Question: As shown in Figure 1, in the Cartesian coordinate system, the linear function $y=-\frac{2}{3}x+4$ intersects the x-axis at point $B$ and the y-axis at point $A$. Point $C$ is the midpoint of segment $AB$. Through point $C$, $DC\perp x$ axis, and the foot of the perpendicular is $D$. Assume point $E$ is on the negative half of the y-axis, and connecting $CE$ intersects the x-axis at point $F$, with $CF=FE$. On the line $CD$, there exists a point $P$ such that $AP+EP$ is minimized. What are the coordinates of point $P$?

Solution:
\textbf{Solution:} Since point C is the midpoint of AB, and given points A(0,4) and B(6,0), it follows that point C(3,2), therefore CD=2, and OD=3; since CD is perpendicular to the x-axis, it follows that $\angle EOF=\angle CDF=90^\circ$. In $\triangle EOF$ and $\triangle CDF$, we have
\[
\left\{
\begin{array}{l}
\angle EOF=\angle CDF\\
\angle OFE=\angle CFD\\
EF=CF
\end{array}
\right.
\]
therefore, $\triangle EOF\cong \triangle CDF$ (by AAS), hence OE=CD=2, and OF=FD=$\frac{1}{2}$OD=$\frac{3}{2}$, therefore point F($\frac{3}{2}$,0), and point E(0,-2); Construct the symmetric point A$^{\prime}$ of point A with respect to line CD, and draw line A$^{\prime}$E intersecting CD at point F, draw line AP, thus point A$^{\prime}$(6,4). Suppose the equation of line A$^{\prime}$E is y=kx+b, then
\[
\left\{
\begin{array}{l}
6k+b=4\\
b=-2
\end{array}
\right.
\]
Solving this yields:
\[
\left\{
\begin{array}{l}
k=1\\
b=-2
\end{array}
\right.
\]
therefore, the equation of A$^{\prime}$E is y=x-2, when x=3, y=1, therefore point P(3,1) is $\boxed{(3,1)}$.


Question: As shown in Figure 1, in a Cartesian coordinate system, the linear function $y=-\frac{2}{3}x+4$ intersects the $x$-axis at point $B$ and the $y$-axis at point $A$. Point $C$ is the midpoint of line segment $AB$. A line is drawn through point $C$ such that $DC\perp x$-axis, with the foot of the perpendicular being $D$. As shown in Figure 2, does there exist a point $Q$ on the line $CD$ such that $\angle ABQ=45^\circ$? Find the coordinates of point $Q$.

Solution:
Solution:When point $Q$ is above the x-axis, $\angle ABQ=45^\circ$. Draw $AM\perp AB$ at point $A$, intersecting $BQ$ at point $M$. Draw $MH\perp y$-axis at point $H$,\\
$\therefore \angle BAM=90^\circ$,\\
$\therefore \triangle BAM$ is an isosceles right triangle,\\
$\therefore AM=AB$,\\
$\because \angle HAM+\angle OAB=90^\circ$, $\angle OAB+\angle ABO=90^\circ$,\\
$\therefore \angle HAM=\angle ABO$,\\
In $\triangle AMH$ and $\triangle ABO$
\[
\left\{\begin{array}{l}
\angle AHM=\angle AOB\\
\angle HAM=\angle ABO\\
AM=AB
\end{array}\right.
\]
$\therefore \triangle AMH\cong \triangle ABO$ (AAS),\\
$\therefore MH=AO=4$, $AH=BO=6$,\\
$\therefore OH=AH+AO=4+6=10$,\\
$\therefore$ the coordinates of point $M$ are $(4,10)$,\\
Let the equation of the line $MB$ be $y=mx+n$
\[
\left\{\begin{array}{l}
4m+n=10\\
6m+n=0
\end{array}\right.
\]
Solving it, we get:
\[
\left\{\begin{array}{l}
m=-5\\
n=30
\end{array}\right.
\]
$\therefore y=-5x+30$,\\
When $x=3$, $y=-15+30=15$,\\
$\therefore$ the coordinates of point $Q$ are $(3,15)$;\\
When point $Q$ is below the x-axis, $\angle ABQ=45^\circ$. Draw $AN\perp AB$ at point $A$, intersecting the extended line of $BQ$ at point $N$. Draw $NG\perp y$-axis at point $G$,\\
Similarly, it can be proved that $\triangle ANG\cong \triangle ABO$,\\
$\therefore NG=AO=4$, $AG=BO=6$,\\
$\therefore OG=6-4=2$,\\
$\therefore$ the coordinates of point $N$ are $(-4,-2)$,\\
Let the equation of $BN$ be $y=ax+c$,
\[
\left\{\begin{array}{l}
-4a+c=-2\\
6a+c=0
\end{array}\right.
\]
Solving it, we get:
\[
\left\{\begin{array}{l}
a=\frac{1}{5}\\
c=-\frac{6}{5}
\end{array}\right.
\]
$\therefore y=\frac{1}{5}x-\frac{6}{5}$,\\
When $x=3$, $y=\frac{1}{5}\times 3-\frac{6}{5}=-\frac{3}{5}$,\\
$\therefore$ the coordinates of point $Q$ are $\left(3, -\frac{3}{5}\right)$ or $(3,15)$.\\
Hence, the coordinates of point $Q$ are $\boxed{\left(3, -\frac{3}{5}\right)}$ or $\boxed{(3,15)}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle A=40^\circ$, $AB=AC$, and the perpendicular bisector $MN$ of $AB$ intersects $AC$ at point D and $AB$ at point E. The perimeter of $\triangle ABC$ is $13cm$, $AB=5cm$. What is the degree measure of $\angle DBC$?

Solution:
\textbf{Solution:} \\
Since $AB=AC$, \\
it follows that $\triangle ABC$ is an isosceles triangle, \\
because $\angle A=40^\circ$, \\
thus $\angle ABC=\angle ACB=\frac{1}{2}\left(180^\circ-40^\circ\right)=70^\circ$, \\
since $MN$ is the perpendicular bisector of $AB$, \\
it follows that $\triangle ABD$ is an isosceles triangle, \\
therefore $\angle ABD=\angle A=40^\circ$, \\
hence $\angle DBC=\angle ABC-\angle ABD=70^\circ-40^\circ=\boxed{30^\circ}$;


Question: As shown in the diagram, in $ \triangle ABC$, $ \angle A=40^\circ, AB=AC$, and the perpendicular bisector of $AB$, $MN$, intersects $AC$ at point D and $AB$ at point E. The perimeter of $ \triangle ABC$ is $13cm$, and $AB=5cm$. What is the perimeter of $ \triangle DBC$?

Solution:
\textbf{Solution:} Since the perimeter of $\triangle ABC$ is $13\text{cm}, AB=5\text{cm}, AB=AC$,\\
it follows that $BC=13-2\times 5=3\text{cm}$,\\
Since $MN$ bisects $AB$ perpendicularly,\\
it implies $AD=BD$,\\
Therefore, the perimeter of $\triangle DBC=BC+CD+BD=BC+CD+AD=BC+AC=5+3=\boxed{8\text{cm}}$.


Question: As shown in the figure, in $\triangle ABC$, $BO$ bisects $\angle ABC$, $CO$ bisects $\angle ACB$, $MN$ passes through point $O$ and is parallel to $BC$, intersecting $AB$ and $AC$ at $M$ and $N$, respectively. Given $AB=12$, $AC=18$, and $MN=BM+CN$, what is the perimeter of $\triangle AMN$?

Solution:
\textbf{Solution:} From (1) we know that $MN = BM + CN$,\\
$\therefore$ the perimeter of $\triangle AMN = AM + AN + MN = AM + AN + BM + CN = AB + AC$,\\
$\because$ $AB = 12$, $AC = 18$,\\
$\therefore$ the perimeter of $\triangle AMN = AB + AC = \boxed{30}$.


Question: As shown in the figure, in $\triangle ABC$, the perpendicular bisectors of sides $AB$ and $AC$ respectively intersect $BC$ at $D$ and $E$. If $BC=5$, what is the perimeter of $\triangle ADE$?

Solution:
\textbf{Solution:}

Given that the perpendicular bisectors of sides AB and AC intersect BC at D and E, respectively,

$\therefore DA=DB$, $EA=EC$,

$\therefore$ the perimeter of $\triangle ADE = AD + DE + AE = DB + DE + EC = BC = \boxed{5}$.


Question: As shown in the figure, in $\triangle ABC$, the perpendicular bisectors of sides $AB$ and $AC$ intersect $BC$ at points $D$ and $E$, respectively. If $\angle BAD + \angle CAE = 60^\circ$, what is the degree measure of $\angle BAC$?

Solution:
\textbf{Solution:} Since $DA=DB$ and $EA=EC$,\\
it follows that $\angle DAB = \angle B$ and $\angle EAC = \angle C$,\\
therefore, $\angle B + \angle C = \angle DAB + \angle EAC = 60^\circ$,\\
hence, $\angle BAC = \boxed{120^\circ}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $\angle D=90^\circ$, $\angle B=60^\circ$, $AD=6$, $AB^2=\frac{100}{3}$, $AB\perp AC$. A point $E$ is chosen on $CD$, and line $AE$ is drawn. Triangle $ADE$ is folded along $AE$ so that point $D$ falls on point $F$ on $AC$. Find: What is the length of $CD$?

Solution:
\textbf{Solution:} Since $AB \perp AC$ and $\angle B = 60^\circ$,\\
it follows that $\angle BCA = 30^\circ$,\\
which implies $BC = 2AB$,\\
hence $BC^2 = 4AB^2 = \frac{400}{3}$,\\
therefore $AC^2 = BC^2 - AB^2 = 100$,\\
leading to $AC = 10$.\\
Since $\angle D = 90^\circ$,\\
in $\triangle ADC$, where $AD = 6, AC = 10$,\\
by the Pythagorean theorem, it is found that $CD = \boxed{8}$.


Question: As shown in the figure, in the quadrilateral $ABCD$, $\angle D = 90^\circ$, $\angle B = 60^\circ$, $AD=6$, $AB^2=\frac{100}{3}$, $AB \perp AC$. A point E is chosen on $CD$, and $AE$ is connected. The triangle $ADE$ is folded along $AE$ such that point D falls on point F on $AC$. Find: What is the length of $DE$?

Solution:
\textbf{Solution:} Let $ED=x$, thus $EF=x, CE=8-x$.\\
From the folding, we have: $AF=AD=6, CF=10-6=4$.\\
In $\triangle CFE$ right-angled at $F$, we have $CF^2+FE^2=CE^2$,\\
$\therefore 4^2+x^2=(8-x)^2$,\\
Solving this gives $x=\boxed{3}$.\\
Therefore, $DE=\boxed{3}$.


Question: As shown in the figure, $AD$ is the altitude of $\triangle ABC$, $CE$ is the angle bisector of $\angle ACB$ in $\triangle ACB$, and $F$ is the midpoint of $AC$, with $\angle ACB=50^\circ$ and $\angle BAD=65^\circ$. What is the degree measure of $\angle AEC$?

Solution:
\textbf{Solution:}

Since $AD$ is the altitude of $\triangle ABC$,\\
therefore $\angle ADB=90^\circ$,\\
Since $\angle BAD=65^\circ$,\\
therefore $\angle ABD=90^\circ-65^\circ=25^\circ$,\\
Since $CE$ is the bisector of $\triangle ACB$, and $\angle ACB=50^\circ$,\\
therefore $\angle ECB=\frac{1}{2} \angle ACB=25^\circ$,\\
therefore $\angle AEC=\angle ABD+\angle ECB=25^\circ+25^\circ=\boxed{50^\circ}$.


Question: As shown in the diagram, $AD$ is the altitude of $\triangle ABC$, $CE$ is the angle bisector of $\angle ACB$ in $\triangle ACB$, and $F$ is the midpoint of $AC$. Given $\angle ACB=50^\circ$ and $\angle BAD=65^\circ$, if the difference in the perimeters of $\triangle BCF$ and $\triangle BAF$ is 3, with $AB=7$ and $AC=4$, then what is the length of $BC$?

Solution: \textbf{Solution:} From the given, we have $BC=\boxed{10}$.


Question: As shown in the figure, in $\triangle ABC$, $AE$ is the altitude on side $BC$. If $AD$ is the median on side $BC$, $AE=3\,\text{cm}$, and $S_{\triangle ABC}=12\,\text{cm}^2$. What is the length of $DC$ in $\text{cm}$?

Solution:
\textbf{Solution:} Given that $AE=3\,\text{cm}$ and $S_{\triangle ABC}=12\,\text{cm}^2$,\\
it follows that $BC=12\times 2\div 3=8\,\text{cm}$,\\
since $AD$ is the median to side $BC$,\\
thus $DC =\frac{1}{2} BC=4\,\text{cm}$;\\
hence, the length of $DC$ is $\boxed{4}\,\text{cm}$.


Question: As shown in the figure, in $\triangle ABC$, AE is the altitude on side BC. If AD is the bisector of $\angle BAC$, $\angle B=40^\circ$, and $\angle C=50^\circ$, what is the measure of $\angle DAE$?

Solution:
\textbf{Solution}:\\
Since $\angle B=40^\circ$ and $\angle C=50^\circ$,\\
it follows that $\angle BAC=90^\circ$,\\
Since $AD$ is the bisector of $\angle BAC$,\\
it results in $\angle BAD=45^\circ$,\\
Since $\angle ADE$ is an exterior angle of $\triangle ABD$,\\
it follows that $\angle ADE=\angle B+\angle BAD=40^\circ+45^\circ=85^\circ$,\\
Therefore, $\angle DAE=90^\circ-85^\circ=\boxed{5^\circ}$.


Question: As shown in the figure, the parabola $y=-x^{2}+2x+c$ intersects the $x$-axis at points $A(-1, 0)$ and $B(3, 0)$, and intersects the $y$-axis at point $C$. Point $D$ is the vertex of the parabola. What are the coordinates of point $D$?

Solution: \textbf{Solution:} The coordinates of point $D$ are $\boxed{(1,4)}$.


Question: As shown in the figure, the parabola $y=-x^2+2x+c$ intersects the $x$-axis at points $A(-1,0)$ and $B(3,0)$, and intersects the $y$-axis at point $C$. $D$ is the vertex of the parabola. The point symmetric to $C$ with respect to the axis of symmetry of the parabola is point $E$. Point $M$ is a point on the axis of symmetry of the parabola, and $\triangle DMB$ is similar to $\triangle BCE$. What are the coordinates of point $M$?

Solution: \textbf{Solution:} The coordinates of point $M$ are $ \boxed{(1,\frac{2}{3}) or (1, -2)}$.


Question: As shown in Figure (1), both quadrilateral ABCD and quadrilateral AEFG are squares. Rotate the square AEFG around point A and connect BE and CF. What is the value of $FC\colon BE$?

Solution: \textbf{Solution:} We have $FC: BE = \boxed{\sqrt{2}}$.


Question: As shown in Figure (1), the quadrilateral ABCD and the quadrilateral AEFG are both squares. Rotate the square AEFG around point A and connect BE and CF. When points G, F, and C are collinear, as shown in Figure (2), if $AB=5$ and $AE=\sqrt{5}$, then what is the length of $BE$?

Solution: \textbf{Solution:} We have $BE=\boxed{\sqrt{10}}$.


Question: As shown in the figure, Xiaohui used two right-angled triangle paper pieces to form a quadrilateral, where $\angle A=90^\circ$, $AD=2\text{cm}$, $AB=3\text{cm}$, and the hypotenuses overlap to form a quadrilateral. Then, points $E$ and $F$ are chosen on $CB$ and $CD$ respectively, and $AE$ and $BF$ are connected such that $AE \perp BF$. If the quadrilateral is formed as shown in the figure, what is the value of $\frac{BF}{AE}$?

Solution: \textbf{Solution:} Thus, $\frac{BF}{AE} = \boxed{\frac{2}{3}}$.


Question: During the comprehensive practice class, Xiaohui used two right-angled triangle paper pieces as shown in the figure: $\angle A=90^\circ$, $AD=2\text{cm}$, $AB=3\text{cm}$, and overlapped the hypotenuses to form a quadrilateral. Then, points $E$ and $F$ were chosen on $CB$ and $CD$, respectively, and lines $AE$ and $BF$ were drawn such that $AE\perp BF$. If the quadrilateral formed is as shown in the figure, then what is the value of $\frac{BF}{AE}$?

Solution: \textbf{Solution:} Therefore, $\therefore \frac{BF}{AE} = \boxed{\frac{13}{12}}$.


Question: Quadrilateral ABCD is a rectangular piece of paper, with point E on side AD. Fold $\triangle ABE$ along BE such that point A falls onto the diagonal BD of the rectangle. Draw line CF. Given that $DE=1$, investigate the following question: As shown in Figure 1, when F is exactly at the midpoint of BD, what is the length of $AE$?

