Question: As shown in the figure, point $C$ is a point on line segment $AB$, and points $M$ and $N$ are the midpoints of line segments $AC$ and $BC$, respectively. If point $P$ is the midpoint of line segment $AB$, $AC=6\,\text{cm}$, and $CP=2\,\text{cm}$, what is the length of line segment $PN$ in cm?

Solution:
\textbf{Solution:} Since $AC=6\text{cm}$, $CP=2\text{cm}$,\\
it follows that $AP=AC+CP=8\text{cm}$,\\
Since $P$ is the midpoint of segment $AB$,\\
it follows that $AB=2AP=16\text{cm}$,\\
Therefore, $CB=AB-AC=16-6=10\text{cm}$,\\
Since $N$ is the midpoint of segment $CB$,\\
it follows that $CN= \frac{1}{2} CB=5\text{cm}$,\\
Therefore, $PN=CN-CP=5-2=\boxed{3}\text{cm}$.


Question: As shown in the figure, it is known that points $A$, $B$, $C$, and $D$ lie on the same straight line, point $C$ is the midpoint of segment $AB$, and point $D$ is on segment $AB$. If $AB=6$ and $BD=\frac{1}{3}BC$, what is the length of segment $CD$?

Solution:
\textbf{Solution:} Since point C is the midpoint of segment AB, and $AB=6$, \\
therefore $BC= \frac{1}{2}AB=3$, \\
since $BD=\frac{1}{3}BC$, \\
thus $BD=\frac{1}{3} \times 3=1$, \\
therefore $CD=BC-BD=3-1=\boxed{2}$.


Question: Given points $A$, $B$, $C$, and $D$ lie on the same straight line, with point $C$ being the midpoint of segment $AB$, and point $D$ lies on segment $AB$. If point $E$ is a point on segment $AB$, and $AE=2BE$, when $AD:BD=2:3$, what is the value of $CD:CE$?

Solution:
\textbf{Solution:} Let $AD=2x$, $BD=3x$, then $AB=5x$.\\
Since point C is the midpoint of line segment AB,\\
it follows that $AC=\frac{1}{2}AB=\frac{5}{2}x$,\\
thus $CD=AC-AD=\frac{1}{2}x$,\\
since $AE=2BE$,\\
it follows that $AE=\frac{2}{3}AB=\frac{10}{3}x$,\\
thus $CE=AE-AC=\frac{10}{3}x-\frac{5}{2}x=\frac{5}{6}x$,\\
thus $CD\colon CE=\frac{1}{2}x\colon \frac{5}{6}x=\boxed{3\colon 5}$.


Question: As shown in the figure, point $B$ is a point on line segment $AC$, and $AB=27\text{cm}$, $BC=\frac{1}{3}AB$. Try to find the length of line segment $AC$?

Solution:
\textbf{Solution:} Given that $AB=27\text{cm}$, and $BC=\frac{1}{3}AB=9\text{cm}$,\\
thus $AC=AB+BC=27+9=\boxed{36}\text{cm}$.


Question: As shown in the figure, point B is a point on line segment AC, and $AB=27\text{cm}$, $BC=\frac{1}{3}AB$. If point O is the midpoint of line segment AC, what is the length of line segment $OB$ in cm?

Solution:
\textbf{Solution:} From (1) we know: $AC=24\,\text{cm}$,\\
since point O is the midpoint of line segment AC,\\
therefore $OC=\frac{1}{2}AC=\frac{1}{2}\times 24=12\,\text{cm}$.\\
Therefore $OB=OC-BC=12-9=\boxed{9}\,\text{cm}$.


Question: As shown in the figure, point C is a point on the line segment AB, where M and N are the midpoints of AB and CB, respectively. Given that $AC=10\,cm$ and $NB=6\,cm$, find the length of CM.

Solution:
\textbf{Solution:} \\
$\because$ N is the midpoint of BC, and NB = 6cm, \\
$\therefore$ BC = 2NB = $2 \times 6$ = 12cm, \\
$\because$ AC = 10cm, \\
$\therefore$ AB = AC + BC = 10 + 12 = 22cm, \\
$\because$ M is the midpoint of AB, \\
$\therefore$ BM = AM = $\frac{1}{2}$AB = $\frac{1}{2} \times 22$ = 11cm, \\
$\therefore$ CM = AM - AC = 11 - 10 = \boxed{1}cm.


Question: As shown in the figure, point C is a point on line segment AB, M and N are the midpoints of AB and CB respectively, with $AC=10\,cm$ and $NB=6\,cm$. Find the length of MN.

Solution:
\textbf{Solution:} From (1) we have $BM=11\,\text{cm}$,

$\because NB=6\,\text{cm}$,

$\therefore MN=BM-NB=11-6=\boxed{5}\,\text{cm}$.


Question: As shown in the figure, it is known that $AB=60\,\text{cm}$, point C is the midpoint of line segment AB, and point D is a point on line segment AB, with the length ratio of $AD$ to $DB$ being $2\colon 1$. What is the length of $BD$ in $\text{cm}$?

Solution:
\textbf{Solution:} Given that $AB=60\text{cm}$ and the ratio of the lengths of AD to DB is $2\colon 1$, \\
it follows that $BD=60\times \frac{1}{3}=\boxed{20}\text{cm}$.


Question: As shown in the figure, it is known that $AB=60\,cm$, point C is the midpoint of line segment AB, and point D is a point on line segment AB with the ratio of the lengths of AD to DB being $2\colon 1$. What is the length of CD in $cm$?

Solution:
\textbf{Solution:} Since $AB=60\,cm$, and point C is the midpoint of line segment AB,\\
therefore, $BC=\frac{1}{2}AB=30\,cm$,\\
therefore, $CD=BC-BD=30-20=\boxed{10}\,cm$.


Question: As shown in the figure, $C$ is the midpoint of line segment $AB$, and point $D$ divides line segment $AB$ into two parts, $AD$ and $DB$, in the ratio of $3\colon 2$. If $CD=1\text{cm}$, what is the length of line segment $AB$?

Solution:
\textbf{Solution:} Let $AD=3x\,(\text{cm})$, and $BD=2x\,(\text{cm})$, thus $AB=5x\,(\text{cm})$ \\
Since $C$ is the midpoint of $AB$, \\
Then $AC=BC=\frac{5}{2}x\,(\text{cm})$ \\
Therefore $CD=BC-BD=\frac{1}{2}x\,(\text{cm})$ \\
Since $CD=1\,\text{cm}$, \\
Then $\frac{1}{2}x=1$, \\
Therefore $x=2\,\text{cm}$, \\
Therefore $AB=5x=\boxed{10\,\text{cm}}$;


Question: As shown in the diagram, it is known that the length of segment $AB=20$. Point $C$ is on segment $AB$, point $D$ is on segment $CB$, and point $E$ is the midpoint of segment $DB$, with $DE=3$. If $CE=8$, what is the length of $AC$?

Solution:
\textbf{Solution:} We have $E$ as the midpoint of $DB$, with $DE=3$, and $CE=8$, \\
thus $DB=2DE=6$, and $CD=CE-DE=5$, \\
therefore, $AC=AB-CD-DB=20-6-5=\boxed{9}$.


Question: As shown in the figure, it is known that segment $AB=20$, $C$ is a point on segment $AB$, $D$ is a point on $CB$, $E$ is the midpoint of $DB$, and $DE=3$. If $C$ is the midpoint of $AB$, what is the length of $CD$?

Solution:
\textbf{Solution:} Since $C$ is the midpoint of $AB$, and $AB=20$,

$\therefore$ $AC=CB=10$. Given also that $DB=6$,

$\therefore$ $CD=CB-DB=10-6=\boxed{4}$.


Question: Point $O$ is the midpoint of line segment $AB$, where $OB=14\,\text{cm}$. Point $P$ divides the line segment $AB$ into two parts such that the ratio of $AP$ to $PB$ is $5:2$. Find the length of line segment $OP$.

Solution:
\textbf{Solution:} Since point O is the midpoint of line segment AB, $OB=14\,\text{cm}$,\\
therefore $AB=2OB=28\,\text{cm}$,\\
since $AP\colon PB=5\colon 2$,\\
therefore $BP= \frac{2}{7}AB=8\,\text{cm}$,\\
therefore $OP=OB-BP=14-8=\boxed{6}\,\text{cm}$.


Question: As shown in the figure, point $O$ is the midpoint of line segment $AB$, with $OB=14\,cm$. Point $P$ divides line segment $AB$ into two parts, with $AP:PB=5:2$. Point $M$ is on line segment $AB$. If the distance from point $M$ to point $P$ is $4\,cm$, what is the length of line segment $AM$ in $cm$?

Solution:
Solution: As shown in Figure 1, when point M is to the left of point P,\\
$AM=AB-(PM+BP)=28-(4+8)=\boxed{16}\,(\text{cm})$,\\
As shown in Figure 2, when point M is to the right of point P,\\
$AM=AB-BM=AB-(BP-PM)=28-(8-4)=\boxed{24}\,(\text{cm})$.\\
In summary, $AM=16\,\text{cm}$ or $24\,\text{cm}$.


Question: Given points $B$ and $D$ on line segment $AC$, as shown in the figure, if $AC=20$, $AB=8$, and point $D$ is the midpoint of line segment $AC$, what is the length of line segment $BD$?

Solution:
\textbf{Solution:} Given that $AC=20$ and point $D$ is the midpoint of segment $AC$, \\
$\therefore AD=DC=\frac{1}{2}AC=\frac{1}{2}\times 20=10$, \\
Since $AB=8$, \\
$\therefore BD=AD-AB=10-8=\boxed{2}$.


Question: As shown in the figure, it is known that points $B$ and $D$ are on the line segment $AC$. If $BD=\frac{1}{3}AB=\frac{1}{4}CD$, $AE=BE$, and $EC=13$, find the length of the line segment $AC$.

Solution:
\textbf{Solution:} Given that $BD=\frac{1}{3}AB=\frac{1}{4}CD$,\\
it follows that $AB=3BD, CD=4BD$,\\
Given $AE=BE$,\\
it then follows $AE=BE=\frac{1}{2}AB=\frac{3}{2}BD$,\\
Since $EC=BE+BD+DC=\frac{3}{2}BD+BD+4BD=\frac{13}{2}BD=13$,\\
then $BD=2$,\\
Thus, $AE=\frac{3}{2}BD=\frac{3}{2}\times 2=3$,\\
Therefore, $AC=AE+EC=3+13=\boxed{16}$.


Question: As shown in the figure, point $C$ is on line segment $AB$, $D$ is the midpoint of $AC$, and $E$ is the midpoint of $BC$. If $AB=9\, \text{cm}$ and $AC=5\, \text{cm}$. Find: What is the length of $AD$ in centimeters?

Solution:
\textbf{Solution:} Given that AC$=5\,\text{cm}$ and D is the midpoint of AC,\\
it follows that AD$=$DC$=\frac{1}{2}$AC$=\boxed{\frac{5}{2}}\,\text{cm}$.


Question: As shown in the figure, $C$ is a point on the line segment $AB$, $D$ is the midpoint of $AC$, and $E$ is the midpoint of $BC$. If $AB=9\,\text{cm}$, $AC=5\,\text{cm}$. Find: the length of $DE$ in centimeters?

Solution:
\textbf{Solution:} Given that $AB=9\text{cm}$, $AC=5\text{cm}$,\\
thus $BC=AB-AC=9-5=4\text{cm}$,\\
since $E$ is the midpoint of $BC$,\\
thus $CE=\frac{1}{2}BC=2\text{cm}$,\\
therefore $DE=CD+CE=\frac{5}{2}+2=\boxed{\frac{9}{2}}\text{cm}$.


Question: As shown in Figure 1, point $E$ is the midpoint of side $BC$ of a square $ABCD$ with side length 2, and $DF \perp AE$ at $F$. What is the length of $DF$?

Solution:
\textbf{Solution:} As shown in the figure, connect $DE$,

since point $E$ is the midpoint of side $BC$ of square $ABCD$ with a side length of 2,

therefore $BE=1$,

thus, in $\triangle ABE$ right-angled at $B$, $AE=\sqrt{1^2+2^2}=\sqrt{5}$,

since $DF\perp AE$,

therefore $\frac{1}{2}\times AE\times DF=\frac{1}{2}\times AD\times AB$,

thus, $DF=\frac{AD\times AB}{AE}=\frac{2\times 2}{\sqrt{5}}=\boxed{\frac{4}{\sqrt{5}}}$.


Question: As shown in the figure, point $E$ is a point on the diagonal $AC$ of the square $ABCD$. Connect $DE$ and draw $EF\perp DE$ to intersect ray $BC$ at point $F$. Construct rectangle $DEFG$ with $DE$ and $EF$ as adjacent sides, and connect $CG$. Rectangle $DEFG$ is a square; if $AB=4$ and $CE=3\sqrt{2}$, what is the length of $CG$?

Solution:
\textbf{Solution:} As shown in the figure, draw EH $\perp$ AB at point H, \\
then $\angle EHB=90^\circ$, since $\angle ENB=90^\circ$, and $\angle B=90^\circ$, thus quadrilateral BNEH is a rectangle, and therefore $EN=BH$. Within square ABCD, $AD=CD$, and $\angle ADC=90^\circ$. In square DEFG, $DE=DG$, and $\angle EDG=90^\circ$, thus $\angle ADE=\angle CDG$. In $\triangle ADE$ and $\triangle CDG$, we have $\left\{ \begin{array}{l} AD=CD \\ \angle ADE=\angle CDG \\ DE=DG \end{array} \right.$, therefore $\triangle ADE \cong \triangle CDG$ (SAS), hence $CG=AE$. Since $CE=3\sqrt{2}$, $\angle ECN=45^\circ$, and $\angle ENC=90^\circ$, therefore $\angle NEC=45^\circ$, thus $NC=NE=3$. Given that $AB=4$, thus $AH=4-3=1$. As $\angle HAE=45^\circ$, therefore $\angle HEA=45^\circ$, thus $AH=HE=1$, hence $AE=\sqrt{1^2+1^2}=\sqrt{2}$, thus $CG=\boxed{\sqrt{2}}$;


Question: As shown in the figure, point $E$ is a point on the diagonal $AC$ of the square $ABCD$. Line $DE$ is drawn, and through point $E$, line $EF\perp DE$ is drawn, which intersects the ray $BC$ at point $F$. A rectangle $DEFG$ is constructed with $DE$ and $EF$ as adjacent sides, and $CG$ is connected. When the angle between line segment $DE$ and one side of square $ABCD$ is $25^\circ$, directly write down the degree measure of $\angle EFC$.

Solution: \textbf{Solution:} We have $\angle EFC = \boxed{115^\circ}$ or $\boxed{25^\circ}$.


Question: As shown in the figure, point E is a point on side AD of quadrilateral ABCD. Extend line EB such that $BF = BE$, and extend line EC such that $CG = CE$. Connect FG. H is the midpoint of FG. Connect DH and AF. If $\angle BAE = 70^\circ$ and $\angle DCE = 20^\circ$, what is the measure of $\angle DEC$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a parallelogram, \\
$\therefore \angle BAE=\angle BCD=70^\circ$, with AD$\parallel$BC, \\
since $\angle DCE=20^\circ$, and given $AB\parallel CD$, \\
$\therefore \angle CDE=180^\circ-\angle BAE=110^\circ$, \\
$\therefore \angle DEC=180^\circ-\angle DCE-\angle CDE=\boxed{50^\circ}$;


Question: As shown in the diagram, point E is located on side AD of quadrilateral $ABCD$. Extend EB such that $BF=BE$, and extend EC such that $CG=CE$, then connect F and G. Let H be the midpoint of FG, and connect DH and AF. Quadrilateral AFHD is a parallelogram; connect EH, intersecting BC at point O. If $OC=OH$, prove that $OE=\frac{1}{2}BC$.

Solution:
\textbf{Solution:} Proof: Connect $BH$ and $CH$, \\
$\because$ $CE=CG$, $FH=HG$, \\
$\therefore$ $CH=\frac{1}{2}EF$, $CH\parallel EF$, \\
$\because$ $EB=BF=\frac{1}{2}EF$, \\
$\therefore$ $BE=CH$, \\
$\therefore$ Quadrilateral $EBHC$ is a parallelogram, \\
$\therefore$ $OB=OC$, $OE=OH$, \\
$\because$ $OC=OH$, \\
$\therefore$ $OE=OB=OC=\boxed{\frac{1}{2}BC}$.


