$DH=DC−CH=2−(3−$\sqrt{3}$)=$\sqrt{3}$−1,\\
Since Quadrilateral DHB$^{\prime}$M is a rectangle\\
$\therefore$DM=B'H=$\sqrt{3}$−1, MB$^{\prime}$=DH=$\sqrt{3}−1$,\\
$\therefore$D'M=DD'−DM=$\sqrt{3}$ AD−DM=2$\sqrt{3}$−($\sqrt{3}$−1)=$\sqrt{3}$+1,\\
$\therefore$D'B'= $\sqrt{M{{B}^{\prime}}^{2}+M{{D}^{\prime}}^{2}}=\sqrt{{(\sqrt{3}−1)}^{2}+{(\sqrt{3}+1)}^{2}}=2\sqrt{2}$.\\
Therefore, the minimum value of DF+DG is $\boxed{2\sqrt{2}}$.


Question: As shown in the figure, in an equilateral triangle $ABC$ with side length of 4, $D$ and $E$ are the midpoints of sides $BC$ and $AC$ respectively. $DF \perp AB$ at point $F$. Connect $EF$. What is the length of $EF$?

Solution:
\textbf{Solution:} Since $D$ and $E$ are the midpoints of sides $BC$ and $AC$ respectively, and the equilateral triangle $ABC$ has a side length of $4$,\\
$\therefore AB=BC=AC=4$, $\angle B=60^\circ=\angle C$, $AE=CE=2$, $BD=CD=2$,\\
$\therefore \triangle DEC$ is an equilateral triangle,\\
$\therefore DE=2$, $\angle EDC=60^\circ$,\\
Since $DF\perp AB$, then $\angle BDF=30^\circ$,\\
$\therefore BF=\frac{1}{2}BD$, $BF\colon DF\colon BD=1\colon \sqrt{3}\colon 2$,\\
$\therefore DF=\sqrt{3}$,\\
Since $\angle BDF=30^\circ$, $\angle EDC=60^\circ$,\\
$\therefore \angle FDE=180^\circ-30^\circ-60^\circ=90^\circ$,\\
$EF=\sqrt{{\left(\sqrt{3}\right)}^{2}+{2}^{2}}=\sqrt{7}.$\\
Therefore, the answer is $EF=\boxed{\sqrt{7}}.$


Question: As shown in the figure, a right-angle triangle plate is stacked in the manner shown. Where $ \angle A=45^\circ$ and $ \angle D=30^\circ$. If $ DF\parallel BC$, then what is the degree measure of $ \angle AGE$?

Solution:
\textbf{Solution}: Since $\triangle ABC$ and $\triangle DFE$ are both right-angled triangles, with $\angle A=45^\circ$ and $\angle D=30^\circ$,\\
thus $\angle B=45^\circ$ and $\angle DFE=90^\circ-30^\circ=60^\circ$,\\
since $DF\parallel BC$ and $\angle D=30^\circ$,\\
thus $\angle FEC=\angle DFE=60^\circ$,\\
thus $\angle GEB=90^\circ-\angle DFE=90^\circ-60^\circ=30^\circ$,\\
thus $\angle AGE=\angle GEB+\angle B=30^\circ+45^\circ=\boxed{75^\circ}$,


Question: As shown in the figure, it is known that $\triangle ABC$ is an equilateral triangle with the height $AH=8\,\text{cm}$. $P$ is a moving point on $AH$, and $D$ is the midpoint of $AB$. What is the minimum value of $PD+PB$ in $\,\text{cm}$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is an equilateral triangle and $AH$ is the height,\\
$\therefore$ point $H$ is the midpoint of $BC$,\\
point $B$ is symmetric to point $C$ with respect to $AH$, connecting $DC$ intersects $AH$ at a point $P$, at this moment there is a minimum value, the minimum value is the length of $DC$,\\
since $D$ is the midpoint of $AB$,\\
$\therefore DC$ is the height from $AB$,\\
$\therefore DC=AH=8$,\\
$\therefore PD+PB=PD+PC=DC=\boxed{8\,\text{cm}}$.


Question: As shown in the figure, point $D$ is inside $\triangle ABC$, with $CD$ bisecting $\angle ACB$, $BE \perp CD$ at point $D$, and intersecting $AC$ at point $E$, with $\angle A = \angle ABE$. If $AC = 7$ and $BC = 4$, what is the length of $BD$?

Solution:
\textbf{Solution:} As shown in the diagram, since $CD$ bisects $\angle ACB$ and $BE\perp CD$,\\
it follows that $\angle BCD=\angle ECD$ and $\angle BDC=\angle EDC=90^\circ$,\\
since $CD=CD$,\\
it follows that $\triangle BCD\cong \triangle ECD\ (ASA)$,\\
hence $BC=CE=4$, $BD=DE=\frac{1}{2}BE$,\\
since $\angle A=\angle ABE$,\\
it follows that $\triangle ABE$ is an isosceles triangle,\\
thus $AE=BE=AC−EC=AC−BC=3$,\\
therefore $BD=DE=\frac{1}{2}BE=\boxed{1.5}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is an axisymmetric figure with line $AC$ as its axis of symmetry. If $\angle BAC = 65^\circ$ and $\angle B = 50^\circ$, what is the measure of $\angle BCD$?

Solution:
\textbf{Solution:} Given that $\angle BAC=65^\circ$ and $\angle B=50^\circ$,\\
thus $\angle BCA=180^\circ-\angle B-\angle BAC=180^\circ-50^\circ-65^\circ=65^\circ$,\\
since quadrilateral ABCD is an axisymmetric figure,\\
therefore $\angle BCA=\angle DCA=65^\circ$,\\
hence $\angle BCD=\angle BCA+\angle DCA=\boxed{130^\circ}$.


Question: As shown in the diagram, $D$ is a point inside $\triangle ABC$, where $CD$ bisects $\angle ACB$, and $BE\perp CD$ with the foot of the perpendicular being $D$, intersecting $AC$ at point $E$. It is given that $\angle A = \angle ABE$. If $AC=7$ and $BC=4$, what is the length of $BD$?

Solution:
\textbf{Solution:} Given that $CD$ bisects $\angle ACB$,\\
thus $\angle BCD=\angle ECD$,\\
since $BE\perp CD$,\\
thus $\angle BDC=\angle EDC=90^\circ$,\\
since $CD=CD$,\\
thus $\triangle BDC\cong \triangle EDC$ (by ASA),\\
therefore $BC=CE=4$, $BD=DE$,\\
also since $\angle A=\angle ABE$,\\
thus $AE=BE$,\\
since $AC=7$, $BC=4$,\\
thus $AE=AC-CE=3$,\\
therefore $BE=AE=3$,\\
therefore $BD=\frac{1}{2}BE=\boxed{\frac{3}{2}}$,


Question: As shown in the figure, in $\triangle ABC$, with $AB=AC=7$, the perpendicular bisector $DE$ of $AB$ intersects $BC$ and $AB$ at points $D$ and $E$, respectively. Line $AD$ is drawn. If $BC=10$, then what is the perimeter of $\triangle ACD$?

Solution:
\textbf{Solution:} Given $DE$ is the perpendicular bisector of $AB$,\\
$\therefore$ $AD=DB$,\\
$\therefore$ the perimeter of $\triangle ADC$ is $AD+DC+AC=DB+DC+AC=BC+AC=10+7=\boxed{17}$


Question: As shown in the figure, in $\triangle ABC$, $\angle A=90^\circ$, $\angle B=30^\circ$, $CD$ bisects $\angle ACB$ and intersects $AB$ at point $D$, $DE\parallel BC$, and intersects $AC$ at point $E$. If $BC=9$, what is the length of $AE$?

Solution:
\textbf{Solution:} In the right-angled triangle $ABC$, $\angle B=30^\circ$, \\
$\therefore AC= \frac{1}{2}BC= \frac{9}{2}$, \\
$\angle ACB=90^\circ-\angle B=60^\circ$, \\
$\because CD$ bisects $\angle ACB$, \\
$\therefore \angle ACD= \frac{1}{2}\angle ACB=30^\circ$, \\
$\because DE\parallel BC$, \\
$\therefore \angle EDC=\angle DCB=\angle ACD=30^\circ$, \\
$\therefore CE=DE$, \\
$\therefore \angle ADC=60^\circ$, \\
$\therefore \angle ADE=\angle ADC-\angle EDC=30^\circ$, \\
$\therefore AE=\frac{1}{2}DE=\frac{1}{2}EC$, \\
$\therefore AE= \frac{1}{3}AC= \frac{1}{3}\times \frac{9}{2}=\boxed{\frac{3}{2}}$


Question: As shown in the figure, $AB \perp CD$, and $AB = CD$, $CE \perp AD$ at $E$, $BF \perp AD$ at $F$. Given $CE = 6$, $BF = 3$, $EF = 2$, what is the length of $AD$?

Solution:
\textbf{Solution:} Since $AB\perp CD$, $CE\perp AD$, and $BF\perp AD$,\\
it follows that $\angle A+\angle D=90^\circ$, $\angle C+\angle D=90^\circ$, and $\angle CED=\angle AFB=90^\circ$,\\
thus, $\angle A=\angle C$,\\
in $\triangle ABF$ and $\triangle CDE$,\\
$\begin{cases}
\angle A=\angle C \\
\angle AFB=\angle CED \\
AB=CD
\end{cases}$,\\
therefore, $\triangle ABF\cong \triangle CDE$ (AAS),\\
hence, $AF=CE=6$, $BF=DE=3$,\\
therefore, $AD=AF-EF+DE=6-2+3=\boxed{7}$.


Question: As shown in the figure, points D and E are the trisection points of $BC$, and $\triangle ADE$ is an equilateral triangle. What is the measure of $\angle BAC$?

Solution:
Solution: Since $E$ is the trisecting point of $BC$, and $\triangle ADE$ is an equilateral triangle,\\
$\therefore BD=DE=EC=AD=AE$, $\angle ADE=\angle AED=60^\circ$,\\
$\therefore \angle B=\angle BAD=\angle C=\angle EAC=30^\circ$,\\
$\therefore \angle BAC=180^\circ-\angle B-\angle C=\boxed{120^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C = 90^\circ$, $\angle A = 20^\circ$, $BD$ is the bisector of $\angle ABC$, then what is the measure of $\angle BDC$ in degrees?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $\angle C=90^\circ$, $\angle A=20^\circ$,\\
thus $\angle ABC=180^\circ-\angle C-\angle A=70^\circ$,\\
since $BD$ is the bisector of $\angle ABC$,\\
thus $\angle DBC=\frac{1}{2}\angle ABC=35^\circ$,\\
in $\triangle DBC$,\\
since $\angle C=90^\circ$, $\angle DBC=35^\circ$,\\
thus $\angle BDC=180^\circ-\angle C-\angle DBC=\boxed{55^\circ}$.


