Question: As shown in the figure, in the equilateral triangle $ABC$, point $E$ is a point on side $CB$ (not coinciding with point $C$), and point $F$ is a point on side $AC$. Given $BC=10$, $EC=6$, and $\angle AEF=60^\circ$, what is the length of $AF$?

Solution:
\textbf{Solution:} Given from the problem, $\triangle ABC$ is an equilateral triangle,
\[
\therefore \angle B = \angle C = 60^\circ, \quad AC = BC = AB = 10,
\]
\[
\because EC = 6,
\]
\[
\therefore BE = 4,
\]
\[
\because \angle AEC = \angle BAE + \angle B,
\]
that is, $\angle AEF + \angle CEF = \angle BAE + \angle B$,
and considering $\angle AEF = 60^\circ$, $\angle B = 60^\circ$,
\[
\therefore \angle BAE = \angle CEF,
\]
\[
\because \angle B = \angle C,
\]
\[
\therefore \triangle ABE \sim \triangle ECF,
\]
\[
\therefore \frac{BE}{CF} = \frac{AB}{EC}, \text{ that is, } \frac{4}{CF} = \frac{10}{6},
\]
\[
\therefore CF = \frac{12}{5},
\]
\[
\therefore AF = AC - CF = 10 - \frac{12}{5} = \boxed{\frac{38}{5}}\].


Question: As shown in the figure, a circular arc in the Cartesian coordinate system passes through the grid points $A(0, 4)$, $B(4, 4)$, and $C(6, 2)$. What are the coordinates of the center of the circle containing this arc?

Solution: \textbf{Solution:} The center of the circle containing the arc has coordinates $\boxed{(2,0)}$.


Question: As shown in the figure, within the rectangular coordinate system, an arc passes through grid points $A\left(0, 4\right)$, $B\left(4, 4\right)$, and $C\left(6, 2\right)$. Find the length of arc ABC.

Solution:
\textbf{Solution:} Connect AC,\\
Based on the grid, we can determine that OP=CQ=2, OA=PQ=4,\\
$\angle AOP=\angle PQC=90^\circ$,\\
By Pythagorean theorem,\\
AP= $\sqrt{OP^2+OA^2}=\sqrt{2^2+4^2}=2\sqrt{5}=PC$,\\
$\because AP^2=2^2+4^2=20$, CP$^2=2^2+4^2=20$, AC$^2=2^2+6^2=40$,\\
$\therefore AP^2+CP^2=AC^2$,\\
$\therefore \angle APC=90^\circ$,\\
$\therefore$ The length of arc ABC is $\frac{90^\circ \pi \times 2\sqrt{5}}{180}=\boxed{\sqrt{5}\pi}$.


Question: As shown in the figure, it's known that $AB$ is the diameter of $\odot O$, $PD$ is tangent to $\odot O$ at point C, and intersects the extension line of $AB$ at point D, and $\angle D=2\angle CAD$. Find the size of $\angle D$.

Solution:
\textbf{Solution:} Connect $OC$.\\
Since $\overset{\frown}{BC}=\overset{\frown}{BC}$,\\
it follows that $ \angle CAD=\frac{1}{2}\angle COB$, i.e., $ \angle COB=2\angle CAD$.\\
Since $ \angle D=2\angle CAD$,\\
it follows that $ \angle COB=\angle D$.\\
Since $ PD$ is tangent to $\odot O$,\\
it follows that $ OC\perp PD$, i.e., $ \angle OCD=90^\circ$.\\
Therefore, $ \angle COB+\angle D=90^\circ$.\\
Therefore, $ 2\angle D=90^\circ$.\\
Therefore, $ \angle D=\angle COB=\boxed{45^\circ}$.


Question: As shown in the figure, it is known that $AB$ is the diameter of the circle $\odot O$, $PD$ is tangent to $\odot O$ at point C, and intersects the extension line of $AB$ at point D, and $\angle D = 2\angle CAD$. If $CD=2$, find the length of the arc $\overset{\frown}{AC}$.

Solution:
Solution: Since $\angle COB=\angle D$, and $CD=2$,
thus $CO=CD=2$.
Since $\angle COB=45^\circ$,
thus $\angle AOC=135^\circ$.
Therefore, the length of the arc $\overset{\frown}{AC}$, $l=\frac{n\pi R}{180}=\frac{135\times 2\times \pi }{180}=\boxed{\frac{3}{2}\pi}$


Question: As shown in the figure, point $E$ is the incenter of $\triangle ABC$. The extension line of $AE$ intersects the circumcircle of $\triangle ABC$ at point $D$. Connect $BE$. If $\angle CBD = 34^\circ$, what is the degree measure of $\angle BEC$?

Solution:
\textbf{Solution}:\\
Since $\angle CBD=34^\circ$,\\
it follows that $\angle CAD=34^\circ$,\\
since point E is the incenter of $\triangle ABC$,\\
it follows that $\angle BAC=2\angle CAD=68^\circ$,\\
therefore $\angle EBC+\angle ECB=(180^\circ-68^\circ)\div 2=56^\circ$,\\
hence, $\angle BEC=180^\circ-56^\circ=\boxed{124^\circ}$.


Question: As shown in the figure, $O$ is the center of the semicircle, and $C$, $D$ are two points on the semicircle. Connect $CD$, $BD$, $AD$, where $CD=BD$. Extend $AC$ to intersect the extension of $BD$ at point $E$. Given that $CD=DE$; if $AC=6$ and the radius $OB=5$, find the length of $BD$.

Solution:
\textbf{Solution:} In right triangle $\triangle ACB$, we have $BC=\sqrt{AB^{2}-AC^{2}}=\sqrt{10^{2}-6^{2}}=8$.\\
Since $CD=DE$ and $CD=BD$,\\
it follows that $BD=ED$.\\
In $\triangle ADB$ and $\triangle ADE$, we have\\
$\left\{\begin{array}{l}
AD=AD \\
\angle ADB=\angle EDA \\
BD=ED
\end{array}\right.$,\\
therefore, $\triangle ADB \cong \triangle ADE \text{ (SAS)}$,\\
thus, $AE=AB=10$,\\
hence, $CE=AE-AC=10-6=4$,\\
In right triangle $\triangle ECB$, $BE=\sqrt{CE^{2}+CB^{2}}=\sqrt{4^{2}+8^{2}}=4\sqrt{5}$,\\
therefore, $BD=\frac{1}{2}BE=2\sqrt{5}=\boxed{2\sqrt{5}}$.


Question: As shown in the figure, we have $AB=AC=2\sqrt{2}$ and $\angle BAC=90^\circ$. There is a moving point $D$ on the line segment $BC$. We connect $AD$. Through point $A$, we draw $AD\perp AE$ where $AD=AE$. We then connect $DE$. What is the length of $BC$?

Solution: \textbf{Solution:} According to the problem, we have $BC=\boxed{4}$.


Question: As shown in the figure, $AB=AC=2\sqrt{2}$, $\angle BAC=90^\circ$. On the line segment $BC$, there is a moving point $D$. Connect $AD$. Through point $A$, construct $AD \perp AE$ and $AD=AE$. Connect $DE$. When $BD=1$, what is the length of $DE$?

Solution: \textbf{Solution:} When $BD = 1$, the length of $DE$ is $\boxed{\sqrt{10}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=75^\circ$, $AE\perp BC$ at point E, $CD\perp AB$ at point D, $AE$ and $CD$ intersect at point F, and $\angle AFC=120^\circ$. Calculate how many degrees is $\angle BCD$?

Solution: \textbf{Solution}: $\angle BCD = \boxed{30^\circ}$.


Question: As shown, in $\triangle ABC$, $\angle ACB=75^\circ$, $AE\perp BC$ at point E, $CD\perp AB$ at point D, $AE$ and $CD$ intersect at point F, and $\angle AFC=120^\circ$. If $BD=1$, $AD=\sqrt{3}$, then how long is $EF$?

Solution: \textbf{Solution:} We have $EF=\boxed{\frac{\sqrt{3}-1}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, CE is the altitude to the hypotenuse AB, and the angle bisector BD intersects CE at point M. $\triangle CDM$ is an isosceles triangle. If AB = 10 and AC = 8, find the length of CM.

Solution:
\textbf{Solution:} Construct $DF\perp AB$ at point $F$, as shown in the figure,\\
$\because \angle DCB=90^\circ$ and $BD$ bisects $\angle ABC$,\\
$\therefore DC=DF$,\\
$\because \angle ACB=90^\circ$, $AB=10$, $AC=8$,\\
$\therefore BC= \sqrt{AB^{2}-AC^{2}}=\sqrt{10^{2}-8^{2}}=6$,\\
$\because S_{\triangle ABC}=S_{\triangle BCD}+S_{\triangle ADB}$,\\
$\therefore \frac{AC\cdot BC}{2}=\frac{BC\cdot CD}{2}+\frac{AB\cdot DF}{2}$,\\
thus $\frac{8\times 6}{2}=\frac{6\cdot CD}{2}+\frac{10\cdot DF}{2}$,\\
solving gives $CD=DF=3$,\\
From (1) we know: $CM=CD$,\\
$\therefore CM=\boxed{3}$,\\
which means the length of $CM$ is $3$.


Question: In the diagram, in $\triangle ABC$, $\angle BAC=90^\circ$, $AB=5\,cm$, $BC=13\,cm$, point $D$ is on line segment $AC$, and $CD=7\,cm$. A moving point $P$ starts from point $E$, which is $15\,cm$ away from point $B$, and moves in the direction of the ray $EA$ at a speed of $2\,cm$ per second, for a time of $t$ seconds. Find the length of $AD$.

Solution:
\textbf{Solution:} According to the Pythagorean theorem, we have $AC=\sqrt{BC^{2}-AB^{2}}=\sqrt{13^{2}-5^{2}}=12\,(\text{cm})$,\\
$\therefore AD=AC-CD=5$,\\
$\therefore$ the length of $AD$ is $\boxed{5}\,\text{cm}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, $AB=5\,cm$, and $BC=13\,cm$. Point $D$ is on line segment $AC$, and $CD=7\,cm$. Moving point $P$ starts from point $E$, which is $15\,cm$ away from point $B$, and moves in the direction of ray $EA$ at a speed of $2\,cm$ per second, for a time of $t$ seconds. Express the length of $AP$ in centimeters as an algebraic expression containing $t$.

Solution:
\textbf{Solution:} Given that $BE=15$ and $AB=5$,\\
therefore $AE=20$,\\
thus when $P$, $A$ coincide, $t=\frac{20}{2}=10$,\\
therefore for $0\le t\le 10$, $AP=\left(20-2t\right)\text{cm}$,\\
when $t>10$, $AP=\left(2t-20\right)\text{cm}$.\\
$\therefore \boxed{AP=\left(20-2t\right)\text{cm} \text{ for } 0\le t\le 10}$,\\
$\boxed{AP=\left(2t-20\right)\text{cm} \text{ for } t>10}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, $AB=5\text{cm}$, $BC=13\text{cm}$. Point $D$ is on segment $AC$, and $CD=7\text{cm}$. A moving point $P$ starts from point $E$, which is $15\text{cm}$ away from point $B$, and moves in the direction of ray $EA$ at a speed of $2\text{cm}$ per second, for a time of $t$ seconds. During the motion, is there a moment when $\triangle ABC$ and $\triangle ADP$ are congruent? If yes, please calculate the value of $t$; if no, please explain the reason.

Solution:
\textbf{Solution:} There exists a moment such that $\triangle ABC$ is congruent to $\triangle ADP$, for the following reasons:\\
$\because$ $\angle DAP=\angle BAC=90^\circ$, and $DA=BA$,\\
$\therefore$ when $AP=AC$, $\triangle ABC\cong \triangle ADP$,\\
$\therefore$ $20-2t=12$ or $2t-20=12$,\\
Solving this, we get $t=\boxed{4}$ or $t=16$,\\
$\therefore$ the values of $t$ that satisfy the conditions are \boxed{4} or \boxed{16}.


Question: As shown in the diagram, point $C$ is a moving point on $AB$. On the same side of $AB$, isosceles right triangles $ACD$ and $CBE$ are constructed with $AC$ and $BC$ as hypotenuses, respectively. Line segment $DE$ is drawn, and point $F$ is located on $DE$ with $CF$ connected and $FD=FC$. Point $F$ is the midpoint of $DE$. A perpendicular line to $AB$ is drawn from point $F$, and point $G$ is the foot of the perpendicular. What is the value of $\frac{FG}{AB}$?

Solution:
\textbf{Solution:} Construct $DM \perp AB$ at $M$ through point $D$, and construct $EN \perp AB$ at $N$ through point $E$,\\
since $\triangle ADC$ and $\triangle BCE$ are isosceles right triangles,\\
therefore $AC=2DM$, $BC=2EN$,\\
thus $AB=AC+BC=2(DM+EN)$,\\
since $FG\perp AB$,\\
thus $DM\parallel FG\parallel EN$,\\
since $DF=EF$,\\
thus $MG=NG$,\\
therefore $FG=\frac{1}{2}(DM+EN)$,\\
therefore $\frac{FG}{AB}=\frac{\frac{1}{2}(DM+EN)}{2(DM+EN)}=\boxed{\frac{1}{4}}$;


Question: As shown in the figure, point $C$ is a moving point on line segment $AB$. Isoceles right triangles $\triangle ACD$ and $\triangle CBE$ are constructed on the same side of $AB$ with $AC$ and $BC$ as their hypotenuses, respectively. Line segment $DE$ is drawn, and point $F$ is located on $DE$. Line segments $CF$ and $FD=FC$ are drawn. Given that $AB=12$, and the sum of the areas of triangles $\triangle ACD$ and $\triangle CEB$ is denoted as $S$, find the minimum value of $S$.

Solution:
\textbf{Solution:} Let $\because AB=12$,\\
$\therefore$let $AC=x$, then $BC=12-x$,\\
$\because DM\perp AB$, $EN\perp AB$,\\
$\therefore DM=\frac{1}{2}AC$, $EN=\frac{1}{2}BC$,\\
$\therefore S=\frac{1}{2}AC\cdot DM+\frac{1}{2}BC\cdot EN=\frac{1}{4}AC^{2}+\frac{1}{4}BC^{2}=\frac{1}{4}x^{2}+\frac{1}{4}(12-x)^{2}$,\\
i.e., $S=\frac{1}{2}x^{2}-6x+36=\frac{1}{2}(x-6)^{2}+18$\\
$\therefore S$'s minimum value is $\boxed{18}$.


Question: As shown in the figure, in the Cartesian coordinate system (with $O$ as the origin), it is known that the line $y=kx+b$ intersects the $x$-axis and the $y$-axis at points $A(-2,0)$ and $B(0,-1)$ respectively, and the coordinates of point $C$ are $(0,2)$. What is the equation of line $AB$?

Solution:
\textbf{Solution:} Since the line $y=kx+b$ intersects the x-axis and y-axis at points $A(-2,0)$ and $B(0,-1)$ respectively,

$\therefore$ $\begin{cases} -2k+b=0\\ b=-1 \end{cases}$,

Solving this yields $\begin{cases} k=-\frac{1}{2}\\ b=-1 \end{cases}$,

$\therefore$ the equation of line $AB$ is: $\boxed{y=-\frac{1}{2}x-1}$.


Question: As shown in the figure, in the Cartesian coordinate system (where $O$ is the origin), it is known that the line $y=kx+b$ intersects the $x$ axis and the $y$ axis at points $A(-2,0)$ and $B(0,-1)$, respectively. The coordinates of point $C$ are $(0,2)$. Suppose point $D$ is a point on line $AB$ and $CD=BD$. Find the coordinates of point $D$.

