Question: As shown in the graph, the line $y=-\frac{3}{4}x+3$ intersects the x-axis at point A and the y-axis at point B. The $\triangle AOB$ is folded along a certain line passing through point B, such that point A falls on point D located on the negative y-axis. The fold line intersects the x-axis at point C. Please find the coordinates of points A, B, and C?

Solution:
\textbf{Solution:} Let's solve it. When $x=0$, then $y=3$; \\
when $y=0$, then $x=4$, \\
therefore, the coordinates of point A are $\boxed{(4,0)}$, and the coordinates of point B are $\boxed{(0,3)}$, \\
therefore, $AB=\sqrt{3^2+4^2}=5$, \\
since BD=AB=5, \\
therefore, the coordinates of point D are $\boxed{(0,-2)}$, \\
let the coordinates of point C be $(a,0)$, according to the problem, we have CD=AC, \\
therefore, $\sqrt{4+a^2}=4-a$, \\
solving it we find $a=\frac{3}{2}$, \\
therefore, the coordinates of point C are $\boxed{\left(\frac{3}{2}, 0\right)}$;


Question: As shown in the figure, the line $y=-\frac{3}{4}x+3$ intersects the x-axis at point A and the y-axis at point B. Triangle $AOB$ is folded along a certain line passing through point B such that point A falls on point D located on the negative y-axis. The fold intersects the x-axis at point C. Find the equation of line BC.

Solution:
\textbf{Solution:} Let the expression for line BC be $y=kx+b$,\\
$\therefore \begin{cases}3=b\\ 0=\frac{3}{2}k+b\end{cases}$,\\
solving yields $\begin{cases}k=-2\\ b=3\end{cases}$,\\
$\therefore$ the expression for line BC is $\boxed{y=-2x+3}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $y=mx$ intersects with the hyperbola $y=\frac{n}{x}$ at points $A(-1, a)$ and $B$. The line segment $BC$ is perpendicular to the $x$-axis, with the foot of the perpendicular being $C$, and the area of $\triangle AOC$ is 1. Find the equation of line $AC$.

Solution:
\textbf{Solution:} Given that the line $y=mx$ intersects the hyperbola $y=\frac{n}{x}$ at points $A(-1, a)$ and $B$,\\
the x-coordinate of point $B$ is 1,\\
since $BC$ is perpendicular to the x-axis with the foot at $C$,\\
we have $C(1, 0)$, and the distance $OC=1$,\\
considering the area of $\triangle AOC$ is 1,\\
we get $\frac{1}{2}\times 1\cdot a=1$,\\
solving this equation yields $a=2$,\\
thus, $A(-1, 2)$,\\
Let the equation of line $AC$ be $y=kx+b$,\\
substituting points $A(-1, 2), C(1, 0)$ into the equation yields: $\left\{\begin{array}{l}-k+b=2 \\ k+b=0 \end{array}\right.$, solving these equations gives $\left\{\begin{array}{l}k=-1 \\ b=1 \end{array}\right.$,\\
therefore, the equation of line $AC$ is $\boxed{y=-x+1}$.


Question: As shown in the diagram, in the Cartesian coordinate system, given points $A(8, 1)$ and $B(0, -3)$, the graph of the inverse proportion function $y=\frac{k}{x}$ $(x>0)$ passes through point A. The moving line $x=t$ $(0<t<8)$ intersects the graph of the inverse proportion function at point M and intersects line $AB$ at point N. What is the value of $k$?

Solution:
\textbf{Solution:} Since the inverse proportion function $y=\frac{k}{x}(x>0)$ passes through point $A\left(8, 1\right)$,\\
thus, substituting the coordinates of point A into $y=\frac{k}{x}$ yields: $1=\frac{k}{8}$,\\
solving for $k$ gives $k=\boxed{8}$;


Question: As shown in the figure, in the Cartesian coordinate system, it is known that points $A(8, 1)$ and $B(0, -3)$. The graph of the inverse proportion function $y=\frac{k}{x}$ $(x>0)$ passes through point A. The moving line $x=t$ $(0<t<8)$ intersects with the graph of the inverse proportion function at point M, and intersects with line $AB$ at point N. When $t=4$, what is the area of $\triangle BMA$?

Solution:
\textbf{Solution:} Since $k=8$\\
thus, the equation of the inverse proportion function is: $y=\frac{8}{x}(x>0)$.\\
Let the equation of line $AB$ be $y=k'x+b$,\\
substituting points $A\left(8, 1\right)$ and $B\left(0, -3\right)$, we get $\left\{\begin{array}{l}8k'+b=1\\ b=-3\end{array}\right.$,\\
solving this yields $\left\{\begin{array}{l}k'=\frac{1}{2}\\ b=-3\end{array}\right.$,\\
thus, the equation of line $AB$ is $y=\frac{1}{2}x-3$,\\
thus, in $y=\frac{1}{2}x-3$, when $x=t=4$, we find $y=-1$,\\
thus, point $N\left(4, -1\right)$,\\
in $y=\frac{8}{x}(x>0)$, when $x=t=4$, we find $y=2$,\\
thus, point $M\left(4, 2\right)$,\\
thus, the distance $MN=2-\left(-1\right)=3$,\\
since points $A\left(8, 1\right)$ and $B\left(0, -3\right)$,\\
thus, the area of $\triangle BMA=\frac{1}{2}MN\times \left({x}_{A}-{x}_{B}\right)=\frac{1}{2}\times 3\times \left(8-0\right)=\boxed{12}$.


Question: The location of rectangle $ABCD$ in the Cartesian coordinate system is shown in the diagram, where $F$ is a point on $AB$. Fold $\triangle BCF$ along $CF$ so that point $B$ precisely falls on the intersection point $E$ of $AD$ and the $y$-axis. Connect $CE$, if the lengths of $AE, AB$ satisfy $\sqrt{AE-4}+(AB-8)^{2}=0$. What are the coordinates of points $A$ and $B$?

Solution:
\textbf{Solution:} Solving from $\sqrt{AE-4}+(AB-8)^{2}=0$, we have:\\
$AE-4=0$ and $AB-8=0$\\
$\therefore AE=4$\\
$AB=8$\\
$\therefore A(\boxed{-4,8})$\\
$B(\boxed{-4,0})$


Question: As shown in the figure, rectangle $ABCD$ is positioned in the Cartesian coordinate system as depicted, with $F$ being a point on $AB$. Fold $\triangle BCF$ along $CF$ making point $B$ precisely fall on the intersection point $E$ of $AD$ and the y-axis. Connect $CE$. If the lengths of $AE$ and $AB$ satisfy $\sqrt{AE-4}+(AB-8)^2=0$, what are the coordinates of point $D$?

Solution:
\textbf{Solution:} Let $AE$ be $x$. According to the Pythagorean theorem, we have:\\
$(8-x)^{2}-x^{2}=4^{2}$\\
Solving this, we get: $x=3$\\
Let $ED$ be $y$. According to the Pythagorean theorem, we have:\\
$y^{2}+8^{2}=(y+4)^{2}$\\
Solving this, we get: $y=6$\\
$\therefore \boxed{D(6,8)}$


Question: As shown in the figure, in the Cartesian coordinate system \(xOy\), the side AB of the rhombus ABCD lies on the x-axis, and the vertex C is on the y-axis. It is known that the coordinates of points A and C are \((-2, 0)\) and \((0, 4)\), respectively. What are the coordinates of points B and D?

Solution:
\textbf{Solution:} According to the problem, we know: $OA=2$, $OC=4$, $AB=BC$,\\
Let $AB=BC=x$, $OB=AB-OA=x-2$,\\
$\therefore$ in right-angled $\triangle OBC$, $BC^2=OB^2+OC^2$,\\
$\therefore$ $x^2=(x-2)^2+4^2$, solving this gives: $x=5$,\\
$\because$ rhombus ABCD, $AB=BC=CD$,\\
$\therefore$ $AB=CD=5$,\\
$\therefore$ $OB=3$,\\
$\therefore$ $B(3, 0)$,\\
$\because$ $CD \parallel x$ axis, $CD=5$, $OC=4$,\\
$\therefore$ point $D(-5, 4)$,\\
Thus, the coordinates of points B and D are respectively \boxed{B(3, 0)} and \boxed{D(-5, 4)}.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the rhombus $ABCD$ has its edge $AB$ on the x-axis and vertex $C$ on the y-axis. It is known that the coordinates of points $A$ and $C$ are $\left(-2, 0\right)$ and $\left(0, 4\right)$, respectively. What is the area of the rhombus $ABCD$?

Solution: \textbf{Solution:} From the problem, we have the area of rhombus $ABCD$, $S_{\text{rhombus} ABCD} = AB \times OC = 5 \times 4 = \boxed{20}$.


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

Solution:
\textbf{Solution:} Since the graph of the direct proportion function $y=2x$ intersects with the graph of the linear function $y=kx+b$ at point $A(m, 2)$,\\
therefore $2m=2$, that is, $m=1$.\\
Substituting $(1, 2)$ and $(-2, -1)$ into $y=kx+b$, we get\\
$\begin{cases}
k+b=2\\
-2k+b=-1
\end{cases}$, which solves to $\begin{cases}
k=1\\
b=1
\end{cases}$,\\
therefore the equation of the linear function is $\boxed{y=x+1}$.


Question: As shown in the diagram, the graph of the direct proportion function $y=2x$ intersects with the graph of the linear function $y=kx+b$ at point $A(m, 2)$. The graph of the linear function passes through point $B(-2, -1)$ and intersects with the $x$-axis at point $C$. What is the area of $\triangle AOC$?

Solution:
\textbf{Solution:} Let $y=0$, then $x+1=0$, thus $x=-1$.\\
Hence, the coordinates of point C are $(-1,0)$,\\
therefore, the area of $\triangle AOC$ is $=\frac{1}{2} \times 1 \times 2 = \boxed{1}$.


Question: The graph of the linear function $y=kx+b$ (where $k$ and $b$ are constants, and $k \neq 0$) is shown in the figure. What are the values of $k$ and $b$?

Solution:
\textbf{Solution:} Since the graph of the linear function $y=kx+b$ passes through the points $(1,0)$ and $(0,2)$,\\
it follows that $\begin{cases} b=2 \\ k+b=0 \end{cases}$,\\
which yields $\begin{cases} k=\boxed{-2} \\ b=\boxed{2} \end{cases}$.


Question: The graph of the linear function $y=kx+b$ ($k$ and $b$ are constants, and $k\neq 0$) is shown in the figure. Using the graph, directly write down the solution set of the inequality $kx+b<2$ with respect to $x$.

Solution: \textbf{Solution:} Let \boxed{x > 0}


Question: As shown in the figure, the graph of the linear function $y=ax+b$ passes through the points $(1,2)$ and $(-1,6)$. Question: What is the analytic expression of this linear function?

Solution:
\textbf{Solution:} According to the problem, \\
when $x=1$, $y=2$; \\
when $x=-1$, $y=6$. \\
Therefore, $\begin{cases}2=a+b\\ 6=-a+b\end{cases}$ \\
Solving this system, we get $\begin{cases}a=-2\\ b=4\end{cases}$ \\
$\therefore$ the linear equation is: $y=-2x+4$. \\
Hence, the answer is $\boxed{y=-2x+4}$.


Question: Given the graph as shown, the graph of the linear function $y=ax+b$ passes through points $(1,2)$ and $(-1,6)$. Find: the area enclosed by the graph of the linear function and the two coordinate axes.

Solution:
\textbf{Solution:} The graph of the linear function intersects the y-axis and x-axis at points A and B, respectively.

Given $y=-2x+4$, we have

Coordinates of point A $(0,4)$, coordinates of point B $(2,0)$,

which means $OA=4$, $OB=2$.

Therefore, the area of $\triangle AOB$ is $\frac{1}{2}OA \cdot OB=\frac{1}{2} \times 4 \times 2=\boxed{4}$.

Thus, the area enclosed by the graph of the linear function and the two coordinate axes is \boxed{4}.


Question: As shown in the figure, $AC$ bisects $\angle BAD$, and $\angle BAD = \angle BCD$. What is the measure of $\angle DBC$?

Solution: \textbf{Solution:} We have $\angle DBC = \boxed{45^\circ}$.


Question: As shown in the figure, $AC$ bisects $\angle BAD$, and $\angle BAD = \angle BCD$. If $AD=6$, $AB=8$, then what is the length of $AC$?

Solution: \textbf{Solution:} We have $AC=\boxed{7\sqrt{2}}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, and the quadrilateral ABCD is inscribed in $\odot O$. OD intersects AC at point E, where OD $\parallel$ BC, and AD = CD; if AC = 8 and DE = 2, find the length of BC.

Solution:
\textbf{Solution:} Given that $\because$ OD$\perp$AC, and AC$=8$,

$\therefore$ AE$= \frac{1}{2}$ AC$=4$,

Let the radius of $\odot$O be r,

$\because$ DE$=2$,

$\therefore$ OE$=$OD$-$DE$=r-2$,

In the $\triangle AEO$, $AE^{2}$ + $OE^{2}$ = $AO^{2}$,

$\therefore$ $16+(r-2)^{2}=r^{2}$,

Solving for r we get r$=5$,

$\therefore$ AB$=2r=10$,

In $\triangle ACB$, $BC= \sqrt{AB^{2}-AC^{2}} = \sqrt{10^{2}-8^{2}} = \boxed{6}$,

$\therefore$ the length of BC is $\boxed{6}$.


Question: As shown in the figure, $AB$ is a chord of the circle $\odot O$, $OD\perp AB$ with the foot of the perpendicular at $E$, intersecting $\odot O$ at points $C$ and $D$. If $\angle AOD=50^\circ$, what is the degree measure of $\angle DOB$?

Solution:
\textbf{Solution:} Since $OD \perp AB$,\\
it follows that $\overset{\frown}{AD} = \overset{\frown}{BD}$,\\
therefore, $\angle DOB = \angle AOD = \boxed{50^\circ}$.


Question: As shown in the figure, $AB$ is a chord of the circle $\odot O$, $OD\perp AB$, the foot of the perpendicular is $E$, intersecting $\odot O$ at points $C$ and $D$. Given that $AB=2\sqrt{5}$ and $ED=1$, what is the length of the radius of $\odot O$?

Solution: \textbf{Solution:} Let the radius be $r$. Since $OD\perp AB$ and $AB=2\sqrt{5}$, it follows that $AE=BE=\frac{1}{2}AB=\sqrt{5}$. In $\triangle AOE$, we have $OE^2+AE^2=OA^2$. Given $ED=1$ and the radius is $r$, we have $OE=r-1$. Thus, $(r-1)^2+(\sqrt{5})^2=r^2$, which solves to $r=\boxed{3}$. Therefore, $OA=\boxed{3}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, and the semicircle $O$ with diameter $AB$ intersects $BC$ and $AC$ at points $D$ and $E$, respectively. Connect $DE$ and $OD$. $\overset{\frown}{BD}=\overset{\frown}{ED}$. When the ratio of the degrees of $\overset{\frown}{AE}$ and $\overset{\frown}{BE}$ is $4\colon 5$, find the degrees of the four interior angles of quadrilateral $ABDE$.

Solution: \textbf{Solution:} Given $\because$ $\overset{\frown}{AE}$ + $\overset{\frown}{BE}$ = $180^\circ$ and the ratio of the degrees of $\overset{\frown}{AE}$ to $\overset{\frown}{BE}$ is $4\colon 5$, $\therefore$ $\overset{\frown}{AE}$ = $80^\circ$, $\overset{\frown}{BE}$ = $100^\circ$. Consequently, $\overset{\frown}{BD}$ = $\overset{\frown}{ED}$ = $50^\circ$, hence $\overset{\frown}{AD}$ = $\overset{\frown}{AE}$ + $\overset{\frown}{ED}$ = $130^\circ$. Therefore, $\angle BAE$ = $\frac{1}{2} \overset{\frown}{BE}$ = $50^\circ$, $\angle B$ = $\frac{1}{2} \overset{\frown}{AD}$ = $65^\circ$, $\because$ $\angle AED$ + $\angle B$ = $180^\circ$ and $\angle BDE$ + $\angle A$ = $180^\circ$, $\therefore$ $\angle AED$ = $115^\circ$, $\angle BDE$ = $130^\circ$, thus $\angle BAE$ = $\boxed{50^\circ}$, $\angle B$ = $\boxed{65^\circ}$, $\angle BDE$ = $\boxed{130^\circ}$, $\angle AED$ = $\boxed{115^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$. Circle $\odot O$ is drawn with $AB$ as the diameter, intersecting $BC$ at point $D$ and $AC$ at point $E$. The arc $\overset{\frown}{BD}=\overset{\frown}{DE}$. If $\angle BAC=50^\circ$, find the degree measure of the arc $\overset{\frown}{AE}$.

