Question: As shown in the figure, given the quadratic function $y=ax^2+bx+c$ with part of its graph as illustrated, if $y>0$, what is the range of $x$?

Solution:
\textbf{Solution:} From the figure, we know that the axis of symmetry of the parabola is the line $x=-1$, and one intersection point with the $x$-axis has an abscissa of $-3$,\\
$\therefore$ the abscissa of the other intersection point of the parabola with the $x$-axis is $1$,\\
$\because y>0$,\\
$\therefore -3<x<1$,\\
Therefore, the answer is: $x \in \boxed{(-3, 1)}$.


Question: As shown in the figure, the parabola $y=\frac{1}{2}x^2-2x+c$ intersects the x-axis at points A and B, and intersects the negative half of the y-axis at point C. The vertex of the parabola is M, and D and E are the midpoints of AB and BM, respectively. If the area ratio of $\triangle DEB$ to $\triangle ACD$ is $9:10$, what is the value of $c$?

Solution:
\textbf{Solution:} Let $S_{\triangle DEB} = \frac{1}{2} \text{DB} \cdot |y_E|$, and $S_{\triangle ADC} = \frac{1}{2} \text{AD} \cdot |y_C|$,
\\
Since D is the midpoint of AB,
\\
Therefore, AD = DB,
\\
Also, since the area ratio of $\triangle DEB$ to $\triangle ACD$ is $9:10$,
\\
Therefore, $\frac{|y_E|}{|y_C|} = \frac{9}{10}$,
\\
Also, since E is the midpoint of BM,
\\
Therefore, $|y_E| = \frac{1}{2} |y_M|$,
\\
Substituting $x = 0$ into the equation yields $y_C = c$,
\\
Therefore, $|y_E| = \left| \frac{9}{10} c \right|$,
\\
Therefore, $|y_M| = \left| \frac{9}{5} c \right|$,
\\
Since $y_M < 0$ and $c < 0$,
\\
Therefore, $y_M = \frac{9}{5} c$,
\\
Since M is the vertex of the parabola,
\\
Therefore, $x_M = -\frac{b}{2a} = -\frac{-2}{2 \times \frac{1}{2}} = 2$,
\\
Substituting $x = 2$ into the equation yields: $y_M = \frac{1}{2} \times 2^2 - 2 \times 2 + c = c - 2 = \frac{9}{5} c$,
\\
Solving this gives $c = \boxed{-\frac{5}{2}}$.


Question: As shown in the figure, the parabola $y=-2x^{2}+8x-6$ intersects the x-axis at points $A$ and $B$. Let the part of the parabola above and on the x-axis be denoted as $C_{1}$. After shifting $C_{1}$ to the right, we obtain $C_{2}$, which intersects the x-axis at points $B$ and $D$. If the line $y=x+m$ has 3 different intersection points with $C_{1}$ and $C_{2}$, what is the range of values for $m$?

Solution:
\textbf{Solution:} Let $y=-2x^{2}+8x-6=0$, \\
that is, $x^{2}-4x+3=0$, \\
solving yields $x=1$ or $3$, \\
thus, points $A(1,0)$, $B(3,0)$, \\
since translating $C_{1}$ rightward by 2 units gives $C_{2}$, \\
then, the equation for $C_{2}$ is $y=-2(x-4)^{2}+2$, $3\leq x\leq 5$, \\
when $y=x+m_{1}$ is tangent to $C_{2}$, \\
let $y=x+m_{1}=-2(x-4)^{2}+2$, \\
that is, $2x^{2}-15x+30+m_{1}=0$, \\
$\Delta=-8m_{1}-15=0$, \\
solving yields $m_{1}=\boxed{-\frac{15}{8}}$, \\
when $y=x+m_{2}$ passes through point $B$, \\
that is, $0=3+m_{2}$, \\
$m_{2}=-3$, \\
when $-3<m<-\frac{15}{8}$, the line $y=x+m$ intersects with $C_{1}$ and $C_{2}$ at 3 distinct points.


Question: As shown in the figure, the parabola $y=-x^2+2x+1$ intersects the x-axis at points A and B, and intersects the y-axis at point C. Point D is the vertex of the parabola, and the point symmetrical to C with respect to the axis of symmetry of the parabola is point E. Points G and F are located on the x-axis and the y-axis, respectively. What is the minimum perimeter of the quadrilateral EDFG?

Solution:
Solution: Let $x=0$, then $y=1$,\\
$\therefore$ $C(0,1)$,\\
$\because$ $y=-x^2+2x+1=-{(x-1)}^2+2$,\\
$\therefore$ $D(1,2)$, the axis of symmetry of the parabola is the line $x=1$,\\
$\because$ the symmetric point of C with respect to the axis of symmetry of the parabola is point E,\\
$\therefore$ $E(2,1)$,\\
$\therefore$ $DE=\sqrt{1^2+1^2}=\sqrt{2}$,\\
Construct the symmetric point ${D}^{\prime }$ of D with respect to the y-axis, and the symmetric point ${E}^{\prime }$ of E with respect to the x-axis, connect ${D}^{\prime }{E}^{\prime }$, intersecting the x-axis at point G and the y-axis at point F, as shown in the figure,\\
Then ${D}^{\prime }(-1,2)$, ${E}^{\prime }(2,-1)$, $FD=F{D}^{\prime }$, $GE=G{E}^{\prime }$,\\
At this time, $DF+FG+EG=F{D}^{\prime }+FG+G{E}^{\prime }={D}^{\prime }{E}^{\prime }$,\\
$\therefore$ At this time, the perimeter of the quadrilateral EDFG is minimal,\\
Extend $D{D}^{\prime }$ and $E{E}^{\prime }$, they intersect at point H, as shown in the figure,\\
Then ${D}^{\prime }H=3$, $H{E}^{\prime }=3$,\\
$\therefore$ ${D}^{\prime }{E}^{\prime }=\sqrt{3^2+3^2}=3\sqrt{2}$,\\
$\therefore$ The minimum perimeter of the quadrilateral EDFG is $DE+DF+FG+EG=DE+{D}^{\prime }{E}^{\prime }=\sqrt{2}+3\sqrt{2}=\boxed{4\sqrt{2}}$.


Question: As shown in the figure, it is known that $AB=8$, point $P$ is a moving point on line segment $AB$, a square $APCD$ is constructed with $AP$ as a side, and an isosceles triangle $\triangle PBE$ is constructed with $PB$ as the base. Connecting $CE$, what is the maximum area of $\triangle PCE$?

Solution:
\textbf{Solution:} Consider a line EH$\perp$AB at point H,\\
Let AP=x, then PB=8−x,\\
$\because$ $\triangle PBE$ is an isosceles triangle with BP as its base,\\
$\therefore$ PH=$\frac{1}{2}$(8−x)=4−$\frac{1}{2}$x,\\
$\because$ square APCD,\\
$\therefore$AP=PC=x,\\
$\therefore$ ${S}_{\triangle PCE}=\frac{1}{2}PC \cdot PH=\frac{1}{2}x(4−\frac{1}{2}x)=-\frac{1}{4}(x−4)^2+4$,\\
$\because$ the parabola opens downward,\\
$\therefore$ when x=4, the maximum value of $s$ is $\boxed{4}$.


Question: As shown in the figure, on the graph of the inverse proportion function $y=\frac{2}{x}$ ($x>0$), there are points $P_1$, $P_2$, $P_3$, $P_4$ whose abscissae are 1, 2, 3, 4, respectively. Perpendiculars are drawn from these points to the $x$-axis and the $y$-axis. The areas of the shaded regions formed in the diagram from left to right are $S_1$, $S_2$, and $S_3$, respectively. What is the value of $S_1+S_2+S_3$?

Solution:
\textbf{Solution:} After translation, as shown in the figure, \\
when $x=4$, $y=\frac{1}{2}$, \\
$\therefore$ the area of rectangle AOCB is $1\times\frac{1}{2}=\frac{1}{2}$; \\
when $x=1$, $y=2$, \\
$\therefore S_{1}+S_{2}+S_{3}+\text{Area of rectangle AOCB}=2$ \\
$\therefore S_{1}+S_{2}+S_{3}=2-\frac{1}{2}=\boxed{\frac{3}{2}}$.


Question: The diagram shows a schematic of an arched bridge, with the points where the arch intersects the deck being O and B, respectively. Taking point O as the origin and the horizontal line OB as the x-axis, we establish a Cartesian coordinate system. The arch of the bridge can be approximated by the parabola $y=-0.01(x-20)^2+4$. The point C, where the arch intersects the pier AC, is precisely at the water's surface, and $AC \perp x$-axis. If OA is 5 meters, what is the height of the deck above the water surface AC?

Solution:
\textbf{Solution:} Since $OA=5$ and $CA\perp x$-axis, \\
it follows that the coordinates of point A are $(-5, 0)$, \\
when $x=-5$, $y=-0.01(-5-20)^{2}+4=-2.25$, \\
thus point C has coordinates $(-5, -2.25)$, \\
hence $AC=|-2.25|=\boxed{2.25}$ meters.


Question: As shown in the figure, the parabola \(y_{1}=ax^{2}+bx+c\) intersects the line \(y_{2}=mx+n\) at points \((3,0)\) and \((0,3)\). If \(ax^{2}+bx+c>mx+n\), what is the range of values for \(x\)?

Solution:
\textbf{Solution:} According to the graph of the functions, \\
when $x<0$ or $x>3$, $y_1>y_2$, \\
thus the solution set for $ax^2+bx+c>mx+n$ is $x<0$ or $x>3$. \\
Therefore, the answer is: $\boxed{x<0$ or $x>3}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertex $A$ of $\square OABC$ lies on the inverse proportional function $y=\frac{1}{x}$, the vertex $B$ lies on the inverse proportional function $y=\frac{5}{x}$, and point $C$ is located on the positive half of the $x$-axis. What is the area of $\square OABC$?

Solution:
\textbf{Solution:} As shown in the figure, draw perpendiculars from points A and B to the x-axis, with the feet of the perpendiculars being M and N, respectively, and connect OB. \\
Since quadrilateral ABCO is a parallelogram, \\
it follows that $OA=BC$ and $AB\parallel OC$, \\
which means $AM=BN$, \\
In $\triangle AMO$ and $\triangle BNC$ right-angled triangles, \\
since $OA=CB$ and $AM=BN$, \\
it leads to $\triangle AMO \cong \triangle BNC$ (HL criterion), \\
Also, since point A lies on the inverse proportion function $y=\frac{1}{x}$, \\
it implies $Area_{\triangle AOM}=\frac{1}{2}\times 1=\frac{1}{2}=Area_{\triangle BNC}$, \\
Since point B lies on the inverse proportion function $y=\frac{5}{x}$, \\
it implies $Area_{\triangle BON}=\frac{1}{2}\times 5=\frac{5}{2}$, \\
Therefore, $Area_{\triangle OBC}=Area_{\triangle BON}-Area_{\triangle BNC}=\frac{5}{2}-\frac{1}{2}=2=\frac{1}{2}Area_{\square ABCO}$, \\
Thus, $Area_{\square ABCO}=\boxed{4}$.


Question: After shifting the parabola \(y=x^{2}-4x+1\) to the left so that its vertex falls on the y-axis, as shown in the figure, what is the area \(S\) of the shape formed by the two parabolas, the line \(y=-3\), and the x-axis (the shaded area in the figure)?

Solution:
\textbf{Solution:} The parabola $y=x^{2}-4x+1=(x-2)^{2}-3$ has its vertex at $C(2,-3)$. Shifting it leftwards to have the vertex lie on the y-axis, we get the vertex $B(0,-3)$. Point $A$ is one of the intersections of the parabola with the $x$-axis. Connecting $OC$, $AB$,\\
as shown in the figure,\\
the shaded area equals the area of $\triangle ABCO$, with $S=2\times 3=\boxed{6}$.


Question: As illustrated in the figure, the line $y=-\frac{1}{2}x+m$ ($m>0$) intersects the $x$-axis at point $C$ and the $y$-axis at point $D$. A rectangle ABCD is constructed with $CD$ as one side, where point $A$ is on the $x$-axis. The hyperbola $y=-\frac{6}{x}$ passes through point $B$ and intersects the line $CD$ at point $E$. What are the coordinates of point $E$?

Solution:
\textbf{Solution:} According to the given conditions, the line $y=-\frac{1}{2}x+m$ intersects the x-axis at C and the y-axis at D, \\
by setting $x=0$, $y=0$, \\
we get $y=m$, $x=2m$, \\
thus $D(0,m)$ and $C(2m,0)$, \\
since AD is perpendicular to CD and passes through D, \\
the equation of line AD is: $y=2x+m$, \\
setting $y=0$, we find $x=-\frac{1}{2}m$, \\
thus $A\left(-\frac{1}{2}m,0\right)$, \\
draw BH perpendicular to AC at H, \\
since quadrilateral ABCD is a rectangle, \\
therefore $AD=BC$ and $\angle DAO=\angle BCH$, \\
in $\triangle AOD$ and $\triangle CHB$ we have \\
$\left\{\begin{array}{l}
\angle DAO=\angle BCH\\
\angle AOD=\angle CHB=90^\circ\\
AD=BC
\end{array}\right.$, \\
thus $\triangle AOD \cong \triangle CHB$ (AAS), \\
hence $BH=OD=m$ and $CH=OA=\frac{1}{2}m$, \\
thus $OH=\frac{3}{2}m$, \\
therefore, the coordinates of point B are $B\left(\frac{3}{2}m,-m\right)$, \\
since B lies on the hyperbola $y=-\frac{6}{x}$, \\
thus $\frac{3}{2}m\cdot (-m)=-6$, \\
solving for $m$ gives $m=\pm 2$, \\
since $m>0$, \\
thus $m=2$, \\
hence, the equation of line CD is $y=-\frac{1}{2}x+2$, \\
solving $\left\{\begin{array}{l}
y=-\frac{6}{x}\\
y=-\frac{1}{2}x+2
\end{array}\right.$, \\
gives $\left\{\begin{array}{l}
x=6\\
y=-1
\end{array}\right.$ and $\left\{\begin{array}{l}
x=-2\\
y=3
\end{array}\right.$, \\
therefore, the coordinates of point $E$ are $\boxed{\left(6,-1\right)}$.


Question: As shown in the figure, point A is on the graph of the inverse proportion function $y = \frac{k}{x}$ (where $k > 0$), and AB is perpendicular to the x-axis at point B. If the area of $\triangle AOB$ is 3, what is the value of $k$?

Solution:
\textbf{Solution:} Since point $A$ lies on the graph of the inverse proportion function $y=\frac{k}{x}$, we have $S_{\triangle AOB}=\frac{1}{2}|k|=3$;\\
Since $k>0$, it follows that $k=\boxed{6}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertex \(A\) of \(\square OABC\) is on the inverse proportion function \(y=\frac{1}{x}\), the vertex \(B\) is on the inverse proportion function \(y=\frac{5}{x}\), and the point \(C\) is on the positive half of the x-axis. What is the area of \(\square OABC\)?

Solution:
\textbf{Solution:} Construct line AF $\perp$ x-axis at point F through point A, and construct line BE $\perp$ x-axis at point E through point B. Extend BA to intersect the y-axis at point G. Thus, both quadrilaterals OFAG and OEBG are rectangles. \\
$\because$ Point A lies on the inverse proportional function $y=\frac{1}{x}$, and vertex B lies on the inverse proportional function $y=\frac{5}{x}$, \\
$\therefore$ Area of rectangle OFAG = 1, Area of rectangle OEBG = 5, \\
$\therefore$Area of $\square OABC$ = Area of rectangle ABEF = Area of rectangle OEBG - Area of rectangle OFAG = \boxed{4}.


Question: As shown in the figure, a rectangular window frame in the shape of the character "目" is made from 48 meters of wood (the crossbars EF and GH are also made of wood). Where AB $\parallel$ EF $\parallel$ GH $\parallel$ CD, to maximize the area of the window frame ABCD, what should be the length of AB?

