Question: In the mathematics activity class, Xiao Dong wants to measure the radius of a circular gear's inner circle. As shown in the figure, Xiao Dong first selects points $A$ and $B$ on the inner circle, then constructs the perpendicular bisector of chord $AB$ with the foot of the perpendicular being $C$ and intersecting $\overset{\frown}{AB}$ at point $D$, and connects $CD$. After measuring, $AB=8\text{cm}$ and $CD=2\text{cm}$. So, what is the radius of this gear's inner circle in centimeters?

Solution:
\textbf{Solution:} Let the center of the circle be O, and connect OB.\\
In $\triangle OBC$, $BC = \frac{1}{2}AB = 4\text{cm}$,\\
According to the Pythagorean theorem, $OC^2 + BC^2 = OB^2$, which is: $(OB - 2)^2 + 4^2 = OB^2$,\\
Solving this, we get: $OB = \boxed{5}$;\\
Therefore, the radius of the wheel is $5\text{cm}$.


Question: As shown in the figure, within circle $\odot O$, $OC \perp AB$, and $\angle ADC=32^\circ$, what is the degree measure of $\angle OBA$?

Solution:
\textbf{Solution:} Since in circle $O$, $OC\perp AB$,\\
it follows that $\overset{\frown}{AC}=\overset{\frown}{BC}$, and $\angle BOC+\angle OBA=90^\circ$,\\
thus $\angle BOC=2\angle ADC=64^\circ$,\\
hence $\angle OBA=90^\circ-\angle BOC=90^\circ-64^\circ=\boxed{26^\circ}$.


Question: As shown in the figure, points $A$, $B$, $P$ are three points on the circle $\odot O$. If $\angle AOB = 50^\circ$, what is the degree measure of $\angle APB$?

Solution:
Given that points A, B, and P lie on the circle centered at O, and $\angle AOB = 50^\circ$, \\
it follows that $\angle APB = \frac{1}{2} \angle AOB = \boxed{25^\circ}.$


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in $\odot O$, with $AD\parallel BC$. $BD$ bisects $\angle ABC$, and $\angle A=126^\circ$. What is the degree measure of $\angle BDC$?

Solution:
\textbf{Solution:} Since $AD \parallel BC$ and $\angle A=126^\circ$,\\
it follows that $\angle ABC=180^\circ-\angle A=180^\circ-126^\circ=54^\circ$,\\
Given that $BD$ bisects $\angle ABC$,\\
we have $\angle DBC=54^\circ\div 2=27^\circ$,\\
Since $AD \parallel BC$,\\
it follows that $\angle ADB=\angle DBC=27^\circ$,\\
Given that quadrilateral $ABCD$ is circumscribed around circle $O$,\\
it follows that $\angle ADC=180^\circ-54^\circ=126^\circ$,\\
Therefore, $\angle BDC=\angle ADC-\angle ADB=126^\circ-27^\circ=\boxed{99^\circ}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in a circle, with extensions of two pairs of opposite sides intersecting at points $E$ and $F$, respectively, and $\angle E=40^\circ$, $\angle F=60^\circ$. What is the measure of $\angle A$ in degrees?

Solution:
\textbf{Solution:} Connect EF, as shown in the figure, \\
$\because$ Quadrilateral ABCD is inscribed in a circle, $\therefore \angle$ECD$=\angle A$, \\
$\because \angle$ECD$=\angle 1+\angle 2$, $\therefore \angle A=\angle 1+\angle 2$, \\
$\because \angle A+\angle 1+\angle 2+\angle E+\angle F=180^\circ$, \\
$\therefore 2\angle A+40^\circ+60^\circ=180^\circ$, \\
$\therefore \angle A=\boxed{40^\circ}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\odot O$, with $E$ being a point on the extension line of diameter $AB$, and $AB \parallel DC$. If $\angle A=70^\circ$, what is the degree measure of $\angle CBE$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in circle $O$,\\
it follows that $\angle A+\angle C=180^\circ$,\\
Since $\angle A=70^\circ$,\\
it follows that $\angle C=110^\circ$,\\
Since $AB\parallel DC$,\\
it follows that $\angle CBE=\angle C=110^\circ$;\\
Therefore, the answer is: $\boxed{110^\circ}$.


Question: As shown in the figure, points A, B, C, and D are all on the circle \( \odot O\), with \( OA\perp BC\), and \( \angle CDA=25^\circ\). What is the degree measure of \( \angle AOB\)?

Solution:
\textbf{Solution:}\\
Since $ AO\perp BC$\\
It follows that $\overset{\frown}{AC}=\overset{\frown}{AB}$\\
Since $\angle CDA=25^\circ$\\
Thus, $\angle AOB=2\angle CDA=50^\circ$\\
Therefore, the answer is: $\boxed{50^\circ}$


Question: As shown in the diagram, the quadrilateral $ABCD$ is an inscribed quadrilateral of the circle $\odot O$, and $\angle BCD=120^\circ$. What is the measure of $\angle BOD$?

Solution:
\textbf{Solution:} Since $\angle BCD = 120^\circ$,\\
it follows that $\angle A = 180^\circ - \angle BCD = 60^\circ$,\\
therefore $\angle BOD = 2\angle A = \boxed{120^\circ}$.


Question: As shown in the figure, points A, B, and C are three points on the circle $\odot O$, with $\angle C = 50^\circ$, what is the degree measure of $\angle AOB$?

Solution:
\textbf{Solution:} Since arc AB = arc AB,\\
it follows that $\angle AOB = 2\angle C = 2\times50^\circ = \boxed{100^\circ}$.


Question: As shown in the figure, it is known that the line $l_1\colon y=kx+b$ intersects with the line $l_2\colon y=mx+n$ at point $P(-3, -2)$, what is the solution set of the inequality $mx+n>kx+b$ with respect to $x$?

Solution:
\textbf{Solution:} Given that line $l_1\colon y=kx+b$ intersects with line $l_2\colon y=mx+n$ at point $P(-3, -2)$.\\
Therefore, the solution set for $mx+n>kx+b$ is $\boxed{x<-3}$.


Question: As shown in the figure, the line $I_1 \colon y=x+1$ intersects with the line $I_2 \colon y=mx+n$ at point $P(a, 2)$. What is the solution set of the inequality $x+1\ge mx+n$ in terms of $x$?

Solution:
\textbf{Solution:} Since $y=x+1$ intersects with the line $I_2 \colon y=mx+n$ at point $P(1, 2)$,\\
substituting $y=2$ into $y=x+1$ gives $x=1$,\\
therefore, the coordinates of point $P$ are $(1, 2)$;\\
From the graph, it is known that when $x \geq 1$, $x+1 \geq mx+n$.\\
Thus, the answer is: $\boxed{x \geq 1}$.


Question: As shown in the figure, the line $y=kx+b$ passes through point $A(-1, -3)$ and point $B(-2, 0)$, and the line $y=3x$ passes through point $A$. What is the solution set of the inequality $3x<kx+b<0$?

Solution:
\textbf{Solution:} According to the problem, we have $y=kx+b$ intersects with $y=3x$ at point $A(-1, -3)$,\\
solving the inequality $3x<kx+b<0$ for its solution set, which refers to the part of the graph between points $A$ and $B$,\\
furthermore, since $B(-2,0)$,\\
the range of the independent variable $x$ is $-2<x<-1$.\\
Therefore, the solution set of the inequality $3x<kx+b<0$ is: $-2<x<-1$.\\
Hence, the answer is: $\boxed{-2<x<-1}$.


Question: Squares $A_{1}B_{1}C_{1}O$, $A_{2}B_{2}C_{2}C_{1}$, $A_{3}B_{3}C_{3}C_{2}$, ..., are placed as shown in the diagram. Points $A_{1}$, $A_{2}$, $A_{3}$, ..., and points $C_{1}$, $C_{2}$, $C_{3}$, ..., are located on the line $y=kx+b$ $(k>0)$ and the $x$-axis, respectively. Given that point $B_{1}$ is at $(1, 1)$ and $B_{2}$ is at $(3, 2)$, what are the coordinates of point $B_{2021}$?

Solution:
\textbf{Solution:} Since the coordinates of point B$_1$ are (1,1), and the coordinates of point B$_2$ are (3,2),\\
therefore, the side length of square A$_1$B$_1$C$_1$O$_1$ is 1, and the side length of square A$_2$B$_2$C$_2$C$_1$ is 2,\\
thus, the coordinates of A$_1$ are (0,1), and the coordinates of A$_2$ are: (1,2),\\
substituting into y=kx+b yields: $\left\{\begin{array}{l}b=1 \\ k+b=2\end{array}\right.$,\\
solving gives: $\left\{\begin{array}{l}k=1 \\ b=1\end{array}\right.$,\\
hence, the equation of the line is: y=x+1.\\
Since A$_1$B$_1$=1, and the coordinates of point B$_2$ are (3,2),\\
thus, the coordinates of point A$_3$ are (3,4),\\
therefore, A$_3$C$_2$=A$_3$B$_3$=B$_3$C$_3$=4,\\
therefore, the coordinates of point B$_3$ are (7,4),\\
thus, the y-coordinate of B$_1$ is: 1=$2^0$, the x-coordinate of B$_1$ is: 1=$2^1$−1,\\
thus, the y-coordinate of B$_2$ is: 2=$2^1$, the x-coordinate of B$_2$ is: 3=$2^2$−1,\\
thus, the y-coordinate of B$_3$ is: 4=$2^2$, the x-coordinate of B$_3$ is: 7=$2^3$−1,\\
thus, the y-coordinate of B$_n$ is: $2^{n-1}$, the x-coordinate is: $2^{n}$−1,\\
therefore, the coordinates of B$_n$ are: ($2^{n}$−1,$2^{n-1}$).\\
therefore, the coordinates of B$_{2021}$ are: $\boxed{(2^{2021}-1,2^{2020})}$.


Question: \textit{As shown in the figure, the intersection point of the lines $y=mx-3m$ and $y=-\frac{1}{2}x+n$ has a coordinate of 5. What is the solution set for the system of inequalities in terms of $x$ $\begin{cases} -\frac{1}{2}x+n>mx-3m \\ mx-3m>0 \end{cases}$?}

Solution:
\textbf{Solution:} Given that the intersection point of the lines $y=mx-3m$ and $y=-\frac{1}{2}x+n$ has a coordinate of 5,\\
it follows from the graph that when $-\frac{1}{2}x+n>mx-3m$, it is solved that $x<5$;\\
Since the graph shows that $y=mx-3m$ increases as $x$ increases,\\
hence $m>0$\\
Therefore, when $mx-3m>0$, it is solved that $x>3$;\\
Thus, $3<x<5$.\\
So, the answer is: $x\in\boxed{(3,5)}$.


Question: As shown in the figure, it is known that the line $l_{1}: y= \frac{3}{2}x+6$ and the line $l_{2}: y= -\frac{5}{2}x-2$ intersect at point $P(-2,3)$. What is the solution set of the inequality $\frac{3}{2}x+6 > -\frac{5}{2}x-2$?

Solution:
\textbf{Solution:} Given that the line $l_{1}\colon y= \frac{3}{2}x+6$ intersects with the line $l_{2}\colon y= -\frac{5}{2}x-2$ at the point (-2, 3), and when $x>-2$, the line $l_{1}$ is above line $l_{2}$,\\
the solution set for the inequality $\frac{3}{2}x+6 > -\frac{5}{2}x-2$ is $x>-2$.\\
Thus, the answer is: $\boxed{x>-2}$.


Question: As shown in the figure, the line passing through the point $(3,0)$: $y=-x+b$ intersects with the line: $y=ax$ at point $P(n,2)$. What is the solution set of the system of inequalities $0<ax\leq -x+b$?

Solution:
\textbf{Solution:} Given the line passing through the point $(3,0)$ can be expressed as $y=-x+b$ and intersects with the line $y=ax$ at point $P(n,2)$,\\
$\therefore$ $-3+b=0$,\\
$\therefore$ $b=3$,\\
$\therefore$ $y=-x+3$,\\
$\therefore$ $2=-n+3$,\\
$\therefore$ $n=1$,\\
$\therefore$ $P(1,2)$,\\
From the graph, the solution set of the system of inequalities $0<ax\leq -x+b$ is $0<x\leq 1$.\\
Hence, the answer is $\boxed{0<x\leq 1}$.


Question: As shown in the figure, the line $y=kx+b$ passes through points $A(m, -2)$ and point $B(-2, 0)$. If the line $y=2x$ passes through point A, what is the solution set of the inequality $2x>kx+b$?

Solution:
\textbf{Solution}: Since the line $y=2x$ passes through point A, \\
hence $-2=2m$, solving yields: $m=-1$. \\
As known from the graph, the solution set of $2x>kx+b$ is to the right side of point A, that is, $\boxed{x>-1}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=x+2$ and the line $y=ax+b$ ($a \neq 0$) intersect at point $P$. According to the graph, what is the value of $x$ that solves the equation $x+2=ax+b$?

Solution:
\textbf{Solution:} Substituting $y=7$ into $y=x+2$ yields $7=x+2$, \\
Solving this gives $x=5$, \\
$\therefore$ The x-coordinate of point P is 5, \\
$\because$ The line $y=x+2$ intersects with the line $y=ax+b$ ($a\neq 0$) at point P, \\
$\therefore$ The solution to the equation $x+2=ax+b$ is $\boxed{x=5}$.


Question: As shown in the figure, the vertices A and C of the square $OABC$ are located on the positive semiaxes of the coordinate axes, respectively, and point B is a point in the first quadrant on the line $y=\frac{1}{2}x+3$. What are the coordinates of point B?

Solution:
\textbf{Solution:} From the problem, we know that point B lies on the line $y=\frac{1}{2}x+3$ and point B is a vertex of the square ABCD,\\
Let $B\left(x, \frac{1}{2}x+3\right)$,\\
$\therefore x=\frac{1}{2}x+3$,\\
Solving yields: $x=6$,\\
$\therefore \frac{1}{2}x+3=6$,\\
$\therefore B\left(6, 6\right)$;\\
Thus, the answer is $B\boxed{\left(6, 6\right)}$.


Question: As shown in the figure, the line $y=kx+b$ ($k<0$) passes through the point $P(2,0)$. What is the range of $x$ when $kx+b>0$?

Solution:
\textbf{Solution:} The line $y=kx+b$ ($k<0$) passes through the point $P(2,0)$.\\
When $kx+b>0$, the range of $x$ is $x<2$,\\
Therefore, the answer is: $\boxed{x<2}$.


Question: The graph of the function $y=kx$ and $y=6-x$ is shown in the figure below. What is the value of $k$?

Solution:
\textbf{Solution:} Since the abscissa of the point of intersection between the linear function $y=6-x$ and $y=kx$ is 2,\\
$\therefore y=6-2=4$,\\
$\therefore$ the coordinates of the intersection point are $(2,4)$,\\
Substituting $(2,4)$ into $y=kx$, we get $2k=4$, which gives: $k=\boxed{2}$.


Question: As shown in the figure, the line $l_1: y=kx+4$ intersects with the line $l_2: y=ax+b$ at point $P(4, m)$. What is the solution set of the inequality $kx+4 \ge ax+b$ with respect to $x$?

Solution:
\textbf{Solution:} To solve: for the line $l_{1}: y=kx+4$ and the line $l_{2}: y=ax+b$ intersect at point $P(4, m)$,\\
to satisfy the inequality $kx+4\ge ax+b$,\\
the graph of $l_{1}: y=kx+4$ must be above that of $l_{2}: y=ax+b$,\\
from the graph, it is known to intersect at point $P(4, m)$,\\
with the x-coordinate being 4,\\
hence, the solution set for the inequality $kx+4\ge ax+b$ is: $\boxed{x\ge 4}$.


Question: As shown in the figure, in the Cartesian coordinate system, points $A_1$, $A_2$, $A_3$ lie on the line $y=\frac{1}{6}x+b$, and points $B_1$, $B_2$, $B_3$ lie on the x-axis; $\triangle O A_1 B_1$, $\triangle B_1 A_2 B_2$, $\triangle B_2 A_3 B_3$ are all isosceles right triangles. Given that point $A_1(1,1)$, what is the y-coordinate of point $A_3$?

