Question: As shown in the figure, $OC \perp AB$ at point O, $OD$ bisects $\angle BOC$, and $OE$ 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, point $C$ is a point on line $AB$, and point $M$ is the midpoint of segment $AC$. If $AB=8$, point $C$ is on segment $AB$, and $AC=3BC$, then what is the length of $AM$?

Solution: \textbf{Solution:} Since the information given in the problem is insufficient, it is not possible to provide a solution. Hence, the length of $AM$ cannot be determined. Therefore, the length of $AM$ is \boxed{\text{undetermined}}.


Question: As shown in the figure, point $C$ is a point on line $AB$, and point $M$ is the midpoint of segment $AC$. If $AB=a$, and $AB-AC=\frac{a}{3}$, find the length of $BM$ (expressed in an algebraic expression with $a$).

Solution:
\textbf{Solution:}

(1) When point $C$ lies between points $A$ and $B$, \\
$\because BC=AB-AC=\frac{a}{3}$, \\
$\therefore AC=AB-BC=a-\frac{a}{3}=\frac{2a}{3}$, \\
$\because$ point $M$ is the midpoint of $AC$, \\
$\therefore CM=\frac{1}{2}AC=\frac{1}{2}\times \frac{2a}{3}=\frac{a}{3}$, \\
$\therefore BM=CM+BC=\frac{a}{3}+\frac{a}{3}=\boxed{\frac{2a}{3}}$, \\
(2) When point $C$ is to the left of point $A$, \\
$\because AC=AB-\frac{a}{3}=\frac{2a}{3}$, \\
Additionally, $\because$ point $M$ is the midpoint of $AC$, \\
$\therefore AM=\frac{1}{2}AC=\frac{1}{2}\times \frac{2a}{3}=\frac{a}{3}$, \\
$\therefore BM=AM+AB=\frac{a}{3}+a=\boxed{\frac{4a}{3}}$.


Question: As shown in Figure 1, it is known that $\angle BOC = 40^\circ$, $OE$ bisects $\angle AOC$, and $OF$ bisects $\angle BOC$. If $AO \perp BO$, then what is the measure of $\angle EOF$ in degrees?

Solution:
\textbf{Solution:} Given $\because$ $\angle BOC=40^\circ$ and $AO\perp BO$,\\
$\therefore$ $\angle AOB=90^\circ, \angle AOC=\angle AOB+\angle BOC=130^\circ$,\\
$\because$ $OE$ bisects $\angle AOC$ and $OF$ bisects $\angle BOC$,\\
$\therefore$ $\angle EOC=\frac{1}{2}\angle AOC=65^\circ, \angle COF=\frac{1}{2}\angle BOC=20^\circ$,\\
$\therefore$ $\angle EOF=\angle EOC−\angle COF=65^\circ−20^\circ=\boxed{45^\circ}$.


Question: As shown in the figure, it is known that the straight lines $AB$ and $CD$ intersect at point $O$, and $OE$ bisects $\angle AOD$. If $\angle AOC = 50^\circ$, find the measure of $\angle BOE$.

Solution:
\textbf{Solution:} Given $\because \angle AOC=50^\circ$, \\
$\therefore \angle BOD=\angle AOC=50^\circ$ \\
$\therefore \angle AOD=180^\circ-\angle AOC=180^\circ-50^\circ=130^\circ$\\
Since $OE$ bisects $\angle AOD$\\
$\therefore \angle DOE=\frac{1}{2}\angle AOD=\frac{1}{2}\times 130^\circ=65^\circ$\\
Therefore, $\angle BOE=\angle BOD+\angle DOE=50^\circ+65^\circ=\boxed{115^\circ}$


Question: As shown in Figure (1), it is known that $AM \parallel CN$, and point B is a point in the plane, with $AB \perp BC$ at point B. A line is drawn through point B such that $BD \perp AM$ at point D, where $\angle BCN = \alpha$. If $\alpha = 30^\circ$, what is the measure of $\angle ABD$ in degrees?

Solution:
\textbf{Solution:} Extend $DB$ to intersect $NC$ at point H, as shown in the diagram,\\
$\because AM\parallel CN$ and $BD\perp AM$,\\
$\therefore DH\perp NC$.\\
$\therefore \angle BHC=90^\circ$.\\
$\because \angle BCN=\alpha =30^\circ$,\\
$\therefore \angle HBC=90^\circ-\angle BCN=60^\circ$.\\
$\because AB\perp BC$,\\
$\therefore \angle ABC=90^\circ$.\\
$\therefore \angle ABD=180^\circ-\angle ABC-\angle HBC=\boxed{30^\circ}$;


Question: As shown in Figure (1), it is known that $AM\parallel CN$, where point B is a point in the plane and $AB\perp BC$ at point B. Through point B, construct $BD\perp AM$ at point D, and let $\angle BCN=\alpha$. As shown in Figure (2), if points E and F are on $DM$, and $BE$, $BF$, and $CF$ are connected such that $BE$ bisects $\angle ABD$ and $BF$ bisects $\angle DBC$, what is the measure of $\angle EBF$ in degrees?

Solution:
\textbf{Solution:} Extend $DB$, intersect $NC$ at point H, as shown,\\
since $AM\parallel CN$, $BD\perp AM$,\\
therefore $DH\perp NC$.\\
Therefore $\angle BHC=90^\circ$.\\
Since $\angle BCN=\alpha $,\\
therefore $\angle HBC=90^\circ-\alpha $.\\
Since $AB\perp BC$,\\
therefore $\angle ABC=90^\circ$.\\
Therefore $\angle ABD=180^\circ-\angle ABC-\angle HBC=\alpha $.\\
Since $BE$ bisects $\angle ABD$,\\
therefore $\angle DBE=\angle ABE=\frac{1}{2}\alpha $.\\
Since $\angle HBC=90^\circ-\alpha $,\\
therefore $\angle DBC=180^\circ-\angle HBC=90^\circ+\alpha $.\\
Since $BF$ bisects $\angle DBC$,\\
therefore $\angle DBF=\angle CBF=\frac{1}{2}\angle DBC=45^\circ+\frac{1}{2}\alpha $.\\
Therefore $\angle EBF=\angle DBF-\angle DBE=45^\circ+\frac{1}{2}\alpha -\frac{1}{2}\alpha =\boxed{45^\circ}$;


Question: As shown in Figure (1), it is known that \( AM\parallel CN \), with B being a point on the plane, and \( AB\perp BC \) at point B. A line is drawn through point B such that \( BD\perp AM \) at point D, and let \( \angle BCN=\alpha \). As in Figure (3), under the conditions of question (2), if \( CF \) bisects \(\angle BCH\), and \(\angle BFC=3\angle BCN\), find the degree measure of \(\angle EBC\).

Solution:
\textbf{Solution:} Given the construction, we have
\begin{align*}
\because \angle BCN &= \alpha, \\
\therefore \angle HCB &= 180^\circ - \angle BCN = 180^\circ - \alpha. \\
\because CF &\text{ bisects } \angle BCH, \\
\therefore \angle BCF &= \angle HCF = \frac{1}{2}\angle HCB = 90^\circ - \frac{1}{2}\alpha. \\
\because AM &\parallel CN, \\
\therefore \angle DFC &= \angle HCF = 90^\circ - \frac{1}{2}\alpha. \\
\because \angle BFC &= 3\angle BCN, \\
\therefore \angle BFC &= 3\alpha. \\
\therefore \angle DFB &= \angle DFC - \angle BFC = 90^\circ - \frac{7}{2}\alpha. \\
From (2) we know: & \angle DBF = 45^\circ + \frac{1}{2}\alpha. \\
\because BD &\perp AM, \\
\therefore \angle D &= 90^\circ. \\
\therefore \angle DBF + \angle DFB &= 90^\circ. \\
\therefore 45^\circ + \frac{1}{2}\alpha + 90^\circ - \frac{7}{2}\alpha &= 90^\circ. \\
Solving, we find: & \alpha = \boxed{15^\circ}. \\
\therefore \angle FBC &= \angle DBF = 45^\circ + \alpha = 60^\circ. \\
\therefore \angle EBC &= \angle FBC + \angle EBF = 60^\circ + 37.5^\circ = \boxed{97.5^\circ}.
\end{align*}


Question: As shown in the figure, it is known that $O$ is a point on the line $AB$, $\angle COD$ is a right angle, $OE$ bisects $\angle BOC$. If $\angle AOC = 30^\circ$, what is the degree measure of $\angle DOE$?

Solution:
\textbf{Solution:}\\
$\because \angle AOC=30^\circ$,\\
$\therefore \angle BOC=180^\circ-30^\circ=150^\circ$,\\
$\because$ OE bisects $\angle BOC$,\\
$\therefore \angle COE= \frac{1}{2} \angle BOC=75^\circ$,\\
$\because \angle COD$ is a right angle,\\
$\therefore \angle COD=90^\circ$,\\
$\therefore \angle DOE=90^\circ-75^\circ=\boxed{15^\circ}$.


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

Solution:
\textbf{Solution:} Given that $\angle AOC = \alpha$ \\
Therefore, $\angle BOC = 180^\circ - \alpha$ \\
Since OE bisects $\angle BOC$ \\
Then, $\angle COE = \frac{1}{2} \angle BOC = 90^\circ - \frac{1}{2} \alpha$ \\
Given that $\angle COD$ is a right angle \\
Therefore, $\angle COD = 90^\circ$ \\
Hence, $\angle DOE = 90^\circ - (90^\circ - \frac{1}{2} \alpha) = \boxed{\frac{1}{2} \alpha}$


Question: It is known that $O$ is a point on line $AB$, $\angle COD$ is a right angle, and $OE$ bisects $\angle BOC$. In the process where $\angle DOC$ in Figure 1 is rotated clockwise around vertex $O$ to the position in Figure 2, with ray $OC$ remaining above line $AB$, at what degree of $\angle AOC$ will $\angle COE = 2\angle DOB$?

Solution:
\textbf{Solution:}

(1) When OD is above the line AB, let $\angle DOB=\alpha$, then $\angle COE=2\alpha$.\\
Since OE bisects $\angle BOC$,\\
it follows that $\angle COE=\angle BOE=2\alpha$.\\
Since $\angle DOE=2\alpha-\alpha=\alpha$,\\
it follows that $\angle COD=90^\circ=\angle COE+\angle DOE=2\alpha+\alpha=3\alpha$.\\
Thus, $\alpha=30^\circ$.\\
Therefore, from (2) we have $\angle AOC=2\angle DOE=60^\circ$.\\
(2) When OD is below the line AB, let $\angle DOB=\alpha$, then $\angle COE=2\alpha$.\\
Since OE bisects $\angle BOC$,\\
it follows that $\angle COE=\angle BOE=2\alpha$.\\
Since $\angle COD=90^\circ=\angle COE+\angle BOE+\angle DOB=5\alpha$,\\
it follows that $\alpha=18^\circ$.\\
Thus, $\angle COE+\angle BOE=4\alpha=72^\circ$.\\
Therefore, $\angle AOC=180^\circ-72^\circ=\boxed{108^\circ}$.


Question: As shown in the figure, point O is a point on line AB, with $\angle BOC : \angle AOC = 1 : 2$. OD bisects $\angle BOC$, and OE is perpendicular to OD at point O. What is the measure of $\angle BOC$ in degrees?

Solution: \textbf{Solution:} Since $\angle AOB = \angle BOC + \angle AOC = 180^\circ$ and $\angle BOC : \angle AOC = 1 : 2$, it follows that $\angle AOC = 2\angle BOC$. Therefore, $\angle BOC + 2\angle BOC = 180^\circ$, which implies $\angle BOC = \boxed{60^\circ}$.


Question: As shown in the figure, it is known that $AB\parallel CD$, $AD\parallel BC$, $\angle DCE=90^\circ$, point E is on segment AB, $\angle FCG=90^\circ$, point F is on line AD, $\angle AHG=90^\circ$. If $\angle ECF=25^\circ$, what is the degree measure of $\angle BCD$?

Solution:
\textbf{Solution:} Since $\angle ECF=25^\circ$ and $\angle DCE=90^\circ$,

$\therefore \angle FCD=65^\circ$. Also, since $\angle BCF=90^\circ$,

$\therefore \angle BCD=65^\circ+90^\circ=\boxed{155^\circ}$.


Question: As shown in the figure, it is known that $AB\parallel CD$, $AD\parallel BC$, $\angle DCE=90^\circ$, point E is on the line segment AB, $\angle FCG=90^\circ$, point F is on the line AD, $\angle AHG=90^\circ$. Given the conditions in (2), point C (point C does not coincide with points B and H) starts from point B and moves in the direction of ray BG, with other conditions unchanged, find the degree measure of $\angle BAF$?

Solution:
\textbf{Solution:} Discuss in two cases:\\
(1) As shown in figure a, when point C is on segment BH, point F is on the extension of DA, then $\angle ECF=\angle DCG=\angle B=25^\circ$.\\
Since AD$\parallel$BC,\\
therefore $\angle BAF=\angle B=25^\circ$;\\
(2) As shown in figure b, when point C is on the extension of BH, point F is on segment AD.\\
Since $\angle B=25^\circ$, AD$\parallel$BC,\\
therefore $\angle BAF=180^\circ-25^\circ=155^\circ$.\\
In summary, the degree of $\angle BAF$ is $\boxed{25^\circ or 155^\circ}$.


Question: As shown in the figure, it is known that lines $AB$ and $CD$ intersect at point $O$, and $\angle COE = 90^\circ$. If $\angle AOC = 37^\circ$, what is the measure of $\angle BOE$?

Solution:
\textbf{Solution:} Given that $\because \angle COE=90^\circ$, $\angle AOC=37^\circ$, \\
$\therefore \angle BOE=180^\circ-\angle AOC-\angle COE$ \\
$=180^\circ-37^\circ-90^\circ$ \\
$=\boxed{53^\circ}$.


Question: As shown in the figure, it is known that lines $AB$ and $CD$ intersect at point $O$, and $\angle COE=90^\circ$. If $\angle BOD:\angle BOC = 3:6$, what is the degree measure of $\angle AOE$?

Solution:
\textbf{Solution:}

Given that $\angle BOD : \angle BOC = 3 : 6$,\\
and $\angle BOD + \angle BOC = 180^\circ$,\\
hence, $\angle BOD = 60^\circ$,\\
Since $\angle BOD = \angle AOC$,\\
it follows that $\angle AOC = 60^\circ$,\\
Given $\angle COE = 90^\circ$,\\
thus, $\angle AOE = \angle COE + \angle AOC = 90^\circ + 60^\circ = \boxed{150^\circ}$.


Question: Understanding Calculation: As shown in Figure (1), $ \angle AOB=80^\circ$, $ \angle AOC=40^\circ$. Ray $ OM$ bisects $ \angle BOC$, $ ON$ bisects $ \angle AOC$, find the degree measure of $ \angle MON$?

Solution:
\textbf{Solution:} Given $\because \angle BOC=\angle AOB+\angle AOC=80^\circ+40^\circ=120^\circ$,\\
ray $OM$ bisects $\angle BOC$,\\
$\therefore \angle COM=\frac{1}{2}\angle BOC=\frac{1}{2}\times 120^\circ=60^\circ$,\\
$\because ON$ bisects $\angle AOC$,\\
$\therefore \angle CON=\frac{1}{2}\angle AOC=\frac{1}{2}\times 40^\circ=20^\circ$,\\
$\therefore \angle MON=\angle COM-\angle CON=60^\circ-20^\circ=\boxed{40^\circ}$.


Question: Extended exploration: As shown in the figure, $ \angle AOB = \alpha $, $ \angle AOC = \beta $ ($\alpha$ and $\beta$ are acute angles). Ray $OM$ bisects $\angle BOC$, and ray $ON$ bisects $\angle AOC$. What is the degree measure of $\angle MON$?