Solution:
\textbf{Solution:} Let the length of AB be $x$ meters, then the length of AD is $\frac{48-4x}{2}$ meters.\\
Using the formula for the area of a rectangle, we get: $S_{\text{rectangle}} = \text{AD} \cdot \text{AB} = x \cdot \frac{48-4x}{2} = -2x^2 + 24x = -2(x-6)^2 + 72$,\\
$\because 48-4x > 0$,\\
$\therefore x < 12$,\\
$\therefore 0 < x < 12$,\\
$\because -2 < 0$,\\
$\therefore$ when $x = \boxed{6}$, the area of the rectangle has its maximum value.


Question: As shown in the figure, the line $y=x+2$ and the hyperbolic function $y=\frac{k}{x}$ intersect at point P in the first quadrant. If $OP=\sqrt{20}$, what is the value of $k$?

Solution:
\textbf{Solution:} Let $P(x, x+2)$,\\
since $OP=\sqrt{20}$,\\
it follows that $x^{2}+(x+2)^{2}=(\sqrt{20})^{2}$,\\
simplifying, we get:\\
$x^{2}+2x-8=0$,\\
thus, $(x+4)(x-2)=0$,\\
hence, $x_{1}=-4, x_{2}=2$,\\
since $P$ is in the first quadrant, then\\
$P(2, 4)$,\\
therefore, $k=xy=2\times 4=\boxed{8}$


Question: As shown in the figure, $AB \perp x$-axis, with $B$ being the foot of the perpendicular. The hyperbola $y = \frac{k}{x}$ (where $x > 0$) intersects the two sides $OA$ and $AB$ of $\triangle AOB$ at points $C$ and $D$, respectively, with $OC = CA$. If the area of $\triangle ACD$ is $3$, what is the value of $k$?

Solution:
\textbf{Solution:} Draw OD, and construct CE $\perp$ x-axis through point C,\\
$\because$ OC = CA,\\
$\therefore$ OE : OB = 1 : 2;\\
Assume the area of $\triangle OBD$ as x, according to the meaning of the inverse ratio function k, the area of triangle OCE is x,\\
$\because$ $\triangle COE \sim \triangle AOB$,\\
$\therefore$ the ratio of the areas of triangle COE to triangle BOA is 1 : 4,\\
$\because$ the area of $\triangle ACD$ is 3,\\
$\therefore$ the area of $\triangle OCD$ is 3,\\
$\therefore$ the area of triangle BOA is 6 + x,\\
That is, the area of triangle BOA is 6 + x = 4x,\\
Solving this gives x = 2,\\
$\therefore$ $\frac{1}{2}|k|=2$,\\
$\because$ k > 0,\\
$\therefore$ k = \boxed{4}.


Question: As shown in the figure, the regular hexagon ABCDEF is inscribed in $\odot O$, where the radius of $\odot O$ is 1. What is the length of the apothem $OM$?

Solution:
\textbf{Solution:} Connect OB, \\
$\because$ hexagon ABCDEF is a regular hexagon inscribed in $\odot$ O, \\
$\therefore \angle$BOM = $\frac{360^\circ}{6\times 2}$ = $30^\circ$, \\
$\therefore$ OM = OB $\cdot \cos \angle$BOM = 1 $\times \frac{\sqrt{3}}{2}$ = $\boxed{\frac{\sqrt{3}}{2}}$;


Question: As shown in the figure, it is known that isosceles $\triangle ABC$, with $AB=BC$, and the circle with diameter $AB$ intersects $AC$ at point D. The tangent of $\odot O$ through point D intersects $BC$ at point E. If $CD=4\sqrt{5}, CE=8$, then what is the radius of $\odot O$?

Solution:
\textbf{Solution:} As shown in the diagram, connect OD and BD,\\
$\because$ DE is a tangent,\\
$\therefore$ OD $\perp$ DE,\\
$\because$ AB is a diameter,\\
$\therefore$ $\angle ADB = 90^\circ$, and AB=BC,\\
$\therefore$ AD=CD=$4\sqrt{5}$, and AO=OB,\\
$\therefore$ DO $\parallel$ BC, and DE $\perp$ OD,\\
$\therefore$ DE $\perp$ EC,\\
$\therefore$ DE$=\sqrt{CD^{2}-CE^{2}}=\sqrt{80-64}=4$,\\
$\because$ $\tan C = \frac{BD}{CD} = \frac{DE}{EC} = \frac{4}{8} = \frac{1}{2}$,\\
$\therefore$ BD=$2\sqrt{5}$,\\
$\therefore$ AB$=\sqrt{AD^{2}+DB^{2}}=10$,\\
$\therefore$ OA=$\boxed{5}$.


Question: As shown in the figure, there is a paper circle $\odot O$ with a radius of 6cm. $\triangle ABC$ is an inscribed triangle in $\odot O$. The paper circle $\odot O$ is folded along the straight lines $AB$ and $AC$, and both arcs $\overset{\frown}{AB}$ and $\overset{\frown}{AC}$ pass through the center $O$. What is the area of the shaded region in the figure?

Solution:
\textbf{Solution:} Draw lines $AO, BO, CO$, and extend $AO$ to intersect $BC$ at point D, as shown in the diagram:\\
Since $\triangle ABC$ is an inscribed triangle of circle $O$ with a radius of $6\,\text{cm}$,\\
therefore, $AD\perp BC$ and $AO=6\,\text{cm}$,\\
therefore, $OD=3\,\text{cm}$,\\
therefore, $CD=\sqrt{OC^{2}-OD^{2}}=3\sqrt{3}$,\\
therefore, $BC=2CD=6\sqrt{3}\,\text{cm}$,\\
From the diagram, it can be seen that the area of the shaded part is the same as the area of $\triangle OBC$,\\
therefore, $S_{\text{shaded}}=S_{\triangle COB}=\frac{1}{2}\times BC\times OD=\frac{1}{2}\times 6\sqrt{3}\times 3=\boxed{9\sqrt{3}\,\text{cm}^2}$.


Question: As shown in the figure, a regular octagon is assembled with four quadrilaterals as depicted in Figure 1 and four diamonds as illustrated in Figure 2 (as in Figure 3). What is the ratio of the area of the shaded part to the area of the blank part in Figure 3?

Solution:
\textbf{Solution:} As shown in the figure, draw $BE\perp AD$ at E through the vertex B of the rhombus; Let the center of the regular octagon be point O, with one side being MN. Connect OM and ON, and draw $MP\perp ON$ at P through point M. Suppose the side length of the regular octagon is $a$, then we have $AB=AD=MN=a$,\\
$\therefore \angle ABC=180^\circ-360^\circ\div 8=135^\circ$, $\angle MON=360^\circ\div 8=45^\circ$,\\
$\because AD\parallel BC$,\\
$\therefore \angle BAE=180^\circ-\angle ABC=45^\circ$,\\
$\therefore BE=\frac{\sqrt{2}}{2}AB=\frac{\sqrt{2}}{2}a$,\\
$\therefore S_{\text{rhombus}ABCD}=AD\cdot BE=\frac{\sqrt{2}}{2}a^{2}$,\\
$\therefore$ the area of the blank part $=4\times \frac{\sqrt{2}}{2}a^{2}=2\sqrt{2}a^{2}$,\\
$\because \angle MON=45^\circ$,\\
$\therefore OP=PM$, let $OP=PM=b$, then $OM=ON=\sqrt{2}b$,\\
$\therefore PN=(\sqrt{2}-1)b$,\\
In $\triangle MPN$, $PM^{2}+PN^{2}=MN^{2}$,\\
$\therefore b^{2}+(\sqrt{2}-1)^{2}b^{2}=a^{2}$,\\
Upon rearranging: $b^{2}=\frac{2+\sqrt{2}}{4}a^{2}$,\\
$\therefore S_{\triangle OMN}=\frac{1}{2}ON\cdot PM=\frac{1}{2}\times \sqrt{2}b\cdot b=\frac{\sqrt{2}}{2}b^{2}=\frac{1+\sqrt{2}}{4}a^{2}$,\\
$\therefore$ the area of the regular octagon is: $8\times \frac{1+\sqrt{2}}{4}a^{2}=2(1+\sqrt{2})a^{2}$,\\
$\therefore$ the area of the shaded part is: $2(1+\sqrt{2})a^{2}-2\sqrt{2}a^{2}=2a^{2}$,\\
$\therefore$ the ratio of the area of the shaded part to the area of the blank part $=\frac{2a^{2}}{2\sqrt{2}a^{2}}=\boxed{\frac{\sqrt{2}}{2}}$.


Question: As shown in the figure, the regular hexagon $ABCDEF$ is inscribed in the circle $\odot O$. Draw $OM \perp BC$ at point $M$. If the radius of $\odot O$ is 4, what is the length of the apothem $OM$?

Solution:
\textbf{Solution:} As shown in the figure, connect OB and OC.\\
Since hexagon ABCDEF is a regular hexagon,\\
it follows that $\angle BOC=60^\circ$, and OB=OC=4,\\
therefore $\triangle OBC$ is an equilateral triangle,\\
hence BC=OB=OC=4,\\
since OM$\perp$BC,\\
it follows that BM=CM=2,\\
in $\triangle OBM$, OM=$\sqrt{OB^{2}-BM^{2}}=\sqrt{4^{2}-2^{2}}=\boxed{2\sqrt{3}}$.


Question: As shown in the diagram, in the Cartesian coordinate system, a regular hexagon $ABCDEF$ with side length 2 has its center coinciding with the origin $O$, and $AB$ is parallel to the x-axis, intersecting the y-axis at point $P$. If $\triangle OAP$ is rotated clockwise around point $O$, rotating $90^\circ$ each time, what are the coordinates of point $A$ after the 2022nd rotation?

Solution:
\textbf{Solution:} Given a regular hexagon ABCDEF with side length 2, and its center coincides with the origin O,\\
\[ \because OA=AB=2, \angle BAO=60^\circ, \]
\[ \because AB\parallel x \text{ axis}, \]
\[ \therefore \angle APO=90^\circ, \]
\[ \therefore \angle AOP=30^\circ, \]
\[ \therefore AP=1, OP=\sqrt{3}, \]
\[ \therefore A(1,\sqrt{3}), \]
\[ \because \text{rotating } \triangle OAP \text { clockwise around point O by } 90^\circ\text{ each time, it is known that point A}_2 \text{ coincides with } D, \]
\[ \text{Since } 360^\circ \div 90^\circ=4, \]
\[ \text{indicating each cycle consists of 4 rotations,} \]
\[ \therefore 2022 \div 4=505\ldots 2, \]
\[ \therefore \text{point A}_{2022} \text{ coincides with point A}_2, \]
\[ \because \text{point A}_2 \text{ is symmetric to point A concerning the origin } O, \]
\[ \therefore A_2(-1,-\sqrt{3}), \]
\[ \therefore \text{after the 2022nd rotation, the coordinates of point A are} \boxed{(-1,-\sqrt{3})}\].


Question: As shown in the figure, it is known that the circle $\odot O$ is the incircle of $\triangle ABC$, and $\angle ABC=50^\circ$, $\angle ACB=80^\circ$. What is the measure of $\angle BOC$?

Solution:
\textbf{Solution:} Since $\odot O$ is the inscribed circle of $\triangle ABC$,\\
it follows that $OB$ and $OC$ are the angle bisectors of $\triangle ABC$,\\
therefore $\angle OBC=\frac{1}{2}\angle ABC=25^\circ$, $\angle OCB=\frac{1}{2}\angle ACB=40^\circ$,\\
hence $\angle BOC=180^\circ-\angle OBC-\angle OCB=\boxed{115^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $BC=5$, and circle $O$ is tangent to the three sides of $\triangle ABC$ at points D, E, and F. If the radius of circle $O$ is 2, what is the perimeter of $\triangle ABC$?

Solution:
\textbf{Solution:} Let us draw lines OE and OF, and suppose AD $= x$. By the Tangent-Secant Theorem, we have AE $= x$. \\
$\because$ Circle O and right triangle $\triangle ABC$ intersect at points D, E, and F, \\
$\therefore$ OE $\perp$ AC, OF $\perp$ BC, \\
$\therefore$ Quadrilateral OECF is a square, \\
$\because$ the radius of circle O is 2, and BC $= 5$, \\
$\therefore$ CE $=$ CF $= 2$, BD $=$ BF $= 3$, \\
$\therefore$ in right triangle $\triangle ABC$, \\
$\because$ $AC^{2} + BC^{2} = AB^{2}$, i.e., $(x + 2)^{2} + 5^{2} = (x + 3)^{2}$, \\
solving this, we get $x = 10$, \\
$\therefore$ the perimeter of $\triangle ABC$ is $12 + 5 + 13 = \boxed{30}$.


Question: As shown in the figure, AB is the diameter of $\odot O$. Point D is on the extension line of AB, and DC is tangent to $\odot O$ at point C. If $\angle A = 20^\circ$, what is the measure of $\angle D$?

Solution:
\textbf{Solution:} Draw line OC, \\
$\because$ DC is tangent to $\odot$O at point C, \\
$\therefore$ $\angle$OCD=$90^\circ$, \\
$\because$ $\angle$A=$20^\circ$, \\
$\therefore$ $\angle$OCA=$20^\circ$, \\
$\therefore$ $\angle$DOC=$40^\circ$, \\
$\therefore$ $\angle$D=$90^\circ-40^\circ=\boxed{50^\circ}$.


Question: As shown in the figure, $AB$ is the tangent to the circle $\odot O$ with $A$ being the point of tangency. The line segments $OA$ and $OB$ are connected. If $\angle B = 35^\circ$, what is the measure of $\angle AOB$?

Solution:
\textbf{Solution:} Since AB is the tangent line of the circle O, \\
it follows that $\angle$OAB=$90^\circ$, \\
Since $\angle$B=$35^\circ$, \\
it follows that $\angle$AOB=$90^\circ-\angle$B=\boxed{55^\circ}.


Question: As shown in the figure, the regular pentagon ABCDE has a side length of 6. A circle is drawn with A as the center and AB as the radius. What is the area of the shaded portion in the figure?

Solution:
\textbf{Solution:} Given that pentagon ABCDE is a regular pentagon with side length 6,\\
it follows that $AB=AE=6$ and $\angle BAE=\frac{180^\circ \times (5−2)}{5}=108^\circ$,\\
thus, the area of the shaded region in the diagram is $\frac{108\pi \times {6}^{2}}{360}=\boxed{\frac{54}{5}\pi}$.


Question: As shown in the figure, the regular pentagon ABCDE is inscribed in $\odot O$. Connecting AC, what is the degree measure of $\angle ACD$?

Solution:
\textbf{Solution:} As shown in the figure, connect OA, OE, OD, \\
since a regular pentagon ABCDE is inscribed in the circle $O$, \\
therefore, $\angle AOE = \angle DOE = \frac{360^\circ}{5} = 72^\circ$, \\
hence, $\angle ACD = \frac{1}{2} \angle AOD = \boxed{72^\circ}$.


Question: As shown in the figure, the radius $OA=2$ of the regular hexagon ABCDEF. What are the coordinates of point B?

Solution:
\textbf{Solution:} Draw OB, let BC intersect the x-axis at M, \\
$\because$ the radius of the regular hexagon ABCDEF, $OA=2$, \\
$\therefore$ OB=OA=AB=2, $\angle$AOB=$\angle$ABO=$\angle$OBC=$60^\circ$, \\
$\therefore$ $\angle$BOM=$30^\circ$, $\angle$BMO=$90^\circ$, \\
$\therefore$ BM=$\frac{1}{2}$OB=1, \\
$\therefore$ OM=$\sqrt{OB^{2}-BM^{2}}=\sqrt{2^{2}-1^{2}}=\sqrt{3}$, \\
$\therefore$ the coordinates of point B are ($-\sqrt{3}$, 1), \\
Hence the answer is: $\boxed{(-\sqrt{3}, 1)}$.


