Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the linear function \(y=kx+b\) intersects the x-axis and y-axis at points A and B, respectively, and it passes through the point \(D(-2, 6)\). It intersects the graph of the direct proportion function \(y=3x\) at point C, where the \(x\)-coordinate of point C is 1. What is the equation of the linear function \(y=kx+b\)?

Solution:
\textbf{Solution:} Solve: When $x=1$, $y=3$. Let the line $y=kx+b$ pass through $(1, 3)$ and $(-2, 6)$\\
$\therefore$ $\begin{cases}6=-2k+b\\ 3=k+b\end{cases}$\\
Solving these, we get: $\begin{cases}k=-1\\ b=4\end{cases}$\\
$\therefore$ The equation of the function is $y=-x+4=\boxed{-x+4}$


Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ intersects the x-axis and y-axis at points A and B respectively, and passes through the point $D\left(-2,6\right)$. It intersects the graph of the direct proportionality function $y=3x$ at point C, where the x-coordinate of point C is 1. What is the area of $\triangle BOC$?

Solution:
\textbf{Solution:} When $x=0$, then $y=4$, \\
$\therefore$ $S_{\triangle BOC}=\frac{1}{2}\times 4\times 1=\boxed{2}$.


Question: As shown in the figure, in a Cartesian coordinate system, the graph of the linear function $y=kx+b$ intersects the x-axis and the y-axis at points A and B, respectively, and passes through point $D(-2, 6)$. It intersects the graph of the direct proportion function $y=3x$ at point C, where the x-coordinate of point C is 1. If point E is a moving point on line $OC$, and the area of $\triangle ADE$ is 6 times the area of $\triangle BOC$, please directly write down the coordinates of point E.

Solution:
\textbf{Solution:}

From given, we have: $6S_{\triangle BOC}=6\times 2=12$,\\
As per the diagram: $S_{\triangle AED}=\frac{1}{2}\times 6\times AE=\frac{1}{2}\times 6\times 4=12$,\\
$\therefore E(0, 0)$,\\
Point E is a variable point on line $OC$. According to symmetry, when point $E$ at the origin is symmetrical about point $C$ and falls on $y=3x$, it also satisfies the condition,\\
Based on symmetry about a point, we obtain: $E(2, 6)$,\\
Therefore, the coordinates that satisfy the condition are: $E=\boxed{(0, 0)}$, $E=\boxed{(2, 6)}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the vertices of rectangle $OABC$ are such that vertex $A$ is on the positive $y$-axis, and point $C$ is on the positive $x$-axis, with $AO=\sqrt{3}$, and $\angle BAC=30^\circ$. Point $D$ is a point on segment $OC$, and triangle $AOD$ is folded along line $AD$ such that point $O$ coincides with point $E$ on the diagonal $AC$ of the rectangle. What are the coordinates of point $B$?

Solution: \textbf{Solution:} Given that quadrilateral $OABC$ is a rectangle, it follows that $BC=AO=\sqrt{3}$. In $\triangle ACB$, with $\angle BAC=30^\circ$, we have $AC=2BC=2\sqrt{3}$. Thus, $AB=\sqrt{AC^{2}-BC^{2}}=3$. Therefore, the coordinates of point $B$ are $\boxed{B\left(3, \sqrt{3}\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertices of rectangle $OABC$ with vertex $A$ on the positive y-axis and point $C$ on the positive x-axis. Given $AO=\sqrt{3}$ and $\angle BAC=30^\circ$, point $D$ is located on line segment $OC$. When $\triangle AOD$ is folded along line $AD$, point $O$ falls on point $E$ on the diagonal $AC$ of the rectangle. What is the equation of line $DE$?

Solution:
\textbf{Solution:} Let $EH\perp y$-axis at $H$, and $EG\perp x$-axis at $G$. Then, quadrilateral EHOG is a rectangle, \\
$\therefore EG=HO$. By folding, $AE=AO=\sqrt{3}$.\\
$\because AC=2\sqrt{3}$,\\
$\therefore E$ is the midpoint of $AC$,\\
$\therefore G$ is the midpoint of $OC$, and $H$ is the midpoint of $AO$,\\
$\therefore HE, EG$ are midlines,\\
$HE=OG=GC=\frac{1}{2}OC=\frac{3}{2}$,\\
$EG=HO=AH=\frac{1}{2}AO=\frac{\sqrt{3}}{2}$,\\
$\therefore$ Point $E\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)$.\\
In $\triangle ADO$, $\angle OAD=\frac{1}{2}\angle OAC=30^\circ$, $AO=\sqrt{3}$,\\
$\therefore OD=1$,\\
$\therefore$ Point $D\left(1, 0\right)$.\\
Let the equation of line $ED$ be $y=kx+b$.\\
Substituting the coordinates of $D$, $E$ gives $\left\{\begin{array}{l}\frac{3}{2}k+b=\frac{\sqrt{3}}{2} \\ k+b=0\end{array}\right.$,\\
Solving gives, $\left\{\begin{array}{l}k=\sqrt{3} \\ b=-\sqrt{3}\end{array}\right.$.\\
Thus, the equation of line $ED$ is: $y=\sqrt{3}x-\sqrt{3}$.\\
Hence, the answer is $\boxed{y=\sqrt{3}x-\sqrt{3}}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertices of rectangle $OABC$ are such that vertex $A$ is on the positive half of the $y$-axis, and point $C$ is on the positive half of the $x$-axis, with $AO=\sqrt{3}$ and $\angle BAC=30^\circ$. Point $D$ is a point on segment $OC$, and triangle $AOD$ is folded along line $AD$, placing point $O$ at point $E$ on the diagonal $AC$ of the rectangle. There exists a point $M$ on line $DE$ and a point $N$ on one of the coordinate axes, such that the quadrilateral formed by vertices $N$, $A$, $M$, and $C$ is a rhombus. Write down the coordinates of point $N$.

Solution:
\textbf{Solution:} Consider two situations:
1. When the diagonal of the rhombus is $AC$, the line $ED$ is the perpendicular bisector of $AC$. The vertex $N$ of the rhombus must be on the line $ED$, and point $N$ only has two opportunities on the coordinate axis:
   (1) When line $ED$ intersects with the $x$-axis, point $N$ coincides with point $D$, forming the rhombus $AN_1CM_1$, as shown in the figure $N_1(1, 0)$;
   (2) When line $ED$ intersects with the $y$-axis, $y=\sqrt{3}x-\sqrt{3}$, let $x=0$, then $y=-\sqrt{3}$, forming the rhombus $AN_2CM_2$, as shown in the figure $N_2(0, -\sqrt{3})$;
2. When the side of the rhombus is $AC$:
   (1) When $N$, $M$ are above side $AC$, point $N$ is on the $y$-axis, $AN=AC=2\sqrt{3}$, $N(0, 3\sqrt{3})$.
   $y=\sqrt{3}x-\sqrt{3}$, let $x=3$, then $y=2\sqrt{3}$.
   $CM_3=2\sqrt{3}$. $M_3(0, 3\sqrt{3})$. $AN$ is parallel and equal to $CM_3$ forming the rhombus $ACM_3N_3$. As shown in the figure $N_3(0, 3\sqrt{3})$.
   (2) When $N$, $M$ are below side $AC$, point $N$ is on the $x$-axis, $ON_4=OC=3$, $N_4(-3, 0)$.
   $y=\sqrt{3}x-\sqrt{3}$, let $x=0$, then $y=-\sqrt{3}$.
   $AO=OM_4=\sqrt{3}$, forming the rhombus $AN_4M_4C$. As shown in the figure $N_4(-3, 0)$.

$\therefore$ The coordinates of $N$ are: \boxed{N(1, 0)}, \boxed{N(0, -\sqrt{3})}, \boxed{N(0, 3\sqrt{3})}, or \boxed{N(-3, 0)}.


Question: As shown in the figure, the line $y=2x+1$ intersects the x-axis at point $A$ and the y-axis at point $B$. What are the coordinates of points $A$ and $B$?

Solution:
\textbf{Solution:} Substituting $x=0$ into $y=2x+1$, we get $y=1$,\\
thus the coordinates of point $B$ are $(0,1)$;\\
Substituting $y=0$ into $y=2x+1$, we get $0=2x+1$,\\
solving for $x$, we find $x=-\frac{1}{2}$,\\
thus the coordinates of point $A$ are $\left(-\frac{1}{2},0\right)$.\\
Hence, the coordinates of points $A$ and $B$ are $\boxed{\left(-\frac{1}{2},0\right)}$ and $\boxed{(0,1)}$, respectively.


Question: As shown in the figure, the line $y=2x+1$ intersects the $x$ axis at point $A$, and the $y$ axis at point $B$. Draw line $BP$ from point $B$ to intersect the $x$ axis at point $P$, and let $OP=2OA$, find the equation of the line $BP$.

Solution:
\textbf{Solution:}

Given that the coordinates of point A is $\left(-\frac{1}{2}, 0\right)$, \\
therefore, OA $= \frac{1}{2}$, \\
thus, OP $= 2\text{OA} = 1$, \\
when point P is on the positive half of the x-axis, then the coordinates of P is $(1,0)$, \\
let the equation of line BP be: $y=kx+b$, \\
substituting P $(1,0)$ and B $(0,1)$ into the equation, we get
\[
\begin{cases}
k+b=0\\
b=1
\end{cases}
\]
Solving this we get:
\[
\begin{cases}
k=-1\\
b=1
\end{cases}
\]
therefore, the equation of line BP is: $y=-x+1$; \\
when point P is on the negative half of the x-axis, then the coordinates of P is $(-1,0)$, \\
let the equation of line BP be $y=mx+n$, \\
substituting P $(-1, 0)$ and B $(0, 1)$, we get
\[
\begin{cases}
-m+n=0\\
n=1
\end{cases}
\]
Solving this we get
\[
\begin{cases}
m=1\\
n=1
\end{cases}
\]
therefore, the equation of line BP is: $y=x+1$; \\
In summary, the equation of line BP is either \boxed{y=x+1} or \boxed{y=-x+1}.


Question: As shown in the figure, line $l_1$ passes through points $A(0, 4)$ and $D(4, 0)$, line $l_2\colon y=\frac{1}{2}x+1$ intersects the $x$-axis at point $C$, and the two lines $l_1$ and $l_2$ intersect at point $B$. Find the equation of line $l_1$ and the coordinates of point $B$.

Solution:
Solution: Let the equation of line $l_1$ be $y=kx+b$,\\
According to the problem, we have $\begin{cases}b=4 \\ 4k+b=0\end{cases}$,\\
Solving these equations gives: $\begin{cases}k=-1 \\ b=4\end{cases}$,\\
Therefore, the equation of line $l_1$ is $y=-x+4$;\\
By combining $\begin{cases}y=-x+4 \\ y=\frac{1}{2}x+1\end{cases}$,\\
We find: $\begin{cases}x=2 \\ y=2\end{cases}$,\\
$\therefore$The coordinates of point \boxed{B(2,2)};


Question: As shown in the figure, the line $l_1$ passes through points A(0,4) and D(4,0), the line $l_2: y=\frac{1}{2}x+1$ intersects the x-axis at point C, and the two lines $l_1$ and $l_2$ intersect at point B. Find the area of $\triangle BCD$.

Solution:
\textbf{Solution:} The line $y=\frac{1}{2}x+1$ intersects the x-axis at point $C$,\\
$\therefore$ the coordinates of point $C$ are $(-2,0)$,\\
$\therefore$ $CD=6$,\\
$\therefore$ the area of $\triangle ABC=S_{\triangle ACD}-S_{\triangle BCD}=\frac{1}{2}\times 6\times 4-\frac{1}{2}\times 6\times 2=6$.\\
$\therefore$ the area of $\triangle BCD=\boxed{6}$.


Question: The graph of the linear function $y=kx+b$ passes through points $A(1, 6)$ and $B(-3, -2)$. Find the equation of this linear function.

Solution:
\textbf{Solution:} Let's substitute points A$(1,6)$ and B$(-3,-2)$ into $y=kx+b$ to get:\\
\[
\begin{cases}
k+b=6\\
-3k+b=-2
\end{cases},
\]
from which we find:
\[
\begin{cases}
k=2\\
b=4
\end{cases},
\]
therefore the expression of the linear function is: $\boxed{y=2x+4}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ passes through the points $A(1, 6)$ and $B(-3, -2)$. What is the area of $\triangle AOB$?

Solution:
\textbf{Solution:} As shown in the diagram, the line intersects the y-axis at point D,\\
for $y=2x+4$, by setting $x=0$, we get $y=4$,\\
$\therefore$ the coordinates of point D are: $(0,4)$,\\
$\therefore$ OD=4,\\
$\therefore$ ${S}_{\triangle AOB}={S}_{\triangle AOD}+{S}_{\triangle BOD}$,\\
$\therefore$ ${S}_{\triangle AOB}=\frac{1}{2}\times 4\times 1+\frac{1}{2}\times 4\times \left|-3\right|=2+6=\boxed{8}$;\\
$\therefore$ the area of $\triangle AOB$ is $\boxed{8}$.


Question: As shown in the diagram, the line BC intersects the x-axis at point C and the y-axis at point B, and it intersects the line $y=ax$ at point A. The x-coordinate of point A is 2, $\angle ACO = 45^\circ$, and the area of $\triangle ABO$ is 1. What is the value of $a$ and the equation of the line BC?

Solution:
\textbf{Solution:} Let the coordinates of point A be $(x_A, y_A)$,

$\because$ $S_{\triangle ABO} = \frac{1}{2} BO \cdot x_A = \frac{1}{2} \times BO \times 2 = 1$,

$\therefore$ $BO = 1$,

$\therefore$ $B(0, 1)$,

$\because$ $\angle ACO = 45^\circ$, $\angle BOC = 90^\circ$,

$\therefore$ $\angle CBO = \angle ACO = 45^\circ$,

$\therefore$ OC = BO = 1

$\therefore$ C$(-1,0)$

Let the analytic expression for BC be $y=kx+b$, where $k\neq 0$

Substituting the coordinates of points B and C into $y=kx+b$, we get: $\left\{ \begin{array}{l} b=1 \\ -k+b=0 \end{array} \right.$

$\therefore$ $\left\{ \begin{array}{l} k=1 \\ b=1 \end{array} \right.$

$\therefore$ The analytic expression for BC is $y=x+1$,

$\because$ when $x=2$, $y=3$,

$\therefore$ $A(2, 3)$,

Also $\because$ A is on the line $y=ax$,

$\therefore$ $a=\boxed{\frac{3}{2}}$;


Question: As shown in the diagram, the line BC intersects the x-axis at point C and the y-axis at point B, and intersects the line $y=ax$ at point A. The x-coordinate of point A is 2, $\angle ACO = 45^\circ$, and the area of $\triangle ABO$ is 1. If the line $y=ax+m$ intersects the y-axis at point D, when the area of $\triangle ABD$ is 4, what is the value of $m$?

Solution:
\textbf{Solution:} Since the line $y=ax+m$ intersects the y-axis at point D, \\
therefore $D(0, m)$, \\
since the area of $\triangle ABD=\frac{1}{2}BD\cdot x_A=\frac{1}{2}BD\times 2=4$, \\
thus $BD=4$, \\
since $B(0, 1)$, \\
therefore the coordinates of point D are $(0, 5)$ or $(0, -3)$, \\
thus $m=\boxed{5}$ or \boxed{-3};


Question: As shown in the figure, the line BC intersects the x-axis at point C, the y-axis at point B, and intersects the line $y=ax$ at point A. The x-coordinate of point A is 2, $\angle ACO = 45^\circ$, and the area of $\triangle ABO$ is 1. Suppose point P is a point on the line BC, and point Q is a point in the coordinate plane. There exist points P and Q that satisfy the conditions making the quadrilateral OAPQ a rhombus. Write down the coordinates of point Q.

