Question: As shown in the figure, in the parallelogram $ABCD$, $CE\perp AD$ at point $E$. Extend $DA$ to point $F$ such that $AF=DE$. Connect $BF$ and $CF$. Quadrilateral $BCEF$ is a rectangle; If $AB=6$, $CF=8$, $DF=10$, what is the length of $EF$?

Solution:
\textbf{Solution:} Since according to the properties of parallelograms we know $AB=CD$, \\
thus $CD=6$, \\
since in $\triangle FCD$, $DF^2=10^2=6^2+8^2=CF^2+DC^2$, \\
therefore $\triangle FCD$ is a right triangle, $\angle FCD=90^\circ$, \\
based on the formula for the area of a triangle: $\frac{1}{2}\times FC\times CD=\frac{1}{2}\times FD\times EC$, \\
then $EC=\frac{24}{5}$, \\
since in rectangle $FBCE$, $\angle FBC=90^\circ$, $EC=BF$, \\
thus $BF=\frac{24}{5}$, \\
given $EF\parallel BC$, it means $\angle FCB=\angle CFD$, \\
therefore $\triangle FCB\sim \triangle DFC$, \\
thus $\frac{BC}{FC}=\frac{FB}{DC}$, \\
thus $BC=\frac{FB}{DC}\times FC=\frac{\frac{24}{5}\times 8}{6}=\frac{32}{5}$, \\
thus $EF=BC=\boxed{\frac{32}{5}}$.


Question: As shown in the diagram, in rectangle $ABCD$, $AD=4$, $AB=8$, diagonals $AC$ and $BD$ intersect at point $O$. Points $E$ and $F$ are located on the extensions of $CD$ and $DA$ respectively, and $DE=4$, $AF=2$. The segments $EF$ and $OE$ are drawn, with $G$ being the midpoint of $EF$. The segment $OE$ intersects $AD$ at point $H$, and segment $GH$ is drawn. Given that $H$ is the midpoint of $OE$, find the length of $GH$.

Solution:
\textbf{Solution:} Connect OF. \\
Since quadrilateral ABCD is a rectangle, \\
it follows that $ \angle ADC=90^\circ$. \\
Since $OM\parallel DC$, \\
it follows that $ \angle FMO=\angle ADC$, \\
therefore $ \angle FMO=90^\circ$. \\
Since $AD=4$ and M is the midpoint of AD, \\
it follows that $AM=\frac{1}{2}AD=2$. \\
Since $AF=2$, \\
it follows that $FM=4$. \\
Therefore, in $ \triangle FOM$, $ \angle FMO=90^\circ$, $ OM=4$, and $ FM=4$. \\
By the Pythagorean theorem, we get $ OF=\sqrt{OM^{2}+FM^{2}}=4\sqrt{2}$. \\
Since G is the midpoint of EF and H is the midpoint of OE, \\
it follows that GH is the midline of $ \triangle FEO$, \\
therefore $GH=\frac{1}{2}OF=2\sqrt{2}$. \\
What is the length of $GH$? $GH=\boxed{2\sqrt{2}}$.


Question: As shown in the figure, take a point $O$ on the line $AB$ and draw a ray $OC$ such that $\angle BOC = 70^\circ$. Place the right angle vertex of a right-angle triangle at point $O$. (Note: $\angle DOE = 90^\circ$) As shown in Figure 1, if one side $OD$ of the right-angle triangle $DOE$ falls on the ray $OB$, then what is the measure of $\angle COE$?

Solution: \textbf{Solution:} Therefore, $\angle COE = \boxed{20^\circ}$.


Question: As shown in the figure, a ray $OC$ is drawn from point $O$ on line $AB$, making an angle $ \angle BOC=70^\circ$. A right-angle triangle ruler is placed with its right-angle vertex at point $O$. (Note: $ \angle DOE=90^\circ$) As shown in figure 2, if $OC$ exactly bisects $ \angle BOE$, what is the degree measure of $ \angle COD$?

Solution:
\textbf{Solution:} Since $OC$ bisects $\angle EOB$ and $\angle BOC=70^\circ$,\\
it follows that $\angle EOB=2\angle BOC=140^\circ$.\\
Given $\angle DOE=90^\circ$,\\
we have $\angle BOD=\angle BOE-\angle DOE=50^\circ$,\\
therefore $\angle COD=\angle BOC-\angle BOD=\boxed{20^\circ}$.


Question: As shown in the figure, put together a set of triangles as demonstrated below. Write down the degrees of $\angle A$, $\angle B$, $\angle BCD$, $\angle D$, and $\angle AED$.

Solution:
\textbf{Solution:} From the problem, we get \\
$\angle A=30^\circ$, \\
$\angle B=90^\circ$, \\
$\angle BCD=150^\circ$, \\
$\angle D=45^\circ$, \\
$\angle AED=135^\circ$.


Question: Given that O is a point on line AB, $\angle COD$ is a right angle, and OE bisects $\angle BOC$. As shown in Figure 1, if $\angle AOC= 30^\circ$, find the measures of $\angle COE$ and $\angle BOD$.

Solution:
\textbf{Solution:} Given that $\angle AOC=30^\circ$ and $\angle COD$ is a right angle,\\
it follows that $\angle BOC= 180^\circ-\angle AOC=150^\circ$ and $\angle COD= 90^\circ$,\\
thus, $\angle BOD= 180^\circ-\angle AOC-\angle COD=60^\circ$.\\
Since OE bisects $\angle BOC$,\\
it implies that $\angle COE=\frac{1}{2}\angle BOC=75^\circ$;\\
therefore, $\angle BOD=\boxed{60^\circ}$ and $\angle COE=\boxed{75^\circ}$.


Question: Given that $O$ is a point on the line $AB$, $\angle COD$ is a right angle, and $OE$ bisects $\angle BOC$. As shown in Figure 1, if $\angle AOC = \alpha$, what is the degree measure of $\angle DOE$ (expressed as an algebraic expression containing $\alpha$)?

Solution:
\textbf{Solution:} Given $\because$ $\angle AOC = \alpha$, $\angle COD$ is a right angle,\\
$\therefore$ $\angle BOC = 180^\circ - \angle AOC = 180^\circ - \alpha$, $\angle COD = 90^\circ$,\\
$\therefore$ $\angle BOD = 180^\circ - \angle AOC - \angle COD = 90^\circ - \alpha$.\\
Since OE bisects $\angle BOC$,\\
$\therefore$ $\angle BOE = \frac{1}{2} \angle BOC = 90^\circ - \frac{1}{2}\alpha$,\\
$\therefore$ $\angle DOE = \angle BOE - \angle BOD = 90^\circ - \frac{1}{2}\alpha - (90^\circ - \alpha) = \boxed{\frac{1}{2}\alpha}$.


Question: As shown in the figure, $\angle AOB$ is a straight angle, and $OM$, $ON$ are the angle bisectors of $\angle AOC$, $\angle BOD$ respectively. Given that $\angle AOC=40^\circ$, $\angle BOD=60^\circ$, what is the degree measure of $\angle MON$?

Solution:
Solution:Given that $\angle AOB$ is a straight angle, $\angle AOC=40^\circ$, $\angle BOD=60^\circ$,\\
thus, $\angle COD=\angle AOB-\angle AOC-\angle BOD=180^\circ-40^\circ-60^\circ=80^\circ$,\\
since $OM$ and $ON$ are the angle bisectors of $\angle AOC$ and $\angle BOD$ respectively,\\
thus, $\angle COM= \frac{1}{2} \angle AOC=20^\circ$, $\angle DON= \frac{1}{2} \angle BOD=30^\circ$,\\
since $\angle MON=\angle COM+\angle COD+\angle DON$,\\
thus, $\angle MON=20^\circ+80^\circ+30^\circ=\boxed{130^\circ}$.


Question: As shown in the figure, $\angle AOB$ is a straight angle, and $OM$ and $ON$ are the angle bisectors of $\angle AOC$ and $\angle BOD$ respectively. Given that $\angle COD=90^\circ$, find the measure of $\angle MON$?

Solution: Solution: Since $\angle COD=90^\circ$, it follows that $\angle AOC+\angle BOD =90^\circ$. Furthermore, since OM and ON are the bisectors of $\angle AOC$ and $\angle BOD$ respectively, we have $\angle MOC= \frac{1}{2} \angle AOC$ and $\angle NOD= \frac{1}{2} \angle BOD$. Thus, $\angle MOC+\angle NOD= \frac{1}{2} (\angle AOC+\angle BOD)=45^\circ$. Also, $\angle MON=\angle MOC+\angle NOD+\angle COD$, yielding $\angle MON=45^\circ+90^\circ=\boxed{135^\circ}$.


Question: As the figure shows, it is known that: $O$ is a point on 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:} Given that $\angle COD$ is a right angle, and $\angle AOC=30^\circ$,\\
it follows that $\angle BOD=180^\circ-90^\circ-30^\circ=60^\circ$,\\
and $\angle COB=180^\circ-30^\circ=150^\circ$. Since OE bisects $\angle BOC$,\\
we have $\angle BOE=\frac{1}{2}\angle BOC=75^\circ$,\\
thus, $\angle DOE=\angle BOE-\angle BOD=75^\circ-60^\circ=\boxed{15^\circ}$.


Question: As shown in the figure, it is known that $ \angle AOB=90^\circ$, $ \angle AOC$ is an acute angle, $ ON$ bisects $ \angle AOC$, and the ray $ OM$ is inside $ \angle AOB$. How many angles in the figure are less than a straight angle?

Solution:
\textbf{Solution:} Let's consider angles that start from edge $OC$: there are $\angle CON, \angle COA, \angle COM, \angle COB$,\\
Angles that start from edge $ON$: $\angle NOA, \angle NOM, \angle NOB$,\\
Angles that start from edge $OA$: $\angle AOM, \angle AOB$,\\
Angles that start from edge $OM$: $\angle MOB$,\\
$\therefore$ In the figure, there are a total of $4+3+2+1=\boxed{10}$ angles that are less than a straight angle.


Question: As shown in the figure, it is known that $\angle AOB=90^\circ$, $\angle AOC$ is an acute angle, $ON$ bisects $\angle AOC$, and ray $OM$ is inside $\angle AOB$. If $\angle AOC=50^\circ$ and $\angle MON=45^\circ$, find the measure of $\angle AOM$.

Solution:
\textbf{Solution:} Since $ON$ bisects $\angle AOC$ and $\angle AOC=50^\circ$,\\
it follows that $\angle AON=\angle CON=\frac{1}{2}\angle AOC=25^\circ$,\\
Since $\angle MON=45^\circ$,\\
therefore, $\angle AOM=\angle MON-\angle AON=45^\circ-25^\circ=\boxed{20^\circ}$.


