Question: As shown in the figure, it is known that the parabola $y=x^2+bx+c$ passes through points A($-3$, $0$) and C($0$, $-3$). Determine the coordinates of another intersection point B of the parabola with the x-axis, and based on the graph, write the range of values for $x$ when $y<0$.

Solution:
\textbf{Solution:} From (1) we know the equation of the parabola is $y = x^2 + 2x - 3$,\\
When $y = 0$, $x^2 + 2x - 3 = 0$,\\
We find: $x_1 = 1$, $x_2 = -3$,\\
$\therefore$ the other intersection point B of the parabola and the x-axis is at $\boxed{(1,0)}$,\\
From the graph, we know that when $y < 0$, $\boxed{-3 < x < 1}$.


Question: As shown in the figure, it is known that the parabola $y=ax^{2}+bx+c\ (a\ne 0)$ has its vertex at point $Q(2, -1)$, it intersects the y-axis at point $C(0, 3)$, and intersects the x-axis at points $A$ and $B$ (with point $A$ to the right of point $B$). Point $P$ is a moving point on this parabola, moving from point $C$ along the parabola towards point $A$ (point $P$ does not coincide with $A$). A line is drawn through point $P$ parallel to the y-axis, intersecting $AC$ at point $D$. What is the equation of this parabola?

Solution:
\textbf{Solution:} Since the vertex of the parabola is Q(2,−1),\\
hence, let $y=a(x-2)^{2}-1$,\\
Substituting C(0, 3) into the above equation, we get $3=a(0-2)^{2}-1$,\\
Solving for $a$, we find that $a=1$,\\
therefore $y=(x-2)^{2}-1$,\\
which simplifies to $\boxed{y=x^{2}-4x+3}$.


Question: As shown in the diagram, it is known that the parabola $y=ax^2+bx+c\ (a\ne 0)$ has its vertex at point Q$(2, -1)$, and it intersects with the y-axis at point C$(0, 3)$ and with the x-axis at points A and B (point A to the right of point B). Point P is a moving point on the parabola, moving from point C towards point A along the parabola (point P does not coincide with A). A line through point P is drawn, $PD \parallel y$-axis, intersecting AC at point D. When $\triangle ADP$ is a right-angled triangle, what are the coordinates of point P?

Solution:
\textbf{Solution:} There are two cases to consider: When point $P_1$ is the right angle vertex, point $P_1$ coincides with point B (as shown in the diagram)\\
    Let $y=0$, then $x^{2}-4x+3=0$\\
    Solving this, we get $x_{1}=1$, $x_{2}=3$\\
    Since point A is to the right of point B,\\
    Therefore, B(1, 0), A(3, 0)\\
    Hence, $P_1$(1, 0)
    When point A is the right angle vertex of $\triangle APD_{2}$ (as shown in the diagram)\\
    Since OA=OC and $\angle AOC=90^\circ$,\\
    Therefore, $\angle OAD_{2}=45^\circ$\\
    When $\angle D_{2}AP_{2}=90^\circ$, $\angle OAP_{2}=45^\circ$,\\
    Hence, AO bisects $\angle D_{2}AP_{2}$\\
    Also, since $P_{2}D_{2}\parallel y$-axis,\\
    Therefore, $P_{2}D_{2}\perp$ AO,\\
    Therefore, $P_{2}$ and $D_{2}$ are symmetric about the $x$-axis\\
    Let the equation of line AC be $y=kx+b$\\
    Substituting A(3, 0), C(0, 3) into the equation, we get\\
    $\left\{\begin{array}{l}
    0=3k+b\\
    3=b
    \end{array}\right.$,\\
    Therefore, $\left\{\begin{array}{l}
    k=-1\\
    b=3
    \end{array}\right.$\\
    Hence, $y=-x+3$\\
    Since $D_{2}$ is on $y=-x+3$ and $P_{2}$ is on $y=x^{2}-4x+3$,\\
    Let $D_{2}$($x$, $-x+3$), $P_{2}$($x$, $x^{2}-4x+3$)\\
    Hence, ($-x+3$) + ($x^{2}-4x+3$) = 0\\
    That is, $x^{2}-5x+6=0$,\\
    Therefore, $x_{1}=2$, $x_{2}=3$ (discard)\\
    Therefore, when $x=2$, $y=x^{2}-4x+3=2^{2}-4\times 2+3=-1$\\
    Hence, the coordinates of $P_{2}$ are $P_{2}$(2, $-1$) (i.e., the vertex of the parabola)\\
    Therefore, the coordinates of point P are \boxed{P_{1}(1, 0)}, \boxed{P_{2}(2, -1)}


Question: As shown in the figure, it is known that the hyperbola $y=\frac{k}{x}$ ($k\ne 0$) and the line $y=mx+n$ intersect at points $A$ and $B$, with $B$ having coordinates $(2, -3)$. The line $AC$ is perpendicular to the $y$-axis at point $C$, and $AC=\frac{3}{2}$. What are the analytical equations of the hyperbola and the line?

Solution:
\textbf{Solution:}

Since $B$ lies on the hyperbola $y=\frac{k}{x}$ ($k\ne 0$), and the coordinates of point $B$ are $(2, -3)$,\\
hence $k=-6$,\\
therefore the equation of the hyperbola is: $y=-\frac{6}{x}$.\\
Since $AC$ is perpendicular to the y-axis at point $C$, and $AC=\frac{3}{2}$,\\
thus the x-coordinate of point $A$ is $-\frac{3}{2}$,\\
substituting into $y=-\frac{6}{x}$ gives $y=-\frac{6}{-\frac{3}{2}}=4$,\\
hence $A\left(-\frac{3}{2}, 4\right)$,\\
substituting the coordinates of A and B into $y=mx+n$, we obtain $\begin{cases}-\frac{3}{2}m+n=4\\ 2m+n=-3\end{cases}$,\\
solving this yields $\begin{cases}m=-2\\ n=1\end{cases}$,\\
therefore, the equation of line $AB$ is $\boxed{y=-2x+1}$.


Question: As shown in the figure, it is known that the hyperbola $y=\frac{k}{x} \ (k \ne 0)$ and the line $y=mx+n$ intersect at points $A$ and $B$, with the coordinates of point $B$ being $(2, -3)$. The line $AC$ is perpendicular to the $y$-axis at point $C$, with $AC = \frac{3}{2}$. If the area of $\triangle AOB$ is $S_{1}$ and the area of $\triangle ACB$ is $S_{2}$, what is the value of $\frac{S_{1}}{S_{2}}$?

Solution:
\textbf{Solution:} In the equation $y=-2x+1$, let $y=0$, then we solve for $x=\frac{1}{2}$,\\
$\therefore$ the coordinates of the intersection point of line $AB$ and the x-axis are $\left(\frac{1}{2}, 0\right)$,\\
$\therefore$ ${S}_{1}=\frac{1}{2}\times \frac{1}{2}\times 4+\frac{1}{2}\times \frac{1}{2}\times 3=\frac{7}{4}$,\\
$\because$ ${S}_{2}=\frac{1}{2}\times \frac{3}{2}\times (4+3)=\frac{21}{4}$\\
$\therefore$ $\frac{{S}_{1}}{{S}_{2}}=\boxed{\frac{1}{3}}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=kx+b$ ($k \neq 0$) intersects with the hyperbola $y=\frac{m}{x}$ ($m \neq 0$) at points $A(2, -3)$ and $B(n, 2)$. What are the expressions for the functions of the line and the hyperbola respectively?

Solution:
\textbf{Solution:} Since the hyperbola $y = \frac{m}{x}$ (where $m \neq 0$) passes through point $A(2, -3)$,
it follows that $m = -6$.
Therefore, the equation of the hyperbola is $y = -\frac{6}{x}$.

Given that point $B(n, 2)$ lies on the hyperbola $y = -\frac{6}{x}$,
the coordinates of point $B$ are $(-3, 2)$.

Since the line $y = kx + b$ passes through points $A(2, -3)$ and $B(-3, 2)$,
we have the system of equations $\begin{cases} 2k + b = -3 \\ -3k + b = 2 \end{cases}$.

Solving this system yields $\begin{cases} k = -1 \\ b = -1 \end{cases}$.

Therefore, the equation of the line is $\boxed{y = -x - 1}$;


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=kx+b$ ($k\neq 0$) intersects with the hyperbola $y=\frac{m}{x}$ ($m\neq 0$) at points $A(2,-3)$ and $B(n,2)$. Directly write out the solution set of the inequality $kx+b>\frac{m}{x}$ in terms of $x$.

Solution: \textbf{Solution:} Solve: \boxed{x < -3} or \boxed{0 < x < 2}.


Question: As shown in the figure, in the Cartesian coordinate system, the line $AB$ intersects the $x$ axis and $y$ axis at points $A$ and $B$, respectively; the line $CD$ intersects the $x$ axis and $y$ axis at points $C$ and $D$, respectively. The lines $AB$ and $CD$ intersect at point $E$. The lengths of the segments $OA$ and $OC$ are the two roots of the quadratic equation $x^{2}-18x+72=0$ (with $OA>OC$). Given that $BE=5$ and $\tan\angle ABO=\frac{3}{4}$, what are the coordinates of points $A$ and $C$?

Solution:
\textbf{Solution:} Since ${x}^{2}-18x+72=0$, \\
we find $x_{1}=6, x_{2}=12$, \\
Since the lengths of line segments $OA$ and $OC$ are the two roots of the quadratic equation ${x}^{2}-18x+72=0$ ($OA>OC$), \\
therefore $OA=12, OC=6$, \\
therefore, the coordinates of point $A$ are $\boxed{(-12,0)}$, and the coordinates of point $C$ are $\boxed{(6,0)}$;


Question: As shown in the figure, in the Cartesian coordinate system, the line $AB$ intersects the $x$-axis and $y$-axis at point $A$ and $B$, respectively, and the line $CD$ intersects the $x$-axis and $y$-axis at point $C$ and $D$, respectively. The lines $AB$ and $CD$ intersect at point $E$. The lengths of segments $OA$ and $OC$ are the two roots of the quadratic equation ${x}^{2}-18x+72=0$ (with $OA>OC$), and $BE=5$, $\tan\angle ABO=\frac{3}{4}$. If the graph of the inverse proportion function $y=\frac{k}{x}$ passes through point $E$, what is the value of $k$?

Solution:
\textbf{Solution:} Draw a line $EH \perp AC$ at point H through point E,\\
because $\tan\angle ABO=\frac{3}{4}$,\\
thus $\frac{OA}{OB}=\frac{3}{4}$, solving gives: $OB=16$,\\
hence $AB=\sqrt{16^2+12^2}=20$,\\
since $BE=5$,\\
thus $AE=15$,\\
because $EH\perp AC$, $OB\perp AC$,\\
thus $EH\parallel OB$,\\
hence $\triangle AEH\sim \triangle AOB$,\\
thus $\frac{EH}{BO}=\frac{AH}{AO}=\frac{AE}{AB}$, that is $\frac{EH}{16}=\frac{AH}{12}=\frac{15}{20}$,\\
solving gives: $EH=12$, $AH=9$,\\
thus $OH=3$,\\
hence point E is (−3,12),\\
since the graph of the inverse proportion function $y=\frac{k}{x}$ passes through point E,\\
thus $12=\frac{k}{-3}$, solving gives: $k=\boxed{-36}$;


Question: Given in the diagram, points $A(-4, 2)$ and $B(n, -4)$ are two intersection points of the graph of the linear function $y=kx+b$ and the graph of the inverse proportion function $y=\frac{m}{x}$. Find the equations of the linear function and the inverse proportion function?

Solution:
\textbf{Solution:} Given that $A(-4, 2)$ lies on the curve of $y=\frac{m}{x}$,\\
$\therefore m=-8$.\\
$\therefore$ The equation of the inverse proportion function is $y=-\frac{8}{x}$.\\
Given that point $B(2, -4)$ is on $y=-\frac{8}{x}$,\\
$\therefore n=2$.\\
$\therefore B(2, -4)$.\\
Given $y=kx+b$ passes through $A(-4, 2)$ and $B(2, -4)$,\\
$\therefore \left\{\begin{array}{l} -4k+b=2 \\ 2k+b=-4 \end{array}\right.$.\\
Solving yields: $\left\{\begin{array}{l} k=-1 \\ b=-2 \end{array}\right.$.\\
$\therefore$ The linear function equation is $\boxed{y=-x-2}$.


Question: As shown in the figure, it is known that points $A(-4, 2)$ and $B(n, -4)$ are two intersection points between the graph of the linear function $y=kx+b$ and the graph of the inverse proportion function $y=\frac{m}{x}$. By observing the graphs, directly write the solution set of the inequality $kx+b-\frac{m}{x}>0$.

Solution: \textbf{Solution:} It follows from the problem statement that the solution set for $kx+b-\frac{k}{x}>0$ is $\boxed{\text{indeterminate without further information}}$.


Question: As shown in the figure, it is known that points $A(-4, 2)$ and $B(n, -4)$ are two intersection points of the graph of the linear function $y=kx+b$ with the graph of the inverse proportion function $y=\frac{m}{x}$. What is the area of $\triangle AOB$?

Solution:
\textbf{Solution:} Since $y=-x-2$,\\
therefore when y=0, x=$-2$.\\
Thus, point C is ($-2$, $0$).\\
Therefore, OC=2.\\
Therefore, $S_{\triangle AOB}=S_{\triangle ACO}+S_{\triangle BCO}=\frac{1}{2}\times 2\times 4+\frac{1}{2}\times 2\times 2=\boxed{6}$.


Question: As shown in the figure, the parabola $y=ax^2+bx-3a$ intersects the negative half of the x-axis at point $A(-1, 0)$, intersects the x-axis at another point $B$, and intersects the positive half of the y-axis at point $C(0, 3)$. The axis of symmetry of the parabola intersects line $BC$ at point $M$, and intersects the x-axis at point $G$. Find the analytical expression of the parabola and the axis of symmetry?

Solution:
Solution: Substituting $A(-1, 0)$ and $C(0, 3)$ into $y=ax^{2}+bx-3a$ yields:
\[
\left\{
\begin{array}{l}
a-b-3a=0 \\
-3a=3
\end{array}
\right.
\]
Solving this, we get:
\[
\left\{
\begin{array}{l}
a=-1 \\
b=2
\end{array}
\right.
\]
Thus, the equation of the parabola is $y=-x^{2}+2x+3$, and its axis of symmetry is $x=-\frac{b}{2a}=\boxed{1}$


Question: As shown in the figure, the parabola $y=ax^2+bx-3a$ intersects the negative half of the x-axis at point $A(-1,0)$, and another intersection point with the x-axis is $B$, and it intersects the positive half of the y-axis at point $C(0,3)$. The axis of symmetry of the parabola intersects line $BC$ at point $M$ and intersects the x-axis at point $G$. There exists a point $P$ on the axis of symmetry of the parabola, and when point $P$ is above the x-axis, it satisfies $\angle APB = \angle ABC$. What is the length of $PG$?

Solution:
\textbf{Solution:} Let $y=0$, we get: $-x^2+2x+3=0$, solving this gives $x_1=-1$, $x_2=3$,\\
$\therefore$OB=OC=3,\\
$\therefore\angle ABC=45^\circ$,\\
$\because\angle APB=\angle ABC=45^\circ$, and PA=PB,\\
$\therefore\angle PBA=\frac{1}{2}(180^\circ-45^\circ)=67.5^\circ$,\\
$\therefore\angle MPB=\frac{1}{2}\angle APB=22.5^\circ$,\\
$\because\angle MBP=67.5^\circ-45^\circ=22.5^\circ$,\\
$\therefore\angle MPB=\angle MBP$,\\
$\therefore$MP=MB, in Rt$\triangle BMG$, we have BG=MG=2, by the Pythagorean theorem, we find: BM=$2\sqrt{2}$,\\
$\therefore$MP=$2\sqrt{2}$,\\
$\therefore$PG=MG+MP=2+$2\sqrt{2}$.\\
Hence, the length of $PG$ is $\boxed{2+2\sqrt{2}}$.