Solution:
\textbf{Solution:} Connect $OE$, as shown in Figure 2:\\
$\because$ $AB$ is the diameter of $\odot O$,\\
$\therefore$ $OA$ is the radius,\\
$\therefore$ $OA=OE$,\\
$\therefore$ $\angle OEA=\angle BAC=50^\circ$,\\
$\therefore$ $\angle AOE=180^\circ-50^\circ-50^\circ=80^\circ$,\\
$\therefore$ The degree measure of $\overset{\frown}{AE}$ is $\boxed{80^\circ}$.


Question: As shown in the figure, in circle $O$, chord $AB=8$, point $C$ is on circle $O$ ($C$ does not coincide with $A$ and $B$). Connect $CA$ and $CB$, and draw $OD\perp AC$ and $OE\perp BC$ through point $O$, where the feet of the perpendiculars are points $D$ and $E$, respectively. Find the length of line segment $DE$.

Solution:
\textbf{Solution:} Since $OD$ passes through the center $O$, and $OD\perp AC$,\\
$\therefore AD=DC$,\\
Similarly, $CE=EB$,\\
$\therefore DE$ is the median of $\triangle ABC$,\\
$\therefore DE=\frac{1}{2}AB$,\\
Since $AB=8$,\\
$\therefore DE=\boxed{4}$.


Question: As shown in the diagram, within circle $O$, chord $AB=8$, and point $C$ is on circle $O$ (and is distinct from $A$ and $B$). Connect $CA$ and $CB$, and draw $OD \perp AC$ and $OE \perp BC$ with perpendicular foot at points $D$ and $E$, respectively. The distance from point $O$ to $AB$ is 3. What is the radius of circle $O$?

Solution:
\textbf{Solution:} Draw $OH\perp AB$ through point $O$, with the foot as point $H$, where $OH=3$, and connect $OA$.\\
Since $OH$ passes through the center $O$,\\
it follows that $AH=BH=\frac{1}{2}AB$,\\
Since $AB=8$,\\
it follows that $AH=4$,\\
In $\triangle AHO$, we have $AH^2+OH^2=AO^2$,\\
therefore $AO=\boxed{5}$,\\
which means the radius of circle $O$ is $\boxed{5}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\odot O$, with $AC$ being the diameter of $\odot O$, and $\angle ADB = \angle CDB$. If $AB = 2\sqrt{2}$ and $AD = 2$, find the length of $CD$.

Solution:
\textbf{Solution:} Since $\triangle ABC$ is an isosceles right-angled triangle,\\
it follows that $BC=AB=2\sqrt{2}$,\\
hence $AC=\sqrt{AB^{2}+BC^{2}}=4$,\\
In $\triangle ADC$, where $\angle ADC=90^\circ$ and $AD=2$,\\
it therefore follows that $CD=\sqrt{AC^{2}-AD^{2}}=\boxed{2\sqrt{3}}$.


Question: As shown in the diagram, in $\triangle ABC$, $AB=AC$, a semicircle with diameter $AB$ intersects $AC$ and $BC$ at points $E$ and $D$, respectively. Connect $ED$. If $\angle BAC=55^\circ$, find the degree measure of arc $AE$.

Solution:
\textbf{Solution:} Connect $BE$, $OE$\\
Since $AB$ is a diameter\\
Therefore $\angle AEB=90^\circ$\\
Since $\angle BAC=55^\circ$\\
Therefore $\angle ABE=90^\circ-\angle BAC=35^\circ$\\
Therefore $\angle AOE=2\angle ABE=70^\circ$;\\
Therefore, the degree measure of arc $AE$ is $\boxed{70^\circ}$;


Question: Given: As shown in the figure, $\triangle ABC$ is inscribed in $\odot O$, with $AB$ being the diameter. The bisector of $\angle CBA$ intersects $AC$ at point $F$ and intersects $\odot O$ at point $D$. $DE \perp AB$ at point $E$, and it intersects $AC$ at point $P$. Connect $AD$. $\angle DAC = \angle DBA$; Connect $CD$. If $CD=6$ and $BD=8$, find the radius of $\odot O$ and the length of $DE$.

Solution:
Solution: As shown in the diagram:\\
$\because \angle CBD=\angle DBA$,\\
$\therefore AD=CD=6$,\\
$\therefore AB=\sqrt{AD^{2}+BD^{2}}=10$,\\
$\therefore$ the radius of $\odot O$ is $\boxed{5}$,\\
$\because \frac{1}{2}AD\cdot BD=\frac{1}{2}AB\cdot DE$,\\
$\therefore \frac{1}{2}\times 6\times 8=\frac{1}{2}\times 10\times DE$,\\
Solving gives: $DE=\boxed{4.8}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\odot O$, with $\angle ADC=2\angle B$, and point $D$ is the midpoint of $\overset{\frown}{AC}$. What is the measure of $\angle B$ in degrees?

Solution:
\textbf{Solution:} Since $\angle ADC=2\angle B$\\
And since $\angle ADC+\angle B=180^\circ$\\
Therefore, $\angle B=\boxed{60^\circ}$


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\odot O$, with $OC=2$, and $AC=2\sqrt{2}$. The quadrilateral $AOCD$ is a rhombus. Find the distance from point $O$ to $AC$.

Solution:
\textbf{Solution:} Connect OA and construct OH $\perp$ AC at H,\\
OA$^{2}$+OC$^{2}=8$, AC$^{2}=8$,\\
$\therefore$ OA$^{2}$+OC$^{2}$=AC$^{2}$,\\
$\therefore$ $\triangle AOC$ is an isosceles right triangle, $\angle AOC=90^\circ$,\\
$\therefore$ OH=$\frac{1}{2}$AC=$\sqrt{2}$, that is, the distance from point O to AC is $\boxed{\sqrt{2}}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in a circle $\odot O$, $OC=2$, and $AC=2\sqrt{2}$. What is the measure of $\angle ADC$ in degrees?

Solution:
\textbf{Solution:} From the given condition, we have\\
$\because \angle AOC=90^\circ,$\\
$\therefore \angle B= \frac{1}{2}\angle AOC=45^\circ,$\\
$\because$ quadrilateral ABCD is inscribed in circle $O$,\\
$\therefore \angle ADC=180^\circ-45^\circ=\boxed{135^\circ}.$


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, point $C$ is a point on $\odot O$, connect $BC$ and $AC$. Point $E$ is the midpoint of $BC$, connect and extend $OE$ to intersect the circle at point $D$. $OD\parallel AC$. If $DE=2$ and $BE=2\sqrt{3}$, find the area of the shaded region.

Solution:
\textbf{Solution:} Connect OC, and set the radius of the circle as $r$, then $OE=r-2$,\\
$\because O{E}^{2}+B{E}^{2}=O{B}^{2}$,\\
$\therefore {\left(r-2\right)}^{2}+{\left(2\sqrt{3}\right)}^{2}={r}^{2}$,\\
Solving this, we get $r=4$,\\
$\therefore OE=2=\frac{1}{2}OB$,\\
$\therefore \angle ABC=30^\circ$,\\
$\therefore \angle COA=60^\circ$,\\
From (1), we have $BE=CE=2\sqrt{3}$,\\
$\therefore BC=4\sqrt{3}$,\\
$\therefore {S}_{\triangle BOC}=\frac{1}{2}OE\cdot BC=4\sqrt{3}$,\\
$\because \triangle BOC$ and $\triangle AOC$ have the same base and height,\\
$\therefore {S}_{\triangle AOC}={S}_{\triangle BOC}=4\sqrt{3}$,\\
$\therefore {S}_{\text{shaded}}={S}_{\text{sector} AOC}-{S}_{\triangle AOC}=\frac{60^\circ \cdot \pi \times {4}^{2}}{360}-4\sqrt{3}=\boxed{\frac{8\pi }{3}-4\sqrt{3}}$.


Question: As shown in Figure 1, $AB$ is the diameter of $\odot O$, and the tangent $BM$ is drawn from point B. The chord $CD$ is parallel to $BM$ and intersects $AB$ at point F, with $\overset{\frown}{DA}=\overset{\frown}{DC}$. The lines $AC$ and $AD$ are connected, and $AD$ is extended to meet $BM$ at point E. $\triangle ACD$ is an equilateral triangle; $OE$ is connected. If $OE=2\sqrt{7}$, find the length of $DE$.

Solution:
\textbf{Solution:} Given the problem statement, since $\triangle ACD$ is an equilateral triangle and $AB \perp CD$, \\
it follows that $\angle BAD = 30^\circ$, \\
connect $BD$, \\
since $AB$ is the diameter of $\odot O$, \\
it follows that $\angle ADB = 90^\circ$, \\
thus $\angle ABD = 60^\circ$, $BD = \frac{1}{2}AB = OB$, \\
since $\angle ABE = 90^\circ$, \\
it follows that $\angle DBE = 30^\circ$, \\
thus $BE = 2DE$, \\
let $DE = x$, then $BE = 2x$, \\
thus $BD = \sqrt{BE^{2} - DE^{2}} = \sqrt{(2x)^{2} - x^{2}} = \sqrt{3}x$, \\
thus $OB = \sqrt{3}x$, \\
thus $OE = \sqrt{BE^{2} + OB^{2}} = \sqrt{(2x)^{2} + (\sqrt{3}x)^{2}} = \sqrt{7}x$, \\
since $OE = 2\sqrt{7}$, \\
it follows that $x = \boxed{2}$, \\
thus $DE = \boxed{2}$.


Question: As shown in the figure, with side $AB$ of $\triangle ABC$ as the diameter, draw $\odot O$, which intersects side $AC$ at point $D$, and $BC$ is a tangent of $\odot O$. A chord $DE \perp AB$ at point $F$, and connect $BE$. It is given that $\angle ABE = \angle C$. If point $F$ is the midpoint of $OB$, and $OF=1$, what is the length of the line segment $ED$?

Solution:
\textbf{Solution:} Connect $OE$,\\
since point $F$ is the midpoint of $OB$,\\
it follows that $OF=\frac{1}{2}OB=\frac{1}{2}OE$,\\
and since $OF=1$,\\
it follows that $OE=2$. By the Pythagorean theorem, $EF=\sqrt{3}$,\\
and since $DE\perp AB$,\\
it follows that $DE=2EF=2\sqrt{3}$,\\
thus, the length of segment $ED$ is $DE=\boxed{2\sqrt{3}}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and chord $CD\perp AB$ at point $E$. A perpendicular line to $DB$ is drawn through point $C$, intersecting the extension of $AB$ at point $G$, with the foot of the perpendicular being point $F$. Connect $AC$, where $\angle A=\angle D$. $AC=CG$; given $CD=EG=8$, what is the radius of $\odot O$?

Solution:
\textbf{Solution:} As shown in the figure, connect $OC$,\\
let the radius of $\odot O$ be $r$, thus $OA=OC=r$,\\
since $CA=CG$, $CD\perp AB$, $CD=EG=8$,\\
therefore $AE=EG=8$, $EC=ED=\frac{1}{2}CD=4$,\\
hence $OE=AE−OA=8−r,$\\
in $\triangle OEC$, by the Pythagorean theorem, we have $OC^{2}=OE^{2}+EC^{2}$, which is $r^{2}=(8−r)^{2}+4^{2}$,\\
solving it, we get $r=\boxed{5}$,\\
therefore, the radius of $\odot O$ is $5$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, $CD$ is a chord of $\odot O$, and $CD \perp AB$ at point $E$. $\angle BCO = \angle D$; Given $BE = 8\,\text{cm}, CD = 6\,\text{cm}$, what is the radius of $\odot O$?

Solution:
\textbf{Solution:}

Because $AB$ is the diameter of the circle $O$, and $CD \perp AB$,

therefore $CE=DE=\frac{1}{2}CD=\frac{1}{2}\times 6=3$,

because $\angle B=\angle D$ and $\angle BEC=\angle DEA$,

therefore $\triangle BCE \sim \triangle DAE$,

therefore $AE: CE=DE: BE$,

therefore $AE: 3=3: 8$,

solving gives $AE=\frac{9}{8}$,

therefore $AB=AE+BE=\frac{9}{8}+8=\frac{73}{8}$,

therefore the radius of circle $O$ is $\boxed{\frac{73}{16}}$ (cm).


Question: As shown in the figure, within circle $O$, $AB$ is a diameter, and point $D$ lies on circle $O$. Point $C$ is the midpoint of arc $\stackrel{\frown}{AD}$. Chord $CM$ is perpendicular to $AB$ at point $F$. Connect $AD$, which intersects $CF$ at point $P$, and connect $BC$. Given that $\angle DAB=30^\circ$, what is the measure of $\angle ABC$?

Solution:
\textbf{Solution}: Draw line BD, since AB is the diameter of $\odot$O,\\
therefore $\angle ADB=90^\circ$,\\
since $\angle DAB=30^\circ$,\\
therefore $\angle ABD=90^\circ-30^\circ=60^\circ$,\\
since C is the midpoint of $\overset{\frown}{AD}$,\\
therefore $\angle ABC=\angle DBC=\frac{1}{2}\angle ABD=30^\circ$;\\
thus, the degree of $\angle ABC$ is $\boxed{30^\circ}$.


Question: As shown in the figure, inside the circle $\odot O$, $AB$ is a diameter, point $D$ is a point on $\odot O$, point $C$ is the midpoint of the arc $\overset{\frown}{AD}$, the chord $CM\perp AB$ at point $F$, connect $AD$, intersecting $CF$ at point $P$, and connect $BC$, with $\angle DAB=30^\circ$. If $CM=8\sqrt{3}$, find the length of the arc $\overset{\frown}{AC}$ (the result should retain $\pi$).

Solution:
\textbf{Solution:} Connect OC, then $\angle AOC=2\angle ABC=60^\circ$,\\
$\because$ CM $\perp$ diameter AB at point F,\\
$\therefore$ CF=$\frac{1}{2}$CM=$4\sqrt{3}$, $\angle CFO=90^\circ$,\\
$\therefore$ in $\triangle COF$, $\angle OCF=30^\circ$,\\
$\therefore$ OC=2OF, $OF^2+CF^2=OC^2$, that is $OF^2+(4\sqrt{3})^2=(2OF)^2$,\\
Solving gives: OF=4, $\therefore$ OC=8,\\
$\therefore$ the length of $\overset{\frown}{AC}$ is $\frac{60^\circ\pi \times 8}{180}=\boxed{\frac{8\pi }{3}}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\odot O$, with $\angle C = 2\angle A$. $DE$ is the diameter of $\odot O$, and $BD$ is connected. What is the degree measure of $\angle A$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in circle $\odot O$,\\
it follows that $\angle C+\angle A=180^\circ$,\\
and since $\angle C=2\angle A$,\\
it results in $\angle A=\boxed{60^\circ}$;


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\odot O$, with $\angle C=2\angle A$. $DE$ is the diameter of circle $\odot O$, and $BD$ is connected. If the diameter of $\odot O$ is 4, what is the length of $BD$?

Solution:
\textbf{Solution:} Connect $BE$,\\
since by the Inscribed Angle Theorem we have $\angle BED=\angle A=60^\circ$,\\
and since $DE$ is the diameter of circle $\odot O$,\\
it follows that $\angle EBD=90^\circ$,\\
thus $\angle BDE=90^\circ-60^\circ=30^\circ$,\\
therefore $BE=\frac{1}{2}DE=\frac{1}{2}\times 4=2$,\\
therefore $BD=\sqrt{DE^{2}-BE^{2}}=2\sqrt{3}$.\\
Hence, the length of $BD$ is $\boxed{2\sqrt{3}}$.