Question: As shown in the figure, the diagonals of rhombus $ABCD$ intersect at point $O$. $DE \parallel AC$ and $CE \parallel BD$. Connect $OE$. Quadrilateral $OCED$ is a rectangle. If $AC=6$ and $BD=8$, find the length of $OE$.

Solution: \textbf{Solution}: Since the quadrilateral $ABCD$ is a rhombus, $AC=6, BD=8$, it follows that $OC=\frac{1}{2}AC=3, OD=\frac{1}{2}BD=4, AC\perp BD$, hence $CD=\sqrt{OC^{2}+OD^{2}}=5$. Since the quadrilateral $OCED$ is a rectangle, it follows that $OE = \boxed{5}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is a square, with $E$ and $F$ being two points on the diagonal $AC$, and $AE=CF$. $\triangle ADE\cong \triangle CBF$; the quadrilateral $BEDF$ forms a rhombus. If $AC=8, AE=2$, what is the perimeter of the quadrilateral $BEDF$?

Solution:
\textbf{Solution:} As shown in the figure, connect $BD$, intersecting $AC$ at point O, \\
since quadrilateral $ABCD$ is a square and $AC=8$, \\
therefore $AC\perp BD, OD=OA=\frac{1}{2}AC=4$, \\
since $AE=2$, \\
therefore $OE=OA-AE=2$, \\
In right triangle $\triangle DOE$, $DE=\sqrt{OD^{2}+OE^{2}}=2\sqrt{5}$, \\
thus, the perimeter of quadrilateral $BEDF$ is $4DE=4\times 2\sqrt{5}=\boxed{8\sqrt{5}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $DE \perp AC$, $BF \perp AC$, with the perpendicular feet being $E$ and $F$ respectively. Extend $DE$ and $BF$ to meet $AB$ at point $H$ and $BC$ at point $G$ respectively. If $AD \parallel BC$ and $AE=CF$, then prove that quadrilateral $ABCD$ is a parallelogram.

Solution:
Proof:

Since $DE \perp AC$ and $BF \perp AC$,

it follows that $\angle AED = \angle CFB = 90^\circ$ and $DE \parallel BF$.

Given $AD \parallel BC$,

it implies $\angle DAE = \angle BCF$.

In $\triangle DAE$ and $\triangle BCF$,

\[
\left\{\begin{array}{l}
\angle DEA = \angle BFC = 90^\circ \\
AE = CF \\
\angle DAE = \angle BCF
\end{array}\right.
\]

Therefore, $\triangle DAE \cong \triangle BCF$ (ASA),

which means $AD = CB$.

Since $AD \parallel BC$,

the quadrilateral $ABCD$ is \boxed{a \ parallelogram}.


Question: Given: In quadrilateral $ABCD$, $DE\perp AC$ and $BF\perp AC$ with perpendicular feet $E$ and $F$ respectively. Extend $DE$ and $BF$ to intersect $AB$ at point $H$ and $BC$ at point $G$, if $AD\parallel BC$ and $AE=CF$. If $\angle DAH=\angle GBA$, $GF=2$, and $CF=4$, find the length of $AD$.

Solution: \textbf{Solution:} Since $DE\perp AC$ and $BF\perp AC$, it follows that $DH\parallel BG$. Consequently, $\angle GBA=\angle DHA$. Given that $\angle DAH=\angle GBA$, we have $\angle DAH=\angle DHA$, which implies $AD=DH$. In $\triangle GCF$, where $GF=2$ and $CF=4$, we find that $CG=\sqrt{CF^2+GF^2}=2\sqrt{5}$. Because quadrilateral $ABCD$ is a parallelogram, we have $DC\parallel AB$ and $AB=CD$, which means $DG\parallel HB$. Given that $DH\parallel GB$, quadrilateral $DHBG$ forms a parallelogram, thus $DG=HB$. Therefore, $AH=CG=2\sqrt{5}$. In $\triangle AEH$, where $AE=CF=4$ and $AH=2\sqrt{5}$, it follows that $EH=2$. In $\triangle ADE$, we have $AD^2=DE^2+AE^2$, which gives us $AD^2=(AD-2)^2+4^2$. Solving this, we find $AD=\boxed{5}$.


Question: As shown in the figure, in the Cartesian coordinate system, it is known that the graph of the linear function $y=\frac{1}{2}x+1$ intersects the x-axis and y-axis at points A and B, respectively. A square ABCD is constructed in the second quadrant with AB as a side. What is the area of the square ABCD?

Solution:
\textbf{Solution:} For the line $y=\frac{1}{2}x+1$, by setting $x=0$, we find $y=1$; by setting $y=0$, we find $x=-2$,\\
$\therefore$A$(-2,0)$, B$(0,1)$,\\
$\therefore$in $\triangle AOB$, $OA=2$, $OB=1$,\\
$\therefore$according to the Pythagorean theorem: $AB=\sqrt{2^2+1^2}=\sqrt{5}$,\\
$\therefore$the area of the square $ABCD$ is $\boxed{5}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the linear function \(y=\frac{1}{2}x+1\) intersects the x-axis and y-axis at points A and B, respectively. A square ABCD is constructed in the second quadrant with AB as one of its sides. What are the coordinates of points C and D?

Solution:
\textbf{Solution:} As shown in the figure, let $CE\perp y$ axis, $DF\perp x$ axis,\\
$\therefore \angle CEB=\angle AFD=\angle AOB=90^\circ$.\\
$\because$ Quadrilateral $ABCD$ is a square,\\
$\therefore BC=AB=AD$, $\angle DAB=\angle ABC=90^\circ$,\\
$\therefore \angle DAF+\angle BAO=90^\circ$, $\angle ABO+\angle CBE=90^\circ$,\\
$\because \angle DAF+\angle ADF=90^\circ$, $\angle BAO+\angle ABO=90^\circ$,\\
$\therefore \angle BAO=\angle ADF=\angle CBE$,\\
$\therefore \triangle BCE\cong \triangle DAF\cong \triangle ABO(AAS)$,\\
$\therefore BE=DF=OA=2$, $CE=AF=OB=1$,\\
$\therefore OE=OB+BE=3$, $OF=OA+AF=3$,\\
$\therefore$ The coordinates of point C are $\boxed{(-1, 3)}$, the coordinates of point D are $\boxed{(-3, 2)}$;


Question: As shown in the figure, in the Cartesian coordinate system, it is known that the graph of the linear function $y=\frac{1}{2}x+1$ intersects the x-axis and y-axis at points A and B, respectively. A square ABCD is constructed in the second quadrant with AB as a side. Is there a point M on the x-axis such that the perimeter of $\triangle MDB$ is minimized? Find the coordinates of point M.

Solution:
\textbf{Solution:} As shown in the figure, find the point $B^{\prime}$, which is the reflection of point B across the x-axis. Connect $B\prime D$. If it intersects the x-axis at point M, then the perimeter of $\triangle BMD$ is minimized at this moment.\\
$\because$B(0,1),\\
$\therefore B^{\prime}$(0,−1)\\
Let the equation of the line $B\prime D$ be $y=kx+b(k\ne 0)$,\\
Substituting the coordinates of $B^{\prime}$ and D gives: $\begin{cases}b=−1\\−3k+b=2\end{cases}$,\\
Solving these, we get: $\begin{cases}k=−1\\b=−1\end{cases}$,\\
$\therefore$The equation of the line $B\prime D$ is $y=−x−1$.\\
For $y=−x−1$, let $y=0$, we find $x=−1$,\\
$\therefore M(\boxed{-1},0)$.


Question: As shown in the figure, in rectangle $ABCD$, point $O$ is the midpoint of diagonal $AC$. Through point $O$, $EF \perp AC$ intersects $BC$ and $AD$ at points $E$ and $F$, respectively. Connect $AE$ and $CF$. Quadrilateral $AECF$ is a rhombus; if $AB = 3$ and $BC = 5$, find the length of $AE$.

Solution:
\textbf{Solution:} Let $AE=CE=x$, so $BE=5-x$, \\
since quadrilateral ABCD is a rectangle, \\
therefore, $\angle B=90^\circ$.\\
In $\triangle ABE$, by the Pythagorean theorem, $AB^2+BE^2=AE^2$,\\
which gives $3^2+(5-x)^2=x^2$,\\
solving this, we get $x=\boxed{3.4}$.


Question: As shown in the figure, in the quadrilateral $ABCD$, $E$ is the midpoint of $AD$, and $F$ is a point on the extension of $BC$, with $CF = \frac{1}{2}BC$. Connect $CE$ and $DF$. Quadrilateral $CEDF$ is a parallelogram; if $AB=4$, $AD=6$, and $\angle B=60^\circ$, find the length of $DF$.

Solution:
\textbf{Solution:} Let the line through $DH$ perpendicular to $BF$ at $H$, \\
since quadrilateral $ABCD$ is a parallelogram, it follows that $AB \parallel CD$ and $CD=AB=4$. Hence, $\angle DCH=\angle B=60^\circ$, thus $\angle CDH=30^\circ$. Therefore, $CH=\frac{1}{2}CD=2$, and thus $DH=\sqrt{CD^{2}-CH^{2}}=\sqrt{4^{2}-2^{2}}=2\sqrt{3}$. Since $CF=\frac{1}{2}AD=\frac{1}{2}BC=3$, it follows that $HF=CF-CH=3-2=1$. Therefore, $DF=\sqrt{DH^{2}+HF^{2}}=\sqrt{12+1}=\boxed{\sqrt{13}}$.


Question: As illustrated in the Cartesian coordinate system, $AB\parallel OC$, with $A(0, 12)$, $B(a, c)$, $C(b, 0)$, and $a, b$ satisfying $\left|a-21\right|+\left(b-16\right)^2=0$. A moving point $P$ departs from point $A$ and travels towards point $B$ along line segment $AB$ at a speed of 2 units per second; another moving point $Q$ starts from point $O$ and moves towards point $C$ along line segment $OC$ at a speed of 1 unit per second. Points $P$ and $Q$ commence their movements from points $A$ and $O$ respectively, simultaneously. When point $P$ reaches point $B$, it stops, and point $Q$ stops accordingly. Let the duration of the movement be $t$ seconds. What are the coordinates of points $B$ and $C$?

Solution:
\textbf{Solution:} We have $\because$ $\left|a-21\right|+\left(b-16\right)^{2}=0$,

$\therefore$ $\left\{\begin{array}{l}
a-21=0\\
b-16=0
\end{array}\right.$, solving this gives us: $a=21$, $b=16$,

$\therefore$ $\boxed{C(16,0)}$, $\because$ $AB\parallel OC$, $OA=12$,

$\therefore$ $\boxed{B(21,12)}$.


Question: As shown in the figure, in the Cartesian coordinate system, $AB\parallel OC$, with $A(0, 12)$, $B(a, c)$, $C(b, 0)$, and $a, b$ satisfying $\left|a-21\right|+\left(b-16\right)^2=0$. A moving point $P$ starts from point $A$ and moves towards point $B$ along the line segment $AB$ at a speed of 2 units per second; a moving point $Q$ starts from point $O$ and moves towards point $C$ along the line segment $OC$ at a speed of 1 unit per second. Points $P$ and $Q$ start moving from points $A$ and $O$ simultaneously, and when point $P$ reaches point $B$, it stops moving, and point $Q$ stops moving as well. Let the movement time be $t$ seconds. At what value of $t$ will the quadrilateral $PQCB$ be a parallelogram? What are the coordinates of points $P$ and $Q$ at this time?

Solution:
\textbf{Solution:} Since $BQ \parallel CP$, when $BQ=CP$, the quadrilateral $PQCB$ is a parallelogram,\\
thus, $21-2t=16-t$, solving this gives: $t=\boxed{5}$,\\
therefore, when $t=5$, the quadrilateral $PQCB$ is a parallelogram,\\
thus, $AP=2t=10$,$OQ=t=5$,\\
$\boxed{P(10,12)}$, $\boxed{Q(5,0)}$.


Question: As shown in the figure, in the Cartesian coordinate system, $AB\parallel OC$, $A(0, 12)$, $B(a, c)$, $C(b, 0)$, and $a, b$ satisfy $\left|a-21\right|+\left(b-16\right)^2=0$. A moving point $P$ starts from point $A$ and moves towards point $B$ along the segment $AB$ at a speed of 2 units per second; another moving point $Q$ starts from point $O$ and moves towards point $C$ along the segment $OC$ at a speed of 1 unit per second. Points $P, Q$ start simultaneously from points $A, O$ respectively, and the movement of point $P$ stops when it reaches point $B$, and consequently, point $Q$ stops moving. Suppose the movement duration is $t$ seconds. At what value of $t$ is $\triangle PQC$ an isosceles triangle with $PQ$ as its equal side? Determine the coordinates of points $P$ and $Q$ at this time.

Solution:
\textbf{Solution:} Given: $Q(t,0)$, $P(2t,12)$,

(1) When $PQ=PC$, constructing $PM \perp x$-axis at point $M$, it follows that: $QM=t$, $CM=16-2t$,

$\therefore QM=CM$ (coincidence of three lines),

$\therefore t=16-2t$,

Solving for $t$ gives $t=\frac{16}{3}$,

$\therefore AP=2t=\frac{32}{3}$, $OQ=\frac{16}{3}$,

$\therefore P\left(\frac{32}{3}, 12\right), Q\left(\frac{16}{3}, 0\right)$;

(2) When $PQ=CQ$, constructing $QN \perp AB$ at point $N$, it follows that: $QN=12$, $PN=2t-t=t$, $CQ=16-t$,

$\because PQ^2=CQ^2$,

$\therefore 12^2+(2t-t)^2=(16-t)^2$,

Solving for $t$ gives $t=\frac{7}{2}$,

$\therefore AP=2t=7$, $OQ=\frac{7}{2}$,

$\therefore P(7, 12), Q\left(\frac{7}{2}, 0\right)$;

In summary, the coordinates of points $P$ and $Q$ are $\boxed{P(7, 12)}, \boxed{Q\left(\frac{7}{2}, 0\right)}$ or $\boxed{P\left(\frac{32}{3}, 12\right)}, \boxed{Q\left(\frac{16}{3}, 0\right)}$.


Question: As shown in the figure, it is known that the diagonals of the rhombus $ABCD$ intersect at point $O$. Extend $AB$ to point $E$ such that $BE=AB$, and connect $CE$. Given $BD=EC$; at what angle $\angle DAB$ will the quadrilateral $BECD$ form a rhombus? Please explain the reason.

Solution:
\textbf{Solution:} When $\angle DAB=60^\circ$, the quadrilateral $BECD$ is a rhombus.\\
The reason is as follows:\\
$\because$ Quadrilateral $ABCD$ is a rhombus,\\
$\therefore$ $AD=AB$,\\
$\because$ $\angle DAB=60^\circ$,\\
$\therefore$ $\triangle ADB$, $\triangle DCB$ are equilateral triangles,\\
$\therefore$ $DC=DB$,\\
$\because$ Quadrilateral $BECD$ is a parallelogram,\\
$\therefore$ Quadrilateral $BECD$ is a rhombus.\\
$\therefore$ \boxed{\angle DAB=60^\circ}


Question: As shown in the figure, in parallelogram $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. $E$ and $F$ are respectively the midpoints of $OB$ and $OD$. Connect $AE$, $AF$, $CE$, and $CF$. Quadrilateral $AECF$ is a parallelogram; if $AB \perp AC$ and $AB = 3$, $BC = 5$. What is the length of $BD$?

Solution:
\textbf{Solution:} Since AB$\perp$AC,\\
it follows that $\angle$BAC$=90^\circ$,\\
hence, AC$= \sqrt{BC^{2}-AB^{2}}=\sqrt{5^{2}-3^{2}}=4$,\\
thus, OA$=\frac{1}{2}AC=2$,\\
In right triangle $\triangle$AOB, by the Pythagorean theorem, we get OB$=\sqrt{AB^{2}+OA^{2}}=\sqrt{3^{2}+2^{2}}=\sqrt{13}$,\\
therefore, BD$=2OB=2\sqrt{13}$. Therefore, the length of BD is $\boxed{2\sqrt{13}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AC$ intersects $BD$ at point $O$, and $AO=CO$. Point $E$ is on segment $BO$, and $\angle CEO=\angle ADO$. Prove that quadrilateral $AECD$ is a parallelogram.