Question: As shown in the figure, the parabola $y=x^{2}+bx+c$ (where $b$ and $c$ are constants) intersects the $x$-axis at points $A(-1,0)$ and $B(3,0)$, and intersects the $y$-axis at point $C$. Let $P$ be a point on the parabola with an abscissa of $m$. Find the equation of this parabola.

Solution:
\textbf{Solution}: Since the parabola $y=x^{2}+bx+c$ (where $b$ and $c$ are constants) intersects the x-axis at points $A(-1, 0)$ and $B(3, 0)$,\\
we have $\begin{cases} 1-b+c=0 \\ 9+3b+c=0 \end{cases}$,\\
which yields $\begin{cases} b=-2 \\ c=3 \end{cases}$.\\
Thus, the equation of the parabola is $\boxed{y=x^{2}-2x+3}$.


Question: As shown in the figure, the parabola $y=x^2+bx+c$ (where $b$ and $c$ are constants) intersects the x-axis at points $A(-1,0)$ and $B(3,0)$, and the y-axis at point $C$. Let $P$ be a point on the parabola with the x-coordinate $m$. Let the area of $\triangle ABP$ be denoted as $S$. Find the range of possible values for $S$ when $0 \leq m \leq \frac{5}{2}$.

Solution:
\textbf{Solution:} Draw $PE\perp AB$ at point E through point P, as shown in the figure,\\
since $0\leq m\leq \frac{5}{2}$,\\
thus P($m$, $m^{2}-2m-3$) is in the fourth quadrant,\\
therefore $PE=-m^{2}+2m+3$.\\
Since A($-1$,0), B(3,0),\\
thus $OA=1$, $OB=3$,\\
therefore $AB=OA+OB=4$.\\
Therefore, the area of $\triangle PAB=\frac{1}{2}AB\cdot PE$\\
$=\frac{1}{2}\times 4\times (-m^{2}+2m+3)$\\
$=-2m^{2}+4m+6$\\
$=-2(m-1)^{2}+8$.\\
Thus, when $m=1$, the area of $\triangle PAB$ reaches its maximum value of $8$.\\
Since $0\leq m\leq \frac{5}{2}$,\\
thus, when $m=\frac{5}{2}$, the area of $\triangle PAB$ reaches its minimum value of $\frac{7}{2}$.\\
Therefore, the range of values for $S$ is: $\boxed{\frac{7}{2}\leq S\leq 8}$.


Question: As shown in the figure, the parabola $y=x^{2}+bx+c$ (where $b$ and $c$ are constants) intersects the x-axis at points $A(-1,0)$ and $B(3,0)$, and intersects the y-axis at point $C$. Let $P$ be a point on the parabola with an abscissa of $m$. When the difference between the maximum and minimum y-coordinates of the section of the parabola between points $C$ and $P$ (including points $C$ and $P$) is $2$, find the value of $m$?

Solution:
\textbf{Solution:} From the given, we have $y=x^{2}-2x-3=(x-1)^{2}-4$,\\
$\therefore$ the vertex of the parabola is $(1,-4)$.\\
Let $x=0$, then $y=-3$,\\
$\therefore$ point $C(0,-3)$ is established.\\
$\because$ the difference between the highest and lowest ordinate (y-coordinate) of the segment between point $C$ and point $P$ (including points $C$ and $P$) is $2$,\\
$\therefore$ point $P$ cannot be below point $C$.\\
$\therefore$ point $P$ is above point $C$.\\
$\therefore$ the ordinate of point $P$ is $-1$,\\
Let $y=-1$, then $m^{2}-2m-3=-1$.\\
Solving yields: $m=1\pm\sqrt{3}$.\\
$\therefore$ the value of $m$ is: $1+\sqrt{3}$ or $1-\sqrt{3}$.\\
Thus, the value of $m$ is $\boxed{1+\sqrt{3}}$ or $\boxed{1-\sqrt{3}}$.


Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\frac{m}{x}$ at points $A$ and $B$. The coordinates of point $A$ are $\left(-2, 1\right)$, and the coordinates of point $B$ are $\left(1, n\right)$. Determine the analytical expressions of the two functions.

Solution:
\textbf{Solution:} Since the graph of the inverse proportion function $y=\frac{m}{x}$ passes through point $A(-2, 1)$,

\[m=-2\times 1=-2,\]

thus, the analytical expression of the inverse proportion function is $y=-\frac{2}{x}$.

Since the graph of the inverse proportion function $y=-\frac{2}{x}$ passes through point $B(1, n)$,

\[-2=1\times n,\]

thus, $n=-2$,

implying $B(1, -2)$.

Since the graph of the linear function $y=kx+b$ passes through points $A$, $B$,

we have the system of equations
\[
\left\{\begin{array}{l}
k+b=-2 \\
-2k+b=1
\end{array}\right.,
\]

solving this system yields
\[
\left\{\begin{array}{l}
k=-1 \\
b=-1
\end{array}\right.,
\]

hence, the analytical expression of the linear function is $y=-x-1$;

Therefore, the answers are \boxed{y=-\frac{2}{x}} and \boxed{y=-x-1}.


Question: As shown in the graph, it is known that the graph of the linear function $y=kx+b$ intersects with that of the hyperbolic function $y=\frac{m}{x}$ at points $A$ and $B$, where point $A$ has the coordinates $\left(-2, 1\right)$ and point $B$ has the coordinates $\left(1, n\right)$. Is there a point $P$ on the positive x-axis such that $\triangle ABP$ is an isosceles triangle? If it exists, find the coordinates of point $P$; if not, explain why.

Solution:
\textbf{Solution:} Exist, \\
Let point $P(m, 0) (m>0)$, \\
$\because A(-2, 1)$, $B(1, -2)$, \\
$\therefore AB^{2}=(-2-1)^{2}+[1-(-2)]^{2}=18$, \\
$AP^{2}=(m+2)^{2}+1^{2}=m^{2}+4m+5$, \\
$BP^{2}=(m-1)^{2}+4=m^{2}-2m+5$, \\
$\because \triangle ABP$ is isosceles, \\
$\therefore$ (1) When $AP=AB$, $AP^{2}=AB^{2}$, \\
$\therefore m^{2}+4m+5=18$, \\
$\therefore m=-2-\sqrt{17}$ (discard) or $m=-2+\sqrt{17}$, \\
$\therefore P(-2+\sqrt{17}, 0)$, \\
(2) When $AP=BP$, $AP^{2}=BP^{2}$, \\
$\therefore m^{2}+4m+5=m^{2}-2m+5$, \\
$\therefore m=0$ (discard), \\
(3) When $AB=BP$, $AB^{2}=BP^{2}$, \\
$\therefore m^{2}-2m+5=18$, \\
$\therefore m=1-\sqrt{14}$ (discard) or $m=1+\sqrt{14}$, \\
$\therefore P(1+\sqrt{14}, 0)$, \\
Hence, the coordinates of point $P$ that meet the conditions are $\boxed{(-2+\sqrt{17}, 0)}$ or $\boxed{(1+\sqrt{14}, 0)}$.


Question: As shown in the figure, the graph of the linear function $y_1=k_1x+b$ intersects with the graph of the inverse proportion function $y_2=\frac{k_2}{x}\ (x<0)$ at points $A$ and $B$. The intersections with the coordinate axes are $\left(-6, 0\right)$ and $\left(0, 6\right)$, respectively, and the x-coordinate of point $B$ is $-4$. Determine the equation of the inverse proportion function?

Solution:
\textbf{Solution:} Let the equation of the linear function be $y=kx+b$,\\
$\because$ The intersections of the linear function with the coordinate axes are $\left(-6, 0\right)$ and $\left(0, 6\right)$,\\
$\therefore$
$\begin{cases}
-6k+b=0, \\
b=6,
\end{cases}$\\
$\therefore$
$\begin{cases}
k=1, \\
b=6,
\end{cases}$\\
$\therefore$ The equation of the linear function is: $y=x+6$,\\
$\therefore$ For point $B\left(-4, 2\right)$,\\
$\therefore$ The equation of the inverse proportional function is: $\boxed{y=\frac{-8}{x}}$;


Question: As shown in the figure, the graph of the linear function $y_1=k_1x+b$ intersects with the graph of the inverse proportion function $y_2=\frac{k_2}{x}$ (where $x<0$) at points $A$ and $B$. The graphs also intersect the coordinate axes at $(-6, 0)$ and $(0, 6)$. The x-coordinate of point $B$ is $-4$. What is the area of $\triangle AOB$?

Solution:
Solution: Since point A and point B are the intersection points of the inverse proportion function and the linear function,

it follows that: $x+6=-\frac{8}{x}$,

Solving this, we get $x=-2$ or $x=-4$,

hence $A(-2, 4)$,

therefore, the area of $\triangle AOB$ is $6\times 6\div 2-6\times 2=\boxed{6}$.


Question: As shown in the figure, a linear function $y_{1}=k_{1}x+b$ intersects with the hyperbolic function $y_{2}=\frac{k_{2}}{x}\left(x<0\right)$ at points $A$ and $B$, and intersects with the coordinate axes at $\left(-6, 0\right)$ and $\left(0, 6\right)$. The x-coordinate of point $B$ is $-4$. Directly write the solution to the inequality $k_{1}x+b>\frac{k_{2}}{x}$.

Solution:
\textbf{Solution:} Solve: $-4<x<-2$.

Directly write the solution of the inequality $k_{1}x+b>\frac{k_{2}}{x}$ as $x \in \boxed{(-4, -2)}$.


Question: As shown in the figure, within the square $OABC$, point $O$ is the origin, point $C(-3, 0)$, point $A$ is on the positive y-axis, points $E$ and $F$ are on $BC$ and $CO$ respectively, with $CE=CF=2$. The graph of the linear function $y=kx+b(k\ne 0)$ passes through points $E$ and $F$, intersecting the y-axis at point $G$. The graph of the reciprocal function $y=\frac{m}{x}(m\ne 0)$ passing through point $E$ intersects $AB$ at point $D$. What are the equations of the reciprocal function and the linear function?

Solution: \textbf{Solution:} Given the problem statement, point $C(-3, 0)$, hence the side length of the square $OABC$ is 3, implying $OA=AB=BC=CO=3$. Since $CE=CF=2$, it follows that $OF=1$. Thus, $E(-3, 2)$ and $F(-1, 0)$. Substituting $E(-3, 2)$ and $F(-1, 0)$ into $y=kx+b$ yields $\begin{cases}-3k+b=2\\ -k+b=0\end{cases}$. Solving this system of equations, we get $\begin{cases}k=-1\\ b=-1\end{cases}$. Therefore, the linear function is $y=-x-1$. Substituting $E(-3, 2)$ into $y=\frac{m}{x}$ gives $2=\frac{m}{-3}$. Solving for $m$ yields $m=-6$. Therefore, the inverse proportionality function is $y=-\frac{6}{x}$. Answer: The inverse proportionality function is $\boxed{y=-\frac{6}{x}}$, and the linear function is $\boxed{y=-x-1}$.


Question: As shown in the figure, in the square OABC, point O is the origin, point $C(-3, 0)$, point A is on the positive y-axis, points E and F are respectively on BC and CO, with $CE=CF=2$. The graph of the linear function $y=kx+b\ (k\ne 0)$ passes through points E and F, intersecting the y-axis at point G. The graph of the inverse proportion function $y=\frac{m}{x}\ (m\ne 0)$ through point E intersects AB at point D. Is there a point P on segment EF such that $S_{\triangle ADP}=S_{\triangle APG}$? If so, find the coordinates of point P; if not, please explain why.

Solution:
\textbf{Solution:} Suppose there exists a point $P$ such that $S_{\triangle ADP}=S_{\triangle APG}$. In the equation $y=-x-1$, let $x=0$ to obtain $y=-1$,\\
therefore $G(0, -1)$,\\
hence $AG=4$,\\
in the equation $y=-\frac{6}{x}$, let $y=3$ to obtain $x=-2$,\\
therefore $D(-2, 3)$,\\
thus $AD=2$,\\
let $P(t, -t-1)$,\\
since $S_{\triangle ADP}=S_{\triangle APG}$,\\
it follows that $\frac{1}{2}\times 2\cdot [3-(-t-1)]=\frac{1}{2}\times 4\times (-t)$,\\
solving this yields $t=-\frac{4}{3}$,\\
therefore \boxed{P\left(-\frac{4}{3}, \frac{1}{3}\right)}.


Question: As shown in the figure, the graph of the inverse proportion function $y_{1}=\frac{k}{x}$ (where $k$ is a constant and $k\neq 0$) and the graph of the linear function $y_{2}=2x+2$ both pass through points $A(1, m)$ and $B(a, b)$. Find the coordinates of point A and the expression for the inverse proportion function.

Solution:
\textbf{Solution:} Substituting point A into $y_2=2x+2$, we get $m=4$,\\
$\therefore$ the coordinates of A are $\boxed{(1,4)}$;\\
$\because$ $y_1=\frac{k}{x}$ passes through point A,\\
$\therefore$ $k=1\times 4=4$,\\
$\therefore$ the expression for the inverse proportion function is $\boxed{y_1=\frac{4}{x}}$.


Question: As shown in the figure, the graph of the inverse proportion function $y_{1}=\frac{k}{x}$ (where $k$ is a constant and $k \neq 0$) and the graph of the linear function $y_{2}=2x+2$ both pass through points $A(1, m)$ and $B(a, b)$. Determine the coordinates of point B and, by referring to the graph, directly write out the range of values for $x$ when $y_{1} > y_{2}$.

Solution:
\textbf{Solution:} Since $y_{1} = \frac{k}{x}$ intersects with $y_{2} = 2x + 2$ at points A and B,\\
it follows that $\frac{4}{x} = 2x + 2$,\\
thus $x = -2$ or $x = 1$,\\
therefore $y = -2$ or $y = 4$,\\
hence, the coordinates of point B are \boxed{(-2, -2)};\\
When $y_{1} > y_{2}$, the range of $x$ is \boxed{x < -2} or \boxed{0 < x < 1}.


Question: As shown in the figure, the graph of the direct proportion function $y_1=x$ intersects with that of the quadratic function $y_2=x^2-bx$ at points $O(0,0)$ and $A(4,4)$. What is the value of $b$?

Solution:
\textbf{Solution:} Substitute point A (4,4) into $y_2=x^2-bx$, we get $16-4b=4$, $4b=12$,\\
solving for $b$, we find $b=\boxed{3}$.


Question: As shown in the figure, the proportional function $y_{1}=x$ and the quadratic function $y_{2}=x^{2}-bx$ intersect at points $O(0,0)$ and $A(4,4)$. When $y_{1} < y_{2}$, directly write the range of $x$.

Solution:
\textbf{Solution:} Solve: When $y_{1}< y_{2}$, directly write the range of $x$ as $x<0$ or $x>4$, \\
hence, the range of $x$ is $\boxed{x<0}$ or $\boxed{x>4}$.


Question: As shown in the figure, the vertex A of the right triangle $\triangle ABO$ is the intersection point in the second quadrant between the hyperbola $y=\frac{k}{x}$ and the line $y=-x-(k+1)$. $\overline{AB} \perp x$-axis at B, and the area of $\triangle ABO$ is $\frac{3}{2}$. What are the equations of these two functions?