Solution:
\textbf{Solution:} Let's set the ordinate of points $A_{2}$ and $A_{3}$ to be $m$ and $n$ respectively, \\
since $A_{1}(1,1)$ lies on the line $y= \frac{1}{6}x+b$, \\
we have $b= \frac{5}{6}$, \\
thus $y= \frac{1}{6}x+ \frac{5}{6}$. \\
Let $A_{2}(m+2, m)$, substituting into $y=\frac{1}{6}x+\frac{5}{6}$, we get: $m=\frac{1}{6}(m+2)+\frac{5}{6}$, solving for $m$ gives: $m=\frac{7}{5}$; \\
Let $A_{3}\left(2+2\times \frac{7}{5}+n, n\right)$, substituting into $y=\frac{1}{6}x+\frac{5}{6}$, we get: $n=\frac{1}{6}\left(\frac{24}{5}+n\right)+\frac{5}{6}$, solving for $n$ gives: $n=\boxed{\frac{49}{25}}$.


Question: As shown in the figure, the rectangle $ABCD$ has its side $BC$ on the $x$-axis, point $E$ is the midpoint of diagonal $BD$, and the graph of the function $y=\frac{2}{x}$ passes through points $A$ and $E$. If $\angle ABD=45^\circ$, what is the function expression corresponding to line $BD$?

Solution:
\textbf{Solution:} Solve: When $ \angle ABD=45^\circ$, then $AB=AD$, \\
$\therefore$ the rectangle $ABCD$ is a square, \\
Let the coordinates of point $A$ be $(a, \frac{2}{a})$, then $B(a, 0)$, $D\left(a+\frac{2}{a}, \frac{2}{a}\right)$, \\
$\because$ $E$ is the midpoint of the diagonal $BD$, \\
$\therefore$ point $E\left(a+\frac{1}{a}, \frac{1}{a}\right)$, \\
$\because$ the graph of the function $y=\frac{2}{x}$ passes through points $A$ and $E$, \\
$\therefore \left(a+\frac{1}{a}\right)\cdot \frac{1}{a}=2$,  \\
Solving, we get $a=1$ (negatives are discarded), \\
$\therefore B(1, 0)$, $D(3, 2)$, \\
Let the equation of line $BD$ be $y=kx+b$, \\
$\therefore$ $\begin{cases}k+b=0\\ 3k+b=2\end{cases}$, solving we get $\begin{cases}k=1\\ b=-1\end{cases}$, \\
$\therefore$ the expression for line $BD$ is $y=x-1$, \\
Hence, the answer is $\boxed{y=x-1}$.


Question: As shown in the figure, the graph of the directly proportional function $y=-x$ intersects with the graph of the linear function $y=kx+ \frac{5}{2}$ (where $k\neq 0$) at point $P$. What is the solution to the equation $-x=kx+ \frac{5}{2}$ in terms of $x$?

Solution:
\textbf{Solution:} Let the coordinates of point P be $(x, 1)$,\\
Substituting $(x, 1)$ into $y=-x$,\\
$\therefore$ $-x=1$,\\
$\therefore$ $x=-1$,\\
Thus, the coordinates of point P are $(-1, 1)$,\\
$\therefore$ the solution to the equation concerning $x$, $-x=kx+ \frac{5}{2}$, is $x=\boxed{-1}$.


Question: Given the graph of the linear function $y=kx+b$ ($k\ne 0$) as shown, what is the solution set of the inequality $kx+b>0$ with respect to $x$?

Solution:
\textbf{Solution:} The graph of the function $y=kx+b$ passes through the point $(4,0)$, and the function value $y$ decreases as $x$ increases,\\
therefore, when $x<4$, the function value is greater than $0$, meaning the solution set for the inequality $kx+b>0$ with respect to $x$ is $\boxed{x<4}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects the x-axis and y-axis at points $A(3, 0)$ and $B(0, 5)$, respectively. What is the solution set of the inequality $kx+b\le 0$?

Solution:
\textbf{Solution:} From the graph, we can observe that when $y \leq 0$, then $x \geq 3$,\\
which means the solution set of the inequality $kx+b \leq 0$ is $\boxed{x \geq 3}$.


Question: In the Cartesian coordinate system $xOy$, the graphs of linear functions $y=kx$ and $y=-x+3$ are shown as in the figure. Then, what is the solution to the system of linear equations $\begin{cases}y=kx\\ y=-x+3\end{cases}$?

Solution:
\textbf{Solution}: Since the graph of the linear function $y=kx$ intersects with the graph of $y=-x+3$ at point $(1, 2)$,
therefore the solution to the system of linear equations $\begin{cases}y=kx\\ y=-x+3\end{cases}$ is $\boxed{\begin{cases}x=1\\ y=2\end{cases}}$.


Question: As shown in the figure, the intersection's x-coordinate of the lines $y = -x + m$ and $y = nx + 4n$ ($n \neq 0$) is $-2$. What is the solution set of the inequality $-x + m > nx + 4n$ with respect to $x$?

Solution:
\textbf{Solution:} From the graph, it can be seen that when $x < -2$, the graph of the line $y = -x + m$ is above the graph of the line $y = nx + 4n$ (where $n \neq 0$).\\
Therefore, the solution set for the inequality $-x + m > nx + 4n$ with respect to $x$ is $x < -2$.\\
Thus, the answer is: $\boxed{x < -2}$.


Question: As shown in the figure, given that the graph of function $y=k_{1}x+b_{1}$ intersects with the graph of function $y=k_{2}x+b_{2}$ at point $A(-2,1)$, what is the solution set of the inequality $k_{1}x+b_{1}>k_{2}x+b_{2}$ with respect to $x$?

Solution:
\textbf{Solution:} Since from the graph of the function, it is known that when $x<-2$, the line $y=k_{1}x+b_{1}$ is above the line $y=k_{2}x+b_{2}$,\\
therefore, the solution set for the inequality $k_{1}x+b_{1}>k_{2}x+b_{2}$ with respect to $x$ is $x<-2$.\\
Thus, the answer is: $\boxed{x<-2}$.


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. When $\triangle AOB$ is folded along a line through point $A$, such that point $B$ falls on point $C$ on the positive half of the $x$-axis, and the fold intersects the $y$-axis at point $D$, what is the equation of the line where the fold lies?

Solution:
\textbf{Solution:} Let $y=0$, then $\frac{4}{3}x+4=0$, solving this we get $x=-3$,\\
$\therefore$ $A(-3, 0)$, $OA=3$,\\
Let $x=0$, then $y=\frac{4}{3}\times 0+4=4$,\\
$\therefore$ $B(0, 4)$, $OB=4$,\\
In $\triangle AOB$, $AB=\sqrt{OA^2+OB^2}=\sqrt{3^2+4^2}=5$,\\
By folding, we get: $AC=AB=5$ and $CD=DB$,\\
$\therefore$ $OC=AC-OA=5-3=2$, $CD=BD=OB-OD=4-OD$,\\
In $\triangle COD$, $OD^2+OC^2=CD^2$, that is $OD^2+2^2=(4-OD)^2$,\\
Solving this gives $OD=\frac{3}{2}$,\\
$\therefore$ $D\left(0,\frac{3}{2}\right)$,\\
Let the equation of the line containing the crease $AD$ be $y=kx+b$,\\
$\because$ $A(-3, 0)$, $D\left(0,\frac{3}{2}\right)$,\\
$\therefore$ $\begin{cases}-3k+b=0\\ b=\frac{3}{2}\end{cases}$,\\
Solving gives: $\begin{cases}k=\frac{1}{2}\\ b=\frac{3}{2}\end{cases}$,\\
$\therefore$ $y=\frac{1}{2}x+\frac{3}{2}$\\
Hence, the answer is $\boxed{y=\frac{1}{2}x+\frac{3}{2}}$


Question: As shown in the figure, the line $y_1=x+b$ intersects with $y_2=kx-1$ at point P. What is the solution set of the inequality $x+b>kx-1$ with respect to $x$?

Solution:
\textbf{Solution:} From the graph, we know that the coordinates of the intersection point of the two lines are $\left(-1, \frac{1}{2}\right)$. Additionally, when $x > -1$, the graph of the function $y = x + b$ is above the graph of the function $y = kx - 1$,\\
$\therefore$ the solution set for the inequality in terms of $x$, $x + b > kx - 1$, is $x > -1$.\\
Therefore, the answer is: $\boxed{x > -1}$.


Question: As shown in the graph of the linear function $y=kx+b$, what is the solution set of the linear inequality $kx+b>0$?

Solution:
\textbf{Solution:} From the graph, it is known that $y$ decreases as $x$ increases, and when $x<2$, the line is above the x-axis,\\
therefore, the solution set for $kx+b>0$ is: $\boxed{x<2}$.


Question: As shown in the figure, the graph of the linear function $y_1=x+b$ intersects with the graph of the linear function $y_2=kx-1$ at point P. Then, what is the solution set of the inequality concerning $x$, $x+b-kx+1>0$?

Solution:
\textbf{Solution:} Given the graph of the function, when the graph of $y_1=x+b$ is above the graph of $y_2=kx-1$, we have $x+b>kx-1$, which leads to $x>-1$; \\
Hence, the answer is: $\boxed{x>-1}$.


Question: As shown in the figure, the line $y=kx+b$ ($k<0$) passes through the point $(2, 1)$. When $kx+b<\frac{1}{2}x$, what is the range of values for $x$?

Solution:
\textbf{Solution:} Let's solve it: when $x=2$, $y=1$, \\
$\therefore$ the line $y=\frac{1}{2}x$ goes through the point $(2,1)$,\\
As can be seen from the graph, \\
when $kx+b<\frac{1}{2}x$, $x>2$\\
Therefore, the answer is: $\boxed{x>2}$.


Question: As shown in the figure, in the Cartesian coordinate system, $O$ is the origin. The coordinate of point $A$ is $(0, 8)$, and the coordinate of point $B$ is $(-4, 0)$. Point $P$ is a moving point on the line $l: x+y=4$. If $\angle PAB = \angle ABO$, what are the coordinates of point $P$?

Solution:
\textbf{Solution:} As illustrated:\\
(1) When point P is on the left side of the y-axis, connect AP, $\because$ $\angle PAB = \angle ABO$, $\therefore$ AP$\parallel$OB, $\therefore$ the y-coordinate of point P is 8. When y=8, x+8=4, solving this gives x=−4, $\therefore$ P(−4, 8);\\
(2) When point P is on the right side of the y-axis, connect AP intersecting the x-axis at point C, $\because$ $\angle PAB = \angle ABO$, $\therefore$ AC=BC, let OC=a, then BC=AC=4+a. In $\triangle AOC$, by Pythagoras' theorem, we get $(4+a)^{2}=a^{2}+8^{2}$, solving for a gives a=6, $\therefore$ C(6,0). Let the linear equation of line AC be y=kx+b. Substituting A(0,8) and C(6,0) into y=kx+b gives $\left\{\begin{array}{l}b=8\\ 6k+b=0\end{array}\right.$, solving this system gives $\left\{\begin{array}{l}k=-\frac{4}{3}\\ b=8\end{array}\right.$, $\therefore$ $y=-\frac{4}{3}x+8$. $\because$ Point P is the intersection of line l and line AC, $\therefore$ $\left\{\begin{array}{l}y=-\frac{4}{3}x+8\\ x+y=4\end{array}\right.$, solving this system gives $\left\{\begin{array}{l}x=12\\ y=-8\end{array}\right.$, $\therefore$ P(12,−8), hence the answer is: $\boxed{(12,−8) \text { or } (−4, 8)}$.


Question: Given the graph, the functions $y=bx$ and $y=ax+4$ intersect at point $A(1,3)$. What is the solution set of the inequality $bx<ax+4$?

Solution:
\textbf{Solution:} \\
From the given condition, the solution set of the inequality $bx<ax+4$ is $x<1$,\\
Therefore, the answer is: $\boxed{x<1}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graphs of the functions $y_1=kx$ and $y_2=ax+3$ intersect at point $A(-1, 2)$. What is the solution set of the inequality $kx>ax+3$ with respect to $x$?

Solution:
\textbf{Solution:} By observing the graph, we can see that when $x<-1$, the line $y_1=kx$ is above the line $y_2=ax+3$. \\
Therefore, the solution set for the inequality $kx>ax+3$ with respect to $x$ is $x<-1$.\\
Hence, the answer is: $\boxed{x<-1}$.


Question: In the Cartesian coordinate system, the vertices of rectangle OABC are as follows: vertex O is the origin of coordinates, vertices A and C are on the x-axis and y-axis respectively, with $OA=4$ and $OC=3$. D is the midpoint of side AB, and E is a moving point on side OA. When the perimeter of $\triangle CDE$ is minimized, what are the coordinates of point E?

Solution:
\textbf{Solution:} Construct point F symmetrical to point D with respect to the x-axis, as shown in the figure,\\
$\because$ Quadrilateral OABC is a rectangle,\\
$\therefore$OC＝BD＝3, the coordinates of point C are $\left(0,3\right)$,\\
$\because$D is the midpoint of side AB,\\
$\therefore$AD＝$\frac{3}{2}$,\\
$\because$OA＝4,\\
$\therefore$The coordinates of point D are $\left(4,\frac{3}{2}\right)$, then the coordinates of point F are $\left(4,-\frac{3}{2}\right)$,\\
According to the property of axial symmetry: EF＝ED,\\
$\therefore$P$_{\triangle CDE}$＝CD+CE+DE＝CD+CE+EF, where CD is a constant,\\
When the sum of CE+EF is minimized, the perimeter of $\triangle CDE$ is minimal, at this time points C, E, F are collinear,\\
Let the equation of line CF be: $y=kx+b\left(k\ne 0\right)$,\\
Substituting $\left(0,3\right)$ and $\left(4,-\frac{3}{2}\right)$ into the equation yields:\\
$\begin{cases}b=3\\ 4k+b=-\frac{3}{2}\end{cases}$, solving gives: $\begin{cases}k=-\frac{9}{8}\\ b=3\end{cases}$,\\
$\therefore$The equation of line CF is: $y=-\frac{9}{8}x+3$,\\
Setting $y=0$, we get: $-\frac{9}{8}x+3=0$,\\
Solving for $x$ gives: $x=\frac{8}{3}$,\\
$\therefore$The coordinates of point E are $\boxed{\left(\frac{8}{3},0\right)}$.


Question: As shown in the graph, the intersection point of the line $y = -x + 2$ and $y = kx + b$ ($k \neq 0$ and $k$, $b$ are constants) is at the coordinates $(3, -1)$. What is the solution set of the inequality $kx + b \geq -x + 2$ with respect to $x$?

Solution:
\textbf{Solution:} From the graph, we obtain that when $x \ge 3$, the graph corresponding to the points of $y = -x + 2$ lies below the graph of the function $y = kx + b$ (where $k \ne 0$ and $k$, $b$ are constants).\\
Therefore, the solution set of the inequality $kx + b \ge -x + 2$ is $\boxed{x \ge 3}$.


Question: Given the lines $y_{1}=x$, $y_{2}=\frac{1}{3}x+1$, $y_{3}=-\frac{4}{5}x+5$ as shown in the figure, if for any value of $x$, $y$ always takes the minimum value among $y_{1}$, $y_{2}$, and $y_{3}$, what is the maximum value of $y$?