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

Solution:
\textbf{Solution:} Draw line OA and extend it to intersect BC at point E,\\
$\because$ AB = AC,\\
$\therefore$ $\overset{\frown}{AB}$ = $\overset{\frown}{AC}$,\\
$\because$ AE passes through the center O,\\
$\therefore$ AE is the perpendicular bisector of BC,\\
$\therefore$ AE bisects $\angle$BAC, and BE = $\frac{1}{2}$BC = $4\sqrt{3}$,\\
$\therefore$ $\angle$BAC = $2\angle$BAE,\\
$\because$ OA = OB,\\
$\therefore$ $\angle$ABD = $\angle$BAE,\\
$\because$ AB = AC and $\angle$ACB = $60^\circ$,\\
$\therefore$ $\angle$ABC = $\angle$ACB = $\angle$BAC = $60^\circ$,\\
$\therefore$ $\angle$ABD = $\frac{1}{2}\angle$BAC = $30^\circ$,\\
$\therefore$ $\angle$CBD = $30^\circ$,\\
$\therefore$ OB = \boxed{8},\\
Thus, the radius of $\odot$O is 8.


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

Solution:
\textbf{Solution:} Let $\angle ABD=x$,\\
from (1) we know that $\angle BAC=2\angle ABD=2x$,\\
$\therefore \angle BDC=3x$,\\
$\triangle BCD$ is an isosceles triangle,\\
(1) If $BD=BC$,\\
then $\angle C=\angle BDC=3x$,\\
$\because AB=AC$,\\
$\therefore \angle ABC=\angle C=3x$,\\
In $\triangle ABC$, $\angle ABC+\angle C+\angle BAC=180^\circ$,\\
$\therefore 3x+3x+2x=180^\circ$,\\
solving for $x$, we get $x=22.5^\circ$,\\
$\therefore \angle BCD=3x=67.5^\circ$,\\
(2) If $BC=CD$, then $\angle BDC=\angle CBD=3x$,\\
$\therefore \angle ABC=\angle ACB=4x$,\\
In $\triangle ABC$, $\angle ABC+\angle C+\angle BAC=180^\circ$,\\
$\therefore 4x+4x+2x=180^\circ$,\\
$\therefore x=18^\circ$,\\
$\therefore \angle BCD=4x=72^\circ$,\\
In summary, $\triangle BCD$ is an isosceles triangle, $\angle BCD$ is either $\boxed{67.5^\circ}$ or $\boxed{72^\circ}$.


Question: As shown in the figure, the running trajectory of the "Tube Cart" water barrel is a circle centered at $O$. It is known that the center $O$ is always above the water surface, and when the chord $AB$ cut by the water surface is $6\, \text{m}$, the maximum depth of the water barrel below the water surface is $1\, \text{m}$ (that is, the maximum distance from a point on the part of the circle below the water surface to the water surface). Find the radius of the circle in meters.

Solution:
\textbf{Solution:} As shown in the figure, draw $OD\perp AB$ at point $E$, intersecting the circle at point $D$, \\
then $AE=\frac{1}{2}AB=3$ meters, $DE=1$ meter, \\
let the radius of the circle be $r$ meters, \\
in the right triangle $\triangle AOE$, we have $AE^{2}+OE^{2}=OA^{2}$, \\
$\therefore 3^{2}+(r-1)^{2}=r^{2}$, \\
solving this equation, we find $r=\boxed{5}$, \\
$\therefore$ the radius of the circle is $\boxed{5}$ meters.


Question: As illustrated, the operating track of the "Tank Car" water tank is a circle centered at axis $O$. It is known that the center $O$ is always above the water surface. When the chord $AB$ intercepted by the water surface is $6\text{m}$, the maximum depth of the water tank below the surface is $1\text{m}$ (i.e., the maximum distance from a point on the circle below the surface to the water surface). If the water level rises causing the chord $AB$ intercepted by the water surface to change from the original $6\text{m}$ to $8\text{m}$, what is the new maximum depth of the water tank below the surface, in meters?

Solution:
\textbf{Solution:} As shown in the figure above, when $AB=8$ meters, $AE=\frac{1}{2}AB=4$,\\
in the right-angled triangle $\triangle AOE$, $AE^2+OE^2=OA^2$,\\
$\therefore 4^2+OE^2=5^2$,\\
$\therefore OE=3$ meters,\\
$\therefore DE=5-3=2$ meters,\\
Therefore, the maximum depth of the water container below the surface is $\boxed{2}$ meters.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $\angle A=30^\circ$, $AB=10$. The circle $\odot O$ with $AB$ as the diameter intersects $BC$ at point $D$ and $AC$ at point $E$. Connect $DE$. Draw $BP$ through point $B$ parallel to $DE$ and intersect $\odot O$ at point $P$. Connect $CP$ and $OP$. Point $D$ is the midpoint of $BC$; find the length of $\overset{\frown}{AP}$.

Solution:
\textbf{Solution:} Given $AB=AC$ and $\angle A=30^\circ$,\\
it follows that $\angle ABC=\frac{1}{2}\left(180^\circ-30^\circ\right)=75^\circ$,\\
since quadrilateral $ABDE$ is inscribed in a circle,\\
we have $\angle BAE+\angle BDE=\angle BDE+\angle CDE=180^\circ$,\\
thus $\angle CDE=\angle BAE=30^\circ$,\\
given $BP\parallel DE$,\\
it follows that $\angle PBC=\angle EDC=30^\circ$,\\
therefore, $\angle ABP=\angle ABC-\angle PBC=75^\circ-30^\circ=45^\circ$,\\
thus, $\angle AOP=2\angle ABP=90^\circ$,\\
hence, the length of $\overset{\frown}{AP}$ is: $\frac{90\pi \times 5}{180}=\boxed{\frac{5}{2}\pi}$.


Question: As shown in the figure, $AB$ is the diameter of the semicircle $O$, and $C$ and $D$ are two points on the semicircle $O$, with $OD \parallel BC$, and $OD$ intersects $AC$ at point $E$. If $\angle B=72^\circ$, find the degree measure of $\angle CAD$.

Solution:
\textbf{Solution:} Given that $AB$ is the diameter of the semicircle $O$,\\
it follows that $\angle C=90^\circ$,\\
since $\angle B=72^\circ$,\\
it follows that $\angle CAB=90^\circ-\angle B=18^\circ$,\\
since $OD\parallel BC$,\\
it follows that $\angle AOD=\angle B=72^\circ$,\\
since $OA=OD$,\\
it follows that $\angle OAD=\angle D=54^\circ$,\\
therefore $\angle CAD=\angle OAD-\angle CAB=\boxed{36^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of the semicircle $O$, and $C$, $D$ are two points on the semicircle $O$, with $OD\parallel BC$, and $OD$ intersects $AC$ at point $E$. If $AB=13$ and $AC=12$, find the length of $DE$.

Solution:
\textbf{Solution:} Given that $AB=13$ and $AC=12$,\\
it follows that $BC=\sqrt{AB^{2}-AC^{2}}=5$.\\
Since $OD\parallel BC$ and $\angle C=90^\circ$,\\
it implies $OD\perp AC$,\\
which means $AE=CE$.\\
Given $OA=OB$,\\
it can be deduced that $OE=\frac{1}{2}BC=2.5$.\\
Since $OD=OA=\frac{1}{2}AB=6.5$,\\
it follows that $DE=OD-OE=\boxed{4}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, $BC$ is the chord of $\odot O$, the radius $OD \perp BC$ with the foot of the perpendicular being $E$. Given that $BC=6\sqrt{3}$ and $DE=3$, find the radius of $\odot O$?

Solution:
\textbf{Solution:} Let the radius of $\odot O$ be $r$,\\
since radius $OD\perp BC$,\\
then it follows that $CE=\frac{1}{2}BC=3\sqrt{3}$,\\
since $OC^{2}=CE^{2}+OE^{2}$,\\
then $r^{2}=(r-3)^{2}+(3\sqrt{3})^{2}$,\\
thus $r=\boxed{6}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, $BC$ is a chord of $\odot O$, and the radius $OD\perp BC$ with the foot of the perpendicular being $E$. If $BC=6\sqrt{3}$ and $DE=3$, find the area of the shaded region (the result should retain $\pi$).

Solution:
\textbf{Solution:} Given that $\angle CEO=90^\circ$ and $CO=2OE$,\\
$\therefore \angle ECO=30^\circ$.\\
Since $AB$ is the diameter of $\odot O$,\\
$\therefore \angle ACB=90^\circ$,\\
$\therefore \angle ACO=60^\circ$.\\
Since $OA=OC$,\\
$\therefore \triangle ACO$ is an equilateral triangle,\\
$\therefore \angle AOC=60^\circ$,\\
$\therefore S_{\text{shaded}}=S_{\text{sector} ACO}-S_{\triangle AOC}=\frac{60\pi \times 6^2}{360}-\frac{1}{2}\times 6\times 6\times \frac{\sqrt{3}}{2}=6\pi -9\sqrt{3}$.\\
$\therefore$ The area of the shaded part is $S_{\text{shaded}}=\boxed{6\pi -9\sqrt{3}}$.


Question: As shown in the figure, it is known that $C$ and $D$ are points on the circle $\odot O$ and are located on either side of the diameter $AB$, with $BC=2\sqrt{3}$ and $\angle ABC=30^\circ$. When $CD\perp AB$ at point $E$, what is the length of $CD$?

Solution:
\textbf{Solution:} Given that $CD \perp AB$ and $AB$ is the diameter of the circle,\\
therefore $CD = 2CE$ and $\angle CEB = 90^\circ$. Also, given that $\angle ABC = 30^\circ$,\\
thus $BC = 2CE$,\\
therefore $CD = BC = 2\sqrt{3}$;\\
Hence, the length of $CD$ is $\boxed{2\sqrt{3}}$.


Question: As shown in the figure, it is known that $C$ and $D$ are points on the circle $\odot O$, and they are located on both sides of the diameter $AB$, with $BC=2\sqrt{3}$, and $\angle ABC=30^\circ$. When $D$ is the midpoint of arc $AB$, what is the length of $CD$?

Solution:
\textbf{Solution:} As shown in the figure, construct $BF\perp CD$ at point F,\\
since D is the midpoint of $\overset{\frown}{AB}$, and $BC=2\sqrt{3}$,\\
thus $\angle BCD=45^\circ$,\\
therefore, $\triangle BCF$ is an isosceles right triangle,\\
thus $BF=CF=\frac{\sqrt{2}BC}{2}=\sqrt{6}$,\\
also, since $\angle BDC=60^\circ$,\\
thus $DF=\frac{\sqrt{3}BF}{3}=\sqrt{2}$,\\
therefore, $CD=\sqrt{6}+\sqrt{2}=\boxed{\sqrt{6}+\sqrt{2}}$.


Question: As shown in the figure, within circle $\odot O$, chord $AB$ perpendicularly bisects radius $OC$. Find the degree measure of $\angle C$?

Solution:
\textbf{Solution:} Since chord $AB$ perpendicularly bisects radius $OC$,

it follows that $OB=BC$,

and since $OB=OC$,

it follows that $\triangle OCB$ is an equilateral triangle,

hence, $\angle C = \boxed{60^\circ}$.


Question: As shown in the figure, within circle $O$, chord $AB$ perpendicularly bisects radius $OC$. If the length of chord $AB$ is $10$, what is the diameter of circle $O$?

Solution:
\textbf{Solution:} Given that AB is a chord, OC is the radius, and $OC\perp AB$ with $AB=10$,\\
hence $BD= \frac{1}{2}AB=5$,\\
since the chord AB bisects the radius OC vertically and $OB=OC$,\\
thus $OD= \frac{1}{2}OC= \frac{1}{2}OB$,\\
in $\triangle ODB$, we have $OD^{2}+BD^{2}=OB^{2}$, that is $\left( \frac{1}{2}OB\right)^{2}+5^{2}=OB^{2}$,\\
solving this, we get: $OB= \frac{10\sqrt{3}}{3}$ (neglecting the negative value),\\
therefore, the diameter of $\odot O$ is $2\times \frac{10\sqrt{3}}{3}=\boxed{\frac{20\sqrt{3}}{3}}$.


Question: As shown in the figure, within a circle $\odot O$ with a radius of $50\,mm$, the chord $AB$ has a length of $50\,mm$. Find the measure of $\angle AOB$ in degrees.

Solution:
\textbf{Solution:} Since OA and OB are radii of the circle O, \\
thus OA = OB = 50mm, \\
and since AB = 50mm, \\
therefore OA = OB = AB, \\
therefore $\triangle$ AOB is an equilateral triangle, \\
therefore $\angle$AOB = \boxed{60^\circ}


Question: As shown in the figure, in circle $\odot O$ with a radius of $50mm$, the chord $AB$ has a length of $50mm$. Question: What is the distance in $mm$ from point $O$ to $AB$?

Solution:
\textbf{Solution:} Draw $OC\perp AB$ at point $O$ with the foot at point $C$, as shown in the figure,\\
by the perpendicular diameter theorem, we have $AC=CB=\frac{1}{2}AB=25\,mm$,\\
in $\triangle OAC$, $OC^2=OA^2-AC^2=50^2-25^2=25^2\times 3$,\\
$\therefore OC=\sqrt{25^2\times 3}=25\sqrt{3}\,mm$,\\
that is, the distance from point $O$ to $AB$ is \boxed{25\sqrt{3}\,mm}.


Question: In rectangle $ABCD$, $E$ is a point on side $AB$. Triangle $ADE$ is folded along $DE$ so that the corresponding point of $A$ is precisely point $F$ on side $BC$. Line $AF$ intersects $DE$ at point $N$, and line $BN$ is drawn. If $AD = 3$ and $BF = \frac{1}{3}BC$, what is the area of rectangle $ABCD$?

Solution: \textbf{Solution:} The area of rectangle $ABCD$ is $3\sqrt{5}=\boxed{3\sqrt{5}}$.


Question: In rectangle $ABCD$, where $E$ is a point on side $AB$, fold $\triangle ADE$ along $DE$ such that the corresponding point of $A$, point $F$, falls exactly on side $BC$. Connect $AF$ and intersect $DE$ at point $N$, then connect $BN$. Given $AD=3$ and $BF=\frac{1}{3}BC$, find the value of $\sin\angle BNF$.

Solution: \textbf{Solution:} We have $\sin\angle BNF = \boxed{\frac{\sqrt{5}}{3}}$.


Question: As shown in Figure 1, $\triangle ABC$ is an equilateral triangle, where D is a point on side $AC$ that does not coincide with point A. Extend $BC$ to point E such that $CE=AD$, and extend $AC$ to F so that $CF=AC$. Connect $EF$ and $BD$. If $\angle ABD=20^\circ$, what are the degrees of $\angle CFE$ and $\angle CEF$?

Solution: Solution: Since $\triangle ABC$ is an equilateral triangle, we have $AB=AC$, and $\angle A=\angle ACB=60^\circ$. Therefore, $\angle ECF=60^\circ$, which means $\angle A=\angle ECF$. Given that $CF=AC$, we find $AB=CF$. Since $CE=AD$, we deduce $\triangle ABD \cong \triangle CFE$, thus $\angle CFE=\angle ABD=20^\circ$. Consequently, $\angle CEF=180^\circ-\angle CFE-\angle ECF=\boxed{100^\circ}$.


Question: As shown in the figure, behind a building \textup{AB} there is an artificial hill with a slope \textup{CD} of $i=1\colon 2$. At point \textup{E} on the slope, there is a pavilion. It is measured that the horizontal distance from the foot of the hill to the building $BC=24$ meters, and the distance to the pavilion $CE=8\sqrt{5}$ meters. Xiao Li measures the angle of depression of point \textup{E} from the top of the building to be $45^\circ$. What is the distance from point \textup{E} to the horizontal ground in meters?