Solution: \textbf{Solution:} We have $AE = \boxed{\frac{1}{2}}$.


Question: Quadrilateral $ABCD$ is a rectangular piece of paper, with point $E$ located on side $AD$. When $\triangle ABE$ is folded along $BE$ such that point $A$ lies on the diagonal $BD$ of the rectangle, and $CF$ is drawn. Given that $DE=1$, explore the following question: As shown in Figure 2, when points $C$, $E$, and $F$ lie on the same straight line, what is the length of $AE$?

Solution: \textbf{Solution:} When points $C$, $E$, and $F$ are on the same line, we have $AE=\boxed{\frac{\sqrt{5}-1}{2}}$.


Question: As shown in Figure 1, in $\triangle ABC$, $\angle BCA=90^\circ$, $AC=3$, and $BC=4$. Point $P$ is located on side $AB$, and a line is drawn through point $P$ such that $PD\perp AC$, intersecting $AC$ and $BC$ at points $D$ and $E$, respectively. If $P$ is the midpoint of $AB$, with $PD\perp AC$ and $PE\perp BC$, what is the value of $\frac{PD}{PE}$?

Solution: \textbf{Solution:} According to the given conditions, we have $ \frac{PD}{PE}=\boxed{\frac{4}{3}}$.


Question: As shown in the figure, in $ \triangle ABC$, $\angle BCA=90^\circ$, $AC=3$, $BC=4$. Point $P$ is a point on hypotenuse $AB$. A ray $PD$ is drawn perpendicular to $PE$, intersecting $AC$ and $BC$ at points $D$ and $E$, respectively. $DE$ is connected. If $\triangle PDE$ is similar to $\triangle ABC$, what is the value of $BP$?

Solution:
\textbf{Solution:} Construct $PM\perp AC$, $PN\perp BC$, intersecting $AC$, $BC$ at points $M$, $N$ respectively.\\
Since $\angle BCA=90^\circ$\\
Therefore, quadrilateral $PMCN$ is a rectangle,\\
Since $\angle DPE=90^\circ$\\
Therefore $\angle DPM=\angle EPN$\\
Since $PM\perp AC$, $PN\perp BC$,\\
Therefore $\angle DMP=\angle ENP=90^\circ$\\
Therefore $\triangle DPM\sim \triangle EPN$\\
(1) If $\triangle DPE\sim \triangle ACB$, it can be obtained that $\frac{PD}{PE}=\frac{AC}{BC}=\frac{3}{4}$\\
Since $\triangle DPM\sim \triangle EPN$\\
Therefore $\frac{PM}{PN}=\frac{PD}{PE}=\frac{3}{4}$\\
Let $PM=3m$, then $PN=4m$\\
Since quadrilateral $PMCN$ is a rectangle, $CN=3m$\\
In $\triangle ACB$, with $\angle C=90^\circ$, it can be obtained that $\tan B=\frac{AC}{BC}=\frac{3}{4}$\\
In $\triangle PNB$, $\tan B=\frac{PN}{BN}=\frac{4m}{BN}=\frac{3}{4}$, solving gives $BN=\frac{16}{3}m$\\
Since $BC=CN+BN$\\
Therefore $4=3m+\frac{16}{3}m$, solving gives $m=\frac{12}{25}$\\
Therefore $PN=\frac{48}{25}$\\
Since $\angle B=\angle B$, $\angle PNB=\angle C=90^\circ$\\
Therefore $\triangle BPN\sim \triangle BAC$\\
Therefore $\frac{BP}{BA}=\frac{PN}{AC}$, that is $\frac{BP}{5}=\frac{\frac{48}{25}}{3}$, solving gives $BP=\boxed{\frac{16}{5}}$\\
(2) If $\triangle DPE\sim \triangle BCA$, it can be obtained that $\frac{PD}{PE}=\frac{BC}{AC}=\frac{4}{3}$\\
Since $\triangle DPM\sim \triangle EPN$\\
Therefore $\frac{PM}{PN}=\frac{PD}{PE}=\frac{4}{3}$\\
Let $PM=4m$, then $PN=3m$,\\
Since quadrilateral $PMCN$ is a rectangle, $CN=4m$\\
In $\triangle ACB$, with $\angle C=90^\circ$, it can be obtained that $\tan B=\frac{AC}{BC}=\frac{3}{4}$\\
In $\triangle PNB$, $\tan B=\frac{PN}{BN}=\frac{3m}{BN}=\frac{3}{4}$, solving gives $BN=4m$\\
Since $BC=CN+BN$\\
Therefore $4=4m+4m$, solving gives $m=\frac{1}{2}$\\
Therefore $PN=\frac{3}{2}$\\
Since $\angle B=\angle B$, $\angle PNB=\angle C=90^\circ$\\
Therefore $\triangle BPN\sim \triangle BAC$\\
Therefore $\frac{BP}{BA}=\frac{PN}{AC}$, that is $\frac{BP}{5}=\frac{\frac{3}{2}}{3}$,\\
Solving gives $BP=\boxed{\frac{5}{2}}$


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=ax^{2}+bx+8\, (a\neq 0)$ passes through points $A(-3, 5)$ and $B(5, 3)$, and intersects the y-axis at point $C$. The circle with diameter $AB$ passes through points $O$ and $C$ and intersects the x-axis at point $D$. Lines $AO$ and $AC$ are drawn. What is the equation of the parabola?

Solution:
\textbf{Solution:} Substitute points $A\left(-3, 5\right)$ and $B\left(5, 3\right)$ into $y=ax^{2}+bx+8\left(a\ne 0\right)$, to get\\
\[
\begin{cases}
9a-3b+8=5\\
25a+5b+8=3
\end{cases},
\]
solving this, we get\\
\[
\begin{cases}
a=-\frac{1}{4}\\
b=\frac{1}{4}
\end{cases},
\]
$\therefore$ the expression of the parabola is $\boxed{y=-\frac{1}{4}x^{2}+\frac{1}{4}x+8}$;


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=ax^{2}+bx+8\left(a\neq 0\right)$ passes through point $A\left(-3, 5\right)$ and $B\left(5, 3\right)$, and intersects the y-axis at point C. The circle with diameter AB passes through points O and C, and intersects the x-axis at point D. Connect AO and AC. What are the coordinates of point D?

Solution:
\textbf{Solution:} Let's connect AD, and assume the coordinates of point D are $(x,0)$,\\
since AB is the diameter of the circle,\\
it follows that $\angle ADB=90^\circ$,\\
hence $AB^2=AD^2+BD^2$,\\
therefore $\left(5+3\right)^2+\left(3-5\right)^2=\left(-3-x\right)^2+5^2+\left(5-x\right)^2+3^2$,\\
solving this gives $x=2$ or $x=0$ (discard),\\
thus, the coordinates of point D are $\boxed{(2,0)}$;


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=ax^2+bx+8\left(a\ne 0\right)$ passes through points $A(-3, 5)$ and $B(5, 3)$, and intersects the y-axis at point C. The circle with diameter AB passes through points O and C, and intersects the x-axis at point D. Lines AO and AC are drawn. Point E is on the x-axis. Lines BD and BE are drawn. When $\triangle BDE$ is similar to $\triangle OAC$, find the length of OE that satisfies this condition.

Solution:
\textbf{Solution:} From the given information, \\
$\because y=-\frac{1}{4}x^2+\frac{1}{4}x+8$ intersects the y-axis at point C, \\
$\therefore$ C(0, 8), \\
Draw AF $\perp$ y-axis through point A and BG $\perp$ x-axis through point B, then F(0, 5), G(5, 0), \\
$\therefore \tan\angle ACF=\frac{AF}{CF}=\frac{3}{8-5}=1$, $\tan\angle BDG=\frac{BG}{DG}=\frac{3}{5-2}=1$, \\
$\therefore \angle ACF=\angle BDG=45^\circ$, \\
(1) When $\triangle BDE\sim \triangle ACO$, \\
then $\frac{AC}{BD}=\frac{AO}{BE}=\frac{CO}{DE}$, \\
$\because AC=\sqrt{3^2+3^2}=3\sqrt{2}$, $BD=\sqrt{3^2+3^2}=3\sqrt{2}$, \\
$\therefore \frac{CO}{DE}=\frac{3\sqrt{2}}{3\sqrt{2}}=1$, \\
$\therefore DE=CO=8$, \\
$\therefore OE=OD+DE=2+8=\boxed{10}$; \\
(2) When $\triangle EBD\sim \triangle AOC$, \\
then $\frac{DE}{CA}=\frac{BD}{OC}=\frac{BE}{OA}$, \\
$\because AC=\sqrt{3^2+3^2}=3\sqrt{2}$, $BD=\sqrt{3^2+3^2}=3\sqrt{2}$, OC=8, \\
$\therefore \frac{DE}{3\sqrt{2}}=\frac{3\sqrt{2}}{8}$, \\
$\therefore DE=\frac{9}{4}$, \\
$\therefore OE=OD+DE=2+\frac{9}{4}=\boxed{\frac{17}{4}}$; \\
In summary, the length of OE is \boxed{10} or \boxed{\frac{17}{4}}.


Question: It is known that the acute triangle $ABC$ is inscribed in $\odot O$. As shown in Figure 1, when point $A$ is the midpoint of $\overset{\frown}{BAC}$,

Solution:
\textbf{Solution:}

Let us connect OB and OC, and assume AB and BC intersect at point P,

$\because$ point A is the midpoint of $\overset{\frown}{BAC}$,

$\therefore \overset{\frown}{AB}=\overset{\frown}{AC}$,

$\therefore AB=AC$,

Furthermore, $\because OB=OC$,

$\therefore AO$ is the perpendicular bisector of BC,

$\therefore AO\perp BC$;

(2) $\because AB=AC$ and $AP\perp BC$,

$\therefore BP=CP=4$,

$\therefore AP=\sqrt{AB^{2}-BP^{2}}=8$,

$\because BO^{2}=OP^{2}+BP^{2}$,

$\therefore BO^{2}=(8-OB)^{2}+16$,

$\therefore BO=\boxed{5}$,

$\therefore$ the radius of the circle $O$ is \boxed{5};


Question: As shown in the figure, it is known that acute triangle $ABC$ is inscribed in circle $\odot O$. When $AB>AC$, connect $BO$ and extend it to intersect side $AC$ at point $D$. If $\angle A=45^\circ$, $\frac{OD}{OB}=\frac{2}{3}$, what is $\frac{AD}{DC}$?

Solution:
\textbf{Solution:}

Extend $BD$ to intersect $\odot O$ at point $H$, and connect $CH$, $CO$, and $AH$. \\

Since $\frac{OD}{OB} = \frac{2}{3}$, \\

Thus, set $BO = 3a = OC = OH$, and $OD = 2a$, \\

Hence, $DH = a$, \\

Since $\angle BAC = 45^\circ$, \\

Thus $\angle BOC = 2\angle BAC = 90^\circ$, \\

Hence, $CD = \sqrt{OD^2 + OC^2} = \sqrt{13}a$, $CH = \sqrt{OH^2 + OC^2} = 3\sqrt{2}a$, \\

Since $BH$ is a diameter, hence $\angle BAH = 90^\circ$, \\

Since $\angle BAC = \angle BHC = 45^\circ$ \\

Thus $\angle CAH = 90^\circ - \angle BAC = 45^\circ = \angle CHD$ \\

Also, since $\angle ACH = \angle DCH$, \\

Therefore, $\triangle ACH \sim \triangle HCD$, \\

Therefore, $\frac{AC}{CH} = \frac{CH}{CD}$, \\

Thus, $\frac{AC}{3\sqrt{2}a} = \frac{3\sqrt{2}a}{\sqrt{13}a}$, \\

Hence, $AC = \frac{18\sqrt{13}a}{\sqrt{13}}$, \\

Thus, $AD = AC - CD = \frac{18\sqrt{13}a}{\sqrt{13}} - \sqrt{13}a = \frac{5\sqrt{13}a}{\sqrt{13}}$, \\

Therefore, $\frac{AD}{CD} = \frac{\frac{5\sqrt{13}a}{\sqrt{13}}}{\sqrt{13}a} = \boxed{\frac{5}{13}}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ is the angle bisector, and point $E$ is a point on side $AC$, satisfying $\angle ADE = \angle B$. $\triangle ADB \sim \triangle AED$; if $AE = 3$ and $AD = 5$, find the length of $AB$.

Solution:
\textbf{Solution:} Since $\triangle ADB \sim \triangle AED$, \\
it follows that $\frac{AD}{AE}=\frac{AB}{AD}$, \\
since $AE=3$, and $AD=5$, \\
thus $\frac{5}{3}=\frac{AB}{5}$, \\
therefore $AB=\boxed{\frac{25}{3}}$.


Question: As shown in the diagram, in quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $F$. Point $E$ is on $BD$, and $\angle BAE = \angle CAD$, $\frac{AB}{AE} = \frac{AC}{AD}$. Prove that: $\triangle ABC \sim \triangle AED$.

Solution:
\textbf{Solution:} Since $\angle BAE = \angle CAD$, \\
it follows that $\angle BAE + \angle EAF = \angle CAD + \angle EAF$, \\
which implies $\angle BAC = \angle DAE$, \\
and thus $\frac{AB}{AE}=\frac{AC}{AD}$. \\
Considering triangles $\triangle ABC$ and $\triangle AED$, where \\
$\left\{\begin{array}{l}
\frac{AB}{AE}=\frac{AC}{AD} \\
\angle BAC = \angle DAE
\end{array}\right.$, \\
it can be concluded that \boxed{\triangle ABC \sim \triangle AED}.


Question: As shown in the diagram, within quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $F$. Point $E$ is on $BD$, and $\angle BAE=\angle CAD$, $\frac{AB}{AE}=\frac{AC}{AD}$. If $\angle BAE=20^\circ$, find the measure of $\angle CBD$.

Solution:
Solution: Since $\triangle ABC\sim \triangle AED$, \\
it follows that $\angle ADB=\angle ACB$, \\
Also, since $\angle AFD=\angle BFC$, by virtue of vertical angles being equal, \\
it implies $\triangle AFD\sim \triangle BFC$, \\
thus $\angle CBD=\angle CAD$, \\
Given $\angle BAE=\angle CAD$, and $\angle BAE=20^\circ$, \\
it results in $\angle CAD=20^\circ$, \\
Therefore, the answer is: $\boxed{20^\circ}$.


Question: As shown in the figure, it is known that D and E are points on sides AC and AB of $\triangle ABC$ respectively, with $\angle AED = \angle C$, $AE=5$, $AC=9$, and $DE=6$. $\triangle ABC \sim \triangle ADE$. Find the length of BC.

Solution:
\textbf{Solution:}

Since $ \triangle ABC\sim \triangle ADE$,\\
$\therefore \frac{BC}{DE}=\frac{AC}{AE}$,\\
$\therefore \frac{BC}{6}=\frac{9}{5}$,\\
$\therefore BC=\boxed{\frac{54}{5}}$.


Question: As shown in Figure 1, given the quadratic function $y=-\frac{4}{3}(x+1)^{2}+\frac{16}{3}$ intersects the x-axis at points $A$ and $B$ (with point $A$ to the left of point $B$), and intersects the y-axis at point $C$. Point $D$ is the vertex of the parabola. Find the coordinates of points $A$ and $C$.

Solution:
\textbf{Solution:} Let $y=0$, we get $-\frac{4}{3}(x+1)^2+\frac{16}{3}=0$,\\
Solving for $x$, we find $x_1=-3$ and $x_2=1$,\\
$\therefore \boxed{A(-3, 0)}, B(1, 0)$,\\
Let $x=0$, we get $y=-\frac{4}{3}(0+1)^2+\frac{16}{3}=4$,\\
$\therefore \boxed{C(0,4)}$.


Question: As shown in Figure 1, it is known that the graph of the quadratic function $y=-\frac{4}{3}(x+1)^2+\frac{16}{3}$ intersects the x-axis at points A and B (with point A to the left of B) and intersects the y-axis at point C. Point D is the vertex of the parabola. As shown in Figure 2, connect AC and DC, and draw $CE\parallel AB$ through point C to intersect the parabola at point E. As shown in Figure 3, $\angle DCE = \angle CAO$; connect BC, and on the extension of EC there is a point P such that the triangle with vertices D, E, P is similar to $\triangle ABC$. What is the length of EP?