Question: As shown in the figure, in triangle $AOB$ and triangle $COD$, we have $\angle AOB = \angle COD$, with $\angle AOB = 90^\circ$. When the two triangles are combined as shown in the figure and $\angle BOD = 30^\circ$, what is the measure of $\angle AOC$?

Solution:
\textbf{Solution:} \\
Since $\angle AOC = \angle AOB + \angle BOC = 90^\circ + \angle BOC$,\\
$\angle DOB = \angle DOC - \angle BOC = 90^\circ - \angle BOC$,\\
Therefore, $\angle AOC + \angle DOB = 90^\circ + \angle BOC + 90^\circ - \angle BOC = 180^\circ$,\\
Since $\angle BOD = 30^\circ$,\\
Therefore, $\angle AOC = \boxed{150^\circ}$.


Question: In triangles $AOB$ and $COD$, $\angle AOB = \angle COD$ and it is known that $\angle AOB = 90^\circ$. When the two triangles are put together as shown in the figure, and $\angle AOC = 2\angle BOD$, what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:}\\
$\because \angle AOC = \angle AOB + \angle BOC = 90^\circ + \angle BOC$,\\
$\angle DOB = \angle DOC - \angle BOC = 90^\circ - \angle BOC$,\\
$\therefore \angle AOC + \angle DOB = 90^\circ + \angle BOC + 90^\circ - \angle BOC = 180^\circ$,\\
$\because \angle AOC = 2\angle BOD$,\\
$\therefore \angle BOD = \boxed{60^\circ}$.


Question: In triangles $AOB$ and $COD$, $\angle AOB = \angle COD$. When $\angle AOB = \alpha$, the two triangles are combined as shown in the diagram. Express $\angle AOC + \angle BOD$ using an algebraic formula involving $\alpha$.

Solution:
\textbf{Solution:} Given:
\[
\because \angle AOC = \angle AOB + \angle BOC = \alpha + \angle BOC,
\]
\[
\angle DOB = \angle DOC - \angle BOC = \alpha - \angle BOC,
\]
\[
\therefore \angle AOC + \angle DOB = \alpha + \angle BOC + \alpha - \angle BOC = \boxed{2\alpha}.
\]


Question: As shown in the figure, $O$ is a point on line $AB$, $\angle AOC=40^\circ$, $OD$ bisects $\angle AOC$, $\angle DOE=90^\circ$. Please answer the following question: How many angles in the figure are less than $180^\circ$?

Solution: \textbf{Solution:} The angles in the figure that are less than $180^\circ$ are: $\angle AOD$, $\angle AOC$, $\angle AOE$, $\angle DOC$, $\angle DOE$, $\angle DOB$, $\angle COE$, $\angle COB$, $\angle EOB$, for a total of $\boxed{9}$ angles.


Question: As shown in the figure, $\angle AOC$ and $\angle BOD$ are both right angles. If $\angle DOC = 35^\circ$, then what is the degree measure of $\angle AOB$?

Solution: \textbf{Solution:} We have $\angle AOB = \boxed{145^\circ}$.


Question: As shown in the figure, the straight lines AB and CD intersect at point O. OE is the bisector of $\angle AOD$. If $\angle AOC=60^\circ$ and $OF\perp OE$, what is the measure of $\angle BOE$?

Solution:
\textbf{Solution:} Since $\angle AOC=60^\circ$,\\
it follows that $\angle BOD=\angle AOC=60^\circ$,\\
$\angle AOD=180^\circ-60^\circ=120^\circ$,\\
since $OE$ is the bisector of $\angle AOD$,\\
it follows that $\angle DOE= \frac{1}{2} \angle AOD=60^\circ$,\\
thus, $\angle BOE=\angle BOD+\angle DOE=60^\circ+60^\circ=\boxed{120^\circ}$.


Question: As shown in the figure, two triangular boards OAB and OCD are positioned with $\angle ABO = 30^\circ$ and $\angle ODC = 45^\circ$. If OM and ON are the angle bisectors of $\angle AOD$ and $\angle BOC$ respectively, what is the degree measure of $\angle MON$?

Solution:
\textbf{Solution:}\\
$\because \angle ABO=30^\circ$ and $\angle ODC=45^\circ$\\
$\therefore \angle COB=90^\circ-45^\circ=45^\circ$, $\angle AOB=90^\circ-30^\circ=60^\circ$\\
$\because$ OM and ON bisect $\angle AOD$ and $\angle BOC$ respectively\\
$\therefore \angle NOB= \frac{1}{2}\angle COB= \frac{1}{2}\times45^\circ=22.5^\circ$\\
$\angle BOM= \frac{1}{2}\angle AOB= \frac{1}{2}\times60^\circ=30^\circ$\\
$\therefore \angle MON=\angle NOB+\angle BOM=\boxed{52.5^\circ}$


Question: As shown in Figure (1), two triangular boards, OAB and OCD, are placed with $\angle ABO=30^\circ$ and $\angle ODC=45^\circ$. If OM and ON bisect $\angle AOD$ and $\angle BOC$ respectively, rotate the triangular board OCD clockwise around point O as shown in Figure 2. If OD bisects $\angle AOB$, what is the rotation angle $\alpha$?

Solution:
\textbf{Solution:} Since OD bisects $\angle AOB$,

$\therefore \angle BOD = \frac{1}{2}\angle AOB = 30^\circ$,

$\therefore$ The angle of rotation $\angle BOD = \alpha = \boxed{30^\circ}$.


Question: A pair of set squares OAB and OCD are placed as shown in the figure, where ($\angle ABO=30^\circ$, $\angle ODC=45^\circ$). If OM and ON bisect $\angle AOD$ and $\angle BOC$ respectively. Rotate set square OCD clockwise around point O in the figure, if $\angle AOC=90^\circ$, find the degree of the rotation angle $\alpha$.

Solution:
\textbf{Solution:} Given $\because \angle AOC=90^\circ$ and $\angle COD=45^\circ$,\\
$\therefore \angle AOD=\angle AOC-\angle COD=45^\circ$,\\
$\therefore$ the angle of rotation $\angle BOD=\angle AOB-\angle AOD=60^\circ-45^\circ=\alpha =\boxed{15^\circ}$.


Question: As shown in the figure, two triangles OAB and OCD are placed with $\angle ABO=30^\circ$ and $\angle ODC=45^\circ$. If OM and ON bisect $\angle AOD$ and $\angle BOC$ respectively. Rotate the triangle OCD clockwise around point O from figure 1. In the process of rotation, is there a scenario where $\angle BOD = \angle COB + \angle AOD$? If it exists, calculate the rotation angle $\alpha$; if not, explain the reason.

Solution:
\textbf{Solution:}\\
Since $\angle BOD = \angle COB + \angle AOD$,\\
it follows that $\angle DOC + \angle AOB = 45^\circ + 60^\circ = 105^\circ$, thus $\angle COB + \angle BOD + \angle AOD + \angle BOD = 105^\circ$,\\
therefore $3\angle BOD = 105^\circ$,\\
The angle of rotation $\angle BOD = \alpha = \boxed{35^\circ}$.


Question: As shown in Figure 1, it is known that $\angle AOB=160^\circ$, $\angle COE$ is a right angle, and $OF$ bisects $\angle AOE$. If $\angle COF=32^\circ$, then what is the degree measure of $\angle BOE$?

Solution: \textbf{Solution:} We have $\angle BOE = \boxed{44^\circ}$.


Question: As shown in Figure 1, point O is a point on the line AB. A ray OC is drawn through point O such that $\angle AOC=60^\circ$. A right-angled triangle plate is placed with its right-angle vertex at point O, one side OM on the ray OB, and the other side ON below the line AB. The triangle plate in Figure 1 is rotated counterclockwise around point O to Figure 2, such that one side OM is inside $\angle BOC$ and exactly bisects $\angle BOC$. What are the measures of $\angle CON$ and $\angle AOM$?

Solution: \textbf{Solution:} Since $\angle AOC = 60^\circ$, it follows that $\angle BOC = 120^\circ$. Since OM bisects $\angle BOC$, we have $\angle COM = \frac{1}{2}\angle BOC = 60^\circ$. Thus, $\angle CON = \angle COM + 90^\circ = 150^\circ$, and $\angle AOM = \angle AOC + \angle COM = 60^\circ + 60^\circ = 120^\circ$; therefore, the measure of $\angle CON$ is $\boxed{150^\circ}$, and the measure of $\angle AOM$ is $\boxed{120^\circ}$.


Question: Place the right-angled vertex O of the right-angled triangle OMN on the line AB, and let the ray OC bisect $\angle AON$. As shown in the figure, if $\angle BON = 60^\circ$, what is the degree measure of $\angle AOM$?

Solution:
\textbf{Solution}: Since $\angle MON=90^\circ$ and $\angle BON=60^\circ$, \\
it follows that $\angle AOM=180^\circ-\angle MON-\angle BON=180^\circ-90^\circ-60^\circ=\boxed{30^\circ}$.


Question: As shown in the figure, the right-angled vertex $O$ of the right triangle $OMN$ is placed on line $AB$, and ray $OC$ bisects $\angle AON$. If $\angle AOM = 2\angle COM$, what is the measure of $\angle AON$ in degrees?

Solution:
\textbf{Solution:} Since line OC bisects $\angle AON$,
\[
\therefore \angle AON=2\angle AOC,
\]
Let $\angle COM=x$, then $\angle AOM=2x$,
\[
\therefore \angle CON=\angle AOC=\angle COM+\angle AOM=x+2x=3x,
\]
Since $\angle COM+\angle CON=90^\circ$,
\[
\therefore x+3x=90^\circ,
\]
Solving this, we find $x=22.5^\circ$;
\[
\therefore \angle AON=6x=6\times22.5^\circ=\boxed{135^\circ}.
\]


Question: As shown in the figure, place the right-angled vertex $O$ of the right triangle $OMN$ on the line $AB$, and the ray $OC$ bisects $\angle AON$. Rotate the right triangle $OMN$ counterclockwise around vertex $O$. During the rotation process: when $\angle BON = 120^\circ$, what is the measure of $\angle COM$?

