Question: As shown in the figure, in $\triangle ABC$, $AB=AC=\sqrt{5}$, $\sin B=\frac{2\sqrt{5}}{5}$. Find the value of $\cos A$?

Solution:
\textbf{Solution:} As shown in the diagram, let us draw the height $BE$ of $\triangle ABC$,\\
$S_{\triangle ABC}=\frac{1}{2}BC\cdot AD=\frac{1}{2}AC\cdot BE$,\\
$\therefore 2\cdot 2=\sqrt{5}\cdot BE$,\\
$\therefore BE=\frac{4\sqrt{5}}{5}$.\\
Since in $Rt\triangle ABE$, $AE=\sqrt{AB^{2}-BE^{2}}=\sqrt{(\sqrt{5})^{2}-\left(\frac{4\sqrt{5}}{5}\right)^{2}}=\frac{3\sqrt{5}}{5}$,\\
$\therefore \cos A=\frac{AE}{AB}=\frac{\frac{3\sqrt{5}}{5}}{\sqrt{5}}=\boxed{\frac{3}{5}}$


Question: As shown in the diagram, there is an advertising board CD at the top of a building. Xiaoming and his classmates measured the elevation angle of the bottom of the advertising board D to be $53^\circ$ at the foot of a hill at point A. Walking up the slope AB to point B, they measured the elevation angle of the top of the advertising board C to be $45^\circ$. Given that the gradient of hill AB is $i=1\colon \sqrt{3}$, $AB=10$ m, and $AE=21$ m. (Ignoring the height of the protractor, and the results are accurate to 0.1 meter. Reference data: $\sqrt{2}\approx1.41$, $\sqrt{3}\approx1.73$, $\sin53^\circ\approx\frac{4}{5}$, $\cos53^\circ\approx\frac{3}{5}$, $\tan53^\circ\approx\frac{4}{3}$). Calculate the height of point B above the horizontal ground AE in meters.

Solution:
\textbf{Solution:} As shown in the diagram, draw BM$\perp$AE and BN$\perp$CE through point B, where M and N are the feet of the perpendiculars, respectively. \\
$\because \angle CBN=45^\circ$, $\angle DAE=53^\circ$, $i=1\colon \sqrt{3}=\frac{BM}{AM}=\tan\angle BAM$, \\
$\therefore \angle BAM=30^\circ$, \\
$\therefore BM=\frac{1}{2}AB=5\text{ (meters)}$, \\
thus, the height of point B from the horizontal ground AE is \boxed{5}\text{ meters};


Question: As shown in the diagram, a billboard CD is erected at the top of a building. Xiaoming and his classmates measure the angle of elevation to the bottom of the billboard, D, to be $53^\circ$ from point A at the base of the slope. They then walk up the slope along AB to point B and measure the angle of elevation to the top of the billboard, C, to be $45^\circ$. It is known that the slope gradient of AB is $i=1\colon \sqrt{3}$, $AB=10$ meters, and $AE=21$ meters. (The height of the protractor is negligible, and the result is accurate to 0.1 meters. Reference data: $\sqrt{2}\approx1.41$, $\sqrt{3}\approx1.73$, $\sin53^\circ\approx\frac{4}{5}$, $\cos53^\circ\approx\frac{3}{5}$, $\tan53^\circ\approx\frac{4}{3}$) What is the height of the billboard CD in meters?

Solution:
\textbf{Solution:}\\
Since $BM \perp AE$, $BN \perp CE$, and $CE \perp AE$,\\
it follows that quadrilateral $BMEN$ is a rectangle,\\
therefore, $NE = BM = 5$ meters, and $BN = ME$,\\
In $\triangle ABM$, $\angle BAM = 30^\circ$,\\
therefore, $AM = AB \cdot \cos30^\circ = \frac{\sqrt{3}}{2}AB = 5\sqrt{3}$ (meters),\\
thus, $ME = AM + AE = 5\sqrt{3} + 21$ (meters) = $BN$,\\
Since $\angle CBN = 45^\circ$,\\
it follows that $CN = BN = 5\sqrt{3} + 21$ meters,\\
thus, $CE = CN + NE = 5\sqrt{3} + 26$ meters,\\
In $\triangle ADE$, $\angle DAE = 53^\circ$, and $AE = 21$ meters,\\
therefore, $DE = AE \cdot \tan53^\circ \approx 21 \times \frac{4}{3} = 28$ (meters),\\
thus, $CD = CE - DE = 5\sqrt{3} + 26 - 28 = 5\sqrt{3} - 2 \approx \boxed{6.7}$ (meters),\\
i.e., the height of the billboard $CD$ is approximately $6.7$ meters.


Question: As illustrated, the radius $OA$ of the arc $\overset{\frown}{AB}$ is 2, $OC \perp AB$ at point $C$, and $\angle AOC = 60^\circ$. What is the length of the chord $AB$?

Solution:
\textbf{Solution:} \\
$\because$ The radius $OA=2$ of arc $\widehat{AB}$, and $OC\perp AB$ at point C, $\angle AOC=60^\circ$,\\
$\therefore AC=OA\cdot\sin60^\circ=2\times \frac{\sqrt{3}}{2}=\sqrt{3}$,\\
$\therefore AB=2AC=2\sqrt{3}$;\\
Thus, the length of chord AB is $AB=\boxed{2\sqrt{3}}$.


Question: As shown in the figure, the radius OA of the arc $\overset{\frown}{AB}$ is 2, OC is perpendicular to AB at point C, and $\angle AOC=60^\circ$. What is the length of the arc $\overset{\frown}{AB}$?

Solution:
\textbf{Solution:} Since $OC\perp AB$ and $\angle AOC=60^\circ$, \\
it follows that $\angle AOB=120^\circ$, \\
Since $OA=2$,\\
the length of the arc $\overset{\frown}{AB}$ is: $\frac{120^\circ\pi \times 2}{180^\circ}=\boxed{\frac{4\pi }{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $\cos\angle BAC=\frac{12}{13}$, $AB=26$, $AC=15$, $BD\perp AC$ with the foot of the perpendicular being D, E is the midpoint of BD, and AE intersects BC at point F. What is the tangent value of $\angle CBD$?

Solution:
\textbf{Solution:} In right-angled $\triangle ADB$, where $AB=26$ and $\cos\angle BAC= \frac{12}{13}$, \\
$\therefore AD=24$, \\
By the Pythagorean theorem, we have $BD= \sqrt{AB^{2}-AD^{2}}=10$, \\
$\because AC=15$, \\
$\therefore CD=9$, \\
$\therefore \tan\angle CBD=\frac{DC}{BD}=\frac{9}{10}$, \\
$\therefore$ the tangent of $\angle CBD$ is $\boxed{\frac{9}{10}}$.


Question: As shown in the figure, in $\triangle ABC$, $\cos\angle BAC=\frac{12}{13}$, $AB=26$, $AC=15$, $BD\perp AC$ with the foot being D, E is the midpoint of BD, and AE intersects BC at point F. Find the value of $\frac{EF}{AE}$?

Solution:
\textbf{Solution:} Construct line EM $\parallel$ DC intersecting BC at M.\\
Since E is the midpoint of BD,\\
Thus EM$= \frac{1}{2}$DC=4.5\\
Since EM$\parallel$CA,\\
Thus $\triangle$EMF$\sim$$\triangle$ACF,\\
Thus $\frac{EF}{AF}=\frac{EM}{AC}=\frac{4.5}{15}=\frac{3}{10}$,\\
Thus $\frac{EF}{AE}=\boxed{\frac{3}{13}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $CD\perp AB$ with foot D, $AF$ bisects $\angle CAB$ and intersects $CD$ at point E and $CB$ at point F. If $AC=6$ and $BC=8$, directly write out the length of $CD$.

Solution: \textbf{Solution:} We have $CD = \boxed{4.8}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, $CD \perp AB$ at D, $AF$ bisects $\angle CAB$, intersects $CD$ at point E, and intersects $CB$ at point F. Given $AC=6$, $BC=8$. Find the length of $CF$ and the value of $\tan\angle BAF$.

Solution:
\textbf{Solution:} As shown in the figure, draw $FG\perp AB$ at point G through point F.\\
$\because$AF bisects $\angle CAB$,\\
$\therefore \angle BAF=\angle CAF$, $CF=FG$.\\
$\because$ ${S}_{\triangle ABC}={S}_{\triangle ACF}+{S}_{\triangle ABF}=\frac{1}{2}AC\cdot CF+\frac{1}{2}AB\cdot FG$,\\
$\therefore \frac{1}{2}AC\cdot CF+\frac{1}{2}AB\cdot CF=\frac{1}{2}AC\cdot BC$, i.e., $6CF+10CF=6\times 8$,\\
Solving yields: $CF=\boxed{3}$.\\
$\therefore \tan\angle BAF=\tan\angle CAF=\frac{CF}{AC}=\frac{3}{6}=\boxed{\frac{1}{2}}$.


Question: As shown in the figure, the mathematics interest group measures the height of the flagpole $CD$ and the academic building $AB$. First, at point E, they use a theodolite $EF$ that is 1.5 meters high to measure the elevation angle $\angle AFH$ of the top of the academic building A as $45^\circ$, at this moment the top of the flagpole D is exactly in line with sight $FA$. Then, moving forward 12 meters to point G (G is on $CD$), they measure the elevation angle $\angle AGH$ of the top of the academic building A again as $60^\circ$. Points B, C, and E are on the same horizontal line. Calculate the height of the flagpole $CD$ in meters.

Solution:
\textbf{Solution:} Given the problem, we have: $\angle FDG = \angle AFH = 45^\circ$, $EF = CG = BH = 1.5$ meters, $GF = CE = 12$ meters,\\
in $\triangle AFH$, since $\angle AFH = 45^\circ$, $\therefore \triangle DFG$ is an isosceles right triangle,\\
$\therefore DG = FG = 12$ meters, $\therefore CD = CG + DG = 1.5 + 12 = \boxed{13.5}$ (meters).


Question: As shown in the figure, the mathematics interest group measures the height of the flagpole $CD$ and the academic building $AB$. Firstly, at point E, they use an altimeter $EF$ of 1.5 meters high to measure the elevation angle of the top of the academic building A, $\angle AFH$, as $45^\circ$. At this moment, the top of the flagpole D is exactly in line with sight $FA$. Then, moving forward 12 meters to point G (G is on $CD$), the elevation angle of the top of the academic building A, $\angle AGH$, is measured as $60^\circ$. Points B, C, and E are on the same horizontal line. Find the height of the academic building $AB$ (express the result as an exact value).

Solution:
\textbf{Solution:} Let GH = x meters. Given that AB$\parallel$DC$\parallel$EF, EF = CG = BH, and $\angle$ABE = $90^\circ$, \\
$\therefore$ the quadrilateral BCGH is a rectangle, $\therefore$ $\angle$AHF = DGF = $90^\circ$. From (1), we have: DG = FG = 12 meters, BH = EF = 1.5 meters, \\
$\because$ $\angle$AFH = $45^\circ$, $\therefore$ $\triangle$AFH is an isosceles right triangle, \\
$\therefore$ AH = FH, $\because$ $\angle$AGH = $60^\circ$, $\therefore$ $\tan$ $\angle$AGH = $\frac{AH}{GH} = \tan60^\circ = \sqrt{3}$, \\
$\therefore$ AH = $\sqrt{3}$GH = $\sqrt{3}x$ (meters), \\
$\because$ FH = GH + FG = (x + 12) meters, $\therefore$ $\sqrt{3}x = x + 12$, \\
Solving gives: x = $6\sqrt{3}+6$, $\therefore$ GH = ($6\sqrt{3}+6$) meters, AH = (18 + $6\sqrt{3}$) meters, \\
$\therefore$ AB = BH + AH = (19.5 + $6\sqrt{3}$) meters, \\
Therefore, the height of the teaching building AB is \boxed{19.5+6\sqrt{3}} meters.


Question: As shown in the figure, in $\triangle ABC$, $AB=5$, $AC=8$, and $\angle A=60^\circ$. Find the length of $BC$.

Solution:
\textbf{Solution:} Construct $CD \perp AB$ at point C, with D being the foot of the perpendicular.\\
In $\triangle ACD$,\\
since $\angle A=60^\circ$ and $AC=8$,\\
then $\angle ACD=30^\circ$,\\
thus $AD= \frac{1}{2}AC=4$,\\
and $CD=\sqrt{8^2-4^2}=4\sqrt{3}$, $BD=AB-AD=1$.\\
Therefore, in $\triangle BCD$, $BC=\sqrt{CD^2+BD^2}=\sqrt{48+1}=\boxed{7}$;


Question: As shown in the figure, in $\triangle ABC$, we have $AB=5$, $AC=8$, and $\angle A=60^\circ$. Find $\sin B$.

Solution:
\textbf{Solution:} In $\triangle BCD$,\\
$\because$ from (1) we know that $CD=4\sqrt{3}$ and $BC=7$,\\
$\therefore$ $\sin B=\frac{CD}{BC}=\frac{4\sqrt{3}}{7}=\boxed{\frac{4\sqrt{3}}{7}}$.


Question: Given the parabola: $y=ax^2-6ax-16a$ (where $a>0$) intersects the x-axis at points $A$ and $B$ (with $A$ to the left of $B$), and intersects the y-axis at point $C$. Point $G$ is the midpoint of segment $AC$. Find the coordinates of points $A$ and $B$, and the axis of symmetry of the parabola.

Solution:
\textbf{Solution:} Given the problem, we have $y=ax^2-6ax-16a=a(x^2-6x-16)$ \\
Setting $y=0$, then $x^2-6x-16=0$, \\
Solving yields $x=-2$, $x=8$ \\
Therefore, $\boxed{A(-2, 0)}$, $\boxed{B(8, 0)}$ \\
Also, since $y=a(x-3)^2-25a$, \\
Therefore, the axis of symmetry is the line $x=\boxed{3}$.


Question: Given the parabola: $y=ax^{2}-6ax-16a$ ($a>0$) intersects the x-axis at points $A$ and $B$ (with $A$ to the left of $B$) and intersects the y-axis at point $C$. Point $G$ is the midpoint of segment $AC$. The line $y=-\frac{3}{2}x$ intersects the parabola at points $M$ and $N$, and $MO=NO$. Find the equation of the parabola.

Solution:
\textbf{Solution:} Solve the system of equations
\[
\begin{cases}
y=-\frac{3}{2}x\\
y=ax^2-6ax-16a
\end{cases}
\]
Upon simplification, we have $ax^2-6ax+\frac{3}{2}x-16a=0$,\\
Thus, $x_M+x_N=6-\frac{3}{2a}$,\\
Since $MO=NO$,\\
It follows that points M and N are symmetric about the origin,\\
Thus, $x_M+x_N=0$,\\
Therefore, $6-\frac{3}{2a}=0$,\\
Thus, $a=\frac{1}{4}$,\\
Therefore, $\boxed{y=\frac{1}{4}x^2-\frac{3}{2}x-4}$.


Question: Given the parabola: $y=ax^2-6ax-16a$ ($a>0$) intersects the $x$-axis at points $A$ and $B$ ($A$ is to the left of $B$), and intersects the $y$-axis at point $C$. The midpoint of $AC$ is point $G$. It is known that point $P$ is a moving point in the fourth quadrant on the parabola, and a perpendicular line to the $x$-axis through point $P$ intersects $BC$ at point $E$ and the $x$-axis at point $F$. If the triangle formed by points $C$, $P$, $E$ is similar to $\triangle AOG$, determine the coordinates of point $P$.