Question: As shown in the figure, in $\triangle ABC$ and $\triangle DEC$, we have $AB=DE$, $AC=DC$, $CE=CB$. Point $E$ is on $AB$. If $\angle ACE=2\angle ECB=50^\circ$, then what is the degree measure of $\angle A$?

Solution:
\textbf{Solution:} Since $\angle ACE = 2\angle ECB = 50^\circ$,\\
it follows that $\angle ECB = 25^\circ$,\\
thus $\angle ACB = \angle ACE + \angle ECB = 50^\circ + 25^\circ = 75^\circ$,\\
since $CE = CB$, $\angle ECB = 25^\circ$,\\
it follows that $\angle B = \angle CEB = \frac{1}{2}(180^\circ - \angle ECB) = \frac{1}{2}(180^\circ - 25^\circ) = 77.5^\circ$,\\
therefore $\angle A = 180^\circ - \angle B - \angle ACB = 180^\circ - 77.5^\circ - 75^\circ = \boxed{27.5^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, point D is the midpoint of the hypotenuse $AB$, $CD=5$, $BC=6$. Connect $CD$, fold $\triangle BCD$ along $CD$ so that point B falls on point E, point F is a point on the right-angle side $AC$, connect $DF$, fold $\triangle ADF$ along $DF$ so that point A coincides with point E. What is the area of $\triangle EFC$?

Solution:
\textbf{Solution:} Given that $\angle ACB=90^\circ$ and point D is the midpoint of hypotenuse $AB$ with $CD=5$ and $BC=6$,\\
$\therefore$ $AB=10$,\\
$\therefore$ $AC=\sqrt{AB^2-BC^2}=8$,\\
From folding, we know that $\angle B=\angle DEC$, $\angle A=\angle DEF$, $CE=BC=6$, and $AF=EF$,\\
$\because$ $\angle A+\angle B=90^\circ$,\\
$\therefore$ $\angle DEF+\angle DEC=90^\circ$, which means $\angle FEC=90^\circ$,\\
$\therefore$ $EF^2+CE^2=CF^2$,\\
Let $AF=EF=x$, then $CF=AC-AF=8-x$,\\
$\therefore$ $x^2+6^2=(8-x)^2$,\\
Solving for $x$ yields $x=\frac{7}{4}$,\\
$\therefore$ $EF=\frac{7}{4}$,\\
$\therefore$ The area of $\triangle EFC$ is $\frac{1}{2}CE\times EF=\frac{1}{2}\times 6\times \frac{7}{4}=\boxed{\frac{21}{4}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=9\text{cm}$, $DE$ is the perpendicular bisector of $AB$, intersecting $AB$ and $AC$ at points $D$ and $E$, respectively. If $BC=5\text{cm}$, what is the perimeter of $\triangle BCE$ in $\text{cm}$?

Solution:
\textbf{Solution:} Given that $DE$ is the perpendicular bisector of $AB$,\\
it follows that $AE=BE$,\\
given $AC=9\text{cm}$, $BC=5\text{cm}$, and $AE=BE$,\\
thus, the perimeter of $\triangle BCE$ is $BC+BE+CE=BC+AE+CE=BC+AC=9\text{cm}+5\text{cm}=\boxed{14\text{cm}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AD$ bisects $\angle BAC$ and intersects $BC$ at point $D$, $DE\perp AB$ with the foot at $E$. If $BC=4$ and $DE=1.6$, what is the length of $BD$?

Solution:
\textbf{Solution:} Since $AD$ bisects $\angle BAC$, $DE\perp AB$, and $\angle C=90^\circ$,\\
it follows that $CD=DE$,\\
Since $DE=1.6$,\\
it follows that $CD=1.6$,\\
therefore $BD=BC-CD=4-1.6=\boxed{2.4}$.


Question: As shown in the figure, three squares form a right-angled triangle, with $64$ and $400$ as the areas of the respective squares. What is the area of the square represented by the letter $M$ in the figure?

Solution:
\textbf{Solution:} According to the relationship between the area and the side length of a square,\\
the square of the short leg of the right-angled triangle in the figure is 64, and the square of the hypotenuse is 400,\\
based on the Pythagorean theorem, the square of the long leg of the right-angled triangle is $400-64=336$,\\
thus the area of the square represented by the letter $M$ is $\boxed{336}$.


Question: As shown in the diagram, $ \triangle ABC \cong \triangle ADE$, and $\angle EAB = 120^\circ$, $\angle B = 30^\circ$, $\angle CAD = 10^\circ$, what is the degree measure of $\angle CFD$?

Solution:
\textbf{Solution:} Since $\triangle ABC \cong \triangle ADE$, \\
it follows that $\angle EAD = \angle CAB$, \\
and since $\angle EAB = 120^\circ$ and $\angle CAD = 10^\circ$, \\
it follows that $\angle EAD = \angle CAB = 55^\circ$, \\
therefore, $\angle CFD = \angle FAB + \angle B = 10^\circ + 55^\circ + 30^\circ = \boxed{95^\circ}$


Question: As shown in the figure, in $\triangle ABC$, the perpendicular bisector of AB intersects AC at point D. Given that $BC=10$ and the perimeter of $\triangle BDC$ is $25$, what is the length of AC?

Solution:
\textbf{Solution:} Since $DE$ is the perpendicular bisector of $AB$,\\
it follows that $DB=DA$,\\
Since the perimeter of $\triangle BDC$ is 25,\\
it follows that $BC+BD+CD=25$,\\
hence $BC+CD+DA=25$,\\
thus $BC+CA=25$,\\
since $BC=10$,\\
it follows that $AC=25-10=\boxed{15}$.


Question: As shown in the figure, in $ \triangle ABC$, $ DE \parallel BC$, $ DF \parallel AC$, $ AD=3$, $ BD=2$, what is the value of $ BF: DE$?

Solution:
\textbf{Solution:} Since $DE \parallel BC$ and $DF \parallel AC$, \\
it follows that $\angle ADE=\angle B$ and $\angle BDF=\angle BAC$, \\
which means $\triangle DBF \sim \triangle ADE$, \\
thus $\frac{BD}{DA}=\frac{BF}{DE}=\frac{2}{3}$, \\
hence $\frac{BF}{DE}=\boxed{\frac{2}{3}}$.


Question: As shown in the figure, within rectangle $ABCD$, the bisector of $\angle BCD$ intersects side $AD$ at point $E$, the bisector of $\angle AEC$ intersects the extended line of side $CB$ at point $G$ and intersects side $AB$ at point $F$. If $AB = 3\sqrt{2}$ and $AF = 2BF$, then what is the length of $GB$?

Solution:
\textbf{Solution:} From rectangle $ABCD$, the angle bisector of $\angle BCD$, CE, intersects AD at E; \\
$\therefore CD=AB=3\sqrt{2}$, $\angle DCE=\angle BCE=45^\circ$, \\
$\therefore CD=DE=3\sqrt{2}$, \\
From right triangle CDE, \\
$\therefore CE=\sqrt{CD^2+DE^2}=6$, \\
Also, the angle bisector of $\angle AEC$, EG, intersects AB at point F, \\
$\therefore \angle AEG=\angle CEG$, \\
Since AD$\parallel$BC, \\
$\therefore \angle G=\angle AEG$, \\
$\therefore \angle CEG=\angle G$, \\
$\therefore CG=CE=6$, \\
Since $\angle G=\angle AEF$, $\angle AFE=\angle BFG$, \\
$\therefore \triangle AEF\sim \triangle BGF$, \\
$\therefore \frac{GB}{AE}=\frac{FG}{EF}=\frac{FB}{AF}=\frac{BF}{2BF}=\frac{1}{2}$, \\
Let $BG=x$, then $AE=2x$, thus $BC=AD=3\sqrt{2}+2x$, \\
Because $CG=BC+BG$, \\
$\therefore 6=3\sqrt{2}+2x+x$, solving for $x$ gives $x=\boxed{2-\sqrt{2}}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $\angle C=90^\circ$, and $\tan A=\frac{5}{12}$. After rotating $\triangle ABC$ $90^\circ$ counterclockwise around point $A$, we obtain $\triangle ADE$, where point $B$ falls on point $D$, and point $C$ falls on point $E$. By connecting $BE$ and $CD$, and drawing the bisector $AN$ of $\angle CAD$, which intersects line segment $BE$ at point $M$ and line segment $CD$ at point $N$, what is the value of $\frac{AM}{AN}$?

Solution:
\textbf{Solution:} Let's establish a Cartesian coordinate system with point C as the origin. Through point N, draw $NF\perp AC$ extending to intersect BP at point P, and $NH\perp AD$ intersecting at point H, and $NG\perp x$-axis intersecting at point G. Through point D, draw $DQ\perp x$-axis intersecting at point Q,\\
$\because$ $\angle C=90^\circ$, $\tan A=\frac{5}{12}$,\\
$\therefore$ let $BC=5k$, $AC=12k$, $AB=13k$,\\
By rotation, we can obtain: $AE=AC=12k$, $DE=BC=5k$, $AB=AD=13k$,\\
$\therefore$ $CQ=12k$, $DQ=7k$,\\
$\therefore$ $B(-5k,0)$, $E(12k,12k)$, $D(12k,7k)$,\\
$\because$ AN is the bisector of $\angle CAD$,\\
$\therefore$ $NF=NH$,\\
$\therefore$ $\frac{S_{\triangle ANC}}{S_{\triangle AND}}=\frac{AC}{AD}=\frac{12k}{13k}=\frac{12}{13}$, hence $\frac{CN}{DN}=\frac{12}{13}$,\\
$\therefore$ $N(\frac{144k}{25},\frac{84k}{25})$,\\
Let the equation of line BE be $y=mx+n$,\\
Substitute $B(-5k,0)$, $E(12k,12k)$ into it: $\left\{\begin{array}{l}-5km+n=0 \\ 12km+n=12k\end{array}\right.$,\\
Solving them, we get: $\left\{\begin{array}{l}m=\frac{12}{17} \\ n=\frac{60k}{17}\end{array}\right.$,\\
$\therefore$ $y=\frac{12}{17}x+\frac{60k}{17}$,\\
When $y=\frac{84k}{25}$, $\frac{12}{17}x+\frac{60k}{17}=\frac{84k}{25}$,\\
Solving for $x$: $x=-\frac{6k}{25}$,\\
$\therefore$ $P(-\frac{6k}{25},\frac{84k}{25})$,\\
$\therefore$ $NP=\frac{144k}{25}-(-\frac{6k}{25})=6k$,\\
$\because$ $NF\perp AC$, $\angle CAE=90^\circ$,\\
$\therefore$ $\angle CFN=\angle CAE=90^\circ$,\\
$\therefore$ $AE\parallel PN$,\\
$\therefore$ $\triangle MAE\sim \triangle MNP$,\\
$\therefore$ $\frac{AM}{NM}=\frac{AE}{NP}=\frac{12k}{6k}=2$,\\
$\therefore$ $\frac{AM}{AN}=\boxed{\frac{2}{3}}$.