Solution:
\textbf{Solution:}

As shown in the figure, we find the coordinates of the intersection points of $y_{1}$, $y_{2}$, and $y_{3}$ as $A\left(\frac{3}{2},\frac{3}{2}\right)$; $B\left(\frac{25}{9},\frac{25}{9}\right)$; $C\left(\frac{60}{17},\frac{37}{17}\right)$

When $x<\frac{3}{2}$, $y=y_{1}$

When $\frac{3}{2}\le x<\frac{25}{9}$, $y=y_{2}$

When $\frac{25}{9}\le x<\frac{60}{17}$, $y=y_{2}$

When $x\ge \frac{60}{17}$, $y=y_{3}$

Since $y$ always takes the minimum value among $y_{1}$, $y_{2}$, and $y_{3}$,

Therefore, the maximum value of $y$ is $\boxed{\frac{37}{17}}$


Question: As shown in the figure, it is known that point $E$ is a point on side $AD$ of rectangle $ABCD$. Quadrilateral $BCDE$ is folded along the line $BE$. After folding, the corresponding point of point $C$ is $C'$, and the corresponding point of point $D$ is $D'$. If point $A$ is on $C'D'$, and $AB=10$, $BC=8$, what is the length of $AE$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
$\therefore$ $\angle D=\angle C=\angle DAB=90^\circ$, AB=DC=10, AD=BC=8,\\
According to the properties of folding, we get: $\angle D'=\angle D=90^\circ$, $\angle C'=\angle C=90^\circ$, BC'=BC=8, D'C'=DC=10,\\
$\therefore$ $AC'=\sqrt{AB^{2}-BC'^{2}}=6$,\\
$\therefore$ AD'=D'C'-AC'=10-6=4,\\
Let DE=D'E=x, then AE=8-x,\\
$\therefore$ $(8-x)^{2}=4^{2}+x^{2}$;\\
$\therefore$ x=3,\\
$\therefore$ AE=$\boxed{5}$.


Question:
\[
In an isosceles \triangle ABC with a right angle at C, a point D is on the side AC. Connect BD and fold \triangle ABC along BD so that point A falls on side BC. Connect AE. Then, what is the value of \frac{{S}_{\triangle BDE}}{{S}_{\triangle ACE}}?
\]

Solution:
Solution:Let's fold $\triangle ABC$ along BD so that point A falls on side BC,\\
hence, $BE=AB=AC$, $\angle BED=\angle BAC=90^\circ$, and $AD=DE$,\\
thus, $\angle CED=90^\circ$,\\
since $\angle C=45^\circ$,\\
it follows that $\triangle DEC$ is an isosceles right triangle,\\
thus, $DE=CE$,\\
let $AD=DE=CE=x$,\\
hence, $CD=\sqrt{2}x$,\\
thus, $BE=AC=x+\sqrt{2}x$,\\
draw EH $\perp$ CD at H from point E,\\
hence, $EH=CH=\frac{\sqrt{2}}{2}CE=\frac{\sqrt{2}}{2}x$,\\
since $S_{\triangle BDE}=\frac{1}{2}BE\cdot DE=\frac{1}{2}x(x+\sqrt{2}x)=\frac{1+\sqrt{2}}{2}x^2$, and $S_{\triangle ACE}=\frac{1}{2}AC\cdot EH=\frac{1}{2}\times \frac{\sqrt{2}}{2}x(x+\sqrt{2}x)=\frac{\sqrt{2}+2}{4}x^2$,\\
therefore, $\frac{S_{\triangle BDE}}{S_{\triangle ACE}}=\frac{\frac{1+\sqrt{2}}{2}x^2}{\frac{\sqrt{2}+2}{4}x^2}=\boxed{\sqrt{2}}$,


Question: As shown in the figure, in the isosceles $\triangle ABC$ with $AB=AC$, $\angle BAC=56^\circ$. The bisector of $\angle BAC$ intersects the perpendicular bisector of $AB$ at point $O$. If point $C$ is folded along $EF$ to coincide with point $O$, what is the measure of $\angle CEF$?

Solution:
\textbf{Solution:} As shown in the figure, connect BO,\\
$\because$ in the isosceles $\triangle ABC$, $AB=AC$, $\angle BAC=56^\circ$,\\
$\therefore \angle ABC=\angle ACB=(180^\circ-56^\circ)÷2=62^\circ$,\\
$\because AO$ is the bisector of $\angle BAC$,\\
$\therefore \angle BAO=28^\circ$,\\
Also $\because OD$ is the perpendicular bisector of AB,\\
$\therefore$ OA=OB,\\
$\therefore \angle OBA=\angle OAB=28^\circ$,\\
$\therefore \angle OBC=\angle OCB=62^\circ-28^\circ=34^\circ$,\\
$\because EF$ is the perpendicular bisector of the segment OC,\\
$\therefore \angle CEF=90^\circ-34^\circ=\boxed{56^\circ}$.


Question: As shown in the figure, point $A$ is inside $\angle MON=45^\circ$ with $OA=6\text{cm}$. Points $B$ and $C$ are moving points on sides $OM$ and $ON$, respectively. What is the minimum perimeter of $\triangle ABC$ in $\text{cm}$?

Solution:
\textbf{Solution:} Let $A'$ be the symmetric point of A with respect to $OM$, and let $A''$ be the symmetric point of A with respect to $ON$. Draw line $A'A''$, intersecting $OM$ at point B and $ON$ at point C. Draw lines $OA'$ and $OA''$. At this moment, the perimeter of $\triangle ABC$ is minimal. \\
$\because$ $AB = A'B$, $AC = A''C$, $OA = OA' = OA'' = 6$, $\angle BOA' = \angle BOA$, $\angle COA'' = \angle COA$, \\
$\because$ $\angle MON = 45^\circ$, \\
$\therefore$ $\angle A''OA' = 90^\circ$, \\
$\therefore$ $A''A' = 6\sqrt{2}$, \\
$\therefore$ The perimeter of $\triangle ABC = AB + BC + AC = A'B + BC + A''C = A''A' = 6\sqrt{2}$, \\
$\therefore$ The minimum perimeter of $\triangle ABC$ is $\boxed{6\sqrt{2}}$ cm.


Question: As shown in the figure, in $\triangle ABC$, $AD$ is the angle bisector of $\triangle ABC$, $DE \perp AB$ with the foot of the perpendicular at $E$, $DF \perp AC$ with the foot of the perpendicular at $F$, if $AB=5$, $AC=3$, and $DF=2$, what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Since AD is the angle bisector of $\triangle ABC$, and DE$\perp$AB, DF$\perp$AC,\\
thus DE=DF=2,\\
therefore, the area of $\triangle ABC$ = $\frac{1}{2} \times 5 \times 2 + \frac{1}{2} \times 3 \times 2 = \boxed{8}$


Question: As shown in the figure, in $\triangle ABC$, $\angle A=80^\circ$. After cutting off $\angle A$, we obtain the quadrilateral $BCDE$. Then, what is the sum of $\angle 1 + \angle 2$?

Solution:
\textbf{Solution:} Given: $\because$ $\angle 1$ and $\angle 2$ are exterior angles of $\triangle ADE$,\\
$\therefore$ $\angle 1 = \angle ADE + \angle A$, $\angle 2 = \angle AED + \angle A$,\\
$\therefore$ $\angle 1 + \angle 2 = \angle ADE + \angle A + \angle AED + \angle A$,\\
Moreover, $\because$ $\angle ADE + \angle A + \angle AED = 180^\circ$,\\
$\therefore$ $\angle 1 + \angle 2 = 180^\circ + \angle A = \boxed{260^\circ}$.


Question: As shown in the figure, $AB \parallel EF$, $\angle ABC = 75^\circ$, $\angle CDF = 135^\circ$. What is the measure of $\angle BCD$ in degrees?

Solution:
\textbf{Solution:} Let point $G$ be the intersection of $BC$ and $EF$, as shown in the diagram below: \\
$\because$ $AB\parallel EF$, $\angle ABC=75^\circ$, $\angle CDF=135^\circ$,\\
$\therefore$ $\angle EGC=\angle ABC=75^\circ$, $\angle EDC=180^\circ-\angle CDF=180^\circ-135^\circ=45^\circ$,\\
Furthermore, $\because$ $\angle EGC=\angle BCD+\angle EDC$,\\
$\therefore$ $\angle BCD=75^\circ-45^\circ=\boxed{30^\circ}$.


Question: As shown in the figure, an isosceles right-angled triangle object $ABC$ is leaning against a corner of the wall ($\angle O = 90^\circ$). If $OA = 50cm$ and $OB = 28cm$, what is the distance in $cm$ of point $C$ from the ground?

Solution:
\textbf{Solution:} Draw $CD \perp OB$ at point D, as shown in the diagram:\\
$\therefore$ $\angle CDB=\angle AOB=90^\circ$\\
$\because$ $\triangle ABC$ is an isosceles right triangle\\
$\therefore$ AB=CB, $\angle ABC=90^\circ$\\
$\therefore$ $\angle ABO+\angle CBD=90^\circ$\\
$\therefore$ $\angle ABO=\angle BCD$,\\
$\therefore$ $\triangle ABO\cong \triangle BCD(AAS)$\\
$\therefore$ CD=BO=\boxed{28\text{cm}}


Question: Rectangle $OABC$ is in the Cartesian coordinate system, with $OA = \sqrt{3}$ and $AB = 1$. Folding it along $OB$, point $A$ falls at point ${A}^{\prime}$. What are the coordinates of ${A}^{\prime}$?

Solution:
\textbf{Solution:} As shown in the figure, construct ${A}^{\prime}D \perp OA$ at D,\\
since in rectangle $OABC$, $OA=\sqrt{3}$, $AB=1$,\\
then $OB=\sqrt{OA^{2}+AB^{2}}=2$, that is $AB=\frac{1}{2}OB$,\\
therefore $\angle AOB=30^\circ$,\\
from the folding, $\angle {A}^{\prime}OB=\angle AOB=30^\circ$, $O{A}^{\prime}=OA=\sqrt{3}$,\\
therefore $\angle {A}^{\prime}OD=60^\circ$,\\
therefore $\angle O{A}^{\prime}D=30^\circ$,\\
therefore $OD=\frac{1}{2}O{A}^{\prime}=\frac{\sqrt{3}}{2}$,\\
therefore ${A}^{\prime}D=\sqrt{O{{A}^{\prime}}^{2}−OD^{2}}=\sqrt{{\left(\sqrt{3}\right)}^{2}−{\left(\frac{\sqrt{3}}{2}\right)}^{2}}=\frac{3}{2}$,\\
therefore, the coordinates of ${A}^{\prime}$ are $\boxed{\left(-\frac{\sqrt{3}}{2},\frac{3}{2}\right)}$.


Question: In the figure, within $\triangle ABC$, we have $AB=AC$, and $D$ is a point on $BC$ such that $CD=AD$ and $AB=BD$. What is the measure of $\angle B$?

Solution:
\textbf{Solution:} We have AB=AC,\\
thus $\angle B=\angle C$,\\
and since CD=DA,\\
it follows that $\angle C=\angle DAC$,\\
also given BA=BD,\\
it implies $\angle BDA=\angle BAD=2\angle C=2\angle B$,\\
furthermore, since $\angle B+\angle BAD+\angle BDA=180^\circ$,\\
we conclude $5\angle B=180^\circ$,\\
hence $\angle B=\boxed{36^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, the perpendicular bisector of BC intersects with the angle bisector of $\angle BAC$ at point D, with the foot of the perpendicular being point P. If $\angle BAC=84^\circ$, then what is the degree measure of $\angle BDC$?

Solution:
\textbf{Solution:} Construct DE $\perp$ AB at point E, extending through AB, and DF $\perp$ AC at F,\\
since AD is the bisector of $\angle$BAC,\\
therefore DE = DF,\\
since DP is the perpendicular bisector of BC,\\
therefore BD = CD,\\
In right triangles $\triangle$DEB and $\triangle$DFC,\\
$\left\{\begin{array}{l}DB=DC\\ DE=DF\end{array}\right.$,\\
therefore right $\triangle$DEB $\cong$ right $\triangle$DFC (HL).\\
therefore $\angle$BDE = $\angle$CDF,\\
therefore $\angle$BDC = $\angle$EDF,\\
since $\angle$DEB = $\angle$DFC = $90^\circ$,\\
therefore $\angle$EAF + $\angle$EDF = $180^\circ$,\\
since $\angle$BAC = $84^\circ$,\\
therefore $\angle$BDC = $\angle$EDF = \boxed{96^\circ},


Question: As shown in the figure, triangle $ABC$ is rotated $22^\circ$ clockwise around point B to obtain triangle $DBE$. If $\angle C = 28^\circ$ and side $DE$ intersects side $BC$ at point F, then what is the measure of $\angle CFE$ in degrees?

Solution:
\textbf{Solution:} Rotate $\triangle ABC$ clockwise around point B by $22^\circ$ to get $\triangle DBE$,\\
$\therefore$ $\angle CBE=22^\circ$.\\
$\because$ $\angle C=28^\circ$,\\
$\therefore$ $\angle E=\angle C=28^\circ$.\\
$\because$ $\angle CFE$ is the external angle of $\triangle BEF$,\\
$\therefore$ $\angle CFE=\angle CBE+\angle E=22^\circ+28^\circ=\boxed{50^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, point D is on side AB. $\triangle CBD$ is folded along CD such that point B falls exactly on point E on side AC. If $\angle A=26^\circ$, what is the degree measure of $\angle ADE$?

Solution:
\textbf{Solution:} From the folding, it can be seen that $\angle ACD=\angle BCD$, $\angle BDC=\angle CDE$,\\
$\because$ $\angle ACB=90^\circ$,\\
$\therefore$ $\angle ACD=45^\circ$,\\
$\because$ $\angle A=26^\circ$,\\
$\therefore$ $\angle BDC=\angle ACD+\angle A=71^\circ$,\\
$\therefore$ $\angle CDE=71^\circ$,\\
$\therefore$ $\angle ADE=180^\circ-71^\circ-71^\circ=\boxed{38^\circ}$.


Question: As shown in the figure, point $B$ is on ray $AN$, an equilateral triangle $\triangle ABC$ is constructed with $AB$ as a side, $M$ is the midpoint of $AN$, and $AN=4$. $P$ is the midpoint of $BC$. When $PM+PN$ is minimized, what is the length of $AB$?

Solution:
\textbf{Solution:} Draw $AP$ and construct the point $D$ as the symmetric point of $M$ with respect to $AP$, and connect $DP$,\\
$\therefore PD=PM$, $AD=AM$,\\
$\therefore PM+PN=PD+PN\ge DN$,\\
thus, when $D$, $P$, $N$ are collinear and $DN\perp AC$, $PM+PN$ is minimized;\\
$\because M$ is the midpoint of $AN$, and $AN=4$,\\
$\therefore AD=AM=\frac{1}{2}AN=2$;\\
$\because P$ is the midpoint of $BC$, and $\triangle ABC$ is an equilateral triangle,\\
$\therefore AP\perp BC$, $\angle CAP=30^\circ$, $\angle C=60^\circ$,\\
in $Rt\triangle ADP$, $AP=2PD$, by the Pythagorean theorem $DP^{2}+2^{2}=(2DP)^{2}$,\\
$\therefore DP=\frac{2\sqrt{3}}{3}$;\\
$\because \angle C=60^\circ$, $DN\perp AC$,\\
$\therefore \angle DPC=30^\circ$,\\
$\therefore PC=2DC$\\
in $Rt\triangle PDC$, by the Pythagorean theorem $DC^{2}+DP^{2}=(2DC)^{2}$,\\
that is $DC^{2}+\left(\frac{2\sqrt{3}}{3}\right)^{2}=(2DC)^{2}$, solving gives: $DC=\frac{2}{3}$,\\
$\therefore AC=AD+DC=2+\frac{2}{3}=\frac{8}{3}$,\\
$\therefore AB=AC=\boxed{\frac{8}{3}}$.


Question: In the paper $\triangle ABC$, $\angle A=65^\circ$, $\angle B=75^\circ$, fold one corner of the paper so that point C falls inside $\triangle ABC$ (as shown in the figure). If $\angle 1=20^\circ$, what is the degree measure of $\angle 2$?

Solution:
\textbf{Solution:} Given in $\triangle ABC$, $\angle A=65^\circ$, $\angle B=75^\circ$,\\
$\therefore$ $\angle C=180^\circ-\angle A-\angle B=180^\circ-65^\circ-75^\circ=40^\circ$,\\
since $\angle 1=20^\circ$,\\
$\therefore$ $\angle CED=\frac{180^\circ-\angle 1}{2}=80^\circ$,\\
in $\triangle CDE$, $\angle CDE=180^\circ-\angle C-\angle CED=180^\circ-40^\circ-80^\circ=60^\circ$,\\
$\therefore$ $\angle 2=180^\circ-2\angle CDE=180^\circ-2\times60^\circ=\boxed{60^\circ}$.