Solution:
\textbf{Solution.} Proof:

Since $\angle CEO = \angle ADO$,

it follows that $CE \parallel AD$.

Given $\angle CEO = \angle ADO$, $\angle COE = \angle AOD$, and $OC = OA$,

we have $\triangle COE \cong \triangle AOD$,

which implies $CO = AO$, $OE = OD$.

Thus, quadrilateral \boxed{AECD} is a parallelogram.


Question: As shown in the figure, in quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $O$, and $AO=CO$. Point $E$ is on line segment $BO$, and $\angle CEO=\angle ADO$. If $AB=BC$, $CD=10$, and $AC=16$, find the area of quadrilateral $AECD$.

Solution: \textbf{Solution:} Given $AB=BC$ and $OA=OC$, it follows that $BO \perp AC$. Within $\triangle COD$, we have $OC=\frac{1}{2}AC=8$, therefore $OD=\sqrt{CD^{2}-OC^{2}}=\sqrt{10^{2}-8^{2}}=6$. Consequently, $DE=2OD=12$. Thus, the area of quadrilateral AECD $=\frac{1}{2}\times DE\times AC=\frac{1}{2}\times 12\times 16=\boxed{96}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BCA=90^\circ$, CD is the median to side AB, and lines parallel to BA and BC are drawn through points C and D, respectively, intersecting at point E, where DE intersects AC at point O, and AE is connected. It is given that AD=CD; if $\angle B=60^\circ$ and BC=3, what is the area of quadrilateral ADCE?

Solution:
\textbf{Solution:} In the right-angled $\triangle ABC$, where $CD$ is the median to side $AB$, and $\angle B=60^\circ$, $BC=3$,\\
$\therefore AD=DB=CD=3$.\\
$\therefore AB=6$, by the Pythagorean theorem, $AC=3\sqrt{3}$.\\
Since quadrilateral $DBCE$ is a parallelogram, $\therefore DE=BC=3$.\\
Since $EC=BD=AD$ and $CE\parallel DB$, $\therefore$ quadrilateral $ADCE$ is a parallelogram.\\
$\therefore ED\parallel BC$.\\
$\therefore \angle AOD=\angle ACB$.\\
Since $\angle ACB=90^\circ$, $\therefore \angle AOD=\angle ACB=90^\circ$.\\
$\therefore$ parallelogram $ADCE$ is a rhombus,\\
$\therefore$ Area of the rhombus $ADCE$ is $\frac{AC\cdot ED}{2}=\frac{3\sqrt{3}\times 3}{2}=\boxed{\frac{9\sqrt{3}}{2}}$.


Question: As shown in the figure, fold the rectangular paper $ABCD$ along $EF$ so that vertex $C$ coincides with vertex $A$, and point $D$ falls at point $G$. If $AB=4$ and $BC=8$, find the area of the overlapping part of the paper after folding.

Solution:
Solution: From the folding, we have: $AE=EC$, $\angle AEF=\angle CEF$. Let $BE=x$, then $AE=EC=8-x$. In $\triangle ABE$, we have $AB^2+BE^2=AE^2$,
\[\therefore x^2+4^2=(8-x)^2\], solving this gives $x=3$,
\[\therefore AE=EC=8-x=5\],
\[\because rectangle ABCD\],
\[\therefore AD\parallel BC\],
\[\therefore \angle AFE=\angle CEF\],
\[\therefore \angle AEF=\angle AFE\],
\[\therefore AE=AF=5\],
Therefore, the area of the overlapped part of the paper after folding, $\triangle AEF$, is $\frac{1}{2}AB\cdot AF=\frac{1}{2}\times 4\times 5=\boxed{10}$


Question: As shown in the figure, rectangle $OABC$ is in a Cartesian coordinate system, with the coordinate of vertex $B$ being $\left(4, n\right)$. Diagonals $OB$ and $AC$ intersect at $D$. The line $y=\frac{3}{4}nx-\frac{3}{2}n$ intersects $OA$, $AC$, and $OB$ at $P$, $M$, and $N$, respectively. Find the length of $DP$ (expressed in terms of $n$).

Solution:
\textbf{Solution:} Since quadrilateral OABC is a rectangle,\\
it follows that point D is the midpoint of OB,\\
Since $O(0,0)$ and $B(4,n)$,\\
it follows that $D(2,\frac{n}{2})$,\\
From $\frac{3}{4}nx-\frac{3}{2}n=0$ we get: $x=2$, therefore $P(2,0)$,\\
thus $DP\parallel$ y-axis,\\
thus $DP=\boxed{\frac{n}{2}}$;


Question: As shown in the figure, rectangle $OABC$ is in the Cartesian coordinate system, where the coordinate of vertex $B$ is $(4, n)$, and the diagonals $OB$ and $AC$ intersect at $D$. The line $y=\frac{3}{4}nx-\frac{3}{2}n$ intersects $OA$, $AC$, and $OB$ at points $P$, $M$, and $N$, respectively. When $CN=2MN$, what is the value of $n$?

Solution:
\textbf{Solution:} Since point M is the midpoint of the line segment PN,\\
$PN=2MN=CN$,\\
Since $CN=\sqrt{3^2+\left(n-\frac{3}{4}n\right)^2}=\sqrt{9+\frac{1}{16}n^2}$,\\
$PN=\sqrt{(2-3)^2+\left(0-\frac{3}{4}n\right)^2}=\sqrt{1+\frac{9}{16}n^2}$,\\
Therefore, $\sqrt{9+\frac{1}{16}n^2} = \sqrt{1+\frac{9}{16}n^2}$,\\
Solving this, we get: $n=4$ or $n=-4$,\\
Since $n>0$,\\
Therefore, $n=\boxed{4}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=9$, $AD=3$, and $M$, $N$ are points on $AB$, $CD$ respectively. When quadrilateral $MBCN$ is folded along $MN$, point $B$ coincides exactly with point $D$ and point $C$ with point $E$. Connect $BN$. Quadrilateral $DMBN$ is a rhombus; determine the length of the line segment $AM$.

Solution:
\textbf{Solution:} Let $AM=x$, then $DM=BM=AB-AM=9-x$,\\
since quadrilateral $ABCD$ is a rectangle,\\
therefore $\angle A=90^\circ$,\\
in $\triangle ADM$, $AD^2+AM^2=DM^2$,\\
therefore $3^2+x^2=(9-x)^2$, solving gives: $x=\boxed{4}$,\\
that is, $AM=\boxed{4}$.


Question: In rectangle $ABCD$, $AB=9$ and $AD=3$, $M$ and $N$ are points on $AB$ and $CD$, respectively. When quadrilateral $MBCN$ is folded along $MN$, point $B$ coincides exactly with point $D$, and point $C$ coincides with point $E$. Connect $BN$. What is the length of the fold $MN$?

Solution:
\textbf{Solution:} As shown in the figure, connect BD, \\
$\because$ Quadrilateral ABCD is a rectangle, \\
$\therefore$ $\angle A=90^\circ$, \\
In $\triangle ABD$, AB=9, AD=3, \\
$\therefore$ $BD=\sqrt{AD^2+AB^2}=3\sqrt{10}$, \\
From (2), we get: AM=4, \\
$\therefore$BM=5, \\
$\because$ ${S}_{\text{rhombus }BMDN}=\frac{1}{2}MN\cdot BD=BM\cdot AD$, \\
$\therefore$ $\frac{1}{2}MN \times 3\sqrt{10}=5\times 3$, \\
Solving this gives: $MN=\boxed{\sqrt{10}}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertices $A$ and $C$ of the rectangle $OABC$ are located on the positive semi-axis of the $x$-axis and $y$-axis, respectively, and the coordinates of point $B$ are $\left(8, 4\right)$. The rectangle is folded along $OB$, with point $A$ corresponding to point $D$. $OD$ intersects $BC$ at point $E$. Given that $EO=EB$, find the coordinates of point $E$?

Solution: \textbf{Solution:} From equation (1), it is known that $EO=EB$, and since the vertices $A$, $C$ of the rectangle $OABC$ are located on the positive halves of the $x$-axis and $y$-axis respectively, with the coordinate of point $B$ being $\left(8, 4\right)$, it follows that $OC=AB=4$, and $OA=CB=8$. Let $OE=x$, then $DE=8-x$, and consequently $BE=OE=x$, $BD=4$. In $\triangle BDE$, we have $BD^2+DE^2=BE^2$, hence $4^2+(8-x)^2=x^2$, leading to $x=5$, thus $BE=5$. Therefore, $CE=BC-BE=8-5=3$, which yields that the coordinate of point $E$ is $\boxed{\left(3, 4\right)}$.


Question: As illustrated in the rectangle $ABCD$, where $DC=6$, there exists a point $E$ on $DC$. When $\triangle ADE$ is folded along the line $AE$, such that point $D$ precisely falls on side $BC$, let this point be $F$. If the area of $\triangle ABF$ is 24, what is the length of $BF$?

Solution: \textbf{Solution:} From the given information, the length of $BF$ is $\boxed{8}$.


Question: As shown in the figure, in rectangle $ABCD$, $DC=6$. There is a point $E$ on $DC$. By folding $\triangle ADE$ along the line $AE$, such that point $D$ falls exactly on the side $BC$, and let this point be $F$. If the area of $\triangle ABF$ is $24$, what is the length of $CE$?

Solution: \textbf{Solution:} We have $CE = \boxed{\frac{8}{3}}$.


Question: As shown in the figure, $\triangle ABC$ is inscribed in $\odot O$, $\angle ABC>90^\circ$, the bisector of its exterior angle $\angle EAC$ intersects $\odot O$ at point D, connect $DB, DC$, and $DB$ intersects $AC$ at point F. If $\angle EAD=75^\circ$, what is the degree measure of $\overset{\frown}{BC}$?

Solution:
\textbf{Solution:} Since $\angle EAD=75^\circ$ and $AD$ bisects $\angle EAC$,\\
it follows that $\angle EAC=2\angle DAE=150^\circ$,\\
thus $\angle BAC=180^\circ-\angle EAC=30^\circ$,\\
therefore, the degree measure of $\overset{\frown}{BC}$ is $30^\circ \times 2=\boxed{60^\circ}$;


Question: As shown in the figure, $ \triangle ABC$ is inscribed in $ \odot O$, $ \angle ABC>90^\circ$. The bisector of its exterior angle $ \angle EAC$ intersects $ \odot O$ at point D. Connect $ DB, DC$, where $ DB$ intersects $ AC$ at point F. Given $ DB=DC$ and $ DA=DF$, when $ \angle ABC=\alpha $, find the measure of $ \angle DFC$ in degrees (expressed in an algebraic formula involving $ \alpha$)?

Solution:
\textbf{Solution:} Given that $DA=DF$,\\
it follows that $\angle DAF=\angle DFA$,\\
hence, $\angle DAF=\angle DFA=\angle CBD=\angle BCD$,\\
therefore, $\triangle DAF\sim \triangle DBC$,\\
which means that $\angle ADF=\angle BDC$,\\
given that $\angle ABC=\alpha$ and knowing that the opposite angles in a cyclic quadrilateral are supplementary,\\
it follows that $\angle ADC=180^\circ-\alpha$,\\
thus, $\angle ADF=90^\circ-\frac{\alpha}{2}$,\\
resulting in $\angle DAF=\angle DFA=\left(180^\circ-\angle ADF\right)\div 2=45^\circ+\frac{\alpha}{4}$,\\
therefore, $\angle DFC=180^\circ-\angle DFA=\boxed{135^\circ-\frac{\alpha}{4}}$.


Question: As shown in Figure 1, triangle $ABC$ is inscribed in circle $O$, with $AB$ as the diameter and point $C$ located above $AB$, such that $\angle CAB > \angle B$. Connect $OC$. $CG$ is the altitude from $C$ to side $AB$. A line through point $O$ and perpendicular to $OC$ intersects the extension of $CG$ at point $D$ and intersects $BC$ at point $E$. It is given that $\angle ACG = \angle BCO$. When $OE = OD$, find the value of $\frac{OG}{OC}$.

Solution:
\textbf{Solution:} Since $DE \perp OC$, it follows that $\angle BCO + \angle CEO = 90^\circ$. Since $\angle ACG = \angle BCO$, we have $\angle CEO = \angle A$.\\
Since $\angle A + \angle ACG = 90^\circ$ and also $\angle BCG + \angle ACG = 90^\circ$, it follows that $\angle BCG = \angle A$, hence $\angle BCG = \angle CEO$,\\
therefore $DC = DE$. Since $OE = OD$ and $CO \perp DE$, we have $DC = CE$, thus $DC = DE = CE$,\\
therefore, $\triangle CDE$ is an equilateral triangle, hence $\angle DCE = 60^\circ$, and thus $\angle OCG = 30^\circ$, therefore $\frac{OG}{OC} = \sin 30^\circ = \boxed{\frac{1}{2}}$;


Question: In the Cartesian coordinate system $xOy$, it is known that the parabola $y=ax^{2}+bx$ passes through the points $A(4,0)$ and $B(1,4)$. Point $P$ lies on the parabola and above the line $AB$. What is the equation of the parabola?

Solution:
\textbf{Solution:} Substituting $A(4, 0)$ and $B(1, 4)$ into $y = ax^2 + bx$,\\
$\therefore$ we find the solution.\\
$\therefore$ The equation of the parabola is: \boxed{y = -x^2 + x}.


Question: In quadrilateral $ABCD$, $E$ is the midpoint of $DC$. Line $AE$ is extended to meet the extended line of $BC$ at point $F$, and $BC = CF$. Let $G$ be a point on $CF$, and line $AG$ intersects $CD$ at point $H$, where $\angle DAF = \angle GAF$. Given $CG = 2$ and $GF = 5$, find the length of $AH$.

Solution:
\textbf{Solution:} Given that $AD\parallel BC$,\\
it follows that $\angle DAF=\angle F$,\\
since $\angle GAF=\angle DAF$,\\
we have $\angle GAF=\angle F$,\\
thus $AG=GF=5$,\\
given that $CG=2$, $GF=5$,\\
it follows that $AD=CF=7$,\\
since $AD\parallel BC$,\\
it implies $\triangle AHD\sim \triangle GHC$,\\
therefore $\frac{AH}{GH}=\frac{AD}{GC}=\frac{7}{2}$,\\
which means $\frac{AH}{AG}=\frac{7}{9}$, i.e., $\frac{AH}{5}=\frac{7}{9}$,\\
$AH=\boxed{\frac{35}{9}}$.


Question: As shown in Figure 1, there is a circular paper with point O as the center and AB as the diameter. Point C is on the circle $\odot$O. Folding the circular paper along the diameter CF, point B falls on point D located on $\odot$O (not coinciding with point A). After unfolding the paper, connect CB, CD, and AD. Let CD intersect with diameter AB at point E. AD$\parallel$OC; as shown in Figure 2, when CD$\perp$AB, if OC＝2, what is the length of BC?

Solution:
\textbf{Solution:} From the fold, it is known that: $\angle ECO = \angle BCO$,\\
$\because OB = OC$,\\
$\therefore \angle OCB = \angle B$,\\
$\therefore \angle ECO = \angle OCB = \angle B$,\\
$\because CD \perp AB$,\\
$\therefore \angle ECO + \angle OCB + \angle B = 90^\circ$,\\
$\therefore \angle ECO = \angle OCB = \angle B = 30^\circ$,\\
$\because OC = OB = 2$,\\
$\therefore OE = 1$,\\
$\therefore CE = \sqrt{3}$,\\
$\therefore BC = 2CE = 2\sqrt{3}$;\\
Hence, the length of BC is $BC = \boxed{2\sqrt{3}}$.


Question: As shown in Fig. 1, there is a circular paper with point O as the center and AB as the diameter. Point C is on the circle $\odot$O. The circular paper is folded along the diameter CF, causing point B to fall on point D on $\odot$O (not coinciding with point A). After restoring the paper, connect CB, CD, AD. Assume CD intersects diameter AB at point E. As shown in Fig. 3, when $AD=DE$, if $BC=2$, what is the length of $AD$?