Solution:
\textbf{Solution:} As shown in the figure, connect DE,\\
$\because A(-3,0), B(1,0), C(0,4), D(-1,\frac{16}{3})$,\\
$\therefore AC=\sqrt{3^2+4^2}=5$, AB=4, $DC=\sqrt{1^2+\left(\frac{4}{3}\right)^2}=\frac{5}{3}$,\\
Using the symmetry property of the parabola, we have: $DE=DC=\frac{5}{3}$, CE=2,\\
$\therefore \angle ECD=\angle DEC$,\\
From (2) we get $\angle ECD=\angle CAO$,\\
$\therefore \angle DEC=\angle CAB$,\\
$\because \triangle DEP$ and $\triangle ABC$ are similar,\\
(1) When $\triangle DEP \sim \triangle CAB$, we have $\frac{DE}{AC}=\frac{EP}{AB}$,\\
$\therefore \frac{\frac{5}{3}}{5}=\frac{EP}{4}$,\\
$\therefore EP=\boxed{\frac{4}{3}}$.\\
(2) When $\triangle DEP \sim \triangle BAC$, we have $\frac{DE}{AB}=\frac{EP}{AC}$,\\
$\therefore \frac{\frac{5}{3}}{4}=\frac{EP}{5}$,\\
$\therefore EP=\frac{25}{12}$.\\
In conclusion, EP=$\boxed{\frac{4}{3}}$ or $\boxed{\frac{25}{12}}$.


Question: As shown in the figure, point D lies on the circle $\odot O$, and point C is on the extension line of diameter BA. Connect CD, and $\angle CDA=\angle CBD$. CD is the tangent of $\odot O$; if $DC=4$ and $AC=2$, what is the length of OC?

Solution:
\textbf{Solution:} Since $\angle CDA=\angle CBD$ and $\angle ACD=\angle DCB$,\\
it follows that $\triangle ACD\sim \triangle DCB$,\\
thus $\frac{CD}{CB}=\frac{AC}{DC}$,\\
i.e., $\frac{4}{CB}=\frac{2}{4}$,\\
hence $CB=8$,\\
therefore $OA=\frac{CB-AC}{2}=\frac{8-2}{2}=3$,\\
thus $OC=OA+AC=3+2=\boxed{5}$.


Question: As shown in the figure, the line segment $CD \parallel AB$, $AD$ and $BC$ intersect at point $E$. $AE \cdot CE = DE \cdot BE$; through point $E$, draw $EF \parallel AB$, intersecting $AC$ at point $F$. If $AB=5$ and $EF=2$, what is the length of $CD$?

Solution:
\textbf{Solution:} Since EF$\parallel$AB,\\
it follows that $\angle$EFC$=\angle$BAC, $\angle$CEF$=\angle$B,\\
therefore, $\triangle$ CEF$\sim$ $\triangle$ CBA,\\
thus, $\frac{CF}{CA}=\frac{EF}{AB}=\frac{2}{5}$,\\
so, $\frac{AF}{CA}=\frac{3}{5}$,\\
Since CD$\parallel$AB and EF$\parallel$AB,\\
it follows that EF$\parallel$CD,\\
therefore, $\angle$EAF$=\angle$DAC, $\angle$EFA$=\angle$DCA,\\
thus, $\triangle$ AEF$\sim$ $\triangle$ ADC,\\
it follows that $\frac{EF}{CD}=\frac{AF}{CA}$,\\
hence, $\frac{2}{CD}=\frac{3}{5}$,\\
therefore, CD$=\boxed{\frac{10}{3}}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, point $F$ is on $\odot O$, $AE$ bisects $\angle CAD$ and intersects $\odot O$ at point $E$. Through point $E$, construct $ED\perp AF$, and extend it to meet the extension of $AF$ at point $D$. Extend $DE$ and $AB$ to meet at point $C$. $CD$ is a tangent to $\odot O$; given that $AB=10, \frac{ED}{AD}=\frac{1}{2}$, find the length of $BC$.

Solution:
\textbf{Solution:} Draw BE. Since AB is the diameter,\\
$\angle AEB=90^\circ=\angle D$, and $\angle DAE=\angle BAE$,\\
thus $\triangle ADE\sim \triangle AEB$,\\
thus $\frac{AD}{AE}=\frac{AE}{AB}=\frac{DE}{BE}$,\\
and $\frac{ED}{AD}=\frac{1}{2}$,\\
thus $\frac{DE}{AD}=\frac{BE}{AE}=\frac{1}{2}$, hence AE=2BE, and AB=10,\\
in $\triangle ABE$, $AE^2+BE^2=AB^2$, that is $(2BE)^2+BE^2=10^2$,\\
solving gives: BE=$2\sqrt{5}$, then AE=$4\sqrt{5}$,\\
thus $\frac{AD}{4\sqrt{5}}=\frac{4\sqrt{5}}{10}=\frac{DE}{2\sqrt{5}}$,\\
solving gives: AD=8, DE=4,\\
since OE$\parallel$AD,\\
$\triangle COE\sim \triangle CAD$,\\
thus $\frac{CO}{CA}=\frac{OE}{AD}$, let BC=x,\\
thus $\frac{x+5}{x+10}=\frac{5}{8}$, solving gives: x=$\boxed{\frac{10}{3}}$,\\
upon verification: x=$\frac{10}{3}$ is the solution to the original equation,\\
therefore, the length of BC is $\boxed{\frac{10}{3}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AD \parallel BC$, $\angle B = \angle ACD = 90^\circ$, $\triangle ABC \sim \triangle DCA$. If $BC=1$, $AC=2$, find the length of $AD$.

Solution:
Solution: Since $\triangle ABC \sim \triangle DCA$, \\
$\therefore \frac{BC}{AC}=\frac{AC}{AD}$, which means $\frac{1}{2}=\frac{2}{AD}$, \\
$\therefore AD=\boxed{4}$.


Question: As shown in the figure, the line $y=-x+3$ intersects the graph of the inverse proportion function $y=\frac{2}{x}\ (x>0)$ at points A and B. What are the coordinates of points A and B?

Solution:
\textbf{Solution}: From the given problem, we have
\[
\left\{
\begin{array}{l}
y=-x+3 \\
y=\frac{2}{x}
\end{array}
\right.
\]
Therefore, $-x+3=\frac{2}{x}$, solving this gives $x_1=1$, $x_2=2$, $y_1=2$, $y_2=1$,
therefore, the points are \boxed{A(1, 2)}, \boxed{B(2, 1)}.


Question: As shown in the figure, the line $y=-x+3$ intersects with the inverse proportion function $y=\frac{2}{x}\ (x>0)$ at points A and B. As shown in Figure 1, point E is a point on segment AC. Connect OE and OA. If $\angle AOE=45^\circ$, what is the value of $\frac{AE}{EC}$?

Solution: \textbf{Solution:} Since $y=-x+3$ intersects the $x$-axis at point $C$, $C(3, 0)$. Since $\angle OCA = \angle AOE = 45^\circ$ and $\angle OAE = \angle CAO$, $\triangle AOE \sim \triangle ACO$. Therefore, $\frac{AO}{AE} = \frac{AC}{AO}$, and thus $AO^2 = AE \cdot AC$. Given $A(1, 2)$ and $C(3, 0)$, we find $AO = \sqrt{2^2 + 1^2} = \sqrt{5}$ and $AC = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$. Therefore, $AE = \frac{AO^2}{AC} = \frac{5}{2\sqrt{2}}$ and $EC = \frac{3}{2\sqrt{2}}$. Hence, $\frac{AE}{EC} = \boxed{\frac{5}{3}}$.


Question: As shown in the figure, the line $y=-x+3$ intersects the graph of the inverse proportion function $y=\frac{2}{x}$ $(x>0)$ at points A and B. As illustrated in figure 2, when line AB is translated m units to the right along the x-axis, it intersects the graph of the inverse proportion function $y=\frac{2}{x}$ $(x>0)$ at points P and Q. By connecting AP and BQ, if the area of the quadrilateral ABQP is exactly equal to $m^{2}$, find the value of m.

Solution:
\textbf{Solution:} Let the equation of the line after translation be $y_{PQ}=-x+3+m$. Draw $DF\perp PQ$ at point F from D, thus $ED=m$ and $DF= \frac{\sqrt{2}m}{2}$.\\
The area of $\triangle ABPQ$ is $\frac{(AB+PQ)\cdot \frac{\sqrt{2}m}{2}}{2}=\frac{\sqrt{2}m(\sqrt{2}+PQ)}{4}=m^{2}$.\\
Therefore, $\sqrt{2}+PQ=2\sqrt{2}m$,\\
Hence, $PQ= 2\sqrt{2}m - \sqrt{2}$.\\
According to the problem, $\begin{cases}y=-x+3+m\\ y=\frac{x}{2}\end{cases}$.\\
Solving, we get $x_{1}=\frac{m+3+\sqrt{m^{2}+6m+1}}{2}$ and $x_{2}=\frac{m+3-\sqrt{m^{2}+6m+1}}{2}$,\\
Thus, $QH=x_{1}-x_{2}= \sqrt{m^{2}+6m+1}$,\\
Therefore, $PQ=\sqrt{2}\sqrt{m^{2}+6m+1}$,\\
Therefore, $\sqrt{2}\sqrt{m^{2}+6m+1} = 2\sqrt{2}m - \sqrt{2}$.\\
Therefore, $m^{2}+6m+1=4m^{2}-4m+1$,\\
Solving we get $m_{1}=0$ (discarded), $m_{2}=\frac{10}{3}$,\\
thus $m=\boxed{\frac{10}{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=90^\circ$, circle $\odot O$ with diameter $AB$ intersects $AC$ at point $D$, $AE$ is perpendicular to the tangent at point $D$, with the foot of the perpendicular being $E$. $AD$ bisects $\angle BAE$; If $CD=\frac{9}{5}$ and $BC=3$, find the length of segment $AB$.

Solution:
\textbf{Solution:} Connect BD, as shown in the figure,

$\because$ AB is a diameter,

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

$\because$ $\angle 2+\angle$ABD$=90^\circ$, $\angle 3+\angle$ABD$=90^\circ$,

$\therefore$ $\angle 2=\angle 3$,

$\because$ $\angle$DCB$=\angle$BCA,

$\therefore$ $\triangle CDB\sim \triangle CBA$,

$\therefore$ $\frac{CD}{CB}=\frac{CB}{AC}$,

$\therefore$ $\frac{\frac{9}{5}}{3}=\frac{3}{AC}$,

$\therefore$ AC$=5$,

$\therefore$ AB$=\sqrt{AC^2-BC^2}=\sqrt{5^2-3^2}=\boxed{4}$


Question: As shown in the diagram, $AB$ is the diameter of $\odot O$, with points $E$ and $F$ on $\odot O$, and $\overset{\frown}{BF} = 2\overset{\frown}{BE}$. Connect $OE$ and $AF$, and draw the tangent to $\odot O$ at point $B$, which intersects the extensions of $OE$ and $AF$ at points $C$ and $D$, respectively. $\angle COB = \angle A$; if $AB = 6$ and $CB = 4$, calculate the length of the line segment $FD$.

Solution:
\textbf{Solution:} Connect $BF$, as shown in the figure, \\
since $CD$ is the tangent to $\odot O$, \\
therefore $AB\perp CD$, \\
hence $\angle OBC=\angle ABD=90^\circ$, \\
since $\angle COB=\angle A$, \\
thus $\triangle OBC\sim \triangle ABD$, \\
thus $\frac{OB}{AB}=\frac{BC}{BD}$, that is $\frac{3}{6}=\frac{4}{BD}$, solving for $BD=8$, \\
in $\triangle ABD$ right-angled at $B$, $AD=\sqrt{AB^{2}+BD^{2}}=\sqrt{6^{2}+8^{2}}=10$, \\
since $AB$ is the diameter of $\odot O$, \\
thus $\angle AFB=90^\circ$, \\
since $\angle BDF=\angle ADB$, \\
thus $\triangle BDF$ right-angled at $B \sim \triangle ADB$ right-angled at $B$, \\
thus $\frac{DF}{DB}=\frac{DB}{DA}$, that is $\frac{DF}{8}=\frac{8}{10}$, \\
solving for: $DF=\boxed{\frac{32}{5}}$.


Question: Let point E be chosen on the side of rectangle ABCD. Fold $ \triangle BCE$ along BE so that point C precisely lands on point F on side AD. If $ \angle CBE=15^\circ$, $ BC=2BA$, as shown in the figure, when $ AB=5$ and $ AF\cdot FD=10$, find the length of BC.

Solution:
\textbf{Solution:} Let's fold $\triangle BCE$ along $BE$ such that point $C$ precisely lands on point $F$ on side $AD$, \\
then $\angle BFE=\angle C=90^\circ$, $CE=EF$, \\
also, since in rectangle $ABCD$, $\angle A=\angle D=90^\circ$, \\
then $\angle AFB+\angle DFE=90^\circ$, $\angle DEF+\angle DFE=90^\circ$, \\
therefore $\angle AFB=\angle DEF$, \\
thus $\triangle FAB\sim \triangle EDF$, \\
thus $\frac{AF}{DE}=\frac{AB}{DF}$, \\
thus $AF\cdot DF=AB\cdot DE$, \\
since $AF\cdot DF=10$, $AB=5$, \\
then $DE=2$, \\
therefore $CE=DC-DE=5-2=3$, \\
thus $EF=3$, \\
therefore $DF=\sqrt{EF^2-DE^2}=\sqrt{3^2-2^2}=\sqrt{5}$, \\
thus $AF=\frac{10}{\sqrt{5}}=2\sqrt{5}$, \\
therefore $BC=AD=AF+DF=2\sqrt{5}+\sqrt{5}=\boxed{3\sqrt{5}}$


Question: Consider a point $E$ on the side of rectangle $ABCD$. Fold $\triangle BCE$ along $BE$ such that point $C$ exactly aligns with point $F$ on side $AD$. As shown in the diagram, extend $EF$ and let it intersect the angle bisector of $\angle ABF$ at point $M$. Let $BM$ intersect $AD$ at point $N$. Given that $NF = AN + FD$, find the value of $\frac{AB}{BC}$.

Solution:
\textbf{Solution:} Draw $NG \perp BF$ at point G,

since $NF=AN+FD$,

thus $NF= \frac{1}{2}AD= \frac{1}{2}BC$,

since $BC=BF$,

thus $NF= \frac{1}{2}BF$,

since $\angle NFG=\angle AFB$, $\angle NGF=\angle BAF=90^\circ$,

thus $\triangle NFG \sim \triangle BFA$,

thus $\frac{NG}{AB}=\frac{FG}{FA}=\frac{NF}{BF}=\frac{1}{2}$,

Let $AN=x$,

since $BN$ bisects $\angle ABF$, $AN \perp AB$, $NG \perp BF$,

thus $AN=NG=x$, $AB=BG=2x$,

Let $FG=y$, then $AF=2y$,

since $AB^2+AF^2=BF^2$,

thus $(2x)^2+(2y)^2=(2x+y)^2$,

solving yields $y=\frac{4}{3}x$,

thus $BF=BG+GF=2x+\frac{4}{3}x=\frac{10}{3}x$,

thus $\frac{AB}{BC}=\frac{AB}{BF}=\frac{2x}{\frac{10}{3}x}=\boxed{\frac{3}{5}}$


Question: As shown in the figure, the parabola $y=-\frac{3}{4}x^{2}+bx+c$ intersects the x-axis at point $A(4,0)$ and the y-axis at point $B(0,3)$. Point $M(m,0)$ is a moving point on segment $OA$. The line passing through point $M$ and perpendicular to the x-axis intersects line $AB$ and the parabola at points $P$ and $N$, respectively. Find the equation of the parabola and write down the coordinates of its axis of symmetry and vertex.

Solution:
\textbf{Solution:} Since the parabola $y=-\frac{3}{4}x^{2}+bx+c$ intersects the $x$ axis at point $A(4,0)$ and the $y$ axis at point $B(0,3)$,\\
we have $\begin{cases}-\frac{3}{4}\times 4^{2}+4b+c=0\\ c=3\end{cases}$,\\
solving this, we get $\begin{cases}b=\frac{9}{4}\\ c=3\end{cases}$,\\
therefore, the equation of the parabola is $y=-\frac{3}{4}x^{2}+\frac{9}{4}x+3$,\\
since $y=-\frac{3}{4}x^{2}+\frac{9}{4}x+3=-\frac{3}{4}(x-\frac{3}{2})^{2}+\frac{75}{16}$,\\
therefore, the axis of symmetry of this parabola is $x=\frac{3}{2}$,\\
and the vertex coordinates are $\boxed{\left(\frac{3}{2}, \frac{75}{16}\right)}$.