Solution:
\textbf{Solution:}

\[
\because \angle BOC = \angle AOB + \angle AOC = \alpha + \beta,
\]
\[
\because \text{ray } OM \text{ bisects } \angle BOC,
\]
\[
\therefore \angle COM = \frac{1}{2}\angle BOC = \frac{1}{2}(\alpha + \beta),
\]
\[
\because \text{ray } ON \text{ bisects } \angle AOC,
\]
\[
\therefore \angle CON = \frac{1}{2}\angle AOC = \frac{1}{2}\beta,
\]
\[
\therefore \angle MON = \angle COM - \angle CON = \frac{1}{2}(\alpha + \beta) - \frac{1}{2}\beta = \boxed{\frac{1}{2}\alpha}.
\]


Question: As shown in the figure, the length of the line segment $AB$ is $a$. Extend the line segment $AB$ to $C$ so that $BC=b$. Let points $M$ and $N$ be the midpoints of $AC$ and $BC$, respectively. Find the length of $MN$.

Solution:
\textbf{Solution:} Given that $AB=a$ and $BC=b$,\\
thus $AC=AB+BC=a+b$,\\
and since points $M$ and $N$ are the midpoints of $AC$ and $BC$ respectively,\\
thus $CM=\frac{1}{2}AC=\frac{1}{2}(a+b)$ and $CN=\frac{1}{2}BC=\frac{1}{2}b$,\\
thus $MN=CM-CN=\frac{1}{2}(a+b)-\frac{1}{2}b=\boxed{\frac{1}{2}a}$.


Question: As shown in the figure, in triangle $ABC$, the angle bisectors of $\angle ABC$ and $\angle ACB$ intersect at point $P$. When $\angle A=60^\circ$, what is the measure of $\angle BPC$ in degrees?

Solution:
\textbf{Solution:} As shown in the diagram, since BP is the bisector of $\angle ABC$,\\
then $\angle PBC = \frac{1}{2}\angle ABC$. (Definition of an angle bisector)\\
Since CP is the bisector of $\angle ACB$,\\
then $\angle PCB = \frac{1}{2}\angle ACB$,\\
thus $\angle PBC + \angle PCB = \frac{1}{2}(\angle ACB + \angle ABC)$,\\
Since $\angle A = 60^\circ$,\\
then $\angle ACB + \angle ABC = 120^\circ$,\\
thus $\angle PBC + \angle PCB = \frac{1}{2}\left(\angle ACB + \angle ABC\right) = 60^\circ$,\\
therefore $\angle BPC = 180^\circ - \angle PBC - \angle PCB = 180^\circ - (\angle PBC + \angle PCB) = 180^\circ - 60^\circ = \boxed{120^\circ}.$


Question: Given $\angle AOB$ and $\angle COD$, the ray $OE$ bisects $\angle AOD$; as shown in Figure 1, given $\angle AOB = 180^\circ$ and $\angle COD = 90^\circ$, if $\angle DOB = 46^\circ$, what is the measure of $\angle COE$ in degrees?

Solution:
\textbf{Solution:} Given $\angle AOB=180^\circ$ and $\angle DOB=46^\circ$, \\
it follows that $\angle AOD=134^\circ$. \\
Since $OE$ bisects $\angle AOD$, \\
we have $\angle EOD=\frac{1}{2}\angle AOD=\frac{1}{2}\times 134^\circ=67^\circ$. \\
Given $\angle COD=90^\circ$, \\
it follows that $\angle COE=\angle COD-\angle EOD=90^\circ-67^\circ=\boxed{23^\circ}$.


Question: Given $ \angle AOB$ and $ \angle COD$, the ray $ OE$ bisects $ \angle AOD$; the positions of $ \angle AOB$ and $ \angle COD$ are as shown in Figure 2. Given that $ \angle COD = \frac{1}{2}\angle AOB$, find the value of $ \angle COE : \angle DOB$.

Solution:
\textbf{Solution:} From the given information, we have\\
$\because OE$ bisects $\angle AOD$,\\
$\therefore \angle EOD=\frac{1}{2}\left(\angle AOB+\angle BOD\right)$\\
$\because \angle COD=\frac{1}{2}\angle AOB$,\\
$\therefore \angle EOC=\angle EOD-\angle COD$\\
$=\frac{1}{2}\angle AOB+\frac{1}{2}\angle BOD-\frac{1}{2}\angle AOB$\\
$=\frac{1}{2}\angle BOD$\\
$\therefore \angle COE\colon \angle DOB=\boxed{1\colon 2}$


Question: As shown in Figure (1), given point $O$ is the intersection point of lines $AB$ and $CD$, $\angle BOD=32^\circ$, $OE$ bisects $\angle AOD$, $\angle EOF=90^\circ$, find the degree measure of $\angle COF$.

Solution:
Solution: $\because$ $\angle BOD=32^\circ$,\\
$\therefore$ $\angle AOD=180^\circ-\angle BOD=180^\circ-32^\circ=148^\circ$.\\
$\because$ $OE$ bisects $\angle AOD$,\\
$\therefore$ $\angle DOE=\frac{1}{2}\angle AOD=74^\circ$.\\
$\because$ $\angle EOF=90^\circ$,\\
$\therefore$ $\angle COF=180^\circ-\angle DOE-\angle EOF=180^\circ-74^\circ-90^\circ=\boxed{16^\circ}$


Question: As shown in the figure, it is known that the length of line segment $AB=10$. Extend $AB$ to point $C$ such that $BC=2AB$. Let points $M$ and $N$ be the midpoints of line segments $AC$ and $AB$, respectively. Find the length of $MN$.

Solution:
\textbf{Solution:} Given that $BC=2AB$ and $AB=10$,\\
then $BC=2\times 10=20$.\\
Therefore, $AC=AB+BC=10+20=30$.\\
Since points $M$ and $N$ are the midpoints of line segments $AC$ and $AB$ respectively,\\
then $AM=\frac{1}{2}AC=\frac{1}{2}\times 30=15$.\\
And $AN=\frac{1}{2}AB=\frac{1}{2}\times 10=5$.\\
Therefore, $MN=AM-AN=15-5=\boxed{10}$.


Question: As shown in the figure, O is a point on the line AB, $\angle AOC=50^\circ$, OD bisects $\angle AOC$, and $\angle DOE=90^\circ$. What is the measure of $\angle COD$?

Solution:
\textbf{Solution:} Since $\angle AOC=50^\circ$ and $OD$ bisects $\angle AOC$,

$\therefore \angle COD=\angle DOA= \frac{1}{2}\angle AOC=\boxed{25^\circ}$,


Question: As shown in the figure, $O$ is a point on the line $AB$, $\angle AOC=50^\circ$, $OD$ bisects $\angle AOC$, and $\angle DOE=90^\circ$. What is the measure of $\angle BOD$ in degrees?

Solution:
\textbf{Solution:} Since $\angle AOD + \angle BOD = 180^\circ$,\\
it follows that $\angle BOD = 180^\circ - \angle AOD = \boxed{155^\circ}$.


Question: Given $\angle AOB=40^\circ$, $OD$ is the bisector of $\angle BOC$. As shown in Figure 1, when $\angle AOB$ and $\angle BOC$ are complementary, what is the degree measure of $\angle COD$?

Solution:
\textbf{Solution:} Given that $\angle AOB$ and $\angle BOC$ are complementary,\\
it follows that $\angle AOB+\angle BOC=180^\circ$,\\
therefore $\angle BOC=180^\circ-40^\circ=140^\circ$,\\
since $OD$ is the bisector of $\angle BOC$,\\
it results in $\angle COD=\frac{1}{2}\angle BOC=\boxed{70^\circ}$;


Question: As shown in the figure, it is known that $\angle AOB = 40^\circ$, and $OD$ is the bisector of $\angle BOC$. When $\angle AOB$ and $\angle BOC$ are complementary, what is the degree measure of $\angle COD$?

Solution:
\textbf{Solution:} Since $\angle AOB$ and $\angle BOC$ are complementary,\\
it follows that $\angle AOB+\angle BOC=90^\circ$,\\
hence $\angle BOC=90^\circ-40^\circ=50^\circ$,\\
since $OD$ bisects $\angle BOC$,\\
it results in $\angle COD=\frac{1}{2}\angle BOC=\boxed{25^\circ}$.


Question: As shown in the figure, OC is the bisector of $\angle AOB$, and $\angle COD=20^\circ$. If $\angle AOD=30^\circ$, what is the measure of $\angle AOB$?

Solution:
\textbf{Solution:} Since $\angle COD=20^\circ$ and $\angle AOD=30^\circ$,\\
$\therefore \angle AOC=\angle COD+\angle AOD=20^\circ+30^\circ=50^\circ$,\\
Since OC is the bisector of $\angle AOB$,\\
$\therefore \angle AOB=2\angle AOC=\boxed{100^\circ}$.


Question: As shown in the figure, OC is the bisector of $\angle AOB$, and $\angle COD = 20^\circ$. If $\angle BOD = 2\angle AOD$, what is the degree measure of $\angle AOB$?

Solution:
\textbf{Solution:} Let $\angle AOD=x$, then $\angle BOD=2x$,\\
thus $\angle AOB=\angle AOD+\angle BOD=3x$,\\
since $OC$ is the bisector of $\angle AOB$,\\
thus $\angle AOC=\frac{1}{2}\angle AOB=\frac{3}{2}x$,\\
thus $\frac{3}{2}x-x=20^\circ$,\\
solving for $x=40^\circ$,\\
thus $\angle AOB=3x=\boxed{120^\circ}$.


Question: As shown in the figure, line AE intersects with CD at point B, and BF $\perp$ AE. If $\angle DBE=60^\circ$, what is the degree measure of $\angle FBD$?

Solution:
\textbf{Solution:} Since BF$\perp$AE,\\
it follows that $\angle$DBF + $\angle$DBE = $90^\circ$,\\
since $\angle$DBE = $60^\circ$,\\
thus $\angle$DBF = $90^\circ$ - $\angle$DBE = $30^\circ$.\\
Hence, the measure of $\angle FBD$ is $\boxed{30^\circ}$.


Question: As shown in the figure, lines AB and CD intersect at point O, $\angle DOE = 90^\circ$, OD bisects $\angle BOF$, $\angle BOE = 50^\circ$. What is the degree measure of $\angle AOC$?

Solution:
\textbf{Solution:} Given that $\angle DOE=90^\circ$, $\angle BOE=50^\circ$,\\
$\therefore \angle BOD=\angle DOE-\angle BOE=40^\circ$, $\angle AOD=180^\circ-\angle BOD=140^\circ$,\\
$\therefore \angle AOC=180^\circ-\angle AOD=\boxed{40^\circ}$.


Question: As shown in the figure, line AB and CD intersect at point O, $ \angle DOE = 90^\circ$, OD bisects $ \angle BOF$, $ \angle BOE = 50^\circ$. What is the degree measure of $ \angle EOF$?

Solution:
\textbf{Solution:} Since $OD$ bisects $\angle BOF$, \\
$\therefore \angle DOF=\angle BOD=40^\circ$, \\
$\therefore \angle EOF=\angle DOF+\angle DOE=40^\circ+90^\circ=\boxed{130^\circ}$.


Question: As shown in the figure, line $AB$ and line $CD$ intersect at point $O$, $\angle DOE = 90^\circ$, $OD$ bisects $\angle BOF$, $\angle BOE = 50^\circ$. What is the measure of $\angle AOF$ in degrees?

Solution:
\textbf{Solution:} From the given conditions, we have
\[
\angle AOF = 180^\circ - \angle AOC - \angle DOF = 180^\circ - 40^\circ - 40^\circ = \boxed{100^\circ}.
\]


Question: As shown in the figure, it is known that points $B$ and $C$ divide the line segment $AD$ into three parts in the ratio $2:4:3$, with $M$ being the midpoint of $AD$. If $CD=6$, find the length of line segment $MC$.

Solution:
\textbf{Solution:} Given that points B and C divide the line segment AD into three parts in the ratio 2:4:3, that is, 2+4+3=9,\\
it follows that $CD=\frac{1}{3}AD$,\\
Since CD equals 6,\\
it follows that AD equals 18,\\
Given that M is the midpoint of AD,\\
it follows that $MD=\frac{1}{2}AD=9$,\\
therefore, $MC=MD-CD=9-6=\boxed{3}$.


Question: As shown in the figure, $ OM $, $ OB $, and $ ON $ are three rays within $ \angle AOC $, where $ OM $ and $ ON $ are the angle bisectors of $ \angle AOB $ and $ \angle BOC $, respectively. Furthermore, $ \angle NOC $ is three times that of $ \angle AOM $, and $ \angle BON $ is $ 20^\circ $ larger than $ \angle MOB $. Determine the degree measure of $ \angle AOC $.

Solution:
Let the measure of $\angle AOM$ be $x$, then the measure of $\angle NOC$ is $3x$,\\
since OM and ON are the bisectors of $\angle AOB$ and $\angle BOC$ respectively,\\
therefore $\angle MOB=\angle AOM=x$ and $\angle BON=\angle NOC=3x$,\\
since $\angle BON$ is $20^\circ$ greater than $\angle MOB$,\\
therefore $3x-x=20^\circ$,\\
therefore $x=10^\circ$,\\
therefore $\angle AOC=\angle AOM+\angle MOB+\angle BON+\angle NOC=8x=\boxed{80^\circ}$.


Question: As shown in the figure, $\angle AOB$ is a right angle, $\angle BOC$ is an acute angle, and $OM$ bisects $\angle AOC$, $ON$ bisects $\angle BOC$. If $\angle BOC=36^\circ$, what is the measure of $\angle MON$?

Solution:
\textbf{Solution}: Since $\angle AOB$ is a right angle and $\angle BOC=36^\circ$,\\
it follows that $\angle AOC=\angle AOB+\angle BOC=90^\circ+36^\circ=126^\circ$,\\
Since OM bisects $\angle AOC$ and ON bisects $\angle BOC$.\\
Thus, $\angle MOC= \frac{1}{2}\angle AOC=63^\circ$, $\angle NOC= \frac{1}{2}\angle BOC=18^\circ$,\\
Therefore, $\angle MON=\angle MOC-\angle NOC=63^\circ-18^\circ=\boxed{45^\circ}$;


Question: As shown in the figure, $\angle AOB$ is a right angle and $\angle BOC$ is an acute angle. Moreover, $OM$ bisects $\angle AOC$ and $ON$ bisects $\angle BOC$. Given $\angle AOB = 90^\circ$ and $\angle BOC = \beta$, with $\alpha$ and $\beta$ being acute angles and all other conditions unchanged, what is the degree measure of $\angle MON$?

Solution:
\textbf{Solution:}
Given: $\angle AOB=a$ and $\angle BOC=\beta$,\\
Thus, $\angle AOC=\angle AOB+\angle BOC= a+\beta$,\\
Since OM bisects $\angle AOC$ and ON bisects $\angle BOC$,\\
Thus $\angle MOC= \frac{1}{2}\angle AOC= \frac{1}{2}(a+\beta)$ and $\angle NOC= \frac{1}{2}\angle BOC= \frac{1}{2}\beta$,\\
Therefore, $\angle MON=\angle MOC-\angle NOC= \frac{1}{2}(a+\beta)- \frac{1}{2}\beta= \frac{1}{2}a$. Hence, the measure of $\angle MON$ is $\boxed{\frac{1}{2}a}$.


Question: Given: As shown in the figure, $OC$ is the bisector of $\angle AOB$. When $\angle AOB = 60^\circ$, what is the measure of $\angle AOC$?

Solution:
\textbf{Solution:} Given that OC bisects $\angle AOB$,\\
it follows that $\angle AOC= \frac{1}{2}\angle AOB$,\\
since $\angle AOB=60^\circ$,\\
we find $\angle AOC=\boxed{30^\circ}$.


Question: As shown in the figure, $OC$ is the bisector of $\angle AOB$. Given the condition (1), that $\angle EOC = 90^\circ$, please complete the figure in the diagram and find the measure of $\angle AOE$?

Solution:
\textbf{Solution:} Since OE $\perp$ OC, it follows that $\angle$EOC$=90^\circ$, as shown in Figure 1, $\angle$AOE = $\angle$COE + $\angle$COA = $90^\circ + 30^\circ = \boxed{120^\circ}$.

As shown in Figure 2, $\angle$AOE = $\angle$COE - $\angle$COA = $90^\circ - 30^\circ = \boxed{60^\circ}$.


