Question: As shown in the figure, the line $y=kx+b$ intersects the $x$-axis at point $B$ and the $y$-axis at point $C(0,3)$, and it intersects $y=2x$ at point $A(a,2a)$. Point $M$ lies on the line $OA$. Is there a point $M$ such that the area of $\triangle OMC$ is equal to the area of $\triangle OAB$? Find the coordinates of point $M$.

Solution:
\textbf{Solution:} Let there be a point $M$ such that the area of $\triangle OMC$ is equal to the area of $\triangle OAB$, for the following reasons:\\
$\because$ Point $C(0,3)$,\\
$\therefore OC=3$, $\therefore OB=OC=3$.\\
$\because$ The area of $\triangle OMC$ is equal to the area of $\triangle OAB$,\\
$\therefore$ The distance from $M$ to the $y$-axis is equal to the $y$-coordinate of point $A$, which is $2$,\\
$\therefore$ The $x$-coordinate of point $M$ is $2$ or $-2$;\\
When the $x$-coordinate of $M$ is $2$, in $y=2x$,\\
when $x=2$, $y=4$, then the coordinates of $M$ are $(2,4)$;\\
When the $x$-coordinate of $M$ is $-2$, in $y=2x$,\\
when $x=-2$, $y=-4$, then the coordinates of $M$ are $(-2,-4)$.\\
In summary: The coordinates of point $M$ are either $\boxed{(2,4)}$ or $\boxed{(-2,-4)}$.


Question: As shown in the figure, it is known that the line $y=kx+5$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$, and the coordinates of point $A$ are $(5,0)$. The line $y=2x-4$ intersects the $x$-axis at point $D$ and intersects the line $AB$ at point $C$. What are the coordinates of point $C$?

Solution:
\textbf{Solution:} Given that the line $y=kx+5$ passes through point $A(5,0)$,\\
thus $5k+5=0$ and we find $k=-1$.\\
Therefore, the equation of line $AB$ is: $y=-x+5$;\\
By solving the system of equations for lines $AB$ and $CD$ simultaneously $\left\{\begin{array}{l}y=-x+5 \\ y=2x-4\end{array}\right.$, we find: $\left\{\begin{array}{l}x=3 \\ y=2\end{array}\right.$,\\
hence, the coordinates of point $C$ are $\boxed{(3,2)}$.


Question: As shown in the figure, it is known that the line $y=kx+5$ intersects the $x$ axis at point $A$ and the $y$ axis at point $B$, and the coordinate of point $A$ is $(5,0)$. The line $y=2x-4$ intersects the $x$ axis at point $D$, and intersects the line $AB$ at point $C$. Based on the graph, what is the solution set of the inequality $2x-4>kx+5$ with respect to $x$?

Solution:
\textbf{Solution:} Observing the graph of the function, we see that when $x>3$, the line $y=2x-4$ is above the line $y=kx+5$. \\
$\therefore$ The solution set of the inequality $2x-4>kx+5$ is $\boxed{x>3}$.


Question: As shown in the figure, it is known that the line $y=kx+5$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$, with the coordinates of point $A$ being $(5,0)$. The line $y=2x-4$ intersects the $x$-axis at point $D$ and intersects line $AB$ at point $C$. What is the area of $\triangle ADC$?

Solution: \textbf{Solution:} Substitute $y=0$ into $y=2x-4$ to obtain $2x-4=0$, solving this yields $x=2$, $\therefore$ the coordinates of point D are $(2,0)$, $\because$ A$(5,0)$, $\therefore$ $AD=3$, $\because$ the coordinates of point C are $C(3,2)$, $\therefore$ $S_{\triangle ADC} = \frac{1}{2} \cdot 3 \cdot 2 = \boxed{3}$.


Question: As shown in the figure, the graph of the function $y=-2x+3$ intersects with the graph of the function $y=-\frac{1}{2}x+m$ at point $P(n, -2)$. The line $y=-2x+3$ intersects with the y-axis at point A, and the line $y=-\frac{1}{2}x+m$ intersects with the y-axis at point B. Determine the values of $m$ and $n$.

Solution:
\textbf{Solution:} Since the graph of the function $y=-2x+3$ passes through the point $P(n, -2)$,\\
it follows that $-2=-2n+3$. Solving this, we find $n=\boxed{\frac{5}{2}}$.\\
Thus, the coordinates of point $P$ are $\left(\frac{5}{2}, -2\right)$.\\
Since the graph of the function $y=-\frac{1}{2}x+m$ passes through point $P\left(\frac{5}{2}, -2\right)$,\\
it follows that $-2=-\frac{1}{2}\times \frac{5}{2}+m$. Solving this, we find $m=\boxed{-\frac{3}{4}}$;


Question: As shown in the figure, the graph of the function $y=-2x+3$ intersects with the graph of $y=-\frac{1}{2}x+m$ at point $P(n, -2)$. The line $y=-2x+3$ intersects with the y-axis at point A, and the line $y=-\frac{1}{2}x+m$ intersects with the y-axis at point B. Directly write out the solution set for the inequality $-\frac{1}{2}x+m > -2x+3$.

Solution: \textbf{Solution:} The solution set of the inequality $-\frac{1}{2}x+m>-2x+3$ is $\boxed{x>\frac{5}{2}}$.


Question: As shown in the figure, the graph of the function $y=-2x+3$ intersects with the graph of the function $y=-\frac{1}{2}x+m$ at point $P(n, -2)$. The line $y=-2x+3$ intersects the y-axis at point A, and the line $y=-\frac{1}{2}x+m$ intersects the y-axis at point B. Calculate the area of $\triangle ABP$.

Solution: \textbf{Solution:} Given that in the equation $y=-2x+3$, when $x=0$, $y=3$, thus point $A(0, 3)$ is found. Also, in the equation $y=-\frac{1}{2}x-\frac{3}{4}$, when $x=0$, $y=-\frac{3}{4}$, hence point $B\left(0, -\frac{3}{4}\right)$ is determined. Consequently, the length of $AB=\frac{15}{4}$. Therefore, the area of the triangle $\triangle ABP$ is calculated as $\frac{1}{2}\times \frac{15}{4}\times \frac{5}{2}=\boxed{\frac{75}{16}}$.


Question: As shown in the figure, the line $l_1: y = -x - b$ intersects the $x$ and $y$ axes at points $A(6, 0)$ and $B$ respectively. The line $l_2$ passing through point $B$ intersects the negative $x$ axis at point $C$, and $OB:OC = 3:1$. Find the coordinates of points $B$ and $C$, and the function expression of line $BC$.

Solution: \textbf{Solution:} Substitute point $A(6,0)$ into the equation of line $AB$ to get: $0=-6-b$, solving this gives: $b=-6$, $\therefore$ the equation of line $AB$ is $y=-x+6$, $\therefore$ the coordinates of point $B$ are: $\boxed{(0,6)}$. Since $OB:OC=3:1$, $\therefore$ $OC=2$, $\therefore$ the coordinates of point $C$ are $\boxed{(-2,0)}$. Let the equation of line $BC$ be $y=kx+6$, $0=-2k+6$, solving this gives: $k=3$, $\therefore$ the equation of line $BC$ is: $\boxed{y=3x+6}$;


Question: As shown in the figure, the line $l_{1}\colon y=-x-b$ intersects the $x$ and $y$ axes at points $A(6, 0)$ and $B$, respectively. The line $l_{2}$ passing through point $B$ intersects the negative semi-axis of the $x$ axis at point $C$, with $OB:OC=3:1$. Find the value of $S_{\triangle AOB}-S_{\triangle BOC}$.

Solution: \textbf{Solution:} Given by the problem, we have $S_{\triangle AOB}-S_{\triangle BOC}=\frac{1}{2}\cdot OA\cdot OB-\frac{1}{2}\cdot OC\cdot OB=\frac{1}{2}\times 6\times 6-\frac{1}{2}\times 6\times 2=\boxed{12}$.


Question: As shown in the figure, the line $l_{1}\colon y=-x-b$ intersects the $x$-axis at $A(6, 0)$ and the $y$-axis at point $B$. Another line $l_{2}$ passing through point $B$ intersects the negative semi-axis of the $x$-axis at point $C$, and $OB:OC = 3:1$. If the point $P(m+1, m-1)$ is inside $\triangle ABC$ (including the boundary), what is the range of $m$?

Solution:
\textbf{Solution:} Therefore, the range of values for $m$ is $1 \leq m \leq 3$,\\
$\therefore$ the range of values for $m$ is $\boxed{1 \leq m \leq 3}$.


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ passes through point $A(-2, 6)$ and intersects with the $x$-axis at point $B$, and intersects with the graph of the direct proportion function $y=3x$ at point $C$, where the $x$-coordinate of point $C$ is 1. What is the analytic expression of the linear function?

Solution:
\textbf{Solution:} Given that the x-coordinate of point C is 1,\\
$\therefore C(1, 3)$,\\
substitute points $A(-2, 6)$ and $C(1, 3)$ into $y=kx+b$,\\
$\therefore \left\{\begin{array}{l}k+b=3 \\ -2k+b=6\end{array}\right.$,\\
solving this yields $\left\{\begin{array}{l}k=-1 \\ b=4\end{array}\right.$,\\
$\therefore \boxed{y=-x+4}$;


Question:
Integrated Exploration

As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ passes through point $A(-2,6)$, intersects with the $x$-axis at point $B$, and intersects with the graph of the direct proportion function $y=3x$ at point $C$, where the $x$-coordinate of $C$ is $1$. If point $D$ is on the $y$-axis and satisfies $S_{\triangle COD} = \frac{1}{3} S_{\triangle BOC}$, what are the coordinates of point $D$?

Solution:
\textbf{Solution:} From the problem statement, we have $y=-x+4$,\\
thus $B(4, 0)$,\\
therefore $OB=4$,\\
thus $\frac{1}{3}S_{\triangle BOC}=\frac{1}{3}\times \frac{1}{2}\times 4\times 3=2$,\\
since $S_{\triangle COD}=\frac{1}{3}S_{\triangle BOC}$,\\
thus $S_{\triangle COD}=2=\frac{1}{2}\times 1\times |OD|$,\\
therefore $OD=4$,\\
hence the coordinates of $D$ are $\boxed{(0, 4)}$ or $\boxed{(0, -4)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ passes through point $A(-2,6)$, and intersects with the x-axis at point $B$, and intersects with the graph of the direct proportion function $y=3x$ at point $C$, where the x-coordinate of point $C$ is $1$. In the coordinate plane, is there a point $P$ such that the quadrilateral with vertices $O$, $B$, $C$, and $P$ is a parallelogram? If it exists, directly write down the coordinates of point $P$.

Solution: \textbf{Solution:} Yes, there exists such a point $P$, and the coordinates of point $P$ are $\boxed{(-3, 3)}$ or $\boxed{(3, -3)}$ or $(5, 3)$.


Question: As shown in the figure, it is known that the graphs of the functions $y_1=2x+b$ and $y_2=ax-3$ intersect at the point $P(-2, -5)$. The graphs of these two functions intersect the x-axis at points A and B, respectively. Determine the analytic expressions of these two functions.

Solution:
\textbf{Solution:} Since substituting point $P(-2, -5)$ into $y_1=2x+b$ gives $-5=2\times (-2)+b$, solving this equation yields $b=-1$.\\
Substituting point $P(-2, -5)$ into $y_2=ax-3$ gives $-5=a\times (-2)-3$, solving this equation yields $a=1$.\\
Therefore, the equations of the two functions are $\boxed{y_1=2x-1}$ and $\boxed{y_2=x-3}$.


Question: As shown in the figure, it is known that the graphs of the functions $y_{1}=2x+b$ and $y_{2}=ax-3$ intersect at point $P(-2, -5)$, while the graphs of these two functions intersect the x-axis at points A and B, respectively. What is the area of $\triangle ABP$?

Solution:
\textbf{Solution:} Since in $y_{1}=2x-1$, by setting $y_{1}=0$, we get $x=\frac{1}{2}$.\\
Hence $A\left(\frac{1}{2}, 0\right)$.\\
Since in $y_{2}=x-3$, by setting $y_{2}=0$, we get $x=3$,\\
Hence $B\left(3, 0\right)$.\\
Therefore, the area $S_{\triangle ABP}=\frac{1}{2}AB\times 5=\frac{1}{2}\times \frac{5}{2}\times 5=\boxed{\frac{25}{4}}$.


Question: As shown in the figure, the graphs of the functions $y_1=2x+b$ and $y_2=ax-3$ intersect at point $P(-2, -5)$. The graphs of these two functions intersect the x-axis at points A and B, respectively. Based on the graph, directly write the range of values for $x$ when $y_1<y_2$.

Solution: \textbf{Solution:} It can be inferred from the graph of the function that, when $x<-2$, \boxed{y_1<y_2}.


Question: As shown in the figure, it is known that the graph of the linear function $y=kx+k+1$ intersects the graph of the linear function $y=-x+4$ at point $A(1, a)$. What are the values of $a$ and $k$?

Solution:
\textbf{Solution:} From the given, we have $a=-1+4=3$,\\
$\therefore A(1,3)$,\\
Substituting the coordinates of point A into the function, we get $3=2k+1$,\\
Solving for $k$ yields $k=\boxed{1}$,\\
$\therefore a=\boxed{3}$, $k=1$.


Question: As shown in the figure, it is known that the graph of the linear function $y=kx+k+1$ intersects with the graph of the linear function $y=-x+4$ at point $A(1, a)$. Based on the graph, write the solution set of the inequality $-x+4>kx+k+1$.

Solution:
\textbf{Solution:} Since the intersection point of the two lines is A$(1,3)$, when $x<1$, the graph of the linear function $y=-x+4$ lies above the graph of the linear function $y=kx+k+1$, \\
thus, the solution set of the inequality $-x+4>kx+k+1$ is $\boxed{x<1}$.


Question: As shown in the figure, it is known that the graph of the linear function $y=kx+k+1$ intersects with the graph of the linear function $y=-x+4$ at point $A(1, a)$. Considering the graph, when $x>2$, find the range of values of $y$ for the linear function $y=-x+4$.

Solution:
\textbf{Solution:} When $x=2$, $y=-x+4=-2+4=2$,\\
since $k=-1<0$,\\
thus, according to this function, as $x$ increases, it decreases,\\
therefore, when $x>2$, $y<2$.\\
Hence, the range of the function value $y$ is $y\in(-\infty, \boxed{2})$.


Question: As shown in the figure, the line $AB\colon y=2x-k$ passes through the point $M\left(k, 2\right)$, and intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. Find the value of $k$.

Solution:
\textbf{Solution:} Since the line $AB: y = 2x - k$ passes through the point $M(k, 2)$,

therefore, $2k - k = 2$. Solving this gives $k = \boxed{2}$.


Question: As shown in the figure, the line $AB: y=2x-k$ passes through point $M(k, 2)$ and intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. The line $AB$ is translated upward by 3 units to obtain line $l$. If $C$ is a point on line $l$, and the area of $\triangle AOC=3$, find the coordinates of point $C$.

Solution: \textbf{Solution:} Since the equation of line AB is $y=2x-2$, when $y=0$, $2x-2=0$. Solving this gives $x=1$, hence point A is $(1,0)$, and thus OA$=1$. Translating line $AB$ upwards by 3 units gives line $l$, with the equation $y=2x+1$. Given that $C$ is a point on line $l$, let $C(m,2m+1)$. Since the area of $\triangle AOC$ is $3$, we have $\frac{1}{2}\times 1\times |2m+1|=3$. Solving this yields $m_1=\frac{5}{2}$ and $m_2=-\frac{7}{2}$. When $m_1=\frac{5}{2}$, $2m+1=6$; and when $m_1=-\frac{7}{2}$, $2m+1=-6$; hence the coordinates of point $C$ are $\boxed{\left(\frac{5}{2}, 6\right)}$ or $\boxed{\left(-\frac{7}{2}, -6\right)}$.


Question: As shown in the figure, the line ${l}_{1}\colon y=x+3$ intersects with the line ${l}_{2}$, which passes through point $A(3, 0)$, at point $C(1, m)$, and intersects with the x-axis at point B. What is the equation of line ${l}_{2}$?