Solution:
\textbf{Solution:} Since $AB \perp x$-axis at B, and the area of $\triangle ABO = \frac{3}{2}$,\\
it follows that $\frac{1}{2}|k| = \frac{3}{2}$. Solving this, we find $k = \pm 3$.\\
Since the graph of the inverse proportionality function lies in the second and fourth quadrants,\\
we have $k < 0$,\\
thus $k = -3$,\\
which means the equation of the inverse proportionality function is $\boxed{y = -\frac{3}{x}}$, and the equation of the linear function is $\boxed{y = -x + 2}$.


Question: As shown in the figure, the vertex A of the right triangle \(Rt\triangle ABO\) is the intersection point in the second quadrant of the hyperbola \(y=\frac{k}{x}\) and the line \(y=-x-(k+1)\). \(AB \perp x\)-axis at B, and the area of the triangle \(S_{\triangle ABO}=\frac{3}{2}\). Determine the coordinates of the two intersection points A and C of the line and the hyperbola.

Solution:
Solution: Combine the analytic expressions of the two functions into a system of equations, $\begin{cases}y=-\frac{3}{x}\\ y=-x+2\end{cases}$. Solving this, we find: $\begin{cases}x_1=-1\\ y_1=3\end{cases}$ and $\begin{cases}x_2=3\\ y_2=-1\end{cases}$,\\
$\therefore$ the coordinates of point A are $\boxed{\left(-1, 3\right)}$, and the coordinates of point C are $\boxed{\left(3, -1\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $AB$ intersects the x-axis at point $B$ and the y-axis at point $C(0, 2)$, and intersects the graph of the inverse proportion function $y=\frac{6}{x}$ within the first quadrant at point $A$. Line $AD$ is drawn perpendicular to the x-axis at point $D$, with $OD=2$. Find the analytic equation of line $AB$?

Solution:
Solution:

Since AD is perpendicular to the x-axis, and OD = 2, the x-coordinate of point D is 2. By substituting x = 2 into $y= \frac{6}{x}$, we get y = 3. Therefore, A(2, 3). Let the equation of line AB be $y=kx+b$ (where $k\neq 0$). Substituting points C(0, 2) and A(2, 3) into $y=kx+b$ gives $\begin{cases}b=2\\ 2k+b=3\end{cases}$. Therefore, $\begin{cases}b=2\\ k=\frac{1}{2}\end{cases}$. Hence, the equation of line AB is $y=\frac{1}{2}x+2$; so, the answer is $\boxed{y=\frac{1}{2}x+2}$.


Question: In the Cartesian coordinate system $xOy$, the line $AB$ intersects the x-axis at point $B$ and the y-axis at point $C(0, 2)$. It also intersects the graph of the inverse proportion function $y=\frac{6}{x}$ within the first quadrant at point $A$. Line $AD$ is perpendicular to the x-axis at point $D$, with $OD=2$. Let point $P$ be a point on the y-axis. If the area of triangle $ACP$ is equal to 4, what are the coordinates of point $P$?

Solution:
\textbf{Solution:} Since point P is on the y-axis and the area of $\triangle ACP$ is equal to 4, with A(2,3),\\
it follows that $S_{\triangle ACP} = \frac{1}{2} \times \text{CP} \times |x_A| = \frac{1}{2} \times \text{CP} \times 2 = 4$,\\
thus CP = 4.\\
Since point C is (0,2) and point P is on the y-axis,\\
it follows that \boxed{P(0,6)} or \boxed{P(0,-2)}.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, line $AB$ intersects the x-axis at point B and intersects the y-axis at point $C(0, 2)$. It also intersects the graph of the inverse proportional function $y=\frac{6}{x}$ within the first quadrant at point A. Draw $AD\perp x$ axis at point $D$, with $OD=2$. Suppose point E is on the x-axis, and $\triangle EBC$ is an isosceles triangle. Directly write down the coordinates of point E.

Solution: \textbf{Solution:} The coordinates of point $E$ are $(2\sqrt{5}-4, 0)$ or $(-2\sqrt{5}-4, 0)$ or $(4, 0)$ or $(-1.5, 0)$. Therefore, the answer is $(2\sqrt{5}-4, 0)$ or $(-2\sqrt{5}-4, 0)$ or $\boxed{(4, 0)}$ or $(-1.5, 0)$.


Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=\frac{1}{2}x+2$ intersects with the x-axis at point A, and intersects with the y-axis at point C. The axis of symmetry of the parabola $y=ax^{2}+bx+c$ is $x=-\frac{3}{2}$, and it passes through points A and C, with another intersection point on the x-axis being point B. What are the coordinates of points A and B?

Solution:
Solution: From the equation of the line, we get $y=\frac{1}{2}x+2$. When $x=0$, $y=2$; when $y=0$, $x=-4$.\\
$\therefore$ $C(0, 2)$ and $A(-4, 0)$. Due to the symmetry of the parabola, point A and point B are symmetric about the line $x=-1.5$,\\
$\therefore$ $B\left(1, 0\right)$.\\
Therefore, the coordinates of points A and B are respectively $A=\boxed{(-4, 0)}$, $B=\boxed{(1, 0)}$.


Question: As shown in the diagram, within the Cartesian coordinate system, the line $y=\frac{1}{2}x+2$ intersects the x-axis at point A and the y-axis at point C. The parabola $y=ax^2+bx+c$ has its axis of symmetry at $x=-\frac{3}{2}$ and passes through points A and C, with another intersection point with the x-axis being point B. Determine the equation of the parabola.

Solution:
Solution: The parabola $y=ax^{2}+bx+c$ passes through the points $A(-4, 0)$, $B(1, 0)$, so we can set the equation of the parabola to $y=a(x+4)(x-1)$. Moreover, the parabola passes through the point $C(0, 2)$, $\therefore$ $-4a=2$, $\therefore$ $a=-0.5$, \\
$\therefore$ \boxed{y=-\frac{1}{2}x^{2}-\frac{3}{2}x+2}.


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=\frac{1}{2}x+2$ intersects the x-axis at point A and the y-axis at point C. The parabola $y=ax^2+bx+c$ has its axis of symmetry at $x=-\frac{3}{2}$, and it passes through points A and C, with another intersection with the x-axis at point B. Suppose point P is a point on the parabola above line AC, and let $PQ\perp x$ axis, intersecting AC at point Q. Find the maximum value of the line segment PQ and the coordinates of point P at this maximum value.

Solution:
\textbf{Solution:} Let $P\left(m, -\frac{1}{2}m^{2}-\frac{3}{2}m+2\right)$, and $PQ\perp x$ axis, intersecting line AC at point Q, \\
$\therefore Q\left(m, \frac{1}{2}m+2\right)$, \\
$\therefore PQ=\left(-\frac{1}{2}m^{2}-\frac{3}{2}m+2\right)-\left(\frac{1}{2}m+2\right)=-\frac{1}{2}m^{2}-2m$, \\
$PQ=-\frac{1}{2}(m+2)^{2}+2$, \\
$\therefore$ when $m=-2$, the maximum value of $PQ$ is $\boxed{2}$, at this time $\boxed{P\left(-2, 3\right)}$.


Question: As shown in the figure, the graph of the linear function $y_{1}=-x-1$ intersects the x-axis at point A and the y-axis at point B. It intersects the graph of the inverse proportion function $y_{2}= \frac{k}{x}$ at point M($-2$, $m$). What is the analytic expression of the inverse proportion function?

Solution:
\textbf{Solution:} Since point M($-2$, $m$) lies on the graph of the linear function $y_{1}=-x-1$,\\
by substitution we get: $m=-(-2)-1=1$,\\
thus, the coordinates of M are ($-2$, $1$),\\
substituting M's coordinates into $y_{2}=\frac{k}{x}$ gives: $k=-2$,\\
therefore, the equation of the inverse proportion function is: $y_{2}=\boxed{-\frac{2}{x}}$.


Question: As shown in the figure, the graph of the linear function $y_{1}=-x-1$ intersects the x-axis at point A and the y-axis at point B. It has a point of intersection, M($-2$, m), with the graph of the inverse proportional function $y_{2}= \frac{k}{x}$. What is the area of $\triangle MOB$?

Solution:
\[ \textbf{Solution:} \text{Given the problem, when } x=0, \text{ we have } y=-1, \text{ meaning the coordinates of B are }(0, -1), \text{ thus } OB=1, \]
\[ \text{Since the coordinates of M are } (-2, 1), \]
\[ \text{Therefore, the distance from point M to line OB is } 2, \]
\[ \text{Therefore, the area of }\triangle MOB \text{ is } \frac{1}{2} \times 1 \times 2 = \boxed{1}. \]


Question: As shown in the figure, it is known that the graph of the linear function $y_1=kx+b$ and the graph of the hyperbolic function $y_2=\frac{m}{x}$ intersect at points $A(6, 1)$ and $B(a, -3)$ in the first and third quadrants, respectively. Lines $OA$ and $OB$ are drawn. What are the analytical expressions for the linear function and the hyperbolic function?

Solution:
\textbf{Solution:} Substitute $A(6, 1)$, $B(-2, -3)$ into ${y}_{2}=\frac{m}{x}$, we get $m=-3a=6\times 1$ $\therefore$ $m=6$, $a=-2$ $\therefore$ ${y}_{2}=\frac{6}{x}$, $B(-2, -3)$. Substitute $A(6, 1)$, $B(-2, -3)$ into ${y}_{1}=kx+b$, we obtain: $\begin{cases} 6k+b=1 \\ -2k+b=-3 \end{cases}$ $\therefore$ $\begin{cases} k=\frac{1}{2} \\ b=-2 \end{cases}$ \\
$\therefore$ The equation of the linear function is $\boxed{{y}_{1}=\frac{1}{2}x-2}$.


Question: As shown in the diagram, it is known that the graph of the linear function $y_1=kx+b$ and the graph of the inverse proportion function $y_2=\frac{m}{x}$ intersect at points A(6,1) and B(a,-3) in the first and third quadrants, respectively. Lines OA and OB are drawn. What is the area of $\triangle AOB$?

Solution:
\textbf{Solution:}  Let the linear function $y=\frac{1}{2}x-2$ intersect the y-axis at point C. Draw $AD\perp y$ axis at point D through point A, and draw $BE\perp y$ axis at point E through point B. In $y=\frac{1}{2}x-2$, let $x=0$, then $y=-2$, thus $OC=2$. Given $A(6, 1)$, $B(-2, -3)$, and $AD\perp y$ axis, $BE\perp y$ axis. Therefore, $AD=6$, $BE=2$ \\
Therefore, the area of $\triangle AOB$ is $\frac{1}{2}OC\cdot AD+\frac{1}{2}OC\cdot BE= \frac{1}{2}OC(AD+BE)=\frac{1}{2}\times 2\times 8 = \boxed{8}$.


Question: As shown in the figure, it is known that the graph of the linear function $y_{1}=kx+b$ and the graph of the inverse proportional function $y_{2}=\frac{m}{x}$ intersect at points A(6,1) and B(a,−3) in the first and third quadrants, respectively. Lines OA and OB are connected. Directly state the range of the independent variable $x$ when $y_{1}<y_{2}$.

Solution: \textbf{Solution:} When $y_{1}<y_{2}$, the range of the independent variable $x$ is \boxed{x<-2} or \boxed{0<x<6}.


Question: As shown in the figure, the graph of the linear function $y = -x + 4$ intersects with the graph of the inverse proportion function $y = \frac{k}{x}$ at points $A(3, a)$ and $B$, and intersects the y-axis at point $C$. What is the expression of the inverse proportion function?

Solution:
\textbf{Solution:} Substituting $A\left(3, a\right)$ into the linear equation $y=-x+4$, we have,\\
$a=-3+4=1$,\\
$\therefore$ the coordinates of point A are $\boxed{\left(3, 1\right)}$.


Question: As shown in the figure, the graph of the linear function $y=-x+4$ intersects with the graph of the inverse proportion function $y=\frac{k}{x}$ at points $A(3, a)$ and $B$, and intersects with the y-axis at point $C$. Find the coordinates of point $B$, and directly write out the area of $\triangle AOB$.

Solution:
\textbf{Solution:} Substitute $(3,1)$ into the inverse proportion function $y=\frac{3}{x}$, to get $k=3$,\\
therefore, the analytical expression of the inverse proportion function is: $y=\frac{3}{x}$,\\
By solving the system of equations $\left\{\begin{array}{l}y=-x+4 \\ y=\frac{3}{x}\end{array}\right.$, we find $\left\{\begin{array}{l}x_1=1 \\ y_1=3\end{array}\right.\left\{\begin{array}{l}x_2=3 \\ y_2=1\end{array}\right.$,\\
therefore, the coordinates of point B are $\boxed{(3,1)}$,\\
The area of $\triangle AOB$ is $\boxed{4}$.


Question: As shown in the figure, the graph of the linear function $y=-x+4$ intersects with the graph of the inverse proportion function $y=\frac{k}{x}$ at points $A(3, a)$ and $B$, and intersects the y-axis at point $C$. Based on the graph, please directly write the solution set of the inequality $-x+4>\frac{k}{x}$.

Solution: \textbf{Solution:} The solution set of the inequality $-x+4>\frac{k}{x}$ is $1<x<3$. Hence, the solution set is $x \in \boxed{(1,3)}$.


Question: As shown in Figure 1, it is an ancient trebuchet capable of attacking from a distance. Figure 2 represents a schematic of the trebuchet during a moment in the stone-throwing process, where $GP$ is the lever, the sling hangs at point $G$, the counterweight is hung at point $P$, point $A$ is the fulcrum, and point $D$ is a point on the horizontal baseboard $BC$, with $AD=AC=3$ meters, and $CD=3.6$ meters. When the trebuchet is prepared, point $G$ exactly coincides with point $B$, at this time $AG$ and $AC$ are perpendicular. Then, how many meters is $BD$?

Solution: \textbf{Solution:} We have $BD=\boxed{1.4}$ meters.


Question: As shown in Figure 1, it illustrates an ancient trebuchet capable of long-range attacks, while Figure 2 depicts a schematic diagram of the trebuchet in action at a given moment. Here, $GP$ stands for the lever, with the sling attached at point $G$ and the counterweight hanging at point $P$. Point $A$ serves as the fulcrum, and point $D$ is located on the horizontal base $BC$, where $AD=AC=3$ meters, and $CD=3.6$ meters. During the operation of the trebuchet, the extension line of $AP$ intersects segment $DC$ at point $E$. Assuming the ratio $DE:CE=5:1$, how many meters above the ground is point $G$?

Solution: \textbf{Solution}: The point $G$ is $\boxed{\frac{8\sqrt{5}+12}{5}}$ meters away from the ground.


Question: As shown in the figure, in parallelogram $ABCD$, point $G$ is on the extension of side $DC$, $AG$ intersects side $BC$ at point $E$, and intersects diagonal $BD$ at point $F$. It is given that $AF^2=EF\cdot FG$; if $EF=\frac{3}{2}$ and $FG=\frac{8}{3}$, find the value of $\frac{BE}{EC}$.

Solution:
\textbf{Solution:} Since $\triangle GDF \sim \triangle ABF$ and $\triangle AFD \sim \triangle EFB$, \\
from (1) we have $AF^2=EF \cdot FG = \frac{3}{2} \times \frac{8}{3} = 4$, \\
thus $AF=2$, \\
hence $\frac{BE}{AD} = \frac{EF}{AF} = \frac{\frac{3}{2}}{2} = \frac{3}{4}$, \\
therefore $\frac{BE}{EC} = \frac{3}{4-3} = \boxed{3}.$


Question: As shown in the figure, in $\triangle ABC$, point D is on side $AB$, and points F, E are on side $AC$, with $DF \parallel BE$, $\frac{AF}{AE}=\frac{AE}{AC}$, and $DE\parallel BC$. Given $\frac{AF}{FE}=\frac{3}{2}$ and the area of $\triangle ABC$ is 12, find the value of the area of $\triangle ADE$.