Question: As shown in the figure, the parabola $y=-\frac{3}{4}x^{2}+bx+c$ intersects the x-axis at point $A(4,0)$ and the y-axis at point $B(0,3)$. Point $M(m,0)$ is a moving point on line segment $OA$, and the line passing through point $M$ and perpendicular to the x-axis intersects line $AB$ and the parabola at points $P$ and $N$, respectively. If the quadrilateral formed by points $P$, $N$, $B$, and $O$ is a parallelogram, find the value of $m$?

Solution:
\textbf{Solution:} Let the equation of line AB be $y=px+q$,\\
Substituting A(4, 0) and B(0, 3) gives $\left\{\begin{array}{l}4p+q=0 \\ q=3\end{array}\right.$,\\
Solving for $p$ and $q$ yields $\left\{\begin{array}{l}p=-\frac{3}{4} \\ q=3\end{array}\right.$,\\
$\therefore$ The equation of line AB is $y=-\frac{3}{4}x+3$,\\
$\because$ M(m, 0), and MN$\perp$x-axis,\\
$\therefore$ N(m, $-\frac{3}{4}m^{2}+\frac{9}{4}m+3$), and P(m, $-\frac{3}{4}m+3$),\\
$\therefore$ NP= $-\frac{3}{4}m^{2}+3m$, OB=3,\\
$\because$ NP$\parallel$OB, and the quadrilateral formed by vertices P, N, B, and O is a parallelogram,\\
$\therefore$ NP= OB, i.e., $-\frac{3}{4}m^{2}+3m=3$,\\
Upon rearrangement: $m^{2}-4m+4=0$,\\
Solving yields: $m=\boxed{2}$.


Question: As shown in the figure, the parabola $y=-\frac{3}{4}x^{2}+bx+c$ intersects the x-axis at point $A(4,0)$ and the y-axis at point $B(0,3)$. Point $M(m,0)$ is a moving point on line segment $OA$, and the line passing through point $M$ and perpendicular to the x-axis intersects line $AB$ and the parabola at points $P$ and $N$, respectively. If the triangle with vertices $B$, $P$, $N$ is similar to $\triangle ABO$, what are the coordinates of point $M$?

Solution:
\textbf{Solution:} Given that $A(4,0)$, $B(0,3)$, and $P(m, -\frac{3}{4}m+3)$,\\
hence $AB= \sqrt{4^2+3^2}=5$, and $BP= \sqrt{m^2+\left(-\frac{3}{4}m+3-3\right)^2}=\frac{5}{4}m$,\\
meanwhile $NP=-\frac{3}{4}m^2+3m$,\\
since $PN\parallel OB$,\\
it follows that $\angle BPN=\angle ABO$,\\
when $\frac{PB}{OB}=\frac{PN}{AB}$, $\triangle BPN\sim \triangle OBA$,\\
that is $\frac{\frac{5}{4}m}{3}=\frac{-\frac{3}{4}m^2+3m}{5}$,\\
after simplification, we get $9m^2-11m=0$, solving this yields $m_1=0$ (discard), $m_2=\frac{11}{9}$,\\
at this point, the coordinates of point M are $\boxed{\left(\frac{11}{9},0\right)}$;\\
when $\frac{PB}{AB}=\frac{PN}{OB}$, $\triangle BPN\sim \triangle ABO$,\\
that is $\frac{\frac{5}{4}m}{5}=\frac{-\frac{3}{4}m^2+3m}{3}$,\\
after simplification, we get $2m^2-5m=0$, solving this yields $m_1=0$ (discard), $m_2=3$,\\
at this point, the coordinates of point M are $\boxed{(3,0)}$;\\
In conclusion, the coordinates of point M are either $\boxed{\left(\frac{11}{9},0\right)}$ or $\boxed{(3,0)}$.


Question: As shown in the figure, in $\triangle ABC$ and $\triangle ADE$, $\triangle ABD \sim \triangle ACE$; $\angle BAC = \angle DAE = 90^\circ$, $\angle ABC = \angle ADE = 30^\circ$, $AC$ and $DE$ intersect at point F. Point D is on side $BC$, $\frac{AD}{BD} = \sqrt{3}$. Find the value of $\frac{DF}{CF}$.

Solution:
\textbf{Solution:} As shown in Figure 1, draw line EC,\\
$\because \angle BAC = \angle DAE = 90^\circ$, $\angle ABC = \angle ADE = 30^\circ$,\\
$\therefore \triangle ABC \sim \triangle ADE$,\\
From (1), we know $\triangle ABD \sim \triangle ACE$,\\
$\therefore \frac{AE}{CE} = \frac{AD}{BD} = \sqrt{3}$, $\angle ACE = \angle ABD = \angle ADE$,\\
In $\triangle ADE$, $\angle ADE = 30^\circ$,\\
$\therefore \frac{AD}{AE} = \sqrt{3}$,\\
$\therefore \frac{AD}{EC} = \frac{AD}{AE} \times \frac{AE}{CE} = \sqrt{3} \times \sqrt{3} = 3$.\\
$\because \angle ADF = \angle ECF$, $\angle AFD = \angle EFC$,\\
$\therefore \triangle ADF \sim \triangle ECF$,\\
$\therefore \frac{DF}{CF} = \frac{AD}{CE} = \boxed{3}$.


Question: As shown in the figure, A, B, and C are three points on circle \(O\) with a radius of 2, where AB is the diameter. The bisector of \(\angle BAC\) intersects the circle at point D. A perpendicular line is drawn from point D to the extended line of AC, intersecting it at point E. The extended line ED intersects the extended line of AB at point F. The line EF is tangent to circle \(O\); if \(DF = 4\sqrt{2}\), find the value of \(\cos\angle EAD\).

Solution:
\textbf{Solution:} In $\triangle \text{ODF}$, we have $OD=2$, $DF=4\sqrt{2}$,\\
$\therefore OF= \sqrt{OD^2+DF^2}=6$,\\
$\because OD\parallel AE$,\\
$\therefore \frac{OD}{AE}=\frac{OF}{AF}$,\\
$\therefore \frac{2}{AE}=\frac{6}{8}$,\\
$\therefore AE=\frac{8}{3}$,\\
$\therefore ED=\sqrt{AF^2-AE^2}=\sqrt{8^2-\left(\frac{8}{3}\right)^2}=\frac{16}{3}\sqrt{2}$,\\
$\therefore AD=\sqrt{AE^2+ED^2}=\sqrt{\frac{64}{9}+\frac{256}{9}}=\frac{4\sqrt{6}}{3}$,\\
$\therefore \cos\angle EAD=\frac{AE}{AD}=\boxed{\frac{\sqrt{6}}{3}}$.


Question: As shown in the figure, in a square $ABCD$ with side length 1, the diagonals $AC$ and $BD$ intersect at point $O$. Points $Q$ and $R$ are located on sides $AD$ and $DC$, respectively. Line $BR$ intersects line segment $OC$ at point $P$, where $QP \perp BP$ and $QP$ intersects $BD$ at point $E$. $\triangle APQ \sim \triangle DBR$; when $\angle QED$ equals $60^\circ$, find the value of $\frac{AQ}{DR}$.

Solution:
\textbf{Solution:} Let $OE=a$. In square $ABCD$, $\angle POE=90^\circ$, $OA=OB=OD$, \\
$\because \angle QED=60^\circ$, \\
$\therefore \angle BEP=60^\circ$, \\
In $\triangle OEP$, \\
$PE=\frac{OE}{\cos 60^\circ}=2a$, $OP=OE\cdot \tan 60^\circ=\sqrt{3}a$, \\
$\because QP\perp BP$, $\angle BEP=60^\circ$, \\
$\therefore \angle PBE=30^\circ$, \\
$\therefore BE=2PE=4a$, $BP=PE\cdot \tan 60^\circ=2\sqrt{3}a$, \\
$\therefore$ OA=OB=BE−OE=3a, \\
$\therefore$ BD=2OB=6a, \\
$\therefore AP=OA+OP=3a+\sqrt{3}a=(3+\sqrt{3})a$, \\
$\because \triangle APQ\sim \triangle DBR$, \\
$\therefore \frac{AQ}{DR}=\frac{AP}{BD}=\frac{(3+\sqrt{3})a}{6a}=\boxed{\frac{3+\sqrt{3}}{6}}$.


Question: As shown in the figure, $AB$ is the tangent to the circle $\odot O$, point $D$ is on $\odot O$, and line $AD$ intersects $\odot O$ at $C$. $CE$ is the diameter of $\odot O$, and line $BC$ is drawn. Given $\angle A=90^\circ$ and $CB$ bisects $\angle ACE$; when $AB=2$ and $AC=1$, what is the radius of $\odot O$?

Solution:
\textbf{Solution:} As shown in the diagram, connect $BE$,\\
in the right triangle $ABC$, $AB=2$, $AC=1$,\\
by the Pythagorean theorem, we have: $BC=\sqrt{AB^2+AC^2}=\sqrt{5}$,\\
since $CE$ is the diameter of $\odot O$,\\
thus $\angle CBE=90^\circ$,\\
therefore $\angle BAC=\angle EBC$,\\
since $\angle ACB=\angle OCB$,\\
thus $\triangle BAC\sim \triangle EBC$,\\
therefore $\frac{AC}{BC}=\frac{BC}{CE}$, that is, $\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{CE}$,\\
solving this gives: $CE=5$,\\
thus, the radius of $\odot O$ is $\boxed{\frac{5}{2}}$.


Question: Given the figure in the Cartesian coordinate system $xOy$, it is known that the parabola $y=x^{2}+bx$ passes through point A $(2, 0)$ and point $B\left(-1, m\right)$, with vertex at point D. What is the equation of line AB?

Solution:
\textbf{Solution:}

Given the parabola $y=x^{2}+bx$ passes through point A(2, 0),

\[
\therefore 2^{2}+2b=0, \text{ solving for } b, \text{ we get } b=-2,
\]

\[
\therefore \text{the equation of the parabola is } y=x^{2}-2x,
\]

When $x=-1$, $y=3$,

\[
\therefore \text{the coordinates of point B are } B(-1, 3),
\]

Let the equation of line AB be $y=kx+m\ (k\neq 0)$,

Substituting A(2, 0) and $B(-1, 3)$ into it, we get:

\[
\begin{cases}
2k+m=0, \\
-k+m=3,
\end{cases}
\]
solving these yields:
\[
\begin{cases}
k=-1, \\
m=2,
\end{cases}
\]

\[
\therefore \text{the equation of line AB is } \boxed{y=-x+2};
\]


Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, it is known that the parabola $y=x^{2}+bx$ passes through point $A(2,0)$ and point $B(-1, m)$, with its vertex at point $D$. What is the value of $\tan(\angle ABD)$?

Solution:
\textbf{Solution:} Connect BD and AD,

\[
\because y=x^{2}-2x=(x-1)^{2}-1,
\]

\[
\therefore \text{the coordinates of point } D \text{ are } D(1, -1),
\]

\[
\because A(2,0), B(-1, 3),
\]

\[
\therefore AB^{2}=(-1-2)^{2}+3^{2}=18, AD^{2}=(2-1)^{2}+(-1)^{2}=2, BD^{2}=(-1-1)^{2}+(-1-3)^{2}=20,
\]

\[
\therefore AB^{2}+AD^{2}=BD^{2},
\]

\[
\therefore \triangle ABD \text{ is a right-angled triangle},
\]

\[
\therefore \tan\angle ABD=\frac{AD}{AB}=\frac{\sqrt{2}}{\sqrt{18}}=\boxed{\frac{1}{3}}
\]


Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, it is known that the parabola $y=x^{2}+bx$ passes through point A $(2,0)$ and point $B(-1, m)$, with its vertex at point D. Suppose the line segment BD intersects the $x$-axis at point P. If point C is on the $x$-axis, and $\triangle ABC$ is similar to $\triangle ABP$, find the coordinates of point C.

Solution:
Solution: Let the equation of line BD be $y=k_1x+b_1\left(k_1\ne 0\right)$.\\
Substituting points $D\left(1, -1\right)$ and $B\left(-1, 3\right)$ yields:\\
$\begin{cases}k_1+b_1=-1\\ -k_1+b_1=3\end{cases}$, solving this gives: $\begin{cases}k_1=-2\\ b_1=1\end{cases}$,\\
$\therefore$ the equation of line BD is $y=-2x+1$,\\
When $y=0$, $x=\frac{1}{2}$,\\
$\therefore$ the coordinates of point P are $P\left(\frac{1}{2}, 0\right)$,\\
When $\triangle ABP\sim \triangle ABC$, $\angle ABC=\angle APB$,\\
As illustrated, draw BQ⊥x-axis at point Q from point B, then BQ=3, OQ=1,\\
$\because \triangle ABP\sim \triangle ABC$,\\
$\therefore \angle ABD=\angle BCQ$,\\
From (2), we know $\tan\angle ABD=\frac{1}{3}$,\\
$\therefore \tan\angle BCQ=\frac{1}{3}$,\\
$\therefore \frac{BQ}{CQ}=\frac{1}{3}$,\\
$\therefore CQ=9$,\\
$\therefore OC=OQ+CQ=10$,\\
$\therefore$ the coordinates of point C are $C\left(-10, 0\right)$;\\
When $\triangle ABP\sim \triangle ABC$, $\angle APB=\angle ACB$, in this case, point C coincides with point P,\\
$\therefore$ the coordinates of point C are $C\left(\frac{1}{2}, 0\right)$,\\
In conclusion, the coordinates of point C are either $C\left(-10, 0\right)$ or $\boxed{C\left(\frac{1}{2}, 0\right)}$.


Question: As shown in Figure 1, in the quadrilateral $ABCD$, $\angle ABD = \angle BCD = 90^\circ$, and $DB$ bisects $\angle ADC$. If $CD = 6$ and $AD = 8$, what is the length of $BD$?

Solution:
\textbf{Solution:} Given that $DB$ bisects $\angle ADC$,\\
thus $\angle ADB=\angle BDC$,\\
since $\angle ABD=\angle BCD=90^\circ$,\\
it follows that $\triangle ADB\sim \triangle BDC$,\\
hence $\frac{AD}{BD}=\frac{BD}{CD}$,\\
given that $CD=6$, $AD=8$,\\
thus $\frac{8}{BD}=\frac{BD}{6}$,\\
solving yields: $BD=\boxed{4\sqrt{3}}$.


Question: As shown in Figure 1, in quadrilateral $ABCD$, $\angle ABD = \angle BCD = 90^\circ$, $DB$ bisects $\angle ADC$, given that $CD = 6$, $AD = 8$. As shown in Figure 2, draw $BM \parallel CD$ through point $B$ to intersect $AD$ at $M$, and connect $CM$ to intersect $DB$ at $N$. Find the length of $DN$.

Solution:
\textbf{Solution:} Given that $BM\parallel CD$ and $DB$ bisects $\angle ADC$,\\
$\therefore \angle MBD=\angle BDC$, $\angle BDC=\angle BDM$,\\
$\therefore \angle MBD=\angle BDM$,\\
$\therefore MB=MD$,\\
Given $\angle MBD+\angle MBA=\angle ABD=90^\circ$ and $\angle BDM+\angle A=90^\circ$,\\
$\therefore \angle MBA=\angle A$,\\
$\therefore MB=MA$,\\
$\therefore MB=\frac{1}{2}AD=4$,\\
Given $BM\parallel CD$,\\
$\therefore \triangle MNB\sim \triangle CND$,\\
$\therefore \frac{BM}{DC}=\frac{BN}{DN}$,\\
$\therefore \frac{BN}{DN}=\frac{4}{6}=\frac{2}{3}$,\\
$\therefore \frac{DN}{DB}=\frac{3}{5}$,\\
Given $BD=4\sqrt{3}$,\\
$\therefore DN=\frac{3}{5}DB=\frac{3}{5}\times 4\sqrt{3}=\boxed{\frac{12\sqrt{3}}{5}}$,\\
i.e., the length of $DN$ is $\frac{12\sqrt{3}}{5}$.


Question: As shown in the figure, in $\triangle ABC$, $D$ is a point on $AC$ and $AC=kAD$. $E$ is a point on $BC$, and $BD$ intersects $AE$ at point $O$. If $\angle AOB + \angle BAC = 180^\circ$. In figure 2, if $BE=2CE$, find the value of $\frac{AO}{AB}$.