Solution:
\textbf{Solution:} Given that $AD=DE$, \\
it follows that $\angle DAE=\angle DEA$, \\
given that $\angle DEA=\angle BEC$ and $\angle DAE=\angle BCE$, \\
it follows that $\angle BEC=\angle BCE$, \\
considering the circle paper folded along the line CF, \\
it follows that $\angle ECO=\angle BCO$, \\
furthermore, given that $OB=OC$, \\
it follows that $\angle OCB=\angle B$, \\
given that $\angle ECO=\angle B$, and $\angle CEO=\angle CEB$, \\
it follows that $\triangle CEO\sim \triangle BEC$, \\
resulting in $\frac{CE}{ED} = \frac{BE}{CE}$, \\
and therefore $CE^2=EO\cdot BE$, \\
let EO=x, EC=OC=OB=a, \\
hence $a^2=x(x+a)$, \\
solving yields, x= $\frac{\sqrt{5}-1}{2}a$ (negative value discarded), \\
therefore, $OE= \frac{\sqrt{5}-1}{2}a$, \\
therefore, $AE=OA-OE=a- \frac{\sqrt{5}-1}{2}a= \frac{3-\sqrt{5}}{2}a$, \\
given that $\angle AED=\angle BEC$, $\angle DAE=\angle BCE$, \\
it follows that $\triangle BCE\sim \triangle DAE$, \\
resulting in $\frac{BC}{AD} = \frac{EC}{AE}$, \\
given that $BC=2$, \\
$\frac{2}{AD} = \frac{a}{\frac{3-\sqrt{5}}{2}a}$, \\
thus, $AD=\boxed{3- \sqrt{5}}$.


Question: As shown in Figure 1, in the intelligent triangle $\triangle ABC$, where $AB=2$, $BC=2\sqrt{2}$, and $AD$ is the median on side $BC$, find the value of $\frac{AD}{AC}$.

Solution:
\textbf{Solution:} Let: $AD$ be the median of $BC$, $AB=2$, $BC=2\sqrt{2}$,\\
Thus, $BD=\frac{1}{2}BC=\sqrt{2}$,\\
Therefore, $\frac{BD}{AB}=\frac{AB}{BC}=\frac{\sqrt{2}}{2}$,\\
Given that, $\angle B=\angle B$,\\
It follows that, $\triangle ABD\sim \triangle CBA$,\\
Therefore, $\frac{AD}{AC}=\frac{BD}{AB}=\boxed{\frac{\sqrt{2}}{2}}$.


Question: As shown in the figure, quadrilaterals ABCD, DCFE, and EFGH are three squares with side length of 1. Connect AC, AF, and AG. It follows that $\triangle ACF \sim \triangle GCA$. Find the degree measure of $\angle 1 + \angle 2 + \angle 3$?

Solution:
\textbf{Solution:} From (1) we have $\angle 1 = \angle \text{FAC}$.\\
Since quadrilateral ABCD is a square,\\
therefore $\angle 3 = 45^\circ$,\\
therefore $\angle 2 + \angle \text{FAC} = \angle 3 = 45^\circ$,\\
therefore $\angle 1 + \angle 2 = 45^\circ$,\\
therefore $\angle 1 + \angle 2 + \angle 3 = \boxed{90^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, point D is located on side AB, and $\angle ADC=\angle ACB$, with $AD=2$ and $BD=6$. Given that $\triangle ACD \sim \triangle ABC$, find the length of side AC.

Solution:
\textbf{Solution:} Given that $AD=2, BD=6$,

$\therefore AB=AD+BD=8$,

From (1) it has been proven that $\triangle ACD\sim \triangle ABC$,

$\therefore \frac{AC}{AB}=\frac{AD}{AC}$, that is $\frac{AC}{8}=\frac{2}{AC}$,

Solving this yields $AC=\boxed{4}$ or $AC=-4<0$ (which is discarded as it does not meet the problem requirements),

Upon verification, $AC=4$ is a solution of the listed fractional equation,

Therefore, the length of side $AC$ is \boxed{4}.


Question: In figure, in $ \triangle ABC$, $AB=AC$, $AD$ is the median to side $BC$, $DE\perp AB$ at point $E$. Triangles $ \triangle BDE \sim \triangle CAD$. If $AC=13$, $BC=10$, find the length of the line segment $BE$.

Solution:
\textbf{Solution:} Since $AD$ is the median line on side $BC$ and $BC=10$,\\
since $BD=CD=\frac{1}{2}BC=5$,\\
since $\triangle BDE\sim \triangle CAD$ and $AC=13$,\\
therefore $\frac{BE}{CD}=\frac{BD}{AC}$,\\
which means $\frac{BE}{5}=\frac{5}{13}$,\\
therefore $BE=\boxed{\frac{25}{13}}$.


Question: In $\triangle ABC$, $AC=BC=10$, $\tan A = \frac{3}{4}$, points $D$ and $E$ are movable points on sides $AB$ and $AC$ respectively, and segment $DE$ is connected. Construct $\triangle ADE$ symmetric to $\triangle A'DE$ about $DE$. As shown in figure 1, when point $A'$ coincides with point $C$, what is the length of $DE$?

Solution:
\textbf{Solution:} Given that $AE=CE$, $\angle AED=90^\circ$,\\
since $AC=10$,\\
it follows that $AE=5$,\\
since $\tan A=\frac{3}{4}=\frac{DE}{AE}$,\\
then $DE=5\times \frac{3}{4}=\boxed{\frac{15}{4}}$.


Question: In $\triangle ABC$, $AC=BC=10$, and $\tan A = \frac{3}{4}$. Points $D$ and $E$ are moving points on sides $AB$ and $AC$, respectively. Connect $DE$ and construct the triangle $\triangle A'DE$ symmetric to $\triangle ADE$ with respect to $DE$. As shown in figure 2, $E$ is the midpoint of side $AC$. When $\triangle A'DE$ is an isosceles triangle, what is the length of $AD$?

Solution:
\textbf{Solution:} Given that $\triangle ADE$ and $\triangle A'DE$ are symmetric about $DE$,\\
then to find the length of $AD$ when $\triangle A'DE$ is an isosceles triangle is equivalent to finding the length of $AD$ when $\triangle ADE$ is an isosceles triangle,\\
since $E$ is the midpoint of side $AC$,\\
then $AE=5$.\\
(1) When $AE=AD$,\\
$AD=AE=5$;\\
(2) When $AE=ED$,\\
as shown in the diagram, draw $EM \perp AD$ intersecting $AD$ at $M$,\\
since $AE=ED$ and $EM \perp AD$,\\
then $EM$ is the perpendicular bisector of $AD$,\\
since $\tan A=\frac{3}{4}$ and $AE=5$,\\
then $AM=4$,\\
thus $AD=8$;\\
(3) When $AD=ED$,\\
as shown in the diagram, draw $DN \perp AE$ intersecting $AE$ at $N$,\\
since $DN \perp AE$ and $AD=ED$,\\
then $DN$ is the perpendicular bisector of $AE$,\\
thus $AN=\frac{5}{2}$,\\
since $\tan A=\frac{3}{4}$,\\
then $AD=\frac{25}{8}$;\\
therefore, the length of $AD$ is $5$, $8$, or \boxed{\frac{25}{8}}.


Question: In $\triangle ABC$, $AC=BC=10$, $\tan A=\frac{3}{4}$. Points $D$ and $E$ are moving points on sides $AB$ and $AC$ respectively, connecting $DE$. Construct the symmetric figure of $\triangle ADE$ with respect to $DE$, denoted as $\triangle A'DE$. As shown in Figure 3, $E$ is the midpoint of side $AC$. Connect $A'B$, let $F$ be the midpoint of $A'B$, and then connect $CF$. During the movement of point $D$, find the maximum length of segment $CF$.

Solution:
\textbf{Solution:} As shown in the figure, connect $BE$, draw $CH\perp AB$ at H, take the midpoint O of $BE$, connect $OF, OH$, and draw $OG\perp CH$ at G,\\
$\because$ $\tan A = \frac{3}{4}$, $AC=10$,\\
$\therefore$ $AH=8$, $HC=6$,\\
$\because$ $E$ is the midpoint of side $AC$,\\
$\therefore$ $AE=CE=5$,\\
$\therefore$ $A'E=5$,\\
$\because$ point F is the midpoint of $A'B$, point O is the midpoint of $BE$,\\
$\therefore$ $FO=\frac{1}{2}A'E=\frac{5}{2}$,\\
$\therefore$ point F moves on a circle with center O and radius $OF$,\\
$\therefore$ when point F is on the extension line of $CO$, $CF$ has its maximum value,\\
$\because$ $AH=BH$, point O is the midpoint of $BE$,\\
$\therefore$ $OH\parallel AE$, $OH=\frac{1}{2}AE=\frac{5}{2}$,\\
$\therefore$ $\angle ACH = \angle CHO$,\\
And $\because$ $\angle AHC = \angle OGH = 90^\circ$,\\
$\therefore$ $\triangle ACH \sim \triangle OHG$,\\
$\therefore$ $\frac{OH}{AC}=\frac{GH}{CH}=\frac{OG}{AH}=\frac{1}{4}$,\\
$\therefore$ $HG=\frac{3}{2}, GO=2$,\\
$\therefore$ $CG=\frac{9}{2}$,\\
In $\triangle GCO$ right-angled at $G$, by Pythagoras theorem, we get: $CO=\sqrt{\frac{81}{4}+4}=\frac{\sqrt{97}}{2}$,\\
$\therefore$ the maximum value of $CF$ is $\boxed{\frac{\sqrt{97}+5}{2}}$.


Question: As shown in the figure, the graph of the linear function $y=-2x+10$ intersects the graph of the inverse proportion function $y=\frac{k}{x}$ at points $A$ and $B$ (with $A$ being to the right of $B$), and the x-coordinate of point $A$ is 4. Determine the equation of the inverse proportion function and the coordinates of point $B$.

Solution:
\textbf{Solution:} Substitute $x=4$ into $y=-2x+10$ to obtain $y=2$, \\
$\therefore A(4,2)$, \\
Substitute $A(4,2)$ into $y=\frac{k}{x}$, to get $k=4\times 2=8$. \\
$\therefore$ The equation of the inverse proportion function is $y=\frac{8}{x}$, \\
Solve the system of equations $\left\{\begin{array}{l}y=-2x+10 \\ y=\frac{8}{x}\end{array}\right.$, to obtain $\left\{\begin{array}{l}x=1 \\ y=8\end{array}\right.$, or $\left\{\begin{array}{l}x=4 \\ y=2\end{array}\right.$\\
$\therefore$ The coordinates of point $B$ are $\boxed{(1,8)}$;


Question: Given: As shown in the figure, the graph of the linear function $y=-2x+10$ intersects with the graph of the inverse proportion function $y=\frac{k}{x}$ at points $A$ and $B$ (with $A$ located to the right of $B$), where the abscissa of point $A$ is $4$. Observing the graph, directly write out the solution set of the inequality $-2x+10-\frac{k}{x}>0$ with respect to $x$.

Solution: \textbf{Solution:} Observing the graph, the solution set of the inequality $-2x+10-\frac{k}{x}>0$ with respect to $x$ is: \boxed{1<x<4} or \boxed{x<0}.


Question: Given: As shown in the figure, the graph of the linear function $y=-2x+10$ intersects with the graph of the hyperbolic function $y=\frac{k}{x}$ at points $A$ and $B$ (with $A$ to the right of $B$), and the x-coordinate of point $A$ is $4$. On the other branch of the hyperbolic function graph, is there a point $P$ such that $\triangle PAB$ is a right triangle with $AB$ as one of its legs? If such point $P$ exists, find the coordinates of all points $P$ that meet the conditions; if it does not exist, please explain why.

Solution:
\textbf{Solution:} The point exists,\\
Reason:\\
(1) If $\angle BAP=90^\circ$,\\
draw $AH\perp OE$ at H from point A, let the intersection of AP and the x-axis be M, as shown in figure 1,\\
for $y=-2x+10$,\\
when $y=0$, $-2x+10=0$, solving gives $x=5$,\\
$\therefore$ point E(5,0), $OE=5$.\\
$\because A(4,2)$,\\
$\therefore OH=4$, $AH=2$,\\
$\therefore HE=5-4=1$.\\
$\because AH\perp OE$,\\
$\therefore \angle AHM=\angle AHE=90^\circ$.\\
Furthermore, $\because \angle BAP=90^\circ$,\\
$\therefore \angle AME+\angle AEM=90^\circ$, $\angle AME+\angle MAH=90^\circ$,\\
$\therefore \angle MAH=\angle AEM$,\\
$\therefore \triangle AHM\sim \triangle EHA$,\\
$\therefore \frac{AH}{EH}=\frac{MH}{AH}$, i.e., $\frac{2}{1}=\frac{MH}{2}$\\
$\therefore MH=4$,\\
$\therefore M(0,0)$,\\
we can set the equation of the line AP as $y=mx$,\\
then $4m=2$, solving gives $m=\frac{1}{2}$,\\
$\therefore$ the equation of line AP is $y=\frac{1}{2}x$,\\
Solve the system of equations $\begin{cases}y=\frac{1}{2}x\\ y=\frac{8}{x}\end{cases}$ to obtain $\begin{cases}x=4\\ y=2\end{cases}$, $\begin{cases}x=-4\\ y=-2\end{cases}$\\
$\therefore$ the coordinates of point P are $(-4,-2)$.\\
(2) If $\angle ABP=90^\circ$,\\
Similarly, we get: the coordinates of point P are $(-16,-\frac{1}{2})$.\\
In summary: The coordinates of point P that meet the conditions are $\boxed{(-4,-2)}$, $\boxed{(-16,-\frac{1}{2})}$.


Question: As shown in the figure, a painting with a rectangular frame is to be drawn on the gable roof of a building for beautification. It is known that $BC=12m$, and the height $AD=8m$. If the frame is a square, and one side of the square lies on $BC$, with the other two vertices on $AB$ and $AC$ respectively, what is the side length of the square?

Solution:
\textbf{Solution:} Let the side length of the square $PN=x$, then $AE=(8-x)$,\\
since quadrilateral $PQMN$ is a square,\\
therefore $PN\parallel QM$,\\
since $QM$ lies on $BC$,\\
therefore $PM\parallel BC$,\\
therefore $\triangle APN\sim \triangle ABC$,\\
therefore $\frac{PN}{BC}=\frac{AE}{AD}$,\\
that is $\frac{x}{12}=\frac{8-x}{8}$,\\
solving for $x$ yields $x=\boxed{4.8}$, which means: the side length of the square is \boxed{4.8}m.


Question: As shown in the figure, the goal is to draw a painting with a rectangular frame on the gable roof of a building to enhance its beauty. It is known that $BC = 12\,m$ and the height $AD = 8\,m$. If the painting's area is to be maximized, and one side of the rectangle is on $BC$, with the other two vertices on $AB$ and $AC$ respectively, find the dimensions of the frame's length and width as well as the maximum area.

Solution:
\textbf{Solution:} Let the length of the rectangle be $PN=x$ and $PQ=y$. Then $AE=8-y$. \\
Since $\triangle APN\sim \triangle ABC$, \\
it follows that $\frac{PN}{BC}=\frac{AE}{AD}$, \\
i.e., $\frac{x}{12}=\frac{8-y}{8}$, \\
yielding $y=8-\frac{2}{3}x$, \\
Therefore, $S=xy=x\left(8-\frac{2}{3}x\right)=-\frac{2}{3}x^2+8x=-\frac{2}{3}\left(x-6\right)^2+24$ \\
Therefore, when the length is $6$ meters and the width is $4$ meters, the maximum area of the rectangle that can be drawn is $\boxed{24}$ square meters.


Question: As shown in the diagram, in $\triangle ABC$, $\angle C=90^\circ$, $AC=6$, $BC=8$, $D$ is the midpoint of side $BC$, $E$ is a moving point on side $AB$, and $\angle DEF=90^\circ$, line $EF$ intersects the ray $BC$ at point $F$. Let $BE=x$, the area of $\triangle BED$ is $y$. Determine the functional relationship of $y$ in terms of $x$, and state the domain of the independent variable $x$.

