Question: As shown in the figure, parallelogram $OABC$ has its side $OA$ on the $x$-axis, and vertex $C$ is on the graph of the inverse proportion function $y=\frac{k}{x}$. $BC$ intersects the $y$-axis at point $D$, and $D$ is the midpoint of $BC$. If the area of the parallelogram $OABC$ is $8$, what is the value of $k$?

Solution:
\textbf{Solution:} Since $D$ is the midpoint of $BC$, and the area of parallelogram $OABC$ is 8,\\
it follows that $S_{\triangle OCD}=\frac{1}{4}\times 8=2$,\\
which means $\frac{1}{2}\left|k\right|=2$,\\
thus $k=\pm 4$,\\
since the graph of the inverse proportion function is in the second quadrant,\\
it follows that $k<0$,\\
hence, $k=\boxed{-4}$.


Question: As shown in the figure, the line $y=kx+b$ intersects the parabola $y=-x^2+2x+3$ at points $A$ and $B$, where point $A$ is at $(0, 3)$ and point $B$ is at $(3, 0)$. The parabola intersects the x-axis at another point $C(-1, 0)$. What is the solution set of the inequality $-x^2+2x+3>kx+b$?

Solution:
\textbf{Solution:} From the graph and the problem statement, it is given that:\\
The solution set for the inequality $-x^{2}+2x+3>kx+b$ is $0<x<3$;\\
Therefore, the answer is $\boxed{0<x<3}$.


Question: As shown in the figure, in $\triangle AOB$, $\angle AOB=90^\circ$, with vertex $A$ on the graph of the inverse proportion function $y=\frac{1}{x}$ $(x>0)$ and vertex $B$ on the graph of $y=\frac{-3}{x}$ $(x<0)$, then what is the degree measure of $\angle BAO$?

Solution:
\textbf{Solution:} Draw $AC \perp x$-axis at $C$, and draw $BD \perp x$-axis at $D$,\\
then $\angle BDO = \angle ACO = 90^\circ$,\\
since the vertex $A$ is on the graph of the inverse proportion function $y=\frac{1}{x}\left(x>0\right)$ and vertex $B$ is on the graph of $y=\frac{-3}{x}\left(x<0\right)$,\\
thus, the area ${S}_{\triangle AOC} = \frac{1}{2} \cdot {x}_{A} \cdot {y}_{A} = \frac{1}{2} \times 1 = \frac{1}{2}$, and the area ${S}_{\triangle BDO} = \frac{1}{2} \cdot |\,{x}_{B} \cdot {y}_{B}| = \frac{1}{2} \times |-3| = \frac{3}{2}$,\\
since $\angle AOB=90^\circ$,\\
thus $\angle BOD + \angle DBO = \angle BOD + \angle AOC = 90^\circ$,\\
therefore, $\angle DBO = \angle AOC$,\\
therefore, $\triangle BDO \sim \triangle OCA$,\\
therefore, $\frac{{S}_{\triangle BOD}}{{S}_{\triangle OAC}} = \left(\frac{OB}{OA}\right)^2 = \frac{\frac{3}{2}}{\frac{1}{2}} = 3$,\\
therefore, $\frac{OB}{OA} = \sqrt{3}$,\\
therefore, $\tan\angle BAO = \frac{OB}{OA} = \sqrt{3}$,\\
therefore, $\angle BAO = \boxed{60^\circ}$.


Question: As shown in the figure, given the quadratic function $y=-x^2+2x+m$ with its partial graph illustrated, what is the solution set of the quadratic inequality $-x^2+2x+m>0$ with respect to $x$?

Solution:
\textbf{Solution:} From the graph, it is known that the axis of symmetry is the line $x=1$, \\
hence, the coordinate of the other intersection point of the quadratic function graph with the x-axis is $(-1, 0)$, \\
According to the graph: The range of $x$ for which the function value is greater than 0 is: $-1<x<3$, \\
therefore, the solution set for $-x^2+2x+m>0$ is $\boxed{-1<x<3}$.


Question: As shown in the figure, it is known that the parabola \(y_1=ax^2+bx+c\ (a\neq 0)\) and the line \(y_2=mx+n\ (m\neq 0)\) intersect at points \(A\) and \(B\), with the \(x\)-coordinates of points \(A\) and \(B\) being \(-2\) and \(\frac{5}{4}\), respectively. What is the solution to the inequality \(ax^2+bx+c<mx+n\)?

Solution:
Solution: From the graph, it is known that the solution to $ax^2+bx+c<mx+n$ occurs when the line is above the parabola, \\
$\therefore$ $-2<x<\frac{5}{4}$, \\
Therefore, the answer is $\boxed{-2<x<\frac{5}{4}}$.


Question: As shown in the figure, it is known that the parabola $C_1$ is centrally symmetric about the origin with the parabola $C_2$. If the equation of the parabola $C_1$ is $y=\frac{3}{4}(x+2)^2-1$, what are the coordinates of the vertex of the parabola $C_2$, and what is its equation?

Solution:
\textbf{Solution:} Given that the equation of parabola $C_1$ is $y=\frac{3}{4}(x+2)^{2}-1$,\\
$\therefore$ the parabola $C_1$ opens upwards, and its vertex is at: $\left(-2,-1\right)$ \\
$\because$ Parabola $C_1$ and parabola $C_2$ are centrally symmetric about the origin.\\
$\therefore$ Parabola $C_2$ opens downwards, and its vertex is at: $\boxed{\left(2,1\right)}$ \\
$\therefore$ The equation of parabola $C_2$ is $\boxed{y=-\frac{3}{4}(x-2)^{2}+1}$


Question: As shown in the figure, the graph of the direct proportion function $y = x$ intersects with the graph of the inverse proportion function $y = \frac{1}{x}$ at points A and C. $AB \perp x$-axis at B, and $CD \perp x$-axis at D. What is the area of the quadrilateral $ABCD$?

Solution:
\textbf{Solution:} Solve the system of equations:
\[\begin{cases}
y=x\\
y=\frac{1}{x}
\end{cases}\]

We obtain points A(1,1) and C(-1,-1).

Therefore, the area of $\triangle ADB = \frac{1}{2} \times (1+|-1|) \times 1 = 1$, the area of $\triangle BCD = \frac{1}{2} \times (1+|-1|) \times |-1| = 1$,

Thus, the area of the quadrilateral $ABCD = S_{\triangle ADB} + S_{\triangle BCD} = 1 + 1 = \boxed{2}$


Question: As shown in the figure, the parabola $y_1=ax^2\ (a\neq 0)$ intersects with the line $y_2=bx+c\ (b\neq 0)$ at two points, whose coordinates are $A(-2, 4)$ and $B(1, 1)$, respectively. What is the range of $x$ when $y_1<y_2$?

Solution:
\textbf{Solution:} From the graph, it is known that when the line $y_2=bx+c\ (b\ne 0)$ is above the parabola $y_1=ax^2\ (a\ne 0)$, the range of $x$ is $-2<x<1$,\\
$\therefore$ when $y_1<y_2$, the range of $x$ is $-2<x<1$.\\
Therefore, the answer is: $x \in \boxed{(-2, 1)}$


Question: As shown in the figure, $O$ is the origin of coordinates, point $A$ lies on the graph of the function $y=\frac{k}{x}$ ($x<0$), line segment $AB$ is perpendicular to the $x$-axis at point $B$, and the area of $\triangle AOB$ is 4. What is the value of $k$?

Solution:
\textbf{Solution:} Since $AB \perp x$-axis, \\
therefore $S_{\triangle ABO} = \frac{1}{2}|k|$, which means $\frac{1}{2}|k| = 4$, \\
Since $k < 0$, \\
therefore $k = \boxed{-8}$.


Question: As shown in the figure, it is known that the graph of the linear function $y=2x+4$ intersects with the graph of the inverse proportion function $y=\frac{k}{x}$ at points $A$ and $B$. The x-coordinate of point $B$ is $1$. A line is drawn from point $A$ perpendicular to the y-axis at point $C$, and line segment $BC$ is connected. What is the area of triangle $ABC$?

Solution:
\textbf{Solution:} Since the graph of the linear function $y=2x+4$ intersects with the graph of the inverse proportion function $y=\frac{6}{x}$ at points A and B, and the x-coordinate of point B is 1,\\
we substitute $x=1$ into $y=2x+4$ to get $y=6$,\\
thus, B$(1,6)$,\\
and therefore $k=1\times6=6$,\\
hence the equation of the inverse proportion function is $y=\frac{6}{x}$,\\
solving $\left\{\begin{array}{l}y=\frac{6}{x}\\ y=2x+4\end{array}\right.$ gives $\left\{\begin{array}{l}x=1\\ y=6\end{array}\right.$ or $\left\{\begin{array}{l}x=-3\\ y=-2\end{array}\right.$,\\
therefore, A$(-3,-2)$,\\
since AC is perpendicular to the y-axis at point C,\\
therefore, AC=3,\\
hence, the area of $\triangle ABC=\frac{1}{2}\times3\times(6+2)=\boxed{12}$.


Question: As shown in the figure, the graph represents the inverse proportion function $y=\frac{k}{x}$ in the second quadrant. If the area of the rectangle $OABC$ in the graph is $4$, what is the value of $k$?

Solution:
\textbf{Solution:} Given the inverse proportion function $y=\frac{k}{x}$, and the area of rectangle OABC is 4,\\
$\therefore |k|=4$, which means $k=\pm 4$,\\
Since the graph of the inverse proportion function $y=\frac{k}{x}$ is in the second quadrant, $k<0$,\\
$\therefore k=\boxed{-4}$.


Question: As shown in the figure, the hyperbola $y=\frac{k}{x}\ (x>0)$ intersects the side BC of the square ABCD at point E, and intersects the side CD at point F. Given $BE=3CE$ and $A(4, 0)$, $B(8, 0)$, what is the length of $CF$?

Solution:
\textbf{Solution:} Given $\because BE=3CE$,

$\therefore CE=a$, $BE=3a$,

$\therefore E(8,3a)$, $C(8,4a)$,

$\because$ point E lies on $y=\frac{k}{x}$,

$\therefore k=24a$,

$\therefore$ the equation of the hyperbola is: $y=\frac{24a}{x}$,

From the coordinates of point C, we find that the y-coordinate of point F is $4a$,

$\because$ point F also lies on $y=\frac{k}{x}$, by substituting the y-coordinate, we find,

$\therefore x=6$,

$\therefore CF=8-6=\boxed{2}$.


Question: As shown in the figure, there is a crescent-shaped pond in a park, with five decorative lamps labeled A, B, C, D, and E along the edge of the pond. To estimate the size of the pond, observers at points A and D noticed that points A, E, C and D, E, B are on the same straight line. Walking in the direction of AD to point F, it was found that $\angle AFC=90^\circ$. It was measured that $AD=9.6$ meters, $AE=DE=8$ meters, $DF=2.4$ meters. What is the radius of the circle containing $\overset{\frown}{AED}$? What is the radius of the circle containing $\overset{\frown}{ACD}$?

Solution: \textbf{Solution:} Draw line EG $\perp$ AD at point G through point E. Thus, the center of the circle passing through points A, E, D must lie on EG. As shown in the figure, take the center as O, and connect AO. Since $AE=DE=8$ meters, AD=9.6 meters, therefore $AG=$ \(\frac{1}{2}AD=4.8\) meters, $EG=$ $\sqrt{AE^{2}-AG^{2}}=\sqrt{8^{2}-4.8^{2}}=6.4$ meters. Let the radius of the circle containing $\overset{\frown}{AED}$ be r meters, then $AO=EO=r$ meters, $OG=6.4−r$ (meters). In $\triangle AOG$, $AG^{2}$+$OG^{2}=AO^{2}$, that is $4.8^{2}+(6.4−r)^{2}=r^{2}$. Solving this yields: $r=\boxed{5}$ meters, that is, the radius is 5 meters. Extending GE to intersect the larger circle at point H, the center of the containing circle must lie on HG. As shown in the figure, let the center of the large circle be $O_{1}$, make $O_{1}M\perp AC$ at point M, and connect $O_{1}C$. Since AD=9.6 meters, DF=2.4 meters, therefore $AF=9.6+2.4=12$ meters. Since $ACF=90^\circ$, therefore $EG\parallel CF$. Therefore $\frac{AG}{AF}=\frac{AE}{AC}$, that is, $\frac{4.8}{12}=\frac{8}{AC}$. Therefore $AC=20$ meters. Therefore $AM=CM=10$ meters, $EM=10−8=2$ meters. Since $\triangle AEG\sim \triangle O_{1}EM$, therefore $\frac{EM}{O_{1}M}=\frac{EG}{AG}$, that is, $\frac{2}{O_{1}M}=\frac{6.4}{4.8}$. Therefore $O_{1}M=\frac{3}{2}$. Therefore $O_{1}C=\sqrt{O_{1}M^{2}+MC^{2}}=\sqrt{\left(\frac{3}{2}\right)^{2}+10^{2}}=\boxed{\frac{\sqrt{409}}{2}}$.


Question: Five rectangular books are placed in a square bookshelf as shown in the diagram, with four of them placed vertically, and the fifth one placed diagonally, where point $G$ precisely aligns with the edge of the bookshelf. Each book has a thickness of $5\,cm$ and a height of $20\,cm$, and the shelf has a width of $40\,cm$. What is the length of $FI$ in $cm$?

Solution:
\textbf{Solution:} From the given information, we have: $FG=5\text{cm}$, $BC=5\times 4=20\text{cm}$, $BI=40\text{cm}$, $EF=20\text{cm}$,\\
$CD\perp BI$, $BI\perp GI$, $EF\perp GF$,\\
$\therefore \angle I=\angle ECF=90^\circ$, $\angle GFI+\angle EFC=\angle FEC+\angle EFC=90^\circ$,\\
$\therefore \angle GFI=\angle FEC$,\\
In $\triangle GFI$ and $\triangle FEC$, $\left\{\begin{array}{l}\angle I=\angle ECF=90^\circ \\ \angle GFI=\angle FEC\end{array}\right.$,\\
$\therefore \triangle GFI\sim \triangle FEC$,\\
$\therefore \frac{FI}{CE}=\frac{FG}{EF}$,\\
Let $FI=x\text{cm}$, then $CF=BI-BC-FI=(20-x)\text{cm}$,\\
$\therefore \frac{x}{CE}=\frac{5}{20}$, solving for $CE$ we get $CE=4x\text{cm}$,\\
In right triangle $Rt\triangle FEC$, $CF^2+CE^2=EF^2$, which gives $(20-x)^2+(4x)^2=20^2$,\\
Solving, we get $x=\frac{40}{17}$ or $x=0$ (not meeting the requirement, hence discarded),\\
Thus, $FI=\boxed{\frac{40}{17}\text{cm}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, $CD \perp AB$ with the foot of the perpendicular as D. If $AD = 2$ and $BD = 4$, then what is the length of $CD$?

Solution:
\textbf{Solution:} In the right $\triangle ACB$, $\angle ACB=90^\circ$, $CD\perp AB$,\\
$\therefore \angle ACD=\angle B=90^\circ-\angle A$,\\
Also, $\because \angle ACD=\angle CDB=90^\circ$,\\
$\therefore \triangle ACD\sim \triangle CBD$,\\
$\therefore CD^2=AD\cdot BD=8$, that is $CD=\boxed{2\sqrt{2}}$.


Question: As shown in the figure, in $ \triangle ABC$, points D and E are located on sides $ AB$ and $ AC$ respectively, and $ DE\parallel BC$. If $ AD=2\,cm$, $ AB=6\,cm$, and $ AE=1.5\,cm$, what is the length of $ EC$ in $cm$?

Solution:
\textbf{Solution:} Since $DE\parallel BC$,\\
it follows that $\frac{AD}{AB}=\frac{AE}{AC}=\frac{2}{6}=\frac{1}{3}$,\\
thus $\frac{1.5}{AC}=\frac{1}{3}$,\\
solving this gives: $AC=4.5\text{cm}$,\\
therefore $EC=AC-AE=4.5-1.5=\boxed{3}\text{cm}$.


Question: As shown in the diagram, a cup filled with water is placed tilted on a horizontal table surface. Its cross-section can be considered a rectangle with width $BC=6\, \text{cm}$ and length $CD=16\, \text{cm}$. When the water surface touches the edge of the cup, half of side $CD$ is above the water surface. What is the height of the water surface in centimeters at this moment?