Solution:
\textbf{Solution:} As shown in Figure 2, construct $EF \parallel BD$ intersecting $AC$ at $F$, let $AD=a$, let $OD=x$, thus $AC=k\cdot a$, $CD=(k-1)\cdot a$,

$\therefore \triangle EFC\sim \triangle BDC$, $\triangle AOD\sim \triangle AEF$

$\therefore \frac{EF}{BD}=\frac{CF}{CD}=\frac{CE}{BC}=\frac{1}{3}$, $\frac{OD}{EF}=\frac{AD}{AF}$,

$\therefore BD=3EF$, $CF=\frac{1}{3}CD=\frac{1}{3}(k-1)\cdot a$,

$\therefore \frac{x}{EF}=\frac{a}{ka-\frac{1}{3}(k-1)a}$,

$\therefore EF=\frac{2k+1}{3}x$,

$\therefore BD=(2k+1)\cdot x$,

$\because \angle CAE=\angle ACD$, $\angle ADO=\angle ADB$,

$\therefore \triangle AOD\sim \triangle BAD$,

$\therefore \frac{AD}{BD}=\frac{OD}{AD}=\frac{AO}{AB}$,

$\therefore \frac{a}{(2k+1)\cdot x}=\frac{x}{a}$,

$\therefore x=\sqrt{\frac{1}{2k+1}}\cdot a$

$\therefore \frac{AO}{AB}=\frac{\sqrt{\frac{1}{2k+1}}}{a}\cdot a=\boxed{\frac{\sqrt{2k+1}}{2k+1}}$.


Question: As shown in the figure, DP is the tangent of the circle $\odot O$, with D being the point of tangency. The chord AB is parallel to DP. Extend BO to intersect $\odot O$ at point C and DP at point E. Extend AC to intersect DP at point F. Connect OD. AF is parallel to OD; If OD = 5 and AB = 8, find the length of the line segment EF.

Solution:
\textbf{Solution:}

Given that OH $\perp$ AB, and AB = 8, \\
$\therefore$ BH = AH = 4, \\
$\therefore$ OH = $\sqrt{OB^{2} - BH^{2}}$ = $\sqrt{5^{2} - 4^{2}}$ = 3, \\
Given that BH $\parallel$ ED, \\
$\therefore$ $\triangle$ BOH $\sim$ $\triangle$ EOD, \\
$\therefore$ $\frac{BH}{ED}$ = $\frac{OH}{OD}$, that is, $\frac{4}{ED}$ = $\frac{3}{5}$, \\
Solving this yields: ED = $\frac{20}{3}$, \\
Given that $\angle$BAC = $90^\circ$, DH $\perp$ AB, DH $\perp$ DP, \\
$\therefore$ Quadrilateral AFDH is a rectangle, \\
$\therefore$ DF = AH = 4, \\
$\therefore$ EF = ED - DF = $\frac{20}{3} - 4$ = $\boxed{\frac{8}{3}}$.


Question: As shown in the figure, the side length of the square is 1. Take a point E on ray AB, connect DE, and rotate $\triangle ADE$ $90^\circ$ around point D. Point E falls on point F, connect EF, which intersects with the line containing diagonal BD at point M, and intersects with ray DC at point N. When $AE=\frac{1}{3}$, what is the value of $\tan\angle EDB$?

Solution:
\textbf{Solution:} Draw $EH \perp BD$ at H,

since the side length of the square is 1, $AE=\frac{1}{3}$,

thus $EB=1-AE=1-\frac{1}{3}=\frac{2}{3}$,

since BD is the diagonal of the square,

thus BD bisects $\angle ABC$,

thus $\angle ABD=45^\circ$,

since $EH \perp BD$,

thus $\angle BEH=180^\circ-\angle EBH-\angle EHB=180^\circ-45^\circ-90^\circ=45^\circ$,

thus $EH=BH$,

thus $EH=BH=BE\sin45^\circ= \frac{2}{3}\times \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{3}$,

thus $BD=1\div\frac{\sqrt{2}}{2}=\sqrt{2}$,

thus $DH=DB-BH= \sqrt{2}-\frac{\sqrt{2}}{3}=\frac{2\sqrt{2}}{3}$,

thus $\tan\angle EDB=\frac{EH}{HD}=\frac{\frac{\sqrt{2}}{3}}{\frac{2\sqrt{2}}{3}}=\boxed{\frac{1}{2}}$;


Question: Given: As shown in the diagram, the side length of the square is 1. Take a point E on ray AB, connect DE, and rotate ADE $90^\circ$ about point D. Point E falls on point F. Connect EF, intersecting the line where diagonal BD is located at point M, and intersecting ray DC at point N. When point E is on segment AB, if $AE=x$ and $FM=y$, find the analytic expression of $y$ as a function of $x$, and state its domain.

Solution:
\textbf{Solution:} Given the problem, we have \\
$\because$AE=x,\\
$\therefore$BE=1−x,\\
$\because$ Rotating $\triangle ADE$ $90^\circ$ around point D gives $\triangle DCF$,\\
$\therefore$CF=AE=x, ED=FD=$\sqrt{AD^{2}+AE^{2}}=\sqrt{1+x^{2}}$,\\
$\therefore$BF=BC+CF=1+x,\\
In $\triangle EBF$, EF=$\sqrt{BE^{2}+BF^{2}}=\sqrt{(1−x)^{2}+(1+x)^{2}}=\sqrt{2+2x^{2}}$,\\
$\because$$\angle EDF=90^\circ$, ED=FD,\\
$\therefore$$\triangle DEF$ is an isosceles right triangle,\\
$\therefore$$\angle DFE=\angle DEF=45^\circ$,\\
$\therefore$$\angle EBM=\angle MFD=45^\circ$,\\
$\because$$\angle EMB=\angle DMF$,\\
$\therefore$$\triangle BEM\sim \triangle FDM$,\\
$\therefore$$\frac{BE}{DF}=\frac{BM}{FM}$, that is, $\frac{1−x}{\sqrt{1+x^{2}}}=\frac{BM}{y}$,\\
$\because$$\angle DEM=\angle FBM=45^\circ$, $\angle EMD=\angle BMF$,\\
$\therefore$$\triangle EMD\sim \triangle BMF$,\\
$\therefore$$\frac{ED}{BF}=\frac{EM}{BM}$, that is, $\frac{\sqrt{1+x^{2}}}{1+x}=\frac{\sqrt{2+2x^{2}}−y}{BM}$,\\
$\therefore$$\frac{1−x}{\sqrt{1+x^{2}}}\times \frac{\sqrt{1+x^{2}}}{1+x}=\frac{BM}{y}\times \frac{\sqrt{2+2x^{2}}−y}{BM}$,\\
$\therefore$$\frac{1−x}{1+x}=\frac{\sqrt{2+2x^{2}}−y}{y}$,\\
$\therefore$$\frac{1−x+1+x}{1+x}=\frac{\sqrt{2+2x^{2}}−y+y}{y}$, that is, $\frac{2}{1+x}=\frac{\sqrt{2+2x^{2}}}{y}$,\\
$\therefore$$\boxed{y=\frac{1}{2}(1+x)\sqrt{2+2x^{2}}}$, with the domain $0\leq x\leq 1$.


Question: As shown in the figure, the side length of the square is 1. A point E is chosen on ray AB, and DE is connected. Then, triangle ADE is rotated $90^\circ$ around point D, and point E falls on point F. Connect EF, which intersects the line containing diagonal BD at point M, and intersects ray DC at point N. Connect AM, and let the line AM intersect line BC at point G. When $BG = \frac{1}{3}$, find the value of AE.

Solution:
\textbf{Solution:} For point G on segment BC, where $BG=\frac{1}{3}$,

since quadrilateral ABCD is a square,

therefore AD$\parallel$BG,

thus $\angle$DAM=$\angle$BGM, $\angle$ADM=$\angle$GBM,

which implies $\triangle$BGM$\sim$$\triangle$DAM,

hence $\frac{BG}{DA}=\frac{BM}{DM}=\frac{\frac{1}{3}}{1}=\frac{1}{3}$,

given from (2) that $\triangle$BEM$\sim$$\triangle$FDM,

thus $\frac{BM}{MF}=\frac{BE}{DF}$,

since DB=$\sqrt{AB^{2}+AD^{2}}=\sqrt{2}$,

then BM=$\frac{\sqrt{2}}{4}$,

hence $\frac{\frac{\sqrt{2}}{4}}{y}=\frac{1-x}{\sqrt{1+x^{2}}}$,

given $y=\frac{1}{2}(1+x)\sqrt{2+2x^{2}}$,

thus $\frac{\frac{\sqrt{2}}{4}}{\frac{1}{2}(1+x)\sqrt{2+2x^{2}}}=\frac{1-x}{\sqrt{1+x^{2}}}$ results in $1-x^{2}=\frac{1}{2}$,

solution $x_{1}=\frac{\sqrt{2}}{2}$, $x_{2}=-\frac{\sqrt{2}}{2}$ discard;

For point G on the extension of CB, where $BG=\frac{1}{3}$, draw ML$\perp$BC intersecting line BC at L,

since GB$\parallel$AD,

thus $\angle$DAM=$\angle$BGM, $\angle$ADM=$\angle$GBM,

which implies $\triangle$BGM$\sim$$\triangle$DAM,

hence $\frac{BG}{DA}=\frac{BM}{DM}=\frac{\frac{1}{3}}{1}=\frac{1}{3}$,

thus BM=$\frac{1}{3}DM$,

hence BM=$\frac{1}{2}BD$,

given $\angle$LBM=$\angle$CBD=$45^\circ$, and ML$\perp$BC,

$\triangle$MLB forms an isosceles right triangle,

since ML$\parallel$CD,

thus $\angle$LMB=$\angle$CDB, $\angle$L=$\angle$DCB,

which implies $\triangle$MLB$\sim$$\triangle$DCB,

thus $\frac{BM}{BD}=\frac{ML}{DC}=\frac{1}{2}$, with CD=1,

hence ML=$\frac{1}{2}$,

since ML$\parallel$BE,

thus $\angle$L=$\angle$FBE, $\angle$LMF=$\angle$BEF,

which implies $\triangle$LMF$\sim$$\triangle$BEF,

thus $\frac{LM}{BE}=\frac{LF}{BF}$,

given BE=AE−AB=x−1, LF=LB+BC+CF=$\frac{1}{2}+1+x=\frac{3}{2}+x$, BF=BC+CF=1+x,

thus $\frac{\frac{1}{2}}{x-1}=\frac{\frac{3}{2}+x}{1+x}$,

rearranging gives: $2x^{2}=4$,

solution $x_{3}=\sqrt{2}$, $x_{4}=-\sqrt{2}$ discard,

therefore, the value of AE is $\boxed{\sqrt{2}}$.


Question: As shown in the figure, both $\triangle ADB$ and $\triangle BCD$ are equilateral triangles. Extend $AD$ to $E$ such that $\angle AEC=90^\circ$, and $AD=5$. A moving point $M$ starts from point $B$ and moves towards $D$ at a speed of $1$ unit per second, while point $N$ moves from $D$ towards $C$ at a speed of $2$ units per second. The movement stops when one of them reaches their endpoint. Connect $AM$, $CM$, $MN$, $NE$. Let the movement time be $t$ ($0\leq t\leq 2.5$). At what value of $t$ is $MN\parallel BC$?

Solution:
\textbf{Solution:} Given that $\triangle ADB$ and $\triangle BCD$ are both equilateral triangles, and $AD=5$,\\
thus $BD=DC=AD=5$,\\
therefore $BM=t$, $DN=2t$,\\
given $MN\parallel BC$;\\
thus $\angle NMD=\angle DBC=60^\circ=\angle MDN$,\\
therefore $\triangle MDN$ is an equilateral triangle,\\
thus $DM=DN$, i.e., $5-t=2t$,\\
solving gives $t=\boxed{\frac{5}{3}}$.


Question: As illustrated, both $\triangle ADB$ and $\triangle BCD$ are equilateral triangles. Extend $AD$ to point $E$ such that $\angle AEC=90^\circ$, with $AD=5$. A moving point $M$ starts from point $B$, travelling towards $D$ at a speed of 1 unit/second, while simultaneously, point $N$ moves from $D$ towards $C$ at a speed of 2 units/second. The movement of both points ceases the moment either reaches its destination. Connect $AM$, $CM$, $MN$, and $NE$. Given the duration of movement as $t$ ($0 \leq t \leq 2.5$), connect $BN$. At what value of $t$ will points $B$, $N$, and $E$ be collinear?

Solution:
\textbf{Solution:} Since $\angle ADE$ is a straight angle, \\
it follows that $\angle CDE=180^\circ-\angle ADB-\angle BDC=180^\circ-60^\circ-60^\circ=60^\circ$,\\
Since $\angle CEA=90^\circ$,\\
it follows that $\angle DCE=180^\circ-\angle CDE-\angle CED=180^\circ-60^\circ-90^\circ=30^\circ$,\\
therefore $DE=\frac{1}{2}CD=\frac{1}{2}\times 5=2.5$, and $CE=\sqrt{CD^{2}-DE^{2}}=\sqrt{5^{2}-2.5^{2}}=\frac{5}{2}\sqrt{3}$\\
Since points B, N, and E are collinear;\\
it follows that $\angle BNC=\angle END$,\\
Since $\angle BCD=\angle CDE=60^\circ$,\\
it follows that BC$\parallel$DE,\\
therefore $\triangle BCN\sim \triangle EDN$,\\
thus $\frac{BC}{ED}=\frac{CN}{DN}$ that is $\frac{5}{2.5}=\frac{5-DN}{DN}$,\\
solving gives $DN=\frac{5}{3}$,\\
hence $2t=\frac{5}{3}$,\\
solving for $t$ yields $t=\boxed{\frac{5}{6}}$,\\
therefore, when $t=\boxed{\frac{5}{6}}$, points B, N, and E are collinear.


Question: As shown in the figure, it is known that in triangle $ABC$, the side $AB$ is tangent to circle $O$ at point $B$. $AC$ passes through the center $O$ and intersects the circle at points $D$ and $C$. A line $CE$ is drawn through $C$ perpendicular to $AB$, intersecting the extension of $AB$ at point $E$, and $CB$ bisects $\angle ACE$; if $BE=3$ and $CE=4$, what is the radius of circle $O$?

Solution:
\textbf{Solution:} As shown in Figure 2, draw line BD, \\
since $CE \perp AB$, \\
hence $\angle E=90^\circ$, \\
thus $BC=\sqrt{BE^2+CE^2}=\sqrt{3^2+4^2}=5$, \\
since CD is the diameter of the circle $O$, \\
hence $\angle DBC=90^\circ$, \\
thus $\angle E=\angle DBC$, \\
hence $\triangle DBC \sim \triangle CBE$, \\
thus $\frac{CD}{BC}=\frac{BC}{CE}$, \\
hence $BC^2=CD \cdot CE$, \\
thus $CD=\frac{5^2}{4}=\frac{25}{4}$, \\
hence $OC=\frac{1}{2}CD=\frac{25}{8}$, \\
thus the radius of circle $O$ is $\boxed{\frac{25}{8}}$.


Question: As shown in the figure, in $\triangle ABC$, D and E are points on AB and AC respectively, $\angle AED = \angle B$, $AD = 2$, $AC = 3$, and the angle bisector of $\triangle ABC$ intersects DE at point G and intersects BC at point F. $\triangle ADE \sim \triangle ACB$; find the value of $\frac{AG}{GF}$.

Solution:
\textbf{Solution:} Given that $\triangle ADE \sim \triangle ACB$,\\
$\therefore \angle ADE = \angle C$,\\
as $AF$ bisects $\angle BAC$,\\
$\therefore \angle DAG = \angle CAF$,\\
$\therefore \triangle ADG \sim \triangle ACF$,\\
$\therefore \frac{AG}{AF} = \frac{AD}{AC}$,\\
given that $AD = 2$ and $AC = 3$,\\
$\therefore \frac{AG}{AF} = \frac{2}{3}$,\\
$\therefore \frac{AG}{GF} = \boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AB=5$, $BC=\frac{10}{3}$, point $O$ is on $AB$, $OB=2$, and a circle $\odot O$ with $OB$ as the radius intersects $BC$ at point $D$. $AC$ is a tangent to $\odot O$; what is the length of $CD$?

Solution:
\textbf{Solution:} Draw OF$\perp$BC at point O, with F as the foot,\\
since $\angle$OEA=$\angle$C=$\angle$OFC=90$^\circ$,\\
thus quadrilateral OFCE is a rectangle,\\
thus OE=CF=2,\\
since BC=$\frac{10}{3}$,\\
thus BF=BC−CF=$\frac{4}{3}$,\\
since OF$\perp$BD,\\
thus BD=2BF=$\frac{8}{3}$,\\
thus CD=BC−BD=$\boxed{\frac{2}{3}}$.


Question: As shown in the figure, point $C$ is on the line segment $AB$. Through point $B$, a line $l \perp AB$ is drawn, and point $P$ is a point on line $l$. Connect $AP$, and point $Q$ is the midpoint of $AP$. Draw $QR \perp AB$ with the foot of the perpendicular being $R$. Connect $CQ$, with $AC=2$, $BC=1$, and $CQ=1$. What is the length of $CR$?