Solution:
\[ \textbf{Solution:} \]
Given \(y=\frac{1}{4}x^{2}-\frac{3}{2}x-4\), we know \(C(0, -4)\),

Suppose the equation of line BC is \(y=kx+b\),

\[
\therefore \left\{\begin{array}{l}
b=-4\\
8k+b=0
\end{array}\right.,
\]

\[
\therefore \left\{\begin{array}{l}
k=\frac{1}{2}\\
b=-4
\end{array}\right.,
\]

\[
\therefore y=\frac{1}{2}x-4,
\]

Let \(P(t, \frac{1}{4}t^{2}-\frac{3}{2}t-4)\), then \(E(t, \frac{1}{2}t-4)\),

\[
\therefore PE=-\frac{1}{4}t^{2}+2t,
\]

\[
\therefore CE=\sqrt{t^{2}+(\frac{1}{2}t-4)^{2}}, \quad CP=\sqrt{t^{2}+\frac{1}{16}t^{2}(t-6)^{2}},
\]

Since point G is the midpoint of AC,

\[
\therefore G(-1, -2),
\]

\[
\therefore AG=\sqrt{5}, \quad GO=\sqrt{5},
\]

\[
\therefore \triangle AOG \text{ is an isosceles triangle},
\]

Since OA=2, OC=4,

\[
\therefore \tan\angle OAC=2,
\]

Since OB=8,

\[
\therefore \tan\angle OCB=2,
\]

\[
\therefore \angle CAO=\angle OCB,
\]

Since \(PE\parallel OC\),

\[
\therefore \angle FEB=\angle OCB,
\]

\[
\therefore \angle FEB=\angle CAO,
\]

(1) When \(CP=PE\), \(\triangle AOG\sim \triangle CEP\),

\[
\therefore PE=PC,
\]

\[
\therefore \sqrt{t^{2}+\frac{1}{16}t^{2}(t-6)^{2}}=-\frac{1}{4}t^{2}+2t,
\]

Solving gives \(t=3\), \(t=0\),

Since point P is in the fourth quadrant,

\[
\therefore t>0,
\]

\[
\therefore P(3, -\frac{25}{4});
\]

(2) When \(PC=CE\), \(\triangle AOG\sim \triangle PEC\),

\[
\therefore \sqrt{t^{2}+\frac{1}{16}t^{2}(t-6)^{2}}=\sqrt{t^{2}+(\frac{1}{2}t-4)^{2}},
\]

Solving gives \(t=4\),

\[
\therefore P(4, -6).
\]

In conclusion, the coordinates of point P are \(\boxed{(4, -6)}\) or \(\boxed{(3, -\frac{25}{4})}\).


Question: As shown in Figure 1, $E$ is a point on the side $BC$ of the equilateral triangle $\triangle ABC$ (not coinciding with point $B$ or $C$). Connect $AE$ and construct an equilateral triangle $\triangle AEF$ to the right using $AE$ as a side, and connect $CF$. It is known that the function relationship between the area ($S$) of $\triangle ECF$ and the length ($x$) of $BE$ is shown in Figure 2 ($P$ is the vertex of the parabola). When the area of $\triangle ECF$ is maximized, what is the degree measure of $\angle FEC$?

Solution:
\textbf{Solution:} Let the side length of the equilateral $\triangle ABC$ be $a$. Draw $FD\perp BC$ at $D$, \\
$\because$ $\triangle ABC$ is equilateral, $\triangle AEF$ is equilateral, \\
$\therefore AB=AC$, $AE=AF$, $\angle BAC=\angle ABC=\angle ACB=\angle EAF=\angle AEF=60^\circ$, \\
$\therefore \angle BAE=\angle CAF$, \\
$\therefore \triangle ABE\cong \triangle ACF$ (SAS), \\
$\therefore CF=BE=x$, $\angle ACF=\angle ABE=60^\circ$, $\angle FCD=180^\circ-\angle ACB-\angle ACF=60^\circ$, \\
$FD=CF\cdot\sin60^\circ=\frac{\sqrt{3}}{2}x$, \\
$S_{\triangle ECF}=\frac{1}{2}CE\times FD = -\frac{\sqrt{3}}{4}x^2+\frac{\sqrt{3}}{4}ax$ \\
$\therefore$ When $S_{\triangle ECF}$ is maximized, $x= \frac{1}{2}a$, i.e., E is the midpoint of BC, the maximum value of $S_{\triangle ECF}$ is $\frac{\sqrt{3}}{16}a^2$ \\
$\because$ E is the midpoint of BC \\
$\therefore AE\perp BC$, $\angle AEB=90^\circ$ \\
$\therefore \angle FEC=180^\circ-\angle AEB-\angle AEF=30^\circ$ \\
Hence, the measure of $\angle FEC$ is $\boxed{30^\circ}$.


Question: As shown in Figure 1, E is a point on side BC of the equilateral $\triangle ABC$ (not coinciding with points B and C). Draw AE, and construct an equilateral $\triangle AEF$ to the right side using AE as its side. Draw CF. It is known that the function relationship between the area ($S$) of $\triangle ECF$ and the length ($x$) of BE is shown in Figure 2 (P is the vertex of the parabola). What is the side length of the equilateral $\triangle ABC$?

Solution:
\textbf{Solution:} It is known from the diagram that the maximum value of $S_{\triangle ECF}$ is $2\sqrt{3}$,\\
$\therefore \frac{\sqrt{3}}{16}a^{2}=2\sqrt{3}$\\
Solving gives $a=4\sqrt{2}$ or $a=-4\sqrt{2}$ (discard)\\
$\therefore$ The side length of equilateral $\triangle ABC$ is $\boxed{4\sqrt{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, point O is on side AC. Circle $\odot O$ passes through point C and is tangent to hypotenuse AB at point D, intersecting side AC at point E. $\angle ACD=\frac{1}{2}\angle B$. Given that $BC=6$ and $AC=8$, what are the lengths of $AD$ and $CD$?

Solution:
\textbf{Solution:} In $\triangle ABC$, we have $AB=\sqrt{6^2+8^2}=10$,\\
since $OC \perp CB$ and $OC$ is the radius,\\
thus $BC$ is the tangent,\\
therefore $BD=BC=6$,\\
thus $AD=4$,\\
let the radius of $\odot O$ be $r$, then $OD=OC=r$, $OA=8-r$,\\
in $\triangle AOD$, we have $r^2+4^2=(8-r)^2$,\\
solving this, we get $r=3$,\\
therefore, $OC=3$,\\
draw $OB$ intersecting $CD$ at $H$,\\
since $OC=OD$ and $BC=BD$,\\
thus $OB$ perpendicularly bisects $CD$,\\
in $\triangle OCB$, we have $OB=\sqrt{3^2+6^2}=3\sqrt{5}$,\\
because $\frac{1}{2}OB \cdot CH=\frac{1}{2}OC \cdot BC$,\\
thus $CH=\frac{3 \times 6}{3\sqrt{5}}=\frac{6\sqrt{5}}{5}$,\\
therefore, $CD=2CH=\boxed{\frac{12\sqrt{5}}{5}}$.


Question: As shown, in $\triangle ABC$, $\angle ACB=90^\circ$, $CB=CA$, $CE\perp AB$ at point $E$, point $F$ is on $CE$, connect $AF$ and extend it to intersect $BC$ at point $D$, $CG\perp AD$ at point $G$, connect $EG$. $CD^2=DG\cdot DA$; as shown in Figure 1, if $CF=2EF$, prove that point $D$ is the midpoint of $BC$.

Solution:
\textbf{Solution:} As shown in Figure 1, draw EH$\parallel$AD through point E and intersecting BC at point H,\\
since HE$\parallel$AD,\\
therefore, BH:HD = BE:EA, and CD:HD = CF:EF,\\
since CB=CA, and $\angle$ACB$=90^\circ$, with CE$\perp$AB,\\
therefore, E is the midpoint of AB,\\
therefore, BE:EA=1,\\
therefore, BH:HD=BE:EA=1,\\
since CF=2EF,\\
therefore, CD:HD=CF:EF=2,\\
therefore, BH=HD, and CD=2HD,\\
therefore, BD=BH+HD=2HD,\\
therefore, BD=CD,\\
therefore, \boxed{D\text{ is the midpoint of }BD}.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, $CB = CA$, $CE \perp AB$ at point $E$, point $F$ is a point on $CE$, extend $AF$ to intersect $BC$ at point $D$, $CG \perp AD$ at point $G$, connect $EG$. As shown in Figure 2, if $GC = 2$ and $GE = 2\sqrt{2}$, find the length of $GD$.

Solution:
\textbf{Solution:} Since $CB=CA$, and $\angle ACB=90^\circ$,\\
it follows that $\angle BAC=45^\circ$,\\
Since $CE\perp AB$, and $CG\perp AD$,\\
it implies that $\angle AGC=\angle AEC=90^\circ$, and $\angle ACE=45^\circ$,\\
Therefore, points A, C, G, E are concyclic,\\
hence $\angle AGE=\angle ACE=45^\circ$,\\
As shown in Figure 2, draw EH$\perp$AD at point H through point E,\\
thus $\triangle EGH$ is an isosceles right triangle,\\
$EH=GH=GE\cdot \sin 45^\circ=2\sqrt{2}\times \frac{\sqrt{2}}{2}=2$,\\
Since $CG=2$,\\
it follows that $CG=EH$,\\
Since $\angle CGF=\angle EHF=90^\circ$, and $\angle CFG=\angle EFH$,\\
therefore $\triangle CFG\cong \triangle EFH$ (by AAS),\\
thus $FG=FH=1$, and $CF=EF$,\\
In $\triangle CFG$, $CF=\sqrt{CG^2+FG^2}=\sqrt{2^2+1^2}=\sqrt{5}$,\\
thus $CE=2CF=2\sqrt{5}$,\\
thus $AC=\frac{CE}{\sin\angle CAE}=\frac{2\sqrt{5}}{\sin 45^\circ}=2\sqrt{10}$,\\
thus $AG=\sqrt{AC^2-CG^2}=\sqrt{(2\sqrt{10})^2-2^2}=6$,\\
Since $\angle CGD=\angle AGC=90^\circ$,\\
it follows that $\angle CAG+\angle ACG=90^\circ$,\\
Since $\angle ACG+\angle DCG=90^\circ$,\\
it implies that $\angle CAG=\angle DCG$,\\
Therefore, $\triangle CAG\sim \triangle DCG$,\\
thus $\frac{DG}{CG}=\frac{CG}{AG}$,\\
thus $DG=\frac{CG^2}{AG}=\frac{2^2}{6}=\boxed{\frac{2}{3}}$.


Question: Given in the figure, it is known that point $P$ is a point inside the equilateral triangle $ABC$, and $PA=6$, $PB=8$, $PC=10$. What is the measure of $\angle APB$ in degrees?

Solution:
\textbf{Solution}: Draw EP, \\
$\therefore$BE=BP=8, AE=PC=10, $\angle$PBE=60$^\circ$\\
$\therefore$ $\triangle BPE$ is an equilateral triangle,\\
$\therefore$PE=PB=8, $\angle$BPE=60$^\circ$,\\
In $\triangle AEP$, AE=10, AP=6, PE=8,\\
$\therefore$AE$^{2}$=PE$^{2}$+PA$^{2}$,\\
$\therefore$ $\triangle APE$ is a right-angled triangle, and $\angle$APE=90$^\circ$,\\
$\therefore$ $\angle APB=90^\circ+60^\circ=\boxed{150^\circ}$.


Question: As shown in the figure, point $O$ is a point inside the equilateral triangle $ABC$, with $\angle BOC = 150^\circ$. Triangle $BOC$ is rotated counterclockwise around point $C$ to obtain triangle $ADC$. Connect $OD$ and $OA$. What is the degree measure of $\angle ODC$?

Solution:
\textbf{Solution:} From the property of rotation, we have $CD=CO$, $\angle ACD=\angle BCO$,\\
since $\angle ACB=60^\circ$,\\
hence $\angle DCO=60^\circ$,\\
therefore, $\triangle OCD$ is an equilateral triangle,\\
thus $\angle ODC=\boxed{60^\circ}$;


Question: As shown in the figure, point $O$ is a point inside the equilateral triangle $ABC$, with $\angle BOC = 150^\circ$. The triangle $BOC$ is rotated counterclockwise around point $C$ to obtain triangle $ADC$, and then lines $OD$ and $OA$ are drawn. If $OB=2$ and $OC=3$, what is the length of $AO$?

Solution:
\textbf{Solution}: From the property of rotation, we have $AD=OB=2$,\\
since $\triangle OCD$ is an equilateral triangle,\\
therefore $OD=OC=3$,\\
since $\angle BOC=150^\circ$, $\angle ODC=60^\circ$,\\
therefore $\angle ADO=90^\circ$,\\
in $\triangle AOD$, by Pythagoras' theorem, we get $AO=\sqrt{AD^2+OD^2}=\sqrt{2^2+3^2}=\boxed{\sqrt{13}}$.


Question: As shown in the figure, $\odot O$ is the circumcircle of quadrilateral $ABCD$, with $AD$ being the diameter of $\odot O$. Connect $BD$. If $\overset{\frown}{AC} = \overset{\frown}{BD}$ and $\angle 1 = \angle 2$, when $AD = 4\sqrt{2}$ and $BC = 4$, what is the area of $\triangle ABD$?

Solution:
\textbf{Solution:} Draw $OE\perp BC$ at point E. \\
$\therefore BE=CE=\frac{1}{2}BC=2$, \\
$\because AD$ is the diameter of $\odot O$, \\
$\therefore OB=\frac{1}{2}AD=2\sqrt{2}$, \\
$\therefore OE=\sqrt{OB^2-BE^2}=\sqrt{(2\sqrt{2})^2-2^2}=2$, \\
$\therefore {S}_{\triangle ABD}=\frac{1}{2}AD\cdot OE=\frac{1}{2}\times 4\sqrt{2}\times 2=\boxed{4\sqrt{2}}$.


Question: As shown in the figure, it is known that the graph of the inverse proportion function $y= \frac{k}{x}$ intersects with the line $y=ax+b$ at point $A\left( \frac{1}{3}, m+9\right)$ and point $B(1, m)$. Determine the analytical expressions for the inverse proportion function $y= \frac{k}{x}$ and the line $y=ax+b$.

Solution:
\textbf{Solution:} Let's solve this. Since the graph of the inverse proportion function $y= \frac{k}{x}$ intersects with the line $y=ax+b$ at points A$\left(\frac{1}{3}m, m+9\right)$ and B$(1, m)$.

Therefore, substituting point A into $y= \frac{k}{x}$ gives: $k=\frac{1}{3}m(m+9)$, and substituting point B gives: $k=m$,

Therefore, $\frac{1}{3}m(m+9)=m$, solving this gives: $m_1=0$ (discard), $m_2=-6$,

Therefore, A$(-2, 3)$, B$(1, -6)$.

Therefore, substituting point A$(-2, 3)$ into $y= \frac{k}{x}$ gives: $k=-6$,

Therefore, the equation of the inverse proportion function $y= \frac{k}{x}$ is $\boxed{y=-\frac{6}{x}}$,

Substituting points A$(-2, 3)$, B$(1, -6)$ into the line $y=ax+b$, gives: $\begin{cases}3=-2a+b\\ -6=a+b\end{cases}$, solving this gives: $\begin{cases}a=-3\\ b=-3\end{cases}$,

Therefore, the equation of the line $y=ax+b$ is $\boxed{y=-3x-3}$.


Question: As shown in the figure, it is known that the graph of the inverse proportion function \(y = \frac{k}{x}\) intersects with the line \(y = ax + b\) at points \(A\left(\frac{1}{3}, m + 9\right)\) and \(B(1, m)\). There is a point \(P\) on the x-axis such that the area of \(\triangle PAB\) is \(18\). Find the coordinates of point \(P\).

Solution:
\textbf{Solution:} Let us draw line $AB$ and construct $AC\perp x$-axis at point $C$ and $BD\perp x$-axis at point $D$. Let $AB$ intersect the $x$-axis at point $E$. When $y=-3x-3=0$, we solve and find $x=-1$. That is, $E(-1, 0)$. The area of $\triangle PAB$ is $\frac{1}{2}PE\cdot AC+\frac{1}{2}PE\cdot DB=\frac{3}{2}PE+\frac{6}{2}PE=\frac{9}{2}PE$, and since the area of $\triangle PAB=18$, we have $\frac{9}{2}PE=18$, $\therefore PE=4$. Thus, when point $P$ is to the right of the origin, $P(3, 0)$; when point $P$ is to the left of the origin, $P(-5, 0)$.

$\therefore \boxed{P(3, 0)}$ and $\boxed{P(-5, 0)}$.


Question: As shown in the figure, $\triangle AOB$ is an isosceles right triangle. If $A(-4, 1)$, what are the coordinates of point $B$?