Question: Given: As shown in the figure, $OC$ is the bisector of $\angle AOB$. When $\angle AOB = \alpha$, $\angle EOC = 90^\circ$, write down the degree measure of $\angle AOE$ directly. (Express in terms of $\alpha$)

Solution: \textbf{Solution:} We have $\angle AOE=90^\circ+\frac{\alpha}{2}$ or $\angle AOE=90^\circ-\frac{\alpha}{2}$. Therefore, the degree measure of $\angle AOE$ is \boxed{90^\circ+\frac{\alpha}{2}} or \boxed{90^\circ-\frac{\alpha}{2}}.


Question: As shown in the figure, it is known that $\angle AOB = 80^\circ$, $OC$ is the bisector of $\angle AOB$, and $OD$ is the bisector of $\angle BOC$. What is the measure of $\angle AOD$?

Solution:
\textbf{Solution:} Given that $\angle AOB=80^\circ$ and OC is the bisector of $\angle AOB$, \\
it follows that $\angle AOC=\angle BOC=0.5\times 80^\circ=40^\circ$. \\
Also, since OD is the bisector of $\angle BOC$, \\
it follows that $\angle COD=\angle BOD=0.5\times 40^\circ=20^\circ$. \\
Therefore, $\angle AOD=\angle AOC+\angle COD=40^\circ+20^\circ=\boxed{60^\circ}$.


Question: As shown in the figure, it is known that $\angle AOB = 80^\circ$, $OC$ is the bisector of $\angle AOB$, $OD$ is the bisector of $\angle BOC$. If $\angle COE = \frac{1}{4}\angle COB$, what is the degree measure of $\angle AOE$?

Solution:
\textbf{Solution:} Given from (1) that $\angle COB=40^\circ$,\\
$\therefore \angle COE=10^\circ$,\\
when OE is above OC, then $\angle AOE=\angle AOC-\angle COE=40^\circ-10^\circ=\boxed{30^\circ}$;\\
when OE is below OC, then $\angle AOE=\angle AOC+\angle COE=40^\circ+10^\circ=\boxed{50^\circ}$.


Question: It is known that there are three rays inside $\angle AOB$. Among them, $OE$ bisects $\angle BOC$, and $OF$ bisects $\angle AOC$. As shown in Figure 1, if $\angle AOB=90^\circ$ and $\angle AOC=30^\circ$, what is the measure of $\angle EOF$ in degrees?

Solution:
\textbf{Solution:}

Given: $\angle BOC=\angle AOB-\angle AOC=90^\circ-30^\circ=60^\circ$,\\
Since OE bisects $\angle BOC$, and OF bisects $\angle AOC$,\\
Then $\angle EOC=\frac{1}{2}\angle BOC=\frac{1}{2}\times 60^\circ=30^\circ$ and $\angle COF= \frac{1}{2}\angle AOC=\frac{1}{2}\times 30^\circ=15^\circ$,\\
Therefore $\angle EOF=\angle EOC+\angle COF=30^\circ+15^\circ=\boxed{45^\circ}$


Question: As shown in the figure, it is known that inside $ \angle AOB$ there are three rays, among which, $OE$ bisects $ \angle BOC$, and $OF$ bisects $ \angle AOC$. If $ \angle AOB = \alpha $, what is the measure of $ \angle EOF$ (expressed in terms of $\alpha$)?

Solution:
\textbf{Solution:} Since OE bisects $\angle BOC$, and OF bisects $\angle AOC$,

$\therefore \angle EOC= \frac{1}{2}\angle BOC$, $\angle COF= \frac{1}{2}\angle AOC$,

$\therefore \angle EOF=\angle EOC+\angle COF= \frac{1}{2}\angle BOC+ \frac{1}{2}\angle AOC= \frac{1}{2}(\angle BOC+\angle AOC)= \frac{1}{2}\angle AOB= \boxed{\frac{1}{2}\alpha}$.


Question: As shown in Figure 1, point $D$ is the midpoint of segment $AC$, and $AB=\frac{2}{3}BC$, $BC=6$. Find the length of segment $BD$.

Solution:
\textbf{Solution:} By the given conditions, \\
since $AB=\frac{2}{3}BC$ and $BC=6$, \\
then $AB=4$ and $AC=10$, \\
since point $D$ is the midpoint of line segment $AC$, \\
thus $AD=\frac{1}{2}AC=5$, \\
therefore $BD=5-4=\boxed{1}$.


Question: Since there was no text provided in the initial query, I am unable to provide a translated version in LaTeX format. If you have specific text in mind, please share it, and I'll be happy to assist with the translation.

Solution:
\textbf{Solution:} Given the problem,\\
\[\because \angle AOD=100^\circ,\text{ and }OB \text{ bisects } \angle AOD,\]
\[\therefore \angle AOB=\frac{1}{2}\angle AOD=50^\circ,\]
\[\because \angle AOC : \angle COB = 3 : 2,\]
\[\therefore \angle COB=\frac{2}{5}\angle AOC=\frac{2}{5}\times 50^\circ=\boxed{20^\circ}.\]


Question: As shown in the diagram, $OC$ is outside $\angle AOB$, and $OM$, $ON$ are the angle bisectors of $\angle AOC$ and $\angle BOC$, respectively. If $\angle AOB=110^\circ$ and $\angle BOC=60^\circ$, what is the measure of $\angle MON$?

Solution:
\textbf{Solution:} Given that $\angle AOB=110^\circ$ and $\angle BOC=60^\circ$,\\
thus $\angle AOC=\angle AOB+\angle BOC=170^\circ$,\\
since OM bisects $\angle AOC$,\\
thus $\angle MOC=\angle MOA=\frac{1}{2}\angle AOC=85^\circ$,\\
since ON bisects $\angle BOC$,\\
thus $\angle BON=\angle CON=30^\circ$,\\
therefore $\angle MON=\angle COM-\angle CON=\boxed{55^\circ}$;


Question: As shown in the figure, OC is outside of $\angle AOB$, and OM, ON are the angle bisectors of $\angle AOC$ and $\angle BOC$, respectively. If $\angle AOB = \alpha$ and $\angle BOC = 40^\circ$, with all other conditions remaining the same, find the value of $\angle MON$ (expressed in terms of $\alpha$).

Solution:
\textbf{Solution:} When $OM$ and $ON$ are on the same side of $OC$,\\
$\because$ $\angle AOB=\alpha$, $\angle BOC=40^\circ$,\\
$\therefore$ $\angle AOC=\angle AOB+\angle BOC=40^\circ+\alpha$,\\
$\because$ $OM$ bisects $\angle AOC$,\\
$\therefore$ $\angle MOC=\angle MOA=\frac{1}{2}\angle AOC=\frac{1}{2}\alpha +20^\circ$,\\
$\because$ $ON$ bisects $\angle BOC$,\\
$\therefore$ $\angle BON=\angle CON=20^\circ$,\\
$\therefore$ $\angle MON=\angle COM-\angle CON=\frac{1}{2}\alpha +20^\circ-20^\circ=\frac{1}{2}\alpha$; When $OM$ and $ON$ are on opposite sides of $OC$,\\
$\because$ $\angle AOB=\alpha$, $\angle BOC=40^\circ$,\\
$\therefore$ $\angle AOC=360^\circ-\angle AOB-\angle BOC=320^\circ-\alpha$,\\
$\because$ $OM$ bisects $\angle AOC$,\\
$\therefore$ $\angle MOC=\angle MOA=\frac{1}{2}\angle AOC=160^\circ-\frac{1}{2}\alpha$,\\
$\because$ $ON$ bisects $\angle BOC$,\\
$\therefore$ $\angle BON=\angle CON=20^\circ$,\\
$\therefore$ $\angle MON=\angle COM+\angle CON=160^\circ-\frac{1}{2}\alpha +20^\circ=180^\circ-\frac{1}{2}\alpha$; In summary, the value of $\angle MON$ is either $\boxed{\frac{1}{2}\alpha}$ or $\boxed{180^\circ-\frac{1}{2}\alpha}$.


Question: As shown in the figure, $OB$ is the bisector of $\angle AOC$, $OD$ is the bisector of $\angle COE$. If $\angle AOB = 42^\circ$, $\angle DOE = 36^\circ$, what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Since $OB$ is the bisector of $\angle AOC$ and $\angle AOB=42^\circ$,\\
and $OD$ is the bisector of $\angle COE$ and $\angle DOE=36^\circ$,\\
$\therefore$ $\angle AOB=\angle BOC=42^\circ$,\\
$\angle COD=\angle DOE=36^\circ$,\\
$\therefore$ $\angle BOD=\angle BOC+\angle DOC=42^\circ+36^\circ=\boxed{78^\circ}$.


Question: As shown in the figure, $OB$ is the bisector of $\angle AOC$, and $OD$ is the bisector of $\angle COE$. If $\angle AOD$ and $\angle BOD$ are complementary, and $\angle DOE=30^\circ$, find the degree measure of $\angle AOC$.

Solution:
\textbf{Solution:} Since $\angle AOD$ and $\angle BOD$ are complementary,\\
we have $\angle AOD+\angle BOD=180^\circ$,\\
From the figure, we know: $\angle AOD=\angle AOC+\angle COD$,\\
and $\angle BOD=\angle BOC+\angle COD$,\\
By the definition of angle bisector, we know: $\angle BOC=\frac{1}{2}\angle AOC$,\\
Thus, $\angle AOC+\angle DOE+\frac{1}{2}\angle AOC+\angle DOE=180^\circ$,\\
which implies $\frac{3}{2}\angle AOC+2\angle DOE=180^\circ$,\\
Since $\angle DOE=30^\circ$,\\
it follows that $\frac{3}{2}\angle AOC+2\times 30^\circ=180^\circ$,\\
hence, $\angle AOC=\boxed{80^\circ}$.


Question: As shown in the figure, $OC$ is the bisector of $\angle AOD$, and $OE$ is the bisector of $\angle BOD$. If $OB$ and $OD$ are perpendicular to each other, and $\angle AOD=30^\circ$, then what is the degree measure of $\angle COE$?

Solution:
Solution: Since $OB$ and $OD$ are perpendicular to each other, \\
$\therefore \angle BOD=90^\circ$, \\
Since $OE$ is the bisector of $\angle BOD$, \\
$\therefore \angle DOE=\frac{1}{2}\angle BOD=\frac{1}{2}\times 90^\circ=45^\circ$, \\
Since $OC$ is the bisector of $\angle AOD$, and $\angle AOD=30^\circ$, \\
$\therefore \angle COD=\frac{1}{2}\angle AOD=\frac{1}{2}\times 30^\circ=15^\circ$, \\
$\therefore \angle COE=\angle DOE+\angle COD=45^\circ+15^\circ=\boxed{60^\circ}$.


Question: As shown in the figure, $OC$ is the bisector of $\angle AOD$, and $OE$ is the bisector of $\angle BOD$. If $\angle AOB=140^\circ$, then what is the degree measure of $\angle COE$?

Solution:
\textbf{Solution:} Since $OC$ is the bisector of $\angle AOD$ and $OE$ is the bisector of $\angle BOD$,\\
$\therefore \angle COD=\frac{1}{2}\angle AOD$, $\angle DOE=\frac{1}{2}\angle BOD$,\\
Given that $\angle AOB=140^\circ$,\\
$\therefore \angle AOD+\angle BOD=140^\circ$,\\
$\therefore \angle COE=\angle COD+\angle DOE=\frac{1}{2}\left(\angle AOD+\angle BOD\right)=\frac{1}{2}\times 140^\circ=\boxed{70^\circ}$.


Question: As shown in the diagram, $AB$ is a straight line, and $OC$ is the bisector of $\angle AOD$. As illustrated in figure (1), if $\angle COE$ is a right angle, and $\angle AOD=70^\circ$, what is the measure of $\angle BOE$?

Solution:
\textbf{Solution:} Since $OC$ is the bisector of $\angle AOD$ and $\angle AOD = 70^\circ$,\\
it follows that $\angle AOC = \angle DOC = \frac{1}{2}\angle AOD = \frac{1}{2}\times 70^\circ = 35^\circ$,\\
Since $\angle COE is a right angle,\\
we have $\angle COE = 90^\circ$,\\
Hence, $\angle BOE = 180^\circ - \angle AOC - \angle COE = 180^\circ - 35^\circ - 90^\circ = \boxed{55^\circ};$


Question: As shown in the figure, AB is a straight line, and OC is the bisector of $\angle AOD$. As shown in Figure (2), if the ratio $\angle DOE : \angle BOD = 2 : 5$ and $\angle COE = 80^\circ$, find the degree measure of $\angle BOE$.

Solution:
\textbf{Solution:} Let $\angle COD=x^\circ$, $\angle DOE=y^\circ$,\\
$\therefore x+y=80$,\\
$\because$ OC is the bisector of $\angle AOD$,\\
$\therefore \angle AOD=2\angle COD=2x^\circ$,\\
$\because \angle DOE\colon \angle BOD=2\colon 5$,\\
$\therefore \angle BOD= \frac{5}{2}\angle DOE=\frac{5}{2}y^\circ$,\\
$\because \angle AOD+\angle BOD=180^\circ$,\\
$\therefore 2x+\frac{5}{2}y=180$,\\
$\therefore$ solving the above equations we get $y=40$,\\
$\therefore x=80-40=40$,\\
$\therefore \angle AOD=2x^\circ=80^\circ$,\\
$\therefore \angle BOD=180^\circ-80^\circ=100^\circ$,\\
$\therefore \angle BOE=\angle BOD-\angle DOE=100^\circ-40^\circ=\boxed{60^\circ}$.


Question: As shown in the figure, it is known that $ OB$ is a ray inside $ \angle AOC$, $ OE$ bisects $ \angle AOC$, and $ OF$ bisects $ \angle BOC$. If $ AO\perp BO$ and $ \angle BOC=60^\circ$, find the measure of $ \angle EOF$.

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


Question: As shown in the figure, in $\triangle ABC$, $BD$ bisects $\angle ABC$, $E$ is a point on $BD$, $EA \perp AB$, and $EB=EC$, $\angle EBC=\angle ECB$. If $\angle ABC=40^\circ$, what is the measure of $\angle DEC$?

Solution:
\textbf{Solution:} Since $\angle ABC=40^\circ$ and BD bisects $\angle ABC$,\\
it follows that $\angle EBC= \frac{1}{2}\angle ABC=20^\circ$,\\
Since $EB=EC$,\\
we have $\angle ECB=\angle EBC=20^\circ$,\\
Because $\angle DEC is an exterior angle of \triangle EBC$,\\
it follows that $\angle DEC=\angle ECB+\angle EBC=\boxed{40^\circ}$;


Question: As shown in the figure, points $C$ and $D$ divide the line segment $AB$ into three parts in the ratio $2\colon 5\colon 3$, with $E$ being the midpoint of $AB$. If the numbers represented by points $A$, $B$, and $D$ are $-10$, $+20$, and $x$ respectively, what is the value of $x$?

Solution:
\textbf{Solution:} Given the problem, the numbers represented by points $A$, $B$, $D$ are respectively $-10$, $+20$, $x$,\\
$\therefore AB=20-(-10)=30$, $BD=20-x$,\\
$\because$ On the number line, points $C$ and $D$ divide segment $AB$ into three parts in the ratio $2\colon 5\colon 3$,\\
$\therefore BD=\frac{3}{2+5+3}AB=\frac{3}{10}\times 30=9$,\\
$\therefore 20-x=9$,\\
$\therefore x=\boxed{11}$.


Question: As shown in the figure, on the number line, points $C$ and $D$ divide the line segment $AB$ into three parts in the ratio $2:5:3$, with $E$ being the midpoint of $AB$. If $ED=3\,cm$, what is the length of the line segment $AB$?

Solution:
\textbf{Solution:} From the given information, \\
since $E$ is the midpoint of $AB$, \\
thus $EB=\frac{1}{2}AB$, \\
since points $C$, $D$ on the number line divide the segment $AB$ into three parts in the ratio $2:5:3$, \\
thus $BD=\frac{3}{2+5+3}AB=\frac{3}{10}AB$, \\
thus $ED=EB-BD=\frac{1}{2}AB-\frac{3}{10}AB=\frac{1}{5}AB$, \\
also, since $ED=3\text{cm}$, \\
thus $\frac{1}{5}AB=3\text{cm}$, \\
thus $AB=\boxed{15\text{cm}}$.