Solution:
\textbf{Solution:} Let us draw $DH \perp BC$ from point $D$, with the perpendicular foot at $H$, as shown in the figure:\\
$\because CD=BD$,\\
$\therefore HC=HB=\frac{1}{2}BC$,\\
$\because B(0, -1), C(0, 2) \therefore BC=3, OC=2$\\
$\therefore CH=\frac{3}{2}$,\\
$\therefore OH=\frac{1}{2}$,\\
$\therefore$ substitute $y=\frac{1}{2}$ into the line equation $y=-\frac{1}{2}x-1$,\\
we get $\frac{1}{2}=-\frac{1}{2}x-1$, solving for $x$, we find $x=-3$\\
$\therefore$ The coordinates of point $D$ are $\boxed{(-3, \frac{1}{2})}$.


Question: As shown in the figure, $AD \parallel BC$, E is the midpoint of $CD$, extend line $AE$ to intersect the extension of line $BC$ at point F. $\triangle ADE \cong \triangle FCE$. If $\angle BAF = 90^\circ$, $AB = 8$, and $EF = 3$, what is the area of quadrilateral $ABCD$?

Solution:
\textbf{Solution:} Given that $ \because \triangle ADE \cong \triangle FCE$,\\
$ \therefore AE=EF=3$, $ {S}_{\triangle ADE}={S}_{\triangle FCE}$\\
$ \therefore AF=AE+EF=6$\\
$ \because \angle BAF=90^\circ$,\\
$ {S}_{\triangle ABF}=\frac{1}{2}AB\cdot AF=\frac{1}{2}\times 8\times 6=24$\\
$ \therefore {S}_{\text{quadrilateral }ABCD}={S}_{\text{quadrilateral }ABCE}+{S}_{\triangle ADE}={S}_{\text{quadrilateral }ABCE}+{S}_{\triangle FCE}={S}_{\triangle ABF}=\boxed{24}$


Question: As shown in the diagram, the line \(l_{1}\colon y=x+3\) intersects with the line \(l_{2}\) passing through point \(A(3,0)\) at point \(C(1,m)\), and intersects with the \(x\)-axis at point \(B\). What is the equation of line \(l_{2}\)?

Solution:
\textbf{Solution:} Substitute $(1, m)$ into $y = x + 3$ to get $m = 4$, therefore $C(1, 4)$. Let the equation of line $l_2$ be $y = kx + b$. Substituting $(1, 4)$ and $(3, 0)$ into the equation, we get
\[
\left\{
\begin{array}{l}
k + b = 4 \\
3k + b = 0
\end{array}
\right.
\]
Solving the system, we find
\[
\left\{
\begin{array}{l}
k = -2 \\
b = 6
\end{array}
\right.
\]
Therefore, the equation of line $l_2$ is $y = -2x + 6$. Hence, the answer is $y = \boxed{-2x + 6}$.


Question: As shown in the figure, the line $l_{1}\colon y=x+3$ intersects with the line $l_{2}$ that passes through the point $A(3,0)$ at point $C(1,m)$, and intersects with the $x$-axis at point $B$. Point $M$ is on the line $l_{1}$, $MN$ is parallel to the $y$-axis, and intersects line $l_{2}$ at point $N$. If $MN=AB$, find the coordinates of point $M$.

Solution:
\textbf{Solution:} In the equation $y = x + 3$, let $y = 0$ to find $x = -3$,

$\therefore B(-3,0)$,

$\therefore AB = 3 - (-3) = 6$,

Let $M(a, a + 3)$. Since $MN$ is parallel to the y-axis, we get $N(a, -2a + 6)$,

$MN = |a + 3 - (-2a + 6)| = |3a - 3|$,

Since $MN = AB$,

$\therefore |3a - 3|= 6$,

Solving this gives $a = 3$ or $a = -1$,

$\therefore \boxed{M(3, 6)} or \boxed{M(-1, 2)}$.


Question: As shown in the figure, the line $l_{1}: y = x + 3$ intersects with the line $l_{2}$ passing through point A$(3, 0)$ at point C$(1, m)$, and with the x-axis at point B. Is there a point P on the x-axis that makes the triangle BCP an isosceles triangle? If such a point exists, please write down the coordinates of point P; if not, please explain why.

Solution:
\textbf{Solution:} As shown in Fig. 2, $\because B(-3,0), C(1,4)$.
$\therefore BC= \sqrt{(1+3)^{2}+(4-0)^{2}}=4\sqrt{2}$.
Let $P(x,0)$,
When $PC=BC$, at this time point $P$ is symmetric to point $B$ about the line $x=1$, then $P_{1}(5,0)$;
When $PC=PB$, $(x+3)^{2}=(x-1)^{2}+(0-4)^{2}$.
Solving this yields $x=1$. At this time, $P_{2}(1,0)$;
When $BP=BC$, $\left|x+3\right|=4\sqrt{2}$,
Solving this yields $x=-3+4\sqrt{2}$ or $x=-3-4\sqrt{2}$.
At this time, $P_{3}(-3+4\sqrt{2},0)$, $P_{4}(-3-4\sqrt{2},0)$.
In summary, the coordinates of point $P$ that meet the conditions are \boxed{P_{1}(5,0)}, \boxed{P_{2}(1,0)}, \boxed{P_{3}(-3+4\sqrt{2},0)}, \boxed{P_{4}(-3-4\sqrt{2},0)}.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, $AC=BC$, $BE\perp CE$ at $E$, and $AD\perp CE$ at $D$. We have $\triangle ADC \cong \triangle CEB$. Given $AD=5\text{cm}$ and $DE=3\text{cm}$, find the length of $BE$ in cm.

Solution:
\textbf{Solution:} \\
From (1), we know that $\triangle ADC \cong \triangle CEB$,\\
hence $AD = CE = 5\,\text{cm}$, and $CD = BE$.\\
Since $CD = CE - DE$,\\
it follows that $BE = AD - DE = 5 - 3 = \boxed{2}\,\text{cm}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=12$, $AC=16$, $BC=20$. If point $P$ is a point on line segment $AC$, connect $BP$, and $BP=CP$, what is the length of $AP$?

Solution:
\textbf{Solution:} Let $AP=x$, then $BP=CP=16-x$.\\
In $\triangle ABP$ right-angled at $B$, since $AB^2+AP^2=BP^2$,\\
hence $12^2+x^2=(16-x)^2$,\\
solving gives $x=\boxed{3.5}$, therefore the length of $AP$ is \boxed{3.5}.


Question: As shown in the figure, in $\triangle ABC$, $AD$ is the altitude on side $BC$, $CE$ is the median on side $AB$, $G$ is the midpoint of $EC$, connecting $DG$, $CD = AE$, $AD = 6$, and $BD = 8$. Find the length of $CD$?

Solution:
\textbf{Solution:} Since $AD$ is the altitude from $A$ to side $BC$, and $AD=6$, \\
$BD=8$, \\
therefore $\angle ADB=90^\circ$, and $AB=\sqrt{AD^{2}+BD^{2}}=\sqrt{6^{2}+8^{2}}=10$, \\
since $CE$ is the median to side $AB$, \\
therefore $BE=AE=\frac{1}{2}AB=5$, \\
since $CD=AE$, \\
therefore $CD=\boxed{5}$


Question: As shown in the figure, in $\triangle ABC$, $D$ is a point on $AB$, $E$ is the midpoint of $AC$, $DE$ is extended to point $F$ such that $EF=ED$, and $CF$ is drawn. $CF\parallel AB$. If $\angle A=70^\circ$, $\angle F=35^\circ$, and $BE\perp AC$, what is the degree measure of $\angle BED$?

Solution:
Solution: From (1) we know that $\angle A = \angle ACF$,\\
$\therefore \angle ACF = \angle A = 70^\circ$,\\
Additionally, since $\angle F = 35^\circ$,\\
$\therefore \angle DEC = \angle F + \angle ACF = 70^\circ + 35^\circ = 105^\circ$\\
Also, since $BE \perp AC$\\
$\therefore \angle BEC = 90^\circ$\\
Thus, $\angle BED = \angle DEC - \angle BEC = 105^\circ - 90^\circ = \boxed{15^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, the perpendicular bisector $OM$ of side $AB$ intersects with the perpendicular bisector $ON$ of side $AC$ at point $O$. These two perpendicular bisectors respectively intersect $BC$ at points $D$ and $E$. Given $\angle ABC=30^\circ$ and $\angle ACB=40^\circ$, what is the measure of $\angle DAE$ in degrees?

Solution:
\textbf{Solution:} Since $\angle ABC=30^\circ$ and $\angle ACB=40^\circ$, \\
it follows that $\angle BAC=180^\circ-\angle ABC-\angle ACB=180^\circ-30^\circ-40^\circ=110^\circ$, \\
Since $DM$ is the perpendicular bisector of segment $AB$, \\
it follows that $DA=DB$, \\
Therefore, $\angle DAB=\angle ABC=30^\circ$, \\
Similarly, $EA=EC$, \\
Therefore, $\angle EAC=\angle ACB=40^\circ$, \\
Therefore, $\angle DAE=\angle BAC-\angle BAD-\angle EAC=110^\circ-30^\circ-40^\circ=\boxed{40^\circ}$;


Question: As shown in the figure, in $\triangle ABC$, the perpendicular bisector of side $AB$, $OM$, and the perpendicular bisector of side $AC$, $ON$, intersect at point $O$. These two perpendicular bisectors intersect $BC$ at points $D$ and $E$, respectively. It is known that the perimeter of $\triangle ADE$ is $7\,\text{cm}$. The lines $OA$, $OB$, and $OC$ are drawn. If the perimeter of $\triangle OBC$ is $15\,\text{cm}$, find the length of $OA$.

Solution:
\textbf{Solution:} Connect OA, OB, OC,\\
$\because$ the perimeter of $\triangle ADE$ is $7\,\text{cm}$,\\
$\therefore$ $AD+DE+EA=7\,\text{cm}$,\\
$\therefore$ $BC=DB+DE+EC=AD+DE+EA=7\,\text{cm}$;\\
$\because$ the perimeter of $\triangle OBC$ is $15$,\\
$\therefore$ $OB+OC+BC=15$,\\
$\because$ $BC=7$,\\
$\therefore$ $OB+OC=8$,\\
$\because$ $OM$ is the perpendicular bisector of $AB$,\\
$\therefore$ $OA=OB$,\\
Similarly, $OA=OC$,\\
$\therefore$ $OA=OB=OC=\boxed{4\,\text{cm}}$.


Question: As shown in the diagram, in $ \triangle ABC$, extend $ BC$ to point D such that $ CD=AB$, draw $ DE\parallel AB$ with $ DE=BC$, and connect $ CE, AE$. $ \angle BAC=\angle DCE$; If $ \angle B=32^\circ$ and $ \angle ACD=58^\circ$, what is the measure of $ \angle CEA$?

Solution:
\textbf{Solution:} Since $\angle B=32^\circ$ and $\angle ACD=58^\circ$,\\
it follows that $\angle DCE=\angle BAC=\angle ACD-\angle B=58^\circ-32^\circ=26^\circ$,\\
therefore, $\angle ACE=\angle ACD+\angle DCE=58^\circ+26^\circ=84^\circ$,\\
since $\triangle ABC\cong \triangle CDE$,\\
it implies $AC=CE$,\\
thus, $\angle CEA=\angle CAE=\frac{1}{2}\left(180^\circ-\angle ACE\right)=\frac{1}{2}\left(180^\circ-84^\circ\right)=\boxed{48^\circ}$.


Question: As shown in the diagram, in $\triangle ABC$, point D is on the extension of line $BC$. A line through point D, $DF\perp AC$ at point F, is extended to intersect line $AB$ at point E and the bisector of $\angle ACB$ at point N. Point M is the intersection of $CN$ and $AB$, with $\angle BMC=80^\circ$ and $\angle B=40^\circ$. What is the measure of $\angle AEF$?

Solution:
\textbf{Solution:} Since $\angle BMC=80^\circ$ and $\angle B=40^\circ$, \\
it follows that $\angle BCM=180^\circ-\angle B-\angle BMC=60^\circ$ and $\angle AMC=180^\circ-\angle BMC=100^\circ$. \\
Since $DF\perp AC$, \\
it implies that $\angle EFC=90^\circ$. \\
Since $CN$ bisects $\angle ACB$, \\
it implies that $\angle ACN=\angle BCN=60^\circ$. \\
Hence, $\angle MEF=360^\circ-\angle AMC-\angle ACM-\angle EFC=110^\circ$, \\
thus, $\angle AEF=180^\circ-\angle MEF=\boxed{70^\circ}$.


Question: As shown in the figure, the line $l_{1}\colon y=ax+b$ (with constants $a<0$, $b>0$) intersects the x-axis and the y-axis at points A and B, respectively, while the line $l_{2}\colon y=ca+d$ (with constants $c>0$, $d>0$) intersects the x-axis and the y-axis at points C and D, respectively. The lines $l_{1}$ and $l_{2}$ intersect at point E, and $\triangle AOB\cong \triangle COD$. $AB\perp CD$. If $a=-2$ and $b=4$, what is the area of $\triangle ADE$?

Solution:
\textbf{Solution}: From the given information, we have $a=-2$, $b=4$, \\
$\therefore$ the line $l_1\colon y=-2x+4$, \\
Let $x=0$, we get $y=4$; let $y=0$, we get $-2x+4=0$, solving this gives $x=2$, \\
$\therefore$ point A is $(2, 0)$, point B is $(0, 4)$, \\
$\therefore OA=2$, $OB=4$, \\
$\because \triangle AOB \cong \triangle COD$, \\
$\therefore OC=OA=2$, $OD=OB=4$, \\
$\therefore$ points C and D are $(0, 2)$ and $(-4, 0)$ respectively, \\
Solving gives: $\begin{cases} c=\frac{1}{2} \\ d=2 \end{cases}$, \\
$\therefore$ the line $l_2\colon y=\frac{1}{2}x+2$, \\
By solving the system of equations $\begin{cases} y=\frac{1}{2}x+2 \\ y=-2x+4 \end{cases}$, we get: $\begin{cases} x=\frac{4}{5} \\ y=\frac{12}{5} \end{cases}$, \\
$\therefore$ the coordinates of point E are $\left(\frac{4}{5}, \frac{12}{5}\right)$, \\
$\therefore$ the area of $\triangle ADE$ is $\frac{1}{2}AD\cdot y_E=\frac{1}{2}\times (2+4)\times \frac{12}{5}=\boxed{\frac{36}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=CB$, $\angle ABC=90^\circ$, $F$ is a point on the extension of $AB$, and point $E$ is on $BC$, with $BE=BF$. $\triangle ABE \cong \triangle CBF$. If $\angle CAE = 20^\circ$, what is the measure of $\angle ACF$?

Solution: \textbf{Solution:} Given that $AB=BC$ and $\angle ABC=90^\circ$, it follows that $\angle CAB=\angle ACB=45^\circ$. Also, given that $\angle BAE=\angle CAB-\angle CAE=45^\circ-20^\circ=25^\circ$ and according to (1), $\triangle ABE\cong \triangle CBF$. Hence, $\angle BCF=\angle BAE=25^\circ$, which means $\angle ACF=\angle BCF+\angle ACB=45^\circ+25^\circ=\boxed{70^\circ}$.


Question: As shown in the figure, in $\triangle ABC$ and $\triangle ECF$, we have $\angle ACB = \angle ECF = 90^\circ$, and also $AC = BC$, $EC = FC$. Connecting $AE$ and $BF$, they intersect at point $O$. $\triangle ACE \cong \triangle BCF$. What is the measure of $\angle AOB$?

Solution:
\textbf{Solution:} In $\triangle BOM$ and $\triangle ACM$, \\
since $\angle AMC = \angle BMO$, $\angle CAE = \angle BEF$, \\
it follows that $\angle BOM = \angle ACM$, \\
since $\angle ACB = 90^\circ$, \\
it implies $\angle AOB = \boxed{90^\circ}$.


Question: As shown in the figure, $CE$ is the angle bisector of the exterior angle $\angle ACD$ of $\triangle ABC$, and $CE$ intersects the extension of $BA$ at point $E$. If $\angle B=35^\circ$ and $\angle E=25^\circ$, what is the measure of $\angle CAE$?