Solution:
\textbf{Solution:} Let's substitute the point $C\left(1, m\right)$ into the line $l_{1}: y=x+3$, we get: $m=1+3=4$,

$\therefore$ the point $C\left(1, 4\right)$,

Let's assume the equation of the line $l_{2}$ is: $y=kx+b\left(k\ne 0\right)$,

Substituting points $A\left(3, 0\right)$ and $C\left(1, 4\right)$,

We obtain $\begin{cases}3k+b=0\\ k+b=4\end{cases}$,

Solving these, we find $\begin{cases}k=-2\\ b=6\end{cases}$,

$\therefore$ the equation of $l_{2}$: \boxed{y=-2x+6}.


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=\frac{1}{2}x+b$ intersects the x-axis at point A, and the line $y=-x+2$ intersects the x-axis at point B. The two lines intersect at point $C(-2, a)$. What are the values of $a$ and $b$?

Solution: \textbf{Solution:} Since point $C$ lies on the line $y=-x+2$, it follows that $a=2+2=\boxed{4}$. Also, since point $C(-2, 4)$ lies on the line $y=\frac{1}{2}x+b$, it follows that $\frac{1}{2}\times(-2)+b=4$, solving for $b$ gives $b=\boxed{5}$; therefore, $a=4$, $b=5$.


Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=\frac{1}{2}x+b$ intersects the x-axis at point A, and the line $y=-x+2$ intersects the x-axis at point B. The two lines intersect at point $C(-2, a)$. What is the area of $\triangle ABC$?

Solution: \textbf{Solution:} For the equation $y=\frac{1}{2}x+5$, when $y=0$, we have $x=-10$, thus $A(-10, 0)$. In the equation $y=-x+2$, when $y=0$, we get $x=2$, hence $B(2, 0)$. Consequently, $AB=12$ and the area of $\triangle ABC$ is given by ${S}_{\triangle ABC}=\frac{1}{2}\times 12\times 4=\boxed{24}$.


Question: As shown in the figure, within the Cartesian coordinate system, the line $y=\frac{1}{2}x+b$ intersects the x-axis at point A, and the line $y=-x+2$ intersects the x-axis at point B. The two lines intersect at point $C(-2, a)$. A point $P(m, 0)$ is to the right of point A, and segment $PC$ is drawn. When $\triangle ACP$ is an isosceles triangle, find the value of $m$.

Solution:
\textbf{Solution:} As shown in the diagram, draw $CD\perp AB$ and connect $PC$,\\
we have $CD=4$, $AD=8$, $PA=m+10$,\\
$\therefore AC=\sqrt{8^2+4^2}=4\sqrt{5}$.\\
When $PC=PA$, let $PC=PA=m+10$, then $PD=-2-m$,\\
in $\triangle PCD$, $PC^2=PD^2+CD^2$,\\
$\therefore (m+10)^2=(-2-m)^2+4^2$, solving this yields: $m=\boxed{-5}$;\\
When $AC=AP$, we have $AP=4\sqrt{5}$,\\
$\therefore m=\boxed{-10+4\sqrt{5}}$;\\
When $CA=CP$, point D is the midpoint of $AP$,\\
$\therefore AP=2AD=16$,\\
$\therefore m=\boxed{6}$.\\
In summary, when $\triangle ACP$ is an isosceles triangle, the value of $m$ is \boxed{-5}, \boxed{-10+4\sqrt{5}}, or \boxed{6}.


Question: As shown in the figure, the equation of the straight line $l$ is $y=kx+b$. The line $l$ intersects the $x$ axis at point $A$ and intersects the $y$ axis at point $B(0,12)$. Additionally, the area of $\triangle AOB$ is $24$. What is the equation of the straight line $l$?

Solution:
\textbf{Solution:} Given $B(0,12)$ and the area of $\triangle AOB = 24$,\\
we have $\frac{1}{2} \cdot OA \cdot 12 = 24$,\\
therefore $OA=4$,\\
thus $A(-4,0)$,\\
Substituting $A(-4,0)$ and $B(0,12)$ into $y=kx+b$ yields: $\left\{\begin{array}{l} -4k+b=0 \\ b=12 \end{array}\right.$,\\
solving this we get $\left\{\begin{array}{l} k=3 \\ b=12 \end{array}\right.$,\\
thus, the equation of line $l$ is: \boxed{y=3x+12};


Question: As shown in the figure, the equation of line $l$ is $y=kx+b$. Line $l$ intersects the $x$-axis at point $A$ and intersects the $y$-axis at point $B(0,12)$, and the area ${S}_{\triangle AOB}=24$. If point $P$ is a point on line $l$ and ${S}_{\triangle AOP}=36$, determine the coordinates of point $P$.

Solution:
\textbf{Solution:} Let $P(m, 3m+12)$, \\
since $S_{\triangle AOP} = 36$, \\
thus $\frac{1}{2} \cdot OA \cdot |y_P| = 36$, that is $\frac{1}{2} \times 4 \times |3m + 12| = 36$, \\
therefore $|3m + 12| = 18$, \\
solving for $m$, we get $m = 2$ or $m = -10$, \\
thus $P(2, 18)$ or $P(-10, -18)$. \\
Hence, the coordinates of point $P$ are $P(2, \boxed{18})$ or $P(-10, \boxed{-18})$.


Question: It is known, as shown in the figure, that the graph of the linear function $y=kx+b$ intersects the $x$-axis and $y$-axis at points $A(3, 0)$ and $B(0, -4)$, respectively. What is the analytical expression of the linear function $y=kx+b$?

Solution:
\textbf{Solution:} Since the graph of the linear function $y=kx+b$ intersects the x-axis at point $A(3,0)$ and the y-axis at point $B(0,-4)$,\\
we have $\begin{cases}3k+b=0\\ b=-4\end{cases}$,\\
thus, $\begin{cases}k=\frac{4}{3}\\ b=-4\end{cases}$,\\
therefore, the equation of the linear function is $\boxed{y=\frac{4}{3}x-4}$.


Question: It is known that, as shown in the diagram, the graph of the linear function $y=kx+b$ intersects the $x$-axis and $y$-axis at points $A(3,0)$ and $B(0,-4)$ respectively. Point $M$ is on the $y$-axis, and the area of $\triangle ABM$ is $7.5$. Write down the coordinates of point $M$ directly.

Solution: \textbf{Solution:} The coordinates of point $M$ are $\boxed{(0,1)}$ or $\boxed{(0,-9)}$.


Question: As shown in the figure, the straight line \(l\) is the graph of the linear function \(y=kx+b\).

Solution:
\textbf{Solution:} Based on the problem, we know that the points $(-2,0)$ and $(2,2)$ lie on the graph of a linear function,\\
Substituting $(-2,0)$ and $(2,2)$ into $y=kx+b$, we get: $\begin{cases}-2k+b=0 \\ 2k+b=2\end{cases}$,\\
Solving this, we find: $\begin{cases}k=\frac{1}{2} \\ b=1\end{cases}$,\\
Therefore, the equation of the linear function is \boxed{y=\frac{1}{2}x+1};


Question: As shown in the figure, the line $l$ is the graph of the linear function $y=kx+b$. Based on the graph of the function, directly write down the range of $x$ when $y<0$.

Solution: \textbf{Solution:} According to the graph of the function, we can directly state that when $y<0$, the range of $x$ is $\boxed{x<-2}$.


Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, line $AB$ intersects the $x$-axis and $y$-axis at point $A$ and point $B$, respectively, and line $CD$ intersects the $x$-axis and $y$-axis at point $C$ and point $D$, respectively. The equation of line $AB$ is $y=-\frac{5}{8}x+5$, and the equation of line $CD$ is $y=kx+b\ (k\neq 0)$. The two lines intersect at point $E\left(m, \frac{10}{3}\right)$, and $OB:OC=5:4$. What is the equation of line $CD$?

Solution:
\textbf{Solution:} Let's substitute point $E\left(\frac{8}{3}, \frac{10}{3}\right)$ into $y=-\frac{5}{8}x+5$, then we get $m=\frac{8}{3}$, hence the coordinates of point $E$ are $\left(\frac{8}{3}, \frac{10}{3}\right)$. Setting $x=0$ in $y=-\frac{5}{8}x+5$, we find $y=5$, hence the coordinates of point $B$ are $(0, 5)$, thus $OB=5$. Since $OB:OC=5:4$, it follows that $OC=4$, hence the coordinates of point $C$ are $(-4, 0)$. Substituting $E\left(\frac{8}{3}, \frac{10}{3}\right)$, $C(-4, 0)$ into $y=kx+b$ (where $k\neq 0$), we obtain
\[
\left\{
\begin{array}{l}
\frac{8}{3}k+b=\frac{10}{3}\\
-4k+b=0
\end{array}
\right.
\]
Solving the system, we find
\[
\left\{
\begin{array}{l}
k=\frac{1}{2}\\
b=2
\end{array}
\right.
\]
Therefore, the equation of line $CD$ is $\boxed{y=\frac{1}{2}x+2}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $AB$ intersects the $x$-axis and $y$-axis at point $A$ and point $B$, respectively, and the line $CD$ intersects the $x$-axis and $y$-axis at point $C$ and point $D$, respectively. The equation of the line $AB$ is $y=-\frac{5}{8}x+5$, and the equation of the line $CD$ is $y=kx+b$, where $k\ne 0$. The two lines intersect at point $E\left(m, \frac{10}{3}\right)$, and $OB:OC=5:4$. The line $CD$ is translated downwards by a certain distance so that the translated line passes through point $A$ and intersects the $y$-axis at point $F$. What is the area of the quadrilateral $AEDF$?

Solution:
\textbf{Solution:} Let $y=-\frac{5}{8}x+5$ with $y=0$, we obtain $-\frac{5}{8}x+5=0$. Solving for $x$ yields $x=8$, $\therefore$ the coordinates of point A are $(8,0)$. Assume the equation of line CD after moving downwards is $y=\frac{1}{2}x+m$. Substituting the coordinates of point A, we find $m=-4$, $\therefore$ the equation of line AF is $y=\frac{1}{2}x-4$, $\therefore$ the coordinates of point F are $(0,-4)$. $\because$ the equation of line CD is $y=\frac{1}{2}x+2$, $\therefore$ the intersection with the y-axis gives the coordinates of point D as $(0,2)$. Connecting OE,\\
$\therefore$ the area of quadrilateral $AEDF = S_{\triangle ODE}+S_{\triangle OAE}+S_{\triangle OAF} = \frac{1}{2}\times 2\times \frac{8}{3}+\frac{1}{2}\times 8\times \frac{10}{3}+\frac{1}{2}\times 8\times 4 = \boxed{32}$


Question: As shown in the figure, in the Cartesian coordinate system, the line \(y=-\frac{4}{3}x+4\) intersects the \(x\) and \(y\) axes at points \(A\) and \(B\), respectively. The coordinates of point \(C\) are \((0, -2)\), and a line is drawn through \(A\) and \(C\). What is the equation of the line \(AC\)?

Solution: \textbf{Solution:} In the equation $y=-\frac{4}{3}x+4$, let $y=0$ to find $x=3$. $\therefore$ The coordinates of point A are $(3,0)$. Assume the equation of line $AC$ is $y=kx+b$. Substituting A$(3,0)$ and C$(0,-2)$ into it, we get: $\left\{\begin{array}{l} 0=3k+b \\ b=-2 \end{array}\right.$, solving this yields $\left\{\begin{array}{l} k=\frac{2}{3} \\ b=-2 \end{array}\right.$. $\therefore$ The equation of line $AC$ is $\boxed{y=\frac{2}{3}x-2}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=-\frac{4}{3}x+4$ intersects with the x-axis and y-axis at points $A$ and $B$ respectively. The coordinate of point $C$ is $(0, -2)$, and a line is drawn through points $A$ and $C$. Given that point $P$ is a moving point on line $AB$, and point $Q$ is a moving point on line $AC$, when the quadrilateral formed by vertices $O$, $A$, $P$, $Q$ is a parallelogram, determine the coordinates of point $P$.

Solution:
\textbf{Solution:} Let P($m$, $-\frac{4}{3}m+4$), Q($n$, $\frac{2}{3}n-2$), and A(3, 0), O(0, 0),

(1) For the parallelogram with diagonals PQ and AO, the midpoints of PQ and AO coincide,
\[
\therefore \left\{\begin{array}{l}m+n=3\\ -\frac{4}{3}m+4+\frac{2}{3}n-2=0\end{array}\right.,
\]
solving we get
\[
\left\{\begin{array}{l}m=2\\ n=1\end{array}\right.,
\]
therefore the coordinates of point P are ($2$, $\frac{4}{3}$);

(2) For the parallelogram with diagonals PA and QO, the midpoints of PA and QO coincide,
\[
\therefore \left\{\begin{array}{l}m+3=n\\ -\frac{4}{3}m+4=\frac{2}{3}n-2\end{array}\right.,
\]
solving we get
\[
\left\{\begin{array}{l}m=2\\ n=5\end{array}\right.,
\]
therefore the coordinates of point P are ($2$, $\frac{4}{3}$);

(3) For the parallelogram with diagonals PO and QA, the midpoints of PO and QA coincide,
\[
\therefore \left\{\begin{array}{l}m=n+3\\ -\frac{4}{3}m+4=\frac{2}{3}n-2\end{array}\right.,
\]
solving we get
\[
\left\{\begin{array}{l}m=4\\ n=1\end{array}\right.,
\]
therefore the coordinates of point P are ($4$, $-\frac{4}{3}$);

In conclusion, the coordinates of point P are ($2$, $\frac{4}{3}$) or ($4$, $-\frac{4}{3}$). Therefore, the coordinates of point P are $\boxed{(2, \frac{4}{3})}$ or $\boxed{(4, -\frac{4}{3})}$.


Question: As shown in the figure, in quadrilateral ABCD, $AB\parallel DC$, $AB=AD$, diagonals AC and BD intersect at point O, AC bisects $\angle BAD$, and a line is drawn through point C such that $CE\perp AB$ and intersects the extended line of AB at point E. Quadrilateral ABCD is a rhombus; if $AC=8$, $BD=6$, find the length of CE.

Solution:
\textbf{Solution:}

Since quadrilateral ABCD is a rhombus, and $BD=6$, $AC=8$,\\
it follows that $OA=OC=\frac{1}{2}AC=\frac{1}{2}\times 8=4$, and since $BD\perp AC$, we have $OB=OD=\frac{1}{2}BD=\frac{1}{2}\times 6=3$,\\
thus, $\angle AOB=90^\circ$,\\
In $\triangle AOB$, by applying the Pythagorean theorem, we find\\
$AB=\sqrt{OA^2+OB^2}=\sqrt{4^2+3^2}=5$,\\
therefore, the area of the rhombus $S=\frac{1}{2}AC\cdot BD=\frac{1}{2}\times 8\times 6=24$,\\
since $CE\perp AB$,\\
the area of the rhombus $S=AB\cdot CE=5CE=24$,\\
therefore, $CE=\boxed{\frac{24}{5}}$.


Question: As shown in the figure, the rectangle $ABCD$ has two sides $AD$, $AB$ with lengths $3$, $8$ respectively, and $E$ is the midpoint of $DC$. The graph of the inverse proportion function $y=\frac{m}{x}$ (where $x<0$) passes through point $E$, and intersects $AB$ at point $F$. Given that the coordinates of point $B$ are $(-6, 0)$, find the value of $m$ and the expression of the linear function that passes through points $A$ and $E$.

Solution:
\textbf{Solution:} Given that the coordinates of point B are $(-6,0)$, $AD=3$, $AB=8$, and $E$ is the midpoint of $CD$, thus $A(-6,8)$ and $E(-3,4)$. Since the graph of the inverse proportion function passes through point $E$, it follows that $m=-3\times 4=\boxed{-12}$. Suppose the expression of the linear function passing through points $A$ and $E$ is $y=kx+b(k\ne 0)$. Then,
\[
\left\{
\begin{array}{l}
-6k+b=8, \\
-3k+b=4,
\end{array}
\right.
\]
from which we can solve
\[
\left\{
\begin{array}{l}
k=-\frac{4}{3}, \\
b=0,
\end{array}
\right.
\]
therefore, the expression of the linear function is $\boxed{y=-\frac{4}{3}x}$.