Question: As shown in the figure, the perimeter of $\triangle ABC$ is $20\text{cm}$, $BC=6\text{cm}$. Circle $O$ is the inscribed circle of $\triangle ABC$. The tangent line $MN$ of circle $O$ intersects $AB$ and $CA$ at points $M$ and $N$, respectively. What is the perimeter of $\triangle AMN$?

Solution:
\textbf{Solution:} Given that circle O is the inscribed circle of $\triangle ABC$, and the tangent of circle O, MN, intersects AB and AC at points M and N, respectively,

$\therefore$BF=BE, CF=CD, DN=NG, EM=GM, AD=AE,

$\because$ the perimeter of $\triangle ABC$ is 20cm, and BC=6cm,

$\therefore$AE=AD=$\frac{AB+AC-BC}{2}$=$\frac{20-BC-BC}{2}$=$\frac{20-12}{2}$=4(cm),

$\therefore$ the perimeter of $\triangle AMN$ is AM+MG+NG+AN=AM+ME+AN+ND=AE+AD=4+4=\boxed{8}(cm).


Question: In isosceles triangle $ABC$, where $AC = BC = 2$, and $D$ is a point on side $AB$. A circle with diameter $AD$, denoted as $\odot O$, is tangent to $BC$ at point $C$. What is the length of $BD$?

Solution:
\textbf{Solution:} Connect OC,

$\because$ AC = BC,

$\therefore$ $\angle$A = $\angle$B,

$\because$ OA = OC,

$\therefore$ $\angle$A = $\angle$ACO,

$\because$ $\angle$COB = $\angle$A + $\angle$ACO = 2$\angle$A,

$\therefore$ $\angle$COB = 2$\angle$B,

$\because$ circle O is tangent to BC at point C,

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

$\therefore$ $\angle$COB + $\angle$B = 2$\angle$B + $\angle$B = 90$^\circ$,

$\therefore$ $\angle$B = 30$^\circ$,

$\therefore$ OC = $\frac{\sqrt{3}}{3}$BC = $\frac{2\sqrt{3}}{3}$,

$\therefore$ OB = 2OC = $\frac{4\sqrt{3}}{3}$,

$\therefore$ BD = OB - OD = $\boxed{\frac{2\sqrt{3}}{3}}$,


Question: As shown in the figure, points A, B, C are on circle $O$, and $\angle ABC=28^\circ$. A tangent line to circle $O$ at point C intersects the extension of OA at point D. What is the measure of $\angle D$?

Solution:
\textbf{Solution:} Given that line CD is tangent to circle O,\\
thus OC $\perp$ CD,\\
therefore $\angle$OCD = $90^\circ$,\\
By the Inscribed Angle Theorem, we know that $\angle$COD = $2\angle$CBA = $56^\circ$,\\
therefore $\angle$D = $90^\circ$ - $\angle$OCD = $90^\circ$ - $56^\circ$ = $\boxed{34^\circ}$


Question: As shown in the figure, within the circle $\odot O$ where the radius is 1, the chord $AB = \sqrt{2}$. Construct an equilateral triangle $\triangle ABC$ inside $\odot O$ with $AB$ as one side. Rotate $\triangle ABC$ counterclockwise around point A. When side $AC$ first touches $\odot O$ tangentially, what is the angle of rotation?

Solution:
\textbf{Solution:} When $AC$ is tangent to $\odot O$, connect $CO$ and extend it to meet $AB$ at point $M$, connect $AO$,\\
since $\triangle ABC$ is an equilateral triangle,\\
thus, $CM\perp AB$,\\
since $AB = \sqrt{2}$,\\
thus, $AM = \frac{\sqrt{2}}{2}$,\\
since $OA = 1$,\\
thus, $OM = \frac{\sqrt{2}}{2}$,\\
therefore, $\angle OAM = 45^\circ$,\\
since $\angle OAC'=90^\circ$,\\
thus, $\angle BAC'=135^\circ$,\\
since $\angle C'AB'=60^\circ$,\\
thus, $\angle BAB'=75^\circ$,\\
therefore, when $AC$ is tangent to $\odot O$ for the first time, the angle of rotation is $\boxed{75^\circ}$;


Question: As shown in the figure, the radius of the circle is 4. What is the perimeter of the shaded area in the figure?

Solution:
\textbf{Solution:} As shown in the diagram, connect OA and OB, and draw OC $\perp$ AB at point C\\
$\angle OCB=90^\circ$, $\angle OBA=30^\circ$, with the radius of the circle being 4,\\
$\therefore$OC=2,\\
$\therefore$BC=$2\sqrt{3}$,\\
$\therefore$AB=2BC=$4\sqrt{3}$,\\
$\therefore$the perimeter of the shaded area is: $6\times 4\sqrt{3}=\boxed{24\sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle A=70^\circ$, and $\odot O$ is the inscribed circle of $\triangle ABC$. What is the value of $\angle BOC$?

Solution:
\textbf{Solution:} Since $\odot O$ is the inscribed circle of $\triangle ABC$,

it follows that OB bisects $\angle ABC$, and OC bisects $\angle ACB$,

therefore, $\angle OBC = \frac{1}{2}\angle ABC$, and $\angle OCB = \frac{1}{2}\angle ACB$,

hence, $\angle OBC + \angle OCB = \frac{1}{2}(\angle ABC + \angle ACB) = \frac{1}{2}(180^\circ - \angle A)$,

thus, $\angle BOC = 180^\circ - (\angle OBC + \angle OCB) = 180^\circ - \frac{1}{2}(180^\circ - \angle A) = 90^\circ + \frac{1}{2}\angle A = 90^\circ + \frac{1}{2} \times 70^\circ = \boxed{125^\circ}$


Question: As shown in the figure, given points A, B, and C are three points on circle $\odot O$, with radius $OC=2$, $\angle ABC=30^\circ$, the tangent line $AP$ intersects the extension line of $OC$ at point $P$, then how long is $AP$?

Solution:
\textbf{Solution:} As shown in the diagram, connect OA.\\
$\because$ $\angle ABC=30^\circ$,\\
$\therefore$ $\angle AOC=2\angle ABC=60^\circ$.\\
$\because$ PA is the tangent of circle $O$,\\
$\therefore$ $\angle OAP=90^\circ$,\\
$\therefore$ $\angle OPA=30^\circ$.\\
$\because$ OA=OC=2,\\
$\therefore$ OP=2OA=4.\\
$\therefore$ $AP=\sqrt{OP^{2}-OA^{2}}=2\sqrt{3}$,\\
Thus, $AP=\boxed{2\sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, if $\angle AOC = 110^\circ$, what is the measure of $\angle B$?

Solution:
\textbf{Solution}: Since circle $O$ is inscribed in $\triangle ABC$ and $\angle AOC=110^\circ$,\\
it follows that $\angle OAC+\angle OCA=180^\circ-\angle AOC=70^\circ$,\\
and $\angle OAC+\angle OCA=\frac{1}{2}(\angle BAC+\angle ACB)$,\\
therefore $\angle BAC+\angle ACB=140^\circ$,\\
thus $\angle B=180^\circ-(\angle BAC+\angle ACB)=180^\circ-140^\circ=\boxed{40^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $BC=8$, the perpendicular bisector of $AB$ intersects $BC$ at $D$, and the perpendicular bisector of $AC$ intersects $BC$ at $E$. Then, what is the perimeter of $\triangle ADE$?

Solution:
\textbf{Solution:} Given that the perpendicular bisector of line segment AB intersects BC at point D,\\
$ \therefore DB=DA$\\
Given that the perpendicular bisector of line segment AC intersects BC at point F,\\
$ \therefore EA=EC$,\\
$ \therefore $ the perimeter of $\triangle ADE = AD+DE+EA = DB+DE+EC = BC = \boxed{8}$


Question: As shown in the isosceles $\triangle ABC$, where $AB=AC$ and $\angle BAC=50^\circ$, the angle bisector of $\angle BAC$ intersects the perpendicular bisector of $AB$ at point $E$. By folding along $FG$ such that point $C$ coincides with point $E$, what is the measure of $\angle CFG$?

Solution:
\textbf{Solution:} As shown in the figure, connect BE,

$\because$ $\angle BAC=50^\circ$ and AE is the bisector of $\angle BAC$,

$\therefore$ $\angle BAE = \frac{1}{2} \times 50^\circ = 25^\circ$.

Moreover, $\because$ AB=AC,

$\therefore$ $\angle ABC=\angle ACB=65^\circ$.

$\because$ DE is the perpendicular bisector of AB,

$\therefore$ EA=EB,

$\therefore$ $\angle ABE=\angle BAE=25^\circ$,

$\therefore$ $\angle EBC=\angle ABC-\angle ABE=65^\circ-25^\circ=40^\circ$.

$\because$ AE is the bisector of $\angle BAC$ and AB=AC,

$\therefore$ The line AE bisects BC perpendicularly,

$\therefore$ EB=EC,

$\therefore$ $\angle ECB=\angle EBC=40^\circ$,

$\because$ $\angle ACB$ is folded along FG,

$\therefore$ EF=CF,

$\therefore$ $\angle CEF=\angle ECF=40^\circ$;

In $\triangle ECF$, $\angle EFC=180^\circ-\angle CEF-\angle ECF=180^\circ-40^\circ-40^\circ=100^\circ$,

$\therefore$ $\angle CFG = \frac{1}{2} \times \angle CFE = \boxed{50^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, and D is the midpoint of BC. If $\angle BAC=50^\circ$, what is the measure of $\angle BAD$?

Solution:
\textbf{Solution:} Since $AB=AC$, and point $D$ is the midpoint of $BC$, $\angle BAC=50^\circ$,

$\therefore AD$ is the angle bisector of $\angle BAC$,

$\therefore \angle BAD=\angle CAD=\frac{1}{2}\angle BAC=\boxed{25^\circ}$


Question: As shown in the figure, in the equilateral $\triangle ABC$, $AD$ is the height from $A$ to side $BC$, points $M$ and $N$ are respectively on $AD$ and $AC$, and $AM=CN$. Connect $BM$ and $BN$. When $BM+BN$ is at its minimum, what is the measure of $\angle MBN$?

Solution:
\textbf{Solution:} As shown in Figure 1, construct $CH\perp BC$ such that $CH=BC$, and connect $NH$, $BH$.\\
Since $\triangle ABC$ is an equilateral triangle, $AD\perp BC$, $CH\perp BC$,\\
it follows that $\angle DAC=\angle DAB=30^\circ$, $AD\parallel CH$,\\
thus $\angle HCN=\angle CAD=\angle BAM=30^\circ$,\\
since $AM=CN$, $AB=BC=CH$,\\
it follows that $\triangle ABM\cong \triangle CHN$ (SAS),\\
hence, $BM=HN$,\\
since $BN+HN\geq BH$,\\
it follows that when $B$, $N$, and $H$ are collinear, the value of $BM+BN=NH+BN$ is minimized,\\
As shown in Figure 2, when $B$, $N$, and $H$ are collinear,\\
since $\triangle ABM\cong \triangle CHN$,\\
it follows that $\angle ABM=\angle CHB=\angle CBH=45^\circ$,\\
since $\angle ABD=60^\circ$,\\
it follows that $\angle DBM=15^\circ$,\\
hence, $\angle MBN=45^\circ-15^\circ=\boxed{30^\circ}$,\\
thus, when the value of $BM+BN$ is minimized, $\angle MBN=30^\circ$.


Question: As shown in the figure, in $\triangle ABC$, with $\angle ACB=90^\circ$, fold $\triangle CBD$ along CD, so that point B precisely lands on point E on side AC. If $\angle A=22^\circ$, then what is the measure of $\angle EDA$?

Solution:
\textbf{Solution:} In $\triangle ABC$, $\angle ACB=90^\circ$, $\angle A=22^\circ$,\\
thus $\angle B=90^\circ-\angle A=68^\circ$,\\
By the property of folding, we have: $\angle CED=\angle B=68^\circ$,\\
therefore $\angle EDA=\angle CED-\angle A=\boxed{46^\circ}$.


Question: As shown in the figure, $\triangle ABC \cong \triangle ADE$, where D is on side $BC$, $\angle E = 35^\circ$, $\angle DAC = 30^\circ$. What is the measure of $\angle BDA$?

Solution:
\textbf{Solution:} Since $\triangle ABC\cong \triangle ADE$\\
$\therefore \angle E=\angle C=35^\circ$\\
Since in $\triangle ADC$, $\angle DAC=30^\circ$, $\angle C=35^\circ$\\
$\therefore \angle BDA=30^\circ+35^\circ=\boxed{65^\circ}$


Question: When a set of triangles is arranged as shown in the figure, what is the value of $ \angle \alpha $?

Solution:
\textbf{Solution:} By the exterior angle property of triangles, we have $ \angle \alpha =45^\circ+30^\circ=75^\circ$\\
Therefore, the answer is: $\boxed{75^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=90^\circ$, squares are constructed externally on the sides AB, BC, and AC, with areas of 225, 400, and $S$ respectively. What is the value of $S$?

Solution:
\textbf{Solution:} By the Pythagorean theorem, we have $AB^{2}+BC^{2}=AC^{2}$,\\
thus $S=AC^{2}=225+400=625$,\\
therefore, $S=\boxed{625}$.


Question: As illustrated, $E$ is a point within $\triangle ABC$, $BE$ bisects $\angle ABC$, $AD \perp BE$ at point $E$, intersecting $BC$ at point $D$, where point $D$ is precisely on the perpendicular bisector of side $AC$, $BC=10$, and $AB=6$. What is the length of $AE$?

Solution:
\textbf{Solution:} Given that BE bisects $\angle$ABC, and AD$\perp$BE,\\
it follows that $\triangle ABD$ is an isosceles triangle,\\
hence, AB=BD=6, and AE=ED,\\
given that BC=10,\\
thus, DC=BC-BD=10-6=4,\\
since point D lies exactly on the perpendicular bisector of side AC,\\
thus, DA=DC=4,\\
therefore, AE=\boxed{2},


Question: As shown in the figure, it is known that $\triangle ABC \cong \triangle ADE$. If $\angle E = 70^\circ$ and $\angle D = 30^\circ$, what is the degree measure of $\angle BAC$?

Solution:
\textbf{Solution:} Since $\angle E=70^\circ$ and $\angle D=30^\circ$,\\
it follows that $\angle EAD=180^\circ-\angle E-\angle D=180^\circ-70^\circ-30^\circ=80^\circ$,\\
Since $\triangle ABC \cong \triangle ADE$,\\
it follows that $\angle BAC=\angle EAD=\boxed{80^\circ}$


Question: As shown in the figure, $\triangle ABC \cong \triangle AED$, with point D on side BC. If $\angle EAB = 50^\circ$, what is the degree measure of $\angle ADE$?

Solution:
\textbf{Solution:} Given that $\triangle ABC \cong \triangle AED$,\\
$\therefore \angle BAC = \angle EAD$,\\
$\therefore \angle EAB + \angle BAD = \angle DAC + \angle BAD$,\\
$\therefore \angle DAC = \angle EAB = 50^\circ$,\\
$\because AD = AC$,\\
$\therefore \angle ADC = \angle C = \angle ADE = \frac{1}{2}(180^\circ - \angle DAC) = \boxed{65^\circ}$.


Question: As shown in the figure, AD is the angle bisector of $\triangle ABC$, $\angle C=28^\circ$, and AB+BD=AC. When $\triangle ABD$ is folded along the line containing AD, the point where B lands on side AC is marked as point E. What is the measure of $\angle AED$?

Solution:
\textbf{Solution:} According to the property of folding, we have \(BD=DE\), \(AB=AE\).

\(\because AB+BD=AC\), \(AC=AE+EC\),

\(\therefore AB+BD=AE+EC\),

\(\therefore DE=EC\),

\(\therefore \angle EDC=\angle C=28^\circ\),

\(\therefore \angle AED=\angle EDC+\angle C=28^\circ+28^\circ=\boxed{56^\circ}\).


Question: As shown in the figure, quadrilateral $OABC$ is a rectangle, where point $A$ is on the $x$-axis, and point $C$ is on the $y$-axis, with the coordinates of point $B$ being $(8, 6)$. If $\triangle OAB$ is folded along $OB$, with the corresponding point of $A$ being $E$, and $OE$ intersects $BC$ at point $D$, what are the coordinates of point $D$?