Solution:
\textbf{Solution:} Since $\angle DCE$ is an exterior angle of $\triangle BCE$ and $\angle B=35^\circ$, $\angle E=25^\circ$,
\[\therefore \angle DCE=\angle B+\angle E=60^\circ,\]
Since $CE$ bisects $\angle ACD$,
\[\therefore \angle ACE=\angle DCE=60^\circ,\]
\[\therefore \angle CAE=180^\circ-\angle ACE-\angle E=\boxed{95^\circ}.\]


Question: As shown in the figure, in $\triangle ABC$, $D$ is a point on side $BC$, and $AB=AD=DC$. If $\angle B=60^\circ$, what is the degree measure of $\angle C$?

Solution:
\textbf{Solution:} Since $AB=AD$, and $\angle B=60^\circ$,\\
thus, $\triangle ABD$ is an equilateral triangle,\\
therefore, $\angle ADB=60^\circ$,\\
Since $AD=CD$,\\
thus $\angle C=\angle DAC=\frac{1}{2}\angle ADB= \boxed{30^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $D$ is a point on side $BC$, and $AB=AD=DC$. If $AC=BC$, what is the degree measure of $\angle C$?

Solution:
\textbf{Solution:} Let $ \angle C=x$\\
Since $AD=DC$\\
Then $\angle DAC=\angle C=x$\\
Thus $\angle ADB=2x$\\
Since $AB=AD$\\
Then $\angle B=\angle ADB=2x$\\
Since $AC=BC$\\
Then $\angle BAC=\angle B=2x$\\
In $ \triangle ABC$,\\
$2x+2x+x=180^\circ$\\
Therefore $x=36^\circ$\\
Thus, the measure of $\angle C$ is $\boxed{36^\circ}$


Question: As shown in the figure, in $ \triangle ABC$, $D$ is a point on side $BC$, with $AB=AD=DC$. If $ \angle B=60^\circ$, find the measure of $ \angle C$.

Solution:
\textbf{Solution:} Since $AB=AD, \angle B=60^\circ$,\\
it follows that $\triangle ABD$ is an equilateral triangle,\\
therefore $\angle ADB=60^\circ$,\\
since $AD=DC$,\\
it implies that $\angle C=\angle DAC=\frac{1}{2}\angle ADB=\boxed{30^\circ}$


Question: As shown in the figure, in $\triangle ABC$, D is a point on side $BC$, and $AB=AD=DC$. If $AC=BC$, find the measure of $\angle C$.

Solution:
\textbf{Solution:} Let $\angle C=x$,\\
since $AD=DC$,\\
thus $\angle DAC=\angle C=x$,\\
therefore $\angle ADB=2x$,\\
since $AB=AD$,\\
thus $\angle B=\angle ADB=2x$,\\
since $AC=BC$,\\
thus $\angle BAC=\angle B=2x$,\\
in $\triangle ABC$, $2x+2x+x=180^\circ$,\\
therefore $x=\boxed{36^\circ}$,\\
therefore the degree of $\angle C$ is $\boxed{36^\circ}$.


Question: As shown in the figure, it is known that $\angle ADC=90^\circ$, $DC\parallel AB$, $BA=BC$, $AE\perp BC$ with the foot of the perpendicular being point E, and point F is the midpoint of $AC$. Is $\angle AFB=90^\circ$?

Solution:
\textbf{Solution:} Proof: \\
Since $BA = BC$, and $F$ is the midpoint of $AC$, \\
therefore $BF \perp AC$. \\
Therefore, $\angle AFB = 90^\circ$. \\
Hence, the answer is $\boxed{90^\circ}$.


Question: As shown in Figure 1, in quadrilateral $ABCD$, $AB \parallel DC$, $AB=DC$, and the diagonals $AC$ and $BD$ intersect at point $O$, with $AC$ bisecting $\angle BAD$. $\angle DAC = \angle DCA$; quadrilateral $ABCD$ is a rhombus. As shown in Figure 2, draw $CE \perp AB$ extended to meet at point $E$, and connect $OE$. If $AB= \sqrt{5}$ and $BD=2$, what is the length of $OE$?

Solution:
\textbf{Solution:}

Since quadrilateral ABCD is a rhombus and $OA=OC$, $OB=OD$, $BD\perp AC$, and $CE\perp AB$, it follows that $OE=OA=OC$. Given that $BD=2$, thus $OB=\frac{1}{2}BD=1$. In $\triangle AOB$, by the Pythagorean theorem, we have: $OA=\sqrt{AB^{2}-OB^{2}}=\sqrt{(\sqrt{5})^{2}-1^{2}}=\boxed{2}$, therefore $OE=OA=\boxed{2}$.


Question: As shown in the diagram, in parallelogram $ABCD$, the heights on the adjacent sides $AD$ and $CD$ are equal, that is, $BE=BF$. Parallelogram $ABCD$ is a rhombus; if $DB=10$ and $AB=13$, what is the area of parallelogram $ABCD$?

Solution:
\textbf{Solution:} Draw line $AC$ intersecting $BD$ at point $O$, \\
since quadrilateral $ABCD$ is a rhombus, it follows that $AC \perp BD$, and $BO = \frac{1}{2}BD = 5$,\\
in $\triangle ABO$, by the Pythagorean theorem, we have: $AO = \sqrt{AB^{2} - BO^{2}} = 12$, \\
thus $AC = 2AO = 24$, \\
therefore, the area of parallelogram $ABCD$ $= \frac{1}{2}AC \times BD = \boxed{120}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AD\parallel BC$, $AB=BC$, and the diagonals $AC$ and $BD$ intersect at point $O$, where $BD$ bisects $\angle ABC$. Quadrilateral $ABCD$ is a rhombus; through point $D$, draw $DE\perp BC$ and extend it to meet the extension of $BC$ at point $E$, then connect $OE$. Given that $OE=3$ and $AC=4$, what is the length of the sides of the rhombus?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rhombus, \\
thus AC$\perp$BD, OB=OD=$\frac{1}{2}$BD, OC=$\frac{1}{2}$AC=2, \\
since DE$\perp$BC, \\
thus $\angle$DEB$=90^\circ$, \\
since OB=OD, \\
thus OE=$\frac{1}{2}$BD, \\
since OD=$\frac{1}{2}$BD, \\
thus OD=OE=3, \\
in $\triangle$OCD, by the Pythagorean theorem, $CD=\sqrt{OD^2+OC^2}=\sqrt{3^2+2^2}=\boxed{\sqrt{13}},$ \\
therefore, the side length of the rhombus is $\boxed{\sqrt{13}}$.


Question: As shown in the figure, the perimeter of square $ABCD$ is 40. Point $P$ is a moving point on the diagonal $AC$ of square $ABCD$. Perpendiculars are drawn from point $P$ to $AB$ and $BC$, with the feet of the perpendiculars being $E$ and $F$, respectively. Prove that the quadrilateral $PEBF$ is a rectangle.

Solution:
\textbf{Solution:} Proof:\\
Since $PE\perp AB$ and $PF\perp BC$,\\
it follows that $\angle PEB=\angle PFB=90^\circ$.\\
Furthermore, since $ABCD$ is a square,\\
it follows that $\angle ABC=90^\circ$.\\
Therefore, the quadrilateral PEBF is a \boxed{rectangle}.


Question: As shown in the figure, the perimeter of square $ABCD$ is $40$. Point $P$ is a moving point on the diagonal $AC$ of square $ABCD$. Perpendiculars are drawn from point $P$ to $AB$ and $BC$ respectively, with the feet of the perpendiculars being $E$ and $F$, respectively. Quadrilateral $PEBF$ is a rectangle, and $PD=EF$. As point $P$ moves, the length of $EF$ also changes. What is the minimum value of $EF$?

Solution:
\textbf{Solution:} Given equation (2), we have $DP=EF$. Thus, the minimum value of $EF$, which is also the minimum value of $DP$, occurs when $DP\perp AC$, meaning $DP$ is perpendicular to $AC$.

Since the perimeter of square ABCD is 40,
\[AD=CD=10.\]

Given $AD=CD$ and $\angle ADC=90^\circ$,
\[AC=\sqrt{10^2+10^2}=10\sqrt{2}.\]

Since $DP\perp AC$,
\[DP=\frac{1}{2}AC=5\sqrt{2}.\]

Therefore, the minimum value of $EF$ is \boxed{5\sqrt{2}}.


Question: As shown in the figure, in rectangle $ABCD$, $AB=12$, $AD=16$, $E$ and $F$ are the midpoints of $AD$ and $BC$, respectively. $G$ and $H$ are two moving points on diagonal $AC$, and they move towards each other from points $A$ and $C$ at a speed of $1$ unit length per second, for $t$ seconds, where $0\le t<10$. If quadrilateral $EGFH$ is a rectangle, what is the value of $t$?

Solution:
\textbf{Solution:} As shown in Figure 1, connect $EF$,\\
since $AE=BF$, $AE\parallel BF$, and $\angle B=90^\circ$,\\
thus quadrilateral $ABFE$ is a rectangle,\\
therefore $EF=AB=12$. When quadrilateral $EGFH$ is a rectangle,\\
thus $GH=EF=12$,\\
in $\triangle ABC$, by the Pythagorean theorem, we get $AC=\sqrt{AB^2+BC^2}=\sqrt{12^2+16^2}=20$,\\
since $AG=CH=t$,\\
thus $12+2t=20$,\\
therefore $t=\boxed{4}$, i.e., when quadrilateral $EGFH$ is a rectangle, $t=\boxed{4}$.


Question: In the rectangle $ABCD$, where $AB=12$ and $AD=16$, $E$ and $F$ are the midpoints of $AD$ and $BC$, respectively. Points $G$ and $H$ are two moving points on the diagonal $AC$, moving towards each other from points $A$ and $C$ at a speed of $1$ unit per second, for a duration of $t$ seconds, with $0\le t<10$. If point $E'$ starts from point $E$ and moves to the right along line $AD$, and point $F'$ starts from point $F$ and moves to the left along line $CB$, both at the same speed as points $G$ and $H$ starting simultaneously, find the value of $t$ when the quadrilateral $E'G{F'}H$ forms a rhombus.

Solution:
\textbf{Solution:} Let's assume at time $t$ seconds, the quadrilateral $E'GF'H$ forms a rhombus, with point E and F moved to positions $E'$ and $F'$ respectively (as shown in Figure 2). Connect $AF'$, $CE'$, $E'F'$, where $E'F'$ intersects AC at point O.\\
$\because$ Quadrilateral $E'GF'H$ is a rhombus, $E'F' \perp AC$, $OG=OH$, $OE'=OF'$,\\
$\therefore OA=OC$, $AE'=CF'$, $\therefore$ Quadrilateral $AE'CF'$ is a rhombus, $\therefore CE'=AE'=AE+EE'=8+t$, $DE'=ED-EE'=8-t$. In $\triangle CDE'$,\\
by Pythagoras' theorem, we have $CD^2+DE'^2=CE'^2$, that is $12^2+(8-t)^2=(8+t)^2$. Solving this gives $t=\boxed{\frac{9}{2}}$, $\therefore$ when $t=\frac{9}{2}$, the quadrilateral $E'GF'H$ forms a rhombus.


Question: As shown in the diagram, in $\triangle ABC$, $D$ and $E$ are the midpoints of $AB$ and $AC$, respectively. A line through point $C$ is drawn such that $CF \parallel BD$, intersecting the extension of $DE$ at point $F$. Quadrilateral $BCFD$ is a parallelogram; if $BC = 6$, find the length of $EF$.

Solution:
\textbf{Solution:} Given that $DE$ is the mid-segment of $\triangle ABC$,\\
hence $DE = \frac{1}{2}BC = 3$,\\
since quadrilateral BCFD is a parallelogram,\\
thus $DF = BC = 6$,\\
therefore $EF = DF - DE = 6 - 3 = \boxed{3}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $O$, $AD\parallel BC$, and $BO=DO$. Quadrilateral $ABCD$ is a parallelogram. A line through point $O$ is drawn to be perpendicular to $BD$ and intersects $BC$ at point $E$, and $DE$ is connected. If $\angle CDE=\angle CBD=15^\circ$, what is the degree measure of $\angle ABC$?

Solution:
\textbf{Solution:} Given that $OE\perp BD$, that is $\angle BOE=\angle DOE=90^\circ$,\\
therefore, in $\triangle BOE$ and $\triangle DOE$, we have $\left\{\begin{array}{l}BO=DO \\ \angle BOE=\angle DOE=90^\circ \\ OE=OE \end{array}\right.$,\\
hence $\triangle BOE\cong \triangle DOE$ (SAS),\\
thus $\angle EBD=\angle EDB$, which means $\angle CBD=\angle EDB$.\\
Since $\angle CDE=\angle CBD=15^\circ$,\\
then $\angle CDE=\angle CBD=\angle EDB=15^\circ$,\\
therefore $\angle BDC=\angle EDB+\angle CDE=30^\circ$.\\
Since quadrilateral $ABCD$ is a parallelogram,\\
then $\angle ABD=\angle BDC=30^\circ$,\\
hence $\angle ABC=\angle ABD+\angle CBD=\boxed{45^\circ}$.


Question: As shown in the diagram, point O is a point inside $\triangle ABC$. Connect OB, OC, and join the midpoints D, E, F, G of AB, OB, OC, AC respectively, to get quadrilateral DEFG. Quadrilateral DEFG is a parallelogram. If $\angle OBC=45^\circ$, $\angle OCB=30^\circ$, and OC$=8$, what is the length of EF?

Solution:
\textbf{Solution:} As shown in the figure, draw OM$\perp$BC at M, \\
In Rt$\triangle$OCM, $\angle$OCM=$30^\circ$, OC=8, \\
$\therefore$OM=$\frac{1}{2}$OC=4, \\
$\therefore$CM=$\sqrt{OC^{2}-OM^{2}}=4\sqrt{3}$, \\
In Rt$\triangle$OBM, \\
$\because$$\angle$OBC=$45^\circ$, \\
$\therefore$$\angle$OBM=$\angle$BOM=$45^\circ$ \\
$\therefore$BM=OM=4, \\
$\therefore$BC=4+$4\sqrt{3}$, \\
$\because$EF=$\frac{1}{2}$BC, \\
$\therefore$EF=$2+2\sqrt{3}$. \\
Hence, the length of EF is $\boxed{2+2\sqrt{3}}$.


Question: As shown in the figure, it is known that the graph of the linear function $y=-\frac{1}{2}x+b$ passes through point $A(0,3)$. Point $P$ is a moving point on this line. Perpendiculars $PM$ and $PN$ are drawn from point $P$ to the $x$-axis at point $M$ and the $y$-axis at point $N$, respectively. On the quadrilateral $PMON$, the following segments are taken: $PC=\frac{1}{3}MP$, $MB=\frac{1}{3}OM$, $OE=\frac{1}{3}ON$, $ND=\frac{1}{3}NP$. What is the value of $b$?

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


Question: As shown in the figure, it is known that the graph of the linear function $y=-\frac{1}{2}x+b$ passes through point $A(0,3)$. Point $P$ is a moving point on this line. Through point $P$, $PM$ is drawn perpendicular to the $x$-axis at point $M$, and $PN$ is drawn perpendicular to the $y$-axis at point $N$. On quadrilateral $PMON$, the following segments are intercepted: $PC=\frac{1}{3}MP$, $MB=\frac{1}{3}OM$, $OE=\frac{1}{3}ON$, and $ND=\frac{1}{3}NP$. Quadrilateral $BCDE$ is a parallelogram. Is there such a point $P$ on the line $y=-\frac{1}{2}x+b$ that makes quadrilateral $BCDE$ a square? If such a point exists, please find the coordinates of all such points $P$; if it does not exist, please explain why.

Solution:
\textbf{Solution:} Let the coordinates of point $P$ be $(x,y)$,

when $\triangle OBE \cong \triangle MCB$, quadrilateral $BCDE$ becomes a square,

$OE=BM$,

when point $P$ is in the first quadrant, i.e., $\frac{1}{3}y=\frac{1}{3}x$, $x=y$.