Solution:
\textbf{Solution:} Construct $BF\perp AF$ at $F$, as shown in the diagram:\\
Since $BC=6$ cm, $CD=16$ cm, and $CE=\frac{1}{2}CD$,\\
Therefore $CE=8$ cm,\\
Since $\angle C=90^\circ$,\\
By the Pythagorean theorem, we have $BE=\sqrt{BC^{2}+CE^{2}}=\sqrt{6^{2}+8^{2}}=10$,\\
Since $\angle BCE=\angle FBE=90^\circ$,\\
Therefore $\angle EBC=\angle ABF$,\\
Since $\angle BCE=\angle BFA=90^\circ$,\\
Therefore $\triangle CBE\sim \triangle FBA$,\\
Therefore $\frac{BE}{AB}=\frac{BC}{BF}$,\\
i.e., $\frac{10}{16}=\frac{6}{BF}$,\\
Therefore $BF=\boxed{\frac{48}{5}}$.


Question: As shown in the figure, in parallelogram $ABCD$, point $E$ lies on side $BC$, and $AE$ intersects $BD$ at point $F$. If $BE=2$, $EC=3$, and $DF=10$, then what is the value of $BF$?

Solution:
\textbf{Solution:} Given that $BE=2, EC=3$, \\
it follows that $BC=BE+CE=2+3=5$, \\
Since quadrilateral $ABCD$ is a parallelogram, \\
it follows that $AD\parallel BC$ and $AD=BC=5$, \\
Therefore, $\triangle BEF\sim \triangle DAF$, \\
which implies $\frac{BF}{DF}=\frac{BE}{AD}$, i.e., $\frac{BF}{10}=\frac{2}{5}$, \\
Thus, $BF=\boxed{4}$.


Question: As shown in the figure, in $\triangle ABC$, points D and E are on sides AB and AC respectively, $AD:DB = 3:2$, $AE:EC = 1:2$. The line ED and the extension of CB intersect at point F. What is the value of $\frac{FB}{FC}$?

Solution:
\textbf{Solution:} Draw BG$\parallel$AC through point B, intersecting EF at point G,\\
since AD$\colon$DB = 3$\colon$2, AE$\colon$EC = 1$\colon$2,\\
thus AE = $\frac{1}{2}$EC,\\
thus $\triangle$BFG$\sim$$\triangle$CEF, $\triangle$BGD$\sim$$\triangle$AED,\\
thus $\frac{BG}{AE}=\frac{BD}{AD}=\frac{2}{3}$,\\
$\frac{BG}{EC}=\frac{FB}{FC}$,\\
thus $\frac{BG}{EC}=\frac{1}{3}$\\
thus $\frac{FB}{FC}=\boxed{\frac{1}{3}}$.


Question: As shown in the figure, $\triangle ABC$ and $\triangle A'B'C'$ are similar, with the center of similarity at point O, $OA=2AA'$, and the area of $\triangle ABC$ is 4. What is the area of $\triangle A'B'C'$?

Solution:
\textbf{Solution:} Since $OA = 2AA'$,\\
it follows that $\frac{OA}{OA'}=\frac{2}{3}$,\\
Since $\triangle ABC$ and $\triangle A'B'C'$ are similar,\\
it implies that $\triangle ABC\sim \triangle A'B'C'$, and $AB\parallel A'B'$,\\
therefore $\triangle OAB\sim \triangle OA'B'$,\\
which leads to $\frac{AB}{A'B'}=\frac{OA}{OA'}=\frac{2}{3}$,\\
thus $\frac{S_{\triangle OAB}}{S_{\triangle OA'B'}}=\left(\frac{2}{3}\right)^2=\frac{4}{9}$,\\
Given that the area of $\triangle ABC$ is 4,\\
it follows that the area of $\triangle A'B'C'$ is $\boxed{9}$.


Question: As shown in the figure, rectangle $ABCD$ is inscribed in circle $O$, with $AB=40$ and $BC=55$. Let $E$ be a point on arc $BC$, and draw lines $AE$ and $CE$, where $CE=20$. What is the length of $AE$? Let $F$ be a moving point on ray $CE$ such that $\angle BFE>90^\circ$. When $BF=AB$, what is the length of $EF$?

Solution:
\textbf{Solution:} Draw $BM\perp CF$ at point $M$, connect $AC$, let $AE$ intersect $BC$ at point $G$,\\
since quadrilateral ABCD is a rectangle,\\
therefore $AB=CD=40$, $AD=BC=55$, $\angle ABC=\angle ADC=90^\circ$,\\
therefore $AC$ is the diameter,\\
therefore $\angle AEC=90^\circ$,\\
in $\triangle ADC$, we have $AC^{2}=AD^{2}+DC^{2}=4625$,\\
since $CE=20$,\\
therefore $AE=\sqrt{AC^{2}-CE^{2}}=\boxed{65}$,\\
in $\triangle ABG$ and $\triangle CEG$,\\
$\left\{\begin{array}{l}
\angle ABG=\angle CEG=90^\circ \\
\angle AGB=\angle CGE
\end{array}\right.$,\\
therefore $\triangle ABG\sim \triangle CEG$,\\
therefore $\frac{BG}{EG}=\frac{AG}{CG}=\frac{AB}{CE}=\frac{40}{20}=2$,\\
therefore $BG=2EG$, $AG=2CG$,\\
since $AG+EG=AE$,\\
therefore $2CG+EG=65$ (1),\\
since $BG+CG=BC$,\\
therefore $2EG+CG=55$ (2),\\
from (1) and (2), we get: $CG=25$, $EG=15$,\\
therefore $BG=2EG=30$, $AG=2CG=50$,\\
in $\triangle CBM$ and $\triangle AGB$,\\
$\left\{\begin{array}{l}
\angle BCM=\angle BAG \\
\angle CMB=\angle ABG=90^\circ
\end{array}\right.$,\\
therefore $\triangle CBM\sim \triangle AGB$,\\
therefore $\frac{BM}{BG}=\frac{CM}{AB}=\frac{BC}{AG}$,\\
therefore $\frac{BM}{30}=\frac{CM}{40}=\frac{55}{50}$,\\
therefore $BM=33$, $CM=44$,\\
since $BM\perp CF$, $BF=AB=40$,\\
therefore $FM=\sqrt{BF^{2}-BM^{2}}=\sqrt{40^{2}-33^{2}}=\sqrt{511}$,\\
therefore $EF=CM-CE-FM=44-20-\sqrt{511}=\boxed{24-\sqrt{511}}$.


Question: As shown in the figure, rectangle $ABCD$, where $AB=8$, $AD=6$, and $F$ is a moving point on side $AB$. Connect $CF$, and draw $DG \perp CF$ at $D$ with the foot of the perpendicular being point $G$, intersecting $BC$ at point $E$. Draw $AH \perp DE$ through $A$ with the foot of the perpendicular being point $H$, and connect $CH$. What is the maximum area of quadrilateral $AGCH$?

Solution:
\textbf{Solution:}

Since quadrilateral $ABCD$ is a rectangle,

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

and since $DG\perp CF$, $AH\perp DE$,

$\therefore$ $\angle AHD=\angle DGC=90^\circ$,

$\therefore$ $\angle HAD+\angle ADH=\angle GDC+\angle ADH=90^\circ$,

$\therefore$ $\angle HAD=\angle GDC$,

and $\angle AHD=\angle DGC=90^\circ$,

$\therefore$ $\triangle AHD\sim \triangle DGC$,

$\therefore$ $\frac{DH}{CG}=\frac{AH}{DG}=\frac{AD}{DC}=\frac{3}{4}$,

Let $AH=3x$, $DG=4x$, $DH=3a$, $CG=4a$,

$\therefore$ $GH=DG-DH=4x-3a$,

$\therefore$ $S_{\text{quadrilateral }AGCH}=S_{\triangle AHG}+S_{\triangle HGC}$,

$=\frac{1}{2}AH\cdot GH+\frac{1}{2}CG\cdot GH$

$=\frac{1}{2}GH(AH+CG)$

$=\frac{1}{2}(4x-3a)(3x+4a)$

$=\frac{1}{2}(12x^2+7ax-12a^2)$

$=-6a^2+\frac{7}{2}ax+6x^2$

$=-6(a-\frac{7}{24}x)^2+\frac{625}{96}x^2$,

$\therefore$ when $a=\frac{7}{24}x$, $S_{\text{quadrilateral }AGCH}$ reaches its maximum value of $\frac{625}{96}x^2$,

Since in $Rt\triangle ADH$,

$AH^2+DH^2=AD^2$, that is $(3x)^2+(3a)^2=36$,

$\therefore$ $x^2+a^2=4$,

Substituting $a=\frac{7}{24}x$ into $x^2+a^2=4$,

we get $x^2=\frac{2304}{625}$,

$\therefore$ $\frac{625}{96}x^2=\frac{625}{96}\times \frac{2304}{625}=\boxed{24}$,

The maximum area of quadrilateral $AGCH$ is: $\boxed{24}$.


Question: In rectangle $ABCD$, $AB=2BC$, and points $F$ and $G$ are on sides $AB$ and $DC$ respectively. The quadrilateral $AFGD$ is folded along $GF$ to obtain quadrilateral $EFGP$, where point $E$ lies on side $BC$, and $EP$ intersects $CD$ at point $H$. If $\sin\angle CGP=\frac{3}{5}$ and $GF=2\sqrt{5}$, what is the length of $CE$?

Solution:
\textbf{Solution:} As shown in the figure, connect $AE$, and draw $GQ\perp AB$ through point $G$,\\
$\because$ by folding,\\
$\therefore AE\perp FG$,\\
$\therefore \angle QGF+\angle AFG=\angle BAE+\angle AFG$,\\
$\therefore \angle QGF=\angle BAE$,\\
$\because \angle B=\angle FQG$,\\
$\therefore \triangle ABE\sim \triangle GQF$,\\
$\because AB=2BC$, $GF=2\sqrt{5}$,\\
$\therefore \frac{GF}{AE}=\frac{QG}{AB}=\frac{1}{2}$,\\
$\therefore AE=4\sqrt{5}$,\\
$\because GP\parallel FE$, $DC\parallel AB$,\\
$\therefore \angle PGH=\angle EFB$,\\
$\because \sin\angle CGP=\frac{3}{5}$,\\
$\therefore \sin\angle BFE=\frac{3}{5}$,\\
Let $BE=3k$, then $EF=5k$,\\
$\therefore FB=4k$,\\
$\therefore AB=AF+BF=EF+BF=5k+4k=9k$,\\
$\therefore AE=\sqrt{AB^2+BE^2}=\sqrt{(9k)^2+(3k)^2}=3\sqrt{10}k$,\\
$\therefore 3\sqrt{10}k=4\sqrt{5}$,\\
Solving, we find $k=\frac{2}{3}\sqrt{2}$,\\
$\because BC=\frac{1}{2}AB=\frac{9}{2}k$, $BE=3k$,\\
$\therefore EC=BC-BE=\frac{9}{2}k-3k=\frac{3}{2}k=\boxed{\sqrt{2}}$


Question: As shown in the diagram, in $\triangle ABC$, $\angle B=90^\circ$, point $E$ is on $AB$, and $CE$ is connected. Folding $\triangle BCE$ along line $CE$, we obtain $\triangle FCE$, with the corresponding point $F$ of point $B$ landing exactly on $AC$. Through point $F$, construct $FD \perp AB$ at point $D$, and point $M$ is on the extension of $DF$, with $CM$ connected. Point $N$ is on $CM$, point $R$ is on $CF$, and a point $Q$ is selected on the extension of $CM$, with $FN$, $RN$, and $RQ$ connected. If $\angle CFN = \angle Q$, $RN=FR=CR$, $FN=NQ$, and $AB=\frac{40}{3}$, $AF=\frac{1}{2}AB$, then what is the length of the segment $RQ$?

Solution:
\textbf{Solution:} Given $AB=\frac{40}{3}$ and $AF=\frac{1}{2}AB$, \\
$\therefore AF=\frac{20}{3}$, \\
by the property of folding, we have $CB=CF$, let $CB=CF=x$, \\
by the Pythagorean theorem, we have $\left(x+\frac{20}{3}\right)^{2}=x^{2}+\left(\frac{40}{3}\right)^{2}$, \\
solving this gives $x=10$, \\
$\therefore CB=CF=10$, \\
since $RN=FR=CR$, \\
$\therefore RN=FR=CR=5$, $\angle FNC=90^\circ$, \\
let the intersection of $FN$ and $RQ$ be O, \\
since $\begin{cases}
\angle CFN=\angle Q \\
FN=QN \\
\angle FNC=\angle QNO
\end{cases}$, \\
$\therefore \triangle FNC\cong \triangle QNO$, \\
$\therefore FC=QO=10$, \\
since $RN=FR=CR$, \\
$\therefore \angle CFN=\angle RNO$, \\
since $\angle CFN=\angle Q$, \\
$\therefore \angle Q=\angle RNO$, \\
since $\angle NRO=\angle QRN$, \\
$\therefore \triangle NRO\sim \triangle QRN$, \\
$\therefore \frac{NR}{RQ}=\frac{OR}{RN}$, \\
solving this gives $OR=5\sqrt{5}-5$ (discarding $OR=-5\sqrt{5}-5$), \\
$\therefore RQ=RO+OQ=5\sqrt{5}-5+10=5\sqrt{5}+5$, \\
thus the answer is: $RQ=\boxed{5\sqrt{5}+5}$.


Question: As shown in the figure, in parallelogram $ABCD$, $AC$ and $BD$ intersect at point $O$. Point $E$ is the midpoint of $OA$. Connect $BE$ and extend it to intersect $AD$ at point $F$. Given that the area of $\triangle AEF$ is $3$, what is the area of $\triangle ABE$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a parallelogram,\\
it follows that $AD\parallel BC$ and $OA=OC$,\\
therefore $\triangle AFE\sim \triangle CBE$,\\
which implies $\frac{AE}{EC}=\frac{EF}{BE}$,\\
since point E is the midpoint of $OA$,\\
it follows that $AE=\frac{1}{3}EC$,\\
hence, $EF=\frac{1}{3}BE$,\\
given that $S_{\triangle AEF}=3$,\\
thus, $S_{\triangle AEB}=\boxed{9}$


Question: As shown in the figure, it is known that the area of $\triangle ABF$ is equal to that of the quadrilateral $CDFE$, with point $E$ on side $BC$, $AE\parallel CD$ intersecting $BD$ at point $F$, $CD=12$, $EF=9$. What is the length of $AF$?

Solution:
\textbf{Solution:}\\
$\because$ The area of $\triangle ABF$ is equal to the area of the quadrilateral $CDFE$,\\
$\therefore$ The area of $\triangle BEA$ is equal to the area of $\triangle BCD$,\\
$\because$ $AE\parallel CD$,\\
$\therefore$ $\triangle BEF\sim \triangle BCD$,\\
$\therefore$ $\frac{S_{\triangle BEF}}{S_{\triangle BCD}}=\left(\frac{EF}{CD}\right)^2$\\
$\because$ $CD=12$ and $EF=9$,\\
$\therefore$ $\frac{S_{\triangle BEF}}{S_{\triangle BCD}}=\left(\frac{9}{12}\right)^2=\frac{9}{16}$,\\
$\therefore$ $\frac{S_{\triangle BEF}}{S_{\triangle BEA}}=\frac{9}{16}$\\
$\therefore$ $\frac{EF}{EA}=\frac{9}{16}$\\
$\therefore$ $\frac{AF}{EF}=\frac{7}{9}$\\
$\therefore$ $AF=\frac{7}{9}EF=\frac{7}{9}\times 9 = \boxed{7}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=6$, and $BC=4$. Rotate $\triangle ABC$ $90^\circ$ clockwise around the right-angle vertex $C$ to get $\triangle DEC$. If point $F$ is the midpoint of $DE$, and connect $AF$, then what is the length of $AF$?

Solution:
\textbf{Solution:} Draw $FH \perp AC$ at H, as shown in the figure,\\
since $\triangle ABC$ is rotated clockwise by $90^\circ$ around the right angle vertex C to obtain $\triangle DEC$,\\
therefore, $CE=BC=4$, $CD=CA=6$, and $\angle ECD=\angle ACB=90^\circ$,\\
thus, $AE=AC-CE=2$, and $FH \parallel CD$,\\
therefore, $\triangle EHF \sim \triangle ECD$,\\
since point F is the midpoint of DE,\\
thus, $\frac{EH}{EC}=\frac{HF}{CD}=\frac{EF}{ED}=\frac{1}{2}$,\\
therefore, $EH=\frac{1}{2}EC=2$, $HF=\frac{1}{2}CD=3$,\\
thus, $AH=AE+EH=4$,\\
in right-angled $\triangle AFH$, $AF=\sqrt{AH^2+HF^2}=\sqrt{4^2+3^2}=\boxed{5}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $\cos A = \frac{1}{3}$, where $BE$ and $CF$ are the altitudes on sides $AC$ and $AB$ respectively. Connecting $EF$, what is the ratio of the perimeters of $\triangle AEF$ and $\triangle ABC$?