Solution:
\textbf{Solution:} When $NO$ is above the line $AB$,

$\because \angle AON = 180^\circ - \angle BON$,

$\therefore \angle AON = 60^\circ$,

$\because OC$ bisects $\angle AON$,

$\therefore \angle CON = \frac{1}{2}\angle AON = 30^\circ$,

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

$\therefore \angle COM = \angle MON - \angle CON = 90^\circ - 30^\circ = 60^\circ$;

When $ON$ is below the line $AB$,

$\because \angle AON = 180^\circ - \angle BON$,

$\therefore \angle AON = 60^\circ$,

$\because OC$ bisects $\angle AON$,

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

$\therefore \angle COM = \angle MON + \angle CON = 90^\circ + 30^\circ = 120^\circ$;

$\therefore$ The measure of $\angle COM$ is either $\boxed{60^\circ}$ or $\boxed{120^\circ}$.


Question:
During our study in the latter half of seventh grade, we delved into the problems concerning the angles between bisectors of interior angles and the angles between external and internal angle bisectors. Throughout this process, Kun found a new category of problems during his independent exploration. His exploration process is as follows:

\textbf{Exploratory Research}: As shown in Figure 1, in $\triangle ABC$, $BM$ bisects $\angle ABC$ and $CM$ bisects $\angle ACB$, with $BM$ and $CM$ intersecting at point $M$. If $\angle A=50^\circ$, then what is the measure of $\angle BMC$?

Solution: \textbf{Solution:} We have $\angle BMC = \boxed{115^\circ}$.


Question: As shown in the figure, within $\triangle ABC$, $BM$ bisects $\angle ABC$, and $CM$ bisects $\angle ACB$, with $BM$ and $CM$ intersecting at point $M$. If $\triangle ABC$ is folded along $DE$ such that point $A$ coincides with point $M$, and given that $\angle 1+\angle 2=120^\circ$, what is the measure of $\angle BMC$?

Solution:
\textbf{Solution:} From the property of folding, we have $\angle AED=\angle MED$, $\angle ADE=\angle MDE$,
\\
$\because \angle 1+\angle ADM=180^\circ$, $\angle 2+\angle AEM=180^\circ$, $\angle 1+\angle 2=120^\circ$,
\\
$\therefore \angle ADM+\angle AEM=240^\circ$,
\\
$\therefore 2\angle AED+2\angle ADE=240^\circ$,
\\
$\therefore \angle AED+\angle ADE=120^\circ$,
\\
$\therefore \angle A=180^\circ-\angle AED-\angle ADE=60^\circ$,
\\
$\therefore$ By the same principle as in (1), we can conclude that $\angle BMC=\boxed{120^\circ}$.


Question: As shown in Figure 1, a right-angled triangle board is placed on line $AD$ (right-angled triangle board $OBC$ and right-angled triangle board $MON$, with $\angle OBC=90^\circ$, $\angle BOC=45^\circ$, $\angle MON=90^\circ$, $\angle MNO=30^\circ$). Keeping the triangle board $OBC$ stationary, the triangle board $MON$ is rotated clockwise around point $O$ at a speed of $6^\circ$ per second until the side $OM$ coincides with the line $AD$ for the first time. The rotation time is recorded as $t$ seconds. At what value of $t$ does $OM$ bisect $\angle AOC$?

Solution: \textbf{Solution}: When $t = \boxed{\frac{15}{4}}$ seconds, $OM$ bisects $\angle AOC$.


Question: We define two triangles as "opposite triangles" if their internal angles are vertically opposite angles. For example, in Figure 1, since the internal angle $\angle AOB$ of $\triangle AOB$ and the internal angle $\angle COD$ of $\triangle COD$ are vertically opposite angles, then $\triangle AOB$ and $\triangle COD$ are considered "opposite triangles". According to the triangle angle sum theorem, opposite triangles have the following property: $\angle A+\angle B=\angle C+\angle D$. Given two triangles defined as "opposite triangles", such as $\triangle AOB$ and $\triangle COD$ in Figure 1, with $\angle AOB=70^\circ$, what is the degree measure of $\angle C+\angle D$?

Solution: \textbf{Solution}: We have $\angle C + \angle D = \boxed{110^\circ}$.


Question: As shown in Figure 2, in $\triangle ABC$, $AD$ and $BE$ bisect $\angle BAC$ and $\angle ABC$ respectively. If $\angle C = 60^\circ$ and $\angle ADE$ is $6^\circ$ larger than $\angle BED$, what is the measure of $\angle BED$?

Solution:
\textbf{Solution:} Given that $AD$ and $BE$ bisect $\angle BAC$ and $\angle ABC$ respectively,

$\therefore \angle 1=\angle 2$, $\angle 3=\angle 4$,

also, given that $\angle C=60^\circ$,

$\therefore \angle BAC+\angle ABC=180^\circ - \angle C=180^\circ - 60^\circ=120^\circ$,

$\therefore \angle 1+\angle 2+\angle 3+\angle 4=120^\circ$,

$\therefore 2\angle 1+2\angle 3=120^\circ$,

$\therefore \angle 1+\angle 3=60^\circ$,

From the figure, it is known that $\triangle ABF$ and $\triangle DEF$ are vertically opposite triangles,

$\therefore \angle 1+\angle 3=\angle ADE+\angle BED=60^\circ$ (1),

also, given that $\angle ADE$ is $6^\circ$ larger than $\angle BED$,

$\therefore \angle ADE-\angle BED=6^\circ$ (2),

Solving (1) and (2) together yields $\begin{cases} \angle ADE+\angle BED=60^\circ \\ \angle ADE-\angle BED=6^\circ \end{cases}$,

Solution obtained is: $\begin{cases} \angle ADE=33^\circ \\ \angle BED=27^\circ \end{cases}$,

$\therefore \angle BED=\boxed{27^\circ}$.


Question: As shown in Figure 3, in $\triangle ABC$, $\angle A=45^\circ$, $\angle ABC=120^\circ$, $BC=2\sqrt{3}$, and $D$ is a point in the plane containing $\triangle ABC$. When the quadrilateral $ABCD$ is a semisymmetric quadrilateral and $AC$ is the diagonal that divides it, what is the area of the quadrilateral $ABCD$?

Solution:
\textbf{Solution:} Construct $CE\perp AB$, intersecting the extension of $AB$ at point $E$, as shown in the diagram,\\
since $CE\perp AB$, $\angle A=45^\circ$, $\angle ABC=120^\circ$,\\
therefore $\angle ACE=45^\circ$, $\angle EBC=60^\circ$,\\
therefore $AE=EC$, $\angle ECB=30^\circ$,\\
therefore $BE=\frac{1}{2}BC=\sqrt{3}$,\\
therefore $EC=\sqrt{BC^{2}-BE^{2}}=\sqrt{(2\sqrt{3})^{2}-(\sqrt{3})^{2}}=3$,\\
therefore $AE=EC=3$,\\
therefore $AC=\sqrt{2}EC=3\sqrt{2}$.\\
therefore $AB=AE-BE=3-\sqrt{3}$.\\
therefore $S_{\triangle ABC}=\frac{1}{2}AB\cdot EC=\frac{1}{2}(3-\sqrt{3})\times 3=\frac{9}{2}-\frac{3\sqrt{3}}{2}$.\\
(1) When $DA=DC$ and $AC$ bisects $\angle BAD$, as shown in the diagram,\\
according to the problem: $\angle DAC=\angle BAC=45^\circ$,\\
therefore $DA=DC$,\\
therefore $\angle DCA=\angle DAC=45^\circ$,\\
therefore $\angle ADC=90^\circ$,\\
therefore $\triangle ADC$ is an isosceles right triangle,\\
therefore $AD=CD=\frac{\sqrt{2}}{2}AC=3$,\\
therefore $S_{\triangle DAC}=\frac{1}{2}AD\cdot CD=\frac{9}{2}$,\\
therefore $S_{\text{quadrilateral} ABCD}=S_{\triangle ABC}+S_{\triangle ADC}=\frac{9}{2}-\frac{3\sqrt{3}}{2}+\frac{9}{2}=9-\frac{3\sqrt{3}}{2}$;\\
(2) When $DA=DC$ and $AC$ bisects $\angle BCD$, as shown in the diagram,\\
according to the problem: $\angle ACD=\angle BCA=15^\circ$,\\
therefore $DA=DC$,\\
therefore $\angle DCA=\angle DAC=15^\circ$,\\
therefore $\angle ADC=150^\circ$,\\
Construct $CF\perp AD$ through point $C$, intersecting the extension of $AD$ at point $F$,\\
then $\angle CDF=30^\circ$,\\
therefore $CF=\frac{1}{2}CD$,\\
therefore $DF=\frac{\sqrt{3}}{2}CD$.\\
Let $CD=x$, then $AD=x$, $CF=\frac{1}{2}x$, $AF=AD+DF=(1+\frac{\sqrt{3}}{2})x$,\\
In $Rt\triangle ACF$,\\
since $AC^{2}=AF^{2}+CF^{2}$,\\
therefore $(3\sqrt{2})^{2}=[(1+\frac{\sqrt{3}}{2})x]^{2}+(\frac{1}{2}x)^{2}$,\\
therefore $x=3+3\sqrt{3}$ (contradicts the problem statement, thus discarded) or $x=3\sqrt{3}-3$,\\
therefore $AD=3\sqrt{3}-3$, $CF=\frac{3\sqrt{3}-3}{2}$.\\
therefore $S_{\triangle ADC}=\frac{1}{2}AD\cdot CF=\frac{18-9\sqrt{3}}{2}$,\\
therefore $S_{\text{quadrilateral} ABCD}=S_{\triangle ABC}+S_{\triangle ADC}=9-\frac{3\sqrt{3}}{2}+\frac{18-9\sqrt{3}}{2}=18-6\sqrt{3}$.\\
In summary, the quadrilateral $ABCD$


Question: Overlay a right-angled triangle board DOE with AOC as shown in figure (1), where $\angle O=90^\circ$, $\angle A=30^\circ$, and $\angle E=45^\circ$, with $OD>OC$. Within the plane containing the two triangle boards, rotate the triangle board DOE clockwise around point O by $\alpha$ degrees $(0^\circ<\alpha<90^\circ)$ to the position $D_{1}OE_{1}$, so that $OD_{1}\parallel AC$, as shown in figure (2). What is the value of $\alpha$?

Solution:
\textbf{Solution:} By the given information, we have: $\alpha =\angle AOD_{1}$\\
Since $OD_{1}\parallel AC$ and $\angle A=30^\circ$\\
Therefore, $\angle AOD_{1}=\angle A=30^\circ$\\
Hence, $\alpha =\boxed{30^\circ}$


Question: Place a triangular board $ABC$ ($\angle ACB=90^\circ$, $\angle A=30^\circ$) inside the acute angle $\angle POQ$ as shown in figure (1), with the right-angle side $BC$ on side $OQ$, and let $\angle POQ=\alpha$. Now, rotate the triangular board $ABC$ counterclockwise around point B at a speed of $m^\circ$ per second for $t$ seconds (with the right-angle side $BC$ moving to the position shown in figure (2), and point A always within $\angle POQ$). Draw $MN\parallel OQ$ intersecting ray $OP$ at point M, and bisect $\angle MAB$ with $AD$ intersecting ray $OQ$ at point D. What value of $m$ minimizes the value of the algebraic expression $\left|10−m\right|+3$? What is the value of $m$?