Solution:
\textbf{Solution:} Given $\because \frac{AF}{FE}=\frac{3}{2}$,\\
$\therefore \frac{AD}{BD}=\frac{AF}{EF}=\frac{3}{2}$,\\
$\because DE\parallel BC$,\\
$\therefore \frac{AE}{CE}=\frac{AD}{BD}=\frac{3}{2}$,\\
$\therefore \frac{AE}{AC}=\frac{3}{5}$.\\
$\because DE\parallel BC$,\\
$\therefore \triangle ADE\sim \triangle ABC$.\\
$\therefore \frac{S_{\triangle ADE}}{S_{\triangle ABC}}=\left(\frac{AE}{AC}\right)^2=\left(\frac{3}{5}\right)^2$, hence $\frac{S_{\triangle ADE}}{S_{\triangle ABC}}=\frac{9}{25}$,\\
$\because S_{\triangle ABC}=12$,\\
$\therefore \frac{S_{\triangle ADE}}{12}=\frac{9}{25}$, hence $S_{\triangle ADE}=\boxed{\frac{108}{25}}$.


Question: In the figure, in $\triangle ABC$, points $D$ and $E$ are located on sides $AB$ and $AC$, respectively, with $DE \parallel BC$. Lines $BE$ and $CD$ intersect at point $O$. Given $AE = 2$, $AD = EC = 3$, $BD = 4.5$, and $DE = 4$. What is the length of the segment $BC$?

Solution:
\textbf{Solution:} Given that $DE\parallel BC$,\\
it follows that $\triangle ADE\sim \triangle ABC$,\\
hence, $\frac{DE}{BC}=\frac{AE}{AC}$,\\
given that $AE=2, EC=3, DE=4$,\\
we get $\frac{4}{BC}=\frac{2}{2+3}$, solving yields: $BC=\boxed{10}$;


Question: As shown in the figure, in $\triangle ABC$, points $D$ and $E$ are located on the sides $AB$ and $AC$ respectively, with $DE \parallel BC$. Lines $BE$ and $CD$ intersect at point $O$. Given $AE=2$, $AD=EC=3$, $BD=4.5$, and $DE=4$. If the area of $\triangle BOC$ is $5$, what is the area of $\triangle CED$?

Solution:
\textbf{Solution:} Since $DE \parallel BC$,
\[\therefore \triangle ODE \sim \triangle OBC,\]
\[\therefore \frac{S_{\triangle DOE}}{S_{\triangle COB}} = \left( \frac{DE}{BC} \right)^2 = \left( \frac{4}{10} \right)^2 = \frac{4}{25},\] \[\frac{OE}{OB} = \frac{DE}{BC} = \frac{2}{5},\]
Since $S_{\triangle BOC} = 5$,
\[\therefore S_{\triangle DOE} = \frac{4}{5},S_{\triangle EOC} = 2,\]
\[\therefore S_{\triangle CED} = S_{\triangle EOD} + S_{\triangle EOC} = \frac{4}{5} + 2 = \boxed{2.8}.\]


Question: Given: As shown in the trapezoid $ABCD$, $AD\parallel BC$, and diagonals $AC$ and $BD$ intersect at point $E$. Point $F$ is on side $AB$, and $EF\parallel BC$. It is known that $AD=3$, $BC=6$. Question: What is the length of $EF$?

Solution:
\textbf{Solution:} Given that $EF\parallel BC$,\\
it follows that $\triangle AFE\sim \triangle ABC$,\\
thus $\frac{EF}{BC}=\frac{AF}{AB}$,\\
Given that $AD\parallel BC$ and $EF\parallel BC$,\\
it implies $EF\parallel AD$,\\
therefore $\triangle BFE\sim \triangle BAD$,\\
hence $\frac{EF}{AD}=\frac{BF}{AB}$,\\
therefore $\frac{EF}{BC}+\frac{EF}{AD}=\frac{AF}{AB}+\frac{BF}{AB}=\frac{AF+BF}{AB}=1$,\\
which means $\frac{EF}{BC}+\frac{EF}{AD}=1$,\\
Given that $AD=3$ and $BC=6$,\\
therefore $\frac{EF}{6}+\frac{EF}{3}=1$,\\
Solving this equation yields $EF=\boxed{2}$;


Question: Given: As shown in the figure, in trapezoid $ABCD$, $AD\parallel BC$, diagonals $AC$ and $BD$ intersect at point $E$, point $F$ is on side $AB$, and $EF\parallel BC$. It is known that $AD=3$, $BC=6$. Fill in the blank: If ${S}_{\triangle AEF}=5$, then ${S}_{\triangle ECD}$ equals what?

Solution: \textbf{Solution:} From the given, it is obtained that the area of $\triangle ECD$ is $\boxed{15}$.


Question: As shown in the figure, $E$ is a point on the side $CB$ of rectangle $ABCD$, and $AF \perp DE$ at point $F$. $\triangle EDC \sim \triangle DAF$. If $AB = 3$, $AD = 2$, and $CE = 1$, find the length of segment $DF$.

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
it follows that DC = AB = 3, and $\angle ADC = \angle C = 90^\circ$.\\
Since CE = 1,\\
then DE = $\sqrt{DC^2 + CE^2} = \sqrt{10}$.\\
Since $\triangle EDC \sim \triangle DAF$,\\
it follows that $\frac{DE}{AD} = \frac{CE}{FD}$, that is $\frac{\sqrt{10}}{3} = \frac{1}{FD}$,\\
therefore, FD = $\boxed{\frac{\sqrt{10}}{5}}$.


Question: As shown in the diagram, in quadrilateral $ABCD$, $AE:EB=2:3$, and $DE$ intersects $AC$ at $F$. Calculate the ratio of the perimeters of $\triangle AEF$ and $\triangle CDF$.

Solution:
\textbf{Solution:} Given that in quadrilateral $ABCD$, \\
$\therefore AB=CD, AB\parallel CD$, \\
$\therefore \triangle AEF\sim \triangle CDF$, \\
$\therefore$ The ratio of the perimeters of $\triangle AEF$ and $\triangle CDF$ is equal to the similarity ratio $AE:CD$, \\
$\because$ $AE:EB=2:3$, $AB=CD$, \\
$\therefore AE:AB=AE:CD=2:5$, \\
$\therefore$ The ratio of the perimeters of $\triangle AEF$ and $\triangle CDF$ is $ \boxed{2:5}$.


Question: As shown in the figure, in quadrilateral $ABCD$, the ratio $AE:EB=2:3$, and $DE$ intersects $AC$ at $F$. If the area of $\triangle CDF$ is $25\,cm^{2}$, what is the area of quadrilateral $ABCD$?

Solution:
\textbf{Solution:} Since $\triangle AEF\sim \triangle CDF$, \\
it follows that $\frac{AF}{CF}=\frac{AE}{CD}=\frac{2}{5}$, \\
thus, $\frac{{S}_{\triangle AFD}}{{S}_{\triangle CDF}}=\frac{AF}{CF}=\frac{2}{5}$, \\
Given that the area of $\triangle CDF$ is $25\,cm^{2}$, \\
hence, ${S}_{\triangle AFD}=\frac{2}{5}{S}_{\triangle CDF}=\frac{2}{5}\times 25=10\,cm^{2}$; \\
Therefore, ${S}_{\triangle ACD}=10+25=35\,cm^{2}$, \\
Consequently, ${S}_{\square ABCD}=2{S}_{\triangle ACD}=2\times 35=70\,cm^{2}$. \\
Hence, the area of quadrilateral $ABCD$ is $\boxed{70\,cm^{2}}$.


Question: As shown in the figure, $AB \perp BC$, $EC \perp BC$, point $D$ is on $BC$, $AB=1$, $BD=2$, $CD=3$, $CE=6$. $\triangle ABD \sim \triangle DCE$; what is the measure of $\angle ADE$ in degrees?

Solution:
Solution: From (1), we know that $\triangle ABD\sim \triangle DCE$, thus $\angle BAD=\angle EDC$.\\
Since $\angle BAD+\angle ADB=90^\circ$,\\
it follows that $\angle ADB+\angle EDC=90^\circ$.\\
Hence, $\angle ADE=180^\circ-\angle ADB-\angle EDC=90^\circ$.\\
Therefore, the measure of $\angle ADE$ is $\boxed{90^\circ}$.


Question: As shown in the figure, $\triangle ADE\sim \triangle ABC$, and $\frac{AC}{AB}=\frac{2}{3}$. Point D is inside $\triangle ABC$, and lines $BD$, $CD$, $CE$ are connected. $\triangle ABD\sim \triangle ACE$. If $CD=CE$, $BD=3$, and $\angle ABD + \angle ACD = 90^\circ$, what is the length of $DE$?

Solution:
\textbf{Solution:} Given $\triangle ABD\sim \triangle ACE$, \\
$\therefore$ $\frac{AC}{AB}=\frac{CE}{BD}=\frac{2}{3}$, $\angle ABD=\angle ACE$, \\
Also given $BD=3$, \\
$\therefore$ $CE=2$, \\
$\therefore$ $CD=CE=2$, \\
Given $\angle ABD+\angle ACD=90^\circ$, \\
$\therefore$ $\angle ACD+\angle ACE=90^\circ$, \\
$\therefore$ $\angle DCE=90^\circ$, \\
$\therefore$ $DE=\sqrt{2}CD=2\sqrt{2}$. \\

Hence, the length of $DE$ is $\boxed{2\sqrt{2}}$.


Question: As shown in the figure, it is known that the line $y=-\frac{4}{3}x+8$ intersects the x-axis at point A and the y-axis at point B. A moving point C starts from point B and moves towards point A at a uniform speed of 1 unit per second along direction $BA$, while another moving point D starts from point A and moves towards point O at a uniform speed of 2 units per second along direction $AO$. When one point stops moving, the other point stops as well. Let the time of movement be t seconds ($t>0$). What is the area of $\triangle AOB$?

Solution:
\textbf{Solution:} Solve: In the equation $y=-\frac{4}{3}x+8$,\\
when $x=0$, $y=8$,\\
when $y=0$, $-\frac{4}{3}x+8=0$,\\
we get: $x=6$,\\
$\therefore A(6,0), B(0,8)$,\\
thus $OA=6, OB=8$,\\
$\therefore {S}_{\triangle AOB}=\frac{1}{2}OA\cdot OB=\frac{1}{2}\times 6\times 8=\boxed{24}$,\\
meaning the area of $\triangle AOB$ is $\boxed{24}$.


Question: As shown in the figure, it is known that the line $y=-\frac{4}{3}x+8$ intersects the x-axis at point A and the y-axis at point B. A moving point C starts from point B and moves towards point A along line $BA$ at a constant speed of 1 unit per second. Simultaneously, another moving point D starts from point A and moves towards point O along line $AO$ at a constant speed of 2 units per second. The motion of one point stops when the other point stops. Let the time of motion be t seconds ($t>0$). Express the coordinates of point C using an algebraic equation involving t.

Solution:
\textbf{Solution:} Construct $CM\perp x$ axis and $CN\perp y$ axis through point C,\\
$\therefore CN\parallel x$ axis and $CM\parallel y$ axis,\\
In $\triangle AOB$, $AB=\sqrt{OA^2+OB^2}=10$,\\
According to the given information, $BC=t$, thus $AC=10-t$,\\
$\therefore \frac{CN}{OA}=\frac{BC}{AB}$, which means $\frac{CN}{6}=\frac{t}{10}$,\\
Solving this gives: $CN=\frac{3}{5}t$,\\
$\frac{CM}{OB}=\frac{AC}{BC}$, which means $\frac{CM}{8}=\frac{10-t}{10}$,\\
Solving this gives: $CM=\frac{40-4t}{5}$,\\
$\therefore$The coordinates of point C are $\boxed{\left(\frac{3}{5}t, \frac{40-4t}{5}\right)}$.


Question: As shown in the figure, it is known that the line $y=-\frac{4}{3}x+8$ intersects the x-axis at point A and the y-axis at point B. A moving point C starts from point B, moving towards point A at a uniform speed of 1 unit per second along direction $BA$. Simultaneously, a moving point D starts from point A, moving towards point O at a uniform speed of 2 units per second along direction $AO$. When one point stops moving, the other point stops as well. Let the time of movement be t seconds ($t>0$). Directly state the value of t when the area of $\triangle ACD$ is $\frac{84}{5}$.

Solution: \textbf{Solution:} We have $t=\boxed{3}$.


Question: As shown in the figure, it is known that the line $y=-\frac{4}{3}x+8$ intersects the x-axis at point A and the y-axis at point B. A moving point C starts from point B and moves at a uniform speed of 1 unit per second in the direction of $BA$. Simultaneously, a moving point D starts from point A and moves at a uniform speed of 2 units per second in the direction of $AO$ towards point O. When one point stops moving, the other point will also stop moving. Let the time of movement be $t$ seconds ($t>0$). Directly state the value of $t$ when $\triangle ACD$ is similar to $\triangle AOB$.

Solution: \textbf{Solution:} From the problem, we know that when $\triangle ACD$ is similar to $\triangle AOB$, the value of $t$ is $\boxed{\frac{30}{13}}$.


Question: As shown in the figure, it is known that in trapezoid $ABCD$, $AB \parallel CD$, and diagonals $AC$ and $BD$ intersect at point $O$, with $CO = \frac{2}{5} AC$. What is the value of $\frac{CD}{AB}$?

Solution:
\textbf{Solution:} Because $CO = \frac{2}{5}AC$, \\
thus $CO: OA = 2:3$, \\
Because $CD \parallel AB$, \\
thus $\frac{CD}{AB} = \frac{OC}{OA} = \boxed{\frac{2}{3}}$.


Question: As illustrated in the rectangle $ABCD$, where $AB=4$ and $BC=5$, $P$ is a moving point on ray $BC$. Construct $PE\perp AP$ with $PE$ intersecting ray $DC$ at point $E$, and ray $AE$ intersecting ray $BC$ at point $F$. Let $BP=x$ and $CF=y$. When $\sin\angle APB=\frac{4}{5}$, what is the length of $CE$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a rectangle,\\
$\therefore \angle B=90^\circ$,\\
In $\triangle APB$, $AB=4$, $\sin\angle APB=\frac{4}{5}$,\\
$\therefore \sin\angle APB=\frac{AB}{AP}=\frac{4}{5}$,\\
$\therefore AP=5$,\\
$\therefore PB=\sqrt{AP^2-AB^2}=\sqrt{5^2-4^2}=3$,\\
$\therefore PC=BC-PB=5-3=2$,\\
Because $PE\perp AP$,\\
$\therefore \angle APE=90^\circ$,\\
$\therefore \angle CPE+\angle APB=90^\circ$,\\
Because $\angle BAP+\angle APB=90^\circ$,\\
$\therefore \angle CPE=\angle BAP$,\\
Because $\angle ECP=\angle B=90^\circ$,\\
$\therefore \triangle PCE\sim \triangle ABP$,\\
$\therefore \frac{CE}{BP}=\frac{PC}{AB}$,\\
i.e., $\frac{CE}{3}=\frac{2}{4}$,\\
Solving for: $CE=\boxed{\frac{3}{2}}$;


Question: As shown in the rectangle $ABCD$, where $AB=4$ and $BC=5$. Let $P$ be a moving point on ray $BC$. Construct $PE\perp AP$ where $PE$ intersects ray $DC$ at point E, and ray $AE$ intersects ray $BC$ at point F. Let $BP=x$ and $CF=y$. As illustrated, when point P is on side $BC$ (excluding points B and C), find the function relationship between $y$ and $x$, and state its domain.