Solution:
\textbf{Solution:} Let $BE$ and $CF$ be the altitudes from $AC$ and $AB$, respectively,\\
Hence, $\angle AEB=\angle AFC=90^\circ$,\\
Since $\angle A=\angle A$,\\
Therefore, $\triangle AEB\sim \triangle AFC$,\\
Therefore, $\frac{AE}{AF}=\frac{AB}{AC}$,\\
Therefore, $\frac{AE}{AB}=\frac{AF}{AC}$,\\
Since $\angle A=\angle A$,\\
Therefore, $\triangle AEF\sim \triangle ABC$,\\
Therefore, the ratio of the perimeters of $\triangle AEF$ and $\triangle ABC$ is $AE:AB$,\\
Since $\cos A=\frac{AE}{AB}=\frac{1}{3}$,\\
Therefore, the ratio of the perimeters of $\triangle AEF$ and $\triangle ABC$ is $AE:AB=\boxed{1:3}$.


Question: As shown in the figure, it is known that in $\triangle ABC$ and $\triangle CDB$, $\angle ACB=\angle CDB=90^\circ$, $AC=BC$, $CD=BD$. Connect $AD$ and intersect $BC$ at point $P$. What is the value of $\tan\angle CAP$?

Solution:
\textbf{Solution:} Since $\angle ACB=\angle CDB=90^\circ$ and $AC=BC$, $CD=BD$,\\
it follows that $\triangle ACB$ and $\triangle CDB$ are isosceles right triangles,\\
therefore $\angle ABC+\angle CBD=90^\circ$\\
thus $\angle ABD+\angle CDB=180^\circ$\\
which means $CD\parallel AB$\\
hence $\triangle CDP\sim \triangle BAP$,\\
let $CD=DB=a$, then $CB=\sqrt{2}a$, $AB=2a$\\
therefore, $\frac{CD}{AB}=\frac{CP}{BP}=\frac{1}{2}$\\
thus, $\frac{CP}{BC}=\frac{1}{3}$\\
Since $AC=BC$\\
it follows that $\tan\angle CAP=\boxed{\frac{1}{3}}$


Question: As shown in the figure, in $\triangle ABC$, points $D$ and $E$ are located on sides $AB$ and $AC$ respectively, and $DE \parallel BC$. If $AD = 2$, $DB = 3$, and $DE = 1$, what is the length of $BC$?

Solution:
\textbf{Solution:} Given $\because$ DE $\parallel$ BC, \\
$\therefore$ $\triangle ADE \sim \triangle ABC$, \\
$\therefore$ $\frac{AD}{AB} = \frac{DE}{BC}$, \\
$\because$ AD = 2, DB = 3, \\
$\therefore$ AB = 5, \\
$\therefore$ $\frac{2}{5} = \frac{1}{BC}$, solving this we get: $BC = \boxed{\frac{5}{2}}$.


Question: As depicted, the rectangle $DEFG$ has its side $DE$ on side $BC$ of $\triangle ABC$, with points $G$ and $E$ respectively on sides $AB$ and $AC$. Given that $BC=6\,cm$, $DE=3\,cm$, $EF=2\,cm$, what is the area of $\triangle ABC$ in $\,cm^{2}$?

Solution:
\textbf{Solution:} Draw $AH\perp BC$ at $H$, intersecting $GF$ at $M$,\\
since $AH\perp BC$ and quadrilateral $DEFG$ is a rectangle,\\
thus, quadrilateral $HEFM$ is a rectangle, so $MH=EF=2\,cm$,\\
since quadrilateral $DEFG$ is a rectangle,\\
thus, $GF\parallel BC$, $DG=EF=2\,cm$, $GF=DE=3\,cm$,\\
therefore, $\triangle AGF\sim \triangle ABC$,\\
thus, $\frac{AM}{AH}=\frac{GF}{BC}$, that is $\frac{AM}{AM+2}=\frac{3}{6}$, solving gives: $AM=2\,cm$,\\
thus, $AH=AM+MH=4\,cm$,\\
therefore, ${S}_{\triangle ABC}=\frac{1}{2}BC\cdot AH=\frac{1}{2}\times 6\times 4=\boxed{12}\,cm^{2}$;


Question: As shown in the figure, points A, C, D are all on the circle $\odot O$, and point B is inside $\odot O$. Furthermore, $AB \perp BC$ at point B, and $BC \perp CD$ at point C. Given that $AB = 6$, $BC = 12$, and $CD = 4$, what is the circumference of $\odot O$?

Solution:
\textbf{Solution:} Extend $CB$ to meet $\odot O$ at point $E$, connect $AE$, $DE$, $AC$, and have $AD$ intersect $CB$ at point $F$.\\
Therefore, $AC=\sqrt{AB^{2}+BC^{2}}=6\sqrt{5}$.\\
Since $AB\perp BC$ and $BC\perp CD$,\\
Therefore, $AB\parallel CD$,\\
Therefore, $\triangle ABF\sim \triangle DCF$,\\
Therefore, $\frac{AB}{CD}=\frac{BF}{CF}$,\\
Therefore, $\frac{BF}{CF}=\frac{6}{4}=\frac{3}{2}$.\\
Since $BC=BF+CF=12$,\\
Therefore, $BF=\frac{3}{5}\times 12=\frac{36}{5}$, $CF=\frac{2}{5}\times 12=\frac{24}{5}$,\\
Therefore, $AF=\sqrt{AB^{2}+BF^{2}}=\sqrt{6^{2}+\left(\frac{36}{5}\right)^{2}}=\frac{6}{5}\sqrt{61}$, $DF=\sqrt{CF^{2}+CD^{2}}=\sqrt{\left(\frac{24}{5}\right)^{2}+4^{2}}=\frac{4}{5}\sqrt{61}$,\\
Therefore, $AD=AF+DF=2\sqrt{61}$.\\
Since $\angle DCE=90^\circ$,\\
Therefore, DE is the diameter,\\
Therefore, $\angle EAD=\angle ABC=90^\circ$.\\
Since $\angle EDA=\angle ACB$,\\
Therefore, $\triangle EDA\sim \triangle ACB$,\\
Therefore, $\frac{AD}{BC}=\frac{DE}{AC}$, that is, $\frac{2\sqrt{61}}{12}=\frac{DE}{6\sqrt{5}}$,\\
Solving gives: $DE=\sqrt{305}$, that is, the diameter of $\odot O$ is $\sqrt{305}$,\\
Therefore, the circumference of $\odot O$ is $\boxed{\sqrt{305}\pi}$.


Question: As shown in the figure, in $\triangle ABC$, $DE \parallel BC$, $AD=2$, $BD=3$, $AC=10$, what is the length of $AE$?

Solution:
\textbf{Solution:} Since $DE\parallel BC$, \\
it follows that $\frac{AD}{BD}=\frac{AE}{EC}=\frac{2}{3}$, \\
thus, $\frac{AE}{AC}=\frac{2}{5}$, \\
therefore, $AE=\frac{2}{5}AC=\frac{2}{5}\times 10=\boxed{4}$.


Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is on $DC$, with $DE:EC = 3:2$. Connecting $AE$ and intersecting $BD$ at point $F$, what is the ratio of $S_{\triangle DEF}$ to $S_{\triangle BAF}$?

Solution:
\textbf{Solution:} Given, DE: EC = 3: 2,\\
Therefore, DE: DC = 3: 5,\\
Given parallelogram ABCD,\\
Therefore, AB $\parallel$ CD and AB = CD,\\
Therefore, $\frac{DE}{AB} = \frac{DE}{CD} = \frac{3}{5}$, $\triangle$ DEF $\sim$ $\triangle$ BAF,\\
Therefore, $\frac{S_{\triangle DEF}}{S_{\triangle BAF}} = \left(\frac{DE}{AB}\right)^2 = \left(\frac{3}{5}\right)^2 = \boxed{\frac{9}{25}}$,


Question: As shown in the figure, points A and B are on the graphs of the inverse proportion functions $y=\frac{1}{x} (x>0)$ and $y=\frac{k}{x} (x>0)$, respectively, and $\angle AOB=90^\circ$, $\sin B=\frac{1}{2}$. What is the value of $k$?

Solution:
\textbf{Solution:} Construct $AC\perp y$-axis through $A$, and $BD\perp y$-axis through $B$, it can be obtained that $\angle ACO=\angle BDO=90^\circ$, \\
$\therefore \angle AOC+\angle OAC=90^\circ$, \\
$\because OA\perp OB$, \\
$\therefore \angle AOC+\angle BOD=90^\circ$, \\
$\therefore \angle OAC=\angle BOD$, \\
$\therefore \triangle AOC\sim \triangle OBD$, \\
$\because$ points $A$ and $B$ lie on the graphs of the inverse proportion functions $y=\frac{1}{x}(x>0)$ and $y=\frac{k}{x}(x>0)$ respectively, \\
$\therefore S_{\triangle AOC}=\frac{1}{2}$, $S_{\triangle OBD}=|\frac{k}{2}|$, \\
$\therefore S_{\triangle AOC}\colon S_{\triangle OBD}=1\colon |k|$, \\
$\therefore \left(\frac{OA}{OB}\right)^2=1\colon |k|$, \\
then, in $\triangle AOB$ with right angle, $\because \sin B=\frac{1}{2}$, \\
$\therefore \angle B=30^\circ$, \\
$\therefore \tan B=\frac{OA}{OB}=\frac{\sqrt{3}}{3}$, \\
$\therefore 1\colon |k|=1\colon 3$, \\
$\therefore |k|=3$, \\
$\because$ the graph of $y=\frac{k}{x}(x>0)$ is in the fourth quadrant, \\
$\therefore k=\boxed{-3}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $CD \perp AB$ at $D$, $AD=9$, $CD=6$, then what is the length of $BD$?

Solution:
\textbf{Solution:} Given that $CD\perp AB$,\\
it follows that $\angle ADC=\angle CDB=90^\circ$, $\angle A+\angle ACD=90^\circ$,\\
since $\angle ACB=90^\circ$,\\
it follows that $\angle ACD+\angle BCD=90^\circ$,\\
therefore $\angle A=\angle BCD$,\\
therefore $\triangle ADC\sim \triangle CDB$,\\
therefore $\frac{AD}{CD}=\frac{CD}{BD}$,\\
since $AD=9, CD=6$,\\
it follows that $\frac{9}{6}=\frac{6}{BD}$,\\
therefore $BD=\boxed{4}$.


Question: As shown in the figure, it is known that in rectangle $ABCD$, $AB=2$. A point $E$ is chosen on $BC$, and $\triangle ABE$ is folded upward along $AE$, causing point $B$ to coincide with point $F$ on $AD$. If quadrilateral $EFDC$ is similar to rectangle $ABCD$, what is the length of $AD$?

Solution:
\textbf{Solution:} Given that in rectangle ABCD, AF is obtained by folding AB,\\
thus ABEF is a square.\\
Also, since AB = 2, thus AF = AB = EF = 2.\\
Let AD = x, then FD = x - 2.\\
Since quadrilateral EFDC is similar to rectangle ABCD,\\
thus $\frac{EF}{FD} = \frac{AD}{AB}$, that is $\frac{2}{x-2} = \frac{x}{2}$\\
Solving this, we get $x_{1}=1+\sqrt{5}$, $x_{2}=1−\sqrt{5}$ (negative value is discarded).\\
Upon verification, $x_{1}=1+\sqrt{5}$ is the solution of the original equation.\\
Therefore, AD $ =\boxed{1+\sqrt{5}}$.


Question: In the figure, within $\triangle ADC$, $\angle C=90^\circ$. Point B is on the extension line of $CD$. Connect $AB$. Point E is on $AC$. Connect $DE$. $AD$ bisects $\angle BAC$. $CE = 2AE$, $DB = DE$, and $CD = 3$. What is the length of $AC$?

Solution:
\textbf{Solution:} As shown in the diagram, draw $DF\perp AB$ at $F$ through point $D$, \\
since $\angle C=90^\circ$ and $AD$ bisects $\angle BAC$, with $CD=3$, \\
therefore $DF=CD=3$, \\
since $DB=DE$, \\
therefore by the HL theorem, $Rt\triangle EDC\cong Rt\triangle BDF$, \\
hence $BF=CE$, \\
let $AE=x$, then $CE=2x=BF$, \\
since $CD=DF$ and $AD=AD$, \\
therefore, by the Pythagorean theorem, we get: $AC=AF=3x$, \\
hence $AB=5x$, \\
since $\angle B=\angle B$ and $\angle BFD=\angle ACB=90^\circ$, \\
therefore $\triangle BFD\sim \triangle BCA$, \\
therefore $\frac{DF}{AC}=\frac{BD}{AB}=\frac{BF}{BC}$, \\
therefore $\frac{BD}{5x}=\frac{3}{3x}$, \\
thus $BD=5$, \\
therefore from $BF^2+DF^2=BD^2$, we obtain: $(2x)^2=5^2-3^2=16$, \\
solving gives: $x=2$, \\
thus $AC=3x=\boxed{6}$.


Question: As shown in the figure, $\triangle ABC$ is inscribed in circle $O$, with $AB$ being the diameter of circle $O$. $\triangle ABC$ is rotated about point $C$ to $\triangle EDC$, where point $E$ is on the circle. Given that $AE=2$ and $\tan D=3$, what is the length of $AB$?

Solution:
\textbf{Solution:} Since AB is the diameter,\\
it follows that $\angle ACB=90^\circ$\\
Because rotating $\triangle ABC$ around point C to $\triangle EDC$,\\
it follows that $\angle ABC=\angle D$, $\angle ACB=\angle ECD=90^\circ$, ED=AB, BC=CD, AC=CE,\\
therefore, $\angle ACE=\angle BCD=\angle ABE$,\\
thus, $\angle ABE+\angle ABC+\angle CBD=\angle BCD+\angle D+\angle CBD=180^\circ$,\\
therefore, points E, B, D lie on the same line,\\
in $\triangle ECD$, $\tan D= \frac{EC}{CD}=3$,\\
Let CD=x, then EC=3x,\\
therefore, $ED=\sqrt{CD^2+EC^2}=\sqrt{x^2+9x^2}=\sqrt{10}x$;\\
Because $\overset{\frown}{AC}=\overset{\frown}{AC}$\\
it follows that $\angle ACE=\angle BCD$, $\angle D=\angle ABC=\angle AEC$,\\
therefore, $\triangle ACE\sim \triangle BCD$,\\
therefore, $\frac{CE}{CD}=\frac{AE}{BD}=3$,\\
$\angle CBD=\angle CAE$,\\
thus, $\frac{2}{BD}=3$ Solving this yields: $BD=\frac{2}{3}$;\\
therefore, $BE=DE−BD=\sqrt{10}x-\frac{2}{3}$\\
In $\triangle AEB$, $AE^2+BE^2=AB^2$\\
thus, solving $2^2+\left(\sqrt{10}x-\frac{2}{3}\right)^2=\left(\sqrt{10}x\right)^2$ yields: $x=\frac{\sqrt{10}}{3}$\\
therefore, $AB=DE=\frac{\sqrt{10}}{3}\times \sqrt{10}=\boxed{\frac{10}{3}}$.


Question: As shown in the figure, it is known that quadrilateral $ABCD$ is inscribed in circle $\odot O$, with the diameter $AC=12$. The diagonals $AC$ and $BD$ intersect at point $E$, and $AB=BD$, $EC=2$. What is the length of $AD$?

Solution:
\textbf{Solution:} Connect BO to intersect AD at point F, and draw OD, \\
$\because$ BA = BD, OA = OD, \\
$\therefore$ BF is the perpendicular bisector of line segment AD, \\
$\therefore$ BF $\perp$ AD, \\
$\because$ AC is the diameter of $\odot$O, \\
$\therefore$ $\angle$ADC = $90^\circ$, i.e., AD $\perp$ DC, \\
$\therefore$ BF $\parallel$ CD, \\
$\therefore$ $\triangle$BOE $\sim$ $\triangle$DCE, \\
$\therefore$ OB : CD = EO : EC, \\
$\because$ AC = 12, EC = 2, \\
$\therefore$ OB = OC = 6, \\
$\therefore$ OE = 4, \\
$\therefore$ 4CD = 12, \\
$\therefore$ CD = 3, \\
$\therefore$ In $\triangle$ADC, with AD = $\sqrt{12^{2}-3^{2}}$ = $\boxed{3\sqrt{5}}$.


Question: As shown in the figure, it is known that $AM$ and $DN$ are the medians of $\triangle ABC$ and $\triangle DCE$ respectively, $B$, $C$, $D$ are on the same straight line, $AB\parallel CD$, $AC\parallel DE$. If $BC=6$, what is the length of $MN$?