Solution:
\textbf{Solution:} Draw line $EF \perp BC$ extending through point $E$ to $F$. In $\triangle CEF$,\\
since the slope of $CD$ is $i=EF:CF=1:2$,\\
it follows that $EF:CF:CE=1:2:\sqrt{5}$,\\
since $CE=8\sqrt{5}$,\\
thus $EF=8$, and $CF=16$ meters,\\
therefore, the distance from point $E$ to the flat ground is $\boxed{8}$ meters.


Question: As shown in the figure, behind the building \(AB\) there is a rockery \(CD\) with a slope of \(i=1:2\). At point \(E\) on the slope, there is a pavilion. It is measured that the horizontal distance from the bottom of the rockery to the building is \(BC=24\) meters, and the distance to the pavilion is \(CE=8 \sqrt{5}\) meters. Xiao Li measures the angle of depression to \(E\) from the top of the building to be \(45^\circ\). What is the height of the building \(AB\) in meters?

Solution: \textbf{Solution:} Construct $EH \perp AB$ at point H. Since $AB \perp BF$ and $EF \perp BF$, it follows that quadrilateral BFEH is a rectangle; thus, $BH=EF=8$ and $HE=BF$. Given $BC=24$ and $CF=16$, we have $HE=BF=BC+CF=24+16=40$. In right triangle $AHE$, since $\angle HAE=90^\circ-45^\circ=45^\circ$, it follows that $AH=HE=40$. Therefore, $AB=AH+HB=40+8=\boxed{48}$ meters.


Question: As shown in the figure, in $\triangle ABC$, $AC=5$, $\cot A=2$, $\cot B=3$. Point D is on side $AB$, and $\angle BDC=45^\circ$. What is the length of segment $BD$?

Solution:
\textbf{Solution:}

Construct $CE\perp AB$ at point E,

since $\angle BDC=45^\circ$,

therefore $\angle DCE=90^\circ-\angle CDE=45^\circ=\angle CDE$,

therefore $DE=CE$,

since $AC=5$, and $\cot A=2$,

let $CE=x$, and $\cot A=\frac{AE}{CE}=2$,

therefore, $AE=2CE=2x$,

in $\triangle ACE$, according to Pythagoras' theorem $AC^2=CE^2+AE^2$ hence $5^2=x^2+(2x)^2$,

solving for $x$ gives $x=\sqrt{5}$,

therefore, $DE=CE=\sqrt{5}$,

in $\triangle CEB$, $\cot B=\frac{EB}{CE}=3$,

therefore, $BE=3CE=3\sqrt{5}$,

therefore, $BD=DE+BE=\sqrt{5}+3\sqrt{5}=4\sqrt{5}$;

Hence, the length of segment $BD$ is $BD=\boxed{4\sqrt{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $AC=5$, $\cot A=2$, $\cot B=3$. Point $D$ lies on side $AB$, and $\angle BDC=45^\circ$. If we let $\overrightarrow{CA}=a$ and $\overrightarrow{CB}=b$, then $\overrightarrow{AB}=\square$, $\overrightarrow{AD}=\square$, $\overrightarrow{CD}=\square$ (expressed in terms of $a$ and $b$).

Solution:
\textbf{Solution:} From the given information, we have $ \overrightarrow{CA}=a, \overrightarrow{CB}=b$,

since $ \overrightarrow{AB}=\overrightarrow{AC}+\overrightarrow{CB}$ and $ \overrightarrow{AC}=-\overrightarrow{CA}$,

thus $ \overrightarrow{AB}=\overrightarrow{CB}-\overrightarrow{CA}=b-a$,

since AD=AE−DE=2DE−DE=DE and BD=4DE,

thus AB=AD+BD=5AD,

thus AD=$ \frac{1}{5}AB$,

thus $ \overrightarrow{AD}=\frac{1}{5}\overrightarrow{AB}=\frac{1}{5}(b-a)=\frac{1}{5}b-\frac{1}{5}a$,

thus $ \overrightarrow{CD}=\overrightarrow{CA}+\overrightarrow{AD}=a+\frac{1}{5}b-\frac{1}{5}a=\frac{4}{5}a+\frac{1}{5}b$.

Therefore, the answers are: $ \boxed{b-a}$, $ \boxed{\frac{1}{5}b-\frac{1}{5}a}$, $ \boxed{\frac{4}{5}a+\frac{1}{5}b}$.


Question: As shown in the diagram, in $\triangle ABC$, $\angle ACB=90^\circ$, $CD\perp AB$ at $D$, draw $CE\parallel AB$ through point $C$, and draw $AE\parallel CD$ through point $A$, with the two lines intersecting at point $E$. Connect $DE$. Quadrilateral $AECD$ is a rectangle. If $BD=4\sqrt{5}$ and $\sin\angle ACE=\frac{2\sqrt{5}}{5}$, what is the length of $DE$?

Solution:
Solution: Given that the quadrilateral AECD is a rectangle,

$\therefore \angle DCE=\angle AEC=90^\circ$, AC=DE,

$\because \angle ACB=90^\circ$,

$\therefore \angle DCB+\angle ACD=90^\circ$,

$\because \angle ACE+\angle ACD=90^\circ$,

$\therefore \angle BCD=\angle ACE$,

$\because \sin\angle ACE=\frac{2\sqrt{5}}{5}$,

$\therefore \sin\angle BCD=\frac{2\sqrt{5}}{5}$,

$\because CD\perp AB$,

$\therefore \angle CDB=90^\circ$,

$\because BD=4\sqrt{5}$,

$\therefore BC=10, CD=2\sqrt{5}$,

$\therefore AE=CD=2\sqrt{5}$,

In $\triangle ACE$,

$\angle AEC=90^\circ, \sin\angle ACE=\frac{2\sqrt{5}}{5}$,

$\therefore AE=2\sqrt{5}, AC=5$,

$\therefore DE=AC=\boxed{5}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ is the altitude on side $BC$, $AE$ is the median on side $BC$, $\angle C=45^\circ$, $\sin B=\frac{2}{3}$, and $AD=4$. What is the length of $BC$?

Solution:
\textbf{Solution:} Given $\triangle ABC$, since AD is the height from A to side BC,\\
thus $\angle ADB = \angle ADC = 90^\circ$.\\
In $\triangle ADC$, since $\angle ADC = 90^\circ$, $\angle C = 45^\circ$, and AD = 4,\\
thus DC = AD = 4.\\
In $\triangle ADB$, since $\angle ADB = 90^\circ$, $\sin B = \frac{2}{3}$, and AD = 4,\\
thus AB = $\frac{AD}{\sin B} = 6$\\
thus BD = $\sqrt{AB^2 - AD^2} = 2\sqrt{5}$,\\
thus BC = BD + DC = $2\sqrt{5} + 4 = \boxed{2\sqrt{5} + 4}$


Question: As illustrated, in $\triangle ABC$, $AD$ is the altitude to side $BC$, $AE$ is the median to side $BC$, $\angle C=45^\circ$, $\sin B=\frac{2}{3}$, and $AD=4$. What is the value of $\tan \angle DAE$?

Solution:
\textbf{Solution:} Since AE is the median to side BC,\\
hence CE = $\frac{1}{2}$BC = $\sqrt{5}+2$,\\
therefore DE = CE - CD = $\sqrt{5} - 2$,\\
thus $\tan \angle DAE = \frac{DE}{AD} = \boxed{\frac{\sqrt{5}-2}{4}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AB=5$, and $BC=3$. Point P starts from point A and moves along $AC-CB-BA$ at a speed of 2 units per second until it stops at point A. When point P does not coincide with the vertices of $\triangle ABC$, a perpendicular line is drawn from point P to the side it lies on, intersecting another side of $\triangle ABC$ at point Q. Let the movement time of point P be t seconds. What is the length of side $AC$?

Solution: \textbf{Solution:} Due to the insufficient information provided by the problem, it is impossible to directly determine the length of side $AC$. Therefore, specific solution steps and results cannot be provided.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AB=5$, and $BC=3$. Point $P$ starts from point $A$ and moves along $AC-CB-BA$ at a speed of 2 units per second until it stops at point $A$. When point $P$ does not coincide with the vertices of $\triangle ABC$, a perpendicular line is drawn from point $P$ to the side it is on, intersecting another side of $\triangle ABC$ at point $Q$. Let the movement time of point $P$ be $t$ seconds. When point $P$ is moving along the right-angle sides of $\triangle ABC$, what is the distance from point $P$ to side $AB$? (Expressed in an algebraic equation containing $t$)

Solution:
\textbf{Solution:} Let the distance from point P to side $AB$ be $h$.
\begin{enumerate}
    \item When point P moves on side $AC$, draw $PH\perp AB$ at H, as shown in figure 1:\\
    $\because \sin A=\frac{PH}{AP}=\frac{BC}{AB}=\frac{3}{5}$, $AP=2t$,\\
    $\therefore PH=h=AP\cdot \sin A=2t\times \frac{3}{5}=\boxed{\frac{6}{5}t}$;
    \item When point P moves on side $BC$, draw $PH\perp AB$ at H, as shown in figure 2:\\
    $\because \sin B=\frac{PH}{PB}=\frac{AC}{AB}=\frac{4}{5}$, $PB=7-2t$,\\
    $\therefore PH=h=PB\cdot \sin B=(7-2t)\times \frac{4}{5}=\boxed{-\frac{8}{5}t+\frac{28}{5}}$;
\end{enumerate}
In conclusion, the distance from point P to side $AB$ is $\boxed{\frac{6}{5}t}$ or $\boxed{-\frac{8}{5}t+\frac{28}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AB=5$, $BC=3$. Point P starts from point A and moves at a speed of 2 units per second along the path $AC-CB-BA$ until it stops at point A. When point P does not coincide with the vertices of $\triangle ABC$, a perpendicular line is drawn from point P to the edge it is on, intersecting another side of $\triangle ABC$ at point Q. Let the movement time of point P be $t$ seconds. When point Q is on the right-angled sides of $\triangle ABC$ and $PQ=\frac{3}{2}$, find the value of $t$.

Solution:
\textbf{Solution}:

From the problem statement, we get:

(1) When point $Q$ is on side $AC$, $AP=12-2t$,

then $PQ=\frac{3}{2}=AP\cdot \tan A$,

which means $\frac{3}{2}=(12-2t)\times \frac{3}{4}$,

solving this yields: $t=\boxed{5}$.

(2) When point $Q$ is on side $BC$, $BP=2t-7$,

then $PQ=\frac{3}{2}=BP\cdot \tan B$,

which means $\frac{3}{2}=(2t-7)\times \frac{4}{3}$,

solving this yields: $t=\boxed{\frac{65}{16}}$;

In summary, if $PQ=\frac{3}{2}$, the value of $t$ is either \boxed{5} or \boxed{\frac{65}{16}}.


Question: As shown in the figure, in $\triangle ABC$, $\angle C = 90^\circ$, $AB = 5$, and $BC = 3$. Point P starts from point A and moves along $AC-CB-BA$ with a velocity of 2 units per second, and stops at point A. When point P does not coincide with the vertices of $\triangle ABC$, a perpendicular line is drawn from point P to the side it is on, intersecting another side of $\triangle ABC$ at point Q. Let the motion time of point P be $t$ seconds. When one vertex of $\triangle APQ$ is equidistant from the hypotenuse and one leg of $\triangle ABC$, write down the value of $t$ directly.

Solution: \textbf{Solution:} The value of $t$ can be directly written as $t=\boxed{\frac{5}{4}}$, $t=\boxed{\frac{8}{3}}$, $t=\boxed{4}$, and $t=\boxed{5}$.


Question: As shown in the figure, the parabola $y=ax^{2}+2ax+c$ intersects the x-axis at points $A$ and $B$ (where point $A$ is to the left of point $B$), and intersects the y-axis at point $C(0, 3)$. The y-coordinate of its vertex $D$ is $4$. What is the expression of this parabola?

Solution:
\textbf{Solution:} Substitute the point $C(0, 3)$ into $y=ax^{2}+2ax+c$ to get: $c=3$\\
$\therefore y=ax^{2}+2ax+3$\\
$\therefore -\frac{b}{2a}=-\frac{2a}{2a}=-1$\\
When $x=-1$, $\therefore y=a-2a=-a+3$\\
Since the ordinate of the vertex D is $4$,\\
$\therefore -a+3=4$\\
$a=-1$\\
$\therefore D(-1, 4)$\\
Therefore, the equation of the parabola is $y=-x^{2}-2x+3=\boxed{-x^{2}-2x+3}$


Question: As shown in the figure, the parabola $y=ax^2+2ax+c$ intersects the x-axis at points $A$ and $B$ (with point $A$ to the left of point $B$) and intersects the y-axis at point $C(0, 3)$. The y-coordinate of its vertex $D$ is $4$. What is the value of the tangent of $\angle ACB$?

Solution:
\textbf{Solution:} Draw $BE\perp AC$ at point E,\\
Let $y=-x^2-2x+3=0$\\
Then $x^2+2x-3=0$\\
$\therefore x_1=-3$, $x_2=1$\\
Hence $A(-3,0)$, $B(1,0)$\\
$\therefore AB=1-(-3)=4$, $OC=3$\\
$\therefore S_{\triangle ABC}=\frac{1}{2}\cdot AB\cdot OC=\frac{1}{2}\times 4\times 3=6$\\
$\because A(-3, 0)$, $C(0, 3)$\\
$\therefore AC=\sqrt{(-3-0)^2+(0-3)^2}=3\sqrt{2}$\\
$\because S_{\triangle ABC}=\frac{1}{2}\cdot AC\cdot BE=\frac{1}{2}\times 3\sqrt{2}\times BE$\\
$\therefore \frac{1}{2}\times 3\sqrt{2}\times BE=6$\\
$\therefore BE=2\sqrt{2}$\\
$\because BC^2=OC^2+OB^2=3^2+1^2=10$\\
$\therefore EC=\sqrt{BC^2-BE^2}=\sqrt{10-8}=\sqrt{2}$\\
$\therefore \tan\angle ACB=\frac{BE}{EC}=\frac{2\sqrt{2}}{\sqrt{2}}=\boxed{2}$


Question: As shown in the figure, the parabola $y=ax^2+2ax+c$ intersects the x-axis at points $A$ and $B$ (with point $A$ on the left side of point $B$), and intersects the y-axis at point $C(0, 3)$. The y-coordinate of its vertex $D$ is $4$. Point $F$ is located on the extension line of segment $CB$, and $\angle AFC = \angle DAB$. What is the length of $CF$?

Solution:
\textbf{Solution:} Draw $DM\perp x$-axis through point D, and draw $AN\perp BC$ through point A,\\
$\because D(-1, 4)$, $\because M(-1, 0)$,\\
$\because A\left(-3, 0\right)$,\\
$\therefore AM=2$, $DM=4$,\\
$\therefore \tan\angle DAB=\frac{DM}{AM}=\frac{4}{2}=2$,\\
$\because AN\perp BC$,\\
$\therefore \angle ANB=90^\circ$,\\
$\therefore \angle ANB=\angle BOC$,\\
$\because \angle ABN=\angle OBC$,\\
$\therefore \triangle OBC\sim \triangle NBA$,\\
$\therefore \frac{OB}{NB}=\frac{OC}{NC}=\frac{BC}{BA}$,\\
$\therefore \frac{1}{NB}=\frac{3}{NC}=\frac{\sqrt{10}}{4}$,\\
$NB=\frac{2\sqrt{10}}{5}$, $NA=\frac{6\sqrt{10}}{5}$,\\
$\because \angle DAB=\angle AFC$,\\
$\therefore \tan\angle DAB=\tan\angle AFC=2$,\\
$\therefore \tan\angle AFC=\frac{AN}{NF}=2$,\\
$\therefore \frac{\frac{6\sqrt{10}}{5}}{NF}=2$,\\
$NF=\frac{3\sqrt{10}}{5}$,\\
When point F is on the extension of CB, F can only be in the fourth quadrant, hence $CF=NF+CN$,\\
$\because BC=\sqrt{10}NB=\frac{2\sqrt{10}}{5}$,\\
$\because CN=BC-NB=\sqrt{10}-\frac{2\sqrt{10}}{5}=\frac{3\sqrt{10}}{5}$,\\
$\therefore CF=\frac{3\sqrt{10}}{5}+\frac{3\sqrt{10}}{5}=\boxed{\frac{6\sqrt{10}}{5}}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $\angle C = 90^\circ$, $AB = 4$, and $AC = \sqrt{7}$. Find the length of $BC$.