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a rectangle, \\
$\therefore EC\parallel AB$, \\
$\therefore \triangle ECF \sim \triangle ABF$, \\
$\therefore \frac{CF}{BF}=\frac{CE}{AB}$, \\
that is, $\frac{y}{5+y}=\frac{CE}{4}$, \\
solving this, we get: $CE=\frac{4y}{5+y}$, \\
$\because \triangle PCE \sim \triangle ABP$, \\
$\therefore \frac{CE}{x}=\frac{5-x}{4}$, \\
solving this, we obtain: $CE=\frac{x(5-x)}{4}$, \\
$\therefore \frac{4y}{5+y}=\frac{x(5-x)}{4}$, \\
which simplifies to: $y=\frac{25x-5x^2}{x^2-5x+16}$, for $0<x<5$. \\
Thus, the functional relationship of $y$ with respect to $x$ is $\boxed{y=\frac{25x-5x^2}{x^2-5x+16}}$, with the domain being $(0,5)$.


Question: As shown in the diagram, in rectangle $ABCD$, $AB=4$, $BC=5$, P is a moving point on the ray $BC$, construct $PE\perp AP$, with $PE$ intersecting the ray $DC$ at point E, and the ray $AE$ intersecting the ray $BC$ at point F. Let $BP=x$, $CF=y$. When $\frac{PE}{AP}=\frac{1}{2}$, find the length of $CF$.

Solution:
\textbf{Solution:} (1) When point $P$ lies on segment $BC$, and $E$ lies on segment $CD$,
\\
$\because \triangle PCE\sim \triangle ABP$,
\\
$\therefore \frac{PC}{AB}=\frac{CE}{BP}=\frac{PE}{AP}=\frac{1}{2}$,
\\
$\therefore PC=\frac{1}{2}AB=2$, $CE=\frac{1}{2}BP$,
\\
$\therefore BP=BC-PC=5-2=3$,
\\
$\therefore CE=\frac{3}{2}$,
\\
$\because AB\parallel CD$,
\\
$\therefore \triangle CEF\sim \triangle BAF$,
\\
$\therefore \frac{CF}{BF}=\frac{CE}{AB}$,
\\
which gives $\frac{CF}{5+CF}=\frac{\frac{3}{2}}{4}$,
\\
Solving this we get: $CF=\boxed{3}$;
\\
(2) When $P$ lies on the extension of $BC$, as shown in the diagram:
\\
$\because \angle EPC+\angle BPA=90^\circ$, $\angle BPA+\angle BAP=90^\circ$,
\\
$\therefore \angle EPC=\angle BAP$,
\\
$\because \angle B=\angle PCE$,
\\
$\therefore \triangle ABP\sim \triangle PCE$,
\\
$\therefore \frac{PC}{AB}=\frac{CE}{BP}=\frac{PE}{AP}=\frac{1}{2}$,
\\
$\therefore PC=\frac{1}{2}AB=2$, $CE=\frac{1}{2}BP$,
\\
$\therefore BP=BC+PC=7$, $CE=\frac{7}{2}$,
\\
$\because \angle AFB=\angle CFE$, $\angle B=\angle FCE$,
\\
$\therefore \triangle ABF\sim \triangle ECF$,
\\
$\therefore \frac{AB}{EC}=\frac{BF}{CF}$,
\\
which gives $\frac{4}{\frac{7}{2}}=\frac{5-CF}{CF}$,
\\
Solving this we get: $CF=\boxed{\frac{7}{3}}$;
\\
In summary, the length of $CF$ can be $3$ or $\boxed{\frac{7}{3}}$.


Question: As shown in the figure, it is known that in quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $O$, $AB\perp AC$, and $CD\perp BD$. $\triangle AOD\sim \triangle BOC$; if $\sin\angle ABO=\frac{2}{3}$ and the area of $\triangle AOD=4$, what is the value of the area of $\triangle BOC$?

Solution:
\textbf{Solution:} In the right triangle $\triangle AOB$, we have $\sin\angle ABO=\frac{2}{3}$,\\
hence $\frac{AO}{BO}=\frac{2}{3}$,\\
since $\triangle AOD\sim \triangle BOC$,\\
it follows that $\frac{{S}_{\triangle AOD}}{{S}_{\triangle BOC}}=\left(\frac{AO}{BO}\right)^{2}=\left(\frac{2}{3}\right)^{2}=\frac{4}{9}$,\\
given that $S_{\triangle AOD}=4$,\\
therefore $S_{\triangle BOC}=\boxed{9}$.


Question: As shown in the figure, it is known that $AB \parallel CD$, $AC$ intersects $BD$ at point $E$, and point $F$ is on line segment $BC$, with $\frac{AB}{CD} = \frac{1}{2}$, $\frac{BF}{CF} = \frac{1}{2}$. $AB \parallel EF$; what is the ratio of $AB : EF : CD$?

Solution:
\textbf{Solution:} Since $AB\parallel EF$,\\
we have $\frac{EF}{AB}=\frac{CF}{BC}=\frac{2}{3}$,\\
Since $AB\parallel CD$,\\
we have $\frac{AB}{CD}=\frac{AE}{EC}=\frac{1}{2}=\frac{3}{6}$,\\
Therefore, $AB: EF: CD=\boxed{3:2:6}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=6\,cm$, $BC=12\,cm$, $\angle B=90^\circ$. Point P starts from point A and moves towards point B along side $AB$ at a speed of $1\,cm/s$, and point Q starts from point B and moves towards point C along side $BC$ at a speed of $2\,cm/s$. If P and Q start moving from A and B at the same time, respectively, and the moving time is denoted as $t\,(s)$. At what value of $t$ is the area of $\triangle PBQ$ equal to $9\,cm^{2}$?

Solution:
\textbf{Solution:} Given the problem, point $P$ starts moving from point $A$ towards point $B$ along side $AB$ at a speed of $1\,cm/s$, and point $Q$ starts moving from point $B$ towards point $C$ along side $BC$ at a speed of $2\,cm/s$,\\
$\therefore AP=t$, $BQ=2t$,\\
$\therefore BP=AB-AP=6-t$;\\
Since the area of $\triangle PBQ$ is $9\,cm^{2}$,\\
$\therefore \frac{1}{2}\cdot 2t\cdot (6-t)=9$,\\
solving yields, $t=\boxed{3}$.\\
That is, when $t=3$, the area of $\triangle PBQ$ is $9\,cm^{2}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=6\,cm$, $BC=12\,cm$, $\angle B=90^\circ$. Point P starts from point A and moves along side $AB$ towards point B at a speed of $1\,cm/s$, and point Q starts from point B and moves along side $BC$ towards point C at a speed of $2\,cm/s$. If P and Q start simultaneously from A and B respectively, and the moving time is denoted as $t\,(s)$. When $t$ is how much, will $\triangle PBQ$ be similar to $\triangle ABC$?

Solution:
\textbf{Solution:} Given the motion, we have $BQ=2t$ and $PB=6-t$.\\
Since $\triangle PQB$ and $\triangle ABC$ are right-angled triangles,\\
therefore, when $\triangle PQB$ is similar to $\triangle ABC$, there are two cases,\\
which are $\frac{BP}{AB}=\frac{BQ}{BC}$ or $\frac{BP}{BC}=\frac{BQ}{AB}$.\\
When $\frac{BP}{AB}=\frac{BQ}{BC}$, we get $\frac{6-t}{6}=\frac{2t}{12}$, solving for $t$ yields $t=\boxed{3}$;\\
When $\frac{BP}{BC}=\frac{BQ}{AB}$, we get $\frac{6-t}{12}=\frac{2t}{6}$, solving for $t$ yields $t=\boxed{1.2}$;\\
Therefore, when $t=1.2$ or $3$, $\triangle PQB$ is similar to $\triangle ABC$.


Question: As shown in the figure, it is known that $AB$ is the diameter of $\odot O$, $BC \perp AB$, connect $OC$, the chord $AD \parallel OC$, and the line $CD$ intersects the extension line of $BA$ at point $E$. The line $CD$ is the tangent line of $\odot O$; if $BC=1$, $DE=2BC$, what is the radius of $\odot O$?

Solution:
\textbf{Solution:} In $\triangle OEC$, since $AD \parallel OC$,\\
$\therefore \frac{ED}{CD} = \frac{EA}{OA}$.\\
Additionally, since $\triangle OBC \cong \triangle ODC$,\\
$\therefore CB = CD$,\\
Given that $BC = 1$, $DE = 2BC$,\\
$\therefore DE = 2BC = 2CD = 2$,\\
$\therefore \frac{ED}{CD} = \frac{EA}{OA} = 2$,\\
$\therefore EA = 2OA$.\\
Let the radius of $\odot O$ be $x$, then $OA = OD = x$, $EA = 2x$,\\
$\therefore OE = OA + EA = x + 2x = 3x$,\\
Since the line $CD$ is a tangent to $\odot O$,\\
$\therefore \angle ODE = 90^\circ$,\\
By the Pythagorean theorem, $OD^2 + DE^2 = OE^2$,\\
$\therefore x^2 + 2^2 = (3x)^2$,\\
Solving yields $x = \frac{\sqrt{2}}{2}$,\\
Hence, the radius of $\odot O$ is $\boxed{\frac{\sqrt{2}}{2}}$.


Question: As shown in the figure, it is known that two chords $AB$ and $CD$ of the circle $\odot O$ intersect at the point $M$ with $AB=CD$. Let the radius of $\odot O$ be $r$. $DM=BM$; if $\angle DMB=100^\circ$, $r=1$, what is the length of the arc $\overset{\frown}{BC}$?

Solution:
Solution: Connect $OB$ and $OC$,\\
since $DM=BM$, $\angle DMB=100^\circ$,\\
thus $\angle MDB=\angle MBD=40^\circ$,\\
therefore $\angle BOC=80^\circ$,\\
since $r=1$,\\
hence the length of arc $\overset{\frown}{BC} = \frac{n\pi r}{180} = \frac{80\pi }{180} = \boxed{\frac{4\pi }{9}}$;


Question: As shown in the figure, it is known that $BD$ is the angle bisector of $\triangle ABC$, and $E$ is a point on the extension of $BD$ with $AE=AB$. $\triangle ADE \sim \triangle CDB$. If $AB=4$, $\frac{DC}{AD}=\frac{1}{2}$, find the length of $BC$.

Solution:
Solution: Since $AE=AB$, and $AB=4$,\\
it follows that $AE=4$.\\
Since $\triangle ADE\sim \triangle CDB$,\\
it follows that $\frac{BC}{AE}=\frac{DC}{AD}=\frac{1}{2}$.\\
Therefore, $BC=\frac{1}{2}AE=\boxed{2}$.


Question: As shown in the figure, in rectangle $ABCD$, $E$ is a point on side $CD$. The triangle $ADE$ is folded along $AE$ so that point $D$ precisely falls on point $F$ on side $BC$. $\triangle ABF \sim \triangle FCE$; if $AB = 3\sqrt{3}$ and $AD = 6$, what is the length of $EC$?

Solution:
\textbf{Solution:} By folding, we have $AF=AD=6$,\\
$\therefore BF=\sqrt{AF^{2}-AB^{2}}=3$,\\
$\therefore FC=6-3=3$,\\
$\because \triangle ABF\sim \triangle FCE$,\\
$\therefore \frac{AB}{FC}=\frac{BF}{EC}$,\\
$\frac{3\sqrt{3}}{3}=\frac{3}{EC}$,\\
$\therefore EC=\boxed{\sqrt{3}}$


Question: In rectangle $ABCD$, $AB=6$, $AD=8$, point $P$ is a moving point on line segment $BD$ (not coinciding with points $B$, $D$). A line is drawn through point $P$ such that $PE \perp BD$, intersecting the ray $DC$ at point $E$, and $BE$ is connected. As shown in Figure 1, when point $E$ coincides with point $C$, what is the length of $BP$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle, with $AB=6$ and $AD=8$,\\
then $CD=6$, $BC=8$, $BD=10$, $\angle A=90^\circ$, and $AD\parallel CB$,\\
hence $\angle ADB=\angle CBP$,\\
since $CP\perp BD$,\\
then $\angle CPB=\angle A=90^\circ$,\\
thus $\triangle CPB\sim \triangle BAD$,\\
therefore $\frac{BP}{AD}=\frac{BC}{DB}$,\\
thus $\frac{BP}{8}=\frac{8}{10}$,\\
therefore $BP=\boxed{\frac{32}{5}}$;


Question: As shown in the figure, in triangles \( \triangle ABC \) and \( \triangle ADE \), we have \( \angle BAC = \angle DAE = 90^\circ \), \( AB = 3\sqrt{3} \), \( AD = \sqrt{3} \), \( BC = 6 \), \( DE = 2 \). \( \triangle ABC \sim \triangle ADE \); find the value of \( \frac{BD}{CE} \).

Solution:
\textbf{Solution:} From (1) we have $\frac{AB}{AC} = \frac{AD}{AE}$, \\
$\therefore \frac{AB}{AD} = \frac{AC}{AE}$, \\
$\because \angle BAD=\angle CAE=90^\circ-\angle CAD$, \\
$\therefore \triangle ABD\sim \triangle ACE$, \\
$\therefore \frac{BD}{CE} = \frac{AB}{AC} = \sqrt{3}$, \\
$\therefore$ the value of $\frac{BD}{CE}$ is $\boxed{\sqrt{3}}$.


Question: As shown in the diagram, in parallelogram $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Let $M$ be the midpoint of $AD$, and connect $CM$ to intersect $BD$ at point $N$, with $ON=1$, and $\triangle DMN \sim \triangle BCN$; what is the length of $BD$?

Solution:
\textbf{Solution:} By the given information, we have:
\begin{align*}
\because \triangle DMN\sim \triangle BCN, \\
\therefore \frac{MD}{CB}=\frac{DN}{BN}, \\
\because M\text{ is the midpoint of }AD, \\
\therefore MD=\frac{1}{2}AD=\frac{1}{2}BC, \\
\text{i.e., } \frac{MD}{CB}=\frac{1}{2}, \\
\therefore \frac{DN}{BN}=\frac{1}{2}, \\
\text{i.e., } BN=2DN, \\
\text{let } OB=OD=x, \text{ thus } BD=2x, BN=OB+ON=x+1, DN=x-1, \\
\therefore x+1=2(x-1), \\
\text{Solving, we find } x=3, \\
\therefore BD=2x=\boxed{6}.
\end{align*}


Question: As shown in the diagram, in quadrilateral $ABCD$, $AB \parallel DC$ and $AB=DC$. The diagonals $AC$ and $BD$ intersect at point $O$. $AC$ bisects $\angle BAD$, and a line is drawn from point $C$ perpendicular to $AB$, intersecting the extension of $AB$ at point $E$. Line $OE$ is then drawn. Quadrilateral $ABCD$ is a rhombus. If $AB = \sqrt{10}$ and $BD = 2$, what is the area of $\triangle AOE$?

Solution: \textbf{Solution:} Thus, the area of $\triangle AOE$ is $\boxed{\frac{27}{10}}$.


Question: As shown in the figure, quadrilateral $ABCD$ is a rhombus, and point $E$ is on the extension line of $AC$, $\angle ACD = \angle ABE$. $\triangle ABC \sim \triangle AEB$; when $AD = 4$ and $AC = 3$, what is the length of $AE$?