Point $P$ lies on the line,

$\left\{\begin{array}{l}
y=\frac{1}{2}x+3 \\
y=x
\end{array}\right.$,

solving we get $\left\{\begin{array}{l}
x=\boxed{2} \\
y=2
\end{array}\right.$,

when point $P$ is in the second quadrant, $-x=y$

$\left\{\begin{array}{l}
y=\frac{1}{2}x+3 \\
y=-x
\end{array}\right.$,

solving we get $\left\{\begin{array}{l}
x=-6 \\
y=6
\end{array}\right.$

There exists such a point $P$ on the line $y=-\frac{1}{2}x+b$ that makes quadrilateral $BCDE$ a square, with $P$'s coordinates being $\boxed{(2,2)}$ or $\boxed{(-6,6)}$.


Question: As shown in the diagram, in quadrilateral $ABCD$, $E$ and $F$ are two points on the diagonal $AC$, and $AE=CF$. Quadrilateral $BEDF$ is a parallelogram; if $AB \perp BF$, $AB=8$, $BF=6$, and $AC=16$. Find the length of the line segment $EF$.

Solution:
\textbf{Solution:} In right triangle $Rt\triangle ABF$, we have $AF=\sqrt{AB^{2}+BF^{2}}=\sqrt{8^{2}+6^{2}}=10$,\\
since $AC=16$,\\
then $CF=AC-AF=16-10=6$,\\
since $AE=CF$, thus $AE=6$,\\
therefore $EF=AF-AE=10-6=\boxed{4}$.


Question: As shown in the figure, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, with $DE \parallel AC$ and $CE \parallel BD$. Quadrilateral $ODEC$ is a rhombus; connect $OE$. If $BC = 2\sqrt{2}$, find the length of $OE$.

Solution:
\textbf{Solution}: As shown in the figure, connect OE, intersecting CD at point F,\\
$\because$ from (1) we know that quadrilateral OCED is a rhombus,\\
$\therefore$ OE $\perp$ CD,\\
$\therefore \angle ADC=\angle OFC=90^\circ$,\\
$\therefore$ AD $\parallel$ OE,\\
$\because$ DE $\parallel$ AC,\\
$\therefore$ quadrilateral AOED is a parallelogram,\\
$\therefore$ OE=AD=BC=$\boxed{2\sqrt{2}}$.


Question: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $E$ is the midpoint of $AD$, and points $F$, $G$ are on $CD$, with $EF\perp CD$ and $OG\parallel EF$. Quadrilateral $OEFG$ is a rectangle; if $AD=10$ and $EF=3$, find the lengths of $OE$ and $CG$.

Solution:
Solution: Since quadrilateral ABCD is a rhombus, \\
$\therefore$ $\angle AOD=90^\circ$, AD=CD=10,\\
also $\because$ point E is the midpoint of AD,\\
$\therefore$ $OE=DE=\frac{1}{2}AD=5$,\\
$\therefore$ $GF=OE=5$,\\
In $\triangle DEF$, $\angle EFD=90^\circ$,\\
$\therefore$ $DF=\sqrt{DE^2-EF^2}=\sqrt{5^2-3^2}=4$,\\
$\therefore$ $CG=CD-GF-DF=10-5-4=\boxed{1}$.


Question: Given: As shown in the figure, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Lines parallel to $AC$ and $BD$ are drawn passing through points $B$ and $C$, respectively, and intersect at point $F$. Quadrilateral $BFCO$ is a rhombus. If $AB=3$ and $BC=4$, what is the perimeter of quadrilateral $BFCO$?

Solution: \textbf{Solution:} Because quadrilateral ABCD is a rectangle, it follows that $\angle ABC=90^\circ$. By the Pythagorean theorem, we have $AC=\sqrt{AB^2+BC^2}=\sqrt{3^2+4^2}=5$. Therefore, $OC=\frac{1}{2}AC=\frac{5}{2}$. From statement (1), we know quadrilateral BFCO is a rhombus. Therefore, $BF=FC=OC=OB$, hence the perimeter of quadrilateral BFCO is $4OC=4\times \frac{5}{2}=\boxed{10}$.


Question: As shown in the figure, it is known that after the line segment AD is moved downward by $m$ units and then moved to the right by $n$ units to the line segment BC, the corresponding points of points A and D are points B and C, respectively. Connect AB, CD, AC, and BD, with AC and BD intersecting at point O. It is given that $OB = OD$. Draw $DM \perp BC$ at M, let N be the midpoint of CD, and connect MN. If $\angle ADB = 45^\circ$ and $OD = 3\sqrt{2}$, with MN equal to 4, what is the value of $\frac{BM}{MC}$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\
it follows that $\angle ADB=\angle CBD$. $OB=OD=3\sqrt{2}$, $BD=6\sqrt{2}$.\\
Since $\angle ADB=45^\circ$, and $DM\perp BC$, \\
it follows that $\triangle BDM$ is an isosceles right triangle, hence $BM=MD=6$.\\
Since $\triangle CMD$ is a right triangle, with $N$ being the midpoint of $CD$, and $MN=4$, it follows that $CD=2MN=8$.\\
In $\triangle DCM$, by the Pythagorean theorem, we find $MC=2\sqrt{7}$, hence $\frac{BM}{MC}=\frac{6}{2\sqrt{7}}=\boxed{\frac{3\sqrt{7}}{7}}$.


Question: Given that line segment $AD$ is translated down by $m$ units and then translated to the right by $n$ units to line segment $BC$, with corresponding points of $A$, $D$ being points $B$ and $C$ respectively, and connecting $AB$, $CD$, $AC$, $BD$, where $AC$ intersects $BD$ at point $O$. Under the given conditions, as $H$ moves on $BC$, when $\triangle CDH$ is an isosceles triangle, write down the length of $HC$.

Solution:
\textbf{Solution:}\\
(1) When $HC=DC=8$, as shown in the figure, an isosceles triangle is formed.\\
(2) When $HD=HC$, as shown in the figure, an isosceles triangle is formed.\\
At this time, let $HM$ be $x$,\\
$\because H{M}^{2}+M{D}^{2}=H{D}^{2}=M{C}^{2}$,\\
$\therefore {x}^{2}+{6}^{2}={\left(x+2\sqrt{7}\right)}^{2}$,\\
Solving this, we get: $x=\frac{2\sqrt{7}}{7}$,\\
$\therefore HC=x+MC= \frac{2\sqrt{7}}{7}+2\sqrt{7}=\frac{16\sqrt{7}}{7}$.\\
(3) When $DH=DC$, as shown in the figure, an isosceles triangle is formed:\\
In this case $HC=2MC= 4\sqrt{7}$.\\
Therefore, the length of $HC$ can be: $8$ or \boxed{\frac{16\sqrt{7}}{7}} or $4\sqrt{7}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=90^\circ$. A line parallel to $AC$ is drawn through point $B$, intersecting the bisector of $\angle BAC$ at point $D$. Point $E$ is a point on $AC$, and $BE\perp AD$ at point $F$. Connect $DE$. The quadrilateral $ABDE$ is a rhombus. Given that $AB=2$ and $\angle ADC=90^\circ$, what is the length of $BC$?

Solution:
\textbf{Solution:}

Given that quadrilateral $ABDE$ is a rhombus, it follows that $DE=AE=AB=2$, hence $\angle EAD=\angle EDA$. Given that $\angle ADC=90^\circ$, it follows that $\angle EDC+\angle EDA=90^\circ$ and $\angle EAD+\angle ECD=90^\circ$. Therefore, $\angle EDC=\angle ECD$, which implies $DE=EC=2$. Consequently, $AC=AE+CE=4$. Considering $\angle ABC=90^\circ$, it can be derived that $BC=\sqrt{AC^{2}-AB^{2}}=\sqrt{4^{2}-2^{2}}=\boxed{2\sqrt{3}}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=4$ and $AD=10$. When the vertex $P$ of the right angle of a right-angled ruler slides along side $AD$ (point $P$ does not coincide with $A$ or $D$), one of the right-angle sides passes through point $C$ and the other right-angle side intersects $AB$ at point $E$. $Rt\triangle AEP \sim Rt\triangle DPC$; when $\angle CPD=30^\circ$, what is the length of $AE$?

Solution:
\textbf{Solution:} In right triangle $\triangle PCD$, $\angle CPD=30^\circ$ and $CD=4$, \\
therefore $PC=8$, \\
therefore $PD=\sqrt{PC^{2}-CD^{2}}=4\sqrt{3}$, \\
therefore $AP=AD-PD=10-4\sqrt{3}$, \\
As right triangles $\triangle AEP \sim \triangle DPC$ indicate: $\frac{PD}{AE}=\frac{CD}{AP}$, thus $\frac{4\sqrt{3}}{AE}=\frac{4}{10-4\sqrt{3}}$, \\
therefore $AE=\boxed{10\sqrt{3}-12}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=4$ and $AD=10$. When the right-angle vertex $P$ of the square slides on $AD$ (point $P$ does not coincide with $A$ or $D$), one right-angle side passes through point $C$, and the other intersects $AB$ at point $E$. Is there such a point $P$ that makes the perimeter of $\triangle DPC$ equal to twice the perimeter of $\triangle AEP$? Find the length of $DP$.

Solution:
\textbf{Solution:} Let's assume that there is a point $P$ that satisfies the given conditions, let $DP=x$, then $AP=10-x$.

Since $\triangle AEP \sim \triangle DPC$, according to the perimeter of $\triangle CDP$ being twice the perimeter of $\triangle PAE$, we get the ratio of similarity between the two triangles is 2,

thus $\frac{CD}{AP}=\frac{4}{10-x}=2$,

solving this equation gives $x=\boxed{8}$,

therefore, there exists a point $P$ that satisfies the conditions, with the length of $DP$ being \boxed{8}.


Question: As shown in the figure, in parallelogram $ABCD$, $E$ and $F$ are points on sides $AD$ and $BC$ respectively, and $\angle ABE = \angle CDF$. Quadrilateral $BEDF$ is a parallelogram; connect $CE$. If $CE$ bisects $\angle DCB$, $CF = 3$, and $DE = 5$, what is the perimeter of parallelogram $ABCD$?

Solution: \textbf{Solution:} Given that CE bisects $\angle BCD$, it follows that $\angle 1 = \angle 2$. Furthermore, since $AD \parallel BC$, it implies $\angle 1 = \angle 3$, which leads to $\angle 2 = \angle 3$, and consequently, $ED = DC = 5$. In quadrilateral $EBFD$, it can be deduced that $BF = ED = 5$. Hence, the perimeter of the quadrilateral $ABCD$ equals $2 \times (5 + 3) + 2 \times 5 = \boxed{26}$.


Question: As shown in the figure, it is known that in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Lines parallel to $AC$ and $BD$ are drawn through points $D$ and $C$ respectively, intersecting at point $E$. Quadrilateral $OCED$ is a rhombus. If $AB=3$ and $BC=4$, what is the area of the rhombus $OCED$?

Solution:
\textbf{Solution:} Given $AB=3$, $BC=4$,\\
thus, the area of rectangle $ABCD$ equals $3\times 4=12$,\\
since $S_{\triangle ODC}=\frac{1}{4}S_{\text{rectangle}ABCD}=3$,\\
therefore, the area of the quadrilateral $OCED$ equals $2S_{\triangle ODC}=\boxed{6}$.


Question: Assemble a set of triangles as shown in the figure, where points B, C, and D lie on the same straight line, $\angle ACB=45^\circ$, $\angle DCE=60^\circ$. If CM and CN bisect $\angle ACB$ and $\angle DCE$ respectively, what is the measure of $\angle MCN$?

Solution: \textbf{Solution:} Hence, $\angle MCN = \boxed{127.5^\circ}$.


Question: Join a set of triangles together as shown in the figure, where points B, C, and D lie on the same straight line, $\angle ACB=45^\circ$, and $\angle DCE=60^\circ$. If CM bisects $\angle BCE$ and CN bisects $\angle DCA$, as shown in Figure 2, what is the measure of $\angle MCN$?

Solution: \textbf{Solution:} We have $\angle MCN = \boxed{52.5^\circ}$.


Question: As shown in the figure, point $A$, $O$, and $B$ are on the same line, and $OC$ bisects $\angle AOB$. If $\angle COD = 28^\circ$, what is the measure of $\angle BOD$?

Solution:
\textbf{Solution:} Since points A, O, and B are on the same line, and OC bisects $\angle AOB$,\\
therefore $\angle AOC = \angle BOC = 90^\circ$,\\
thus $\angle BOD = \angle BOC - \angle COD = 90^\circ - 28^\circ = \boxed{62^\circ}$;


Question: As shown in the figure, points A, O, and B are on the same line, and OC bisects $\angle AOB$. If $\angle COD=28^\circ$ and OE bisects $\angle BOD$, what is the measure of $\angle COE$ in degrees?

Solution:
\textbf{Solution:} Given that OE bisects $\angle$BOD,\\
it follows that $\angle$DOE = $\frac{1}{2} \angle$BOD = $31^\circ$,\\
hence, $\angle$COE = $\angle$DOE + $\angle$COD = $31^\circ + 28^\circ = \boxed{59^\circ}$.


Question: As shown in the figure, $OB$ is a ray inside $\angle AOC$, $OM$ is a ray inside $\angle AOB$, and $ON$ is a ray inside $\angle BOC$. As shown in Figure 1, $OM$ and $ON$ are the angle bisectors of $\angle AOB$ and $\angle BOC$, respectively. Given that $\angle AOB=30^\circ$ and $\angle MON=70^\circ$, what is the degree measure of $\angle BOC$?

Solution:
\textbf{Solution:} Given that $OM$ is the bisector of $\angle AOB$ and $\angle AOB=30^\circ$,\\
it follows that $\angle BOM=\frac{1}{2}\angle AOB=15^\circ$,\\
Given that $\angle MON=70^\circ$,\\
it follows that $\angle BON=\angle MON-\angle BOM=55^\circ$,\\
Since $ON$ is the bisector of $\angle BOC$,\\
it follows that $\angle BOC=2\angle BON=\boxed{110^\circ}$.


Question: As shown in the figure, $OB$ is a ray inside $\angle AOC$, $OM$ is a ray inside $\angle AOB$, and $ON$ is a ray inside $\angle BOC$. In Figure 2, if $\angle AOC=140^\circ$, $\angle AOM=\angle NOC=\frac{1}{4}\angle AOB$, and $\angle BOM:\angle BON=3:2$, what is the degree measure of $\angle MON$?

Solution:
\textbf{Solution:} Let $\angle AOM=\angle NOC=x$, thus $\angle AOB=4x$,\\
$\therefore \angle BOM=\angle AOB-\angle AOM=3x$,\\
$\because \angle BOM\colon \angle BON=3\colon 2$,\\
$\therefore \angle BON=2x$,\\
$\therefore \angle AOC=\angle AOB+\angle BON+\angle NOC=7x=140^\circ$,\\
$\therefore x=20^\circ$,\\
$\therefore \angle MON=\angle BOM+\angle BON=5x=\boxed{100^\circ}$.


Question: As shown in the figure, the direction of ray $OA$ is $15^\circ$ east of north, and the direction of ray $OB$ is $40^\circ$ west of north. Ray $OA$ is the angle bisector of $\angle BOC$. $OF$ is the extension line in the opposite direction of $OB$. Find: the direction of ray $OC$.

Solution:
\textbf{Solution:} From the diagram, it is known that: $\angle AOB=15^\circ+40^\circ=55^\circ$,\\
since $OA$ is the angle bisector of $\angle BOC$,\\
thus $\angle AOC=55^\circ$,\\
therefore $\angle NOC=\angle NOA+\angle AOC$\\
$=15^\circ+55^\circ=70^\circ$,\\
hence, the ray $OC$ is in the direction of \boxed{70^\circ} East of North.


Question: As shown in the figure, the direction of ray $OA$ is $15^\circ$ east of north, and the direction of ray $OB$ is $40^\circ$ west of north. $OA$ is the angle bisector of $\angle BOC$. $OF$ is the straight extension in the opposite direction of $OB$. What is the degree measure of $\angle COF$?

