Question: As shown in the figure, straight lines AB and CD intersect at point O, OE bisects $\angle AOD$, $OF\perp CD$, and  $\angle AOE=70^\circ$, find the degree measure of $\angle BOF$.

Solution:
\textbf{Solution:} Given that OE bisects $\angle AOD$ and $\angle AOE = 70^\circ$, \\
thus $\angle AOD = 2\angle AOE = 140^\circ$, \\
since A, O, B are collinear, \\
thus $\angle BOD = \angle AOC = 180^\circ - \angle AOD = 180^\circ - 140^\circ = 40^\circ$, \\
given that $OF \perp CD$, \\
thus $\angle FOD = 90^\circ$, \\
thus $\angle BOF = \angle FOD - \angle BOD = 90^\circ - 40^\circ = \boxed{50^\circ}$


Question: As shown in the figure, straight lines AB and CD intersect at point O, EO $\perp$ AB with the foot of the perpendicular at O, and $\angle EOC = 35^\circ$. What is the measure of $\angle BOD$?

Solution:
\textbf{Solution:} Given that $EO\perp AB$,\\
it follows that $\angle AOE=90^\circ$,\\
and given $\angle EOC=35^\circ$,\\
it results in $\angle AOC=\angle AOE-\angle EOC=55^\circ$,\\
therefore, $\angle BOD=\angle AOC=\boxed{55^\circ}$.


Question: As shown in the figure, $\angle AOC = 90^\circ$, OC bisects $\angle DOB$, and $\angle DOC = 25^\circ 35'$. Find the degree measure of $\angle BOA$?

Solution:
Solution: Since $OC$ bisects $\angle DOB$ and $\angle DOC=25^\circ35'$, \\
it follows that $\angle DOC=\angle BOC=25^\circ35'$. \\
Since $\angle AOC=90^\circ$, \\
it follows that $\angle BOA=90^\circ-25^\circ35'=64^\circ25'$. \\
Hence, the answer is $\boxed{64^\circ25'}$.


Question: As shown in the figure, $C$ is a point on line segment $AB$, and $D$ is the midpoint of line segment $BC$. If $AB=10$ and $AC=6$, find the length of $AD$.

Solution:
\textbf{Solution:} Since $C$ is a point on line segment $AB$, $AB=10$, $AC=6$,\\
thus $BC=AB-AC=10-6=4$,\\
since $D$ is the midpoint of line segment $BC$,\\
thus $BD=\frac{1}{2}BC=2$,\\
therefore $AD=AB-BD=10-2=\boxed{8}$.


Question: As shown in the diagram, $AO \perp BO$, $CO \perp DO$, where $O$ is the foot of the perpendiculars, $\angle BOC = 50^\circ$. What is the degree measure of $\angle AOD$?

Solution:
\textbf{Solution:} Given $AO\perp BO$, $CO\perp DO$,\\
therefore, $\angle AOB = \angle COD = 90^\circ$,\\
since $\angle AOB + \angle BOC + \angle COD + \angle AOD = 360^\circ$, and $\angle BOC = 50^\circ$,\\
thus, $90^\circ + 50^\circ + 90^\circ + \angle AOD = 360^\circ$,\\
we find that $\angle AOD = \boxed{130^\circ}$.


Question: As shown in the figure, lines $AB$ and $CD$ intersect at point $O$. $OE$ bisects $\angle BOC$, $OF\perp OE$, and $\angle AOD=66^\circ$. What is the degree measure of $\angle BOF$?

Solution:
\textbf{Solution:} Given that $\angle AOD = 66^\circ$,\\
hence $\angle BOC = \angle AOD = 66^\circ$,\\
since OE bisects $\angle BOC$,\\
therefore $\angle BOE = \frac{1}{2} \angle BOC = 33^\circ$,\\
since OF is perpendicular to OE,\\
therefore $\angle EOF = 90^\circ$,\\
thus $\angle BOF = 90^\circ - 33^\circ = \boxed{57^\circ}$.