Solution: \textbf{Solution:} We have $m=\boxed{10}$.


Question: As shown in the figure, a right-angled triangular board $ABC$ (where $\angle ACB=90^\circ$, $\angle A=30^\circ$) is placed inside the acute angle $\angle POQ$, making the right-angled side $BC$ lie on the side $OQ$, and let $\angle POQ=\alpha$. Now the triangular board $ABC$ is rotated counterclockwise around point $B$ at a speed of $m^\circ$ per second for $t$ seconds (the right-angled side $BC$ rotates to the position shown in the figure, and point $A$ always remains within $\angle POQ$). Through point $A$, draw $MN \parallel OQ$ intersecting the ray $OP$ at point $M$, and $AD$ bisects $\angle MAB$ intersecting the ray $OQ$ at point $D$, where the value of $m$ is such that the algebraic expression $\left|10−m\right|+3$ is minimized. When $t=5$ seconds, determine the degree measure of $\angle NAC$.

Solution:
\textbf{Solution:} Given $t=5$, then $\angle CBQ=50^\circ$,\\
since $\angle ACB=90^\circ, \angle BAC=30^\circ$,\\
thus $\angle ABC=90^\circ-\angle BAC=60^\circ$,\\
which means $\angle ABQ=\angle ABC+\angle CBQ=110^\circ$,\\
given that $MN\parallel OQ$,\\
it follows that $\angle NAB=180^\circ-\angle ABQ=70^\circ$,\\
thus $\angle NAC=\angle NAB-\angle BAC=40^\circ$,\\
therefore the measure of $\angle NAC$ is $\boxed{40^\circ}$.


Question: As shown in the figure, $ \angle POQ$ is an acute angle, and point A is a point on side $ OP$, with ray $ AB\parallel OQ$. It is known that the two acute angle vertices of the triangular plate $ CDE$ are located on $ AB$ and $ OQ$ respectively, $ DE\parallel OP$, $ \angle E=90^\circ$, and $ \angle CDE=60^\circ$. As shown in Figure 1, if $ \angle ACD=112^\circ$, what is the degree measure of $ \angle POQ$?

Solution:
\textbf{Solution:} Given $\because AB\parallel OQ, \angle ACD=112^\circ$,\\
$\therefore \angle CDQ=\angle ACD=112^\circ$.\\
$\because \angle CDE=60^\circ$,\\
$\therefore \angle EDQ=\angle CDQ-\angle CDE=112^\circ-60^\circ=52^\circ$.\\
Furthermore, $\because DE\parallel OP$,\\
$\therefore \angle POQ=\boxed{52^\circ}$.


Question: As shown in the figure, the right-angled vertex $P$ of the triangle ruler $ABP$ lies on the line $CD$, with points $A$, $B$ on the same side of line $CD$. As shown in fig. (1), if $\angle APC=40^\circ$, find the degree measure of $\angle BPD$.

Solution:
\textbf{Solution:} By the given conditions,\\
$ \because \angle APB=90^\circ$, $ \angle APC=40^\circ$\\
$ \therefore \angle BPD=180^\circ-\angle APB-\angle APC=180^\circ-90^\circ-40^\circ=\boxed{50^\circ}$


Question: As shown in the figure, the right-angle vertex $P$ of the set square $ABP$ is on the line $CD$, and points $A$ and $B$ are on the same side of line $CD$. As shown in Figure 2, if $PM$ bisects $\angle APC$ and $PN$ bisects $\angle BPD$, find the degree measurement of $\angle MPN$.

Solution:
\textbf{Solution:} Given that $PM$ bisects $\angle APC$, and $PN$ bisects $\angle BPD$,\\
it follows that $\angle APM=\angle CPM$ and $\angle BPN=\angle DPN$.\\
Since $\angle APB=90^\circ$,\\
we have $2\angle APM+2\angle BPN=90^\circ$,\\
which leads to $\angle APM+\angle BPN=45^\circ$,\\
thus, $\angle MPN=\angle APM+\angle APB+\angle BPN=45^\circ+90^\circ=\boxed{135^\circ}$.


Question: As shown in the figure, the right-angle vertex P of the right-angled triangle \( ABP \) lies on the line \( CD \), with points A and B on the same side of line \( CD \). Rotate the right-angled triangle \( ABP \) about point P so that points A and B are on opposite sides of line \( CD \), as shown in figure (3). When \( \angle APC = 4\angle BPD \), find the degree measure of \( \angle BPC \).

Solution:
\textbf{Solution:} Let $\angle BPD=x$, then $\angle APC=4x$,\\
since $\angle APB=90^\circ$,\\
therefore $\angle APD=90^\circ-x$,\\
according to the problem, we have $4x+(90^\circ-x)=180^\circ$,\\
yielding $3x=90^\circ$,\\
solving for $x$, we get $x=30^\circ$,\\
thus $\angle BPC=180^\circ-\angle BPD=180^\circ-30^\circ=\boxed{150^\circ}$.


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

Solution:
\textbf{Solution:} Since $\angle AOC = 40^\circ$, \\
thus $\angle BOC = 180^\circ - \angle AOC = 140^\circ$, \\
since $\angle COD$ is a right angle, \\
thus $\angle COD = 90^\circ$, \\
therefore $\angle BOD = \angle BOC - \angle COD = 140^\circ - 90^\circ = 50^\circ$, \\
since $OE$ bisects $\angle BOC$, \\
thus $\angle BOE = \frac{1}{2} \angle BOC = 70^\circ$, \\
therefore $\angle DOE = \angle BOE - \angle BOD = 70^\circ - 50^\circ = \boxed{20^\circ}$;


Question: As shown in the figure, it is known that: $O$ is a point on line $AB$, $\angle COD$ is a right angle, and $OE$ bisects the obtuse angle $\angle BOC$. As shown in Figure 2, $OF$ bisects $\angle BOD$. What is the degree measure of $\angle EOF$?

Solution:
\textbf{Solution:} From the construction, we have $OE$ bisect $\angle BOC$, $OF$ bisect $\angle BOD$,\\
$\therefore \angle BOE=\frac{1}{2}\angle BOC$, $\angle BOF=\frac{1}{2}\angle BOD$,\\
$\therefore \angle EOF=\angle BOE-\angle BOF=\frac{1}{2}\left(\angle BOC-\angle BOD\right)=\frac{1}{2}\angle COD$,\\
$\because \angle COD=90^\circ$,\\
$\therefore \angle EOF=\boxed{45^\circ}$;


Question: As shown in Figure 1, point O is on line segment $AB$. Ray $OC$ is drawn from point O such that $\angle BOC = 120^\circ$. A right-angle triangle ruler is placed with its right-angle vertex at point O, one side $OM$ on ray $OB$, and the other side $ON$ below line $AB$. The triangle ruler in Figure 1 is rotated clockwise around point O to the position in Figure 3, so that side $OM$ is inside $\angle BOC$ and exactly bisects $\angle BOC$. What is the measure of $\angle CON$?

Solution:
\textbf{Solution:} Given $\because \angle BOC = 120^\circ$, and $OM$ bisects $\angle BOC$,\\
$\therefore \angle COM = \angle BOM = 60^\circ$,\\
$\because \angle MON = 90^\circ$,\\
$\therefore \angle CON = \angle MON + \angle COM = 90^\circ + 60^\circ = \boxed{150^\circ}$;


Question: When two acute angle vertices of a set of triangular boards coincide, $\angle AOB=45^\circ$, $\angle COD=30^\circ$, $OM$ and $ON$ are the angle bisectors of $\angle AOC$ and $\angle BOD$ respectively. As shown in Figure 1, when $OB$ coincides with $OC$, what is the magnitude of $\angle MON$?

Solution: \textbf{Solution:} We have $\angle MON = \boxed{37.5^\circ}$.


Question: When two acute angles of a set of set squares coincide, $\angle AOB=45^\circ$, $\angle COD=30^\circ$, and $OM$, $ON$ are the angle bisectors of $\angle AOC$, $\angle BOD$ respectively. When $\angle COD$ is rotated around point $O$ to the position shown in Figure 2, and $\angle BOC=10^\circ$, what is the degree measure of $\angle MON$?

Solution:
\textbf{Solution:} Given $\because$ when $\angle BOC=10^\circ$, $\angle AOC=35^\circ$, $\angle BOD=20^\circ$, and OM, ON are the bisectors of $\angle AOC$, $\angle BOD$, respectively,\\
$\therefore \angle BON= \frac{1}{2} \angle BOD=10^\circ$,\\
$\therefore \angle MOC= \frac{1}{2} \angle AOC=17.5^\circ$,\\
$\therefore \angle MON=\angle MOC+\angle BON+\angle BOC=17.5^\circ+10^\circ+10^\circ=\boxed{37.5^\circ}.


Question: Given in the figure, two acute vertices of a set squares are coincident, with $\angle AOB=45^\circ$, $\angle COD=30^\circ$. $OM$ and $ON$ are the bisectors of $\angle AOC$ and $\angle BOD$ respectively. When $\angle COD$ is rotated around point $O$ to the position shown in the figure, and $\angle BOC=n^\circ$, what is the degree measure of $\angle MON$?

Solution:
\textbf{Solution:} Given $\because \angle BOC = n^\circ$, then $\angle AOC = 45^\circ + n^\circ$ and $\angle BOD = 30^\circ + n^\circ$. Furthermore, OM and ON are the angle bisectors of $\angle AOC$ and $\angle BOD$ respectively, \\
$\therefore \angle BON = \frac{1}{2} \angle BOD = \frac{1}{2} (30^\circ + n^\circ) = 15^\circ + \frac{1}{2} n^\circ$, \\
$\angle MOB = \frac{1}{2} \angle AOC - \angle BOC = \frac{1}{2} (45^\circ + n^\circ) - n^\circ = 22.5^\circ - \frac{1}{2} n^\circ$, \\
$\therefore \angle MON = \angle MOB + \angle BON = 15^\circ + \frac{1}{2} n^\circ + 22.5^\circ - \frac{1}{2} n^\circ = \boxed{37.5^\circ}$.


Question: As shown in the figure, it is known that $OB$ bisects $\angle AOC$, and the complement of $\angle AOC$ is $30^\circ$ less than $\angle BOC$. What is the measure of $\angle AOB$?