Solution:
\textbf{Solution:} Given that, $QR\perp AB$ and $l\perp AB$, \\
$\therefore QR\parallel BP$, \\
$\therefore \frac{AQ}{AP}=\frac{AR}{AB}$, \\
As point Q is the midpoint of AP, \\
$\therefore \frac{AQ}{AP}=\frac{AR}{AB}=\frac{1}{2}$, \\
Also given that $AC=2$ and $BC=1$, \\
$\therefore AB=3$, \\
$\therefore AR=\frac{3}{2}$, \\
$\therefore CR=AC-AR=\boxed{\frac{1}{2}}$.


Question: As shown in the figure, point $C$ is on line segment $AB$. A line $l \perp AB$ passes through point $B$, with point $P$ on line $l$. Connect $AP$, let point $Q$ be the midpoint of $AP$, and draw $QR \perp AB$ with the foot at $R$. Connect $CQ$, with $AC=2$, $BC=1$, and $CQ=1$. $\triangle RCQ \sim \triangle QCA$. What is the measure, in degrees, of $\angle AQC$?

Solution:
\textbf{Solution:} Since $\triangle RCQ\sim \triangle QCA$, \\
it follows that $\angle AQC=\angle QRC=90^\circ=\boxed{90^\circ}$.


Question: In the figure, point D is on side AB of $\triangle ABC$, with $\angle ABC = \angle ACD$. $\triangle ABC \sim \triangle ACD$; if $AD = 2$ and $AB = 8$, find the length of $AC$.

Solution:
\textbf{Solution:} Since $\triangle ABC\sim \triangle ACD$,\\
it follows that $\frac{AC}{AD}=\frac{AB}{AC}$,\\
Given $AD=2$, $AB=8$,\\
we have $\frac{AC}{2}=\frac{8}{AC}$,\\
therefore $AC= \boxed{4}$.


Question: As shown in the diagram, in $\triangle ABC$, where $AB=AC=10$ and $BC=12$, points $D$ and $E$ are located on sides $BC$ and $AC$ respectively (with point $D$ being distinct from endpoints $B$ and $C$), and it is given that $\angle ADE = \angle B$. $\triangle ABD \sim \triangle DCE$; Let $BD=x$ and $CE=y$, determine the value of $x$ for which $y$ attains its maximum value. What is the maximum value of $y$?

Solution:
Solution: Let $BD=x$, $CE=y$, \\
Therefore, $DC=BC-BD=12-x$, \\
From (1), we know that $\triangle ABD \sim \triangle DCE$, \\
Therefore, $\frac{AB}{CD}=\frac{BD}{CE}$, \\
Therefore, $\frac{10}{12-x}=\frac{x}{y}$, \\
Therefore, $y=-\frac{1}{10}x^2+\frac{6}{5}x=-\frac{1}{10}(x-6)^2+\frac{18}{5}$, \\
Since $0<x<12$ and $-\frac{1}{10}<0$, \\
Therefore, when $x=\boxed{6}$, $y$ reaches its maximum value, which is $\boxed{\frac{18}{5}}$;


Question: As shown in the figure, in $\triangle ABC$, we have $AB=AC=10$ and $BC=12$. Let points $D$ and $E$ be on the sides $BC$ and $AC$, respectively (point $D$ does not coincide with the endpoints $B$ and $C$), and suppose $\angle ADE = \angle B$. When $\triangle ADE$ is an isosceles triangle, find the length of $BD$.

Solution:
\textbf{Solution:}

(1) When $DA=DE$,\\
$\therefore \angle DAE=\angle AED$,\\
$\because AB=AC=10$,\\
$\therefore \angle B=\angle C=\angle ADE$.\\
$\because \angle AED=\angle EDC+\angle C=\angle EDC+\angle ADE$,\\
$\therefore \angle DAE=\angle EDC+\angle ADE$,\\
$\therefore \angle EAD=\angle ADC$,\\
$\therefore CD=AC=10$,\\
$\therefore x=BD=BC-CD=12-10=2$,\\
Therefore, when $x=2$, $\triangle ADE$ is an isosceles triangle,\\
(2) When $AE=DE$, $\triangle ADE$ is an isosceles triangle, i.e., $\angle DAE=\angle ADE=\angle B$\\
Also, $\angle ACD=\angle BCA$,\\
$\therefore \triangle ADC\sim \triangle BAC$,\\
$\therefore \frac{AC}{BC}=\frac{CD}{AC}$,\\
$\therefore DC\cdot BC=AC^{2}$,\\
$\therefore DC= \frac{25}{3}$,\\
$\therefore x=BD=12-DC=12-\frac{25}{3}=\frac{11}{3}$;\\
In summary: $\triangle ADE$ is an isosceles triangle when $x=2$ or $x=\frac{11}{3}$.\\
$\therefore \boxed{2}$ or $\boxed{\frac{11}{3}}$.


Question: Given: As shown in the figure, in $\triangle ABC$, $AD$ is the altitude on side $BC$. There is an inscribed rectangle $PQMN$ within the triangle, with one side of the rectangle on $BC$, and the other two vertices on $AB$ and $AC$, respectively. If $BC=60$ and $AD=40$, what is the length of $PQ$ when $PQ=PN$?

Solution:
\textbf{Solution:} Since quadrilateral PQMN is a rectangle,\\
it follows that PQ$\parallel$BC,\\
which implies $\triangle APQ\sim \triangle ABC$,\\
also, since AD$\perp$BC,\\
this leads to AE$\perp$PQ,\\
thus, quadrilateral PEDN forms a rectangle,\\
resulting in PN=DE,\\
from $\triangle APQ\sim \triangle ABC$ we get $ \frac{PQ}{BC}=\frac{AE}{AD}$,\\
since PQ=PN,\\
it follows that $ \frac{PQ}{60}=\frac{40-PQ}{40}$,\\
solving this yields PQ$=\boxed{24}$;


Question: Given: As shown in the figure, in $\triangle ABC$, $AD$ is the altitude from $A$ to side $BC$. There is an inscribed rectangle $PQMN$ within this triangle, one side of the rectangle lies on $BC$, and the other two vertices are on $AB$ and $AC$, respectively. If $BC=100$, $AD=40$, and $PQ=PN$ with $\angle BAC=90^\circ$, directly provide the value of $BN\cdot CM$.

Solution: \textbf{Solution:} We have $BN \cdot CM = \boxed{\frac{40000}{49}}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ is the altitude to side $BC$. Within this triangle, there is an inscribed rectangle $PQMN$, with one side lying on $BC$, and the other two vertices respectively on $AB$, $AC$. If $BC=60$ and $AD=40$, find the lengths of the sides of rectangle $PQMN$ when the area of $PQMN$ is maximized.

Solution:
\textbf{Solution:} Let $PN=x$, \\
then $DE=PN=x$, $AE=AD-DE=40-x$, \\
since $\triangle APQ \sim \triangle ABC$, \\
we have $\frac{AE}{AD}=\frac{PQ}{BC}$, i.e., $\frac{40-x}{40}=\frac{PQ}{60}$, \\
thus $PQ=60-\frac{3}{2}x$, \\
therefore, the area of rectangle $PQMN$ is $PQ \cdot PN= \left(60-\frac{3}{2}x\right)x$ \\
$= -\frac{3}{2}x^{2}+60x$ \\
$= -\frac{3}{2}(x-20)^{2}+600$, \\
when $x=20$, the area of rectangle $PQMN$ reaches its maximum value of $600$, \\
at this point, $PN=20, PQ=60-\frac{3}{2}x=60-\frac{3}{2}\times 20=30$. \\
Therefore, the dimensions of the rectangle are $\boxed{30}$ and $\boxed{20}$.


Question: As shown in the figure, the side length of square paper $ABCD$ is 12, where $E$ is a point on side $CD$. Connect $AE$ and fold the paper so that point $A$ falls exactly on point $G$ on $AE$, creating a crease $BF$ that intersects with $AD$ at point $F$. When $E$ is the midpoint of $CD$, what is the length of $AF$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a square, \\
it follows that $AB=AD$, and $\angle BAD=\angle D=90^\circ$, \\
Since $\angle ABF+\angle BAG=90^\circ$, and $\angle DAE+\angle BAG=90^\circ$, \\
it follows that $\angle ABF=\angle DAE$, \\
Therefore, $\triangle ABF\cong \triangle DAE$ (by ASA), \\
Hence, $AP=DE$, $BF=AE$ \\
Since E is the midpoint of CD, it follows that $DC=12$, therefore $AF=DE=\boxed{6}$.


Question: As shown in the figure, the side length of the square paper $ABCD$ is $12$. Point $E$ is on side $CD$, and line $AE$ is connected. The paper is folded such that point $A$ exactly coincides with point $G$ on $AE$, resulting in the crease $BF$ intersecting $AD$ at point $F$. If $DE=5$, what is the length of $GE$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a square,\\
we have AB=AD=12 and $\angle$BAD=$\angle$D=90$^\circ$,\\
by the property of folding and axial symmetry, $\triangle$ABF$\cong$$\triangle$GBF, BF bisects AG perpendicularly,\\
thus BF$\perp$AE, and AH=GH,\\
which means $\angle$BAH+$\angle$ABH=90$^\circ$,\\
also, since $\angle$FAH+$\angle$BAH=90$^\circ$,\\
we get $\angle$ABH=$\angle$FAH,\\
thus, $\triangle$ABF$\cong$$\triangle$DAE (ASA),\\
therefore, AF=DE=5,\\
in right triangle $\triangle$ABF,\\
BF=$\sqrt{AB^{2}+AF^{2}}=\sqrt{12^{2}+5^{2}}=13$,\\
$Area_{\triangle ABF}=\frac{1}{2}$AB$\cdot$AF=$\frac{1}{2}$BF$\cdot$AH,\\
therefore, 12$\times$5=13AH,\\
thus AH=$\frac{60}{13}$,\\
therefore, AG=2AH=$\frac{120}{13}$,\\
since AE=BF=13,\\
thus, GE=AE−AG=13−$\frac{120}{13}=\boxed{\frac{49}{13}}$.


Question: As shown in the figure, in $\triangle ABC$, $D$ is a point on $AB$ such that $AD=DC$. $DE$ bisects $\angle ADC$ and intersects $AC$ at point $E$. $DF$ bisects $\angle BDC$ and intersects $BC$ at point $F$, where $\angle DFC=90^\circ$. Quadrilateral $CEDF$ is a rectangle; if $\angle B=30^\circ$ and $AD=2$, connect $BE$ and find the length of $BE$.

Solution:
\textbf{Solution:} From (1), it is known that quadrilateral CEDF is a rectangle,\\
$\therefore \angle CED = \angle ECF = 90^\circ$,\\
$\therefore \angle A = 90^\circ - \angle B = 90^\circ - 30^\circ = 60^\circ$, DE$\perp$AC,\\
$\because AD = DC$,\\
$\therefore CE = AE$, $\triangle ACD$ is an equilateral triangle,\\
$\therefore \angle ACD = 60^\circ$, AC = AD = 2,\\
$\therefore AE = CE = 1$,\\
$\therefore DE = \sqrt{AD^2 - AE^2} = \sqrt{3}$,\\
$\angle DCB = \angle ECF - \angle ACD = 90^\circ - 60^\circ = 30^\circ$,\\
$\therefore \angle DCB = \angle B$,\\
$\therefore DB = DC = AD$,\\
$\therefore DE$ is the median of $\triangle ABC$,\\
$\therefore BC = 2DE = 2\sqrt{3}$,\\
In $\triangle BCE$, by the Pythagorean theorem: $BE = \sqrt{CE^2 + BC^2} = \sqrt{13}$,\\
Hence, the length of BE is $\boxed{\sqrt{13}}$.


Question: As shown in the figure, points A and B are on the graph of the linear function $y=kx+1$, where point A is in the first quadrant and point B is on the x-axis. Point D is on the positive half of the x-axis, and the coordinates of point C are $(1, -2)$. Quadrilateral OADC is a rhombus. What is the value of $k$?

Solution:
\textbf{Solution:} Given the coordinates of point C as $\left(1, -2\right)$, and using the properties of a rhombus,\\
$\therefore$ the coordinates of point A are $\left(1, 2\right)$, and of D $\left(2, 0\right)$,\\
substituting point A $\left(1, 2\right)$ into $y=kx+1$ gives $k+1=2$,\\
solving for $k$ yields $k=\boxed{1}$;\\
$\therefore$ the equation of line AB is $y=x+1$.


Question: As shown in the figure, points A and B lie on the graph of the linear function $y=kx+1$, with point A in the first quadrant and point B on the x-axis. Point D is located on the positive half of the x-axis, and point C has coordinates $(1, -2)$. The quadrilateral OADC is a rhombus. What is the area of $\triangle ABD$?

Solution:
\textbf{Solution:} From (1), the linear equation can be expressed as $y=x+1$,\\
when $y=0$, then $x=-1$,\\
$\therefore$ point B$(-1,0)$,\\
which means $OB=1$,\\
$\therefore$ $BD=1+2=3$,\\
$\therefore$ the area of $\triangle ABD$ is $\frac{1}{2} \times 3 \times 2=\boxed{3}$.


Question: As shown in the figure, for the linear function $y=kx+1$, there are points A and B on its graph, with point A in the first quadrant and point B on the x-axis. Point D is on the positive half of the x-axis, and the coordinates of point C are $(1, -2)$. The quadrilateral OADC is a rhombus. Let point P be a moving point on line AB, and $S_{\triangle AOP}=\frac{1}{2}S_{\text{rhombus} OADC}$. Find the coordinates of point P.

Solution:
\textbf{Solution:} Given by the problem, $A(1, 2), D(0, 2)$,\\
$\therefore S_{\text{rhombus} AOCD}=2\times \frac{1}{2}\times 2\times 2=4$,\\
As shown in the diagram, let the intersection of AB with the y-axis be Q,\\
when $x=0$, then $y=1$, i.e., Q$(0, 1)$,\\
Let point P be $(x, x+1)$,\\
$\therefore S_{\triangle AOP}=S_{\triangle AOQ}+S_{\triangle POQ}=\frac{1}{2}\times OQ\times |x-1|=\frac{1}{2}|x-1|$,\\
Since $S_{\triangle AOP}= \frac{1}{2}S_{\text{rhombus} AOCD}$,\\
$\therefore \frac{1}{2}|x-1|=2$,\\
Solving gives $x=5$ or $x=-3$,\\
When $x=5$, $y=6$, when $x=-3$, $y=-2$,\\
$\therefore$ The coordinates of point P are $\boxed{(5, 6)}$ or $\boxed{(-3, -2)}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $BC\parallel AD$, $\angle ADC=90^\circ$, $BC=8\,\text{cm}$, $AD=CD=10\,\text{cm}$. Point $E$ starts from point $B$ and moves along direction $BC$ at a uniform speed of $2\,\text{cm/s}$; at the same time, point $F$ starts from point $D$ and moves along direction $DA$ at a uniform speed of $3\,\text{cm/s}$. Draw $EH\perp AD$ at point $E$ with the perpendicular foot being $H$, and $EH$ intersects $AC$ at point $G$. Connect $FG$. Let the movement time be $t\,\text{s}$ $\left(0<t<\frac{10}{3}\right)$. Solve the following problem: Find the length of $DH$ (expressed in an algebraic expression involving $t$).

Solution:
\textbf{Solution:} From the given, we have $BE=2t$, $DF=3t$, \\
$\therefore$ $CE=8-2t$, $AF=10-3t$, \\
$\because$ $CD\perp AD$, $EH\perp AD$, \\
$\therefore$ $\angle D=\angle EHA=90^\circ$, \\
$\therefore$ $CD\parallel EH$, \\
Additionally, $\because$ $CB\parallel AD$, \\
$\therefore$ the quadrilateral $CEHD$ is a parallelogram, \\
$\therefore$ the quadrilateral $CEHD$ is a rectangle, \\
$\therefore$ $DH=CE=8-2t$; \\
Thus, the length of $DH$ is $\boxed{8-2t}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $BC\parallel AD$, $\angle ADC=90^\circ$, $BC=8\,cm$, and $AD=CD=10\,cm$. Point $E$ moves uniformly along the direction of $BC$ from point $B$ at a speed of $2\,cm/s$; simultaneously, point $F$ moves uniformly along the direction of $DA$ from point $D$ at a speed of $3\,cm/s$. A line through point $E$ is drawn such that $EH\perp AD$ with the foot of the perpendicular at $H$, and $EH$ intersects $AC$ at point $G$, and $FG$ is connected. Let the movement time be $t(s)\left(0<t<\frac{10}{3}\right)$, answer the following question: When $\triangle CEG\cong \triangle AHG$, what is the value of $t$?