Question: As shown in the diagram, in parallelogram $ABCD$, $AB=8$, $DE\perp AB$ with the foot of the perpendicular at E, $BE=6$. Let $CE$ be connected. If $\angle CEB=\angle ADE$, then what is the length of $DE$?

Solution:
\textbf{Solution:}

Given: In quadrilateral $ABCD$, $AB=8$, $BE=6$,\\
thus $AB\parallel CD$, $CD=AB=8$, $AE=2$,\\
thus $\angle CDE=\angle AED, \angle DCE=\angle CEB$,\\
given that $\angle CEB=\angle ADE$,\\
thus $\angle DCE=\angle ADE$,\\
therefore, $\triangle ADE \sim \triangle ECD$,\\
thus $\frac{AE}{DE}=\frac{DE}{CD}$,\\
thus $\frac{2}{DE}=\frac{DE}{8}$,\\
thus $DE^2=16$,\\
thus $DE=\pm 4$,\\
Upon verification: $DE=4$ is consistent with the problem, $DE=-4$ is not consistent and is discarded,\\
therefore, $DE=\boxed{4}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=14$, $AC=10$, point $D$ is on $BC$, and point $M$ is on the extension of $BA$. Given that $\tan\angle CAM=\frac{4}{3}$ and $\angle DAB=45^\circ$, what is the length of $AD$?

Solution:
\textbf{Solution:} Draw $CE\parallel AD$ through $C$ intersecting ray $AM$ at point $E$, and draw $CF\perp AM$ with the foot at $F$, \\
since $AC=10$ and $\tan\angle CAM=\frac{4}{3}$, $CF\perp AM$, \\
therefore, $\tan\angle CAM=\frac{CF}{AF}=\frac{4}{3}$, \\
let $CF=4m$ and $AF=3m$, \\
according to the Pythagorean theorem: $CF^2+AF^2=AC^2$, that is $16m^2+9m^2=100$, \\
we find: $m=2$ (discard $m=-2$), \\
thus, $CF=8$ and $AF=6$, \\
since $CE\parallel AD$ and $\angle DAB=45^\circ$, \\
thus, $\angle CEF=\angle DAB=45^\circ$, \\
thus, $EF=CF=8$, $CE=8\sqrt{2}$, \\
since $\angle CEF=\angle DAB=45^\circ$ and $\angle B=\angle B$, \\
thus, $\triangle DBA\sim \triangle CBE$, \\
thus, $\frac{AB}{BE}=\frac{AD}{CE}$, that is $\frac{14}{AF+EF+AB}=\frac{14}{28}=\frac{AD}{8\sqrt{2}}$, \\
therefore, $AD=\boxed{4\sqrt{2}}$.


Question: As shown in the diagram, through the centroid \(G\) of \(\triangle ABC\), construct \(ED\parallel AB\) meeting sides \(AC\) and \(BC\) at points \(E\) and \(D\) respectively. Connect \(AD\). If \(AD\) bisects \(\angle BAC\) and \(AB=6\), what is the length of \(EC\)?

Solution:
\textbf{Solution}:
Draw and extend CG to meet AB at H,\\
since G is the centroid of $\triangle ABC$\\
therefore $\frac{CG}{GH}=2$\\
therefore $\frac{CG}{CH}=\frac{2}{3}$\\
since $ED\parallel AB$\\
therefore $\frac{CG}{CH}=\frac{2}{3}=\frac{EC}{AC}$, $\angle ADE=\angle BAD$, $\triangle ECD\sim \triangle ACB$\\
therefore $\frac{EC}{AC}=\frac{DE}{AB}=\frac{2}{3}$\\
therefore $DE=\frac{2}{3}AB=4$\\
since AD bisects $\angle BAC$\\
therefore $\angle EAD=\angle BAD$\\
therefore $\angle EAD=\angle ADE$\\
therefore $DE=AE=4$\\
therefore $\frac{2}{3}=\frac{EC}{AC}=\frac{EC}{EC+AE}$,\\
therefore $EC=\boxed{8}$


Question: As shown in the figure, in $\triangle ABC$, the medians $AD$ and $BE$ intersect at point $O$. If the area of $\triangle AOE$ is 4, what is the area of the quadrilateral $OECD$?

Solution:
\textbf{Solution:} As shown in the figure, connect DE,\\
$\because$ AD, BE are medians on sides BC, AC respectively,\\
$\therefore$ D, E are the midpoints of BC, AC respectively,\\
$\therefore$ DE is the midline of $\triangle ABC$,\\
$\therefore$ DE $=\frac{1}{2}AB$, DE $\parallel$ AB,\\
$\therefore$ $\triangle ABO\sim \triangle DEO$, $\triangle CDE\sim \triangle CBA$,\\
$\therefore$ $\frac{OE}{OB}=\frac{DE}{AB}=\frac{1}{2}$,\\
$\therefore$ $\frac{S_{\triangle ABO}}{S_{\triangle AOE}}=\frac{BO}{EO}=2$,\\
$\therefore$ $S_{\triangle ABO}=8$,\\
$\therefore$ $S_{\triangle ABE}=S_{\triangle ABO}+S_{\triangle AOE}=12$,\\
$\therefore$ $S_{\triangle ABC}=2S_{\triangle ABE}=24$,\\
$\because$ $\frac{S_{\triangle DEO}}{S_{\triangle ABO}}=\left(\frac{DE}{AB}\right)^2=\frac{1}{4}$,\\
$\therefore$ $S_{\triangle DEO}=2$,\\
$\because$ $\frac{S_{\triangle CDE}}{S_{\triangle ABC}}=\left(\frac{DE}{AB}\right)^2=\frac{1}{4}$,\\
$\therefore$ $S_{\triangle CDE}=6$,\\
$\therefore$ $S_{\text{quadrilateral} OECD}=S_{\triangle DEO}+S_{\triangle CDE}=\boxed{8}$,


Question: We call a rectangle whose width to length ratio is the golden ratio $\left(\frac{\sqrt{5}-1}{2}\right)$ a golden rectangle. As shown in the figure, in the golden rectangle $ABCD$, where $AB<BC$, $BC=2$, and the bisector of $\angle ABC$ intersects side $AD$ at point $E$, what is the length of $DE$?

Solution:
\textbf{Solution:} Given a golden rectangle $ABCD$, \\
the ratio of width to length is the golden ratio $\left(\frac{\sqrt{5}-1}{2}\right)$, and since AD$\parallel$BC, AD=BC=2, \\
thus, AB=$\left(\frac{\sqrt{5}-1}{2}\right)$BC=$\frac{\sqrt{5}-1}{2}\times 2=\sqrt{5}-1$, \\
because BE is the bisector of $\angle ABC$, \\
it follows that $\angle$ABE=$\angle$CBE, \\
and since AD$\parallel$BC, \\
it also follows that $\angle$AEB=$\angle$CBE=$\angle$ABE, \\
hence, AE=AB=$\sqrt{5}-1$, \\
therefore, DE=AD−AE=2−($\sqrt{5}-1$)=$\boxed{3-\sqrt{5}}$．


Question: As shown in the figure, it is known that $AB \parallel CD \parallel EF$, which intersect lines $l_1, l_2$ at points $A, D, F$ and $B, C, E$ respectively. If $\frac{AD}{DF} = \frac{2}{3}$ and $BE=20$, what is the length of segment $BC$?

Solution:
\textbf{Solution:} Since $AB\parallel CD\parallel EF$,\\
it follows that $\frac{BC}{CE}=\frac{AD}{DF}=\frac{2}{3}$,\\
thus $CE=\frac{3}{2}BC$,\\
Given $BC+CE=BE=20$,\\
we have $\frac{3}{2}BC+BC=20$,\\
solving this gives $BC=\boxed{8}$.


Question: As shown in the figure, in $ \triangle ABC$, D and E are the midpoints of sides $AB$ and $AC$, respectively. What is the ratio of the perimeters of $ \triangle ADE$ and $ \triangle ABC$?

Solution:
\textbf{Solution:} Given that point D and point E are the midpoints of side AB and AC, respectively, \\
it follows that DE is the mid-segment of $\triangle ABC$, \\
thus, DE$\parallel$BC, and the ratio DE$\colon$BC = 1$\colon$2, \\
therefore, $\triangle ADE\sim \triangle ABC$, \\
hence, the ratio of the perimeters of $\triangle ADE$ to $\triangle ABC$ is $\boxed{1\colon 2}$.


Question: As shown in the figure, $AB\parallel CD\parallel EF$, $AF$ intersects with $BE$ at point $G$, and $AG=4$, $GD=2$, $DF=8$. What is the value of $\frac{BC}{CE}$?

Solution:
\textbf{Solution:} Since $AB\parallel CD\parallel EF$, \\
it follows that $\frac{BC}{CE}=\frac{AD}{DF}=\frac{AG+GD}{DF}$, \\
given that $AG=4, GD=2, DF=8$, \\
we have $\frac{BC}{CE}=\frac{AD}{DF}=\frac{AG+GD}{DF}=\frac{4+2}{8}=\boxed{\frac{3}{4}}$.


Question: Given rectangle $ABCD$, with point $E$ on side $AD$, $BE$ and $BD$ are connected, $\angle BED=2\angle BDC$, $BE=25$, $BC=32$, what is the length of $CD$?

Solution:
\textbf{Solution:} Construct $EF\perp BD$ at point $F$,\\
since quadrilateral $ABCD$ is a rectangle,\\
thus $\angle C=\angle ADC=90^\circ$,\\
since $\angle BED=2\angle BDC$,\\
thus $180^\circ-\angle AEB=2\left(90^\circ-\angle ADB\right)$,\\
thus $\angle AEB=2\angle EDB$,\\
since $\angle AEB=\angle ADB+\angle EBD$,\\
thus $\angle EDB=\angle EBD$,\\
thus $BE=DE$,\\
thus $BF=\frac{1}{2}BD$,\\
since $AD\parallel BC$,\\
thus $\angle ADB=\angle DBC$,\\
thus $\angle EBF=\angle DBC$,\\
thus $\triangle EBF\sim \triangle DBC$,\\
thus $\frac{BE}{BD}=\frac{BF}{BC}$,\\
thus $BD^2=2BC\cdot BE=2\times 25\times 32=40^2$,\\
thus $BD=40$,\\
thus $CD=\sqrt{BD^2-BC^2}=\boxed{24}$.


Question: As shown in the figure, in an equilateral triangle $ABC$ with side length $9$, where $BD=3$, and $\angle ADE=60^\circ$, what is the length of $AE$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is an equilateral triangle, we have $\angle B=\angle C=60^\circ$, and $AB=BC$.\\
Therefore, $CD=BC-BD=9-3=6$, and $\angle BAD+\angle ADB=120^\circ$.\\
Since $\angle ADE=60^\circ$, $\angle ADB+\angle EDC=120^\circ$. Thus, $\angle DAB=\angle EDC$.\\
Moreover, since $\angle B=\angle C=60^\circ$, we conclude that $\triangle ABD\sim \triangle DCE$.\\
Thus, $\frac{AB}{BD}=\frac{DC}{CE}$, which gives $\frac{9}{3}=\frac{6}{CE}\Rightarrow CE=2$.\\
Therefore, $AE=AC-CE=9-2=\boxed{7}$.