Question: As shown in the figure, for the rectangle $ABCD$, the lengths of sides $AD$ and $AB$ are $3$ and $8$, respectively. Point $E$ is the midpoint of $DC$. The graph of the inverse proportional function $y=\frac{m}{x}$ ($x<0$) passes through point $E$ and intersects $AB$ at point $F$. If $AF-AE=2$, find the expression of the inverse proportional function.

Solution:
\textbf{Solution:} Given $\because AD=3, DE=4$, \\
$\therefore AE=\sqrt{AD^2+DE^2}=5$ \\
$\because AF-AE=2$ \\
$\therefore AF=7, BF=1$ \\
Let the coordinates of point $E$ be $(a, 4)$, then the coordinates of point $F$ are $(a-3, 1)$,\\
$\because E, F$ are on the graph of the function $y=\frac{m}{x} (x<0)$,\\
$\therefore 4a=a-3$, solving this gives $a=-1$,\\
$\therefore E(-1, 4)$ \\
$\therefore m=-1\times 4=-4$ \\
$\therefore \boxed{y=-\frac{4}{x}} (x<0)$.


Question: As shown in the figure, it is known that quadrilateral ABCD is a rhombus, AE $\perp$ BC at point E, and AF $\perp$ CD at point F. AE=AF. If $\angle B=70^\circ$, find the degree measure of $\angle EAF$.

Solution: \textbf{Solution:} Since $AE\perp BC$ at point $E$, and $\angle B=70^\circ$, it follows that $\angle BAE=20^\circ$. Because $\triangle ABE\cong \triangle ADF$, it follows that $\angle BAE=\angle DAF=20^\circ$, thus $\angle EAF=180^\circ-\angle B-\angle BAE-\angle DAF=\boxed{70^\circ}$.


Question: As shown in the figure, it is known that point $G$ is on the diagonal $AC$ of the square $ABCD$, $GE \perp BC$ with the foot of the perpendicular as point $E$, and $GF \perp CD$ with the foot of the perpendicular as point $F$. Quadrilateral $CEGF$ is a square. Find the value of $\frac{AG}{BE}$.

Solution: \textbf{Solution:} Since quadrilateral $CEGF$ is a square, it follows that $\angle CEG=\angle B=90^\circ$ and $\angle ECG=45^\circ$. Consequently, $GC=\sqrt{2}EC$ and $AC=\sqrt{2}BC$. Therefore, $AG=AC-GC=\sqrt{2}(BC-EC)=\sqrt{2}(m-n)$ and $BE=m-n$. Thus, $\frac{AG}{BE}=\boxed{\sqrt{2}}$.


Question: In quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, where $AO=CO=10$, $BO=DO$, and $AB=12$, $BC=16$. Quadrilateral $ABCD$ is a rectangle. If $\angle ADF : \angle FDC=3:2$, and $DF\perp AC$ at point $E$, what is the measure of $\angle BDF$?

Solution:
\textbf{Solution:}
Given that $\angle ADC=90^\circ$ and $\angle ADF : \angle FDC = 3 : 2$,
it follows that $\angle FDC=36^\circ$.
Given also that $DF \perp AC$,
it implies that $\angle DCO=90^\circ - 36^\circ = 54^\circ$.
Since quadrilateral ABCD is a rectangle,
it follows that $OC=OD$,
which means $\angle ODC=54^\circ$.
Therefore, $\angle BDF=\angle ODC - \angle FDC=\boxed{18^\circ}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=4$, $AD=10$, point $E$ is on side $AD$, it is known that points $B$ and $E$ are symmetrical with respect to line $l$, line $l$ intersects sides $AD$ and $BC$ at points $M$ and $N$, respectively. Connect $BM$ and $NE$. Quadrilateral $BMEN$ is a rhombus. If $DE=2$, what is the length of $NC$?

Solution: \textbf{Solution:} Let the side length of the rhombus be $x$, then $AM=8-x$. In $\triangle ABM$, we have $4^2+(8-x)^2=x^2$. Solving for $x$, we obtain $x=5$. $\therefore NC=\boxed{5}$.


Question: As shown in the figure, in the square $ABCD$, a rhombus $BDFE$ is constructed with the diagonal $BD$ as its side, such that points $B$, $C$, and $E$ lie on the same line. Connect $BF$ and let it intersect $CD$ at point $G$. It is given that $CG=CE$. If the side length of the square is $4$, what is the area of rhombus $BDFE$?

Solution:
\textbf{Solution:} Since the side length of the square $BC=4$,\\
thus $BD=\sqrt{2}BC=4\sqrt{2}$, and $DC=BC=4$.\\
The area of the rhombus $BDFE$ is $S=4\sqrt{2}\times 4=16\sqrt{2}$,\\
therefore, the area of the rhombus $BDFE$ is $S=\boxed{16\sqrt{2}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AD \parallel BC$, $\angle B=90^\circ$, $AB=8\, \text{cm}$, $AD=12\, \text{cm}$, $BC=18\, \text{cm}$. Point $P$ starts moving from point $A$ towards point $D$ at a speed of $1\, \text{cm/s}$; simultaneously, point $Q$ starts moving from point $C$ towards point $B$ at a speed of $2\, \text{cm/s}$. The movement of point $P$ stops when point $Q$ reaches point $B$. Let the time during which points $P$ and $Q$ are moving be $t\, \text{(s)}$. At what value of $t$ will quadrilateral $PQBA$ be a rectangle?

Solution:
\paragraph{Solution:} Since point P starts from point A and moves toward point D at a speed of $1\,\text{cm/s}$, and point Q starts from point C at the same time, moving toward point B at a speed of $2\,\text{cm/s}$,\\
it follows that $AP=t$, $CQ=2t$, $BQ=18-2t$,\\
Since rectangle $PQBA$,\\
it follows that $AP=BQ$,\\
Thus, $t=18-2t$,\\
Solving this gives $t=\boxed{6}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AD\parallel BC$, $\angle B=90^\circ$, $AB=8\,cm$, $AD=12\,cm$, $BC=18\,cm$. Point $P$ starts moving from point $A$ towards point $D$ at a speed of $1\,cm/s$; simultaneously, point $Q$ starts moving from point $C$ towards point $B$ at a speed of $2\,cm/s$. When point $Q$ reaches point $B$, point $P$ also stops moving. Let the time taken for points $P$ and $Q$ to move be $t(s)$. There exists a value of $t$ during the entire motion such that quadrilateral $PQCD$ is a rhombus. Find the value of $t$.

Solution:
\textbf{Solution:} Since quadrilateral PQCD is a rhombus, \\
therefore CQ=CD, \\
therefore $2t=10$, \\
Solving this, we get $t=\boxed{5}$. At this time, $DP=AD-AP=12-5=7$, \\
therefore $DP \neq CD$, \\
therefore, the quadrilateral PQCD is not a rhombus.


Question: In the rectangle \(ABCD\), where \(AB=6\) and \(BC=8\). The rectangle is folded along \(BD\) so that point \(A\) coincides with point \(E\) (as shown in Figure 1). \(DE\) and \(BC\) intersect at point \(F\), and \(ABDF\) forms an isosceles triangle. Determine the length of \(BF\).

Solution:
\textbf{Solution:} Given that when the rectangle is folded along BD, making point A coincide with point E,\\
thus AD$\parallel$BC and $\angle$ADB = $\angle$EDB,\\
hence $\angle$ADB = $\angle$DBF = $\angle$EDB,\\
therefore BF=DF,\\
implying that $\triangle$BDF is an isosceles triangle;\\
Let BF=DF=x, then CF=8−x,\\
in $\triangle$DCF, we have CD$^{2}$+CF$^{2}$=DF$^{2}$, that is $6^{2}$+(8−x)$^{2}$=x$^{2}$,\\
solving this, we get: x= \boxed{\frac{25}{4}},\\
therefore, the length of BF is \boxed{\frac{25}{4}}.


Question: In rectangle $ABCD$, $AB=6$, and $BC=8$. The rectangle is folded so that point $B$ coincides with point $D$ (as shown in Figure 2). What is the length of $DG$?

Solution:
\textbf{Solution:}
\[DG = \boxed{\frac{25}{4}}.\]


Question: As shown in the figure, it is known that the side length of the square $ABCD$ is $1$, the area of the square $CEFG$ is $S_1$, point $E$ is on side $CD$, and point $G$ is on the extension of side $BC$. Suppose the area of the rectangle with sides $AD$ and $DE$ as adjacent edges is $S_2$ and $S_1=S_2$. What is the length of the line segment $CE$?

Solution:
\textbf{Solution:} Let the side length of the square $CEFG$ be $a$,\\
since the side length of the square $ABCD$ is $1$, therefore $DE=1-a$, \\
since $S_1=S_2$, therefore $a^2=1\times(1-a)$, \\
therefore $a=\frac{-1+\sqrt{5}}{2}$ or $a=\frac{-1-\sqrt{5}}{2}$ (discard), \\
therefore $CE=\boxed{\frac{-1+\sqrt{5}}{2}}$;


Question: As shown in the figure, within rectangle $ABCD$, point $E$ is on side $CD$. Fold $\triangle BCE$ along $BE$ so that point $C$ falls on point $F$ on side $AD$. Draw $FG \parallel CD$ intersecting $BE$ at point $G$, and then connect $CG$. Quadrilateral $CEFG$ is a rhombus. If $AB = 6$ and $AD = 10$, find the area of quadrilateral $CEFG$.

Solution:
\textbf{Solution:} Given that the figure is folded, \\
we have $BF=BC=AD=10$, thus $AF= \sqrt{BF^2-AB^2}=8$, and $DF=AD-AF=2$. Let $DE=x$, then $EC=CD-DE=6-x=EF$. In $\triangle EDF$, since $DE^2+DF^2=EF^2$, we have $x^2+2^2=(6-x)^2$. Solving this equation gives $x= \frac{8}{3}$, hence $EC=6-DE=6- \frac{8}{3}=\frac{10}{3}$. Therefore, the area of the rhombus CEFG is $EC \times FD= \frac{10}{3} \times 2= \boxed{\frac{20}{3}}$.


Question: As shown in the diagram, in quadrilateral $ABCD$, diagonal $BD$ bisects $\angle ABC$. Quadrilateral $ABCD$ is a rhombus. A line through point $A$ is drawn such that $AE \parallel BD$, intersecting the extension of line $CD$ at point $E$. A line through point $E$ is drawn such that $EF \perp BC$, intersecting the extension of line $BC$ at point $F$. If $\angle ABC=45^\circ$ and $BC=2$, what is the length of $EF$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\
it follows that AB$\parallel$DE, \\
since AE$\parallel$BD, \\
it follows that quadrilateral ABDE is a parallelogram, \\
hence, AB=DE, \\
since quadrilateral ABCD is a rhombus, \\
it follows that CD=AB=BC=2, \\
thus, DE=AB=2, \\
therefore, CE=CD+DE=4, \\
since $\triangle$CFE is an isosceles right triangle, \\
it follows that EF=$\frac{\sqrt{2}}{2}$CE=$\boxed{2\sqrt{2}}$.


Question: As shown in the figure, in rhombus $ABCD$, $E$ is the midpoint of $AB$, and $DE\perp AB$, with $AE=2$. What is the measure of $\angle ABC$ in degrees?

Solution:
\textbf{Solution:} Given that $E$ is the midpoint of $AB$, and $DE \perp AB$,\\
$\therefore AD = BD$.\\
Also, in rhombus $ABCD$, $AD = AB$,\\
$\therefore AD = AB = BD$,\\
$\therefore \triangle ABD$ is an equilateral triangle,\\
$\therefore \angle ABD = 60^\circ$,\\
$\therefore \angle ABC = \boxed{120^\circ}$.


Question: As shown in the diagram, in rhombus $ABCD$, $E$ is the midpoint of $AB$, and $DE \perp AB$, with $AE=2$. What are the lengths of the diagonals $AC$ and $BD$?

Solution:
\textbf{Solution:} Since $\triangle ABD$ is an equilateral triangle and $AE=2$,

$\therefore AB=BD=AD=4$,

$\therefore DE=AO=\sqrt{AD^{2}-AE^{2}}=\sqrt{4^{2}-2^{2}}=2\sqrt{3}$,

$\therefore AC=\boxed{4\sqrt{3}}$.


Question: As shown in the figure, in rhombus $ABCD$, $AB=4$, $E$ is the midpoint of $BC$, $AE \perp BC$ at point $E$, $AF \perp CD$ at point $F$, $CG \parallel AE$, $CG$ intersects $AF$ at point $H$, and intersects $AD$ at point $G$. What is the area of the rhombus $ABCD$?

Solution: \textbf{Solution:} As shown in the diagram, connect $AC$. Since $E$ is the midpoint of $BC$ and $AE\perp BC$, it follows that $AB=AC$. Furthermore, since quadrilateral $ABCD$ is a rhombus, we have $AB=BC$. Therefore, $\triangle ABC$ is an equilateral triangle. Hence, $BE=\frac{1}{2}BC=2$, and $AE=\sqrt{AB^{2}-BE^{2}}=\sqrt{4^{2}-2^{2}}=2\sqrt{3}$. Consequently, the area of rhombus $ABCD$ is given by $BC\cdot AE=4\times 2\sqrt{3}=\boxed{8\sqrt{3}}$.


Question: As shown in the figure, in rhombus $ABCD$, $AB=4$, $E$ is the midpoint of $BC$, $AE\perp BC$ at point $E$, $AF\perp CD$ at point $F$, $CG\parallel AE$, $CG$ intersects $AF$ at point $H$, and intersects $AD$ at point $G$. What is the degree measure of $\angle CHA$?

Solution:
\textbf{Solution:} In equilateral triangle ABC,\\
$\because AE\perp BC$,\\
$\therefore \angle CAE=\frac{1}{2}\angle BAC=\frac{1}{2}\times 60^\circ=30^\circ.$\\
Similarly, $\angle CAF=30^\circ$.\\
$\therefore \angle EAF=\angle CAE+\angle CAF=30^\circ+30^\circ=60^\circ$\\
$\because AE\parallel CG$\\
$\therefore \angle CHA=180^\circ-\angle EAF=180^\circ-60^\circ=\boxed{120^\circ}$


Question: As shown in the diagram, extend the side $DC$ of the quadrilateral $ABCD$ to point $E$ such that $CE=DC$. Connect $AE$, which intersects $BC$ at point $F$, and connect $AC$, $BE$. Given that $BF=CF$; if $AB=2$, $AD=4$, and $\angle AFC=2\angle D$, what is the area of the quadrilateral $ABCD$?

Solution: \textbf{Solution:} From (1), we know that quadrilateral ABEC is a parallelogram, $\therefore$ FA=FE, FB=FC. $\because$ Quadrilateral ABCD is a parallelogram $\therefore$ $\angle ABC=\angle D$. Also $\because$ $\angle AFC=2\angle D$, $\therefore$ $\angle AFC=2\angle ABC$. $\because$ $\angle AFC=\angle ABC+\angle BAF$, $\therefore$ $\angle ABC=\angle BAF$, $\therefore$ FA=FB, $\therefore$ FA=FE=FB=FC, $\therefore$ AE=BC, $\therefore$ Quadrilateral ABEC is a rectangle, $\therefore$ $\angle BAC=90^\circ$ $\therefore$ BC=AD=4, $\therefore$ $AC=\sqrt{BC^2-AB^2}=\sqrt{4^2-2^2}=2\sqrt{3}$, $\therefore$ ${S}_{\square ABCD}=AB\cdot AC=2\times 2\sqrt{3}=\boxed{4\sqrt{3}}$.