Solution: \textbf{Solution:} The coordinates of point $Q$ are $\boxed{\left(-1, 1\right)}$ or $\left(\frac{\sqrt{26}}{2}, \frac{\sqrt{26}}{2}\right)$ or $\left(-\frac{\sqrt{26}}{2}, -\frac{\sqrt{26}}{2}\right)$ or $\left(\frac{13}{10}, \frac{13}{10}\right)$.


Question: As shown in the figure, two lines $l_1$ and $l_2$ passing through point $A(2, 0)$ intersect the $y$-axis at points $B$ and $C$, respectively, where point $B$ is above the origin and point $C$ is below the origin. It is known that $AB = \sqrt{13}$. What are the coordinates of point $B$?

Solution:
\textbf{Solution:} Given that $\because A(2,0)$, \\
$\therefore OA=2$, \\
In the right-angled triangle $Rt\triangle AOB$, $\angle AOB=90^\circ$, \\
$OB=\sqrt{AB^{2}-OA^{2}}=\sqrt{13-4}=3$, \\
$\therefore$ the coordinates of point B are $\boxed{(0,3)}$.


Question: As shown in the figure, two lines $l_{1}$ and $l_{2}$ pass through point A(2,0) and intersect the y-axis at points B and C, respectively, where point B is above the origin and point C is below the origin. It is known that $AB=\sqrt{13}$. If $BC=4$, find the equation of line $l_{2}$.

Solution:
\textbf{Solution:} Since $BC=4$,\\
the coordinates of point C are (0, -1),\\
let the equation of $l_{2}$ be $y=kx+b$,\\
substituting A(2, 0), C(0, -1) into it, we get:\\
$\therefore \left\{\begin{array}{l}2k+b=0 \\ b=-1 \end{array}\right.$\\
Solving, we find: $\left\{\begin{array}{l}k=\frac{1}{2} \\ b=-1 \end{array}\right.$\\
Hence, the equation of $l_{2}$ is $y=\frac{1}{2}x-1=\boxed{y=\frac{1}{2}x-1}$.


Question: As shown in the figure, in rectangle $ABCO$, point $C$ is on the x-axis, point $A$ is on the y-axis, and the coordinates of point $B$ are $(-9, 12)$. Rectangle $ABCO$ is folded along line $BD$ so that point $A$ falls at point $E$ on the diagonal $OB$, and line $BD$ intersects $OA$ and the x-axis at points $D$ and $F$, respectively. What is the length of segment $BO$?

Solution:
\textbf{Solution:} As shown in Figure 3, since quadrilateral $ABCO$ is a rectangle, \\
thus $\angle BCO=90^\circ$. \\
In $\triangle BCO$ right-angled at $C$, \\
$BO^{2}=BC^{2}+OC^{2}$, $BO^{2}=12^{2}+9^{2}$, \\
hence, $BO=\boxed{15}$.


Question: As shown in the diagram, in rectangle $ABCO$, point $C$ is on the x-axis, and point $A$ is on the y-axis, with the coordinates of point $B$ being $(-9, 12)$. The rectangle $ABCO$ is folded along line $BD$ such that point $A$ falls on point $E$ on the diagonal $OB$, and line $BD$ intersects $OA$ and the x-axis at points $D$ and $F$, respectively. What is the area of $\triangle OBD$?

Solution:
\textbf{Solution:} Let $OD=x$, \\
since quadrilateral $ABCO$ is a rectangle, \\
therefore $\angle BAD=90^\circ$. \\
According to the problem, $\triangle BAD\cong \triangle BED$, \\
therefore $BE=BA=9$, $AD=ED=12-x$, $\angle BED=\angle BAD=90^\circ$, \\
thus $\angle OED=90^\circ$, $EO=BO-BE=15-9=6$. \\
In $\triangle DEO$, $OD^2=OE^2+DE^2$, $x^2=6^2+(12-x)^2$, \\
$x=\frac{15}{2}$, i.e., $OD=\frac{15}{2}$, \\
therefore, the area of $\triangle OBD=\frac{1}{2}OD\cdot AB=\boxed{\frac{135}{4}}$.


Question: As shown in the figure, within the rectangle $ABCO$, point $C$ is on the $x$-axis, and point $A$ is on the $y$-axis, with point $B$ having the coordinates $(-9, 12)$. The rectangle $ABCO$ is folded along line $BD$, such that point $A$ falls at point $E$ on the diagonal $OB$, and line $BD$ intersects lines $OA$ and the $x$-axis at points $D$ and $F$, respectively. On the $x$-axis, there exists a point $M$, such that the quadrilateral formed by vertices $A$, $B$, $F$, and $M$ is a parallelogram. If such a point $M$ exists, find the coordinates of point $M$ that satisfy this condition.

Solution:
\textbf{Solution:} From equation (2), we have $OD=\frac{15}{2}$, thus $D\left(0,\frac{15}{2}\right)$.\\
Let the equation of line BD be $y=kx+b$,\\
Since $B(-9,12)$, $D\left(0,\frac{15}{2}\right)$,\\
Then $\begin{cases}-9k+b=12,\\ b=\frac{15}{2},\end{cases}$ yields $\begin{cases}k=-\frac{1}{2},\\ b=\frac{15}{2},\end{cases}$\\
Therefore, the equation of $BD$ is: $y=-\frac{1}{2}x+\frac{15}{2}$.\\
When $y=0$, $x=15$,\\
Therefore, $OF=15$.\\
Also, since $AB=9$,\\
Therefore, the coordinates of the point M that satisfy the conditions are $\boxed{(6,0)}$ or $\boxed{(24,0)}$.


Question: In the Cartesian coordinate system $xOy$, the linear function $y=-4x-2$ intersects with the $y$-axis at point $C$ and intersects with the line $AB: y=kx+2$ at point $D$, and the area of $\triangle ACD$ is $\frac{16}{7}$. What are the coordinates of point $D$?

Solution:
\textbf{Solution:} Start by considering the line $y=-4x-2$ which intersects the y-axis at point C, \\
thus the coordinates of point C are $(0, -2)$, \\
From the line $AB: y=kx+2$, \\
we get the coordinates of point A as $(0, 2)$, \\
$\therefore AC=4$, \\
$\because$ the area of $\triangle ACD=\frac{16}{7}$, \\
$\therefore$ the height of $\triangle ACD$ from side $AC=\frac{8}{7}$, \\
$\therefore$ the x-coordinate of point D is $-\frac{8}{7}$, \\
When $x=-\frac{8}{7}$, $y=-4x-2=\frac{18}{7}$ \\
$\therefore$ the coordinates of point D are $\boxed{\left(-\frac{8}{7}, \frac{18}{7}\right)}$.


Question: In the Cartesian coordinate system $xOy$, the linear function $y=-4x-2$ intersects the $y$-axis at point $C$ and intersects the line $AB: y=kx+2$ at point $D$, and the area of $\triangle ACD$ is $\frac{16}{7}$. Connect $BC$. What is the area of $\triangle BCD$?

Solution:
\textbf{Solution:} Substituting the coordinates of point D into the equation of line $AB: y=kx+2$,\\
we get $-\frac{8}{7}k+2=\frac{18}{7}$,\\
$\therefore k=-\frac{1}{2}$,\\
$\therefore$ the equation of line $AB$ is $y=-\frac{1}{2}x+2$;\\
$\because$ B is the intersection point of line AB and the x-axis,\\
$\therefore$ the coordinates of point B are (4, 0),\\
$\therefore S_{\triangle ABC}=\frac{1}{2}AC\cdot OB=\frac{1}{2}\times4\times4=8$,\\
$\therefore$ the area of $\triangle BCD$ $=S_{\triangle ACD}+S_{\triangle ABC}=\frac{16}{7}+8=\boxed{\frac{72}{7}}$;


Question: In the Cartesian coordinate system $xOy$, the linear function $y=-4x-2$ intersects with the $y$ axis at point $C$ and intersects with the line $AB: y=kx+2$ at point $D$, and the area of $\triangle ACD$ is $\frac{16}{7}$. Point $P$ lies on the $y$ axis. When $\triangle ABP$ is an isosceles triangle, find the coordinates of point $P$.

Solution:
\textbf{Solution:}
Given the equation of line $AB$ as $y=-\frac{1}{2}x+2$,\\
we find the coordinates of point $B$ to be $(4, 0)$,\\
$\therefore AB=2\sqrt{5}$,\\
When $\triangle ABP$ is an isosceles triangle, consider three cases:\\
(1) Taking A as the center and the length of $AB$ as the radius, we draw a circle, so $AB=AP=2\sqrt{5}$,\\
$\therefore$ the coordinates of point $P_1$ are $(0, 2+2\sqrt{5})$, and of $P_2$ are $(0, 2-2\sqrt{5})$;\\
(2) Taking B as the center and the length of $AB$ as the radius, we draw a circle, so $AB=BP=2\sqrt{5}$,\\
From the property of the isosceles triangle that all lines converge, we get $OP=OA=2$,\\
$\therefore$ the coordinates of point $P_3$ are $(0, -2)$;\\
(3) Drawing the perpendicular bisector of $AB$, which intersects the y-axis at point P, then $AP_4=BP_4$,\\
Let $OP_4=x$, so $AP_4=BP_4=x+2$,\\
Since $OB=4$, in $\triangle OBP_4$,\\
$OP_4^2+OB^2=BP_4^2$, i.e., $x^2+4^2=(x+2)^2$,\\
Solving this gives: $x=3$,\\
$\therefore$ the coordinates of point $P_4$ are $(0, -3)$,\\
$\therefore$ the coordinates of point $P$ are $\boxed{(0, 2+2\sqrt{5})}$ or $\boxed{(0, 2-2\sqrt{5})}$ or $\boxed{(0, -2)}$ or $\boxed{(0, -3)}$.


Question: Given the line $y=kx+b$ passes through points $A(6, 0)$ and $B(2, 2)$. Find the analytical expression of line $AB$.

Solution:
\textbf{Solution}: Since the line $y=kx+b$ passes through points $A(6, 0)$ and $B(2, 2)$, \\
then $\left\{\begin{array}{l}
6k+b=0 \\
2k+b=2
\end{array}\right.$. Solving this system of equations, we find $\left\{\begin{array}{l}
k=-\frac{1}{2} \\
b=3
\end{array}\right.$. \\
Therefore, the equation of line $AB$ is: \boxed{y=-\frac{1}{2}x+3}.


Question: As shown in the diagram, it is known that the line $y=kx+b$ passes through points $A\left(6, 0\right)$ and $B\left(2, 2\right)$. If the line $y=x-3$ intersects with line $AB$ at point $C$, find the coordinates of point $C$.

Solution:
\textbf{Solution}: Given the problem, if the line $y=x-3$ intersects with line $AB$ at point $C$,\\
then we have the system of equations $\left\{\begin{array}{l}y=-\frac{1}{2}x+3 \\ y=x-3\end{array}\right.$, solving this gives $\left\{\begin{array}{l}x=4\\ y=1\end{array}\right.$\\
$\therefore$ The coordinates of point $C$ are $\boxed{(4, 1)}$.


Question: Given the line $y=kx+b$ passes through the points $A(6, 0)$, $B(2, 2)$. According to the graph, write the solution set for the inequality $x-3>kx+b$ with respect to $x$.

Solution: \textbf{Solution:} From the graph, we can conclude that $\boxed{x>4}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, point $A$ is on the positive half of the $y$-axis, and point $B$ is on the positive half of the $x$-axis, with $OA = OB = 10$. What is the equation of line $AB$?

Solution:
\textbf{Solution:} Since point $A$ is on the positive half of the y-axis, and point $B$ is on the positive half of the x-axis, with $OA=OB=10$, \\
thus $A\left(0, 10\right)$, $B\left(10, 0\right)$. \\
Let the equation of line $AB$ be $y=kx+b$, \\
hence $\begin{cases}10k+b=0\\ b=10\end{cases}$, solving gives: \\
$\begin{cases}k=-1\\ b=10\end{cases}$, \\
thus the equation of line $AB$ is $\boxed{y=-x+10}$.


Question: As shown in the diagram, in the Cartesian coordinate system xOy, point A lies on the positive half of the y-axis, and point B lies on the positive half of the x-axis, with $OA=OB=10$. If point P is a moving point on line AB, and $S_{\triangle OBP} = \frac{1}{4} S_{\triangle OAP}$, find the coordinates of point P.

Solution:
Solution: Given equation (1) and the conditions, we can assume point $P\left(m, -m+10\right)$. Thus, for $\triangle OBP$ with OB as the base, the absolute value of the y-coordinate of point P is its height, and for $\triangle OAP$ with OA as the base, the absolute value of the x-coordinate of point P is its height.\\
Since $S_{\triangle OBP} = \frac{1}{4} S_{\triangle OAP}$,\\
it follows that $\frac{1}{2}\cdot OB\cdot \left|-m+10\right|=\frac{1}{4}\cdot \frac{1}{2}\cdot OA\cdot \left|m\right|$,\\
Given $OA=OB=10$,\\
it implies $\left|-m+10\right|=\frac{1}{4}\left|m\right|$,\\
Solving this gives: $m=8$ or $m=\frac{40}{3}$,\\
Therefore, the coordinates of point P are $\boxed{\left(8, 2\right)}$ or $\boxed{\left(\frac{40}{3}, -\frac{10}{3}\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, point $A$ is on the positive half of the $y$-axis, and point $B$ is on the positive half of the $x$-axis, with $OA=OB=10$. Line $AB$ is translated downwards by 10 units to obtain line $l$, and points $M$ and $N$ are moving points on line $l$ (the abscissas of $M$ and $N$ are $x_{M}$ and $x_{N}$, respectively, and $x_{M}<x_{N}$). Given $MN=4\sqrt{2}$, find the minimum perimeter of quadrilateral $ABNM$ and explain the reasoning.

Solution:
\textbf{Solution:} From the given problem, we can construct as shown in the figure: Draw point C as the symmetric point of A with respect to MN, draw MD $\parallel$ BN, and connect MC and CD,\\
$\therefore$ $AM=MC$, $AC\perp MN$,\\
$\because$ MN $\parallel$ AB,\\
$\therefore$ quadrilateral $MNBD$ is a parallelogram, $\angle CAD=90^\circ$,\\
$\therefore$ $MN=BD$, $MD=BN$,\\
$\because$ MN $=4\sqrt{2}$,\\
$\therefore$ $BD=4\sqrt{2}$,\\
$\because$ $OA=OB=10$,\\
$\therefore$ $\triangle AOB$ is an isosceles right-angled triangle,\\
$\therefore$ $AB=\sqrt{2}OA=10\sqrt{2}$, the distance from line l to AB is $\frac{1}{2}AB=5\sqrt{2}$,\\
$\therefore$ based on the property of axis symmetry, $AC=10\sqrt{2}$,\\
$\therefore$ the perimeter of quadrilateral $ABNM$ is $AB+BN+MN+AM=AB+MD+MN+CM=14\sqrt{2}+MD+MC$,\\
To minimize the perimeter of quadrilateral $ABNM$, it is necessary to make $MC+MD$ as small as possible, which is achieved when points C, M, D are collinear. At this time, $MC+MD=CD$,\\
$\therefore$ in $\triangle ADC$, $AC=10\sqrt{2}$, $AD=AB−BD=6\sqrt{2}$, by Pythagoras' theorem, we have:\\
$CD=\sqrt{AC^2+AD^2}=4\sqrt{17}$,\\
$\therefore$ $MC+MD=4\sqrt{17}$,\\
$\therefore$ The minimum perimeter of quadrilateral $ABNM$ is $14\sqrt{2}+4\sqrt{17}=\boxed{14\sqrt{2}+4\sqrt{17}}$.