Solution:
\textbf{Solution:} Let $\angle BOC = \angle AOB = x$, then $\angle AOC = 2x$\\
From the problem, we get: $90^\circ - 2x = x - 30^\circ$\\
$3x = 120^\circ$\\
$x = 40^\circ$\\
Thus, the measure of $\angle AOB$ is $\boxed{40^\circ}$.


Question: As shown in the figure, it is known that $OB$ bisects $\angle AOC$, and the complement of $\angle AOC$ is $30^\circ$ smaller than $\angle BOC$. A ray $OD$ is drawn through point O such that $\angle AOD = \frac{1}{5}\angle AOC$. Can you find the measure of $\angle COD$ in degrees?

Solution:
\textbf{Solution:} From (1) we have $\angle AOC = 2x = 2\times 40^\circ = 80^\circ$\\
Then, $\angle AOD = \frac{1}{5}\angle AOC = \frac{1}{5}\times 80^\circ = 16^\circ$.\\
When ray $OD$ is below $OA$,\\
$\angle COD = \angle AOC + \angle AOD = 80^\circ + 16^\circ = \boxed{96^\circ}$\\
When ray $OD$ is above $OA$,\\
$\angle COD = \angle AOC - \angle AOD = 80^\circ - 16^\circ = \boxed{64^\circ}$\\
In summary: The measure of $\angle COD$ is either $96^\circ$ or $64^\circ$.


Question: As shown in the figure, a set of triangles (ignoring thickness) is placed on the table, with the right angle $\angle DAE$ and the $60^\circ$ angle $\angle BAC$ vertices coinciding. The triangle $ADE$ can rotate around point A. When $AD$ bisects $\angle BAC$, find the degree of $\angle 1$.

Solution:
\textbf{Solution:} Since AD bisects $\angle BAC$, and $\angle BAC=60^\circ$,\\
it follows that $\angle DAC=\frac{1}{2}\angle BAC=30^\circ$.\\
Since $\angle DAE=90^\circ$,\\
it follows that $\angle 1=\angle DAE-\angle DAC=\boxed{60^\circ}$.


Question: As shown in the figure, a set of triangles is placed on a table (ignoring thickness), with the right angle $(\angle DAE)$ and the $60^\circ$ angle $(\angle BAC)$ having coincident vertices. The triangle $ADE$ can rotate around point $A$. During the rotation process, keeping $AD$ inside $\angle BAC$, what is the degree measure of $\angle 1 - \angle 2$?

Solution:
\textbf{Solution:} During the rotation process, when $AD$ is inside $\angle BAC$, we have $\angle 1 = \angle DAE - \angle DAC = 90^\circ - \angle DAC$, and $\angle 2 = \angle BAC - \angle DAC = 60^\circ - \angle DAC$.

$\therefore \angle 1 - \angle 2 = (90^\circ - \angle DAC) - (60^\circ - \angle DAC) = \boxed{30^\circ}$.


Question: As shown in the figure, one corner of the rectangular notebook loose-leaf sheet is folded over so that the vertex $A$ falls at $A'$, with $BC$ being the crease. If $\angle ACB = 35^\circ$, find the degree measure of $\angle A'CD$.

Solution:
\textbf{Solution:} Given that $\because \angle ABC=35^\circ$, \\
$\therefore$ by folding we get $\angle A^{\prime}BC=\angle ABC=35^\circ$, \\
$\therefore \angle A^{\prime}CD=180^\circ-\angle ACB-\angle A^{\prime}CB \\
=180^\circ-35^\circ-35^\circ \\
=\boxed{110^\circ};$


Question: As shown in the figure, we have already learned the concept of angle bisectors. So, can you use it to solve related problems? Assuming another angle is also folded over such that side $CD$ coincides with $CA'$, with the crease being $CE$, what are the measures of $\angle 1$ and $\angle BCE$?

Solution:
\textbf{Solution}: From the conclusion of (1), we have $\angle A'CD=110^\circ$,\\
due to the property of folding, we obtain $\angle 1= \frac{1}{2} \angle A'CD= \frac{1}{2} \times 110^\circ=55^\circ$,\\
by the property of angle bisector, we get $\angle 2=\angle ACB=35^\circ$;\\
$\therefore \angle BCE=\angle 1+\angle 2=35^\circ+55^\circ=\boxed{90^\circ}$;


Question: Place the right-angled vertex $O$ of the right-angled triangle $OMN$ on the line $AB$, and let the ray $OC$ bisect $\angle AON$. As shown in the figure, if $\angle BON=60^\circ$, what is the degree measure of $\angle COM$?

Solution:
\textbf{Solution:}\\
$\because \angle BON=60^\circ$,\\
$\therefore \angle AON=120^\circ$,\\
$\because OC$ bisects $\angle AON$,\\
$\therefore \angle CON=60^\circ$,\\
$\because \angle MON=90^\circ$,\\
$\therefore \angle COM=90^\circ-60^\circ=\boxed{30^\circ}$.


Question: As shown in the figure, place the right angle vertex $O$ of the right-angled triangle $OMN$ on the line $AB$, with the ray $OC$ bisecting the angle $\angle AON$. Rotate the right-angled triangle $OMN$ counterclockwise about vertex $O$. During the rotation process:

Solution:
\textbf{Solution:}

(1) When ON is above line AB,

$\because \angle BON=140^\circ$,

$\therefore \angle AON=40^\circ$,

$\because OC$ bisects $\angle AON$,

$\therefore \angle CON=20^\circ$,

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

$\therefore \angle COM=70^\circ$.

When ON is below line AB,

$\because \angle BON=140^\circ$,

$\therefore \angle AON=40^\circ$,

$\because OC$ bisects $\angle AON$,

$\therefore \angle CON=20^\circ$,

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

$\therefore \angle COM=\boxed{110^\circ}$.

(2) $\angle BON=2\angle COM$ or $\angle BON+2\angle COM=360^\circ$


Question: As shown in the figure, point $O$ is a point on line $AB$, and the right-angled vertex of a right-angled triangle $OMN$ is placed at point $O$. The ray $OC$ bisects $\angle MOB$. As shown in Figure 1, if $\angle AOM = 30^\circ$, what is the degree measure of $\angle CON$?

Solution:
\textbf{Solution}: From the given, we find that \angle BOM=180^\circ-\angle AOM=150^\circ,\\
Furthermore, since \angle MON is a right angle and $OC$ bisects \angle BOM,\\
it follows that \angle CON=\angle MON-\frac{1}{2}\angle BOM \\
=90^\circ-\frac{1}{2}\times 150^\circ,\\
= \boxed{15^\circ}


Question: As shown in the figure, point $O$ is on line $AB$. Place the right-angled vertex of the right-angled triangle $OMN$ at point $O$. The ray $OC$ bisects $\angle MOB$. Rotate the right-angled triangle $OMN$ in Figure 1 clockwise around the vertex $O$ to the position in Figure 2, with one side $OM$ above the ray $OB$ and the other side $ON$ below line $AB$.

Solution:
\textbf{Solution:} Let $ \angle AOM=\alpha $, then $ \angle BOM=180^\circ-\alpha $,

(1) $\angle AOM=2\angle CON$, the reason is as follows:

Since $OC$ bisects $ \angle BOM$,

$\therefore \angle MOC=\frac{1}{2}\angle BOM=\frac{1}{2}\left(180^\circ-\alpha \right)$

$=90^\circ-\frac{1}{2}\alpha $,

Since $\angle MON=90^\circ$,

$\therefore \angle CON=\angle MON-\angle MOC=90^\circ-\left(90^\circ-\frac{1}{2}\alpha \right)=\frac{1}{2}\alpha $,

$\therefore \angle AOM=2\angle CON$,

(2) From (1), we know $ \angle BON=\angle MON-\angle BOM=90^\circ-\left(180^\circ-\alpha \right)=\alpha -90^\circ$,

$ \angle AOC=\angle AOM+\angle MOC=\alpha +90^\circ-\frac{1}{2}\alpha =90^\circ+\frac{1}{2}\alpha $,

Since $\angle AOC=3\angle BON$,

$\therefore 90^\circ+\frac{1}{2}\alpha =3\left(\alpha -90^\circ\right)$, solving this gives $ \alpha =\boxed{144^\circ}$,

Thus, $\angle AOM=144^\circ$.


Question: Take point O on line AB as the endpoint to draw ray OC so that $\angle BOC=40^\circ$. Place the right-angled vertex of a right-angle set square at point O, i.e., $\angle DOE=90^\circ$. As shown in Figure 1, if one side OE of the right-angle set square DOE lies on the ray OA, what is the value of $\angle COD$?

Solution: \textbf{Solution:} Hence, $\angle COD = \boxed{50^\circ}$.


Question: Given point O on line AB, construct ray OC such that $\angle BOC=40^\circ$. Place the right-angled vertex of a right triangle at O, that is, $\angle DOE=90^\circ$. As shown in figure 2, rotate the right triangle DOE clockwise around point O to a certain position. If OE exactly bisects $\angle AOC$, then what is the measure of $\angle COD$?

Solution: \textbf{Solution:} We have $\angle COD = \boxed{20^\circ}$.


Question: As shown in Figure 1, two right-angled triangles are stacked on point $O$ with two of their vertices overlapping, where the vertex $C$ of one triangle is on the side $OA$ of the other triangle. It is known that $\angle ABO=\angle DCO=90^\circ$, $\angle AOB=45^\circ$, $\angle COD=60^\circ$. The bisector of $\angle AOD$ intersects side $CD$ at point $E$. What is the degree measure of $\angle BOE$?

Solution:
\textbf{Solution:} Given that $\because \angle AOD=60^\circ$, and $OE$ bisects $\angle AOD$,\\
$\therefore \angle AOE=\frac{1}{2}\angle AOD=30^\circ$\\
$\because \angle AOB=45^\circ$\\
$\therefore \angle BOE=\angle AOE+\angle AOB=\boxed{75^\circ}$


Question: As shown in Figure 1, two right-angle triangles are overlapped at point $O$ with their vertices coinciding, where the vertex $C$ of one triangle falls on the edge $OA$ of the other triangle. It is known that $\angle ABO=\angle DCO=90^\circ$, $\angle AOB=45^\circ$, $\angle COD=60^\circ$. Draw the bisector of $\angle AOD$ intersecting side $CD$ at point $E$. As shown in Figure 2, if point $C$ does not fall on side $OA$, when $\angle COE=15^\circ$, find the degree measure of $\angle BOD$.