Question: As shown in the figure, line $AB$ intersects line $CD$ at point $O$, $OE \perp CD$ and $OE$ bisects $\angle BOF$. If $\angle BOD$ is $10^\circ$ larger than $\angle BOE$, what is the degree measure of $\angle COF$?

Solution:
\textbf{Solution:}
Since $OE \perp CD$,\\
it follows that $\angle DOE = 90^\circ$,\\
which means $\angle DOB + \angle EOB = 90^\circ$,\\
Since $\angle DOB = \angle EOB + 10^\circ$,\\
it follows that $\angle EOB = 40^\circ$,\\
Since $OE$ bisects $\angle BOF$,\\
it follows that $\angle EOF = \angle BOE = 40^\circ$,\\
Therefore, $\angle COF = 90^\circ - \angle EOF = \boxed{50^\circ}$.


Question: As shown in the figure, it is known that lines AB and CD intersect at point O, with OE $\perp$ CD. If $\angle AOC = 42^\circ$, what is the degree measure of $\angle BOE$?

Solution:
\textbf{Solution:} Since $\angle AOC=42^\circ$,\\
it follows that $\angle BOD=42^\circ$,\\
since $OE \perp CD$,\\
it follows that $\angle BOE=90^\circ-42^\circ=\boxed{48^\circ}$.


Question: As shown in the figure, it is known that lines AB and CD intersect at point O, with OE $\perp$ CD. If $\angle BOD : \angle BOC = 2 : 7$ and OF bisects $\angle AOD$, what is the measure of $\angle EOF$?

Solution:
\textbf{Solution:}\\
Since $\angle$BOD : $\angle$BOC = 2 : 7,\\
then $\angle$BOC = $180^\circ \times \frac{7}{7+2} = 140^\circ$,\\
thus $\angle$AOD = $140^\circ$,\\
Since OF bisects $\angle$AOD,\\
then $\angle$DOF = $\frac{1}{2}\angle$AOD = $70^\circ$,\\
Since OE$\perp$CD,\\
then $\angle$EOF = $90^\circ + 70^\circ = \boxed{160^\circ}$.


Question: As shown in the figure, point O is on line AB, $\angle BOD$ and $\angle COD$ are complementary, and $\angle BOC = n\angle EOC$. If $\angle AOD = 24^\circ$ and $n = 3$, what is the degree measure of $\angle DOE$?

Solution:
\textbf{Solution:} Given that,\\
since $\angle BOD$ and $\angle COD$ are complementary,\\
thus $\angle BOD + \angle AOD = 180^\circ$,\\
therefore, $\angle COD = \angle AOD$,\\
since $\angle AOD = 24^\circ$, $n=3$, $\angle BOC = n\angle EOC$,\\
thus $\angle COD = 24^\circ$, $\angle BOC = 3\angle EOC$,\\
therefore, $\angle BOC = 180^\circ - 24^\circ - 24^\circ = 132^\circ$, $\angle EOC = 44^\circ$,\\
therefore, $\angle DOE = 24^\circ + 44^\circ = \boxed{68^\circ}$.


Question: As shown in the figure, point $O$ is on line $AB$, $\angle BOD$ and $\angle COD$ are complementary, and $\angle BOC = n\angle EOC$. If $DO \perp OE$, what is the value of $n$?

Solution:
\textbf{Solution:} Given that, since $\angle BOD$ and $\angle COD$ are complementary,\\
we have $\angle BOD + \angle AOD = 180^\circ$,\\
thus $\angle COD = \angle AOD$,\\
Let $\angle COE = x$,\\
then $\angle BOE = nx - x$,\\
since $OD \perp OE$,\\
it follows that $\angle COD + \angle COE = 90^\circ = \angle AOD + \angle BOE$,\\
thus $\angle COE = \angle BOE$,\\
hence $nx - x = x$,\\
and given $x \ne 0$,\\
we find $n = \boxed{2}$.


Question: As shown in the figure, point $O$ lies on line $AB$, $\angle BOD$ and $\angle COD$ are complementary, $\angle BOC = n\angle EOC$. If $n=4$, let $\angle AOD = \alpha$, find the degree measure of $\angle DOE$ (express the degree measure of $\angle DOE$ in terms of $\alpha$).

Solution:
\textbf{Solution:} Given the problem statement,\\
\(\because \angle BOD\) and \(\angle COD\) are complementary, \\
\(\therefore \angle BOD + \angle AOD = 180^\circ\),\\
\(\therefore \angle COD = \angle AOD = \alpha\),\\
Let \(\angle COE = y\),\\
and \(n = 4\),\\
then \(\angle BOC = 4y\), \(\angle BOE = 3y\),\\
\(\therefore 4y + 2\alpha = 180^\circ\),\\
\(\therefore y = 45^\circ - \frac{1}{2}\alpha\),\\
\(\therefore \angle DOE = \alpha + 45^\circ - \frac{1}{2}\alpha = 45^\circ + \frac{1}{2}\alpha\).\\
Hence, the measure of \(\angle DOE\) is \(\boxed{45^\circ + \frac{1}{2}\alpha}\).


Question: Given the diagram, $OC\perp AB$ at point $O$, and $\angle COD= \frac{1}{4}\angle BOD$, with $OE$ bisecting $\angle BOD$. Find the measures of $\angle COE$ and $\angle AOE$?

Solution: \textbf{Solution:} Since OC $\perp$ AB, it follows that $\angle BOC = \angle AOC = 90^\circ$. Since $\angle COD = \frac{1}{3}\angle BOC = 30^\circ$, it follows that $\angle BOD = 120^\circ$. Since OE bisects $\angle BOD$, it follows that $\angle BOE = \angle DOE = 60^\circ$. Therefore, $\angle COE = \angle BOC - \angle BOE = 90^\circ - 60^\circ = \boxed{30^\circ}$, $\angle AOE = 180^\circ - \angle BOE = 180^\circ - 60^\circ = \boxed{120^\circ}$.


Question: As shown in the figure, $OC\perp AB$ at point $O$, $\angle COD=\frac{1}{4}\angle BOD$, and $OE$ bisects $\angle BOD$. A ray $OF$ is drawn through point $O$, and if $OF\perp OE$, what is the measure of $\angle BOF$?

Solution:
Solution: As shown in the figure, when OF is above line AB,

$\because$ OF$\perp$OE,

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

$\because$ $\angle BOE=60^\circ$,

$\therefore$ $\angle BOF=\angle BOE+\angle EOF=60^\circ+90^\circ=150^\circ$;

When OF is below line AB,

$\because$ OF$\perp$OE,

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

$\because$ $\angle BOE=60^\circ$,

$\therefore$ $\angle BOF=\angle EOF-\angle BOE=90^\circ-60^\circ=\boxed{30^\circ}$,

Therefore, the measure of $\angle BOF$ is $150^\circ$ or $\boxed{30^\circ}$.


Question: As shown in the figure, points $C$, $D$, and $E$ are on the line segment $AB$, with $AB=12$ and $AC=4$. $D$ and $E$ are the midpoints of $AB$ and $CB$ respectively. Find the length of $DE$.

Solution:
\textbf{Solution:} Given $\because AB=12$, $AC=4$, \\
$\therefore BC=AB-AC=12-4=8$, \\
$\because D$ and $E$ are the midpoints of $AB$ and $CB$, respectively, \\
$\therefore BD= \frac{1}{2}AB= \frac{1}{2}\times 12=6$, $BE=CE= \frac{1}{2}BC= \frac{1}{2}\times 8=4$, \\
$\therefore DE=BD-BE=6-4=\boxed{2}$.


Question: As shown in the figure, it is known that $OC$ bisects $\angle AOD$, $\angle BOC=30^\circ$, and $\angle BOC$ is complementary to $\angle AOD$. Find the measures of $\angle AOD$ and $\angle BOD$.

Solution:
\textbf{Solution:}\\
Since $\angle$BOC and $\angle$AOD are complementary,\\
$\therefore$ $\angle$BOC + $\angle$AOD = 180$^\circ$,\\
Since $\angle$BOC = 30$^\circ$,\\
$\therefore$ $\angle$AOD = 150$^\circ$,\\
Since OC bisects $\angle$AOD,\\
$\therefore$ $\angle$DOC = $\frac{1}{2}\angle$AOD = $\frac{1}{2} \times 150^\circ = 75^\circ$,\\
$\therefore$ $\angle$BOD = $\angle$DOC + $\angle$BOC\\
= 75$^\circ$ + 30$^\circ$\\
= \boxed{105^\circ},\\
that is, $\angle$AOD = \boxed{150^\circ}, $\angle$BOD = \boxed{105^\circ}.


Question: As shown in the figure, $OC$ is a ray inside $\angle AOB$, $OD$ bisects $\angle BOC$, and $OE$ bisects $\angle AOC$. If $\angle AOB=130^\circ$, find the degree measure of $\angle DOE$.

Solution:
\textbf{Solution:} When $\angle AOB = 130^\circ$,\\
we have $\angle DOE = \frac{1}{2}\angle AOB$\\
$= \frac{1}{2}\times 130^\circ$\\
$= 65^\circ$.\\
Therefore, the measure of $\angle DOE$ is $\boxed{65^\circ}$.


Question: As shown in the figure, $AB\parallel CD$, and point E is a point between the two lines. If $\angle BAE=35^\circ$ and $\angle DCE=20^\circ$, then what is the degree measure of $\angle AEC$?

Solution: \textbf{Solution:} We have $\angle AEC = \boxed{55^\circ}$.


Question: As shown in the figure, point $O$ is a point on line $AB$, $OM$ and $ON$ are on opposite sides of line $AB$, and $\angle MON=90^\circ$, $OE$ bisects $\angle MOB$, $OF$ bisects $\angle AON$. Given $\angle BOM=150^\circ$, what are the degrees of $\angle BOE$ and $\angle NOF$?

Solution:
\textbf{Solution:} \\
Since $OE$ bisects $\angle MOB$ and $\angle BOM=150^\circ$,\\
we have $\angle BOE=\frac{1}{2}\angle BOM=\frac{1}{2}\times 150^\circ=\boxed{75^\circ}$;\\
Since $\angle AOM=180^\circ-\angle BOM=180^\circ-150^\circ=30^\circ$,\\
and since $\angle MON=90^\circ$,\\
we get $\angle AON=\angle MON-\angle AOM=90^\circ-30^\circ=60^\circ$,\\
Since $OF$ bisects $\angle AON$,\\
therefore, $\angle NOF=\frac{1}{2}\angle AON=\frac{1}{2}\times 60^\circ=\boxed{30^\circ}$.


Question: As shown in the figure, point $O$ is on line $AB$, with $OM$ and $ON$ on opposite sides of line $AB$ and $\angle MON=90^\circ$. $OE$ bisects $\angle MOB$, and $OF$ bisects $\angle AON$. Let $\angle AOF = \theta$, express $\angle MOE$ in terms of $\theta$.

Solution:
\textbf{Solution:}\\
Since $\angle AOF=\theta$,\\
it follows that $\angle AON=2\angle AOF=2\theta$,\\
and since $\angle MON=90^\circ$,\\
we have $\angle AOM=90^\circ-2\theta$,\\
thus $\angle BOM=180^\circ-\left(90^\circ-2\theta \right)=90^\circ+2\theta$,\\
therefore $\angle MOE=\frac{1}{2}\angle BOM=\frac{1}{2}\left(90^\circ+2\theta \right)=45^\circ+\theta$.\\
Hence, $\angle MOE$ can be expressed in a formula containing $\theta$ as $\angle MOE=\boxed{45^\circ+\theta}$.


Question: As shown in the figure, $OA \perp OB$, $\angle COD = 60^\circ$. If $OC$ bisects $\angle AOD$, what is the degree measure of $\angle BOC$?

Solution:
\textbf{Solution:} Since $OC$ bisects $\angle AOD$, $\angle COD=60^\circ$,\\
thus $\angle AOC=\angle COD=60^\circ$,\\
since $OA\perp OB$,\\
thus $\angle AOB=90^\circ$,\\
therefore $\angle BOC=\angle AOB-\angle AOC=90^\circ-60^\circ=\boxed{30^\circ}$.


Question: As shown in the figure, $OA \perp OB$, $\angle COD = 60^\circ$. If $\angle BOC = \frac{3}{7}\angle AOD$, what is the degree measure of $\angle AOD$?

Solution:
\textbf{Solution:}
Given: $\because$ $\angle AOB=90^\circ$,\\
$\therefore$ $\angle BOD=\angle AOD-\angle AOB=\angle AOD-90^\circ$,\\
$\because$ $\angle COD=60^\circ$,\\
$\therefore$ $\angle BOD=\angle COD-\angle BOC=60^\circ-\angle BOC$,\\
$\therefore$ $60^\circ-\angle BOC=\angle AOD-90^\circ$,\\
$\because$ $\angle BOC=\frac{3}{7}\angle AOD$,\\
$\therefore$ $60^\circ-\frac{3}{7}\angle AOD=\angle AOD-90^\circ$,\\
Solving, we get: $\angle AOD=\boxed{105^\circ}$.


Question: As shown in the figure, point $D$ is the midpoint of segment $AC$, and $AB = \frac{2}{3} BC$, $BC = 6$. Find the length of segment $BD$.

Solution:
\textbf{Solution:} Given that $AB= \frac{2}{3} BC$ and $BC=6$,
\[
\therefore AB= \frac{2}{3} \times 6=4,
\]
\[
\therefore AC=AB+BC=10,
\]
Since point D is the midpoint of segment AC,
\[
\therefore AD= \frac{1}{2} AC=5,
\]
\[
\therefore BD=AD-AB=5-4=\boxed{1}.
\]


Question: As shown in the figure, it is known that $OB$ bisects $\angle AOD$, $\angle BOC = \frac{2}{3} \angle AOC$, and if $\angle AOD=100^\circ$, find the degree measure of $\angle BOC$.

Solution:
\textbf{Solution:} Given that OB bisects $\angle AOD$, and $\angle AOD = 100^\circ$, \\
$\therefore \angle AOB = \frac{1}{2} \angle AOD = 50^\circ$, \\
Given that $\angle BOC + \angle AOC = \angle AOB$, and $\angle BOC = \frac{2}{3} \angle AOC$, \\
$\therefore \frac{2}{3} \angle AOC + \angle AOC = 50^\circ$, \\
$\therefore \angle AOC = 30^\circ$, \\
$\therefore \angle BOC = \frac{2}{3} \angle AOC = \boxed{20^\circ}$.


Question: Given the diagram, place the right-angled vertex \(O\) of the right-angled triangle \(MON\) on the line \(AB\), and let the ray \(OC\) bisect \(\angle AON\). As shown in Figure 1, if \(\angle MOC = 28^\circ\), what is the degree measure of \(\angle BON\)?

Solution:
\textbf{Solution:} As shown in Figure 1, since $\angle MOC=28^\circ$ and $\angle MON=90^\circ$,\\
it follows that $\angle NOC=90^\circ-28^\circ=62^\circ$,\\
and since $OC$ bisects $\angle AON$,\\
it follows that $\angle AOC=\angle NOC=62^\circ$,\\
therefore $\angle BON=180^\circ-2\angle NOC=180^\circ-62^\circ\times 2=\boxed{56^\circ}$.


Question: As shown in the figure, it is known that the right-angled vertex $O$ of the right-angled triangle $MON$ is placed on the line $AB$, and the ray $OC$ bisects $\angle AON$. If $\angle MOC = m^\circ$, what is the measure of $\angle BON$ in degrees?

Solution: \textbf{Solution:} We have $\angle BON = \boxed{2m^\circ}$.


Question: As shown in the figure, point $O$ is on line $AB$, $CO \perp AB$, and $OE$ bisects $\angle BOD$, with $OF \perp OE$. If $\angle 2 - \angle 1 = 20^\circ$, what is the degree measure of $\angle BOE$?

Solution:
\textbf{Solution:} Given $CO \perp AB$,

it follows that $\angle COA = 90^\circ$, which means $\angle 2 = 90^\circ - \angle 1$.