Solution:
\textbf{Solution:} Since $\triangle ABC\sim \triangle AEB$, \\
it follows that $\frac{AB}{AE}=\frac{AC}{AB}$, \\
Since $ABCD$ is a rhombus, we have $AB=AD=4$, $AC=3$, \\
thus $\frac{4}{AE}=\frac{3}{4}$, \\
hence $AE=\boxed{\frac{16}{3}}$.


Question: As shown in the figure, within the Cartesian coordinate system, the quadrilateral \(OABC\) is a square. Point \(B\) has the coordinates \((2\sqrt{2}, 0)\). Point \(D\) is a moving point on ray \(OB\) that does not coincide with point \(O\). Segment \(CD\) is rotated \(90^\circ\) clockwise around point \(D\) to obtain \(ED\), and segments \(AD\) and \(AE\) are drawn. \(DA=DE\); As in figure 2, by connecting \(AC\) and \(BE\), when triangles \(CDA\) and \(DBE\) are similar, what is the length of \(BD\)?

Solution:
\textbf{Solution:} As shown in Figure 2, let AC intersect OB at point J, and draw DK $\perp$ BC at point K. \\
$\because$ DA=DC=DE, \\
$\therefore$ $\angle$DAC=$\angle$DCA, $\angle$DAE=$\angle$DEA, \\
$\because$ $\triangle ADC$ is similar to $\triangle EBD$, \\
$\therefore$ $\angle$ACD=$\angle$DEB, \\
$\therefore$ $\angle$CAD=$\angle$DAE, \\
$\because$ $\angle$CAB=$\angle$ACB=$45^\circ$, \\
$\therefore$ $\angle$CAD=$\angle$DAB=$\angle$ACD=$\angle$BCD=$22.5^\circ$, \\
$\because$ DJ$\perp$CJ, DK$\perp$CB, \\
$\therefore$ DJ=DK, \\
$\because$ B$(2\sqrt{2},0)$, \\
$\therefore$ OJ=JB=$\sqrt{2}$, \\
$\therefore$ DJ+$\sqrt{2}$DJ=$\sqrt{2}$, \\
$\therefore$ DJ=$2-\sqrt{2}$, \\
$\therefore$ OD=2, \\
$\therefore$ BD=$2\sqrt{2}-2$; \\
Therefore, the length of BD is $BD=\boxed{2\sqrt{2}-2}$.


Question: As shown in the figure, in the plane Cartesian coordinate system, the quadrilateral OABC is a square. The coordinates of point B are $(2\sqrt{2}, 0)$. Point D is a moving point on ray OB that does not coincide with point O. The line segment CD is rotated clockwise by $90^\circ$ around point D to obtain ED. Connect AD and AE. When the symmetric point $A'$ of point A with respect to line ED falls on the edge of the square, find the coordinates of point D.

Solution:
\textbf{Solution:} As shown in Figure 3-1, when point $A^{\prime}$ falls on BC,\\
due to symmetry, it can be known that $\angle CDE=\angle ADA^{\prime}=90^\circ$, $\angle EDA=\angle EDA^{\prime}=45^\circ$,\\
$\therefore \angle ADC=135^\circ$,\\
$\therefore \angle BDC=\angle BDA=67.5^\circ$,\\
$\because \angle DBC=45^\circ$,\\
$\therefore \angle BCD=\angle BDC=67.5^\circ$,\\
$\therefore BD=BC=2$,\\
$\therefore OD=2\sqrt{2}-2$,\\
$\therefore D(2\sqrt{2}-2,0)$;\\
When point D is on AC, $A^{\prime}$ coincides with point C, satisfying the condition, at this time $D(\sqrt{2},0)$;\\
As shown in Figure 3-2, when point $A^{\prime}$ falls on AC, by the same method, it can be proved that $OD=OC=2$, so we get $D(2,0)$.\\
When point D coincides with B, A coincides with $A^{\prime}$, satisfying the condition, at this time $D(2\sqrt{2},0)$.\\
In summary, the coordinates of point D that meet the condition are $\boxed{(2\sqrt{2}-2,0)}$ or $\boxed{(\sqrt{2},0)}$ or $\boxed{(2,0)}$ or $\boxed{(2\sqrt{2},0)}$.


Question: As shown in the diagram, it is known that $AD \parallel BE \parallel CF$, and they intersect lines $l_{1}$ and $l_{2}$ at points $A$, $B$, $C$ and $D$, $E$, $F$ respectively. If $AB = 6$, $BC = 8$, and $DF = 7$, what is the length of $EF$?

Solution:
\textbf{Solution:} Since $AD \parallel BE \parallel CF$, \\
it follows that $\frac{AB}{BC} = \frac{DE}{EF} = \frac{3}{4}$, \\
given that $DE=7$, \\
thus $EF=\boxed{4}$.


Question: As shown in the figure, it is known that $AD \parallel BE \parallel CF$, and they intersect lines $l_{1}$ and $l_{2}$ at points $A$, $B$, $C$, and $D$, $E$, $F$, respectively. If $AB:AC = 2:5$ and $EF=9$, and segment $x$ is the mean proportional of segments $DE$ and $DF$, what is the value of $x$?

Solution:
\textbf{Solution:} Given that $AD\parallel BE\parallel CF$,\\
thus, $\frac{AB}{AC} = \frac{DE}{DF} = \frac{2}{5}$,\\
therefore, $\frac{DE}{EF}=\frac{2}{3}$,\\
given that EF=9,\\
therefore, DE=6, DF=15,\\
since $x^{2} = DE \times DF = 90$, and $x>0$,\\
thus, $x= \boxed{3\sqrt{10}}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertices of rectangle $OABC$ are such that vertex $O$ coincides with the origin of the coordinates, vertices $A$ and $C$ are respectively on the x-axis and y-axis, and the coordinates of point $B$ are $\left(2, 3\right)$. The graph of the inverse proportion function $y=\frac{k}{x}$ passes through the midpoint $D$ of $BC$ and intersects $AB$ at point $E$. Connect $DE$. Find the value of $k$ and the coordinates of point $E$.

Solution:
\textbf{Solution:} In rectangle $OABC$,\\
since the coordinates of point $B$ are $(2, 3)$,\\
then the coordinates of midpoint $D$ on side $BC$ are $(1, 3)$,\\
and since the inverse proportion function $y=\frac{k}{x}$ passes through point $D(1, 3)$,\\
then $3=\frac{k}{1}$,\\
therefore $k=\boxed{3}$,\\
since point $E$ is on $AB$,\\
then the x-coordinate of $E$ is 2,\\
and since $y=\frac{3}{x}$ passes through point $E$,\\
then the y-coordinate of $E$ is $\frac{3}{2}$,\\
therefore, the coordinates of point $E$ are $\boxed{\left(2, \frac{3}{2}\right)}$,


Question: As shown in the figure, in the Cartesian coordinate system, the rectangle $OABC$ has its vertex O coinciding with the origin, vertices A and C respectively on the x-axis and y-axis, and the coordinate of point B is $\left(2, 3\right)$. The graph of the inverse proportion function $y=\frac{k}{x}$ passes through the midpoint D of BC, and intersects AB at point E, with DE being joined. If point F is a point on the side OC, and $\triangle FBC\sim \triangle DEB$, find the equation of the straight line FB.

Solution:
\textbf{Solution:} From (1) we have $BD=1$, $BE=\frac{3}{2}$, $CB=2$,

since $\triangle FBC\sim \triangle DEB$,

then $\frac{BD}{CF}=\frac{BE}{CB}$, which is $\frac{1}{CF}=\frac{\frac{3}{2}}{2}$,

therefore $CF=\frac{4}{3}$,

thus $OF=\frac{5}{3}$, meaning the coordinates of point F are $\left(0, \frac{5}{3}\right)$,

Assume the equation of line $FB$ is $y=k_1x+b\ (k_1\ne 0)$, and line $FB$ passes through $B(2, 3)$, $F\left(0, \frac{5}{3}\right)$,

therefore $\begin{cases}3=2k_1+b \\ \frac{5}{3}=b \end{cases}$,

hence $k_1=\frac{2}{3}, b=\frac{5}{3}$,

therefore, the equation of line $FB$ is $\boxed{y=\frac{2}{3}x+\frac{5}{3}}$;


Question: As shown in the diagram, in the Cartesian coordinate system, the vertices of the rectangle $OABC$ are such that vertex $O$ coincides with the origin, and vertices $A$ and $C$ are respectively on the x-axis and y-axis. The coordinates of point $B$ are $(2, 3)$. The graph of the inverse proportion function $y=\frac{k}{x}$ passes through the midpoint $D$ of $BC$, and intersects with $AB$ at point $E$. Connect $DE$. If point $P$ is on the y-axis, and the area of $\triangle OPD$ is equal to the area of quadrilateral $BDOE$, find the coordinates of point $P$.

Solution:
\textbf{Solution:} Given that $S_{\text{quadrilateral} BDOE}=S_{\text{rectangle} OABC}-{S}_{\triangle AOE}-{S}_{\triangle COD}=2\times 3-\frac{1}{2}\times 2\times \frac{3}{2}-\frac{1}{2}\times 3\times 1=3$,\\
according to the problem, we have $\frac{1}{2}OP\cdot DC=3, DC=1$,\\
$\therefore OP=6$,\\
$\therefore$ the coordinates of point P are $\boxed{\left(0, 6\right)}$ or $\left(0, -6\right)$.


Question: In the figure, in the isosceles $\triangle ABC$, where $AB=AC$, the circle $\odot O$ with diameter $AB$ intersects $BC$ at point $D$, $DE\perp AC$ with the foot of the perpendicular at $E$, and the extension of $ED$ intersects the extension of $AB$ at point $F$. $EF$ is a tangent to $\odot O$; If the radius of $\odot O$ is $\frac{7}{2}$ and $BD=3$, find the length of $CE$.

Solution:
\textbf{Solution:} As shown in the diagram, connect $AD$,\\
since $AB$ is the diameter of $\odot O$,\\
thus $AD\perp BC$.\\
Since $DE\perp AC$,\\
thus $\angle ADC=\angle DEC=90^\circ$.\\
Since $\angle C=\angle C$,\\
thus $\triangle CDE\sim \triangle CAD$,\\
thus $\frac{CD}{CA}=\frac{CE}{CD}$.\\
Since $AB=AC=7$,\\
thus $DC=DB=3$.\\
Therefore, $\frac{3}{7}=\frac{CE}{3}$\\
Therefore, $CE=\boxed{\frac{9}{7}}$.


Question: As shown in Figure 1, it is known that $AB$ is the diameter of $\odot O$, $\triangle ABC$ is inscribed in $\odot O$, $AB=10$, $BC=6$, and point $D$ is a moving point on $\odot O$ (point $D$ does not coincide with points $A$ or $B$). $OC \parallel AD$, what is the length of $AD$?

Solution:
\textbf{Solution:} Given that $AB=10$ and $BC=6$,\\
thus $OC=AO=OB=5$,\\
since $\angle BHO=\angle BHC=90^\circ$,\\
it follows that $OB^2-OH^2=CB^2-CH^2=BH^2$,\\
thus $5^2-OH^2=6^2-(5-OH)^2$,\\
therefore $OH=1.4$,\\
since $\overset{\frown}{BC}=\overset{\frown}{CD}$,\\
it follows that $DH=BH$,\\
hence $AD=2OH=\boxed{2.8}$;


Question:
As shown in Figure 1, it is known that $AB$ is the diameter of $\odot O$, $\triangle ABC$ is inscribed in $\odot O$, $AB=10$, $BC=6$, and point $D$ is a moving point on $\odot O$ (point $D$ does not coincide with points $A$ or $B$).

As shown in Figure 2, if $CD$ bisects $\angle ACB$, and $AD$ is connected, determine the area of $\triangle ACD$.

Solution:
\textbf{Solution:} Draw line BD and let AG $\perp$ CD at point G,\\
since AB is the diameter,\\
it follows that $\angle ACB=\angle ADB=90^\circ$.\\
Since CD bisects $\angle ACB$,\\
we have $\angle ACD=\angle BCD=45^\circ$,\\
which implies $\overset{\frown}{AD}=\overset{\frown}{BD}$,\\
hence $AD=BD=\frac{\sqrt{2}}{2}AB=5\sqrt{2}$, $AC=\sqrt{AB^{2}-BC^{2}}=8$,\\
thus $AG=CG=\frac{\sqrt{2}}{2}AC=4\sqrt{2}$,\\
and therefore $DG=\sqrt{AD^{2}-AG^{2}}=3\sqrt{2}$,\\
which means $CD=DG+CG=7\sqrt{2}$,\\
so $S_{\triangle ACD}=\frac{1}{2}CD\cdot AG=\frac{1}{2}\times 7\sqrt{2}\times 4\sqrt{2}=\boxed{28}$.


Question: As shown in Figure 1, it is known that $AB$ is the diameter of $\odot O$, $\triangle ABC$ is inscribed in $\odot O$, $AB=10$, $BC=6$, and point $D$ is a moving point on $\odot O$ (point $D$ does not coincide with points $A$ or $B$). When is $AD$ such that $\triangle ACD$ is an isosceles triangle?

Solution:
\textbf{Solution:}
(1) When $AC=AD=8$, $\triangle ACD$ is an isosceles triangle,\\
(2) When $AC=CD=8$, $\triangle ACD$ is an isosceles triangle,\\
As shown in the figure, draw $CF\perp AD$ at $F$,\\
$\therefore AD=2DF$, $\angle CFD=90^\circ$,\\
$\because \angle B=\angle D$, $\angle CFD=\angle ACB=90^\circ$,\\
$\therefore \triangle CDF\sim \triangle ABC$,\\
$\therefore \frac{CD}{AB}=\frac{DF}{BC}$,\\
$\therefore \frac{8}{10}=\frac{DF}{6}$,\\
$\therefore DF=\frac{24}{5}$,\\
$\therefore AD=\frac{48}{5}$;\\
(3) When $AD=CD$, $\triangle ACD$ is an isosceles triangle,\\
As shown in the figure, when point $D$ is on the major arc $ADC$, connect $DO$ and extend it to meet $AC$ at $M$,\\
then $AM=CM=\frac{1}{2}AC=4$, $DM\perp AC$,\\
$\because AO=BO$,\\
$\therefore OM=\frac{1}{2}BC=3$,\\
$\therefore DM=OM+OD=8$,\\
$\therefore AD=\sqrt{AM^2+DM^2}=\sqrt{4^2+8^2}=4\sqrt{5}$,\\
When point $D$ is on the minor arc $AD$, by similar reasoning, we get $AD=2\sqrt{5}$,\\
In summary, when $AD$ is $8$ or $\boxed{\frac{48}{5}}$ or $\boxed{4\sqrt{5}}$ or $\boxed{2\sqrt{5}}$, $\triangle ACD$ is an isosceles triangle.


Question: As shown in the figure, in rectangle $ABCD$, $AB=8$, $BC=12$. Point $E$ is on $AB$ with $AE=5$. $P$ is a point on $AD$. The rectangle is folded along $PE$ making point $A$ fall on point $A'$. $AC$ is joined and intersects $PE$ at point $F$. Let $AP=x$. What is the length of $AC$?

Solution: \textbf{Solution:} We have $AC=\boxed{4\sqrt{13}}$.


Question: As shown in the diagram, within rectangle $ABCD$, $AB=8$, $BC=12$. Point $E$ is on $AB$, with $AE=5$. $P$ is a point on $AD$, and the rectangle is folded along $PE$ such that point $A$ falls on point ${A}^{\prime}$. Connect $AC$ and let it intersect $PE$ at point $F$, with $AP=x$. If point ${A}^{\prime}$ lies on the bisector of $\angle BAC$, what is the length of $FC$?