Solution:
Solution: Since $\angle COD=60^\circ$ and $\angle COE=15^\circ$, \\
therefore $\angle DOE=\angle COD-\angle COE=45^\circ$\\
Since $OE$ bisects $\angle AOD$,\\
therefore $\angle AOD=2\angle DOE=90^\circ$\\
Since $\angle AOB=45^\circ$\\
therefore $\angle BOD=\angle AOD+\angle AOB=\boxed{135^\circ}$


Question: As shown in the figure, it is known that point $O$ is a point on line $AB$. A right-angled triangle ruler $MON$ is placed as shown, with the right-angled vertex at point $O$. Ray $OC$ is drawn inside $\angle MON$, and $OC$ exactly bisects $\angle MOB$. If $\angle CON=20^\circ$, what is the degree measure of $\angle AOM$?

Solution:
\textbf{Solution:} Given, \\
$\because \angle MON=90^\circ$ and $\angle CON=20^\circ$, \\
$\therefore \angle MOC=90^\circ-\angle CON=70^\circ$, \\
$\because OC$ bisects $\angle MOB$, \\
$\therefore \angle BOM=2\angle MOC=140^\circ$, \\
$\therefore \angle AOM=180^\circ-\angle BOM=\boxed{40^\circ}$;


Question: Given that point $O$ is a point on line $AB$, a right-angled triangle ruler $MON$ is placed as shown in the figure, with the right-angle vertex at point $O$. A ray $OC$ is drawn inside $\angle MON$, and $OC$ exactly bisects $\angle MOB$. If $\angle BON = 2 \angle NOC$, what is the measure of $\angle AOM$ in degrees?

Solution:
\textbf{Solution:} Since $\angle BON=2\angle NOC$ and $OC$ bisects $\angle MOB$,\\
it follows that $\angle MOC=\angle BOC=3\angle NOC$\\
Since $\angle MOC+\angle NOC=\angle MON=90^\circ$,\\
then $3\angle NOC+\angle NOC=90^\circ$\\
Thus, $4\angle NOC=90^\circ$,\\
which implies $\angle BON=2\angle NOC=45^\circ$,\\
Therefore, $\angle AOM=180^\circ−\angle MON−\angle BON=180^\circ−90^\circ−45^\circ=\boxed{45^\circ}$.


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

Solution:
\textbf{Solution:} Given the diagram, we have: $\angle AOB=15^\circ+40^\circ=55^\circ$,\\
since $OA$ is the bisector of $\angle BOC$,\\
thus $\angle AOC=55^\circ$,\\
therefore $\angle NOC=\angle NOA+\angle AOC$\\
$=15^\circ+55^\circ=70^\circ$,\\
hence, ray $OC$ points in the direction of $70^\circ$ east of north.


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

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


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

Solution:
\textbf{Solution:} Since $\angle MON=90^\circ$ and $\angle BOC=65^\circ$, \\
it follows that $\angle MOC=\angle MON-\angle BOC=90^\circ-65^\circ=\boxed{25^\circ}$;


Question: Let point O be a point on line AB. A ray OC is drawn through point O such that $\angle BOC = 65^\circ$. Place the right-angled vertex of a right-angled triangle ruler at point O. As shown in Figure 2, rotate the triangle ruler MON counterclockwise around point O by a certain angle, such that OC is the bisector of $\angle MOB$. What are the measures of $\angle BON$ and $\angle CON$?

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


Question: Let $O$ be a point on line $AB$. A ray $OC$ is drawn from point $O$ such that $\angle BOC=65^\circ$. Place the right-angled vertex of a right-angled triangle plate at point $O$. When the triangle plate $MON$ is rotated counterclockwise around point $O$ to position 3, $\angle NOC=\frac{1}{4}\angle AOM$, what is the degree measure of $\angle NOB$?

Solution:
\textbf{Solution:} Given that, \\
$\because \angle NOC = \frac{1}{4}\angle AOM$,\\
$\therefore \angle AOM = 4\angle NOC$.\\
$\because \angle BOC = 65^\circ$,\\
$\therefore \angle AOC = \angle AOB - \angle BOC = 180^\circ - 65^\circ = 115^\circ$,\\
$\because \angle MON = 90^\circ$,\\
$\therefore \angle AOM + \angle NOC = \angle AOC - \angle MON = 115^\circ - 90^\circ = 25^\circ$,\\
$\therefore 4\angle NOC + \angle NOC = 25^\circ$,\\
$\therefore \angle NOC = 5^\circ$,\\
$\therefore \angle NOB = \angle NOC + \angle BOC = \boxed{70^\circ}$.


Question: Given that point O is a point on line AB, place the right-angled triangle MON as shown in the figure, with the right-angled vertex at O. Draw a ray OC inside $ \angle MON$ such that OC exactly bisects $ \angle BOM$. If $ \angle CON=24^\circ$, find the degree measure of $ \angle AOM$.

Solution:
\textbf{Solution:} Since $\angle MON=90^\circ$ and $\angle CON=24^\circ$, \\
it follows that $\angle MOC=90^\circ-\angle CON=66^\circ$. \\
Since OC bisects $\angle MOB$, \\
it follows that $\angle BOM=2\angle MOC=132^\circ$, \\
hence $\angle AOM=180^\circ-\angle BOM=\boxed{48^\circ}$.


Question: As shown in the figure, it is known that point O is on the line AB. The right-angled triangle plate MON is placed as shown, with the right-angle vertex at point O. A ray OC is drawn inside $\angle MON$, and OC exactly bisects $\angle BOM$. If $\angle BON = 2\angle CON$, find the measure of $\angle AOM$.

Solution:
\textbf{Solution:} Given $\because \angle BON = 2\angle NOC$, OC bisects $\angle MOB$,\\
$\therefore \angle MOC = \angle BOC = 3\angle NOC$,\\
$\because \angle MOC + \angle NOC = \angle MON = 90^\circ$,\\
$\therefore 3\angle NOC + \angle NOC = 90^\circ$,\\
$\therefore 4\angle NOC = 90^\circ$,\\
$\therefore \angle BON = 2\angle NOC = 45^\circ$,\\
$\therefore \angle AOM = 180^\circ - \angle MON - \angle BON = 180^\circ - 90^\circ - 45^\circ = \boxed{45^\circ}$;


Question: As shown in Figure 1, it is known that $\angle AOB = 120^\circ$, ray $OC$ is the angle bisector of $\angle AOB$, point $D$ is a point inside $\angle BOC$, and point $D$ does not lie on the bisector of $\angle BOC$. When $\angle BOD = 20^\circ$, calculate the degree measure of $\angle COD$?

Solution:
\textbf{Solution:} Since the ray $OC$ is the angle bisector of $\angle AOB$ and $\angle AOB=120^\circ$,\\
it follows that $\angle BOC=\frac{1}{2}\angle AOB=\frac{1}{2}\times 120^\circ=60^\circ$,\\
therefore $\angle COD=\angle BOC-\angle BOD=60^\circ-20^\circ=\boxed{40^\circ}$;


Question: As shown in the figure, it is known that $OE$ is the angle bisector of $\angle AOC$, and $OD$ is the angle bisector of $\angle BOC$. If $\angle AOC=120^\circ$ and $\angle BOC=30^\circ$, what is the degree measure of $\angle DOE$?

Solution:
\textbf{Solution.} Since OE is the bisector of $\angle AOC$ and OD is the bisector of $\angle BOC$, with $\angle AOC=120^\circ$ and $\angle BOC=30^\circ$,
therefore, $\angle EOC=60^\circ$ and $\angle DOC=15^\circ$,
hence $\angle DOE=\angle EOC-\angle DOC=60^\circ-15^\circ=\boxed{45^\circ}$.


Question: As shown in the figure, it is known that $OE$ is the angle bisector of $\angle AOC$, and $OD$ is the angle bisector of $\angle BOC$. If $\angle AOB=90^\circ$ and $\angle BOC=\alpha$, what is the degree measure of $\angle DOE$?

Solution:
\textbf{Solution:} Given $\because$ $\angle AOB=90^\circ$, $\angle BOC=\alpha$ \\
$\therefore$ $\angle AOC = 90^\circ+\alpha$ \\
$\because$ OE is the bisector of $\angle AOC$, OD is the bisector of $\angle BOC$. \\
$\therefore$ $\angle AOE= \frac{1}{2} \angle AOC$, $\angle COD= \frac{1}{2}\angle BOC$ \\
$\angle AOE= \frac{1}{2}(90^\circ+\alpha)$, $\angle COD= \frac{1}{2}\alpha$ \\
$\angle DOE=\angle AOC - \angle AOE-\angle COD$ \\
$= (90^\circ+\alpha) - \frac{1}{2}(90^\circ+\alpha) - \frac{1}{2}\alpha = \boxed{45^\circ}$


Question: As shown in Figure 1, $E$ is the midpoint of line segment $AB$, point $C$ is on line segment $AB$, and $F$ is the midpoint of $AC$. If $EF = 5\,cm$ and $AC = 6\,cm$, find the lengths of line segments $CE$ and $AB$.

Solution:
Solution: Since F is the midpoint of AC and $AC=6\,cm$, \\
thus $CF=AF=\frac{1}{2}AC=3\,cm$, \\
Since $EF=5\,cm$, \\
thus $CE=EF-CF=5-3=\boxed{2}\,(cm)$, \\
thus $AE=AC+CE=6+2=8\,(cm)$, \\
Since E is the midpoint of the line segment AB, \\
thus $AB=2AE=\boxed{16}\,cm$.


Question: As shown in the figure, it is known that $O$ is a point on line $AB$, $\angle AOE = \angle COD$, ray $OC$ bisects $\angle BOE$, and $\angle EOC = 50^\circ$. What is the measure of $\angle DOE$ in degrees?

Solution:
\textbf{Solution:} Since OC bisects $\angle BOE$ and $\angle EOC=50^\circ$, \\
$\therefore \angle BOE=2\angle EOC=100^\circ$,\\
$\therefore \angle AOE=180^\circ-\angle BOE=80^\circ$,\\
Since $\angle AOE=\angle COD$, \\
$\therefore \angle COD=80^\circ$,\\
$\therefore \angle DOE=\angle COD-\angle COE=80^\circ-50^\circ=\boxed{30^\circ}$.


Question: As shown in Figure 1, it is known that $\angle AOB = 160^\circ$, $\angle COE$ is a right angle, and $OF$ bisects $\angle AOE$. If $\angle COF = 32^\circ$, what is the measure of $\angle BOE$ in degrees?

Solution: \textbf{Solution:} We have $\angle BOE = \boxed{44^\circ}$.


Question: As shown in the figure, point O is a point on line AB, $\angle COD$ is a right angle, $OE$ bisects $\angle BOC$. As shown in Figure (1), if $\angle AOC = 40^\circ$, what is the degree measure of $\angle DOE$?