Question: As shown, $\triangle ABE\cong \triangle ADC\cong \triangle ABC$, if $\angle 1=150^\circ$, what is the measure of $\angle \alpha$?

Solution:
\textbf{Solution:} Given that $\triangle ABE \cong \triangle ADC \cong \triangle ABC$,\\
it follows that $\angle E = \angle DCA$, and $\angle BAE = \angle 1 = 150^\circ$,\\
since $\angle EFD = \angle CFA$, and $\angle \alpha + \angle E + \angle EFD = \angle EAC + \angle DCA + \angle CFA$\\
it implies that $\angle EAC = \angle \alpha$;\\
Given that $\angle EAC = 360^\circ - \angle BAE - \angle 1 = 360^\circ - 150^\circ - 150^\circ = 60^\circ$,\\
thus, $\angle \alpha = \boxed{60^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, squares $AGFB$, $BCDE$, and $ACMN$ are constructed upwards on sides $AB$, $BC$, and $AC$ respectively, with point $E$ on $FG$. If $AC=2$ and $BC=\sqrt{13}$, what is the area of the shaded region in the diagram?

Solution:
\textbf{Solution:} As shown in the diagram, we connect $DM$, and draw $DQ\perp CG$ through point $D$,
\[\therefore \angle DQC=90^\circ,\]
\[\because\] Quadrilateral $BCDE$ is a square,
\[\therefore BC=CD,\; \angle ACB+\angle QCD=90^\circ,\]
Also \[\because \angle BAC=90^\circ,\]
\[\therefore \angle ACB+\angle ABC=90^\circ\]
\[\therefore \angle ABC=\angle QCD\]
In $\triangle ABC$ and $\triangle QCD$: \[\left\{\begin{array}{l}\angle BAC=\angle DQC \\ \angle ABC=\angle QCD \\ BC=CD \end{array}\right.\]
\[\therefore \triangle ABC \cong \triangle QCD \text{ (AAS)}\]
\[\therefore DQ=AC\]
Also, \[\because ACMN\] is a square,
\[\therefore CM=AC,\; \angle QCM=90^\circ,\; MN\parallel AC\]
\[\therefore QD\parallel CM,\; DQ=CM\]
\[\therefore CMDQ\] is a parallelogram,
\[\therefore MD\parallel AC\]
\[\therefore\] Points $D$, $N$, and $M$ lie on a straight line,
Therefore, $\triangle DCM$ is also a right triangle and $\triangle ABC \cong \triangle DCM$, given that the quadrilateral $BCDE$ is a square, $AGFB$ is a Square, $ACMN$ is a square, $\triangle DCM$, $\triangle ABC$ are congruent triangles, analogously to Zhao Shuang's chord diagram it is known $\triangle ABC \cong \triangle DCM \cong \triangle FBE$, thus proving $\triangle DNK \cong \triangle EGP$ (the proof is omitted here)\\
Then: $S_{\text{shaded}}=S_{\triangle FBE}+S_{\triangle EGP}+S_{CMNK}=S_{\triangle FBE}+S_{\triangle DNK}+S_{CMNK}=S_{\triangle FBE}+S_{\triangle DCM}=2S_{\triangle ABC}$\\
\[\because AC=2,\; BC=\sqrt{13}\]
\[\therefore AB=\sqrt{BC^2-AC^2}=\sqrt{(\sqrt{13})^2-2^2}=3\]
\[\therefore S_{\text{shaded}}=2S_{\triangle ABC}=2\times \frac{1}{2}AB\cdot AC=2\times \frac{1}{2}\times 2\times 3=\boxed{6}\]


Question: As shown in the figure, in $ \triangle ABC$, $AD$ is the altitude from $A$ to side $BC$, and $BE$ is the altitude from $B$ to side $AC$, with $AD$ and $BE$ intersecting at point $F$. If $BF=AC$, $CD=3$, and $BD=8$, what is the length of line segment $AF$?

Solution:
\textbf{Solution:} Given that $AD$ is the altitude from $A$ on side $BC$ and $BE$ is the altitude from $B$ on side $AC$,\\
therefore $\angle ADC=\angle BDF=\angle AEB=90^\circ$,\\
thus, $\angle DAC+\angle C=90^\circ$, $\angle C+\angle DBF=90^\circ$,\\
which means $\angle DAC=\angle DBF$,\\
in $\triangle ADC$ and $\triangle BDF$,\\
given that $\left\{\begin{array}{l}\angle ADC=\angle BDF \\ \angle DAC=\angle DBF \\ AC=BF \end{array}\right.$,\\
we can conclude that $\triangle ADC\cong \triangle BDF$ (AAS),\\
hence, $CD=FD=3$, $AD=BD=8$,\\
thus, $AF=AD-FD=8-3=\boxed{5}$.


Question: As shown in the figure, points $B$ and $D$ are on ray $AM$, and points $C$ and $E$ are on ray $AN$, with $AB=BC=CD=DE$. Given that $\angle EDM=84^\circ$, what is the degree measure of $\angle A$?

Solution:
\textbf{Solution:} Given that $AB = BC = CD = DE$,\\
it follows that $\angle A = \angle BCA$, $\angle CBD = \angle BDC$, and $\angle ECD = \angle CED$,\\
Based on the exterior angle property of triangles, $\angle A + \angle BCA = \angle CBD$, $\angle A + \angle CDB = \angle ECD$, and $\angle A + \angle CED = \angle EDM$,\\
Also, since $\angle EDM = 84^\circ$,\\
it follows that $\angle A + 3\angle A = 84^\circ$,\\
Solving for $\angle A$, we find that $\angle A = \boxed{21^\circ}$.


Question: As shown in the figure, $ \angle 1 : \angle 2 : \angle 3 = 1 : 3 : 6$, what is the value of $ \angle 4$?

Solution:
\textbf{Solution:} Given that $\angle 2\colon \angle 3=3\colon 6$ and $\angle 2+\angle 3=180^\circ$,\\
therefore, $\angle 2=\frac{3}{9}\times 180^\circ=60^\circ$,\\
hence, $\angle 1=60^\circ\times \frac{1}{3}=20^\circ$,\\
thus, $\angle 4=180^\circ-20^\circ-80^\circ=\boxed{80^\circ}$


Question: As shown in the figure, two triangular plates of the same size containing $45^\circ$, triangular plate ACF and triangular plate CFB are placed as shown. Point D is on side AC, and point E is on side BC, with $\angle CFE=13^\circ$ and $\angle CFD=32^\circ$. What is the measure of $\angle DEC$ in degrees?

Solution:
\textbf{Solution:}

Construct $FH \perp FE$, intersecting $AC$ at point $H$.\\
$\because$ $\angle AFC=\angle EFH=90^\circ$\\
Also, $\because$ $\angle AFC=\angle AFH+\angle CFH$ and $\angle HFE=\angle CFE+\angle CFH$\\
$\therefore$ $\angle AFH=\angle CFE=13^\circ$\\
$\because$ $\angle A=\angle FCE=45^\circ$ and $FA=CF$\\
$\therefore$ $\triangle FAH\cong \triangle FCE$ (ASA)\\
$\therefore$ $FH=FE$\\
$\because$ $\angle DFE=\angle DFC+\angle EFC=32^\circ+13^\circ=45^\circ$\\
$\because$ $\angle DFH=\angle HFE-\angle DFE=90^\circ-45^\circ=45^\circ$\\
$\therefore$ $\angle DFE=\angle DFH$\\
Also, $\because$ $DF=DF$\\
$\therefore$ $\triangle HDF\cong \triangle EDF$ (SAS)\\
$\therefore$ $\angle DHF=\angle DEF$\\
$\because$ $\angle DHF=\angle A+\angle HFA=45^\circ+13^\circ=58^\circ$\\
$\therefore$ $\angle DEF=58^\circ$\\
$\because$ $\angle CFE+\angle CEF+\angle FCE=180^\circ$\\
$\therefore$ $\angle CEF=180^\circ-\angle CFE-\angle FCE=180^\circ-13^\circ-45^\circ=122^\circ$\\
$\therefore$ $\angle DEC=\angle CEF-\angle DEF=122^\circ-58^\circ=\boxed{64^\circ}$


Question: As shown in the diagram, $\triangle ABC$ is folded along $EF$ such that point $A$ falls on point ${A}^{\prime}$. $BP$ and $CP$ are respectively the angle bisectors of $\angle ABD$ and $\angle ACD$. If $\angle P=30^\circ$ and $\angle {A}^{\prime}EB=20^\circ$, what is the degree measure of $\angle {A}^{\prime}FC$?

Solution:
\textbf{Solution:} As shown in the figure,\\
since $BP$ and $CP$ are the angle bisectors of $\angle ABD$ and $\angle ACD$ respectively,\\
it follows that $\angle PBD=\frac{1}{2}\angle ABD$ and $\angle BCP=\frac{1}{2}\angle BCA$.\\
Also, since $\angle PBD=\angle P+\angle PCB$,\\
we have $\angle P=\angle PBD−\angle PCB=\frac{1}{2}\angle ABD−\frac{1}{2}\angle BCA=\frac{1}{2}(\angle ABD−\angle ACB)$,\\
and since $\angle ABD=\angle A+\angle ACB$,\\
it follows that $\angle ABD−\angle ACB=\angle A$,\\
thus $\angle P=\frac{1}{2}\angle A$,\\
and hence $\angle A=2\angle P=2\times 30^\circ=60^\circ$,\\
Given that: $\angle A\prime =\angle A=60^\circ$,\\
it follows that $\angle 1=\angle A\prime +\angle A\prime EB=60^\circ+20^\circ=80^\circ$,\\
thus $\angle A\prime FC=\angle A+\angle 1=60^\circ+80^\circ=\boxed{140^\circ}$,\\


Question: As shown in the figure, $\angle ACD=90^\circ$, $\angle D=15^\circ$, and point $B$ is on the perpendicular bisector of $AD$. If $AC=4$, what is the length of $AB$?

Solution:
\textbf{Solution:} Since point B is on the perpendicular bisector of AD,\\
it follows that $BA=BD$,\\
which means $\angle D=\angle BAD=15^\circ$\\
Thus, $\angle ABC=\angle D+\angle BAD=30^\circ$\\
Since $\angle ACD=90^\circ$ and $AC=4$,\\
it follows that $AB=2AC=\boxed{8}$


Question: As shown in the figure, in $\triangle ABC$, the perpendicular bisector $MN$ of $AB$ intersects $AC$ at point $D$, and $BD$ is drawn. If $AC=9$ and $BC=5$, what is the perimeter of $\triangle BDC$?

Solution:
\textbf{Solution:} Since the perpendicular bisector $MN$ of $AB$ intersects $AC$ at point D,\\
it follows that $AD=BD$,\\
Since $AC=9$ and $BC=5$,\\
the perimeter of $\triangle BDC$ is $BD+CD+BC=AD+CD+BC=AC+BC=9+5=\boxed{14}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, the perpendicular bisector of $AB$ intersects side $AB$ at point $D$ and side $AC$ at point $E$. If the perimeters of $\triangle ABC$ and $\triangle EBC$ are 15 and 9, respectively, what is the length of $BC$?

Solution:
\textbf{Solution:} Since $AB=AC$, the perimeter of $\triangle ABC$ is 15. \\
Therefore, $2AC+BC=15$. \\
Since $E$ is on the perpendicular bisector of $AB$, \\
we have $AE=BE$. \\
Since the perimeter of $\triangle BCE$ is $BC+CE+BE=BC+CE+AE=BC+AC=9$, \\
we find $AC=15-9=6$, \\
thus $BC=9-6=\boxed{3}$.


Question: As shown in the figure, there is an observation tower $EF$ at the top $E$ of a mountain. It is known that the angle $\angle EAB$ between the mountain slope and the horizontal plane is $30^\circ$, the height of the mountain $EB$ is 120 meters, point $C$ is 180 meters away from the base of the mountain $A$, $CD\parallel AB$, and intersects $EB$ at point $D$. The elevation angle $\angle FCD$ of the top end $F$ of the observation tower measured at point $C$ is $60^\circ$. What is the height of the observation tower $EF$ in meters?

Solution:
\textbf{Solution:} According to the question, we have: $\angle ABE=90^\circ$, \\
$\because$ $\angle EAB=30^\circ$, \\
$\therefore$ $\angle AEB=60^\circ$, $AE=2EB$, \\
$\because$ $EB=120$, \\
$\therefore$ $AE=2EB=240$, \\
$\because$ $AC=180$, \\
$\therefore$ $CE=AE-AC=60$, \\
$\because$ $CD\parallel AB$, \\
$\therefore$ $\angle CDE=\angle ABE=90^\circ$, $\angle ECD=\angle EAB=30^\circ$, \\
$\because$ $\angle FCD=60^\circ$, $\angle FCE=\angle FCD-\angle ECD=30^\circ$, \\
$\therefore$ $\angle FCE=\angle CFE=30^\circ$, \\
$\therefore$ $EF=CE=60$, \\
thus, the height of the tower is \boxed{60} meters.


Question: As shown in the figure, $DE$ is the perpendicular bisector of side $AC$ of $\triangle ABC$, with the foot of the perpendicular at point $E$, intersecting $AB$ at point $D$. Connect $CD$, $\angle B=30^\circ$, $CD\perp BC$, $CD=3$, then what is the length of $AB$?

Solution:
\textbf{Solution:} Since $DE$ is the perpendicular bisector of side $AC$ of $\triangle ABC$, and $CD=3$, \\
it follows that $AD=CD=3$, \\
Because $\angle B=30^\circ$ and $CD\perp BC$, \\
it follows that $BD=2CD=6$, \\
Hence, $AB=AD+BD=\boxed{9}$.


Question: As shown in the figure, in an acute-angled $\triangle ABC$, where $\angle A=75^\circ$, $DE$ and $DF$ are the perpendicular bisectors of sides $AB$ and $AC$, respectively. What is the degree measure of $\angle DBC$?

Solution:
\textbf{Solution:} Draw DA, DC, \\
since $\angle BAC=75^\circ$, \\
thus $\angle ABC + \angle ACB = 180^\circ - 75^\circ = 105^\circ$, \\
since DE and DF are the perpendicular bisectors of AB and AC respectively, \\
thus DA=DB, DA=DC, \\
thus DB=DC, $\angle DBA=\angle DAB$, $\angle DAC=\angle DCA$, \\
thus $\angle DBA+\angle DCA=\angle DAB+\angle DAC=75^\circ$, \\
thus $\angle DBC=\angle DBC= \frac{1}{2} \times (105^\circ - 75^\circ) = \boxed{15^\circ}$, \\


Question: As shown in the figure, $\angle AOB = 30^\circ$, point $P$ is located inside $\angle AOB$ with $OP = 3$. Points $M$ and $N$ are moving points on the rays $OA$ and $OB$, respectively. What is the minimum perimeter of $\triangle PMN$ when its perimeter is at its minimum?

Solution:
\textbf{Solution:} As shown in the figure, construct the symmetric points $E$ and $D$ of point $P$ with respect to $OA$ and $OB$, respectively, and connect $DN$, $EM$, $DE$, $OD$, $OE$,\\
then $OA$ perpendicularly bisects $PE$ and $OB$ perpendicularly bisects $PD$,\\
$\therefore EM=PM$, $DN=PN$, $OD=OP=OE=3$,\\
$\therefore$ the perimeter of $\triangle PMN$ is $PM+MN+PN=EM+MN+DN$,\\
Knowing that the length of the segment between two points is the shortest, when points $E$, $M$, $N$, $D$ are collinear, the value of $EM+MN+DN$ is minimal, and the minimum value is equal to the length of $DE$,\\
$\because OE=OP$, $OA\perp PE$,\\
$\therefore \angle AOE=\angle AOP$ (the three lines of an isosceles triangle coincide),\\
Similarly, it can be derived: $\angle BOD=\angle BOP$,\\
$\because \angle AOB=30^\circ$,\\
$\therefore \angle BOP+\angle AOP=30^\circ$,\\
$\therefore \angle DOE=\angle BOD+\angle BOP+\angle AOP+\angle AOE=2\left(\angle BOP+\angle AOP\right)=60^\circ$,\\
Also, $\because OD=OE=3$,\\
$\therefore \triangle DOE$ is an equilateral triangle,\\
$\therefore DE=OD=3$,\\
$\therefore$ the minimum value of $EM+MN+DN$ is 3,\\
$\therefore$ the minimum perimeter of $\triangle PMN$ is $\boxed{3}$,\\


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, point D is on side AB, and $\triangle CBD$ is folded along CD such that point B coincides exactly with point E on side AC. If $\angle A=25^\circ$, then what is the measure of $\angle CDE$?