Solution:
\textbf{Solution:} Given that $AB\parallel CD$, $AC\parallel DE$,\\
it follows that $\angle B = \angle DCE$, $\angle E = \angle ACB$,\\
thus, $\triangle ABC \sim \triangle DCE$,\\
since $AM$ and $DN$ are the medians of $\triangle ABC$ and $\triangle DCE$ respectively,\\
it implies that $\frac{BC}{CE} = \frac{AM}{DN} = 2$, $CM = \frac{1}{2}BC = 3$, $CN = \frac{1}{2}CE$,\\
thus, $\frac{6}{CE} = 2$,\\
therefore $CE = 3$,\\
hence $CN = 1.5$,\\
consequently $MN = CM + CN = 3 + 1.5 = \boxed{4.5}$.


Question: As shown in the figure, in the square $ABCD$, where $AB=6$, $E$ is the midpoint of $BC$, and $F$ is a point on side $CD$. The triangle $CEF$ is folded along $EF$ to obtain the triangle $GEF$. Connect $CG$ and extend it to intersect $EF$ and $AB$ at points $O$ and $H$, respectively. If $G$ is the midpoint of $OH$, what is the length of $CF$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a square,\\
it follows that $\angle B=90^\circ$,\\
since $AB=6$ and E is the midpoint of $BC$,\\
then $CE=BE=\frac{1}{2}AB=3$,\\
since $\triangle CEF$ is folded along $EF$ to obtain $\triangle GEF$,\\
it follows that $EF$ bisects $CG$ perpendicularly,\\
thus $OG=OC$, $\angle COE=90^\circ$\\
Given G is the midpoint of $OH$,\\
then $OG=GH$,\\
therefore $OG=GH=OC=\frac{1}{3}CH$,\\
let $OC=x$, then $CH=3x$,\\
since $\angle COE=\angle B=90^\circ$, $\angle OCE=\angle BCH$,\\
it follows that $\triangle COE\sim \triangle CBH$,\\
which means $\frac{CO}{BC}=\frac{CE}{CH}$,\\
thus $\frac{x}{6}=\frac{3}{3x}$,\\
solving gives $x=\sqrt{6}$,\\
therefore $CO=\sqrt{6}$,\\
hence $OE=\sqrt{CE^2-CO^2}=\sqrt{3}$,\\
therefore $\sin\angle OCE=\frac{OE}{CE}=\frac{\sqrt{3}}{3}$,\\
since $\angle CFO+\angle OCF=\angle OCE+\angle OCF=90^\circ$,\\
it follows that $\angle CFO=\angle OCE$,\\
thus $\sin\angle CFO=\sin\angle OCE=\frac{OC}{CF}=\frac{\sqrt{3}}{3}$,\\
hence $CF=\sqrt{3}OC=\sqrt{3}\times \sqrt{6}=\boxed{3\sqrt{2}}$


Question: As shown in the figure, $AB=AC$, $A\left(0, \sqrt{15}\right)$, $C(1, 0)$, D is a point on the ray AO, and a moving point P starts from A, moving along the path $A-D-C$. The speed on AD is $4$ units per second, and the speed on CD is $1$ unit per second. What are the coordinates of D when the total moving time is minimal?

Solution:
\textbf{Solution:} As shown in the figure, draw $DH\perp AB$ at H, and $CM\perp AB$ at $M$, intersecting AO at ${D}^{\prime}$.\\
Since the travel time $t=\frac{AD}{4}+CD$,\\
and since $AB=AC$, $AO\perp BC$,\\
it follows that $BO=OC=1$,\\
given $A(0,\sqrt{15})$, $C(1,0)$, $AB=AC$, $AO\perp BC$,\\
it follows that $AB=AC=\sqrt{OA^2+OB^2}=\sqrt{15+1}=4$,\\
Since $\angle DAH=\angle BAO$, $\angle DHA=\angle AOB=90^\circ$,\\
it follows that $\triangle AHD\sim \triangle AOB$,\\
therefore $\frac{AD}{AB}=\frac{DH}{OB}$,\\
hence $DH=\frac{1}{4}AD$,\\
therefore $\frac{1}{4}AD+CD=CD+DH$,\\
hence, when C, D, H are collinear and coincide with CM, the travel time is minimized,\\
thus $\frac{1}{2}BC\cdot AO=\frac{1}{2}AB\cdot CM$,\\
hence $CM=\frac{\sqrt{15}}{2}$,\\
therefore $AM=\sqrt{AC^2-CM^2}=\sqrt{4^2-\left(\frac{\sqrt{15}}{2}\right)^2}=\frac{7}{2}$,\\
Given $AD^{\prime}=4MD^{\prime}$, let $MD^{\prime}=m$, then $AD^{\prime}=4m$,\\
we get: $16m^2-m^2=\frac{49}{4}$\\
Hence $m=\frac{7\sqrt{15}}{30}$ or $-\frac{7\sqrt{15}}{30}$ (discard latter),\\
Therefore $AD^{\prime}=\frac{14\sqrt{15}}{15}$\\
Thus $D\left(0,\frac{\sqrt{15}}{15}\right)$,\\
Therefore, the answer is $\boxed{\left(0,\frac{\sqrt{15}}{15}\right)}$.


Question: As shown in the figure, in $\triangle ABC$, points D and E are on sides $AB$ and $AC$, respectively. If $DE \parallel BC$ and $\frac{AD}{DB} = \frac{2}{3}$, $DE = 6\,\text{cm}$, then what is the length of $BC$ in cm?

Solution:
\textbf{Solution:} Since $\frac{AD}{DB}=\frac{2}{3}$,\\
it follows that $\frac{AD}{AB}=\frac{2}{5}$,\\
Since $DE\parallel BC$,\\
it implies $\triangle ADE\sim \triangle ABC$,\\
thus $\frac{AD}{AB}=\frac{DE}{BC}$, and given $DE=6$,\\
it means $\frac{6}{BC}=\frac{2}{5}$,\\
therefore $BC=\boxed{15}$, which is verified to be in accordance with the problem statement;


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, $AB=3$, and $AC=4$. Let point $P$ be any point on $BC$. Connect $PA$, and construct parallelogram $PAQC$ with $PA$ and $PC$ as adjacent sides. Connect $PQ$. What is the minimum value of $PQ$?

Solution:
\textbf{Solution:} Given that $\angle BAC=90^\circ$ and $AB=3$, $AC=4$,\\
$\therefore$ $BC=\sqrt{AC^2+AB^2}=5$,\\
Since quadrilateral $APCQ$ is a parallelogram,\\
$\therefore$ $PO=QO$, $CO=AO$,\\
Since the shortest distance is $PQ$ which equates to the shortest distance $PO$,\\
$\therefore$ draw the perpendicular line from $O$ to $BC$, labelled as $OP^{\prime }$,\\
Since $\angle ACB=\angle P^{\prime }CO$ and $\angle CP^{\prime }O=\angle CAB=90^\circ$,\\
$\therefore$ $\triangle CAB\sim \triangle CP^{\prime }O$,\\
$\therefore$ $\frac{CO}{BC}=\frac{OP^{\prime }}{AB}$,\\
$\therefore$ $\frac{2}{5}=\frac{OP^{\prime }}{3}$,\\
$\therefore$ $OP^{\prime }=\frac{6}{5}$,\\
$\therefore$ the minimum value of $PQ$ is $2OP^{\prime }=\boxed{\frac{12}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=15$, $\sin\angle A=\frac{4}{5}$. Points D and E are located on sides AB and AC, respectively, with $AD=2DB$. The $\triangle ADE$ is folded along line DE to obtain $\triangle DEF$. If ray $EF\perp BC$, then what is the length of $AE$?

Solution:
\textbf{Solution:}

Construct $DM \perp AC$ at $M$, and construct $BH \perp AC$ at $H$. \\
Since $AB = AC = 15$, $\sin\angle A = \frac{4}{5}$, and $AD = 2DB$, \\
then $AD = 10$, $DM = 8$, $AM = 6$, $BH = 12$, $AH = 9$. \\
Therefore, $CH = AC - AH = 6$. \\
Thus, $\tan\angle C = \frac{BH}{CH} = 2$, and $BC = \sqrt{BH^2 + CH^2} = 6\sqrt{5}$. \\
Construct $DG \perp EF$ intersecting $EF$ at $N$, and intersecting $AC$ at $G$. \\
Since triangle $ADE$ is folded along the straight line $DE$ to form $\triangle DEF$, \\
then $DE$ bisects $\angle AEF$. \\
Hence, $DM = DN = 8$, $EM = EN$. \\
Since $EF \perp BC$, \\
then $DG \parallel BC$. \\
Thus, $\frac{DG}{BC} = \frac{AD}{AB} = \frac{2}{3}$, and $\angle C = \angle NGE$. \\
Therefore, $DG = \frac{2}{3}BC = 4\sqrt{5}$. \\
Thus, $NG = DG - DN = 4\sqrt{5} - 8$. \\
Since $\tan\angle C = \tan\angle NGE = 2 = \frac{EN}{NG}$, \\
then $EM = EN = 2NG = 8\sqrt{5} - 16$. \\
Therefore, $AE = AM + EM = \boxed{8\sqrt{5} - 10}$.


Question: As shown in the figure, $AB=20$, point $C$ is the midpoint of segment $AB$, and point $D$ is the midpoint of segment $CB$. Find the length of $AD$.

Solution:
\textbf{Solution:} Since point $C$ is the midpoint of segment $AB$, and $AB=20$, \\
therefore $AC=BC=\frac{1}{2}AB=10$. \\
Since point $D$ is the midpoint of segment $BC$, \\
therefore $CD=\frac{1}{2}BC=5$. \\
Therefore, $AD=AC+CD=10+5=\boxed{15}$.


Question: As shown in the figure, $A$, $B$, and $C$ are three consecutive points on line $L$, with $M$ being the midpoint of $AB$ and $N$ being the midpoint of $MC$. Given that $AB=6\,\text{cm}$ and $NC=8\,\text{cm}$, what is the length of $BC$ in centimeters?

Solution:
Solution: Since M is the midpoint of AB,\\
it follows that AM = BM = $\frac{1}{2}$AB = 3,\\
Since N is the midpoint of MC,\\
it follows that MN = NC = 8.\\
Therefore, BN = MN - BM = 5,\\
Therefore, BC = BN + NC = 5 + 8 = \boxed{13} (cm)


Question: As shown in the figure, it is known that the line segment $AB=10\,\text{cm}$, and point C is a point on the line segment AB. If M is the midpoint of AC and $AM=2\,\text{cm}$, find the length of the line segment BC.

Solution:
\textbf{Solution:} Given that $M$ is the midpoint of $AC$, and $AM=2\,\text{cm}$, \\
thus $AC=2AM=4\,\text{cm}$, \\
since $AB=10\,\text{cm}$, \\
therefore $BC=AB-AC=\boxed{6}\,\text{cm}$.


Question: As shown in the figure, points B and C are on the line segment AD, and the ratio of AB:BC:CD is 2:3:4. Let M be the midpoint of line segment AC, and let N be the midpoint of line segment CD, with MN=9. Find the length of BD.

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

$\because M$ is the midpoint of segment $AC$, and $N$ is the midpoint of segment $CD$,

$\therefore CM = \frac{1}{2}AC$, $CN = \frac{1}{2}CD$,

$\therefore MN = CM + CN = \frac{1}{2}(AC + CD) = \frac{1}{2}AB = x = 9$,

$\therefore AD = 18$.

$\because AB : BC : CD = 2 : 3 : 4$,

$\therefore BC = 3x$, $CD = 4x$,

$\therefore BD = \frac{3 + 4}{2 + 3 + 4} \times AD = \boxed{14}$.


Question: As shown in the figure, $D$ is the midpoint of segment $AC$, and $E$ is the midpoint of segment $AB$. Given that $AD=6\,cm$ and $BC=3\,cm$, find the lengths of segment $AB$ and $EC$ in centimeters.

Solution:
\textbf{Solution:} Since $D$ is the midpoint of segment $AC$, and $E$ is the midpoint of segment $AB$\\
Thus, $AD=CD=\frac{1}{2}AC$, $AE=EB=\frac{1}{2}AB$\\
Since $AD=6cm$, $BC=3cm$\\
Thus, $AB = AC + CB = 2AD + BC = 12 + 3 = \boxed{15cm}$\\
$EC = EB - BC = \frac{1}{2}AB - BC = \frac{15}{2} - 3 = \frac{9}{2} = \boxed{4.5cm}$


Question: As shown in the figure, P is a point on segment AB, and Q is a point on the extension of AB. If $AB=10$ and $\frac{AP}{PB} = \frac{AQ}{QB} = \frac{3}{2}$. Find the length of segment PQ.

Solution:
\textbf{Solution:} Given $\because AP:PB=3:2$,\\
$\therefore$ let AP=3a, and PB=2a,\\
$\because P$ is a point on segment AB, and AB=10,\\
$\therefore 3a+2a=10$, solving for $a$ we find $a=2$,\\
$\therefore AP=6$, PB=4. Let the length of BQ be $x$, then $AQ=AB+BQ=10+x$,\\
$\because \frac{AQ}{BQ}=\frac{3}{2}$,\\
$\therefore \frac{10+x}{x}=\frac{3}{2}$, solving for $x$ yields $x=20$, i.e., $BQ=20$,\\
$\therefore PQ=PB+BQ=\boxed{24}$.


Question: As shown in the figure, it is known: the length of line segment $AB=20\text{cm}$, points $C$ and $D$ are two points on line segment $AB$, and $BC=\frac{1}{5}AB$, $AD=\frac{1}{4}AB$. Points $M$ and $N$ are the midpoints of line segments $AC$ and $BD$, respectively. Question: what is the length of line segment $MN$?

Solution:
\textbf{Solution:} Since $AB=20cm$, $BC=\frac{1}{5}AB$, $AD=\frac{1}{4}AB$,\\
hence $BC=4cm$, $AD=5cm$,\\
thus $AC=AB-BC=16cm$, $BD=AB-AD=15cm$,\\
as points $M$ and $N$ are the midpoints of line segments $AC$ and $BD$ respectively,\\
thus $AM=\frac{1}{2}AC=8cm$, $BN=\frac{1}{2}BD=\frac{15}{2}cm$,\\
therefore $MN=AB-AM-BN=20-8-\frac{15}{2}=\boxed{\frac{9}{2}cm}$.


Question: As shown in the figure, point C is a point on segment AB, point M is the midpoint of segment AC, and point N is the midpoint of segment BC. It is known that $AM = \frac{5}{4}BC = 5\,\text{cm}$. What is the length of MN in centimeters?

Solution:
\textbf{Solution:} Since $M$ is the midpoint of segment $AC$,\\
$\therefore AM=CM=5\text{cm}$, $BC=4\text{cm}$.\\
Also, since $N$ is the midpoint of segment $BC$,\\
$\therefore CN=\frac{1}{2}BC=2\text{cm}$,\\
$\therefore MN=MC+CN=\boxed{7}\text{cm}$.


Question: As shown in the figure, $AB=24$, point $C$ is the midpoint of $AB$, and point $D$ is on the line segment $AC$ with $AD = \frac{1}{3}CB$. Find the lengths of line segments $CD$ and $BD$.

Solution:
\textbf{Solution:} Since point C is the midpoint of AB,\\
it follows that $AC=BC=\frac{1}{2}AB=12$,\\
since $AD=\frac{1}{3}CB$,\\
it follows that $AD=\frac{1}{3}\times 12=4$,\\
therefore $CD=AC-AD=\boxed{8}$,\\
thus, $BD=BC+CD=20$


Question: As shown in the figure, points C and D are the trisection points of line segment $AB$, and point E is the midpoint of line segment $DB$. Given $AB=12\,\text{cm}$, find the length of line segment $CE$.

Solution:
\textbf{Solution:} Since points C and D are the trisection points of $AB$ and $AB=12cm$,\\
it follows that $CD=DB=\frac{1}{3}AB=4cm$,\\
Since point E is the midpoint of the line segment $DB$,\\
it follows that $DE=\frac{1}{2}DB=2cm$,\\
therefore, $CE=CD+DE=\boxed{6cm}$


Question: As shown in the figure, $C$, $D$, $E$ are points on the line segment $AB$, with $AC=5$, $DB=3$. Points $C$ and $E$ are the midpoints of line segments $AD$ and $BD$, respectively. Find the length of $CE$.

Solution:
\textbf{Solution:} Since points C and E are the midpoints of segments $AD$ and $BD$, respectively, \\
therefore $CD=AC=5$, $DE=\frac{1}{2}BD=\frac{1}{2}\times 3=1.5$, \\
therefore $CE=CD+DE=5+1.5=\boxed{6.5}$.


Question: Unfortunately, you haven't provided any specific text to translate. Please provide the text you'd like translated into English along with the required LaTeX formatting, and I'd be more than happy to assist you.