Solution:
\textbf{Solution:} Given that $\angle AOC + \angle BOC = 180^\circ$,\\
we have $\angle BOC = 180^\circ - \angle AOC = 180^\circ - 40^\circ = 140^\circ$,\\
Since $OE$ bisects $\angle BOC$,\\
we get $\angle COE = \frac{1}{2} \angle BOC = \frac{1}{2} \times 140^\circ = 70^\circ$,\\
Given $\angle COD$ is a right angle,\\
thus $\angle COE + \angle DOE = 90^\circ$,\\
leading to $\angle DOE = 90^\circ - \angle COE = 90^\circ - 70^\circ = \boxed{20^\circ};$


Question: As shown in the figure, point O is a point on the line AB, $\angle COD$ is a right angle, and $OE$ bisects $\angle BOC$. As shown in figure (2), if $\angle COE = \angle DOB$, what is the degree measure of $\angle AOC$?

Solution:
\textbf{Solution:} \\
$\because$ OE bisects $\angle BOC$, \\
$\therefore \angle COE = \angle BOE$, \\
$\because \angle COE = \angle BOD$, \\
$\therefore \angle COE = \angle BOE = \angle DOB$, \\
$\because \angle COD = 90^\circ$, \\
$\therefore \angle COE = \angle BOE = \frac{1}{3} \times 90^\circ = 30^\circ$, \\
$\therefore \angle AOC = 180^\circ - 30^\circ - 30^\circ = \boxed{120^\circ}.$


Question: As shown in the figure, right-angled triangle $ABC$ has its right-angled vertex $C$ on line $DE$, and $CF$ bisects $\angle BCD$. As shown in Figure (1), if $\angle BCE=30^\circ$, find the degree measure of $\angle ACF$.

Solution:
\textbf{Solution:} Given the problem statement, since $\angle ACB=90^\circ,\angle BCE=30^\circ$,\\
$\therefore \angle ACD=180^\circ-90^\circ-30^\circ=60^\circ,\angle BCD=180^\circ-30^\circ=150^\circ$.\\
Also, since $CF$ bisects $\angle BCD$,\\
$\therefore \angle DCF=\angle BCF=\frac{1}{2}\angle BCD=75^\circ$,\\
$\therefore \angle ACF=\angle DCF-\angle ACD=75^\circ-60^\circ=\boxed{15^\circ}$.


Question: As shown in the figure, the right-angled vertex $C$ of the right-angled triangle $ABC$ is on the line $DE$, and $CF$ bisects $\angle BCD$. When the triangle $ABC$ in figure (1) is rotated about the vertex $C$ to the position in figure (2), if $\angle BCE=140^\circ$, calculate the degrees of $\angle ACF$ and $\angle ACE$.

Solution: \textbf{Solution}: As shown in the diagram (2), since $\angle BCE=140^\circ$, it follows that $\angle BCD=40^\circ$. Since $CF$ bisects $\angle BCD$, we have $\angle BCF=\angle FCD=20^\circ$, hence $\angle ACF=90^\circ-\angle BCF=\boxed{70^\circ}$, and $\angle ACD=90^\circ-\angle BCD=50^\circ$, thus $\angle ACE=180^\circ-\angle ACD=\boxed{130^\circ}$.


Question: Given that $O$ is a point on line $AB$, place the right-angle vertex of the right-angled triangle $OMN$ at point $O$. Ray $OC$ bisects $\angle MOB$. As shown in Figure 1, if $\angle AOM = 30^\circ$, what is the measure of $\angle CON$?

Solution:
\textbf{Solution:}
Given $\angle BOM = 180^\circ - \angle AOM = 150^\circ$,\\
since $\angle MON$ is a right angle, and $OC$ bisects $\angle BOM$,\\
thus, $\angle CON = \angle MON - \frac{1}{2} \angle BOM = 90^\circ - \frac{1}{2} \times 150^\circ = \boxed{15^\circ}$.


Question: It is known that $O$ is a point on line $AB$, and the right-angle vertex of a right-angled triangle $OMN$ is placed at point $O$. Ray $OC$ bisects $\angle MOB$. Rotate the right-angled triangle $OMN$ in Figure 1 clockwise about vertex $O$ to the position in Figure 2. When $\angle AOC = 3\angle BON$, what is the measure of $\angle AOM$?

Solution:
\textbf{Solution:} Let $\angle AOM = x$, then $\angle BOM = 180^\circ - x$,

$\because$ OC bisects $\angle BOM$,

$\therefore \angle MOC = \frac{1}{2}\angle BOM = \frac{1}{2}(180^\circ - x) = 90^\circ - \frac{1}{2}x$,

$\therefore \angle AOC = \angle AOM + \angle MOC = x + 90^\circ - \frac{1}{2}x = 90^\circ + \frac{1}{2}x$,

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

$\therefore \angle BON = \angle MON - \angle BOM = 90^\circ - (180^\circ - x) = x - 90^\circ$,

$\because \angle AOC = 3\angle BON$,

$\therefore 90^\circ + \frac{1}{2}x = 3(x - 90^\circ)$,

Solving gives $x = \boxed{144^\circ}$,

$\therefore \angle AOM = 144^\circ$.


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

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


Question: Given point O on line AB, draw ray OC through point O such that $\angle BOC=65^\circ$. Place the right angle vertex of a right-angled triangle ruler at point O. As shown in the diagram, rotate the triangle ruler MON about point O anticlockwise by a certain angle, such that OC is the angle bisector of $\angle MOB$. Find the degrees of $\angle BON$ and $\angle CON$?

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


Question: Point O is a point on line AB. A ray OC is drawn through point O such that $\angle BOC = 65^\circ$. Place the right-angled vertex of a right-angled triangle ruler at point O. When the triangle ruler MON is rotated counterclockwise around point O to the position shown in figure (3), $\angle NOC = \frac{1}{4}\angle AOM$, what is the degree measure of $\angle NOB$?

Solution:
\textbf{Solution:} According to the given problem, \\
$\because \angle NOC = \frac{1}{4}\angle AOM$, \\
$\therefore \angle AOM = 4\angle NOC$, \\
$\because \angle BOC = 65^\circ$, \\
$\therefore \angle AOC = \angle AOB - \angle BOC = 180^\circ - 65^\circ = 115^\circ$, \\
$\because \angle MON = 90^\circ$, \\
$\therefore \angle AOM + \angle NOC = \angle AOC - \angle MON = 115^\circ - 90^\circ = 25^\circ$, \\
$\therefore 4\angle NOC + \angle NOC = 25^\circ$, \\
$\therefore \angle NOC = 5^\circ$, \\
$\therefore \angle NOB = \angle NOC + \angle BOC = \boxed{70^\circ}$.


Question: As shown in Figure 1, a set of triangular plates are pieced together, with sides $OA$ and $OC$ coinciding with line $EF$, where $\angle AOB=45^\circ$ and $\angle COD=60^\circ$. What is the degree measure of $\angle BOD$ in Figure 1?

Solution:
\textbf{Solution:} Given that $\because$ $\angle AOB=45^\circ$ and $\angle COD=60^\circ$,\\
$\therefore$ $\angle BOD=180^\circ-\angle AOB-\angle COD=180^\circ-45^\circ-60^\circ=\boxed{75^\circ}$.


Question: As shown in Figure 1, a set of triangular plates are combined together, with sides $OA$, $OC$ coinciding with line $EF$, where $\angle AOB=45^\circ$, $\angle COD=60^\circ$. In Figure 2, the triangular plate $COD$ is fixed, and the triangular plate $AOB$ is rotated clockwise around point $O$ by an angle. During the rotation, triangular plate $AOB$ always stays inside $\angle EOD$, let $\angle EOA=\alpha$. If $OB$ bisects $\angle EOD$, what is the value of $\alpha$?

Solution:
\textbf{Solution:}\\
(1) Since $\angle COD=60^\circ$,\\
then $\angle EOD=180^\circ-\angle COD=180^\circ-60^\circ=120^\circ$,\\
Since $OB$ bisects $\angle EOD$,\\
then $\angle EOB=\frac{1}{2}\angle EOD=\frac{1}{2}\times 120^\circ=60^\circ$,\\
Since $\angle AOB=45^\circ$,\\
then $\alpha =\angle EOB-\angle AOB=60^\circ-45^\circ=\boxed{15^\circ}$;\\
(2) Since $\angle AOB=45^\circ$ and $\angle COD=60^\circ$,\\
then $\angle AOC=\angle AOB+\angle BOD+\angle COD=45^\circ+\angle BOD+60^\circ=105^\circ+\angle BOD$,\\
Since $\angle AOC=4\angle BOD$,\\
then $105^\circ+\angle BOD=4\angle BOD$,\\
Solving yields: $\angle BOD=35^\circ$,\\
then $\angle AOC=4\angle BOD=4\times 35^\circ=140^\circ$,\\
then $\alpha =180^\circ-\angle AOC=180^\circ-140^\circ=\boxed{40^\circ}$.


Question: As shown in the figure, it is known that $OE$ is the angle bisector of $\angle AOC$, and $OD$ is the angle bisector of $\angle BOC$. If $\angle AOE=70^\circ$ and $\angle COD=20^\circ$, what is the degree measure of $\angle AOB$?

Solution:
\textbf{Solution:} Since $OD$ is the angle bisector of $\angle BOC$ and $\angle COD=20^\circ$,\\
$\therefore \angle BOC=2\angle COD=40^\circ$.\\
Since $OE$ is the angle bisector of $\angle AOC$,\\
$\therefore \angle AOC=2\angle AOE=140^\circ$.\\
$\therefore \angle AOB=\angle AOC-\angle BOC=140-40^\circ=\boxed{100^\circ}$.


Question: As shown, it is known that $OE$ is the angle bisector of $\angle AOC$, and $OD$ is the angle bisector of $\angle BOC$. If $\angle DOE=45^\circ$, and $\angle AOC + \angle BOC = 180^\circ$, what is the degree measure of $\angle COD$?

Solution:
\textbf{Solution:} Since $OD$ is the angle bisector of $\angle BOC$,\\
let $\angle COD=\angle BOD=x$.\\
Since $\angle DOE=45^\circ$,\\
then $\angle BOE=45^\circ-x$, $\angle COE=45^\circ+x$.\\
Since $OE$ is the angle bisector of $\angle AOC$,\\
then $\angle AOE=\angle COE=45^\circ+x$.\\
Since $\angle AOC+\angle BOC=180^\circ$,\\
then $2(45^\circ+x)+2x=180^\circ$,\\
thus $x=\boxed{22.5^\circ}$, that is $\angle COD=22.5^\circ$.