Solution:
\textbf{Solution:} Given $\because \angle ACB=90^\circ$ and $\angle A=25^\circ$,\\
$\therefore \angle B=65^\circ$,\\
From the folding, we have: $\angle BCD=\angle ACD= \frac{1}{2}\angle ACB=45^\circ$ and $\angle CED=\angle B=65^\circ$,\\
$\therefore \angle CDE=180^\circ-\angle CED-\angle ACD=\boxed{70^\circ}$.


Question: As shown in the figure, in the isosceles triangle $ABC$, where $AB=AC$, $DE$ is the perpendicular bisector of $AB$, intersecting $AB$ at point $E$ and $AC$ at point $D$. If $\angle ADE=40^\circ$, what is the measure of $\angle CBD$?

Solution:
\textbf{Solution:} Given $DE$ bisects $AB$ perpendicularly,\\
hence $AD=BD$, $\angle AED=90^\circ$,\\
given $\angle ADE=40^\circ$,\\
thus $\angle ABD=\angle A=50^\circ$,\\
therefore, $\angle ABC=\angle C=\frac{1}{2}\times(180^\circ-50^\circ)=65^\circ$,\\
thus, $\angle CBD=65^\circ-50^\circ=\boxed{15^\circ}$.


Question: As shown in the figure, in $ \triangle ABC$, the bisectors of $ \angle ABC$ and $ \angle ACB$ intersect at point $D$. Through point $D$, a line $EF\parallel BC$ is drawn, intersecting $AB$ at point $E$ and $AC$ at point $F$. If $AB=10$, $BC=7$, and $AC=8$, what is the perimeter of $ \triangle AEF$?

Solution:
\textbf{Solution:} Since $EF\parallel BC$,\\
it follows that $\angle EDB = \angle DBC$,\\
Since $BD$ bisects $\angle ABC$,\\
it implies $\angle ABD = \angle DBC$,\\
Therefore, $\angle EBD = \angle EDB$,\\
which means $ED = EB$,\\
Similarly, it can be proven that $DF = FC$,\\
Therefore, $AE + AF + EF = AE + EB + AF + FC = AB + AC = 10 + 8 = 18$,\\
Hence, the perimeter of $\triangle AEF$ is $\boxed{18}$.


Question: As shown in the figure, the side length of each small square is 1. What is the degree measure of $\angle ABC$?

Solution:
\textbf{Solution:} Connect AC,\\
by the Pythagorean theorem, we have $AC^{2}=2^{2}+1^{2}=5$,\\
$BC^{2}=2^{2}+1^{2}=5$,\\
$AB^{2}=1^{2}+3^{2}=10$,\\
$\therefore AC^{2}+BC^{2}=5+5=10=AB^{2}$,\\
$\therefore \triangle ABC$ is an isosceles right triangle, $\angle ACB=90^\circ$,\\
$\therefore \angle ABC=\boxed{45^\circ}$.


Question: As shown in the figure, if $\triangle ABC \cong \triangle DEF$, $AC = 4$, $AB = 3$, $EF = 5$, then what is the perimeter of $\triangle ABC$?

Solution:
\textbf{Solution:} Since $\triangle ABC \cong \triangle DEF$, \\
\[
\therefore AB = DE = 3, \quad AC = DF = 4, \quad BC = EF = 5,
\]
\[
\therefore \text{the perimeter of} \ \triangle ABC = AB + AC + BC = 3 + 4 + 5 = \boxed{12}
\]


Question: As shown in the figure, in $\triangle ACB$, $\angle ACB = 90^\circ$, $AC = BC$, the coordinates of point $C$ are $(-2, 0)$, and the coordinates of point $A$ are $(-8, 3)$. What are the coordinates of point $B$?

Solution:
\textbf{Solution:} As shown in the figure, extend $AD \perp OC$ at $D$, and $BE \perp OC$ at $E$, such that $\angle 1=\angle 2=\angle 3=90^\circ$,\\
$\therefore$ $\angle$DAC+$\angle$ACD=$\angle$ACD+$\angle$ECB=$90^\circ$,\\
$\therefore$ $\angle DAC=\angle ECB$,\\
$\therefore$ $\angle ACD=\angle CBE$,\\
In $\triangle ACD$ and $\triangle CEB$,\\
$\because$ $\left\{\begin{array}{l}\angle DAC=\angle ECB\hfill \\ AC=CB\hfill \\ \angle ACD=\angle CBE\hfill \end{array}\right.$,\\
$\therefore$ $\triangle ADC\cong \triangle CEB$ (ASA),\\
$\therefore$ $DC=BE$, $AD=CE$,\\
$\because$ the coordinates of point $C$ are $(-2,0)$, and the coordinates of point $A$ are $(-8,3)$,\\
$\therefore$ $OC=2$, $AD=CE=3$, $OD=8$,\\
$\therefore$ $CD=OD-OC=6$, $OE=CE-OC=1$,\\
$\therefore$ $BE=6$,\\
$\therefore$ the coordinates of point $B$ are $\boxed{(1,6)}$.


Question: As shown in the figure, point $P$ is a point inside $\angle AOB$. Construct the symmetrical points $P_{1}$ and $P_{2}$ of point $P$ with respect to $OB$ and $OA$ respectively. Connect $P_{1}P_{2}$, let it intersect $OB$ at $M$, and intersect $OA$ at $N$. If $\angle AOB=40^\circ$, what is the measure of $\angle MPN$?

Solution:
\textbf{Solution:} Since the symmetric point of $P$ about $OB$ is ${P}_{1}$ and the symmetric point of $P$ about $OA$ is ${P}_{2}$, \\
therefore, $PM={P}_{1}M$, $PN={P}_{2}N$, $\angle {P}_{2}=\angle {P}_{2}PN$, $\angle {P}_{1}=\angle {P}_{1}PM$, $PP_{1}\perp OB$, $PP_{2}\perp OA$, \\
since $\angle AOB=40^\circ$, \\
therefore, $\angle {P}_{1}PP_{2}=140^\circ$, \\
therefore, $\angle {P}_{1}+\angle {P}_{2}=180^\circ-\angle {P}_{1}PP_{2}=40^\circ$, \\
since $\angle PMN=\angle {P}_{1}+\angle MP{P}_{1}=2\angle {P}_{1}$, $\angle PNM=\angle {P}_{2}+\angle NP{P}_{2}=2\angle {P}_{2}$, \\
therefore, $\angle PMN+\angle PNM=2(\angle {P}_{1}+\angle {P}_{2})=2\times 40^\circ=80^\circ$, \\
therefore, $\angle MPN=180^\circ-(\angle PMN+\angle PNM)=180^\circ-80^\circ=\boxed{100^\circ}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $BC=18$, $E$ and $F$ are the trisection points of $BC$, $AE=10$, $AF=8$, $G$ and $H$ are the midpoints of $AC$ and $AB$ respectively. What is the perimeter of the quadrilateral $EFGH$?

Solution:
\textbf{Solution:} Since $E$ and $F$ are the trisection points of $BC$,\\
we have $EF=\frac{1}{3}BC=6$, and $BE=EF=FC$.\\
Since $G$ and $H$ are the midpoints of $AC$ and $AB$ respectively,\\
$GH$ is the midline of $\triangle ABC$, and $GH=\frac{1}{2}BC=9$.\\
$HE$ is the midline of $\triangle ABF$, and $HE=\frac{1}{2}AF=4$.\\
$GF$ is the midline of $\triangle ACE$, and $GF=\frac{1}{2}AE=5$.\\
Therefore, the perimeter of the quadrilateral $EFGH=6+9+4+5=\boxed{24}$.


Question: As shown in the figure, it is known that $AB=3$, $AC=1$, and $\angle D=90^\circ$. If $\triangle DEC$ is centrally symmetric to $\triangle ABC$ with respect to point $C$, what is the length of $AE$?

Solution:
\textbf{Solution:} Given that $\triangle DEC$ and $\triangle ABC$ are centrally symmetric with respect to point C,\\
therefore $\triangle DEC \cong \triangle ABC$,\\
since $AB=3$ and $AC=1$,\\
then $DE=3$ and $CD=1$,\\
thus $AD=2$,\\
also, since $\angle D=90^\circ$,\\
it follows that $AE= \sqrt{DE^2+AD^2}=\sqrt{3^2+2^2}=\boxed{\sqrt{13}}$.


Question: As shown in the figure, vertex $C$ of $ABCD$ is on the side $BF$ of the equilateral triangle $\triangle BEF$, with point $E$ on the extension of $AB$, and $G$ as the midpoint of $DE$. Connecting $CG$. If $AD=5$, $AB=CF=3$, then what is the length of $CG$?

Solution:
Solution: As shown in the figure, extend DC to meet EF at point H, \\
$\because$ $\square ABCD$, AD=5, AB=CF=3, \\
$\therefore$ AD$\parallel$BC, BC=5, CD=CF=3, \\
$\therefore$ BF=BC+CF=5+3=8, $\angle$FCH=$\angle$EBF, \\
$\because$ equilateral $\triangle BEF$, \\
$\therefore$ $\angle F=\angle BEF=\angle EBF=\angle FCH=60^\circ$, EF=BF=8, \\
$\therefore$ $\triangle FCH$ is an equilateral triangle, \\
$\therefore$ CH=CF=FH=3, \\
$\therefore$ HE=EF−HF=8−3=5, \\
also $\because$ G is the midpoint of DE, \\
$\therefore$ CG is the midline of $\triangle DHE$, \\
$\therefore$ CG=$\frac{1}{2}$HE=$\frac{1}{2} \times 5=\boxed{\frac{5}{2}}$.


Question: As shown in the figure, a set of triangular plates is placed as in Figure 1, with $AB=CD= \sqrt{6}$, and vertex $E$ coinciding. When $\triangle DEC$ is rotated around its vertex $E$, as in Figure 2, during the rotation, when $\angle AED=75^\circ$, and $AD$ and $BC$ are connected, what is the area of $\triangle ADE$?

Solution:
\textbf{Solution:} As shown in Figure 2, draw $EG \parallel CD$ through point E, extend CE to meet AB at point H, \\
$\therefore \angle EDC=\angle DEG$, \\
Given that, we know: $\angle EDC=30^\circ$, $\angle DEC=60^\circ$, $\angle ECD=\angle AEB=90^\circ$, $\angle ABE=\angle BAE=45^\circ$, \\
$\because \angle AED=75^\circ$, \\
$\therefore \angle AEG=\angle AED-\angle DEG=75^\circ-30^\circ=45^\circ$, \\
$\therefore EG\parallel AB\parallel CD$, \\
$\therefore HC\perp AB$, \\
$\therefore HE=BH=AH=\frac{1}{2}AB$, \\
And since $AB=CD$, \\
$\therefore$ Quadrilateral ABCD is a parallelogram, \\
$\because AB=CD=\sqrt{6}$, \\
$\therefore HE=BH=AH=\frac{\sqrt{6}}{2}$, EC=$\frac{CD}{\sqrt{3}}=\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}$, \\
HC=HE+EC=$\frac{\sqrt{6}}{2}+\sqrt{2}=\frac{\sqrt{6}+2\sqrt{2}}{2}$, \\
$\therefore S_{\triangle ABE}=\frac{1}{2} \times \sqrt{6} \times \frac{\sqrt{6}}{2}=\frac{3}{2}$, $S_{\triangle DEC}=\frac{1}{2} \times \sqrt{6} \times \sqrt{2}=\sqrt{3}$, $S_{\triangle CEB}=\frac{1}{2} \times \sqrt{2} \times \frac{\sqrt{6}}{2}=\frac{\sqrt{3}}{2}$, \\
$S_{\square ABCD}=\sqrt{6} \times \frac{\sqrt{6}+2\sqrt{2}}{2}=3+2\sqrt{3}$, \\
$\therefore S_{\triangle ADE}=S_{\square ABCD}-S_{\triangle ABE}-S_{\triangle DEC}-S_{\triangle CEB}=3+2\sqrt{3}-\frac{3}{2}-\sqrt{3}-\frac{\sqrt{3}}{2}=\boxed{\frac{3+\sqrt{3}}{2}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $\angle BAD=120^\circ$. Points E and F are on the extensions of CD and BC, respectively, with AE $\parallel$ BD and EF $\perp$ BC. Given that EF $= 5\sqrt{3}$, what is the length of AB?

Solution: \textbf{Solution:} Since quadrilateral ABCD is a parallelogram, and $\angle BAD=120^\circ$, it follows that AB$\parallel$DC and AB=CD, and $\angle BAD=\angle BCD=120^\circ$. Hence, $\angle FCE=180^\circ-120^\circ=60^\circ$. As AE$\parallel$BD, quadrilateral ABDE is also a parallelogram. Therefore, AB=DE, and hence AB=DE=CD, making D the midpoint of CE. Given that EF$\perp$BC, we have $\angle EFC=90^\circ$. In $\triangle CEF$, $\angle CEF=30^\circ$. Let CF=x, and CE=2x, thus EF$^{2}$+CF$^{2}$=CE$^{2}$ leads to ($5\sqrt{3}$)$^{2}$+x$^{2}$=4x$^{2}$. Solving this, we find x=5 (taking the positive value). Therefore, CE=5×2=10, which makes CD=AB=$\frac{1}{2}$CE=5. Thus, the answer is: $\boxed{5}$.


Question: As shown in the figure, in rhombus $ABCD$, $AB=10$, diagonals $AC$ and $BD$ intersect at point $O$. If $E$ is the midpoint of side $AD$ and $\angle AEO = 32^\circ$, what is the length of $OE$? What is the degree measure of $\angle ADO$?

Solution:
\textbf{Solution:} Given: In diamond ABCD, AC and BD intersect at point O,\\
it follows that OA=OC, $\angle$ADB=$\angle$CDB, and CD=AB=10,\\
since E is the midpoint of side AD,\\
it follows AE=DE,\\
thus OE is the median line of $\triangle$ADC,\\
hence OE $\parallel$ CD, and OE=$\frac{1}{2}$CD=$\frac{1}{2} \times 10=\boxed{5}$,\\
therefore $\angle$ADC=$\angle$AEO=32$^\circ$,\\
hence $\angle$ADO=$\frac{1}{2}\angle$ADC=$\frac{1}{2} \times 32^\circ=\boxed{16^\circ}$,


Question: As shown in the figure, fold side AD of the rectangle so that point D falls on point F on side BC. If $ AB=8\,cm$ and $BC=10\,cm$, what is the length of EC in cm?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
hence AB=CD=8cm, AD=BC=10cm, $\angle$B=$\angle$C=$90^\circ$,\\
By the property of folding: AF=AD=10cm, DE=EF,\\
In $\triangle$ABF, BF=$\sqrt{AF^{2}-AB^{2}}=6cm$,\\
thus CF=BC-BF=4cm,\\
Let EC=x, then DE=EF=8-x,\\
In $\triangle$EFC, since EF$^{2}$=EC$^{2}$+CF$^{2}$,\\
thus (8-x)$^{2}$=x$^{2}$+4$^{2}$,\\
thus x=\boxed{3}cm,


Question: As shown in the figure, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. If $\angle ADB=30^\circ$, what is the measure of $\angle AOB$?

Solution:
\textbf{Solution:} Since ABCD is a rectangle,\\
it follows that $\angle$DAB = 90$^\circ$.\\
Since $\angle$ADB = 30$^\circ$,\\
it follows that $\angle$ABD = 90$^\circ - \angle$ADB = 60$^\circ$.\\
Since OA = OB,\\
it follows that $\triangle$ABO is an equilateral triangle,\\
therefore, $\angle$AOB = \boxed{60^\circ}.