Question: As shown in the figure, lines AB, CD, and EF intersect at point O. If $\angle COF = 120^\circ$ and $\angle AOD = 100^\circ$, what is the measure of $\angle AOF$ in degrees?

Solution:
\textbf{Solution:} Since $\angle COF=120^\circ$,\\
it follows that $\angle COE=180^\circ-120^\circ=60^\circ$,\\
thus $\angle DOF=\angle COE=60^\circ$,\\
and since $\angle AOD=100^\circ$,\\
it follows that $\angle AOF=\angle AOD-\angle DOF=100^\circ-60^\circ=\boxed{40^\circ}$.


Question: In $\triangle ABC$, $AD$ is the altitude from $A$ to side $BC$, $\angle C=45^\circ$, $\sin B=\frac{1}{3}$, and $AD=1$. Find the length of $BC$?

Solution:
\textbf{Solution:} Given $\tan C = \frac{AD}{DC}$, it follows that $\tan 45^\circ = \frac{1}{DC} = 1$,\\
thus, DC = 1.\\
Since $\sin B = \frac{AD}{AB} = \frac{1}{3}$, this implies $\frac{1}{AB} = \frac{1}{3}$,\\
thus, AB = 3.\\
In right triangle $ABD$, we have $AD^2 + BD^2 = AB^2$,\\
therefore, $BD = \sqrt{AB^2 - AD^2} = \sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}$.\\
Hence, BC = BD + DC = $2\sqrt{2} + 1 = \boxed{2\sqrt{2} + 1}$.


Question: As illustrated, in $\triangle ABC$, $\angle ACB = 90^\circ$, $CD$ and $CH$ are the median and the altitude on side $AB$, respectively, $BC = \sqrt{14}$, $\cos\angle ACD = \frac{3}{4}$, calculate the lengths of $AB$ and $CH$.

Solution:
\textbf{Solution:} Draw DE $\perp$ AC at E, then $\angle AED = \angle CED = 90^\circ$,\\
since $\angle ACB = 90^\circ$,\\
it follows that $\angle AED = \angle ACB$,\\
hence, DE $\parallel$ BC,\\
since CD is the median of $\triangle ABC$,\\
it follows that AD = BD,\\
hence, CE = AE, i.e., AC = 2CE\\
since BC = $\sqrt{14}$,\\
it follows that DE = $\frac{1}{2}$BC = $\frac{\sqrt{14}}{2}$,\\
since $\cos\angle ACD = \frac{CE}{CD} = \frac{3}{4}$\\
let CE = 3x, CD = 4x,\\
by the Pythagorean theorem, we get: $DE = \sqrt{CD^2 - CE^2} = \sqrt{(4x)^2 - (3x)^2} = \sqrt{7}x$\\
hence, $\sqrt{7}x = \frac{\sqrt{14}}{2}$, i.e., x = $\frac{\sqrt{2}}{2}$\\
hence, AE = CE = 3x = $\frac{3\sqrt{2}}{2}$\\
hence, AC = AE + CE = $3\sqrt{2}$\\
since $\cos\angle ACD = \frac{AC}{AB} = \frac{3}{4}$, i.e., $\frac{3\sqrt{2}}{AB} = \frac{3}{4}$\\
hence, AB = $4\sqrt{2}$\\
since $S_{\triangle ABC} = \frac{1}{2} \times AC \times BC = \frac{1}{2} \times AB \times CH$\\
hence, $\frac{1}{2} \times 3\sqrt{2} \times \sqrt{14} = \frac{1}{2} \times 4\sqrt{2} \times CH$, solving this gives: CH = $\frac{3\sqrt{14}}{4}$.\\
hence, the length of CH is $\boxed{\frac{3\sqrt{14}}{4}}$, and the length of AB is $\boxed{4\sqrt{2}}$.