Question: As shown in the figure, it is known that in quadrilateral $ABCD$, $AB \parallel CD$, $BC=AD=4$, $AB=CD=10$, $\angle DCB=90^\circ$. $E$ is a point on the side $CD$, with $DE=7$. A moving point $P$ starts from point $A$ and moves towards point $B$ along side $AB$ at a speed of 1 unit per second. Line $PE$ is drawn, and the time taken for point $P$ to move is $t$ seconds. What is the length of $BE$?

Solution:
\textbf{Solution:} Since $CD=10$ and $DE=7$, \\
thus $CE=10-7=3$, \\
Since $\angle DCB=90^\circ$, \\
hence, in $\triangle CBE$, $BE=\sqrt{BC^2+CE^2}=\sqrt{4^2+3^2}=\boxed{5}$;


Question: Given the quadrilateral $ABCD$, where $AB \parallel CD$, $BC=AD=4$, $AB=CD=10$, and $\angle DCB=90^\circ$. Let $E$ be a point on the side $CD$, with $DE=7$. A moving point $P$ starts from point $A$, moving towards point $B$ along side $AB$ at a speed of 1 unit per second. Connect $PE$. Let the time taken for point $P$ to move be $t$ seconds. If $\triangle BPE$ is a right-angled triangle, find the value of $t$.

Solution:
\textbf{Solution:}

Since $BC=AD=4$ and $AB=CD=10$,

it follows that quadrilateral ABCD is a parallelogram.

Since $\angle DCB=90^\circ$,

ABCD is a rectangle.

Therefore $\angle A=\angle D=90^\circ$.

When $\angle BPE=90^\circ$, that is $\angle APE=90^\circ$,

quadrilateral APED is a rectangle,

thus $AP=DE=7$,

so $t=7\div 1=\boxed{7}$ seconds,

When $\angle BEP=90^\circ$, draw $PF\perp CD$ at F through point P,

since $\angle A=\angle D=\angle DFP=90^\circ$,

quadrilateral ADFP is a rectangle,

hence $DF=AP=t$, $PF=AD=4$,

thus $PE^2=PF^2+EF^2=4^2+(7-t)^2$,

since in $\triangle BEP$, $BP^2=BE^2+PE^2$,

it follows that $(10-t)^2=5^2+4^2+(7-t)^2$

Solving yields: $t=\frac{5}{3}$,

Therefore, when $t=7$ or $t=\boxed{\frac{5}{3}}$, $\triangle BPE$ is a right-angled triangle.


Question: As shown in the figure, rectangle $ABCD$ is folded along $EF$, causing point $D$ to coincide with point $B$. If $\angle AEB = 40^\circ$, what is the degree measure of $\angle BFE$?

Solution:
\textbf{Solution:} Given the method,\\
$\because \angle AEB=40^\circ$,\\
$\therefore \angle BED=140^\circ$,\\
$\therefore$ From the property of folding: $\angle BEF=\angle DEF=70^\circ$,\\
$BG=CD$, $GF=CF$, $\angle G=\angle D=90^\circ$,\\
$\because$ Quadrilateral ABCD is a rectangle,\\
$\therefore AD\parallel BC$, $CD=AB$, $BC=AD$,\\
$\therefore \angle BFE=\angle DEF=\boxed{70^\circ}$.


Question: As shown in the figure, rectangle $ABCD$ is folded along $EF$ so that point $D$ coincides with point $B$. Given $AB = 6$ and $AD = 18$, find the length of $CF$.

Solution:
\textbf{Solution:} Given that $BC=AD=18$, \\
hence $BF=18-CF$, \\
since $\angle G=90^\circ$, \\
it follows that $BG^2+GF^2=BF^2$, \\
given that $BG=CD=AB=6$ and $GF=CF$, \\
we get $6^2+CF^2=(18-CF)^2$, \\
thus, $CF=\boxed{8}$.


Question: As shown in the figure, in the rhombus $ABCD$, $CE \perp AB$ at point $E$. If $CE=4$ and $AE=2BE$, what is the perimeter of the rhombus $ABCD$?

Solution:
\textbf{Solution:} Given that $AE=2BE$, \\
it follows that $AB=3BE$. \\
Since quadrilateral $ABCD$ is a rhombus, \\
we have $AB=BC=CD=AD=3BE$. \\
In $\triangle BCE$, we have $BE^2+CE^2=BC^2$, \\
therefore $BE^2+4^2=(3BE)^2$, solving this gives $BE=\sqrt{2}$ (neglect the negative value), \\
thus $AB=BC=CD=AD=3BE=3\sqrt{2}$, \\
hence the perimeter of rhombus $ABCD$ is $3\sqrt{2}\times 4=\boxed{12\sqrt{2}}$.


Question: As shown in the figure, within quadrilateral $ABCD$, point $F$ is the midpoint of side $AD$, and side $BC$ is extended to point $E$ such that $CE = \frac{1}{2}BC$. When connecting $DE$ and $CF$, the quadrilateral $CEDF$ forms a parallelogram. Given that $AB = 4$, $AD = 6$, and $\angle B = 60^\circ$, what is the length of $DE$?

Solution: \textbf{Solution:} As shown in the diagram, draw $DH\perp BE$ at point H. In quadrilateral $ABCD$, since $\angle B=60^\circ$ and $AB\parallel DC$, it follows that $\angle B=\angle DCE$, hence $\angle DCE=60^\circ$. Since $AB=4$, it follows that $CD=AB=4$, thus $CH=\frac{1}{2}CD=2$, and $DH=2\sqrt{3}$. In quadrilateral $CEDF$, $CE=DF=\frac{1}{2}AD=3$, thus $EH=1$. Therefore, in $\triangle DHE$, according to the Pythagorean theorem, $DE=\sqrt{(2\sqrt{3})^2+1^2}=\sqrt{13}$. Hence, the length of $DE$ is $\boxed{\sqrt{13}}$.


Question: As shown in the figure, it is known that quadrilateral $ABCD$ is a rhombus, $E$ is a point moving on diagonal $AC$, $DE$ is extended to intersect the extension line of $AB$ at point $F$, and $BE$ is connected. $\angle AFD = \angle EBC$. If $DE = EC$ and $BE \perp AF$, find the degree measure of $\angle DAB$?

Solution: \textbf{Solution:} Given rhombus ABCD, it follows that $\angle DAB = \angle DCB$ and DC$\parallel$AB implies $\angle CBF = \angle DCB$. From (1) it has been proven that $\triangle DCE \cong \triangle BCE$, thus $\angle EDC = \angle EBC$ and $\angle DCE = \angle BCE$. Since DE=EC, it results in $\angle EDC = \angle ECD$, hence $\angle EDC = \angle ECD = \angle EBC = \angle BCE$. Let $\angle EDC = \angle ECD = \angle EBC = \angle BCE = x$, thus $\angle CBF = \angle DCB = 2x$. Given BE$\perp$AF, it leads to $\angle CBF + \angle EBC = 2x + x = 90^\circ$, consequently $x = 30^\circ$, which means $\angle DAB = \angle DCB = 2x = \boxed{60^\circ}$.


Question: Given that quadrilateral $ABCD$ is a rhombus, $E$ is a moving point on the diagonal $AC$. Extend line $DE$ to intersect the extension of line $AB$ at point $F$, and connect $BE$. If $\angle DAB=90^\circ$ and $\triangle BEF$ is an isosceles triangle, find the measure of $\angle EFB$.

Solution:
\textbf{Solution:}

(1) When $F$ is on the extension line of $AB$, as shown in the figure, if $\angle DAB=90^\circ$, the rhombus $ABCD$ is a square, $\therefore \angle DAB=\angle ADC=90^\circ$, $DC\parallel AB$, $\therefore \angle CDF=\angle EFB$, $\because \triangle BEF$ is an isosceles triangle, $\therefore \angle EFB=\angle FEB=\angle CDF$, let $\angle EFB=\angle FEB=\angle CDF=y$, $\therefore \angle ABE=2y$, $\angle CDF=y$. Combining the conclusions from (1), it is easy to prove that $\triangle ADE\cong \triangle ABE$, $\therefore \angle ADE=\angle ABE$, $\therefore 90^\circ-y=2y$, $\therefore \boxed{y=30^\circ}$, $\therefore \angle EFB=30^\circ$;

(2) When $F$ is on the line segment $AB$, as shown in the figure, if $\angle DAB=90^\circ$, the rhombus $ABCD$ is a square, $\therefore \angle DAB=\angle ADC=90^\circ$, $DC\parallel AB$, $\therefore \angle CDF=\angle AFD$, $\because \triangle BEF$ is an isosceles triangle, $\therefore \angle EBF=\angle BEF$, let $\angle EFB=\angle FEB=\angle CDF=y$, $\therefore \angle CBE=90^\circ-y$, $\angle CDF=\angle AFD=2y$. Combining the conclusions from (1), $\triangle CDE\cong \triangle CBE$, $\therefore \angle CDE=\angle CBE=90^\circ-y$, $\therefore 2y=90^\circ-y$, $\therefore y=30^\circ$, $\therefore \angle EFB=180^\circ-30^\circ-30^\circ=\boxed{120^\circ}$,

In summary, $\angle EFB=30^\circ$ or $\boxed{120^\circ}$.


Question: As shown in the figure, it is known that the area of the rhombus ABCD is $2\sqrt{6}$, and the length of the diagonal AC is $2\sqrt{3}$. What is the length of the diagonal BD?

Solution:
\textbf{Solution:} Since the area of the rhombus $ABCD$ is $2\sqrt{6}$ and the length of diagonal $AC$ is $2\sqrt{3}$, \\
\therefore $\frac{1}{2}\times 2\sqrt{3}\times BD=2\sqrt{6}$, solving this we get $BD=\boxed{2\sqrt{2}}$.


Question: As shown in the figure, it is known that the area of the rhombus ABCD is $2\sqrt{6}$, and the length of the diagonal AC is $2\sqrt{3}$. Find the perimeter of the rhombus ABCD.

Solution:
\textbf{Solution:} \\
Since quadrilateral ABCD is a rhombus, \\
it follows that $AC\perp BD$, $AO=CO$, $BO=DO$\\
Since $AC=2\sqrt{3}$, $BD=2\sqrt{2}$ \\
it follows that $AO=\sqrt{3}$, $BO=\sqrt{2}$\\
Therefore, $AB=\sqrt{BO^2+AO^2}=\sqrt{5}$\\
Thus, the perimeter of the rhombus ABCD is $4\sqrt{5}=\boxed{4\sqrt{5}}$


Question: As shown in the figure, in $\triangle ABC$, where $AB=AC=4$, $D$ and $E$ are the midpoints of $AB$ and $AC$, respectively. Connect $CD$, and draw $EF \parallel DC$ through $E$ to intersect the extension of $BC$ at $F$; $DE=CF$; If $\angle B=60^\circ$, find the length of $EF$.

Solution:
\textbf{Solution:} Given $\because AB=AC=4$, $\angle B=60^\circ$,\\
$\therefore BC=AB=AC=4$,\\
Also, $\because D$ is the midpoint of $AB$,\\
$\therefore CD\perp AB$,\\
$\therefore$ in $\triangle BCD$,\\
$BD= \frac{1}{2}AB=2$,\\
$\therefore CD= \sqrt{BC^{2}-BD^{2}}=2\sqrt{3}$,\\
$\because$ quadrilateral $CDEF$ is a parallelogram,\\
$\therefore EF=CD=2\sqrt{3}$,\\
Thus, the length of $EF$ is $\boxed{2\sqrt{3}}$.


Question: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Line $DE$ is drawn through point $D$ such that $DE \parallel AC$ and $DE = OC$, and $CE$ is then connected. Quadrilateral $OCED$ is a rectangle. Line $AE$ is drawn and intersects $OD$ at point $F$. If the side length of rhombus $ABCD$ is $6$ and $\angle ABC = 60^\circ$, find the length of $AE$.

Solution:
\textbf{Solution:}\\
$\because$ The side length of rhombus ABCD is 6,\\
$\therefore AB=BC=CD=AD=6$, $BD\perp AC$, $AO=CO=\frac{1}{2}AC$\\
$\because \angle ABC=60^\circ$, $AB=BC$\\
$\therefore \triangle ABC$ is an equilateral triangle\\
$\therefore AC=AB=6$\\
$\because$ In $\triangle AOD$, $BD\perp AC$, $AD=6$, $AO=3$,\\
$\therefore OD=\sqrt{AD^2-AO^2}=3\sqrt{3}$\\
$\because$ Quadrilateral OCED is a rectangle,\\
$\therefore CE=OD=3\sqrt{3}$\\
$\because$ In $\triangle ACE$, $AC=6$, $CE=3\sqrt{3}$,\\
$\therefore AE=\sqrt{AC^2+CE^2}=\sqrt{6^2+(3\sqrt{3})^2}=\boxed{3\sqrt{7}}$


Question: Fold the rectangle $ABCD$ as shown in the diagram, with $BE$, $EG$, and $FG$ being the fold lines. If vertices $A$, $C$, $D$ all fall on point $O$, and points $B$, $O$, $G$ are on the same straight line, while points $E$, $O$, $F$ are on another straight line, what is the value of $\frac{EG}{BE}$?

Solution: \textbf{Solution:} From the context, we know that the value of $\frac{EG}{BE}$ is \boxed{\frac{\sqrt{2}}{2}}.


Question: Fold rectangle $ABCD$ as shown in the figure, with $BE$, $EG$, $FG$ being the creases. If vertices $A$, $C$, $D$ all coincide at point $O$, and points $B$, $O$, $G$ lie on the same line, while points $E$, $O$, $F$ lie on another line. If $AD=4\sqrt{2}$, what is the area of quadrilateral $BEGF$?

Solution: \textbf{Solution:} Hence, the area of the quadrilateral $BEGF$ is $\boxed{9\sqrt{2}}$.


Question: As shown in the figure, points A, B, C, and D are on the circle, where $\overset{\frown}{AB}=\overset{\frown}{AD}=\overset{\frown}{CD}$. The lines BD and AC intersect at point E. As shown in figure 1, if AE=4 and CE=8, what is the length of AB?

Solution:
\textbf{Solution:} Given that $\because \overset{\frown}{AB}=\overset{\frown}{AD}=\overset{\frown}{CD}$, $\therefore \angle ABD=\angle CBD=\angle ACD=\angle ACB$, \\
Furthermore, $\because \angle BAE=\angle CAB$, \\
$\therefore \triangle ABE \sim \triangle ACB$, \\
$\therefore \frac{AE}{AB}=\frac{AB}{AC}$, that is, $\frac{4}{AB}=\frac{AB}{12}$ \\
$\therefore AB=\boxed{4\sqrt{3}}$


Question: As shown in the figure, points A, B, C, and D are on the circle, with $ \overset{\frown}{AB}=\overset{\frown}{AD}=\overset{\frown}{CD}$. Line BD intersects AC at point E. As shown in figure 2, the bisector of $\angle ACB$ intersects BD at point M and AB at point N. If point N is the midpoint of AB, and BE equals DM, what is the value of $\frac{DM}{BM}$?

Solution:
\textbf{Solution:} \\
Since CN bisects $\angle ACB$, it follows that NH=NF, \\
Since AN=BN, it follows that Rt$\triangle ANH \cong$ Rt$\triangle BNF$, \\
Hence, $\angle BAC = \angle ABC$, \\
Therefore, $\overset{\frown}{BC} = \overset{\frown}{AC} = 2\overset{\frown}{CD} = 2\overset{\frown}{AD} = 2\overset{\frown}{AB}$, \\
Therefore, $\angle ABD = \angle CBD = \angle ACB = 36^\circ$, \\
$\angle BAC = \angle BDC = 72^\circ$, \\
Therefore, $\angle DCM = 54^\circ$, $\angle DMC = 54^\circ$, \\
Thus, CD=DM, \\
Since $\angle BDC = 72^\circ$, $\angle ACD = 36^\circ$, it follows that $\angle DEC = 72^\circ$, \\
Hence, CD=CE, \\
Since $\angle CBD = \angle ACB$, it follows that BE=CE, \\
Therefore, BE=CE=CD=DM. \\
Let DE=1, CD=x, then BE=CE=CD=x. \\
It is easy to see that $\triangle DCE \sim \triangle DBC$, hence $\frac{DE}{CD} = \frac{CD}{BD}$, \\
Therefore, $\frac{1}{x} = \frac{x}{1+x}$, \\
Solving this obtains $x=\frac{1+\sqrt{5}}{2}$ ($x>0$). \\
Since BE=DM, it follows that BM=DE. \\
Therefore, $\frac{DM}{BM}=\frac{DM}{DE}=\frac{CD}{DE}=\boxed{\frac{1+\sqrt{5}}{2}}$.