Question: In the figure, in the Cartesian coordinate plane, point $A(2, m)$ and point $B(6, 2)$ lie on the graph of an inverse proportionality function. What is the equation of the line $AB$?

Solution:
\textbf{Solution:} Let the inverse proportion function be $y=\frac{k}{x}$,\\
$\because$ point $A(2, m)$ and point $B(6, 2)$ lie on the graph of $y=\frac{k}{x}$,\\
$\therefore k=2m=6\times 2$,\\
Solving this gives $m=6$,\\
$\therefore$ the coordinates of point $A$ are $(2, 6)$,\\
Let the expression for line $AB$ be $y=ax+b$,\\
Substituting $A(2, 6)$ and $B(6, 2)$ into it gives $\left\{\begin{array}{l}2a+b=6 \\ 6a+b=2\end{array}\right.$,\\
Solving this yields $\left\{\begin{array}{l}a=-1 \\ b=8\end{array}\right.$,\\
$\therefore$ the expression for line $AB$ is $y=-x+8$, hence the expression for line $AB$ is $\boxed{y=-x+8}$.


Question: As shown in the figure, in the Cartesian coordinate plane, point $A(2, m)$ and point $B(6, 2)$ are both on the graph of an inverse proportion function. Find the area of $\triangle AOB$ and the distance $AH$ from point $A$ to line $OB$.

Solution: \textbf{Solution:} Let the point of intersection between line $AB$ and the $x$-axis be $C$. Within the line equation $AB$ as $y=-x+8$, by setting $y=0$ we get $x=8$. Therefore, $C(8, 0)$. Thus, the area of $\triangle AOB$ can be calculated as ${S}_{\triangle AOB}={S}_{\triangle AOC}-{S}_{\triangle BOC}=\frac{1}{2}\times 8\times 6-\frac{1}{2}\times 8\times 2=\boxed{16}$. Given $B(6, 2)$, it follows that $OB=\sqrt{6^2+2^2}=2\sqrt{10}$. Since ${S}_{\triangle AOB}=\frac{1}{2}OB\cdot AH=16$, it concludes that $AH=\frac{16\times 2}{2\sqrt{10}}=\boxed{\frac{8\sqrt{10}}{5}}$.


Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, points A and B are respectively the intersections of the line $y=\frac{3}{4}x+3$ with the x-axis and the y-axis. Given that point C is in the first quadrant, if quadrilateral $ABCO$ is a parallelogram, what are the coordinates of point C?

Solution:
\textbf{Solution:} Let $x=0$, then $y=3$; let $y=0$, then $\frac{3}{4}x+3=0$, \\
$\therefore x=-4$, \\
$\because$ Points A and B are the intersections of the line $y=\frac{3}{4}x+3$ with the $x$ and $y$ axes, respectively, \\
$\therefore$ A$(-4,0)$, B$(0,3)$, \\
$\because$ Point C is in the first quadrant, if quadrilateral ABCO is a parallelogram, \\
$\therefore$ AB$\parallel$CD, and AB=CD, \\
Based on coordinates and translations, we can derive that C$(4,3)$. \\
Hence, the coordinates of point C is $\boxed{(4,3)}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, points A and B are the intersections of the line $y=\frac{3}{4}x+3$ with the x-axis and y-axis, respectively. Given the condition in (1), point D is on the x-axis. If the quadrilateral $ABCD$ is an isosceles trapezoid, find the equation of line $CD$.

Solution:
Solution: As shown in the figure, draw CM $\perp$ x-axis at M, then $\angle AOB=\angle DMC=90^\circ$, M(4,0),\\
$\because$ quadrilateral ABCD is an isosceles trapezoid,\\
$\therefore \angle BAO=\angle CDM$, AB=DC; in $\triangle ABO$ and $\triangle DCM$,\\
$\left\{\begin{array}{l}
\angle AOB=\angle DMC \\
\angle BAO=\angle CDM \\
AB=DC
\end{array}\right.$,\\
$\therefore \triangle ABO \cong \triangle DCM$ (AAS),\\
$\therefore AO=DM=4$,\\
$\therefore OD=OM+MD=8$,\\
$\therefore D(8,0)$,\\
Let the equation of line CD be $y=k(x-8)$, substituting point C(4,3),\\
then $k= -\frac{3}{4}$,\\
$\therefore$ the equation of line CD is $y=-\frac{3}{4}x+6$. Therefore, the equation of line CD is $\boxed{y=-\frac{3}{4}x+6}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $l_1$ passes through points $A(0, 1)$ and $B(2, 2)$. The line $l_1$ is translated downward by $m$ units to obtain the line $l_2$, which is known to pass through point $(-1, -2)$ and intersects the x-axis at point C. What is the equation of the line $l_1$?

Solution:
\textbf{Solution:} Let the equation of line $l_1$ be $y=kx+b$,\\
since line $l_1$ passes through points A(0, 1) and B(2, 2),\\
thus $\begin{cases}b=1\\ 2k+b=2\end{cases}$, solving yields: $\begin{cases}k=\frac{1}{2}\\ b=1\end{cases}$,\\
therefore, the equation of line $l_1$ is $y= \frac{1}{2}x+1=\boxed{\frac{1}{2}x+1}$.


Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, the line $l_{1}$ passes through points $A(0, 1)$ and $B(2, 2)$. The line $l_{1}$ is translated downwards by $m$ units to obtain the line $l_{2}$. It is known that the line $l_{2}$ passes through point $(-1, -2)$ and intersects the x-axis at point C. Calculate the value of $m$ and the coordinates of point C.

Solution:
\textbf{Solution:} Let's move the line $l_{1}$ downwards by $m$ units to get the line $l_{2}$, thus the equation of line $l_{2}$ is $y= \frac{1}{2}x+1-m$, \\
$\because$ The line $l_{2}$ passes through the point $(-1,-2)$, \\
$\therefore -2= \frac{1}{2} \times (-1)+1-m$, solving this gives $m= \boxed{\frac{5}{2}}$, \\
$\therefore$ The equation of line $l_{2}$ is $y= \frac{1}{2}x- \frac{3}{2}$, \\
Let $y=0$, then solving gives $x=3$, \\
$\therefore$ The coordinates of point C are $\boxed{(3,0)}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, line $l_{1}$ passes through points $A(0, 1)$ and $B(2, 2)$. Line $l_{1}$ is translated downwards by $m$ units to obtain line $l_{2}$. It is known that line $l_{2}$ passes through point $(-1, -2)$ and intersects the x-axis at point $C$. Point $D$ is a point on line $l_{2}$. If points $A$, $B$, $C$, and $D$ can form a parallelogram, find the coordinates of point $D$.

Solution:
Solution: From the problem, we know that AB$\parallel$CD. When the four points A, B, C, and D form the parallelogram ABDC, \\
$\because$A (0,1), B (2,2), C (3,0), \\
$\therefore$point A moves 3 units to the right and then 1 unit down to coincide with point C, \\
$\therefore$point B moves 3 units to the right and then 1 unit down to coincide with point D, at this time the coordinates of D are $\boxed{(5,1)}$; \\
$\because$AB$\parallel$CD, when points A, B, C, and D form the parallelogram ABCD, \\
$\because$A (0,1), B (2,2), C (3,0), \\
$\therefore$point B moves 1 unit to the right and then 2 units down to coincide with point C, \\
$\therefore$point A moves 1 unit to the right and then 2 units down to coincide with point D, at this time the coordinates of D are $\boxed{(1,-1)}$; \\
Overall, if points A, B, C, and D can form a parallelogram, the coordinates of point D are either \boxed{(5,1)} or \boxed{(1,-1)}.


Question: In the Cartesian coordinate system xOy, it is known that the line $y=-\frac{1}{2}x+2$ intersects the x-axis and y-axis at points A and B, respectively. The graph of the line $y=kx-1$ intersects the y-axis at point C and intersects the given line at point D, where the x-coordinate of point D is 2. Find the analytical expression of the line $y=kx-1$?

Solution:
\textbf{Solution:} Given the problem, \\
since the graph of the line $y=kx-1$ intersects with the given line at point $D$, and the x-coordinate of $D$ is 2, \\
therefore, when $x=2$, $y=-\frac{1}{2}x+2=1$, \\
thus, the coordinates of $D$ are $(2, 1)$, \\
substituting the coordinates of $D$ into the line $y=kx-1$, we get $k=1$, \\
therefore, the equation of the line $y=kx-1$ is $\boxed{y=x-1}$.


Question: In the Cartesian coordinate system $xOy$, it is known that the line $y=-\frac{1}{2}x+2$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively. The graph of the line $y=kx-1$ intersects the $y$-axis at point $C$ and intersects the given line at point $D$, where the $x$-coordinate of point $D$ is $2$. If the graph of the line $y=kx-1$ is translated upwards or downwards, intersecting the given line at point $E$, and assuming the intercept of the resulting function graph after translation is $b$, if the intersection point $E$ always falls on the line segment $AB$, what is the range of values for $b$?

Solution:
\textbf{Solution:} Let $x=0$, then $y=-\frac{1}{2}x+2=2$,\\
Let $y=0$, then $-\frac{1}{2}x+2=0$,\\
$\therefore x=4$,\\
$\because$ the line $y=-\frac{1}{2}x+2$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively,\\
$\therefore$ the coordinates of $A$ are $(4, 0)$, and the coordinates of $B$ are $(0, 2)$,\\
Assume the equation of the line $y=kx-1$ after translation is $y=x+b$,\\
When the line $y=x+b$ passes through point $A$, $b=-4$,\\
When the line $y=x+b$ passes through point $B$, $b=2$,\\
From the graph, when the intersection point $E$ always falls on the segment $AB$, the range of $b$ is $-4\leq b\leq 2$,\\
Therefore, the range of $b$ is $\boxed{-4\leq b\leq 2}$.


Question: In the Cartesian coordinate system $xOy$, it is known that the line $y=-\frac{1}{2}x+2$ intersects the $x$ axis and the $y$ axis at points $A$ and $B$, respectively. The graph of the line $y=kx-1$ intersects the $y$ axis at point $C$, and intersects the known line at point $D$, where the $x$-coordinate of point $D$ is $2$. Is there a point $P$ on the $x$ axis such that the quadrilateral formed by points $P$, $A$, $B$, and $C$ is a trapezoid? If so, find the coordinates of point $P$.

Solution:
\textbf{Solution:}\\
Given the graph of the line $y=kx-1$ intersects the $y$ axis at point $C$,\\
$\therefore$ when $x=0$, $y=-1$,\\
$\therefore$ the coordinates of $C$ are $(0, -1)$.\\
(1) When $CP\parallel AB$, the quadrilateral $ABPC$ forms a trapezoid,\\
$\therefore$ the equation of line $CP$ is $y=-\frac{1}{2}x-1$,\\
Let $y=0$, then $x=-2$,\\
$\therefore \boxed{P(-2, 0)}$.\\
(2) When $BP\parallel AC$, the quadrilateral $ABPC$ forms a trapezoid,\\
Let the equation of line $AC$ be $y=mx-1$, substituting point $A(4, 0)$ gives $m=\frac{1}{4}$,\\
$\therefore$ the equation of line $AC$ is $y=\frac{1}{4}x-1$,\\
$\because BP\parallel AC$,\\
$\therefore$ the equation of line $BP$ is $y=\frac{1}{4}x+2$,\\
Let $y=0$, then $x=-8$,\\
$\therefore \boxed{P(-8, 0)}$.\\
Thus, such a point $P$ exists, making the quadrilateral formed by points $P$, $A$, $B$, and $C$ a trapezoid, with coordinates $(-2, 0)$ or $(-8, 0)$.


Question: As shown in the figure, in the Cartesian coordinate system, $A(0,4)$, $B(2,4)$, and $C(4,2)$. What are the coordinates of the center $M$ of the circle containing the arc passing through points $A$, $B$, and $C$?

Solution: \textbf{Solution:} Let $M$ be the center of the circle that passes through points $A$, $B$, and $C$. The coordinates of $M$ are $\boxed{(1,1)}$.


Question: As shown in the figure, in the Cartesian coordinate system, $A(0,4)$, $B(2,4)$, $C(4,2)$. What is the radius of this circle?

Solution: \textbf{Solution:} The radius of the circle is $\boxed{\sqrt{10}}$.


Question: As shown in the Cartesian coordinate system, let $A(0,4)$, $B(2,4)$, $C(4,2)$. What is the position of point $D(3,-1)$ relative to the circle $\odot M$? (Fill in: inside, outside, or on).

Solution: \textbf{Solution:} Point $D$ is \boxed{inside} circle $\odot M$.


Question: As shown in the figure, $\triangle ABC$ is inscribed in $\odot O$, with $AB$ being the diameter of $\odot O$ and $AB=5$, $AC=3$. What is the value of $\tan A$?

Solution:
\textbf{Solution:} Since AB is the diameter of the circle $\odot$O,\\
therefore $\angle ACB=90^\circ$,\\
since AB=5, and AC=3,\\
therefore BC=4,\\
hence $\tan A=\boxed{\frac{4}{3}}$.


Question: As illustrated in the figure, $\triangle ABC$ is inscribed in $\odot O$, with $AB$ being the diameter of $\odot O$, and $AB=5$, $AC=3$. If $D$ is the midpoint of $\overset{\frown}{AB}$, and $CD$ and $BD$ are connected, what is the length of chord $CD$?

Solution:
\textbf{Solution:} Draw $BE\perp CD$ at E passing through point B, \\
since D is the midpoint of $\overset{\frown}{AB}$, \\
it follows that $\overset{\frown}{AD}=\overset{\frown}{BD}$, \\
thus, $\angle ACD=\angle BCD=45^\circ$, \\
given that BC=4, \\
in $\triangle BCE$, we have $BE=CE=2\sqrt{2}$, \\
since $\angle A=\angle D$, \\
it follows that $\tan D=\tan A=\frac{4}{3}$, \\
in $\triangle BDE$, \\
$DE=\frac{BE}{\tan D}=\frac{2\sqrt{2}}{\frac{4}{3}}=\frac{3\sqrt{2}}{2}$, \\
therefore, $CD=CE+DE=\boxed{\frac{7\sqrt{2}}{2}}$.


Question: As shown in the diagram, $AB$ is a chord of $\odot O$. Through point $O$, construct $OC\perp OA$, with $OC$ intersecting $AB$ at point $P$, and $CP=BC$. Point $Q$ is a point on the arc $\overset{\frown}{AmB}$. Prove that: $BC$ is a tangent to $\odot O$.

Solution:
\textbf{Solution} Proof: Connect OB,\\
since OA = OB,\\
it follows that $\angle OAB = \angle OBA$,\\
since PC = CB,\\
it follows that $\angle CPB = \angle PBC$,\\
since $\angle APO = \angle CPB$,\\
it follows that $\angle APO = \angle CBP$,\\
since OC $\perp$ OA,\\
it follows that $\angle AOP = 90^\circ$,\\
thus, $\angle OAP + \angle APO = 90^\circ$,\\
thus, $\angle CBP + \angle ABO = 90^\circ$,\\
thus, $\angle CBO = 90^\circ$,\\
thus, \boxed{BC \text{ is the tangent to the circle } O}.