Solution:
\textbf{Solution:} As shown in the diagram, draw AE$\perp$x-axis at E through point A, and draw BF$\perp$x-axis at F through point B,\\
$\therefore \angle$AEO=$\angle$OFB=$90^\circ$,\\
$\therefore \angle$AOE+$\angle$OAE=$90^\circ$,\\
Also, $\because \angle$AOB=$90^\circ$,\\
$\therefore \angle$AOE+$\angle$BOF=$90^\circ$,\\
$\therefore \angle$OAE=$\angle$BOF,\\
$\because$AO=OB,\\
$\therefore \triangle$OAE$\cong \triangle$BOF (AAS),\\
$\therefore$OF=AE, BF=OE,\\
$\because$the coordinates of point A are (-4, 1),\\
$\therefore$OF=AE=1, BF=OE=4,\\
$\therefore$the coordinates of point B are $\boxed{(1,4)}$;


Question: As shown in the figure, $\triangle AOB$ is an isosceles right triangle. $AN \perp$ the y-axis, with the foot of the perpendicular being N, and $BM \perp$ the y-axis, with the foot of the perpendicular being point M. Point P is the midpoint of AB. Connect PM and find the degree measure of $\angle PMO$?

Solution:
\textbf{Solution:} As shown in the diagram, we extend $MP$ to intersect $AN$ at $H$,\\
since $AH \perp y$-axis, and $BM \perp y$-axis,\\
it follows that $BM \parallel AN$,\\
hence, $\angle MBP = \angle HAP$, and $\angle AHP = \angle BMP$,\\
since point $P$ is the midpoint of $AB$,\\
it implies that $AP = BP$,\\
therefore, $\triangle APH \cong \triangle BPM$ (AAS),\\
thus, $AH = BM$,\\
since the coordinates of point $A$ are $(-4,1)$, and the coordinates of point $B$ are $(1,4)$,\\
it follows that $AN = 4$, $OM = 4$, $BM = 1$, $ON = 1$,\\
thus, $HN = AN - AH = AN - BM = 3$, $MN = OM - ON = 3$,\\
hence, $HN = MN$,\\
therefore, $\angle NHM = \angle NMH = 45^\circ$, that is $\angle PMO = \boxed{45^\circ}$.


Question: As shown in the figure, in a right-angled triangle paper with $\angle C=90^\circ$, $AB=10$, $BC=8$, and $AC=6$, the triangle is folded along a line passing through point $B$ such that point $C$ falls on point $E$ on side $AB$, with the fold line being $BD$. What is the perimeter of $\triangle ADE$?

Solution:
\textbf{Solution:} From the properties of folding, we know that $BE=BC=8$, $DE=CD$,\\
$\therefore AE=AB-BE=2$,\\
$\therefore$ the perimeter of $\triangle ADE$ $=AD+AE+DE=AD+DE+AE=AC+AE=\boxed{8}$.


Question: As shown in the figure, in a right-angled triangle paper, where $\angle C=90^\circ$, $AB=10$, $BC=8$, and $AC=6$, fold the triangle along the line passing through point $B$ such that point $C$ falls on point $E$ on side $AB$, with the crease being $BD$. Find the length of $DE$.

Solution:
\textbf{Solution:} Let $CD=DE=x$, then $AD=AC-CD=6-x$. By the property of folding, we know that $\angle DEB=\angle C=90^\circ$, $\therefore \angle DEA=90^\circ$, $\therefore AD^2=AE^2+DE^2$, $\therefore (6-x)^2=2^2+x^2$, solving this, we find $x=\frac{8}{3}$, \\
$\therefore DE=\boxed{\frac{8}{3}}$.


Question: Consider the diagram, where we have an equilateral triangle $\triangle ABC$ with point $M$ on side $BC$, and $MN\perp AC$ at point $N$, intersecting $AB$ at point $P$. Prove that: $BP=BM$.

Solution:
\textbf{Solution:} Proof:

Since $\triangle ABC$ is an equilateral triangle,

$\therefore \angle ABC=\angle C=60^\circ$,

Since $MN \perp AC$ at $N$,

$\therefore \angle BMP=\angle CMN=90^\circ-60^\circ=30^\circ$,

$\therefore \angle P+\angle BMP=\angle ABC=60^\circ$,

$\therefore \angle P=30^\circ$, that is, $\angle P=\angle BMP=30^\circ$,

$\therefore \boxed{BP=BM}$.


Question: Given an equilateral triangle $\triangle ABC$, $M$ is on side $BC$, $MN\perp AC$ at $N$, and intersects $AB$ at point $P$. If $E$, $F$ are respectively on $AB$, $AC$, and $\triangle MEF$ is an equilateral triangle, when the value of $\frac{S_{\triangle MEF}}{S_{\triangle ABC}}$ is minimal, what is the value of $\frac{BM}{BC}$?

Solution: \textbf{Solution:} When the value of $\frac{S_{\triangle MEF}}{S_{\triangle ABC}}$ is minimal, $\frac{BM}{BC}=\boxed{\frac{1}{2}}$.


Question: As shown in the figure, a high school in Liupanshui City has an irregular quadrilateral space ABCD. The school plans to lay a suspended floor on the space. After measurement, $\angle ABC=90^\circ$, $BC=6\text{m}$, $AB=8\text{m}$, $AD=26\text{m}$, $CD=24\text{m}$. What is the area of the space ABCD in square meters?

Solution:
\textbf{Solution:} Connect AC,\\
In the right-angled triangle ABC,\\
$\because \angle ABC=90^\circ$, BC=6m, AB=8m,\\
$\therefore AC=\sqrt{BC^2+AB^2}=10$m,\\
$\because AC^2+CD^2=10^2+24^2=676=AD^2$,\\
$\therefore \angle ACD=90^\circ$,\\
$\therefore S_{\text{quadrilateral }ABCD}=S_{\triangle ABC}+S_{\triangle ACD}=\frac{1}{2}\times 6\times 8+\frac{1}{2}\times 10\times 24=\boxed{144}$m$^2$.


Question: As shown in the figure, there is an irregular quadrilateral plot of land, ABCD, at a certain middle school in Liupanshui City. The school plans to pave the space with a floating floor. After measurement, $\angle ABC=90^\circ$, $BC=6\text{m}$, $AB=8\text{m}$, $AD=26\text{m}$, and $CD=24\text{m}$. If it costs $120\text{ yuan}$ to pave $1\text{ square meter}$ of floating floor, how much in total will need to be invested?

Solution: \textbf{Solution:} Solve: $144\times 120=\boxed{17280}$ (yuan).


Question: Given that the length of segment $AB=15\text{cm}$, and point $C$ is on segment $AB$ with $AC:CB=3:2$. Find the lengths of segments $AC$ and $CB$.

Solution:
\textbf{Solution:} Given that $AB=15\text{cm}$, and the ratio of $AC$ to $CB$ is $3:2$,

it follows that $AC=\frac{3}{3+2}AB=\frac{3}{5}\times 15=\boxed{9\text{cm}}$, and $CB=\frac{2}{3+2}AB=\frac{2}{5}\times 15=\boxed{6\text{cm}}$;


Question: As shown in the figure, it is known that the length of segment $AB=15\text{cm}$, point $C$ is on segment $AB$, and $AC:CB=3:2$. If point $P$ is the midpoint of segment $AB$, and point $M$ is the midpoint of segment $AP$, find the length of segment $MC$.

Solution:
\textbf{Solution:} Since point P is the midpoint of line segment AB,\\
it follows that $AP=BP=\frac{1}{2}AB=\frac{15}{2}$,\\
Furthermore, since point M is the midpoint of line segment AP,\\
it follows that $MP=\frac{1}{2}AP=\frac{15}{4}$,\\
Also, since $BC=6$,\\
it follows that $PC=PB−BC=\frac{15}{2}−6=\frac{3}{2}$,\\
Therefore, $MC=MP+PC=\frac{15}{4}+\frac{3}{2}=\boxed{\frac{21}{4}}cm$.


Question: As shown in the figure, the hexagon $ABCDEF$ is a regular hexagon inscribed in circle $\odot O$. Let the area of the circle $\odot O$ be $S_1$ and the area of the hexagon $ABCDEF$ be $S_2$. Find the value of $\frac{S_1}{S_2}$.

Solution:
\textbf{Solution:} Draw OG $\perp$ DE at G, and connect OE,\\
Let the radius of circle O be r,\\
$\therefore$ EF=BC=ED=r, AD=2r,\\
In the regular hexagon ABCDEF,\\
$\angle$OED=$\angle$ODE=60$^\circ$,\\
$\therefore$ $\angle$EOG=30$^\circ$,\\
$\therefore$ EG=$ \frac{1}{2}$r,\\
$\therefore$ OG=$ \sqrt{OE^{2}-EG^{2}}$=$ \frac{\sqrt{3}}{2}$r,\\
$\therefore$ The area of regular hexagon ABCDEF=$ 6\times \frac{1}{2}\times r\times \frac{\sqrt{3}}{2}r$=$ \frac{3\sqrt{3}}{2}r^{2}$,\\
The area of circle O=$ \pi r^{2}$,\\
$\therefore$ $ \frac{{S}_{1}}{{S}_{2}}$=$ \frac{\pi r^{2}}{\frac{3\sqrt{3}}{2}r^{2}}$=$ \boxed{\frac{2\sqrt{3}\pi }{9}}$.


Question: As shown in the figure, in $\triangle ABC$, the perpendicular bisector DM of AC intersects AC at D, and the perpendicular bisector EN of BC intersects BC at E, where DM and EN intersect at point F. If the perimeter of $\triangle CMN$ is 16 cm, what is the length of AB in cm?

Solution:
Solution: Since $DM$ is the perpendicular bisector of side $AC$,\\
it follows that $MA=MC$,\\
and since $EN$ is the perpendicular bisector of side $BC$,\\
it follows that $NB=NC$,\\
Thus, $AB=AM+MN+NB=MC+MN+NC=$ the perimeter of $\triangle CMN = \boxed{16}$cm;


Question: As shown in the figure, in $\triangle ABC$, the perpendicular bisector DM of side AC intersects AC at D, and the perpendicular bisector EN of side CB intersects BC at E. DM and EN intersect at point F. If $\angle MFN=70^\circ$, what is the measure of $\angle MCN$?

Solution:
\textbf{Solution:} Given that MD$\perp$AC and NE$\perp$BC, \\
it follows that $\angle ACB=180^\circ-\angle MFN=110^\circ$, \\
which leads to $\angle A+\angle B=70^\circ$, \\
Given MA=MC and NB=NC, \\
it follows that $\angle MCA=\angle A$ and $\angle NCB=\angle B$, \\
Hence, $\angle MCN=\angle ACB -\angle MCA-\angle NCB =\angle ACB -(\angle A+\angle B)=\boxed{40^\circ}$.


Question: In the figure, suppose $\triangle AED$ where $\angle AED=90^\circ$, $AB=AC=AD$, $EC=2$, and $BE=8$. If $AB=\sqrt{41}$, what is the value of $AE$?

Solution: \textbf{Solution:} We have $AE = \boxed{5}$.


Question: In the figure, in $\triangle AED$, $\angle AED=90^\circ$, $AB=AC=AD$, $EC=2$, $BE=8$, then what is the value of $DE$?

Solution: \textbf{Solution:} We have $DE=\boxed{4}$.


Question: As shown in the figure, $\triangle ABC$ is an isosceles right-angled triangle, with $\angle ACB=90^\circ$, $AC=BC=6$. Point $D$ is on segment $BC$, and $E$ is a point on segment $AD$. Now taking $CE$ as the right angle side, with $C$ as the right angle vertex, an isosceles right-angled triangle $\triangle ECF$ is constructed below $CE$, and $BF$ is connected. $\triangle AEC \cong \triangle BFC$. When points $A$, $E$, and $F$ are collinear, as shown in figure 2, if $AF=8$, what is the length of $BF$?

Solution:
Solution: As shown in Figure 2,\\
$\because CA=CB=6$, $\angle ACB=90^\circ$,\\
$\therefore AB=6\sqrt{2}$,\\
$\because \triangle ACE \cong \triangle BCF$,\\
$\therefore \angle CAD=\angle DBF$,\\
$\because \angle ADC=\angle BDF$,\\
$\therefore \angle ACD=\angle DFB=90^\circ$,\\
$\therefore BF=\sqrt{AB^{2}-AF^{2}}=\sqrt{(6\sqrt{2})^{2}-8^{2}}=\boxed{2\sqrt{2}}$.


Question: As shown in the figure, $\triangle ABC$ is an isosceles right triangle with $\angle ACB=90^\circ$ and $AC=BC=6$. Point $D$ is on the line segment $BC$, and point $E$ is on the line segment $AD$. Now, taking $CE$ as the right angle side and $C$ as the right angle vertex, an isosceles right triangle $\triangle ECF$ is constructed below $CE$, and $BF$ is connected. As shown in figure 3, if $\angle BAD=15^\circ$ and $DF$ is connected, when $E$ moves to make $\angle ACE=30^\circ$, what is the area of $\triangle CDF$?

Solution:
\textbf{Solution:} Let $FH \perp BC$ at $H$.\\
Since $\angle ACE = \angle CAE = 30^\circ$,\\
it follows that $AE = EC$,\\
since $\triangle ACE \cong \triangle BCF$,\\
it follows that $BF = AE$ and $CF = CE$,\\
thus $CF = BF$ and $\angle FCB = \angle CBF = 30^\circ$,\\
since $FC = FB$ and $FH \perp BC$,\\
it follows that $CH = BH = 3$ and $FH = \sqrt{3}$, $CF = BF = 2\sqrt{3}$,\\
since $\angle CED = \angle CAE + \angle ACE = 60^\circ$ and $\angle ECD = 90^\circ - 30^\circ = 60^\circ$,\\
thus $\triangle ECD$ is an equilateral triangle,\\
thus $EC = CF = CD = 2\sqrt{3}$,\\
therefore, the area of $\triangle CDF$ is $\frac{1}{2} \times CD \times FH = \frac{1}{2} \times 2\sqrt{3} \times \sqrt{3} = \boxed{3}$.


Question: As shown in the figure, in $\triangle ABC$, point $D$ is on side $BC$, and $AD=AB$, $AE\parallel BC$, $\angle BAD=\angle CAE$. Connect $DE$ and intersect $AC$ at point $F$. If $\angle B=70^\circ$, what is the degree measure of $\angle C$?

Solution:
\textbf{Solution:} Since $ AD=AB$ and $ \angle B=70^\circ$,\\
it follows that $ \angle B=\angle ADB=70^\circ$,\\
which means $ \angle BAD=180^\circ-\angle B-\angle ADB=40^\circ$,\\
thus $ \angle BAD=\angle CAE=40^\circ$,\\
and since $ AE\parallel BC$,\\
it follows that $ \angle C=\boxed{40^\circ}$.


Question: In the figure, if $\triangle ACB$ and $\triangle DCE$ are both isosceles right triangles with $\angle ACB = \angle DCE = 90^\circ$, points A, D, and E lie on the same straight line, and $CM$ is the altitude from $C$ to the side $DE$ in $\triangle DCE$, with $BE$ connected. $\triangle ACD \cong \triangle BCE$; if $CM = 2$ and $BE = 3$, find the length of $AE$. The core problem of this LaTeX is {to find the length of $AE$.}

Solution: \textbf{Solution:} Since $\triangle ACD\cong \triangle BCE$, it follows that $AD=BE=3$. Since $CD=CE$ and $CM\perp DE$, it follows that $DM=ME$, and $\angle MCD=\angle MCE=45^\circ$. Since $\angle CDE=\angle CED=45^\circ$, it follows that $DM=ME=CM$. Therefore, $DE=2CM=4$. Thus, $AE=AD+DE=\boxed{7}$.


Question: As shown in the figure, $AD$ is the altitude of $\triangle ABC$, $E$ is a point on $AC$, $BE$ intersects $AD$ at point $F$, and it is given that $BF=AC$, $FD=CD$. $\triangle BFD \cong \triangle ACD$; what is the measure of $\angle ABD$?