Solution:
\textbf{Solution:} As shown in Figure 1, draw $EH\perp CB$ at $H$ through point $E$.\\
$\therefore EH\parallel AC$\\
In $\triangle ABC$, $\angle C=90^\circ$, $AC=6$, $BC=8$,\\
$\therefore AB=\sqrt{AC^2+BC^2}=10$,\\
$\because E$ can be a moving point on $AB$ that coincides with $A$ or $B$,\\
when $x=5$, $E$ is the midpoint of $AB$, $ED\parallel AC$, $\angle DEF=90^\circ$, at this time $EF\parallel BC$, $EF$ has no intersection with $BC$,\\
Let the distance from $C$ to $AB$ be $h$,\\
then $h=\frac{AC\times CB}{AB}=\frac{24}{5}$\\
When $ED\perp AB$, $DE=\frac{1}{2}h=\frac{12}{5}$,\\
at this time $BE=\sqrt{DB^2-DE^2}=\frac{16}{5}$, according to the figure, when $\frac{16}{5}<BE<5$, $EF$ has no intersection with $BC$,\\
$\therefore 0\le x\le \frac{16}{5}$ or $5<x\le 10$\\
$\because EH\parallel AC$\\
$\therefore \triangle BEH\sim \triangle BAC$\\
$\therefore \frac{EH}{AC}=\frac{BE}{AB}$\\
$EH=\frac{AC\cdot BE}{AB}=\frac{3}{5}x$\\
$\therefore {S}_{\triangle DEB}=\frac{1}{2}BD\cdot EH=\frac{1}{2}\times 4\times \frac{3}{5}x=\frac{6}{5}x$,\\
$\therefore \boxed{y=\frac{6}{5}x}(0<x\le \frac{16}{5}$ or $5<x\le 10)$


Question: As shown in the figure, in $\triangle ABC$, $\angle C = 90^\circ$, $AC = 6$, $BC = 8$, $D$ is the midpoint of side $BC$, $E$ is a moving point on side $AB$. Construct $\angle DEF = 90^\circ$, and let $EF$ intersect the extension of line $BC$ at point $F$. Let $BE = x$, and the area of $\triangle BED$ be $y$. If the triangle formed by vertices $B$, $E$, $F$ is similar to $\triangle BED$, find the area of $\triangle BED$.

Solution:
\textbf{Solution}: From the problem, we know that $ \angle BEF\ne 90^\circ$, so we can consider two cases.\\
(1) As shown in Figure 2, when $ \angle BEF$ is acute,\\
from the known fact that the triangle with vertices $B, E, F$ is similar to $\triangle BED$ and knowing that $\angle EBF=\angle DBE$ and $\angle BEF<\angle BED$, it follows that $\angle BEF=\angle BDE$.\\
Draw $DM\perp BA$ at $M$ through point $D$, and draw $EH\perp BC$ at $H$ through $E$.\\
$\therefore$ $ \angle MDE=\angle HDE$,\\
$\therefore$ $ EM=EH$.\\
Since $ EM=MB-EB=\frac{16}{5}-x$\\
and $\because$ $ EH=\frac{3}{5}x$\\
$\therefore$ $ \frac{16}{5}-x=\frac{3}{5}x$,\\
we find $ x=2$\\
$\therefore$ $ y=\frac{6}{5}\times 2=\frac{12}{5}$\\
(2) As shown in Figure 3, when $\angle BEF$ is obtuse,\\
similarly, we find $ x-\frac{16}{5}=\frac{3}{5}x$,\\
$\therefore$ $ x=8$.\\
$\therefore$ $ y=\frac{6}{5}\times 8=\frac{48}{5}$\\
In summary, the area of $\triangle BED$ is $\boxed{\frac{12}{5}}$ or $\boxed{\frac{48}{5}}$.


Question: As shown in the figure, $D$ is a point on $AC$, and $DE \parallel AB$, with $\angle B = \angle DAE$. $\triangle ABC \sim \triangle DAE$; If $AB = 4$, $AD = 3$, and $AE = 6$, find the length of $BC$.

Solution:
\textbf{Solution:} Since $\triangle ABC\sim \triangle DAE$,\\
it follows that $\frac{BC}{AE}=\frac{BA}{AD}$,\\
given $AB=4$, $AD=3$, and $AE=6$,\\
we have $\frac{BC}{6}=\frac{4}{3}$,\\
thus, $BC=\boxed{8}$.


Question: As shown in the figure, $\odot O$ is the circumcircle of $\triangle ABC$, where the diameter $BD$ intersects $AC$ at point E. A tangent to $\odot O$ through point D intersects the extension of $BC$ at point F, with $\angle F=\angle BAC$. Given $DF\parallel AC$, $AB = 8$, and $CF = 2$, what is the length of $AC$?

Solution:
\textbf{Solution:} Draw $CD$,\\
$\because DF\parallel AC$ and $\angle ODF=90^\circ$,\\
$\therefore \angle BEC=\angle ODF=90^\circ$,\\
$\therefore$ diameter $BD\perp AC$ at E,\\
$\therefore AE=CE=\frac{1}{2}AC$,\\
$\therefore AB=BC=8$,\\
$\because BD$ is the diameter of $\odot O$,\\
$\therefore \angle BCD=90^\circ$,\\
$\therefore \angle DBC+\angle BDC=90^\circ$,\\
$\because \angle DBC+\angle F=90^\circ$,\\
$\therefore \angle BDC=\angle F$,\\
$\because \angle BCD=\angle FCD=90^\circ$,\\
$\therefore \triangle BCD\sim \triangle DCF$,\\
$\therefore \frac{BC}{DC}=\frac{DC}{CF}$, i.e., $\frac{8}{DC}=\frac{DC}{2}$,\\
$\therefore DC=4$,\\
$\therefore BD=\sqrt{BC^2+CD^2}=\sqrt{8^2+4^2}=4\sqrt{5}$,\\
$\because$ in $\triangle BCD$, $S_{\triangle BCD}=\frac{1}{2}BC\cdot CD=\frac{1}{2}BD\cdot CE$,\\
$\therefore \frac{1}{2}\times 8\times 4=\frac{1}{2}\times 4\sqrt{5}\cdot CE$,\\
$\therefore CE=\frac{8}{5}\sqrt{5}$,\\
$\therefore AC=2CE=\boxed{\frac{16}{5}\sqrt{5}}$.


Question: As shown in the figure, in $ \triangle ABC$, with $ \odot O$ passing through vertices A and C of $ \triangle ABC$ and intersecting side AB at point D. A line is drawn through point D such that $ DF\parallel BC$, intersecting $ AC$ at point E and $ \odot O$ at point F. Line $ DC$ is drawn, with point C being the midpoint of arc $ DF$, and $ BC$ being the tangent of $ \odot O$. If the radius of $ \odot O$ is 3, and $ DF=4\sqrt{2}$, what is the value of $ CE\cdot CA$?

Solution:
\textbf{Solution:} Draw line $OC$ intersecting $DF$ at $M$, and connect $OD$, $AF$, as shown in the figure:\\
Since $C$ is the midpoint of arc $DF$,\\
$\therefore$ $OC\perp DF$,\\
$\therefore$ $DM=MF=\frac{1}{2}DF=2\sqrt{2}$,\\
Since the radius of circle $O$ is 3,\\
$\therefore$ $OM=\sqrt{OD^{2}-DM^{2}}=\sqrt{3^{2}-\left(2\sqrt{2}\right)^{2}}=1$,\\
$\therefore$ $CM=OC-OM=3-1=2$,\\
$\therefore$ $DC^{2}=DM^{2}+CM^{2}=\left(2\sqrt{2}\right)^{2}+2^{2}=12$,\\
Since $\overset{\frown}{DC}=\overset{\frown}{FC}$,\\
$\therefore$ $\angle DAC=\angle CAF$,\\
Since $\angle CDF=\angle CAF$,\\
$\therefore$ $\angle CDF=\angle DAC$,\\
Since $\angle DCE=\angle ACD$,\\
$\therefore$ $\triangle DCE\sim \triangle ACD$,\\
$\therefore$ $\frac{CD}{AC}=\frac{CE}{CD}$,\\
$\therefore$ $CD^{2}=CE\cdot CA$,\\
$\therefore$ $CE\cdot CA=\boxed{12}$.


Question: As shown in Figure 1, there exists a point $C$ on the circle $\odot O$ with diameter $AB$. Connect $BC$, and draw the bisector $BD$ of $\angle ABC$ that intersects $\odot O$ at point $D$. Through point $D$, draw the extension line $DE \perp BC$ at point $E$. $DE$ is the tangent to the circle $\odot O$; if $BE=9$ and $DE=6$, find the radius of $\odot O$.

Solution:
\textbf{Solution:} In right triangle $\triangle BDE$, by the Pythagorean theorem, we have $BD = \sqrt{BE^2 + DE^2} = \sqrt{9^2 + 6^2} = 3\sqrt{13}$,\\
since $AB$ is the diameter,\\
then $\angle ADB = 90^\circ$,\\
thus $\angle ADB = \angle E$,\\
since $\angle ABD = \angle DBE$,\\
it follows that $\triangle ABD \sim \triangle DBE$,\\
hence $\frac{AB}{BD} = \frac{BD}{BE}$,\\
thus $\frac{AB}{3\sqrt{13}} = \frac{3\sqrt{13}}{9}$,\\
therefore $AB = 13$,\\
hence the radius of circle $O$ is $\boxed{\frac{13}{2}}$.


Question: As shown in the figure, it is known that point $D$ lies on the extension of the diameter $AB$ of circle $O$, and point $C$ lies on circle $O$. A line is drawn through $D$ perpendicular to $AD$ and intersects with the extension of $AC$ at $E$. $CD$ is a tangent to circle $O$; $AB=2$ and $AE=3$. $CD=DE$; find the length of $BD$?

Solution:
\textbf{Solution:} Method One: \\
Since $AB=2$, \\
it follows that $OA=OB=OC=1$, \\
Since $OC\perp CD$, \\
by the Pythagorean theorem, we have $CD^{2}=(1+BD)^{2}-1^{2}$, \\
Since $ED\perp AD$, \\
by the Pythagorean theorem, we have $DE^{2}=3^{2}-(2+BD)^{2}$, \\
Since $CD=DE$, \\
it follows that $(1+BD)^{2}-1^{2}=3^{2}-(2+BD)^{2}$, \\
Solving this gives: $BD=\frac{-3+\sqrt{19}}{2}$ or $\frac{-3-\sqrt{19}}{2}$ (discard this result),\\
thus $BD=\boxed{\frac{-3+\sqrt{19}}{2}}$. \\

Method Two: \\
By the chord tangent theorem, we have $\angle DCB=\angle DAC$, \\
Since $\angle CDB=\angle ADC$, \\
it follows that $\triangle CDB\sim \triangle ADC$, \\
hence $\frac{CD}{AD}=\frac{BD}{CD}$, that is, $CD^{2}=AD\cdot BD=(2+BD)\cdot BD$, \\
Since $ED\perp AD$, \\
by the Pythagorean theorem, we have $DE^{2}=3^{2}-(2+BD)^{2}$, \\
Since $CD=DE$, \\
it follows that $(2+BD)\times BD=3^{2}-(2+BD)^{2}$, \\
Solving this gives $BD=\frac{-3+\sqrt{19}}{2}$ or $\frac{-3-\sqrt{19}}{2}$ (discard this result),\\
thus $BD=\boxed{\frac{-3+\sqrt{19}}{2}}$.


Question: As shown in the diagram, it is known that point $D$ lies on the extension of the diameter $AB$ of circle $\odot O$, point $C$ is on circle $\odot O$. A line through $D$ is drawn perpendicular to $AD$ at $E$, intersecting the extension of $AC$ at $E$, and $CD$ is a tangent to circle $\odot O$. It is given that $AB=2$ and $AE=3$. If the bisector of $\angle ACB$ intersects circle $\odot O$ at point $F$ and $P$ is the incenter of $\triangle ABC$, determine the length of $PF$.

Solution:
\textbf{Solution:} Draw $BF$, $PB$, and $AF$, \\
since $CF$ bisects $\angle ACB$, \\
thus $\overset{\frown}{A}F=\overset{\frown}{B}F$, \\
hence $AF=BF$, \\
since $AB$ is the diameter and $AB=2$, \\
thus $AF=BF=\sqrt{2}$, \\
since $P$ is the incenter of $\triangle ABC$, \\
thus $\angle 1=\angle 2$, $\angle CBP=\angle ABP$, \\
since $\angle 1=\angle 3$, \\
thus $\angle 2=\angle 3$, \\
hence $\angle 2+\angle CBP=\angle 3+\angle ABP$, \\
thus $\angle FPB=\angle FBP$, \\
hence $FP=FB=\boxed{\sqrt{2}}$.


Question: In $\triangle ABC$ and $\triangle ADE$, where point $E$ is on $BC$. It is known that $\angle B = \angle D$, $\angle DAB = \angle EAC$. Thus, $\triangle ABC \sim \triangle ADE$. If $AC \parallel DE$ and $\angle AEC = 45^\circ$, what is the degree measure of $\angle C$?

Solution:
\textbf{Solution:} Given that $AC \parallel DE$,\\
$\therefore \angle AED = \angle EAC$,\\
Since $\triangle ABC \sim \triangle ADE$,\\
$\therefore \angle AED = \angle C$,\\
$\therefore \angle EAC = \angle C$,\\
Given that $\angle AEC = 45^\circ$,\\
$\therefore \angle C = (180^\circ - 45^\circ) \div 2 = 67.5^\circ$,\\
Hence, the measure of $\angle C$ is $\boxed{67.5^\circ}$.


Question: As shown in the figure, it is known that $AB$ is the diameter of circle $\odot O$, $C$ is a point on circle $\odot O$, and the bisector of $\angle OCB$ intersects circle $\odot O$ at point $D$. A tangent to circle $\odot O$ passing through point $D$ intersects the extension line of $CB$ at point $E$. $CE\perp DE$; If $AB=10$, $\tan A = \frac{1}{3}$, find the length of $DE$.

Solution:
\textbf{Solution:} Given that $AB$ is the diameter of $\odot O$,\\
it follows that $\angle ADB=90^\circ$,\\
since $\tan A=\frac{BD}{AD}=\frac{1}{3}$,\\
let $BD=x$, then $AD=3x$,\\
hence $A{D}^{2}+B{D}^{2}=A{B}^{2}$, which leads to ${\left(3x\right)}^{2}+{x}^{2}=10^{2}$,\\
solving for $x$ yields $\sqrt{10}$,\\
since $\angle ODE=\angle ADB=90^\circ$,\\
it follows that $\angle BDE=\angle ADO$,\\
since $OA=OD$,\\
it follows that $\angle ADO=\angle OAD$,\\
hence $\angle BDE=\angle OAD$,\\
since $\angle E=\angle ADB=90^\circ$,\\
it follows that $\triangle BDE\sim \triangle BAD$,\\
hence $\frac{DE}{BD}=\frac{AD}{AB}$, which means $\frac{DE}{\sqrt{10}}=\frac{3\sqrt{10}}{10}$,\\
yielding $DE=\boxed{3}$.


Question: To effectively carry out labor education on campus, the Dongfanghong School utilized a piece of land with an area of unit 1 denoted by $\triangle ABC$, located on the southeast side of the school next to the wall. This land was divided into three parts for the seventh, eighth, and ninth grades, as shown in the diagram. Within $\triangle ABC$, point P is any point on side $BC$ (excluding points B and C), and the vertices F and E of rectangle $AFPE$ are respectively on sides $AB$ and $AC$. The area for the seventh grade corresponds to rectangle $AFPE$, while the eighth and ninth grades ocupy areas $\triangle PEC$ and $\triangle BPF$, respectively. If $BP: PC = 2:3$, what is the area of $\triangle BPF$?

Solution:
\textbf{Solution:} Since quadrilateral $AFPE$ is a rectangle, \\
hence $PF\parallel CA$, \\
hence $\triangle BFP\sim \triangle BAC$, \\
since $BP\colon PC=2\colon 3$, \\
hence $BP\colon BC=2\colon 5$, \\
hence $\frac{{S}_{\triangle BFP}}{{S}_{\triangle BAC}}=\frac{4}{25}$, \\
since ${S}_{\triangle ABC}=1$, \\
hence ${S}_{\triangle BFP}=\boxed{\frac{4}{25}}$;


Question: As shown in the figure, in $\triangle ABC$, point $P$ is an arbitrary point on side $BC$ (point $P$ does not coincide with points B and C), and the vertices $F$ and $E$ of rectangle $AFPE$ are on $AB$ and $AC$, respectively. The seventh grade corresponds to the area of rectangle $AFPE$, while the eighth and ninth grades correspond to the areas of $\triangle PEC$ and $\triangle BPF$, respectively. Given that $BC=2$ and $S_{\triangle ABC}=1$. Let $BP=x$, and the area of rectangle $AFPE$ be $y$, find the function relationship between $y$ and $x$.