Question: As shown in the figure, in $ \square ABCD$, the diagonals AC, BD intersect at point O, with OA=OB. If AD=4 and $\angle AOD=60^\circ$, what is the length of AB?

Solution:
\textbf{Solution:} Since $ABCD$ is a square and $OA=OB$,\\
it follows that $AO=BO=DO=OC$,\\
which implies $AC=BD$,\\
therefore, quadrilateral ABCD is a rectangle,\\
hence $\angle DAB=90^\circ$,\\
since $\angle AOD=60^\circ$,\\
it means $\triangle ADO$ is an equilateral triangle,\\
thus $DO=AD=4$,\\
therefore $BD=OD+OB=8$,\\
in $\triangle ABD$, $AB=\sqrt{BD^{2}-AD^{2}}=\sqrt{64-16}=4\sqrt{3}$.\\
Hence, the answer is: $AB=\boxed{4\sqrt{3}}$.


Question: As shown in the figure, in parallelogram $ABCD$, $\angle B=60^\circ$, $AB=4$, point $H$ and point $G$ are moving points on side $DC$ and $BC$, respectively, where point $H$ does not coincide with point $C$. Connect $AH$ and $HG$. Point $E$ is the midpoint of $AH$ and point $F$ is the midpoint of $GH$. Connect $EF$. What is the minimum value of $EF$?

Solution:
\textbf{Solution:} As shown in the figure, connect AG,\\
because point E is the midpoint of AH, and point F is the midpoint of GH,\\
therefore EF=$\frac{1}{2}AG$, thus the minimum value of EF,\\
can only be achieved when AG is at its minimum value, which is the vertical segment AG,\\
draw AM$\perp$BC through point A, with the foot at M,\\
because $\angle B=60^\circ$, $AB=4$,\\
therefore BM=2,\\
AM=$\sqrt{4^2-2^2}=2\sqrt{3}$,\\
therefore the minimum value of EF is $\frac{1}{2}AG=\boxed{\sqrt{3}}$.


Question: As shown in the diagram, in rhombus $ABCD$, $O$ is the intersection point of the two diagonals. Three lines passing through point $O$ divide the rhombus into shaded and unshaded areas. When the side length of the rhombus is $5$ and one of the diagonals is $8$, what is the area of the shaded region?

Solution:
\textbf{Solution:} Draw lines AC and BD as shown in the figure: \\
$\because$ Quadrilateral ABCD is a rhombus, \\
$\therefore$ AB=5, OA=OC=$\frac{1}{2}$AC=4, OB=OD, and AC$\perp$BD, \\
$\therefore$ OB=$\sqrt{AB^{2}-OA^{2}}=\sqrt{5^{2}-4^{2}}=3$, \\
$\therefore$ BD=2OB=6, \\
$\therefore$ The area of rhombus ABCD= $\frac{1}{2}$AC$\times$BD= $\frac{1}{2}\times6\times8=24$, \\
$\because$ O is the intersection point of the diagonals of the rhombus, and a rhombus is a centrally symmetric figure, \\
$\therefore$ The area of the shaded part=$\frac{1}{2}\times24=\boxed{12}$;


Question: As shown in the figure, in parallelogram $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, with $OA=3$. To make parallelogram $ABCD$ a rectangle, what should the length of $BD$ be?

Solution:
\textbf{Solution:} Suppose parallelogram \(ABCD\) is a rectangle, \\
\(\therefore\) \(OA=OC=OB=OD\), \\
\(\because\) \(OA=3\), \\
\(\therefore\) \(BD=2OB=\boxed{6}\).


Question: In rectangle $ABCD$, where $AB = 6$ and $AD = 8$, $P$ is a point on segment $AC$. If $\triangle PCD$ is an isosceles triangle, then what is the length of $AP$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\
$\therefore$ $\angle$ADC=$90^\circ$, DC=AB=6, AD=8,\\
$\therefore$ AC=$\sqrt{AD^{2}+DC^{2}}=\sqrt{8^{2}+6^{2}}=10$,\\
Consider three cases:\\
(1) When CP=CD=6, AP=AC−CP=10−6=4;\\
(2) When PD=PC, $\angle$PDC=$\angle$PCD,\\
$\because$ $\angle$PCD+$\angle$PAD=$\angle$PDC+$\angle$PDA=$90^\circ$,\\
$\therefore$ $\angle$PAD=$\angle$PDA,\\
$\therefore$ PD=PA,\\
$\therefore$ PA=PC,\\
$\therefore$ AP=$\frac{1}{2}$AC=5;\\
(3) When DP=DC, as illustrated, draw DQ$\perp$AC at Q,\\
then PQ=CQ,\\
$\because$ Area$_{\triangle ADC}$=$\frac{1}{2}$AD$\cdot$DC=$\frac{1}{2}$AC$\cdot$DQ,\\
$\therefore$ DQ=$\frac{AD\cdot DC}{AC}=\frac{8\times 6}{10}=\frac{24}{5}$,\\
$\therefore$ CQ=$\sqrt{DC^{2}-DQ^{2}}=\sqrt{6^{2}-\left(\frac{24}{5}\right)^{2}}=\frac{18}{5}$,\\
$\therefore$ PC=2CQ=$\frac{36}{5}$,\\
$\therefore$ AP=AC−PC=10−$\frac{36}{5}=\boxed{\frac{14}{5}}$;\\
In summary, if $\triangle$PCD is an isosceles triangle, the length of AP could be 4, 5, or $\frac{14}{5}$.


Question: As shown in the figure, in rhombus ABCD, the diagonals AC and BD intersect at point O, DE $\perp$ AB at point E, and OE is connected. If DE $= \sqrt{3}$, and BE $= 1$, what is the degree measure of $\angle ABD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\
$\therefore$ AC $\perp$ BD, DO=BO, \\
Since DE $\perp$ AB, DE=$\sqrt{3}$, BE=1, \\
$\therefore$ BD=$\sqrt{DE^2+BE^2}=2$, \\
$\therefore$ DO=BO=1, \\
Because DE $\perp$ BA, DO=BO, \\
$\therefore$ EO=DO=BO=1, \\
$\therefore$ BE=BO=EO=1, \\
$\therefore$ $\triangle$BEO is an equilateral triangle, \\
$\therefore$ $\angle$ABD=$\boxed{60^\circ}$,


Question: As shown in the figure, in rectangle $ABCD$, it is known that $AB=6$, $BC=8$. The perpendicular bisector of $BD$ intersects $AD$ at point $E$ and intersects $BC$ at point $F$. What is the length of $BF$?

Solution:
\textbf{Solution:}
Given that quadrilateral ABCD is a rectangle,\\
$\therefore \angle A=90^\circ$, AB=6, AD=BC=8,\\
$\therefore$BD=$\sqrt{AB^{2}+AD^{2}}=10$,\\
Additionally, since EF is the perpendicular bisector of BD,\\
$\therefore$OB=OD=5, $\angle$BOF=$90^\circ$,\\
Also, $\because \angle C=90^\circ$,\\
$\therefore \triangle BOF\sim \triangle BCD$,\\
$\therefore \frac{BO}{BC}=\frac{BF}{BD}$, that is: $\frac{5}{8}=\frac{BF}{10}$, solving this yields: $BF=\boxed{\frac{25}{4}}$.


Question: As shown in the figure, the quadrilateral ABCD is a rhombus, with diagonals AC and BD intersecting at point O. We have $AC=4\sqrt{3}$ and $BD=4$. Point P is a moving point on AC, and point E is the midpoint of AB. What is the minimum value of $PD+PE$?

Solution:
\textbf{Solution:} As shown in the figure, connect DE,\\
since quadrilateral ABCD is a rhombus, with diagonals AC and BD intersecting at point O, and AC = $4\sqrt{3}$, BD = 4,\\
thus AO = $\frac{1}{2}$AC = $2\sqrt{3}$, BO = $\frac{1}{2}$BD = 2, and AC $\perp$ BD,\\
thus AB = $\sqrt{AO^{2}+BO^{2}} = \sqrt{(2\sqrt{3})^{2}+2^{2}} = 4$,\\
thus AB = AD = BD, meaning $\triangle ABD$ is an equilateral triangle,\\
since point E is the midpoint of AB,\\
thus DE $\perp$ AB,\\
thus DE = $\sqrt{AD^{2}-AE^{2}} = \sqrt{4^{2}-2^{2}} = 2\sqrt{3}$,\\
since DP+PE $\geq$ DE,\\
thus the minimum value of PD+PE is the length of DE,\\
i.e., the minimum value of PD+PE is $2\sqrt{3}$,\\
hence, the answer is: the minimum value of $PD+PE$ is $\boxed{2\sqrt{3}}$.


Question: As shown in the figure, in rectangle $ABCD$, with $AB=6$ and $AD=8$, points $E$ and $F$ are located on $BC$ and $AD$ respectively. Connect $EF$. Quadrilateral $CDFE$ is folded along the line on which $EF$ lies, so that point $C$ coincides exactly with point $A$, and point $D$ falls at the position labeled $D'$. What is the length of $AE$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
it follows that BC=AD=8. Since quadrilateral CDFE is folded along the line containing EF, making point C coincide exactly with point A,\\
it follows that AE=CE. Let AE=CE=x, then BE=8-x,\\
in $\triangle ABE$, $AB^2+BE^2=AE^2$,\\
thus, $6^2+(8-x)^2=x^2$,\\
solving for x, we get $x=\boxed{\frac{25}{4}}$.


Question: As shown in the figure, parallelogram $ABCD$ has its diagonal $BD$ bisecting $\angle ABC$. Given $BC = 6$, $\angle ABC = 45^\circ$, and a moving point $P$ on diagonal $BD$ and another moving point $Q$ on side $BC$ such that $PQ + PC$ reaches its minimum, what is this minimum value?

Solution:
\textbf{Solution:} Let's start. Since quadrilateral ABCD is a parallelogram, \\
it follows that AD $\parallel$ BC, \\
thus $\angle ADB = \angle CBD$, \\
since BD bisects $\angle ABC$, \\
it follows that $\angle ABD = \angle CBD$, \\
therefore, $\angle ABD = \angle ADB$, \\
thus AB=AD, \\
hence, parallelogram ABCD is a rhombus, \\
therefore, points A and C are symmetric with respect to BD, \\
Draw AQ $\perp$ BC at Q, intersecting BD at P, \\
then the minimum value of PQ+PC is AQ, \\
since $\angle ABC = 45^\circ$, \\
it follows that $\triangle ABQ$ is an isosceles right triangle, \\
since AB=BC=6, \\
it follows that AQ=$\frac{\sqrt{2}}{2}$AB=$3\sqrt{2}$, \\
thus, this minimal value is $\boxed{3\sqrt{2}}$.


Question: As shown in the figure, in rectangle $ABCD$, $BC=6$, $AB=3$, $R$ is on side $CD$, and $CR=2$. Let $P$ be a moving point on $BC$, and $E$, $F$ be the midpoints of $AP$, $RP$, respectively. As $P$ moves from $B$ to $C$, what is the length of segment $EF$?

Solution:
\textbf{Solution}: As shown in the figure, connect AR.\\
$\because$ Quadrilateral ABCD is a rectangle,\\
$\therefore \angle D=90^\circ$,\\
$\because BC=6, AB=3, CR=2$,\\
$\therefore AD=6, DR=1$,\\
$\therefore AR=\sqrt{6^2+1^2}=\sqrt{37}$,\\
$\because$E and F are the midpoints of AP and RP respectively,\\
$\therefore EF=\frac{1}{2}AR=\frac{1}{2}\times \sqrt{37}=\boxed{\frac{\sqrt{37}}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, with $BC=4$, D and E are respectively the midpoints of $AB$ and $AC$, G and H are respectively the midpoints of $AD$ and $AE$, then what is the length of $GH$?

Solution:
\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively,\\
hence $DE$ is the median of $\triangle ABC$,\\
thus $DE=\frac{1}{2}BC=\frac{1}{2}\times 4=2$,\\
since $G$ and $H$ are the midpoints of $AD$ and $AE$ respectively,\\
hence $GH$ is the median of $\triangle ADE$,\\
thus $GH=\frac{1}{2}DE=\frac{1}{2}\times 2=\boxed{1}$.


Question: As shown in the diagram, in rectangle $ABCD$, points $E$ and $F$ are located on sides $BC$ and $AD$ respectively. The rectangle is folded along $EF$ such that point $B$ falls exactly on point $H$ on side $AD$. Given that the area of rectangle $ABCD$ is $16\sqrt{3}$ and $FH = 2HD$, what is the length of the crease $EF$?

Solution:
\textbf{Solution:} As shown in the diagram,\\
since folding the rectangle along EF makes point B precisely overlap with point H on edge AD, \\
thus, $AF=FG$, $\angle EHG=\angle B=\angle A=\angle G=90^\circ$, $HG=AB$, $\angle BEF=\angle FEH$,\\
since $\angle BEF+\angle FEH+\angle HEC=180^\circ$, $\angle HEC=60^\circ$,\\
thus, $\angle BEF=\angle FEH=60^\circ$,\\
since $AD\parallel BC$,\\
thus, $\angle FHE=\angle HEC=60^\circ$,\\
thus, $\triangle EFH$ is an equilateral triangle,\\
thus, $\angle FHE=60^\circ$,\\
thus, $\angle GHF=30^\circ$,\\
thus, $FG=\frac{1}{2}FH$,\\
since $FH=2HD$,\\
thus, $DH=FG=AF$,\\
let $DH=FG=AF=x$,\\
thus, $AB=HG=\sqrt{3}x$, $AD=4x$,\\
since the area of rectangle ABCD is $16\sqrt{3}$,\\
thus, $\sqrt{3}x \cdot 4x = 16\sqrt{3}$,\\
thus, $x=2$ (negative value is discarded),\\
thus, $AB=2\sqrt{3}$,\\
Draw $FM\perp BC$ at $M$ through F,\\
then, $FM=AB=2\sqrt{3}$,\\
thus, $EF=\frac{2\sqrt{3}}{3}FM=\boxed{4}$,\\


Question: As shown in the diagram, in square $ABCD$, $AB=2$. Extend $AD$ to point $E$ so that $DE=1$, $EF \perp AE$, and $EF=AE$. Connect $AF$ and $CF$ respectively. Let $M$ be the midpoint of $CF$. What is the length of $AM$?

Solution:
\textbf{Solution:} Extend FE and BC to meet at H, and connect $AC$,\\
In square ABCD: $\angle ADC=\angle BCD=90^\circ$, $\angle DAC=45^\circ$, $AB=CD=AD=2$,\\
$\therefore \angle EDC=\angle HCD=90^\circ$,\\
$\because EF\perp AE$,\\
$\therefore \angle AEF=\angle AEH=90^\circ$,\\
$\therefore \angle EDC=\angle HCD=\angle AEH=90^\circ$,\\
$\therefore$ quadrilateral $DCHE$ is a rectangle,\\
$\therefore CD=EH=2$, $DE=CH=1$,\\
$\because EF=AE$, $\angle AEF=90^\circ$,\\
$\therefore \angle FAE=45^\circ$,\\
$\because DE=1$, $AD=2$,\\
$\therefore AE=EF=3$,\\
$\therefore FH=5$,\\
According to the Pythagorean theorem: $CF=\sqrt{CH^2+FH^2}=\sqrt{1^2+5^2}=\sqrt{26}$,\\
$\because \angle CFA=\angle FAE+\angle EAF=90^\circ$,\\
$\therefore \triangle CAF$ is a right triangle,\\
$\because M$ is the midpoint of $CF$,\\
According to the fact that the median to the hypotenuse of a right triangle is half the length of the hypotenuse: $AM=\frac{1}{2}CF=\boxed{\frac{\sqrt{26}}{2}}$.


Question: As shown in the figure, it is known that the side length of rhombus ABCD is $6\sqrt{3}$, and point M is a moving point on diagonal AC, with $\angle ABC=120^\circ$. What is the degree measure of $\angle DAC$? What is the minimum value of $MA+MB+MD$?