Solution:
\textbf{Solution:} As shown in Figure 1,\\
$\because AA'$ bisects $\angle BAC$,\\
$\therefore \angle EAA' = \angle FAA'$,\\
From the folding, it is known that $AA' \perp EF$,\\
$\therefore \angle EAA' + \angle AEF = 90^\circ$, $\angle AFE + \angle FAA' = 90^\circ$,\\
$\therefore \angle AEF = \angle AFE$,\\
$\therefore AE = AF = 5$,\\
$\therefore CF = AC - AF = 4\sqrt{13} - 5 = \boxed{4\sqrt{13} - 5}$;


Question: As shown in the diagram, within rectangle $ABCD$, where $AB=8$ and $BC=12$, point E is on $AB$ with $AE=5$. Point P is located on $AD$. The rectangle is folded along $PE$ so that point A lands on point ${A}^{\prime}$. Line $AC$ is drawn and intersects $PE$ at point F, with $AP=x$. Find the minimum distance between points ${A}^{\prime}$ and D, and calculate the value of $\frac{AE}{AP}$ under these conditions;

Solution:
\textbf{Solution:} Connect $DE$ and $D{A}^{\prime}$.

In $\triangle ADE$, $\angle EAD=90^\circ$, $AE=5$, $AD=BC=12$,

$\therefore DE=13$,

$\because$ According to the property of folding: $EA=E{A}^{\prime}=5$,

$\therefore$ In $\triangle D{A}^{\prime}E$, $D{A}^{\prime}>DE-E{A}^{\prime}=8$,

When ${A}^{\prime}$ is on $DE$, $D{A}^{\prime}=DE-E{A}^{\prime}=8$,

That is: $D{A}^{\prime}\ge 8$,

$\therefore$ The minimum value of $D{A}^{\prime}$ is $\boxed{8}$,

At this time, ${A}^{\prime}$ is on $DE$, that is, E, ${A}^{\prime}$, D are collinear,

According to the property of folding: $\angle P{A}^{\prime}E=\angle PAE=90^\circ$,

That is: $\angle P{A}^{\prime}E=\angle P{A}^{\prime}D=90^\circ$,

Therefore, $\triangle P{A}^{\prime}D$ is a right triangle, thus $PD^2={A}^{\prime}P^2+{A}^{\prime}D^2$,

According to $PA=P{A}^{\prime}=x$, we have $PD=AD-PA=12-x$,

Thus, $(12-x)^2=x^2+8^2$,

Solving gives $x=\frac{10}{3}$,

$\therefore \frac{AE}{AP}=\frac{5}{\frac{10}{3}}=\boxed{\frac{3}{2}}$;


Question: As shown in the figure, in rectangle $ABCD$, $AB=8$, $BC=12$, point E is on $AB$ with $AE=5$, and P is a point on $AD$. The rectangle is folded along $PE$ such that point A falls on point ${A}^{\prime}$, and $AC$ is connected, intersecting with $PE$ at point F. Let $AP=x$. If point ${A}^{\prime}$ is inside $\triangle ABC$, directly write down the range of $x$.

Solution:
\textbf{Solution:} As shown in Figure 3-1, when point $A'$ falls on $AC$, \\
$\because$ $\angle AEP+\angle EAC=90^\circ$, $\angle ACB+\angle EAC=90^\circ$, \\
$\therefore$ $\angle AEP=\angle ACB$, \\
Also, $\because$ in rectangle $ABCD$, we have $\angle EAP=\angle ABC=90^\circ$, \\
$\therefore$ $\triangle AEP\sim \triangle BCA$, \\
$\therefore$ $\frac{AP}{AE}=\frac{AB}{BC}$, \\
i.e., $AP=\frac{AB}{BC}\times AE=\frac{8}{12}\times 5=\frac{10}{3}$; \\
In Figure 3-2, when point $A'$ falls on $CB$, draw $PH\perp BC$ at H, \\
It is easy to prove that quadrilateral $APHB$ is a rectangle, \\
Then $PH=AB=8$, $PA=BH$. \\
In $\triangle BEA'$, $BE=3$, $EA'=EA=5$, \\
$\therefore$ $BA'=4$, \\
$\because$ $\angle B=\angle EA'P=\angle PHA'=90^\circ$, \\
$\therefore$ $\angle BA'E+\angle PA'H=90^\circ$, $\angle PA'H+\angle A'PH=90^\circ$, \\
$\therefore$ $\angle BA'E=\angle A'PH$, \\
$\therefore$ $\triangle BA'E\sim \triangle HPA'$, \\
$\therefore$ $\frac{A'H}{HP}=\frac{BE}{BA'}$, \\
$\because$ $BE=3$, $PH=AB=8$, $BA'=4$, \\
$\therefore$ $A'H=\frac{BE}{BA'}\times HP=\frac{3}{4}\times 8=6$, \\
$\therefore$ $AP=BH=BA'+A'H=10$. \\
Thus, when $\frac{10}{3}<x<10$, point $A'$ is inside $\triangle ABC$. Hence, the range of $x$ is $\boxed{\frac{10}{3}<x<10}$.


Question: In square $ABCD$, $E$ and $F$ are points on sides $AD$ and $CD$ respectively, where $AE=ED$ and $DF=\frac{1}{4}DC$. Extend line $EF$ to intersect the extension of $BC$ at point $G$. $\triangle ABE\sim \triangle DEF$; if the side length of the square is $6$, what is the length of $BG$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a square,\\
it follows that $ED\parallel BG$,\\
therefore $\angle DEF=\angle G$,\\
since $\angle DFE=\angle GFC$, $\triangle EDF\sim \triangle GCF$,\\
it follows that $\frac{ED}{CG}=\frac{DF}{CF}$,\\
also, given that $DF=\frac{1}{4}DC$ and the side length of the square is 6,\\
it follows that $ED=3$, $\frac{ED}{CG}=\frac{DF}{CF}=\frac{1}{3}$,\\
therefore $CG=9$,\\
thus $BG=BC+CG=\boxed{15}$.


Question: As shown in the diagram, in $\triangle ABC$, the perpendicular bisectors of sides $AB$ and $AC$ intersect $BC$ at points $D$ and $E$, respectively. If $BC=6$, what is the perimeter of $\triangle ADE$?

Solution:
\textbf{Solution:} In triangle $ \triangle ABC$, the perpendicular bisectors of sides $AB$ and $AC$ intersect $BC$ at points $D$ and $E$, respectively.\\
$\therefore AD=BD$, $CE=AE$,\\
Given $BC=6$,\\
$\therefore$ the perimeter of $ \triangle ADE$ $ =AD+DE+EA=BD+DE+EC=BC=\boxed{6}$.


Question: As shown in the figure, in $\triangle ABC$, the perpendicular bisectors of sides $AB$ and $AC$ respectively meet $BC$ at $D$ and $E$. If $\angle BAC=130^\circ$, find the measure of $\angle DAE$.

Solution:
\textbf{Solution:} Since $\angle BAC=130^\circ$,\\
then $\angle B+\angle C=50^\circ$,\\
Since $AD=BD$, $CE=AE$,\\
then $\angle BAD=\angle B$, $\angle EAC=\angle C$,\\
thus $\angle BAD+\angle EAC=50^\circ$,\\
hence $\angle DAE=\angle BAC-(\angle BAD+\angle EAC)=130^\circ-50^\circ=\boxed{80^\circ}$.


Question: As shown in the figure, $AD$ is the altitude from $A$ to the side $BC$ of $\triangle ABC$, with $E$ being a point on $AD$, and it's given that $BD=AD$ and $DE=DC$. If $AE=5$ and $CD=2$, what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Given that $Rt\triangle BDE\cong Rt\triangle ADC$ and $CD=2$,
it follows that $DE=CD=2$ and $BD=AD$.
Since $AE=5$,
we have $AD=AE+DE=5+2=7$.
Thus, $BC=BD+CD=7+2=9$.
Therefore, the area of $\triangle ABC$ is $\frac{1}{2}BC\cdot AD=\frac{1}{2}\times 9\times 7=\boxed{\frac{63}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle A=60^\circ$, point D is on side AB, and point E is on the extension line of side AC, with $BD=CE$. Line DE intersects BC at point F, and point F is the midpoint of DE. Point G is on BC, and $DG\parallel AC$. It is given that $DG=CE$; $\triangle ABC$ is an equilateral triangle; if $ED\perp AB$ and $BD=2$, what is the length of AB?

Solution:
\textbf{Solution:} Given that $\because \triangle ABC$ is an equilateral triangle,\\
$\therefore AB=BC, \angle B=60^\circ$.\\
$\because ED\perp AB$,\\
$\therefore \angle BFD=90^\circ-\angle B=30^\circ, \angle E=90^\circ-\angle A=30^\circ$,\\
$\therefore BF=2BD=4$.\\
$\because \angle CFE=\angle BFD=30^\circ$,\\
$\therefore \angle CFE=\angle E$,\\
$\therefore CF=CE=BD=2$,\\
$\therefore AB=BC=BF+CF=\boxed{6}$.


Question: As shown in the figure, in $\triangle ABC$, it is known that $AB=AC$, $BC$ bisects $\angle ABD$, and $AC\parallel BD$; if $\angle A=100^\circ$, find the measure of $\angle 1$.

Solution:
\textbf{Solution}: Since $\angle A=100^\circ$,\\
it follows that $\angle ABC=\angle ACB=\frac{180^\circ-100^\circ}{2}=40^\circ$,\\
therefore $\angle 1=\boxed{40^\circ}$.


Question: As shown in the figure of quadrilateral $ABCD$, it is known that $AB=3$, $BC=4$, $CD=12$, $AD=13$, and $\angle B=90^\circ$. Find: the length of $AC$?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $AB=3$, $BC=4$, and $\angle B=90^\circ$,\\
it follows that $AC=\sqrt{AB^2+BC^2}=\boxed{5}$;


Question: As shown in the figure, the quadrilateral $ABCD$ is given with $AB=3$, $BC=4$, $CD=12$, $AD=13$, and $\angle B=90^\circ$. Find: The area of the quadrilateral $ABCD$.

Solution:
\textbf{Solution:} For $\triangle ABC$, we have $S_{\triangle ABC}=\frac{1}{2}AB\cdot BC=\frac{1}{2}\times 3\times 4=6$, \\
Since, in $\triangle ACD$, $CD=12$, $AD=13$, and $AC=5$, \\
Thus, $CD^2+AC^2=12^2+5^2=13^2=AD^2$, \\
Therefore, $\triangle ACD$ is a right-angled triangle, with $\angle ACD=90^\circ$, \\
Hence, $S_{\triangle ACD}=\frac{1}{2}AC\cdot CD=\frac{1}{2}\times 5\times 12=30$. \\
Therefore, $S_{\text{Quadrilateral} ABCD}=S_{\triangle ABC}+S_{\triangle ACD}=6+30=\boxed{36}$.


Question: In quadrilateral $ABCD$, $\triangle OAB$ and $\triangle OCD$ share a common vertex $O$, and $\triangle OAB \cong \triangle OCD$. As shown in Figure 1, $O$ is a point on side $BC$. Given that $AO=DO$, and if $\angle AOD=80^\circ$, connect $AC$ and $BD$, with the intersection point being $E$. Find the measure of $\angle DEC$.

Solution:
\textbf{Solution:} As shown, \\
$\because$ $\triangle OAB \cong \triangle OCD$, \\
$\therefore$ $\angle BOA = \angle COD = \frac{180^\circ - 80^\circ}{2} = 50^\circ, OA = OC, OB = OD$, \\
$\therefore$ $\angle AOC = \angle DOB = 130^\circ$, \\
$\therefore$ $\angle ACO = \angle DBO = \frac{180^\circ - 130^\circ}{2} = 25^\circ$, \\
$\therefore$ $\angle DEC = \boxed{50^\circ}$.


Question: In quadrilateral \(ABCD\), triangles \(OAB\) and \(OCD\) share the common vertex \(O\), and \( \triangle OAB \cong \triangle OCD\). As shown in Figure 2, points \(B\), \(O\), and \(C\) are not on the same line, and \(\angle AOB=90^\circ\), given that \(AD^2+BC^2=50\), \(AO=3\), find the area of \(\triangle OAB\).

Solution:
\textbf{Solution:} As shown in the figure, connect $AC, BD$ and they intersect at point Q, \\
$\because$ $\triangle OAB \cong \triangle OCD$, \\
$\therefore$ $\angle COD = \angle AOB = 90^\circ$, $OA = OC, OB = OD$, $AB = CD$, \\
$\therefore$ $\angle AOC = \angle BOD$, \\
$\therefore$ $\angle OAC = \angle OCA = \angle OBD = \angle ODB$. \\
$\because$ $\angle OBC + \angle OCB = 180^\circ - \angle BOC$, $\angle AOD = 180^\circ - \angle BOC$, \\
$\therefore$ $\angle OBC + \angle OCB = \angle AOD$, \\
$\therefore$ $\angle DQC = \angle QBO + \angle OBC + \angle QCO + \angle OCB$ \\
$= \angle OAC + \angle OCA + \angle AOD$ \\
$= 180^\circ - 90^\circ = 90^\circ$. \\
$\therefore$ $AD^2 = AQ^2 + DQ^2$, $BC^2 = BQ^2 + CQ^2$, \\
$\therefore$ $AD^2 + BC^2 = AQ^2 + DQ^2 + BQ^2 + CQ^2$ \\
$= AB^2 + CD^2$, \\
$\because$ $AB = CD$, \\
$\therefore$ $AD^2 + BC^2 = 2AB^2 = 50$, \\
$\therefore$ $AB = 5$, \\
$\because$ $AO = 3$, \\
$\therefore$ $BO = 4$, \\
$\therefore$ the area of $\triangle OAB = \boxed{6}$.


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

Solution:
\textbf{Solution}: Given the problem, \\
$\because DA=DC$, \\
$\therefore \angle DAC=\angle C$, \\
$\because \angle ADB$ is an exterior angle of $\triangle ADC$, \\
$\therefore \angle ADB=\angle DAC+\angle C=2\angle C$, \\
$\because AB=AD$, \\
$\therefore \angle B=\angle ADB=2\angle C$, \\
$\because \angle B+\angle ADB+\angle BAD=180^\circ$, \\
$\therefore 2\angle C+20^\circ+2\angle C=180^\circ$, \\
$\therefore \angle C=\boxed{40^\circ}$.


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

Solution: \textbf{Solution:} Since the given information is insufficient, it is impossible to provide a solution.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $\angle BAC=110^\circ$, $D$ is a point on $BC$, connect $AD$, make $\angle ADE=35^\circ$, and $DE$ intersects line segment $AC$ at point $E$. Write down the sizes of $\angle B$ and $\angle C$ directly.

Solution: \textbf{Solution}: Let $ \angle B = \angle C = \boxed{30^\circ}$


Question: As shown in the figure, in $ \triangle ABC$, we have $AB=AC$ and $\angle BAC=110^\circ$. Let $D$ be a point on $BC$. Connect $AD$ and construct $\angle ADE=35^\circ$, with $DE$ intersecting line segment $AC$ at point $E$. If $AE=DE$, what is the size of $\angle BDA$?

Solution:
Solution: Since $AE=DE$,\\
it follows that $\angle DAE=\angle ADE=35^\circ$,\\
Since $\angle C=35^\circ$,\\
it follows that $\angle BDA=\angle DAE+\angle C=35^\circ+35^\circ=\boxed{70^\circ}$.


Question: As shown in the figure, it is known that $\triangle ABD \cong \triangle CFD$, $AD \perp BC$ at point $D$, and $CE \perp AB$; given that $BC=7$ and $AD=5$, find the length of $AF$.