Question: The lookout tower $AB$ has a height of $20m$. From the base of the lookout tower $B$, the angle of elevation to the top of the opposite tower $C$ is measured to be $60^\circ$, and from the top of the lookout tower $A$, the angle of elevation to the tower top $C$ is measured to be $45^\circ$. Given that the terrain height of the lookout tower and tower $CD$ is the same, what is the height of tower $CD$ in meters?

Solution:
\textbf{Solution:} Let the height of the tower $CD$ be $x$, thus $BD=\frac{\sqrt{3}}{3}x$,\\
from $BD \cdot \tan 60^\circ - BD \cdot \tan 45^\circ = AB$, substituting $BD=\frac{\sqrt{3}}{3}x$,\\
we get: $x-\frac{\sqrt{3}}{3}x=20$,\\
solving this gives: $x=\boxed{30+10\sqrt{3}}$.


Question: As shown in the figure, $\angle CAB=\angle CBD$, $AB=4$, $AC=6$, $BD=7.5$, $BC=5$, find the length of $CD$.

Solution:
\textbf{Solution:} Given $AB=4$, $AC=6$, $BD=7.5$, $BC=5$,\\
$\therefore \frac{AC}{BD}=\frac{AB}{BC}=\frac{4}{5}$,\\
$\because \angle CAB=\angle CBD$,\\
$\therefore \triangle ABC \sim \triangle BCD$,\\
$\therefore \frac{BC}{CD}=\frac{4}{5}$,\\
$\therefore CD=\frac{5}{4}BC=\frac{5}{4}\times 5=\boxed{\frac{25}{4}}$.


Question: As shown in the figure, in $\triangle APB$, $C$ and $D$ are respectively points on $AP$ and $BP$. Moreover, $\angle CDP = \angle A$, $AB = 8\text{cm}$, $\frac{BP}{CP} = \frac{4}{3}$. Find the length of $CD$.

Solution:
\textbf{Solution:} Since $\left\{\begin{array}{l}
\angle CDP=\angle A \\
\angle P=\angle P
\end{array}\right.$,\\
it follows that $\triangle PCD\sim \triangle PBA$,\\
hence $\frac{BP}{CP}=\frac{AB}{CD}$,\\
given that $AB=8\text{cm}$ and $\frac{BP}{CP}=\frac{4}{3}$,\\
we have $\frac{4}{3}=\frac{8}{CD}$,\\
therefore, $CD=\boxed{6}\text{cm}$.


Question: As shown in the figure, $D,E$ are points on the sides $AB$ and $BC$ of $\triangle ABC$, respectively, with $DE\parallel AC$. If the ratio of the areas ${S}_{\triangle BDE}:{S}_{\triangle ADE}=1:3$, what is the value of ${S}_{\triangle DOE}:{S}_{\triangle AOC}$?

Solution:
\textbf{Solution:} From the construction, we have
\[
\because S_{\triangle BDE} : S_{\triangle ADE} = 1 : 3,
\]
\[
\therefore BD : AD = 1 : 3,
\]
\[
\therefore BD : BA = 1 : 4,
\]
\[
\because DE \parallel AC,
\]
\[
\therefore \triangle DOE \sim \triangle AOC,
\]
\[
\therefore \frac{DE}{AC} = \frac{BD}{BA} = \frac{1}{4},
\]
\[
\therefore S_{\triangle DOE} : S_{\triangle AOC} = \boxed{1 : 16}.
\]


Question: As shown in the figure, points E and F are located on sides BC and CD of rectangle ABCD, respectively, and $CE = \frac{2}{3}BC$, $CF = \frac{2}{3}CD$. If the perimeter of $\triangle ABC$ is 39, what is the perimeter of $\triangle CEF$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\
$\therefore \angle B=\angle ECF=90^\circ$, $AB=CD$\\
$\because CE=\frac{2}{3}BC$, $CF=\frac{2}{3}CD$\\
$\therefore \frac{CE}{BC}=\frac{CF}{CD}=\frac{CF}{AB}=\frac{2}{3}$\\
$\therefore \triangle CEF \sim \triangle BCA$, with a similarity ratio of $\frac{2}{3}$.\\
$\because$ The perimeter of $\triangle ABC$ is 39, $\therefore$ the perimeter of $\triangle CEF$ is $39\times \frac{2}{3}=\boxed{26}$.