Solution:
\textbf{Solution:} Given that $Rt\triangle BFD\cong Rt\triangle ACD$,\\
it follows that $BD=AD$,\\
since $\angle ADB=90^\circ$,\\
it results in $\angle ABD=\angle BAD=\frac{1}{2}(180^\circ-\angle ADB)=\frac{1}{2}\times (180^\circ-90^\circ)=45^\circ$,\\
hence, the measure of $\angle ABD$ is $\boxed{45^\circ}$.


Question: As shown in the figure, it is known that in an equilateral triangle $ABC$, $D$ is the midpoint of $AC$, $E$ is a point on the extension line of $BC$, and $CD = CE$, $M$ is the midpoint of $BE$. What is the measure of $\angle E$ in degrees?

Solution:
\textbf{Solution:} Given that $\triangle ABC$ is an equilateral triangle, \\
$\therefore \angle ACB=60^\circ$, \\
since $CD=CE$, \\
$\therefore \angle CDE=\angle E$, \\
Also, since $\angle ACB=\angle CDE+\angle E$, \\
$\therefore \angle E=\boxed{30^\circ}$;


Question: As shown in the figure, $DE\perp AB$ at $E$, $DF\perp AC$ at $F$. Given that $BD=CD$ and $BE=CF$, $\triangle BDE \cong \triangle CDF$. It's known that $AC=12$ and $BE=2$, find the length of $AB$.

Solution:
\textbf{Solution:} From (1) it is known that: $Rt\triangle BDE \cong Rt\triangle CDF$,\\
$\therefore DE=DF$, $BE=CF=2$,\\
Also, since $AC=12$, $\therefore AF=10$,\\
Because $\angle E=\angle DFA=90^\circ$ and $AD=AD$,\\
$\therefore$ $Rt\triangle AED \cong Rt\triangle AFD$ (HL),\\
$\therefore AE=AF=10$,\\
$\therefore AB=AE-BE=10-2=\boxed{8}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle A=40^\circ$, and the bisector of the exterior angle $\angle CBD$ of $\triangle ABC$, labeled as $BE$, intersects the extension of $AC$ at point $E$. Point $F$ is located on the extension of $AC$, and $DF$ is connected. What is the measure of $\angle CBE$ in degrees?

Solution:
\textbf{Solution:} Since $\angle ACB=90^\circ$ and $\angle A=40^\circ$,
\[\therefore \angle CBD=\angle A+\angle ACB=130^\circ,\]
Since $BE$ bisects $\angle CBD$,
\[\therefore \angle CBE=\frac{1}{2}\angle CBD=\boxed{65^\circ}.\]


Question: As shown in the figure, $AE$ and $BD$ are two altitudes of $\triangle ABM$, where $AE$ and $BD$ intersect at point $C$, and $AE=BE$. $\triangle AME \cong \triangle BCE$; what is the measure of $\angle MDE$ in degrees?

Solution:
\textbf{Solution:} Draw $EF\perp ED$ through point E and intersect $BC$ at point F,\\
since $\angle DEF=\angle AEB$,\\
therefore $\angle DEA=\angle BEF$,\\
in $\triangle AED$ and $\triangle BEF$,\\
$\left\{\begin{array}{l}
\angle DEA=\angle BEF \\
AE=BE \\
\angle DAE=\angle FBE
\end{array}\right.$,\\
therefore $\triangle AED\cong \triangle BEF$ (ASA),\\
thus $ED=EF$,\\
therefore $\angle EDF=\angle EFD=45^\circ$,\\
since $\angle BDE=90^\circ$,\\
thus $\angle MDE=\boxed{45^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=4$, $\angle B=\angle C=40^\circ$. Point D moves on line segment $BC$ (D does not coincide with B or C). Connect $AD$ and construct $\angle ADE=40^\circ$, with $DE$ intersecting line segment $AC$ at $E$. When $\angle BDA=100^\circ$, what is the value of $\angle DEC$?

Solution:
\textbf{Solution:} Given that in $\triangle BAD$, $\angle B=\angle C=40^\circ$, $\angle BDA=100^\circ$, \\
thus, $\angle BAD=180^\circ-\angle B-\angle BDA=180^\circ-40^\circ-100^\circ=40^\circ$;\\
$\angle EDC=180^\circ-\angle ADB-\angle ADE=180^\circ-100^\circ-40^\circ=40^\circ$.\\
$\angle DEC=180^\circ-\angle C-\angle EDC=180^\circ-40^\circ-40^\circ=\boxed{100^\circ}$;


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=4$ and $\angle B=\angle C=40^\circ$. Point $D$ moves on line segment $BC$ (not coinciding with $B$ or $C$), and $AD$ is connected. Let $\angle ADE=40^\circ$, and let $DE$ intersect line segment $AC$ at $E$. When does $DC$ equal such that $\triangle ABD\cong \triangle DCE$? Please state your reasons.

Solution:
\textbf{Solution:} When $DC=4$, $\triangle ABD\cong \triangle DCE$ for the following reason:\\
$\because \angle C=40^\circ$,\\
$\therefore \angle DEC+\angle EDC=140^\circ$,\\
Also, $\because \angle ADE=40^\circ$,\\
$\therefore \angle ADB+\angle EDC=140^\circ$,\\
$\therefore \angle ADB=\angle DEC$,\\
Also, $\because AB=DC=4$,\\
In $\triangle ABD$ and $\triangle DCE$, we have $\left\{\begin{array}{l}\angle ADB=\angle DEC \\ \angle B=\angle C \\ AB=DC\end{array}\right.$,\\
$\therefore \triangle ABD\cong \triangle DCE$ (AAS);\\
Thus, $DC=\boxed{4}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=4$, $\angle B=\angle C=40^\circ$. Point D moves along the segment $BC$ (D does not coincide with B or C). Connect $AD$ and construct $\angle ADE=40^\circ$, where $DE$ intersects segment $AC$ at $E$. During the movement of point D, is $\triangle ADE$ an isosceles triangle? Calculate the degree measure of $\angle BDA$.

Solution:
\textbf{Solution:} Solve: When the degree of $\angle BDA$ is $110^\circ$ or $80^\circ$, the shape of $\triangle ADE$ is an isosceles triangle,\\
$\because$ when $\angle BDA = 110^\circ$,\\
$\therefore$ $\angle ADC = 70^\circ$,\\
$\because$ $\angle C = 40^\circ$,\\
$\therefore$ $\angle DAE = 70^\circ$,\\
$\therefore$ $\angle AED = 180^\circ - 70^\circ - 40^\circ=70^\circ$\\
$\therefore$ the shape of $\triangle ADE$ is an isosceles triangle;\\
$\because$ when the degree of $\angle BDA$ is $80^\circ$,\\
$\therefore$ $\angle ADC = 100^\circ$,\\
$\because$ $\angle C = 40^\circ$,\\
$\therefore$ $\angle DAE = 40^\circ$,\\
$\therefore$ $\angle DAE = \angle ADE$,\\
$\therefore$ the shape of $\triangle ADE$ is an isosceles triangle.\\
Therefore, the degree of $\angle BDA$ is $\boxed{110^\circ}$ or $\boxed{80^\circ}$.


Question: As shown in the diagram, in $\triangle ABC$, $D$ is a point on $AB$, and $E$ is the midpoint of $AC$. Connect $DE$ and extend it to point $F$ such that $EF=ED$, and connect $CF$. Given $CF\parallel AB$ and $\angle ABC=50^\circ$, connect $BE$. $BE$ bisects $\angle ABC$, and $AC$ bisects $\angle BCF$. Find the measure of $\angle A$.

Solution:
\textbf{Solution:} Given $AC$ bisects $\angle BCF$,\\
thus, $\angle ACB = \angle ACF$,\\
since $\angle A = \angle ACF$,\\
thus, $\angle A = \angle ACB$,\\
since $\angle A + \angle ABC + \angle ACB = 180^\circ$, $\angle ABC = 50^\circ$,\\
thus, $2\angle A = 130^\circ$,\\
therefore, $\angle A = \boxed{65^\circ}$.


Question: As shown in the figure, $AD$ and $BC$ intersect at point $O$, with $AD=BC$, and $\angle C=\angle D=90^\circ$. We have $AC=BD$; if $\angle ABC=35^\circ$, what is the degree measure of $\angle CAO$?

Solution:
\textbf{Solution:} In $\triangle ACB$, $\angle ABC=35^\circ$, \\
$\therefore \angle CAB=90^\circ-35^\circ=55^\circ$, \\
From (1) it is known $\triangle ACB \cong \triangle BDA$, \\
$\therefore \angle BAD=\angle ABC=35^\circ$, \\
$\therefore \angle CAO=\angle CAB-\angle BAD=55^\circ-35^\circ=\boxed{20^\circ}$.


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle, point D is a point on side AB, an equilateral $\triangle CDE$ is constructed upwards with CD as its side, and AE is connected. $\triangle BCD \cong \triangle ACE$. If $AE=1$, $AB=3$, what is the length of $AD$?

Solution:
\textbf{Solution:} Since $\triangle BCD \cong \triangle ACE$, $AE=1$, $AB=3$,\\
$\therefore$ $BD=AE=1$,\\
$\therefore$ $AD=AB-BD=3-1=\boxed{2}$,\\
$\therefore$ the length of $AD$ is \boxed{2}.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$. An arc is drawn with point B as the center and BC as the radius, intersecting segment AB at point D; an arc is drawn with point A as the center and AD as the radius, intersecting segment AC at point E. Connect CD. If $\angle A=24^\circ$, what is the degree measure of $\angle ACD$?

Solution:
\textbf{Solution:} Given $\because \angle ACD=90^\circ$ and $\angle A=24^\circ$,\\
$\therefore \angle B=66^\circ$.\\
Since $BD=BC$,\\
$\therefore \angle BCD=\angle BDC=\frac{180^\circ-66^\circ}{2}=57^\circ$.\\
Thus, $\angle ACD=90^\circ-\angle BCD=90^\circ-57^\circ=\boxed{33^\circ};$


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, an arc is drawn with point B as the center and the length of BC as the radius, intersecting line segment AB at D; an arc is drawn with point A as the center and AD as the radius, intersecting line segment AC at point E, and CD is connected. If $BC=5$ and $AC=12$, what is the length of $AD$?

Solution:
\textbf{Solution:} Since $\angle ACB=90^\circ$ and $BC=5$, $AC=12$,\\
by the Pythagorean theorem, we have $AB=\sqrt{AC^2+BC^2}=\sqrt{12^2+5^2}=13$,\\
Since $AB=AD+BD$ and $BD=BC=5$,\\
it follows that $AD=13-5=\boxed{8}$.


Question: As shown in the figure, in $\triangle ABC$, points D and E are located on sides AB and AC, respectively. CD and BE are connected, and it is given that BD = BC = BE. If $\angle A = 30^\circ$ and $\angle ACB = 70^\circ$, what are the measures of $\angle BDC$ and $\angle ACD$?

Solution:
\textbf{Solution:} Given $\because \angle A + \angle ACB + \angle ABC = 180^\circ$, with $\angle A = 30^\circ$ and $\angle ACB = 70^\circ$,\\
$\therefore \angle ABC = 80^\circ$.\\
In $\triangle BDC$, where $BD = BC$,\\
$\therefore \angle BDC = \angle BCD = \frac{180^\circ - 80^\circ}{2} = 50^\circ$,\\
$\therefore \angle ACD = \angle BDC - \angle A = 20^\circ$.\\
Hence, $\angle BDC = \boxed{50^\circ}$ and $\angle ACD = \boxed{20^\circ}$.


Question: As shown in Figure 1, in $\triangle ABC$, $BO \perp AC$ at point $O$, $AO=BO=3$, $OC=1$. A line is drawn through point $A$ such that $AH \perp BC$ at point $H$, intersecting $BO$ at point $P$. What is the length of line segment $OP$?

Solution:
\textbf{Solution:} Since $BO \perp AC$ and $AH \perp BC$,\\
$\therefore$ $\angle AOP=\angle BOC=\angle AHC=90^\circ$,\\
$\therefore$ $\angle OAP+\angle C=\angle OBC+\angle C=90^\circ$,\\
$\therefore$ $\angle OAP=\angle OBC$,\\
In $\triangle OAP$ and $\triangle OBC$,\\
$\left\{\begin{array}{l}
\angle AOP=\angle BOC \\
AO=BO \\
\angle OAP=\angle OBC
\end{array}\right.$,\\
$\therefore$ $\triangle OAP \cong \triangle OBC$ (ASA),\\
$\therefore$ OP=OC=\boxed{1};


Question: As shown in Figure 1, in $\triangle ABC$, $BO\perp AC$ at point $O$, $AO=BO=3$, $OC=1$. A line is drawn through point $A$ such that $AH\perp BC$ at point $H$, and it intersects $BO$ at point $P$. Connect $OH$. What is the degree measure of $\angle AHO$?

Solution:
\textbf{Solution:} Draw $OM \perp CB$ at point M and $ON \perp HA$ at point N, as shown in Figure 1:\\
In quadrilateral $OMHN$, $\angle MON = 360^\circ - 3 \times 90^\circ = 90^\circ$,\\
$\therefore \angle COM = \angle PON = 90^\circ - \angle MOP$.\\
In $\triangle COM$ and $\triangle PON$,\\
$\left\{\begin{array}{l}
\angle COM = \angle PON \\
\angle OMC = \angle ONP = 90^\circ \\
OC = OP
\end{array}\right.$,\\
$\therefore \triangle COM \cong \triangle PON$ (AAS),\\
$\therefore OM = ON$.\\
Because $OM \perp CB$ and $ON \perp HA$,\\
$\therefore HO$ bisects $\angle CHA$,\\
$\therefore \angle AHO = \frac{1}{2} \angle AHC = \boxed{45^\circ}$;


Question: As shown in the figure, in an acute-angled $\triangle ABC$, point E is a point on side AB, with $BE=CE$, $AD\perp BC$ at point D, and $AD$ intersects $EC$ at point G. $\triangle AEG$ is an isosceles triangle. If $BE=10$, $CD=3$, and $G$ is the midpoint of $CE$, what is the length of $AG$?

Solution:
\textbf{Solution:} Draw EF$\perp$AG at point E with F as the foot, \\
$\therefore \angle EFG=90^\circ$, \\
$\because EA=EG$ and EF$\perp$AG, \\
$\therefore AG=2FG$, \\
$\because G$ is the midpoint of $CE$, \\
$\therefore EG=GC=\frac{1}{2}EC$, \\
$\because EB=EC=10$, \\
$\therefore GC=\frac{1}{2}EC=5$, \\
$\because \angle EFG=\angle CDG=90^\circ$ and $\angle EGF=\angle CGD$, \\
$\therefore \triangle EFG \cong \triangle CDG$ (AAS), \\
$\therefore FG=DG$, \\
In $\triangle CDG$, $CD=3$, \\
$\therefore DG=\sqrt{CG^2-CD^2}=\sqrt{5^2-3^2}=4$, \\
$\therefore FG=DG=4$, \\
$\therefore AG=2FG=8$, \\
$\therefore$ The length of $AG$ is $\boxed{8}$.


Question: As shown in the figure, $\triangle ABC$ is an isosceles triangle, with $AB=BC=5$ and $AC=6$. Point $O$ is the midpoint of $AC$, and a line is drawn through $O$ such that $OD \perp AB$ at point $D$. Point $P$ lies on ray $BO$, and point $Q$ lies on line segment $BC$ (not coinciding with points $B$ or $C$). Connect $PQ$. Find the lengths of $BO$ and $OD$.

Solution:
\textbf{Solution:} Given that $AB=BC=5$, $AC=6$, and $OD\perp AB$,\\
it follows that $AO=CO=3$,\\
in $\triangle ABO$, we have $BO=\boxed{4}$,\\
since the area $S_{\triangle ABO}=\frac{1}{2}\times 5\times OD=\frac{1}{2}\times 3\times 4$,\\
thus, $OD=\boxed{\frac{12}{5}}$;


Question: As shown in the diagram, $\triangle ABC$ is an isosceles triangle with $AB=BC=5$ and $AC=6$. Point $O$ is the midpoint of $AC$, and a line is drawn from $O$ perpendicular to $AB$ at point $D$. Point $P$ is located on the ray $BO$, and point $Q$ is located on the line segment $BC$ (not coinciding with points $B$ or $C$), with $PQ$ connected. Given $BQ=4$, point $P$ is rotated $90^\circ$ counterclockwise around point $Q$ to obtain point $E$. When point $E$ lies on one side of $\triangle BPQ$, find the length of $BP$.