Question: As shown in the figure, AB is a chord of $\odot O$. Draw OC $\perp$ OA, where OC intersects AB at P, and CP = BC. Point Q is a point on the arc $\overset{\frown}{AmB}$. BC is a tangent of $\odot O$; given that $\angle BAO = 25^\circ$, what is the degree measure of $\angle AQB$?

Solution:
\textbf{Solution:} Since $\angle BAO=25^\circ$,\\
it follows that $\angle ABO=25^\circ$, $\angle APO=65^\circ$,\\
thus $\angle POB=\angle APO-\angle ABO=40^\circ$,\\
therefore $\angle AQB=\frac{1}{2}(\angle AOP+\angle POB)=\frac{1}{2}\times 130^\circ=\boxed{65^\circ}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, C and D are points on $\odot O$, BD bisects $\angle ABC$, $DE \perp BE$, and DE intersects the extension of BC at point E. DE is a tangent to $\odot O$; if $CE=1$ and $AC=2\sqrt{7}$, what is the radius $r$ of $\odot O$?

Solution:
\textbf{Solution:} Set OD intersects AC at point M, as shown in the diagram:\\
$\because$ AB is the diameter of $\odot O$,\\
$\therefore \angle ACB=\angle ACE=90^\circ$,\\
From (1), it is known that $\angle ODE=90^\circ$,\\
$\therefore \angle ACE=\angle E=\angle ODE=90^\circ$,\\
$\therefore$ Quadrilateral DECM is a rectangle,\\
$\therefore EC=DM=1$,\\
$\because MO\parallel CB$ and O is the midpoint of AC,\\
$\therefore MO$ is the midline of $\triangle ABC$, and $\angle AMO=\angle ACB=90^\circ$,\\
$\therefore AM=MC=\frac{1}{2}AC=\sqrt{7}$,\\
Let the radius of the circle be $r$, then $MO=DO-DM=r-1$,\\
In $\triangle AMO$, by Pythagorean theorem: $AO^2=AM^2+MO^2$, substituting the values:\\
$r^2=(\sqrt{7})^2+(r-1)^2$,\\
Solving gives: $r=\boxed{4}$,\\
Thus, the radius of the circle $\odot O$ is \boxed{4}.


Question: As shown in the figure, a circle with center $O$ on side $BC$ of $\triangle ABC$ passes through points $A$ and $B$, intersecting side $BC$ at point $E$, with $DO\perp BE$ at point $O$. Line $AD$ is drawn to intersect $BC$ at point $F$, given that $AC=FC$. $AC$ is the tangent of circle $O$: Given $BF=8$ and $DF= 2\sqrt{10}$, what is the radius of circle $O$?

Solution:
\textbf{Solution:} Let the radius of $\odot O$ be $r$, \\
$\therefore BO=DO=r$, \\
$\because BF=8$, \\
$\therefore OF=8-r$, \\
$\because \angle DOF=90^\circ$, \\
$\therefore$ in $\triangle ODF$ right, by the Pythagorean theorem, we have $OF^2+OD^2=DF^2$, \\
$\because DF=2\sqrt{10}$, \\
$\therefore (8-r)^2+r^2=(2\sqrt{10})^2$, \\
Solving, we get: $r=6$ or $r=2$ (discarded as it does not meet the problem statement), \\
Hence, the radius of $\odot O$ is $\boxed{6}$.


Question: As shown in the figure, there is a circle with a point $O$ on the side $BC$ of $\triangle ABC$ as the center. This circle goes through points $A$ and $B$ and intersects with side $BC$ at point $E$. $DO \perp BE$ at point $O$, and $AD$ is connected intersecting $BC$ at $F$, given $AC=FC$. If $\angle ADB= 60^\circ$ and $BD=1$, calculate the area of the shaded part (leave the result in radical form).

Solution:
\textbf{Solution:} Since $BO=DO$, and $BD=1$, $\angle DOB=90^\circ$\\
Therefore, in $\triangle BOD$ right at $O$, by the Pythagorean theorem $BO^2+OD^2=BD^2$\\
Thus, we find $BO=DO=\frac{\sqrt{2}}{2}$\\
That is, the radius of the circle $O$ is $\frac{\sqrt{2}}{2}$\\
Since $\angle ADB=60^\circ$\\
Therefore $\angle AOB=2\angle ADB=120^\circ$\\
Therefore $\angle AOC=180^\circ-\angle AOB=60^\circ$\\
Since $OA\perp AC$\\
Therefore $\angle OAC=90^\circ$\\
Therefore, in $\triangle OAC$ right at $O$, $\tan AOC=\tan 60^\circ=\frac{AC}{OA}=\sqrt{3}$．\\
Since $OA=\frac{\sqrt{2}}{2}$\\
Therefore $AC=\sqrt{3}OA=\frac{\sqrt{6}}{2}$\\
Therefore, the area of $\triangle OAC=\frac{1}{2}OA\cdot AC=\frac{1}{2}\times \frac{\sqrt{2}}{2}\times \frac{\sqrt{6}}{2}=\frac{\sqrt{3}}{4}$, and the area of the sector $OAE=\frac{60^\circ\pi (\frac{\sqrt{2}}{2})^2}{360^\circ}=\frac{\pi }{12}$\\
Therefore, the area of the shaded region $=S_{\triangle OAC}-S_{\text{sector }OAE}=\boxed{\frac{\sqrt{3}}{4}-\frac{\pi }{12}}$.


Question: As shown in the figure, in right $\triangle ABC$, $\angle ABC=90^\circ$, point $E$ is the midpoint of $BC$, and $DE$ is drawn. $DE$ is a tangent to the semicircle $\odot O$. If $\angle BAC=30^\circ$ and $DE=2$, find the length of $AD$.

Solution:
\textbf{Solution:}

Given that \\
$DE=2$, \\
it follows that $BC=2ED=4$, \\
Given $\angle BAC=30^\circ$, \\
it follows that $AC=2BC=8$, \\
Thus, $AB=\sqrt{8^2-4^2}=4\sqrt{3}$, \\
So, $BD=\frac{1}{2}AB=2\sqrt{3}$, \\
Therefore, $AD=\sqrt{(4\sqrt{3})^2-(2\sqrt{3})^2}=\boxed{6}$.


Question: As shown in the figure, $\triangle ABC$ is inscribed in circle $\odot O$, and $AB=AC$. Point $P$ is on arc $\overset{\frown}{AC}$, and $\angle BAC=30^\circ$. What is the degree measure of $\angle APC$?

Solution:
\textbf{Solution:} Given that $AB=AC$, $\angle BAC=30^\circ$,\\
hence, $\angle ABC=\angle ACB= \frac{1}{2}\left(180^\circ-\angle BAC\right)=\frac{1}{2}\left(180^\circ-30^\circ\right)=75^\circ$,\\
since quadrilateral ABCP is a cyclic quadrilateral,\\
it follows that $\angle ABC+\angle APC=180^\circ$,\\
thus, $\angle APC=180^\circ-\angle ABC=180^\circ-75^\circ=\boxed{105^\circ}$,


Question: As shown in the figure, $\triangle ABC$ is inscribed in $\odot O$, and $AB=AC$. Let $P$ be a point on the arc $\overset{\frown}{AC}$, and $\angle BAC=30^\circ$. If the radius of $\odot O$ is 6, find the length of the arc $\overset{\frown}{APC}$ (keep $\pi$ in the result).

Solution:
\textbf{Solution:} Draw the lines OA and OC.

$\because \angle ABC=75^\circ$,

$\therefore \angle AOC=2\angle ABC=2\times 75^\circ=150^\circ$,

$\therefore L_{\overset{\frown}{APC}} = \frac{150\pi R}{180}=\frac{5}{6}\pi \times 6=\boxed{5\pi}$.


Question: As shown in the figure, the circumcircle of $\triangle ABC$ has its center at $O$, with $AB$ being the diameter. $AB \perp CD$ at point $E$, and $P$ is a point on the extension of $AB$, and $\angle BCP = \angle BCD$. Prove that $CP$ is a tangent to $\odot O$.

Solution:
\textbf{Solution:} Proof: Connect OC.\\
$\because$OB=OC,\\
$\therefore \angle OBC=\angle OCB$,\\
$\because$AB$\perp$CD at point E,\\
$\therefore \angle CEB=90^\circ$,\\
$\therefore \angle OBC+\angle BCD=90^\circ$,\\
$\therefore \angle OCB+\angle BCD=90^\circ$,\\
$\because \angle BCP=\angle BCD$,\\
$\therefore \angle OCB+\angle BCP=90^\circ$,\\
$\therefore$OC$\perp$CP,\\
$\therefore \boxed{CP \text{ is the tangent to the circle } O}$.


Question: As shown in the figure, $O$ is the circumcircle of $\triangle ABC$, $AB$ is the diameter of $O$, $AB \perp CD$ at point $E$, and $P$ is a point on the extension line of $AB$, with $\angle BCP = \angle BCD$. Extend $DO$ to intersect $AC$ at point $F$ and intersect $O$ at point $G$, and connect $GC$. Given that the radius of $O$ is $5$ and $OE=3$, find the lengths of $GC$ and $OF$.

Solution:
\textbf{Solution:} Since $AB\perp CD$ at point E, \\
hence E is the midpoint of CD, \\
Since O is the midpoint of GD, \\
hence OE is the median of $\triangle DCG$, \\
hence $GC=2OE=6$, $OE\parallel GC$, \\
Since $AO\parallel GC$, \\
hence $\triangle GCF\sim \triangle OAF$, \\
hence $\frac{GC}{OA}=\frac{GF}{OF}$, \\
that is $\frac{6}{5}=\frac{GF}{OF}$, \\
Since $GF+OF=5$, \\
hence $OF=\boxed{\frac{25}{11}}$.


Question: Given that in $\triangle ABC$, we have $AB=AC=5$ and $BC=8$. Point $E$ is a moving point on ray $CA$, and point $O$ is a moving point on side $BC$, with $OC=OE$. Ray $OE$ intersects ray $BA$ at point $D$. As shown in Figure 1, if $OC=2$, find the value of $\frac{{S}_{\triangle ADE}}{{S}_{\triangle ODB}}$.

Solution:
\textbf{Solution:} Given $AB=AC$,\\
it follows $\angle B=\angle C$,\\
since $OC=OE$,\\
it follows $\angle OEC=\angle C$,\\
hence, $\angle B=\angle OEC$,\\
thus, $\triangle ABC\sim \triangle OEC$,\\
implying $\frac{AC}{OC}=\frac{BC}{CE}$,\\
which gives $\frac{5}{2}=\frac{8}{CE}$,\\
therefore $CE=3.2$,\\
thus, $AE=1.8$;\\
Since $\angle AED=\angle OEC=\angle B, \angle D=\angle D$,\\
it implies $\triangle OBD\sim \triangle AED$,\\
giving $\frac{AE}{OB}=\frac{1.8}{6}=0.3$,\\
therefore $\frac{S_{\triangle ADE}}{S_{\triangle ODB}}=\boxed{0.09}$.


Question: Given that in $\triangle ABC$, $AB=AC=5, BC=8$, point $E$ is a moving point on ray $CA$, and point $O$ is a moving point on side $BC$, with $OC=OE$, ray $OE$ intersects ray $BA$ at point $D$. Connect $AO$. If $\triangle AEO$ is an isosceles triangle with $AE$ as the leg, what is the length of the line segment $OC$?

Solution:
\textbf{Solution:}\\
Since $\triangle AEO$ is an isosceles triangle with $AE$ as its leg,\\
thus $AE=OE$,\\
since $OC=OE$,\\
thus, let $AE=OE=OC=x$,\\
from (1), we get $\triangle ABC \sim \triangle OEC$,\\
thus, $\frac{AC}{OC}=\frac{BC}{CE}$,\\
thus, $\frac{5}{x}=\frac{8}{5-x}$,\\
solving this, we obtain $x=\boxed{\frac{25}{13}}$, upon verification, $x=\frac{25}{13}$ is a solution to the original equation;\\
hence, the length of $OC$ is \boxed{\frac{25}{13}}.


Question: Given that, in $\triangle ABC$, $AB=AC=5$, $BC=8$. Point $E$ is a moving point on ray $CA$, and point $O$ is a moving point on side $BC$, with $OC=OE$. Ray $OE$ intersects ray $BA$ at point $D$. When point $E$ is on side $AC$, connecting $BE$ and $CD$, $\angle DBE = \angle CDO$, find the length of segment $OC$.

Solution:
\textbf{Solution:} From equation (1), we have $\angle B = \angle OEC$,\\
$\because \angle OEC + \angle OEA = 180^\circ$,\\
$\therefore \angle B + \angle OEA = 180^\circ$,\\
$\therefore$ Points A, B, O, and E are concyclic,\\
$\therefore \angle DBE = \angle AOD$,\\
$\because \angle DBE = \angle CDO$,\\
$\therefore \angle AOD = \angle CDO$,\\
$\therefore AO \parallel DC$,\\
$\therefore \triangle AOE \sim \triangle CDE, \triangle ABO \sim \triangle DBC$,\\
$\therefore \frac{AO}{DC} = \frac{BO}{CB}$, $\frac{AO}{DC} = \frac{AE}{CE}$,\\
$\therefore \frac{AE}{EC} = \frac{BO}{CB}$,\\
Let OC=x, OB=8-x,\\
$\because \triangle ABC \sim \triangle OEC$,\\
$\therefore \frac{AC}{OC} = \frac{BC}{CE}$,\\
$\therefore \frac{5}{x} = \frac{8}{CE}$,\\
Solving, $CE=1.6x$,\\
$\therefore AE=5-1.6x$\\
$\therefore \frac{5-1.6x}{1.6x} = \frac{8-x}{8}$,\\
Solving, $x_1=8-\sqrt{39}$, $x_2=8+\sqrt{39}$ (discard),\\
Hence, the length of OC is $\boxed{8-\sqrt{39}}$.


Question: As shown in the figure, $ \triangle ABC$ is inscribed in $ \odot O$, with chords AE and BC intersecting at point D. Line BO is drawn such that $ \angle CAE = \angle ABO$ and $ AE \perp BC$. If $ ED = \sqrt{3} CD$, find the degree measure of $ \angle ABC$.

Solution:
Solution: As shown in the figure, connect CE,\\
$\therefore \angle ABC=\angle AEC$,\\
$\because ED=\sqrt{3}CD$, AE$\perp$BC,\\
$\therefore \tan\angle AEC=\frac{CD}{DE}=\frac{\sqrt{3}}{3}$,\\
$\therefore \angle AEC=30^\circ$,\\
$\therefore \angle ABC=\boxed{30^\circ}$;


Question: As shown in the figure, $\triangle ABC$ is inscribed in circle $\odot O$, with chords AE and BC intersecting at point D. BO is connected, and $\angle CAE = \angle ABO$. Through point O, construct $OH \perp BC$ at point H, and extend HO to intersect AB at point P. If $HO = 1$ and $AP = 6$, find the length of the radius of circle $\odot O$.