Also, given $\angle 2 - \angle 1 = 20^\circ$,

it results in $\angle 2 = 20^\circ + \angle 1$,

hence $90^\circ - \angle 1 = 20^\circ + \angle 1$.

Solving this, we find $\angle 1 = 35^\circ$,

therefore $\angle 2 = 55^\circ$,

and thus $\angle BOD = 180^\circ - \angle 2 = 125^\circ$.

Since $OE$ bisects $\angle BOD$,

it means $\angle BOE = \angle DOE = \frac{1}{2} \angle BOD = \boxed{62.5^\circ}$.


Question: As shown in the figure, $ \angle AOB=40^\circ$, $OB$ is the bisector of $ \angle AOC$, and $OD$ is the bisector of $ \angle COE$. If $ \angle DOE=10^\circ$, find the degree measure of $ \angle BOD$.

Solution:
\textbf{Solution:} Since $OB$ is the bisector of $\angle AOC$, and $OD$ is the bisector of $\angle COE$,\\
$\therefore \angle BOC=\angle AOB=40^\circ$, $\angle COD=\angle DOE=10^\circ$,\\
$\therefore \angle BOD=\angle BOC+\angle COD=40^\circ+10^\circ=\boxed{50^\circ}$.


Question: As shown in the diagram, $ \angle AOB=40^\circ$, $ OB$ is the bisector of $ \angle AOC$, and $ OD$ is the bisector of $ \angle COE$. If $ \angle AOD$ and $ \angle BOD$ are complementary angles, find the degree measure of $ \angle COE$?

Solution:
\textbf{Solution:}

Given that $OB$ is the bisector of $\angle AOC$, and $OD$ is the bisector of $\angle COE$,\\
$\therefore \angle BOC=\angle AOB=40^\circ$,\\
Let $\angle COD=\angle DOE=x^\circ$,\\
$\therefore \angle BOD=\angle BOC+\angle COD=(40+x)^\circ$, $\angle AOD=\angle AOB+\angle BOC+\angle COD=(80+x)^\circ$,\\
Since $\angle AOD$ and $\angle BOD$ are complementary,\\
$\therefore \angle AOD+\angle BOD=(40+x)^\circ+(80+x)^\circ=180^\circ$,\\
$\therefore x=\boxed{30}$,\\
$\therefore \angle COD=\angle DOE=30^\circ$,\\
$\therefore \angle COE=2\angle COD=60^\circ$.


Question: As shown in the figure, it is known that the line segment \( AC=12\,cm \), and point \( B \) is on line segment \( AC \), satisfying \( BC= \frac{1}{2} AB \). What is the length of \( AB \)?

Solution:
\textbf{Solution:} Since $BC=\frac{1}{2}AB$, \\
it follows that $AB=\frac{2}{3}AC=\frac{2}{3}\times 12=\boxed{8\text{cm}}$.


Question: As shown in the figure, it is known that the length of line segment $AC=12\, \text{cm}$, and point $B$ is on line segment $AC$, satisfying $BC= \frac{1}{2} AB$. If $D$ is the midpoint of $AB$, and $E$ is the midpoint of $AC$, find the length of $DE$.

Solution:
\textbf{Solution:}\\
Since $D$ is the midpoint of $AB$,\\
thus $AD=\frac{1}{2}AB=4\text{cm}$,\\
Since $E$ is the midpoint of $AC$,\\
thus $AE=\frac{1}{2}AC=6\text{cm}$,\\
Therefore, $DE=AE-AD=\boxed{2\text{cm}}$.


Question: As shown in the figure, $O$ is a point on the straight line $AB$, $\angle DOB=90^\circ$, $\angle EOC=90^\circ$. If $\angle DOE=50^\circ$, what is the degree measure of $\angle BOC$?

Solution:
\textbf{Solution:} From the given conditions, we have \\
$\because \angle BOC + \angle COD = 90^\circ$, \\
$\angle COD + \angle DOE = 90^\circ$, \\
$\therefore \angle BOC = \angle DOE = \boxed{50^\circ}$.


Question: As shown in the figure, $O$ is a point on line $AB$, $\angle DOB = 90^\circ$, $\angle EOC = 90^\circ$. If $OE$ bisects $\angle AOD$, what is the degree measure of $\angle BOE$?

Solution:
Solution:
\[
\because OE \text{ bisects } \angle AOD,
\]
\[
\therefore \angle DOE=\frac{1}{2}\angle AOD=45^\circ,
\]
\[
\text{Also, } \because \angle BOE = \angle BOD + \angle DOE,
\]
\[
\therefore \angle BOE = \boxed{135^\circ}.
\]


Question: As shown in the figure, it is known that there is a line $AE$, and $O$ is a point on line $AE$. $OB$ is the bisector of $\angle AOC$, $OD$ is the bisector of $\angle COE$. If $\angle AOB=30^\circ$, what is the degree measure of $\angle COE$?

Solution:
\textbf{Solution:} Since OB is the bisector of $\angle AOC$ and $\angle AOB=30^\circ$,\\
it follows that $\angle BOC=\angle AOB=30^\circ$,\\
thus $\angle AOC=2\angle AOB=60^\circ$.\\
Since $\angle AOC+\angle COE=180^\circ$,\\
we have $\angle COE=180^\circ-\angle AOC=180^\circ-60^\circ=\boxed{120^\circ}$;


Question: As shown in the figure, it is known that $AE$ is a straight line, and $O$ is a point on the straight line $AE$. $OB$ is the bisector of $\angle AOC$, and $OD$ is the bisector of $\angle COE$. Given that $\angle AOB=30^\circ$, what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Since $OD$ is the bisector of $\angle COE$,\\
it follows that $\angle COD= \frac{1}{2}\angle COE=60^\circ$,\\
therefore, $\angle BOD=\angle COD+\angle BOC=60^\circ+30^\circ=\boxed{90^\circ}$.


Question: As shown in the figure, with point O as the endpoint, rays OA, OB, OC, OD, and OE are drawn successively in clockwise direction. Moreover, OB is the bisector of $\angle AOC$, and OD is the bisector of $\angle COE$. If $\angle AOB=50^\circ$ and $\angle DOE=30^\circ$, what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Given that $OB$ is the bisector of $\angle AOC$,\\
hence $\angle BOC = \angle AOB = 50^\circ$;\\
since $OD$ is the bisector of $\angle COE$,\\
thus $\angle COD = \angle DOE = 30^\circ$,\\
therefore $\angle BOD = \angle BOC + \angle COD = 50^\circ + 30^\circ = \boxed{80^\circ}$.


Question: As shown in the figure, rays OA, OB, OC, OD, and OE are drawn sequentially in a clockwise direction from point O as the endpoint. Moreover, OB is the bisector of $\angle AOC$, and OD is the bisector of $\angle COE$. If $\angle AOD = 110^\circ$ and $\angle BOE = 100^\circ$, what is the measure of $\angle AOE$?

Solution: \textbf{Solution:} Given $OD$ bisects $\angle COE$, let $\angle EOD = \angle DOC = x$, $OB$ bisects $\angle AOC$, hence $\angle AOB = \angle COB$. Given $\angle BOE = 100^\circ$, we have: $\angle AOB = \angle COB = \angle BOE - \angle COE = 100^\circ - 2x$. From $\angle AOD = 110^\circ$, we have: $\angle AOD = \angle COD + \angle COB + \angle AOB = 110^\circ$, thus $x + (100^\circ - 2x) + (100^\circ - 2x) = 110^\circ$. Solving this gives $x = 30^\circ$, i.e., $\angle EOD = \angle DOC = 30^\circ$. Therefore, $\angle AOE = \angle AOD + \angle DOE = 110^\circ + 30^\circ = \boxed{140^\circ}$.


Question: As shown in the figure, with point O as the endpoint, make rays OA, OB, OC, OD, and OE in clockwise order. Moreover, OB is the bisector of $\angle AOC$ and OD is the bisector of $\angle COE$. When $\angle AOD = n^\circ$, then $\angle BOE = (150 - n)^\circ$. What is the degree measure of $\angle BOD$?

Solution: \textbf{Solution:} Let $\angle EOD = \angle DOC = x^\circ$ and $\angle AOB = \angle BOC = y^\circ$. According to the problem statement, we have $\angle AOD = \angle AOB + \angle BOC + \angle COD = y^\circ + y^\circ + x^\circ = n^\circ$ and $\angle BOE = \angle BOC + \angle COD + \angle DOE = y^\circ + x^\circ + x^\circ = (150 - n)^\circ$. Hence, $\angle AOD + \angle BOE = 3x^\circ + 3y^\circ = 150^\circ$, therefore $x^\circ + y^\circ = 50^\circ$. Since $\angle BOD = \angle BOC + \angle COD = x^\circ + y^\circ$, it follows that $\angle BOD = \boxed{50^\circ}$.


Question: In quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $O$, with $AD \parallel BC$ and $AO=CO$. Quadrilateral $ABCD$ is a parallelogram. Through point $O$, draw $OE \perp BD$ intersecting $BC$ at point $E$, and connect $DE$. If $\angle CDE = \angle CBD = 15^\circ$, what is the measure of $\angle ABC$ in degrees?

Solution: \textbf{Solution:} Given that OB = OD, and OE $\perp$ BD, it follows that BE = ED. Consequently, $\angle CBD = \angle BDE = 15^\circ$. Since $\angle CDE = 15^\circ$, we find that $\angle BDC = 30^\circ$. Given that quadrilateral ABCD is a parallelogram, it follows that AB $\parallel$ CD, implying $\angle ABD = \angle BDC = 30^\circ$. Therefore, $\angle ABC = \angle ABD + \angle CBD = 30^\circ + 15^\circ = \boxed{45^\circ}$.


Question: As shown in the figure, points $D$, $E$, and $F$ are the midpoints of the sides $AB$, $BC$, and $AC$ of $\triangle ABC$, respectively. Extend $DE$ to point $G$ such that $DE=EG$, and connect $AE$, $FG$. Quadrilateral $AEGF$ is a parallelogram. If $\angle BAC=90^\circ$ and $AD=AC=3$, find the length of $FG$.

Solution: \textbf{Solution:} From (1), we know: $DE=\frac{1}{2}AC$, $DE\parallel AC$, since $\angle BAC=90^\circ$, therefore $\angle EAD=90^\circ$. Also, since $AC=AD=3$, therefore $DE=\frac{3}{2}$, hence $AE=\sqrt{DE^2+AD^2}=\sqrt{\left(\frac{3}{2}\right)^2+3^2}=\frac{3\sqrt{5}}{2}$. From (1), it is known that quadrilateral AEGF is a parallelogram, therefore $FG=\boxed{\frac{3\sqrt{5}}{2}}$.


Question: As shown in the figure, in parallelogram \(ABCD\), \(\angle B = 60^\circ\), \(BC = 6\), and point \(E\) is the midpoint of side \(BC\). Triangle \(ABE\) is folded to the right along \(AE\), with point \(B\) falling on \(B'\). Connect \(CB'\) and extend it to intersect \(AD\) at point \(F\). Prove that quadrilateral \(AECF\) is a parallelogram.

Solution:
Proof:

Because $\triangle ABE$ is folded along $AE$ with point $B$ landing on $B'$,

it follows that $\triangle ABE \cong \triangle AB'E$,

thus $\angle AEB = \angle AEB' = \frac{1}{2}\angle BEB'$, and $EB = EB'$,

since $E$ is the midpoint of side $BC$,

we have $EB = EC = EB'$,

therefore, $\angle EB'C = \angle ECB' = \frac{1}{2}\angle BEB'$,

hence, $\angle BEA = \angle ECB'$,

which implies $AE \parallel CF$,

given that quadrilateral $ABCD$ is a parallelogram,

we have $AD \parallel BC$,

thus, quadrilateral \boxed{AECF} is a parallelogram.

This completes the proof.


Question: Given, as shown in the diagram, in parallelogram $ABCD$, $\angle B=60^\circ$, $BC=6$, and point $E$ is the midpoint of side $BC$. Triangle $ABE$ is folded to the right along $AE$ so that point $B$ falls at point $B'$. Line $CB'$ is drawn and extended to intersect $AD$ at point $F$. When $AB' \perp CD$, what is the length of $AE$?

Solution:
\textbf{Solution:} Given $AB \parallel CD$ and $AB' \perp CD$, \\
it follows that $AB' \perp AB$, which means $\angle BAB' = 90^\circ$, \\
by folding, we have $\angle BAE = \angle EAB' = 45^\circ$, \\
as shown in the diagram, draw $EH \perp AB$ at point H, \\
in $\triangle BEH$, $\angle B = 60^\circ$ and $BE = 3$, \\
thus $BH = \frac{3}{2}$, $HE = \sqrt{BE^2 - BH^2} = \sqrt{3^2 - \left(\frac{3}{2}\right)^2} = \frac{3}{2}\sqrt{3}$, \\
in $\triangle AEH$, $\angle HAE = \angle AEH = 45^\circ$, $AH = EH = \frac{3}{2}\sqrt{3}$, \\
therefore $AE = \sqrt{2}HE = \boxed{\frac{3}{2}\sqrt{6}}$.


Question: As shown in the figure, the diagonals $AC$ and $BD$ of the quadrilateral $\square ABCD$ intersect at point $Q$, with $E$ and $F$ being the midpoints of $OB$ and $OD$, respectively. Lines $AE$, $AF$, $CE$, and $CF$ are connected. Quadrilateral $AECF$ is a parallelogram. If $AB \perp AC$ and $AB=3$, $BC=5$, what is the length of $BD$?

Solution: \textbf{Solution:} Given that $AB \perp AC$, it follows that $\angle BAC = 90^\circ$. In right triangle $\triangle ABC$, we have $AC = \sqrt{BC^2 - AB^2} = \sqrt{5^2 - 3^2} = 4$. Thus, $OA = \frac{1}{2}AC = 2$. In right triangle $\triangle AOB$, $BO = \sqrt{AB^2 + OA^2} = \sqrt{3^2 + 2^2} = \sqrt{13}$, hence $BD = 2BO = 2\sqrt{13}$. Therefore, the length of $BD$ is $\boxed{2\sqrt{13}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $BD=AD$. Extend $CB$ to point $E$ such that $BE=BD$, and connect $AE$. Quadrilateral $AEBD$ is a rhombus; connect $DE$ and intersect $AB$ at point $F$. If $DC=\sqrt{10}$ and the ratio $DC:DE=1:3$, what is the length of $AD$?

Solution:
\textbf{Solution}: As shown in the diagram, extend DE to meet AB at F, \\
$\because$ Quadrilateral AEBD is a rhombus,\\
$\therefore$ AB$\perp$DE,\\
$\therefore$ $\angle$EFB$=90^\circ$. \\
$\because$ Quadrilateral ABCD is a parallelogram,\\
$\therefore$ AB$ \parallel $DC.\\
$\therefore$ $\angle$EDC $=\angle$EFB$=90^\circ$.\\
$\because$ DC$= \sqrt{10}$ and the ratio DC$\colon$ DE$=1\colon 3$,\\
$\therefore$ DE$= 3\sqrt{10}$.\\
In $\triangle$EDC, by the Pythagorean theorem, we have $EC=\sqrt{ED^{2}+DC^{2}}=10$\\
$\therefore$ AD$=\boxed{5}$.


Question: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, $DE \parallel AC$, and $AE \parallel BD$. Quadrilateral $AODE$ is a rectangle; if $AB=10$ and $\angle ABC=60^\circ$, what is the area of quadrilateral $AODE$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rhombus,\\
$\therefore$ $AD=AB=BC=10$, $OA=OC$, $AC\perp BD$,\\
Since $\angle ABC=60^\circ$,\\
$\therefore$ $\triangle ABC$ is an equilateral triangle,\\
$\therefore$ $AC=AB=10$,\\
$\therefore$ $OA=\frac{1}{2}AC=5$,\\
In $Rt\triangle AOD$, by the Pythagorean theorem, $OD=\sqrt{AD^{2}-OA^{2}}=\sqrt{10^{2}-5^{2}}=5\sqrt{3}$,\\
From (1), quadrilateral $AODE$ is a rectangle,\\
$\therefore$ the area of quadrilateral $AODE$ $=OA\cdot OD=5\times 5\sqrt{3}=\boxed{25\sqrt{3}}$.