Question: As shown in the figure, within $\triangle ABC$, points E and F are located on side AD and diagonal AC respectively. Line segment EF is drawn, and $\angle AEF = \angle CAB$. Prove that: $\triangle AEF \sim \triangle ABC$.

Solution:
\textbf{Solution:} Proof: \\
$\because$ Quadrilateral ABCD is a parallelogram, \\
$\therefore$ AB $\parallel$ CD, \\
$\therefore$ $\angle$CAB $=$ $\angle$ACD, \\
$\because$ $\angle$AEF $=$ $\angle$CAB, \\
$\therefore$ $\angle$AEF $=$ $\angle$ACD, \\
$\because$ $\angle$EAF $=$ $\angle$CAD, \\
$\therefore$ \boxed{$\triangle AEF \sim \triangle ACD$}.


Question: As shown in the diagram, in square $ABCD$, points $E$ and $F$ are respectively on side $AD$ and diagonal $AC$. Connect $EF$, and $\angle AEF = \angle CAB$, $\triangle AEF \sim \triangle ACD$; if $AF = 2CF$, $AE = 4$, $DE = 5$, find the length of $AC$?

Solution:
\textbf{Solution:} Given the problem statement,\\
$\because \triangle AEF\sim \triangle ACD$,\\
$\therefore \frac{AE}{AC}=\frac{AF}{AD}$.\\
$\because AF=2CF$,\\
$\therefore AF=\frac{2}{3}AC$.\\
$\because AE=4, DE=5$,\\
$\therefore \frac{4}{AC}=\frac{\frac{2}{3}AC}{4+5}$.\\
$\therefore \frac{2}{3}AC^2=36$.\\
Solving gives $AC=\boxed{3\sqrt{6}}$ (Negative values discarded).


Question: As shown in the figure, $AF$ and $AG$ are the altitudes of $\triangle ABC$ and $\triangle ADE$, respectively, with $\angle BAF = \angle DAG$. $\triangle ABC \sim \triangle ADE$; If $DE = 3$ and $\frac{AD}{AB} = \frac{2}{5}$, find the length of $BC$.

Solution:
\textbf{Solution:} Since $\triangle ABC\sim \triangle ADE$,\\
$\therefore \frac{AD}{AB}=\frac{DE}{BC}$,\\
and given $\frac{AD}{AB}=\frac{2}{5}$, $DE=3$,\\
$\therefore \frac{2}{5}=\frac{3}{BC}$,\\
$\therefore BC=\boxed{\frac{15}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=8$, $\angle B=90^\circ$, point $P$ is a moving point on side $AB$, draw $PD\parallel AC$ intersecting $BC$ at $D$, and draw $PE\parallel BC$ intersecting $AC$ at $E$. When point $P$ is the midpoint of $AB$, $PE=3$. What is the length of side $BC$?

Solution:
\textbf{Solution:} Given $\because PE\parallel BC$, and point P is the midpoint of AB\\
$\therefore \frac{AE}{EC}=\frac{AP}{PB}=1$\\
$\therefore AE=EC$\\
This means E is the midpoint of AC\\
$\therefore$ PE is the median\\
$\therefore BC=2PE=2\times 3=\boxed{6}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=8$, $\angle B=90^\circ$. Point $P$ is a moving point on side $AB$, and through point $P$, $PD \parallel AC$ intersects $BC$ at $D$, and $PE \parallel BC$ intersects $AC$ at $E$. When point $P$ is the midpoint of $AB$, $PE=3$. $\triangle APE \sim \triangle PBD$; find the value of $AP$ for which the quadrilateral $PDCE$ is a rhombus.

Solution:
\textbf{Solution:} Let $AP=x$, then $BP=8-x$\\
Since $PE\parallel BC$, $PD\parallel AC$\\
Therefore, quadrilateral PDCE is a parallelogram\\
Therefore, $CD=PE$\\
Since $\triangle APE\sim \triangle PBD$\\
Therefore, $\frac{AP}{PB}=\frac{PE}{BD}$\\
i.e., $\frac{x}{8-x}=\frac{PE}{6-CD}$\\
$\frac{x}{8-x}=\frac{PE}{6-PE}$\\
Therefore, $PE=\frac{3}{4}x$\\
Therefore, $CD=PE=\frac{3}{4}x$\\
Therefore, $BD=BC-CD=6-\frac{3}{4}x$\\
When $PE=PD$, parallelogram PDCE becomes a rhombus.\\
By the Pythagorean theorem, $PB^{2}+BD^{2}=PD^{2}$\\
Therefore, $(8-x)^{2}+(6-\frac{3}{4}x)^{2}=(\frac{3}{4}x)^{2}$\\
Simplify to get $x^{2}-25x+100=0$\\
Solving this equation gives $x_{1}=5$, $x_{2}=20$\\
$x=20>8$ is not feasible, disregard it.\\
Therefore, when $AP=5$, quadrilateral PDCE is a rhombus.\\
Solutions 2, 3, and 4 are omitted, as they are identical to Solution 1.\\
Therefore, when $AP=\boxed{5}$, quadrilateral PDCE is a rhombus.


Question: As shown in the figure, it is known that $AB$ is the diameter of $\odot O$, the bisector of the acute angle $\angle DAB$ intersects $\odot O$ at point $C$. Draw $CD \perp AD$ at $D$, and the extension of line $CD$ intersects the extension of $AB$ at point $E$. The line $CD$ is the tangent of $\odot O$; when $AB=2BE$, and $CE=\sqrt{3}$, find the length of $AD$.

Solution:
\textbf{Solution}: Since diameter AB $= 2$BE,\\
it follows that OE $= 2$OC,\\
In $\triangle EOC$, let CO $= x$, thus OE $= 2x$,\\
By the Pythagorean theorem: CE $= \sqrt{3}x$,\\
Also, since CE $= \sqrt{3}$, it follows that $x = 1$, hence OC $= 1$,\\
Since OC $\parallel$ AD, it follows that $\triangle EOC \sim \triangle EAD$,\\
Hence $\frac{OC}{AD} = \frac{OE}{AE}$, that is, $\frac{1}{AD} = \frac{2}{3}$,\\
Solving gives $AD = \boxed{\frac{3}{2}}$.


Question: As shown in the figure, rectangle $ABCD$, $BF \perp AC$ intersects $CD$ at point $E$, and intersects the extension line of $AD$ at point $F$. $AB^{2} = BC \cdot AF$. When $\frac{BC}{AB} = \frac{2}{3}$ and $DF = 5$, find the length of $AC$.

Solution:
\textbf{Solution:}
Given that,
\begin{align*}
\because \frac{BC}{AB} &= \frac{2}{3}, \\
\therefore \text{let} BC &= 2x, AB = 3x, \\
\because AB^2 &= BC \cdot AF, \\
\therefore (3x)^2 &= 2x(2x + 5), \\
\therefore x_1 &= 0, x_2 = 2, \\
\therefore BC &= 4, AB = 6, \\
\therefore AC &= \sqrt{6^2 + 4^2} = 2\sqrt{13}.
\end{align*}
Thus, the length of $AC$ is $\boxed{2\sqrt{13}}$.


Question: As shown in Figure 1, it is known that $\triangle ABC$, $\angle CAB = 45^\circ$, $AB = 7$, $AC = 3\sqrt{2}$, and $CD \perp AB$ at point D. Point E is a moving point on side BC, a circle $\odot O$ is drawn with DE as the diameter, intersecting BC at F, and intersecting AB at G. Connect DF and FG. $\angle BCD = \angle FDB$. When point E is on segment BF, and $\triangle DFG$ is an isosceles triangle, find the length of DG.

Solution:
\textbf{Solution:}

(i) When $DF = DG$,

$\because \angle CAB = 45^\circ$, $CD \perp AB$, $AC = 3\sqrt{2}$,

$\therefore AD = CD = 3$,

$\because AB = 7$,

$\therefore BD = 7 - 3 = 4$,

$\therefore BC = \sqrt{3^2 + 4^2} = 5$,

$\therefore DF = \frac{3 \times 4}{5} = \frac{12}{5}$,

$\therefore DG = \frac{12}{5}$,

(ii) When $DF = FG$, draw $FH \perp BD$ intersecting $BD$ at point $H$,

$\therefore \triangle DFH \sim \triangle CBD$,

$\therefore \frac{DH}{CD} = \frac{DF}{CB}$,

$\therefore DH = DF \times \frac{3}{5} = \frac{12}{5} \times \frac{3}{5} = \frac{36}{25}$,

$\therefore DG = 2DH = \frac{72}{25}$,

(iii) When $FG = DG$, $\angle 1 = \angle 2$,

$\because \angle 1 + \angle 3 = \angle 2 + \angle 4 = 90^\circ$,

$\therefore \angle 3 = \angle 4$,

$\therefore FG = GB = DG$,

$\therefore DG = \frac{1}{2}BD = 2$,

Hence, the lengths of $DG$ are $\boxed{\frac{12}{5}}$, $\boxed{\frac{72}{25}}$, $\boxed{2}$.


Question: As shown in Figure 1, given $\triangle ABC$ with $\angle CAB=45^\circ$, $AB=7$, and $AC=3\sqrt{2}$, $CD\perp AB$ at point D. E is a moving point on side BC. A circle $\odot O$ with diameter DE intersects BC at point F and AB at point G. Lines DF and FG are drawn. As shown in Figure 2, another intersection point of $\odot O$ and CD is P. If ray AP passes through point F, find the value of $\frac{AP}{DE}$.

Solution:
\textbf{Solution:}\\
(1) Since quadrilateral PDEF is a cyclic quadrilateral inside circle O,\\
therefore $\angle APD = \angle DEF$,\\
since $\angle APD + \angle PAD = \angle DEF + \angle EDF = 90^\circ$,\\
therefore $\angle PAD = \angle EDF$.\\
(2) Connect PG,\\
since $\angle PAD = \angle EDF$,\\
$\angle ADP = \angle DFE = 90^\circ$,\\
therefore $\triangle ADP \sim \triangle DFE$,\\
therefore $\frac{AP}{DE} = \frac{AD}{DF} = 3 \times \frac{5}{12} = \boxed{\frac{5}{4}}$.\\
Since $\angle PDG = 90^\circ$,\\
therefore PG is a diameter,\\
therefore $\angle PFG = 90^\circ$,\\
since $\angle FPG = \angle FDG = \angle BCD$,\\
therefore $\triangle CDB \sim \triangle PFG$,\\
therefore $\frac{FG}{FG} = \frac{CB}{DB} = \frac{5}{4}$,\\
therefore $\frac{DE}{FG} = \frac{CB}{DB} = \frac{5}{4}$,\\
therefore $\frac{AP}{FG} = \frac{AP}{DE} \cdot \frac{DE}{FG} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$.


Question: In $\triangle ABC$, $\angle CAB=90^\circ$, $AC=AB$. If point $D$ lies on $AC$, and $BD$ is connected and rotated $90^\circ$ clockwise around point $B$ to obtain $BE$, and $CE$ is connected and intersects $AB$ at point $F$, as shown in Figure 1. If $\angle ABE=75^\circ$ and $BD=4$, what is the length of $AC$?

Solution:
\textbf{Solution:} Draw DG$\perp$BC from D with the foot as G, as shown in Fig. 1:\\
$\because$ Rotating BD $90^\circ$ clockwise around point B yields BE,\\
$\therefore$ $\angle$EBD$=90^\circ$,\\
$\because$ $\angle$ABE$=75^\circ$,\\
$\therefore$ $\angle$ABD$=15^\circ$,\\
$\because$ $\angle$ABC$=45^\circ$,\\
$\therefore$ $\angle$DBC$=30^\circ$,\\
$\therefore$ in right $\triangle$BDG, we have $DG=\frac{1}{2}BD=2$, $BG=\sqrt{3}DG=2\sqrt{3}$,\\
$\because$ $\angle$ACB$=45^\circ$,\\
$\therefore$ in right $\triangle$DCG, CG$=DG=2$,\\
$\therefore$ BC$=BG+CG=2+2\sqrt{3}$,\\
$\therefore$ AC$=\frac{\sqrt{2}}{2}BC=\sqrt{2}+\sqrt{6}$;\\
Thus, the length of AC is $\boxed{\sqrt{2}+\sqrt{6}}$.


Question: In $\triangle ABC$, $\angle CAB=90^\circ$, and $AC=AB$. Let point $D$ be a point on $AC$, and connect $BD$. Rotate $BD$ $90^\circ$ clockwise around point $B$ to get $BE$, and connect $CE$ to intersect $AB$ at point $F$. As shown in figure 3, if $AB=4$ and $D$ is the midpoint of $AC$, rotate $\triangle ABD$ around point $B$ to get $\triangle A'BD'$. Connect $A'C$ and $A'D$. When $A'D+ \frac{\sqrt{2}}{2} A'C$ is minimized, find $S_{\triangle A'BC}$.

Solution:
\textbf{Solution:} Let $AB=a$, then $BC=\sqrt{2}a$. Take the midpoint $N$ of $BC$, connect $A'D$, $A'C$, and $A'N$, also connect $DN$. By rotation, we know $A'B=AB=a$, \\
$\because \frac{A'B}{BN}=\frac{a}{\frac{\sqrt{2}}{2}a}=\sqrt{2}$, $\frac{BC}{A'B}=\frac{\sqrt{2}a}{a}=\sqrt{2}$, \\
$\therefore \frac{A'B}{BN}=\frac{BC}{A'B}=\sqrt{2}$. Also, $\angle A'BN=\angle CBA'$, \\
$\therefore \triangle A'BN\sim \triangle CBA'$, \\
$\therefore \frac{A'N}{A'C}=\frac{A'B}{BC}=\frac{\sqrt{2}}{2}$, \\
$\therefore A'N=\frac{\sqrt{2}}{2}A'C$. According to rotation and the principle that the shortest distance between two points is a straight line, $A'D+\frac{\sqrt{2}}{2}A'C$ is minimized, which means $A'D+A'N$ is minimized. At this point, $D$, $A'$, $N$ are collinear, that is, $A'$ is on the line segment $DN$. Assume at this moment $A'$ falls at the position $A''$. Draw $A''F\perp AB$ at $F$, connect $AA''$, \\
$\because D$, $N$ are the midpoints of $AC$, $BC$ respectively, \\
$\therefore DN$ is the median of $\triangle ABC$, \\
$\therefore DN\parallel AB$, \\
$\because AB\perp AC$, \\
$\therefore DN\perp AC$, \\
$\because \angle A=\angle A''FA=\angle A''DA=90^\circ$, \\
$\therefore$ quadrilateral $A''FAD$ is a rectangle, \\
$\therefore AF=A''D$, $A''F=AD=2$, also $A''B=AB=4$. Let $AF=x$, in the right triangle $A''FB$, $A''B^2=A''F^2+BF^2$, \\
$\therefore 16=4+(4-x)^2$, solving this gives $x=4-2\sqrt{3}$. \\
$\therefore$ at this moment, $S_{\triangle A''BC}=S_{\triangle ABC}-S_{\triangle AA''B}-S_{\triangle A''AC}=\frac{1}{2}AB\cdot AC-\frac{1}{2}AB\cdot A''F-\frac{1}{2}AC\cdot A''D=4\sqrt{3}-4$. \\
Hence, $S_{\triangle A'BC}=\boxed{4\sqrt{3}-4}$.