Solution:
\textbf{Solution:} As shown in the figure, draw $OF$ perpendicular to $AB$ at $F$, \\
$\therefore BF = AF$, \\
Let $FP = x$, \\
$\therefore BF = AF = AP + PF = 6 + x$, \\
$\therefore BP = BF + PF = 6 + 2x$, \\
Since $\angle ABC = 30^\circ$, and $PH \perp BC$, \\
$\therefore \angle BPH = 60^\circ$, and $BP = 2PH$, \\
Also, since $\angle OF \perp AB$, \\
$\therefore \angle OFP = 90^\circ$, \\
$\therefore \angle POF = 30^\circ$, \\
$\therefore OP = 2FP = 2x$, \\
$\therefore PH = OP + OH = 1 + 2x$, \\
$\therefore BP = 2 + 4x$, \\
$\therefore 2 + 4x = 6 + 2x$, \\
Solving gives $x = 2$, \\
$\therefore PF = 2$, $BF = 8$, $PO = 4$, \\
$\therefore OF = \sqrt{PO^{2} - PF^{2}} = 2\sqrt{3}$, \\
$\therefore OB = \sqrt{OF^{2} + BF^{2}} = \boxed{2\sqrt{19}}$, \\
$\therefore$ The radius of circle $O$ is $2\sqrt{19}$.


Question: As shown in the diagram, in $\triangle ABC$, $AB=AC$, and $\angle BAC=54^\circ$. The circle $\odot O$ with diameter $AB$ intersects $AC$ and $BC$ at points $D$ and $E$, respectively. A tangent to $\odot O$ passing through point $B$ intersects the extension of $AC$ at point $F$. Given $BE=CE$; find the measure of $\angle CBF$ in degrees.

Solution:
\textbf{Solution:} Since $\angle BAC=54^\circ$ and $AB=AC$, it follows that $\angle ABC=63^\circ$.\\
Also, since $BF$ is the tangent to the circle $O$, it follows that $\angle ABF=90^\circ$.\\
Therefore, $\angle CBF=\angle ABF-\angle ABC=\boxed{27^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $\angle BAC=54^\circ$. The circle $\odot O$ with diameter $AB$ intersects $AC$ and $BC$ at points $D$ and $E$, respectively. A tangent line to $\odot O$ passing through point $B$ intersects the extension of $AC$ at point $F$. Given $AB=6$, what is the length of $\overset{\frown}{AD}$?

Solution:
\textbf{Solution:} Draw line OD, \\
since $OA = OD$, $\angle BAC = 54^\circ$, \\
therefore $\angle BOD = 72^\circ$, $\angle AOD = 72^\circ$. \\
Also, since $AB = 6$, \\
therefore $OA = 3$. \\
Therefore, $\overset{\frown}{AD} = \frac{72\pi \times 3}{180} = \boxed{\frac{6\pi }{5}}$.


Question: As shown in the figure, in parallelogram $ABCD$, point $O$ is on side $AD$, and a circle $\odot O$ with $O$ as the center and $OA$ as the radius passes exactly through point $B$ and intersects side $AD$ at point $E$. The line containing side $CD$ is tangent to $\odot O$ at point $H$. Lines $AH$ and $EH$ are connected. If $2\angle HAB + \angle C = 180^\circ$, $CB$ is tangent to $\odot O$; if $DE = 1$, what is the radius of $\odot O$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a parallelogram, \\
it follows that $AD\parallel BC$ and $AB\parallel CH$. \\
Since $\angle OBC=90^\circ$, \\
it follows that $\angle BOA=90^\circ$, \\
thus, $\triangle AOB$ is an isosceles right triangle, \\
therefore, $\angle BAD=\angle ABO=45^\circ$, \\
thus, $\angle ODH=\angle BAD=45^\circ$. \\
Since $\angle CHO=90^\circ$, which means $\angle DHO=90^\circ$, \\
therefore, $\triangle DHO$ is an isosceles right triangle, \\
therefore, $OH=DH=OE$. \\
Let $OH=x$, then $OE=DH=x$, $OD=OE+DE=x+1$. \\
Since in $\triangle DHO$, $OH^2+DH^2=OD^2$, which means $x^2+x^2=(x+1)^2$ \\
Solving gives: $x_1=1+\sqrt{2}$, $x_2=1-\sqrt{2}$ (discard) \\
Therefore, the radius of $\odot O$ is $\boxed{1+\sqrt{2}}$.


Question: As shown in the diagram, it is known that $AB$ is the diameter of the circle $\odot O$, point $C$ lies on circle $\odot O$, and the tangent line to $\odot O$ passing through point $C$ intersects the extended line of $AB$ at point $P$. Point $D$ is a point on arc $AC$. Connect $BD$, $BC$, and $DC$. If $\angle D = 26^\circ$, what is the measure of $\angle PCB$?

Solution:
\textbf{Solution:} Let's connect CO,\\
$\because$ $\angle D=26^\circ$,\\
$\therefore$ $\angle COB=52^\circ$,\\
$\because$ $OB=OC$,\\
$\therefore$ $\angle BCO=\frac{180^\circ-\angle COB}{2}=64^\circ$,\\
$\because$ PC is tangent to $\odot O$,\\
$\therefore$ $\angle PCO=90^\circ$,\\
$\therefore$ $\angle PCB=\angle PCO-\angle BCO=26^\circ$;\\
Hence, the measure of $\angle PCB$ is $\boxed{26^\circ}$.


Question: Given that $AB$ is the diameter of the circle $\odot O$, and point $C$ is a point on the circle $\odot O$. A tangent line to $\odot O$ passing through point $C$ intersects the extension line of $AB$ at point $P$. Let $D$ be a point on arc $AC$, and connect $BD$, $BC$, and $DC$. If quadrilateral $CDBP$ is a parallelogram, find the measures of $\angle PCB$ and $\angle ADC$?

Solution:
\textbf{Solution:} Connect CO, AC,

since quadrilateral CDBP is a parallelogram,

it follows that $\angle P=\angle CDB$,

since AB is the diameter of $\odot O$,

it follows that $\angle ACB=\angle ADB=90^\circ$, i.e., $\angle ACO+\angle OCB=90^\circ$,

since PC is tangent to $\odot O$,

it follows that $\angle PCO=90^\circ$, i.e., $\angle BCP+\angle OCB=90^\circ$,

therefore, $\angle ACO=\angle BCP$,

since $\angle OAC=\angle BDC$,

it follows that $\angle OAC=\angle OCA=\angle BCP=\angle P=\angle CDB$,

hence $\angle COB=2\angle OAC$, $\angle OBC=2\angle P$,

therefore, $\angle COB=\angle OBC$,

since $OB=OC$,

it follows that $\angle OCB=\angle OBC$,

therefore, $\angle OCB=\angle OBC=\angle COB=60^\circ$,

therefore, $\angle PCB=\frac{1}{2}\angle CBO=30^\circ$,

therefore, $\angle OAC=\angle OCA=\angle BCP=\angle P=\angle CDB=30^\circ$,

therefore, $\angle ADC=\angle ADB+\angle BDC=90^\circ+30^\circ=\boxed{120^\circ}$;

therefore, $\angle PCB=\boxed{30^\circ}$, $\angle ADC=120^\circ$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, with chord $CD \perp AB$ at point $E$. Connect $AC$, $BC$, $BD$, and $OF \perp AC$ at point $F$, where $OF = 1$. What is the length of $BD$?

Solution:
\textbf{Solution:} Since $OF \perp AC$,\\
it follows that $AF=FC$,\\
Since $OA=OB$,\\
it follows that $OF$ is the median of $\triangle ABC$,\\
Therefore, $BC=2OF=2$,\\
Since $AB \perp CD$,\\
it follows that $\overset{\frown}{BC}=\overset{\frown}{BD}$,\\
Therefore, $BD=BC=2$;\\
Hence, the length of $BD$ is $\boxed{2}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and chord $CD\perp AB$ at point $E$. Connect $AC$, $BC$, $BD$, $OF\perp AC$ at point $F$, and $OF=1$. When $\angle D = 30^\circ$, find the length of $\overset{\frown}{AC}$ and the area of the shaded region (keep the results in terms of square roots and $\pi$).

Solution:
\textbf{Solution:} Connect $OC$ as shown below:\\
$\because \angle CAB=\angle D=30^\circ$, $OA=OC$,\\
$\therefore \angle OAC=\angle OCA=30^\circ$,\\
$\therefore \angle AOC=120^\circ$,\\
In $Rt\triangle ABC$, $\because \angle ACB=90^\circ$, $BC=2$, $\angle CAB=30^\circ$,\\
$\therefore AB=2BC=4$, $AC=\sqrt{3}BC=2\sqrt{3}$,\\
$\therefore$ the length of $\overset{\frown}{AC}$ $=\frac{120\pi \cdot 2}{180}=\boxed{\frac{4\pi }{3}}$,\\
The area of the shaded region $=\frac{120\pi \cdot{2}^{2}}{360}-\frac{1}{2}\times 2\sqrt{3}\times 1=\boxed{\frac{4\pi }{3}-\sqrt{3}}$.


Question: As shown in the figure, $AB$ is the diameter of circle $O$, and points $C$ and $D$ are on circle $O$, satisfying: point $C$ is the midpoint of arc $AD$, and $AD$ intersects $OC$ at point $E$. Given $AD=8$ and $EC=2$, what is the radius of circle $O$?

Solution:
\textbf{Solution:} Since point C is the midpoint of arc AD,\\
it follows that $OC\perp AD$,\\
thus $AE=\frac{1}{2}AD=4$,\\
Let the radius of circle O be $r$, then $OA=OC=r, OE=r-2$,\\
Since in $\triangle AOE$, $OE^2+AE^2=OA^2$,\\
it follows that $(r-2)^2+4^2=r^2$,\\
Solving for $r$ gives $r=\boxed{5}$,\\
therefore, the radius of circle O is \boxed{5}.


Question: As shown in the diagram, $AB$ is the diameter of circle $O$, and points $C$ and $D$ are on circle $O$ with the condition that point $C$ is the midpoint of arc $AD$, and $AD$ intersects $OC$ at point $E$. Given $AD=8$, $EC=2$. A line parallel to $AB$ passing through point $C$ intersects chord $AD$ at point $F$. What is the length of segment $EF$?

Solution:
\textbf{Solution}: Since $CF\parallel AB$,\\
it follows that $\frac{CE}{OE}=\frac{EF}{AE}$,\\
thus $\frac{2}{3}=\frac{EF}{4}$,\\
solving this gives $EF=\boxed{\frac{8}{3}}$.


Question: As shown in the figure, it is known that $AB$ is the diameter of $\odot O$, $CD$ is a chord, and $AB\perp CD$ at point $E$. $G$ is any point on the arc $AC$ (not coinciding with $A$ or $C$), and $AD$ and $GD$ are connected. If $EB=4\,\text{cm}$ and $CD=16\,\text{cm}$, what is the radius of $\odot O$?

Solution:
\textbf{Solution:} Connect $OD$, as shown in the figure:\\
Since $EB=4\,\text{cm}$ and $CD=16\,\text{cm}$, and $AB$ is the diameter of $\odot O$ with $AB\perp CD$,\\
Therefore, $DE=\frac{1}{2}CD=8\,\text{cm}$,\\
Let $OD=OB=r\,\text{cm}$, then $OE=(r-4)\,\text{cm}$,\\
Therefore, in $\triangle OED$, we have $r^2=(r-4)^2+8^2$,\\
Solving this, we find: $r=\boxed{10}\,\text{cm}$,\\
Therefore, the radius of $\odot O$ is \boxed{10}\,\text{cm}.


Question: As shown in the figure, it is known that $AB$ is the diameter of the circle $\odot O$ and $CD$ is a chord, with $AB \perp CD$ at point $E$. $G$ is any point on the arc $AC$ (not coinciding with $A$ or $C$), and segments $AD$ and $GD$ are connected. When $G$ moves to the position where points $O$, $D$, and $G$ are collinear, find the length of $AG$ at this moment.

Solution:
\textbf{Solution:} Given from (2), we know that the radius of $ \odot O$ is 10cm,\\
therefore $GD=20cm$, $AE=16cm$, $DE=8cm$,\\
therefore $AD^{2}=DE^{2}+AE^{2}=320cm^2$,\\
since $GD$ is the diameter of $ \odot O$,\\
therefore $\angle GAD=90^\circ$,\\
therefore $AG=\sqrt{GD^{2}-AD^{2}}=\boxed{4\sqrt{5}}$.


Question: As shown in the figure, in $\triangle ABC$, we have $AB=AC$, and $BD$ bisects $\angle ABC$ and intersects $AC$ at point $D$. Point $E$ is on the extension line of $BC$, and $BD=DE$. Point $G$ is the midpoint of the line segment $BC$. We draw line $AG$ and it intersects $BD$ at point $F$. A line is drawn through point $D$ such that $DH\perp BC$ with $H$ as the foot of the perpendicular. $\triangle DCE$ is an isosceles triangle; if $\angle CDE=22.5^\circ$ and $DC=\sqrt{2}$, find the length of $GH$.

Solution:
\textbf{Solution:}\\
Since $\angle CDE=22.5^\circ$ and $CD=CE=\sqrt{2}$,\\
it follows that $\angle DCH=45^\circ$ and $DH\perp BC$,\\
thus $\angle HDC=\angle DCH=45^\circ$,\\
implying $DH=CH$,\\
since $DH^2+CH^2=DC^2=2$,\\
we have $DH=CH=1$,\\
since $\angle ABC=\angle DCH=45^\circ$,\\
$\triangle ABC$ is an isosceles right triangle.\\
Furthermore, since point G is the midpoint of BC,\\
$AG\perp BC$ and $AG=GC=BG$,\\
and since $BD=DE$ and $DH\perp BC$,\\
it follows $BH=HE=\sqrt{2}+1$,\\
since $BH=BG+GH=CG+GH=CH+GH+GH=\sqrt{2}+1$,\\
thus $1+2GH=\sqrt{2}+1$,\\
yielding $GH=\boxed{\frac{\sqrt{2}}{2}}$.


Question: As shown in the diagram, point $C$ is any point on segment $AB$ (point $C$ is not coincident with points $A$ and $B$). Equilateral triangles $\triangle ACD$ and $\triangle BCE$ are constructed on the same side of line $AB$ with $AC$ and $BC$ as sides, respectively. $AE$ and $CD$ intersect at point $M$, $BD$ and $CE$ intersect at point $N$, and $AE$ and $BD$ intersect at point $F$. $\triangle ACE \cong \triangle DCB$; what is the measure of $\angle AFC$?

Solution:
\textbf{Solution:} Given $\triangle ACE \cong \triangle DCB$, we have $\angle MAC=\angle NDC$,\\
also, $\angle ACM=60^\circ$, and $\angle AMC=\angle DMF$,\\
therefore, $\angle DFM=\angle ACM=60^\circ$,\\
thus, $\angle AFB=180^\circ-\angle DFM=120^\circ$,\\
draw $CG \perp AE$ at point G, and $CH \perp BD$ at point H from point C.\\
Since $\triangle ACE \cong \triangle DCB$,\\
thus, $DB=AE$, and $S_{\triangle ACE}=S_{\triangle DCB}$,\\
therefore, $\frac{1}{2}DB \cdot CH=\frac{1}{2}AE \cdot CG$,\\
thus, $CG=CH$,\\
therefore, $CF$ bisects $\angle AFB$.\\
Thus, $\angle AFC=\frac{1}{2}\angle AFB=\frac{1}{2}\times 120^\circ=\boxed{60^\circ}$.


Question: As shown in the figure, within $ \triangle ABC$, $ AD$ is the altitude on side $ BC$, $ CE$ is the median on side $ AB$, $ DG \perp CE$ at $G$, $CD = \frac{1}{2}AB$, and $DE$ is connected. Given that $BC = 13$, $CD = 5$, and $CG = EG$; what is the area of $\triangle BEC$?