Solution:
\textbf{Solution:} Since $AB = 18$, and point D is the midpoint of line segment AB, \\
thus $BD = 18 \div 2 = 9$; \\
Since $BD = 3BC$, \\
thus $BC = 9 \div 3 = 3$, \\
therefore $AC = AB + BC = 18 + 3 = \boxed{21}$.


Question: As shown in the figure, the segment \(AC=6\,cm\), the segment \(BC=15\,cm\), point M is the midpoint of \(AC\), and a point N is taken on \(CB\) such that \(CN=\frac{1}{2}NB\). What is the length of \(MN\)?

Solution:
\textbf{Solution:} Given that $M$ is the midpoint of $AC$, and $AC=6\,\text{cm}$,\\
it follows that $MC=\frac{1}{2}AC=\frac{1}{2}\times 6=3\,\text{cm}$,\\
since $CN=\frac{1}{2}NB$ and $BC=15\,\text{cm}$,\\
then $CN=\frac{1}{3}BC=\frac{1}{3}\times 15=5\,\text{cm}$,\\
therefore, $MN=MC+NC=3\,\text{cm}+5\,\text{cm}=\boxed{8\,\text{cm}}$.


Question: As shown in the figure, point $C$ is the midpoint of segment $AB$, point $E$ is a point on segment $AB$, and point $D$ is the midpoint of segment $AE$. If $AB=16$ and $CE=5$, find the length of segment $AD$.

Solution:
\textbf{Solution:} Since point C is the midpoint of segment AB, and $AB=16$, \\
$\therefore BC=\frac{1}{2}AB=8$, \\
because $CE=5$, \\
$\therefore BE=BC-CE=8-5=3$, \\
hence $AE=AB-BE=16-3=13$, \\
because point D is the midpoint of segment AE, \\
$\therefore AD=\frac{1}{2}AE=\boxed{6.5}$.


Question: Given: Line segment $AB = 6$, point $C$ is the midpoint of line segment $AB$, extend line segment $AB$ to $D$, such that $BD = 3BC$. Find the length of line segment $AD$?

Solution:
\textbf{Solution:} Since point C is the midpoint of line segment AB,\\
therefore $BC=\frac{1}{2}AB$,\\
since AB = 6,\\
therefore BC = 3,\\
since BD= 3BC,\\
therefore BD= 9,\\
therefore AD=AB+BD=6+9=\boxed{15}


Question: As shown in the figure, point $C$ is a point on segment $AB$ with $AC>BC$. Point $M$ is the midpoint of segment $AB$, and point $N$ is the midpoint of segment $BC$. Given $AM=8$ and $CN=3$, find the length of segment $MN$.

Solution:
\textbf{Solution:} Since point M is the midpoint of line segment AB, $AM=8$,\\
hence $BM=AM=8$,\\
and since point N is the midpoint of BC, $CN=3$,\\
hence $BN=CN=3$,\\
therefore $MC=BM-BN-CN=8-3-3=2$,\\
therefore $MN=MC+CN=2+3=\boxed{5}$.


Question: As shown in the figure, it is known that points $C$, $D$, and $E$ lie on line $AB$, and point $D$ is the midpoint of segment $AC$, point $E$ is the midpoint of segment $DB$. If point $C$ is on segment $EB$ and $DB=6$, $CE=1$, what is the length of segment $AB$?

Solution:
\textbf{Solution:} Since point $E$ is the midpoint of line segment $DB$ and $DB=6$,\\
it follows that $DE=\frac{1}{2}DB=\frac{1}{2}\times 6=3$,\\
since $EC=1$,\\
then $DC=DE+EC=3+1=4$,\\
since point $D$ is the midpoint of line segment $AC$,\\
it follows that $AD=DC=4$,\\
therefore $AB=AD+DB=4+6=\boxed{10}$.


Question: As shown in the figure, point $C$ is on the line segment $AB$, and point $D$ is the midpoint of the line segment $AB$. If $AC=6\,cm$ and $BC=3\,cm$, what is the length of the line segment $CD$ in $cm$?

Solution:
\textbf{Solution:} Since $AC=6\,cm$, $BC=3\,cm$,

it follows that $AB=AC+BC=6+3=9\,cm$,

Since point D is the midpoint of segment AB,

it follows that $AD=\frac{1}{2}AB=\frac{1}{2}\times9=4.5\,cm$,

thus $CD=AC-AD=6-4.5=\boxed{1.5}\,cm$


Question: As shown in the figure, the midpoint of line segment AB, which has a length of 18, is M. Point C divides line segment MB in the ratio MC:CB=1:2. Find the length of line segment AC.

Solution:
\textbf{Solution:} Given that the midpoint of line segment AB, which has a length of 18, is M, \\
$ \therefore AM=BM=\frac{1}{2}AB=9, $ \\
$ \because $ the ratio of MC to CB is 1:2, \\
$ \therefore MC=\frac{1}{3}\times 9=3, $ \\
$ \therefore AC=AM+MC=9+3=\boxed{12}.$


Question: As shown in the figure, the length of segment $AB$ is $10\, \text{cm}$. Point $C$ is a point on segment $AB$ with $BC=3\, \text{cm}$. Points $D$ and $E$ are the midpoints of segments $AC$ and $AB$, respectively. Find the length of segment $DE$.

Solution:
\textbf{Solution:} Given that point $C$ lies on the segment $AB$, with segment $AB$ measuring $10cm$, and $BC=3cm$,\\
therefore, $AC=AB−BC=10−3=7cm$,\\
since point $D$ is the midpoint of $AC$,\\
thus $AD=\frac{1}{2}AC=\frac{1}{2}\times 7=\frac{7}{2}$cm;\\
since point $E$ is the midpoint of $AB$,\\
thus $AE=\frac{1}{2}AB=\frac{1}{2}\times 10=5$cm,\\
hence $DE=AE−AD=5−\frac{7}{2}=\boxed{\frac{3}{2}}$cm.


Question: As shown in the figure, it is known that point C is the midpoint of line segment AB, and point D is the midpoint of line segment BC. Given AB = 16 cm, find the length of line segment AD.

Solution:
Solution: Since $AB=16\,cm$, and $C$ is the midpoint of $AB$, \\
therefore $AC=BC=\frac{1}{2}AB=8\,cm$, \\
Since $D$ is the midpoint of $BC$, \\
therefore $CD=\frac{1}{2}BC=4\,cm$ \\
thus $AD=AC+CD=\boxed{12}\,cm$.


Question: As shown in the figure, point \(C\) is on line segment \(AB\), \(D\) is the midpoint of line segment \(AC\), and \(E\) is the midpoint of line segment \(BC\).

Solution:
\textbf{Solution:}

(1) Since $D$ is the midpoint of segment $AC$, and $AC=8$,\\
it follows that $DC= \frac{1}{2}AC=\frac{1}{2}\times 8=4$\\
Since $E$ is the midpoint of segment $BC$, and $BC=3$,\\
it follows that $CE= \frac{1}{2}BC=\frac{1}{2}\times 3=\frac{3}{2}$,\\
therefore, $DE=DC+CE= 4+\frac{3}{2}=\boxed{5\frac{1}{2}}$;\\
(2) Since point $C$ is on segment $AB$, $D$ is the midpoint of segment $AC$, and $E$ is the midpoint of segment $BC$.\\
it follows that $AC=2DC$, $BC=2CE$,\\
therefore, $AB=AC+BC=2DC+2CE=2(DC+CE)=2DE=2\times 5=\boxed{10}$.


Question: As shown in the figure, the length of line segment $AB$ is $12$, $C$ is a point on line segment $AB$, $AC=4$, $M$ is the midpoint of $AB$, $N$ is the midpoint of $AC$. Find the length of line segment $MN$.

Solution:
\textbf{Solution:} Given: $AM = \frac{1}{2}AB = 6$, $AN = \frac{1}{2}AC = 2$\\
$\therefore MN = AM - AN = \boxed{4}$


Question: As shown in the figure, in $\triangle ABC$, $AD \perp BC$ at $D$, $AE$ bisects $\angle DAC$, $\angle BAC = 80^\circ$, $\angle B = 60^\circ$, what is the measure of $\angle DAE$?

Solution:
Solution: In $\triangle ABC$, since $\angle BAC + \angle B + \angle C = 180^\circ$,\\
then $\angle C = 180^\circ - 80^\circ - 60^\circ = 40^\circ$,\\
since $AD \perp BC$ at D,\\
then $\angle ADC = 90^\circ$,\\
in $\triangle ADC$, $\angle DAC = 90^\circ - \angle C = 90^\circ - 40^\circ = 50^\circ$,\\
since $AE$ bisects $\angle DAC$,\\
then $\angle DAE = \frac{1}{2} \angle DAC = \boxed{25^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $AD$ is the altitude, $AE$ and $BF$ are the angle bisectors, and they intersect at point $O$. Given $\angle ABC=50^\circ$ and $\angle C=70^\circ$, find the measures of $\angle DAE$ and $\angle BOA$.

Solution:
\textbf{Solution:} Since $AD \perp BC$,\\
it follows that $\angle ADC = 90^\circ$,\\
Since $\angle C = 70^\circ$,\\
it follows that $\angle CAD = 180^\circ - 90^\circ - 70^\circ = 20^\circ$,\\
Since $\angle ABC = 50^\circ$, \\
it follows that $\angle BAC = 180^\circ - \angle ABC - \angle C = 60^\circ$,\\
Since AE is the bisector of $\angle BAC$,\\
it follows that $\angle EAC = \angle BAE = 30^\circ$,\\
therefore $\angle EAD = \angle EAC - \angle CAD = 30^\circ - 20^\circ = \boxed{10^\circ}$,\\
Since BF is the bisector of $\angle ABC$,\\
it follows that $\angle ABO = 25^\circ$,\\
therefore $\angle BOA = 180^\circ - \angle BAO - \angle ABO = 180^\circ - 30^\circ - 25^\circ = \boxed{125^\circ}$,\\
Hence, the measures of $\angle DAE$ and $\angle BOA$ are $10^\circ$ and $125^\circ$, respectively.


Question: As shown in the figure, line segment $AC$ intersects line segment $BD$ at point $O$. If $\angle A = 70^\circ$, $\angle B = 30^\circ$, $\angle C = 60^\circ$, what is the degree measure of $\angle D$?

Solution:
\textbf{Solution:} Given that $\angle AOB = \angle COD$,\\
it follows that $\angle A + \angle B = \angle C + \angle D$,\\
given $\angle A = 70^\circ$, $\angle B = 30^\circ$, and $\angle C = 60^\circ$,\\
thus $70^\circ + 30^\circ = 60^\circ + \angle D$,\\
therefore, $\angle D = \boxed{40^\circ}$.


Question: As shown in the figure, $AD$ and $BE$ are the altitude and angle bisector of $\triangle ABC$, respectively, with $\angle BAC=86^\circ$ and $\angle C=58^\circ$. Find the measure of $\angle AOB$?

Solution:
\textbf{Solution:} Given $\angle BAC=86^\circ$, $\angle C=58^\circ$\\
Therefore, $\angle ABC=180^\circ-86^\circ-58^\circ=36^\circ$\\
Since $AD$ and $BE$ are the altitude and angle bisector of $\triangle ABC$ respectively,\\
Therefore, $\angle OBD=\frac{1}{2}\angle ABC=18^\circ$, $\angle ADB=90^\circ$\\
Hence, $\angle AOB=\angle OBD+\angle ADB=18^\circ+90^\circ=\boxed{108^\circ}$


Question: As shown in the figure, Xiaoying and her classmate are swinging. When the swing AB is at rest, the lower end B' is 0.6m above the ground. When the swing moves to the position AB, the horizontal distance from the lower end B to the rest position EB is 2.4m, and it is 1.4m above the ground. Find the length of the swing AB.

Solution:
\textbf{Solution:} Let the length of AB=AB$^{\prime }$ be $x$ meters. According to the problem, we can deduce that B$^{\prime }$E = 1.4 - 0.6 = 0.8 (m),\\
hence, AE = AB - 0.8,\\
in $\triangle AEB$, since AE$^{2}$ + BE$^{2}$ = AB$^{2}$,\\
therefore, (x-0.8)$^{2}$ + 2.4$^{2}$ = x$^{2}$\\
Solving this, we find x = \boxed{4},\\
hence, the length of the swing AB is 4m.


Question: As shown in the diagram, in $\triangle ABC$, $AB=AC$, $AD$ bisects $\angle BAC$. Given $BC=10$, $AD=12$, find the length of $AC$.

Solution:
\textbf{Solution:} Since $AB=AC$ and $AD$ bisects $\angle BAC$,\\
it follows that $AD\perp BC$, $BD=CD=\frac{1}{2}BC=5$,\\
Given that $AD=12$,\\
we have $AC=\sqrt{AD^{2}+CD^{2}}=\sqrt{12^{2}+5^{2}}=\boxed{13}$


Question: As shown in the figure, $AE$ and $AD$ are respectively the altitude and the angle bisector of $\triangle ABC$, and $\angle B=40^\circ$, $\angle C=60^\circ$, find the degree measure of $\angle DAE$.

Solution:
\textbf{Solution:} Since $\angle B = 40^\circ$ and $\angle C = 60^\circ$, \\
it follows that $\angle BAC = 180^\circ - \angle B - \angle C = 80^\circ$.\\
Since $AD$ is the angle bisector of $\triangle ABC$, \\
it follows that $\angle BAD = \angle DAC = \frac{1}{2} \angle BAC = 40^\circ$, \\
therefore $\angle ADE = \angle B + \angle BAD = 80^\circ$, \\
therefore $\angle DAE = 90^\circ - \angle ADE = 90^\circ - 80^\circ = \boxed{10^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, D is the midpoint of side $AB$, E is a point on side $AC$, and line $BF$ is drawn parallel to $AC$ through point B and intersects the extension of $ED$ at point F. If $AD=6$ and $BF=9$, find the length of $CE$.

Solution:
\textbf{Solution:} Given that $BF\parallel AC$, \\
it follows that $\angle F=\angle AED$, \\
since $D$ is the midpoint of $AB$, \\
it implies $AD=BD$, \\
considering $\triangle ADE$ and $\triangle BDF$, \\
we have
$\left\{\begin{array}{l}
\angle AED=\angle F \\
\angle ADE=\angle BDF \\
AD=BD
\end{array}\right.$, \\
thus, $\triangle ADE\cong \triangle BDF$ (by AAS), \\
which means $AE=BF=9$, \\
given $AB=AC$, \\
it results in $AC=2AD=12$, \\
therefore, $CE=AC-AE=12-9=\boxed{3}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=45^\circ$, $AB=6\sqrt{2}$, $AC=8$, $BC>6$. Points $E$ and $F$ are located on sides $BC$ and $AC$, respectively, and $AF=CE$. What is the minimum value of $AE+BF$?

Solution:
\textbf{Solution:} Construct line $AG \parallel BC$ through point $A$ such that $AG=AC$, and connect $FG$, $BG$. \\
Since $AG \parallel BC$, \\
then $\angle GAF=\angle ACE$. \\
Since $AF=CE$, $AG=AC$, \\
then $\triangle ACE$ is congruent to $\triangle GAF$, \\
thus $AE=GF$, \\
hence $AE+BF=GF+BF$. \\
When points $B$, $F$, $G$ are collinear, $BF+GF$ is minimized. \\
Construct $AI \perp BC$ through point $A$, intersecting $BC$ at point $I$, and construct $GH \perp BC$ through point $G$, intersecting the extended line of $BC$ at point $H$. \\
Since $\angle ABC=45^\circ$, \\
then $AI=BI=6$. \\
It is easy to see that quadrilateral $AIHG$ is a rectangle, thus $GH=AI=6$, $IH=AG=8$, \\
therefore $BH=BI+IH=6+8=14$. \\
In $\triangle BGH$, $BG= \sqrt{BH^2+GH^2}=\sqrt{14^2+6^2}=2\sqrt{58}$. \\
Thus, the minimum value of $AE+BF$ is $\boxed{2\sqrt{58}}$.


Question: As shown in the figure, in $ \triangle ABC$, $CD$ is the altitude of $ \triangle ABC$, $AE$ is the angle bisector of $ \triangle ABC$, and $CD$ intersects $AE$ at point $G$. Given $\angle BCD = 50^\circ$ and $\angle BEA = 110^\circ$, find the measure of $\angle ACD$.

Solution:
\textbf{Solution:} By the given information, $CD$ is the height of $\triangle ABC$,\\
hence $\angle BDC=\angle ADC=90^\circ$.\\
Therefore, $\angle B=90^\circ-\angle BCD=90^\circ-50^\circ=40^\circ$.\\
In $\triangle ABE$, $\angle BAE=180^\circ-\angle B-\angle BEA=180^\circ-40^\circ-110^\circ=30^\circ$.\\
Since $AE$ is the angle bisector of $\triangle ABC$,\\
it follows that $\angle BAC=2\angle BAE=60^\circ$.\\
Therefore, $\angle ACD=90^\circ-\angle BAC=\boxed{30^\circ}$.