Question: As shown in the figure, in right triangle $ABC$, $\angle C=90^\circ$, $AC=8\,\text{cm}$, $BC=6\,\text{cm}$. Now, a moving point $P$ starts from point $B$, moving towards point $A$ along segment $BA$, and another moving point $Q$ starts from point $A$, moving towards the end point along the polyline $AC-CB$. If the speed of point $P$ is $1\,\text{cm/s}$ and the speed of point $Q$ is $1\,\text{cm/s}$. They start at the same time, and when one point reaches the end point, the other point also stops moving. Let the time of motion be $t$ seconds. As shown in Figure 1, $Q$ is on $AC$, at what time $t$ in seconds, will the triangle formed by points $A$, $P$, $Q$ be similar to $\triangle ABC$?

Solution:
Solution: As shown in Figure 1, when $\angle AQP=90^\circ$, $\triangle AQP\sim \triangle ABC$,
\[\therefore \frac{AQ}{AC}=\frac{AP}{AB}.\]
In $\triangle ABC$, by the Pythagorean theorem, we have
\[AB= \sqrt{AC^2+BC^2}=\sqrt{8^2+6^2}=10 \text{ (cm)}.\]
\[\because BP=t,\ AQ=t,\]
\[\therefore PA=10-t,\]
\[\therefore \frac{t}{8}=\frac{10-t}{10},\]
\[\therefore t= \boxed{\frac{40}{9}};\]
As shown in Figure 2, when $\angle APQ=90^\circ$, $\triangle APQ\sim \triangle ABC$,
\[\therefore \frac{AQ}{AB}=\frac{AP}{AC},\]
\[\therefore \frac{t}{10}=\frac{10-t}{8},\]
\[t= \boxed{\frac{50}{9}}.\]
In conclusion, when $t= \boxed{\frac{40}{9}} or t= \boxed{\frac{50}{9}}, the triangle with vertices A, P, and Q is similar to $\triangle ABC$;


Question: As shown in the figure, in right-angled triangle $ABC$, $\angle C=90^\circ$, $AC=8\text{cm}$, and $BC=6\text{cm}$. Now, a moving point $P$ starts from point $B$ and moves towards the endpoint $A$ along line segment $BA$, and a moving point $Q$ starts from point $A$ and moves towards the endpoint along the polyline $AC-CB$. If the speed of point $P$ is $1\text{cm/s}$, and the speed of point $Q$ is $1\text{cm/s}$. They start at the same time, and when one point reaches the endpoint, the other point also stops moving. Let the time of movement be $t$ seconds. As shown in figure 2, $Q$ is on $CB$. Is there a moment that makes the triangle with vertices $B$, $P$, $Q$ similar to $\triangle ABC$? If it exists, find the value of $t$; if not, please explain the reason.

Solution:
\textbf{Solution:} As shown in Figure 3, when $\triangle BPQ\sim \triangle BAC$, \\
$\therefore \frac{BP}{AB}=\frac{BQ}{BC}$. \\
$\because BQ=14-t$ and $BP=t$,\\
$\therefore \frac{t}{10}=\frac{14-t}{6}$,\\
$\therefore t= \boxed{\frac{35}{4}}$,\\
When $\triangle BQP\sim \triangle BAC$,\\
$\therefore \frac{BP}{BC}=\frac{BQ}{BA}$,\\
$\therefore \frac{t}{6}=\frac{14-t}{10}$,\\
$\therefore t= \frac{21}{4}$,\\
$\because Q$ is on $CB$,\\
$\therefore t>8$,\\
$\therefore t= \frac{21}{4}$ is discarded,\\
$\therefore when $t= \frac{35}{4}$, $Q$ is on $CB$, and the triangle formed by points $B$, $P$, and $Q$ is similar to $\triangle ABC$.


Question: As illustrated in the diagram, in $\triangle ABC$, where $AB=AC$, the circle $\odot O$ with diameter $AB$ intersects $BC$ at point $D$. A line $EF$ passing through point $D$ intersects $AC$ at point $F$ and the extension of $AB$ at point $E$, and $\angle BAC=2\angle BDE$. $DF$ is a tangent to $\odot O$; when $CF=4$ and $BE=6$, find the length of $AF$.

Solution: \textbf{Solution:} Since $AB=AC$ and $AD\perp BC$, we have $BD=CD$. Since $BO=AO$, it follows that $OD$ is the median line of triangle $ABC$, thus $OD\parallel AC$ and $AC=2OD$. Therefore, $\triangle EOD\sim \triangle EAF$, which implies $\frac{OD}{AF}=\frac{EO}{EA}$. Let $OD=x$, then $AC=2x$. Since $CF=4$ and $BE=6$, we have $OA=OB=x$, $AF=AC-CF=2x-4$, $EO=x+6$, and $EA=2x+6$. So, $\frac{x}{2x-4}=\frac{x+6}{2x+6}$. Solving this, we find $x=12$. Checking, we find $x=12$ is the solution of the given fractional equation. Therefore, $AF=2x-4=\boxed{20}$.


Question: Given that $AC$ and $EC$ are the diagonals of the rectangles $ABCD$ and $EFCG$ respectively, point $E$ is inside $\triangle ABC$, and $\frac{AB}{BC}=\frac{EF}{FC}=k$, $\angle CAE+\angle CBE=90^\circ$. $\triangle CAE\sim \triangle CBF$; if $BE=2$, $AE=4$, $CE=6$, find the value of $k$.

Solution:
\textbf{Solution:}\\
Since $\triangle CAE\sim \triangle CBF$,\\
we have $\frac{AC}{BC}=\frac{EF}{FC}=k$,\\
thus $\angle CBF=\angle CAE$,\\
and since $\angle CAE+\angle CBE=90^\circ$,\\
it follows that $\angle EBF=90^\circ$,\\
let $BC=a$, then $AB=ka$, let $FC=b$, then $EF=kb$,\\
thus $AC=\sqrt{AB^2+BC^2}=a\sqrt{k^2+1}$,\\
and $EC=\sqrt{EF^2+FC^2}=b\sqrt{k^2+1}$,\\
hence $\frac{AE}{BF}=\frac{AC}{BC}=\sqrt{k^2+1}$,\\
given that $AE=4$,\\
it follows that $BF=\frac{4}{\sqrt{k^2+1}}$,\\
since $\angle EBF=90^\circ$,\\
it follows that $EF^2=BE^2+BF^2=4+\frac{16}{k^2+1}=k^2b^2$,\\
since $CE^2=EF^2+CF^2$, i.e., $6^2=(kb)^2+b^2$,\\
we get $b^2=\frac{36}{k^2+1}$, hence $\frac{36k^2}{k^2+1}=4+\frac{16}{k^2+1}$,\\
solving for $k$ yields $k=\boxed{\frac{\sqrt{10}}{4}}$ or $k=-\frac{\sqrt{10}}{4}$ (discard as irrelevant).\\
Therefore, the value of $k$ is $\boxed{\frac{\sqrt{10}}{4}}$.


Question: As shown in the figure, $D$ is a point on the circle $\odot O$, and point $C$ is on the extension line of diameter $BA$, with $CD^{2}=CA\cdot CB$; $CD$ is the tangent of $\odot O$; a tangent to $\odot O$ is drawn through point $B$ and intersects the extension line of $CD$ at point $E$. If $BC=10$ and $\tan\angle CDA=\frac{3}{5}$, what is the length of $BE$?

Solution:
\textbf{Solution:} From the given information, $BE$ and $CE$ are tangent to the circle $O$, \\
$\therefore ED=EB$, \\
$\because \triangle DCA\sim \triangle BCD$, \\
$\therefore \angle DBA=\angle CDA$, \\
$\therefore \frac{DC}{BC}=\frac{DA}{BD}=\tan\angle DBA=\tan\angle CDA=\frac{3}{5}$, \\
$\therefore CD=\frac{3}{5}BC=6$, \\
Let $BE=x$, then $DE=x$, $CE=x+6$, \\
In $\triangle CBE$, $(x+6)^{2}=x^{2}+10^{2}$, \\
Solving this gives: $x=\frac{16}{3}$, \\
$\therefore BE=\boxed{\frac{16}{3}}$.


Question: As shown in the figure, square DEFG has a side length of 4 cm, and all four vertices are on the sides of $\triangle ABC$. In right $\triangle ABC$, with $\angle A=90^\circ$, where $BD > EC$, and $BC=14$ cm, $\triangle BDG \sim \triangle FEC$; find the length of $BD$ in cm?

Solution:
\textbf{Solution:} Given that the side length of the square DEFG is 4cm,\\
thus DE=DG=FE=4cm. Let BD=xcm.\\
Given that BC=14cm,\\
thus $EC=BC−BD−DE=(10−x)cm$.\\
Given that $\triangle BDG\sim \triangle FEC$,\\
thus $\frac{BD}{FE}=\frac{DG}{EC}$.\\
Therefore, $\frac{x}{4}=\frac{4}{10−x}$. Solving gives $x_{1}=8$, $x_{2}=2$. Upon verification, $x=8$ and $x=2$ are solutions to the aforementioned fractional equation,\\
thus BD=8cm or BD=2cm. When BD=8cm, EC=2cm, which satisfies BD>EC. When BD=2cm, EC=8cm, does not satisfy BD>EC.\\
Therefore, BD=\boxed{8cm}.


Question: As shown in the figure, the parabola intersects the x-axis at $A(5, 0)$ and $B(-1, 0)$, and intersects the positive semi-axis of the y-axis at point $C$. Lines $BC$ and $AC$ are drawn, and it is known that $\sin\angle BAC=\frac{\sqrt{2}}{2}$. What is the analytical equation of the parabola?

Solution:
\textbf{Solution:} Given $A\left(5, 0\right)$, it is known that $OA=5$. In $\triangle AOC$, $\sin\angle BAC=\frac{\sqrt{2}}{2}$, \\
$\therefore \angle BAC=45^\circ$, \\
$\therefore OA=OC=5$, i.e., point C is (0, 5), \\
According to the problem, let $y=a\left(x-5\right)\left(x+1\right)$, substituting point C into it results in: \\
$-5a=5$, \\
Solving this gives: $a=-1$, \\
$\therefore$ The equation of the parabola is $\boxed{y=-\left(x-5\right)\left(x+1\right)=-x^{2}+4x+5}$.


Question: As shown in the figure, the parabola intersects the x-axis at $A(5,0)$ and $B(-1,0)$, and intersects the positive half-axis of the y-axis at point $C$. Lines $BC$ and $AC$ are drawn. It is known that $\sin\angle BAC=\frac{\sqrt{2}}{2}$. The line $y=kx$ ($k>0$) intersects segment $AC$ at point $M$. When the triangle with vertices $A$, $O$, and $M$ is similar to $\triangle ABC$, find the value of $k$ and then find the coordinates of point $M$.

Solution:
\textbf{Solution:} From (1) we obtain: $C(0,5)$, $A(5,0)$. Assume the equation of line AC is $y=k_1x+b$. Substituting the coordinates of points A and C, we get:
\[
\begin{cases}
b=5\\
5k_1+b=0
\end{cases}
\]
Solving, we find:
\[
\begin{cases}
b=5\\
k_1=-1
\end{cases}
\]
$\therefore$ The equation of line AC is $y=-x+5$,

$\because$ Line $y=kx$ ($k>0$) intersects segment AC at point M, then let $M\left(m, -m+5\right)$,

$\therefore k=\frac{-m+5}{m}$,

From (1) it is known $OA=OC=5$, $OB=1$,

$\therefore AC=\sqrt{(0-5)^2+(5-0)^2}=5\sqrt{2}$, $AB=6$,

According to the problem, we can divide into two cases:

(1) When $\triangle AOM\sim \triangle ABC$,

$\therefore \frac{AO}{AB}=\frac{AM}{AC}=\frac{5}{6}$,

$\therefore AM=\frac{5}{6}AC=\frac{25\sqrt{2}}{6}$,

$\therefore$ By the distance formula between two points:

$(m-5)^2+(m-5)^2=\frac{625}{18}$,

Solving, we find: $m_1=\frac{5}{6}$, $m_2=\frac{55}{6}$,

$\because 0\le m\le 5$,

$\therefore m=\frac{5}{6}$,

$\therefore M\left(\frac{5}{6}, \frac{25}{6}\right)$, $k=\boxed{5}$;

(2) When $\triangle AOM\sim \triangle ACB$,

$\therefore \frac{AO}{AC}=\frac{AM}{AB}=\frac{5}{5\sqrt{2}}=\frac{\sqrt{2}}{2}$,

$\therefore AM=\frac{\sqrt{2}}{2}AB=3\sqrt{2}$,

$\therefore$ By the distance formula between two points:

$(m-5)^2+(m-5)^2=18$,

Solving, we find: $m_1=2$, $m_2=8$ (discarding as it does not meet the problem criteria),

$\therefore M\left(2, 3\right)$, $k=\boxed{\frac{3}{2}}$;


Question: As shown in the diagram, the vertices A and C of $\triangle ABC$ are located on circle O, and circle O intersects with side AB at point D. Through point D, $DF \parallel BC$ is drawn, intersecting AC at point E and circle O at point F. DC is connected, and point C is the midpoint of arc DF. BC is the tangent to circle O; if the radius of circle O is 3 and $DF=4\sqrt{2}$, find the value of $CE\cdot CA$.

Solution:
\textbf{Solution:} Connect CO and extend it to meet $\odot O$ at G, DF and CG intersect at point M, connect DO, as shown in the figure:\\
Since $OC\perp BC$ and $DF\parallel BC$,\\
it follows that $OM\perp DF$\\
Therefore, $DM=FM=\frac{1}{2}DF=2\sqrt{2}$\\
Since the radius of $\odot O$ is 3,\\
it follows that $OD=OC=OG=3$\\
Therefore, $OM=\sqrt{OD^{2}-DM^{2}}=1$, and $CG=OC+OG=6$\\
Therefore, $MC=OC-OM=2$\\
Since CG is a diameter,\\
it follows that $\angle CAG=90^\circ$\\
Since $\angle MCE=\angle ACG$\\
it follows that $\triangle MCE\sim \triangle ACG$\\
Therefore, $\frac{MC}{CA}=\frac{CE}{CG}$\\
Therefore, $\frac{2}{CA}=\frac{CE}{6}$\\
Therefore, $CE\cdot CA=\boxed{12}$;


Question: As shown in the figure, the vertices A and C of $\triangle ABC$ are on circle O, and circle O intersects side AB at point D. Through point D, draw $DF \parallel BC$, intersecting AC at point E and circle O at point F. Join DC, with point C being the midpoint of arc DF. Given these conditions, connect AF. If $BD = AF$, find the length of AD.

Solution:
\textbf{Solution:} Let's extend CO and intersect $\odot O$ at G, and let DF and CG intersect at point M, then connect DO, AF, CF,\\
according to the conclusion of (2), we have $OM\perp DF$, $MC=2$, $DM=FM=2\sqrt{2}$\\
$\therefore CD=CF=\sqrt{MC^{2}+MF^{2}}=2\sqrt{3}$\\
$\because$DF$\parallel$BC,\\
$\therefore \angle ADF=\angle B$, $\angle BCD=\angle CDF$\\
$\because \angle ADF=\angle ACF$, $\angle CAF=\angle CDF$\\
$\therefore \angle B=\angle ACF$, $\angle BCD=\angle CAF$\\
$\therefore \triangle BCD\sim \triangle CAF$\\
$\therefore \frac{BD}{CF}=\frac{CD}{AF}$\\
$\therefore \frac{BD}{2\sqrt{3}}=\frac{2\sqrt{3}}{AF}$\\
$\because BD=AF$\\
$\therefore BD=AF=2\sqrt{3}$ or $BD=AF=-2\sqrt{3}$ (discard)\\
$\therefore BD=AF=CF=CD=2\sqrt{3}$\\
$\because$ point C is the midpoint of the arc DF.\\
$\therefore \angle CFD=\angle CAF$\\
$\because \angle ECF=\angle FCA$\\
$\therefore \triangle ECF\sim \triangle FCA$\\
$\therefore \frac{EF}{AF}=\frac{CE}{CF}$\\
$\because AF=CF=2\sqrt{3}$\\
$\therefore EF=CE$\\
Let $CE=x$\\
$\therefore ME=FM-EF=FM-CE=2\sqrt{2}-x$\\
$\because CE^{2}=MC^{2}+ME^{2}$\\
$\therefore x^{2}=4+{(2\sqrt{2}-x)}^{2}$\\
$\therefore x=\frac{3\sqrt{2}}{2}$\\
$\therefore CA=\frac{12}{CE}=4\sqrt{2}$\\
$\therefore AE=CA-CE=\frac{5}{2}\sqrt{2}$\\
$\because \angle ADF=\angle B$, $\angle DAE=\angle BAC$\\
$\therefore \triangle ADE\sim \triangle ABC$\\
$\therefore \frac{AD}{AB}=\frac{AE}{AC}=\frac{5}{8}$\\
$\therefore \frac{AD}{AD+BD}=\frac{AD}{AD+2\sqrt{3}}=\frac{5}{8}$\\
$\therefore AD=\boxed{\frac{10\sqrt{3}}{3}}$.