Solution:
\textbf{Solution:} Since $\triangle ABD \cong \triangle CFD$, \\
$\therefore BD=DF$,\\
Since $BC=7$, $AD=DC=5$,\\
$\therefore BD=BC-CD=2$,\\
$\therefore AF=AD-DF=5-2=\boxed{3}$.


Question: In $\triangle ABC$, $\angle ABM = 45^\circ$, $AM \perp BM$ with the foot of the perpendicular at $M$, and point $C$ lies on the extension of $BM$. Connect $AC$. If $AB = \sqrt{18}$ and $BC = 5$, what is the length of $AC$?

Solution:
\textbf{Solution:} Since $\angle ABM=45^\circ$ and $AM\perp BM$,\\
it follows that $AM=BM=\frac{\sqrt{2}}{2}AB=\frac{\sqrt{2}}{2}\times 3\sqrt{2}=3$,\\
thus $CM=BC-BM=5-3=2$,\\
therefore $AC=\sqrt{AM^2+CM^2}=\sqrt{3^2+2^2}=\sqrt{13}$;\\
hence, the length of $AC$ is $\boxed{\sqrt{13}}$.


Question: As shown in the figure, $\triangle ABC$ and $\triangle ADE$ are symmetric with respect to line $MN$, and the intersection point $F$ of $BC$ and $DE$ lies on line $MN$. If $\angle BAC=100^\circ$ and $\angle CAD=30^\circ$, what is the degree measure of $\angle EAF$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ and $\triangle ADE$ are symmetrical with respect to line MN, \\
$\therefore \triangle ABC \cong \triangle ADE$, $\angle CAF = \angle EAF$,\\
$\therefore \angle DAE = \angle BAC = 100^\circ$,\\
$\because \angle CAD = 30^\circ$,\\
$\therefore \angle CAE = 100^\circ - 30^\circ = 70^\circ$,\\
$\therefore \angle EAF = \frac{70^\circ}{2} = \boxed{35^\circ}$.


Question: As shown in the figure, $\triangle ABC$ and $\triangle ADE$ are symmetric with respect to line $MN$, and the intersection $F$ of $BC$ and $DE$ lies on line $MN$. If $BC \parallel AD$, $AE$ bisects $\angle BAM$, and $\angle BFE + \angle C = 81^\circ$, what is the degree measure of $\angle EAF$?

Solution:
\textbf{Solution:} Given $BC\parallel AD$,\\
it follows that $\angle C = \angle CAD$,\\
given $\angle DAC = \angle BAE$, and $\angle EAF = \angle CAF$,\\
and since $AE$ bisects $\angle BAM$,\\
it follows $\angle DAC = \angle CAF = \angle EAF = \angle BAE$,\\
given $\angle BFE + \angle C = 81^\circ$,\\
it follows $\angle D + \angle DAC = 81^\circ$,\\
therefore $\angle CAF + \angle EAF + \angle E = 180^\circ - 81^\circ = 99^\circ$,\\
given $\angle C = \angle E$,\\
it follows $3\angle EAF = 99^\circ$,\\
hence $\angle EAF = \boxed{33^\circ}$.


Question: As shown in the figure, given: In $\triangle ABC$, $\angle ABC=60^\circ$, $\angle ACB=40^\circ$, $BD$ bisects $\angle ABC$, and $CD$ bisects $\angle ACB$. What is the measure of $\angle BDC$?

Solution:
\textbf{Solution:}

Since BD bisects $\angle ABC$, \\
it follows that $\angle DBC = \frac{1}{2}\angle ABC = \frac{1}{2} \times 60^\circ = 30^\circ$, \\
Since CD bisects $\angle ACB$, \\
it follows that $\angle DCB = \frac{1}{2}\angle ACB = \frac{1}{2} \times 40^\circ = 20^\circ$, \\
Therefore, $\angle BDC = 180^\circ - \angle DBC - \angle DCB = 180^\circ - 30^\circ - 20^\circ = \boxed{130^\circ}$;


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=60^\circ$, $\angle ACB=40^\circ$, $BD$ bisects $\angle ABC$, $CD$ bisects $\angle ACB$, connect $AD$, construct $DE\perp AB$, $DE=2$, and $AC=4$. What is the area of $\triangle ADC$?

Solution:
\textbf{Solution:} Let $DF\perp AC$ at $F$, and $DH\perp BC$ at $H$, as shown in Figure 2,\\
since $BD$ bisects $\angle ABC$, $DE\perp AB$, and $DH\perp BC$,\\
therefore $DH=DE=2$,\\
since $CD$ bisects $\angle ACB$, $DF\perp AC$, and $DH\perp BC$,\\
therefore $DF=DH=2$,\\
thus, the area of $\triangle ADC$ is $\frac{1}{2}DF\cdot AC=\frac{1}{2}\times 2\times 4=\boxed{4}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=45^\circ$, D is a point on BC, $\angle ADC=60^\circ$, $AE\perp BC$ at point E, $CF\perp AD$ at point F, $AE$ and $CF$ intersect at point G, $\angle CAE=15^\circ$. What is the degree measure of $\angle ACF$?

Solution:
\textbf{Solution:} Given that $\because \angle ADC=60^\circ$ and $AE \perp BC$,\\
$\therefore \angle DAE=30^\circ$,\\
$\because \angle CAE=15^\circ$,\\
$\therefore \angle CAF=\angle CAE+\angle DAE=45^\circ$,\\
$\because CF \perp AD$,\\
$\therefore \angle ACF=90^\circ-\angle CAF=45^\circ$.

Thus, the answer is $\angle ACF=\boxed{45^\circ}$.


Question: In the figure, within $\triangle ABC$, $\angle BAC=90^\circ$, $\angle ABC=60^\circ$, where $AD$ and $CE$ bisect $\angle BAC$ and $\angle ACB$, respectively. What is the degree measure of $\angle AOE$?

Solution:
\textbf{Solution:} Since $\angle BAC=90^\circ$ and $\angle ABC=60^\circ$, \\
it follows that $\angle ACB=30^\circ$, \\
Because $AD$ bisects $\angle BAC$ and $CE$ bisects $\angle BAC$, \\
it follows that $\angle CAD=\frac{1}{2}\angle BAC=45^\circ$ and $\angle ACE=\frac{1}{2}\angle ACB=15^\circ$, \\
Since $\angle AOE is the exterior angle of \triangle AOC$, \\
it follows that $\angle AOE=\angle CAD+\angle ACE=\boxed{60^\circ}$;


Question: As shown in the figure, the line $y = \frac{4}{3}x + 4$ intersects the x-axis and the y-axis at points A and B, respectively. What are the coordinates of points A and B?

Solution: \textbf{Solution:} For $y=\frac{4}{3}x+4$, setting $y=0$ gives $y=\frac{4}{3}x+4=0$. Solving this, we find $x=-3$. Setting $x=0$, we get $y=4$. Therefore, the coordinates of points A and B are \boxed{(-3,0)} and \boxed{(0,4)}, respectively.


Question: As shown in the figure, the line $y=\frac{4}{3}x+4$ intersects the x-axis and the y-axis at points $A$ and $B$ respectively. A line is drawn through point $B$ intersecting the x-axis at point $P$. If the area of $\triangle ABP$ is $8$, find the coordinates of point $P$.

Solution:
\textbf{Solution:} Let the point $P(x,0)$, \\
then the area of $\triangle ABP = \frac{1}{2} \times AP \times OB = \frac{1}{2} \times 4 \times |x+3| = 2|x+3|$, solving for $x$ gives $x=1$ or $x=-7$, \\
$\therefore$ the coordinates of point $P$ are $\boxed{(1,0)}$ or $\boxed{(-7,0)}$;


Question: As shown in the figure, the line $y=\frac{4}{3}x+4$ intersects the x-axis and y-axis at points A and B, respectively. Point M is a point on OB, and if $\triangle ABM$ is folded along AM, point B exactly falls on point $B_{1}$ on the x-axis. Find the coordinates of point M.

Solution:
\textbf{Solution:} Given the coordinates of points $A$ and $B$, we know that $OA=3$, $BO=4$. Therefore, $AB=\sqrt{OA^2+BO^2}=5=AB_1$,\\
thus the coordinates of point $B_1$ are $(2,0)$,\\
let the coordinates of point $M$ be $(0,m)$,\\
according to the problem, $MB=MB_1$, which means $m^2+4=(m-4)^2$. Solving this, we find $m=1.5$,\\
therefore, the coordinates of point $M$ are $\boxed{(0,1.5)}$.


Question: As shown in the figure, the line $y = \frac{4}{3}x + 4$ intersects the x-axis and y-axis at points A and B, respectively. Point C is on the y-axis, and line AC is drawn. If $\triangle ABC$ is an isosceles triangle with AB as its base, please write down the coordinates of point C directly.

Solution: \textbf{Solution:} The coordinates of point C are $\boxed{(0,9)}$ or $(0,-1)$ or $(0,-4)$.


Question: As shown in the figure, in rectangle $ABCD$, we have $AB=10$ and $AD=6$. Let point E be on side $BC$, and fold $\triangle CDE$ along $DE$ such that point C falls onto point F on side $AB$. What is the length of $AF$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
it follows that $\angle A=90^\circ$, and AB=CD=10,\\
by the properties of folding, we know that DF=CD=10,\\
therefore in $\triangle ADF$, by the Pythagorean theorem, we have $AF=\sqrt{DF^{2}-AD^{2}}=\boxed{8}$.


Question: As shown in the figure, in rectangle $ABCD$, we have $AB=10$ and $AD=6$. Let E be a point on side $BC$. Fold $\triangle CDE$ along $DE$ so that point C falls on point F on side $AB$. What is the length of $CE$?

Solution:
\textbf{Solution:} Given the property of folding, we can determine that $CE=EF$,\\
from the properties of a rectangle, we have $\angle B=90^\circ$, $BC=AD=6$,\\
let $CE=EF=x$, then $BE=6-x$, $BF=AB-AF=2$,\\
in $\triangle BEF$, by the Pythagorean theorem, we have $EF^2=BE^2+BF^2$,\\
$\therefore$ $2^2+(6-x)^2=x^2$,\\
solving for $x$, we get $x=\frac{10}{3}$,\\
$\therefore$ $CE=\boxed{\frac{10}{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=2\sqrt{10}$, and $AD$ is the altitude on side $BC$, with $DE\parallel AC$ intersecting $AB$ at point $E$. $\triangle ADE$ is an isosceles triangle; connect $CE$ and intersect $AD$ at point $H$. If $\angle DCE=45^\circ$, find the length of $EH$.

Solution:
\textbf{Solution:} Draw $EG\parallel BC$, intersecting $AD$ at $G$,\\
$\because AE=BE$,\\
$\therefore AG=DG$,\\
$\therefore EG= \frac{1}{2} BD= \frac{1}{2} CD$,\\
$\because EG\parallel BC$,\\
$\therefore \frac{GH}{DH}=\frac{EH}{CH}=\frac{EG}{CD} = \frac{1}{2}$,\\
$\therefore GH= \frac{1}{2} DH$, $EH= \frac{1}{2} CH$,\\
$\because AD\perp BC$, $\angle DCE=45^\circ$,\\
$\therefore \triangle CDH$ is an isosceles right triangle,\\
$\therefore DH=DC$,\\
$\therefore AD=3DC$,\\
$\because AB=AC=2\sqrt{10}$, $AC^2=AD^2+DC^2$,\\
$\therefore 40=9DC^2+DC^2$,\\
$\therefore DC=2$,\\
$\therefore DH=DC=2$,\\
$\therefore CH= \sqrt{2^2+2^2} =2\sqrt{2}$,\\
$\therefore EH= \frac{1}{2} CH= \boxed{\sqrt{2}}$,\\


Question: As shown in the figure, $\triangle ABC$ is an arbitrary triangle. Equilateral triangles $ABD$ and $ACE$ are constructed externally on sides $AB$ and $AC$, respectively. Lines $CD$ and $BE$ are drawn and intersect at point $P$. $\triangle DAC \cong \triangle BAE$; what is the measure of $\angle BPC$?

Solution:
\textbf{Solution:} From (1) we have $\angle ADC=\angle ABE$, \\
$\therefore \angle BPC=\angle PBD+\angle PDB=\angle ABD+\angle ABE+\angle PDB=\angle ABD+\angle ADC+\angle PDB=\angle ABD+\angle ADB$, \\
$\because \angle ABD=\angle ADB=60^\circ$, \\
$\therefore \angle BPC=\boxed{120^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $D$ is the midpoint of $BC$, $DE\perp AB$, and $DF\perp AC$ with the perpendicular feet being $E$ and $F$ respectively, and $BE=CF$. $AD$ is the angle bisector of $\triangle ABC$. If $AB=8$ and the area of $\triangle ABC$ is $36$, what is the length of $DE$?

Solution:
\textbf{Solution:} From (1), we have $\triangle DEB \cong \triangle DFC$,\\
$DE=DF$,\\
$\therefore \angle B=\angle C$,\\
$\therefore AB=AC=8$,\\
$\because S_{\triangle ABC}=36$,\\
$\therefore \frac{AB\cdot DE}{2}+\frac{AC\cdot DF}{2}=36$,\\
$\therefore \frac{8DE}{2}+\frac{8DE}{2}=36$,\\
From this, we find $DE=\boxed{\frac{9}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, $\angle A = 36^\circ$. The bisector of the external angle $\angle CBD$ of $\triangle ABC$ intersects the extension of line $AC$ at point $E$. Point $F$ is located on the extension of line $AC$, and line $DF$ is drawn. Find the degree measure of $\angle CBE$?

Solution:
\textbf{Solution:} Given $\because \angle ACB=90^\circ$ and $\angle A=36^\circ$. Since $\angle CBD$ is an exterior angle of $\triangle ABC$,\\
$\therefore \angle CBD=\angle ACB+\angle A=126^\circ$.\\
Since $BE$ bisects $\angle CBD$,\\
$\therefore \angle CBE= \frac{1}{2} \angle CBD=\boxed{63^\circ}$.


Question: As shown in the figure, $AB=AC$, $\angle BAC=120^\circ$, and the perpendicular bisector of $AB$ intersects $BC$ at point $D$. What is the degree measure of $\angle ADC$?

Solution:
\textbf{Solution:} Given $AB=AC$ and $\angle BAC=120^\circ$,\\
it follows that $\angle B=\angle C= \frac{1}{2} (180^\circ-\angle BAC)= \frac{1}{2} (180^\circ-120^\circ)=30^\circ$,\\
since $DE$ is the perpendicular bisector of $AB$,\\
it implies that $AD=BD$,\\
therefore $\angle BAD=\angle B=30^\circ$,\\
thus $\angle ADC=\angle B+\angle BAD=30^\circ+30^\circ=\boxed{60^\circ};$


Question: As shown in Figure 1, in the Cartesian coordinate system, the linear function $y=-\frac{2}{3}x+4$ intersects the x-axis at point B and the y-axis at point A. Point C is the midpoint of segment AB. A line is drawn through point C such that $DC \perp x$-axis, with D being the foot of the perpendicular. What are the coordinates of points A and B?

Solution:
\textbf{Solution:}

Since the linear function $y=-\frac{2}{3}x+4$ intersects with the x-axis at point B, and with the y-axis at point A,\\
therefore, when $x=0$, $y=4$,\\
thus, the coordinates of point A are $\boxed{(0,4)}$;\\
when $y=0$, solving $-\frac{2}{3}x+4=0$ yields $x=6$,\\
therefore, the coordinates of point B are $\boxed{(6,0)}$.