Solution:
\textbf{Solution}: From (1) we have: $\triangle BFP \sim \triangle BAC$, \\
$\because BP=x, BC=2$, \\
$\therefore \frac{{S}_{\triangle BFP}}{{S}_{\triangle BAC}}=\left(\frac{x}{2}\right)^{2}$, \\
$\because {S}_{\triangle ABC}=1$, \\
$\therefore {S}_{\triangle BFP}=\frac{{x}^{2}}{4}$, \\
Similarly, ${S}_{\triangle PEC}=\left(\frac{2-x}{2}\right)^{2}=\frac{4-4x+{x}^{2}}{4}$, \\
$\therefore y=1-\frac{{x}^{2}}{4}-\frac{4-4x+{x}^{2}}{4}=-\frac{1}{2}{x}^{2}+x$, \\
$\therefore$ the function relationship between y and x is $\boxed{y=-\frac{1}{2}{x}^{2}+x}$.


Question: As shown in the figure, in rectangle $ABCD$, $E$ is the midpoint of side $BC$, and $DF\perp AE$ at point $F$. It is given that $\frac{AF}{BE} = \frac{AD}{AE}$. Knowing that $AB=8, BC=12$, what is the length of $AF$?

Solution:
\textbf{Solution:} Given, $\because E$ is the midpoint of $BC$,\\
$\therefore BE = \frac{1}{2}BC = 6$,\\
$\therefore AE = \sqrt{AB^{2} + BE^{2}} = 10$.\\
$\because \frac{AF}{BE} = \frac{AD}{AE}$,\\
$\therefore \frac{AF}{6} = \frac{12}{10}$,\\
$\therefore AF = \boxed{7.2}$.


Question: As shown in the figure, in $ \triangle ABC$, $ \angle A=55^\circ$, $ \angle B=45^\circ$, points $ D$ and $ E$ are on the sides $ AB$ and $ AC$ of $ \triangle ABC$, respectively, and $ \angle ADE=80^\circ$. $ \triangle AED\sim \triangle ABC$. If $ AD=4$, $ BD=6$, and $ AE=5$, find the length of $ CE$.

Solution:
\textbf{Solution:} From (1) we have $\triangle AED\sim \triangle ABC$,\\
$\therefore \frac{AD}{AC}=\frac{AE}{AB}$,\\
$\because AD=4$, $BD=6$,\\
$\therefore AB=10$,\\
$\because AD=4$, $AB=10$, $AE=5$,\\
$\therefore AC=8$,\\
$\therefore CE=AC-AE=8-5=\boxed{3}$.


Question: The graph of the quadratic function $y=ax^2+bx+4$ (where $a\neq 0$) passes through points $A(-4,0)$ and $B(1,0)$, and intersects the y-axis at point $C$. Point $P$ is a point on the parabola in the second quadrant. Lines $BP$ and $AC$ intersect at point $Q$. A line is drawn from point $P$ perpendicular to the x-axis, meeting it at point $D$. Find the expression of the quadratic function.

Solution:
\textbf{Solution:} From the given conditions, we have:
\[
\left\{
\begin{array}{l}
a\cdot (-4)^{2}+b\cdot (-4)+4=0\\
a+b+4=0
\end{array}
\right.
\]
Solving the system, we get:
\[
\left\{
\begin{array}{l}
a=-1\\
b=-3
\end{array}
\right.
\]
Therefore, the expression for the quadratic function is \boxed{y=-x^{2}-3x+4}.


Question: The graph of the quadratic function $y=ax^2+bx+4$ ($a\neq 0$) passes through points $A(-4,0)$ and $B(1,0)$, and intersects the $y$-axis at point $C$. Point $P$ is a point on the parabola in the second quadrant. Lines $BP$ and $AC$ are drawn and intersect at point $Q$. A perpendicular line $PD$ is drawn from point $P$ to the $x$-axis at point $D$. Line $BC$ is drawn. When $\angle DPB=2\angle BCO$, find the equation of line $BP$.

Solution:
\textbf{Solution}: Let $BP$ intersect the $y$-axis at point $E$, \\
since $PD \parallel y$-axis, \\
$\therefore \angle DPB = \angle OEB$, \\
since $\angle DPB = 2\angle BCO$, \\
$\therefore \angle OEB = 2\angle BCO$, \\
$\therefore \angle ECB = \angle EBC$, \\
$\therefore BE = CE$, let $OE = a$, then $CE = 4-a$, \\
$\therefore BE = 4-a$, \\
in $\triangle BOE$, by Pythagoras theorem, we have $BE^2 = OE^2 + OB^2$, \\
$\therefore (4-a)^2 = a^2 + 1^2$, solving this gives $a = \frac{15}{8}$, \\
$\therefore E\left(0, \frac{15}{8}\right)$, \\
let the equation of the line containing $BE$ be $y = kx + e (k \neq 0)$, \\
$\therefore \left\{\begin{array}{l}k \cdot 0 + e = \frac{15}{8}, \\ k \cdot 1 + e = 0.\end{array}\right.$\\
Solving this gives $\left\{\begin{array}{l}k = -\frac{15}{8}, \\ e = \frac{15}{8}.\end{array}\right.$\\
$\therefore$ The equation of line $BP$ is $\boxed{y = -\frac{15}{8}x + \frac{15}{8}}$.


Question: As shown in the figure, the graph of the quadratic function $y=ax^2+bx+4$ ($a\neq 0$) passes through points $A(-4,0)$, $B(1,0)$, and intersects the y-axis at point $C$. Point $P$ is a point on the parabola in the second quadrant. Connect $BP$ and $AC$, intersecting at point $Q$, and draw $PD\perp x$ axis at point $D$. Determine whether $\frac{PQ}{QB}$ has a maximum value? If so, find the coordinates of point $P$ when $\frac{PQ}{QB}$ is maximized; if not, explain why.

Solution: \textbf{Solution}: Set point $N$ where line segment $PD$ intersects with line segment $AC$. Draw a line parallel to the y-axis through point $B$ that intersects line segment $AC$ at point $M$. Given that the coordinates of points $A$ and $C$ are $(-4,0)$ and $(0,4)$, respectively, the equation of line $AC$ is $y=x+4$. Therefore, the coordinates of point $M$ are $(1,5)$, and the length of $BM$ is $5$. Since $BM \parallel PN$, triangles $\triangle PNQ$ and $\triangle BMQ$ are similar. Thus, $\frac{PQ}{QB}=\frac{PN}{BM}=\frac{PN}{5}$. Let $P\left(a_{0}, -a_{0}^{2}-3a_{0}+4\right)\left(-4<a_{0}<0\right)$, then $N\left(a_{0}, a_{0}+4\right)$. Consequently, $\frac{PQ}{QB}=\frac{-a_{0}^{2}-3a_{0}+4-(a_{0}+4)}{5}=\frac{-a_{0}^{2}-4a_{0}}{5}=\frac{-\left(a_{0}+2\right)^{2}+4}{5}$. Therefore, when $a_{0}=-2$, $\frac{PQ}{QB}$ reaches its maximum value of $0.8$. At this point, the coordinates of point $P$ are $(-2, 6)$. Hence, the maximum value of $\frac{PQ}{QB}$ is $\boxed{0.8}$, and the coordinates of point $P$ at this moment are $\boxed{(-2, 6)}$.


Question: In $\triangle ABC$, $AB=AC=6$, $\angle BAC=108^\circ$, point $D$ is on side $BC$, $\angle BAD=36^\circ$. $\triangle BAD\sim \triangle BCA$; what is the length of $AD$?

Solution: \textbf{Solution:} Since $\angle B=\angle BAD$, it follows that $BD=AD=x$. Because $\angle ADC=\angle B+\angle BAD=72^\circ$ and $\angle DAC=108^\circ-36^\circ=72^\circ$, we have $\angle ADC=\angle DAC$, thus $DC=AC=6$. Since $\triangle BAD\sim \triangle BCA$, we have $\frac{AD}{AB}=\frac{AB}{AC}$, that is $\frac{x}{6}=\frac{6}{x+6}$. Solving this, we get: $x=3\sqrt{5}-3$. Therefore, $AD=\boxed{3\sqrt{5}-3}$.


Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, rectangle $OABC$ has its sides $OA$ and $OC$ along the $x$-axis and $y$-axis, respectively, and point $B$ has coordinates $(3, 2)$. Let $D$ be a moving point on side $OA$, and construct rectangle $BDEF$ to the right side of $BD$ with $DE:BD=2:3$. Connect $CD$ and $AF$. Given $3AF=2CD$, when the coordinates of point $D$ are $(1, 0)$, find the coordinates of point $F$.

Solution:
\textbf{Solution:} Draw a perpendicular line from $F$ to the $x$-axis, with $H$ being the foot of the perpendicular.\\
$\because$ $AB\perp x$-axis,\\
$\therefore$ $FH\parallel AB$.\\
$\therefore$ $\angle AFH=\angle BAF$.\\
$\because$ $CB\parallel OA$,\\
$\therefore$ $\angle CDO=\angle BCD$.\\
By $(1)$, $\angle BAF=\angle BCD$,\\
$\therefore$ $\angle AFH=\angle CDO$.\\
Moreover, $\because$ $\angle AHF=\angle COD=90^\circ$,\\
$\therefore$ $\triangle AHF\sim \triangle COD$.\\
$\therefore$ $\frac{AH}{CO}=\frac{HF}{OD}=\frac{AF}{CD}=\frac{2}{3}$.\\
$\therefore$ $AH=\frac{2}{3}CO=\frac{4}{3}$, $HF=\frac{2}{3}OD=\frac{2}{3}$.\\
$\because$ $OH=OA+AH=\frac{13}{3}$.\\
$\therefore$ The coordinates of point $F$ are $\boxed{\left(\frac{13}{3}, \frac{2}{3}\right)}$.


Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, rectangle $OABC$ has its sides $OA$ and $OC$ along the $x$-axis and $y$-axis, respectively, and the coordinates of point $B$ are $(3, 2)$. Let $D$ be a moving point on side $OA$, and a rectangle $BDEF$ is constructed to the right side with $BD$ as one of its sides, where the ratio of $DE$ to $BD$ is $2:3$. Connect $CD$ and $AF$. The range of motion of point $E$ is a line segment, and the length of this line segment is to be found.

Solution:
\textbf{Solution:} Join $OB$ and $BH$.\\
Since $\frac{AB}{AH}=\frac{AO}{AB}=\frac{2}{3}$, and $\angle BAH=\angle OAB=90^\circ$.\\
Therefore, $\triangle BAH\sim \triangle OAB$.\\
Thus, $\angle ABH=\angle AOB$,\\
Therefore, $\angle OBA+\angle ABH=\angle OBA+\angle AOB=90^\circ$.\\
That is, $OB\perp BH$, and $\frac{BH}{OB}=\frac{2}{3}$.\\
Construct rectangle $OBHG$ with $OB$ and $BH$ as adjacent sides.\\
Then, when point $D$ coincides with $O$, the position of point $E$ is point $G$; when point $D$ coincides with $A$, the position of point $E$ is point $H$.\\
Join $BE$.\\
Since $\frac{BA}{AH}=\frac{BD}{DE}$, and $\angle BAH=\angle BDE=90^\circ$.\\
Therefore, $\triangle BAH\sim \triangle BDE$.\\
Thus, $\frac{BH}{BA}=\frac{BE}{BD}$, and $\angle ABH=\angle DBE$.\\
Therefore, $\angle EBH=\angle DBA$.\\
Thus, $\triangle BHE\sim \triangle BAD$.\\
Therefore, $\angle BHE=\angle BAD=90^\circ$, that is, $EH\perp BH$.\\
Therefore, the range of point $E$ is the line segment $GH$.\\
$GH=OB=\sqrt{OA^2+AB^2}=\boxed{\sqrt{13}}$.


Question: As shown in the figure, in parallelogram $ABCD$, $E$ is a point on side $DC$, and $\angle EAB = \angle EBC$. If $BE=2$ and $\triangle ABE \sim \triangle BEC$; what is the value of $AB \cdot CE$?

Solution:
\textbf{Solution}: Since $\triangle ABE \sim \triangle BEC$,\\
it follows that $\frac{AB}{EB}=\frac{EB}{EC}$,\\
thus $BE^2=AB \cdot CE$,\\
and since $BE=2$,\\
we have $AB \cdot CE=\boxed{4}$.


Question: As shown in the figure, an equilateral triangle $ABC$ is folded such that point $A$ falls on point $D$ on side $BC$ (not coinciding with $B$ or $C$), with the crease being $EF$. We have $\triangle BDE \sim \triangle CFD$; if $BD=6$ and $DC=2$, calculate the perimeters of $\triangle BDE$ and $\triangle CFD$ respectively.

Solution:
\textbf{Solution:} Given the problem, \\
$\because BD=6$ and $DC=2$, \\
$\therefore BC=BD+DC=8$, \\
$\because$ triangle $ABC$ is equilateral, \\
$\therefore AB=AC=BC=8$, \\
$\because$ triangle $ABC$ is folded such that point $A$ falls on point $D$ on side $BC$, \\
$\therefore AE=ED$ and $AF=FD$, \\
$\therefore$ the perimeter of $\triangle BDE$ is: $BD+DE+BE=BD+AE+BE=BD+AB=6+8=\boxed{14}$, \\
the perimeter of $\triangle CFD$ is: $CD+DF+CF=CD+AF+FC=CD+AC=2+8=\boxed{10}$.


Question: As illustrated, fold the equilateral triangle $ABC$ such that point $A$ falls on point $D$ on side $BC$ (not coinciding with $B$ or $C$), with the fold line being $EF$. Under the condition mentioned in (2), find the length of $BE$.

Solution:
\textbf{Answer:} Given $\because \triangle BDE\sim \triangle CFD$, and the perimeter of $\triangle BDE$ is 14, and the perimeter of $\triangle CFD$ is 10,\\
$\therefore \frac{BE}{CD}=\frac{14}{10}$,\\
$\because DC=2$,\\
$\therefore \frac{BE}{2}=\frac{14}{10}$,\\
$\therefore BE=\boxed{2.8}$.


Question: As shown in the diagram, in quadrilateral $ABCD$, $AB\parallel CD$, $AD\parallel BC$, $AC$ bisects $\angle DAB$. Line $BD$ intersects $AC$ at point $O$, and a line through point $C$ and perpendicular to $AB$ intersects the extension of line $AB$ at point $E$. Quadrilateral $ABCD$ is a rhombus; if $OA=4$ and $OB=3$, find the length of $CE$.

Solution:
\textbf{Solution:} From equation (1), we have $AO\perp OB$ and $AC=2OA=8$,\\
thus $AB=\sqrt{OA^2+OB^2}=\sqrt{4^2+3^2}=5$,\\
since $CE\perp AB$ and $\angle OAB=\angle EAC$,\\
it follows that $\triangle OAB\sim \triangle EAC$,\\
hence $\frac{OB}{AB}=\frac{CE}{AC}$, that is $\frac{3}{5}=\frac{CE}{8}$,\\
thus $CE=\boxed{\frac{24}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $AC=BC=10\text{cm}$, $CD\perp AB$, $CD=8\text{cm}$. Point P moves from B to A at a speed of $1\text{cm/s}$, simultaneously, points Q and R move from C to B and A to C, respectively, at a speed of $2\text{cm/s}$. The movement of all points stops when one of them stops. The movement time is $t$ (unit: $s$, $0<t<5$). At what value of $t$ is $PR\parallel BC$?