Solution:
\textbf{Solution:} As shown in the figure, draw $ME\perp AB$ at point E and connect BD.\\
Since in rhombus ABCD, $\angle ABC=120^\circ$,\\
it follows that $\angle DAB=60^\circ$, $AD=AB=DC=BC$, AC perpendicularly bisects BD,\\
thus $\triangle ADB$ is an equilateral triangle, DM=BM,\\
hence $\angle DAC=\angle BAC=30^\circ$,\\
thus AM=2ME,\\
since MD=MB,\\
it follows that MA+MB+MD=2ME+2DM,\\
thus when points D, M, E are collinear, ME+DM is minimized, i.e., $MA+MB+MD$ is minimized,\\
since the side length of rhombus ABCD is $6\sqrt{3}$, $\angle DAE=60^\circ$, $\angle AED=90^\circ$,\\
it follows that AE=$\frac{1}{2}AD=3\sqrt{3}$,\\
thus DE=$\sqrt{AD^2-AE^2}=\sqrt{(6\sqrt{3})^2-(3\sqrt{3})^2}=9$,\\
hence $2DE=18$.\\
Therefore, the minimum value of $MA+MB+MD$ is $\boxed{18}$.\\
Therefore, the answer is: $\angle DAC=\boxed{30^\circ}$, the minimum value of $MA+MB+MD$ is $\boxed{18}$.


Question: As shown in the figure, $E$ and $F$ are two movable points on the side $AD$ of the square $ABCD$, satisfying $AE=DF$. Connect $CF$ and intersect $BD$ at point $G$, connect $BE$ and intersect $AG$ at point $H$. If the side length of the square is $2$, what is the minimum value of the length of segment $DH$?

Solution:
\textbf{Solution:} Take the midpoint O of AB, and connect OH and OD,\\
since it is a square ABCD,\\
therefore AB＝AD＝CD, $\angle BAD＝\angle CDA$, $\angle ADG＝\angle CDG$,\\
in $\triangle ABE$ and $\triangle DCF$,\\
\[
\left\{
\begin{array}{l}
AB=CD\\
\angle BAD=\angle CDA\\
AE=DF
\end{array}
\right.
\]\\
therefore $\triangle ABE \cong \triangle DCF$ (SAS),\\
therefore $\angle 1＝\angle 2$,\\
in $\triangle ADG$ and $\triangle CDG$,\\
\[
\left\{
\begin{array}{l}
AD=CD\\
\angle ADG=\angle CDG\\
DG=DG
\end{array}
\right.
\]\\
therefore $\triangle ADG \cong \triangle CDG$ (SAS),\\
therefore $\angle 2＝\angle 3=\angle 1$,\\
since $\angle BAH+\angle 3＝\angle BAD＝90^\circ$,\\
therefore $\angle 1+\angle BAH＝90^\circ$,\\
therefore $\angle AHB＝180^\circ−90^\circ＝90^\circ$,\\
since OH is the median of Rt$\triangle ABH$,\\
therefore OH＝AO＝$\frac{1}{2}$AB＝1,\\
in Rt$\triangle AOD$,\\
$OD=\sqrt{AO^{2}＋AD^{2}}=\sqrt{1^{2}+2^{2}}=\sqrt{5}$\\
since OH＋DH＞OD,\\
therefore when points O, D, and H are collinear, the length of DH is minimal,\\
The minimum value of DH ＝ OD−OH ＝ $\boxed{\sqrt{5}−1}$.


Question: As shown in the figure, it is known that the side length of the square $ABCD$ is 1. Point $E$ is a moving point on side $AB$. Connect $ED$, and rotate $ED$ clockwise $90^\circ$ around point $E$ to $EF$. Connect $DF$ and $CF$. What is the minimum value of $DF+CF$?

Solution:
\textbf{Solution:} Connect BF, through point F draw $FG\perp AB$ intersecting the extension of AB at point G, as shown in the figure,\\
$\because$ Rotating $ED$ $90^\circ$ clockwise around point $E$ to EF,\\
$\therefore EF\perp DE$, $EF=DE$,\\
$\therefore \angle EDA=\angle FEG$,\\
In $\triangle AED$ and $\triangle GFE$,\\
$\left\{\begin{array}{l}
\angle A=\angle FGE\\
\angle EDA=\angle FEG\\
DE=EF
\end{array}\right.$,\\
$\therefore \triangle AED\cong \triangle GFE (AAS)$,\\
$\therefore FG=AE$,\\
$\therefore$ Point F moves along the ray BF,\\
Construct the symmetric point ${C}^{\prime }$ of point C with respect to BF,\\
$\because EG=DA$, $FG=AE$,\\
$\therefore AE=BG$,\\
$\therefore BG=FG$,\\
$\therefore \angle FBG=45^\circ$,\\
$\therefore \angle CBF=45^\circ$,\\
$\therefore BF$ is the angle bisector of $\angle CBC^{\prime}$, hence point $F$ moves on the angle bisector of $\angle CBC^{\prime}$,\\
$\therefore$ Point ${C}^{\prime }$ is on the extension line of AB,\\
When points D, F, ${C}^{\prime }$ are collinear, DF + CF= D${C}^{\prime }$ is minimal,\\
In right $\triangle$AD${C}^{\prime }$, where AD= 1, A${C}^{\prime }$= 2,\\
$\therefore D{C}^{\prime }=\sqrt{AD^{2}+AC^{2}}=\sqrt{1^{2}+2^{2}}=\boxed{\sqrt{5}}$.


Question: As shown in the diagram, in parallelogram $ABCD$, $\angle ABC=60^\circ$. Points $E$ and $F$ lie on the extensions of $CD$ and $BC$, respectively. $AE \parallel BD$, $EF \perp BC$, and $CF=2$. What is the length of $AB$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a parallelogram,\\
it follows that $AB \parallel DC$ and $AB = CD$.\\
Given that $AE \parallel BD$,\\
it follows that quadrilateral $ABDE$ is a parallelogram.\\
Therefore, $AB = DE = CD$,\\
thus $D$ is the midpoint of $CE$.\\
Given that $EF \perp BC$,\\
it follows that $\angle EFC = 90^\circ$.\\
Given that $AB \parallel CD$,\\
it follows that $\angle DCF = \angle ABC = 60^\circ$.\\
Therefore, $\angle CEF = 30^\circ$.\\
Therefore, $CF = \frac{1}{2}CE$,\\
Given that $CF = 2$,\\
it follows that $CE = 4$\\
Therefore, $AB = CD = \boxed{2}$.


Question: As shown in the figure, within the rectangle $ABCD$, $E$ is the midpoint of $AD$. After folding $\triangle ABE$ along $BE$, we obtain $\triangle GBE$. Extending $BG$ intersects $CD$ at point $F$. If $CF=2$ and $FD=4$, what is the length of $BC$?

Solution:
\textbf{Solution:} Connect EF, \\
$\because$ $CF=2$, $FD=4$, \\
$\therefore$ $CD=CF+FD=2+4=6$, \\
$\because$ the quadrilateral ABCD is a rectangle, \\
$\therefore$ $AB=CD=6$, \\
As known from the folding, $EG=EA$, \\
Also, $\because$ E is the midpoint of AD, \\
$\therefore$ $ED=EA$, \\
$\therefore$ $ED=EG$, \\
$\because$ $EF=EF$, \\
$\therefore$ $\triangle EGF \cong \triangle EDF$ (HL), \\
$\therefore$ $GF=DF=4$, \\
By the property of folding: $BG=AB=6$, \\
$\therefore$ $BF=BG+GF=6+4=10$, \\
$\therefore$ $BC=\sqrt{BF^{2}-CF^{2}}=\sqrt{10^{2}-2^{2}}=\boxed{4\sqrt{6}}$.


Question: As shown in the diagram, if $A$ represents the area of the square, what is the value of $A$?

Solution:
\textbf{Solution:} Let the side length of the square represented by A be $x$,\\
by the Pythagorean theorem we have $x^{2}+3^{2}=5^{2}$,\\
$\therefore x^{2}=16,$ which means the area of the square represented by A is $\boxed{16}$.


Question: Given the diagram, $P$ is a point inside the right-angled triangle $ABC$ where $\angle BAC=90^\circ$, and $PA=3$, $PB=7$, $PC=9$. What is the maximum value of $BC$?

Solution:
\textbf{Solution:} As shown in the figure, draw PD$\perp$AB at D, PF$\perp$AC at F, then draw BG$\perp$PF at G through B, and CE$\perp$PD at E through C, intersecting BG at H, connect AH, PH\\
$\because$ $\angle$BAC$=90^\circ$,\\
$\therefore$ the quadrilaterals ADPF, ADEC, ABGF, PGHE, and ABHC are all rectangles\\
$\therefore$ PD=BG, PF=AD=EC, PG=HE, BC=AH\\
$\therefore$ $PA^{2}=AD^{2}+PD^{2}=PF^{2}+PD^{2}$\\
$PC^{2}=EC^{2}+PE^{2}=PF^{2}+PE^{2}$\\
$PH^{2}=PE^{2}+HE^{2}=PE^{2}+PG^{2}$\\
$PB^{2}=PG^{2}+BG^{2}=PG^{2}+PD^{2}$\\
$\therefore$ $PB^{2}+PC^{2}=PA^{2}+PH^{2}=PF^{2}+PE^{2}+PG^{2}+PD^{2}$\\
$\because$PA=3, PB=7, PC=9\\
$\therefore$ $7^{2}+9^{2}=3^{2}+PH^{2}$,\\
Solving gives $PH=11$\\
$\because$ in $\triangle PAH$, PA=3, $PH=11$\\
$\therefore$ $8<AH<14$\\
$\because$ when points A, P, H are collinear, $AH=14$\\
$\therefore$ when points A, P, H are collinear, $BC=AH=14$ is the maximum\\
Hence, the answer is: $BC=\boxed{14}$.


Question: As shown in the figure, points B and C are located on the lines $y=2x$ and $y=kx$, respectively. Points A and D are two points on the x-axis. If quadrilateral ABCD is a rectangle, and the ratio of $AB$ to $AD$ is $1:3$, what is the value of $k$?

Solution:
\textbf{Solution:} Let point A be (a, 0),\\
since point B lies on y=2x,\\
thus point B is (a, 2a),\\
therefore AB=2a,\\
since the ratio AB : AD = 1 : 3,\\
thus AD=3AB=$3\times2a$=6a,\\
therefore point D is (7a, 0),\\
since quadrilateral ABCD is a rectangle,\\
thus point C is (7a, 2a),\\
Substituting point C (7a, 2a) into y=kx gives 7ak=2a,\\
therefore k=$\boxed{\frac{2}{7}}$.


Question: As shown in the figure, if the leg $AF=6$ and the chord $AD=10$, then what is the area of the small square $EFGH$?

Solution:
\textbf{Solution:} As shown in the diagram, the right-angled triangles in the "Zhao Shuang Chord Diagram" are all congruent triangles, that is $AE=BF$:\\
$\because$ leg $AF=6$, chord $AD=AB=10$,\\
$\therefore$ leg $BF=\sqrt{AB^{2}-AF^{2}}=\sqrt{100-36}=\sqrt{64}=8$,\\
$\therefore$ the side length of the small square $EF=AF-AE=AF-BF=8-6=2$,\\
$\therefore$ the area of the small square $=2^{2}=\boxed{4}$.


Question: As shown in the figure, it is known that the graph of the linear function $y=2x+4$ intersects with the graph of the inverse proportion function $y=\frac{k}{x}$ at points $A$ and $B$, where the x-coordinate of point $B$ is $1$. A line is drawn from point $A$ such that $AC \perp y$ axis at point $C$, and line $BC$ is connected. What is the area of $\triangle ABC$?

Solution:
\textbf{Solution:}

Since the graph of the linear function $y=2x+4$ intersects the graph of the inverse proportionality function $y=\frac{6}{x}$ at points A and B, and the x-coordinate of point B is 1,

Substituting $x=1$ into $y=2x+4$ gives $y=6$,

Therefore, point B is (1, 6),

Therefore, $6=\frac{6}{1}$, solving for $k$ gives $k=6$,

Therefore, the equation of the inverse proportionality function is $y=\frac{6}{x}$,

Solving $\left\{\begin{array}{l}y=\frac{6}{x}\\ y=2x+4\end{array}\right.$ yields: $\left\{\begin{array}{l}x=1\\ y=6\end{array}\right.$ or $\left\{\begin{array}{l}x=-3\\ y=-2\end{array}\right.$,

Therefore, point A is (-3, -2),

Since AC is perpendicular to the y-axis at point C,

Therefore, AC equals 3,

Therefore, the area of $\triangle ABC=\frac{1}{2}\times3\times(6+2)=\boxed{12}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertices $A$ and $C$ of rectangle $ABCD$ are located on the negative half-axis of the $x$-axis and the positive half-axis of the $y$-axis, respectively. The $y$-axis bisects side $AB$. The coordinates of point $A$ are $(-2,0)$, and $AB=5$. What is the expression of the inverse proportion function passing through point $D$?

Solution:
\textbf{Solution:} As shown in the diagram, draw $DE\perp x$-axis at point $E$, and let the intersection point of $AB$ and the $y$-axis be point $F$,

since quadrilateral $ABCD$ is a rectangle,

then $AD=BC$, $\angle BAD=\angle B=90^\circ$,

since the $y$-axis bisects side $AB$, and $AB=5$,

then $AF=BF=\frac{5}{2}$,

since $A(-2,0)$,

then $OA=2$,

therefore, in $Rt\triangle OAF$, $OF=\sqrt{AF^{2}-OA^{2}}=\frac{3}{2}$,

in $\triangle BCF$ and $\triangle OAF$, $\left\{\begin{array}{l}\angle BFC=\angle OFA \\ \angle B=\angle AOF=90^\circ \end{array}\right.$,

thus $\triangle BCF\sim \triangle OAF$,

thus $\frac{BC}{OA}=\frac{BF}{OF}$, that is, $\frac{BC}{2}=\frac{\frac{5}{2}}{\frac{3}{2}}$,

solving gives $BC=\frac{10}{3}$,

thus $AD=\frac{10}{3}$,

since $\angle BAD=90^\circ$, $DE\perp x$-axis,

thus $\angle ADE+\angle DAE=\angle FAO+\angle DAE=90^\circ$,

thus $\angle ADE=\angle FAO$,

in $\triangle ADE$ and $\triangle FAO$, $\left\{\begin{array}{l}\angle ADE=\angle FAO \\ \angle AED=\angle FOA=90^\circ \end{array}\right.$,

thus $\triangle ADE\sim \triangle FAO$,

thus $\frac{DE}{OA}=\frac{AE}{OF}=\frac{AD}{AF}$, that is, $\frac{DE}{2}=\frac{AE}{\frac{3}{2}}=\frac{\frac{10}{3}}{\frac{5}{2}}$,

solving gives $DE=\frac{8}{3}$, $AE=2$,

thus $OE=OA+AE=4$,

thus $D(-4,\frac{8}{3})$,

let the expression for the inverse proportion function passing through point $D$ be $y=\frac{k}{x}$,

substituting point $D(-4,\frac{8}{3})$ gives: $k=-4\times \frac{8}{3}=-\frac{32}{3}$,

then the expression for the inverse proportion function passing through point $D$ is $y=-\frac{32}{3x}$,

hence, the answer is: $\boxed{y=-\frac{32}{3x}}$.


Question: In the Cartesian coordinate system, for the right triangle $\triangle ABC$ with vertices $A$ and $C$ having coordinates $(0,2)$ and $(2,0)$ respectively, and $\angle ACB = 90^\circ$, where $AC = 2BC$. If the graph of the function $y = \frac{k}{x}$ ($k>0$, $x>0$) passes through point $B$, what is the value of $k$?