Question: As shown in the figure, it is known that the perimeter of $\triangle ACD$ is $14$, $AB-AC=2$, and the perpendicular bisector of $BC$ intersects $AB$ at point $D$ and intersects $BC$ at point $E$. Find the lengths of $AB$ and $AC$.

Solution:
\textbf{Solution}: Since the perimeter of $\triangle ACD$ is 14,\\
$\therefore AC+AD+CD=14$,\\
because $DE$ is the perpendicular bisector of $BC$,\\
$\therefore BD=CD$,\\
$\therefore AC+AD+BD=14$,\\
$\therefore AC+AB=14$,\\
since $AB-AC=2$,\\
$\therefore AB=\boxed{8}$, $AC=\boxed{6}$.


Question: As shown in the figure, $AE \perp BD$, $CD \perp BD$, $AB = BC$, $BE = CD$. What is the measure of $\angle ABC$ in degrees?

Solution:
\textbf{Solution:} Given $\because AE\perp BD$, $CD\perp BD$,\\
$\therefore \angle AEB=\angle BDC=90^\circ$,\\
In $Rt\triangle AEB$ and $Rt\triangle BDC$\\
$\because \left\{\begin{array}{l}AB=BC\\ BE=CD\end{array}\right.$,\\
$\therefore Rt\triangle AEB\cong Rt\triangle BDC$,\\
$\therefore \angle A=\angle CBD$,\\
$\because \angle AEB=90^\circ$,\\
$\therefore \angle A+\angle ABE=90^\circ$,\\
$\therefore \angle ABC=\angle ABE+\angle CBD=\angle ABE+\angle A=90^\circ$.\\
Hence, the measure of $\angle ABC$ is $\boxed{90^\circ}$.


Question: As shown in the figure, within $\triangle ABC$, $AD$ is the angle bisector of $\angle BAC$, $BE$ is the altitude from $B$ to side $AC$, and $AD$ and $BE$ intersect at point $O$. If $\angle AOE=70^\circ$, what is the measure of $\angle ABE$ in degrees?

Solution:
\textbf{Solution:} Since $BE$ is the altitude on side $AC$,\\
it follows that $\angle AEB=90^\circ$.\\
Therefore, $\angle OAE=90^\circ-\angle AOE=20^\circ$.\\
Since $AD$ is the angle bisector of $\angle BAC$,\\
it follows that $\angle BAC=2\angle OAE=40^\circ$.\\
Therefore, $\angle ABE=90^\circ-\angle BAC=\boxed{50^\circ}$.


Question: As shown in the figure, it is known that in the right-angled $\triangle ABC$, $\angle ACB=90^\circ$, $AC=9$, $BC=12$, the perpendicular bisector of $AB$ intersects $AB$ at point $D$ and intersects $BC$ at point $E$. Connect $AE$. Find the length of $BE$.

Solution:
\textbf{Solution:} In right-angled $\triangle ABC$, by the Pythagorean theorem, we have

\[AB = \sqrt{AC^{2} + BC^{2}} = \sqrt{9^{2} + 12^{2}} = 15,\]

since $DE$ is the perpendicular bisector of $AB$,

thus $AE = BE$.

Let $BE = AE = x$, then $CE = 12 - x$.

In right-angled $\triangle ACE$, by the Pythagorean theorem, we get

\[AE^{2} = AC^{2} + CE^{2},\]

which implies

\[x^{2} = 9^{2} + (12-x)^{2},\]

solving this, we find $x = \boxed{\frac{75}{8}}$,

thus, the length of $BE$ is $\frac{75}{8}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=100^\circ$, $AD\perp BC$ at point D, and $AE$ bisects $\angle BAC$ and intersects $BC$ at point E. If $\angle C=25^\circ$, find the degree measure of $\angle DAE$.

Solution:
\textbf{Solution:} Since AE bisects $\angle BAC$, \\
it follows that $\angle EAC = \frac{1}{2}\angle BAC = \frac{1}{2}\times 100^\circ = 50^\circ$, \\
Since $\angle C = 25^\circ$, \\
it follows that $\angle AED = \angle C + \angle EAC = 25^\circ + 50^\circ = 75^\circ$, \\
Since AD is perpendicular to BC, \\
it follows that $\angle ADE = 90^\circ$, \\
therefore, $\angle DAE = 90^\circ - 75^\circ = \boxed{15^\circ}$.


Question: As shown in the figure, in triangle $ABC$, $AB=AC$, $\angle C=25^\circ$, point $D$ is on the extension line of segment $CA$, and $DA=AC$. What is the measure of $\angle ABD$ in degrees?

Solution:
\textbf{Solution:} Since AB = AC, and $\angle C = 25^\circ$, \\
hence $\angle ABC = 25^\circ$, \\
therefore $\angle BAD = \angle C + \angle ABC = 50^\circ$, \\
since DA = AC, \\
hence AB = AD, \\
therefore $\angle ABD =  \frac{1}{2} (180^\circ - \angle BAD) = \boxed{65^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ is the altitude on side $BC$, and $CE$ is an angle bisector that intersects with $AD$ at point $P$. Given that $\angle APE=55^\circ$ and $\angle AEP=80^\circ$, find the measure of $\angle BAC$.

Solution:
\textbf{Solution:} Since $AD\perp BC$,\\
it follows that $\angle PDC=90^\circ$,\\
Given $\angle CPD=\angle APE=55^\circ$,\\
it implies $\angle PCD=90^\circ-55^\circ=35^\circ$,\\
Since $CE$ is an angle bisector,\\
it follows $\angle ACE=\angle PCD=35^\circ$,\\
Hence, $\angle BAC=180^\circ-35^\circ-80^\circ=\boxed{65^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, point D is the midpoint of $BC$, point E is on $AC$, $AD=AE$, if $\angle BAD=50^\circ$, what is the degree measure of $\angle CED$?

Solution:
\textbf{Solution:} Since $AB=AC$ and point D is the midpoint of $BC$, \\
it follows that triangle $ABC$ is an isosceles triangle, making $AD$ a median of $BC$ and the bisector of $\angle BAC$, hence $\angle CAD=\angle BAD=50^\circ$, \\
Since $AD=AE$, \\
it follows that $\angle ADE=\angle AED$, \\
Since $\angle ADE+\angle AED+\angle CAD=180^\circ$, \\
it follows that $\angle ADE=\angle AED=\frac{1}{2}(180^\circ-50^\circ)=65^\circ$, \\
Therefore, $\angle CED=180^\circ-65^\circ=\boxed{115^\circ}$.


Question: As shown in the figure, $\angle B = \angle C = 90^\circ$, $M$ is the midpoint of $BC$, $DM$ bisects $\angle ADC$, and $\angle ADC = 110^\circ$, find the degree measure of $\angle MAB$?

Solution:
\textbf{Solution:} As shown in the figure, draw $MN \perp AD$ at N,\\
$\because$ $\angle B = \angle C = 90^\circ$,\\
$\therefore$ $AB \parallel CD$,\\
$\therefore$ $\angle DAB = 180^\circ - \angle ADC = 70^\circ$,\\
$\because$ $DM$ bisects $\angle ADC$, $MN \perp AD$, $MC \perp CD$,\\
$\therefore$ $\angle DMN = \angle DMC$, $\angle DNM = \angle DCM = 90^\circ$,\\
In $\triangle DMN$ and $\triangle DMC$,\\
$\left\{\begin{array}{l}
\angle DMN = \angle DMC \\
\angle DNM = \angle DCM \\
DM = DM
\end{array}\right.$\\
$\therefore$ $\triangle DMN \cong \triangle DMC$ (AAS),\\
$\therefore$ $MN = MC$.\\
$\because$ M is the midpoint of $BC$,\\
$\therefore$ $MC = MB$,\\
$\therefore$ $MN = MB$,\\
And since $MN \perp AD$, $MB \perp AB$,\\
$\therefore$ $\angle MAB = \frac{1}{2}\angle DAB = \boxed{35^\circ}$.


Question: As shown in the diagram, it is known that $\triangle ABC \cong \triangle DBE$, with point D on $AC$, and $BC$ intersects $DE$ at point P. If $\angle ABE=160^\circ$ and $\angle DBC=30^\circ$, what is the degree measure of $\angle CBE$?

Solution:
\textbf{Solution:} Given that $\triangle ABC \cong \triangle DBE$,\\
it follows that $\angle ABC = \angle DBE$, that is $\angle ABD + \angle DBC = \angle DBC + \angle CBE$,\\
therefore $\angle ABD = \angle CBE$,\\
since $\angle ABE = 160^\circ$ and $\angle DBC = 30^\circ$,\\
it follows that $\angle ABD + \angle DBC + \angle CBE = \angle ABE = 160^\circ$,\\
therefore $\angle ABD = \angle CBE = \frac{1}{2}(\angle ABE - \angle DBC) = \frac{1}{2}(160^\circ - 30^\circ) = \boxed{65^\circ}$.


Question: As shown in the figure, in $ \triangle ABC$, point $D$ is on side $BC$, $BD=AD=AC$, $E$ is the midpoint of $CD$, and $ \angle CAE=20^\circ$. What is the measure of $ \angle BAC$?

Solution:
\textbf{Solution:} Because $AD=AC$, and point E is the midpoint of $CD$,
\[
\therefore AE\perp CD,
\]
\[
\therefore \angle AEC=90^\circ,
\]
Because $\angle CAE=20^\circ$,
\[
\therefore \angle C=90^\circ-\angle CAE=70^\circ, \quad \angle DAC=2\angle CAE=40^\circ,
\]
Because $AD=AC$,
\[
\therefore \angle ADC=\angle C=70^\circ,
\]
Because $AD=BD$,
\[
\therefore \angle B=\angle BAD,
\]
Because $\angle B+\angle BAD=\angle ADC=70^\circ$,
\[
\therefore \angle B=\angle BAD=\frac{1}{2}\angle ADC=35^\circ,
\]
\[
\therefore \angle BAC=\angle DAC+\angle BAD=40^\circ+35^\circ=\boxed{75^\circ}.
\]


Question: As shown in the figure, $AD$ is the altitude of $\triangle ABC$ on side $BC$, $AE$ bisects $\angle BAC$. Given that $\angle B=40^\circ$, $\angle C=72^\circ$, find the measures of $\angle AEC$ and $\angle DAE$.

Solution:
\textbf{Solution:} Given $\because \angle BAC + \angle B + \angle C = 180^\circ$, $\angle B = 40^\circ$, $\angle C = 72^\circ$,\\
$\therefore \angle BAC = 68^\circ$,\\
$\because$ AE bisects $\angle BAC$,\\
$\therefore \angle BAE = \angle CAE = \frac{1}{2} \angle BAC = 34^\circ$,\\
$\therefore \angle AEC = \angle B + \angle BAE = 74^\circ$,\\
$\because$ AD$\perp$BC,\\
$\therefore \angle ADE = 90^\circ$,\\
$\therefore \angle DAE = 90^\circ - \angle AEC = \boxed{16^\circ}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $AD$ bisects $\angle BAC$ and intersects $BC$ at $D$, $AE \perp BC$ at $E$. If $\angle ADE = 80^\circ$, $\angle EAC = 20^\circ$, what is the degree measure of $\angle B$?

Solution:
\textbf{Solution:} Since $AE \perp BC$ and $\angle EAC = 20^\circ$,\\
it follows that $\angle C = 70^\circ$,\\
hence $\angle BAC + \angle B = 110^\circ$.\\
Since $\angle ADE = \angle B + \angle BAD = \frac{1}{2}(\angle BAC + \angle B) + \frac{1}{2}\angle B$,\\
it follows that $\angle B = \boxed{50^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $DE\parallel BC$ and $AD: DB = 2: 1$. The area of $\triangle ABC$ is $27$. What is the area of $\triangle ADE$?

Solution:
\textbf{Solution:} Since $AD:DB = 2:1$, \\
it follows that $\frac{AD}{AB} = \frac{2}{3}$. \\
Since $DE \parallel BC$, \\
it implies $\triangle ADE \sim \triangle ABC$. \\
Therefore, $\frac{S_{\triangle ADE}}{S_{\triangle ABC}} = \left( \frac{AD}{AB} \right)^2$, that is $\frac{S_{\triangle ADE}}{27} = \frac{4}{9}$, \\
Solving this we get: $S_{\triangle ADE} = \boxed{12}$, \\
Thus, the area of $\triangle ADE$ is $12$.


Question: As shown in the figure, AD and BC intersect at point P. AC and BD are connected, and $\angle 1 = \angle 2$, $AC=3$, $CP=2$, $DP=1$. Find the length of BD.

Solution:
Solution: Since $\angle 1 = \angle 2$ and $\angle APC = \angle BPD$,\\
$\therefore \triangle APC \sim \triangle BPD$,\\
$\therefore \frac{AC}{BD} = \frac{CP}{DP}$,\\
$BD = \frac{DP \cdot AC}{CP} = \frac{1 \times 3}{2} = \boxed{1.5}$.


Question: As shown in the figure, $D$ and $E$ are points on the sides $AB$ and $AC$ of $\triangle ABC$ respectively, with $\angle AED=\angle B$, $AD=3$, $AB=8$, and $AE=4$. What is the length of $AC$?

Solution:
\textbf{Solution:} Given that $\angle AED = \angle B$, and $\angle A = \angle A$,\\
$\therefore \triangle ADE \sim \triangle ACB$,\\
$\therefore \frac{AD}{AC} = \frac{AE}{AB}$,\\
Given that $AD = 3$, $AB = 8$, $AE = 4$,\\
$\therefore \frac{3}{AC} = \frac{4}{8}$,\\
$\therefore AC = \boxed{6}$.


Question: As shown in the figure, in $\odot O$, $CP=2$, $PD=6$, $AP=5$, and chord $CD\perp AB$ with the foot of the perpendicular at point $P$. Find the length of $OP$.

Solution:
\textbf{Solution:} Draw $OE\perp CD$ at point E, and $OF\perp AB$ at point F, connect $OA$ and $OD$,\\
since $CD\perp AB$,\\
therefore, quadrilateral $OEPF$ is a rectangle,\\
thus $PF=OE$, $OF=PE$,\\
given $CP=2$, $PD=6$,\\
thus $CD=CP+DP=8$,\\
since $CD\perp OE$,\\
thus $DE=\frac{1}{2}CD=4$,\\
thus $OF=PE=PD-DE=2$,\\
let $OE=x$, then $PF=x$, $AF=5-x$,\\
in $\triangle AOF$, $AF^2+OF^2=OA^2$,\\
in $\triangle DOE$, $OE^2+DE^2=OD^2$,\\
since $OA=OD$,\\
thus $AF^2+OF^2=OE^2+DE^2$, i.e., $(5-x)^2+2^2=x^2+4^2$,\\
solving for $x$ gives $x=\frac{13}{10}$,\\
in $\triangle POE$, $OP=\sqrt{OE^2+PE^2}=\sqrt{\left(\frac{13}{10}\right)^2+2^2}=\boxed{\frac{\sqrt{569}}{10}}$.


Question: As shown in the figure, the line $y=-\frac{1}{2}x+2$ intersects the $x$-axis and the $y$-axis at points $B$ and $A$, respectively. Let $Q$ be a variable point (not coinciding with $A$ or $B$) on segment $AB$. When point $Q$ is rotated clockwise by $90^\circ$ about point $P(1, 0)$, point $Q'$ is obtained. What is the minimum value of $OQ'$ when connecting $OQ'$?

Solution:
\textbf{Solution}: Construct $QM \perp x$-axis at point $M$, $Q'N \perp x$-axis at $N$,\\
$\because$ $\angle PMQ=\angle PNQ'=\angle QPQ'=90^\circ$,\\
$\therefore$ $\angle QPM+\angle NPQ'=\angle PQ'N+\angle NPQ'$,\\
$\therefore$ $\angle QPM=\angle PQ'N$,\\
In $\triangle PQM$ and $\triangle Q'PN$,\\
$\left\{\begin{array}{l}
\angle PMQ=\angle PNQ' \\
\angle QPM=\angle PQ'N \\
PQ=PQ'
\end{array}\right.$,\\
$\therefore$ $\triangle PQM\cong \triangle Q'PN$ (AAS),\\
$\therefore$ $PN=QM$, $Q'N=PM$,\\
Let $Q\left(m, -\frac{1}{2}m+2\right)$,\\
$\therefore$ $PM=\left|m-1\right|$, $QM=\left|-\frac{1}{2}m+2\right|$,\\
$\therefore$ $ON=\left|3-\frac{1}{2}m\right|$,\\
$\therefore$ $Q'\left(3-\frac{1}{2}m, 1-m\right)$,\\
$\therefore$ $OQ'^2=\left(3-\frac{1}{2}m\right)^2+\left(1-m\right)^2=\frac{5}{4}m^2-5m+10=\frac{5}{4}\left(m-2\right)^2+5$,\\
When $m=2$, $OQ'^2$ reaches its minimum value of 5,\\
$\therefore$ The minimum value of $OQ'$ is $\boxed{\sqrt{5}}$.