Solution:
\textbf{Solution:} Given $\because BC=13, CD=5$,\\
$\therefore AB=10, BD=BC-CD=13-5=8$,\\
$\therefore AD=\sqrt{AB^{2}-BD^{2}}=\sqrt{10^{2}-8^{2}}=6$,\\
$\therefore S_{\triangle ABC}=\frac{1}{2}BC\cdot AD=\frac{1}{2}\times 13\times 6=39$,\\
$\because CE$ is the median to side $AB$,\\
$\therefore S_{\triangle BEC}=\frac{1}{2}S_{\triangle ABC}=\boxed{19.5}$.


Question: Given in the figure, points B, D, and C lie on a straight line, with AB = AD, AC = AE, and $\angle BAD = \angle CAE$. If $\angle B = 65^\circ$ and BC = DE, what is the measure of $\angle CDE$?

Solution:
\textbf{Solution:} Given that $AB=AD$ and $\angle B=65^\circ$, \\
it follows that $\angle ADB=\angle B=65^\circ$, \\
since $\triangle BAC \cong \triangle DAE$, \\
it implies that $\angle ADE=\angle B=65^\circ$, \\
therefore, $\angle CDE=180^\circ-\angle ADE-\angle ADB \\
=180^\circ-2\times65^\circ \\
=\boxed{50^\circ}.$


Question: As shown in the figure, in the isosceles $\triangle ABC$ where $AB=AC$ and $\angle A<90^\circ$, $CD$ is the height of $\triangle ABC$, and $BE$ is the angle bisector of $\triangle ABC$, with $CD$ and $BE$ intersecting at point $P$. As the size of $\angle A$ changes, the shape of $\triangle EPC$ also changes. When $\angle A=36^\circ$, what is the degree measure of $\angle BPD$?

Solution:
\textbf{Solution:} Given $AB=AC$ and $\angle A=36^\circ$,\\
it follows that $\angle ABC=\angle ACB=\left(180-36\right)^\circ \div 2=72^\circ$,\\
since $CD\perp AB$,\\
thus $\angle BDC=90^\circ$,\\
given that $BE$ bisects $\angle ABC$,\\
hence $\angle ABE=\angle CBE=\frac{1}{2}\angle ABC=36^\circ$,\\
therefore $\angle BPD=90^\circ-36^\circ=\boxed{54^\circ}$;


Question: As shown in the figure, in the isosceles $\triangle ABC$, $AB=AC$, $\angle A<90^\circ$, $CD$ is the altitude of $\triangle ABC$, $BE$ is the angle bisector of $\triangle ABC$, and $CD$ intersects $BE$ at point $P$. As the size of $\angle A$ changes, the shape of $\triangle EPC$ also changes. When $\triangle EPC$ is an isosceles triangle, what is the degree measure of $\angle ACB$?

Solution:
\textbf{Solution:} Let $ \angle A=\alpha $ and $ \angle EPC=\beta $. \\
(1) If $ EP=EC$, \\
then $ \angle ECP=\angle EPC=\beta $, \\
and $ \angle ABC=\angle ACB=90^\circ-\frac{\alpha }{2}$, $ \angle ABC+\angle BCD=90^\circ$, \\
thus, $ 90^\circ-\frac{\alpha }{2}+\left(90^\circ-\frac{\alpha }{2}-\beta \right)=90^\circ$, \\
from (2), we know $ \beta =45^\circ+\frac{\alpha }{4}$, \\
$ \therefore 90^\circ-\frac{\alpha }{2}+90^\circ-\frac{\alpha }{2}-\left(45^\circ+\frac{\alpha }{4}\right)=90^\circ$, \\
solving gives: $ \alpha =36^\circ$, \\
$ \therefore \angle ACB=90^\circ-\frac{\alpha }{2}=\boxed{72^\circ}$; \\
(2) If $ PC=PE$, \\
then $ \angle PCE=\angle PEC=\left(180^\circ-\beta \right)\div 2=90^\circ-\frac{\beta }{2}$, \\
from (1), we have: $ \angle ABC+\angle BCD=90^\circ$, \\
$ \therefore 90^\circ-\frac{\alpha }{2}+90^\circ-\frac{\alpha }{2}-\left(90^\circ-\frac{\beta }{2}\right)=90^\circ$, \\
since $ \beta =45^\circ+\frac{\alpha }{4}$, \\
$ \therefore 90^\circ-\frac{\alpha }{2}+90^\circ-\frac{\alpha }{2}-\left(90^\circ-\frac{45^\circ+\frac{\alpha }{4}}{2}\right)=90^\circ$, \\
solving gives: $ \alpha =\frac{180^\circ}{7}$, \\
$ \therefore \angle ACB=90^\circ-\frac{\alpha }{2}=\boxed{\frac{540^\circ}{7}}$; \\
(3) If $ CP=CE$, \\
then $ \angle EPC=\angle PEC=\beta$, $ \angle PCE=180^\circ-2\beta$, \\
from (1), we have: $ \angle ABC+\angle BCD=90^\circ$, \\
$ \therefore 90^\circ-\frac{\alpha }{2}+90^\circ-\frac{\alpha }{2}-\left(180^\circ-2\beta \right)=90^\circ$, \\
since $ \beta =45^\circ+\frac{\alpha }{4}$, \\
$ \therefore 90^\circ-\frac{\alpha }{2}+90^\circ-\frac{\alpha }{2}-\left[180^\circ-2\left(45^\circ+\frac{\alpha }{4}\right)\right]=90^\circ$, \\
solving gives: $ x=0$, which does not make sense, \\
In summary: when $ \triangle EPC$ is an isosceles triangle, the degree of $ \angle ACB$ is either $ \boxed{72^\circ}$ or $ \boxed{\frac{540^\circ}{7}}$.


Question: As shown in the figure, $AD$ and $CE$ are respectively the median and the angle bisector of $\triangle ABC$, with $AB = AC$. If the area of $\triangle ABC$ is $20$ and $BC = 4$, what is the length of $AD$?

Solution:
\textbf{Solution:} Given that $AD$ is the median of $\triangle ABC$, and $AB=AC$. \\
$\therefore AD \perp BC$,\\
Since the area of $\triangle ABC$ is 20, and $BC=4$,\\
$\therefore \frac{1}{2}BC\cdot AD=20$,\\
$\therefore \frac{1}{2}\times 4\times AD=20$,\\
$\therefore AD=\boxed{10}$.


Question: As shown in the figure, $AD$ and $CE$ are respectively the median and angle bisector of $\triangle ABC$, with $AB=AC$. If $\angle CAD=20^\circ$, find the measure of $\angle ACE$.

Solution:
\textbf{Solution:} Given that, since $AD$ is the median of $\triangle ABC$, $AB=AC$, $\angle CAD=20^\circ$, \\
thus $\angle CAB=2\angle CAD=40^\circ$, $\angle B=\angle ACB=\frac{1}{2}\left(180^\circ-\angle CAB\right)=70^\circ$.\\
Since $CE$ is the angle bisector of $\triangle ABC$,\\
thus $\angle ACE=\frac{1}{2}\angle ACB=\boxed{35^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ and $AE$ are the median and altitude to side $BC$, respectively. If $AE = 5\,\text{cm}$ and the area of $\triangle ABC = 30\,\text{cm}^2$, what is the length of $DC$ in cm?

Solution:
\textbf{Solution:} Let $AD$ and $AE$ be the median and the height to the side $BC$ respectively, with $AE=5\,\text{cm}$ and $S_{\triangle ABC}=30\,\text{cm}^{2}$,\\
therefore $S_{\triangle ADC}=15\,\text{cm}^{2}$,\\
thus $\frac{1}{2} \times AE \times CD = 15$,\\
hence $\frac{1}{2} \times 5 \times CD = 15$,\\
solving this gives: $CD=\boxed{6}\,(\text{cm})$;


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

Solution:
\textbf{Solution:} Since $\angle B=40^\circ$ and $\angle C=50^\circ$, \\
therefore, $\angle BAC=90^\circ$, \\
and since AD is the median, \\
thus AD$= \frac{1}{2}$ BC$=$BD, \\
hence, $\angle ADE=2\angle B=80^\circ$, \\
and since AE$\perp$BC, \\
therefore, $\angle DAE=\boxed{10^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AD=DC$. If $\angle BAD=20^\circ$, what is the measure of $\angle C$ in degrees?

Solution:
\textbf{Solution:}

Given:
\[ \because AB=AD \]
\[ \therefore \angle ABD=\angle ADB=\frac{180^\circ-\angle BAD}{2}=\frac{180^\circ-20^\circ}{2}=80^\circ \]
\[ \because DA=DC \]
\[ \therefore \angle C=\angle DAC \]
Since $\angle ADB$ is an exterior angle of $\triangle ADC$,
\[ \therefore \angle ADB=\angle C+\angle DAC \]
\[ \therefore 80^\circ=\angle C+\angle C \]
Solving, we find $\angle C=\boxed{40^\circ}$


Question: As shown in the figure, in $\triangle ABC$, we have $AB=AD=DC$. If $\angle BAC=78^\circ$, then what is the measure of $\angle C$ in degrees?

Solution: \textbf{Solution:} We have $\angle C = \boxed{34^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ is the altitude from $A$ to side $BC$, $AE$ and $BF$ are the angle bisectors of $\angle BAC$ and $\angle ABC$ respectively, $\angle BAC=56^\circ$, $\angle C=70^\circ$. What is the measure of $\angle DAE$?

Solution:
\textbf{Solution:} Given that $AE$ bisects $\angle BAC$, \\
we have $\angle EAC = \frac{1}{2}\angle BAC = \frac{1}{2}\times 56^\circ = 28^\circ$. \\
Since $AD$ is the height from $A$ to $BC$, \\
it follows that $\angle ADC = 90^\circ$. \\
Hence, in $\triangle ADC$, $\angle DAC = 90^\circ - \angle C = 70^\circ = 20^\circ$. \\
Therefore, $\angle DAE = \angle EAC - \angle DAC = 28^\circ - 20^\circ = \boxed{8^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ is the altitude to side $BC$, $AE$ and $BF$ are the angle bisectors of $\angle BAC$ and $\angle ABC$ respectively, $\angle BAC=56^\circ$, $\angle C=70^\circ$. What is the measure of $\angle BOA$?

Solution:
\textbf{Solution}: In $\triangle ABC$, we have $\angle ABC=180^\circ-\angle C-\angle BAC=180^\circ-70^\circ-56^\circ=54^\circ$ \\
Since $BF$ bisects $\angle ABC$ \\
Therefore, $\angle 1=\frac{1}{2}\angle ABC=\frac{1}{2}\times 54^\circ=27^\circ$ \\
Therefore, $AE$ bisects $\angle BAC$ \\
Therefore, $\angle 2=\frac{1}{2}\angle BAC=\frac{1}{2}\times 56^\circ=28^\circ$ \\
Therefore, in $\triangle AOB$, $\angle AOB=180^\circ-(\angle 1+\angle 2)=180^\circ-(27^\circ+28^\circ)=\boxed{125^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $AD$ is the altitude on the side $BC$, $CE$ is the median on the side $AB$, $DG \perp CE$ at $G$, and $CD = AE$. Given that $BD = 6$, $CD = 5$, and $CG = EG$; find the area of $\triangle CDG$.

Solution:
\textbf{Solution:} Given $CE$ is the median of side $AB$,\\
hence $AE=BE$,\\
since $BD=6, CD=5$,\\
it follows that $AB=10$,\\
thus $AD=\sqrt{10^{2}-6^{2}}=8$,\\
therefore, the area of $\triangle ABC=\frac{1}{2}\times 11\times 8=44$, the area of $\triangle ABD=\frac{1}{2}\times 6\times 8=24$,\\
since $CE$ is the median of side $AB$,\\
it follows that the area of $\triangle BEC=\frac{1}{2}$ area of $\triangle ABC=22$,\\
since $DE$ is the median of side $AB$,\\
it follows that the area of $\triangle BDE=\frac{1}{2}$ area of $\triangle ABD=12$,\\
thus the area of $\triangle EDC$ is area of $\triangle BEC$ - area of $\triangle EDB=22-12=10$,\\
additionally, since $CG=EG$,\\
the area of $\triangle CDG=\frac{1}{2}$ area of $\triangle EDC=\boxed{5}$.


Question: As shown in Figure 1, it is known that point P lies on segment \(AC\), \(BP = DP\), \(AB = 3\), and \(CD = 7\). If \(\angle A = \angle C = \angle BPD = 90^\circ\), what is the length of \(AC\)?

Solution:
\textbf{Solution:} Since $\angle A=\angle C=\angle BPD=90^\circ$\\
Therefore, $\angle APB=\angle CDP=90^\circ-\angle CPD$ \\
Since $BP=DP$\\
Therefore, $\triangle APB\cong \triangle CDP$ \\
Therefore, $AB=PC=3$ and $AP=CD=7$ \\
Therefore, $AC=\boxed{10}$.


Question: Given: Point $P$ is a point on segment $AC$, $BP=DP$, $AB=3$, $CD=7$. As shown in Figure 2, if $\angle A = \angle C = \angle BPD \neq 90^\circ$, find the length of $AC$?

Solution:
\textbf{Solution:}\\
Since $\angle A=\angle C=\angle BPD$, $\angle A+\angle B=\angle BPD+\angle CPD$\\
Thus, $\angle B=\angle CPD$\\
Since $BP=DP$\\
Therefore, $\triangle APB \cong \triangle CDP$\\
Thus, $AB=PC=3$, $AP=CD=7$\\
Therefore, $AC=\boxed{10}$.


Question: As shown in the diagram, in $\triangle ABC$, $\angle B=40^\circ$, $\angle C=80^\circ$. What is the measure of $\angle BAC$ in degrees?

Solution:
\textbf{Solution:} Because $\angle B + \angle BAC + \angle C = 180^\circ$, \\
it follows that $\angle BAC = \boxed{60^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B = 40^\circ$, $\angle C = 80^\circ$. $AE$ bisects $\angle BAC$ and intersects $BC$ at $E$, $AD \perp BC$ at $D$. Find the degree measure of $\angle EAD$.

Solution:
\textbf{Solution:}\\
Since $AD \perp BC$,\\
it follows that $\angle ADC = 90^\circ$,\\
Since $\angle DAC = 180^\circ - \angle ADC - \angle C$,\\
it follows that $\angle DAC = 10^\circ$,\\
Since $AE$ bisects $\angle BAC$,\\
it follows that $\angle BAE = \angle CAE = 30^\circ$,\\
Therefore, $\angle EAD = \angle CAE - \angle DAC = \boxed{20^\circ}$.


Question: Given in $\triangle ABC$, $\angle C = 3\angle B$, $AD$ bisects $\angle BAC$ and intersects $BC$ at $D$. As shown in figure 1, if $AE \perp BC$ at $E$, $\angle C = 75^\circ$, what is the measure of $\angle DAE$?

Solution:
\textbf{Solution:} Given that $\because \angle C = 3\angle B$ and $\angle C = 75^\circ$,\\
$\therefore \angle B = 25^\circ$,\\
$\therefore \angle BAC = 180^\circ - \angle B - \angle C = 80^\circ$,\\
$\because AD$ bisects $\angle BAC$,\\
$\therefore \angle BAD = \frac{1}{2}\angle BAC = 40^\circ$,\\
$\therefore \angle ADE = \angle BAD + \angle B = 65^\circ$,\\
$\because AE \perp BC$, $\therefore \angle AED = 90^\circ$,\\
$\therefore \angle DAE = 90^\circ - \angle ADE = 90^\circ - 65^\circ = \boxed{25^\circ}$;


Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the inverse proportion function $y=\frac{k_2}{x}$ at points $A(1, 2)$ and $B(-2, n)$. What are the analytical expressions of the linear function and the inverse proportion function?