Solution:
\textbf{Solution:} Given that quadrilateral OABC is a rectangle, with point A on the x-axis and point C on the y-axis, and the coordinates of point B are $(8,6)$

$\therefore$ OC=AB=6, and BC=OA=8, with $\angle OCB=90^\circ$, and BC$\parallel$OA

$\therefore$$\angle AOB = \angle OBC$

Since triangle $\triangle OAB$ is folded along OB, the corresponding point for A is E

$\therefore$$\angle EOB=\angle AOB$

$\therefore$$\angle OBC=\angle EOB$

$\therefore$OD=BD

Let CD=x, then

$OD=DB=BC-CD=8-x$

In $\triangle OCD$, a right triangle,

$OC^{2}+CD^{2}=OD^{2}$

$\therefore x^{2}+6^{2}=(8-x)^{2}$

Solving gives: $x=\frac{7}{4}$

$\therefore$ The coordinates of point D are $\boxed{(\frac{7}{4},6)}$


Question: As shown in the figure, the perpendicular bisectors of line segments $AB$ and $BC$, denoted as $ {l}_{1}$ and $ {l}_{2}$, intersect at point $O$. If $\angle 1=40^\circ$, what is the measure of $\angle AOC$?

Solution:
\textbf{Solution:} Draw line BO and extend BO to point P, \\
$\because$ the perpendicular bisectors $l_{1}$ and $l_{2}$ of line segments AB and BC intersect at point O, and $l_{1}$, $l_{2}$ intersect AB, BC at points D, E respectively, \\
$\therefore$ AO=OB=OC, $\angle$BDO=$\angle$BEO=$90^\circ$, \\
$\therefore$ $\angle$DOE + $\angle$ABC=$180^\circ$, \\
$\because$ $\angle$DOE + $\angle$1=$180^\circ$, \\
$\therefore$ $\angle$ABC=$\angle$1=$40^\circ$, \\
$\because$ OA=OB=OC, \\
$\therefore$ $\angle$A=$\angle$ABO, $\angle$OBC=$\angle$C, \\
$\because$ $\angle$AOP=$\angle$A+$\angle$ABO, $\angle$COP=$\angle$C+$\angle$OBC, \\
$\therefore$ $\angle$AOC=$\angle$AOP+$\angle$COP=$\angle$A+$\angle$ABC+$\angle$C=2$\angle$ABC=2$\times$40$^\circ$=$\boxed{80^\circ}$;


Question: As shown in the figure, $AB\parallel CD$, $\angle A=45^\circ$, $\angle C=\angle E$, what is the measure of $\angle C$?

Solution:
\textbf{Solution:} Since $AB\parallel CD$ and $\angle A=45^\circ$,\\
it follows that $\angle DOE=\angle A=45^\circ$\\
Given that $\angle C=\angle E$ and $\angle C+\angle E=\angle DOE$,\\
we have $\angle C=\frac{1}{2}\angle DOE=\frac{1}{2}\times 45^\circ=\boxed{22.5^\circ}$


Question: As shown in the figure, it is known that $AB=AD$, $AE=AC=BC$, $\angle 1=\angle 2$, $\angle C=40^\circ$. What is the measure of $\angle ADE$ in degrees?

Solution:
\textbf{Solution:} Since $\angle 1=\angle 2$,\\
it follows that $\angle 1+\angle DAC=\angle 2+\angle DAC$,\\
which means $\angle BAC=\angle DAE$,\\
In $\triangle ABC$ and $\triangle ADE$,\\
since $\left\{\begin{array}{l}AB=AD\\ \angle BAC=\angle DAE\\ AC=AE\end{array}\right.$,\\
it follows that $\triangle ABC\cong \triangle ADE$ (SAS),\\
hence $\angle C=\angle E=40^\circ$, $AE=BC=DE$,\\
therefore $\angle ADE=\angle EAD=\frac{1}{2}\left(180^\circ-\angle E\right)=\frac{1}{2}\left(180^\circ-40^\circ\right)=\boxed{70^\circ}$.


Question: As shown in the figure, BD bisects $\angle ABC$ and intersects AC at point D. If $\angle C - \angle A = 20^\circ$, then what is the value of $\angle ADB$?

Solution:
\textbf{Solution:} Given $BD$ bisects $\angle ABC$ at point D,\\
it follows that $\angle ABD=\angle DBC$,\\
given $\angle C-\angle A=20^\circ$ hence $\angle C=\angle A+20^\circ$,\\
also, since $\angle ADB=\angle C+\angle DBC$,\\
it follows that $\angle ADB=\angle A+20^\circ+\angle DBC$,\\
since $\angle A+\angle ABD+\angle ADB=180^\circ$\\
which means $\angle A+\angle ABD=180^\circ-\angle ADB$,\\
it follows that $\angle ADB=180^\circ-\angle ADB+20^\circ$,\\
thus, $2\angle ADB=200^\circ$,\\
therefore, $\angle ADB=\boxed{100^\circ}$.


Question: As shown in the figure, fold the triangular paper ABC along AD so that point C falls on point E on side BD. If $\angle C = 45^\circ$, $\angle B = 30^\circ$, and AD = 2, what is the value of $AB^2 - AC^2$?

Solution:
\textbf{Solution:} Given that the triangle paper ABC is folded along AD, making point C fall on point E on side BD,\\
thus, $\angle ADC=\angle ADE=90^\circ$,\\
since $\angle C=45^\circ$, AD=2,\\
thus, AC= $\sqrt{2}$ AD= $2\sqrt{2}$,\\
since $\angle B=30^\circ$,\\
thus, AB=2AD=$2\times2$=4,\\
therefore, AB$^{2}$−AC$^{2}$=4$^{2}$−($2\sqrt{2}$)$^{2}$=\boxed{8},


Question: As shown in the figure, in $\triangle ABC$, $\angle B=90^\circ$, arcs are drawn with centers at $A$ and $C$ and radii greater than the length of $AC$, intersecting at points $M$ and $N$. Line $MN$ is drawn, intersecting $AC$ and $BC$ at $D$ and $E$ respectively. Connect $AE$. If $AB=6$ and $AC=10$, what is the perimeter of $\triangle ABE$?

Solution:
\textbf{Solution:} From the construction, we have $ED$ perpendicular bisector of $AC$,\\
$\therefore EA=EC$,\\
In $\triangle ABC$, $BC= \sqrt{AC^{2}-AB^{2}}=\sqrt{10^{2}-6^{2}}=8$,\\
$\therefore$ The perimeter of $\triangle ABE$ $=AB+BE+AE=AB+BE+CE=AB+BC=6+8=\boxed{14}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, the perpendicular bisector of $AC$ intersects $AB$ and $AC$ at $D$ and $E$, respectively. Connect $CD$. If $\angle B=70^\circ$, what is the measure of $\angle DCB$?

Solution:
\textbf{Solution:} Given that $\angle B=70^\circ$ and AB=AC,\\
it follows that $\angle A=180^\circ-2\times70^\circ=40^\circ$,\\
furthermore, since DE is perpendicular bisector of AC,\\
it implies that DA=DC,\\
thus, $\angle ACD=\angle A=40^\circ$,\\
therefore, $\angle BCD=\angle ACB-\angle ACD=70^\circ-40^\circ=\boxed{30^\circ}$.


Question: As shown in the figure, a ruler is placed together with a set square that includes a $30^\circ$ angle. If $\angle 1 = 24^\circ$, what is the measure of $\angle 2$?

Solution:
\textbf{Solution:} As shown in the figure, \\
since $\angle$BEF is an exterior angle of $\triangle AEF$, $\angle 1=24^\circ$, $\angle F=30^\circ$, \\
therefore $\angle$BEF=$\angle 1+\angle F=54^\circ$, \\
since AB$\parallel$CD, \\
therefore $\angle 2=\angle$BEF=$\boxed{54^\circ}$.


Question: As shown in the figure, it is known that points $D$ and $E$ are the midpoints of sides $BC$ and $AB$ in an equilateral triangle $ABC$, respectively, with $AD=6$. Point $F$ is a moving point on segment $AD$. What is the minimum value of $BF+EF$?

Solution:
\textbf{Solution:} Draw line CE intersecting AD at point F, and connect BF. \\
$\because$ $\triangle ABC$ is an equilateral triangle and D is the midpoint of BC, \\
$\therefore$ BF = CF, \\
$\therefore$ BF + EF = CF + EF = CE, \\
At this point, the value of BF + EF is minimal, and the minimum value is CE, \\
$\because$ D and E are respectively the midpoints of sides BC and AB of the equilateral $\triangle ABC$, \\
$\therefore$ AD = CE, \\
$\because$ AD = 6, \\
$\therefore$ CE = 6, \\
$\therefore$ The minimum value of BF + EF is \boxed{6},


Question: As shown in the figure, $ \angle AOB=30^\circ$, P is a point on the bisector of $\angle AOB$, $PM \perp OB$ at point M, $PN \parallel OB$ intersects OA at point N, if PM=1, what is the length of PN?

Solution:
\textbf{Solution:} Draw $PD\perp AO$ at point D, \\
since P lies on the angle bisector of $\angle AOB$ and $PM\perp OB$ at point M, \\
therefore $\angle AOP=\angle BOP=15^\circ$, and $PD=PM=1$, \\
since $PN\parallel OB$, \\
therefore $\angle NPO=\angle BOP=15^\circ$, \\
thus $\angle DNP=30^\circ$, \\
hence $PN=2PD=\boxed{2}$.


Question: As shown in the figure, $ \triangle ACB \cong \triangle ACB^{\prime }$, $ \angle BCB^{\prime } = 30^{\circ}$, what is the measure of $ \angle ACA^{\prime }$?

Solution:
\textbf{Solution:} Since $\triangle ACB\cong \triangle A'C'B'$,\\
it follows that $\angle ACB=\angle A'C'B'$, which means $\angle ACA'+\angle A'CB=\angle B'CB+\angle A'CB$,\\
therefore, $\angle ACA'=\angle B'CB$,\\
and given that $\angle B'CB=30^\circ$,\\
it follows that $\angle ACA'=\boxed{30^\circ}$.


Question: As shown in the diagram, in $\triangle ABC$, DE is the perpendicular bisector of AB. If $AE=4$ and $EC=2$, what is the length of BC?

Solution:
\textbf{Solution:} Since DE is the perpendicular bisector of AB, we have AE=4,\\
therefore, BE=AE=4,\\
since EC=2,\\
therefore, BC=BE+EC=\boxed{6}


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, and through point $A$, $DA\perp AC$ intersects $BC$ at point $D$. If $\angle B=2\angle BAD$, what is the degree measure of $\angle BAD$?

Solution:
\textbf{Solution:} Since AB = AC, \\
it follows that $\angle B = \angle C = 2\angle BAD$, \\
hence $\angle ADC = \angle BAD + \angle B$, \\
since $\angle B = 2\angle BAD$, \\
it implies $\angle ADC = 3\angle BAD$, \\
given DA $\perp$ AC, \\
thus $\angle DAC = 90^\circ$, \\
therefore $\angle ADC + \angle C = 90^\circ$, \\
it follows that $3\angle BAD + 2\angle BAD = 90^\circ$, \\
hence $\angle BAD = \boxed{18^\circ}$.


Question: As shown in the diagram, within $\triangle ABC$, $\angle C = 90^\circ$, $BE$ bisects $\angle ABC$, $ED$ perpendicularly bisects $AB$ at foot $D$. If $AC=12$, what is the length of $AE$?

Solution:
\textbf{Solution:} Given that $BE$ bisects $\angle ABC$,\\
hence, $\angle CBE = \angle ABE$,\\
and since $ED$ perpendicularly bisects $AB$ at $D$,\\
it follows that $EA=EB$,\\
thus $\angle A = \angle ABE$,\\
therefore $\angle CBE = \angle A = \angle ABE = \frac{1}{3}\left(180^\circ-\angle C\right)= \frac{1}{3}\left(180^\circ-90^\circ\right)=30^\circ$,\\
hence, $BE=2EC$, that is, $AE=2EC$,\\
and given $AE+EC=AC=12$,\\
it follows that $AE=\boxed{8}$.


Question: As shown in the figure, $\angle AOB = 30^\circ$, point $P$ is a fixed point inside $\angle AOB$ and $OP=3$. If points $M$ and $N$ are moving points on rays $OA$ and $OB$ different from point $O$, respectively, what is the minimum perimeter of $\triangle PMN$?

Solution:
\textbf{Solution:} As shown in the diagram, let $E$ and $D$ be the symmetric points of point $P$ with respect to $OA$ and $OB$, respectively. Draw lines $DN$, $EM$, $DE$, $OD$, and $OE$,\\
then $OA$ perpendicularly bisects $PE$, and $OB$ perpendicularly bisects $PD$,\\
$\therefore EM=PM$, $DN=PN$, $OD=OP=OE=3$,\\
$\therefore$ the perimeter of $\triangle PMN$ is $PM+MN+PN=EM+MN+DN$,\\
As the shortest distance between two points is a straight line, when points $E$, $M$, $N$, $D$ are collinear, the value of $EM+MN+DN$ is minimized, and the minimum value is the length of $DE$,\\
$\because OE=OP$ and $OA\perp PE$,\\
$\therefore \angle AOE=\angle AOP$ (angle-side-angle congruence in isosceles triangle),\\
Similarly, it can be obtained: $\angle BOD=\angle BOP$,\\
$\because \angle AOB=30^\circ$,\\
$\therefore \angle BOP+\angle AOP=30^\circ$,\\
$\therefore \angle DOE=\angle BOD+\angle BOP+\angle AOP+\angle AOE=2\left(\angle BOP+\angle AOP\right)=60^\circ$,\\
And since $OD=OE=3$,\\
$\therefore \triangle DOE$ is an equilateral triangle,\\
$\therefore DE=OD=3$,\\
$\therefore$ the minimum value of $EM+MN+DN$ is 3,\\
$\therefore$ the minimum perimeter of $\triangle PMN$ is $\boxed{3}$,


Question: As shown in the figure, $\angle ACD$ is the exterior angle of $\triangle ABC$, and $CE$ bisects $\angle ACD$. If $\angle A=70^\circ$ and $\angle B=40^\circ$, what is the measure of $\angle ECD$?

Solution:
\textbf{Solution:}\\
Since $\angle A=70^\circ$ and $\angle B=40^\circ$,\\
it follows that $\angle ACD=\angle A+\angle B=110^\circ$,\\
and since CE bisects $\angle ACD$,\\
we have $\angle ECD= \frac{1}{2}\angle ACD=\boxed{55^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=128^\circ$, $P_{1}$ is the intersection point of the bisector of the interior angle $\angle ABC$ of $\triangle ABC$, which is $BP_{1}$, and the bisector of the exterior angle $\angle ACE$, which is $CP_{1}$. $P_{2}$ is the intersection point of the bisector of the interior angle $\angle P_{1}BC$ of $\triangle BP_{1}C$, which is $BP_{2}$, and the bisector of the exterior angle $\angle P_{1}CE$, which is $CP_{2}$. $P_{3}$ is the intersection point of the bisector of the interior angle $\angle P_{2}BC$ of $\triangle BP_{2}C$, which is $BP_{3}$, and the bisector of the exterior angle $\angle P_{2}CE$, which is $CP_{3}$. Continuing in this manner, what is the measure of $\angle P_{6}$?