Solution:
\textbf{Solution:} When point $E$ is on line $PE$, as shown in Figure 1,\\
draw $QH\perp OC$ at point $H$ through point $Q$, let the intersection of $PQ$ and $CO$ be $G$,\\
$\because \angle PQE=90^\circ$, $QE=QP$,\\
$\therefore \angle QPE=45^\circ$,\\
$\therefore \angle OGP=45^\circ$,\\
$\because BQ=4$, $BC=5$,\\
$\therefore CQ=1$,\\
$\because QH\parallel BO$,\\
$\therefore \frac{CQ}{BC}=\frac{CH}{OC}=\frac{QH}{BO}$, i.e., $\frac{1}{5}=\frac{CH}{3}=\frac{QH}{4}$,\\
$\therefore QH=\frac{4}{5}$, $CH=\frac{3}{5}$,\\
$\because \angle QGH=45^\circ$,\\
$\therefore QH=HG=\frac{4}{5}$,\\
$\because CO=3$,\\
$\therefore OG=3-\frac{3}{5}-\frac{4}{5}=\frac{8}{5}$,\\
$\therefore OP=\frac{8}{5}$,\\
$\therefore BP=4+\frac{8}{5}=\frac{28}{5}$;\\
When point $E$ is on line $BQ$,\\
$\because \angle BQP=\angle BOC=90^\circ$, $BQ=BO=4$,\\
$\therefore \triangle BCO\cong \triangle BPQ$ (ASA),\\
$\therefore BP=BC=5$;\\
In summary: The length of $BP$ is $\boxed{5}$ or $\boxed{\frac{28}{5}}$;


Question: As shown in the diagram, $\triangle ABC$ is an isosceles triangle with $AB=BC=5$ and $AC=6$. Point $O$ is the midpoint of $AC$, and a line is drawn from $O$ perpendicular to $AB$ at point $D$. Point $P$ is located on ray $BO$, and point $Q$ is located on line segment $BC$ (not coinciding with points $B$ or $C$). Line $PQ$ is drawn. When point $P$ coincides with point $O$, a point $F$ is chosen on line segment $AB$ such that points $C$ and $F$ are symmetrical about $PQ$. What is the distance $h$ from point $F$ to $AC$?

Solution:
\textbf{Solution:}

Let $O$ be the origin, the line where $AC$ lies be the $x$-axis, and the line where $OB$ lies be the $y$-axis to establish a Cartesian coordinate system. Connect $CF$,\\
since $C(3, 0)$, $B(0, 4)$, $A(3, 0)$,\\
Let the equation of line $BC$ be $y=kx+b$,\\
thus $\left\{\begin{array}{l}b=4 \\ -3k+b=0\end{array}\right.$,\\
solving yields $\left\{\begin{array}{l}b=4 \\ k=\frac{4}{3}\end{array}\right.$,\\
therefore $y=\frac{4}{3}x+4$,\\
Similarly, the equation of line $AB$ can be found as $y=-\frac{4}{3}x+4$,\\
Let $Q\left(m, \frac{4}{3}m+4\right)$, $F\left(n, -\frac{4}{3}n+4\right)$,\\
The equation of line $OQ$ can be found as $y=\left(\frac{4}{3}+\frac{4}{m}\right)x$,\\
since points $C$ and $F$ are symmetric about $PQ$,\\
thus the midpoint $K$ of $CF$ is $\left(\frac{n-3}{2}, -\frac{2}{3}n+2\right)$,\\
Substituting point $K$ into $y=\left(\frac{4}{3}+\frac{4}{m}\right)x$,\\
yields $-\frac{2}{3}n+2=\left(\frac{4}{3}+\frac{4}{m}\right)\cdot \left(\frac{n-3}{2}\right)$,\\
solving for $m$ gives $m=-\frac{3}{2}$,\\
thus $Q\left(-\frac{3}{2}, 2\right)$,\\
since $CQ=QF$,\\
thus $\left(\frac{3}{2}\right)^{2}+4=\left(n+\frac{3}{2}\right)^{2}+\left(2-\frac{4}{3}n\right)^{2}$,\\
solving yields $n=0$ (discarded) or $n=\frac{21}{25}$,\\
thus $F\left(\frac{21}{25}, \frac{72}{25}\right)$,\\
therefore, the distance $h$ from point $F$ to $AC$ is $\boxed{\frac{72}{25}}$.


Question: As shown in the figure, a regular hexagon and a regular pentagon are placed as depicted, with the common vertex $O$, and the side $AB$ of the hexagon and the side $DE$ of the pentagon lying on the same straight line. What is the measure of $\angle ABO$?

Solution: \textbf{Solution}: From what is given in the problem, we have $\angle ABO = \frac{1}{6} \times (6-2) \times 180^\circ = \boxed{120^\circ}$.


Question: As shown in the figure, a regular hexagon and a regular pentagon are placed in the manner shown, with a common vertex O, and side AB of the hexagon and side DE of the pentagon lying on the same straight line. What is the measure of $\angle BOE$?

Solution:
\textbf{Solution:} Similarly, we can obtain: $\angle DEO=108^\circ$,\\
$\therefore \angle OEB=180^\circ-108^\circ=72^\circ$, $\angle OBE=180^\circ-120^\circ=60^\circ$,\\
$\therefore \angle BOE=180^\circ-72^\circ-60^\circ=\boxed{48^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AD\perp BC$ and $CE\perp AB$ with perpendicular feet at points D and E respectively, $AD$ and $CE$ intersect at point H, and $AE=CE$. Since $\triangle BEC\cong \triangle HEA$, if $BE=8$ and $CH=3$, what is the length of the line segment $AB$?

Solution:
Solution: From Equation (1), we know that $\triangle BEC \cong \triangle HEA$,\\
hence, BE = EH = 8,\\
since CH = 3,\\
thus, CE = AE = 3 + 8 = 11,\\
therefore, AB = AE + BE = 11 + 8 = \boxed{19}.


Question: As shown in the figure, in $\triangle ABC$, $AD$ is the altitude from $A$ to side $BC$, and $BE$ is an angle bisector, intersecting at point $P$. Given that $\angle EPD=120^\circ$, what is the degree measure of $\angle BAD$?

Solution:
\textbf{Solution:} Let $\angle BAD=x^\circ$\\
Given the problem, we have: $\angle BDA=90^\circ$, $\angle ABE=\angle CBE=\frac{1}{2}\angle ABD$,\\
$\therefore$ $\angle ABD=90^\circ-x^\circ$, $\angle ABE=\angle CBE=45^\circ-\frac{x^\circ}{2}$,\\
By the exterior angle property of triangles, we get $\angle EBD+\angle BDA=\angle DPE$,\\
that is $45^\circ-\frac{x^\circ}{2}+90^\circ=120^\circ$, solving this gives $x=\boxed{30}$,\\
hence $\angle BAD=\boxed{30^\circ}$;


Question: As shown in the diagram, in $\triangle ABC$, $AD$ is the altitude on side $BC$, and $BE$ is an angle bisector, intersecting at point P. Given that $\angle BAC=75^\circ$ and $BD=\sqrt{3}$, find the length of $AC$.

Solution:
\textbf{Solution:} From (1) we have $ \angle BAD=30^\circ$, hence $ \angle DAC=45^\circ$,\\
$\therefore$ $ \angle DAC=\angle C=45^\circ$,\\
$\therefore$ $ AD=CD$,\\
In $\triangle ABD$, $ \angle BAD=30^\circ$, $ BD=\sqrt{3}$,\\
$\therefore$ $ AB=2BD=2\sqrt{3}$,\\
By the Pythagorean theorem, we have: $ AD=\sqrt{AB^{2}-BD^{2}}=3$,\\
$ AC=\sqrt{AD^{2}+CD^{2}}=3\sqrt{2}$,\\
i.e., $ AC=\boxed{3\sqrt{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $AD\perp BC$, $AE$ bisects $\angle BAC$, $\angle B=70^\circ$, and $\angle C=30^\circ$. What is the measure of $\angle BAE$?

Solution: \textbf{Solution:} Therefore, $\angle BAE = \boxed{40^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AD \perp BC$, $AE$ bisects $\angle BAC$, $\angle B = 70^\circ$, $\angle C = 30^\circ$. What is the measure of $\angle DAE$?

Solution:
\textbf{Solution:} Since $AD \perp BC$, \\
it follows that $\angle ADC=90^\circ$, \\
which implies $\angle DAC=90^\circ - \angle C=60^\circ$, \\
thus $\angle DAE=\angle DAC - \angle EAC=\boxed{20^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=BC$, and point E is a point inside $\angle ACB$. Connect $CE$, draw $AD\perp CE$ and $BE\perp CE$, with the perpendicular feet being points D and E respectively. $\triangle CAD\cong \triangle BCE$, if $AD=15\,cm$ and $BE=6\,cm$, find the length of $DE$.

Solution:
\textbf{Solution:} Since $\triangle CAD\cong \triangle BCE$,\\
it follows that $AD=CE=15\text{cm}$, and $BE=CD=6\text{cm}$,\\
thus $DE=CE-CD=\boxed{9}\text{cm}$.


Question: As shown in the figure, in $\triangle ABC$, $CD$ bisects $\angle ACB$, $\angle A=68^\circ$, $\angle ACD=30^\circ$. Find: the measure of $\angle BDC$?

Solution:
\textbf{Solution:} Since $\angle BDC=\angle A+\angle ACD$, and $\angle A=68^\circ$, $\angle ACD=30^\circ$,\\
therefore, $\angle BDC=68^\circ+30^\circ=\boxed{98^\circ}$;


Question: As shown in the figure, in $\triangle ABC$, $CD$ bisects $\angle ACB$, $\angle A=68^\circ$, $\angle ACD=30^\circ$. Find: the degree measure of $\angle B$?

Solution:
\textbf{Solution:} Given that $CD$ bisects $\angle ACB$ and $\angle ACD=30^\circ$, \\
$\therefore \angle ACB=2\angle ACD=60^\circ$, \\
$\therefore \angle B=180^\circ-\angle A-\angle ACB=180^\circ-68^\circ-60^\circ=\boxed{52^\circ}$.


Question: As shown in the figure, $A$, $B$, and $D$ are consecutively on the same straight line, $\angle A=\angle D=\angle CBE=90^\circ$, $AC=BD$, $CB=BE$. If $AC=2$ and $AD=6$, what is the length of $CE$?

Solution:
Solution: Since $AC=BD=2$ and $AD=6$,\\
then $AB=AD-BD=6-2=4$,\\
thus $CB^{2}=AC^{2}+AB^{2}=2^{2}+4^{2}=20$,\\
which means $CB^{2}=BE^{2}=20$,\\
therefore $CE= \sqrt{CB^{2}+BE^{2}}=\sqrt{20+20}=2\sqrt{10}$,\\
hence the length of $CE$ is $CE=\boxed{2\sqrt{10}}$.


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle, with $P$ and $Q$ located on sides $AC$ and $BC$ respectively, such that $AP=CQ$. Lines $AQ$ and $BP$ intersect at point $O$. Since $\triangle ACQ \cong \triangle BAP$, we are asked to find the measure of $\angle BOQ$.

Solution:
\textbf{Solution:} Since $\triangle ACQ\cong \triangle BAP$,\\
it follows that $\angle CAQ=\angle ABP$,\\
thus $\angle BOQ=\angle BAO+\angle ABP=\angle BAO+\angle CAQ=BAC=60^\circ$;\\
therefore, the measure of $\angle BOQ$ is $\boxed{60^\circ}$.


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle, with points $P$ and $Q$ located on sides $AC$ and $BC$ respectively, and $AP=CQ$. Lines $AQ$ and $BP$ intersect at point $O$. When $\angle CBP=45^\circ$ and $BQ=4\sqrt{6}$, what is the length of $OB$?

Solution:
\textbf{Solution:} As shown in the diagram, construct $QH\perp BP$ at H, \\
since $\angle CBP=45^\circ$ and $QH\perp BP$, \\
it follows that $\angle PBQ=\angle BQH=45^\circ$, \\
hence $BH=HQ$, \\
given $BQ=4\sqrt{6}$, \\
it follows that $BH=HQ=4\sqrt{3}$, \\
since $\angle BOQ=60^\circ$, \\
it follows that $\angle AQH=30^\circ$, \\
therefore $HQ=\sqrt{3}OH$, \\
hence $OH=4$, \\
thus $OB=OH+BH=4\sqrt{3}+4=\boxed{4\sqrt{3}+4}$.


Question: As shown in the figure, in $ \triangle ABC$, $\angle ACB=90^\circ$, $AM=AC$, $BN=BC$. When $\angle A=30^\circ$, what is the measure of $\angle MCN$ in degrees?

Solution:
\textbf{Solution:} Given that $\angle A=30^\circ$ and $AC=AM$,\\
it follows that $\angle AMC=\frac{1}{2}\times \left(180^\circ-30^\circ\right)=75^\circ$,\\
Since $\angle B=60^\circ$ and $BC=BN$,\\
it implies $\angle CNB=\frac{1}{2}\times \left(180^\circ-60^\circ\right)=60^\circ$,\\
Thus, $\angle MCN=180^\circ-60^\circ-75^\circ=\boxed{45^\circ}$;


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AM=AC$, $BN=BC$. When the degree of $\angle A$ changes, does the degree of $\angle MCN$ change? If it does not change, find the degree of $\angle MCN$.

Solution:
\textbf{Solution:}\\
Since $\angle AMC=\frac{1}{2}\left(180^\circ-\angle A\right), \angle CNB=\frac{1}{2}\left(180^\circ-\angle B\right)$,\\
it follows that $\angle MCN=180^\circ-\angle AMC-\angle BNC=\frac{1}{2}\left(\angle A+\angle B\right)=45^\circ$,\\
therefore, as the measure of $\angle A$ changes, the measure of $\angle MCN$ remains constant, $\angle MCN=\boxed{45^\circ}$.


Question: As shown in the figure, in $\triangle ACB$, $\angle ACB=90^\circ$, point M is the midpoint of side $AB$, point E is on segment $AM$, $EF\perp AC$ at point F, connect $CM$, $CE$. It is known that $\angle A=50^\circ$, $\angle ACE=30^\circ$, and $CE=CM$. If $AB=4$, find the length of segment $FC$.

Solution:
\textbf{Solution:} Since $AB=4$,\\
it follows that $CE=CM=\frac{1}{2}AB=2$,\\
since $EF\perp AC$ and $\angle ACE=30^\circ$,\\
it follows that $EF=\frac{1}{2}CE=1$,\\
therefore $FC=\sqrt{CE^{2}-EF^{2}}=\sqrt{2^{2}-1^{2}}=\boxed{\sqrt{3}}$.


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle with side length 2, $D$ is a point on the extension line of $CA$, an equilateral triangle $BDE$ is constructed with $BD$ as a side, and $AE$ is connected. Find the degree measure of $\angle EAD$.

Solution:
\textbf{Solution:} Given that $\triangle ABC$ and $\triangle BDE$ are equilateral triangles,\\
$\therefore AB=BC=AC=2$, $BD=BE$, $\angle ABC=\angle C=\angle BAC=\angle DBE=60^\circ$,\\
$\therefore \angle ABC+\angle ABD=\angle DBE+\angle ABD$,\\
which means $\angle CBD=\angle ABE$,\\
In $\triangle CBD$ and $\triangle ABE$,\\
$ \left\{\begin{array}{l}
BC=AB \\
\angle CBD=\angle ABE \\
BD=BE
\end{array}\right.$\\
$\therefore \triangle CBD\cong \triangle ABE$ (SAS),\\
$\therefore \angle BAE=\angle BCD=60^\circ$,\\
$\therefore \angle EAD=180^\circ-60^\circ-60^\circ=\boxed{60^\circ}$;


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle with side length 2. Point $D$ is on the extension of side $CA$. An equilateral triangle $BDE$ is constructed with $BD$ as a side. Connect $AE$. Find the value of $AE-AD$.