Question: As shown in the figure, there is a street lamp pole $AB$ on the opposite side of the river. Under the illumination of the lamp, Xiao Ming is at point $D$, and his shadow is $DF=4m$ long. He moves in the direction of $BD$ to reach point $F$, and then measures his shadow to be $FG=5m$ long. If Xiao Ming's height is $1.6m$, calculate the height of the street lamp pole $AB$.

Solution:
\textbf{Solution:} Given that $CD\parallel EF\parallel AB$,\\
we can infer that $\triangle ABF\sim \triangle CDF$ and $\triangle ABG\sim \triangle EFG$,\\
hence, $\frac{AB}{CD}=\frac{BF}{DF}$ and $\frac{AB}{EF}=\frac{BG}{FG}$,\\
since $CD=EF$,\\
it follows that $\frac{BF}{DF}=\frac{BG}{FG}$\\
Given $DF=4$, $FG=5$, $BF=BD+DF=BD+4$, $BG=BD+DF+FG=BD+9$,\\
we get $\frac{4+BD}{4}=\frac{9+BD}{5}$,\\
solving for $BD$ yields $BD=16, BF=16+4=20$,\\
therefore $\frac{AB}{1.6}=\frac{20}{4}$,\\
solving gives $AB=\boxed{8}$.


Question: As shown in the figure, in rectangle $ABCD$, $E$ is the midpoint of $BC$, $DF \perp AE$ at $F$, with $AB=6$ and $BC=4$. What are the lengths of $AE$ and $DF$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\
we have $\angle BAE+\angle DAE=90^\circ$, and $\angle ADF+\angle DAF=90^\circ$,\\
thus, $\angle BAE=\angle ADF$,\\
and since $\angle AFD=\angle B=90^\circ$,\\
we obtain $\triangle ADF\sim \triangle EAB$,\\
which implies $\frac{AD}{AE}=\frac{DF}{AB}$,\\
knowing that $E$ is the midpoint of $BC$ and $BC=4$,\\
we find $BE=2$,\\
thus, $AE=\sqrt{6^2+2^2}=2\sqrt{10}$,\\
leading to $\frac{4}{2\sqrt{10}}=\frac{DF}{6}$,\\
solving this, we find: $DF=\boxed{\frac{6\sqrt{10}}{5}}$.


Question: In the figure, for trapezoid $ABCD$, where $AD\parallel BC$, point $E$ is on line segment $AD$, $CE$ and $BD$ intersect at point $H$, and $CE$ intersects the extension line of $BA$ at point $G$. Given $DE:AE=2:3$, $BC=4DE$, and $CE=10$ cm, find the lengths of $EH$ and $GE$.

Solution:
\textbf{Solution:}

Since $AD \parallel BC$,

$\therefore \frac{DE}{BC}=\frac{EH}{HC}$.

Since $BC=4DE$,

$\therefore \frac{DE}{BC}=\frac{1}{4}$.

Since $CE=10$,

$\therefore HC=10-EH$

$\therefore \frac{1}{4}=\frac{EH}{10-EH}$,

$\therefore EH=\boxed{2}$.

Since $BC=4DE$, $DE:AE=2:3$,

$\therefore \frac{AE}{BC}=\frac{3}{8}$.

Since $AD \parallel BC$,

$\therefore \triangle AGE \sim \triangle BGC$,

$\therefore \frac{AE}{BC}=\frac{GE}{GC}$.