Solution:
\textbf{Solution:} Given that the internal angle bisector BP and the external angle bisector CP1 of $\triangle ABC$ intersect at $P_{1}$,\\
hence, $\angle P_{1}BC = \frac{1}{2}\angle ABC$, $\angle P_{1}CE = \frac{1}{2}\angle ACE$,\\
since $\angle ACE = \angle A + \angle ABC$, $\angle P_{1}CE = \angle P_{1}BC + \angle P_{1}$,\\
thus, $\frac{1}{2}(\angle A + \angle ABC) = \angle P_{1}BC + \angle P_{1} = \frac{1}{2}\angle ABC + \angle P_{1}$,\\
therefore, $\angle P_{1} = \frac{1}{2}\angle A = \frac{1}{2} \times 128^\circ = 64^\circ$,\\
similarly, $\angle P_{2} = \frac{1}{2}\angle P_{1} = 32^\circ$,\\
hence, $\angle P_{6} = \boxed{2^\circ}$,


Question: As shown in the figure, in $ \triangle ABC$, $ \angle C=90^\circ$, and $ AD$ bisects $ \angle BAC$. Given $ AB=10$ and $ CD=3$, what is the area of $ \triangle ABD$?

Solution:
\textbf{Solution}: As shown in the figure, draw DE$\perp$AB at point E through point D,\\
since $\angle C=90^\circ$ and AD bisects $\angle$BAC,\\
it follows that $DE=CD=3$,\\
thus, the area of $\triangle ABD$ $=\frac{1}{2}AB\cdot DE=\frac{1}{2}\times 10\times 3=\boxed{15}$.


Question: As shown in the figure, in the isosceles triangle $ABC$, where $AB=AC$ and $BD\perp AC$, with $\angle ABC=70^\circ$, what is the value of $\angle DBC$?

Solution:
\textbf{Solution:} \\
Since AB = AC, \\
thus $\angle$ACB = $\angle$ABC = 70°, \\
Since BD $\perp$ AC, \\
thus $\angle$BDC = 90°, \\
hence $\angle$DBC = 90° - $\angle$ACB = \boxed{20°}.


Question: As shown in the figure, in $\triangle ABC$, $\angle B = \angle C$, and M, N, P are points on sides AB, AC, BC, respectively, with $BM=CP$ and $CN=BP$. If $\angle MPN = 44^\circ$, what is the measure of $\angle A$?

Solution:
\textbf{Solution:}

In $\triangle MBP$ and $\triangle PCN$, we have
\[
\left\{\begin{array}{l}
BM=PC\\
\angle B=\angle C\\
BP=NC
\end{array}\right.,
\]
hence, $\triangle MBP\cong \triangle PCN$ (SAS),

therefore, $\angle BMP=\angle NPC$,

since $\angle MPN=44^\circ$,

therefore, $\angle BPM+\angle CPN=180^\circ-\angle MPN=136^\circ$,

therefore, $\angle BPM+\angle BMP=136^\circ$,

thus, $\angle B=180^\circ-136^\circ=44^\circ$,

since $\angle A+\angle B+\angle C=180^\circ$,

therefore, $\angle A=\boxed{92^\circ}$.


Question: As shown in the figure, in $ \triangle ABC$, $ \angle ACB=90^\circ$, $AC=3$, and $BC=4$. Fold side $AC$ along $CE$, making point $A$ fall onto point $D$ on $AB$; then fold side $BC$ along $CF$, making point $B$ fall onto point $B'$ on the extension line of $CD$. The two creases intersect hypotenuse $AB$ at points $E$ and $F$, respectively. What is the length of segment $B'F$?

Solution:
\textbf{Solution:} Let's consider right-angled triangle $\triangle ABC$ where $\angle ACB=90^\circ$, $AC=3$, and $BC=4$,\\
hence $AB=5$,\\
based on the properties of folding, we know $AC=CD$, $\angle A=\angle CDE$, $CE\perp AB$,\\
thus $B'D=BC-CD=4-3=1$, $\angle DCE+\angle B'CF=\angle ACE+\angle BCF$,\\
since $\angle ACB=90^\circ$,\\
thus $\angle ECF=45^\circ$,\\
hence $\triangle ECF$ is an isosceles right triangle,\\
thus $EF=CE$, $\angle EFC=45^\circ$,\\
thus $\angle BFC=\angle B'FC=135^\circ$,\\
thus $\angle B'FD=90^\circ$,\\
since $S_{\triangle ABC}=\frac{1}{2}AC\cdot BC=\frac{1}{2}AB\cdot CE$,\\
thus $AC\cdot BC=AB\cdot CE$,\\
thus $CE=\frac{12}{5}$,\\
thus $EF=\frac{12}{5}$, $ED=AE=\sqrt{AC^2-CE^2}=\frac{9}{5}$,\\
thus $DF=EF-ED=\frac{3}{5}$,\\
thus $B'F=\sqrt{B'^2D^2-DF^2}=\boxed{\frac{4}{5}}$.


Question: As shown in the Cartesian coordinate system, the coordinate of point A is $(-3, 0)$, and the coordinate of point B is $(0, 4)$. Point C is a point on OB. If $\triangle ABC$ is folded along AC, so that point B lands exactly on the x-axis at point $B'$, what is the coordinate of point C?

Solution:
\textbf{Solution:} By the problem statement, we have $A{B}^{\prime }=AB$ and $C{B}^{\prime }=CB$,\\
Since the coordinates of point A are $(-3,0)$ and the coordinates of point B are $(0,4)$,\\
Therefore, OA=3 and OB=4,\\
Therefore, $AB=\sqrt{OA^{2}+OB^{2}}=5$,\\
Therefore, $A{B}^{\prime }=5$,\\
Therefore, $O{B}^{\prime }=2$,\\
In $\triangle {B}^{\prime }OC$ right-angled at $B'$, $O{C}^{2}+O{{B}^{\prime }}^{2}={B}^{\prime }{C}^{2}$,\\
Therefore, $O{C}^{2}+2^{2}={(4-OC)}^{2}$,\\
Solving, we get: $OC=\frac{3}{2}$,\\
Therefore, the point $C\left(0,\frac{3}{2}\right)$.\\
Hence, the answer is: The coordinates of point C are $\boxed{\left(0,\frac{3}{2}\right)}$.


Question: As shown in the figure, in $\triangle AED$, $\angle AED=90^\circ, AB=AC=AD, EC=3, BE=11$, then what is the value of $ED$?

Solution:
\textbf{Solution:} As the figure shows: draw $AF\perp BC$ with foot F \\
$\because$ $EC=3$, $BE=11$ \\
$\therefore$ $BC=BE+EC=11+3=14$ \\
$\because$ $AB=AC$ \\
$\therefore$ $BF=CF=\frac{1}{2}BC=7$ \\
$\therefore$ $EF=FC-EC=7-3=4$ \\
In $\triangle ADE$, by Pythagoras' theorem, $DE^2=AD^2-AE^2$ \\
In $\triangle AEF$, by Pythagoras' theorem, $AE^2=AF^2+EF^2$ \\
And $\because$ $AB=AD$ \\
$\therefore$ $DE^2=AB^2-(AF^2+EF^2)$ \\
In $\triangle ABF$, by Pythagoras' theorem, $AB^2=AF^2+BF^2$ \\
$\therefore$ $DE^2=AF^2+BF^2-(AF^2+EF^2)=BF^2-EF^2=7^2-4^2=33$ \\
$\therefore$ $DE=\sqrt{33}$ \\
Thus, the answer is: $DE=\boxed{\sqrt{33}}$


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=90^\circ$, $\angle A=60^\circ$, D is a point on side $AC$, $\angle DBC=30^\circ$, $AC=12\, \text{cm}$, what is the perimeter of $\triangle ABD$?

Solution:
\textbf{Solution:} Since $\triangle ABC\cong \triangle ADE$,\\
it follows that $\angle E=\angle C$,\\
In $\triangle ABC$, $\angle BAC=80^\circ$, $\angle B=40^\circ$,\\
therefore, $\angle E=\angle C=180^\circ-80^\circ-40^\circ=\boxed{60^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, arcs are drawn with points A and C as the centers and with radii longer than $AC$, intersecting at points M and N; line $MN$ intersects $BC$ and $AC$ at points D and E, respectively. If $AE=6\,\text{cm}$ and the perimeter of $\triangle ABD$ is $26\,\text{cm}$, what is the perimeter of $\triangle ABC$?

Solution:
\textbf{Solution:} Given that: $MN$ is the perpendicular bisector of $AC$,\\
$\therefore$ $AD=CD$, $AC=2AE=12cm$,\\
$\because$ the perimeter of $\triangle ABD$ is 26cm,\\
which means $AB+AD+BD=26cm$,\\
$\therefore$ $AB+CD+BD=AB+BC=26cm$,\\
$\therefore$ the perimeter of $\triangle ABC$ is $AB+BC+AC=26+12=\boxed{38cm}$;


Question: As shown in the figure, $ \angle B=35^\circ$, and $CD$ is the perpendicular bisector of $AB$, then what is the value of $ \angle ACE$?

Solution:
\textbf{Solution:} Since CD is the perpendicular bisector of AB,\\
it follows that $BC=AC$,\\
hence $\angle A=\angle B=35^\circ$,\\
therefore, $\angle ACE=\angle A+\angle B=2\times 35^\circ=70^\circ$,\\
thus, $\angle ACE = \boxed{70^\circ}$


Question: As shown in the figure, point E is on the bisector of $\angle BAC$ on the line AP, with AB = 4, and the area of $\triangle ABE$ is 12. What is the distance from point E to the line AC?

Solution:
\textbf{Solution:}
Given that $AB = 4$ and the area of $\triangle ABE$ is $12$,\\
it follows that the distance from point E to line AB $= \frac{2 \times 12}{4} = 6$,\\
since E is a point on the bisector of $\angle BAC$,\\
therefore, the distance from point E to line AC $=$ \boxed{6}.


Question: As shown in the figure, in the right-angled triangle $ABC$, $\angle ACB=90^\circ$, $\angle ABC=30^\circ$. Three squares are constructed externally on the sides AC, BC, and AB of $\triangle ABC$, and the areas of these three squares are denoted as $S_1$, $S_2$, and $S_3$, respectively. Given that $S_1=9$, what is the value of $S_3$?

Solution: \textbf{Solution:} Since $S_{1}=AC^{2}=9$, it results that $AC=3$. In $\triangle ABC$, $\angle ACB=90^\circ$ and $\angle ABC=30^\circ$, hence $AB=2AC=6$. Therefore, $S_{3}=AB^{2}=\boxed{36}$.


Question: As shown in the figure, in the right-angled triangle $ABC$, where $\angle C = 90^\circ$, $AD$ bisects $\angle BAC$ and intersects $BC$ at point $D$. Line $DE \parallel AB$ intersects $AC$ at point $E$. Given that $CE = 3$ and $CD = 4$, what is the length of $AD$?

Solution:
\textbf{Solution:} In right $\triangle ABC$, $\angle C=90^\circ$, $CE=3$, $CD=4$, \\
$\therefore DE^{2}=CD^{2}+CE^{2}=9+16=25$, \\
$\therefore DE=5$, \\
$\because DE\parallel AB$, \\
$\therefore \angle BAD=\angle ADE$, \\
$\because AD$ bisects $\angle BAC$, \\
$\therefore \angle BAD=\angle CAD$, \\
$\therefore \angle CAD=\angle ADE$, \\
$\therefore AE=DE=5$, \\
$\therefore AC=AE+EC=5+3=8$, \\
$\therefore$ in right $\triangle ACD$, \\
$\therefore AD^{2}=AC^{2}+CD^{2}=64+16=80$, \\
$\therefore AD=\boxed{4\sqrt{5}}$.


Question: As shown in the figure, $DE$ is the perpendicular bisector of side $BC$ of $\triangle ABC$, intersecting sides $AB$ and $BC$ at points $D$ and $E$, respectively, and $AB=9$, $AC=6$. What is the perimeter of $\triangle ACD$?

Solution:
\textbf{Solution:} Since DE is the perpendicular bisector of side BC in $\triangle ABC$, \\
it follows that BD = CD, \\
and since the perimeter of $\triangle ACD$ is AD + CD + AC = AD + BD + AC = AB + AC = 9 + 6 = \boxed{15}.


Question: In a right-angled triangle $ABC$, with $\angle ABC=90^\circ$, $AC=13$, and $AB=12$, what is the sum of the perimeters of the five small right-angled triangles in the figure?

Solution:
\textbf{Solution:} In $\triangle ABC$, $\angle ABC=90^\circ$, $AC=13$, and $AB=12$. By the Pythagorean theorem, $BC= \sqrt{AC^2-AB^2}=\sqrt{13^2-12^2}=5$. Based on the properties of translation, we know that the sum of the two legs of the five small right-angled triangles in the figure equals $AB+BC=12+5=17$,

$\therefore$ The perimeter of the five small right-angled triangles in the figure is $5 \times (AB+BC) = 5 \times 17 = \boxed{85}$.


Question: As shown in the figure, in $\triangle ABC$, D is a point on the extension of BC, $\angle B=50^\circ$, $\angle ACD=120^\circ$. What is the value of $\angle A$?

Solution:
\textbf{Solution:} It is known from the properties of the exterior angles of a triangle that,
\[
\angle A = \angle ACD - \angle B = 70^\circ,
\]
thus, \(\angle A = \boxed{70^\circ}\).


Question: As shown in the figure, in the right triangle $ABC$, where $\angle C=90^\circ$, the bisector of $\angle ABC$, $BD$, intersects $AC$ at $D$. If $CD = 4\, \text{cm}$, what is the distance from point $D$ to $AB$, denoted as $DE$?

Solution:
\textbf{Solution:} Given that $\angle C=90^\circ$ and BD bisects $\angle ABC$, with $DE \perp AB$,\\
it follows that $DE=CD$,\\
Since $CD=4\text{cm}$,\\
the distance from point D to AB, DE, is $\boxed{4\text{cm}}$.


Question: As shown in the figure, in $\triangle ABC$, it is known that $S_{\triangle ABD} : S_{\triangle ACD} = 2 : 1$, point E is the midpoint of AB, and the area of $\triangle ABC$ is $9\,\text{cm}^2$. What is the area of $\triangle AED$?

Solution:
\textbf{Solution:} Given that $S_{\triangle ABD} : S_{\triangle ACD} = 2 : 1$,\\
it follows that $S_{\triangle ACD} = \frac{1}{2}S_{\triangle ABD}$,\\
since point E is the midpoint of AB,\\
thus $S_{\triangle ADE} = S_{\triangle BDE} = \frac{1}{2}S_{\triangle ABD}$,\\
hence $S_{\triangle ADE} = \frac{1}{3}S_{\triangle ABC} = \frac{1}{3} \times 9 = \boxed{3\text{cm}^2}$.


Question: As shown in the figure, a triangular paper piece with a $60^\circ$ angle is cut to remove this $60^\circ$ angle, resulting in a quadrilateral. What is the degree measurement of $\angle 1+\angle 2$?

Solution:
\textbf{Solution}: As shown in the figure, \\
since $\angle C=60^\circ$, \\
therefore $\angle A+\angle B=180^\circ-\angle C=180^\circ-60^\circ=120^\circ$, \\
therefore $\angle 1+\angle 2=360^\circ-(\angle A+\angle B)=360^\circ-120^\circ=\boxed{240^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, it is known that points $D$, $E$, and $F$ are the midpoints of $BC$, $AD$, and $CE$, respectively. If the area of $\triangle ABC$ is $60\, \text{cm}^2$, what is the area of the shaded region?

Solution:
\textbf{Solution:} Point D is the midpoint of BC,\\
$\therefore S_{\triangle ABD}=S_{\triangle ADC}=\frac{1}{2}S_{\triangle ABC}=30$,\\
$\because$ point E is the midpoint of AD,\\
$\therefore S_{\triangle BDE}=S_{\triangle CDE}=\frac{1}{2}S_{\triangle ADC}=15$;\\
$\therefore S_{\triangle BEC}=2S_{\triangle CDE}=30$,\\
$\because$ point F is the midpoint of EC,\\
$\therefore S_{\triangle BEF}=\frac{1}{2}S_{\triangle BEC}=15$;\\
Therefore, the answer is: $\boxed{15}$


Question: As shown in the figure, in $\triangle ABC$, $BD = \frac{1}{5}BC$, $AE = \frac{1}{4}AD$, and $CF = \frac{1}{2}CE$. Given that the area of $\triangle ABC = 30$, what is the area of $\triangle DEF$?