Question: As shown in the figure, in $\triangle ABC$, point $D$ is on side $AB$, and point $E$ is on side $AC$. The angle $\angle ADE$ is equal to $\angle C$, and the area of quadrilateral $DBCE$ is three times the area of $\triangle ADE$. If $DE = 3$, what is the length of $BC$?

Solution:
\textbf{Solution:} Since $\angle A = \angle A$ and $\angle ADE = \angle C$,\\
it follows that $\triangle ADE \sim \triangle ACB$,\\
thus, $\frac{S_{\triangle ADE}}{S_{\triangle ABC}} = \left(\frac{DE}{BC}\right)^2$.\\
Given that the area of quadrilateral DBCE is three times the area of $\triangle ADE$,\\
we get $\frac{S_{\triangle ADE}}{S_{\triangle ABC}} = \frac{1}{4}$,\\
hence $\left(\frac{DE}{BC}\right)^2 = \frac{1}{4}$.\\
Since $DE = 3$,\\
it follows that $BC = \boxed{6}$.


Question: As shown in the figure, in $\triangle ABC$, if $DE\parallel BC$, $\frac{AD}{BD}=\frac{1}{2}$, and $DE=4\,\text{cm}$, then what is the length of $BC$ in $\,\text{cm}$?

Solution:
\textbf{Solution:} Given $DE\parallel BC$,\\
therefore $\triangle ADE \sim \triangle ABC$,\\
thus $\frac{DE}{BC} = \frac{AD}{AB}$,\\
also, given $\frac{AD}{DB} = \frac{1}{2}$,\\
thus $\frac{AD}{AB}=\frac{1}{3}$,\\
hence $\frac{4}{BC}=\frac{1}{3}$,\\
therefore $BC=\boxed{12}\,\text{cm}$.


Question: As shown in the diagram, within $ \square ABCD$, where E is on DC, if $ DE: EC=1: 2$, what is the value of $ \frac{{S}_{\triangle ABF}}{{S}_{\triangle CEF}}$?

Solution:
\textbf{Solution:} Given $\because DE:EC=1:2$,\\
$\therefore \frac{CD}{CE}=\frac{3}{2}$,\\
$\because$ quadrilateral ABCD is a parallelogram,\\
$\therefore$ AB $\parallel$ CD, AB=CD,\\
$\therefore \triangle ABF\sim \triangle CEF$,\\
$\therefore \frac{S_{\triangle ABF}}{S_{\triangle CEF}}=\left(\frac{AB}{CE}\right)^2=\left(\frac{3}{2}\right)^2=\boxed{\frac{9}{4}}$.


Question: As shown in the diagram, it is known that $\odot O$ is the circumscribed circle of quadrilateral $ABCD$ with $O$ as the center of the circle. If $\angle BCD=120^\circ$ and $AB=AD=2\, \text{cm}$, then what is the length of the radius of $\odot O$?

Solution:
\textbf{Solution:} Draw BD and construct OE $\perp$ AD at E, and draw OD.\\
Since $\odot O$ is the circumcircle of quadrilateral ABCD, and $\angle BCD=120^\circ$,\\
it follows that $\angle BAD=60^\circ$.\\
Given AD=AB=2cm,\\
triangle ABD is an equilateral triangle.\\
Hence, DE= $\frac{1}{2}$ AD=1cm, and $\angle ODE= \frac{1}{2} \angle ADB=30^\circ$,\\
therefore, OD= $\frac{DE}{\cos 30^\circ}$ = $\boxed{\frac{2\sqrt{3}}{3}}$ cm.


Question: As shown in the diagram, points $A$, $B$, and $C$ are three points on the circle $\odot O$, with $AC=BC$. Point $M$ is another point on $\odot O$, and $CE$ is perpendicular to $AM$, with the foot of the perpendicular being point $E$. Given that $AE=2\sqrt{3}$, $BM=\sqrt{3}$, and $CM=\sqrt{7}$, what is the length of arc $\overset{\frown}{CM}$?

Solution:
\textbf{Solution:} Let's cut $AG=BM$ on $AE$, and connect $CG$,\\
since $AC=BC$ and $\angle A=\angle B$,\\
thus $\triangle ACG \cong \triangle BCM$ (SAS),\\
hence $CG=CM=\sqrt{7}$,\\
since $AE=2\sqrt{3}$ and $AG=BM=\sqrt{3}$,\\
thus $GE=\sqrt{3}$,\\
since $CE\perp AM$,\\
thus $CE = \sqrt{CG^{2} - GE^{2}} = \sqrt{(\sqrt{7})^2 - (\sqrt{3})^2} = 2$,\\
thus $\tan\angle A = \frac{CE}{AE} = \frac{\sqrt{3}}{3}$,\\
thus $\angle A = 30^\circ$,\\
thus $\angle COM = 60^\circ$,\\
connect $OM$ and $CO$,\\
since $OC=OM$,\\
thus $\triangle COM$ is an equilateral triangle,\\
thus $OC=\sqrt{7}$,\\
thus the length of $\overset{\frown}{CM} = \frac{60\pi \times \sqrt{7}}{180} = \boxed{\frac{\sqrt{7}}{3}\pi}$.


Question: As shown in the rectangle $ABCD$, where $AB=8$ and $BC=6$, a circular arc $BH$ is drawn with vertex $C$ as the center and the length of $BC$ as the radius. A tangent $PE$ to the arc $BH$ is drawn passing through the midpoint $P$ of $AB$, with $E$ as the point of tangency. Connect $AE$ and extend it to intersect $CD$ at point $F$. What is the numerical value of $\tan\angle DAF$?

Solution:
\textbf{Solution:} As shown in the diagram, connect CE, CP, and BE. \\
Since PE and BC are the tangents to circle C, \\
and PE = PB, \\
and BC = BE, \\
therefore, CP is the perpendicular bisector of BE, \\
therefore, CP $\perp$ BE, \\
Since PA = PB = PE, \\
therefore, $\angle$AEB$=90^\circ$, that is, BE $\perp$ AF, \\
therefore, AF $\parallel$ BC, \\
since AP $\parallel$ FC, \\
therefore, the quadrilateral APCF is a parallelogram, \\
thus, FC = AP = 2, \\
therefore, FD = CD - FC = 8 - 4 = 4, \\
therefore, $\tan\angle$DAF$= \frac{FD}{AD}=\frac{4}{6}=\boxed{\frac{2}{3}}$.


Question: As shown in the figure, each small square in the grid has a side length of \(1\), and the vertices of \(\triangle ABC\) are all located at grid points. What is the sine value of \(\angle ABC\)?

Solution:
\textbf{Solution:} As shown in the figure, \\
$\because AC=\sqrt{2^2+4^2}=2\sqrt{5}$, $AB=\sqrt{2^2+4^2}=2\sqrt{5}$, \\
$\therefore AC=AB$, \\
$\because CD=BD=\sqrt{1^2+1^2}=\sqrt{2}$, \\
$\therefore AD\perp BC$, \\
In $\triangle ABD$ with a right angle, \\
$AD=\sqrt{AB^2−BD^2}=3\sqrt{2}$, \\
$\therefore \sin\angle ABC=\sin\angle ABD=\frac{AD}{AB}=\frac{3\sqrt{2}}{2\sqrt{5}}=\boxed{\frac{3\sqrt{10}}{10}}$.


Question: As shown in the figure, in square $ABCD$, point $E$ is on $BC$, and $CE=\frac{1}{2}BC$. Draw $BG\perp DE$ at $G$, and extend $DC$ to intersect with $BG$ at $H$. Connect $AG$ and extend it to intersect the extended line of $DC$ at $F$. Then, what is the value of $\tan\angle CBH$? What is the value of $\frac{CF}{DF}$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a square,\\
$\therefore \angle BCD = \angle BCH = 90^\circ$, and CD=CB,\\
since $BG \perp DG$,\\
$\therefore \angle BGE = \angle DCE = 90^\circ$,\\
since $\angle BEG = \angle CED$,\\
$\therefore \angle CDE = \angle CBH$,\\
$\therefore \triangle DCE \cong \triangle BCH$ (ASA),\\
$\therefore \angle CBH = \angle CDE$,\\
since CE = $\frac{1}{2}BC$,\\
$\therefore \tan\angle CBH = \tan\angle CDE = \frac{CE}{CD} = \boxed{\frac{1}{2}}$,\\
Let BE=EC=m, then BC=CD=2m,\\
$\therefore EG = \frac{\sqrt{5}}{5}m$, $BG = \frac{2\sqrt{5}}{5}m$, $DE = \sqrt{CD^2 + EC^2} = \sqrt{5}m$,\\
$\therefore DH = DC + CH = 3m$, $DG = DE + EG = \frac{6\sqrt{5}}{5}m$,\\
$\therefore GH = \sqrt{DH^2 - DG^2} = \sqrt{(3m)^2 - \left(\frac{6\sqrt{5}}{5}m\right)^2} = \frac{3\sqrt{5}}{5}m$,\\
since AB $\parallel$ FH,\\
$\therefore \triangle FHG \sim \triangle ABG$,\\
$\therefore \frac{FH}{2m} = \frac{\frac{3\sqrt{5}}{5}m}{\frac{2\sqrt{5}}{5}m}$,\\
$\therefore FH = 3m$,\\
$\therefore DF = 6m$, CF = 4m,\\
$\therefore \frac{CF}{DF} = \frac{4m}{6m} = \boxed{\frac{2}{3}}$.


Question: Given $\angle AOB=60^\circ$ and a circle $\odot P$ with a radius of $3\text{cm}$ moves parallel from the right to the left along side $OA$, the tangent point on side $OA$ is marked as point $C$. $\odot P$ moves to intersect side $OB$ at points $E$ and $F$, given that $EF=4\sqrt{2}\text{cm}$, what is the length of $OC$ in $\text{cm}$?

Solution:
\textbf{Solution:} There are two cases to consider:

(1) As shown in Figure 1, when P is inside $\angle$AOB, draw PE and PC. Through point P, draw PM$\perp$EF at point M and extend CP to intersect OB at point N,

$\because$EF $=$ 4$\sqrt{2}$ cm,

$\therefore$EM $=$ 2$\sqrt{2}$ cm,

In the right triangle $\triangle$EPM, PM $=$ $\sqrt{3^{2}-(2\sqrt{2})^{2}}=1$ cm,

$\because$$\angle$AOB $=$ $60^\circ$,

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

$\therefore$PN $=$ 2PM $=$ 2cm,

$\therefore$NC $=$ PN + PC $=$ 5cm,

In the right triangle $\triangle$OCN, OC $=$ NC$\times\tan30^\circ$ $=$ 5$\times\frac{\sqrt{3}}{3}$ $=$ $\boxed{\frac{5\sqrt{3}}{3}}$ cm.