Solution:
\textbf{Solution:} As shown in the figure, draw $BE\perp OC$ at $E$, \\
since the coordinates of A and C are respectively $(0, 2)$ and $(2, 0)$, \\
therefore, $OA=2$, $OC=2$, \\
therefore, $\angle OAC=\angle ACO=45^\circ$, and $AC=2\sqrt{2}$, \\
since $AC=2BC$ and $\angle ACB=90^\circ$, \\
therefore, $BC=\sqrt{2}$, and $\angle BCE=45^\circ$, \\
therefore, $\angle BCE=\angle CBE=45^\circ$, and $BE=CE=1$, \\
therefore, $B(3,1)$, \\
therefore, $k=xy=3\times 1=\boxed{3}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the vertices $A$ and $C$ of rectangle $ABCD$ are located on the negative half of the $x$-axis and the positive half of the $y$-axis, respectively. The $y$-axis bisects side $AB$, and the coordinate of point $A$ is $(-2,0)$, with $AB=5$. What is the expression of the inverse ratio function that passes through point $B$?

Solution:
\textbf{Solution:} Construct BF $\perp$ x-axis through point B, with F being the foot of the perpendicular. Assume AB intersects with the y-axis at point E,\\
$\because$ The coordinates of point A are $(-2, 0)$,\\
$\therefore$ OA = 2,\\
$\because$ The y-axis bisects side AB, and AB = 5,\\
$\therefore$ AE = BE = $\frac{1}{2}$AB = $\frac{5}{2}$,\\
$\because$ BF $\parallel$ y-axis,\\
$\therefore$ $\angle$AOE = $\angle$AFB, $\angle$AEO = $\angle$ABF,\\
$\therefore$ $\triangle$AOE $\sim$ $\triangle$AFB,\\
$\therefore$ $\frac{AO}{AF} = \frac{AE}{AB} = \frac{1}{2}$,\\
$\therefore$ AF = 2AO = 4,\\
$\therefore$ OF = AF - OA = 4 - 2 = 2,\\
$\therefore$ BF = $\sqrt{AB^{2} - AF^{2}} = \sqrt{5^{2} - 4^{2}} = 3$,\\
$\therefore$ B(2, 3),\\
Assume the expression for the inverse proportion function passing through point B is $y = \frac{k}{x}$,\\
Substituting B(2, 3) into $y = \frac{k}{x}$ yields:\\
3 = $\frac{k}{2}$,\\
$\therefore$ k = 6,\\
$\therefore$ The expression for the inverse proportion function passing through point B is: $\boxed{y=\frac{6}{x}}$.


Question: As shown in the figure, points A and B are on the graphs of the inverse proportion functions $y=\frac{2}{x}$ ($x>0$) and $y=\frac{4}{x}$ ($x>0$) respectively, and AB is parallel to the x-axis. If C is any point on the x-axis, then what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} As shown in the figure, extend BA to intersect the y-axis at point M, and connect OA, OB,\\
since line AB is parallel to the x-axis,\\
since $S_{\triangle AOM}=1$, $S_{\triangle BOM}=2$,\\
therefore $S_{\triangle ABC}=S_{\triangle ABO}=S_{\triangle BOM}-S_{\triangle AOM}=2-1=\boxed{1}$


Question: As shown in the figure, point A is any point on the graph of the inverse proportion function $y=\frac{7}{x}\left(x>0\right)$. Line segment $AB$ is parallel to the x-axis and intersects the graph of the inverse proportion function $y=-\frac{9}{x}$ at point B. Taking $AB$ as a side, a quadrilateral $\square ABCD$ is formed with points C and D on the x-axis. What is the area of $\square ABCD$?

Solution:
\textbf{Solution:} Draw lines OA, OB, and extend AB to meet the y-axis at E, as shown in the diagram,\\
$\because$ AB$\parallel$x-axis,\\
$\therefore$ AB$\perp$y-axis,\\
$\therefore$ Area$_{\triangle OEA}=\frac{1}{2}\times7=\frac{7}{2}$, Area$_{\triangle OBE}=\frac{1}{2}\times9=\frac{9}{2}$,\\
$\therefore$ Area$_{\triangle OAB}=\frac{7}{2}+\frac{9}{2}=8$,\\
$\because$ The quadrilateral ABCD is a parallelogram,\\
$\therefore$ Area$_{\text{parallelogram ABCD}}=2\times$Area$_{\triangle OAB}=\boxed{16}$.


Question: As shown in the figure, the graph of the inverse proportion function $y=\frac{k}{x}$ passes through the vertex $P$ of the rectangle $AOBP$, and the two sides $AO$, $BO$ of the rectangle are respectively on the x-axis and y-axis. If the area of the rectangle is $7$, what is the value of $k$?

Solution:
\textbf{Solution}: Since the area of rectangle $OAPB$ is 7\\
it follows that $\left|k\right|=7$\\
Given that the graph is in the second quadrant, it implies $ k=-7$\\
Therefore, the answer is: $k=\boxed{-7}$


Question: As shown in the figure, it is known that point A lies on the graph of the inverse proportion function $y=\frac{k}{x}$ ($x<0$). A line is drawn through point A perpendicular to the y-axis, and the foot of the perpendicular is B. If the area of $\triangle OAB$ is 1, what is the value of $k$?

Solution:
\textbf{Solution:} Since $AB \perp$ y-axis, \\
therefore, we have $S_{\triangle OAB} = \frac{1}{2}|k| = 1$, \\
and given $k < 0$, \\
therefore, we find $k = \boxed{-2}$.


Question: In the Cartesian coordinate system, a part of the graph of the inverse proportion function $y=\frac{k}{x}$ is shown in the figure. $AB\perp y$-axis at point B, and point P is on the x-axis. If the area of $\triangle ABP$ is 2, what is the value of $k$?

Solution:
\textbf{Solution:} Let the equation of the inverse proportion function be: $y=\frac{k}{x}$. Assume the coordinates of point A are (m, n).\\
Therefore, AB = -m, OB = n, and mn = k.\\
Since the area of $\triangle$ABP is 2,\\
it follows that $\frac{1}{2}$AB$\cdot$OB=2, i.e., -$\frac{1}{2}$mn=2\\
Thus, mn=-4,\\
resulting in $k=mn=\boxed{-4}$.


Question: Given point $A$ is on the graph of $y= \frac{k}{x}$ ($x>0$), and point $B$ is on the negative half of the x-axis. Connect $AB$, intersecting the y-axis at point $C$. If $AC=BC$ and $S_{\triangle BOC}=1$, what is the value of $k$?

Solution:
\textbf{Solution:}\\
Construct AD $\perp$ y-axis through point A, with D being the foot of the perpendicular,\\
$\therefore \angle ADC=\angle BOC=90^\circ$,\\
Also, $\because \angle ACD=\angle BCO$ and AC=BC,\\
$\therefore \triangle BOC \cong \triangle ADC$,\\
$\therefore S_{\triangle BOC}=S_{\triangle ADC}$\\
Connect OA, then $S_{\triangle ACO}= S_{\triangle BOC}=1$,\\
$\therefore S_{\triangle ADO}=2$,\\
$\therefore k=2\times 2=\boxed{4}$


Question: As shown, it is known that the radius of $\odot P$ is 1, and the center P moves on the parabola $y=\frac{1}{2}x^{2}-x-\frac{1}{2}$. When $\odot P$ is tangent to the x-axis, what are the coordinates of the center P?

Solution:
\textbf{Solution:} Let point P$(x, y)$, where $y=\frac{1}{2}x^{2}-x-\frac{1}{2}$,\\
since $\odot P$ is tangent to the x-axis,\\
thus $|y|=1$,\\
therefore $y=\pm 1$.\\
(1) When $y=1$, $1=\frac{1}{2}x^{2}-x-\frac{1}{2}$,\\
we get: $x_{1}=3$, $x_{2}=-1$\\
thus, points P$(3,1)$, $(-1,1)$\\
(2) When $y=-1$, $-1=\frac{1}{2}x^{2}-x-\frac{1}{2}$,\\
we get: $x=1$\\
thus, point P$(1,-1)$\\
Therefore, the answers are $\boxed{(3,1)}$ or $\boxed{(-1,1)}$ or $\boxed{(1,-1)}$.


Question: As shown in the right triangle $ABC$, with $\angle ACB=90^\circ$, $AC=BC=2$, point $P$ is a moving point on segment $AB$. Connect $CP$, and rotate segment $CP$ $90^\circ$ clockwise around point $C$ to obtain segment $CQ$. Connect $PQ$ and $AQ$. What is the maximum area of $\triangle PAQ$?

Solution:
\textbf{Solution:} As shown in the figure, rotate the line segment CP clockwise around point C to obtain line segment CQ, \\
$\therefore \angle PCQ=90^\circ$, CP=CQ,\\
$\therefore \angle ACP+\angle ACQ=90^\circ$,\\
Also, $\because \angle ACB=90^\circ$,\\
$\therefore \angle BCP+\angle ACP=90^\circ$,\\
$\therefore \angle BCP=\angle ACP$,\\
$\because AC=BC$,\\
$\therefore \triangle BPC\cong \triangle AQC$ (SAS),\\
$\therefore \angle B=\angle CAQ$, BP=AQ,\\
$\because BC=AC=2$,\\
$\therefore \angle B=\angle CAQ=\angle BAC=45^\circ$,\\
$\therefore \angle PAQ=\angle BAC+\angle CAQ=90^\circ$,\\
In $\text{Rt}\triangle ABC$, by the Pythagorean theorem AB= $\sqrt{BC^2+AC^2}=\sqrt{2^2+2^2}=2\sqrt{2}$,\\
Let BP=AQ=x, then $AP=2\sqrt{2}-x$,\\
$\therefore S_{\triangle PAQ}=\frac{1}{2}x(2\sqrt{2}-x)=-\frac{1}{2}(x^2-2\sqrt{2}x)=-\frac{1}{2}(x-\sqrt{2})^2+1$,\\
$\because a=-\frac{1}{2}<0$, the function opens downward, and the function has a maximum value,\\
When $x=\sqrt{2}$, $S_{\triangle PAQ_{\max}}=\boxed{1}$.


Question: As shown in the figure, point A is on the curve $y_1=\frac{2}{x}$ ($x>0$), and point B is on the hyperbola $y_2=\frac{k}{x}$ ($x<0$), with AB parallel to the x-axis. Point C is a point on the x-axis, and lines AC and BC are connected. If the area of $\triangle ABC$ is 6, what is the value of $k$?

Solution:
\textbf{Solution:} Let the coordinates of point $A$ be $\left(a,\frac{2}{a}\right)$, then the coordinates of point $B$ are $\left(\frac{ak}{2},\frac{2}{a}\right)$\\
Thus, ${S}_{\triangle ABC}=\frac{1}{2}\times \left|a-\frac{ak}{2}\right|\times \frac{2}{a}=6$\\
We find $k=-10$, $k=14$ (discard)\\
Therefore, the answer is: $k=\boxed{-10}$.


Question: As shown in the Cartesian coordinate system, point O is the origin, and the vertices of the parallelogram OABC, with vertex A lying on the reciprocal function $y = \frac{1}{x}$ and vertex B on the reciprocal function $y = \frac{4}{x}$, and point C on the positive half of the x-axis, what is the area of the parallelogram OABC?

Solution:
\textbf{Solution:} As shown in the diagram,\\
draw AF $\perp$ x-axis at point F passing through point A, draw BE $\perp$ x-axis at point E passing through point B, extend BA to intersect the y-axis at point G, then\\
quadrilateral OFAG and quadrilateral OEBG are rectangles,\\
$\because$ point A is on the inverse proportion function $y = \frac{1}{x}$, point B is on the inverse proportion function $y = \frac{4}{x}$,\\
$\therefore$ area of rectangle OFAG = 1, area of rectangle OEBG = 4,\\
$\therefore$ area of quadrilateral OABC = area of rectangle ABEF = area of rectangle OEBG - area of rectangle OFAG = 4 - 1 = \boxed{3}.


Question: As shown in the figure, the side length of the square $ABCD$ is $3$, with the side $AD$ on the negative x-axis. The graph of the inverse proportional function $y = \frac{k}{x}$ (where $x < 0$) passes through point $B$ and the midpoint $E$ of side $CD$. What is the value of $k$?

Solution:
Let $B(m, 3)$, then $C(m-3, 3)$, \\
$\because$ Point E is the midpoint of CD, \\
$\therefore$ The coordinates of E are $\left(m-3, \frac{3}{2}\right)$. \\
$\because$ B and E both lie on the graph of $y=\frac{k}{x}$, \\
$\therefore$ $(m-3)\cdot\frac{3}{2}=m\cdot3$, \\
Solving for m gives $m=\boxed{-3}$, \\
$\therefore$ $B(-3, 3)$, \\
$\therefore$ $k=-3\times3=\boxed{-9}$.


Question: As shown in the figure, rhombus OABC is in the first quadrant, with $\angle AOC = 60^\circ$, and the graph of the inverse proportion function $y = \frac{k}{x}$ ($k > 0$) passes through point A and intersects side BC at point D. If the area of $\triangle AOD$ is $\sqrt{3}$, what is the value of $k$?

Solution:
\textbf{Solution:} Join AC, through point A draw AE$\perp$OC at E, \\
$\because$ Quadrilateral ABCO is a rhombus, \\
$\therefore$AO$\parallel$CB, OA=OC, and $\angle$AOC$=60^\circ$, \\
$\therefore$$\triangle$AOC is an equilateral triangle, and AE$\perp$OC, \\
$\therefore$S$_{\triangle AOE}=\frac{1}{2}$S$_{\triangle AOC}$, \\
$\because$OA$\parallel$BC, \\
$\therefore$S$_{\triangle OAD}$=S$_{\triangle OAC}=\sqrt{3}$, \\
$\therefore$S$_{\triangle AOE}=\frac{1}{2}$S$_{\triangle AOC}=\frac{\sqrt{3}}{2}=\frac{k}{2}$, \\
$\therefore$k=$\boxed{\sqrt{3}}$,


Question: If the point $P(12, a)$ lies on the graph of the inverse proportion function $y=\frac{60}{x}$, then what is the value of $\cos\angle POH$?

Solution:
\textbf{Solution:} Given that point P lies on the graph of the inverse proportion function $y=\frac{60}{x}$,\\
thus $a=\frac{60}{12}=5$,\\
hence, the coordinates of point P are (12, 5),\\
therefore, $OH=12$ and $PH=5$,\\
in $\triangle OHP$, by the Pythagorean theorem, $OP=\sqrt{OH^2+PH^2}$,\\
thus, $OP=\sqrt{12^2+5^2}=13$,\\
therefore, $\cos\angle POH=\frac{OH}{OP}=\boxed{\frac{12}{13}}$.


Question: The cross-section of the arch of Zhaozhou Bridge is approximately parabolic, as illustrated in the figure, with the equation given by $y=-\frac{1}{25}x^{2}$. When the height $DO$ from the water surface to the top of the arch is $4m$, what is the width $AB$ of the water surface in meters?

Solution: \textbf{Solution:} According to the problem, we have $-4 = -\frac{1}{25}x^{2}$. Solving this yields $x = \pm 10$. Therefore, the coordinates of point A are $(-10, -4)$ and the coordinates of point B are $(10, -4)$. At this moment, the width of the water surface $AB$ is $\boxed{20}$ meters.


Question: As shown in the figure, point \(A\) is on the graph of the inverse proportion function \(y=\frac{1}{x}\left(x>0\right)\), point \(B\) is on the graph of the inverse proportion function \(y=\frac{3}{x}\left(x>0\right)\), and \(AB\) is parallel to the \(x\) axis. Points \(C\) and \(D\) are on the \(x\) axis. If the quadrilateral \(ABCD\) is a rectangle, what is its area?

Solution:
\textbf{Solution:} Extend BA to intersect the y-axis at point E, \\
$\because$ Quadrilateral ABCD is a rectangle, and AB$\parallel$x-axis, with points C, D on the x-axis, \\
$\therefore$ AE$\perp$y-axis, \\
$\therefore$ Quadrilaterals ADOE and BCOE are rectangles, \\
$\because$ point A is on the graph of the inverse proportion function y=$\frac{1}{x}$, and point B is on the graph of the inverse proportion function y=$\frac{3}{x}$, \\
$\therefore$ S$_{\text{rectangle}}$ADOE=1, S$_{\text{rectangle}}$BCOE=3, \\
$\therefore$ S$_{\text{rectangle}}$ABCD=S$_{\text{rectangle}}$BCOE−S$_{\text{rectangle}}$ADOE=3−1=\boxed{2}.