Solution:
\textbf{Solution:}\\
$\because BD= \frac{1}{5} BC$,\\
$\therefore CD= \frac{4}{5} BC$,\\
$\therefore S_{\triangle ACD}= \frac{4}{5} S_{\triangle ABC}= \frac{4}{5} \times 30 = 24$,\\
$\because AE= \frac{1}{4} AD$,\\
$\therefore DE= \frac{3}{4} AD$,\\
$\therefore S_{\triangle CDE}= \frac{3}{4} S_{\triangle ACD}= \frac{3}{4} \times 24 = 18$,\\
$\because CF= \frac{1}{2} CE$,\\
$\therefore EF= \frac{1}{2} CE$,\\
$\therefore S_{\triangle DEF}= \frac{1}{2} S_{\triangle CDE}= \frac{1}{2} \times 18 = \boxed{9}$.


Question: As shown in the figure, rectangle $ABCD$ is folded along the diagonal $AC$, making point $D$ fall on point $M$, with $AM$ intersecting $BC$ at point $E$, and $AB=4$, $AD=8$. What is the length of $BE$?

Solution:
\textbf{Solution:} Given the properties of the rectangle and the fold, we have: $AB=CD=MC=4$, $AD=BC=AM=8$, $\angle AMC=90^\circ$,\\
In $\triangle ABE$ and $\triangle CME$,\\
$\left\{\begin{array}{l}
\angle ABC=\angle CME\\
\angle AEB=\angle CEM\\
AB=CM
\end{array}\right.$\\
$\therefore \triangle ABE\cong \triangle CME$ (AAS),\\
$\therefore BE=ME$,\\
Let $BE=x$, then $ME=x$, $AE=8-x$,\\
In $\triangle ABE$, by the Pythagorean theorem: $AB^2+BE^2=AE^2$, i.e., $4^2+x^2=(8-x)^2$,\\
Solving this, we get: $x=\boxed{3}$, hence $BE=3$.


Question: As shown in the figure, the area of square $ABCD$ is $100 \text{cm}^2$. Triangle $\triangle ABP$ is a right-angled triangle with $\angle P = 90^\circ$, and $PB = 6\text{cm}$. What is the length of $AP$?

Solution:
\textbf{Solution:} Given that the area of square ABCD is $100\,\text{cm}^{2}$,\\
it follows that AB=10cm,\\
since \(\triangle\)ABP is a right triangle with \(\angle P=90^\circ\) and PB=6cm,\\
therefore AP=$\sqrt{AB^{2}-BP^{2}}=\sqrt{10^{2}-6^{2}}=\boxed{8}\,\text{cm}$.


Question: As shown in the figure, in the rectangle ABCD, AB = 4 cm, and BC = 8 cm. Now it is folded along EF, making point C coincide with point A. What is the length of AF?

Solution:
\textbf{Solution:} Let $AF=x$ cm, then $DF=(8-x)$ cm, \\
since in the rectangle ABCD, $AB=4$ cm, $BC=8$ cm, and it is folded along EF such that point C coincides with point A, \\
therefore $DF=D^\prime F$, $AD^\prime=CD=AB=4$, \\
in $\triangle AD^\prime F$, since $AF^2=A{D^\prime}^2+D^\prime F^2$, \\
therefore $x^2=4^2+(8-x)^2$, \\
Solving yields: $x=\boxed{5}$ (cm).


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, point D is on side $AC$ and $AD=BD$, $M$ is the midpoint of $BD$. If $AC=8$ and $BC=4$, what is the length of $CM$?

Solution:
\textbf{Solution:} Since $\angle ACB=90^\circ$ and M is the midpoint of BD,\\
$\therefore CM=\frac{1}{2}BD$,\\
Let $CM=x$, then $BD=AD=2x$,\\
Since $AC=8$,\\
$\therefore CD=AC-AD=8-2x$,\\
In $\triangle BCD$, according to the Pythagorean theorem,\\
$BC^2+CD^2=BD^2$,\\
i.e., $4^2+(8-2x)^2=(2x)^2$,\\
Solving for $x$, we get $x=\boxed{\frac{5}{2}}$.


Question: As shown in the figure, it is known that the area of $\triangle ABC$ is 48, with $AB=AC=8$. Point D is on side $BC$, and lines $DE \perp AB$ at E, and $DF \perp AC$ at F are drawn. If $DF=2DE$, then what is the length of $DF$?

Solution:
\textbf{Solution:} Construct line segment AD, \\
since $S_{\triangle ABD} + S_{\triangle ACD} = S_{\triangle ABC}$, AB = AC = 8, \\
therefore $\frac{1}{2} \times 8 \cdot DF + \frac{1}{2} \times 8 \cdot DE = 48$, \\
therefore DE + DF = 12, \\
since DF = 2DE, \\
therefore DF = \boxed{8}.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, D is a point on $BC$, $DE\perp AB$ at point E, $AE=AC$, connect $AD$, if $BC=8$, what is the value of $BD+DE$?

Solution:
\textbf{Solution:}\\
Since $DE \perp AB$,\\
it follows that $\angle AED = \angle C = 90^\circ$,\\
Since $AE = AC$, $AD = AD$,\\
it follows that $\triangle ADE \cong \triangle ADC\ (HL)$,\\
therefore $DE = DC$,\\
thus $BD + DE = BC + DC = BC = \boxed{8}$.


Question: In the given figure, within $\triangle ABC$, $AD \perp BC$, $AE$ is the angle bisector of $\angle BAC$. If $\angle B = 72^\circ$ and $\angle C = 38^\circ$, what is the measure of $\angle DAE$?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $\angle B=72^\circ$ and $\angle C=38^\circ$,\\
thus $\angle BAC=180^\circ-\angle B-\angle C=180^\circ-72^\circ-38^\circ=70^\circ$,\\
since $AD\perp BC$,\\
thus $\angle ADC=90^\circ$,\\
hence $\angle DAC=90^\circ-\angle C=90^\circ-38^\circ=52^\circ$;\\
since $AE$ bisects $\angle BAC$,\\
thus $\angle EAC= \frac{1}{2}\angle BAC=35^\circ$;\\
therefore $\angle DAE=\angle DAC-\angle EAC=52^\circ-35^\circ=\boxed{17^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=40^\circ$, $\angle C=30^\circ$, and point D is a point on side BC. After folding $\triangle ADC$ along line AD, point C falls onto point E. If $DE\parallel AB$, then what is the measure of $\angle ADE$?

Solution:
\textbf{Solution:} In $\triangle ABC$, $\angle B=40^\circ$, $\angle C=30^\circ$,\\
thus $\angle BAC=180^\circ-\angle B-\angle C=180^\circ-40^\circ-30^\circ=110^\circ$;\\
Because after folding $\triangle ADC$ along the line AD, point C falls on point E,\\
thus $\angle C=\angle E=30^\circ$, $\angle EAD=\angle CAD$, $\angle ADE=\angle ADC$,\\
Because DE$\parallel$AB,\\
thus $\angle BAE=\angle E=30^\circ$,\\
thus $\angle EAD=\angle CAD= \frac{1}{2}(\angle BAC-\angle BAE)= \frac{1}{2}(110^\circ-30^\circ)=40^\circ$;\\
thus $\angle ADE=\angle ADC=180^\circ-\angle C-\angle CAD=180^\circ-30^\circ-40^\circ=\boxed{110^\circ}$.


Question: A ruler and a right-angled triangle ruler containing a $30^\circ$ angle are placed as shown in the figure. If $\angle 1 = 22^\circ$, what is the degree measure of $\angle 2$?

Solution:
\textbf{Solution:} As shown in the figure, draw line CE through point C such that CE$\parallel$AB, \\
$\because$AB$\parallel$GH, $\therefore$AB$\parallel$GH$\parallel$CE, $\therefore$$\angle$1=$\angle$DCE=$22^\circ$, $\angle$2=$\angle$ECF, $\therefore$$\angle$2=$\angle$ECF=$\angle$DCF−$\angle$DCE=$60^\circ−22^\circ=\boxed{38^\circ}$.


Question: As shown in the figure, $AB$, $AC$, $BD$ are the tangents to circle $\odot O$ with the points of tangency at $P$, $C$, and $D$, respectively. If $AB=5$ and $AC=3$, what is the length of $BD$?

Solution:
\textbf{Solution:} Since $AB$, $AC$, and $BD$ are tangents to the circle $\odot O$,\\
it follows that AP=AC and BP=BD,\\
and since $AB=5$ and $AC=3$,\\
it follows that AP=3,\\
thus, BD=BP=AB-AP=\boxed{2}.


Question: As shown in the figure, when a circular card with a radius of \(2 \, \text{cm}\) is folded such that the arc exactly passes through the center of the circle, what is the length of the crease?

Solution:
\textbf{Solution:} As shown in the figure, connect OA, and connect point O with the symmetrical point E with respect to AB, intersecting AB at point D,\\
By folding, we get $OD=DE=\frac{1}{2}OE=1\text{ cm}$, $OD\perp AB$,\\
$\therefore AD=BD=\frac{1}{2}AB$,\\
In $\triangle AOD$, $OD^2+AD^2=OA^2$,\\
$\therefore AD=\sqrt{OA^2-OD^2}=\sqrt{2^2-1^2}=\sqrt{3}\text{ cm}$,\\
$\therefore AB=2AD=2\sqrt{3}\text{ cm}$,\\
Hence, the length of the crease is $AB=\boxed{2\sqrt{3}\text{ cm}}$.


Question: As shown in the Cartesian coordinate system, point $P$ is in the first quadrant. Circle $\odot P$ is tangent to both the x-axis and y-axis, and it passes through the vertex $C$ of rectangle $AOBC$, intersecting with $BC$ at point $D$. Given that the radius of $\odot P$ is 5 and the coordinates of point $A$ are $(0, 8)$, what are the coordinates of point $D$?

Solution:
\textbf{Solution:}\\
Let the points of tangency be $G$, $E$, respectively. Connect $PG$, $PE$, $PC$, $PD$, and extend $EP$ to intersect $BC$ at $F$. Then $PG=PE=PC=5$, and quadrilateral $OBFE$ is a rectangle.\\
$\because$OA=8,\\
$\therefore$CF=8-5=3,\\
$\therefore$PF=4,\\
$\therefore$OB=EF=5+4=9.\\
$\because$PF passes through the circle's center,\\
$\therefore$DF=CF=3,\\
$\therefore$BD=8-3-3=2,\\
$\therefore$The coordinates of $D$ are $\boxed{(9, 2)}$.


Question: As shown in the figure, PA and PB are tangent lines to the circle $O$, with A and B being the points of tangency. AC is a diameter of the circle $O$. Given that $\angle P = 50^\circ$, what is the measure of $\angle ACB$?

Solution:
\textbf{Solution:} Connect OB, as shown in the diagram,\\
since PA and PB are tangents to the circle centered at O,\\
it follows that OA $\perp$ PA, and OB $\perp$ PB,\\
thus, $\angle$OAP=$\angle$OBP=90$^\circ$,\\
hence, $\angle$AOB=360$^\circ$-90$^\circ$-90$^\circ$-50$^\circ$=130$^\circ$,\\
since OB=OC,\\
it follows that $\angle$OCB=$\angle$OBC,\\
and since $\angle$AOB=$\angle$OCB+$\angle$OBC,\\
thus, $\angle$OCB= $\frac{1}{2} \times 130^\circ=65^\circ$,\\
that is, $\angle$ACB=$\boxed{65^\circ}$.


Question: As shown in the diagram, a right-angled triangle ruler containing a $30^\circ$ angle is positioned as shown together with a straight ruler. If $\angle 1 = 62^\circ$, what is the degree measure of $\angle 2$?

Solution:
\textbf{Solution:} As shown in the diagram:\\
$\because \angle 1=62^\circ$,\\
$\therefore$ by the property that parallel lines have equal alternate interior angles, we have $\angle 3=62^\circ$,\\
$\therefore$ by the exterior angle theorem of a triangle, we have $\angle 4=\angle 3-30^\circ=32^\circ$,\\
$\therefore \angle 2=\angle 4=\boxed{32^\circ}$.


Question: As shown in the figure, fold the rectangular paper ABCD upwards along AE, so that point B falls at point F on side DC. If $\angle FEC = 28^\circ$, what is the value of $\angle EAB$ in degrees?

Solution:
\textbf{Solution:} By folding, we obtain $\angle BEA = \angle FEA$,\\
$\because \angle FEC = 28^\circ$,\\
$\therefore \angle AEB = (180^\circ - 28^\circ) \div 2 = 76^\circ$,\\
$\therefore \angle EAB = 90^\circ - \angle AEB = \boxed{14^\circ}$.


Question: As shown in the figure, in rhombus $ABCD$, $E$ and $F$ are the midpoints of $AB$ and $AC$, respectively. If $EF=2$, what is the perimeter of rhombus $ABCD$?

Solution:
\textbf{Solution:} Given that in rhombus $ABCD$, $E$ and $F$ are the midpoints of $AB$ and $AC$ respectively, and $EF=2$,\\
it follows that $BC=2EF=2\times 2=4$. Therefore, $AB=BC=CD=AD=4$.\\
Hence, the perimeter of the rhombus is $4BC=4\times 4=\boxed{16}$.


Question: As shown in the diagram, a rectangular sheet of paper $ABCD$ is folded along $EF$, such that vertices $C$ and $D$ respectively fall on points $C'$ and $D'$. The line $CE$ intersects $AF$ at point $G$, and $\angle CEF = 70^\circ$. What is the value of $\angle GFD'$?

Solution:
\textbf{Solution:} Since $AD \parallel BC$,\\
it follows that $\angle DFE + \angle CEF = 180^\circ$, $\angle AFE = \angle CEF$,\\
given that $\angle CEF = 70^\circ$,\\
thus $\angle DFE = 180^\circ - \angle CEF = 180^\circ - 70^\circ = 110^\circ$, $\angle AFE = 70^\circ$,\\
From the fold, it is known that: $\angle D'FE = \angle DFE = 110^\circ$, therefore $\angle GFD' = \angle D'FE - \angle AFE = 110^\circ - 70^\circ = \boxed{40^\circ}$;


Question: As shown in the figure, in rectangle $ABCD$, $AD = 9$, $AB = 3$. The paper is folded to make point $D$ coincide with point $B$, and the crease is $EF$. What is the length of the crease $EF$?

Solution:
\textbf{Solution:} Construct $FH \perp AD$ meeting AD at point H.\\
Since quadrilateral $EFCB$ is obtained by folding quadrilateral $EFCD$ along EF,\\
therefore ED=BE, CF=$CF$, $BC'=CD=3$\\
Since ED=BE, DE=AD−AE=9−AE\\
therefore $BE=9−AE$\\
Since in $\triangle ABE$, AB=3, BE=9−AE\\
thus $(9−AE)^{2}=3^{2}+AE^{2}$\\
therefore $AE=4$\\
therefore $DE=5$\\
therefore $CF=BC−BF=9−BF$\\
Since in $\triangle BCF$, BC=3, CF=9−BF\\
thus $(9−BF)^{2}+3^{2}=BF^{2}$\\
therefore $BF=5$, EH=1\\
Since in $\triangle EFH$, HF=3, EH=1\\
therefore $EF=\sqrt{EH^{2}+HF^{2}}=\sqrt{3^{2}+1^{2}}=\boxed{\sqrt{10}}$


Question: As shown in the figure, rotating $\triangle ABC$ about point $C(0, -1)$ by $180^\circ$ results in $\triangle A'B'C'$. If the coordinates of point $A$ are $(-4, -3)$, what are the coordinates of point $A'$?

Solution:
\textbf{Solution:} Let $A'$E$\perp$y-axis at point E, and AD$\perp$y-axis at point D, then $\angle A'$EC=$\angle$ADC, \\
$\because$ $\angle A'$CE=$\angle$ACD, AC=$A'$C, \\
$\therefore$ $\triangle A'$EC$\cong \triangle$ADC (AAS), \\
$\therefore$ AD=$A'$E, CE=CD, \\
$\because$ the coordinates of point A are (-4, -3), OC=1, \\
$\therefore$ AD=4, OD=3, \\
$\therefore$ AD=$A'$E=4, CD=2, \\
$\therefore$ CE=2, \\
$\therefore$ OE=1, \\
$\therefore$ the coordinates of point $A'$ are $\boxed{(4, 1)}$.