Solution:
\textbf{Solution:} From the problem, we know that the inverse proportion function $y=\frac{k_2}{x}$ passes through $A(1, 2)$,
\[ \therefore k_2=1\times 2=2, \]
\[ \therefore \] the inverse proportion function is $y=\frac{2}{x}$,
Moreover, since $B(-2, n)$ lies on the graph of the inverse proportion function $y=\frac{2}{x}$,
\[ \therefore n=\frac{2}{-2}=-1, \]
\[ \therefore B(-2, -1), \]
Also, since the line $y=k_1x+b$ passes through $A(1, 2)$, $B(-2, -1)$,
\[ \therefore \left\{\begin{aligned} k_1+b&=2 \\ -2k_1+b&=-1 \end{aligned}\right., \] solving this yields $\left\{\begin{aligned} k_1&=1 \\ b&=1 \end{aligned}\right.$,
\[ \therefore \] the linear function is described by $y=x+1$,
The inverse proportion function is described by $y=\frac{2}{x}$.
Thus, the formula for the linear function is $\boxed{y=x+1}$, and the formula for the inverse proportion function is $\boxed{y=\frac{2}{x}}$.


Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the inverse proportion function $y=\frac{k_2}{x}$ at points $A(1, 2)$ and $B(-2, n)$. Based on the graph, directly write the range of values of $x$ for which $k_1x+b>\frac{k_2}{x}$.

Solution: \textbf{Solution:} Observing the graph, the range of $x$ for $k_1x+b>\frac{k_2}{x}$ is \boxed{-2<x<0} or \boxed{x>1}.


Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects the graph of the inverse proportion function $y=\frac{k_2}{x}$ at points $A(1, 2)$ and $B(-2, n)$. If point $P$ is on segment $AB$, and the area ratio $S_{\triangle AOP} : S_{\triangle BOP} = 1 : 4$, find the coordinates of point $P$.

Solution:
\textbf{Solution:} Let $P(x, x+1)$,

$\because$ the ratio of areas $S_{\triangle AOP} : S_{\triangle BOP} = 1 : 4$,

$\therefore$ the ratio of segments $AP : PB = 1 : 4$,

which means $PB = 4PA$,

$\therefore (x+2)^2+(x+1+1)^2=16[(x-1)^2+(x+1-2)^2]$,

solving yields $x_1 = \frac{2}{5}$, $x_2 = 2$ (discard),

$\therefore$ the coordinates of point $P$ are $\boxed{\left( \frac{2}{5}, \frac{7}{5} \right)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the rectangle $OABC$ has its vertex $O$ at the origin. Side $OC$ is on the positive half of the $x$-axis, and side $OA$ is on the positive half of the $y$-axis, with $OA=3$ and $AB=4$. The graph of the inverse proportion function $y=\frac{k}{x}$ ($k>0$) intersects sides $AB$ and $BC$ at points $D$ and $E$, respectively, and $BD=2AD$. Find the coordinates of point $D$ and the value of $k$?

Solution:
\textbf{Solution:} Given that $AB=4$ and $BD=2AD$,\\
it follows that $AB=AD+BD=AD+2AD=3AD=4$,\\
hence $AD=\frac{4}{3}$.\\
Also, given that $OA=3$,\\
it follows that $D\left(\frac{4}{3}, 3\right)$.\\
Since point D lies on the hyperbola $y=\frac{k}{x}$,\\
it follows that $k=\frac{4}{3}\times 3=\boxed{4}$.


Question: As shown in the figure, in the Cartesian coordinate system, the rectangle OABC has its vertex O at the origin. Side OC is on the positive x-axis, and side OA is on the positive y-axis, with OA = 3 and AB = 4. The graph of the inverse proportion function $y=\frac{k}{x}$ (where $k>0$) intersects the sides AB and BC of the rectangle at points D and E, respectively, and BD = 2AD. Connect OD, OE, and DE. Find the area of $\triangle DOE$.

Solution:
\textbf{Solution:} From the given problem, we have \\
$\because AD=\frac{4}{3}$, $OA=3$, \\
$\therefore S_{\triangle AOD}=\frac{1}{2}AD\cdot AO=\frac{1}{2}\times \frac{4}{3}\times 3=2$. \\
$\because k=4$, \\
$\therefore$ the inverse proportion function can be expressed as $y=\frac{4}{x}$. \\
$\because$ in rectangle ABCD, $BC=OA=3$, $OC=AB=4$, \\
also, $\because$ point E lies on the graph of the inverse proportion function $y=\frac{4}{x}$, \\
$\therefore E(4, 1)$. \\
$\therefore CE=1$, \\
$\therefore BE=2$, \\
$\therefore S_{\triangle OCE}=\frac{1}{2}OC\cdot CE=2$. \\
$\because AB=4$, $AD=\frac{4}{3}$. \\
$\therefore BD=\frac{8}{3}$, \\
$\therefore S_{\triangle BDE}=\frac{1}{2}BD\cdot BE=\frac{8}{3}$, \\
$\because S_{\text{rectangle} AOCB}=AO\cdot OC=3\times 4=12$, \\
$\therefore S_{\triangle DOE}=S_{\text{rectangle} AOCB}-S_{\triangle AOD}-S_{\triangle OCE}-S_{\triangle BDE}=12-2-2-\frac{8}{3}=\boxed{\frac{16}{3}}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertices of rectangle $OABC$ are such that vertex $O$ is at the origin, side $OC$ is on the positive half-axis of the $x$-axis, and side $OA$ is on the positive half-axis of the $y$-axis, with $OA=3$, $AB=4$. The graph of the inverse proportion function $y=\frac{k}{x}$ (where $k>0$) intersects the sides $AB$ and $BC$ of the rectangle at points $D$ and $E$, respectively, and it is given that $BD=2AD$. For a point $P$ on the segment $OC$, does there exist a point $P$ such that $\angle APE=90^\circ$? If such a point exists, find the coordinates of point $P$; if it does not exist, please explain why.

Solution:
\textbf{Solution:} Exist. Assume there exists the required point $P$ with coordinates $(m,0)$,\\
$\therefore OP=m$, $CP=4-m$.\\
$\because \angle APE=90^\circ$,\\
$\therefore \angle APO+\angle EPC=90^\circ$,\\
Also $\because \angle APO+\angle OAP=90^\circ$,\\
$\therefore \angle EPC=\angle OAP$,\\
Also $\because \angle AOP=\angle PCE=90^\circ$,\\
$\therefore \triangle AOP\sim \triangle PCE$,\\
$\therefore \frac{OA}{CP}=\frac{OP}{CE}$,\\
$\because OA\cdot CE=OP\cdot CP$.\\
$\therefore m(4-m)=3$,\\
Solving: $m=1$ or $m=3$.\\
$\therefore$ There exists the required point $P$, with coordinates $\boxed{(1, 0)}$ or $\boxed{(3, 0)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=-x^2+bx+c$ passes through points $A(-3,0)$ and $B(1,0)$ and intersects the $y$-axis at point $C$. Line segment $BC$ is drawn, and point $P$ is a moving point on the parabola, which is above the $x$-axis. A perpendicular line $PE$ is drawn from $P$ to the $x$-axis, with $E$ being the foot of the perpendicular. What is the equation of the parabola?

Solution:
\textbf{Solution:} By substituting A ($-3$, $0$) and B ($1$, $0$) into $y=-x^{2}+bx+c$, we get:
\[
\left\{
\begin{array}{l}
0=-9+3b+c \\
0=-1+b+c
\end{array}
\right.
\]
Solving this, we find:
\[
\left\{
\begin{array}{l}
b=-2 \\
c=3
\end{array}
\right.
\]
$\therefore$ The equation of the parabola is $y=-x^{2}-2x+3=\boxed{-x^{2}-2x+3}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the parabola $y=-x^2+bx+c$ passes through points $A(-3,0)$ and $B(1,0)$, and intersects the $y$-axis at point $C$. Line segment $BC$ is drawn, and point $P$ is a moving point on the parabola above the $x$-axis. A perpendicular line from $P$ to the $x$-axis is drawn, with the foot of the perpendicular being point $E$. Is there a point $P$ such that the triangle formed by $A$, $P$, and $E$ is similar to $\triangle BOC$? If such a point exists, find the coordinates of point $P$ and explain the reasoning.

Solution:
\textbf{Solution:} Let $A(-3,0)$, $B(1,0)$, $C(0,3)$,\\
therefore $OC=3$, $OB=1$,\\
let $P(m,-m^2-2m+3)$,\\
thus $PE=-m^2-2m+3$, $AE=m+3$,\\
Based on the given conditions, we have $\frac{m+3}{-m^2-2m+3}=\frac{1}{3}$,\\
Solving this, we find $m_1=-2$, $m_2=-3$ (discard),\\
thus $-m^2-2m+3=-(-2)^2-2\times(-2)+3=3$,\\
thus $P_1(-2,3)$,\\
or $\frac{m+3}{-m^2-2m+3}=\frac{3}{1}$,\\
Solving this, we find $m_1=\frac{2}{3}$, $m_2=-3$ (discard),\\
thus $-m^2-2m+3=-\left(\frac{2}{3}\right)^2-2\times\frac{2}{3}+3=\frac{11}{9}$,\\
thus $P_2\left(\frac{2}{3},\frac{11}{9}\right)$,\\
In conclusion, the coordinates of point $P$ are $\boxed{P_1(-2,3)}$ or $\boxed{P_2\left(\frac{2}{3},\frac{11}{9}\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=-x^{2}+bx+c$ passes through points $A(-3,0)$ and $B(1,0)$ and intersects the $y$-axis at point $C$. The segment $BC$ is drawn. Point $P$ is a moving point on the parabola above the $x$-axis, and a line through $P$ is drawn perpendicular to the $x$-axis, with the foot of the perpendicular being point $E$. Is there a point $P$ that maximizes the area of quadrilateral $ABCP$? If such a point exists, find the coordinates of point $P$ and provide the reasoning.

Solution:
\textbf{Solution:} To solve, we connect $AC$ and intersect it with $PE$ at point $H$,\\
from point $A(-3,0)$ and point $C(0,3)$, we derive the equation of line $AC$ as: $y=x+3$,\\
let $P(m,-m^2-2m+3)$, then $H(m,m+3)$,\\
thus, $PH=-m^2-3m$,\\
therefore, the area of $\triangle PAC$ is $\frac{1}{2}\cdot(-m^2-3m)\times3$,\\
hence, the area of quadrilateral $ABCP$ is $S_{\triangle PAC}+S_{\triangle ABC}=-\frac{3}{2}m^2-\frac{9}{2}m+6=-\frac{3}{2}\left(m+\frac{3}{2}\right)^2+\frac{75}{8}$,\\
when $m=-\frac{3}{2}$, the maximum area $S_{\text{max}}=\boxed{\frac{75}{8}}$, at this moment, the coordinate of point $P$ is $\boxed{\left(-\frac{3}{2},\frac{15}{4}\right)}$.


Question: As shown in the figure, the graph of the direct proportion function $y_{1}=kx$ intersects with the graph of the inverse proportion function $y_{2}=\frac{m}{x}$ at points A$(2,4)$ and point B. The coordinates of point C are $(4,0)$, and point D lies on the graph of the inverse proportion function $y_{2}=\frac{m}{x}$. What is the expression of the inverse proportion function?

Solution:
\textbf{Solution:} Since the graph of the inverse proportion function $y=\frac{m}{x}$ passes through point A$(2,4)$,\\
then $m=2\times 4=8$,\\
thus, the expression of the inverse proportion function is $\boxed{y=\frac{8}{x}}$.


Question: As shown in the figure, the graph of the direct proportion function $y_1=kx$ intersects with the graph of the inverse proportion function $y_2=\frac{m}{x}$ at points $A(2,4)$ and point $B$. The coordinates of point $C$ are $(4,0)$, and point $D$ is on the graph of $y_2=\frac{m}{x}$. Suppose point $E$ is on the x-axis, and $\angle AEB = 90^\circ$, find the coordinates of point $E$.

Solution:
\textbf{Solution:} As shown in Figure 1, draw $AG\perp x$-axis at $G$, and $BH\perp x$-axis at $H$,\\
since the graph of the direct proportion function $y_{1}=kx$ intersects with the graph of the inverse proportion function $y_{2}=\frac{8}{x}$ at points $A(2,4)$ and $B$,\\
therefore $B(-2,-4)$,\\
hence $OG=OH=2$, $AG=BH=4$,\\
let the coordinate of $E$ be $(x,0)$ (where $|x|>2$), then $EG=|2-x|$, $EH=|x+2|$,\\
since $\angle AEG+\angle BEH=\angle AEB=90^\circ$, $\angle BEH+\angle EBH=90^\circ$,\\
therefore $\angle AEG=\angle EBH$,\\
since $\angle AGE=\angle EHB=90^\circ$,\\
therefore $\triangle EBH\sim \triangle AEG$,\\
therefore $\frac{EH}{AG}=\frac{BH}{EG}$, that is $\frac{|x+2|}{4}=\frac{4}{|x-2|}$,\\
rewriting, we get $x^2-4=16$,\\
solving, $x=\pm 2\sqrt{5}$,\\
therefore the coordinates of point $E$ are $\boxed{(2\sqrt{5},0)}$ or $\boxed{(-2\sqrt{5},0)}$;


Question: As shown in the figure, the graph of the direct proportion function $y_1=kx$ intersects with the graph of the inverse proportion function $y_2=\frac{m}{x}$ at points $A(2,4)$ and point $B$. The coordinate of point $C$ is $(4,0)$, and point $D$ lies on the graph of $y_2=\frac{m}{x}$. Let point $M$ be on the $x$-axis, and point $N$ be within the Cartesian coordinate system. When the quadrilateral $CDNM$ is a square, directly write down the coordinates of point $M$.

Solution: \textbf{Solution:} When the quadrilateral $CDNM$ is a square, the coordinates of the point $M$ are $\boxed{(2,0)}$ or $\boxed{(6,0)}$.


Question: "The Joy of Winter Olympics: Mingxi Enthusiastically Welcomes the Basketball Match" has begun with athletes attacking and defending, every second counts. In this event, the basketball player Xiaoming performs a long throw from position O. The ball is thrown from point A, which is 1 meter above the ground. It reaches its highest point at B, 6 meters away from O, with the highest point C being 4 meters above the ground. The ball bounces again and lands at point E, with the distance between E and F being 2 meters. According to experiments, the trajectory of the basketball's second bounce on the court has the same shape as the first but its maximum height is reduced to half of the original. Taking Xiaoming's standing position O as the origin, a Cartesian coordinate system is established as shown in the figure (all calculated results are to be rounded to the nearest integer, $ \sqrt{3} \approx 1.75$; $ \sqrt{6} \approx 2.5$). What is the equation of the parabola ACD?

Solution:
\textbf{Solution:} Let us assume that the equation of the parabola describing the basketball's trajectory from its initial launch to its first landing is $y=a(x-h)^2+k$,\\
where $h=6$, $k=4$,\\
thus $y=a(x-6)^2+4$.\\
From the given information, when $x=0$, $y=1$,\\
which implies $1=36a+4$,\\
therefore $a=-\frac{1}{12}$,\\
hence the equation is $\boxed{y=-\frac{1}{12}(x-6)^2+4}$.


Question: As shown in the figure, the basketball player Xiao Ming throws the ball from point O. The ball is thrown from point A, which is 1 meter above the ground. The ball reaches its highest point, B, 6 meters away from point O, with the highest point C being 4 meters above the ground. It bounces again and lands at point E, with the distance between E and F being 2 meters. According to experiments, the parabola described by the basketball on its second bounce on the court has the same shape as the first, but the maximum height is reduced to half of the original maximum height. Taking the standing position of Xiao Ming at point O as the origin, a Cartesian coordinate system is established as shown in the figure. (All calculated results are to be rounded to the nearest integer, with $ \sqrt{3}\approx1.75$ and $ \sqrt{6}\approx2.5$). What is the distance from point E, where the basketball lands for the second time, to point O?