(2) As shown in Figure 2, when P is outside $\angle$AOB, draw PF and PC, where PC intersects EF at point N. Through point P, draw PM$\perp$EF at point M,

From (1) we know, PN $=$ 2cm,

$\therefore$NC $=$ PC - PN $=$ 1cm,

In the right triangle $\triangle$OCN, OC $=$ NC$\times\tan30^\circ$ $=$ 1$\times\frac{\sqrt{3}}{3}$ $=$ $\boxed{\frac{\sqrt{3}}{3}}$ cm.

In conclusion, the length of OC is either $\frac{\sqrt{3}}{3}$ cm or $\frac{5\sqrt{3}}{3}$ cm.


Question: As shown in the figure, $\triangle ABC$ is inscribed in $\odot O$, where $AB=AC$, and $AO$ and $CO$ are connected. If $BC=6$ and $\sin\angle BAC=\frac{3}{5}$, what is the radius of $\odot O$?

Solution:
\textbf{Solution:} As shown in the figure, draw and extend BO to intersect circle O at D, then BD is the diameter of circle O,\\
$\therefore$ $\angle BCD=90^\circ$, $\angle BAC=\angle BDC$,\\
$\therefore$ $\sin\angle BAC=\sin\angle BDC=\frac{BC}{BD}=\frac{3}{5}$,\\
$\therefore$ $BD=\frac{5}{3}BC=10$,\\
$\therefore$ the radius of circle O is $\boxed{5}$.


Question: As shown in the diagram, in $\triangle ABC$, $\angle ACB=90^\circ$, $AB=10$, $AC=8$, D is the midpoint of AB, M is a moving point on the segment AC, connect DM, construct a right triangle DEM with DM as the right angle side such that $\angle DEM=30^\circ$, the hypotenuse DE intersects ray MC at point F. If the area of $\triangle MDF$ is $\sqrt{3}$ times the area of $\triangle MEF$, what is the length of CM?

Solution:
\textbf{Solution:} As shown in the figure, draw DG $\perp$ AC at G, and draw EH $\perp$ AC at H, \\
then $\angle$DGM=$\angle$MHE=$90^\circ$, \\
in $\triangle ABC$, $BC=\sqrt{AB^{2}-AC^{2}}=\sqrt{10^{2}-8^{2}}=6$, \\
since D is the midpoint of AB, \\
thus AD=$\frac{1}{2}$AB=5, \\
since $\angle$AGD=$\angle$ACB=$90^\circ$, $\angle$DAG=$\angle$BAC, \\
thus $\triangle ADG\sim \triangle ABC$, \\
thus $\frac{DG}{BC}=\frac{AG}{AC}=\frac{AD}{AB}$, that is $\frac{DG}{6}=\frac{AG}{8}=\frac{5}{10}$, \\
thus DG=3, AG=4, \\
thus CG=AC−AG=8−4=4, \\
since the area of $\triangle MDF$ is $\sqrt{3}$ times the area of $\triangle MEF$, \\
thus $\frac{1}{2}$FM•DG=$\sqrt{3} \times \frac{1}{2}$FM•EH, \\
thus DG=$\sqrt{3}$EH, that is EH=$\frac{\sqrt{3}}{3}$DG=$\sqrt{3}$, \\
in $\triangle DEM$, $\angle$DME=$90^\circ$, $\angle$DEM=$30^\circ$, \\
thus $\frac{DM}{ME}=\tan\angle DEM=\tan30^\circ=\frac{\sqrt{3}}{3}$, \\
since $\angle$DMG+$\angle$MDG=$90^\circ$, $\angle$DMG+$\angle$EMH=$\angle$DME=$90^\circ$, \\
thus $\angle$MDG=$\angle$EMH, \\
thus $\triangle DMG\sim \triangle MEH$, \\
thus $\frac{MG}{EH}=\frac{DM}{ME}$, \\
thus $\frac{MG}{\sqrt{3}}=\frac{\sqrt{3}}{3}$, \\
thus MG=1, \\
thus CM=CG+MG=4+1=\boxed{5},


Question: As shown in the figure, within the square $ABCD$, where $AB=3$, point $M$ is on side $CD$, and $DM=1$. The triangle $AEM$ is symmetrical to triangle $ADM$ about the line on which $AM$ is located. By rotating triangle $ADM$ $90^\circ$ clockwise around point $A$, we obtain triangle $ABF$. Connecting $EF$, what is the value of $\cos\angle EFC$?

Solution:
\textbf{Solution:} As shown in the figure, draw HG $\perp$ AB at point H, intersecting CD at point G, then $\angle AHE = \angle EGM = 90^\circ$, and quadrilateral AHGD is a rectangle. \\
By symmetry, $\angle AEM = \angle D = 90^\circ$, AE = AD = 3, EM = DM = 1, \\
$\therefore \angle AEH + \angle MEG = 90^\circ$, \\
$\because \angle AEH + \angle HAE = 90^\circ$, \\
$\therefore \angle MEG = \angle HAE$, \\
$\therefore \triangle AHE \sim \triangle EGM$, \\
$\therefore \frac{AH}{EG} = \frac{HE}{GM} = \frac{AE}{EM}$, \\
Let MG = x, then AH = MG + MD = x + 1, \\
$\therefore \frac{x + 1}{EG} = \frac{HE}{x} = \frac{3}{1}$, \\
$\therefore HE = 3x$, EG = $\frac{x + 1}{3}$, \\
In $\triangle AHE$, $AH^2 + HE^2 = AE^2$, \\
$\therefore (x + 1)^2 + (3x)^2 = 3^2$, \\
Solving this gives: x = $\frac{4}{5}$ or x = -1 (discard), \\
$\therefore AH = \frac{9}{5}$, HE = $\frac{12}{5}$, EG = $\frac{3}{5}$, \\
Draw EN $\perp$ BC at point N through point E, then $\angle ENF = 90^\circ$, and quadrilateral ENBH is a rectangle, \\
$\therefore EN = BH = AB - AH = 3 - \frac{9}{5} = \frac{6}{5}$, BN = HE = $\frac{12}{5}$, \\
By rotation, BF = DM = 1, \\
$\therefore FN = FB + BN = 1 + \frac{12}{5} = \frac{17}{5}$, \\
$\therefore EF = \sqrt{EN^2 + FN^2} = \sqrt{\left(\frac{6}{5}\right)^2 + \left(\frac{17}{5}\right)^2} = \sqrt{13}$, \\
$\therefore \cos\angle EFN = \frac{FN}{EF} = \frac{\frac{17}{5}}{\sqrt{13}} = \boxed{\frac{17\sqrt{13}}{65}}$.


Question: As shown in the figure, in $\triangle ABC$, $BD$ is the median. If $AB=2\sqrt{10}$, $BC=10$, and $BD=3\sqrt{5}$, what is the value of $\tan\angle ABC$?

Solution:
\textbf{Solution:} Construct a line from point A such that $AE \perp BC$ at E, and $DF \perp BC$ at F,\\
therefore $AE \parallel DF$, $\angle AEB = \angle DFB = 90^\circ$\\
therefore $\triangle AEC \sim \triangle DFC$,\\
$AB^2 - BE^2 = AE^2$,\\
$BD^2 - BF^2 = DF^2$,\\
therefore $\frac{DF}{AE} = \frac{CF}{CE} = \frac{CD}{AC}$,\\
since BD is the median,\\
therefore $CD = AD = \frac{1}{2}AC$,\\
therefore $\frac{DF}{AE} = \frac{CF}{CE} = \frac{CD}{AC} = \frac{1}{2}$,\\
therefore $AE = 2DF$, $CE = 2CF$,\\
therefore $EF = CF$, let $EF = CF = x$, then $BE = BC - CE = 10 - 2x$, $BF = BC - CF = 10 - x$,\\
since $AB^2 - BE^2 = AE^2$, $BD^2 - BF^2 = DF^2$,\\
therefore $AE^2 = (2\sqrt{10})^2 - (10 - 2x)^2 = 40x - 60 - 4x^2$,\\
$DF^2 = (3\sqrt{5})^2 - (10 - x)^2 = 20x - 55 - x^2$,\\
since $AE = 2DF$,\\
therefore $AE^2 = 4DF^2$,\\
therefore $40x - 60 - 4x^2 = 80x - 220 - 4x^2$, solving for $x = 4$,\\
therefore $BE = 10 - 2x = 2$, $AE = \sqrt{AB^2 - BE^2} = 6$,\\
therefore $\tan \angle ABC = \frac{AE}{BE} = \boxed{3}$


Question: In rectangle $ABCD$, where $AB=5$ and $BC=5\sqrt{3}$, point $P$ is a moving point on $BC$. Connect $AP$ and construct an equilateral triangle $APE$ with $AP$ as one of its sides. Connect $DE$. What is the minimum value of segment $DE$ as point $P$ moves?

Solution:
\textbf{Solution:} Construct an equilateral triangle ABF with AB as its side, draw a line connecting FE, and draw line DH perpendicular to the extension of FE at point H, as shown in the figure,\\
since quadrilateral ABCD is a rectangle, it follows that $\angle ABP=\angle BAD=90^\circ$,\\
since $\triangle APE$ and $\triangle ABF$ are equilateral triangles, it follows that $\angle BAF=\angle PAE=60^\circ$, $BA=FA$, PA=AE, thus $\angle BAP=\angle FAE$,\\
therefore $\triangle BAP\cong \triangle FAE$, hence $\angle ABP=\angle AFE=90^\circ$,\\
since $\angle FAG=90^\circ-60^\circ=30^\circ$, it follows that $\angle AGE=90^\circ-30^\circ=60^\circ$,\\
since AB=AF=5, it follows that $AG=AF\div\cos30^\circ=\frac{10\sqrt{3}}{3}$,\\
therefore, point E moves on the ray FH, since AD=BC=$5\sqrt{3}$, it follows that DG=AD−AG=$\frac{5\sqrt{3}}{3}$,\\
since DH is perpendicular to FE, $\angle AGE=\angle HGD=60^\circ$, it follows that $DH=DG\times \sin60^\circ=\frac{5}{2}$,\\
also, since the perpendicular segment is the shortest, it follows that when point E coincides with point H, DE has its minimum value, and the minimum value is $\boxed{\frac{5}{2}}$.


Question: As shown in the figure, the graph of the linear function $y= -\frac{\sqrt{3}}{3}x+b$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. Taking the line segment $AB$ as a side, an equilateral triangle $ABC$ is constructed inside the first quadrant. The graph of the hyperbolic function $y= \frac{k}{x}$ (where $k\neq 0$) just passes through the midpoint $D$ of side $BC$ and intersects side $AC$ at point $E$. If the area of $\triangle ODE$ is $\sqrt{3}$, what is the value of $k$?