Question: As shown in the figure, $\triangle DEF$ is obtained by rotating $\triangle ABC$ around a certain point. What are the coordinates of this point?

Solution:
\textbf{Solution:} As shown in the figure, \\
we connect AD, BE and construct the perpendicular bisectors of line segments AD, BE, \\
the intersection of these lines is the center of rotation $O^{\prime}$. Its coordinates are $\boxed{(0,1)}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $BD$ is the bisector of $\angle ABC$, $DE \perp AB$, and the foot of the perpendicular is E. If $AC=5$ and $DE=2$, what is the length of $AD$?

Solution:
\textbf{Solution:} Since BD is the bisector of $\angle ABC$, $DE \perp AB$, and $\angle C = 90^{\circ}$,\\
it follows that $DE = CD = 2$,\\
Since $AC = 5$,\\
it follows that $AD = AC - CD = 5 - 2 = \boxed{3}$


Question: As shown in the figure, the small tree AB casts a shadow BC under the illumination of the street light O. If the height of the tree AB is 2m, the length of the shadow BC is 3m, and the horizontal distance between the tree and the street light BP is 4.5m, what is the height of the street light OP?

Solution:
Solution: Under the same lighting conditions, the ratio of the height of any object to its shadow remains constant:\\
Since the height of the tree, AB = 2m, and the shadow of the tree, BC = 3m, and BP = 4.5m\\
Therefore, $\frac{OP}{PC}=\frac{AB}{BC}$, substituting the values gives: $\frac{OP}{7.5}=\frac{2}{3}$\\
Therefore, $OP=\boxed{5}$m


Question: As shown in the figure, in the circle $\odot O$, the chord $AC\parallel$ to the radius $OB$, $\angle BOC=48^\circ$, what is the degree measure of $\angle OAB$?

Solution:
\textbf{Solution:} Given that $AC \parallel OB$,\\
it follows that $\angle BOC = \angle ACO = 48^\circ$,\\
since $OA = OC$,\\
then $\angle OAC = \angle ACO = 48^\circ$,\\
since $\angle CAB = \frac{1}{2} \angle BOC = 24^\circ$,\\
hence $\angle BAO = \angle OAC - \angle CAB = \boxed{24^\circ}$.


Question: As shown in the figure, the lines $a$, $b$, and $c$ are intersected by lines $l_1$ and $l_2$, with the intersection points being A, C, E and B, D, F, respectively. It is known that $a \parallel b \parallel c$, and $AC = 3$, $CE = 4$. What is the value of $\frac{BD}{BF}$?

Solution:
\textbf{Solution:} Since $a\parallel b\parallel c$,\\
it follows that $\frac{AC}{CE}=\frac{BD}{DF}=\frac{3}{4}$,\\
therefore, $\frac{BD}{BF}=\frac{3}{3+4}=\boxed{\frac{3}{7}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle A=60^\circ$, and $AC=6$. The triangle $\triangle ABC$ is rotated counterclockwise around point C to obtain $\triangle A'B'C'$, where point A' happens to be on side AB. What is the distance between point B' and point B?

Solution:
\textbf{Solution:} Draw line $B'B$.\\
$\because$ Triangle $ABC$ is rotated counterclockwise around point $C$ to obtain $\triangle A'B'C$,\\
$\therefore$ $AC=A'C$, $AB=A'B$, $\angle A=\angle CA'B'=60^\circ$,\\
$\therefore$ $\triangle AA'C$ is an equilateral triangle,\\
$\therefore$ $\angle AA'C=60^\circ$,\\
$\therefore$ $\angle B'A'B=180^\circ-60^\circ-60^\circ=60^\circ$,\\
$\because$ Triangle $ABC$ is rotated counterclockwise around point $C$ to obtain $\triangle A'B'C$,\\
$\therefore$ $\angle ACA'=\angle BAB'=60^\circ$, $BC=B'C$, $\angle CB'A'=\angle CBA=90^\circ-60^\circ=30^\circ$,\\
$\therefore$ $\triangle BCB'$ is an equilateral triangle,\\
$\therefore$ $\angle CB'B=60^\circ$,\\
$\because$ $\angle CB'A'=30^\circ$,\\
$\therefore$ $\angle A'B'B=30^\circ$,\\
$\therefore$ $\angle B'BA'=180^\circ-60^\circ-30^\circ=90^\circ$,\\
$\because$ $\angle ACB=90^\circ$, $\angle A=60^\circ$, $AC=6$,\\
$\therefore$ $AB=12$,\\
$\therefore$ $A'B=AB−AA'=AB−AC=6$,\\
$\therefore$ $B'B=6\sqrt{3}$,\\
Therefore, the answer is: \boxed{6\sqrt{3}}.


Question: As shown in the figure, it is known that the points A, B, C, D, and E are on the same straight line, with point D being the midpoint of segment AB, and point E being the midpoint of segment BC. If the length of segment AC is 12, what is the length of segment DE?

Solution:
\textbf{Solution:} Given that $D$ is the midpoint of the line segment $AB$, it follows that $AD=BD$,\\
Since $E$ is the midpoint of the line segment $BC$, it follows that $BE=CE$,\\
Given that $AC=12$, it follows that $AD+CD=12$, hence $BD+CD=12$,\\
Additionally, since $BD=2CE+CD$, it follows that $2CE+CD+CD=12$,\\
which simplifies to $2(CE+CD)=12$, hence $CE+CD=6$,\\
meaning the length of the segment $DE$ equals $\boxed{6}$.


Question: As shown in the figure, Xiaofang and her father are taking a walk. Her father is 1.8m tall and his shadow on the ground is 2.1m long. If Xiaofang is only 1.2m tall, how long is her shadow?

Solution:
\textbf{Solution:} Let the shadow length of Xiaofang be $x$ meters.\\
According to the proportionality of height to shadow length at the same moment, we get: $\frac{1.8}{2.1}=\frac{1.2}{x}$,\\
Solving this, we obtain: $x=\boxed{1.4}$.\\
Upon verification, this is consistent with the problem statement.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=46^\circ$, $\angle C=54^\circ$, $AD$ bisects $\angle BAC$, intersects $BC$ at $D$, and $DE\parallel AB$, intersecting $AC$ at $E$. What is the measure of $\angle ADE$?

Solution:
\textbf{Solution:} Since $\angle B=46^\circ$ and $\angle C=54^\circ$, \\
it follows that $\angle BAC=180^\circ-\angle B-\angle C=180^\circ-46^\circ-54^\circ=80^\circ$, \\
Since AD bisects $\angle BAC$, \\
it follows that $\angle BAD=\frac{1}{2}\angle BAC=40^\circ$, \\
Since DE $\parallel$ AB, \\
it follows that $\angle ADE=\angle BAD=\boxed{40^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $CM$ bisects $\angle ACB$ and intersects $AB$ at point $M$. A line through point $M$ is drawn such that $MN \parallel BC$ and intersects $AC$ at point $N$, and $MN$ bisects $\angle AMC$. If $AN=1$, what is the length of $BC$?

Solution:
Solution:In a right-angled triangle $ABC$, let $CM$ bisect $\angle ACB$ and intersect $AB$ at point $M$. Construct $MN \parallel BC$ through point $M$ to intersect $AC$ at point $N$, and let $MN$ bisect $\angle AMC$,\\
$\therefore \angle AMN = \angle NMC = \angle B$, $\angle NCM = \angle BCM = \angle NMC$,\\
$\therefore \angle ACB = 2\angle B$, $NM = NC$,\\
$\therefore \angle B = 30^\circ$,\\
Since $AN = 1$,\\
$\therefore MN = 2$,\\
$\therefore AC = AN + NC = 3$,\\
$\therefore BC = \boxed{6}$.


Question: As shown in the figure, to measure the height of a building, at point A, which is 30 meters away from the base of the building at point O, the angle of elevation to the top of the building at point B is measured to be $\angle OAB = 65^\circ$. What is the height of this building?

Solution:
\textbf{Solution:} As shown in the figure, in the right-angled triangle $\triangle$ABO,\\
since $\angle$AOB$=90^\circ$ and $\angle$A$=65^\circ$ with AO$=30$m,\\
then $\tan 65^\circ = \frac{OB}{AO}$,\\
thus, OB$=30\cdot\tan 65^\circ$ meters.\\
Therefore, the answer is: $BO=\boxed{30\cdot\tan 65^\circ}$ meters.


Question: In a square grid, the position of $\triangle ABC$ is shown as in the figure, where points A, B, and C are all on the grid points. What is the value of $\cos B$?

Solution:
\textbf{Solution:} As shown in the figure, draw AD $\perp$ BC and extend BC to meet point D,\\
$\because$ AD=BD=4, $\angle$ADB=90$^\circ$,\\
$\therefore$ $\triangle$ABD is an isosceles right triangle,\\
$\therefore$ $\angle$B=45$^\circ$\\
$\therefore$ $\cos B=\frac{\sqrt{2}}{2}$\\
Hence, the answer is: $\boxed{\frac{\sqrt{2}}{2}}$.


Question: As shown in the figure, quadrilateral ABCD is a rhombus, with $AC=8$ and $\tan \angle DAC= \frac{3}{4}$. If $DH \perp AB$ at H, what is the distance from point D to side AB?

Solution:
\textbf{Solution:} Let the intersection of AC and BD be O,\\
$\because$ Quadrilateral ABCD is a rhombus,\\
$\therefore$ AC$\perp$BD, $AO=\frac{1}{2}AC=4$, $BD=2OD$, AB=AD,\\
$\therefore$ $\tan\angle DAC=\frac{OD}{OA}=\frac{3}{4}$,\\
$\therefore$ $OD=3$,\\
$\therefore$ $AD=\sqrt{AO^{2}+DO^{2}}=5$, $BD=6$,\\
$\therefore$ $AB=AD=5$,\\
$\because$ $S_{\text{rhombus}ABCD}=\frac{1}{2}AC\cdot BD=AB\cdot DH$,\\
$\therefore$ $DH=\frac{AC\cdot BD}{2AB}=\boxed{\frac{24}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=3$, and $BC=4$, what is the value of $\sin A$?

Solution:
\textbf{Solution:} In $Rt\triangle ABC$, where $\angle C=90^\circ$, $AC=3$, and $BC=4$,\\
by the Pythagorean theorem, we have $AB=\sqrt{AC^2+BC^2}=5$,\\
therefore $\sin A=\frac{BC}{AB}=\boxed{\frac{4}{5}}$


Question: Given a right-angled $\triangle ABC$, with $\angle C=90^\circ$, $\angle A=50^\circ$, and $AB=2$, what is the value of $AC$?

Solution:
\textbf{Solution}: We have $\angle B=90^\circ-\angle A=90^\circ-50^\circ=40^\circ$,\\
Furthermore, since $\sin B= \frac{AC}{AB}$,\\
it follows that $AC=AB \cdot \sin 40^\circ = \boxed{2\sin 40^\circ}$.


Question: As shown in the cross-section of a certain dam, the slope ratio $i$ of the slope AB is $1\colon 2$, and the slope ratio $i$ of the water-facing slope CD is $1\colon 1$. If the length of AB is $6\sqrt{5}$ meters, what is the length of the slope CD?

Solution:
\textbf{Solution:}

Let lines BE and CF be perpendicular to AD, passing through B and C respectively,

$\therefore$ Quadrilateral BEFC is a rectangle,

The slope ratio of ramp AB is $i=1\colon 2$, that is $\frac{BE}{AE}=\frac{1}{2}$. Assuming $BE=x$, then $AE=2x$,

In $\triangle ABE$, according to the Pythagorean theorem, we have: $BE^2+AE^2=AB^2$,

$x^2+(2x)^2=(6\sqrt{5})^2$, solving this gives $x=6$ or $x=-6$ (which is inconsistent with the problem, thus discarded),

Moreover, since the slope ratio of the uphill slope CD is $i=1\colon 1$,

$\therefore$ $CF=DF=6$,

$\therefore$ In $\triangle CFD$, according to the Pythagorean theorem, we have:

$CD=\sqrt{CF^2+DF^2}=\sqrt{6^2+6^2}=\boxed{6\sqrt{2}}$


Question: In the figure, the quadrilateral $ABCD$ is inscribed in circle $\odot O$, $\angle ADC=60^\circ$, and the radius of $\odot O$ is $3\,cm$. What is the length of the chord $AC$?

Solution:
\textbf{Solution:} Draw $OA$ and $OC$, and construct $OE \perp AC$ at $E$,\\
since $\angle ADC=60^\circ$,\\
thus $\angle AOC=2\angle ADC=120^\circ$,\\
since $OA=OC$,\\
thus $\angle AOE=\angle COE=60^\circ$, $AC=2AE$,\\
thus $\sin\angle AOE=\frac{AE}{AO}$,\\
thus $AE=AO\cdot \sin\angle AOE$,\\
thus $AE=3\sin60^\circ=\frac{3\sqrt{3}}{2}$,\\
thus $AC=2AE=\boxed{3\sqrt{3}}$,


Question: Place a right triangular plate as shown in the figure, with point $C$ on the extension of $FD$, $AB\parallel CF$, $\angle F = \angle ACB = 90^\circ$, $\angle E = 30^\circ$, $\angle A = 45^\circ$, $AC = 6\sqrt{2}$. What is the length of $CD$?

Solution:
\textbf{Solution:} As shown in the diagram, construct BM$\perp$FD at M,\\
In $\triangle ACB$, $\angle ACB=90^\circ$, $\angle A=45^\circ$, $AC=6\sqrt{2}$,\\
$\therefore BC=AC=6\sqrt{2}$,\\
$\because AB\parallel CF$,\\
$\therefore BM=BC\times\sin45^\circ=6\sqrt{2}\times \frac{\sqrt{2}}{2}=6$,\\
$\therefore CM=BM=6$,\\
In $\triangle EFD$, $\angle F=90^\circ$, $\angle E=30^\circ$,\\
$\therefore \angle EDF=60^\circ$,\\
$\therefore MD=\frac{BM}{\tan60^\circ}=\frac{6}{\sqrt{3}}=2\sqrt{3}$,\\
$\therefore CD=CM-MD=6-2\sqrt{3}$.\\
Thus, the answer is: $CD=\boxed{6-2\sqrt{3}}$.


Question: As shown in the figure, in the square $ABCD$ with a side length of $4$, where point $E$ is a point on side $CD$, and point $F$ is the reflection of point $D$ across line $AE$. Connecting $AF$ and $BF$, if $\tan \angle ABF = 2$, what is the length of $DE$?

Solution:
\textbf{Solution:} Draw a line FN $\perp$ AB at point N, and extend NF to intersect CD at point M,\\
since AB$\parallel$CD,\\
thus MN$\perp$CD,\\
thus $\angle$FME $= 90^\circ$,\\
since $\tan\angle$ABF $= 2$,\\
thus $\frac{FN}{BN} = 2$,\\
let BN $= x$, then FN $= 2x$,\\
thus AN $= 4 - x$,\\
since point F is symmetric to point D with respect to line AE,\\
thus DE $= EF$, DA $= AF = 4$,\\
since AE $= AE$,\\
thus $\triangle$ADE $\cong$ $\triangle$AFE (SSS),\\
thus $\angle$D $= \angle$AFE $= 90^\circ$,\\
since AN$^2$ + NF$^2$ = AF$^2$,\\
thus $(4 - x)^2 + (2x)^2 = 4^2$,\\
thus $x_1 = 0$ (discard), $x_2 = \frac{8}{5}$,\\
thus AN $= 4 - x = 4 - \frac{8}{5} = \frac{12}{5}$, MF $= 4 - 2x = 4 - \frac{16}{5} = \frac{4}{5}$,\\
since $\angle$EFM + $\angle$AFN $= \angle$AFN + $\angle$FAN $= 90^\circ$,\\
thus $\angle$EFM $= \angle$FAN,\\
thus $\cos\angle$EFM $= \cos\angle$FAN,\\
thus $\frac{FM}{EF} = \frac{AN}{AF}$, thus $\frac{\frac{4}{5}}{EF} = \frac{\frac{12}{5}}{4}$,\\
thus EF $= \frac{4}{3}$,\\
thus DE $= EF = \boxed{\frac{4}{3}}$.