Question: As shown in the figure, given that $\angle AOB=90^\circ$, draw rays $OC$, and $OD$ is the angle bisector of $\angle AOC$, $OE$ is the angle bisector of $\angle BOC$. As shown in Figure (1), when $\angle BOC=70^\circ$, find the degree measure of $\angle DOE$.

Solution:
\textbf{Solution:} Since $\angle AOB=90^\circ$ and $\angle BOC=70^\circ$,\\
it follows that $\angle AOC=90^\circ-\angle BOC=20^\circ$,\\
Since lines $OD$ and $OE$ bisect $\angle AOC$ and $\angle BOC$ respectively,\\
it follows that $\angle COD=\frac{1}{2}\angle AOC=10^\circ$ and $\angle COE=\frac{1}{2}\angle BOC=35^\circ$,\\
therefore $\angle DOE=\angle COD+\angle COE=\boxed{45^\circ}$.


Question: Given: $\angle AOB = 90^\circ$, draw rays $OC$, $OD$ is the bisector of $\angle AOC$, $OE$ is the bisector of $\angle BOC$. As shown in figure (2), if ray $OC$ rotates around point $O$ within $\angle AOB$, when $\angle BOC = a$, what is the degree measure of $\angle DOE$?

Solution:
\textbf{Solution:} The size of $\angle DOE$ remains constant. The reason is as follows:

Since $\angle AOB=90^\circ$, $\angle BOC=\alpha$,

Therefore, $\angle AOD=90^\circ-\alpha$.

Also, since $OE$, $OD$ are the angle bisectors of $\angle BOC$ and $\angle AOC$ respectively,

Therefore, $\angle EOC=\frac{1}{2}\alpha$, $\angle COD=\frac{1}{2}\left(90^\circ-\alpha \right)$.

Therefore, $\angle DOE=\angle EOC+\angle COD$

$=\frac{1}{2}\alpha +\frac{1}{2}\left(90^\circ-\alpha \right)=\boxed{45^\circ}$.


Question: Given: $\angle AOB=90^\circ$, construct rays $OC$, $OD$ is the angle bisector of $\angle AOC$, $OE$ is the angle bisector of $\angle BOC$. If ray $OC$ rotates around point $O$ outside $\angle AOB$ and $\angle AOC$ is an obtuse angle, find the degree measure of $\angle DOE$.

Solution:
\textbf{Solution:} Consider two cases:

As shown in Figure 3, since $OD$ and $OE$ bisect $\angle AOC$ and $\angle BOC$ respectively,\\
it follows that $\angle COD=\frac{1}{2}\angle AOC$, $\angle COE=\frac{1}{2}\angle BOC$,\\
thus $\angle DOE=\angle COD-\angle COE=\frac{1}{2}\left(\angle AOC-\angle BOC\right)=45^\circ$;

As shown in Figure 4, since $OE$ and $OD$ are the bisectors of $\angle BOC$ and $\angle AOC$ respectively,\\
it follows that $\angle EOC=\angle BOE$, $\angle COD=\angle AOD$,\\
and since $\angle AOB=90^\circ$,\\
it follows that $\angle AOD+\angle DOC+\angle COE+\angle EOB=270^\circ$,\\
thus $2\angle DOC+2\angle COE=270^\circ$,\\
therefore $\angle DOC+\angle COE=135^\circ$,\\
leading to $\angle DOE=\boxed{135^\circ}$.


Question: As shown in the figure, lines AB and CD intersect at point O. It is known that $\angle BOD=75^\circ$, and OE divides $\angle AOC$ into two angles, with $\angle AOE: \angle EOC = 2: 3$. What is the degree measure of $\angle AOE$?

Solution:
\textbf{Solution:} From the given information, we have $\angle AOC=\angle BOD=75^\circ$,\\
$\therefore \angle AOE=75^\circ \times \frac{2}{5}=\boxed{30^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle A=40^\circ$. The bisector of the exterior angle $\angle CBD$ of $\triangle ABC$ intersects the extension of $AC$ at point $E$. What is the measure of $\angle CBE$ in degrees?

Solution:
\textbf{Solution:}
Given $\triangle ABC$ where $\angle ACB=90^\circ$ and $\angle A=40^\circ$,\\
thus $\angle ABC=90^\circ-\angle A=50^\circ$.\\
Thus, $\angle CBD=130^\circ$.\\
Since $BE$ bisects $\angle CBD$,\\
thus, $\angle CBE= \frac{1}{2} \angle CBD=\boxed{65^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, $\angle A = 40^\circ$, the bisector of the external angle $\angle CBD$ of $\triangle ABC$ intersects the extension line of $AC$ at point $E$. Through point $D$, draw $DF \parallel BE$, which intersects the extension line of $AC$ at point $F$. What is the measure of $\angle F$ in degrees?

Solution:
\textbf{Solution:} Since $\angle ACB= 90^\circ$ and $\angle CBE=65^\circ$, \\
it follows that $\angle CEB=90^\circ- 65^\circ= 25^\circ$. \\
Since $DF \parallel BE$, it follows that $\angle F=\boxed{25^\circ}$.


Question: As shown in Figure 1, it is known that $\angle AOB=100^\circ$. The rays $OE$ and $OF$ are the angle bisectors of $\angle AOC$ and $\angle COB$, respectively. If the ray $OC$ is inside the angle $\angle AOB$ and $\angle AOC=30^\circ$, what is the degree measure of $\angle EOF$?

Solution:
\textbf{Solution:} By construction,\\
since $OE$ is the bisector of $\angle AOC$, and $\angle AOC=30^\circ$,\\
then $\angle COE=\frac{1}{2}\angle AOC=15^\circ$,\\
since $\angle AOB=100^\circ$,\\
then $\angle COB=\angle AOB-\angle AOC=70^\circ$,\\
since $OF$ is the bisector of $\angle COB$,\\
then $\angle COF=\frac{1}{2}\angle COB=35^\circ$,\\
thus $\angle EOF=\angle COE+\angle COF=15^\circ+35^\circ=\boxed{50^\circ}$.


Question: Given $\angle AOB = 100^\circ$, rays $OE$ and $OF$ are the angle bisectors of $\angle AOC$ and $\angle COB$, respectively. If ray $OC$ rotates around point $O$ within $\angle AOB$, what is the degree measure of $\angle EOF$?

Solution:
\textbf{Solution:} Given that,
\[
\because \angle AOB=100^\circ,
\]
\[
\therefore \angle AOC+\angle COB=100^\circ,
\]
\[
\because OE \text{ is the bisector of } \angle AOC, OF \text{ is the bisector of }\angle COB,
\]
\[
\therefore \angle COE=\frac{1}{2}\angle AOC, \angle COF=\frac{1}{2}\angle COB,
\]
\[
\therefore \angle EOF=\angle COE+\angle COF=\frac{1}{2}\left(\angle AOC+\angle COB\right)=\boxed{50^\circ}.
\]


Question: As illustrated, it is known that the line $AB$ intersects with $CD$ at point $O$, $OE\perp CD$, and $\angle AOC=40^\circ$, with $OF$ being the angle bisector of $\angle AOD$. What is the degree measure of $\angle EOB$?

Solution:
\textbf{Solution:} Since $OE \perp CD$,\\
$\therefore \angle EOD=90^\circ$,\\
and since $\angle BOD=\angle AOC=40^\circ$,\\
$\therefore \angle EOB=\angle EOD-\angle BOD=50^\circ$.\\
Therefore, the measure of $\angle EOB$ is $\boxed{50^\circ}$.


Question: As shown in the figure, it is known that line $AB$ intersects with line $CD$ at point $O$, $OE \perp CD$, $\angle AOC = 40^\circ$, and $OF$ is the bisector of $\angle AOD$. What is the degree measure of $\angle EOF$?

Solution:
\textbf{Solution:} Since the straight lines $AB$ and $CD$ intersect at the point $O$,\\
it follows that $\angle AOC=\angle BOD=40^\circ$,\\
thus $\angle AOD=180^\circ-\angle BOD=140^\circ$,\\
since $OF$ is the bisector of $\angle AOD$,\\
it follows that $\angle AOF=\angle FOD=70^\circ$,\\
therefore $\angle EOF=\angle EOD+\angle FOD=\boxed{160^\circ}$.


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

Solution:
\textbf{Solution:} Since $\angle AOC=40^\circ$,\\
it follows that $\angle BOC=140^\circ$,\\
also, since OE bisects $\angle BOC$,\\
it implies $\angle COE= \frac{1}{2} \times 140^\circ=70^\circ$,\\
since $\angle COD=90^\circ$,\\
it follows that $\angle DOE=90^\circ-70^\circ=\boxed{20^\circ}$.


Question: As shown in the diagram, point $O$ is on line $AB$, $\angle COD$ is a right angle, and $OE$ bisects $\angle BOC$. As shown in figure 2, if $\angle COE = \angle DOB$, find the measure of $\angle AOC$.

Solution:
\textbf{Solution:} Given that $OE$ bisects $\angle BOC$,\\
thus $\angle COE = \angle BOE$,\\
and since $\angle COE = \angle DOB$ and $\angle COD = 90^\circ$,\\
it follows that $\angle COE = \angle DOB = \angle BOE = 30^\circ$,\\
thus $\angle BOC = 60^\circ$,\\
therefore $\angle AOC = 180^\circ - 60^\circ = \boxed{120^\circ}$.


Question: As shown in the figure, line AB intersects with line CD at point O, $\angle BOE = 90^\circ$, $\angle AOD = 30^\circ$, and OF is the angle bisector of $\angle BOD$. What is the degree measure of $\angle EOC$?

Solution:
\textbf{Solution:} Since $\angle BOC = \angle AOD = 30^\circ$ and $\angle BOE = 90^\circ$,
\[
\therefore \angle COE = 90^\circ - 30^\circ = \boxed{60^\circ}.
\]


Question: Given, as shown in the figure, line $AB$ intersects with line $CD$ at point $O$. $\angle BOE = 90^\circ$, $\angle AOD = 30^\circ$, and $OF$ is the angle bisector of $\angle BOD$. Find the measure of $\angle EOF$.

Solution:
\textbf{Solution:} Given that $\angle BOC=30^\circ$,\\
it follows that $\angle BOD=180^\circ-30^\circ=150^\circ$.\\
Since OF is the angle bisector of $\angle BOD$,\\
we have $\angle BOF= \frac{1}{2} \angle BOD= \frac{1}{2} \times 150^\circ=75^\circ$.\\
Therefore, $\angle EOF=\angle EOC+\angle BOC+\angle BOF=60^\circ+30^\circ+75^\circ=\boxed{165^\circ}$.