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


Question: Known in $\triangle ABC$, $\angle C=90^\circ$, $AB=4$, $AC=\sqrt{7}$. Find $\sin A$.

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


Question: As shown in Figure (1), $BC$ is the diameter of $\odot O$, and points $D, F$ are on $\odot O$. Connect $CD, BF$ and extend them to intersect at point $A$, with $AD=2CD=8$, $AF=3BF$. $\triangle AFD\sim \triangle ACB$. Find: $\cos\angle ADF$?

Solution:
\textbf{Solution:} Connect $CF$,\\
since $BC$ is the diameter of $\odot O$,\\
therefore $\angle BFC=90^\circ$,\\
from (1) we know $\triangle AFD\sim \triangle ACB$,\\
therefore $\frac{AF}{AC}=\frac{AD}{AB}$,\\
since $AD=2CD=8$,\\
therefore $CD=4$,\\
since $AF=3BF$,\\
therefore $AB=AF+BF=3BF+BF=4BF$,\\
thus $\frac{3BF}{12}=\frac{8}{4BF}$,\\
therefore $BF=2\sqrt{2}$,\\
therefore $AF=6\sqrt{2}$,\\
in $Rt\triangle ACF$, $CF=\sqrt{AC^{2}-AF^{2}}=\sqrt{12^{2}-(6\sqrt{2})^{2}}=6\sqrt{2}$,\\
in $Rt\triangle BCF$, $BC=\sqrt{BF^{2}+CF^{2}}=\sqrt{(2\sqrt{2})^{2}+(6\sqrt{2})^{2}}=4\sqrt{5}$,\\
therefore $\cos\angle ADF=\cos\angle B=\frac{BF}{BC}=\frac{2\sqrt{2}}{4\sqrt{5}}=\boxed{\frac{\sqrt{10}}{10}}$


Question: As shown in the figure, $BC$ is the diameter of $\odot O$, points $D$ and $F$ are on $\odot O$. Connect $CD$ and $BF$ and extend them to intersect at point $A$, where $AD=2CD=8$, $AF=3BF$. If point $E$ is the midpoint of arc $BC$, connect $BE$ and $DE$. Find the value of $EG\cdot EF$?

Solution:
\textbf{Solution:} \\
Since $E$ is the midpoint of the arc $BC$, \\
hence $\overset{\frown}{BE}=\overset{\frown}{CE}$, \\
therefore $\angle BFE=\angle CBE$, \\
also since $\angle GEB=\angle BEF$, \\
hence $\triangle EBG\sim \triangle EFB$, \\
thus $\frac{BE}{EF}=\frac{EG}{BE}$, \\
that is $EG\cdot EF=BE^{2}$, \\
since $\overset{\frown}{BE}=\overset{\frown}{CE}$ again, \\
hence $BE=CE$, \\
since $BE^{2}+CE^{2}=BC^{2}$, \\
that is $2BE^{2}=BC^{2}$, \\
therefore $BE^{2}=\frac{1}{2}BC^{2}=\frac{1}{2}\times \left(4\sqrt{5}\right)^{2}=40$, \\
therefore $EG\cdot EF=\boxed{40}$.


Question: As shown in the figure, $AB$ is the diameter of the circle $\odot O$, $C$ is a point on $\odot O$, line $AD$ is perpendicular to the line passing through point $C$ with the foot of the perpendicular being $D$, and $AC$ bisects $\angle DAB$. $DC$ is a tangent line of $\odot O$; If the radius of $\odot O$ is 2, and $AC=2\sqrt{3}$, what is the length of the line segment $AD$?

Solution:
\textbf{Solution:} Connect $BC$,

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

Therefore, $\angle ACB=90^\circ$,

Since $\angle DAC=\angle OAC$, $\angle ADC=\angle ACB=90^\circ$,

Therefore, $\triangle ADC\sim \triangle ACB$,

Therefore, $\frac{AD}{AC}=\frac{AC}{AB}$,

Since the radius of $\odot O$ is 2, $AC=2\sqrt{3}$,

Therefore, $\frac{AD}{2\sqrt{3}}=\frac{2\sqrt{3}}{4}$,

Therefore, $AD=\boxed{3}$;


Question: As shown in the diagram, in $ \triangle ABC$, $ \angle ACB=90^\circ$, $CD\perp AB$ with the foot of the perpendicular at D, $AC=4$, $BC=3$. What is the length of $BD$?

Solution:
Solution: In right-angled triangle $ABC$, we have $\angle ACB=90^\circ$, $AC=4$, and $BC=3$,\\
therefore, $AB=\sqrt{AC^2+BC^2}=5$,\\
since $CD\perp AB$,\\
and $\cos B=\frac{BD}{BC}=\frac{BC}{AB}$,\\
it follows that $\frac{BD}{3}=\frac{3}{5}$,\\
thus, $BD=\boxed{\frac{9}{5}}$;


Question: As illustrated in the diagram, in $\triangle ABC$, $\angle ACB=90^\circ$, $CD\perp AB$ with the foot of the perpendicular being D, $AC=4$, and $BC=3$. Find the tangent value of $\angle ACD$.

Solution:
Solution: Since $\angle ACD + \angle BCD = 90^\circ$ and $\angle BCD + \angle B = 90^\circ$,\\
it follows that $\angle ACD = \angle B$,\\
thus $\tan \angle ACD = \tan \angle B = \frac{AC}{BC} = \boxed{\frac{4}{3}}$.


Question: In the Cartesian coordinate system, the graph of the linear function $y=ax+b$ ($a\neq 0$) intersects the graph of the inverse proportion function $y=\frac{k}{x}$ ($k\neq 0$) at points $A$ and $B$ within the second and fourth quadrants, and intersects the $y$-axis at point $C$. A perpendicular line is drawn from point $A$ to the $y$-axis, with $H$ as the foot, where $OH=3$, $\tan\angle AOH =\frac{4}{3}$, and the coordinates of point $B$ are ($m$, $-2$). Calculate the perimeter of $\triangle AHO$.

Solution:
\textbf{Solution:} Given that $\because$ OH $=$ 3, $\tan\angle AOH =\frac{4}{3}$, and $AH\perp$ y-axis\\
$\therefore$ $\frac{AH}{OH}=\frac{4}{3}$ \\
$\therefore$ $AH=\frac{4OH}{3}=4$ \\
$\therefore$ $AO=\sqrt{AH^{2}+OH^{2}}=5$ \\
$\therefore$ the perimeter of $\triangle AHO$ is $AH+OH+AO=4+3+5=\boxed{12}$;


Question: In the Cartesian coordinate system, the graph of the linear function $y=ax+b$ ($a\neq 0$) intersects with the graph of the inverse proportion function $y=\frac{k}{x}$ ($k\neq 0$) at points $A$ and $B$ within the second and fourth quadrants, respectively, and with the $y$-axis at point $C$. Drawing a line perpendicular to the $y$-axis from point $A$ to point $H$ such that $AH \perp y$-axis and the foot of the perpendicular is $H$, with $OH=3$ and $\tan\angle AOH =\frac{4}{3}$. The coordinates of point $B$ are ($m$, $-2$). Determine the analytical expressions of the inverse proportion function and the linear function.

Solution:
\textbf{Solution:} Let $\because$OH$=3$, $AH=4$, and point A is in the second quadrant\\
$\therefore$ $A\left(-4, 3\right)$ \\
$\because$ the graph of the inverse proportion function $y=\frac{k}{x}$ (where $k\neq 0$) passes through point A\\
$\therefore$ $3=\frac{k}{-4}$\\
$\therefore$ $k=-12$ \\
$\therefore$ the equation of the inverse proportion function is: $y=-\frac{12}{x}$;\\
$\because$ the graph of the inverse proportion function passes through point B, and the coordinates of point B are $(6, -2)$\\
$\therefore$ $-2=-\frac{12}{6}$\\
$\therefore$ $m=6$ \\
Upon verification, $m=6$ is a solution to $-2=-\frac{12}{m}$;\\
$\therefore$ $B\left(6, -2\right)$ \\
$\because$ the graph of the linear function $y=ax+b$ (where $a\neq 0$) passes through points A and B\\
$\therefore$ $\left\{\begin{array}{l}-4a+b=3 \\ 6a+b=-2\end{array}\right.$ \\
$\therefore$ $\left\{\begin{array}{l}a=-\frac{1}{2} \\ b=1\end{array}\right.$ \\
$\therefore$ the equation of the linear function is: $\boxed{y=-\frac{1}{2}x+1}$.


Question: As shown in the figure, it is known that the diameter $AB$ of $\odot O$ is perpendicular to the chord $CD$, with the foot of the perpendicular being point $E$. Also, $AD=5$, and $\cos\angle BCD=\frac{4}{5}$. What is the length of the chord $CD$?

Solution:
\textbf{Solution:} Since $AB \perp CD$,\\
thus $CE = DE = \frac{1}{2}CD$,\\
Since $\cos\angle BCD = \frac{4}{5}$,\\
thus $\cos\angle A= \cos\angle BCD = \frac{4}{5}$,\\
In $\triangle ADE$, $AD=5$, and $\cos\angle A= \frac{4}{5}$,\\
thus $AE=4$,\\
thus $DE=3$, then $CD=\boxed{6}$.


Question: As shown in the figure, it is known that the diameter $AB$ of the circle $\odot O$ and chord $CD$ are perpendicular to each other at point $E$. Also, $AD=5$ and $\cos\angle BCD=\frac{4}{5}$. Find the radius of $\odot O$.

Solution:
\textbf{Solution:} Let us connect $OD$, with the radius being $r$, thus $OE = 4 - r$, $OD = r$.\\
In $\triangle BCE$, we have $OE^{2} + DE^{2} = OD^{2}$,\\
$\therefore$ $(4-r)^{2} + 9 = r^{2}$,\\
$\therefore$ $r = \boxed{\frac{25}{8}}$.


Question: In $\triangle ABC$, $AB=6$, $BC=4$, and $\angle B$ is an acute angle with $\cos B=\frac{1}{2}$. What is the measure of $\angle B$ in degrees?

Solution:
\textbf{Solution:} Since $\angle B$ is an acute angle and $\cos B = \frac{1}{2}$,\\
it follows that $\angle B = \boxed{60^\circ}$;


Question: In $\triangle ABC$, $AB=6$, $BC=4$, and $\angle B$ is an acute angle with $\cos B=\frac{1}{2}$. Find the tangent of $\angle C$.

Solution:
\textbf{Solution:} Let $AD \perp BC$ at $D$, as shown in the figure:\\
Since $\cos B = \frac{1}{2}$,\\
it follows that $\frac{BD}{AB} = \frac{1}{2}$,\\
Given $AB = 6$,\\
then $BD = \frac{1}{2}AB = 3$,\\
thus $AD = \sqrt{AB^{2} - BD^{2}} = 3\sqrt{3}$,\\
Given $BC = 4$, and $BD = 3$,\\
then $CD = BC - BD = 1$,\\
therefore, $\tan C = \frac{AD}{CD} = \frac{3\sqrt{3}}{1} = \boxed{3\sqrt{3}}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $\angle ABC=90^\circ$, the circle $\odot O$ with diameter $AB$ intersects $AC$ at point $D$, and point $E$ is the midpoint of $BC$, connect $BD$, $DE$. If $\frac{AD}{AB}=\frac{1}{3}$, find $\sin C$.

Solution:
\textbf{Solution:} From the given information, we know that $AB$ is a diameter,\\
$\therefore \angle ADB=90^\circ$,\\
$\therefore \angle ABD + \angle BAD=90^\circ$,\\
$\because \angle ABC=90^\circ$,\\
$\therefore \angle C + \angle BAC=90^\circ$,\\
$\therefore \angle C = \angle ABD$,\\
$\because \frac{AD}{AB}=\frac{1}{3}$,\\
$\therefore \sin \angle ABD=\frac{1}{3}$,\\
$\therefore \sin C = \boxed{\frac{1}{3}}$;


Question: Given that, in the Cartesian coordinate system, for square $ABCD$, point $A(0,10)$, point $B(8,4)$, and point $C$ is in the first quadrant. Point $P$ starts from vertex $A$ and moves counterclockwise along the sides of the square for one round. Construct $\angle OPQ = \angle OAB$ on the right side of $OP$ intersecting the $x$-axis at point $Q$. Find the coordinates of point $C$.

Solution: \textbf{Solution:} As shown in Figure 1, since quadrilateral ABCD is a square, it follows that $\angle ABC=90^\circ$, and AB=BC. Therefore, $\angle ABE+\angle CBF=90^\circ$. Draw $BE\perp OA$ at E, and draw $CF\perp BE$ at F. Thus, $\angle AEB=\angle ABC=\angle F=90^\circ$, which implies that $\angle EAB+\angle ABE=90^\circ$. Therefore, $\angle EAB=\angle CBF$. Consequently, $\triangle ABE\cong \triangle BCF$ (by AAS). Therefore, BF=AE, CF=BE. Given that point A is at (0,10) and point B is at (8,4), we deduce AE=6, BE=8, OE=4. Therefore, BF=6, CF=8. Hence, EF=BE+BF=14 and CF+OE=12. Thus, the coordinates of point C are \boxed{(14,12)}.


Question: Given, in the Cartesian coordinate system, square $ABCD$, with point $A(0,10)$, point $B(8,4)$, and point $C$ in the first quadrant. Point $P$ starts from vertex $A$ and moves counterclockwise along the edges of the square for one round. On the right side of $OP$, make $\angle OPQ = \angle OAB$ intersecting the x-axis at point $Q$. When point $P$ coincides with point $B$, find the length of $OQ$.

Solution:
\textbf{Solution:} As shown in Figure 2, draw $BE\perp OQ$ at $E$, and $QF\perp OB$ at $F$, \\
since $B(8,4)$, \\
therefore $OB=\sqrt{8^2+4^2}=4\sqrt{5}$, $\tan\angle BOE=\frac{BE}{OE}=\frac{1}{2}$, \\
$\tan\angle OBQ=\tan\angle OAB=\frac{BG}{AG}=\frac{8}{6}=\frac{4}{3}$, \\
thus $\frac{QF}{BF}=\frac{4}{3}$, $\frac{QF}{OF}=\frac{1}{2}$. Let $FQ=4k$, $BF=3k$, $OF=8k$, since $OF+BF=OB$, \\
thus $8k+3k=4\sqrt{5}$, \\
thus $k=\frac{4\sqrt{5}}{11}$, \\
thus $OF=\frac{32\sqrt{5}}{11}$, $FQ=\frac{16\sqrt{5}}{11}$, \\
thus $OQ=\sqrt{OF^2+FQ^2}=\boxed{\frac{80}{11}}$;


Question: Since you have not provided a specific text to translate, I'm unable to fulfill this request accurately. Please provide the text you want translated to English in standard LaTeX format, and I would be happy to assist.