Solution:
\textbf{Solution:} Given the linear function $y= -\frac{\sqrt{3}}{3}x+b$ intersects with the x-axis and y-axis at points $A$ and $B$, respectively.\\
Therefore, when $y=0$, it holds that $-\frac{\sqrt{3}}{3}x+b=0$,\\
solving this gives $x= \sqrt{3}b$,\\
thus, the coordinate of point $A$ is ($\sqrt{3}b$, $0$), and $OA=\sqrt{3}b$,\\
in the linear function $y= -\frac{\sqrt{3}}{3}x+b$, when $x=0$, we have $y=b$,\\
thus, point $B$ is at $(0, b)$, and $OB=b$,\\
hence, $\tan\angle BAO= \frac{OB}{OA}=\frac{b}{\sqrt{3}b}=\frac{\sqrt{3}}{3}$, and $AB= \sqrt{(\sqrt{3}b)^{2}+b^{2}}=2b$,\\
therefore, $\angle BAO=30^\circ$,\\
given that $\triangle ABC$ is an equilateral triangle,\\
it follows that $\angle CAB=60^\circ$, and $AC=AB=2b$,\\
thus, $\angle CAB+\angle BAO=\angle CAO=90^\circ$, which implies $AC\perp$ the x-axis,\\
therefore, point $C$ is located at ($\sqrt{3}b$, $2b$), and point $E$ at ($\sqrt{3}b$, $\frac{\sqrt{3}k}{3b}$), with the area of $\triangle OAE$ being $\frac{1}{2}k$,\\
given that point $D$ is the midpoint of $BC$,\\
thus, the coordinates of point $D$ are ($\frac{\sqrt{3}}{2}b$, $\frac{3}{2}b$),\\
given that the inverse proportion function $y= \frac{k}{x}$ (with $k\neq 0$) exactly passes through the midpoint $D$ of $BC$,\\
it follows that $\frac{3}{2}b\times \frac{\sqrt{3}}{2}b=k$ (1),\\
draw $DF\perp$ x-axis at point $F$,\\
thus, $OF=\frac{\sqrt{3}}{2}b$ and $DF=\frac{3}{2}b$, with the area of $\triangle ODF$ being $\frac{1}{2}k$,\\
given that the area of $\triangle ODE$ equals the area of trapezoid $AEDF$ plus the area of $\triangle ODF$ minus the area of $\triangle AOE$, which is equal to the area of trapezoid $AEDF$,\\
thus, $\frac{1}{2}(DF+AE)\times AF=\sqrt{3}$, i.e., $\frac{1}{2}\left(\frac{3}{2}b+\frac{\sqrt{3}k}{3b}\right)\times \left(\sqrt{3}b-\frac{\sqrt{3}}{2}b\right)=\sqrt{3}$,\\
which implies $\frac{1}{2}\left(\frac{3}{2}b+\frac{\sqrt{3}k}{3b}\right)\times \frac{\sqrt{3}}{2}b=\sqrt{3}$,\\
thus, $\frac{1}{2}\times \frac{3}{2}b\times \frac{\sqrt{3}}{2}b+\frac{1}{2}\times \frac{\sqrt{3}k}{3b}\times \frac{\sqrt{3}}{2}b=\sqrt{3}$,\\
therefore, $\frac{1}{2}k+\frac{1}{4}k=\sqrt{3}$,\\
hence, $k=\boxed{\frac{4\sqrt{3}}{3}}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=\frac{3}{4}x+3$ intersects the x-axis at point A and the y-axis at point B. The parabola $y=-x^2+2x+1$ intersects the y-axis at point C. If point E moves along the axis of symmetry of the parabola, and point F moves along line AB, then what is the minimum value of $CE+EF$?

Solution:
\textbf{Solution:} As shown in the figure, construct the point $C_1$, which is the reflection of point C across the axis of symmetry. Draw $C_1F \perp AB$ at point F, intersecting the axis of symmetry of the parabola at point E. Extend line $C_1C$ to intersect AB at point G, and connect CE,\\
$\therefore CE = C_1E$,\\
$\therefore CE+EF = C_1E+EF = C_1F$ because the perpendicular is the shortest,\\
$\therefore$ The minimum value of $CE+EF$ is the length of $C_1F$;\\
$y = -(x-1)^2 + 2$ $\therefore$ the axis of symmetry of the parabola is the line $x=1$, when $y=0$, $0=\frac{3}{4}x+3$ solving it gives $x=-4$ $\therefore$ point $A(-4, 0)$ When $x=0$, $y=3$, $\therefore$ point $B(0, 3)$ $\therefore$ $AO=4$, $OB=3$ $\therefore$ $AB=\sqrt{3^2 + 4^2}=5$ When $x=0$, $y=-(0-1)^2 + 2=1$ $\therefore$ $OC=1$, $\therefore$ $BC=3-1=2$ because $GC_1 \parallel x$-axis, $\therefore$ $\frac{BC}{OC}=\frac{CG}{OA}$ that is $\frac{2}{3}=\frac{CG}{4}$ $\therefore$ $CG=\frac{8}{3}$, because the axis of symmetry of the parabola is the line $x=1$, points C and $C_1$ are symmetric about the line $x=1$, $\therefore CC_1=2$ $\therefore$ $C_1G=\frac{8}{3}+2=\frac{14}{3}$ $\therefore$ since $GC_1 \parallel x$-axis, $\therefore$ $\angle FGC_1 = \angle BAO$, $\therefore$ $\sin\angle FGC_1 = \sin\angle BAO = \frac{3}{5} = \frac{FC_1}{GC_1}$ $\therefore$ $FC_1=\frac{3}{5}\times \frac{14}{3}=\boxed{\frac{14}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, it is known that $AB=3\sqrt{2}$, $\angle B=45^\circ$, $\angle C=30^\circ$, then what is the length of $AC$?

Solution:
\textbf{Solution:} As shown in the figure, draw $AD \perp BC$ at point D.\\
$\therefore$ $\angle ADB = \angle ADC = 90^\circ$,\\
$\because$ $\angle B = 45^\circ$,\\
$\therefore$ $\angle BAD = \angle B = 45^\circ$,\\
$\therefore$ AD = BD,\\
$\because$ $AB^{2} = AD^{2} + BD^{2}$, AB = $3\sqrt{2}$,\\
$\therefore$ AD = BD = 3,\\
Also, $\because$ $\angle C = 30^\circ$,\\
$\therefore$ AC = 2AD = \boxed{6}


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=3$, and $AB=5$, what is the value of $\sin B$?

Solution:
\textbf{Solution:} In the right-angled triangle $ABC$, $\angle C=90^\circ$, $AC=3$, $AB=5$, \\
$\therefore$ $\sin B=\frac{AC}{AB}=\boxed{\frac{3}{5}}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, with points $C$ and $D$ located on opposite sides of diameter $AB$ on $\odot O$. Connecting $AC$, $AD$, $BD$, $CD$, if the radius of $\odot O$ is $5$ and $BD = 8$, then what is the value of $\sin\angle ACD$?

Solution:
\textbf{Solution:} Given that $AB$ is the diameter of circle $O$,\\
it follows that $\angle ADB=90^\circ$\\
therefore, $\angle ACD=\angle ABD$\\
Given that the radius of circle $O$ is $5$ and $BD=8$,\\
it follows that $\sin\angle ACD=\sin\angle ABD=\frac{6}{10}=\boxed{\frac{3}{5}}$


Question: As shown in the figure, PA and PB respectively tangent to the circle O at points A and B, point E is a point on the circle O, and $ \angle E=50^\circ$, then what is the degree measure of $ \angle P$?

Solution:
\textbf{Solution:} Connect AO, BO,\\
$ \because $ PA and PB are tangents to circle O at points A and B, respectively,\\
$ \therefore \angle PAO=\angle PBO=90^\circ$\\
$ \because \angle E=50^\circ$\\
$ \therefore \angle AOB=2\angle E=100^\circ$\\
$ \therefore \angle P=360^\circ−\angle PAO−\angle PBO−\angle AOB=360^\circ−90^\circ−90^\circ−100^\circ=\boxed{80^\circ}$


Question: As shown in the figure, points $A$ and $D$ lie on the circle $\odot O$, and $BC$ is a diameter. If $\angle ADC=30^\circ$, then what is the measure of $\angle OCA$ in degrees?

Solution:
\textbf{Solution:} Connect AB, as shown in the figure,\\
since BC is a diameter,\\
hence $\angle BAC=90^\circ$,\\
since $\angle B=\angle ADC=30^\circ$,\\
thus $\angle ACB=90^\circ-\angle B=90^\circ-30^\circ=60^\circ$,\\
that is, $\angle OCA=\boxed{60^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle B=36^\circ$. With $C$ as the center and $CA$ as the radius, the circle intersects $AB$ at point $D$ and $BC$ at point $E$. What is the degree measure of the central angle opposite to the arc $DE$?

Solution:
\textbf{Solution:} Connect CD, as shown in the diagram: \\
$\because$ $\angle ACB=90^\circ$, $\angle B=36^\circ$,\\
$\therefore$ $\angle A=90^\circ-\angle B=54^\circ$,\\
$\because$ CA=CD,\\
$\therefore$ $\angle CDA=\angle A=54^\circ$,\\
$\therefore$ $\angle ACD=180^\circ-54^\circ-54^\circ=72^\circ$,\\
$\therefore$ $\angle DCE=90^\circ-\angle ACD=18^\circ$,\\
Hence, the answer is: $\boxed{18^\circ}$.


Question: As shown in the figure, points $I$ and $O$ are the incenter and circumcenter of $\triangle ABC$, respectively. If $\angle AIB = 125^\circ$, then what is the degree measure of $\angle AOB$?

Solution:
\textbf{Solution:} Construct the circumscribed circle $\odot O$ of $\triangle ABC$, and the inscribed circle $\odot I$ of $\triangle ABC$,\\
Since point I is the incenter of $\triangle ABC$,\\
Therefore, AI bisects $\angle CAB$, BI bisects $\angle ABC$,\\
Therefore, $\angle IAB=\frac{1}{2}\angle CAB$, $\angle IBA=\frac{1}{2}\angle CBA$,\\
Since $\angle AIB=125^\circ$,\\
Therefore, $\angle IAB+\angle IBA=180^\circ-\angle AIB=55^\circ$,\\
Therefore, $\angle CAB+\angle CBA=2(\angle IAB+\angle IBA)=110^\circ$,\\
Therefore, $\angle ACB=180^\circ-(\angle CAB+\angle CBA)=70^\circ$,\\
Since point O is the circumcenter of $\triangle ACB$,\\
Therefore, $\angle AOB=2\angle ACB=140^\circ$,\\
Hence, the answer is: $\boxed{140^\circ}$.


Question: As shown in the figure, in rhombus $ABCD$, $\angle A=60^\circ$, points $E$ and $F$ are respectively on sides $BC$ and $CD$, and $CE=DF$, $BF$ intersects $DE$ at point $G$. If $BG=3$ and $DG=5$, what is the area of quadrilateral $ABGD$?