Question: As shown in the figure, in $\triangle ABC$, the circle $\odot O$ with diameter $AB$ intersects $BC$ at point $D$ and the extension line of $CA$ at point $E$. The tangent $DF$ of $\odot O$ is perpendicular to $AC$ at $F$. $AB=AC$. If $CF=2AF$ and $AE=4$, find the radius of $\odot O$.

Solution:
\textbf{Solution:} Connect BE and AD as shown in the figure, \\
$\because$ AB is the diameter of $\odot O$, \\
$\therefore$ AD$\perp$BC, BE$\perp$EC, \\
$\because$ AB=AC, \\
$\therefore$ BD=DC, \\
$\because$ DF$\perp$AC, BE$\perp$EC, \\
$\therefore$ DF$\parallel$BE, \\
$\therefore$ $\frac{CD}{BD}=\frac{CF}{EF}$, \\
$\therefore$ CF=EF, \\
$\because$ $CF=2AF$, AE=4, \\
Let AF=x, then CF=2x, EF=4+x, AC=3x, \\
$\therefore$ 2x=4+x, solving gives: x=4, \\
$\therefore$ AC=12, \\
$\therefore$ AB=AC=12, \\
$\therefore$ The radius of $\odot O$ is $\boxed{6}$.


Question: As shown in Figure 1, in the Cartesian coordinate system, the parabola $y=ax^2-2ax-3a$ ($a<0$) intersects the x-axis at points A and B (A is to the left of B), and intersects the y-axis at point C. Point D is the vertex of the parabola, and the axis of symmetry intersects the x-axis at point E, with $OB=OC$. Find the analytical expression of the parabola?

Solution:
\textbf{Solution:} To solve for the parabola $y=ax^{2}-2ax-3a$, let $y=0$ to get $ax^{2}-2ax-3a=0$. Solving this yields $x=-1$ or $3$, \\
\[\therefore \boxed{A(-1,0)}, \boxed{B(3,0)},\] \\
\[\therefore OA=1, OB=OC=3,\] \\
\[\therefore \boxed{C(0,3)},\] \\
\[\therefore -3a=3,\] \\
\[\therefore a=-1,\] \\
\[\therefore\] the equation of the parabola is \boxed{y=-x^{2}+2x+3}.


Question: As shown in Figure 1, in the Cartesian coordinate system, the parabola $y = ax^2 - 2ax - 3a$ ($a < 0$) intersects the x-axis at points A and B (with A to the left of B), and intersects the y-axis at point C. Point D is the vertex of the parabola, and the axis of symmetry intersects the x-axis at point E, with $OB = OC$. Connect BD. Is there a point F on the parabola such that $\angle FBO = \angle BDE$? If such a point exists, find the coordinates of point F; if not, explain why.

Solution:
\textbf{Solution:} From the given information, we know that $A(-1, 0)$, $B(3, 0)$, $D(1, 4)$,\\
$\therefore DE=4$, $BE=2$,\\
Let $F(x, -x^2+2x+3)$,\\
When $F$ is at ${F}_{1}$, draw ${F}_{1}H\perp$ to the x-axis,\\
$\because \angle BDE=\angle {F}_{1}BH$, $\angle {F}_{1}HB=\angle BED=90^\circ$,\\
$\therefore \triangle BDE\sim \triangle {F}_{1}BH$,\\
$\therefore \frac{BE}{DE}=\frac{{F}_{1}H}{BH}$, which means $\frac{1}{2}=\frac{-x^2+2x+3}{3-x}$,\\
$\therefore 2x^2-5x-3=0$,\\
Solve to get: $x=3$ or $x=-\frac{1}{2}$,\\
$\because x<0$,\\
$\therefore x=-\frac{1}{2}$, at this time $y=\frac{7}{4}$,\\
Thus, ${F}_{1}\left(-\frac{1}{2}, \frac{7}{4}\right)$;\\
When $F$ is at ${F}_{2}$, draw ${F}_{2}K\perp$ to the x-axis,\\
$\because \angle BDE=\angle {F}_{2}BK$, $\angle {F}_{2}KB=\angle BED=90^\circ$,\\
$\therefore \triangle BDE\sim \triangle {F}_{2}BK$,\\
$\therefore \frac{BE}{DE}=\frac{{F}_{2}K}{BK}$, which means $\frac{1}{2}=\frac{-(-x^2+2x+3)}{3-x}$,\\
$\therefore 2x^2-3x-9=0$,\\
Solve to get: $x=3$ or $x=-\frac{3}{2}$,\\
$\because x<0$,\\
$\therefore x=-\frac{3}{2}$, at this time $y=-\frac{9}{4}$,\\
$\therefore$ the coordinates of point $F$ are $\boxed{\left(-\frac{3}{2}, -\frac{9}{4}\right)}$ or $\left(-\frac{1}{2}, \frac{7}{4}\right)$;


Question: As shown in Figure 1, in the Cartesian coordinate system, the parabola $y=ax^2-2ax-3a$ (where $a<0$) intersects the x-axis at points A, B (with A to the left of B), and intersects the y-axis at point C. Point D is the vertex of the parabola, and its axis of symmetry intersects the x-axis at point E. Also, OB=OC. As shown in Figure 2, point P is a moving point on the line $y=5$ (point P is not on the axis of symmetry of the parabola). There are two lines $l_1$ and $l_2$ passing through point P, each having only one common point with the parabola, and neither is parallel to the y-axis. Lines $l_1$ and $l_2$ intersect the axis of symmetry of the parabola at points M and N, respectively. Point G is a point on the axis of symmetry below points M and N, and it always satisfies $GP^2=GM\cdot GN$. Find the coordinates of point G.

Solution:
\textbf{Solution:}

Let $P(p, 5)$, and let $y=5$ intersect with the axis of symmetry at $H$,

then the equation of the line passing through point $P$ is $y=k(x-p)+5$,

\[
\therefore \left\{
\begin{array}{l}
y=-x^{2}+2x+3\\
y=k(x-p)+5
\end{array}
\right.,
\]

\[
\therefore -x^{2}+(2-k)x+kp-2=0,
\]

let $\Delta =b^{2}-4ac=(2-k)^{2}+4(kp-2)=0$,

\[
\therefore k^{2}+(4p-4)k-4=0,
\]

\[
\therefore k_{1}+k_{2}=-\frac{b}{a}=4-4p, \quad k_{1}k_{2}=\frac{c}{a}=-4,
\]

when $x=1$, $y=k(1-p)+5$,

thus $y_{M}=k_{1}(1-p)+5=y_{1}$, $y_{N}=k_{2}(1-p)+5=y_{2}$,

let $G(1, t)$,

\[
\because GP^{2}=GM\cdot GN,
\]

\[
\therefore GH^{2}+PH^{2}=GM\cdot GN, \text{ i.e., } (5-t)^{2}+(p-1)^{2}=(y_{1}-t)(y_{2}-t),
\]

\[
\therefore (5-t)^{2}+(p-1)^{2}=(y_{1}-t)(y_{2}-t),
\]

\[
\therefore (5-t)^{2}+(p-1)^{2}=y_{1}y_{2}-t(y_{1}+y_{2})+t^{2},
\]

\[
\therefore (5-t)^{2}+(p-1)^{2}=[k_{1}(1-p)+5][k_{2}(1-p)+5]-t[k_{1}(1-p)+5+k_{2}(1-p)+5]+t^{2},
\]

\[
\therefore (5-t)^{2}+(p-1)^{2}=k_{1}k_{2}(1-p)^{2}+5(k_{1}+k_{2})(1-p)+25-t(k_{1}+k_{2})(1-p)-10t+t^{2},
\]

\[
\therefore (5-t)^{2}+(p-1)^{2}=-4(1-p)^{2}+20(1-p)^{2}+25-4t(1-p)^{2}-10t+t^{2},
\]

\[
\therefore 25-10t+t^{2}+(p-1)^{2}=-4(1-p)^{2}+20(1-p)^{2}+25-4t(1-p)^{2}-10t+t^{2},
\]

\[
\therefore (15-4t)(1-p)^{2}=0,
\]

\[
\because \text{ The equation always holds, so it is independent of } p,
\]

\[
\therefore 15-4t=0,
\]

Solving for $t$, we get $t=\frac{15}{4}$,

The coordinates of point $G$ are $\left(1, \boxed{\frac{15}{4}}\right)$.


Question: As shown in the figure, point \(P\) is a point on the side \(AB\) of the square \(ABCD\) (not coinciding with points \(A\) or \(B\)), connect \(PD\) and rotate the line segment \(PD\) \(90^\circ\) clockwise around point \(P\) to get the line segment \(PE\), which intersects side \(BC\) at point \(F\), connect \(BE\), \(DF\). If \(\angle ADP = 32^\circ\), what is the measure of \(\angle FPB\)?

Solution:
\textbf{Solution:} Since the quadrilateral ABCD is a square,\\
it follows that $\angle A=\angle PBC=90^\circ$ and AB=AD,\\
therefore, $\angle ADP+\angle APD=90^\circ$,\\
since $\angle DPE=90^\circ$,\\
it follows that $\angle APD+\angle EPB=90^\circ$,\\
therefore, $\angle ADP=\angle FPB=\boxed{32^\circ}$;


Question: As shown in the figure, point $P$ is a point on the side $AB$ of the square $ABCD$ (not coinciding with points $A$, $B$). Connect $PD$ and rotate the line segment $PD$ clockwise by $90^\circ$ around point $P$ to obtain the line segment $PE$, where $PE$ intersects side $BC$ at point $F$. Connect $BE$ and $DF$. If $AP=\sqrt{3}$, what is the length of $BE$?

Solution:
\textbf{Solution:} Draw $EQ \perp AB$ through point E to intersect the extended line of AB at point Q, hence $\angle EQP = \angle A = 90^\circ$,\\
In $\triangle PAD$ and $\triangle EQP$,\\
$\left\{\begin{array}{l}
\angle A = \angle EQP \\
\angle ADP = \angle EPB \\
PD = PE
\end{array}\right.$,\\
$\therefore \triangle PAD \cong \triangle EQP$ (AAS),\\
$\therefore EQ = AP = \sqrt{3}$, AD = AB = PQ,\\
$\therefore AP = EQ = BQ$,\\
$\therefore \angle CBE = \angle EBQ = 45^\circ$;\\
$\therefore BE = \sqrt{2}EQ = \boxed{\sqrt{6}}$.


Question: As shown in the figure, point $P$ is on the side $AB$ of the square $ABCD$ (not coinciding with points $A$, $B$). Connect $PD$ and rotate the line segment $PD$ $90^\circ$ clockwise around point $P$ to obtain the line segment $PE$, where $PE$ intersects side $BC$ at point $F$. Connect $BE$, $DF$. If $\triangle PFD\sim \triangle BFP$, find $\frac{AP}{AB}$?

Solution:
\textbf{Solution:} From the given information,

$\because \triangle PFD \sim \triangle BFP$,

$\therefore \frac{PB}{BF} = \frac{PD}{PF}$,

$\therefore PD \cdot BF = PB \cdot PF$,

$\because \angle ADP = \angle EPB, \angle CBP = \angle A = 90^\circ$,

$\therefore \triangle DAP \sim \triangle PBF$,

$\therefore \frac{PD}{PF} = \frac{PA}{BF}$,

$\therefore PD \cdot BF = AP \cdot PF$,

$\therefore PB \cdot PF = AP \cdot PF$,

$\therefore PA = PB$,

$\because AB = PA + PB = 2PA$,

$\therefore \frac{AP}{AB} = \boxed{\frac{1}{2}}$.


Question: As shown in Figure 1, given $\triangle ABC$, $\angle CAB=45^\circ$, $AB=7$, $AC=3\sqrt{2}$, and $CD\perp AB$ at point D. E is a moving point on side BC. A circle $\odot O$ is drawn with DE as the diameter, intersecting BC at point F and AB at point G. Lines DF and FG are drawn. $\angle BCD=\angle FDB$. When point E is on segment BF, and $\triangle DFG$ is an isosceles triangle, what is the length of DG?

Solution:
\textbf{Solution:}

(i) When $DF=DG$, as shown in the figure:

Since $\angle CAB=45^\circ$, $CD\perp AB$, $AC=3\sqrt{2}$

Therefore $AD=CD=3$

Since $AB=7$

Therefore $BD=7-3=4$

Therefore $BC=\sqrt{3^2+4^2}=5$

Therefore $DF=\frac{3\times 4}{5}=\frac{12}{5}$

Therefore $DG=\boxed{\frac{12}{5}}$

(ii) As shown in the figure:

When $DF=FG$, draw $FH\perp BD$ intersecting $BD$ at point $H$,

Therefore $\triangle DFH\sim \triangle CBD$

Therefore $\frac{DH}{CD}=\frac{DF}{CB}$

Therefore $DH=DF\times \frac{3}{5}=\frac{12}{5}\times \frac{3}{5}=\frac{36}{25}$

Therefore $DG=2DH=\boxed{\frac{72}{25}}$

(iii) As shown in the figure:

When $FG=DG$, $\angle 1=\angle 2$

Since $\angle 1+\angle 3=\angle 2+\angle 4=90^\circ$

Therefore $\angle 3=\angle 4$

Therefore $FG=GB=DG$

Therefore $DG=\frac{1}{2}BD=\boxed{2}$


Question: As shown in Figure 1, given $ \triangle ABC$ with $\angle CAB=45^\circ$, $AB=7$, and $AC=3\sqrt{2}$. $CD\perp AB$ at point D. E is a moving point on side BC. A circle $\odot O$ is drawn with DE as its diameter, intersecting BC at F and AB at point G. Connect DF and FG. As depicted in Figure 2, the other intersection point of $\odot O$ and CD is P. If ray AP passes through point F, find the value of $\frac{AP}{DE}$.

Solution:
\textbf{Solution:} As shown in the figure:\\
$\because$ Quadrilateral PDEF is a cyclic quadrilateral inside circle $O$\\
$\therefore \angle APD=\angle DEF$\\
$\because \angle APD+\angle PAD=\angle DEF+\angle EDF=90^\circ$\\
$\therefore \angle PAD=\angle EDF$\\
Connect PG\\
$\because \angle PAD=\angle EDF$\\
$\angle ADP=\angle DFE=90^\circ$\\
$\therefore \triangle ADP\sim \triangle DFE$\\
$\therefore \frac{AP}{DE}=\frac{AD}{DF}=3\times \frac{5}{12}=\boxed{\frac{5}{4}}$


Question: As shown in the figure, in quadrilateral $ABCD$, draw $BE\perp CD$ at point $E$, connect $AE$, and let $F$ be a point on $AE$ with $\angle AFB=\angle D$. $\triangle ABF\sim \triangle EAD$. If $AB=4\sqrt{3}$, $AD=6$, and $\angle BAE=30^\circ$, find the length of $BF$.