Solution:
\textbf{Solution:} Let point P be on AB, as shown in Figure 3. When $PO=PQ$, draw $PE\perp x$-axis at E, draw $PF\perp OA$ at F\\
$\therefore \angle OPE= \frac{1}{2}\angle OPQ = \frac{1}{2}\angle OAB$,\\
$\because PE\parallel OA$,\\
$\therefore \angle AOP=\angle OPE= \frac{1}{2} \angle OAB$,\\
Construct $\angle OPG=\angle AOP$,\\
$\therefore \angle APG=2\angle AOP=\angle OAB$,\\
$\because \tan\angle OAB= \frac{PF}{AF} = \frac{4}{3}$,\\
$\therefore$ let $PF=4a$, $AF=3a$, $OG=PG=AP=5a$,\\
$\therefore OF=OG+FG=8a$,\\
$\therefore P(4a,8a)$, $B(8,4)$,\\
$\therefore$ the equation of line AB is: $y= -\frac{3}{4}x+10$,\\
$\therefore -\frac{3}{4} \times 4a+10=8a$,\\
$\therefore a= \frac{10}{11}$,\\
$\therefore PF=4a= \frac{40}{11}$,\\
$\therefore OQ=2PF= \boxed{\frac{80}{11}}$,\\
As shown in Figure 4, when $OQ=PQ$, draw $PE\perp OP$ at E,\\
$\therefore \angle POQ=\angle OPQ=\angle OAB$, $OE= \frac{1}{2} OP$,\\
$\because \angle AOP+\angle POQ=90^\circ$,\\
$\therefore \angle OAB+\angle AOP=90^\circ$,\\
$\therefore \angle APO=90^\circ$,\\
$\because \tan\angle OAB= \frac{4}{3}$,\\
$\therefore$ let $AP=3a$, $OP=4a$,\\
$\therefore OA=5a=10$,\\
$\therefore a=2$,\\
$\therefore OP=4a=8$,\\
$\therefore OE=4$,\\
$\because \frac{OQ}{OE} = \frac{5}{3}$,\\
$\therefore OQ= \boxed{\frac{20}{3}}$,\\
When P is on CD, as in Figure 5, when $OQ=PQ$, draw $QE\perp OP$ at E,\\
From above, we know: $OP\perp CD$, $OP=8+10=18$,\\
$\therefore OE=9$,\\
$\therefore OQ=9\times \frac{5}{3} =\boxed{15}$,\\
As in Figure 6, when $PQ=PO$, draw $PE\perp OQ$ at E,\\
$\because$ line AB: $y= -\frac{3}{4}x+10$, $CD\parallel AB$, $C(14,12)$,\\
$\therefore$ the equation of line CD is: $y= -\frac{3}{4}x+ \frac{45}{2}$,\\
Let $P(a,2a)$,\\
$\therefore -\frac{3}{4} \times a+ \frac{45}{2} =2a$,\\
$\therefore 2a= \frac{180}{11}$,\\
$\therefore OQ=2a= \boxed{\frac{180}{11}}$,\\
Overall, during the entire movement of point P, when $\triangle OPQ$ is an isosceles triangle with $PQ$ as the base, $OQ= \boxed{\frac{80}{11}}$ or $\boxed{\frac{20}{3}} or \boxed{15} or \boxed{\frac{180}{11}}$.


Question: During a project learning session, the teacher asked the students to collaborate on creating questions. Inspired by the Chord Diagram of Zhao Shuang, a study group came up with the following problem for you to solve:

Solution:
\textbf{Solution:} Let $AE=x$, then $BE=DG=x+1$. In $\triangle BEF$, $\angle FEB=45^\circ$,\\
$\therefore$ BE=FB,\\
$\because$ BF=DH=x+1,\\
$\therefore$ AH=x+1+1=x+2,\\
Also, $\because$ $\tan\angle AEH=2$,\\
$\therefore$ $\frac{AH}{AE}=2$,\\
$\therefore$ AH=2AE=2x,\\
$\therefore$ 2x=x+2,\\
Solving, we get $x=2$,\\
$\therefore$ AE=\boxed{2}.


Question: As shown in the figure, semicircle $O$ with diameter $AB$ of $\triangle ABC$ intersects sides $AC$ and $BC$ at points $E$ and $D$, respectively, and $D$ is the midpoint of arc $\overset{\frown}{BE}$. If $\angle A=80^\circ$, what is the measure of $\angle DBE$?

Solution:
\textbf{Solution:} Draw $OD$, intersect $BE$ at point $H$, connect $ED$,

$\because D$ is the midpoint of $\overset{\frown}{BE}$,

$\therefore \overset{\frown}{DE}=\overset{\frown}{DB}$,

$OD\perp BE$, $BH=EH$,

$\therefore DE=DB$, $\angle DEB=\angle DBE$,

$\because AB$ is the diameter, then $\angle AEB=90^\circ$,

$\therefore \angle BEC=90^\circ$,

$\therefore \angle CBE+\angle BCE=90^\circ$,

$\angle BED+\angle CED=90^\circ$,

$\therefore \angle CBE+\angle CED=90^\circ$,

$\therefore \angle C=\angle CED$,

$\therefore DC=DE$,

$\therefore DC=DB$,

$\because OA=OB$,

$\therefore OD$ is the median of $\triangle ABC$,

$\therefore OD \parallel AC$,

$\therefore \angle DOB=\angle CAB$,

$\angle ODB=\angle C$,

$\because \angle CAB=80^\circ$, then $\angle DOB=80^\circ$,

$\therefore \angle ODB=\angle OBD=\frac{1}{2}(180^\circ-80^\circ)=50^\circ$,

Also, $\angle OBE=90^\circ-80^\circ=10^\circ$,

Then, $\angle DBE=\angle OBD-\angle OBE=50^\circ-10^\circ=\boxed{40^\circ}$.


Question: As shown in the diagram, with the side $AB$ of $\triangle ABC$ as the diameter, the semicircle $O$ intersects sides $AC$ and $BC$ at points $E$ and $D$, respectively, and $D$ is the midpoint of the arc $\overset{\frown}{BE}$. Given $AB=AC$, if the radius of $\odot O$ is $5cm$ and $BC=12cm$, what is the length of segment $BE$ in $cm$?

Solution:
\textbf{Solution:} Connect $AC$,\\
$\because AB=AC$,\\
$\therefore \triangle ABC$ is an isosceles triangle,\\
$\because AB$ is the diameter of $\odot O$,\\
$\therefore \angle ADB=90^\circ$, which means $AD\perp BC$,\\
$\because BC=12cm$,\\
$\therefore CD=BD=\frac{1}{2}BC=6cm$,\\
Also, $AC=2OD=10cm$,\\
By the Pythagorean theorem, $AD=\sqrt{AC^{2}-CD^{2}}=\sqrt{10^{2}-6^{2}}=8cm$,\\
$\therefore \sin C=\frac{AD}{AC}=\frac{8}{10}=\frac{4}{5}$,\\
$\therefore \sin\angle ODB=\sin C=\frac{4}{5}$,\\
$\therefore \frac{BH}{BD}=\frac{4}{5}$,\\
$\therefore BH=\frac{4}{5}BD=\frac{24}{5}$,\\
$\therefore BE=2BH=\boxed{\frac{48}{5}cm}$.


Question: As shown in the figure, in $\triangle ABC$, with a circle having diameter $BC$ intersecting $AB$ at point $D$, and $\angle ACD = \angle ABC$. If point $E$ is on $BC$, given $BE = 6$, $\cos\angle ABC = \frac{3\sqrt{13}}{13}$, $\tan\angle AEC = \frac{5}{3}$, and $CA$ is a tangent to the circle. What is the diameter of the circle?

Solution: \textbf{Solution:} In $\triangle AEC$, $\tan \angle AEC = \frac{5}{3}$, thus $\frac{AC}{EC} = \frac{5}{3}$, and $EC = \frac{3}{5} AC$. In $\triangle ABC$, $\cos \angle ABC = \frac{3\sqrt{13}}{13}$, thus $\frac{BC}{AB} = \frac{3\sqrt{13}}{13}$, and $\frac{AC}{BC} = \frac{2}{3}$, thus $BC = \frac{3}{2} AC$. Since $BC - EC = BE$ and $BE = 6$, it follows that $\frac{3}{2} AC - \frac{3}{5} AC = 6$. Solving this gives $AC = \frac{20}{3}$, thus $BC = \frac{3}{2} AC = \frac{3}{2} \times \frac{20}{3} = 10$. The diameter of the circle is $\boxed{10}$.


Question: As shown in the figure, in rectangle $ABCD$, $E$ is a point on $BC$, and $DF \perp AE$ at point $F$, with $\frac{AD}{AE} = \lambda$ ($\lambda > 0$). If $\lambda = 1$, prove that $CE=FE$.

Solution:
\textbf{Solution.} Proof: Draw $DE$ as shown in the figure:\\
Since quadrilateral $ABCD$ is a rectangle,\\
it follows that $\angle C=90^\circ$, and $AD\parallel BC$,\\
thus, $\angle ADE=\angle CED$,\\
since $DF\perp AE$,\\
it follows that $\angle DFE=90^\circ$,\\
thus, $\angle DFE=\angle C$,\\
since $\frac{AD}{AE}=\lambda=1$,\\
it follows that $AD=AE$,\\
thus, $\angle ADE=\angle FED$,\\
thus, $\angle FED=\angle CED$,\\
in $\triangle DFE$ and $\triangle DCE$,\\
$\left\{\begin{array}{l}
\angle DFE=\angle C \\
\angle FED=\angle CED \\
DE=DE
\end{array}\right.$,\\
it follows that $\triangle DFE\cong \triangle DCE$ (AAS),\\
thus, $CE=FE=\boxed{FE}$;


Question: As shown in the figure, in rectangle $ABCD$, $E$ is a point on $BC$, $DF\perp AE$ at point $F$, and let $\frac{AD}{AE}=\lambda$ ($\lambda>0$). If $AB=3$, $AD=4$, and $D$, $B$, $F$ are on the same straight line, find the value of $\lambda$.

Solution:
\textbf{Solution:} When points D, B, and F are on the same line as depicted:\\
Since quadrilateral $ABCD$ is a rectangle,\\
it follows that $\angle BAD=\angle ABC=90^\circ$,\\
In $\triangle ADB$, where $AB=3$ and $AD=4$,\\
it follows that $\tan\angle ABD=\frac{AD}{AB}=\frac{4}{3}$,\\
Since $DF\perp AE$,\\
it follows that $\angle BFE=90^\circ$,\\
Since $\angle ABD+\angle DBC=90^\circ$ and $\angle DBC+\angle FEB=90^\circ$,\\
it follows that $\angle FEB=\angle ABD$,\\
Hence $\frac{AB}{BE}=\tan\angle FEB=\tan\angle ABD=\frac{4}{3}$,\\
Since $AB=3$,\\
it follows that $BE=\frac{9}{4}$,\\
In $\triangle ABE$, by the Pythagorean theorem, $AE=\sqrt{3^2+\left(\frac{9}{4}\right)^2}=\frac{15}{4}$,\\
Thus, $\lambda=\frac{AD}{AE}=\frac{4}{\frac{15}{4}}=\boxed{\frac{16}{15}}$.


Question: As shown in the diagram, in rectangle $ABCD$, point $M$ lies on diagonal $BD$. A circle passing through points $A$, $B$, and $M$ intersects $BC$ at point $E$. If $AM=4$, $EB=EM=3$, find the length of $BM$.

Solution:
\textbf{Solution:} Let $AE$ intersect $BD$ at point $G$. Consider the circle passing through points $A$, $B$, and $M$, denoted as $\odot O$. In rectangle $ABCD$, we have $\angle ABC=\angle BAD=90^\circ$, \\
since $\angle ABC=90^\circ$, \\
thus, chord $AE$ is the diameter of $\odot O$,\\
therefore, $\angle AME=90^\circ=\angle ABC$,\\
since $EB=EM=3$, $AE=AE$,\\
therefore, $Rt\triangle AEB\cong Rt\triangle AEM$,\\
therefore, $AB=AM$,\\
since $EB=EM=3$,\\
therefore, $\angle EAB=\angle EAM$, that is, $AG$ bisects $\angle BAM$,\\
therefore, by the "angle bisector theorem", $AG\perp BM$, $BG=GM$,\\
in $Rt\triangle AEM$, $EM=3$, $AM=4$,\\
therefore, $AE=\sqrt{AM^{2}+EM^{2}}=5$,\\
since $\text{Area}_{\triangle AME}=\frac{1}{2}\times AM\times EM=\frac{1}{2}\times AE\times GM$,\\
therefore, $GM=\frac{3\times 4}{5}=\frac{12}{5}$,\\
since $BG=GM$,\\
therefore, $BM=BG+GM=\boxed{\frac{24}{5}}$;


Question: As shown in the figure, in rectangle $ABCD$, point $M$ is on diagonal $BD$. The circle passing through points $A$, $B$, and $M$ intersects $BC$ at point $E$. Given that $AB = 6$, $BC = 8$, and the ratio of $AM$ to $ME$. If $BM = 7$, what is the length of $BE$?

Solution:
\textbf{Solution:} In $(1)$ we have already proved that chord $AE$ is the diameter of $\odot O$,\\
which gives $\angle AME=90^\circ$,\\
therefore, in $\triangle AEM$ right angled at $M$, $AM: ME=\tan\angle AEM$,\\
since $\angle AEM=\angle ABM$,\\
in $\triangle ABD$ right angled at $B$, $\tan\angle ABM=\frac{AD}{AB}$,\\
since in the rectangle $ABCD$, $AB=CD$, $AD=BC$, $AB=6$, $BC=8$\\
therefore, $\tan\angle ABM=\frac{AD}{AB}=\frac{8}{6}=\frac{4}{3}$,\\
therefore, $AM: ME=\tan\angle AEM=\tan\angle ABD=\frac{4}{3}$;\\
$(2)$ Let $AD$ intersect $\odot O$ at point $H$, connect $HM$ and $HE$, as shown in the diagram,\\
According to the property that opposite angles supplement each other in a cyclic quadrilateral,\\
we have: $\angle AHE=\angle ABE=90^\circ=\angle BEH=\angle BAH$, $\angle BMH=\angle BAH=90^\circ$,\\
therefore, quadrilateral $ABEH$ is a rectangle, hence $AH=BE$,\\
since $\angle BMH=\angle BAH=90^\circ$,\\
therefore, $\angle DMH=\angle BAD=90^\circ$,\\
since $\angle HDM=\angle BDA$,\\
therefore, $\triangle DMH\sim \triangle DAB$,\\
therefore, $\frac{DH}{DM}=\frac{DB}{DA}$, $DH=\frac{DB\times DM}{DA}$,\\
since $AB=6$, $AD=BC=8$,\\
therefore, $BD=\sqrt{AD^2+AB^2}=\sqrt{8^2+6^2}=10$,\\
since $BM=7$,\\
therefore, $DM=BD−BM=3$,\\
therefore, $DH=\frac{DB\times DM}{DA}=\frac{10\times 3}{8}=\frac{15}{4}$,\\
therefore, $AH=AD−DH=8−\frac{15}{4}=\frac{17}{4}$,\\
therefore, $BE=AH=\boxed{\frac{17}{4}}$.


Question: As shown in the figure, in $\triangle ABC$, circle $\odot O$ with diameter $BC$ intersects $AC$ at point $E$. A line through point $E$ is drawn such that $EF\perp AB$ at point $F$, and extending $EF$ to intersect the extension of $CB$ at point $G$, with $\angle ABG=2\angle C$. If $\sin\angle EGC=\frac{3}{5}$, and the radius of $\odot O$ is $3$, find the area of the shaded region in the figure. ($\sin 36^\circ=\frac{3}{5}$)

Solution: \textbf{Solution:} Let's consider Right triangles $\triangle EOG$ and $\triangle BFG$, where $\sin\angle EGC=\frac{OE}{OG}=\frac{BF}{BG}=\frac{3}{5}$. This means $\frac{3}{OG}=\frac{3}{5}$. Solving this, we get: $OG=5$, hence $BG=OG-OB=5-3=2$, implying $\frac{BF}{2}=\frac{3}{5}$, hence $BF=\frac{6}{5}$, and thus $EG=\sqrt{OG^2-OE^2}=\sqrt{5^2-3^2}=4$. Let $EF=x$, then $GF=4-x$. Since $BF\parallel OE$, we have $\frac{GB}{OB}=\frac{GF}{EF}$, that is $\frac{2}{3}=\frac{4-x}{x}$. Solving this, we get: $x=EF=\frac{12}{5}$. Given $\sin\angle EGC=\frac{3}{5}$, we find $\angle G=36^\circ$, hence $\angle EOB=90^\circ-36^\circ=54^\circ$. The area of the shaded region $S_{\text{shaded}}=S_{\text{trapezoid} BFEO}-S_{\text{sector} EOB}=\frac{1}{2}\left(\frac{6}{5}+3\right)\times \frac{12}{5}-\frac{54^\circ\pi \times 3^2}{360}=\boxed{\frac{126}{25}-\frac{27}{20}\pi}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, and D, E are two points on $\odot O$ located on opposite sides of AB. Extend BD to point C such that CD=BD, connect AC which intersects $\odot O$ at point F, and connect AE, DE, DF. $\angle E=\angle C$; If $\angle E=55^\circ$, what is the degree measure of $\angle BDF$?