Question: As shown in the figure, in $\square ABCD$, $\triangle ACE$ is constructed with $AC$ as the hypotenuse, and $\angle BED$ is a right angle. Prove that: $\square ABCD$ is a rectangle.

Solution:
\textbf{Solution.} Proof: Connect OE\\
$\because$ quadrilateral ABCD is a parallelogram,\\
$\therefore$ AO=CO, BO=DO.\\
$\because$ $\angle$AEC=$\angle$BED=90$^\circ$,\\
$\therefore$ AC=2OE, BD=2OE,\\
$\therefore$ AC=BD,\\
$\therefore$ $\square ABCD$ is a \boxed{rectangle}.


Question: As shown in the figure, within the square paper ABCD, E is the midpoint of CD. The square paper is folded so that point B falls on point G on segment AE, with the crease being AF. If AD = 4 cm, what is the length of CF in centimeters?

Solution:
\textbf{Solution:}

Given square ABCD, where point E is the midpoint of CD,\\
thus $\angle B = \angle D = 90^\circ$, AD = CD = 4, DE = CE = $\frac{1}{2}$CD = 2,\\
in $\triangle ADE$,\\
$AE = \sqrt{AD^2 + DE^2} = \sqrt{4^2 + 2^2} = 2\sqrt{5}$,\\
because folding the square paper such that point B falls on point G on segment AE, with the fold line being AF,\\
thus DE = EF = 2, AG = AB = 4, BF = FG,\\
thus GE = $2\sqrt{5} - 4$, let BF = x, then CF = 4 - x,\\
in $\triangle GEF$, $EF^2 = GE^2 + GF^2 = (2\sqrt{5} - 4)^2 + x^2$,\\
thus in $\triangle CEF$, $EF^2 = CE^2 + CF^2$,\\
thus $(2\sqrt{5} - 4)^2 + x^2 = (4 - x)^2 + 2^2$, solving it:\\
$x = 2\sqrt{5} - 2$,\\
thus $FC = 4 - (2\sqrt{5} - 2) = \boxed{6 - 2\sqrt{5}}$.


Question: As shown in the figure, $\angle ABC=\angle FAC=90^\circ$, the length of $BC$ is $3\text{cm}$, the length of $AB$ is $4\text{cm}$, and the length of $AF$ is $12\text{cm}$. What is the area of the square $CDEF$ in $\text{cm}^2$?

Solution:
\textbf{Solution:} Given in right triangle $\triangle ABC$, $AC= \sqrt{AB^{2}+BC^{2}}=\sqrt{3^{2}+4^{2}}=5\,\text{cm}$,
in right triangle $\triangle ACF$, $CF= \sqrt{AF^{2}+AC^{2}}=\sqrt{12^{2}+5^{2}}=13\,\text{cm}$,
$\therefore$ the area of square $CDEF$ is $13\times13=\boxed{169}\,\text{cm}^{2}$


Question: As shown in the figure, $\angle ABC=\angle FAC=90^\circ$, the length of $BC$ is 3m, the length of $AB$ is 4cm, and the length of $AF$ is 12cm. Calculate the area of the square $CDEF$.

Solution:
Solution: In right triangle $ABC$,\\
$\because \angle ABC=90^\circ, BC=3\text{cm}, AB=4\text{cm}$\\
$\therefore AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{3^{2}+4^{2}}=5\text{cm}$\\
In right triangle $ACF$, $\because \angle FAC=90^\circ, AF=12\text{cm}$\\
$\therefore CF=\sqrt{AF^{2}+AC^{2}}=\sqrt{12^{2}+5^{2}}=13\text{cm}$\\
$\therefore$ The area of the square $CDEF$ $=13\times 13=\boxed{169}\text{cm}^{2}$.


Question: As shown in the figure, $BD$ is a diagonal of rectangle $ABCD$. Extend $BC$ to point $E$ such that $CE=BD$. Connect $AE$. Given $AB=1$ and $\angle AEB=15^\circ$, find the length of $AD$.

Solution: \textbf{Solution:} As shown in the figure, connect AC and intersect BD at point O. Since quadrilateral ABCD is a rectangle, it follows that AD$\parallel$BE and AC=BD, thus $OA=\frac{1}{2}AC$ and $OD=\frac{1}{2}BD$, therefore $OA=OD$. Consequently, $\angle ADB=\angle CAD$ and $\angle E=\angle DAE$. Also, since $BD=CE$, it follows that $CE=CA$ and thus $\angle E=\angle CAE$. Since $\angle CAD=\angle CAE+\angle DAE=30^\circ$, it follows that $\angle ADB=30^\circ$ and hence $BD=2AB=2$, leading to $AD=\sqrt{BD^2−AB^2}=\boxed{\sqrt{3}}.$


Question: As shown in the figure, it is known that points $B$ and $C$ divide the line segment $AD$ into three parts in the ratio $2:5:3$. Let $M$ be the midpoint of $AD$, and $BM=6$. Find the length of $AD$.

Solution:
\textbf{Solution:} Since points B and C divide line segment AD into three parts in the ratio 2:5:3,\\
it follows that AB=$\frac{1}{5}$AD, and CD=$\frac{3}{10}$AD.\\
Since M is the midpoint of AD,\\
we have AM=$\frac{1}{2}$AD.\\
Since BM=AM−AB,\\
we find that $\frac{1}{2}$AD−$\frac{1}{5}$AD=6.\\
Solving this gives: AD=\boxed{20}.


Question: As shown in the figure, given two lines $AB$ and $CD$ intersect at point $O$, $OE$ bisects $\angle BOD$, and $OF$ bisects $\angle COB$. If $\angle BOE = 36^\circ$, what is the degree measure of $\angle AOF$?

Solution:
\textbf{Solution:} Given that $\angle DOE + \angle EOB + \angle BOF + \angle FOC = 180^\circ$, and that $OE$ bisects $\angle BOD$ and $OF$ bisects $\angle COB$, \\
thus $2\angle EOB + 2\angle BOF = 180^\circ$, \\
hence, $\angle BOE + \angle BOF = 90^\circ$, \\
since $\angle BOE = 36^\circ$, \\
therefore $\angle BOF = 90^\circ - 36^\circ = 54^\circ$, \\
given that $\angle AOF + \angle BOF = 180^\circ$, \\
thus $\angle AOF = 180^\circ - 54^\circ = \boxed{126^\circ}$.


Question: As shown in the figure, point $O$ lies on line $AB$, $OE$ bisects $\angle AOC$, $\angle DOC=90^\circ$, and $\angle EOC=65^\circ$. Find the measure of $\angle BOD$.

Solution:
\textbf{Solution:} Since $OE$ bisects $\angle AOC$,\\
we have $\angle AOC=2\angle EOC=2\times 65^\circ=130^\circ$,\\
thus $\angle BOC=180^\circ-\angle AOC=180^\circ-130^\circ=50^\circ$,\\
given that $\angle DOC=90^\circ$,\\
it follows $\angle BOD=\angle DOC-\angle BOC=90^\circ-50^\circ=\boxed{40^\circ}$.


Question: As shown in the figure, points $B$ and $C$ divide the line segment $AD$ into three parts in the ratio $2:3:4$. $M$ and $N$ are the midpoints of $AD$ and $AB$, respectively. Given that $CD=8\,\text{cm}$, find the length of $MN$.

Solution:
\textbf{Solution:} Given that points B and C sequentially divide segment AD into three parts in the ratio 2:3:4,\\
let AB = 2x, BC = 3x, CD = 4x, and AD = 2x + 3x + 4x = 9x.\\
Since CD = 8cm,\\
then 4x = 8,\\
thus x = 2,\\
hence AD = 9x = 18, and AB = 4.\\
Since M is the midpoint of AD, and N is the midpoint of CD,\\
then AM = $\frac{1}{2}$AD = 9, and AN = $\frac{1}{2}$AB = 2,\\
therefore MN = 9 - 2 = \boxed{7}.


Question: As shown in the figure, $C$ is a point on line segment $AB$, and $P$, $Q$ are the midpoints of line segments $AC$ and $BC$ respectively. If $PQ = 12$, find the length of $AB$.

Solution:
\textbf{Solution:} Since $P$ and $Q$ are the midpoints of line segments $AC$ and $BC$ respectively,\\
it follows that $AP=PC=\frac{1}{2}AC$ and $CQ=QB=\frac{1}{2}CB$,\\
since $PQ=PC+CQ=\frac{1}{2}AC+\frac{1}{2}CB=\frac{1}{2}AB=12$,\\
thus $AB=\boxed{24}$.


Question: As shown in the figure, the line AB intersects with line MN at point O, with OC $\perp$ AB, and OA bisects $\angle MOD$. If $\angle BON = 25^\circ$, what is the degree measure of $\angle COD$?

Solution:
\textbf{Solution:} Given $\angle BON = 25^\circ$, \\
then $\angle AOM = 25^\circ$, \\
since OA bisects $\angle MOD$, \\
then $\angle AOD = \angle MOA = 25^\circ$, \\
since OC $\perp$ AB, \\
then $\angle AOC = 90^\circ$, \\
therefore $\angle COD = 90^\circ - 25^\circ = \boxed{65^\circ}$.


Question: As shown in the figure, the lines $AB$ and $CD$ intersect at point O, and $OE \perp CD$ at O. $OD$ bisects $\angle AOF$, and $\angle AOE = 55^\circ$. What is the degree measure of $\angle BOF$?

Solution:
\textbf{Solution:} According to the construction,\\
since $OE \perp CD$ and $\angle AOE = 55^\circ$,\\
it follows that $\angle AOD = 90^\circ - \angle AOE = 35^\circ$,\\
since $OD$ bisects $\angle AOF$,\\
it yields $\angle DOF = \angle AOD = 35^\circ$,\\
since lines $AB$ and $CD$ intersect at point O,\\
thus $\angle BOC = \angle AOD = 35^\circ$,\\
therefore $\angle BOF = 180^\circ - \angle BOC - \angle DOF = \boxed{110^\circ}$.


Question: As shown in the figure, it is known that $\angle AOB = 90^{\circ}$, $\angle EOF = 60^{\circ}$, $OE$ bisects $\angle AOB$, and $OF$ bisects $\angle BOC$. What are the measures of $\angle AOC$ and $\angle COB$?

Solution:
\textbf{Solution:} Since OE bisects $\angle AOB$ and $\angle AOB=90^\circ$,
\[
\therefore \angle BOE = \angle AOB = 45^\circ,
\]
Also, since $\angle EOF=60^\circ$,
\[
\therefore \angle BOF = \angle EOF - \angle BOE= 15^\circ,
\]
And since OF bisects $\angle BOC$,
\[
\therefore \angle BOC=2\angle BOF=30^\circ,
\]
Thus, $\angle AOC=\angle AOB + \angle BOC=120^\circ$,
Hence, $\angle AOC=\boxed{120^\circ}$, $\angle COB=\boxed{30^\circ}$.


Question: As shown in the diagram, $ \angle AOC=90^\circ$, $OD$ bisects $ \angle AOB$, $ \angle COD=35^\circ$, what is the degree measure of $ \angle BOC$?

Solution:
\textbf{Solution:} Because $OD$ bisects $\angle AOB$, \\
therefore $\angle AOD = \angle BOD = \frac{1}{2} \angle AOB$, \\
because $\angle AOC = 90^\circ$, $\angle COD = 35^\circ$, \\
therefore $\angle BOD = \angle AOD = 90^\circ - 35^\circ = 55^\circ$, \\
therefore $\angle BOC = \angle BOD - \angle COD = 55^\circ - 35^\circ = \boxed{20^\circ}$.


Question: As shown in the figure, the point $O$ is a point on the line $AB$, $OD$ bisects $\angle BOC$, $\angle COE=90^\circ$, if $\angle AOC=40^\circ$, what is the measure of $\angle DOE$ in degrees?

Solution:
\textbf{Solution:} Given that $O$ is a point on the line $AB$ and $\angle AOC=40^\circ$, \\
it follows that $\angle BOC=180^\circ-\angle AOC=140^\circ$.\\
Since $OD$ bisects $\angle BOC$, \\
we have $\angle COD=\frac{1}{2}\angle BOC=70^\circ$.\\
Given $\angle COE=90^\circ$, \\
it follows that $\angle DOE=\angle COE-\angle COD=\boxed{20^\circ}$.


Question: As shown in the figure, the lines $AB$ and $CD$ intersect at point $O$, with $OE$ bisecting $\angle AOD$ and $\angle FOC = 90^\circ$, $\angle 1 = 40^\circ$. Find the measure of $\angle 2$.

Solution:
\textbf{Solution:} Given that $ \because \angle FOC=90^\circ$ and $ \angle 1=40^\circ$,\\
$ \therefore \angle AOC=180^\circ-\angle FOC-\angle 1=180^\circ-90^\circ-40^\circ=50^\circ$,\\
$ \therefore \angle AOD=180^\circ-\angle AOC=180^\circ-50^\circ=130^\circ$,\\
$ \because OE$ bisects $\angle AOD$,\\
$ \therefore \angle 2=\frac{1}{2}\angle AOD=\frac{1}{2}\times 130^\circ=\boxed{65^\circ}$.


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

Solution:
\textbf{Solution:} Let $AB=2x\,cm$, $BC=5x\,cm$, $CD=3x\,cm$,\\
thus, $AD=AB+BC+CD=10x\,cm$\\
Given that $M$ is the midpoint of $AD$,\\
thus, $AM=MD=\frac{1}{2}AD=\frac{1}{2}\times 10x=5x$, and $AB=2x\,cm$,\\
thus, $BM=AM−AB=5x−2x=3x\,cm$,\\
Given that $BM=6\,cm$,\\
thus, $3x=6$, $x=2$,\\
Given that $CM=MD−CD$,\\
thus, $CM=5x−3x=2x=2\times 2=\boxed{4\,cm}$.


Question: As shown in the figure, point C is a point on the line segment AB, with $AC=12\, \text{cm}$ and $AC=\frac{3}{2}CB$. D and E are the midpoints of AC and AB, respectively. Find the length of DE.

Solution:
\textbf{Solution:} Given that $AC=12\,cm$ and $AC=\frac{3}{2}CB$,\\
it follows that $CB=\frac{2}{3}AC=\frac{2}{3}\times 12=8\,cm$,\\
thus $AB=AC+CB=12+8=20\,cm$,\\
since $D$ and $E$ are the midpoints of $AC$ and $AB$ respectively,\\
we have $AD=\frac{1}{2}AC=6\,cm$ and $AE=\frac{1}{2}AB=10\,cm$\\
therefore $DE=AE-AD=10-6=\boxed{4\,cm}$.


Question: As shown in the diagram, it is known that $ \angle AOC : \angle BOC = 1 : 5$, $OD$ bisects $\angle AOB$, and $\angle COD = 36^\circ$. What is the degree measure of $\angle AOB$?

Solution:
\textbf{Solution:} Let $\angle AOC=x$, then $\angle BOC=5x$,\\
since $OD$ bisects $\angle AOB$,\\
thus $\angle AOD=\frac{1}{2}\angle AOB=\frac{1}{2}(x+5x)=3x$,\\
also, since $\angle COD=\angle AOD-\angle AOC$,\\
thus $3x-x=36^\circ$,\\
solving this, we get $x=18^\circ$,\\
therefore $\angle AOB=\angle AOC+\angle BOC=x+5x=\boxed{108^\circ}$.


Question: As shown in the figure, the ratio of the measures of $\angle AOC$ to $\angle BOC$ is $5\colon 2$. $OD$ bisects $\angle AOB$. If $\angle COD=15^\circ$, what is the measure of $\angle AOB$?

Solution:
\textbf{Solution:} Given that the ratio of the measures of $\angle AOC$ to $\angle BOC$ is $5\colon 2$,\\
let $\angle AOC=5x$ and $\angle BOC=2x$, then $\angle AOB=7x$,\\
since OD bisects $\angle AOB$,\\
it follows that $\angle BOD= \frac{1}{2} \angle AOB= \frac{7}{2} x$,\\
since $\angle COD=\angle BOD-\angle BOC$ and $\angle COD=15^\circ$,\\
then $\frac{7}{2} x-2x=15^\circ$,\\
solving for $x$ gives $x=10^\circ$,\\
therefore, $\angle AOB=7\times 10^\circ=\boxed{70^\circ}$.