Solution:
\textbf{Solution:} Given the problem, \\
since $AD=10$ and $DH=8-2t$, \\
thus $AH=AD-DH=10-(8-2t)=2+2t$, \\
since $\triangle CEG\cong \triangle AHG$, \\
thus $CE=AH$, \\
thus $8-2t=2+2t$, \\
solving for $t$ yields $t=\boxed{1.5}$. \\
Therefore, when $\triangle CEG\cong \triangle AHG$, the value of $t$ is \boxed{1.5}.


Question: As shown in the diagram, in quadrilateral $ABCD$, $BC\parallel AD$, $\angle ADC=90^\circ$, $BC=8\text{cm}$, $AD=CD=10\text{cm}$. Point $E$ starts from point $B$ and moves uniformly along the direction of $BC$ at a speed of $2\text{cm/s}$; at the same time, point $F$ starts from point $D$ and moves uniformly along the direction of $DA$ at a speed of $3\text{cm/s}$. Draw $EH\perp AD$ through point $E$ with $H$ as the foot of the perpendicular. $EH$ intersects $AC$ at point $G$, and $FG$ is connected. Let the movement time be $t(s)\left(0<t<\frac{10}{3}\right)$. Solve the following question: During the movement, is there a moment when the quadrilateral formed by points $B$, $E$, $F$, $H$ becomes a parallelogram? If such a moment exists, please write down the value of $t$ directly.

Solution:
\textbf{Solution:} Exists; \\
$\because CB\parallel AD$, \\
$\therefore$ when $BE=FH$, the quadrilateral with vertices $B$, $E$, $F$, $H$ is a parallelogram, \\
when point $F$ is on segment $DH$, $FH=DH-DF=8-2t-3t=8-5t$, \\
we get $2t=8-5t$, solving we find $t=\boxed{\frac{8}{7}}$, \\
when point $F$ is on segment $AH$, $FH=DF-DH=3t-(8-2t)=5t-8$, \\
we get $2t=5t-8$, solving we find $t=\boxed{\frac{8}{3}}$, \\
In conclusion, when $t=\boxed{\frac{8}{3}}$ or $t=\boxed{\frac{8}{7}}$, the quadrilateral with vertices $B$, $E$, $F$, $H$ is a parallelogram.


Question: As shown in the figure, it is known that $BD$ is the angle bisector of $\triangle ABC$, points $E$ and $F$ are respectively on sides $AB$ and $BC$, and $BE=CF$, $ED\parallel BC$. Quadrilateral $EFCD$ is a parallelogram; $\angle ABC=60^\circ$, $\angle ADB=100^\circ$, what is the degree measure of $\angle AEF$?

Solution:
\textbf{Solution:} Given that $BD$ is the angle bisector of $\triangle ABC$, and $\angle ABC=60^\circ$, \\
it follows that $\angle ABD=\frac{1}{2}\angle ABC=30^\circ$, \\
since $\angle ADB=100^\circ$, \\
then $\angle A=180^\circ-\angle ABD-\angle ADB=50^\circ$, \\
Given that quadrilateral EFCD is a parallelogram, \\
it implies $EF\parallel AC$, \\
thus, $\angle AEF=180^\circ-\angle A=\boxed{130^\circ}$.


Question: As shown in the figure, it is known that O is the origin, the coordinates of point A are $(6, 0)$, and point B is a moving point on the positive half-axis of the y-axis. A rectangle OBCA is formed with OB and OA as sides, OC is the diagonal of rectangle OBCA, OE bisects $\angle BOC$ and meets BC at point E, CF bisects $\angle ACO$ and meets OA at point F. When the quadrilateral OECF is a rhombus, what are the coordinates of point B?

Solution:
\textbf{Solution:} Given that the quadrilateral OECF is a rhombus, \\
it follows that $EO=EC$.\\
Hence, $\angle EOC=\angle ECO$.\\
Moreover, since $\angle EOC=\angle BOE$,\\
it follows that $\angle EOB=\angle EOC=\angle ECO=30^\circ$.\\
Also, given that the coordinates of point A are $(6, 0)$,\\
it follows that $OA=6$.\\
Therefore, $BC=6$.\\
Let $BE=x$, then $OE=EC=6-x$. In the right triangle $OBC$, $OE=2BE=2x$.\\
Hence, $6-x=2x$, yielding $x=2$,\\
and therefore $OE=4$.\\
Then, $OB=\sqrt{OE^{2}-BE^{2}}=\sqrt{4^{2}-2^{2}}=2\sqrt{3}$.\\
Thus, the coordinates of point B are \boxed{(0, 2\sqrt{3})}.


Question: As shown in the figure, it is known that O is the origin. The coordinate of point A is $(6, 0)$. Point B is a moving point on the positive half-axis of the y-axis. Rectangle OBCA is formed with OB and OA as sides. OC is the diagonal of rectangle OBCA. OE bisects $\angle BOC$ and intersects BC at point E. CF bisects $\angle ACO$ and intersects OA at point F. Through point E, draw $EG \perp OC$, with G as the foot of the perpendicular. Through point F, draw $FH \perp OC$, with H as the foot of the perpendicular. When points G and H trisect the diagonal OC, find the coordinates of point B.

Solution:
\textbf{Solution:} Given that $\because$OE bisects $\angle BOC$, $EG\perp OC$, $EB\perp OB$,\\
$\therefore \angle BOE=\angle GOE$, $\angle OBE=\angle OGE=90^\circ$,\\
Also $\because OE=OE$,\\
$\therefore \triangle OEB\cong \triangle OEG$,\\
$\therefore OG=OB$. Similarly, $CH=CA$. Since $OB=CA$,\\
$\therefore OG=CH$,\\
When point G is between points O and H, as shown in figure 5:\\
$\because$Points G, H trisect the diagonal OC,\\
$\therefore AC=OG=GH=HC$. Let $AC=m$, then $OC=3m$,\\
In $\triangle OAC$ with a right angle at $A$, $OA=6$,\\
$\because AC^2+OA^2=OC^2$,\\
$\therefore m^2+6^2=(3m)^2$, solving this gives $m=\frac{3\sqrt{2}}{2}$,\\
$\therefore OB=AC=\frac{3\sqrt{2}}{2}$,\\
$\therefore$The coordinates of point B are $\left(0, \frac{3\sqrt{2}}{2}\right)$;\\
When point H is between points O and G, as shown in figure 6,\\
Similarly, it follows that $OG=CH=AC$. Let $OH=n$, then $AC=HC=2n$,\\
In $\triangle OAC$ with a right angle at $A$, $OA=6$,\\
$\because AC^2+OA^2=OC^2$,\\
$\therefore (2n)^2+6^2=(3n)^2$, solving this gives $n=\frac{6\sqrt{5}}{5}$,\\
$\therefore AC=OB=\frac{12\sqrt{5}}{5}$,\\
$\therefore$The coordinates of point B are $\left(0, \frac{12\sqrt{5}}{5}\right)$,\\
$\therefore$The coordinates of point B that satisfy the conditions are $\boxed{\left(0, \frac{3\sqrt{2}}{2}\right)}$ or $\boxed{\left(0, \frac{12\sqrt{5}}{5}\right)}$.


Question: As shown in the diagram, in the plane Cartesian coordinate system $xOy$, it is known that the graph of the linear function $y=x+m$ intersects the x-axis and y-axis at points A and B, respectively. It intersects the graph of the linear function $y=-3x+n\ (n>m)$ at point P. Also, the graph of the linear function $y=-3x+n$ intersects the x-axis and y-axis at points C and D, respectively. Express the coordinates of point P in an algebraic expression involving $m$ and $n$.

Solution:
\textbf{Solution:} Given that the graph of the linear function $y=x+m$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Therefore, $A(-m,0)$ and $B(0,m)$, which implies $OA=OB$. Hence, $\angle PAC=\angle ABO=45^\circ$. Since the graph of the linear function $y=x+m$ intersects with the graph of the linear function $y=-3x+n$ at point $P$, by solving the system of equations
\[\left\{\begin{array}{l}
y=x+m\\
y=-3x+n
\end{array}\right.,\]
we get
\[\left\{\begin{array}{l}
x=\frac{n-m}{4}\\
y=\frac{n+3m}{4}
\end{array}\right..\]
Therefore, $\boxed{P\left(\frac{n-m}{4},\frac{n+3m}{4}\right)}$, which leads to $\angle PAC=45^\circ$, $\boxed{P\left(\frac{n-m}{4},\frac{n+3m}{4}\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, it is known that the graph of the linear function $y=x+m$ intersects the x-axis and y-axis at points A and B, respectively. It also intersects with the graph of the linear function $y=-3x+n\ (n>m)$ at point P. Furthermore, the graph of the linear function $y=-3x+n$ intersects the x-axis and y-axis at points C and D, respectively. If the area of $\triangle ABO$ is equal to the area of $\triangle BDP$, and the area of quadrilateral BOCP is 4, what are the expressions of these two linear functions?

Solution:
\textbf{Solution:}
Given a linear function $y=-3x+n$, its graph intersects the x-axis and y-axis at points $C$ and $D$, respectively.

Therefore, $C\left(\frac{n}{3},0\right)$ and $D(0,n)$.

Since the area of $\triangle ABO$ is equal to the area of $\triangle BDP$,

it follows that $\frac{1}{2}m^2=\frac{1}{2}(n-m)\cdot\frac{n-m}{4}$,

which leads to $4m^2=(n-m)^2$.

Consequently, $n=3m$,

resulting in $C(m,0)$, $D(0,3m)$, $P\left(\frac{m}{2},\frac{3m}{2}\right)$.

Given that the area of quadrilateral $OBCP$ is 4,

we have the area of $\triangle OCD$ minus the area of $\triangle BDP$ equals 4,

yielding $\frac{1}{2}m\cdot 3m-\frac{1}{2}\cdot 2m\cdot\frac{m}{2}=4$,

from which we find $m=2$,

thus, $n=6$.

Hence, the expressions for the two linear functions are $\boxed{y=x+2}$ and $\boxed{y=-3x+6}$.


Question: As shown in the diagram, within the Cartesian coordinate system $xOy$, it is known that the graph of the linear function $y=x+m$ intersects the x-axis and y-axis at points A and B, respectively. It intersects the graph of the linear function $y=-3x+n\ (n>m)$ at point P. Additionally, the graph of the linear function $y=-3x+n$ intersects the x-axis and y-axis at points C and D, respectively. Given the conditions in (2), there exists a point M on line AB and a point N in the plane such that the quadrilateral formed by vertices D, P, M, N is a rhombus. Please directly provide the coordinates of point N.

Solution:
\textbf{Solution:} Given that $m=2, n=6$, we have $D(0,6)$ and $P(1,3)$, thus $OD=6$ and $DP= \sqrt{(0-1)^{2}+(6-3)^{2}}=\sqrt{10}$. Since $\angle PAC=45^\circ$ and $\angle AOB=90^\circ$, it follows that $\angle ABO=45^\circ$. We discuss in three scenarios:

(1) Draw a line from point N such that $NE\perp y$-axis at point E. Since quadrilateral DPMN is a rhombus, it follows that $DN\parallel PM$, and hence $\angle EDN=\angle ABO=45^\circ$, with $DN=DP= \sqrt{10}$. Therefore, $EN=DE=\sqrt{5}$, and thus $OE=6+\sqrt{5}$, which gives $N(\sqrt{5},6+\sqrt{5})$;

(2) Draw a line from point N such that $NF\perp y$-axis at point F. Since quadrilateral DPMN is a rhombus, it follows that $DN\parallel PM$, and hence $\angle FDN=\angle ABO=45^\circ$, with $DN=DP= \sqrt{10}$. Therefore, $EF=DF=\sqrt{5}$, and thus $OE=6-\sqrt{5}$, which gives $N(-\sqrt{5},6-\sqrt{5})$;

(3) Connect MN to intersect DP at point G. Since quadrilateral DMPN is a rhombus, it follows that $DN\parallel PM$, $DG=PG$, and $DP\perp MN$. Given the equation of line AB as $y=x+2$, with $D(0,6)$, the equation for DN becomes $y=x+6$. Suppose $N(n,n+6)$, and since $D(0,6)$, $P(1,3)$, and $DG=PG$, we have $G\left(\frac{1}{2},\frac{9}{2}\right)$. Considering $DN^{2}=DG^{2}+NG^{2}$, we find $(n-0)^{2}+(n+6-6)^{2}=\left(0-\frac{1}{2}\right)^{2}+\left(6-\frac{9}{2}\right)^{2}+\left(n-\frac{1}{2}\right)^{2}+\left(n+6-\frac{9}{2}\right)^{2}$, which gives $n=-\frac{5}{2}$. Therefore, $N\left(-\frac{5}{2},\frac{7}{2}\right)$, leading to $N_1\left(\sqrt{5}, \sqrt{5}+6\right)$, $N_2\left(-\sqrt{5}, 6-\sqrt{5}\right)$, and $N_3\left(-\frac{5}{2}, \frac{7}{2}\right)$.

Therefore, the answer is $N_1\left(\boxed{\sqrt{5}, \sqrt{5}+6}\right)$, $N_2\left(\boxed{-\sqrt{5}, 6-\sqrt{5}}\right)$, $N_3\left(\boxed{-\frac{5}{2}, \frac{7}{2}}\right)$.


Question: As shown in the diagram, in rhombus $ABCD$, $ \angle A=60^\circ$. Points $E$ and $F$ are located on sides $AB$ and $AD$ respectively, and satisfy $ \angle BCE=\angle DCF$, $EF$ is connected. $ \angle CFE=\angle CEF$. If $AF=2\sqrt{3}$ cm, what is the area of $\triangle AEF$?

Solution:
\textbf{Solution:} Construct $FM \perp AB$ at $M$, \\
since $\triangle CDF \cong \triangle CBE$, \\
hence $FD=BE$, \\
since quadrilateral $ABCD$ is a rhombus, \\
hence $AB=AD$, \\
thus $AB-BE=AD-FD$, \\
therefore $AF=AE$, \\
also, since $\angle A=60^\circ$, \\
hence $\triangle AEF$ is an equilateral triangle, \\
thus $AE=AF=2\sqrt{3}$, \\
hence $AM=\sqrt{3}$, \\
therefore $FM=\sqrt{AF^2-AM^2}=3$, \\
hence the area of $\triangle AEF$ = $\frac{1}{2} \times AE \times FM=\frac{1}{2} \times 3 \times 2\sqrt{3}=\boxed{3\sqrt{3}}$.


Question: As shown in the diagram, the diagonals $AC$ and $BD$ of square $\square ABCD$ intersect at point $O$. A line $EF\perp AC$ passing through point $O$ intersects $AB$ and $DC$ at points $E$ and $F$, respectively. Lines $AF$ and $CE$ are connected. If $OE= \frac{3}{2}$, what is the length of $EF$?

Solution: \textbf{Solution:} From the given, since $\square ABCD$, we have AO=OC, AB$\parallel$DC, therefore $\angle$FCO=$\angle$EAO, $\angle$CFO=$\angle$AEO, hence $\triangle FCO\cong \triangle EAO$ (AAS), thus FC=AE, therefore quadrilateral AECF is a parallelogram, thus OE=OF=$\frac{1}{2}$EF, since OE=$\frac{3}{2}$, therefore EF=\boxed{3}.


Question: As shown in the figure, it is known that quadrilateral $ABCD$ is a rhombus, with points $E$ and $F$ being the midpoints of sides $AB$ and $BC$, respectively. Lines $DE$, $EF$, and $DF$ are drawn. $\triangle DEF$ is an isosceles triangle; if $AD=10$ and $EF=8$, what is the area of rhombus $ABCD$?

Solution:
\textbf{Solution:} Connect $AC$ and $BD$ intersecting at point $O$. \\
Since $ABCD$ is a rhombus, \\
then $AC \perp BD$, $OA=OC$, $OB=OD$. \\
Since points $E$ and $F$ are the midpoints of sides $AB$ and $BC$, respectively, \\
then $AC=2EF=16$, thus $OA=8$. \\
Since $AD=10$, \\
then $OD=\sqrt{10^2-8^2}=6$, \\
thus $BD=12$. \\
The area of rhombus $ABCD$ is $\frac{1}{2}AC \cdot BD=\boxed{96}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AD \parallel BC$, $\angle ABC = \angle BCD = 90^\circ$. The diagonals $AC$ and $BD$ intersect at point $O$. Line $DE$ bisects $\angle ADC$ and meets $BC$ at point $E$. Connect $OE$. Quadrilateral $ABCD$ is a rectangle, if $CD = 2$ and $\angle DBC = 30^\circ$, find the area of $\triangle BED$.