Solution:
\textbf{Solution:} Given that $AC=BC=10\text{cm}$ and $CD\perp AB$,\\
it follows that $AD=BD=\sqrt{10^2-8^2}=6\text{cm}$,\\
thus $AB=12\text{cm}$.\\
According to the problem, let $BP=t\text{cm}$ and $AR=2t\text{cm}$,\\
thus $AP=AB-BP=(12-t)\text{cm}$.\\
Since $PR\parallel BC$,\\
it follows that $\frac{AR}{AC}=\frac{AP}{AB}$, i.e., $\frac{2t}{10}=\frac{12-t}{12}$,\\
solving this yields: $t=\boxed{\frac{60}{17}}$;


Question: In $\triangle ABC$, $AC = BC = 10\text{cm}$, $CD \perp AB$, $CD = 8\text{cm}$. Point $P$ moves from $B$ towards $A$ at a speed of $1\text{cm/s}$, simultaneously, points $Q$ and $R$ move towards $B$ and $C$ from $C$ and $A$ respectively, at a speed of $2\text{cm/s}$. The motion of all points stops when any one of them stops. The motion time is $t$ ($t$ is in seconds, $0 < t < 5$). At what value of $t$ will point $P$ be on the perpendicular bisector of $BQ$?

Solution:
\textbf{Solution:} As shown in the figure, point $P$ is on the perpendicular bisector of $BQ$, \\
$\therefore PQ=PB$, $PE\perp BQ$, $BE=QE=\frac{1}{2}BQ$. \\
$\because CQ=2t\,\text{cm}$, \\
$\therefore BQ=BC-CQ=(10-2t)\,\text{cm}$, \\
$\therefore BE=QE=(5-t)\,\text{cm}$. \\
$\because \angle PBE=\angle CBD$, $\angle PEB=\angle CDB=90^\circ$, \\
$\therefore \triangle PBE\sim \triangle CBD$, \\
$\therefore \frac{BE}{BD}=\frac{PB}{BC}$, namely $\frac{5-t}{6}=\frac{t}{10}$, \\
Solving, we get: $t=\boxed{\frac{25}{8}}$;


Question: \textit{"Does there exist a value of $t$ such that $\frac{{S}_{\triangle BQP}}{{S}_{\text{quadrilateral } CDPQ}} = 1:4$? If so, determine the value of $t$."}

Solution:
\textbf{Solution:} As shown in the figure, draw $PF\perp BC$ through point $P$,\\
since the ratio $\frac{{S}_{\triangle BQP}}{{S}_{\text{quadrilateral} CDPQ}}=1:4$,\\
thus the ratio $\frac{{S}_{\triangle BQP}}{{S}_{\triangle BCD}}=1:5$.\\
From (2), it is known that $\triangle PBF\sim \triangle CBD$,\\
thus $\frac{BF}{BD}=\frac{PB}{BC}=\frac{PF}{CD}$, that is $\frac{t}{10}=\frac{PF}{8}$,\\
hence $PF=\frac{4t}{5}\text{cm}$,\\
thus ${S}_{\triangle BPQ}=\frac{1}{2}BQ\cdot PF=\frac{1}{2}\times(10-2t)\times \frac{4}{5}t=\left(-\frac{4}{5}t^2+4t\right)\text{cm}^2$.\\
Since ${S}_{\triangle BCD}=\frac{1}{2}BD\cdot CD=\frac{1}{2}\times 6\times 8=24\text{cm}^2$,\\
therefore $\left(-\frac{4}{5}t^2+4t\right)=\frac{1}{5}\times 24$,\\
solving gives: $t_{1}=\boxed{2}$, $t_{2}=\boxed{3}$,\\
thus the value of $t$ is \boxed{2} or \boxed{3}.


Question: As shown in the figure, in right triangle $ABC$, the right-angle sides $AC=3\text{cm}$ and $BC=4\text{cm}$. Let $P$ and $Q$ be points moving along $AB$ and $BC$, respectively. Point $P$ moves from point $A$ towards point $B$ at a constant speed, while point $Q$ moves from point $B$ towards point $C$ at the same constant speed of $1\text{cm}$ per second. The movement of $P$ stops as soon as $Q$ reaches point $C$. Let the movement time of $P$ and $Q$ be $t$ seconds. For what value of $t$ is $\triangle PBQ$ an isosceles triangle with $\angle B$ as the vertex angle?

Solution:
\textbf{Solution:} Given that the leg $AC=3\text{cm}$ and $BC=4\text{cm}$,\\
thus, by the Pythagorean theorem, we have $AB=\sqrt{AC^2+BC^2}=\sqrt{3^2+4^2}=5$,\\
therefore, $AP=t$ and $BP=5-t$, $BQ=t$,\\
since $\triangle PBQ$ is an isosceles triangle with $\angle B$ as the vertex angle,\\
thus, $BP=BQ$, that is, $5-t=t$, solving this gives $t=\boxed{\frac{5}{2}}$ seconds,\\
therefore, when $t=\boxed{\frac{5}{2}}$ seconds, $\triangle PBQ$ forms an isosceles triangle with $\angle B$ as the vertex angle.


Question: As shown in the figure, in the right-angled triangle $ABC$, the perpendicular sides $AC=3\text{cm}$, $BC=4\text{cm}$. Let $P$ and $Q$ be points moving on $AB$ and $BC$ respectively. Point $P$ moves from point $A$ towards $B$ along $AB$ uniformly, and point $Q$ moves from point $B$ towards $C$ along $BC$ uniformly, with both their speeds being $1\text{cm}$ per second. When point $Q$ reaches point $C$, point $P$ stops moving. Let the time for $P$ and $Q$ to move be $t$ seconds. $\triangle PBQ$ is similar to the right-angled triangle $ABC$, what is the value of $t$?

Solution:
Solution: When $\triangle PBQ \sim \triangle ABC$, we have $\frac{BQ}{BC} = \frac{BP}{AB}$, that is, $\frac{t}{4} = \frac{5-t}{5}$, solving this we get: $t = \boxed{\frac{20}{9}}$; \\
When $\triangle PBQ \sim \triangle CBA$, we have $\frac{BQ}{AB} = \frac{BP}{BC}$, that is, $\frac{t}{5} = \frac{5-t}{4}$, solving this we get: $t = \frac{25}{9}$, \\
Therefore, when $t = \boxed{\frac{20}{9}}$ or $\boxed{\frac{25}{9}}$, $\triangle PBQ$ is similar to the right-angled triangle ABC.


Question: As shown in the figure, in circle $\odot O$, through the midpoint $C$ of the radius $OD$, draw $AB\perp OD$ intersecting $\odot O$ at points $A$ and $B$, and $AB=2\sqrt{3}$. What is the length of $OD$?

Solution:
\textbf{Solution:} Since $AB \perp OD$,\\
$\therefore \angle OCB=90^\circ$,\\
$AC=BC= \frac{1}{2} AB= \sqrt{3}$,\\
Since point C is the midpoint of OD,\\
$\therefore OC= \frac{1}{2} OB$,\\
Since $\cos\angle COB= \frac{OC}{OB} = \frac{1}{2}$,\\
$\therefore \angle COB=60^\circ$,\\
$\therefore OC= \frac{\sqrt{3}}{3} BC= \frac{\sqrt{3}}{3} \times \sqrt{3} =1$,\\
$\therefore OB=2OC=2$,\\
$\therefore OD=OB=\boxed{2}$.


Question: As shown in the figure, within circle $\odot O$, through the midpoint $C$ of radius $OD$, construct $AB\perp OD$ intersecting $\odot O$ at points $A$ and $B$, and $AB=2\sqrt{3}$. What is the area of the shaded region?

Solution:
\[
\textbf{Solution:} \text{The area of the shaded part} = S_{\text{sector } BOD} - S_{\triangle COB} = \frac{60^\circ \times \pi \times 2^2}{360^\circ} - \frac{1}{2} \times \sqrt{3} \times 1 = \boxed{\frac{2}{3} \pi - \frac{\sqrt{3}}{2}}.
\]


Question: As shown in the figure, AB is the diameter of the circle $\odot O$, and the chord CD is perpendicular to AB at point E. Connect AD and BD, and $\angle ADC = \angle ABD$. Construct OF $\perp$ AD at point F. If the radius of $\odot O$ is 5 and OE = 3, find the length of OF.

Solution:
\textbf{Solution:} As shown in the diagram, connect OD.\\
In $\triangle OED$, $DE = \sqrt{OD^2 - OE^2} = \sqrt{5^2 - 3^2} = 4$,\\
In $\triangle ADE$, $AD = \sqrt{AE^2 + DE^2} = \sqrt{8^2 + 4^2} = 4\sqrt{5}$,\\
$\because \sin\angle A = \frac{OF}{OA} = \frac{DE}{AD}$,\\
$\therefore \frac{OF}{5} = \frac{4}{4\sqrt{5}}$,\\
$\therefore OF = \boxed{\sqrt{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, circle $\odot O$ is the circumcircle of $\triangle ABC$, and the extension of $BO$ intersects side $AC$ at point D. If $\angle ACB=60^\circ$ and $BC=8\sqrt{3}$, find the radius of $\odot O$.

Solution:
\textbf{Solution:} Construct line $AO$ and extend $AO$ to meet $BC$ at point $E$. Draw a line from point $O$ such that $OH\perp AB$, intersecting $AB$ at point $H$, as shown in the diagram.\\
Since $AE$ passes through the center and $AB=AC$,\\
it follows that $AE$ perpendicularly bisects $BC$,\\
and thus $AE$ bisects $\angle BAC$.\\
Given that in $\triangle ABC$, $\angle C=60^\circ$,\\
it implies that $\triangle ABC$ is an equilateral triangle.\\
Therefore, $\angle BAO=\frac{1}{2}\angle BAC=30^\circ$, and $AB=BC=8\sqrt{3}$.\\
In right triangle $AOH$, $AH=\frac{1}{2}AB=4\sqrt{3}$, with $\angle HAO=30^\circ$,\\
since $\cos\angle HAO=\frac{AH}{AO}$,\\
it follows that $AO=\frac{AH}{\cos\angle HAO}=8$.\\
Therefore, the radius of circle $O$ is $\boxed{8}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, and $\odot O$ is the circumcircle of $\triangle ABC$. The extension of line $BO$ intersects side $AC$ at point $D$. When $\triangle BCD$ is an isosceles triangle, find the measure of $\angle BCD$.

Solution:
\textbf{Solution}: Let $\angle ABD=\alpha$, \\
since in $\triangle ABC$, $BO=AO$, \\
therefore $\angle ABO=\angle BAO$, \\
therefore $\angle BAC=2\angle ABD=2\alpha$, \\
therefore $\angle BDC=3\alpha$. \\
Furthermore, since $\triangle BCD$ is an isosceles triangle, \\
(1) If $BD=BC$, then $\angle C = \angle BDC = 3\alpha$. \\
Since $AB=AC$, \\
therefore $\angle ABC = \angle C = 3\alpha$. \\
Thus, in $\triangle ABC$, $3\alpha + 3\alpha + 2\alpha = 180^\circ$. \\
Solving gives $\alpha = 22.5^\circ$. \\
therefore $\angle BCD = 67.5^\circ$. \\
(2) If $BC=CD$, then $\angle BDC = \angle CBD = 3\alpha$. \\
Since in $\triangle ABC$, $\angle ABC + \angle BCA + \angle BAD = 180^\circ$, \\
therefore $4\alpha + 4\alpha + 2\alpha = 180^\circ$. \\
Solving gives $\alpha = 18^\circ$. \\
therefore $\angle BCD = 4\alpha = 72^\circ$. \\
(3) If $DB=DC$, then point D coincides with A, and this case does not exist. \\
To summarize, the value of $\angle BCD$ is either $67.5^\circ$ or $\boxed{72^\circ}$.


Question: As shown in the figure, in parallelogram $ABCD$, $BG$ is perpendicular to $CD$, and $AB=BG=BE$, $AE$ intersects $BG$ at point $F$. If $AB=3$ and $\angle BAD=60^\circ$, what is the length of $CE$?

Solution:
Solution: As shown in the figure, in the parallelogram $ABCD$, $\angle BAD=\angle C=60^\circ$.\\
Since $BG$ is perpendicular to $CD$,\\
$\therefore \angle BGC=90^\circ$,\\
$\therefore BC=\frac{BG}{\sin 60^\circ}$.\\
Also, since $AB=BG=BE=3$,\\
$\therefore BC=\frac{3}{\frac{\sqrt{3}}{2}}=2\sqrt{3}$,\\
$\therefore CE=BC-BE=BC-BG=2\sqrt{3}-3$;\\
Thus, the length of $CE$ is $\boxed{2\sqrt{3}-3}$.


Question: As shown in the diagram, $DE$ is the angle bisector of the exterior angle $\angle FDC$ of $\triangle DBC$, intersecting the extension of $BC$ at point $E$. The extension of $DE$ intersects the circumscribed circle of $\triangle DBC$ at point $A$. $AB=AC$; if $\angle DCB=90^\circ$, $\sin E=\frac{\sqrt{5}}{5}$, and $AD=4$, find the length of $BD$.

Solution:
\textbf{Solution:} Given the problem, \\
$\because \angle DCB=90^\circ$, \\
$\therefore \angle DCE=\angle BAD=90^\circ$, \\
$\therefore \angle E+\angle CDE=\angle ABD+\angle ADB=90^\circ$, \\
$\because \angle ADB=\angle FDE=\angle CDE$, \\
$\therefore \angle ABD=\angle E$, \\
$\because \sin E=\frac{\sqrt{5}}{5}$, \\
$\therefore \sin\angle ABD=\frac{AD}{BD}=\frac{\sqrt{5}}{5}$, \\
$\because AD=4$, \\
$\therefore BD=\boxed{4\sqrt{5}}$.


Question: As shown in the figure, in parallelogram $ABCD$, $DE\perp AB$ at point $E$, and $DF\perp BC$ at point $F$. The perimeter of parallelogram $ABCD$ is $28$, and its area is $40$. The ratio of $AB:AD$ is $4:3$. Find: What is the length of $DE$?

Solution:
\textbf{Solution:} In parallelogram $ABCD$, where $AB=CD$ and $AD=BC$, \\
since the perimeter of the parallelogram $ABCD$ is 28, \\
it follows that $AB+AD=14$, \\
and given $AB\colon AD=4\colon 3$, \\
thus, $AB=8$ and $AD=6$, \\
furthermore, the area of parallelogram $ABCD$ is given by $AB \times DE = 40$, \\
therefore, $DE=\boxed{5}$.


Question: As shown in the figure, in the parallelogram $ABCD$, $DE \perp AB$ at point $E$, $DF \perp BC$ at point $F$, the perimeter of the parallelogram $ABCD$ is 28, and the area is 40, with $AB:AD = 4:3$. Find the value of $\sin\angle EDF$.

Solution:
\textbf{Solution:} In quadrilateral $DEBF$, we have $\angle EDF+\angle DEB+\angle B+\angle BFD=360^\circ$,\\
since $DE\perp AB$ and $DF\perp BC$,\\
$\therefore \angle EDF+\angle B=180^\circ$,\\
Moreover, in parallelogram $ABCD$, $\angle A+\angle B=180^\circ$,\\
$\therefore \angle EDF=\angle A$,\\
In $\triangle ADE$, $\sin A=\frac{DE}{AD}=\frac{5}{6}$,\\
$\therefore \sin\angle EDF=\sin A=\boxed{\frac{5}{6}}$.


Question: As shown in the figure, it is known that $AB$ is the diameter of semicircle $O$, $C$ is a point on the semicircle, $P$ is the midpoint of $BC$, and $PD\parallel BC$ intersects the extension of $AB$ at point $D$. Prove that $PD$ is the tangent to $\odot O$.

Solution:
\textbf{Solution.} Proof: Connect $OP$,\\
$\because$ Point $P$ is the midpoint of arc $BC$,\\
$\therefore$ $OP\perp BC$,\\
$\because$ $PD\parallel BC$,\\
$\therefore$ $PD\perp OP$,\\
$\therefore$ \boxed{PD\text{ is the tangent to circle }O}.


Question: As shown in the figure, it is known that $AB$ is the diameter of the semicircle $O$, $C$ is a point on the semicircle arc, $P$ is the midpoint of $BC$, and $PD\parallel BC$ intersects the extension of $AB$ at point $D$. If $AB=10$ and $\cos\angle D=\frac{4}{5}$, what is the value of $PC$?