Solution:
\textbf{Solution:} Since $\triangle CBD \cong \triangle ABE$,\\
it follows that $CD=AE$,\\
thus $AE-AD=CD-AD=AC=\boxed{2}$.


Question: As shown in the figure, it is a "Gougu Circle and Square Diagram" (also known as "Zhao Shuang's Chord Diagram") created by ancient Chinese mathematician Zhao Shuang. It consists of four congruent right-angled triangles and a small square EFGH seamlessly put together to form a larger square ABCD. If $\angle ABE=30^\circ$ and $EF=\sqrt{3}-1$, find the length of AB.

Solution: \textbf{Solution:} Given $\because \triangle ABE \cong \triangle BCF$, $\therefore AE=BF$, $\because \angle ABE=30^\circ$, let $AE=x$, $\therefore \tan \angle ABE = \frac{AE}{BE} = \frac{x}{x+\sqrt{3}-1} = \frac{\sqrt{3}}{3}$, solving this, we get $x=1$, $\therefore AE=1$, $BE=BF+EF= \sqrt{3}$, $\therefore AB=\sqrt{AE^2+BE^2}=\boxed{2}$.


Question: As shown in the figure, it is a "Gougu Circle and Square Diagram" (also known as "Zhao Shuang's Chord Diagram") created by the ancient Chinese mathematician Zhao Shuang. It consists of four congruent right-angled triangles and a smaller square EFGH seamlessly pieced together to form a larger square ABCD. Point M is on FG, with $AB\parallel EM$ and $AB=2EM$. Find the ratio of the perimeter of square ABCD to that of square EFGH.

Solution: \textbf{Solution:} Given that $AB\parallel EM$, it follows that $\angle ABE = \angle FEM$ and $\angle AEB = \angle EFM = 90^\circ$. Therefore, $\triangle ABE \sim \triangle EMF$, which implies $\frac{EM}{AB}=\frac{EF}{BE}=\frac{1}{2}$, so $BF=EF$. Thus, $AE=BF=EF$. Let EF=1, then $AB=\sqrt{AE^2+BE^2}=\sqrt{5}$. Therefore, the perimeter ratio of square ABCD to square EFGH is $AB\colon EF = \boxed{\sqrt{5}\colon 1}$.


Question: As shown in the figure, two straight lines $l_{1}$ and $l_{2}$ passing through point A intersect the y-axis at points B and C, respectively, where point B is above the origin, and point C is below the origin. Given that AB = $\sqrt{13}$ and B(0, 3), what are the coordinates of point A?

Solution:
\textbf{Solution}: Since $B(0,3)$,\\
it follows that $OB=3$,\\
In $\triangle AOB$, $OA= \sqrt{AB^{2}-OB^{2}}=\sqrt{13-3^{2}}=2$,\\
thus \boxed{A(2,0)}.


Question: As shown in the figure, two lines $l_{1}$ and $l_{2}$ passing through point A intersect the y-axis at points B and C, respectively, where point B is above the origin and point C is below the origin. It is known that $AB = \sqrt{13}$ and $B(0,3)$. If the area of $\triangle ABC$ is 4, what is the equation of the line $l_{2}$?

Solution:
\textbf{Solution:} Let $\because S_{\triangle ABC}=\frac{1}{2}BC \cdot OA$,\\
$\therefore 4=\frac{1}{2} \cdot BC \times 2$, solving this gives $BC=4$,\\
$\therefore OC=BC-OB=4-3=1$,\\
$\therefore C(0, -1)$,\\
Let the equation of the line $l_{2}$ be $y=kx+b$,\\
Substituting $A(2, 0)$, $C(0, -1)$ into $y=kx+b$, we get:\\
$\begin{cases}
0=2k+b\\
-1=b
\end{cases}$, solving this gives $\begin{cases}
k=\frac{1}{2}\\
b=-1
\end{cases}$,\\
$\therefore$ the equation of line $l_{2}$ is $\boxed{y=\frac{1}{2}x-1}$;


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle, with $D$ and $E$ located on $AB$ and $BC$ respectively, and $AD=BE$, $AE$ and $CD$ intersect at point $F$. What is the measure of $\angle CFE$ in degrees?

Solution:
\textbf{Solution:} As shown in Figure 1,\\
Figure 1\\
$\because$ $\triangle ABC$ is an equilateral triangle\\
$\therefore$ $AB=BC=CA$, and $\angle ABC=\angle BCA=\angle CAB=60^\circ$\\
In $\triangle ABE$ and $\triangle CAD$,\\
$\because$ $\left\{\begin{array}{l}
AB=CA\\
\angle ABE=\angle CAD\\
BE=AD
\end{array}\right.$\\
$\therefore$ $\triangle ABE \cong \triangle CAD$ (SAS)\\
$\therefore$ $\angle ACD=\angle BAE$\\
In $\triangle AFC$, $\angle CFE=\angle ACD+\angle CAF=\angle BAE+\angle CAF=\angle BAC=60^\circ$\\
$\therefore$ $\angle CFE=\boxed{60^\circ}$


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle, with D and E located on $AB$ and $BC$ respectively, and $AD=BE$, $AE$ intersects $CD$ at point F. As shown in Figure 2, draw $CH\perp AE$ at point H through point C. If $AE=5$ and $HF=2$, what is the length of $DF$?

Solution:
\textbf{Solution:} As shown in Figure 2,\\
$\because CH\perp AE$\\
$\therefore \angle FHC=90^\circ$\\
In $\triangle FHC$, $\angle FCH=90^\circ-\angle CFH=30^\circ$\\
$\because HF=2$\\
$\therefore FC=2FH=4$\\
$\because \triangle ABE\cong \triangle CAD$\\
$\therefore AE=CD=5$\\
$\therefore DF=CD-CF=5-4=\boxed{1}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AC \perp BC$, $AB=13$, $BC=12$, $CD=3$, $AD=4$. What is the length of $AC$?

Solution:
\textbf{Solution:} In $\triangle ABC$,\\
since $AC \perp BC$,\\
it follows that $AC = \sqrt{AB^{2} - BC^{2}} = \sqrt{169 - 144} = \boxed{5}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AC \perp BC$, $AB=13$, $BC=12$, $CD=3$, $AD=4$. What is the area of quadrilateral $ABCD$?

Solution:
\textbf{Solution:} Given that $\because A{D}^{2}+C{D}^{2}=16+9=25=A{C}^{2}$,\\
$\therefore \angle D=90^\circ$,\\
$\therefore {S}_{\text{quadrilateral }ABCD}={S}_{\triangle ABC}+{S}_{\triangle ACD}=\frac{12\times 5}{2}+\frac{4\times 3}{2}=\boxed{36}$.


Question: As shown in the figure, in the right-angled triangle $ABC$, where $\angle C=90^\circ$, the bisector of $\angle CAB$ intersects $BC$ at point $D$, and $DE$ is the perpendicular bisector of $AB$ with the foot of the perpendicular being $E$. What is the measure of $\angle CAD$?

Solution:
\textbf{Solution:} Since DE is the perpendicular bisector of AB,\\
it follows that AD=BD,\\
hence, $\angle B=\angle EAD$,\\
moreover, since AD is the bisector of $\angle CAB$,\\
it follows that $\angle CAD=\angle EAD$,\\
let $\angle CAD=x$, then $3x=90^\circ$,\\
therefore, $x=30^\circ$,\\
thus, $\angle CAD=\boxed{30^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, the bisector of $\angle CAB$ intersects $BC$ at point $D$, and $DE$ is the perpendicular bisector of $AB$, with $E$ being the foot of the perpendicular. If $BC=3$, what is the length of $DE$?

Solution:
\textbf{Solution:} Since $AD$ is the bisector of $\angle CAB$, and $DC \perp AC$, $DE \perp AB$,\\
it follows that $DC=DE$,\\
Let $DC=y$, then $DE=y$, and $BD=3-y$,\\
Furthermore, since $\angle B=30^\circ$,\\
we have $y=\frac{3-y}{2}$,\\
Solving this, we get $\boxed{y=1}$,\\
thus $DE=\boxed{1}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle A=90^\circ$, $BD$ bisects $\angle ABC$ and intersects $AC$ at point $D$, with $AB=4$, $BC=12$, $AD=3$. If point $P$ moves along $BC$, what is the minimum value of the length of segment $DP$?

Solution:
\textbf{Solution:} When $DP \perp BC$, the length of the line segment $DP$ is minimized, \\
$\because BD$ bisects $\angle ABC$ and $\angle A = 90^\circ$, \\
when $DP \perp BC$, $DP = AD$, \\
$\because AD = 3$, \\
$\therefore$ the minimal value of $DP$ is $\boxed{3}$;


Question: As shown in the figure, in $\triangle ABC$, $\angle A=90^\circ$, BD bisects $\angle ABC$ and intersects AC at point D, AB=4, BC=12, and AD=3. If point P moves on BC, when DP is minimized, what is the area of $\triangle CDP$?

Solution:
\textbf{Solution}: Since $\angle A=90^\circ$, \\
it follows that BD=$\sqrt{AB^{2}+AD^{2}}=\sqrt{4^{2}+3^{2}}=5$, \\
When DP is at its minimum, DP=3, and DP$\perp$BC, \\
thus $\angle$DPB=$\angle$DPC=$90^\circ$, \\
so PB=$\sqrt{BD^{2}-DP^{2}}=\sqrt{5^{2}-3^{2}}=4$, \\
thus CP=BC-PB=12-4=8, \\
therefore, the area of $\triangle CDP$=$\frac{1}{2}$CP$\times$DP=$\frac{1}{2}\times8\times3=\boxed{12}$, \\
i.e., when DP is at its minimum, the area of $\triangle CDP$ is \boxed{12}.


Question: As shown in the figure, it is known that point $C$ is a point on $AB$, $AC=30\,cm$, $BC=\frac{2}{5}AC$, and $D$, $E$ are the midpoints of $AC$, $AB$ respectively. How many line segments are there in the figure in total?

Solution: \textbf{Solution:} In the figure, there are \boxed{10} line segments.


Question: As shown in the figure, it is known that point $C$ is a point on $AB$, $AC=30\,cm$, $BC=\frac{2}{5}AC$, and $D$, $E$ are the midpoints of $AC$, $AB$ respectively. What is the length of $DE$ in $cm$?

Solution:
\textbf{Solution:} Given that $AC=30\,cm$, $BC=\frac{2}{5}AC$,\\
hence $BC=\frac{2}{5}\times 30=12\,cm$, $AB=AC+BC=42\,cm$,\\
also, since D and E are the midpoints of $AC$ and $AB$, respectively,\\
thus $AD=DC=\frac{1}{2}AC=15\,cm$, $AE=BE=\frac{1}{2}AB=21\,cm$\\
therefore $DE=AE-AD=21-15=\boxed{6}\,cm$


Question: As shown in the figure, point $O$ is a point on line $AB$, and ray $OC$ passes through point $O$ such that $\angle BOC=65^\circ$. A right-angled triangle ruler is placed with its right-angled vertex at point $O$. As shown in Figure 1, when one side $ON$ of the triangle ruler $MON$ coincides with ray $OB$, then what is the degree measure of $\angle MOC$?

Solution: \textbf{Solution:} We have $\angle MOC = \boxed{25^\circ}$.


Question: As shown in the diagram, point $O$ is located on line segment $AB$. A ray $OC$ is drawn from point $O$ such that $\angle BOC = 65^\circ$. A right-angled triangle ruler is placed with its right angle vertex at point $O$. As shown in Diagram 2, the triangle ruler $MON$ is rotated counterclockwise around point $O$ by a certain angle, such that ray $OC$ bisects angle $\angle MOB$. What are the measures of $\angle BON$ and $\angle CON$?

Solution:
\textbf{Solution}: Since $\angle BOC=65^\circ$ and $OC$ is the angle bisector of $\angle MOB$,\\
it follows that $\angle MOB=2\angle BOC=130^\circ.$\\
Hence, $\angle BON=\angle MOB-\angle MON=130^\circ-90^\circ=40^\circ.$\\
$\angle CON=\angle COB-\angle BON=65^\circ-40^\circ=25^\circ.$\\
Thus, $\angle BON=\boxed{40^\circ}, \angle CON=\boxed{25^\circ}$.


Question: As shown in the figure, $OM$ is the bisector of $\angle AOC$, and $ON$ is the bisector of $\angle BOC$. As in Figure 1, when $\angle AOB$ is a right angle and $\angle BOC=60^\circ$, what is the degree measure of $\angle MON$?

Solution:
\textbf{Solution:} By the given problem, \\
$\because \angle AOB=90^\circ, \angle BOC=60^\circ$, \\
$\therefore \angle AOC=90^\circ+60^\circ=150^\circ$, \\
$\because OM$ bisects $\angle AOC, ON$ bisects $\angle BOC$, \\
$\therefore \angle MOC=\frac{1}{2}\angle AOC=75^\circ, \angle NOC=\frac{1}{2}\angle BOC=30^\circ$, \\
$\therefore \angle MON=\angle MOC-\angle NOC=\boxed{45^\circ}$.


Question: The point $P$ is a point on the line segment $AB$, and the points $M$ and $N$ are the midpoints of the line segments $AP$ and $PB$, respectively. As shown in the figure, if the point $P$ is the midpoint of the line segment $AB$, and $MP=4\, \text{cm}$, what is the length of the line segment $AB$ in cm?

Solution:
\textbf{Solution:} Given that point $M$ and $N$ are the midpoints of line segments $AP$ and $PB$ respectively,\\
$\therefore AP=2MP$, $BP=2PN$.\\
$\because MP=4\, \text{cm}$, $\therefore AP=8\, \text{cm}$.\\
$\because P$ is the midpoint of $AB$, $\therefore AB=2AP=\boxed{16}\, \text{cm}$.


Question: As shown in the figure, point $P$ is a point on the line segment $AB$, and points $M$ and $N$ are the midpoints of line segments $AP$ and $PB$, respectively. If point $P$ is any point on the line segment $AB$, and $AB=12\, \text{cm}$, what is the length of the line segment $MN$ in $\text{cm}$?

Solution:
\textbf{Solution:} Since points M and N are the midpoints of the line segments AP and PB, respectively,\\
$\therefore AP=2MP$, $BP=2PN$.\\
$\therefore AP+BP=2MP+2PN=2MN$, that is $AB=2MN$.\\
$\because AB=12\ \text{cm}$,\\
$\therefore MN=\boxed{6}\ \text{cm}$.


Question: As shown in the figure, it is known that ship A is in the direction of 30 degrees north of east from lighthouse P, ship B is in the direction of 70 degrees south of east from lighthouse P, and ship C is on the bisector of $\angle APB$. What is the degree measure of $\angle APB$?

Solution:
\textbf{Solution:} Given that $\angle APN=30^\circ$ and $\angle BPS=70^\circ$, \\
$\therefore \angle APB=180^\circ-\angle APN-\angle BPS=180^\circ-30^\circ-70^\circ=80^\circ$.\\
$\therefore$ The measure of $\angle APB$ is $\boxed{80^\circ}$.


Question: As shown in the figure, it is known that ship A is in the direction $30^\circ$ north of east from lighthouse P, ship B is in the direction $70^\circ$ south of east from lighthouse P, and ship C is on the bisector of $\angle APB$. How many degrees north of east is ship C from lighthouse P?

Solution:
\textbf{Solution:} Since $PC$ bisects $\angle APB$ and $\angle APB=80^\circ$,
\[\therefore \angle APC= \frac{1}{2}\angle APB= \frac{1}{2}\times 80^\circ=40^\circ.\]
\[\therefore \angle NPC=\angle APN+\angle APC=30^\circ+40^\circ=\boxed{70^\circ}.\]
Therefore, the ship C is in the direction of 70$^\circ$ north by east from the lighthouse P.


Question: As shown in the figure, point C is on the line segment AB, with $AC=12\,cm$ and $CB=9\,cm$. M and N are the midpoints of AC and BC, respectively. What is the length of the line segment MN in cm?