Solution:
\textbf{Solution:} Since $\angle BOC = \angle AOB + \angle AOC$\\
$=55^\circ \times 2 = 110^\circ$,\\
then $\angle COF = 180^\circ - \angle BOC$\\
$=180^\circ - 110^\circ$\\
$=70^\circ$.\\
Therefore, the measure of $\angle COF$ is $\boxed{70^\circ}$.


Question: As shown in the figure, $\angle AOB=90^\circ$, $OM$ bisects $\angle AOC$, and $ON$ bisects $\angle BOC$. If $\angle BOC=30^\circ$, what is the measure of $\angle MON$?

Solution:
\textbf{Solution}: Given that, \\
$\because \angle AOB=90^\circ, \angle BOC=30^\circ$,\\
$\therefore \angle AOC=\angle AOB+\angle BOC=120^\circ$.\\
$\because OM$ bisects $\angle AOC, ON$ bisects $\angle BOC$,\\
$\therefore \angle MOC=\frac{1}{2}\angle AOC=120^\circ \times \frac{1}{2}=60^\circ, \angle NOC=\frac{1}{2}\angle BOC=\frac{1}{2}\times 30^\circ=15^\circ$.\\
$\therefore \angle MON=\angle MOC-\angle NOC=\boxed{45^\circ}$.


Question: As shown in the figure, $\angle AOB=90^\circ$, $OM$ bisects $\angle AOC$, and $ON$ bisects $\angle BOC$. Let $\angle BOC=(2x)^\circ$, find the measure of $\angle MON$?

Solution:
\textbf{Solution:} Given that, $\angle AOB=90^\circ$, \\
$\therefore \angle AOC=\angle AOB+\angle BOC=90^\circ+(2x)^\circ$, \\
$\because OM$ bisects $\angle AOC$ and $ON$ bisects $\angle BOC$,\\
$\therefore \angle MOC=\frac{1}{2}\angle AOC=45^\circ+x^\circ$, $\angle NOC=\frac{1}{2}\angle BOC=\frac{1}{2}\times (2x)^\circ=x^\circ$, \\
$\therefore \angle MON=\angle MOC-\angle NOC=\boxed{45^\circ}$.


Question: As shown in the diagram, $OC \perp AB$ 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 $\angle BOD=\angle COD=\frac{1}{2}\angle BOC=45^\circ$,\\
Therefore, the measure of $\angle BOD$ is $\boxed{45^\circ}$.


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 $\angle COE$.

Solution:
\textbf{Solution:} From (1), we have
\[
\angle AOD = \angle AOC + \angle COD = 90^\circ + 45^\circ = 135^\circ,
\]
\[
\because OE \text{ bisects } \angle AOD,
\]
\[
\therefore \angle AOE = \frac{1}{2}\angle AOD = 67.5^\circ
\]
\[
\therefore \angle COE = 90^\circ - 67.5^\circ = \boxed{22.5^\circ}
\]


Question: As shown in the figure, the lines $EF$ and $CD$ intersect at the point $O$, with $\angle AOB = 90^\circ$, and $OC$ bisects $\angle AOF$. If $\angle AOE = 40^\circ$, find the degree measure of $\angle BOD$.

Solution:
\textbf{Solution:} Given $\because \angle AOE+\angle AOE=180^\circ$, $\angle AOE=40^\circ$,\\
$\therefore \angle AOF=180^\circ-\angle AOE=140^\circ$\\
Since $OC$ bisects $\angle AOF$,\\
$\therefore \angle AOC=\frac{1}{2}\angle AOF=\frac{1}{2}\times 140^\circ=70^\circ$\\
Given $\angle AOB=90^\circ$\\
$\therefore \angle BOD=180^\circ-\angle AOC-\angle AOB$\\
$=180^\circ-70^\circ-90^\circ$\\
$=\boxed{20^\circ}$


Question: As shown in the figure, line $EF$ intersects line $CD$ at point $O$, $\angle AOB=90^\circ$, $OC$ bisects $\angle AOF$. If $\angle AOE=30^\circ$, please write directly the measure of $\angle BOD$ in degrees.

Solution: \textbf{Solution:} We have $\angle BOD = \boxed{15^\circ}$.


Question: As shown in the figure, line AB and CD intersect at point O, with OA bisecting $\angle EOC$. If $\angle EOC = 70^\circ$, what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} \\
$\because$ OA bisects $\angle EOC$, \\
$\therefore \angle AOC= \frac{1}{2}\angle EOC$, \\
$\because \angle EOC=70^\circ$, \\
$\therefore \angle AOC= \frac{1}{2}\times 70^\circ=35^\circ$, \\
$\because$ Straight lines AB and CD intersect at point O, \\
$\therefore \angle BOD=\angle AOC=\boxed{35^\circ}$.


Question: As shown in the figure, lines AB and CD intersect at point O, with OA bisecting $\angle EOC$. If $\angle EOC = \frac{2}{3}\angle EOD$, find the degree measure of $\angle BOD$.

Solution:
\textbf{Solution}: Since $\angle EOC = \frac{2}{3}\angle EOD$ and $\angle EOC + \angle EOD = 180^\circ$,\\
thus $\frac{2}{3}\angle EOD + \angle EOD = 180^\circ$,\\
which implies $\frac{5}{3}\angle EOD = 180^\circ$,\\
therefore $\angle EOD = 108^\circ$,\\
thus $\angle EOC = \frac{2}{3} \times 108^\circ = 72^\circ$,\\
since OA bisects $\angle EOC$,\\
it follows that $\angle AOC = \frac{1}{2}\angle EOC = \frac{1}{2} \times 72^\circ = 36^\circ$,\\
since lines AB and CD intersect at point O,\\
thus $\angle BOD = \angle AOC = \boxed{36^\circ}$;


Question: As shown in the figure, it is known that $O$ is a point on line $AB$. Three rays $OC$, $OD$, and $OE$ are drawn upward from point $O$ to line $AB$, and $OC$ bisects $\angle AOD$, with $\angle 2=3\angle 1$. Suppose $\angle 1=18^\circ$, what is the degree measure of $\angle COE$?

Solution:
\textbf{Solution:} From the given conditions, we have \\
$\because \angle 1=18^\circ$ and $\angle 2=3\angle 1$, \\
$\therefore \angle 2=54^\circ$, \\
$\therefore \angle AOD=180^\circ-\angle 1-\angle 2=108^\circ$. \\
$\because OC$ bisects $\angle AOD$, \\
$\therefore \angle 3=54^\circ$, \\
$\therefore \angle COE=\angle 1+\angle 3=\boxed{72^\circ}$.


Question: As shown in the diagram, it is known that $O$ is a point on line $AB$, and three rays $OC$, $OD$, and $OE$ are drawn upward from point $O$ to line $AB$, with $OC$ bisecting $\angle AOD$ and $\angle 2 = 3\angle 1$. If $\angle COE = 70^\circ$, what is the degree measure of $\angle 2$?

Solution:
\textbf{Solution:} Let $ \angle 1=x^\circ$, $OC$ bisects $ \angle AOD$, $\angle COE=\angle 1+\angle 3=70^\circ$, \\
$\therefore \angle 3=\angle 4=70^\circ-x^\circ$. \\
Moreover, since $\angle 1+\angle 2+\angle 3+\angle 4=180^\circ$, \\
$\therefore x^\circ+\angle 2+2(70^\circ-x^\circ)=180^\circ$, \\
$\therefore \angle 2=40^\circ+x^\circ$. \\
Given $\angle 2=3\angle 1$, that is $40^\circ+x^\circ=3x^\circ$, \\
solving this gives $x=20$, \\
$\therefore \angle 2=3\angle 1=3\times 20^\circ=\boxed{60^\circ}$.


Question: As shown in the figure, the lines AB and CD intersect at point O, with OM $\perp$ AB, and ON $\perp$ CD. If $\angle 1 = \frac{1}{3}\angle BOC$, find the degree measure of $\angle BOD$.

Solution:
\textbf{Solution:} Given: $\because \angle 1 = \frac{1}{3} \angle BOC$,\\
$\therefore \angle BOM = 2\angle 1 = 90^\circ$,\\
We find: $\angle 1 = 45^\circ$,\\
$\therefore \angle BOD = 90^\circ - 45^\circ = \boxed{45^\circ}$.


Question: As shown in Figure 1, segment $AC=6\,\text{cm}$, segment $BC=15\,\text{cm}$, point $M$ is the midpoint of $AC$, and a point $N$ is taken on $CB$ such that $CN\colon NB=1\colon 2$, find the length of $MN$.

Solution:
\textbf{Solution:} Since $M$ is the midpoint of $AC$, and $AC=6$cm,\\
it follows that $MC=\frac{1}{2}AC=6\times\frac{1}{2}=3$cm.\\
Also, since $CN:NB=1:2$, and $BC=15$cm,\\
it follows that $CN=15\times\frac{1}{3}=5$cm,\\
therefore $MN=MC+CN=3+5=\boxed{8}$cm.


Question: As shown in the figure, points B and C divide line segment AD into three parts in the ratio 2:3:4, and $CD = 20$ cm. What is the length of line segment AD in centimeters?

Solution:
\textbf{Solution:}
Let $AB=2x$, $BC=3x$, and $CD=4x$\\
Since $CD=20$\\
$\therefore 4x=20$\\
$\therefore x=5$\\
Then $AD=AB+BC+CD=2x+3x+4x=9x=9\times 5=\boxed{45}$ cm.


Question: As shown in the figure, points B and C divide the line segment AD into three parts in the ratio 2:3:4, and $CD=20$. If point P is the midpoint of AD, and point Q is the midpoint of CD, what is the length of line segment PQ?

Solution:
\textbf{Solution:} Since point P is the midpoint of AD,\\
we have $PD=\frac{1}{2}AD=\frac{45}{2}$.\\
Since point Q is the midpoint of CD,\\
we have $QD=\frac{1}{2}CD=10$.\\
Therefore, $PQ=PD−QD=\frac{45}{2}−10=\boxed{\frac{25}{2}}$.


Question: As shown in the figure, line AB and line CD intersect at point O, and OB bisects $\angle EOD$. If $\angle EOC = 110^\circ$, what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Given $\because \angle DOC = 180^\circ$ and $\angle EOC = 110^\circ$, \\
$\therefore \angle DOE = 180^\circ - \angle EOC = 180^\circ - 110^\circ = 70^\circ$, \\
Since $OB$ bisects $\angle EOD$, \\
$\therefore \angle BOD = \frac{1}{2} \angle DOE = \frac{1}{2} \times 70^\circ = \boxed{35^\circ}$.


Question: As shown in the figure, line AB and line CD intersect at point O, and OB bisects $\angle EOD$. If $\angle DOE : \angle EOC = 2 : 3$, find the degree measure of $\angle AOC$.

Solution:
Solution:\\
Since $\angle DOE \colon \angle EOC = 2 \colon 3$\\
Therefore, $\angle DOE = 180^\circ \times \frac{2}{5} = 72^\circ$\\
Since $OB$ bisects $\angle EOD$\\
Therefore, $\angle DOB = \frac{1}{2} \times 72^\circ = 36^\circ$\\
Therefore, $\angle AOC = \angle DOB = \boxed{36^\circ}$.


Question: As shown in the figure, it is known that $AB \parallel CD$, $\angle BAE = 30^\circ$, $\angle DCE = 60^\circ$, and $EF$, $EG$ trisect $\angle AEC$ (that is, $\angle AEF = \angle FEG = \angle GEC$). What is the measure of $\angle AEF$ in degrees?

Solution:
\textbf{Solution:} Construct $EF' \parallel AB$ through point $E$, then $\angle BAE=\angle F'EA=30^\circ$,\\
because $AB\parallel CD$,\\
therefore $EF' \parallel CD$,\\
thus $\angle DCE=\angle F'EC=60^\circ$\\
Also, because $\angle AEC=\angle F'EA+\angle F'EC$,\\
thus $\angle AEC=90^\circ$,\\
because $EF, EG$ trisect $\angle AEC$ (i.e., $\angle AEF=\angle FEG=\angle GEC$)\\
therefore $\angle AEF=\boxed{30^\circ}$.


Question: As shown in the figure, $OM$ is the bisector of $\angle AOC$, and $ON$ is the bisector of $\angle BOC$. As shown 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:} Given that $\angle AOB=90^\circ$ and $\angle BOC=60^\circ$, \\
thus $\angle AOC=\angle AOB+\angle BOC=90^\circ+60^\circ=150^\circ$. \\
Since $OM$ bisects $\angle AOC$ and $ON$ bisects $\angle BOC$, \\
thus $\angle MOC=\frac{1}{2}\angle AOC=75^\circ$ and $\angle NOC=\frac{1}{2}\angle BOC=30^\circ$. \\
Therefore, $\angle MON=\angle MOC-\angle NOC=75^\circ-30^\circ=\boxed{45^\circ}$.


Question: As shown in the diagram, it is known that $\angle COD$ is inside $\angle AOB$, and $\angle AOB=160^\circ$, $\angle COD=\frac{1}{4}\angle AOB$, ray $OE$ bisects $\angle AOD$, $\angle COE=18^\circ$. Find the measure of $\angle AOD$ in degrees.

Solution:
\textbf{Solution:} Given: $\because$ $\angle AOB=160^\circ$ and $\angle COE=18^\circ$,\\
$\therefore$ $\angle COD=\frac{1}{4}\angle AOB=\frac{1}{4}\times 160^\circ=40^\circ$,\\
$\therefore$ $\angle EOD=\angle EOC+\angle COD=18^\circ+40^\circ=58^\circ$,\\
$\because$ ray $OE$ bisects $\angle AOD$,\\
$\therefore$ $\angle AOE=\angle EOD=\frac{1}{2}\angle AOD$,\\
$\therefore$ $\angle AOD=2\angle EOD=2\times 58^\circ=\boxed{116^\circ}$.


Question: As shown in the figure, given: $\angle COD$ is inside $\angle AOB$, and $\angle AOB=160^\circ$, $\angle COD=\frac{1}{4}\angle AOB$, ray $OE$ bisects $\angle AOD$, $\angle COE=18^\circ$. Find: What is the degree measure of $\angle BOC$?

Solution:
\textbf{Solution:} From equation (1), we have $\angle AOE=\angle EOD=58^\circ$,\\
since $\angle COE=18^\circ$ and $\angle AOB=160^\circ$,\\
therefore $\angle AOC=\angle AOE+\angle COE=58^\circ+18^\circ=76^\circ$,\\
thus $\angle BOC=\angle AOB-\angle AOC=160^\circ-76^\circ=\boxed{84^\circ}$.


Question: As shown in the figure, points $A$, $O$, and $B$ are on the same line, $OC$ bisects $\angle AOB$. If $\angle COD = 28^\circ$, what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Since points A, O, and B lie on the same line, and $OC$ bisects $\angle AOB$,\\
$\therefore$ $\angle AOC=\angle BOC=90^\circ$,\\
$\therefore$ $\angle BOD=\angle BOC-\angle COD=90^\circ-28^\circ=\boxed{62^\circ}$;


Question: As shown in the figure, points $A$, $O$, $B$ lie on the same line, and $OC$ bisects $\angle AOB$. If $\angle COD = 28^\circ$, and $OE$ bisects $\angle BOD$, find the degree measure of $\angle COE$.

Solution:
\textbf{Solution:} Since $OE$ bisects $\angle BOD$,\\
we have $\angle DOE = \frac{1}{2}\angle BOD = 31^\circ$,\\
hence, $\angle COE = \angle DOE + \angle COD = 31^\circ + 28^\circ = \boxed{59^\circ}$.


Question: Given points $A$, $B$, $C$, and $D$ are on the same line, where point $C$ is the midpoint of segment $AB$, and point $D$ is on segment $AB$. If $AB=6$ and $BD=\frac{1}{3}BC$, what is the length of segment $CD$?