Solution: \textbf{Solution:} Given that $AB$ is the diameter, then $\angle ACB=90^\circ$. Since $PD\parallel BC$, it follows that $\angle D=\angle ABC$. Therefore, $\cos\angle D = \cos\angle ABC = \frac{BC}{AB} = \frac{4}{5}$. Solving this gives $BC=8$. Since $OP\perp CB$, it follows that $CE=BE=\frac{1}{2}BC=4$. In $\triangle BOE$, we have $OE=\sqrt{OB^2-BE^2}=\sqrt{5^2-4^2}=3$. Thus, $PE=OP-OE=5-3=2$. In $\triangle CEP$, we find that $PC=\sqrt{CE^2+PE^2}=\sqrt{4^2+2^2}=\boxed{2\sqrt{5}}$.


Question: As shown in the figure, a billboard CD is erected at the top of a building. Xiaoming and his classmates measured the elevation angle of the bottom of the billboard D to be $53^\circ$ from point A at the foot of the hill. Walking up the slope AB to point B, they measured the elevation angle of the top of the billboard C to be $45^\circ$. It is known that the slope $i$ of hill AB is $1\colon \sqrt{3}$, $AB=12$ meters, $AE=24$ meters. (The height of the theodolite is negligible, and the result should be accurate to 0.1 meters. Reference data: $\sqrt{2}\approx 1.41$, $\sqrt{3}\approx 1.73$, $\sin 53^\circ\approx \frac{4}{5}$, $\cos 53^\circ\approx \frac{3}{5}$, $\tan 53^\circ\approx \frac{4}{3}$). How many meters is point B above the horizontal ground AE?

Solution:
\textbf{Solution:} As shown in the diagram, draw $BM\perp AE$ and $BN\perp CE$ through point $B$, with the perpendiculars touching points $M$ and $N$ respectively,\\
it is known from the problem that $\angle CBN=45^\circ$, $\angle DAE=53^\circ$, $i=1\colon \sqrt{3}$, $AB=12$ meters, $AE=24$ meters,\\
since $i=1\colon \sqrt{3}=\frac{BM}{AM}=\tan\angle BAM$,\\
it follows that $\angle BAM=30^\circ$,\\
thus $BM=\frac{1}{2}AB=6$ (meters),\\
which means the height of point $B$ from the horizontal ground $AE$ is $\boxed{6}$ meters;


Question: As shown in the figure, a billboard CD stands on top of a building. Xiao Ming and his classmates measured the angle of elevation to the bottom of the billboard, D, as $53^\circ$ from point A at the foot of the hill. Walking up the slope AB to point B, they measured the angle of elevation to the top of the billboard, C, as $45^\circ$. It is known that the slope of the hill, AB, has a gradient of $i=1\colon \sqrt{3}$ and $AB=12$ meters, $AE=24$ meters. (The height of the protractor is negligible, and the result should be accurate to 0.1 meter, reference data: $\sqrt{2}\approx 1.41$, $\sqrt{3}\approx 1.73$, $\sin 53^\circ\approx \frac{4}{5}$, $\cos 53^\circ\approx \frac{3}{5}$, $\tan 53^\circ\approx \frac{4}{3}$). How tall is the billboard CD, in meters?

Solution:
\textbf{Solution:} In right triangle $ABM$,\\
$\therefore NE=BM=\frac{1}{2}AB=6$ (meters),\\
$AM=\frac{\sqrt{3}}{2}AB=6\sqrt{3}$ (meters),\\
$\therefore ME=AM+AE=\left(6\sqrt{3}+24\right)$ meters,\\
$\because \angle CBN=45^\circ$,\\
$\therefore CN=BN=ME=\left(6\sqrt{3}+24\right)$ meters,\\
$\therefore CE=CN+NE=\left(6\sqrt{3}+30\right)$ meters,\\
In right triangle $ADE$, $\angle DAE=53^\circ$, $AE=24$ meters,\\
$\therefore DE=AE\cdot\tan53^\circ\approx 24\times \frac{4}{3}=32$ (meters),\\
$\therefore CD=CE-DE$\\
$=6\sqrt{3}+30-32$\\
$=6\sqrt{3}-2$\\
$\approx \boxed{8.4}$ (meters)


Question: As shown in the figure, Xiao Ming wants to measure the height of the tower $CD$. He looks up at the top of the tower from point $A$ and measures the angle of elevation to be $30^\circ$. Then, he moves $50$ meters towards the tower to point $B$, where he measures the angle of elevation to be $60^\circ$. It is given that $AB=BD$; find the height of the tower $CD$. (Ignore Xiao Ming's height, and leave the result in square root form.)

Solution:
\textbf{Solution:} Given: $BD=AB=50$ meters,\\
In $\triangle BCD$, $\angle C=90^\circ$,\\
$\therefore \sin \angle DBC = \frac{DC}{BD}$,\\
$\therefore DC=BD \cdot \sin 60^\circ = 50 \times \frac{\sqrt{3}}{2} = \boxed{25\sqrt{3}}$ (meters).


Question: As shown in the figure, $AB$ is the diameter of the circle $\odot O$. The chord $CD \perp AB$ at point $E$, $\angle CDB = 30^\circ$, and $CD = 2\sqrt{3}$. Find the length of the diameter $AB$.

Solution:
\textbf{Solution:} Since $CD\perp AB$ and AB is a diameter,\\
we have $CE=DE=\sqrt{3}$.\\
Since $\angle COB=2\angle CDB=2\times 30^\circ=60^\circ$,\\
we find $OC=\frac{\sqrt{3}}{\sin 60^\circ}=2$.\\
Therefore, the diameter $AB=\boxed{4}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and the chord $CD \perp AB$ at point $E$, $\angle CDB=30^\circ$, $CD=2\sqrt{3}$. Find the area of the shaded figure.

Solution:
Solution: As shown in the figure, connect $BC$\\
$\because \angle CDB=30^\circ$\\
$\therefore \angle COB=60^\circ$\\
$\because OC=OB$\\
$\therefore \triangle OBC$ is an equilateral triangle\\
$\because CE\perp OB$\\
$\therefore OE=BE$\\
$\because CE=DE$, $\angle OEC=\angle BED$\\
$\therefore \triangle OCE\cong \triangle BDE$\\
$\therefore {S}_{\text{shaded}}={S}_{\text{sector} OBC}=\frac{n\pi {r}^{2}}{360}=\frac{60^\circ \cdot \pi \cdot {2}^{2}}{360}=\boxed{\frac{2\pi }{3}}$


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and chord $CD\perp AB$ at point $E$, with $AD, BD$ connected. $\angle ADC=\angle ABD$; draw $OF\perp AD$ at point $F$. If the radius of $\odot O$ is $5$ and $OE=3$, find the length of $OF$.

Solution:
\textbf{Solution:} As shown in the figure, connect $OD$.\\
In $Rt\triangle OED$, we have $DE=\sqrt{OD^{2}-OE^{2}}=\sqrt{5^{2}-3^{2}}=4$,\\
In $Rt\triangle ADE$, $AD=\sqrt{AE^{2}+DE^{2}}=\sqrt{8^{2}+4^{2}}=4\sqrt{5}$,\\
$\because \sin\angle A=\frac{OF}{OA}=\frac{DE}{AD}$,\\
$\therefore \frac{OF}{5}=\frac{4}{4\sqrt{5}}$,\\
$\therefore OF=\boxed{\sqrt{5}}$.


Question: As shown in the figure, in rectangle $ABCD$, where $M$ is a point on $BC$, $EM\perp AM$ intersects the extension of $AD$ at point $E$. $\triangle ABM \sim \triangle EMA$. If $AB=4$, $BM=3$, what is the value of $\sin E$?

Solution:
\textbf{Solution:} Since $\triangle ABM \sim \triangle EMA$,\\
it follows that $\angle E = \angle BAM$,\\
in $\triangle ABM$, $AM = \sqrt{BM^2+AB^2} = \sqrt{3^2+4^2} = 5$,\\
thus $\sin \angle BAM = \frac{BM}{AM} = \frac{3}{5}$,\\
hence $\sin E = \boxed{\frac{3}{5}}$.


Question: As shown in the figure, in acute-angled triangle $ABC$, we have $AB=AC$. Circle $O$ with diameter $BC$ intersects $AB$ and $AC$ at points $D$ and $E$ respectively. Connect $OD, OE$, and $DE$. If $\angle A=50^\circ$, what is the degree measure of $\overset{\frown}{DE}$?

Solution:
\textbf{Solution:} As shown in the figure, draw $BE$.\\
Since $BC$ is the diameter,\\
it follows that $\angle BEC=90^\circ$,\\
thus $\angle AEB=90^\circ$,\\
Since $\angle BAC=50^\circ$,\\
it follows that $\angle ABE=90^\circ-50^\circ=40^\circ$,\\
thus $\angle DOE=2\angle DBE=80^\circ$,\\
Therefore, the degree of $\overset{\frown}{DE}$ is $\boxed{80^\circ}$;


Question: As shown in the figure, in the acute-angled triangle $ABC$, where $AB=AC$, the circle $\odot O$ with $BC$ as the diameter intersects $AB$ and $AC$ at points $D$ and $E$ respectively. Lines $OD$, $OE$, and $DE$ are connected. $DE\parallel BC$. If the radius of $\odot O$ is $m$, and $\tan\angle ABC=2$, what is the area of the quadrilateral $ADOE$ (expressed in an algebraic expression involving $m$)?

Solution:
\textbf{Solution:} Given that $AB=AC$ and $BO=OC$,\\
we have $AO\perp BC$,\\
thus $\tan\angle ABC=\frac{AO}{OC}=2$,\\
since $OB=m$,\\
it follows that $OA=2m$,\\
since $\tan\angle C=\frac{BE}{EC}=2$,\\
we find $EC=\frac{2\sqrt{5}}{5}m$,\\
given $AB=AC=\sqrt{5}m$,\\
thus $AE=AC-EC=\sqrt{5}m-\frac{2\sqrt{5}}{5}m=\frac{3\sqrt{5}}{5}m$,\\
since $DE\parallel CB$,\\
it follows that $\frac{DE}{BC}=\frac{AE}{AC}$,\\
thus $\frac{DE}{2m}=\frac{\frac{3\sqrt{5}}{5}m}{\sqrt{5}m}$,\\
consequently, $DE=\frac{6}{5}m$,\\
since $AO\perp BC$ and $DE\parallel BC$,\\
it follows that $AO\perp DE$,\\
hence the area of quadrilateral $ADOE$ is $\frac{1}{2}\cdot AO\cdot DE=\frac{1}{2}\times 2m\times \frac{6}{5}m=\boxed{\frac{6}{5}m^2}$.


Question: As shown in Figure 1, the graph of the inverse proportion function $y = \frac{k}{x}$ passes through points $A(1, m)$ and $B(2, 1)$. The coordinates of point $P$ are $(6, 1)$. What is the expression of the inverse proportion function?

Solution:
\textbf{Solution:} Since the graph of the inverse proportion function $y=\frac{k}{x}$ passes through point $B(2,1)$,\\
hence $k=2\times 1=2$,\\
therefore, the equation of the inverse proportion function is $\boxed{y=\frac{2}{x}}$.


Question: As shown in Figure 1, the graph of the inverse proportion function $y = \frac{k}{x}$ passes through points $A(1, m)$ and $B(2, 1)$. The coordinates of point $P$ are $(6, 1)$. Connect $PA$ and $PB$. What is the value of $\tan \angle P$?

Solution:
\textbf{Solution:} Since the graph of the inverse proportion function $y = \frac{2}{x}$ passes through point $A(1, 2)$,\\
it follows that $m = 2$,\\
Draw a perpendicular from point $A$ to line $BP$, extending to point $M$ on the extension of $PB$,\\
then $AM = 2 - 1 = 1$, and $MP = 6 - 1 = 5$,\\
In $\triangle PAM$, $\tan \angle P = \frac{AM}{MP} = \frac{1}{5}$;\\
therefore, $\tan \angle P = \boxed{\frac{1}{5}}$.


Question: As shown in Figure 1, the graph of the inverse proportion function $y=\frac{k}{x}$ passes through points $A(1, m)$ and $B(2, 1)$. The coordinates of point $P$ are $(6, 1)$. As shown in Figure 2, the coordinates of points $C$ and $D$ are $(a, 0)$ and $(0, a)$ ($0<a\leq 6$), respectively. When the area of $\triangle PCD$ is $3$, what is the value of $a$?

Solution:
\textbf{Solution:} Construct a perpendicular $PN$ from point P to the x-axis, with N being the foot of the perpendicular, \\
$\because S_{\triangle PCD}=S_{\text{trapezoid}} PNOD-S_{\triangle COD}-S_{\triangle PCN}$, \\
$\therefore \frac{1}{2}(1+a)\times 6-\frac{1}{2}a^{2}-\frac{1}{2}(6-a)\times 1=3$, \\
we obtain $a_{1}=\boxed{1}$, $a_{2}=6$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, D is the midpoint of AB, $AE\parallel CD$, and $CE\parallel AB$. Quadrilateral ADCE is a rhombus. If $BC=6$ and $\tan B=\frac{4}{3}$, what is the perimeter of quadrilateral ADCE?

Solution:
\textbf{Solution:} In right-angled $\triangle ABC$, where $BC=6$ and $\tan B=\frac{AC}{BC}=\frac{4}{3}$,\\
$\therefore AC=\frac{4}{3}BC=\frac{4}{3}\times 6=8$,\\
$\therefore AB=\sqrt{AC^2+BC^2}=\sqrt{8^2+6^2}=10$,\\
$\therefore CD=\frac{1}{2}AB=5$,\\
$\because$ Quadrilateral $ADCE$ is a rhombus,\\
$\therefore CD=DA=AE=EC=5$,\\
$\therefore$ The perimeter of rhombus $ADCE$ is: $5\times 4=\boxed{20}$.


Question: As shown in the figure, it is known that $AB$ is the diameter of $\odot O$, with $AB=6$, and $\sin C= \frac{1}{3}$. Find the length of the chord $AD$.

Solution:
\textbf{Solution:} Draw line BD, \\
$\therefore \angle C=\angle B$, \\
$\therefore \sin B= \frac{1}{3}$, \\
$\because$ AB is the diameter, \\
$\therefore \angle ADB=90^\circ$, \\
$\therefore \sin B =\frac{AD}{AB}=\frac{1}{3}$, \\
$\therefore AD=\boxed{2}$.


Question: As shown in the figure, it is known that $AB$ is the diameter of the circle $O$, with $AB=6$, and $\sin C= \frac{1}{3}$. A line is drawn through point $D$ such that $DE \perp AB$, with $E$ as the foot of the perpendicular, and the extension of $DE$ intersects circle $O$ at point $F$. Find the length of $DF$.

Solution: \textbf{Solution:} Given that $AB$ is the diameter and $DE \perp AB$, it follows that $DF = 2DE$. Connect $DO$ and let $OE = x$. Thus, we have $3^{2} - x^{2} = 2^{2} - (3 - x)^{2}$. This implies $x = \frac{7}{3}$. By the Pythagorean theorem, we obtain $DE = \frac{4\sqrt{2}}{3}$. Therefore, $DF = \boxed{\frac{8\sqrt{2}}{3}}$.


Question: Given that in $\triangle ABC$, $\angle C = 90^\circ$, $AB=4$, and $AC=\sqrt{7}$. Find $BC$?

Solution:
\textbf{Solution:} In triangle $\triangle ABC$, $\angle C=90^\circ$, $AB=4$, $AC=\sqrt{7}$.\\
$BC=\sqrt{AB^2-AC^2}=\sqrt{4^2-(\sqrt{7})^2}=\boxed{3}$.


Question: Given in $\triangle ABC$, $\angle C=90^\circ$, $AB=4$, and $AC=\sqrt{7}$. What is $\sin A$?

Solution: \textbf{Solution:} We have $\sin A = \frac{BC}{AB} = \boxed{\frac{3}{4}}$


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=\sqrt{5}$, and $\sin B=\frac{2\sqrt{5}}{5}$. What is the length of side $BC$?

Solution:
\textbf{Solution}: As shown in the figure, draw the height AD of $\triangle ABC$, \\
since $AB=AC$, \\
therefore, point D is the midpoint of BC, which means $BD=CD$. \\
Since in $\triangle ABD$, $\sin B = \frac{AD}{AB} = \frac{2\sqrt{5}}{5}$, we get $\frac{AD}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$, \\
thus $AD=2$, \\
therefore $BD = \sqrt{AB^{2} - AD^{2}} = \sqrt{(\sqrt{5})^{2} - {2}^{2}} = 1$, \\
therefore, $BC = 2BD = \boxed{2}$.