Solution:
\textbf{Solution:}

Since the quadrilateral AEDF is inscribed in circle O,
\[\therefore \angle AFD=180^\circ-\angle E\]
Since \(\angle CFD=180^\circ-\angle AFD\),
\[\therefore \angle CFD=\angle E=55^\circ\]
From (1), we have \(\angle E=\angle C=55^\circ\),
\[\therefore \angle BDF=\angle C+\angle CFD=55^\circ+55^\circ=\boxed{110^\circ}\]


Question: As shown in the figure, $AB$ is the diameter of circle $O$, and $D,E$ are two points on circle $O$ on opposite sides of $AB$. Connect $BD$ and extend it to point $C$ such that $CD=BD$. Connect $AC$ and let it intersect circle $O$ at point $F$. Connect $AE$, $DE$, $DF$. Suppose $DE$ intersects $AB$ at point $G$, and if $AB=10$ and $E$ is the midpoint of arc $AEB$, what is the value of $EG\cdot ED$?

Solution:
\textbf{Solution:} Since $E$ is the midpoint of $\overset{\frown}{AEB}$, it follows that $AE=BE$\\
Therefore, $\angle EAB=\angle EBA=45^\circ$\\
Also, since $\angle EAB$ and $\angle EDB$ are the angles subtended by arc $\widehat{BE}$ on the circle, it follows that $\angle EAB=\angle EDB=45^\circ$\\
Thus, $\angle EDB=\angle EBA=45^\circ$\\
Additionally, since $\angle DEB=\angle BEG$\\
It follows that $\triangle DEB \sim \triangle BEG$\\
Therefore, $\frac{EG}{EB}=\frac{EB}{ED}$\\
$EG\cdot ED=EB^2$\\
That is, in $\triangle AEB$, $BE=AB\cos45^\circ=5\sqrt{2}$\\
Thus, $EG\cdot ED=\boxed{25}$


Question: As shown in the figure, a supermarket's storage center has a slope $AB$, with a gradient of $i=1\colon 2$, the height at its top $AC=4m$, and points B, C are on the same horizontal ground. What is the length of the slope $AB$?

Solution:
\textbf{Solution:} Given that the slope ratio is $i=1\colon 2$ and $AC=4m$, \\
therefore $i=\frac{AC}{BC}=\frac{1}{2}$, \\
thus $BC=2AC=4\times 2=8m$, \\
In the right triangle $\triangle ABC$, according to the Pythagorean theorem, the slope is: \\
$AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{4^{2}+8^{2}}=\boxed{4\sqrt{5}m}$, \\
hence, the length of the slope $AB$ is $4\sqrt{5}m$.


Question: As shown in the figure, the storage center of a supermarket has a slope $AB$ with a gradient of $1:2$. The height at the top, $AC=4\text{m}$, and $B$, $C$ are on the same horizontal ground. Rectangle $DEFG$ represents the side view of a rectangular container, which is being transported up the slope. At this moment, $GB=6\text{m}$, and Xiao Ming, who is $1.5\text{m}$ tall, stands at point $B$ and sees a chandelier directly above point $D$ at a height of $1.5\text{m}$. Find the height of point $D$ from the ground and calculate the tangent value of Xiao Ming's elevation angle $\alpha$.

Solution:
\textbf{Solution:} Let line $DS \perp BC$ at point S and intersect with $AB$ at point H, \\
$\because \angle DGH=\angle BSH=90^\circ$, $\angle GHD=\angle BHS$, \\
$\therefore \angle GDH=\angle SBH$, \\
$\therefore \tan\alpha =\tan\angle GDH=\tan\angle SBH=\frac{AC}{BC}=\frac{GH}{DG}=\frac{1}{2}$, \\
$\because DG=EF=2\text{m}$, \\
$\therefore GH=1\text{m}$, \\
$\therefore DH=\sqrt{GH^2+DG^2}=\sqrt{1^2+2^2}=\sqrt{5}\text{m}$, $BH=GB-GH=6-1=5\text{m}$, \\
Let $HS=x\text{m}$, then $BS=2x\text{m}$, by the Pythagorean theorem: $HS^2+BS^2=BH^2$, \\
hence, $x^2+(2x)^2=5^2$, \\
Solving gives: $x=\sqrt{5}$ (negative value discarded), \\
$\therefore DS=DH+HS=\sqrt{5}+\sqrt{5}=2\sqrt{5}\text{m}$, thus the height of point D from the ground is $\boxed{2\sqrt{5}\text{m}}$, \\
$\therefore BS=2x=2\sqrt{5}\text{m}$, \\
Let the angle of elevation for Xiaoming be $\alpha$, \\
then $\tan\alpha =\frac{DS+1.5-1.5}{BS}=\frac{2\sqrt{5}+1.5-1.5}{2\sqrt{5}}=\boxed{1}$.


Question: As shown in the figure, there is a schematic diagram of a simulated rocket launch setup. A carrier rocket is launched from the ground at point L. When the rocket reaches point A, the distance $AR$ measured from the radar station located on the ground at point R is $5km$, with an elevation angle of $39^\circ$; about $1.5s$ later, when the rocket reaches point B, the measured elevation angle is $45^\circ$ (Reference data: $\sin 39^\circ \approx 0.63$, $\cos 39^\circ \approx 0.78$, $\tan 39^\circ \approx 0.8$). Determine the horizontal distance from the ground radar station R to the launch site L.

Solution:
\textbf{Solution:} In $\triangle ARL$, we have $RL=AR\cdot \cos 39^\circ \approx 5\times 0.78=3.90$ ($km$),\\
Thus, the horizontal distance from the radar station to the launch site is $RL=\boxed{3.90km}$;


Question: Please provide the text you need translated into English and formatted in standard LaTeX.

Solution:
\textbf{Solution:} In $Rt\triangle ARL$, $AL=AR\cdot \sin 39^\circ \approx 5\times 0.63=3.15$ ($km$),\\
In $Rt\triangle BRL$, $BL=RL\approx 3.90$ ($km$),\\
$\therefore AB=BL-AL=3.90-3.15\approx 0.75$ ($km$),\\
$\therefore$ the average velocity of the rocket from A to B is $0.75\div 1.5=0.5$ ($km/s$),\\
\textbf{Answer:} The average velocity of the rocket from A to B is $\boxed{0.5\ km/s}$.


Question: As shown in the figure, $AB$ is the diameter of the circle $O$, and the chord $CD\perp AB$ with the foot of the perpendicular being $E$. Connect $AC$ and $BC$. If $\angle BAC=30^\circ$ and $CD=6\,\text{cm}$. What is the degree measure of $\angle BCD$?

Solution:
\textbf{Solution:} Since diameter $AB \perp CD$, \\
it follows that $\overset{\frown}{BC} = \overset{\frown}{BD}$,\\
therefore, $\angle BCD = \angle CAB = 30^\circ$,\\
hence, $\angle BCD = \boxed{30^\circ}$.


Question: As shown in the figure, AB is the diameter of the circle $\odot O$. Chord CD is perpendicular to AB at point E. Lines AC and BC are connected. If $\angle BAC=30^\circ$ and $CD=6\text{cm}$, what is the diameter of $\odot O$ in cm?

Solution:
\textbf{Solution:} Since diameter AB$\perp$CD and CD = 6cm,\\
it follows that CE = 3cm,\\
In $\triangle ACE$, with $\angle A=30^\circ$,\\
it follows that AC = 6cm,\\
Since AB is a diameter,\\
it follows that $\angle ACB=90^\circ$,\\
In $\triangle ACB$, AB = $\frac{AC}{\cos\angle A}=\frac{6}{\cos 30^\circ}=4\sqrt{3}$ (cm).\\
Therefore, the diameter of $\odot O$ is $\boxed{4\sqrt{3}}$ cm.


Question: Given in the Cartesian coordinate system $xOy$ (as shown in the figure), the line $y=2x+2$, intersects the x-axis and y-axis at points A and B respectively, and the coordinates of point C are $(3, 2)$. Connecting AC and intersecting the y-axis at point D. Find the length of segment AB.

Solution:
\textbf{Solution:} Let $x=0$, then $y=2$,\\
$\therefore B(0,2)$,\\
$\therefore OB=2$,\\
Let $y=0$, then $x=-1$,\\
$\therefore A(-1,0)$,\\
$\therefore OA=1$,\\
$\therefore AB=\boxed{\sqrt{5}}$;


Question: Given in the Cartesian coordinate system $xOy$ (as shown in the figure), the line $y=2x+2$, intersects the x-axis and y-axis at points A and B, respectively, and the coordinates of point C are $\left(3, 2\right)$. Connect AC and intersect the y-axis at point D. Find the coordinates of point D.

Solution:
\textbf{Solution:} Let the equation of line $AC$ be $y=kx+b$,\\
$\therefore \left\{\begin{array}{l}
-k+b=0 \\
3k+b=2
\end{array}\right.$,\\
Solving, we get $\left\{\begin{array}{l}
k=\frac{1}{2} \\
b=\frac{1}{2}
\end{array}\right.$,\\
$\therefore y=\frac{1}{2}x+\frac{1}{2}$,\\
Let $x=0$, then $y=\frac{1}{2}$,\\
$\therefore \boxed{D\left(0,\frac{1}{2}\right)}$;


Question: As shown in the figure, in $ \triangle ABC$, $ \angle A=90^\circ$, $\sin\angle ABC=\frac{1}{2}$. $\odot O$ is the inscribed circle of $ \triangle ABC$, which is tangent to $ AC$, $ AB$, $ BC$ at points $ F$, $ P$, and $ E$ respectively. What is the value of $\angle EPF$ in degrees?

Solution: \textbf{Solution:} We have $\angle EPF = \boxed{60^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle A=90^\circ$, $\sin\angle ABC=\frac{1}{2}$. Circle $O$ is the inscribed circle of $\triangle ABC$, tangent to $AC$, $AB$, $BC$ at points $F$, $P$, $E$, respectively. If $BC=4$, then what is the length of $AP$?

Solution: \textbf{Solution:} We have $AP = \boxed{\sqrt{3} - 1}$.


Question: As shown in the figure, it is known that $AB$ is the diameter of $\odot O$, $C$ is the midpoint of $\overset{\frown}{BD}$, and $AD$ is perpendicular to the line passing through point $C$ at point $E$. $CE$ is the tangent to $\odot O$. If $\angle BAD=60^\circ$ and $AB=4\sqrt{3}$, what is the length of $CE$?

Solution:
\textbf{Solution:} Draw $BC$.\\
Since $AB$ is the diameter of circle $O$,\\
it follows that $\angle ACB=90^\circ$.\\
Since $\angle BAD=60^\circ$,\\
it follows that $\angle CAD=\angle BAC=30^\circ$.\\
Since $AB=4\sqrt{3}$,\\
it follows that $AC=AB\cos\angle BAC=4\sqrt{3}\times \frac{\sqrt{3}}{2}=6$,\\
thus $CE=AC\sin\angle CAD=6\times \frac{1}{2}=\boxed{3}$.


Question: As shown in the figure, in rectangle $ABCD$, point $F$ is on $CD$, and line $BF$ intersects $AC$ at point $E$. If $\frac{BC}{AB}=\frac{2}{3}$ and point $F$ is the midpoint of $CD$, connecting $AF$, find the value of $\sin\angle CAF$.

Solution:
\textbf{Solution:}
Let a perpendicular be drawn from point F to AC at H, \\
Let $BC = 2a$, then $AB = CD = 3a$, \\
By the Pythagorean theorem, $AC = \sqrt{BC^{2} + CD^{2}} = \sqrt{13}a$, \\
Since point F is the midpoint of $CD$, \\
Thus, $DF = \frac{3}{2}a$, \\
Then $AF = \sqrt{AD^{2} + DF^{2}} = \frac{5}{2}a$, \\
Since the area of $\triangle AFC = \frac{1}{2} AC \cdot FH = \frac{1}{2} CF \cdot AD$, \\
Therefore, $\frac{1}{2} \times \sqrt{13}a \cdot FH = \frac{1}{2} \times \frac{3}{2} \times a \times 2a$, \\
Solving this, we find $FH = \frac{3\sqrt{13}}{13}a$, \\
Thus, $\sin\angle CAF = \frac{FH}{AF} = \frac{\frac{3\sqrt{13}}{13}a}{\frac{5}{2}a} = \boxed{\frac{6\sqrt{13}}{65}}$


Question: Given: As shown in the figure, in the circle $\odot O$ with a radius of 4, $AB$ and $CD$ are two diameters, $M$ is the midpoint of $OB$, the extension line of $CM$ intersects $\odot O$ at point $E$, and $EM>MC$. Connect $DE$, where $DE=\sqrt{15}$. It is known that $AM\cdot MB=EM\cdot MC$; find the length of $EM$.

Solution:
\textbf{Solution}: From the given problem, we know that\\
since $DC$ is the diameter of the circle $O$,\\
thus $\angle DEC = 90^\circ$,\\
therefore $DE^2 + EC^2 = DC^2$,\\
since $DE = \sqrt{15}$ and $CD = 8$, and $EC$ is a positive number,\\
thus $EC = 7$,\\
since $M$ is the midpoint of $OB$,\\
thus $BM = 2$, $AM = 6$,\\
since $AM \cdot BM = EM \cdot CM = EM(EC - EM) = EM(7 - EM) = 12$, and $EM > MC$,\\
therefore $EM = \boxed{4}$.


Question: Given: As shown in the figure, in the circle $\odot O$ with a radius of $4$, $AB$ and $CD$ are two diameters, $M$ is the midpoint of $OB$, and the extension line of $CM$ intersects $\odot O$ at point $E$, with $EM > MC$. Connect $DE$, and $DE = \sqrt{15}$. Find the value of $\sin\angle EOB$.

Solution:
\textbf{Solution:} Draw $EF \perp AB$ through point $E$, with the foot of the perpendicular at point $F$.

Since $OE=4$ and $EM=4$,

it follows that $OE=EM$,

thus $OF=FM=1$,

so $EF=\sqrt{4^2-1^2}=\sqrt{15}$,

therefore, $\sin\angle EOB=\boxed{\frac{\sqrt{15}}{4}}$.


Question: Given: As shown in the figure, within $\triangle ABC$, $AD$ is the altitude from $A$ to side $BC$, $E$ is the midpoint of side $AC$, $BC=14$, $AD=12$, and $\sin B=\frac{4}{5}$. Find: The length of segment $DC$.

Solution:
\textbf{Solution:} In $\triangle ABC$, since AD is the altitude on side BC,\\
therefore AD $\perp$ BC.\\
therefore $\sin B= \frac{AD}{AB}=\frac{4}{5}$.\\
since AD＝12,\\
therefore $AB=\frac{5}{4}AD=15$.\\
In $\triangle ABD$, since $BD=\sqrt{AB^{2}−AD^{2}}=\sqrt{15^{2}−12^{2}}=9$,\\
therefore CD＝BC−BD＝14−9＝\boxed{5}.


Question: Given: As shown in the figure, in $\triangle ABC$, $AD$ is the altitude on the side $BC$, $E$ is the midpoint of side $AC$, $BC=14$, $AD=12$, $\sin B=\frac{4}{5}$. Find the value of $\tan\angle EDC$.

Solution:
\textbf{Solution:} In $\triangle ADC$, E is the midpoint of AC, \\
$\therefore DE=EC$, \\
$\therefore \angle EDC=\angle C$, \\
$\therefore \tan\angle EDC=\tan\angle C=\frac{AD}{CD}=\boxed{\frac{12}{5}}$.