Solution:
\textbf{Solution:} Given that point C is the midpoint of segment AB, and $AB=6$,\\
$\therefore$ $BC=\frac{1}{2}AB=3$,\\
$\therefore$ $BD=\frac{1}{3}BC$,\\
$\therefore$ $BD=\frac{1}{3}\times 3=1$,\\
$\therefore$ $CD=BC-BD=3-1=\boxed{2}$.


Question: As shown in the figure, it is known that points $A$, $B$, $C$, $D$ are on the same line, with point $C$ being the midpoint of segment $AB$, and point $D$ on segment $AB$. If point $E$ is a point on segment $AB$, and $AE=2BE$, when $AD:BD=2:3$, what is the value of $CD:CE$?

Solution:
\textbf{Solution:} Let $AD=2x$ and $BD=3x$, hence $AB=5x$, \\
since point C is the midpoint of segment AB, \\
therefore $AC=\frac{1}{2}AB=\frac{5}{2}x$, \\
therefore $CD=AC-AD=\frac{1}{2}x$, \\
since $AE=2BE$, \\
therefore $AE=\frac{2}{3}AB=\frac{10}{3}x$, \\
therefore $CE=AE-AC=\frac{10}{3}x-\frac{5}{2}x=\frac{5}{6}x$, \\
therefore $CD: CE=\frac{1}{2}x: \frac{5}{6}x=3:5$, \\
thus, the ratio of segment $CD:CE$ is $\boxed{3:5}$.


Question: As shown in the figure, $O$ is a point on the line $MN$, $\angle MOC=130^\circ$, $OA$ bisects $\angle MOC$, $\angle AOB=90^\circ$. What is the measure of $\angle BOC$?

Solution:
\textbf{Solution:} Since $\angle AOB=90^\circ$,\\
$\angle AOC=65^\circ$,\\
$\angle AOB=\angle AOC+\angle BOC$,\\
therefore, $\angle BOC=\angle AOB-\angle AOC=25^\circ$.\\
Hence, $\angle BOC=\boxed{25^\circ}$.


Question: As shown in the figure, it is known that $O$ is a point on the line $AB$, $\angle COD=90^\circ$, $OE$ bisects $\angle BOC$. If $\angle AOC=40^\circ$, what is the degree measure of $\angle DOE$?

Solution: \textbf{Solution}: Thus, we have $\angle DOE = \boxed{20^\circ}$.


Question: As shown in the figure, it is known that $O$ is a point on the line $AB$, $\angle COD = 90^\circ$, and $OE$ bisects $\angle BOC$. If $\angle AOC = \alpha$, what is the degree measure of $\angle DOE$ (expressed in an algebraic expression involving $\alpha$)?

Solution:
\textbf{Solution:} Given the problem, \\
$\because \angle AOC=\alpha$, \\
$\therefore \angle BOC=180^\circ-\alpha$, \\
$\because OE$ bisects $\angle BOC$, \\
$\therefore \angle COE=\frac{1}{2}\angle BOC=90^\circ-\frac{\alpha}{2}$, \\
$\because \angle COD=90^\circ$, \\
$\therefore \angle DOE=90^\circ-(90^\circ-\frac{\alpha}{2})=\boxed{\frac{\alpha}{2}}$.


Question: As shown in the figure, $OD$ bisects $\angle AOB$, $OE$ bisects $\angle BOC$, and $\angle COD=20^\circ$, $\angle AOB=140^\circ$. Find the degree measure of $\angle BOC$?

Solution:
\textbf{Solution:} Since $OD$ bisects $\angle AOB$, \\
it follows that $\angle DOB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 140^\circ = 70^\circ$, \\
therefore, $\angle BOC = \angle BOD - \angle COD = 70^\circ - 20^\circ = \boxed{50^\circ}$.


Question: As shown in the figure, $OD$ bisects $\angle AOB$, $OE$ bisects $\angle BOC$, $\angle COD=20^\circ$, $\angle AOB=140^\circ$. What is the measure of $\angle DOE$ in degrees?

Solution:
\textbf{Solution:} Since $OE$ bisects $\angle BOC$,\\
it follows that $\angle COE = \frac{1}{2}\angle BOC = \frac{1}{2} \times 50^\circ = 25^\circ$,\\
therefore, $\angle DOE = \angle COE + \angle COD = 25^\circ + 20^\circ = \boxed{45^\circ}$.


Question: As shown in the figure, $OB$ is the bisector of $\angle AOC$, and $OD$ is the bisector of $\angle COE$. If $\angle AOB=50^\circ$ and $\angle DOE=35^\circ$, then what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Since OB is the bisector of $\angle AOC$, and OD is the bisector of $\angle COE$, \\
therefore $\angle COD = \angle DOE = 35^\circ$, $\angle COB = \angle BOA = 50^\circ$. \\
Thus, $\angle BOD = \angle COD + \angle COB = \boxed{85^\circ}$.


Question: As shown in the figure, $OB$ is the bisector of $\angle AOC$, and $OD$ is the bisector of $\angle COE$. If $\angle AOE=160^\circ$ and $\angle COD=25^\circ$, then what is the measure of $\angle AOB$ in degrees?

Solution:
\textbf{Solution}: Since $OD$ is the bisector of $\angle COE$,\\
it follows that $\angle COE=2\angle COD=2\times 25^\circ=50^\circ$,\\
thus $\angle AOC=\angle AOE-\angle COE=160^\circ-50^\circ=110^\circ$,\\
also, since $OB$ is the bisector of $\angle AOC$,\\
it follows that $\angle AOB=\frac{1}{2}\angle AOC=\frac{1}{2}\times 110^\circ=\boxed{55^\circ}$.


Question: As shown in the figure, point C is on the line segment $AD$, and point B is the midpoint of $CD$, with $AD=9\,cm$, $BC=2\,cm$. What is the length of $AC$ in cm?

Solution:
Solution: Since $B$ is the midpoint of $CD$, and $BC=2\,cm$, \\
it follows that $CD=2BC=4\,cm$, \\
Given $AD=9\,cm$, \\
thus $AC=AD-CD=\boxed{5}\,cm$.


Question: As shown in the figure, point $C$ is on line segment $AD$, point $B$ is the midpoint of $CD$, and $AD=9\,cm$, $BC=2\,cm$. If point $E$ is on line segment $AD$ and $AE=3\,cm$, what is the length of $BE$?

Solution:
\textbf{Solution:} \\
Since $AB = AC + BC = 7\,cm$, \\
$AE = 3\,cm$, \\
Therefore $BE = AB - AE = 4\,cm$, \\
hence the length of $BE$ is \boxed{4\,cm}.


Question: As shown in the figure, it is known that point $C$ is on the straight line $AB$, and points $M$ and $N$ are the midpoints of $AC$ and $BC$, respectively. If $C$ is on the line segment $AB$ with $AC=6$ cm and $MB=10$ cm, find the lengths of line segments $BC$ and $MN$.

Solution:
\textbf{Solution:} Given that $M$ is the midpoint of $AC$,\\
it follows that $MC=\frac{1}{2}AC=3\,\text{cm}$,\\
thus $BC=MB-MC=7\,\text{cm}$,\\
and since $N$ is the midpoint of $BC$,\\
it then follows that $CN=\frac{1}{2}BC=3.5\,\text{cm}$,\\
therefore $MN=MC+NC=\boxed{6.5\,\text{cm}}$;


Question: Given that point $C$ is on line $AB$, and points $M$ and $N$ are the midpoints of $AC$ and $BC$, respectively. If point $C$ is on the extension of segment $AB$, and it satisfies $AC-BC=a$ cm, please draw a diagram according to the given conditions, and calculate the length of $MN$ (express the result in terms of $a$).

Solution:
\textbf{Solution:} According to the problem, the position of point $M$ can be in two situations:

(1) The position of point $M$ is to the left of point $B$, as shown in the diagram:\\
Since $M$ is the midpoint of $AC$,\\
Therefore, $CM=\frac{1}{2}AC$,\\
Since $N$ is the midpoint of $BC$,\\
Therefore, $CN=\frac{1}{2}BC$,\\
Therefore, $MN=CM-CN=\frac{1}{2}AC-\frac{1}{2}BC=\frac{1}{2}(AC-BC)=\boxed{\frac{1}{2}a}\,cm$;\\
(2) The position of point $M$ is to the right of point $B$, as shown in the diagram:\\
Since $M$ is the midpoint of $AC$,\\
Therefore, $CM=\frac{1}{2}AC$,\\
Since $N$ is the midpoint of $BC$,\\
Therefore, $CN=\frac{1}{2}BC$,\\
Therefore, $MN=CM-CN=\frac{1}{2}AC-\frac{1}{2}BC=\frac{1}{2}(AC-BC)=\boxed{\frac{1}{2}a}\,cm$.


Question: As shown in the figure, it is known that $AB\parallel CD$, and the line $MN$ intersects $AB$ and $CD$ at points $E$ and $F$ respectively, and $\angle NEB=58^\circ$. What is the degree measure of $\angle CFE$?

Solution:
\textbf{Solution:} Given that $\angle NEB$ and $\angle AEF$ are vertical angles, and $\angle NEB=58^\circ$,\\
it follows that $\angle AEF=\angle NEB=58^\circ$.\\
Since $AB \parallel CD$,\\
we have $\angle AEF + \angle CFE = 180^\circ$,\\
therefore $\angle CFE = 180^\circ - 58^\circ = \boxed{122^\circ}$.


Question: As shown in the figure, it is known that $AB \parallel CD$, and the line $MN$ intersects $AB$ and $CD$ at points $E$ and $F$, respectively, with $\angle NEB = 58^\circ$. If $EG \perp AB$, and $EH$ bisects $\angle NEG$ and intersects $CD$ at point $H$, what is the measure of $\angle BEH$ in degrees?

Solution:
\textbf{Solution:} Since $EG\perp AB$, it follows that $\angle BEG=90^\circ$, \\
hence, $\angle BEH=90^\circ-\angle GEH$, \\
since $EH$ bisects $\angle NEG$, \\
it follows that $\angle GEH= \frac{1}{2}\angle NEG=\frac{1}{2}(180^\circ-\angle MEG)$, \\
from (1) we know $\angle AEM=58^\circ$, and since $EG\perp AB$, \\
thus, $\angle MEG=90^\circ-\angle AEM=90^\circ-58^\circ=32^\circ$, \\
hence, $\angle GEH=\frac{1}{2}(180^\circ-32^\circ)=74^\circ$, \\
thus, $\angle BEH=90^\circ-74^\circ=\boxed{16^\circ}$.


Question: As shown in the figure, it is known that $\angle AOB=30^\circ$, $\angle AOE=140^\circ$, $OB$ bisects $\angle AOC$, and $OD$ bisects $\angle AOE$. What is the measure of $\angle AOC$?

Solution:
\textbf{Solution:} From the given, \\
since $\angle AOB = 30^\circ$ and OB bisects $\angle AOC$, \\
it follows that $\angle AOC = 2\angle AOB = 2 \times 30^\circ = \boxed{60^\circ}$.


Question: As shown in the figure, it is known that $ \angle AOB=30^\circ$, $ \angle AOE=140^\circ$, $OB$ bisects $ \angle AOC$, and $OD$ bisects $ \angle AOE$. Find the degree measure of $ \angle COD$.

Solution:
\textbf{Solution:}

Since $\angle AOE = 140^\circ$ and $\angle AOC = 60^\circ$,

it follows that $\angle COE = \angle AOE - \angle AOC = 140^\circ - 60^\circ = 80^\circ$.

Also, since OD bisects $\angle AOE$,

we have $\angle DOE = \frac{1}{2} \angle AOE = \frac{1}{2} \times 140^\circ = 70^\circ$.

Therefore, $\angle COD = \angle COE - \angle DOE = 80^\circ - 70^\circ = \boxed{10^\circ}$.


Question: As shown in the figure, point $O$ is a point on the line $AB$, with $\angle BOC=40^\circ$, and $OD$ bisects $\angle AOC$. What is the degree measure of $\angle AOD$?

Solution:
\textbf{Solution:} Since $\angle BOC=40^\circ$,\\
it follows that $\angle AOC=180^\circ-\angle BOC=140^\circ$,\\
since OD bisects $\angle AOC$,\\
thus $\angle AOD= \frac{1}{2} \angle AOC= \boxed{70^\circ}$;


Question: As shown in the figure, point $O$ is a point on the line segment $AB$, with $ \angle BOC=40^\circ$, and $OD$ bisects $ \angle AOC$. Construct ray $OE$ such that $ \angle BOE=\frac{2}{3}\angle COE$. Find the degree measure of $ \angle COE$.

Solution:
\textbf{Solution:}

(1) As shown in Figure 1, when ray OE is above AB, $\angle BOE = \frac{2}{3} \angle COE$,

$\because \angle BOE + \angle COE = \angle BOC$,

$\therefore \frac{2}{3} \angle COE + \angle COE = 40^\circ$,

$\therefore \angle COE = \boxed{24^\circ}$;

(2) As shown in Figure 2, when ray OE is below AB, $\angle BOE = \frac{2}{3} \angle COE$,

$\because \angle COE - \angle BOE = \angle BOC$,

$\therefore \angle COE - \frac{2}{3} \angle COE = 40^\circ$,

$\therefore \angle COE = \boxed{120^\circ}$;

In conclusion: The degree of $\angle COE$ can be either \boxed{24^\circ} or \boxed{120^\circ}.


Question: As shown in the figure, $\angle AOD = 130^\circ$, $\angle BOC : \angle COD = 1 : 2$, $\angle AOB$ is $\frac{1}{3}$ of the complement of $\angle COD$. How many degrees is $\angle COD$?

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


Question: As shown in the figure, $\angle AOD = 130^\circ$, $\angle BOC : \angle COD = 1 : 2$, and $\angle AOB$ is $\frac{1}{3}$ of the supplement of $\angle COD$. In the plane, the ray OM satisfies $\angle AOM = 2\angle DOM$. What is the measure of $\angle AOM$?

Solution:
\textbf{Solution:} When ray OM is inside $\angle AOD$,\\
$\because \angle AOD = 130^\circ$, and $\angle AOM = 2\angle DOM$,\\
$\therefore \angle AOM = \frac{2}{3}\angle AOD = \boxed{\left(\frac{260}{3}\right)^\circ}$.\\
When ray OM is outside $\angle AOD$,\\
$\because \angle AOD = 130^\circ$, and $\angle AOM = 2\angle DOM$, $\angle AOD + \angle AOM + \angle DOM = 360^\circ$,\\
$\therefore \angle AOM = \boxed{\left(\frac{460}{3}\right)^\circ}$.


Question: Given the figure, $\angle AOD = 130^\circ$, $\angle BOC : \angle COD = 1 : 2$, $\angle AOB$ is $\frac{1}{3}$ of the complement of $\angle COD$. Fix $\angle COD$ and rotate rays OA and OB clockwise at a speed of $2^\circ/s$ until OA overlaps with OD. During the rotation, if ray OP bisects $\angle AOB$ and OQ bisects $\angle COD$, when $\angle POQ + \angle AOD = 50^\circ$, find the range of the rotation time $t$ (in seconds).

Solution:
\textbf{Solution:} From (1) we have: $\angle AOB=40^\circ$, $\angle COD=60^\circ$\\
Since ray OP is the bisector of $\angle AOB$, and OQ is the bisector of $\angle COD$,\\
therefore $\angle AOP=\frac{1}{2}\angle AOB=20^\circ$, $\angle DOQ=\frac{1}{2}\angle COD=30^\circ$,\\
It can be observed that: $\angle AOP+\angle DOQ=50^\circ$,\\
To find the range of time $t$, we need to find the critical condition,\\
When $OP$ coincides with $OQ$, at this moment $\angle AOD=\angle AOP+\angle DOQ=50^\circ$, $\angle POQ=0^\circ$,\\
It is observed that, if $OP$ and $OQ$ do not coincide, there must be $\angle AOD>\angle AOP+\angle DOQ=50^\circ$ which does not satisfy $\angle POQ+\angle AOD=50^\circ$, hence the time $t$ exactly reaches its minimum value at this moment,\\
According to the problem statement: $OA$ rotates a total of $130^\circ-\angle AOD=80^\circ$, thus time $t=80\div 2=40s$,\\
$\therefore t\ge 40s$,\\
When $OA$ coincides with $OD$, at this moment $\angle AOD=0^\circ$, $\angle POQ=\angle AOP+\angle DOQ=50^\circ$,\\
If $OA$ continues to rotate downwards at this moment, there must be $\angle POQ>\angle AOP+\angle DOQ=50^\circ$, which does not satisfy $\angle POQ+\angle AOD=50^\circ$, hence the time $t$ exactly reaches its maximum value at this moment,\\
According to the problem statement: $OA$ rotates a total of $130^\circ-\angle AOD=130^\circ$, thus time $t=130\div 2=65s$,\\
$\therefore t\le 65s$,\\
In conclusion: $40s\le t\le 65s$, hence the range of rotation time $t$ is $t\in\boxed{[40s, 65s]}$.