Solution:
Solution:Since quadrilateral $ABCD$ is a rhombus,\\
$\therefore AB=AD$,\\
Since $\angle BAD=60^\circ$,\\
$\therefore \triangle ABD$ is an equilateral triangle,\\
$\therefore \angle BAD=\angle ABD=60^\circ$,\\
$\therefore \angle DBC=\angle BDF=\angle C=60^\circ$,\\
In $\triangle CDE$ and $\triangle DBF$,\\
$\left\{\begin{array}{l}CD=DB\\ \angle C=\angle BDF\\ CE=DF\end{array}\right.$,\\
$\therefore \triangle CDE \cong \triangle DBF\left(SAS\right)$,\\
$\therefore \angle CDE=\angle DBF$,\\
$\therefore \angle GBE=\angle BDE$,\\
$\therefore \angle DBF+\angle GBE=\angle DBF+\angle BDE=\angle BGE=\angle DGF=60^\circ=\angle BAD$,\\
$\therefore$ Quadrilateral $ABGD$ is a cyclic quadrilateral,\\
$\therefore \angle BGD=120^\circ$,\\
As shown in the figure, draw $AM\perp DE$ and $AN\perp BF$ from point $A$, with the feet $M$ and $N$ respectively,\\
Since $AG$ bisects $\angle BGD$,\\
$\therefore AN=AM$,\\
In $Rt\triangle ABN$ and $Rt\triangle ADM$,\\
$\left\{\begin{array}{l}AB=AD\\ AN=AM\end{array}\right.$,\\
$\therefore Rt\triangle ABN \cong Rt\triangle ADM\left(HL\right)$,\\
$\therefore BN=DM$,\\
$\therefore GN+GM=BG+DG=3+5=8$,\\
Connect $AG$,\\
In $Rt\triangle AGN$ and $Rt\triangle AGM$,\\
$\left\{\begin{array}{l}AG=AG\\ AN=AM\end{array}\right.$,\\
$\therefore Rt\triangle AGN \cong Rt\triangle AGM\left(HL\right)$,\\
$\therefore NG=MG=\frac{1}{2}\left(BG+DG\right)=4$, $\angle AGN=\frac{1}{2}\angle BGD=60^\circ$,\\
$\therefore AN=NG\cdot \tan\angle AGN=4\sqrt{3}$,\\
$\therefore S_{\text{quadrilateral }ABGD}=S_{\text{quadrilateral }ANGM}$,\\
$S_{\text{quadrilateral }ABGD}=2S_{\triangle AGN}=2\times \frac{1}{2}\times NG\times AN=4\times 4\sqrt{3}=\boxed{16\sqrt{3}}$.


Question: As shown in the figure, in the circle $\odot O$ with radius $r=10$, and a chord $AB=16$, $P$ is a moving point on chord $AB$. What is the minimum length of segment $OP$?

Solution:
\textbf{Solution:} Draw $OH \perp AB$ at point H, and connect OB.\\
Therefore, $AH=BH=\frac{1}{2}AB=\frac{1}{2}\times 16=8$,\\
In $\triangle BOH$, $OH=\sqrt{OB^{2}-BH^{2}}=\sqrt{10^{2}-8^{2}}=\boxed{6}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle BAC=30^\circ$, and $BC=2$. The line segment $BC$ rotates around point $B$ to $BD$. Connect $AD$, where $E$ is the midpoint of $AD$. Connect $CE$. What is the maximum value of $CE$?

Solution:
\textbf{Solution:} Construct the midpoint O of AB and connect OE as shown in the diagram: \\
According to the problem, $BD=BC=2$, \\
Since point E is the midpoint of AD, and point O is the midpoint of AB, \\
it follows that $OE=\frac{1}{2}BD=1$, \\
thus, the trajectory of point E is a circle centered at O with a radius of 1, \\
hence, when point E is on the extension line of CO, CE is maximum, \\
since $\angle ACB=90^\circ$, $\angle BAC=30^\circ$, $BC=2$ \\
it follows that $AB=4$, \\
since point O is the midpoint of AB, \\
it follows that $OC=\frac{1}{2}AB=2$, \\
thus, the maximum value of $CE$ is $OC+OE=2+1=\boxed{3}$.


Question: As shown in the figure, $PA$, $PB$ are tangent to the circle $\odot O$ at points $A$, $B$ respectively, and point $C$ is on the circle $\odot O$. If $\angle C = 70^\circ$, what is the degree measure of $\angle P$?

Solution:
\textbf{Solution:} Connect OA and OB, \\
$\because PA$ and $PB$ are tangents to $\odot O$, \\
$\therefore PA\perp OA$, $PB\perp OB$, \\
$\therefore \angle PAO=\angle PBO=90^\circ$, \\
$\because \angle P+\angle PAO+\angle AOB+\angle PBO=360^\circ$, \\
$\therefore \angle P=180^\circ-\angle AOB$, \\
$\because \angle C=70^\circ$, \\
$\therefore \angle AOB=2\angle C=140^\circ$, \\
$\therefore \angle P=180^\circ-140^\circ=\boxed{40^\circ}$.


Question: As shown in the figure, in a grid diagram where each small square has a side length of 1, a circular arc passes through lattice points $A$, $B$, $C$. The line connecting lattice points $C$ and $D$ intersects $\arc{BC}$ at point $E$. What is the length of $\arc{EC}$?

Solution:
\textbf{Solution:} As shown in the figure, connect AE, AC, and AD, \\
$\because \angle ABC=90^\circ$, \\
$\therefore AC$ is the diameter, \\
$\therefore \angle ABC=\angle AEC=90^\circ$, \\
According to the grid figure, we know: $AC=AD=\sqrt{13}$, $CD=\sqrt{26}$, \\
$\therefore AC^{2}+AD^{2}=CD^{2}=26$, \\
$\therefore \triangle ACD$ is an isosceles right-angled triangle, \\
$\therefore \angle CAD=90^\circ$, $\angle ACE=45^\circ$, \\
$\therefore \angle EAC=45^\circ$, \\
$\therefore$ the central angle corresponding to $\overset{\frown}{EC}$ is $90^\circ$, \\
$\therefore$ the length of $\overset{\frown}{EC}$ is $\frac{1}{4}$ of the circumference of the circle with diameter AC, \\
That is, the length of $\overset{\frown}{EC}$ is: $\frac{1}{4}\times \pi \times \sqrt{13}=\boxed{\frac{\sqrt{13}\pi }{4}}$.


Question: As shown in the figure, ABCD is a quadrilateral inscribed in circle O, and $AC \perp BD$. If $AB=10$ and $CD=8$, what is the area of circle O?

Solution:
\textbf{Solution:} As shown in the diagram, connect $AO$ and extend it to meet circle $O$ at point $E$, and connect $EB$, $EC$.\\
Then $AB\perp BE$, $AC\perp CE$.\\
Since $AC\perp BD$,\\
Therefore $BD$ $\parallel$ $EC$,\\
Therefore $\overset{\frown}{CD}=\overset{\frown}{BE}$\\
Therefore $BE=CD$,\\
Since $CD=8$\\
Therefore $EB=CD=8$.\\
In $\triangle ABE$, $AB=10$, $EB=8$\\
So, by the Pythagorean theorem, $AE=\sqrt{AB^{2}+BE^{2}}=\sqrt{10^{2}+8^{2}}=2\sqrt{41}$\\
Therefore $OA=\frac{1}{2}AE=\sqrt{41}$.\\
Thus, the area of circle $O$ is $ \pi \times OA^{2}=\boxed{41\pi}$.


Question: As shown in the figure, $\triangle ABC$ is an inscribed triangle of circle $O$ with $OB$ and $OC$ connected. If the length of chord $BC$ is $2\sqrt{3}$, what is the measure of $\angle BAC$ in degrees?

Solution:
\textbf{Solution:} Construct OE $\perp$ BC at E. \\
Since OE $\perp$ BC, \\
then BE=EC= $\sqrt{3}$, and $\angle$BOE=$\angle$COE, \\
thus OE= $\sqrt{OB^{2}-BE^{2}}=\sqrt{2^{2}-(\sqrt{3})^{2}}=1$, \\
therefore OB=2OE, \\
hence $\angle$OBE=30$^\circ$, \\
thus $\angle$BOE=$\angle$COE=60$^\circ$, \\
therefore $\angle$BOC=120$^\circ$, \\
which implies $\angle$BAC=\boxed{60^\circ}.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, with chord $CD\perp AB$ at $H$, $\angle A=30^\circ$, $OH=1$, then what is the radius of $\odot O$?

Solution:
\textbf{Solution:} Draw $OC$, \\
$\because$ $\angle A=30^\circ$, \\
$\therefore$ $\angle COH=2\angle A=60^\circ$, \\
$\because$ chord $CD \perp AB$ at $H$, \\
$\therefore$ $\angle OHC=90^\circ$, \\
$\therefore$ $\angle OCH=30^\circ$, \\
$\because$ $OH=1$, \\
$\therefore$ $OC=2OH=\boxed{2}$.


Question: As shown in the figure, $CD$ is the diameter of the circle $\odot O$, and $AB$ is a chord. $CD\perp AB$ at point $E$. If $OA=5$ and $AB=8$, what is the length of $AD$?

Solution:
\textbf{Solution:} Given that CD is the diameter of circle O and AB is a chord with CD $\perp$ AB, and AB = 8, \\
$\therefore$ AE=BE=$\frac{1}{2}AB=\frac{1}{2}\times 8=4$, \\
In $\triangle OEB$, according to the Pythagorean theorem, OE=$\sqrt{OA^{2}-EA^{2}}=\sqrt{5^{2}-4^{2}}=3$, \\
$\therefore$ CD=OD+OE=5+3=8, \\
In $\triangle AED$, AD=$\sqrt{AE^{2}+ED^{2}}=\sqrt{4^{2}+8^{2}}=\boxed{4\sqrt{5}}$.


Question: As shown in the figure, the diameter $AB=2$ of the circle $\odot O$, and $C$ is a moving point on $\odot O$. Connect $CB$ and rotate $CB$ $90^\circ$ counterclockwise around point $C$ to get $CD$. Connect $OD$. What is the maximum value of $OD$?

Solution:
\textbf{Solution:} Construct line EO$\perp$AB intersecting circle $\odot$O at point E, with O being the perpendicular foot. Connect BE, ED, OC, and BD. \\
In circle $\odot$O, OB=OE, $\angle BOE=90^\circ$, \\
$\therefore$ $BE^2=OB^2+OE^2=2OB^2$, \\
$\therefore$ $BE=\sqrt{2}OB$, \\
In $\triangle BCD$, BC=CD, $\angle BCD=90^\circ$, \\
$\therefore$ $BD=\sqrt{2}BC$, \\
$\therefore$ $\frac{BE}{OB}=\frac{BD}{BC}=\sqrt{2}$, \\
$\because$ $\angle OBE= \angle CBD= 45^\circ$, \\
$\therefore$ $\angle OBE+\angle EBC = \angle EBC +\angle CBD$, that is, $\angle OBC= \angle EBD$, \\
$\therefore$ $\triangle OBC\sim \triangle EBD$, \\
$\therefore$ $\frac{DE}{OC}=\frac{BE}{OB}=\sqrt{2}$, \\
$\because$ $OC=\frac{1}{2}AB=1$, \\
$\therefore$ $DE=\sqrt{2}$, \\
Hence, point D traces a circle centered at E with radius $\sqrt{2}$. The maximum length of OD, when points O, E, and D are collinear, is $OD=DE+OE=\sqrt{2}+1$. \\
Therefore, the answer is: $\boxed{\sqrt{2}+1}$.