Question: As shown in the figure, quadrilateral ABCD is a rhombus, with $BE \perp AD$ and $BF \perp CD$, where the feet of the perpendiculars are points E and F, respectively. $BE = BF$; When the diagonals of the rhombus ABCD satisfy $AC = 8$ and $BD = 6$, what is the length of $BE$?

Solution:
\textbf{Solution:} Since the diagonals of rhombus ABCD are $AC=8$ and $BD=6$,\\
it follows that $OA=OC=\frac{1}{2}AC=4$, $OB=OD=\frac{1}{2}BD=3$, and $\angle AOD=90^\circ$,\\
thus $AD=\sqrt{OA^{2}+OD^{2}}=5$,\\
Given that the area of rhombus ABCD is $S_{\text{rhombus} ABCD}=\frac{1}{2}AC\cdot BD=\frac{1}{2}AD\cdot BE$,\\
it follows that $BE=\frac{AC\cdot BD}{2AD}=\boxed{\frac{24}{5}}$.


Question: As shown in the figure, in the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, with $DE \parallel AC$ and $AE \parallel BD$. Quadrilateral $AODE$ is a rectangle; if $AB=2\sqrt{3}$ and $\angle BCD=120^\circ$, connect $CE$. What is the length of $CE$?

Solution:
\textbf{Solution:} Given that $\angle BCD=120^\circ$ and quadrilateral ABCD is a rhombus,\\
$\therefore \angle BAD=\angle BCD=120^\circ$, $\angle CAB=\angle CAD=60^\circ$, AB=BC,\\
$\therefore \triangle ABC$ is an equilateral triangle,\\
$\therefore AC=AB=2\sqrt{3}$, OB=OD=AE=3,\\
In $\triangle AEC$, $EC=\sqrt{AC^2+AE^2}=\sqrt{(2\sqrt{3})^2+3^2}=\boxed{\sqrt{21}}$.


Question: As shown in the figure, in quadrilateral ABCD, $AD\parallel BC$, $\angle A=90^\circ$, $BD=BC$, point E is the midpoint of CD, ray BE intersects the extension of AD at point F, connect CF. Prove that the quadrilateral BCFD is a rhombus.

Solution:
\textbf{Solution.} Proof: Since AF$\parallel$BC,\\
it follows that $\angle$DCB=$\angle$CDF, $\angle$FBC=$\angle$BFD,\\
since point E is the midpoint of CD,\\
it follows that DE=EC,\\
in $\triangle$ BCE and $\triangle$ FDE,\\
since $\left\{\begin{array}{l}
\angle FBC=\angle BFD \\
\angle DCB=\angle CDF \\
DE=EC
\end{array}\right.$,\\
it follows that $\triangle$ BCE$\cong$ $\triangle$ FDE,\\
thus, DF=BC,\\
and since DF$\parallel$BC,\\
it follows that quadrilateral BCFD is a parallelogram,\\
since BD=BC,\\
it follows that \boxed{quadrilateral\ BCFD\ is\ a\ rhombus};


Question: As shown in the figure, in quadrilateral $ABCD$, $AD \parallel BC$, $\angle A=90^\circ$, $BD=BC$. Point $E$ is the midpoint of $CD$, the ray $BE$ extends and intersects the extension line of $AD$ at point $F$, and $CF$ is connected. If $AD=1$, $BC=2$, find the length of $BF$.

Solution:
\textbf{Solution:} Since quadrilateral BCFD is a rhombus,\\
it follows that BD=DF=BC=2,\\
in $\triangle BAD$, AB=$\sqrt{BD^{2}-AD^{2}}=\sqrt{3}$,\\
Since AF=AD+DF=1+2=3, in $\triangle BAF$, BF=$\sqrt{AB^{2}+AF^{2}}=\boxed{2\sqrt{3}}$.


Question: Given that AP is a ray outside of square ABCD, and $B'$ is the reflection point of B about line AP, connect $B'D$. As shown in Figure 1. If $\angle BAP=20^\circ$, find the magnitude of the angle $\angle ADB'$.

Solution:
\textbf{Solution:} Draw $A{B}^{\prime}$, as shown in figure 1,\\
$\because$ ${B}^{\prime}$ is the symmetric point of B with respect to the line AP,\\
$\therefore$ $AB=A{B}^{\prime}$,\\
$\therefore$ $\angle BAP=\angle {B}^{\prime}AP=20^\circ$\\
$\because$ quadrilateral ABCD is a square,\\
$\therefore$ $A{B}^{\prime}=AB=AD$,\\
$\therefore$ $\angle A{B}^{\prime}D=\angle AD{B}^{\prime}$,\\
$\because$ $\angle {B}^{\prime}AD=\angle {B}^{\prime}AB+\angle BAD=90^\circ+40^\circ=130^\circ$,\\
$\therefore$ $\angle AD{B}^{\prime}=\boxed{25^\circ}$;


Question: As shown in the figure, the diagonals AC and BD of the rhombus ABCD intersect at point O, and $DE\parallel OC$, $CE\parallel OD$. The quadrilateral OCED is a rectangle; Given that $AC=4\sqrt{5}$ and $BD=2\sqrt{5}$, find the perimeter and area of the rectangle OCED.

Solution:
\textbf{Solution:} \[\quad \text{Since in the rhombus } ABCD, \; OC=\frac{1}{2}AC=2\sqrt{5}, \; OD=\frac{1}{2}BD=\sqrt{5}, \\
\therefore \text{the perimeter of the rectangle } OCED \text{ is: } (2\sqrt{5}+\sqrt{5})\times 2=\boxed{6\sqrt{5}}, \text{ and the area of } OCED \text{ is: } 2\sqrt{5}\times \sqrt{5}=\boxed{10}\].


Question: As shown in the figure, in rectangle $ABCD$, $E$ is the midpoint of $AD$. After folding $\triangle ABE$ along $BE$, $\triangle GBE$ is obtained, and point $G$ is inside the rectangle $ABCD$. Extending $BG$ intersects $DC$ at point $F$ such that $GF=DF$; If $DC=9$ and $DE=2CF$, what is the length of $AD$?

Solution:
\textbf{Solution:} \\
Since quadrilateral ABCD is a rectangle, \\
therefore AB=CD=BG=9, AD=BC, \\
let CF=x, then DF=9-x, DE=2CF=2x, BC=AD=2DE=4x, \\
since GF=DF, \\
therefore GF=9-x, \\
therefore BF=BG+GF=9+9-x=18-x, \\
in $\triangle BCF$, $BC^2+CF^2=BF^2$, \\
therefore $(4x)^2+x^2=(18-x)^2$, \\
upon solving, we get $x=\frac{9\sqrt{17}-9}{8}$ or $x=\frac{9-9\sqrt{17}}{8}$ (discard), \\
therefore AD=4x=$\boxed{\frac{9\sqrt{17}-9}{2}}$;


Question: As shown in the diagram, it is known that in the equilateral $\triangle ABC$, $BE$ and $CD$ are medians on sides $AC$ and $AB$, respectively, and $BE$ and $CD$ intersect at point $O$. Points $M$ and $N$ are the midpoints of line segments $OB$ and $OC$, respectively. Quadrilateral $DENM$ is a rectangle; if the side length of the equilateral $\triangle ABC$ is $12$, find the area of rectangle $DENM$.

Solution:
\textbf{Solution:} Given $\triangle ABC$ with side length 12, and BE as the median,\\
thus $BC=AC=12$, $\angle BEC=90^\circ$, CE=6,\\
thus $DE=MN=\frac{1}{2}BC=6$,\\
using the Pythagorean theorem we get $BE=\sqrt{BC^{2}-EC^{2}}=6\sqrt{3}$,\\
from (1) we know, OM=OE, with M being the midpoint of OB,\\
thus $BM=OM=OE$,\\
thus $OM=OE=\frac{1}{3}BE=2\sqrt{3}$,\\
since quadrilateral DENM is a rectangle, and $\angle DEO=30^\circ$,\\
thus $\angle OEN=60^\circ$, implying $\triangle OEN$ is an equilateral triangle, hence EN=OE=2$\sqrt{3}$,\\
thus the area of rectangle DENM $=EN \times DE=2\sqrt{3} \times 6=\boxed{12\sqrt{3}}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=3\text{cm}$, and $BC=6\text{cm}$. Point $P$ starts moving from point $D$ towards point $A$ and stops upon reaching point $A$; simultaneously, point $Q$ starts moving from point $B$ towards point $C$ and stops upon reaching point $C$. The speeds of points $P$ and $Q$ are both $1\text{cm/s}$. Connect $PQ$, $AQ$, and $CP$. Let the time for points $P$ and $Q$ to move be $t$ seconds. For what value of $t$ is quadrilateral $ABQP$ a rectangle?

Solution:
\textbf{Solution:} From the given information, we have $BQ=DP=t, \, AP=CQ=6-t$,\\
In rectangle $ABCD$, $\angle B=90^\circ$, $AD\parallel BC$,\\
When $BQ=AP$, quadrilateral $ABQP$ forms a rectangle,\\
$\therefore t=6-t$, solving for $t$ yields $t=\boxed{3}$,\\
Therefore, when $t=3$, quadrilateral $ABQP$ is a rectangle.


Question: As shown in the figure, in rectangle $ABCD$, $AB=3\text{cm}$, and $BC=6\text{cm}$. Point $P$ starts moving from point $D$ towards point $A$ and stops upon reaching point $A$; simultaneously, point $Q$ starts moving from point $B$ towards point $C$ and stops upon reaching point $C$. The speeds of points $P$ and $Q$ are both $1\text{cm/s}$. Connect $PQ$, $AQ$, $CP$. Let the time during which points $P$ and $Q$ move be $t$ seconds. When $t$ equals what value, is quadrilateral $AQCP$ a rhombus?

Solution:
\textbf{Solution:} Since $AP=CQ$ and $AP\parallel CQ$, \\
$\therefore$ quadrilateral $AQCP$ is a parallelogram, \\
$\therefore$ when $AQ=CQ$, quadrilateral $AQCP$ becomes a rhombus, \\
i.e., when $\sqrt{3^2+t^2}=6-t$, quadrilateral $AQCP$ is a rhombus, solving this gives $t=\boxed{\frac{9}{4}}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB = 3\,cm$, $BC = 6\,cm$. Point $P$ starts moving from point $D$ towards point $A$ and stops upon reaching $A$; simultaneously, point $Q$ starts moving from point $B$ towards point $C$ and stops upon reaching $C$. The speeds of points $P$ and $Q$ are $1\,cm/s$ each. Connect $PQ$, $AQ$, $CP$. Let the time taken for points $P$ and $Q$ to move be $t$ seconds. When $t=\frac{9}{4}$ seconds, the quadrilateral $AQCP$ forms a rhombus; calculate the perimeter and area of the rhombus $AQCP$?

Solution:
\textbf{Solution:} When $t=\frac{9}{4}$, $AQ=\frac{15}{4},\ CQ=\frac{15}{4}$, \\
then the perimeter is: $4AQ=4\times \frac{15}{4}=\boxed{15}\ (\text{cm})$, \\
and the area is: $CQ\cdot AB=\frac{15}{4}\times 3=\boxed{\frac{45}{4}}\ (\text{cm}^2)$.


Question: As shown in the figure, the quadrilateral $ABCD$ is a rectangle, with two diagonals $AC$ and $BD$ intersecting at point $O$, and $AE \parallel BD$, $DE \parallel AC$. If $BC = 4$ and $AC = 6$, what is the area of the quadrilateral $AODE$?

Solution:
\textbf{Solution:}

Since $BC=4$ and $AC=6$,

we have $AB=\sqrt{AC^{2}-BC^{2}}=\sqrt{36-16}=2\sqrt{5}$.

Given that quadrilateral ABCD is a rectangle,

it follows that $OB=OD$,

implying $Area_{\triangle AOD}=\frac{1}{2}Area_{\triangle ABD}$.

Since quadrilateral AODE is a rhombus,

it follows that $Area_{\triangle AOD}=\frac{1}{2}Area_{\text{rhombus} AODE}$.

Therefore, $Area_{\text{rhombus} AODE}=Area_{\triangle ABD}=\frac{1}{2}AB\cdot AD=\frac{1}{2}\times 2\sqrt{5}\times 4=\boxed{4\sqrt{5}}$.


Question: As shown in the figure, $O$ is the intersection point of the diagonals of rectangle $ABCD$, with $DE \parallel AC$ and $CE \parallel BD$. Quadrilateral $OCED$ is a rhombus. If $AB=3$ and $BC=4$, what is the area of the quadrilateral $OCED$?

Solution:
\textbf{Solution:} Given that $AB=3$, $BC=4$, \\
Therefore, the area of rectangle $ABCD$ is equal to $3\times 4=12$, \\
Since the area of $\triangle ODC=\frac{1}{4}$ of the area of rectangle $ABCD=3$, \\
Hence, the area of quadrilateral $OCED$ equals $2$ times the area of $\triangle ODC=\boxed{6}$.


Question: As shown in the figure, P is a point on the diagonal AC of the rhombus ABCD, $PE\perp AB$ at point E, and $PF\perp AD$ at point F. Given that $\angle BAD=60^\circ$, $PE=1$, find the length of AE.

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus and $\angle BAD=60^\circ$,\\
it follows that $\angle PAE=30^\circ$\\
Since $PE \perp AB$,\\
it follows that $AP=2PE=2$,\\
therefore $AE=\sqrt{AP^{2}-PE^{2}}=\boxed{\sqrt{3}}$;


Question: As shown in Figure 1, the quadrilateral $ABCD$ is a square, and point $G$ is any point on side $BC$. $DE \perp AG$ at point $E$, and $BF \parallel DE$ intersecting $AG$ at point $F$. $DE = AF$; If $AB=4$ and $BG=3$, what is the length of $AF$?

Solution:
\textbf{Solution:} In right triangle $\text{Rt}\triangle ABG$, $AB=4$, $BG=3$. According to the Pythagorean theorem, $AG=5$.\\
$\because \frac{1}{2}AB \cdot BG = \frac{1}{2}AG \cdot BF$,\\
$\therefore BF = \frac{AB \cdot BG}{AG} = \frac{12}{5}$,\\
In right triangle $\text{Rt}\triangle ABF$, $AF = \sqrt{AB^2 - BF^2} = \boxed{\frac{16}{5}}$;


Question: As shown in the figure, within the square $ABCD$, $P$ is a point on the diagonal $BD$, and point $E$ is on the extension of $AB$, such that $PC=PE$ and $PA=PE$; $AE=\sqrt{2}DP$; given that the side length of the square is $2$, what is the minimum value of $CP+\frac{1}{2}BP$?

Solution:
\textbf{Solution:} As shown in the figure, rotate segment BD $30^\circ$ counterclockwise around point B, intersecting AD at G. Draw $PH\perp BG$ at H through point P, and connect AC, intersecting BD at O,\\
$\because \angle PBH=30^\circ$,\\
$\therefore PH=\frac{1}{2}PB$,\\
$\therefore$ when points C, P, H are collinear, $PC+\frac{1}{2}PB$ has the minimum value CH,\\
$\because$ the side length of the square is 2,\\
$\therefore OC=OB=\sqrt{2}$,\\
$\because \angle PBH+\angle BPH=90^\circ$, $\angle OCP+\angle OPC=90^\circ$, $\angle OPC=\angle BPH$,\\
$\therefore \angle OCP=\angle PBH=30^\circ$,\\
$\therefore OP=\frac{1}{2}CP$,\\
In $\triangle OPC$, $PC^2=OP^2+OC^2$, that is $4OP^2=OP^2+2$,\\
$\therefore OP=\frac{\sqrt{6}}{3}$ (discard the negative value),\\
$\therefore CP=\frac{2\sqrt{6}}{3}$,\\
$\therefore PB=\sqrt{2}-\frac{\sqrt{6}}{3}$, $PH=\frac{1}{2}PB=\frac{\sqrt{2}}{2}-\frac{\sqrt{6}}{6}$,\\
$\therefore CH=CP+PH=\frac{2\sqrt{6}}{3}+\frac{\sqrt{2}}{2}-\frac{\sqrt{6}}{6}=\boxed{\frac{\sqrt{6}+\sqrt{2}}{2}}$,\\
$\therefore CP+\frac{1}{2}BP$'s minimum value is $\boxed{\frac{\sqrt{6}+\sqrt{2}}{2}}$.