Question: As shown in the figure, it is known that the parabola $y=-\frac{1}{2}x^2+bx+c$ intersects the x-axis at points $A(-1, 0)$ and $B(4, 0)$, and intersects the y-axis at point C. What is the expression of this parabola?

Solution:
\textbf{Solution}: Let us consider the parabola $y=-\frac{1}{2}x^2+bx+c$ intersects the x-axis at points $A(-1, 0)$, and $B(4, 0)$. Thus,
\[
\begin{cases}
-\frac{1}{2}(-1)^2+b(-1)+c=0\\
-\frac{1}{2}(4)^2+4b+c=0
\end{cases}
\]
Solving the system of equations, we find:
\[
\begin{cases}
b=\frac{3}{2}\\
c=2
\end{cases}
\]
Therefore, the equation of the parabola is $y=-\frac{1}{2}x^2+\frac{3}{2}x+2=\boxed{-\frac{1}{2}x^2+\frac{3}{2}x+2}$


Question: As shown in the figure, it is known that the parabola $y=-\frac{1}{2}x^2+bx+c$ intersects the x-axis at points $A(-1, 0)$ and $B(4, 0)$, and intersects the y-axis at point $C$. If point $P$ is a point on the parabola and $\tan\angle BAP=1$, what are the coordinates of point $P$?

Solution:
\textbf{Solution:} As shown in the diagram:

(1) When point P is above the x-axis, draw $PE\perp x$ axis at point E,

$\because$ $\tan\angle BAP=\frac{PE}{AE}=1$,

$\therefore$ $PE=AE$,

$\because$ $A(-1, 0)$,

$\therefore$ $OA=1$,

$\because$ $AE=PE=OE+OA=OE+1$,

Let $OE=m$, $PE=m+1$,

$\therefore$ the coordinates of point P are $(m, m+1)$,

$\because$ point P is on the parabola,

$\therefore$ $-\frac{1}{2}m^{2}+\frac{3}{2}m+2=m+1$,

Solving this, we get: $m_{1}=2$, $m_{2}=-1$ (irrelevant, discard),

$\therefore$ the coordinates of point P are $\boxed{(2, 3)}$;

(2) When point $P'$ is below the x-axis, draw $P'F\perp x$ axis at point F,

$\because$ $\tan\angle BAP=\frac{P'F}{AF}=1$,

$\therefore$ $P'F=AF$,

$\because$ $OA=1$,

$\therefore$ $AF=P'F=OF+OA=OF+1$,

Let $OF=n$, then $P'F=n+1$,

$\therefore$ the coordinates of point $P'$ are $(n, -n-1)$,

$\because$ point $P'$ is on the parabola,

$\therefore$ $-\frac{1}{2}n^{2}+\frac{3}{2}n+2=-n-1$,

Solving this, we get: $n_{1}=6$, $n_{2}=-1$ (irrelevant, discard),

$\therefore$ the coordinates of point $P'$ are $\boxed{(6, -7)}$,

In conclusion: the coordinates of point P are $\boxed{(2, 3)}$ or $\boxed{(6, -7)}$.


Question: As shown in the figure, it is known that the parabola $y=-\frac{1}{2}x^2+bx+c$ intersects the x-axis at points $A(-1, 0)$ and $B(4, 0)$, and intersects the y-axis at point C. Let the expression of line $BC$ be $y=mx+c$. If the first-degree equation in $x$, $x^2-2(b-m)x+2n=0$, has two positive real roots, directly write the range of values for $n$.

Solution: \textit{Solution:} We have $\boxed{0<n\le 2}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in the circle $\odot O$, $\angle ABC=135^\circ$, $OE\perp AC$, and $\angle AOE=\angle D$; if $AC=6$, what is the radius of $\odot O$?

Solution:
\textbf{Solution}: From (1), we know that $\triangle AOC$ is an isosceles right triangle, which means $\angle OAC=\angle OCA=45^\circ$, \\
Because $AC=6$, \\
Therefore, $OA=OC=6\times \sin 45^\circ=3\sqrt{2}$, \\
Hence, the radius of circle $O$ is $\boxed{3\sqrt{2}}$.


Question: In $\triangle ABC$, $AB=6$, $\angle B$ is an acute angle and $\cos B=\frac{1}{2}$, $\tan C=3\sqrt{3}$. Find the measure of $\angle B$ in degrees.

Solution:
\textbf{Solution:} Since $\angle B$ is an acute angle and $\cos B = \frac{1}{2}$, \\
it follows that $\angle B = \boxed{60^\circ}$.


Question: In $\triangle ABC$, $AB=6$, $\angle B$ is an acute angle and $\cos B=\frac{1}{2}$, $\tan C=3\sqrt{3}$. What is the length of $BC$?

Solution:
\textbf{Solution:} Draw $AH\perp BC$ at H through point A,\\
since $\cos B=\frac{1}{2}$,\\
therefore $\frac{BH}{AB}=\frac{1}{2}$,\\
since $AB=6$,\\
therefore $BH=3$,\\
in $Rt\triangle ABH$, $AH=\sqrt{AB^{2}-BH^{2}}=\sqrt{6^{2}-3^{2}}=3\sqrt{3}$,\\
since $\tan C=3\sqrt{3}$,\\
therefore $\frac{AH}{CH}=3\sqrt{3}$,\\
thus $\frac{3\sqrt{3}}{CH}=3\sqrt{3}$,\\
solving gives $CH=1$,\\
therefore $BC=BH+CH=3+1=\boxed{4}$


Question: In $\triangle ABC$, $AB=6$, $\angle B$ is an acute angle and $\cos B=\frac{1}{2}$, $\tan C=3\sqrt{3}$. Find the area of $\triangle ABC$.

Solution: \textbf{Solution:} Let $S_{\triangle ABC}=\frac{1}{2}BC\cdot AH=\frac{1}{2}\times 4\times 3\sqrt{3}=\boxed{6\sqrt{3}}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, and the chord CD $\perp$ AB at point E. Connect AC and BC. $\angle ACO = \angle BCD$; If CD $= 6$, $\angle A = 30^\circ$, find the area of the shaded region.

Solution:
\textbf{Solution:} From the given problem, \\
$\because \angle A=30^\circ$, \\
$\therefore \angle BOC=60^\circ$, \\
$\therefore \angle AOC=120^\circ$, \\
$\because$ AB is the diameter of circle $O$, and chord CD $\perp$ AB, \\
$\therefore$ CE= $\frac{1}{2}$ CD=3, \\
In $\triangle COE$, OC= $\frac{CE}{\sin 60^\circ}$ =2$\sqrt{3}$, \\
$\therefore$ the area of sector OAC (shaded region) = $\frac{120^\circ \times \pi \times (2\sqrt{3})^2}{360^\circ}$ =4$\pi$, \\
Answer: The area of the shaded region is $\boxed{4\pi}$.


Question: As shown in the figure, it is known that the distance from the rooftop to the ground $GE$ is $18.5$ meters. When Xiao Liang stands at point $B$ in front of the building, the elevation angle of the top end $G$ of the banner measured at point $A$, which is directly above point $B$, is $37^\circ$. Then, moving $15$ meters towards the direction of the teaching building to reach point $D$ (with the bottom part of the building point $E$ and points $B$, $D$ on the same straight line), the elevation angle of the bottom end $F$ of the banner measured at point $C$, which is directly above point $D$, is $42^\circ$. Assuming $AB$, $CD$ are both $1.7$ meters (i.e., quadrilateral $ABCD$ is a rectangle), please help Xiao Liang calculate the distance $BE$ from point $B$ to the teaching building in meters?

Solution:
\textbf{Solution:} Extend $AC$ to intersect $GE$ at $H$,\\
then $AB=CD=EH=1.7$ meters, $AC=BD$, $AH=BE$,\\
since $GE=18.5$ meters,\\
thus $HG=EG-HE=18.5-1.7=16.8$ (meters),\\
in $\triangle AGH$, $\angle GAH=37^\circ$,\\
thus $\tan 37^\circ =\frac{GH}{AH}=\frac{16.8}{15+CH}\approx 0.75$,\\
thus $CH=7.4$,\\
thus $BE=AH=15+7.4=\boxed{22.4}$ (meters).


Question: As shown in the diagram, it is known that the distance from the rooftop to the ground $GE$ is $18.5$ meters. When Xiaoliang stands at point $B$ in front of the building, measuring from point $A$, which is directly above point $B$, the angle of elevation to the top of the banner $G$ is $37^\circ$. Then, moving $15$ meters towards the direction of the teaching building to point $D$ (points $E$, $B$, and $D$ are on a straight line and at the base of the building), and from point $C$, which is directly above point $D$, the angle of elevation to the bottom end $F$ of the banner is $42^\circ$. If $AB$ and $CD$ are both $1.7$ meters (meaning quadrilateral $ABCD$ is a rectangle), please help Xiaoliang calculate the length of the banner $GF$. (The result should be accurate to $0.1m$, reference data: $\sin37^\circ\approx 0.60$, $\cos37^\circ\approx 0.80$, $\tan37^\circ\approx 0.75$, $\sin42^\circ\approx 0.67$, $\cos42^\circ\approx 0.74$, $\tan42^\circ\approx 0.90$)

Solution:
\textbf{Solution:} Given from (1) that \(CH = 7.4\) meters, in the right triangle \(\triangle FCH\), since \(\angle FCH = 42^\circ\),\\
therefore, \(\tan 42^\circ = \frac{FH}{CH} = \frac{FH}{7.4} \approx 0.90\),\\
hence, \(FH = 6.66\),\\
thus, \(FG = GH - FH = 16.8 - 6.66 \approx \boxed{10.1}\) (meters).


Question: As shown in the diagram, within square $ABCD$, $\triangle BPC$ is an equilateral triangle. The extensions of $BP$ and $CP$ intersect $AD$ at points $E$ and $F$, respectively. Connect $BD$, $DP$, and $BD$ intersects $CF$ at point $H$. Given that $\triangle BDE \sim \triangle DPE$, what is the value of $\frac{FP}{PH}$?

Solution:
\textbf{Solution:} Given: $PC=CD$, $\angle PCD=30^\circ$,\\
$\therefore \angle PDC=75^\circ$,\\
$\therefore \angle FDP=15^\circ$,\\
$\because \angle DBA=45^\circ$,\\
$\therefore \angle PBD=15^\circ$,\\
$\therefore \angle FDP=\angle PBD$,\\
$\because \angle DFP=\angle BPC=60^\circ$,\\
$\therefore \triangle DFP\sim \triangle BPH$,\\
$\therefore \frac{PF}{PH}=\frac{DF}{PB}=\frac{DF}{CD}=\boxed{\frac{\sqrt{3}}{3}}$;


Question: As shown in the figure, in square $ABCD$, $\triangle BPC$ is an equilateral triangle. The extensions of $BP$ and $CP$ intersect $AD$ at points $E$ and $F$, respectively. Connect $BD$, $DP$, and $CF$. $BD$ and $CF$ intersect at point $H$. What is the value of $\tan\angle DBE$?

Solution:
\textbf{Solution:} As shown in the diagram, draw $PM\perp CD$ at $M$, $PN\perp BC$ at $N$,\\
Assume the side length of square $ABCD$ is $4a$, $\triangle BPC$ is an equilateral triangle,\\
$\therefore \angle PBC=\angle PCB=60^\circ$, $PB=PC=BC=CD=4a$,\\
$\therefore \angle PCD=30^\circ$,\\
$\therefore CM=PN=PB\cdot \sin 60^\circ=4a\times \frac{\sqrt{3}}{2}=2\sqrt{3}a$,\\
$PM=PC\cdot \sin 30^\circ=2a$,\\
$\because DE\parallel PM$,\\
$\therefore \angle EDP=\angle DPM$,\\
$\therefore \angle DBE=\angle DPM$,\\
$\therefore \tan\angle DBE=\tan\angle DPM=\frac{DM}{PM}=\frac{4a-2\sqrt{3}a}{2a}=2-\sqrt{3}$.\\
Hence, the value of $\tan\angle DBE$ is $\boxed{2-\sqrt{3}}$.


Question: As shown in the figure, point $F$ lies on diagonal $AC$ of the parallelogram $ABCD$. Lines parallel to $AB$ and $AC$ are drawn passing through points $F$ and $B$ respectively, intersecting at point $E$. Line $BF$ is drawn, with given $\angle ABF = \angle FBC + \angle DAC$. Quadrilateral $ABEF$ is a rhombus; if $BE=6$, $AD=10$, and $\tan \angle CBE = \frac{1}{2}$, find the length of $AC$.

Solution:
\textbf{Solution:} Construct DH$\perp$AC at point H, as shown in the diagram:\\
Since BE$\parallel$AC,\\
then $\angle 1=\angle CBE$,\\
Since AD$\parallel$BC,\\
then $\angle 2=\angle 1$,\\
thus $\angle 2=\angle CBE$,\\
thus $\tan \angle 2=\tan \angle CBE=\frac{1}{2}$,\\
In $\triangle ADH$, $\tan \angle 2=\frac{DH}{AH}=\frac{1}{2}$,\\
Let DH$=x$ (where $x>0$), then AH$=2x$,\\
By Pythagoras' theorem, $x^{2}+(2x)^{2}=10^{2}$,\\
Solving this gives: $x=2\sqrt{5}$,\\
therefore AH$=2x=4\sqrt{5}$, DH$=2\sqrt{5}$,\\
Since quadrilateral ABEF is a rhombus,\\
then CD$=$AB$=$BE$=6$,\\
In $\triangle CDH$, CH$=\sqrt{CD^{2}-DH^{2}}=\sqrt{6^{2}-(2\sqrt{5})^{2}}=4$,\\
therefore AC$=$AH$+$CH$=4\sqrt{5}+4=\boxed{4\sqrt{5}+4}$.


Question: As shown in the diagram, $ \triangle ABC$ is inscribed in circle $\odot O$, with D and F respectively on arcs $\overset{\frown}{AC}$ and $\overset{\frown}{AB}$, and $\overset{\frown}{BF} = \overset{\frown}{DA}$. Extend AF to intersect the extension of CB at point E, and connect AD, CD. $\triangle AEB \sim \triangle CAD$; if $\triangle ABC$ is an equilateral triangle with side length 6, and $AD = 3, AE = 8$. Find $\tan\angle DAC$?

Solution:
\textbf{Solution:} Given that $\because \triangle AEB\sim \triangle CAD$,\\
$\therefore \frac{AD}{EB}=\frac{AC}{EA}$, $\angle DAC=\angle AEC$,\\
$\because AD=3, AE=8, AB=AC=BC=6$,\\
$\therefore \frac{3}{EB}=\frac{6}{8}$,\\
$\therefore EB=4$,\\
$\therefore EC=EB+BC=4+6=10$,\\
$\therefore AE^2+AC^2=8^2+6^2=10^2=EC^2$,\\
$\therefore \triangle AEC$ is a right-angled triangle,\\
$\therefore \tan\angle DAC=\tan\angle AEC=\frac{AC}{AE}=\frac{6}{8}=\boxed{\frac{3}{4}}$.


Question: As shown in the figure, there are two islands C and D in the sea. A fishing boat at point A in the sea measures that island D is in the northeast direction, and the distance is $30\sqrt{2}$ nautical miles. The fishing boat then sails from the west to the east for a while and reaches point B. At this point, it's measured that island C is exactly to the north of point B and the distance is $75$ nautical miles. It's also measured that the distance between point B and island D is $30\sqrt{5}$ nautical miles. Find the value of $\sin\angle ABD$.

Solution:
\textbf{Solution:} Construct $DE \perp AB$ at $E$,\\
In $\triangle AED$, $AD=30\sqrt{2}$, $\angle DAE=45^\circ$,\\
$\therefore DE=30\sqrt{2}\times\sin45^\circ=30$,\\
In $\triangle BED$, $BD=30\sqrt{5}$,\\
$\therefore \sin\angle ABD=\frac{ED}{BD}=\frac{30}{30\sqrt{5}}=\boxed{\frac{\sqrt{5}}{5}}$;