Question: As shown in the figure, point C is on the circle $\odot O$. If $\angle ACB=24^\circ$, what is the degree measure of $\angle AOB$?

Solution:
\textbf{Solution:} From the Inscribed Angle Theorem, it can be derived that $ \angle AOB=2\angle ACB=48^\circ$\\
Therefore, the answer is: $ \boxed{48^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and point $C$ is a point on $\odot O$. Given $OA=3$, and $\angle OCA=40^\circ$, what is the area of the shaded region?

Solution:
Solution: Since $OA=OC$,\\
it follows that $\angle OAC=\angle OCA=40^\circ$,\\
therefore, $\angle COB=80^\circ$,\\
thus, the area of the shaded region is: $ \frac{\pi \times {3}^{2}\times 80^\circ}{360^\circ}=\boxed{2\pi}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$. Point $D$ is on $\odot O$, and $\angle DAB=45^\circ$, $BC\parallel AD$, and $CD\parallel AB$. If the radius of $\odot O$ is 1, what is the area of the shaded region? (Keep $\pi$ in the result).

Solution:
\textbf{Solution:} Connect $DO$,\\
$\because \angle DAB=45^\circ$, $\overset{\frown}{DB}=\overset{\frown}{DB}$\\
$\therefore \angle DOB=90^\circ$ meaning $DO\perp AB$\\
$\because$ the radius of $\odot O$ is 1\\
$\therefore AB=2$\\
$\therefore {S}_{\square ABCD}=AB\cdot OD=2\times 1=2$\\
${S}_{\triangle ADO}=\frac{1}{2}AO\cdot OD=\frac{1}{2}\times 1\times 1=\frac{1}{2}$\\
$S_{\text{sector}}{OBD}=\frac{90^\circ}{360^\circ}\pi \cdot OA^{2}=\frac{\pi }{4}$\\
$\therefore S_{\text{shaded area}}={S}_{\square ABCD}-{S}_{\triangle ADO}-S_{\text{sector}}{OBD}$\\
$=2-\frac{1}{2}-\frac{\pi }{4}$\\
$=\boxed{\frac{3}{2}-\frac{\pi }{4}}$


Question: As shown in the figure, point A is inside circle $O$ with a radius of 5, and $OA = \sqrt{3}$. Let P be a moving point on circle $O$. When $\angle OPA$ reaches its maximum value, what is the length of PA?

Solution:
\textbf{Solution:} As shown in the figure:\\
$\because$ OA and OP are constants,\\
$\therefore$ when PA $\perp$ OA, $\angle$OPA is the maximum,\\
In the right-angled triangle OPA, OA $= \sqrt{3}$, OP $= 5$,\\
$\therefore$ PA $= \sqrt{OP^2 - OA^2}$ $= \boxed{\sqrt{22}}$.


Question: As shown in the figure, it is known that point $E$ is a point inside rectangle $ABCD$, satisfying $\angle AEB=90^\circ$. Extend $DE$ to intersect the semicircle with diameter $CD$ at point $F$. When $AE=20$, $BE=15$, and $DF=24$, what is the length of the side $AD$ of the rectangle?

Solution:
\textbf{Solution:} As shown in the diagram, extend a line from point E to draw EH $\perp$ CD at H, intersecting AB at point T, and connect CF.\\
Since quadrilateral ABCD is a rectangle,\\
therefore AB=CD, AB$\parallel$CD,\\
since $\angle$AEB=90$^\circ$, AE=20, BE=15,\\
therefore AB=$\sqrt{AE^2+BE^2}=\sqrt{20^2+15^2}=25$,\\
since ET$\perp$AB,\\
therefore $\frac{1}{2}AE \cdot EB= \frac{1}{2}AB \cdot ET$,\\
therefore ET=$\frac{20\times 15}{25}=12$,\\
therefore AT=DH=$\sqrt{AE^2-ET^2}=\sqrt{20^2-12^2}=16$,\\
since CD is the diameter,\\
therefore $\angle$CFD=90$^\circ$,\\
therefore CF=$\sqrt{CD^2-DF^2}=\sqrt{25^2-24^2}=7$,\\
since $\angle$EDH=$\angle$CDF, $\angle$EHD=$\angle$CFD=90$^\circ$,\\
therefore $\triangle$DHE$\sim$$\triangle$DFC,\\
therefore $\frac{DH}{DF}=\frac{EH}{CF}$,\\
therefore $\frac{16}{24}=\frac{EH}{7}$,\\
therefore EH=$\frac{14}{3}$,\\
therefore AD=BC=HT=ET+EH=12+$\frac{14}{3}=\boxed{\frac{50}{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, point $D$ is on line $AC$, connecting $BD$. If $\angle DBC=45^\circ$, what is the maximum value of $\frac{AD}{CD}$?

Solution:
\textbf{Solution:} As shown in Figure 1, construct the circumcircle of $\triangle BCD$, denoted as $\odot O$. Construct $OE \perp AC$ at point E, and $OF \perp AB$ at point F, and connect OB,\\
$\because \angle AEO = \angle AFO = \angle BAC = 90^\circ$,\\
$\therefore$ Quadrilateral OFAE is a rectangle,\\
$\therefore$ AE=OF,\\
$\because DE=CE= \frac{1}{2}CD$,\\
$\therefore$ CD=2CE,\\
The lengths of CD, CE remain constant. When OF is at its maximum, then AE is at its maximum, and thus AD is at its maximum,\\
$\therefore$ the value of $\frac{AD}{CD}$ is also at its maximum,\\
$\because OF \leq OB$,\\
$\therefore$ when OF=OB, OF is at its maximum,\\
At this time, point B coincides with point F,\\
As shown in Figure 2, when point B coincides with point F, connect OD and OC,\\
Let DE=CE=m, then CD=2m,\\
$\because \angle DBC=45^\circ$,\\
$\therefore \angle DOC=2\angle DBC=90^\circ$,\\
$\because OD=OC$,\\
$\therefore OE= \frac{1}{2}CD=DE=m$,\\
$\therefore AE=OB=OD= \sqrt{DE^2+OE^2}=\sqrt{m^2+m^2}=\sqrt{2}m$,\\
$\therefore AD=AE−DE= \sqrt{2}m−m$,\\
$\therefore \frac{AD}{CD}=\frac{\sqrt{2}m−m}{2m}=\boxed{\frac{\sqrt{2}−1}{2}}$.


Question: As shown in the figure, both $ \triangle ABC$ and $ \triangle CDE$ are equilateral triangles, with $ AB>CD$, $AB=6$. Fixing $ \triangle ABC$ and rotating $ \triangle CDE$ around point C by any angle, we connect AD and BE. Let the lines containing AD and BE intersect at point O. During the rotation process, it is always true that $AD=BE$, and the magnitude of $\angle AOB$ remains constant. What is the maximum distance from point O to line AB in this scenario?

Solution:
\textbf{Solution:}

Given that both $\triangle ABC$ and $\triangle CDE$ are equilateral triangles,

thus, $AC = BC$, $CD = CE$, $\angle BAC = \angle ABC = \angle ACB = \angle DCE=60^\circ$,

therefore, $\angle ACE + \angle DCE = \angle ACE + \angle ACB$,

which means $\angle ACD = \angle BCE$,

thus, $\triangle ACD \cong \triangle BCE$ (SAS),

hence, $\angle CAD = \angle CBE$,

so, $\angle AOB = \angle ACB=60^\circ$,

Construct the circumcircle of $\triangle ABC$, denoted as $\odot M$, as shown in the diagram:

Point O is on $\odot M$,

Construct $OF \perp AB$ at point F,

thus, when point O coincides with point C, the distance from point O to line AB is maximized, and the maximum distance is the length of segment CF,

In $\triangle ACF$, $\angle ACF=30^\circ$

$AF = BF = \frac{1}{2}AB=3$,

$CF = \sqrt{3}AF = 3\sqrt{3}$,

hence, the maximum distance from point O to line AB is $\boxed{3\sqrt{3}}$.


Question: As shown in the figure, in circle $O$, if $\angle BAC=24^\circ$ and $\angle ACB=42^\circ$, then what is the degree measure of $\angle ACO$?

Solution:
\textbf{Solution:} Given that $\angle BAC = 24^\circ$ and $\angle ACB = 42^\circ$,\\
it follows that $\angle BOC = 2\angle BAC = 48^\circ$ and $\angle AOB = 2\angle ACB = 84^\circ$,\\
hence $\angle AOC = \angle BOC + \angle AOB = 132^\circ$.\\
Since OA = OC,\\
it implies that $\angle ACO = \angle OAC = \frac{1}{2}(180^\circ - 132^\circ) = \boxed{24^\circ}$.


Question: As shown in the figure, there is a flask with a spherical bottom, where the radius of the sphere is $5\,cm$. Originally, the maximum depth of the liquid inside the flask, denoted as $CD$, is $4\,cm$. After some of the liquid evaporated, the maximum depth of the liquid inside the flask decreased to $2\,cm$. How much does the length of the chord $AB$ in the cross-sectional circle decrease by (keep the result in square root form)?

Solution:
\textbf{Solution:} Let $A^{\prime}B^{\prime}$ intersect $OD$ at $E$, \\
According to the problem: $OA=O{A}^{\prime}=OD=5\text{cm}$, $OD\perp AB$, $OD\perp A^{\prime}B^{\prime}$, \\
$\therefore AC=BC$, $A^{\prime}E=B^{\prime}E$, \\
$\because CD=4\text{cm}$, \\
$\therefore OC=OD-CD=1\text{cm}$, \\
$\therefore AC=\sqrt{OA^{2}-OC^{2}}=\sqrt{5^{2}-1^{2}}=2\sqrt{6}\text{cm}$, \\
$\therefore AB=2AC=4\sqrt{6}\text{cm}$, \\
$\because DE=2\text{cm}$, \\
$\therefore OE=OD-DE=3\text{cm}$, \\
$\therefore A^{\prime}D=\sqrt{O{A^{\prime}}^{2}-OE^{2}}=\sqrt{5^{2}-3^{2}}=4\text{cm}$, \\
$\therefore A^{\prime}B^{\prime}=2A^{\prime}D=8\text{cm}$, \\
$\therefore AB-A^{\prime}B^{\prime}=(4\sqrt{6}-8)\text{cm}$, \\
Thus, the length of the chord $AB$ in the circular section decreased by $\boxed{4\sqrt{6}-8}\text{cm}$.