Question:
As shown in the figure, point $O$ is on line $AB$, and $OC$, $OD$ bisect $\angle AOE$, $\angle BOE$ respectively. It can be deduced that $\angle COD=90^\circ$. (No solution required)

Given that point $O$ is on line $AB$, and $\angle COD=90^\circ$, $OE$ bisects $\angle BOC$. As shown in the figure, if $\angle AOC=50^\circ$, what is the degree measure of $\angle DOE$?

Solution: \textbf{Solution:} Given $\because$ $\angle COD=90^\circ$, $\therefore$ $\angle AOC+\angle BOD=90^\circ$. $\because$ $\angle AOC=50^\circ$, $\therefore$ $\angle BOD=40^\circ$. $\therefore$ $\angle COB=\angle COD+\angle BOD=90^\circ+40^\circ=130^\circ$. $\because$ $OE$ bisects $\angle BOC$, $\therefore$ $\angle COE=\frac{1}{2}\angle BOC=\frac{1}{2}\times 130^\circ=65^\circ$. $\therefore$ $\angle DOE=\angle COD-\angle COE=90^\circ-65^\circ=\boxed{25^\circ}$.


Question: As shown in the figure, point $O$ is a point on the line $AB$, $CO \perp DO$, $OE$ bisects $\angle BOC$. If $\angle AOC=50^\circ$, what is the degree measure of $\angle DOE$?

Solution:
\textbf{Solution:} Because $\angle AOB$ is a straight angle,\\
therefore $\angle BOC=180^\circ-\angle AOC=180^\circ-50^\circ=130^\circ$,\\
therefore $\angle EOC=\frac{1}{2}\angle BOC=\frac{1}{2}\times 130^\circ=65^\circ$,\\
because $CO\perp DO$,\\
therefore $\angle DCC=90^\circ$,\\
therefore $\angle DOE=\angle DOC=\angle EOC=90^\circ-65^\circ=\boxed{25^\circ}$.


Question: As shown in the figure, point $O$ is a point on line $AB$, with $CO \perp DO$, and $OE$ bisects $\angle BOC$. If $\angle COE = \frac{1}{3}\angle DOB$, determine the measure of $\angle AOC$.

Solution:
\textbf{Solution:} Let $\angle COE=x^\circ$, then $\angle DOB=3x^\circ$ \\
$\because OE$ bisects $\angle BOC$ \\
$\therefore \angle BOE=\angle COE=x^\circ$ \\
$\because CO\perp DO$ \\
$\therefore \angle DOC=90^\circ$ \\
$\therefore x^\circ+x^\circ+3x^\circ=90^\circ$ \\
$\therefore x=18$ \\
$\therefore \angle BOE=\angle COE=18^\circ$ \\
$\therefore \angle AOC=\angle AOB-\angle BOE-\angle COE=180^\circ-18^\circ-18^\circ=\boxed{144^\circ}$.


Question: Given that $O$ is a point on line $AB$, and $OC \perp OE$. As shown in Figure (1), if $\angle COA = 34^\circ$, what is the measure of $\angle BOE$?

Solution:
\textbf{Solution:} Since $OC\perp OE$ and $\angle COA = 34^\circ$, \\
it follows that $\angle BOE = 180^\circ-90^\circ-34^\circ = \boxed{56^\circ}$;


Question: Given that O is a point on line AB, and OC $\perp$ OE. As shown in the diagram, when ray OC is below line AB, OF bisects $\angle AOE$, $\angle BOE = 130^\circ$, find the degree measure of $\angle COF$.

Solution:
\textbf{Solution:} Given $\angle BOE = 130^\circ$,\\
it follows that $\angle AOE = 180^\circ - \angle BOE = 50^\circ$\\
Since OF bisects $\angle AOE$,\\
we have $\angle EOF = \frac{1}{2} \angle AOE = 25^\circ = \angle AOF$\\
Since $OC \perp OE$,\\
it follows that $\angle COF = 90^\circ - \angle EOF = \boxed{65^\circ}$;


Question: Given that $O$ is a point on line $AB$, and $OC \perp OE$. Under these conditions, as shown in the figure, draw ray $OM$ inside $\angle BOE$, such that $\angle COM + \frac{17}{10}\angle AOE = 2\angle BOM + \angle FOM$, what is the degree measure of $\angle BOM$?

Solution:
\textbf{Solution:} Given $\because$OC$\perp$OE,\\
$\therefore \angle AOC=90^\circ-\angle AOE=40^\circ$\\
Let the measure of $\angle BOM$ be $x$\\
$\therefore \angle COM=\angle AOC+\angle AOM=40^\circ+180^\circ-x=220^\circ-x$, and $\angle FOM=\angle AOM-\angle AOF=180^\circ-x-25^\circ=155^\circ-x$\\
$\because \angle COM+ \frac{17}{10}\angle AOE=2\angle BOM+\angle FOM$,\\
$\therefore 220^\circ-x+ \frac{17}{10}\times 50^\circ=2x+155^\circ-x$\\
Solving gives $x=\boxed{75^\circ}$\\
$\therefore$ The measure of $\angle BOM$ is $\boxed{75^\circ}$.


Question: As shown in the figure, it is known that $\angle AOB=120^\circ$, and OC is a ray inside $\angle AOB$, and $\angle AOC : \angle BOC=1 : 2$. Find the measures of $\angle AOC$ and $\angle BOC$.

Solution:
\textbf{Solution:} Since $\angle AOC : \angle BOC = 1 : 2$ and $\angle AOB = 120^\circ$,

thus $\angle AOC = \frac{1}{3} \angle AOB = \frac{1}{3} \times 120^\circ = \boxed{40^\circ}$,

and $\angle BOC = \frac{2}{3} \angle AOB = \frac{2}{3} \times 120^\circ = \boxed{80^\circ}$;


Question: As shown in the figure, it is known that $\angle AOB=120^\circ$. OC is a ray inside $\angle AOB$, and $\angle AOC: \angle BOC=1: 2$. Draw ray OM to bisect $\angle AOC$. Inside $\angle BOC$, draw ray ON such that $\angle CON: \angle BON=1: 3$. What is the measure of $\angle MON$ in degrees?

Solution:
\textbf{Solution:} Given that OM bisects $\angle$AOC,\\
$\therefore \angle$COM$= \frac{1}{2}\angle$AOC$= \frac{1}{2}\times 40^\circ=20^\circ$,\\
Since $\angle$CON$\colon \angle$BON$=1\colon 3$,\\
$\therefore \angle$CON$= \frac{1}{4}\angle$BOC$= \frac{1}{4}\times 80^\circ=20^\circ$,\\
Hence, $\angle$MON$=\angle$COM$+\angle$CON$=20^\circ+20^\circ=\boxed{40^\circ}$;


Question: As shown in the figure, it is known that $\angle AOB=120^\circ$, and $OC$ is a ray inside $\angle AOB$, with $\angle AOC : \angle BOC = 1 : 2$. A ray $OD$ passes through point $O$. If $2\angle AOD = 3\angle BOD$, what is the measure of $\angle COD$ in degrees?

Solution:
\textbf{Solution:} As shown in the figure, when OD is inside $\angle AOB$,\\
let $\angle BOD=x^\circ$,\\
since $2\angle AOD=3\angle BOD$,\\
then $\angle AOD= \frac{3}{2}x^\circ$,\\
since $\angle AOB=120^\circ$,\\
then $x+ \frac{3}{2}x=120$,\\
solving this gives: $x=48$,\\
thus $\angle BOD=48^\circ$,\\
therefore $\angle COD=\angle BOC-\angle BOD=80^\circ-48^\circ=32^\circ$,\\
As shown in the figure, when OD is outside $\angle AOB$,\\
let $\angle BOD=y^\circ$,\\
since $2\angle AOD=3\angle BOD$,\\
then $\angle AOD= \frac{3}{2}y^\circ$,\\
since $\angle AOB=120^\circ$,\\
then $\frac{3}{2}y+y+120^\circ=360^\circ$,\\
solving this gives: $y=96^\circ$,\\
thus $\angle COD=\angle BOD+\angle BOC=96^\circ+80^\circ=176^\circ$,\\
In conclusion, the degree measure of $\angle COD$ is $\boxed{32^\circ}$ or $\boxed{176^\circ}$.


Question: Given that segment $ AD=80$, points B and C are both on segment AD. As shown in Figure 1, $ BC: AB: CD=1:3:4$, point E is the midpoint of segment AC, and point F is the midpoint of segment BD. Find the length of EF.

Solution:
\textbf{Solution:} Let $BC\colon AB\colon CD=1\colon 3\colon 4$,\\
Thus, let BC=x, then AB=3x, CD=4x,\\
Therefore, AD=3x+x+4x=80, which means x=10,\\
Hence, AB=30, BC=10, CD=40,\\
Since point E is the midpoint of segment AC, and point F is the midpoint of segment BD,\\
Thus, AE=20, DF=20,\\
Therefore, EF=80−20−20=\boxed{40};


Question: Given that the line segment $AD=80$, and points B and C are on the line segment AD. If $AB=10$, $BC=20$, and point P moves from B towards D at a speed of 3 units per second, taking $t$ seconds, where point Q is the midpoint of CD, find the value of $t$ when $AP=AQ$.

Solution:
\textbf{Solution:} Given that $AD=80$, $AB=10$, and $BC=20$, \\
hence, $CD=80-10-20=50$, \\
since $Q$ is the midpoint of $CD$, \\
thus, $CQ=\frac{1}{2}CD=25$, and $AQ=10+20+25=55$, \\
after $t$ seconds, $AP=10+3t$, \\
when $AP=AQ$, $10+3t=55$, \\
therefore, $t=\boxed{15}$, \\
when $AP=AQ$, the value of $t$ is \boxed{15}.


Question: As shown in Figure 1, it is known that $\angle AOB = 90^\circ$, OE and OD bisect $\angle AOB$ and $\angle BOC$ respectively. If $\angle EOD = 64^\circ$, what is the degree measure of $\angle BOC$?

Solution: \textbf{Solution:} It is known from the problem that $\angle BOC = \boxed{38^\circ}$.


Question: Given that $\angle AOB = 90^\circ$, as shown in figure 2, $OE$ and $OD$ bisect $\angle AOC$ and $\angle BOC$ respectively. If $\angle BOC = 40^\circ$, find the measure of $\angle EOD$.

Solution:
\textbf{Solution:} Given that $\angle EOD = \angle EOC - \angle COD$ \\
= $\frac{1}{2}\angle AOC - \frac{1}{2}\angle BOC$ \\
= $\frac{1}{2}(\angle AOB + \angle BOC) - \frac{1}{2}\angle BOC$ \\
= $\frac{1}{2}(90^\circ + 40^\circ) - \frac{1}{2} \times 40^\circ$ \\
= $65^\circ - 20^\circ$ \\
= $\boxed{45^\circ}$


Question: As shown in the figure, it is known that $\angle C=\angle D$, $OC=OD$. If $\angle O=72^\circ$, $\angle C=20^\circ$, and $\triangle OAD \cong \triangle OBC$; what is the degree measure of $\angle DAC$?

Solution:
\textbf{Solution:} Given that $ \triangle OAD \cong \triangle OBC$,\\
$\therefore \angle C=\angle D=20^\circ$,\\
$\therefore \angle DAC=\angle O+\angle D=20^\circ+72^\circ=\boxed{92^\circ}$.


Question: As shown in the diagram, in $\triangle ABC$, $D$ is a point on side $BC$, $AB=DB$, $BE$ bisects $\angle ABC$, and intersects side $AC$ at point $E$, with $DE$ connected. $\triangle ABE \cong \triangle DBE$; if $\angle A=105^\circ$, $\angle C=50^\circ$, what is the degree measure of $\angle DEC$?

Solution:
\textbf{Solution:} Given: $\because \angle A=105^\circ$, $\angle C=50^\circ$, $\triangle ABE \cong \triangle DBE$,\\
$\therefore \angle BDE=\angle A=105^\circ$,\\
$\therefore \angle EDC=180^\circ-\angle BDE=180^\circ-105^\circ=75^\circ$,\\
$\therefore \angle DEC=180^\circ-\angle EDC-\angle C=180^\circ-75^\circ-50^\circ=\boxed{55^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=CB$, and $\angle ABC=90^\circ$. Point F lies on the extension of side AB, and point E is on side BC, with $AE=CF$. $Rt\triangle ABE \cong Rt\triangle CBF$. If $\angle CAE=25^\circ$, what is the measure of $\angle CFA$?

Solution:
\textbf{Solution:} Since $AB=CB$ and $\angle ABC=90^\circ$,

it follows that $\angle BAC=\angle BCA=45^\circ$.

Given $\angle CAE=25^\circ$,

then $\angle BAE=45^\circ-25^\circ=20^\circ$.

Since Rt$\triangle ABE \cong$ Rt$\triangle CBF$,

it implies that $\angle BCF=\angle BAE=20^\circ$,

therefore, $\angle CFA=90^\circ-20^\circ=\boxed{70^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, and $CD$ is the altitude to side $AB$. Fold side $AC$ such that the crease is $EF$, and connect $CE$. If $CD$ bisects $\angle BCE$, find the measure of $\angle A$.

Solution:
\textbf{Solution.} From the given problem, $EF$ is the axis of symmetry of $AC$,\\
hence $FA=FC$, $EA=EC$,\\
hence $\angle ECA=\angle A$.\\
Moreover, since $CD$ is the height from $AB$,\\
hence $\angle CDE=\angle CDB=90^\circ$.\\
Also, since $CD$ bisects $\angle BCE$,\\
hence $\angle DCE=\angle DCB$.\\
Also, since $CD=CD$,\\
hence $\triangle CDE\cong \triangle CDB$ (ASA),\\
hence $\angle CED=\angle CBD$.\\
Moreover, since $\angle CED=\angle A+\angle ECA$, $\angle ECA=\angle A$,\\
hence $\angle CBD=2\angle A$.\\
Also, since $\angle ACB=90^\circ$,\\
hence $\angle A+\angle B=90^\circ$,\\
hence $3\angle A=90^\circ$,\\
hence $\angle A=\boxed{30^\circ}$.


Question: In the figure, in $\triangle ABC$, $AB=AC$, $AD\perp BC$ at point D, and point E is on side $AB$, with $EF\parallel AC$ intersecting the extension of $AD$ at point F. If $\angle C=40^\circ$, what is the degree measure of $\angle AEF$?

Solution:
\textbf{Solution:} Given the problem,\\
$\because AB=AC$,\\
$\angle C=40^\circ$,\\
$\therefore \angle B=\angle C=40^\circ$,\\
$\therefore \angle BAC=180^\circ-\angle B-\angle C=100^\circ$,\\
$\because EF\parallel AC$,\\
$\therefore \angle BAC+\angle AEF=180^\circ$,\\
$\therefore \angle AEF=\boxed{80^\circ}$.


Question: As shown in the figure, $AB=AD$, $AC=AE$, $BC=DE$, with point $E$ on $BC$. If $\angle EAC=\angle BAD$ and $\angle EAC=42^\circ$, what is the degree measure of $\angle DEB$?