Solution:
\textbf{Solution:} \\
Since $BE\perp CD$, \\
therefore, $\angle BEC=90^\circ$, \\
since $AB\parallel CD$, \\
therefore, $\angle ABE=\angle BEC=90^\circ$, \\
since $\angle BAE=30^\circ$, \\
therefore, $\cos30^\circ=\frac{AB}{AE}$, \\
thus, $AE=8$, \\
since $\triangle ABF\sim \triangle EAD$, \\
therefore, $\frac{BF}{AD}=\frac{AB}{AE}$, \\
thus, $\frac{BF}{6}=\frac{4\sqrt{3}}{8}$, \\
we find: $BF=\boxed{3\sqrt{3}}$.


Question: As shown in the figure, $AB$ is a chord of the circle $\odot O$, $OP\perp OA$ intersects $AB$ at point $P$, and the line passing through point $B$ intersects the extension of $OP$ at point $C$, and $BC$ is a tangent to $\odot O$. If $OA=6$, $OP=2$, find the length of $CB$.

Solution:
\textbf{Solution}: Let $BC=x$, then $PC=x$,\\
in $\triangle OBC$, $OB=OA=6$, $OC=CP+OP=x+2$,\\
since $OB^2+BC^2=OC^2$,\\
thus $6^2+x^2=(x+2)^2$,\\
we find $x=8$,\\
that is, the length of $BC$ is $\boxed{8}$;


Question: As illustrated in the figure, AB is a chord of the circle $\odot O$, $OP\perp OA$ intersects AB at point P, and a line passing through point B intersects the extension of $OP$ at point C, with $BC$ being a tangent of $\odot O$. Let the area of $\triangle AOP$ be $S_1$ and the area of $\triangle BCP$ be $S_2$, with $\frac{S_1}{S_2}=\frac{2}{5}$. Given that the radius of $\odot O$ is $6$ and $BP=4\sqrt{5}$, find $\tan\angle APO$?

Solution:
\textbf{Solution:} As shown in the figure, construct $CD\perp BP$ at $D$,\\
$\because PC=CB$,\\
$\therefore PD=BD=\frac{1}{2}PB=2\sqrt{5}$,\\
$\because \angle PDC=\angle AOP=90^\circ$, $\angle APO=\angle CPD$,\\
$\therefore \triangle AOP\sim \triangle PCD$,\\
$\because \frac{{S}_{1}}{{S}_{2}}=\frac{2}{5}$,\\
$\therefore \frac{{S}_{\triangle AOP}}{{S}_{\triangle PCD}}=\frac{4}{5}$,\\
$\therefore \frac{OA^{2}}{CD^{2}}=\frac{4}{5}$,\\
$\because OA=6$,\\
$\therefore CD=3\sqrt{5}$,\\
$\therefore \tan\angle APO=\tan\angle CBP=\frac{CD}{BD}=\frac{3\sqrt{5}}{2\sqrt{5}}=\boxed{\frac{3}{2}}$.


Question: As shown in the figure, the line $y=2x$ passing through the origin intersects the inverse proportion function $y_{1}=\frac{2}{x}$ at point $B$, and intersects another inverse proportion function $y_{2}=\frac{k}{x}$ at point $C$, with $OB=BC$. The abscissa of point $A$ is $4$ and is a point on $y_{1}=\frac{2}{x}$. A line is drawn from point $B$ such that $BD\perp x$ axis, with the foot of the perpendicular at point $D$. Find the analytical expression of the inverse proportion function $y_{2}=\frac{k}{x}$ and the line $AD$.

Solution:
\textbf{Solution:} Solving the system of equations, we get
\begin{align*}
\begin{cases}
y=2x \\
y=\frac{2}{x}
\end{cases}
\end{align*}
which yields
\begin{align*}
\begin{cases} x_{1}=1 \\ y_{1}=2 \end{cases}, \quad
\begin{cases} x_{2}=-1 \\ y_{2}=-2 \end{cases}
\end{align*}
\(\therefore B(1, 2)\)

\(\because\) BD \(\perp\) x-axis,

\(\therefore D(1, 0)\)

\(\therefore BD=2, OD=1.\)

When \(x=4\), \(y_{1}=\frac{2}{4}=\frac{1}{2}\)

\(\therefore A(4, \frac{1}{2})\)

Through point C, construct CE \(\perp\) x-axis at point E,

\(\therefore\) BD \(\parallel\) CE

\(\therefore \triangle BOD \sim \triangle COE\)

\(\therefore \frac{OB}{OC}=\frac{BD}{CE}=\frac{OD}{OE}\)

\(\because\) OB = BC, BD = 2

\(\therefore \frac{2}{CE}=\frac{1}{OE}=\frac{1}{2}\)

\(\therefore\) CE = 4, OE = 2

\(\therefore C(2, 4)\)

Substituting into \(y_{2}=\frac{k}{x}\), we get \(\frac{k}{2}=4\)

\(\therefore k=8\)

\(\therefore y_{2}=\frac{8}{x}\)

Assume the equation of line AD is \(y=kx+b\)

Substituting \((1, 0)\) and \((4, \frac{1}{2})\), we get
\begin{align*}
\begin{cases}
k+b=0 \\
4k+b=\frac{1}{4}
\end{cases}
\end{align*}
Solving, we find
\begin{align*}
\begin{cases}
k=\frac{1}{6} \\
b=-\frac{1}{6}
\end{cases}
\end{align*}
\(\therefore\) The equation of line AD is \(y=\frac{1}{6}x-\frac{1}{6}\), thus the answer is \(\boxed{y=\frac{1}{6}x-\frac{1}{6}}\).


Question: As shown in the figure, the line passing through the origin $y=2x$ intersects the inverse proportional function $y_{1}=\frac{2}{x}$ at point $B$, and intersects another inverse proportional function $y_{2}=\frac{k}{x}$ at point $C$, with $OB=BC$. The point $A$ has an abscissa of $4$ and lies on the curve $y_{1}=\frac{2}{x}$. A perpendicular line $BD \perp x$-axis is drawn from point $B$, with $D$ being the foot of the perpendicular. Is there a point $P$ on the graph of the inverse proportional function $y_{2}=\frac{k}{x}$ within the first quadrant such that $S_{\triangle ABP}=\frac{5}{11}S_{\text{quadrilateral} ADBP}$? If such a point exists, find the coordinates of point $P$; if not, please state the reason.

Solution:
\textbf{Solution:} From the given problem, we have $A(4, \frac{1}{2}), B(1, 2), D(1, 0)$,\\
$\therefore S_{\triangle ABD}=\frac{1}{2}BD\cdot |x_{A}-x_{B}|$, $AB=\sqrt{(4-1)^{2}+(\frac{1}{2}-2)^{2}}=\frac{3\sqrt{5}}{2}$\\
Let the equation of line AB be $y=mx+n$. Substituting $A(4, \frac{1}{2}), B(1, 2)$, we get $\left\{\begin{array}{l}m+n=2 \\ 4m+n=\frac{1}{2}\end{array}\right.$, solving this gives, $\left\{\begin{array}{l}m=-\frac{1}{2} \\ n=\frac{5}{2}\end{array}\right.$\\
$\therefore$ The equation of line AB is $y=-\frac{1}{2}x+\frac{5}{2}$\\
Construct PR$\perp$x-axis from P, meeting AB at point F,\\
$\therefore S_{\triangle ABP}=S_{\triangle AFP}+S_{\triangle BFP}$\\
$=\frac{1}{2}PF\cdot |x_{P}-x_{B}|+\frac{1}{2}PF\cdot |x_{A}-x_{P}|$\\
$=\frac{1}{2}PF\cdot |x_{A}-x_{B}|$\\
$\because S_{\triangle ABP}=\frac{5}{11}S_{\text{quadrilateral}ADBP}$,\\
$\therefore S_{\triangle ABP}=\frac{5}{6}S_{\triangle ABD}$\\
$\therefore PF=\frac{5}{6}BD=\frac{5}{6}\times 2=\frac{5}{3}$\\
Let $P(x, \frac{8}{x}), F(x, -\frac{1}{2}x+\frac{5}{2})$,\\
$\therefore PF=\frac{8}{x}-\left(-\frac{1}{2}x+\frac{5}{2}\right)=\frac{8}{x}+\frac{1}{2}x-\frac{5}{2}=\frac{5}{3}$\\
Rearranging gives: $3x^{2}-25x+48=0$\\
Solving, we find $x_{1}=\frac{16}{3}, x_{2}=3$.\\
Upon verification, $x_{1}=\frac{16}{3}, x_{2}=3$ are the roots of the original equation\\
$\therefore \boxed{P_{1}\left(\frac{16}{3}, \frac{3}{2}\right)}, \boxed{P_{2}\left(3, \frac{8}{3}\right)}$


Question: As shown in the figure, the line passing through the origin $y=2x$ intersects the inverse proportion function $y_{1}=\frac{2}{x}$ at point $B$, and intersects another inverse proportion function $y_{2}=\frac{k}{x}$ at point $C$, with $OB=BC$. Point $A$ has an abscissa of $4$ and is a point on $y_{1}=\frac{2}{x}$. A perpendicular line from point $B$ to the x-axis is drawn, and the foot of the perpendicular is point $D$. Suppose a moving point $Q$ is on the graph of $y_{2}=\frac{k}{x}$. Is there a point $H$ in the plane such that points $A$, $B$, $Q$, $H$ can form a rectangle with $AB$ as one side? If it exists, please write down the coordinates of point $H$ directly; if not, please explain the reason.

Solution: \textbf{Solution:} There exists a point $\boxed{H\left(5, \frac{5}{2}\right)}$ such that points $A$, $B$, $Q$, and $H$ can form a rectangle with $AB$ as a side.


Question: In the Cartesian coordinate system shown in the figure, $O$ is the origin. The parabola $y=-x^{2}+6x+3$ intersects the $y$-axis at point $A$. A line $AB$ is drawn parallel to the $x$-axis from point $A$, and it intersects the parabola at point $B$. The line segment $OB$ is connected. A point $P$ is located above segment $AB$ on the parabola, and a line $PA$ is drawn. Perpendicular $PQ$ is dropped from $P$ to $AB$, with the foot of the perpendicular being $H$, and it intersects $OB$ at point $Q$. Find the length of $AB$.

Solution:
\textbf{Solution:} Given $y=-x^{2}+6x+3$, let $x=0$, then we have $y=3$. Thus, point $A(0,3)$, \\
let $y=-x^{2}+6x+3=3$, solving this gives $x=0$ or $x=6$. Hence, point $B(6,3)$, \\
$\therefore AB=\boxed{6}$.


Question: As shown in the Cartesian coordinate system, the point $O$ is the origin. The parabola $y=-x^2+6x+3$ intersects the $y$-axis at point $A$. A line segment $AB$ is drawn parallel to the $x$-axis, intersecting the parabola at point $B$, and $OB$ is connected. Point $P$ lies above $AB$ on the parabola, and $PA$ is connected. A perpendicular $PQ$ is drawn to $AB$ at the foot point $H$, intersecting $OB$ at point $Q$. When $\angle APQ = \angle B$, what are the coordinates of point $P$?

Solution:
\textbf{Solution:} Let $P(m, -m^{2}+6m+3)$,\\
since $\angle P = \angle B$ and $\angle AHP = \angle OAB = 90^\circ$,\\
it follows that $\triangle ABO \sim \triangle HPA$, hence $\frac{HP}{AB}=\frac{AH}{AO}$,\\
thus $\frac{-m^{2}+6m}{6}=\frac{m}{3}$,\\
solving for $m$ gives $m=4$.\\
Therefore, $P(4,11)$. Thus, the coordinates of point $P$ are $\boxed{(4,11)}$.


Question: As shown in the Cartesian coordinate system, let $O$ be the origin. The parabola $y=-x^2+6x+3$ intersects the $y$-axis at point $A$. A line segment $AB$ is drawn parallel to the $x$-axis from $A$ and intersects the parabola at point $B$. The segment $OB$ is then connected. Let $P$ be a point on the parabola above segment $AB$, and draw $PA$ and $PQ \perp AB$ with the foot of the perpendicular on $AB$ at $H$ and intersecting $OB$ at point $Q$. When the area of $\triangle APH$ is twice the area of the quadrilateral $AOQH$, find the coordinates of point $P$.

Solution:
\textbf{Solution:} When the area of $\triangle APH$ is twice the area of quadrilateral $AOQH$,\\
then $2(AO+HQ)=PH$,\\
$\therefore 2\left(3+ \frac{6-m}{2}\right)=-m^2+6m$,\\
solving this gives: $m_1=4$, $m_2=3$,\\
$\therefore \boxed{P(4,11)} or \boxed{P\left(3,12\right)}$.


Question: As illustrated, it is known in $\triangle ABC$ that $\angle BAC=30^\circ$ and $\angle C=90^\circ$. The coordinates of point A are $(-1,0)$ and those of point B are $(3,0)$. The parabola $y_1$ has its vertex labeled as Q and passes through the three vertices A, B, and C of $\triangle ABC$ (with point A to the left of B and point C below the x-axis). Another parabola $y_2$ also intersects the x-axis at points A and B, with its vertex labeled as P. The task is to find the coordinates of point C and the coordinates of the vertex Q of the parabola $y_1$.

Solution: Solution: Since $\angle BAC=30^\circ$ and $\angle C=90^\circ$, with A located at $(-1,0)$ and B at $(3,0)$, it follows that $AC=2\sqrt{3}$. Consequently, $y_C=-\sqrt{3}$. Therefore, the coordinates of point C are $(2, -\sqrt{3})$. Given that the parabola $y_1$ passes through points A, B, and C, let the equation of the parabola $y_1$ be $y_1=a(x+1)(x-3)$. It follows that $-\sqrt{3}=-3a$, hence $a=\frac{\sqrt{3}}{3}$. Therefore, $y_1=\frac{\sqrt{3}}{3}(x+1)(x-3)=\frac{\sqrt{3}}{3}(x-1)^2-\frac{4\sqrt{3}}{3}$. Hence, the coordinates of the vertex Q are $\boxed{(1, -\frac{4\sqrt{3}}{3})}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $\angle BAC=30^\circ$, $\angle C=90^\circ$, the coordinate of point A is $(-1, 0)$, and the coordinate of point B is $(3, 0)$. The vertex of the parabola $y_1$ is denoted as Q, and it passes through the three vertices A, B, and C of $\triangle ABC$ (with point A to the left of point B, and point C below the x-axis). The parabola $y_2$ also intersects the x-axis at points A and B, and its vertex is P. When the value of $BP+CP$ is minimized, find the equation of the parabola $y_2$.

Solution:
Solution: Since the parabola $y_{1}$ and the parabola $y_{2}$ intersect the $x$-axis at the same points, the axis of symmetry of the parabola $y_{2}$ is $x=1$. Since point A is symmetrical to point B about the line $x=1$, the value of $BP+CP$ is minimized when $P$ is the intersection of line $AC$ and the axis of symmetry. Let the equation of line $AC$ be: $y=kx+b$. Then we have
\[
\left\{
\begin{array}{l}
-k+b=0\\
2k+b=-\sqrt{3}
\end{array}
\right.
\]
Solving this, we get
\[
\left\{
\begin{array}{l}
k=-\frac{\sqrt{3}}{3}\\
b=-\frac{\sqrt{3}}{3}
\end{array}
\right.
\]
Therefore, the equation of line $AC$ is: $y=-\frac{\sqrt{3}}{3}x-\frac{\sqrt{3}}{3}$. Thus, the coordinates of point $P$ are $\left(1，-\frac{2\sqrt{3}}{3}\right)$. Let $y_{2}=m(x+1)(x-3)$, then $-4m=-\frac{2\sqrt{3}}{3}$, therefore $m=\frac{\sqrt{3}}{6}$, hence the equation of the parabola $y_{2}$ is: $y_{2}=\frac{\sqrt{3}}{6}(x+1)(x-3)=\frac{\sqrt{3}}{6}x^{2}-\frac{\sqrt{3}}{3}x-\frac{\sqrt{3}}{2}$. Therefore, the answer is $y_{2}=\boxed{\frac{\sqrt{3}}{6}x^{2}-\frac{\sqrt{3}}{3}x-\frac{\sqrt{3}}{2}}$.