Question: As shown in the figure, in square ABCD, point F is on CD, and BF intersects AC at point E. What is the measure of $\angle ACB$ in degrees?

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


Question: As shown in the figure, in the square $ABCD$, point $F$ is on the side $CD$, and $BF$ intersects $AC$ at point $E$. $\triangle ABE \cong \triangle ADE$, If $\angle CBF = 20^\circ$, then what is the measure of $\angle AED$?

Solution: By the given information, we have $\angle AED = \boxed{65^\circ}$.


Question: As shown in the figure, within the square $ABCD$, $P$ is a point on the diagonal $AC$, and point $E$ lies on the extension of $BC$, with $PE=PB$. When $PC=CE$, what is the degree measure of $\angle CDP$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a square,\\
we have AB=BC=CD=AD, $\angle$BCP=$\angle$DCP=$45^\circ$, $\angle$BCD=$\angle$DCE=$90^\circ$,\\
thus, $\angle$PCE=$45^\circ+90^\circ=135^\circ$,\\
In $\triangle$BCP and $\triangle$DCP,\\
$\left\{\begin{array}{l}
BC=DC\\
\angle BCP=\angle DCP\\
CP=CP
\end{array}\right.$,\\
thus, $\triangle$BCP $\cong$ $\triangle$DCP (SAS),\\
therefore, BP=DP, $\angle$CBP=$\angle$CDP,\\
since PE=PB, PC=CE,\\
thus, PD=PE, $\angle$CBP=$\angle$PEB=$\angle$CPE= $\frac{1}{2}(180^\circ - 135^\circ)=22.5^\circ$,\\
therefore, $\angle$CDP=$\boxed{22.5^\circ}$.


Question: As shown in the figure, in square $ABCD$, $E$ is the midpoint of $BC$, and $F$ is a point on $CD$ with $CF=\frac{1}{4}CD$. If the length of $CF$ is 1, what are the lengths of $EF$ and $AE$ respectively?

Solution:
\textbf{Solution}: Given by the problem, $ABCD$ is a square, \\
$\therefore AB=BC=CD=AD$. \\
$\because E$ is the midpoint of $BC$, \\
$\therefore BE=EC$. \\
Also, $\because CF=\frac{1}{4}CD$, \\
$\therefore AB=2BE=2EC=4FC$. \\
$\therefore AB=4$, $BE=EC=2$. \\
In $\triangle ABE$, \\
$AE=\sqrt{AB^2+BE^2}=\sqrt{4^2+2^2}=2\sqrt{5}$. \\
$\therefore$ Similarly, $EF=\sqrt{EC^2+CF^2}=\sqrt{2^2+1^2}=\sqrt{5}$. \\
$\therefore$ The lengths of $EF$ and $AE$ are $\boxed{\sqrt{5}}$ and $\boxed{2\sqrt{5}}$, respectively.


Question: In the square $ABCD$, $E$ is the midpoint of $BC$, and $F$ is a point on $CD$ such that $CF = \frac{1}{4}CD$. Prove what the degree measure of $\angle AEF$ is.

Solution:
\textbf{Solution:} Let the length of $CF$ be $a$ \\
Then $EF=\sqrt{5}a$, $AE=2\sqrt{5}a$ \\
Connect $AF$, thus $DF=3a$, and $AD=4a$, \\
In $Rt\triangle ADF$, $AF=\sqrt{AD^{2}+DF^{2}}=5a$ \\
In $\triangle AEF$, \\
$\because AE^{2}+EF^{2}=AF^{2}$ \\
$\therefore \boxed{\angle AEF=90^\circ}$.


Question: As shown in the figure, a right-angled triangle paper $ABO$ is placed in the Cartesian coordinate system, with point $A(\sqrt{3}, 0)$, point $B(0, 1)$, and point $O(0, 0)$. $P$ is a moving point on side $AB$ (point $P$ does not coincide with points $A$ or $B$), and the paper is folded along $OP$ to obtain the corresponding point $A'$ of point $A$. As shown in Figure 1, when point $A'$ is in the first quadrant and satisfies $A'B \perp OB$, what are the coordinates of point $A'$?

Solution:
\textbf{Solution:} Let point $A\left(\sqrt{3}, 0\right)$ and point $B\left(0, 1\right)$. Hence, $OA=\sqrt{3}$ and $OB=1$. According to the property of folding, we have $OA'=OA=\sqrt{3}$. Since $A'B\perp OB$, it follows that $\angle A'BO=90^\circ$. In right triangle $\triangle A'OB$, we find that $A'B=\sqrt{OA'^{2}-OB^{2}}=\sqrt{2}$,\\
thus the coordinates of point $A'$ are $\boxed{\left(\sqrt{2}, 1\right)}$.


Question: As shown in the figure, a right-angled triangle paper $ABO$ is placed in the Cartesian coordinate system. It is known that point $A\left(\sqrt{3}, 0\right)$, point $B\left(0, 1\right)$, and point $O\left(0, 0\right)$. $P$ is a moving point on side $AB$ (point $P$ does not coincide with points $A$ or $B$). By folding the paper along $OP$, we obtain the corresponding point $A'$ of point $A$. As shown in Figure 2, when $P$ is the midpoint of $AB$, what is the length of $A'B$?

Solution:
\textbf{Solution:} In right triangle $Rt\triangle AOB$, we have $OA=\sqrt{3}$ and $OB=1$, therefore $AB=\sqrt{OA^{2}+OB^{2}}=2$. Since $P$ is the midpoint of $AB$, we have $AP=BP=1$ and $OP=\frac{1}{2}AB=1$. Thus, $OP=OB=BP$, which means $\triangle BOP$ is an equilateral triangle, therefore $\angle OBP=\angle BOP=\angle BPO=60^\circ$. Consequently, $\angle OPA=180^\circ-\angle BPO=120^\circ$. According to the property of reflection, $\angle OPA'=\angle OPA=120^\circ$ and $PA'=PA=1$, which gives $\angle BPA'=\angle OBP=60^\circ$. Therefore, $OB\parallel PA'$. Moreover, since $OB=PA'=1$, the quadrilateral $OPA'B$ is a parallelogram,\\
hence, $A'B=OP=\boxed{1}$.


Question: As shown in the figure, a right-angled triangle paper $ABO$ is placed in the Cartesian coordinate system, where point $A(\sqrt{3}, 0)$, point $B(0, 1)$, and point $O(0, 0)$. $P$ is a moving point on side $AB$ (point $P$ does not coincide with points $A$ or $B$), and the paper is folded along $OP$ to obtain the corresponding point $A'$ of point $A$. When $\angle BPA' = 30^\circ$, directly write down the coordinates of point $P$.

Solution: \textbf{Solution:} When $\angle BPA'=30^\circ$, the coordinates of point $P$ are $\boxed{\left(\frac{3-\sqrt{3}}{2}, \frac{3-\sqrt{3}}{2}\right)}$ or $\boxed{\left(\frac{2\sqrt{3}-3}{2}, \frac{\sqrt{3}}{2}\right)}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=4$, $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively. Line $CD$ is drawn, and through point $E$, line $EF\parallel DC$ intersects the extension of $BC$ at point $F$. $DE=CF$, if $\angle B=60^\circ$, what is the length of $EF$?

Solution:
\textbf{Solution:} Since $AB=AC$ and $\angle B=60^\circ$ \\
it follows that $\triangle ABC$ is an equilateral triangle \\
Hence, $BC=AB=4$ \\
Given $D$ is the midpoint of $AB$ \\
it follows $\angle BDC=90^\circ$ and $BD=\frac{1}{2}AB$ \\
Thus, $BD=2$ \\
Therefore, $CD=\sqrt{BC^{2}-BD^{2}}=\sqrt{4^{2}-2^{2}}=2\sqrt{3}$ \\
Since quadrilateral $DCFE$ is a parallelogram \\
it follows that $EF=CD=2\sqrt{3}$ \\
Thus, the length of $EF$ is $\boxed{2\sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $D$ is any point on side $AB$, $F$ is the midpoint of $AC$, and line $CE \parallel AB$ is drawn through point $C$ intersecting the extension of $DF$ at point $E$. $AE$ and $CD$ are connected. Prove that the quadrilateral $ADCE$ is a parallelogram.

Solution:
Proof: Since $AB \parallel CE$, \\
it follows that $\angle CAD = \angle ACE$ and $\angle ADE = \angle CED$. \\
Since $F$ is the midpoint of $AC$, \\
then $AF = CF$. \\
In $\triangle AFD$ and $\triangle CFE$, \\
we have $\left\{\begin{array}{l}
\angle CAD = \angle ACE \\
\angle ADE = \angle CED \\
AF = CF
\end{array}\right.$, \\
thus $\triangle AFD \cong \triangle CFE$ (AAS), \\
therefore $DF = EF$, \\
hence \boxed{quadrilateral \, ADCE \, is \, a \, parallelogram}.


Question: As shown in the figure, in $\triangle ABC$, D is an arbitrary point on side AB, F is the midpoint of AC, and a line is drawn through point C such that $CE \parallel AB$ intersects the extension of DF at point E. Line segments AE and CD are then drawn. If $\angle B=30^\circ$, $\angle CAB=45^\circ$, and $AC=\sqrt{2}$, find the length of AB.

Solution:
\textbf{Solution:} Construct $CG\perp AB$ at point G,\\
since $\angle CAB=45^\circ$,\\
it follows that $AG=CG$,\\
in $\triangle ACG$, $\angle AGC=90^\circ$,\\
thus $AG^2+CG^2=AC^2$,\\
since $AC=\sqrt{2}$,\\
it follows that $CG=AG=1$,\\
since $\angle B=30^\circ$,\\
it follows that $CG=\frac{1}{2}BC$,\\
thus $BC=2$,\\
in $\triangle BCG$, $BG=\sqrt{BC^2-CG^2}=\sqrt{4-1}=\sqrt{3}$,\\
thus $AB=AG+BG=1+\sqrt{3}=\boxed{1+\sqrt{3}}$.


Question: As shown in the figure, point $O$ is the intersection point of the diagonals of rhombus $ABCD$, with $CE\parallel BD$ and $EB\parallel AC$. Line $OE$ is drawn and intersects $BC$ at point $F$. Quadrilateral $OBEC$ is a rectangle. Line $DG$ is drawn perpendicular to the extension of line $BA$ at point $G$, and line $OG$ is connected. If $OC:OB=1:2$ and $OE=4\sqrt{5}$, find the length of $OG$.

Solution:
\textbf{Solution:} \text{ In rhombus } ABCD, DO = OB,
\[
\text{and since } DG \perp BA,
\]
\[
\therefore \angle DGB = 90^\circ,
\]
\[
\therefore OG = \frac{1}{2}BD = OB,
\]
\[
\text{In rectangle } OBEC, OE = BC = 4\sqrt{5},
\]
\[
\text{and since } OC:OB = 1:2,
\]
\[
\therefore \text{let } OC = x, OB = 2x,
\]
\[
\text{By the Pythagorean theorem: } OC^2 + OB^2 = BC^2, \text{ namely } x^2 + (2x)^2 = (4\sqrt{5})^2,
\]
\[
\text{Solving, we get } x = 4, \text{ thus } OB = 8,
\]
\[
\therefore OG = \boxed{8}
\]


Question: As shown in the figure, the rectangle \(ABCD\) is rotated counterclockwise around point \(A\) to obtain rectangle \(AEFG\), such that point \(E\) falls on the diagonal \(BD\). Connect \(DG\) and \(DF\). If \(\angle BAE = 50^\circ\), what is the degree measure of \(\angle DGF\)?

Solution:
\textbf{Solution:} Given by rotation, $AB=AE$, $AD=AG$, $\angle BAD=\angle EAG=\angle AGF=90^\circ$,\\
$\therefore \angle BAE=\angle DAG=50^\circ$,\\
$\therefore \angle AGD=\angle ADG= \frac{180^\circ-50^\circ}{2}=65^\circ$,\\
$\therefore \angle DGF=90^\circ-65^\circ=\boxed{25^\circ};$


Question: As shown in the figure, in quadrilateral $ABCD$, point $E$ is on side $AC$, and point $F$ is on side $AD$, with $AF=CE$. $EF$ intersects diagonal $BD$ at point $O$. $O$ is the midpoint of $BD$. If $EF\perp BD$ and the perimeter of quadrilateral $ABCD$ is 24, when connecting $BF$, what is the perimeter of $\triangle ABF$?

Solution: \textbf{Solution:} Given the insufficient information from the problem statement, it is not possible to determine the perimeter of $\triangle ABF$ directly. Additional information is required to solve this problem.


Question: As shown in the rectangle $ABCD$, the lengths of sides $AB$ and $BC$ ($AB < BC$) are the two roots of the equation ${x}^{2}-7x+12=0$. Point $P$ starts from point $A$ and moves along the edge $ABC$ in the direction $A\to B\to C\to A$ at a speed of $1$ unit per second, with the movement time being $t$ seconds. What are the lengths of $AB$ and $BC$?

Solution:
\textbf{Solution:} Since $x^{2}-7x+12=0$,\\
then $(x-3)(x-4)=0$,\\
therefore $x_{1}=3$, $x_{2}=4$.\\
Hence $AB=\boxed{3}$, $BC=\boxed{4}$;


Question: As shown in figure, within the rectangle $ABCD$, the lengths of sides $AB$ and $BC$ $(AB<BC)$ are the two roots of the equation ${x}^{2}-7x+12=0$. Point $P$ starts from point $A$ and moves along the perimeter of $\triangle ABC$ in the direction of $A\to B\to C\to A$ at a speed of $1$ unit per second, and the movement time is $t$ seconds. When point $P$ moves onto side $BC$, try to find the value of the movement time $t$, so that the length of $AP$ is $\sqrt{10}$.

Solution:
\textbf{Solution:} By the given, we have $3^2+(t-3)^2=10$,\\
$\therefore t_1=4$, $t_2=2$ (discarded),\\
thus, for $t=\boxed{4}$, we have $AP=\sqrt{10}$;


Question: As shown in the figure, in rectangle $ABCD$, the lengths of sides $AB$ and $BC$ ($AB<BC$) are the two roots of the equation ${x}^{2}-7x+12=0$. Point $P$ starts from point $A$ and moves along the edge $A\to B\to C\to A$ of $\triangle ABC$ at a speed of $1$ unit per second. The movement lasts for $t$ seconds. When point $P$ moves to side $AC$, there exists a point $P$ such that $\triangle CDP$ is an isosceles triangle. What is the value of the movement time $t$?

Solution:
\textbf{Solution:} There exists a point $P$ such that $\triangle CDP$ is an isosceles triangle.\\
(1) When $PC=CD=3$, $t=\frac{3+4+3}{1}=10$ seconds;\\
(2) When $PD=PC$ (i.e., $P$ is the midpoint of the diagonal $AC$), $AB=3$, $BC=4$.\\
$\therefore AC=\sqrt{AB^2+BC^2}=5$, $CP=\frac{1}{2}AC=2.5$,\\
$\therefore t=\frac{3+4+2.5}{1}=9.5$ seconds;\\
(3) When $PD=CD=3$, construct $DQ\perp AC$ at $Q$,\\
$DQ=\frac{\frac{1}{2}\times 3\times 4}{\frac{1}{2}\times 5}=\frac{12}{5}$, $PQ=\sqrt{3^2-\left(\frac{12}{5}\right)^2}=\frac{9}{5}$,\\
$\therefore PC=2PQ=\frac{18}{5}$,\\
$\therefore t=\frac{3+4+\frac{18}{5}}{1}=\frac{53}{5}$ seconds,\\
It is known that when $t$ is $10$ seconds, $9.5$ seconds, or $\frac{53}{5}$ seconds, $\triangle CDP$ is an isosceles triangle. Therefore, the possible values for the duration $t$ are $10$ seconds, $9.5$ seconds, and $\boxed{\frac{53}{5}}$ seconds.