Question: As shown in the figure, in the circle $O$, the radius $OC$ is perpendicular to the chord $AB$ at $D$, point $E$ is on the circle $O$, $\angle E=22.5^\circ$, and $AB=2$. What is the length of the radius $OB$?

Solution:
\textbf{Solution:} By the problem statement, radius $OC$ is perpendicular to chord $AB$ at point $D$,\\
$\therefore \overset{\frown}{AC}=\overset{\frown}{BC}$,\\
$\therefore \angle E=\frac{1}{2}\angle BOC=22.5^\circ$,\\
$\therefore \angle BOD=45^\circ$,\\
$\therefore \triangle ODB$ is an isosceles right triangle,\\
$\because AB=2$,\\
$\therefore DB=OD=1$,\\
$\therefore OB=\sqrt{1^2+1^2}=\boxed{\sqrt{2}}$.


Question: As shown in the figure, M is the midpoint of CD, EM$\perp$CD. If CD$=4$ and EM$=6$, what is the radius of the circle containing arc $\overset{\frown}{CED}$?

Solution:
\textbf{Solution:} Connect OC, \\
$\because$ M is the midpoint of CD, and EM $\perp$ CD, \\
$\therefore$ EM passes through the center point O of $\odot$O, CM = $\frac{1}{2}$CD = 2, \\
Let the radius be x, \\
$\because$ EM = 6, \\
$\therefore$ OM = EM - OE = 6 - x, \\
In Rt$\triangle$OCM, we have OM$^{2}$ + CM$^{2}$ = OC$^{2}$, which gives (6-x)$^{2}$ + $2^{2}$ = x$^{2}$, \\
Solving for x, we get x = $\frac{10}{3}$. \\
$\therefore$ The radius of the circle that contains the arc $CED$ is $\boxed{\frac{10}{3}}$.


Question: As shown in the figure, $AB$ is a chord of $\odot O$, $C$ is a point on $\odot O$, and $\angle ACB=60^\circ$, $OD\perp AB$ at point $E$, intersecting $\odot O$ at point $D$. If the radius of $\odot O$ is $6$, how long is the chord $AB$?

Solution:
\textbf{Solution:} As shown in the diagram, connect OB,\\
then $\angle AOB = 2\angle ACB = 120^\circ$,\\
since OD$\perp$AB,\\
thus $\angle AOE = \frac{1}{2}\angle AOB = 60^\circ$,\\
since AO=6,\\
thus in $\triangle AOE$, $OE = \frac{1}{2} OA = 3$,\\
$AE = \sqrt{OA^2 - OE^2} = 3\sqrt{3}$,\\
thus AB=2AE$=\boxed{6\sqrt{3}}$.


Question: As shown in the figure, $AB$ is a chord of $\odot O$, $OC\perp AB$ at point $M$, and intersects $\odot O$ at point $C$. If the radius of $\odot O$ is $10$, and the ratio $OM:MC = 3:2$, what is the length of $AB$?

Solution:
\textbf{Solution:} As shown in the diagram, connect OA. \\
Since OM:MC=3:2, and OC=10, \\
thus OM=$\frac{3}{5}OC=\frac{3}{5}\times 10=6$. \\
Since OC$\perp$AB, \\
therefore $\angle$OMA$=90^\circ$, and AB=2AM. \\
In $\triangle$AOM, with AO=10, and OM=6, \\
therefore AM$=\sqrt{AO^{2}-OM^{2}}=\sqrt{10^{2}-6^{2}}=8$. \\
Thus, AB=2AM$=\boxed{16}$.


Question: As shown in the figure, $AD$ is the diameter of the circumcircle $\odot O$ of $\triangle ABC$. If $\angle ACB = 50^\circ$, what is the degree measure of $\angle BAD$?

Solution:
\textbf{Solution:} As shown in the figure, connect BD,\\
since AD is the diameter of the circumcircle $\odot O$ of $\triangle ABC$,\\
therefore $\angle ABD=90^\circ$,\\
since $\angle D=\angle C=50^\circ$,\\
therefore $\angle BAD=90^\circ-\angle D=90^\circ-50^\circ=\boxed{40^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $\angle A=50^\circ$, and the perpendicular bisector of $AB$, $DE$, intersects $AC$ and $AB$ at points $D$ and $E$, respectively. What is the degree measure of $\angle CBD$?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $AB=AC$ and $\angle A=50^\circ$ \\
$\therefore$ $\angle ABC=\angle ACB=(180^\circ-50^\circ)÷2=65^\circ$ \\
Since $DE$ is the perpendicular bisector of $AB$ \\
$\therefore$ $DA=DB$ \\
$\therefore$ $\angle ABD=\angle A=50^\circ$ \\
$\therefore$ $\angle CBD=\angle ABC-\angle ABD=65^\circ-50^\circ=\boxed{15^\circ}$


Question: In $\triangle ABC$, $\angle C=90^\circ$, $\angle B=15^\circ$, and $DE$ is the perpendicular bisector of $AB$, with $BE=5$. Find the length of $AC$.

Solution:
\textbf{Solution:} Connect AE, \\
$\because$ DE is the perpendicular bisector of AB, \\
$\therefore$ BE = AE, \\
$\therefore$ $\angle B = \angle EAB = 15^\circ$, \\
$\therefore$ $\angle AEC = 30^\circ$, \\
$\because$ $\angle C = 90^\circ$, \\
$\therefore$ AC = $\frac{1}{2}$ AE = $\frac{1}{2}$ BE = \boxed{2.5}


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC = \angle BAC = 45^\circ$, point $P$ lies on $AB$, $AD \perp CP$, $BE \perp CP$, with the feet of the perpendiculars being $D$ and $E$ respectively. Given that $DC = 2$, what is the length of $BE$?

Solution:
\textbf{Solution:} \\
Given that $\angle ABC=\angle BAC=45^\circ$, \\
thus, $\angle ACB=90^\circ$ and AC=BC, \\
since $\angle DAC+\angle ACD=90^\circ$ and $\angle BCE+\angle ACD=90^\circ$, \\
it follows that $\angle DAC=\angle BCE$, \\
In $\triangle ACD$ and $\triangle CEB$, \\
hence, $\triangle ACD \cong \triangle CEB$ (AAS), \\
therefore, $BE=CD=\boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, and the bisectors of $\angle BAC$ and $\angle ACB$ intersect at point $D$. Given that $\angle ADC=130^\circ$, find the measure of $\angle BAC$?

Solution:
\textbf{Solution:} Given that $AB=AC$, and $AE$ bisects $\angle BAC$,\\
it follows that $AE\perp BC$ (for an isosceles triangle, three lines coincide),\\
since $\angle ADC=130^\circ$,\\
it follows $\angle CDE=50^\circ$,\\
thus $\angle DCE=90^\circ-\angle CDE=40^\circ$, and since $CD$ bisects $\angle ACB$,\\
it follows $\angle ACB=2\angle DCE=80^\circ$.\\
Since $AB=AC$,\\
it follows $\angle B=\angle ACB=80^\circ$,\\
thus $\angle BAC=180^\circ-(\angle B+\angle ACB)=\boxed{20^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, extend $BC$ to point $D$ and connect $AD$. Let $E$ be a point on $AD$. It is known that $\angle B=50^\circ$, $\angle CAE=\angle D$, $\angle DCE=\angle BAC=20^\circ$. What is the degree measure of $\angle CED$?

Solution:
\textbf{Solution:} Since $\angle ACD = \angle B + \angle BAC$, and $\angle DCE = \angle BAC = 20^\circ$, \\
$\therefore \angle ACD = 50^\circ + 20^\circ = 70^\circ$, \\
$\therefore \angle CAE = \angle D = \frac{180^\circ - \angle ACD}{2} = 55^\circ$, \\
$\therefore \angle CED = 180^\circ - \angle DCE - \angle D = 180^\circ - 20^\circ - 55^\circ = \boxed{105^\circ}$


Question: As shown in the figure, within $\triangle ABC$, $AD$ is the height of $\triangle ABC$, $AE$ and $BF$ are the angle bisectors of $\triangle ABC$, and $AE$ intersects $BF$ at point $O$. Given that $\angle BOA = 125^\circ$, what is the measure of $\angle DAC$ in degrees?

Solution:
\textbf{Solution:} Since $\angle OAB + \angle OBA + \angle AOB = 180^\circ$ and $\angle AOB = 125^\circ$,
\[
\therefore \angle OAB + \angle OBA = 180^\circ - 125^\circ = 55^\circ,
\]
Since AE and BF are the angle bisectors of $\triangle ABC$,
\[
\therefore \angle OAB = \frac{1}{2} \angle BAC, \angle OBA = \frac{1}{2} \angle ABC,
\]
\[
\therefore \frac{1}{2} \angle BAC + \frac{1}{2} \angle ABC = 55^\circ,
\]
\[
\therefore \angle BAC + \angle ABC = 110^\circ,
\]
Since $\angle BAC + \angle ABC + \angle ACB = 180^\circ$,
\[
\therefore \angle ACB = 70^\circ,
\]
Since AD is the altitude of $\triangle ABC$,
\[
\therefore \angle ADC = 90^\circ,
\]
\[
\therefore \angle DAC = 90^\circ - 70^\circ = \boxed{20^\circ}.
\]


Question: In the figure, in $ \triangle ABC$, $ \angle C=90^\circ$, $ \angle ABC=60^\circ$, $BD$ bisects $ \angle ABC$, if $AD=8$cm, what is the length of $CD$ in cm?

Solution:
\textbf{Solution}: Since $\angle C=90^\circ$ and $\angle ABC=60^\circ$\\
Therefore, $\angle A=30^\circ$\\
Since $BD$ bisects $\angle ABC$\\
Therefore, $\angle ABD=\angle CBD=30^\circ$\\
Therefore, $\angle A=\angle ABD$\\
Therefore, $DB=AD=8$\\
Since $\angle C=90^\circ$ and $\angle CBD=30^\circ$\\
Therefore, $CD=\frac{1}{2}DB=4$\\
Hence, the length of $CD$ is $\boxed{4}$cm.


Question: In $\triangle ABC$, $D$ is a point on $BC$, $AC=10$, $CD=6$, $AD=8$, $AB=17$. Find the length of $BC$.

Solution:
\textbf{Solution:} Given $\because CD=6, AD=8$,\\
$\therefore CD^{2}+AD^{2}=6^{2}+8^{2}=100$,\\
$\because AC^{2}=10^{2}=100$,\\
$\therefore CD^{2}+AD^{2}=AC^{2}$,\\
$\therefore \angle ADC=90^\circ$,\\
$\therefore \angle ADB=90^\circ$,\\
$\therefore BD=\sqrt{AB^{2}-AD^{2}}=\sqrt{17^{2}-8^{2}}=15$,\\
$\therefore BC=BD+CD=15+6=\boxed{21}$.


Question: As shown in the figure, E is a point on the diagonal BD of the square ABCD. Connect AE and CE, and extend CE to meet AD at point F. If $\angle AEC = 140^\circ$, what is the measure of $\angle DFE$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a square,\\
it follows that AB = CB, and $\angle ABE = \angle CBE = \angle ADB = 45^\circ$.\\
In $\triangle ABE$ and $\triangle CBE$, we have\\
\[
\left\{
\begin{array}{l}
AB = CB\\
\angle ABE = \angle CBE\\
BE = BE
\end{array}
\right.,
\]
thus $\triangle ABE \cong \triangle CBE$,\\
which implies $\angle AEB = \angle CEB$.\\
Also, since $\angle AEC = 140^\circ$,\\
it follows that $\angle CEB = 70^\circ$.\\
Because $\angle DEC + \angle CEB = 180^\circ$,\\
thus $\angle DEC = 180^\circ - \angle CEB = 110^\circ$.\\
Since $\angle DFE + \angle ADB = \angle DEC$,\\
it follows that $\angle DFE = \angle DEC - \angle ADB = 110^\circ - 45^\circ = \boxed{65^\circ}.$


Question: A student performed the following operations on a rectangular piece of paper ABCD: as shown in the figure, first fold along the line AG so that point B falls on point P on the diagonal AC, then fold along the line CH so that point D falls on point Q on the diagonal AC. If $AB=5$ and $BC=12$, what is the area of the quadrilateral $AGCH$?

Solution: \textbf{Solution:} Given that quadrilateral ABCD is a rectangle, with $AB=5$ and $BC=12$. Thus, $AB=CD=5$ and $\angle B=90^\circ$. Therefore, $AC=\sqrt{AB^2+BC^2}=13$. From the property of folding, we know that AP=AB=5, BG=PG, and $\angle B=\angle APG=90^\circ$. Similarly, CQ=CD=5, thus CP=8 and $\angle CPG=90^\circ$. Let CG=x, then BG=PG=12-x. Therefore, by the Pythagorean theorem, we have $(12-x)^2+8^2=x^2$. Solving this, we get $x=\frac{26}{3}$, i.e., CG$=\frac{26}{3}$. Similarly, AH$=\frac{26}{3}$, hence CG=AH. Since quadrilateral ABCD is a rectangle, we have AD$\parallel$BC. Consequently, quadrilateral AGCH is a parallelogram. Therefore, the area of quadrilateral AGCH is $S_{\text{quadrilateral} AGCH}=CG\cdot AB=\boxed{\frac{130}{3}}$.


Question: As shown in the figure, it is known that rectangle $ABCD$ is folded along line $BD$, such that point $C$ falls on point $C'$, $BC$ intersects $AD$ at $E$, $AD=8$, and $AB=4$. What is the length of $DE$?

Solution:
\textbf{Solution:} By the property of folding, we have $\angle EBD = \angle CBD$,\\
$\because$ in rectangle ABCD, AD$\parallel$BC,\\
$\therefore \angle ADB = \angle CBD$,\\
$\therefore \angle EBD = \angle EDB$,\\
$\therefore DE = BE$,\\
Let DE = x, then BE = x, AE = 8 - x,\\
$\because$ in $\triangle ABE$, by the Pythagorean theorem we have $BE^2 = AB^2 + AE^2$,\\
$\therefore x^2 = 4^2 + (8 - x)^2$, solving for $x$ gives: $x = \boxed{5}$,\\
$\therefore DE = \boxed{5}$.


Question: In the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, with $AC=8$, $BD=6$, and $OE\perp BC$ with the foot of the perpendicular being point $E$. Find the length of $OE$.

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\
it follows that AC $\perp$ BD, OB = OD = $\frac{1}{2}$BD = 3, OA = OC = $\frac{1}{2}$AC = 4,\\
in $\triangle OBC$, since OB = 3, OC = 4,\\
thus BC = $\sqrt{3^2 + 4^2} = 5$,\\
since OE $\perp$ BC,\\
it follows that $\frac{1}{2}$OE $\cdot$ BC = $\frac{1}{2}$ OB $\cdot$ OC,\\
thus OE = $\frac{3 \times 4}{5} = \boxed{\frac{12}{5}}$.


Question: As shown in the figure, points A, B, and C are on the same straight line, and BD bisects $\angle ABE$, $\angle EBC=30^\circ$. What is the degree measure of $\angle DBC$?

Solution:
\textbf{Solution:} Since $\angle EBC=30^\circ$,\\
therefore $\angle ABE=180^\circ-30^\circ=150^\circ$,\\
since BD bisects $\angle ABE$,\\
thus $\angle DBE= \frac{1}{2}\angle ABE=75^\circ$,\\
therefore $\angle DBC=\angle DBE+\angle EBC=\boxed{105^\circ}$.


Question: As shown in the figure, point $O$ is on the line $AC$, $OD$ bisects $\angle AOB$, $\angle BOE = \frac{1}{2}\angle EOC$, and $\angle DOE = 70^\circ$. Find the measure of $\angle EOC$.

Solution:
\textbf{Solution:} Let $\angle EOC = \alpha$ \\
Since $\angle BOE = \frac{1}{2} \angle EOC$ \\
Thus $\angle BOE = \frac{1}{2} \alpha$ \\
Since $\angle DOE = 70^\circ$ \\
Thus $\angle BOD = \angle DOE - \angle BOE = 70^\circ - \frac{1}{2} \alpha$ \\
Since $OD$ bisects $\angle AOB$ \\
Thus $\angle AOD = \angle BOD = 70^\circ - \frac{1}{2} \alpha$ \\
Since point O is on line AC, \\
Thus $\angle AOC = 180^\circ$ \\
Therefore $\angle EOC + \angle DOE + \angle AOD = 180^\circ$ \\
Therefore $\alpha + 70^\circ + 70^\circ - \frac{1}{2}\alpha = 180^\circ$ \\
Therefore $\alpha = 80^\circ$ \\
Therefore $\angle EOC = \boxed{80^\circ}$