Solution:
\textbf{Solution:} Let $y=0$, $-\frac{1}{12}(x-6)^2+4=0$,\\
$\therefore (x-6)^2=48$,\\
We find: $x_1=4\sqrt{3}+6\approx 13$, $x_2=-4\sqrt{3}+6$,\\
$\therefore$ the coordinates of point D are $(13,0)$.\\
$\therefore OD=13\text{m}$\\
As shown in the diagram, draw $MG\parallel x$ axis from the vertex M of the parabola DE, intersecting the parabola ACD at points G and H.\\
$\because$ The trajectory of the basketball after the second bounce on the court forms a parabola identical in shape to the first, but its maximum height is reduced to half of the first\\
$\therefore$ The parabola DME is equivalent to the parabola ACD moved two units downward and then moved to the right.\\
$\therefore GH=DE$, the $y$-coordinates of points G, H, and M are 2.\\
When $y=2$, $-\frac{1}{12}(x-6)^2+4=2$\\
$x_1=6-2\sqrt{6}$, $x_2=6+2\sqrt{6}$\\
$\therefore DE=GH=x_2-x_1=4\sqrt{6}\approx 10$\\
$OE=OD+DE$\\
$=13+10$\\
$=\boxed{23\text{m}}$\\
$\therefore$ The distance from point E, the basketball's second landing point, to point O is $23\text{m}$.


Question: As shown in the figure, Xiaoming stands at point $O$ and throws a ball. The ball is thrown from point $A$, which is $1$m above the ground. The basketball reaches its highest point at point $B$, $6$m away from point $O$. The highest point, $C$, is $4$m above the ground. Then it bounces again and falls at point $E$. The distance between $E$ and $F$ is $2$m. According to experiments, the parabola described by the basketball on its second bounce on the court is the same shape as the first parabola, but its maximum height is reduced to half of the original maximum height. Taking the position where Xiaoming stands, $O$, as the origin, a Cartesian coordinate system is established. (All calculated results are rounded to the nearest integer, $\sqrt{3}\approx1.75$; $\sqrt{6}\approx2.5$) If Xiaoming wants to make a basket on his first throw, how far should he walk forward?

Solution:
\textbf{Solution:} As illustrated in the figure, when $y=3$, $-\frac{1}{12}(x-6)^2+4=3$,\\
we find: $x_1=-2\sqrt{3}+6 \approx 3$; $x_2=2\sqrt{3}+6 \approx 9$\\
Thus, the point $OR=9\,\text{m}$,\\
since $EF=2\,\text{m}$,\\
therefore $OF=OE+EF=23+2=25\,\text{m}$\\
If Xiaoming wants to make the basket on his first throw, the distance he should walk forward is:\\
$RF=OF-OR=25-9=\boxed{16}\,\text{m}$


Question: As shown in the figure, the parabola $y=ax^2+bx-3$ passes through three points $A$, $B$, and $C$, where point $A(-3,0)$, point $C(1,0)$, and point $B$ is on the $y$-axis. Point $P$ is a moving point on the parabola below the line $AB$ (not coinciding with $A$ or $B$). What is the analytical expression of this parabola?

Solution:
\textbf{Solution:} Let's substitute points A $(-3,0)$ and C $(1,0)$ into $y=ax^{2}+bx-3$, to get
\[
\left\{
\begin{array}{l}
0=9a-3b-3 \\
0=a+b-3
\end{array}
\right.
\]
Solving this system, we find
\[
\left\{
\begin{array}{l}
a=1 \\
b=2
\end{array}
\right.
\]
$\therefore$ The equation of the parabola is $\boxed{y=x^{2}+2x-3}$.


Question: As shown in the figure, the parabola $y=ax^2+bx-3$ passes through three points $A$, $B$, and $C$, with points $A(-3,0)$, $C(1,0)$, and point $B$ on the $y$-axis. Point $P$ is a moving point on the parabola below line $AB$ (not coinciding with $A$, $B$). A perpendicular line is drawn from point $P$ to the $x$-axis, the foot of the perpendicular is $D$, and it intersects line $AB$ at point $E$. When is $PE$ maximum, and what is the coordinate of point $P$ at this time?

Solution:
\textbf{Solution:} Let $P(x, x^2+2x-3)$, and the equation of line AB is $y=kx+b$,\\
From the parabolic equation $y=x^2+2x-3$,\\
Let $x=0$, then $y=-3$,\\
$\therefore B(0, -3)$,\\
Substituting $A(-3, 0)$ and $B(0, -3)$ into $y=kx+b$, we get
\[
\left\{
\begin{array}{l}
0=-3k+b \\
-3=b
\end{array}
\right.
\]
Solving, we obtain
\[
\left\{
\begin{array}{l}
k=-1 \\
b=-3
\end{array}
\right.
\]
$\therefore$ The equation of line AB is $y=-x-3$,\\
$\because PE\perp x$-axis,\\
$\therefore E(x, -x-3)$,\\
$\because P$ is below the line AB,\\
$\therefore PE=-x-3-(x^2+2x-3)=-x^2-3x=-\left(x+\frac{3}{2}\right)^2+\frac{9}{4}$,\\
When $x=-\frac{3}{2}$, $y=x^2+2x-3=-\frac{15}{4}$,\\
$\therefore$ When $PE$ is maximum, the coordinates of point $P$ are $\boxed{\left(-\frac{3}{2}, -\frac{15}{4}\right)}$.


Question: As shown in the figure, the parabola $y=ax^2+bx-3$ passes through points $A$, $B$, and $C$, where $A(-3,0)$, $C(1,0)$, and point $B$ is on the $y$-axis. Point $P$ is a moving point on the parabola below the line $AB$ (not coinciding with $A$ or $B$). Point $Q$ is a moving point on the axis of symmetry of the parabola. Is there a point $Q$ for which the triangle with vertices $A$, $B$, and $Q$ is a right-angled triangle? If such a point exists, find the coordinates of point $Q$; if it does not exist, please explain why.

Solution:
\textbf{Solution:} The point $Q$ exists, and the reasons are as follows,\\
$\because x=-\frac{2}{2\times 1}=-1$,\\
$\therefore$ the axis of symmetry of the parabola is the line $x=-1$,\\
Let $Q(-1,a)$,\\
$\because B(0,-3), A(-3,0)$,\\
(1) When $\angle QAB=90^\circ$, it holds that $AQ^2+AB^2=BQ^2$,\\
$\therefore 2^2+a^2+3^2+3^2=1^2+(3+a)^2$,\\
Solving this, we find: $a=2$,\\
$\therefore Q_1(-1,2)$,\\
(2) When $\angle QBA=90^\circ$, it holds that $BQ^2+AB^2=AQ^2$,\\
$\therefore 1^2+(3+a)^2+3^2+3^2=2^2+a^2$,\\
Solving this, we find: $a=-4$,\\
$\therefore Q_2(-1,-4)$,\\
(3) When $\angle AQB=90^\circ$, it holds that $BQ^2+AQ^2=AB^2$,\\
$\therefore 1^2+(3+a)^2+2^2+a^2=3^2+3^2$,\\
Solving this, we find: $a_1=\frac{-3+\sqrt{17}}{2}$ or $a_1=\frac{-3-\sqrt{17}}{2}$,\\
$\therefore Q_3(-1,\frac{-3+\sqrt{17}}{2})$, $Q_4(-1,\frac{-3-\sqrt{17}}{2})$,\\
To conclude: the coordinates of point $Q$ are $\boxed{(-1,2)}$, $\boxed{(-1,-4)}$, $\boxed{(-1,\frac{-3+\sqrt{17}}{2})}$, or $\boxed{(-1,\frac{-3-\sqrt{17}}{2})}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y = -2x$ intersects with the graph of the inverse proportion function $y = \frac{k}{x}$ at a point $A(-1, n)$. What is the equation of the inverse proportion function $y = \frac{k}{x}$?

Solution:
\textbf{Solution:} Since point A $(-1, 2)$ lies on the graph of the linear function $y=-2x$,\\
$\therefore n=-2\times(-1)=2$\\
$\therefore$ The coordinates of point A are $(-1, 2)$\\
Since point A lies on the graph of the inverse proportion function,\\
$\therefore k=-2$\\
$\therefore$ The equation of the inverse proportion function is $\boxed{y=-\frac{2}{x}}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y=-2x$ intersects with the graph of the inverse proportion function $y=\frac{k}{x}$ at point $A(-1, n)$. If $P$ is a point on the coordinate axis and satisfies $PA=OA$, write down the coordinates of point $P$ directly.

Solution: \textbf{Solution:} The coordinates of point $P$ are \boxed{(-2, 0)} or \boxed{(0, 4)}.


Question: As shown in the diagram, the line $y=kx+b$ and the inverse proportion function $y=\frac{8}{x}$ (where $x>0$) intersect at points $A(m, 4)$ and $B(8, n)$, and intersect the coordinate axes at points $C$ and $D$ respectively. What is the analytical expression of line $AB$?

Solution:
\textbf{Solution:} From the given information, point $A\left(2, 4\right)$ and point $B\left(8, 1\right)$ lie on the graph of $y=\frac{8}{x}$,\\
$\therefore m=\frac{8}{4}=2$, $n=\frac{8}{8}=1$,\\
that is, $A\left(2, 4\right)$, $B\left(8, 1\right)$.\\
Substituting points $A\left(2, 4\right)$ and $B\left(8, 1\right)$ into $y=kx+b$ yields\\
$\left\{\begin{array}{l}
4=2k+b \\
1=8k+b
\end{array}\right.$, solving for which we get: $\left\{\begin{array}{l}
k=-\frac{1}{2} \\
b=5
\end{array}\right.$,\\
Therefore, the equation of line $AB$ is: $y=-\frac{1}{2}x+5=\boxed{-\frac{1}{2}x+5}$.


Question: The line \(y=kx+b\) and the inverse proportion function \(y=\frac{8}{x}\) (\(x>0\)) intersect at points \(A(m, 4)\) and \(B(8, n)\), respectively, and intersect the coordinate axes at points \(C\) and \(D\). By observing the graph, when \(x>0\), directly write the solution set of \(kx+b>\frac{8}{x}\).

Solution: \textbf{Solution:} When $x>0$, the solution set for $kx+b>\frac{8}{x}$ is $x\in(2,8)$. Therefore, the answer is $x\in\boxed{(2,8)}$.


Question: The straight line $y=kx+b$ and the inverse proportion function $y=\frac{8}{x}$ $(x>0)$ intersect at points $A(m, 4)$ and $B(8, n)$, respectively, and intersect the coordinate axes at points $C$ and $D$, respectively. If $P$ is a moving point on the $x$-axis, and the area of $\triangle ADP$ is $6$, find the coordinates of point $P$.

Solution:
\[ \text{Solution: From (1), the equation of line AB is } y=-\frac{1}{2}x+5, \]
\[ \text{when } y=0, \text{ we have } x=10, \]
\[ \therefore \text{ the coordinates of point } D \text{ are } (10, 0) \]
\[ \text{Let the coordinates of point } P \text{ be } (a, 0), \text{ then } PD=|10-a| \]
\[ \because \text{ the area of } \triangle ADP \text{ is } 6 \]
\[ \therefore \frac{1}{2} \times 4 \times PD = 6 \]
\[ \therefore PD = 3 \]
\[ \therefore |10-a| = 3 \]
\[ \text{Solving this gives } a=7 \text{ or } 13 \]
\[ \therefore \text{The coordinates of } P \text{ are } (\boxed{7}, 0) \text{ or } \boxed{(13, 0)} \]


Question: As shown in the figure, in the Cartesian coordinate system, point O is the origin, and the coordinates of vertex A of the rhombus OABC are (3, 4). Find the equation of the inverse proportion function $y = \frac{k}{x}$ that passes through point B.

Solution:
\textbf{Solution:} Construct AE$\perp$x-axis through point A, and BF$\perp$x-axis through point B, where E and F are the feet of the perpendiculars, as shown in the diagram,\\
since A(3,4),\\
thus OE=3, AE=4,\\
therefore AO$= \sqrt{OE^{2}+AE^{2}}$=5,\\
since quadrilateral OABC is a rhombus,\\
therefore AO=AB=OC=5, AB$\parallel$x-axis,\\
therefore EF=AB=5,\\
therefore OF=OE+EF=3+5=8,\\
therefore B(8,4),\\
since the inverse proportion function passing through point B is $y=\frac{k}{x}$,\\
substituting the coordinates of point B gives $k=32$,\\
therefore, the expression for the inverse proportion function is $\boxed{y=\frac{32}{x}}$;


Question: As shown in the figure, in the Cartesian coordinate system, point O is the origin, and the vertex A of the rhombus OABC has coordinates (3, 4). Connect OB and AC to intersect at point H. Draw line $BD \parallel AC$ passing through point B to intersect the x-axis at point D. Find the equation of line BD.

Solution:
\textbf{Solution:} Since quadrilateral OABC is a rhombus, \\
it follows that OB$\perp$AC, \\
and since BD$\parallel$AC, \\
it also follows that OB$\perp$BD, that is $\angle$OBD$=90^\circ$, \\
therefore, $\angle$OBF+$\angle$DBF$=90^\circ$, \\
and since $\angle$DBF+$\angle$BDF$=90^\circ$, \\
it follows that $\angle$OBF$=\angle$BDF, \\
furthermore, since $\angle$OFB$=\angle$BFD$=90^\circ$, \\
it follows that $\triangle$OBF$\sim$$\triangle$BDF, \\
thus, $\frac{OF}{BF}=\frac{BF}{DF}$, \\
leading to $\frac{8}{4}=\frac{4}{DF}$, \\
from which we solve that DF$=2$, \\
therefore, OD$=$OF$+$DF$=8+2=10$, \\
thus, D(10,0). \\
Suppose the equation of the line on which BD lies is $y=k_{1}x+b$, \\
substituting B(8,4), D(10,0) into the equation gives $\begin{cases}8k_{1}+b=4\\ 10k_{1}+b=0\end{cases}$, \\
solving this yields $\begin{cases}k_{1}=-2\\ b=20\end{cases}$. \\
Therefore, the equation of line BD is $\boxed{y=-2x+20}$.


Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ intersects the $x$ axis and the $y$ axis at points $A$ and $B$ respectively, and intersects the graph of the inverse proportion function $y= \frac{m}{x}$ in the first quadrant at point $C$, with $CD$ perpendicular to the $x$ axis, and the foot of the perpendicular being $D$. If $OA=OB=OD=1$, find: the coordinates of points $A$, $B$, and $C$?

Solution:
\textbf{Solution:} Since $OA=OB=OD=1$, \\
it follows that $A(-1,0)$, $B(0,1)$, substituting into the linear equation of line AB $y=kx+b$, \\
we obtain $\begin{cases}0=-k+b \\ 1=0+b\end{cases}$, solving this gives $\begin{cases}k=1 \\ b=1\end{cases}$, \\
hence, the equation of line AB is $y=x+1$. Substituting $x=1$ into $y=x+1$ yields $y=2$, \\
thus $C(1,2)$.\\
Therefore, the coordinates of points $A$, $B$, and $C$ are $\boxed{A(-1,0)}$, $\boxed{B(0,1)}$, and $\boxed{C(1,2)}$ respectively.