Solution:
\textbf{Solution:} Given that M and N are the midpoints of AC and BC, respectively,

\[
\therefore MC=\frac{1}{2}AC, \quad CN=\frac{1}{2}CB,
\]

\[
\therefore MC+CN=\frac{1}{2}AC+\frac{1}{2}CB=\frac{1}{2}(AC+CB),
\]

which means $MN=\frac{1}{2}(AC+CB)=\frac{1}{2}\times(12+9)=\boxed{\frac{21}{2}}\text{cm}.$


Question: As shown in the figure, points $A$, $B$, and $C$ are on the same straight line, point $D$ is the midpoint of segment $AC$, and point $E$ is the midpoint of segment $BC$. As shown in Figure 1, point $C$ is on the segment $AB$; if $AC=6$ and $BC=4$, how long is segment $DE$?

Solution:
\textbf{Solution:} Since $D$ is the midpoint of line segment $AC$, and $E$ is the midpoint of line segment $BC$,\\
therefore $DC=\frac{1}{2}AC=3$, and $CE=\frac{1}{2}BC=2$,\\
thus $DE=DC+CE=3+2=\boxed{5}$;


Question: As shown in the figure, points A, B, and C are on the same straight line. Point $D$ is the midpoint of line segment $AC$, and point $E$ is the midpoint of line segment $BC$. As shown in Figure 2, point $C$ is on the extension of line segment $AB$. If $DE=5$, find the length of line segment $AB$.

Solution:
\textbf{Solution:} Since $D$ is the midpoint of the line segment $AC$ and $E$ is the midpoint of the line segment $BC$, \\
thus $DC = \frac{1}{2} AC$, $CE = \frac{1}{2} BC$, \\
thus $DE = DC - CE = \frac{1}{2} (AC - BC) = \frac{1}{2} AB = 5$, \\
thus $AB = \boxed{10}$.


Question: As shown in the figure, it is known that $O$ is a point on the line $AB$. Rays $OC$, $OD$, and $OM$ are drawn such that $OM$ bisects $\angle AOC$. When $OC$, $OD$, and $OM$ are all above $AB$, given that $\angle COD=60^\circ$ and $\angle BOC=10^\circ$, what is the degree measure of $\angle DOM$?

Solution:
\textbf{Solution}: From the given information, we have $\angle AOC=180^\circ-10^\circ=170^\circ$, \\
since $OM$ bisects $\angle AOC$, \\
thus $\angle MOC=\frac{170^\circ}{2}=85^\circ$, \\
therefore $\angle MOD=85^\circ-60^\circ=\boxed{25^\circ}$.


Question: As shown in the diagram, it is known that $O$ is a point on line $AB$, with rays $OC$, $OD$, and $OM$ drawn such that $OM$ bisects $\angle AOC$. When $OC$ and $OM$ are above $AB$, and $OD$ is below $AB$, given that $\angle COD=90^\circ$ and the ratio of $\angle BOC$ is $1:10$, what is the degree measure of $\angle DOM$?

Solution:
\textbf{Solution:} Let $\angle AOD=a$, then $\angle BOC=10a$, $\angle AOC=90^\circ-a$,\\
from the given problem we have: $\angle AOC+\angle BOC=90^\circ-a+10a=180^\circ$,\\
solving this gives us $a=10^\circ$.\\
Therefore, $\angle AOC=90^\circ-10^\circ=80^\circ$, $\angle BOC=100^\circ$.\\
Since $OM$ bisects $\angle AOC$,\\
it follows that $\angle AOM=\frac{1}{2}\angle AOC=80^\circ \div 2=40^\circ$.\\
Therefore, $\angle DOM=40^\circ+10^\circ=\boxed{50^\circ}$.


Question: As shown in the figure, it is known that the length of line segment AB is $a$. Extend BA to point C such that $AC = \frac{1}{2} AB$. Point D is the midpoint of line segment BC. What is the length of CD?

Solution:
\textbf{Solution:} Since $AB = a$, \\
therefore $AC = \frac{1}{2}AB = \frac{1}{2}a$, \\
thus $BC = AC + AB = a + \frac{1}{2}a = \frac{3}{2}a$, \\
since D is the midpoint of the line segment BC, \\
thus $CD = BD = \frac{1}{2}BC = \frac{3}{4}a = \boxed{\frac{3}{4}a}$.


Question: As shown in the figure, it is known that the length of line segment $AB$ is $a$. Extend $BA$ to point $C$, making $AC=\frac{1}{2}AB$. Let $D$ be the midpoint of line segment $BC$. If $AD=3\, \text{cm}$, what is the value of $a$?

Solution:
Solution: Since $AD=AB-BD=a-\frac{3}{4}a=3\text{cm}$, \\
it follows that $a=\boxed{12}\text{cm}$.


Question: As shown in the figure, the lines AB and CD intersect at point O. OE is the bisector of $\angle AOC$, $OF\perp CD$, $OG\perp OE$, and $\angle BOD=52^\circ$. What is the measure of $\angle AOF$?

Solution:
\textbf{Solution:} Since OF$\perp$CD, \\
$\therefore$ $\angle$COF$=90^\circ$, \\
Also, since $\angle$AOC and $\angle$BOD are vertical angles, \\
$\therefore$ $\angle$AOC$=\angle$BOD$=52^\circ$, \\
$\therefore$ $\angle$AOF=$\angle$COF-$\angle$AOC=$90^\circ-52^\circ$=$\boxed{38^\circ}$;


Question: As shown in the figure, it is known that $\angle AOB=2\angle BOC$, and $OD$ is the bisector of $\angle AOC$. If $\angle AOB=60^\circ$, what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Given $\because \angle AOB=2\angle BOC$ and $\angle AOB=60^\circ$,\\
$\therefore \angle BOC= \frac{1}{2}\angle AOB=30^\circ$,\\
$\therefore \angle AOC=\angle AOB+\angle BOC=90^\circ$,\\
$\because$ OD is the bisector of $\angle AOC$,\\
$\therefore \angle COD= \frac{1}{2}\angle AOC=45^\circ$,\\
$\therefore \angle BOD=\angle COD-\angle BOC=45^\circ-30^\circ=\boxed{15^\circ}$.


Question: As shown in the figure, it is known that $\angle BOC = 60^\circ$, $OE$ bisects $\angle AOC$, and $OF$ bisects $\angle BOC$. If $AO \perp BO$, then what is the measure of $\angle EOF$ in degrees?

Solution: \textbf{Solution:} Given $\because \angle BOC=60^\circ$, and OF bisects $\angle BOC$, $\therefore \angle COF= \frac{1}{2}\angle BOC=30^\circ$, $\because AO\perp BO$, $\therefore \angle AOB=90^\circ$, hence $\angle AOC=90^\circ+60^\circ=150^\circ$, $\because OE$ bisects $\angle AOC$, $\therefore \angle COE= \frac{1}{2}\angle AOC=75^\circ$, thus $\angle EOF=\angle COE-\angle COF=\boxed{45^\circ};$


Question: As shown in the figure, it is known that $\angle BOC=60^\circ$, $OE$ bisects $\angle AOC$, and $OF$ bisects $\angle BOC$. If $\angle AOC + \angle EOF = 210^\circ$, then how many degrees is $\angle EOF$?

Solution:
\textbf{Solution:} Given that $\angle BOC=60^\circ$ and OF bisects $\angle BOC$,\\
it follows that $\angle COF= \frac{1}{2}\angle BOC=30^\circ$,\\
Given that OE bisects $\angle AOC$,\\
it follows that $\angle AOC=2\angle COE$,\\
Given that $\angle AOC+\angle EOF=210^\circ$,\\
it follows that $2\angle COE+\angle COE-\angle COF=210^\circ$,\\
which simplifies to $3\angle COE-30^\circ=210^\circ$,\\
and therefore $\angle COE=80^\circ$,\\
consequently, $\angle EOF=\angle COE-\angle COF=\boxed{50^\circ}$.


Question: As shown in Figure 1, it is known that $\angle AOB= 40^\circ$. If $\angle AOC= \frac{1}{3}\angle BOC$, what is the degree measure of $\angle BOC$?

Solution:
\textbf{Solution:} As shown in the diagram, when OC is outside of $\angle AOB$,\\
$\because \angle AOC= \frac{1}{3}\angle BOC= \frac{1}{3}(\angle AOB+\angle AOC)$,\\
$\therefore \angle AOC= \frac{1}{2}\angle AOB= \frac{1}{2}\times 40^\circ=20^\circ$,\\
$\therefore \angle BOC=\angle AOB+\angle AOC=40^\circ+ 20^\circ= 60^\circ$;\\
As shown in the diagram, when OC is inside of $\angle AOB$,\\
$\because \angle AOC= \frac{1}{3}\angle BOC= \frac{1}{3} (\angle AOB-\angle AOC)$,\\
$\therefore \angle AOC= \frac{1}{4}\angle AOB= \frac{1}{4}\times 40^\circ=10^\circ$,\\
$\therefore \angle BOC=\angle AOB-\angle AOC=40^\circ- 10^\circ= 30^\circ$.\\
Hence, the degree of $\angle BOC$ is $\boxed{60^\circ}$ or $\boxed{30^\circ}$.


Question: As shown in the figure, it is known that $\angle AOB= 40^\circ$. As shown in figure 2, $\angle AOC= 30^\circ$, $OM$ is a straight line inside $\angle AOB$, $ON$ is the quadrisection line of $\angle MOC$, and $3\angle CON=\angle NOM$, what is the value of $4\angle AON+\angle COM$?

Solution:
\textbf{Solution:} Let $\angle CON = \alpha$, \\
since ON is the angle bisector of $\angle MOC$, \\
thus $\angle MOC = 4\alpha$. \\
Since $3\angle CON = \angle NOM$, \\
thus $\angle MON = 3\angle CON = 3\alpha$ \\
Since $\angle AOC = 30^\circ$, \\
thus $\angle AON = \angle AOC - \angle CON = 30^\circ - \alpha$, \\
therefore $4\angle AON + \angle COM = 4(30^\circ-\alpha) + 4\alpha = \boxed{120^\circ}$.


Question: As shown in the figure, it is known that the length of segment $AB=4$. Extend $AB$ to $C$ such that $BC=\frac{1}{2}AB$. Extend $AB$ in the opposite direction to $D$ such that $AD=\frac{1}{2}AC$. What is the length of segment $CD$?

Solution:
\textbf{Solution:} Given the problem,\\
$\because AB=4$, $BC=\frac{1}{2}AB$,\\
$\therefore BC=\frac{1}{2}\times 4=2$.\\
$\therefore AC=AB+BC=4+2=6$.\\
$\because AD=\frac{1}{2}AC$,\\
$\therefore AD=\frac{1}{2}\times 6=3$.\\
$\therefore CD=AD+AC=3+6=\boxed{9}$.


Question: As shown in the figure, it is known that the line segment $AB=4$. Extend $AB$ to $C$ so that $BC=\frac{1}{2}AB$. Extend $AB$ in the opposite direction to $D$ so that $AD=\frac{1}{2}AC$. If $Q$ is the midpoint of $AB$, and $P$ is a point on the line segment $CD$, and $BP=\frac{1}{2}BC$, what is the length of the line segment $PQ$?

Solution:
\textbf{Solution}: Since $AB=4$, and $Q$ is the midpoint of $AB$, \\
$\therefore QB=\frac{1}{2}AB=\frac{1}{2}\times 4=2$.\\
Since $BC=2$.\\
$\therefore BP=\frac{1}{2}BC=\frac{1}{2}\times 2=1$.\\
When point $P$ is to the right of point $B$, $PQ=QB+BP=2+1=3$;\\
When point $P$ is to the left of point $B$, $PQ=QB-BP=2-1=1$.\\
Thus, the length of $PQ$ is either $1$ or \boxed{3}.


Question: As shown in the figure, line AB and line CD intersect at point O, with OM$\perp$AB at point O. If $\angle BOC=2\angle AOC$, what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution}:
Given that $\because \angle BOC = 2\angle AOC$, and $\angle BOC + \angle AOC = 180^\circ$,\\
$\therefore 2\angle AOC + \angle AOC = 180^\circ$,\\
$\therefore 3\angle AOC = 180^\circ$,\\
$\therefore \angle AOC = 60^\circ$,\\
$\therefore \angle BOD = \angle AOC = \boxed{60^\circ}$.


Question: Lines AB and CD intersect at point O, and OE bisects $\angle BOD$. As shown in Figure 1, if $\angle BOC=130^\circ$, what is the degree measure of $\angle AOE$?

Solution:
\textbf{Solution:} Since $\angle BOC= 130^\circ$,\\
it follows that $\angle AOD=\angle BOC=130^\circ$,\\
$\angle BOD=180^\circ-\angle BOC=50^\circ$.\\
Since OE bisects $\angle BOD$,\\
it follows that $\angle DOE=25^\circ$,\\
therefore $\angle AOE=\angle AOD+\angle DOE=\boxed{155^\circ}$.


Question: Line AB intersects line CD at point O, and OE bisects $\angle BOD$. As shown in Figure 2, ray OF is inside $\angle AOD$.

Solution:
\textbf{Solution:}

(1) OF is the bisector of $\angle$AOD, for the following reason:

Since OF $\perp$ OE, therefore $\angle$EOF $=90^\circ$,

thus $\angle$BOE + $\angle$AOF = $90^\circ$.

Since OE bisects $\angle$BOD, therefore $\angle$BOE = $\angle$DOE,

thus $\angle$DOE + $\angle$AOF = $90^\circ$.

Therefore, $\angle$DOE + $\angle$DOF = $90^\circ$,

therefore, $\angle$AOF = $\angle$DOF,

therefore, OF is the bisector of $\angle$AOD.

(2) Since $\angle AOF = \frac{5}{3}\angle DOF$,

let $\angle$DOF = 3x, then $\angle$AOF = 5x.

Since OF bisects $\angle$AOE,

thus $\angle$AOF = $\angle$EOF = 5x, hence $\angle$DOE = 2x.

Since OE bisects $\angle$BOD, therefore $\angle$BOD = 4x,

therefore 5x + 3x + 4x = $180^\circ$, therefore x = $15^\circ$, therefore $\angle$BOD = 4x = $\boxed{60^\circ}$.


Question: As shown in the figure, it is known that $OB$, $OC$, and $OD$ are three rays within $\angle AOE$, with $OB$ bisecting $\angle AOE$ and $OD$ bisecting $\angle COE$. If $\angle AOB=70^\circ$ and $\angle DOE=20^\circ$, what is the degree measure of $\angle BOC$?

Solution:
\textbf{Solution:} Because $OB$ bisects $\angle AOE$ and $OD$ bisects $\angle COE$, \\
therefore $\angle BOE = \angle AOB = 70^\circ$, $\angle COE = 2\angle DOE = 40^\circ$, \\
thus $\angle BOC = \angle BOE - \angle COE = 70^\circ - 40^\circ = \boxed{30^\circ}$.


Question: As shown in the figure, it is known that $OB$, $OC$, and $OD$ are three rays within $\angle AOE$, $OB$ bisects $\angle AOE$, $OD$ bisects $\angle COE$. If $\angle AOE=136^\circ$ and $AO\perp CO$, what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Given the construction,\\
$\because$ OB bisects $\angle AOE$, OD bisects $\angle COE$,\\
$\therefore \angle BOE=\frac{1}{2}\angle AOE$, $\angle DOE=\frac{1}{2}\angle COE$.\\
$\because \angle BOD=\angle BOE-\angle DOE$,\\
$\therefore \angle BOD=\angle BOE-\angle DOE=\frac{1}{2}\left(\angle AOE-\angle COE\right)=\frac{1}{2}\angle AOC$,\\
$\because$ AO$\perp$CO,\\
$\therefore \angle AOC=90^\circ$,\\
$\therefore \angle BOD=\boxed{45^\circ}$.