Solution:
\textbf{Solution:} Since $AC=AE$, and $\angle EAC=42^\circ$, \\
therefore $\angle AEC=\angle C= \frac{1}{2} \times(180^\circ-42^\circ)=69^\circ$. \\
Since $\triangle ABC \cong \triangle ADE$, \\
therefore $\angle AED=\angle C=69^\circ$, \\
thus $\angle DEB=180^\circ-\angle AED-\angle C=180^\circ-69^\circ-69^\circ=\boxed{42^\circ}$.


Question: Please provide the text you wish to have translated into English in a format suitable for LaTeX. Without the original text, I'm unable to assist with your request.

Solution:
\textbf{Solution:} Given that $AB=AC$ and $\angle A=39^\circ$, \\
it follows that $\angle ABC=\angle ACB=70.5^\circ$, \\
since $DM$ is the perpendicular bisector of $AB$, \\
it implies $DA=DB$ and thus $\angle A=\angle DBA=39^\circ$, \\
therefore, $\angle DBC=\angle ABC-\angle DBA=\boxed{31.5^\circ}$;


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=10$, $BC=8$, $\angle A=39^\circ$. The perpendicular bisector $MN$ of $AB$ intersects $AC$ at $D$ and $AB$ at $M$. $BD$ is connected. Find: What is the perimeter of $\triangle BDC$?

Solution:
\textbf{Solution:} Since $DB=AD$, $AC=10$, $BC=8$, \\
hence $DB+DC=AD+DC=AC=10$. \\
Therefore, the perimeter of $\triangle BDC$ is $DB+DC+BC=18=\boxed{18}$.


Question: As shown in the figure, the line $l_{1}: y=-2x+4$ intersects the x-axis at point D and the y-axis at point A, while the line $l_{2}: y=x+1$ intersects the x-axis at point C. The two lines $l_{1}$ and $l_{2}$ intersect at point B. Connect A to C. Find the coordinates of point B and the equation of the line AC.

Solution: \textbf{Solution:} Since lines $l_1$ and $l_2$ intersect at point B, by solving the system of equations $\begin{cases}y=-2x+4\\ y=x+1\end{cases}$, we obtain $\begin{cases}x=1\\ y=2\end{cases}$. Therefore, the coordinates of point B are $\boxed{(1,2)}$. Since line $l_1: y=-2x+4$ intersects the y-axis at point A, by setting $x=0$, we get $y=4$. Therefore, the coordinates of point A are $\boxed{(0,4)}$. Since line $l_2: y=x+1$ intersects the x-axis at point C, by setting $y=0$, we solve $x+1=0$ to get $x=-1$. Therefore, the coordinates of point C are $\boxed{(-1,0)}$. Let the equation of line AC be $y=kx+b$. Substituting the coordinates, we have $\begin{cases}0=-k+b\\ b=4\end{cases}$, which gives us $\begin{cases}k=4\\ b=4\end{cases}$. Therefore, the equation of line AC is $\boxed{y=4x+4}$.


Question: As shown in the figure, the line $l_{1}\colon y=-2x+4$ intersects with the x-axis at point D and with the y-axis at point A, the line $l_{2}\colon y=x+1$ intersects with the x-axis at point C, the two lines $l_{1}$ and $l_{2}$ intersect at point B, and AC is connected. What is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Let the line $l_{2}\colon y=x+1$ intersect the y-axis at point E. By setting $x=0$, $y=1$,\\
$\therefore$ point E$(0,1)$,\\
$\therefore AE=4-1=3$,\\
$\therefore S_{\triangle ABC}= S_{\triangle AEC}+S_{\triangle ABE}=\frac{1}{2}AE\cdot OC+\frac{1}{2}AE\cdot x_{B}=\frac{1}{2}\times 3\times 1+\frac{1}{2}\times 3\times 1=\boxed{3}$.


Question: As shown in the figure, point $D$ is on side $BC$ of $\triangle ABC$, with $AC \parallel BE$, $BC = BE$, and $\angle ABC = \angle E$. $\triangle ABC \cong \triangle DEB$; if $BE = 9$ and $AC = 4$, find the length of $CD$.

Solution:
\textbf{Solution:} Since $\triangle ABC\cong \triangle DEB$,\\
therefore $AC=DB=4$,\\
therefore $CD=BC-BD=9-4=\boxed{5}$.


Question: As shown in the figure, in an equilateral $\triangle ABC$, points $E$ and $D$ are located on sides $AC$ and $BC$, respectively, with $AE=CD$. Lines $AD$ and $BE$ intersect at point $O$. $\triangle ABE \cong \triangle CAD$; If $\angle OBD=45^\circ$, find the measure of $\angle ADC$.

Solution:
\textbf{Solution:} From (1) we have $\triangle ABE \cong \triangle CAD$,\\
$\therefore \angle ABE = \angle CAD$,\\
$\therefore \angle BOD = \angle ABE + \angle BAO = \angle BAC = 60^\circ$,\\
$\therefore \angle ADC = \angle OBD + \angle BOD = 45^\circ + 60^\circ = \boxed{105^\circ}$.


Question: As shown in Figure (1), it is known that the line $y=-2x+4$ intersects the $x$-axis and the $y$-axis at points $A$ and $C$ respectively. A rectangle $OABC$ is constructed in the first quadrant with $OA$ and $OC$ as sides. What are the coordinates of points $A$ and $C$?

Solution: \textbf{Solution:} The coordinates of point $A$ are \boxed{(2,0)}; the coordinates of point $C$ are \boxed{(0,4)}.


Question: As shown in the figure, it is known that the line $y=-2x+4$ intersects the $x$-axis and $y$-axis at points $A$ and $C$ respectively. A rectangle $OABC$ is constructed in the first quadrant with $OA$ and $OC$ as sides. $\triangle ABC$ is folded so that point $A$ coincides with point $C$, and the fold intersects $AB$ at point $D$. Find the equation of line $CD$.

Solution: \textbf{Solution:} Given from the fold: $CD=AD$. Let $AD=x$, then $CD=x$, $BD=4-x$. According to the given conditions, we have: $(4-x)^{2}+2^{2}=x^{2}$. Solving this gives: $x=\frac{5}{2}$. At this point, $AD=\frac{5}{2}$, $D\left(2, \frac{5}{2}\right)$. Let the equation of line $CD$ be $y=kx+b$, substituting $D\left(2, \frac{5}{2}\right)$ into it yields $\frac{5}{2}=2k+b$. Solving this gives: $k=-\frac{3}{4}$, $b=4$. $\therefore$ The equation of line $CD$ is $\boxed{y=-\frac{3}{4}x+4}$.


Question: As shown in Figure 1, the straight line $AB$ intersects the coordinate axes at points $A(0, -3)$ and $B(-5, 0)$, respectively. Point $C$ is the midpoint of line segment $AB$, and point $P$ is a point on the y-axis. Connecting $CP$, and drawing a perpendicular line from point $C$ to line segment $BO$ intersecting at point $Q$. What is the analytic equation of line $AB$?

Solution: \textbf{Solution:} Let the equation of line AB be $y=kx+b$ ($k\neq 0$). Given $A(0, -3)$, $B(-5, 0)$, we have $\begin{cases}b=-3\\ -5k+b=0\end{cases}$. Solving these equations yields $\begin{cases}k=-\frac{3}{5}\\ b=-3\end{cases}$. Therefore, the equation of line AB is $\boxed{y=-\frac{3}{5}x-3}$.


Question: As shown in Figure 1, line AB intersects the coordinate axes at points A $(0, -3)$ and B $(-5, 0)$, respectively. Point C is the midpoint of line segment AB, and point P is a point on the y-axis. Connect CP, and draw a perpendicular from C to line segment BO intersecting at point Q. As shown in Figure 2, when point Q coincides with point B, connect PQ. Find the length of PO.

Solution:
\[ \text{Solution: From the given, point } C \text{ is the midpoint of segment } AB, \]
\[ \therefore C\left( -\frac{5}{2}, -\frac{3}{2} \right), \]
\[ \text{Let } P(0,t), \]
\[ \because CP\perp CQ \text{ and point } Q \text{ coincides with point } B, \]
\[ \therefore CP \text{ is the perpendicular bisector of segment } AB, \]
\[ \therefore PA=PB, \]
\[ \text{In } \triangle PBO, PB^2=OB^2+OP^2, OB=5, OP=t, \]
\[ \therefore PB^2=5^2+t^2=25+t^2, \]
\[ \because PA=OP+OA=t+3, \]
\[ \therefore (t+3)^2=25+t^2, \]
\[ \text{Solving, we get: } t= \frac{8}{3}, \]
\[ \therefore P\left(0, \frac{8}{3}\right), \]
\[ \therefore \text{The length of } PO \text{ is } \boxed{\frac{8}{3}}. \]


Question: As shown in the figure, the line $AB$ intersects the coordinate axes at points $A(0, -3)$ and $B(-5, 0)$, respectively. Point $C$ is the midpoint of segment $AB$, and point $P$ is a point on the $y$-axis. Connecting $CP$, a perpendicular line to $CP$ is drawn from point $C$ intersecting segment $BO$ at point $Q$. Let $AP=m$ and $BQ=n$. Find the expression of $m$ as a function of $n$.

Solution:
\textbf{Solution:} Let line $CE\perp OA$ at point E, and $CF\perp OB$ at point F,\\
then $E(0,-\frac{3}{2})$, $F(-\frac{5}{2},0)$, $\angle CEP=\angle CFQ=90^\circ$,\\
because $AP=m$, $BQ=n$,\\
thus $OQ=5-n$, $OP=3-m$,\\
therefore $QF=OQ-OF=5-n-\frac{5}{2}=\frac{5}{2}-n$, $PE=OE-OP=\frac{3}{2}-(3-m)=m-\frac{3}{2}$,\\
because $\angle CEP+\angle AOB=180^\circ$,\\
therefore $CE\parallel OB$,\\
therefore $\angle ECF=\angle CFQ=90^\circ$,\\
therefore $\angle PCE+\angle PCF=90^\circ$,\\
because $\angle QCF+\angle PCF=90^\circ$,\\
therefore $\angle PCE=\angle QCF$,\\
therefore $\triangle PCE\sim \triangle QCF$,\\
therefore $\frac{PE}{CE}=\frac{QF}{CF}$, i.e., $\frac{m-\frac{3}{2}}{\frac{5}{2}}=\frac{\frac{5}{2}-n}{\frac{3}{2}}$,\\
therefore $m=-\frac{5}{3}n+\frac{17}{3}$, hence the function expression of $m$ in terms of $n$ is $m=\boxed{-\frac{5}{3}n+\frac{17}{3}}$.


Question: As shown in the figure, $C$ is the midpoint of $\overset{\frown}{ACB}$, $\angle AOC=4\angle B$, and $OC=4$. What is the degree measure of $\angle A$?

Solution:
\textbf{Solution:} As shown in the figure, connect $AC, BC$\\
$\because \overset{\frown}{AB}=\overset{\frown}{AB}$,\\
$\therefore \angle AOB=2\angle ACB$,\\
$\because OA=OB$,\\
$\therefore \angle OAB=\angle OBA$,\\
$\therefore \angle AOB=180^\circ-2\angle B$,\\
$\because C$ is the midpoint of $\overset{\frown}{ACB}$,\\
$\therefore \overset{\frown}{AC}=\overset{\frown}{BC}$,\\
$\therefore \angle AOC=\angle BOC$,\\
$\because \angle AOC=4\angle B$,\\
$\therefore \angle AOC+\angle BOC+\angle AOB=360^\circ$,\\
that is, $4\angle B+4\angle B+180^\circ-2\angle B=360^\circ$,\\
$\therefore 6\angle B=180^\circ$,\\
$\therefore \angle A=\angle B=30^\circ$;\\
Thus, the measure of $\angle A$ is $\boxed{30^\circ}$.


Question: As shown in the figure, $C$ is the midpoint of $\overset{\frown}{ACB}$, $\angle AOC = 4\angle B$, and $OC = 4$. What is the length of line segment $AB$?

Solution:
\textbf{Solution:} As shown in the figure, extend $CO$ to intersect $AB$ at point D,\\
$\because C$ is the midpoint of $\overset{\frown}{ACB}$,\\
$\therefore \overset{\frown}{AC}=\overset{\frown}{BC}$, $CD\perp AB$, $AD=DB$,\\
$\because OC=4$, $\angle A=30^\circ$,\\
$\therefore AO=4, OD=2$,\\
$\therefore AD=\sqrt{AO^{2}-OD^{2}}=\sqrt{4^{2}-2^{2}}=2\sqrt{3}$,\\
$\therefore AB=2AD=4\sqrt{3}=\boxed{4\sqrt{3}}$.


Question: As shown in the diagram, quadrilateral $ABCD$ is inscribed in circle $O$, $\angle ABC=135^\circ$, $OE \perp AC$, $\angle AOE = \angle D$. If $AC=4$, what is the radius of circle $O$?

Solution:
\textbf{Solution:} From (1), we know that $\triangle AOC$ is an isosceles right triangle, \\
because $AC=4$, \\
therefore $OA=OC=2\sqrt{2}$, \\
which means the radius of the circle $O$ is $\boxed{2\sqrt{2}}$.


Question: As shown in the figure, within $\odot O$, points B and C divide $\overset{\frown}{AD}$ into three equal parts. The chords $AC$ and $BD$ intersect at point E. $AC=BD$; $AB$ is connected. If $\angle BAC=25^\circ$, what is the degree measure of $\angle BEC$?

Solution:
\textbf{Solution:} Given $\because \angle BAC=25^\circ$, $\overset{\frown}{AB}=\overset{\frown}{BC}=\overset{\frown}{CD}$, \\
$\therefore \angle BAC=\angle CAD=\angle BDA=25^\circ$, \\
$\because \angle AED+\angle CAD+\angle BDA=180^\circ$, \\
$\therefore \angle AED=180^\circ-\angle CAD-\angle BDA=130^\circ$, \\
$\therefore \angle BEC=\angle AED=\boxed{130^\circ}$.


Question: As shown in the figure, $AD$ is the angle bisector of the exterior angle $\angle EAC$ of $\triangle ABC$, intersecting the circumcircle $\odot O$ of $\triangle ABC$ at point D, with $\angle EAC=120^\circ$. What is the degree measure of arc $\overset{\frown}{BC}$?

Solution:
\textbf{Solution:} Draw lines OB and OC,

since $\angle EAC = 120^\circ$,

therefore $\angle BAC = 180^\circ - \angle EAC = 60^\circ$,

therefore $\angle BDC = \angle BAC = 60^\circ$,

therefore $\angle BOC = 2\angle BDC = 2 \times 60^\circ = 120^\circ$,

thus, the degree measure of arc $\overset{\frown}{BC}$ is $\boxed{120^\circ}$;


Question: As shown in the figure, AB is the diameter of $\odot O$, and CD is a chord with AB $\perp$ CD at point E. Connect AC, OC, and BC. If AE = 2cm and BE = 8cm, what is the length of chord CD in cm?

Solution:
\textbf{Solution:} Given $AE=2\text{cm}$ and $BE=8\text{cm}$,\\
it follows that $OA=5\text{cm}$ and $OE=3\text{cm}$,\\
In $\triangle OCE$, $CE= \sqrt{5^2-3^2}=4\text{cm}$,\\
Since $AB\perp CD$,\\
it implies $CE=DE$,\\
Therefore, $CD=2CE=\boxed{8\text{cm}}$;


Question: As shown in the figure, it is known that $AB$ is the diameter of $\odot O$, $\angle ACD$ is the angle subtended by arc $AD$ on the circumference, $\angle ACD=30^\circ$. What is the measure of $\angle DAB$ in degrees?

Solution:
\textbf{Solution:} Connect BD, \\
$\because$ AB is the diameter of the circle, \\
$\therefore \angle ADB=90^\circ$, \\
$\because \angle ACD=\angle B=30^\circ$, \\
$\therefore \angle DAB=90^\circ-\angle B=90^\circ-30^\circ=\boxed{60^\circ}$.