Solution:
Solution: In the right-angled triangle $BCD$, we have $\angle BCD=90^\circ$, $\angle DBC=30^\circ$, and $CD=2$. \\
Therefore, $BD=2CD=4$, $BC=\sqrt{BD^2-CD^2}=2\sqrt{3}$, \\
As proved in (1), the quadrilateral $ABCD$ is a rectangle, \\
Therefore, $\angle ADC=90^\circ$, \\
Because $DE$ bisects $\angle ADC$, \\
Therefore, $\angle CDE=\frac{1}{2}\angle ADC=45^\circ$, \\
Therefore, $\angle CED=90^\circ-\angle CDE=45^\circ=\angle CDE$, \\
Therefore, $CE=CD=2$, \\
Therefore, $BE=BC-CE=2\sqrt{3}-2$, \\
Thus, the area of $\triangle BED$ is $\frac{1}{2}BE\cdot CD=\frac{1}{2}\times (2\sqrt{3}-2)\times 2=\boxed{2\sqrt{3}-2}$.


Question: As shown in the diagram, within quadrilateral $ABCD$, $AC \perp AD$. Construct $\angle ECA = \angle ACD$, where $CE$ intersects $AB$ at point $O$ and extends to intersect the extended line of $DA$ at point $E$, and connect $BE$. Quadrilateral $ACBE$ is a rectangle; connect $OD$. If $AB=4$ and $\angle ACD=60^\circ$, what is the length of $OD$?

Solution:
\textbf{Solution:} Draw $OF \perp AE$ at point F. In $\triangle ACD$, $\angle ADC = 90^\circ - \angle ACD = 90^\circ - 60^\circ = 30^\circ$,\\
since parallelogram ABCD,\\
therefore $\angle ADC = \angle ABC = 30^\circ$,\\
since rectangle ABCD,\\
therefore $AO = CO = 2$, $\angle ACB = 90^\circ$,\\
therefore $\angle BAC = 90^\circ - \angle ABC = 90^\circ - 30^\circ = 60^\circ$,\\
therefore $\triangle AOC$ is an equilateral triangle,\\
therefore $AC = 2$;\\
therefore $BC = AE = AD = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - 2^2} = 2\sqrt{3}$,\\
since $AC \perp AE$,\\
therefore $OF \parallel AC$,\\
therefore $OF$ is the midline of $\triangle ACE$,\\
therefore $OF = \frac{1}{2}AC = 1$, AF$=\frac{1}{2}AE=\sqrt{3}$;\\
therefore $DF = AF + AD = 2\sqrt{3} + \sqrt{3} = 3\sqrt{3}$,\\
in $\triangle OFD$,\\
$OD=\sqrt{OF^2+DF^2}=\sqrt{1^2+(3\sqrt{3})^2}=\boxed{2\sqrt{7}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB = AC$, $AD$ is drawn through points $A$ and $C$ parallel to $BC$, and $CD$ is drawn parallel to $AB$ intersecting at point $D$. Extend $DC$ to point $E$ such that $DC = CE$, and connect $BE$. Quadrilateral $ACEB$ is a rhombus; If $AB = 4$, $BC = 6$, what is the area of the quadrilateral $ACEB$?

Solution:
\textbf{Solution:} Construct line AE intersecting BC at point O, \\
since quadrilateral ACEB is a rhombus, \\
thus AE$\perp$BC, \\
since AB＝4, BC＝6, \\
thus OB＝$\frac{1}{2}$BC＝3, \\
thus OA＝$\sqrt{AB^{2}−OB^{2}}=\sqrt{4^{2}−3^{2}}=\sqrt{7}$, \\
thus AE＝2OA＝2$\sqrt{7}$, \\
thus the area of quadrilateral $ACEB$, $S=\frac{1}{2}AE\cdot BC=\frac{1}{2}\times 6\times 2\sqrt{7}=\boxed{6\sqrt{7}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AB \parallel DC$, $AB=AD$, and the diagonals $AC$ and $BD$ intersect at point $O$, with $AC$ bisecting $\angle BAD$. Quadrilateral $ABCD$ is a rhombus; through point $C$, draw $CE \perp AB$ to the extended line of $AB$ at point $E$, given that $AB=13$ and $BD=10$, what is the length of $CE$?

Solution:
\textbf{Solution:} In the rhombus ABCD, we have OA = OC, OB = OD, and AC $\perp$ BD, \\
$\therefore$ OB = $\frac{1}{2}$BD = 5, $\angle$AOB = 90$^\circ$, \\
In $\triangle AOB$, AO = $\sqrt{AB^2−AO^2}$ = $\sqrt{13^2−5^2}$ = 12, \\
$\therefore$ AC = 2AO = 24; \\
$\because$ The area of the rhombus ABCD = $\frac{1}{2} \cdot AC \cdot BD = AB \cdot CE$, \\
$\therefore$ $\frac{1}{2} \times 24 \times 10 = 13 \times CE$, \\
$\therefore$ CE = $\boxed{\frac{120}{13}}$.


Question: As shown in the figure, it is known that both quadrilaterals $ABCD$ and $CEFG$ are squares. Point $K$ is on $BC$, and $CD$ is extended to point $H$ such that $DH=BK=CE$. Connect $AK, KF, HF, AH$. $AK=AH$; Quadrilateral $AKFH$ is a square; If the area of quadrilateral $AKFH$ is $10$, and $CE=1$, what is the distance between points $A$ and $E$?

Solution:
\textbf{Solution:} Since the area of quadrilateral $AKFH$ is 10, \\
and from (2) we know that quadrilateral $AKFH$ is a square,\\
therefore $KF=\sqrt{10}$,\\
since $EF=CE=1$,\\
therefore $KE=\sqrt{KF^{2}-EF^{2}}=\sqrt{10-1}=3$,\\
thus $AB=KE=3$,\\
since $BK=EF=1$,\\
therefore $BE=BK+KE=4$,\\
thus $AE=\sqrt{AB^{2}+BE^{2}}=\sqrt{3^{2}+4^{2}}=\boxed{5}$,\\
hence the distance between points $A$ and $E$ is \boxed{5}.


Question: As shown in the figure, in $\triangle ABC$, points D and E are the midpoints of sides BC and AC, respectively, with $BE = 2DE$. Line $AF$ is drawn parallel to $BE$ and intersects the extension of line DE at point F. Quadrilateral ABEF is a rhombus. If $\angle ABE = 45^\circ$ and $AB = 4$, what is the area of quadrilateral ABDF?

Solution:
\textbf{Solution:} Draw $EM\perp AB$ at point M, \\
$AB=2DE$, $AB=4$, \\
$DE=2$, \\
Because quadrilateral ABEF is a rhombus, \\
Therefore $BE=EF=AB=4$, \\
Therefore $DF=DE+EF=2+4=6$, \\
In $\triangle EBM$, $\angle ABE=45^\circ$, \\
Therefore $\triangle EBM$ is an isosceles right triangle, \\
Thus $EM^2+BM^2=BE^2$, \\
Therefore $EM=\frac{\sqrt{2}}{2}BE=\frac{\sqrt{2}}{2}\times 4=2\sqrt{2}$, \\
Therefore, the area of quadrilateral ABDE is: $\frac{1}{2}(AB+DF)\cdot EM = \frac{1}{2}\times (4+6)\times 2\sqrt{2} = \boxed{10\sqrt{2}}$.


Question: As shown in the figure, within square $ABCD$, point $E$ is on side $CD$. Connect $AE$, and draw $AF\perp AE$ intersecting the extension of $CB$ at point $F$. Connect $EF$, and $AG$ bisects $\angle FAE$, with $AG$ intersecting $BC$ and $EF$ at points $G$ and $H$, respectively. Connect $EG$ and $DH$. Given $AF=AE$; if $\angle BAG=10^\circ$, what is the degree measure of $\angle EGC$?

Solution:
\textbf{Solution:} Given that $AG$ bisects $\angle EAF$,\\
it follows that $\angle GAF = \angle GAE$,\\
given that $AF=AE$, and $AG=AG$,\\
it follows that $\triangle AGF \cong \triangle AGE$ (SAS),\\
thus $\angle AGF = \angle AGE = 90^\circ - \angle BAG$,\\
since $\angle BAG = 10^\circ$,\\
it follows that $\angle AGF = \angle AGE = 80^\circ$,\\
thus $\angle EGC = 180^\circ - \angle AGF - \angle AGE = \boxed{20^\circ}$;


Question: As shown in the figure, in the square $ABCD$, with point $E$ on side $CD$, connect $AE$. Construct $AF\perp AE$, intersecting the extension of $CB$ at point $F$, and connect $EF$. Let $AG$ bisect $\angle FAE$, with $AG$ intersecting $BC$ and $EF$ at points $G$ and $H$, respectively. Connect $EG$ and $DH$. If $DE=CE$, find the value of $CE:CG:EG$?

Solution:
\textbf{Solution:}
Given (1), we have $\triangle ADE \cong \triangle ABF$,\\
$\therefore BF=DE$,\\
From (2), we have $\triangle AGF \cong \triangle AGE$,\\
$\therefore GE=GF$,\\
Let $DE=EC=a$, $BG=x$, thus $BF=a$, $BC=CD=2a$,\\
$\therefore FG= a+x$,\\
$\therefore EG=a+x$, $GC=2a-x$,\\
In $\triangle ECG$, $EG^2=EC^2+CG^2$,\\
$\therefore (x+a)^2=a^2+(2a-x)^2$, solving gives $x=\frac{2}{3}a$,\\
$\therefore CG=\frac{4}{3}a$, $EG=\frac{5}{3}a$,\\
$\therefore CE:CG:EG=a:\frac{4}{3}a:\frac{5}{3}a=3:4:5$.\\

Therefore, the answer is $CE:CG:EG=\boxed{3:4:5}$.


Question: As shown in the diagram, it is known that in rectangle $ABCD$, $AC$ and $BD$ intersect at point $O$. Through point $O$, draw $EF\perp BD$ intersecting $AB$ and $CD$ at $E$ and $F$, respectively. Quadrilateral $BEDF$ is a rhombus. If $BC=3$ and $CD=5$, find the area of the rhombus $BEDF$.

Solution:
\textbf{Solution:} Let $CF = x$, then $DF = CD - CF = 5 - x$, \\
since quadrilateral $BEDF$ is a rhombus, \\
therefore $BF = DF = 5 - x$, \\
in $\triangle BFC$, $BF^2 = BC^2 + CF^2$, \\
therefore $(5-x)^2 = 3^2 + x^2$, \\
solving for $x$ yields $x = \frac{8}{5}$, \\
thus $DF = 5 - \frac{8}{5} = \frac{17}{5}$, \\
therefore the area of rhombus $BEDF$ is $DF \cdot BC = \frac{17}{5} \times 3 = \boxed{\frac{51}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, D and E are the midpoints of sides AC and AB, respectively. Lines CE and DE are connected. A line passing through point D and parallel to CE intersects the extension of BC at point F. If $AB = 13$ cm and $AC = 5$ cm, what is the perimeter of quadrilateral DECF in cm?

Solution:
\textbf{Solution:} In right triangle $ABC$, by the Pythagorean theorem, we have: $BC=\sqrt{AB^{2}-AC^{2}}=\sqrt{13^{2}-5^{2}}=12$, \\
since $DE$ is the midline of $\triangle ABC$, \\
thus $DE=\frac{1}{2}BC=\frac{1}{2}\times 12=6$, \\
since the quadrilateral $DECF$ is a parallelogram, \\
thus $DE=CF=6$, and $DF=CE$, \\
since $D$ is the midpoint of the side $AC$, \\
thus $CD=\frac{1}{2}AC=\frac{1}{2}\times 5=\frac{5}{2}$, \\
since $\angle ACB=90^\circ$ and $CF$ is the extension of $BC$, \\
thus $\angle DCF=90^\circ$, \\
in right triangle $DCF$, by the Pythagorean theorem, we get: $DF=\sqrt{CD^{2}+CF^{2}}=\sqrt{\left(\frac{5}{2}\right)^{2}+6^{2}}=\frac{13}{2}$, \\
thus the perimeter of the quadrilateral $DECF$ $=2(\text{DE}+\text{DF})=2\times\left(6+\frac{13}{2}\right)=\boxed{25}$.


Question: As shown in the diagram, in parallelogram $ABCD$, draw $DE\perp AB$ at point $E$, with point $F$ on side $CD$ and $FC=AE$. Connect $AF$ and $BF$. Quadrilateral $DEBF$ is a rectangle; if $AF$ bisects $\angle DAB$, $FC=3$, and $DF=5$, find the length of $BF$.

Solution:
\textbf{Solution:} Given that $\because AF$ bisects $\angle DAB$,\\
$\therefore \angle DAF=\angle BAF$,\\
$\because CD\parallel AB$,\\
$\therefore \angle DFA=\angle BAF$,\\
$\therefore \angle DFA=\angle DAF$,\\
$\therefore AD=DF=5$,\\
In $\triangle AED$, $AE=FC=3$,\\
By the Pythagorean theorem, we have $DE=\sqrt{AD^{2}-AE^{2}}=\sqrt{5^{2}-3^{2}}=4$,\\
From (1), it is deduced that the quadrilateral DEBF is a rectangle,\\
$\therefore BF=DE=\boxed{4}$.


Question: As shown in the figure, on a square sheet of paper ABCD, there is a point P, with $PA=1$, $PD=2$, and $PC=3$. Now, cut out $\triangle PCD$ and place it as shown in the figure (where C coincides with A, P coincides with G, and D coincides with D). Find the length of the segment PG.

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a square, \\
it follows that AD=CD, and $\angle ADC=90^\circ$, \\
since PA=1, PD=2, PC=3, \\
by cutting $\triangle PCD$ and pasting it as shown in the figure (C coincides with A, P with G, and D with D), \\
therefore PD=GD=2, $\angle CDP=\angle ADG$, AG=PC=3, \\
therefore $\angle PDG=\angle ADC=90^\circ$, \\
thus, $\triangle PDG$ is an isosceles right triangle, \\
hence $\angle GPD=45^\circ$, PG=$\sqrt{2}$PD=$\boxed{2\sqrt{2}}$;


Question: As shown in the figure, on the square paper $ABCD$, there is a point $P$ such that $PA=1$, $PD=2$, and $PC=3$. Now, cut $\triangle PCD$ and paste it to the position as shown in the figure (where $C$ coincides with $A$, $P$ coincides with $G$, and $D$ coincides with $D$). Find: What is the degree measure of $\angle APD$?

Solution:
\textbf{Solution:} Given equation (1), we know that $\angle GPD=45^\circ$ and $PG=\sqrt{2}PD=2\sqrt{2}$,\\
because $AG=PC=3$ and $AP=1$,\\
therefore, $1^2+(2\sqrt{2})^2=3^2$,\\
thus, $AP^2+PG^2=AG^2$,\\
which means $\angle GPA=90^\circ$,\\
hence $\angle APD=90^\circ+45^\circ=\boxed{135^\circ}$.


Question: As shown in the figure, let $ABCD$ be a square with side length $2$. Let points $E$ and $F$ be the midpoints of $AB$ and $CD$, respectively. Connect $CE$ and $AF$, and extend a line from $D$ perpendicular to $AF$ at point $G$. Extend $DG$ to intersect with $CE$ at point $H$. What is the length of $DG$?

Solution: \textbf{Solution:} Given a square ABCD, we have AD=CD=AB=2 and $\angle ADC=90^\circ$. Since point F is the midpoint of CD, it follows that DF=$\frac{1}{2}$CD=1. In $\triangle ADF$, we find $AF=\sqrt{AD^2+DF^2}=\sqrt{2^2+1^2}=\sqrt{5}$. Considering $S_{\triangle ADF}=\frac{1}{2}AD\cdot DF=\frac{1}{2}AF\cdot DG$ and DG$\perp$AF, we deduce that $2\times 1=\sqrt{5}DG$. Solving this, we get $DG=\boxed{\frac{2\sqrt{5}}{5}}$.


Question: As shown in the figure, square $ABCD$ has a side length of $2$. Points $E$ and $F$ are the midpoints of $AB$ and $CD$, respectively. Connect $CE$ and $AF$. Draw $DG \perp AF$ through point $D$ with $G$ as the foot of the perpendicular. Extend $DG$ to intersect $CE$ at point $H$. Find the length of $GH$.

Solution:
\textbf{Solution:} Since points E and F are the midpoints of AB and CD respectively, \\
therefore $AE=\frac{1}{2}AB$, $CF=\frac{1}{2}CD$, \\
hence $AE=CF$, \\
since $AE\parallel CF$, \\
thus quadrilateral AFCE is a parallelogram, \\
therefore $AF\parallel CE$, \\
since point F is the midpoint of DC, \\
thus point G is the midpoint of DH, \\
therefore $DG=GH=\boxed{\frac{2\sqrt{5}}{5}}$.