Question: As shown in the figure, it is known that $OB$, $OC$, $OD$ are three rays within $\angle AOE$, with $OB$ bisecting $\angle AOE$, and $OD$ bisecting $\angle COE$. If $\angle DOE = 20^\circ$ and $\angle AOE + \angle BOD = 220^\circ$, what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} From the given information, we have\\
$\because$ OB bisects $\angle AOE$,\\
$\therefore \angle AOE=2\angle BOE$.\\
$\because \angle AOE+\angle BOD=220^\circ$,\\
$\therefore 2\angle BOE+\angle BOD=220^\circ$.\\
$\because \angle BOE-\angle BOD=\angle DOE=20^\circ$,\\
$\therefore 2\angle BOE-2\angle BOD=40^\circ$, i.e., $2\angle BOE=40^\circ+2\angle BOD$,\\
$\therefore 2\angle BOE+\angle BOD=40^\circ+3\angle BOD=220^\circ$,\\
$\therefore 3\angle BOD=180^\circ$,\\
$\therefore \angle BOD=\boxed{60^\circ}$.


Question: As shown in the figure, it is known that lines $AB$ and $CD$ intersect at point $O$. $OE$ and $OF$ are rays, $\angle AOE = 90^\circ$, and $OF$ bisects $\angle BOC$. If $\angle EOF = 30^\circ$, what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Given the construction, we have\\
$\because \angle AOE= 90^\circ$,\\
$\therefore \angle EOB= 180^\circ-\angle AOE= 90^\circ$.\\
$\because \angle EOF= 30^\circ$,\\
$\therefore \angle FOB=\angle EOB-\angle EOF= 60^\circ$.\\
$\because$ OF bisects $\angle BOC$,\\
$\therefore \angle BOC=2\angle FOB= 120^\circ$,\\
$\therefore \angle BOD=180^\circ-\angle BOC=\boxed{60^\circ}$.


Question: As shown in the figure, it is known that point $C$ is on the line segment $AB$, and $M$, $N$ are the midpoints of $AC$, $BC$ respectively. If the length of segment $AC=12$ and $BC=8$, find the length of segment $MN$.

Solution:
\textbf{Solution:} By the given information, we have \(AC=12\) and \(BC=8\), with \(M\), \(N\) being the midpoints of \(AC\), \(BC\) respectively, \\
\(\therefore MC= \frac{1}{2}AC\), \(NC= \frac{1}{2}BC\), \\
\(\therefore MN= \frac{1}{2}AC+ \frac{1}{2}BC= \frac{1}{2} \times 12+ \frac{1}{2} \times 8=\boxed{10}.\)


Question: As shown in the figure, it is known that point $C$ is on the line segment $AB$, and $M$, $N$ are the midpoints of $AC$, $BC$ respectively. Let $AB=a$, find the length of the line segment $MN$.

Solution:
\textbf{Solution:} From (1) it is known that $MN= \frac{1}{2}AC+ \frac{1}{2}BC= \frac{1}{2}(AC+BC)= \frac{1}{2}AB$,\\
since $AB=a$,\\
therefore $MN= \frac{1}{2}a=\boxed{\frac{1}{2}a}$.


Question: As shown in the figure, it is known that point C is on the line segment AB, and M and N are the midpoints of AC and BC, respectively. Given the line segment DE, extend DE to F such that EF= $ \frac{1}{2}$DE; extend ED to G such that DG=2DE, with P and Q being the midpoints of EF and DG, respectively. If PQ=18 cm, what is the length of DE in cm?

Solution:
\textbf{Solution:} Let $DE=x$ cm, then $EF= \frac{1}{2}x$ cm, $EP= \frac{1}{2}EF= \frac{1}{4}x$ cm, $DG=2x$ cm, and $DQ= \frac{1}{2}DG=x$ cm.\\
Since $PQ=18$ cm, we have $x+x+\frac{1}{4}x=18$,\\
solving this yields $x=\boxed{8}$,\\
thus $DE=\boxed{8}$ cm.


Question: As shown in the figure, line AB and line CD intersect at point O, with OM $\perp$ AB. If $\angle 1=40^\circ$ and $\angle 2= 30^\circ$, what is the degree measure of $\angle$NOD?

Solution:
\textbf{Solution:} Since OM$\perp$AB,\\
it follows that $\angle$AOM= $90^\circ$.\\
Since $\angle$1=$40^\circ$,\\
it follows that $\angle$AOC=$\angle$AOM−$\angle$1= $90^\circ−40^\circ= 50^\circ$,\\
Since $\angle$2=$30^\circ$,\\
it follows that $\angle$NOD=$\angle$DOC−$\angle$AOC−$\angle$2= $\boxed{100^\circ}$.


Question: As shown in the figure, the lines AB, CD, and EF intersect at point O. Write the vertical angles of $\angle AOC$ and $\angle BOF$.

Solution: \textbf{Solution:} From the given, we understand that the vertical angle of $\angle AOC$ is \boxed{\angle BOD}, and the vertical angle of $\angle BOF$ is \boxed{\angle AOE}.


Question: As shown in the figure, lines AB, CD, and EF intersect at point O. If $\angle AOC=70^\circ$ and $\angle BOF=20^\circ$, what are the measures of $\angle BOC$ and $\angle DOE$?

Solution:
\textbf{Solution}: Given the problem,\\
since $\angle AOC=70^\circ$, and $\angle AOC + \angle BOC= 180^\circ$,\\
thus $\angle BOC= \boxed{110^\circ}$.\\
Since $\angle BOF= 20^\circ$,\\
thus $\angle COF=\angle BOC-\angle BOF= 90^\circ$,\\
therefore $\angle DOE=\boxed{90^\circ}$.


Question: As shown in the figure, lines MD and CN intersect at point O. OA is a ray inside $\angle MOC$, and OB is a ray inside $\angle NOD$. $\angle MON=70^\circ$. If $\angle BOD=\frac{1}{2}\angle COD$, find the degree measure of $\angle BON$.

Solution:
\textbf{Solution:}

Since $\angle MON=70^\circ$,\\
it follows that $\angle COD= \angle MON=70^\circ$,\\
thus $\angle BOD= \frac{1}{2}\angle COD= \frac{1}{2}\times70^\circ=35^\circ$,\\
therefore $\angle BON= 180^\circ-\angle MON-\angle BOD= 180^\circ-70^\circ- 35^\circ=\boxed{75^\circ}$.


Question: As shown in the figure, line MD and CN intersect at point O. OA is a ray inside $\angle MOC$, and OB is a ray inside $\angle NOD$. $\angle MON=70^\circ$. If $\angle AOD=2\angle BOD$ and $\angle BOC=3\angle AOC$, what is the measure of $\angle BON$?

Solution:
\textbf{Solution:} Let $\angle AOC=x^\circ$, then $\angle BOC=3x^\circ$,\\
since $\angle COD=\angle MON=70^\circ$,\\
thus $\angle BOD=\angle BOC-\angle COD= 3x^\circ- 70^\circ$.\\
Therefore, $\angle AOD=\angle AOC+\angle COD=x^\circ+70^\circ$.\\
Since $\angle AOD=2\angle BOD$,\\
therefore $x+70= 2(3x-70)$,\\
solving this, we get $x=42$,\\
thus $\angle BOD=3x^\circ- 70^\circ=3\times42^\circ- 70^\circ=56^\circ$,\\
therefore, $\angle BON= 180^\circ-\angle MON-\angle BOD= 180^\circ-70^\circ-56^\circ=\boxed{54^\circ}$.


Question: As shown in the figure, $EF \parallel CD$, $\angle 1 + \angle 2 = 180^\circ$. If $CD$ bisects $\angle ACB$, $DG$ bisects $\angle CDB$, and $\angle A = 40^\circ$, what is the measure of $\angle ACB$?

Solution:
\textbf{Solution:} Given from the previous question, we know that $GD\parallel CA$,\\
$\therefore \angle GDB = \angle A = 40^\circ$,\\
$\because DG$ bisects $\angle CDB$, and $\angle CDB$ is the exterior angle of triangle $ADC$,\\
$\therefore \angle CDB = \angle A + \angle ACD = 40^\circ + \angle ACD = 80^\circ$,\\
$\therefore \angle ACD = 40^\circ$,\\
Moreover, since $CD$ bisects $\angle ACB$,\\
$\therefore \angle ACB = 40^\circ \times 2 = \boxed{80^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=80^\circ$, $\angle ACB=50^\circ$. What is the measure of $\angle A$ in degrees?

Solution:
\textbf{Solution}: Since $\angle ABC=80^\circ$ and $\angle ACB=50^\circ$, \\
it follows that $\angle A=180^\circ - \angle ABC - \angle ACB=\boxed{50^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=80^\circ$, $\angle ACB=50^\circ$. $BP$ bisects $\angle ABC$, $CP$ bisects $\angle ACB$. Find the measure of $\angle BPC$ in degrees?

Solution:
\textbf{Solution:} Given that $BP$ bisects $\angle ABC$ and $CP$ bisects $\angle ACB$,\\
it follows that $\angle PBC= \frac{1}{2}\angle ABC$ and $\angle PCB= \frac{1}{2}\angle ACB$,\\
hence $\angle PBC+\angle PCB= \frac{1}{2}(\angle ABC+\angle ACB)=65^\circ$,\\
therefore, $\angle BPC=180^\circ-(\angle PBC+\angle PCB)=\boxed{115^\circ}$.


Question: As shown in Figure 1, it is known that point $O$ is on line $AB$, with rays $OD$ and $OC$ located respectively above and below line $AB$, and $\angle COD = 80^\circ$. Ray $OE$ is always the bisector of $\angle AOD$. If $\angle AOE = 10^\circ$, what is the degree measure of $\angle COE$?

Solution:
\textbf{Solution:} Since $OE$ bisects $\angle AOD$,\\
it follows that $\angle AOE=\angle DOE=10^\circ$,\\
hence $\angle COE-\angle COD-\angle DOE=80^\circ-10^\circ-70^\circ=\boxed{0^\circ}$.


Question: As shown in Figure 1, it is known that point O is on line AB, with rays OD and OC on opposite sides of line AB, and $ \angle COD=80^\circ$. Ray OE is always the bisector of $ \angle AOD$. As shown in Figure 2, let $ \angle AOD=n^\circ$. It is known that $ \angle DOE=2\angle AOC$. Find the value of $n$.

Solution:
\textbf{Solution:} Given that,\\
$\because OE$ bisects $\angle AOD$,\\
$\therefore \angle DOE=\frac{n^\circ}{2}$,\\
$\because \angle COD=80^\circ$,\\
$\therefore \angle AOC=80^\circ-n^\circ$,\\
$\because \angle DOE=2\angle AOC$,\\
$\therefore \frac{n}{2}=2(80-n)$,\\
$\therefore n=\boxed{64}$.


Question: As shown in Figure 1, it is known that point O is on line AB, and rays OD and OC are on the upper and lower sides of line AB, respectively, with $\angle COD=80^\circ$. OE is always the bisector of $\angle AOD$. As illustrated in Figure 3, under the condition of (2), ray OP starts rotating counterclockwise from OB around point O at a speed of $4^\circ$ per second, and ray OQ starts rotating clockwise from OE around point O at a speed of $2^\circ$ per second. Rays OP and OQ start rotating simultaneously, and the rotation time is denoted as t seconds $\left(0\leq t\leq 45\right)$. When $\angle DOQ$ and $\angle EOP$ are complementary, find the value of the rotation time t.

Solution:
\textbf{Solution:} When $n=64$, we have $\angle EOD=\angle AOE=32^\circ$ and $\angle EOB=180^\circ-32^\circ=148^\circ$.

(1) When $0\leq t \leq 16$

Since $\angle QOD=32^\circ-2t^\circ$ and $\angle EOP=148^\circ-4t^\circ$,

it follows that $32-2t+148-4t=90$.

Therefore, $t=\boxed{15}$.

(2) For $16 < t < 37$

Since $\angle QOD=2t^\circ-32^\circ$ and $\angle EOP=148^\circ-4t^\circ$,

it follows that $2t-32+148-4t=90$.

Therefore, $t=13$ (discarded).

(3) For $37 \leq t \leq 45$

$\angle QOD=2t^\circ-32^\circ$ and $\angle EOP=4t^\circ-148^\circ$,

it follows that $2t-32+4t-148=90$.

Therefore, $t=\boxed{45}$.

In summary, $t=\boxed{15}$ seconds or $\boxed{45}$ seconds.


Question: As shown in the figure, $AB$ is a straight line, $OD$ and $OE$ are respectively the bisectors of $\angle AOC$ and $\angle BOC$. Given $\angle BOC=72^\circ 20'$, find the measures of $\angle 1$, $\angle 2$, and $\angle DOE$.

Solution:
\textbf{Solution}: Since $AB$ is a straight line, and $OD, OE$ are the bisectors of $\angle AOC, \angle BOC$ respectively, and $\angle BOC=72^\circ 20'$,\\
thus $\angle 1=\angle EOB=\frac{1}{2}\angle BOC=36^\circ 10'$,\\
thus $\angle DOC=\angle AOD=\frac{1}{2}\angle AOC=\frac{1}{2}(180^\circ-\angle BOC)=\frac{1}{2}(180^\circ-72^\circ 20')=53^\circ 50'$\\
thus $\angle DOE=\angle 1+\angle 2=36^\circ 10'+53^\circ 50'=\boxed{90^\circ}$;


Question: As shown in the figure, $AB$ is a straight line, and $OD$, $OE$ are the angle bisectors of $\angle AOC$, $\angle BOC$, respectively. If $\angle BOC = \alpha$, what is $\angle DOE$?

Solution:
\textbf{Solution:} Since $AB$ is a straight line, and $OD, OE$ are the bisectors of $\angle AOC, \angle BOC$ respectively,\\
thus $\angle 1=\angle EOB=\frac{1}{2}\angle BOC$\\
thus $\angle DOC=\angle AOD=\frac{1}{2}\angle AOC$,\\
thus $\angle DOE=\angle 1+\angle 2=\frac{1}{2}\angle AOC+\frac{1}{2}\angle BOC=90^\circ=\boxed{90^\circ}$.


Question: As shown in the figure, lines $AB$ and $CD$ intersect at point $O$, $EO\perp CD$ at point $O$, and $OF$ bisects $\angle BOC$. If $\angle AOC = 58^\circ$, what is the degree measure of $\angle EOF$?

Solution:
\textbf{Solution:} From the construction, we have\\
$\because \angle AOC + \angle BOC = 180^\circ$, $\angle AOC = 58^\circ$,\\
$\therefore \angle BOC = 122^\circ$,\\
$\because OF$ bisects $\angle BOC$,\\
$\therefore \angle COF = \frac{1}{2}\angle BOC = 61^\circ$,\\
$\because EO \perp CD$,\\
$\therefore \angle COE = 90^\circ$,\\
$\therefore \angle EOF = \angle COE - \angle COF = 90^\circ - 61^\circ = \boxed{29^\circ}$;


Question: As shown in the figure, points $A$, $O$, and $B$ lie on the same straight line, and $\angle AOC = \angle BOD$. Lines $OM$ and $ON$ are the angle bisectors of $\angle AOC$ and $\angle BOD$, respectively. If $\angle COD = 80^\circ$, what is the measure of $\angle MON$?

Solution:
\textbf{Solution:} Since $OM$ and $ON$ are the angle bisectors of $\angle AOC$ and $\angle BOD$ respectively, \\
hence, $\angle MOC = \frac{1}{2}\angle AOC$ and $\angle NOD = \frac{1}{2}\angle BOD$, \\
therefore, $\angle MON = \frac{1}{2}(\angle AOC + \angle BOD) + \angle COD$ \\
$= \frac{1}{2}(180^\circ - \angle COD) + \angle COD$ \\
$= 90^\circ + \frac{1}{2}\angle COD$, \\
since $\angle COD = 80^\circ$, \\
therefore, $\angle MON = 90^\circ + \frac{1}{2} \times 80^\circ = \boxed{130^\circ}$;


Question: As shown in the figure, $OC \perp AB$ at point O, $OD$ bisects $\angle BOC$, $OE$ bisects $\angle AOD$. Find the degree measure of $\angle BOD$.

Solution:
\textbf{Solution:} Since $OC \perp AB$,\\
it follows that $\angle BOC = \angle AOC = 90^\circ$,\\
since $OD$ bisects $\angle BOC$,\\
it follows that $\angle BOD = \angle COD = \frac{1}{2}\angle BOC = 45^\circ = \boxed{45^\circ}$.