Question: As shown in the figure, PA and PB respectively tangent to circle $O$ at points A and B, $\angle APB=60^\circ$, and the radius of circle $O$ is 2. What is the length of PB?

Solution:
\textbf{Solution:} Draw OB and OP, \\
$\because$ PA and PB are tangents to circle $O$, and $\angle APB = 60^\circ$,\\
$\therefore$ $\angle OBP = 90^\circ$, $\angle BPO = \frac{1}{2}\angle APB = 30^\circ$,\\
$\because$ the radius of circle $O$ is 2, i.e., $OB=2$,\\
$\therefore$ $\tan\angle BPO = \frac{OB}{PB} = \frac{2}{PB} = \tan30^\circ = \frac{\sqrt{3}}{3}$,\\
$\therefore$ $PB = \boxed{2\sqrt{3}}$.


Question: As shown in the figure, the vertices of $\triangle ABC$ are all grid points in a square grid. What is the value of $\sin\angle BAC$?

Solution:
\textbf{Solution:} Construct $CD\perp AB$ at D,\\
From the figure, it is known that $BC=2$,\\
By the Pythagorean theorem, $AC=\sqrt{3^2+1^2}=\sqrt{10}$, $AB=\sqrt{3^2+3^2}=3\sqrt{2}$,\\
By the formula for the area of a triangle, $\frac{1}{2}\times 2\times 3=\frac{1}{2}\times 3\sqrt{2}\times CD$,\\
Solving, we find $CD=\sqrt{2}$,\\
$\therefore \sin\angle BAC=\frac{CD}{AC}=\frac{\sqrt{2}}{\sqrt{10}}=\frac{\sqrt{5}}{5}$,\\
Therefore, the answer is: $\boxed{\frac{\sqrt{5}}{5}}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $\angle C=90^\circ$, $AB=10$, and $AC=8$. What is the value of $\tan B$?

Solution:
\textbf{Solution:} In the right-angled $\triangle ABC$, by the Pythagorean theorem, we have $BC=\sqrt{AB^{2}-AC^{2}}=\sqrt{10^{2}-8^{2}}=6$, \\
$\therefore \tan B=\frac{AC}{BC}=\frac{8}{6}=\boxed{\frac{4}{3}}$.


Question: As shown in the figure, fold the rectangular paper ABCD in the manner shown, so that vertex C falls at the position of $C^{\prime}$. If $AB = 4$ and $DE = 8$, what is the value of $\sin\angle C^{\prime}ED$?

Solution:
\textbf{Solution:} Given that ABCD is a rectangle,\\
thus CD=AB=4, and $\angle$C$=90^\circ$,\\
by the properties of paper folding, we have C$^{\prime }$D=CD=4, $\angle$C$^{\prime }$=$\angle$C$=90^\circ$,\\
therefore, $\sin\angle {C}^{\prime }ED=\frac{{C}^{\prime }D}{ED}=\frac{4}{8}=\boxed{\frac{1}{2}}$.


Question: As shown in the figure, a village plans to plant trees on a hillside with a slope angle of $ \alpha $. It is required that the horizontal distance between two adjacent trees be $m$ (m). What is the distance AB between these two trees on the slope?

Solution:
\textbf{Solution:} From the given information, we have $\cos\alpha =\frac{m}{AB}$,\\
thus $AB=\frac{m}{\cos\alpha}=\boxed{\frac{m}{\cos\alpha}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $\angle ABC=30^\circ$, and D is the midpoint of AC, then what is the value of $\tan \angle DBC$?

Solution:
\textbf{Solution:} In $\triangle ABC$, $\angle C=90^\circ$, and $\angle ABC=30^\circ$,\\
hence $\tan\angle ABC=\frac{AC}{BC}=\tan30^\circ=\frac{\sqrt{3}}{3}$. If we let $AC=\sqrt{3}a$, then $BC=3a$,\\
since D is the midpoint of AC,\\
thus $DC=\frac{1}{2}AC=\frac{\sqrt{3}}{2}a$,\\
therefore $\tan\angle DBC=\frac{DC}{BC}=\frac{\frac{\sqrt{3}}{2}a}{3a}=\boxed{\frac{\sqrt{3}}{6}}$.


Question: As shown in the figure, in $\triangle ABC$ where $\angle C=90^\circ$, $BC=5$, and $AC=12$, what is the value of $\tan B$?

Solution:
Solution: In the right triangle $\triangle ABC$, $\angle C=90^\circ$, $BC=5$, and $AC=12$, \\
thus $\tan B=\frac{AC}{BC}=\boxed{\frac{12}{5}}$.


Question: In the $8\times 8$ grid shown in the figure, where the side length of each small square is $1$, and points $A$, $B$, $C$, and $D$ are all on the grid points. If $AB$ and $CD$ intersect at point $E$, what is the value of the tangent of $\angle AED$?

Solution:
\textbf{Solution:} As shown in the figure, select the lattice point K, and draw lines AK and BK.\\
Observing the figure, it is clear that $AK \perp BK$, $BK=2AK$, and $BK \parallel CD$,\\
$\therefore \angle AED=\angle ABK$,\\
$\therefore \tan \angle AED=\tan \angle ABK= \frac{AK}{BK}=\boxed{\frac{1}{2}}$


Question: As shown in the diagram, in the sector $AOB$, $\angle AOB=90^\circ$, taking point A as the center, and the length of OA as the radius, let $\overset{\frown}{OC}$ intersect $\overset{\frown}{AB}$ at point C. If $OA=2$, what is the area of the shaded region?

Solution:
\textbf{Solution:} Connect OC, AC, and construct CD $\perp$ OA at D,\\
$\because$OA = OC = AC,\\
$\therefore$ $\triangle$AOC is an equilateral triangle,\\
$\therefore$ $\angle$OAC = $60^\circ$,\\
$\because$ CD $\perp$ OA, $\angle$CDO = $90^\circ$, OD = AD = $\frac{1}{2}OA = 1$,\\
$\therefore$ CD = OD$\times\tan60^\circ = 1 \times \sqrt{3} = \sqrt{3}$,\\
$S\triangle$OAC = $\frac{1}{2}OA \cdot CD = \frac{1}{2} \times 2 \times \sqrt{3} = \sqrt{3}$,\\
$\therefore$ $\angle$BOC = $30^\circ$,\\
$S_{\text{sector } OBC} = \frac{30\pi \times 2^2}{360} = \frac{1}{3}\pi$,\\
$S_{\text{sector } OAC} = \frac{60\pi \times 2^2}{360} = \frac{2}{3}\pi$,\\
thus, the area of the shaded part equals $\frac{1}{3}\pi - (\frac{2}{3}\pi - \sqrt{3}) = \boxed{\sqrt{3} - \frac{1}{3}\pi}$.


Question: As shown in the figure, both $\triangle ABC$ and $\triangle PQR$ are equilateral triangles, and $AD=BE=CF=\frac{1}{4}AB$. When $\angle AFR=\angle BDP=\angle CEQ=30^\circ$, the area of $\triangle PQR$ is $\sqrt{3}$. What is the side length of $\triangle ABC$?

Solution:
\textbf{Solution:} As illustrated in the figure, extend FP to intersect AB at point O,\\
since $\triangle PQR$ is an equilateral triangle,\\
it follows that $\angle QPR=60^\circ$\\
since $\angle BDP=30^\circ$\\
it follows that $\angle DOP=90^\circ$\\
Let $AD=BE=CF=a(a>0)$\\
therefore, $AB=4a$, $AF=3a$\\
since $\angle AFP=30^\circ$\\
it follows that $AO=\frac{1}{2}AF=\frac{3}{2}a$, $OF=\sqrt{AF^{2}-AO^{2}}=\frac{3\sqrt{3}}{2}a$,\\
thus $DO=AO-AD=\frac{3}{2}a-a=\frac{1}{2}a$\\
since $\angle BDP=30^\circ$\\
it follows that $DP=\frac{OD}{\cos 30^\circ}=\frac{\frac{1}{2}a}{\frac{\sqrt{3}}{2}}=\frac{\sqrt{3}}{3}a$\\
therefore $OP=\frac{1}{2}DP=\frac{1}{2}\times \frac{\sqrt{3}}{3}a=\frac{\sqrt{3}}{6}a$\\
therefore $PR=OF-OP-FR=\frac{3\sqrt{3}}{2}a-\frac{\sqrt{3}}{6}a-\frac{\sqrt{3}}{3}a=\sqrt{3}a$\\
since the area of the equilateral triangle $\triangle PQR$ is $\sqrt{3}$,\\
it follows that $\frac{1}{2}\times PR\times \frac{\sqrt{3}}{2}PR=\frac{1}{2}\times \sqrt{3}a\times \frac{\sqrt{3}}{2}\times \sqrt{3}a=\frac{3\sqrt{3}}{4}a^{2}=\sqrt{3}$\\
thus $a^{2}=\frac{4}{3}$\\
therefore $a=\frac{2\sqrt{3}}{3}$\\
therefore $AB=4a=4\times \frac{2\sqrt{3}}{3}=\boxed{\frac{8\sqrt{3}}{3}}$


Question: As shown in the figure, $\odot O$ is a circle centered at the coordinate origin $O$ with a radius of $4\sqrt{2}$. The coordinate of point $P$ is $(2, 2)$, and the chord $AB$ passes through point $P$. What is the minimum area of the shaded region in the figure?

Solution:
\textbf{Solution:} From the given conditions, when OP$\perp$AB, the area of the shaded region is minimized. \\
Since P(2,2), \\
thus OP=2$\sqrt{2}$, \\
Since OA=OB=4$\sqrt{2}$, \\
thus PA=PB=2$\sqrt{6}$, \\
thus $\tan\angle POB=\frac{PB}{PO}=\sqrt{3}$, \\
thus $\angle AOP=\angle BOP=60^\circ$, \\
thus $\angle AOB=120^\circ$, \\
thus the area of the shaded region $S_{\text{shaded}}=S_{\text{sector} AOB}-S_{\triangle AOB}=\frac{120\times \pi \times {\left(4\sqrt{2}\right)}^{2}}{360}-\frac{1}{2}\times 4\sqrt{6}\times 2\sqrt{2}=\boxed{\frac{32\pi }{3}-8\sqrt{2}}$.


Question: In right triangle $ABC$, where $\angle C=90^\circ$, if $AC=6$ and $BC=8$, then what is the value of $\cos A$?

Solution:
\textbf{Solution: }Since $\angle C=90^\circ$,\\
it follows that AB=$ \sqrt{AC^{2}+BC^{2}}=10$,\\
therefore, $\cos A= \frac{AC}{AB}=\frac{6}{10}=\boxed{\frac{3}{5}}$.


Question: As shown in the figure, $A$, $B$, $C$ are the vertices of small squares, and the side length of each small square is $1$. What is the value of $\sin\angle ABC$?

Solution:
\textbf{Solution:} Connect AC, \\
AC $= \sqrt{5}$, BC $= \sqrt{5}$, AB $= \sqrt{10}$, \\
$\therefore \triangle ABC$ is a right-angled triangle, \\
$\therefore \sin\angle ABC = \frac{AC}{AB} = \frac{\sqrt{5}}{\sqrt{10}} = \boxed{\frac{\sqrt{2}}{2}}$;


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=4$, and $\tan A=\frac{3}{4}$. A circle with point $C$ as the center and the length of $CB$ as the radius intersects $AB$ at point $D$. What is the length of $AD$?

Solution:
\textbf{Solution:} Connect CD, and through point C draw $CE \perp AB$ at point E,\\
Within $\triangle ABC$, $\tan \angle A = \frac{BC}{AC} = \frac{3}{4}$, solve it to get: $BC=3$,\\
$\therefore AB = \sqrt{AC^2 + BC^2} = \sqrt{3^2 + 4^2} = 5$,\\
$\because S_{\triangle ABC} = \frac{1}{2}AC \cdot BC = \frac{1}{2}AB \cdot CE$,\\
$\therefore 3 \times 4 = 5CE$, solve it to get: $CE = \frac{12}{5}$,\\
$\therefore BE = \sqrt{BC^2 - CE^2} = \sqrt{3^2 - \left(\frac{12}{5}\right)^2} = \frac{9}{5}$,\\
$\therefore BD = 2BE = \frac{18}{5}$,\\
$\therefore AD = AB - BD = 5 - \frac{18}{5} = \boxed{\frac{7}{5}}$.


Question: As shown in the figure, it is known that $\odot O$ is the circumcircle of $\triangle ABC$, $AD$ is the diameter of $\odot O$, and $CD$ is connected. If $AD=3$, $AC=2$, then what is the value of $\cos B$?

Solution:
\textbf{Solution:} Given that $AD$ is the diameter of the circle $\odot O$,\\
thus $\angle ACD=90^\circ$,\\
therefore, by the Pythagorean theorem, $CD=\sqrt{AD^{2}-AC^{2}}=\sqrt{3^{2}-2^{2}}=\sqrt{5}$,\\
since $\angle B=\angle D$,\\
thus, $\cos B=\cos D=\frac{CD}{AD}=\frac{\sqrt{5}}{3}=\boxed{\frac{\sqrt{5}}{3}}$.


Question: As shown in the figure, $ABCD$ is a square with side length $1$, and $\triangle BPC$ is an equilateral triangle. Then,what is the area of $\triangle BPD$.

Solution:
\textbf{Solution}: As shown in the figure, \\
draw PE $\perp$ CD and PF $\perp$ BC, \\
since the side length of square ABCD is 1, and $\triangle BPC$ is an equilateral triangle, \\
therefore $\angle PBC=\angle PCB=60^\circ$, PB=PC=BC=CD=1, \\
therefore $\angle PCE=30^\circ$, \\
thus PF=PB$\cdot \sin60^\circ=1\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$, PE=FC=$\frac{1}{2}$, \\
$S_{\triangle BPD}=S_{\text{quadrilateral} PBCD}-S_{\triangle BCD}=S_{\triangle PBC}+S_{\triangle PDC}-S_{\triangle BCD}$ \\
=$\frac{1}{2}\times\frac{\sqrt{3}}{2}\times1+\frac{1}{2}\times\frac{1}{2}\times1-\frac{1}{2}\times1\times1=\boxed{\frac{\sqrt{3}-1}{4}}$;


Question: As shown in the figure, in $\triangle ABC$, it is known that $\angle C = 90^\circ$, $AC = 1$, and $BC = 2$, then what is the value of $\sin B$?

Solution:
\textbf{Solution:} Since $AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{5}$,\\
therefore $\sin B=\frac{AC}{AB}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}=\boxed{\frac{\sqrt{5}}{5}}$.


Question: As shown in the figure, within a grid composed of small squares with a side length of 1, points A, B, and C are located on the grid points, and point D is on the circumcircle of $ \triangle ABC$. What is the value of $\sin\angle ADC$?

Solution:
\textbf{Solution:} Given $\because \overset{\frown}{AC}=\overset{\frown}{AC}$,\\
$\therefore \angle ABC=\angle ADC$,\\
$\because AB^{2}=3^{2}+1^{2}=10, AC^{2}=3^{2}+1^{2}=10, BC^{2}=2^{2}+4^{2}=20$,\\
$\therefore AB^{2}+AC^{2}=BC^{2}$, which means $\triangle ABC$ is an isosceles right triangle,\\
$\therefore \angle ABC=\angle ACB=\angle ADC=45^\circ$,\\
$\therefore \sin\angle ADC=\sin45^\circ=\boxed{\frac{\sqrt{2}}{2}}$