Question: As shown in the diagram, within rectangle ABCD, E is a movable point on BC. After folding $\triangle ABE$ along AE, we obtain $\triangle AFE$, where point F is within rectangle ABCD. Extend AF to intersect CD at point G, with AB = 3 and AD = 4. As shown in Figure 1, when $\angle DAG=30^\circ$, find the length of BE.

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
it follows that $\angle BAD=90^\circ$,\\
and since $\angle DAG=30^\circ$,\\
it follows that $\angle BAG=60^\circ$.\\
From the folding, we know that $\angle BAE=\frac{1}{2}\angle BAG=30^\circ$,\\
In $\triangle BAE$, where $\angle BAE=30^\circ$ and $AB=3$,\\
it follows that $BE=\boxed{\sqrt{3}}$.


Question: As shown in the figure, in rectangle ABCD, E is a moving point on BC, and $\triangle ABE$ is folded along AE to obtain $\triangle AFE$, where point F is inside rectangle ABCD. Extend AF to intersect CD at point G, with $AB=3$ and $AD=4$. As shown in Figure 2, when point E is the midpoint of BC, find the length of line segment GC.

Solution:
\textbf{Solution:} As shown in Diagram 4, connect GE,\\
$\because$ E is the midpoint of BC,\\
$\therefore$ BE=EC,\\
$\because$ $\triangle ABE$ is folded along AE to obtain $\triangle AFE$,\\
$\therefore$ BE=EF,\\
$\therefore$ EF=EC,\\
$\because$ in rectangle ABCD,\\
$\therefore$ $\angle C=90^\circ$,\\
$\therefore$ $\angle EFG=90^\circ$,\\
$\because$ in $\triangle GFE$ and $\triangle GCE$,\\
$\left\{\begin{array}{l}EG=EG \\ EF=EC\end{array}\right.$ \\
$\therefore$ $\triangle GFE \cong \triangle GCE$ (HL),\\
$\therefore$ GF=GC;\\
Let GC=x, then AG=3+x, DG=3-x,\\
in $\triangle ADG$, $4^2+(3-x)^2=(3+x)^2$,\\
solving gives $x=\boxed{\frac{4}{3}}$.


Question: As shown in the figure, in rectangle $ABCD$, $E$ is a moving point on $BC$, and $\triangle ABE$ is folded along $AE$ to obtain $\triangle AFE$. Point $F$ is inside the rectangle $ABCD$, and extending $AF$ intersects $CD$ at point $G$, with $AB=3$ and $AD=4$. As shown in figure 3, when the perimeter of $\triangle CFE$ is at its minimum during the movement of point $E$, write down the length of $BE$.

Solution: \textbf{Solution:} We have $BE = \boxed{\frac{3}{2}}$.


Question: As shown in Figure 1, it is known that square $ABCD$ and isosceles triangle $\triangle EFG$, with $\angle EGF=90^\circ$, points $E$ and $F$ slide along sides $AB$ and $BC$ respectively, and point $G$ is inside the square. The distances from point $G$ to $AB$ and $BC$ are equal. If $AB=4$ and $EF=\sqrt{10}$, as shown in Figure 2, when point $F$ is the midpoint of side $BC$, what is the length of $DG$?

Solution:
\textbf{Solution} Extend $MG$ to intersect $CD$ at point $P$, and connect $DG$,

$\because$ $GM=GN$,

$\therefore$ Rectangle $BNGM$ is a square,

$\therefore MG=GN=BN$,

$\because$ Point $F$ is the midpoint of side $BC$, $AB=4$,

$\therefore BF=2$,

Let $FN=t$, then $EM=t$, $MG=GN=2+t$,

In $\triangle GFN$, by the Pythagorean theorem, we have $GF^2=t^2+(t+2)^2$.

In $\triangle EFG$, by the Pythagorean theorem, we have $EF^2=EG^2+GF^2$,

$\therefore t^2+(t+2)^2+t^2+(t+2)^2=10$,

Solving this yields $t=-1+\frac{\sqrt{6}}{2}$ or $t=-1-\frac{\sqrt{6}}{2}$ (discard this solution),

$\therefore MG=GN=2+t=1+\frac{\sqrt{6}}{2}$, $GP=4-MG=3-\frac{\sqrt{6}}{2}$,

$\therefore DP=3-\frac{\sqrt{6}}{2}$,

In $\triangle DGP$, by the Pythagorean theorem, we get $DG=\sqrt{2}GP=3\sqrt{2}-\sqrt{3}$.

Hence, the length of $DG$ is $\boxed{3\sqrt{2}-\sqrt{3}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, points E and F are on sides BC and AD respectively, and $BE=DF$. Quadrilateral $AECF$ is a parallelogram; If $AE$ bisects $\angle BAD$ and $\angle B=70^\circ$, what is the measure of $\angle DFC$?

Solution:
\textbf{Solution:} Since AD $\parallel$ BC and $\angle B=70^\circ$, \\
it follows that $\angle B+\angle BAD=180^\circ$, \\
hence $\angle BAD=110^\circ$, \\
moreover, since AE bisects $\angle BAD$, \\
it follows that $\angle BAE=\angle EAD=55^\circ$, \\
from (1) we have: quadrilateral AECF is a parallelogram, \\
thus AE $\parallel$ FC, \\
therefore, $\angle DFC=\angle DAE=55^\circ$, \\
hence, $\angle DFC=\boxed{55^\circ}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=3$, $AD=5$, point $E$ is on $BC$, and point $P$ is a moving point inside the rectangle, connecting $PD$, $PB$; when point $P$ moves along segment $BD$, what is the minimum value of $PE+DE$?

Solution:
\textbf{Solution:} As shown in the diagram, construct the symmetric point $D'$ of point D with respect to BC, and connect $ED'$. The minimum value is achieved when points $P'$, $E'$, and $D'$ are collinear and perpendicular to BD. In the rectangle ABCD, $\angle BCD=90^\circ$. In $\triangle D'P'D$, $\angle D'P'D=90^\circ$.\\
Therefore, $\triangle BCD \sim \triangle D'P'D$\\
Therefore, $\frac{DD'}{DB} = \frac{D'P'}{BC}$\\
Therefore, $D'P' = \frac{DD' \cdot BC}{BD}$\\
Also, since $DD' = DC + CD' = 3 + 3 = 6$, and $BD = \sqrt{BC^2 + CD^2} = \sqrt{34}$\\
Therefore, $D'P' = \frac{6 \cdot BC}{\sqrt{34}}$\\
By symmetry, $D'E' = DE'$\\
Therefore, $P'E'+DE' = D'E + P'E' = D'P'$\\
Therefore, the minimum value of $PE+DE$ is $\boxed{\frac{15\sqrt{34}}{17}}$


Question: In rectangle $ABCD$, where $AB=3$ and $AD=5$, point $E$ is on line $BC$, and point $P$ is a moving point inside the rectangle. Lines $PD$ and $PB$ are connected as shown in Figure 3. If point $Q$ is the midpoint of $AD$, and $P$ is a moving point inside the rectangle, with lines $PQ$, $PB$, and $PC$ connected, the question is whether $PQ+\sqrt{2}PB+PC$ has a minimum value. If it does, please write the answer directly; if not, please explain why.

Solution:
\textbf{Solution:} As shown in the figure, rotate triangle $BPC$ clockwise by $90^\circ$ to get the new triangle $B{P}^{\prime }{C}^{\prime }$, and then draw ${C}^{\prime }N$ $\perp$ AD at N through point ${C}^{\prime }$, and connect ${C}^{\prime }Q$ intersecting BC at M.\\
According to the properties of rotation, we know that $BP=B{P}^{\prime }$, $BC=B{C}^{\prime }$, $\angle PB{P}^{\prime }=\angle CB{C}^{\prime }=90^\circ$.\\
$\therefore P{P}^{\prime }=\sqrt{BP^{2}+{P}^{\prime }{B}^{2}}=\sqrt{2}BP$.\\
$\therefore PQ+\sqrt{2}PB+PC=PQ+P{P}^{\prime }+{P}^{\prime }{C}^{\prime }$.\\
Hence, when points Q, P, ${P}^{\prime }$, ${C}^{\prime }$ are collinear, $PQ+\sqrt{2}PB+PC=PQ+P{P}^{\prime }+{P}^{\prime }{C}^{\prime }$ reaches its minimum value.\\
$\because \angle CB{C}^{\prime }=90^\circ$,\\
$\therefore$ points A, B, ${C}^{\prime }$ are collinear,\\
In right triangle ${C}^{\prime }AQ$, ${C}^{\prime }Q=\sqrt{A{{C}^{\prime }}^{2}+A{Q}^{2}}$.\\
$\therefore {C}^{\prime }A=AB+B{C}^{\prime }=AB+BC=8$.\\
$\because$ point $Q$ is the midpoint of $AD$,\\
$\therefore AQ=\frac{5}{2}$,\\
$\therefore {C}^{\prime }Q=\sqrt{A{{C}^{\prime }}^{2}+A{Q}^{2}}=\frac{\sqrt{281}}{2}$.\\
$\therefore$ The minimum value of $PQ+\sqrt{2}PB+PC$ is $\boxed{\frac{\sqrt{281}}{2}}$.


Question: As shown in the figure, the vertices $E$, $G$ of the rectangle $EFGH$ are located on the sides $AD$, $BC$ of the rhombus $ABCD$ respectively, and the vertices $F$, $H$ are on the diagonal $BD$ of the rhombus $ABCD$. Given that $BG=DE$ and $E$ is the midpoint of $AD$, with $FH=4$ cm, what is the perimeter of the rhombus $ABCD$?

Solution:
\textbf{Solution:} Connect $EG$. In the rhombus $ABCD$, $AD\parallel BC$, $AD=BC$.\\
Since $E$ is the midpoint of $AD$,\\
it follows that $AE=ED$,\\
since $BG=DE$,\\
it follows that $AE\parallel BG$ and $AE=BG$.\\
Therefore, in quadrilateral $EFGH$, $EG=FH=4$.\\
Therefore, $AB=4$,\\
hence, the perimeter of the rhombus is $\boxed{16}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=8$ and $BC=4$. A line passing through the midpoint $O$ of diagonal $BD$ intersects $AB$ and $CD$ at points $E$ and $F$, respectively. When quadrilateral $BEDF$ is a rhombus, find the area of the rhombus $BEDF$.

Solution:
\textbf{Solution:} When the quadrilateral $BEDF$ is a rhombus, let $BE=x$, then $DE=x$, $AE=8-x$. In $\triangle ADE$, we have $DE^2=AD^2+AE^2$, therefore
\[
AE=AB-BE=8-x, \quad AD=4,
\]
\[
\therefore (8-x)^2+16=x^2
\]
Solving gives: $x=5$, that is $BE=5$,
\[
\therefore S_{\text{rhombus}BEDF}=5\times 4=\boxed{20}.
\]


Question: As shown in the figure, in square $ABCD$, $E$ is a point on the diagonal $AC$, $FH \perp AC$ at point $E$, intersecting $AD$, $AB$ at points $F$, $H$ respectively. $CF=CH$; if $AH= \frac{1}{3} CH$ and $AB=4$, what is the length of $AH$?

Solution: \textbf{Solution:} Given that quadrilateral ABCD is a square, it follows that $BC=AB=4$ and $\angle B=90^\circ$. Let $AH=x$, then $BH=AB-AH=4-x$. Since $AH=\frac{1}{3}CH$, it follows that $CH=3x$. In $\triangle CBH$, by the Pythagorean theorem, we have $CH^{2}-BH^{2}=BC^{2}$, which simplifies to $(3x)^{2}-(4-x)^{2}=16$. This reduces to $x^{2}+x-4=0$. Solving for $x$52251693, we find $x_{1}=\frac{-1+\sqrt{17}}{2}$, $x_{2}=\frac{-1-\sqrt{17}}{2}$ (discard this solution). Therefore, $x=\boxed{\frac{-1+\sqrt{17}}{2}}$, meaning the length of AH is $\boxed{\frac{-1+\sqrt{17}}{2}}$.


Question: As shown in the figure, within quadrilateral $ABCD$, points $E$ and $F$ are the trisection points of the diagonal $BD$. Connect $AE$, $CF$, $AF$, and $CE$. The quadrilateral $AECF$ forms a parallelogram. If quadrilateral $AECF$ is a rhombus, and $AE=BE$, find the measure of $\angle BAD$.

Solution: \textbf{Solution:} Since quadrilateral AECF is a rhombus, it follows that AE=AF=CF=CE. Also, since AE=BE, it follows that AE=BE=EF=AF=DF, which means that $\angle EAB=\angle EBA$, $\angle EAF=\angle EFA$, and $\angle FAD=\angle FDA$. Therefore, $\triangle AEF$ is an equilateral triangle, which implies that $\angle EAF=\angle AFE=\angle AEF=60^\circ$. Hence, $\angle BAE=30^\circ$ and $\angle FAD=30^\circ$, leading to $\angle BAD=\boxed{120^\circ}$.


Question: As shown in the figure, in the square $ABCD$, point $E$ is a point on the diagonal $AC$. A line through point $E$ is drawn perpendicular to $AC$, intersecting sides $AD$ and $AB$ at points $F$ and $H$, respectively. Lines $CF$ and $CH$ are then drawn. It is given that $CF=CH$; if the side length of the square $ABCD$ is $1$, when the areas of triangles $\triangle AFH$ and $\triangle CDF$ are equal, what is the length of $AE$?

Solution:
\textbf{Solution:} Let $AE=x$, then $AF=\sqrt{2}x$, $DF=1-\sqrt{2}x$, $FH=2AE=2x$,\\
since the area of $\triangle AFH$ is equal to the area of $\triangle CDF$,\\
therefore $\frac{1}{2}\cdot 2x\cdot x=\frac{1}{2}\times 1\cdot (1-\sqrt{2}x)$,\\
which simplifies to $2x^2+\sqrt{2}x-1=0$, solving this we get $x_1=\frac{-\sqrt{2}+\sqrt{10}}{4}$, $x_2=\frac{-\sqrt{2}-\sqrt{10}}{4}$ (discard this solution),\\
therefore $AE=\boxed{\frac{-\sqrt{2}+\sqrt{10}}{4}}$.


Question: As shown in the figure, within the quadrilateral $ABCD$, points E and F are located on sides AD and BC, respectively, with $DE=BF$. Here, line segments $CE$ and $AF$ are drawn. Quadrilateral $AECF$ is a parallelogram; Given that E is the midpoint of AD and $CE \perp AD$, when $CE=4$ and $AB=5$, what is the area of quadrilateral $ABCD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram,\\
it follows that CD = AB = 5,\\
since CE $\perp$ AD,\\
it follows that $\angle$DEC = $90^\circ$,\\
thus DE = $\sqrt{CD^{2} - CE^{2}} = \sqrt{5^{2} - 4^{2}} = 3$,\\
since E is the midpoint of AD,\\
it follows that AD = 2DE = 6,\\
therefore, the area of quadrilateral ABCD = AD × CE = 6 × 4 = $\boxed{24}$.


Question: As shown in the figure, in rectangle $ABCD$, point $O$ is the midpoint of diagonal $AC$, and a line $EF\perp AC$ is drawn through point $O$, intersecting $BC$ at point $E$ and $AD$ at point $F$. Lines $AE$ and $CF$ are connected. The quadrilateral $AECF$ forms a rhombus. When $OB$ is connected, given $AB=4$ and $AF=5$, find the length of $OB$.

Solution:
\textbf{Solution:} As shown in the figure, connect $BO$, \\
$\because AB=4$, $AF=AE=EC=5$, \\
$\therefore BE= \sqrt{AE^{2}-AB^{2}}=\sqrt{25-16}=3$, \\
$\therefore BC=8$, \\
$\therefore AC= \sqrt{AB^{2}+BC^{2}}=\sqrt{16+64}=4\sqrt{5}$, \\
$\because AO=CO$, $\angle ABC=90^\circ$, \\
$\therefore BO= \frac{1}{2} AC= 2\sqrt{5}$, hence the length of $OB$ is $\boxed{2\sqrt{5}}$.