Since $CE=10$,

$\therefore GC=10+GE$,

$\therefore \frac{3}{8}=\frac{GE}{10+GE}$,

$\therefore GE=\boxed{6}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=4$, $AB=5$. Point $D$ is on $AC$ and $AD=3$, $DE\perp AB$ at point $E$. Find the length of $AE$.

Solution:
\textbf{Solution:} Since DE$\perp$AB at point E, and $\angle C=90^\circ$,\\
it follows that $\angle$AED＝$\angle$C,\\
since $\angle$A＝$\angle$A,\\
it follows that $\triangle$ ADE$\sim$ $\triangle$ ABC,\\
thus, $\frac{AD}{AB}=\frac{AE}{AC}$,\\
$\frac{3}{5}=\frac{AE}{4}$,\\
AE＝$\boxed{\frac{12}{5}}$.


Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is the midpoint of $AB$. $CE$ and $BD$ intersect at point $O$. If $S_{\triangle EOB} = 1$, find the area of quadrilateral $AEOD$.

Solution:
\textbf{Solution:} Let's consider parallelogram ABCD, where point E is the midpoint of AB,\\
hence CD$\parallel$AB, and CD=AB=2BE,\\
therefore, $\triangle DOC\sim \triangle BOE$,\\
thus, $\frac{OC}{OE}=\frac{CD}{BE}=2$,\\
therefore, $\frac{S_{\triangle BOC}}{S_{\triangle EOB}}=\frac{OC}{OE}=2$, $ \frac{S_{\triangle DOC}}{S_{\triangle EOB}}=\left(\frac{OC}{OE}\right)^2=4$\\
Since $S_{\triangle EOB}=1$,\\
thus, $S_{\triangle BOC}=2$, $S_{\triangle DOC}=4$,\\
therefore, $S_{\triangle BCD}=6$,\\
therefore, $S_{\triangle DAB}=6$,\\
thus, the area of quadrilateral AEOD is: $S_{\triangle DAB}-S_{\triangle EOB}=6-1=\boxed{5}$.


Question: As shown in the figure, in $\triangle ABC$, it is known that $AB=6$, $AC=10$, and $AD$ bisects $\angle BAC$, with $BD\perp AD$ at point $D$, and $E$ is the midpoint of $BC$. Find the length of $DE$.

Solution:
\textbf{Solution:} As shown in the figure, extend $BD$ to meet $AC$ at point $F$.\\
Since $AD$ bisects $\angle BAC$,\\
it follows that $\angle BAD=\angle FAD$.\\
Since $BD\perp AD$,\\
it follows that $\angle ADB=\angle ADF=90^\circ$.\\
Hence, $\angle ABD=\angle AFD$.\\
Therefore, $AB=AF$ and $BD=DF$.\\
Hence, $D$ is the midpoint of $BF$.\\
Since $AC=10$ and $AB=6$,\\
it follows that $FC=AC-AF=AC-AB=10-6=4$.\\
Since $E$ is the midpoint of $BC$,\\
thus $DE$ is the midline of $\triangle BFC$.\\
Therefore, $DE=\frac{1}{2}FC=\frac{1}{2}\times 4=\boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, $D$, $E$, and $F$ are the midpoints of $AB$, $BC$, and $CA$, respectively, with $AC=20$ and $BC=18$. What is the perimeter of the quadrilateral $DECF$?

Solution:
\textbf{Solution:} Since D, E, F are the midpoints of AB, BC, and CA, respectively, and given that $AC=20$, $BC=18$,\\
it follows that $DE=CF=\frac{1}{2}AC=10$, $DF=CE=\frac{1}{2}BC=9$,\\
therefore, the perimeter of the quadrilateral DECF $=DE+CF+DF+CE=10+10+9+9=\boxed{38}$.


Question: As shown in the diagram, four triangular paper pieces $\triangle ABC$, $\triangle AB_{1}C_{1}$, $\triangle AB_{2}C_{2}$, and $\triangle AB_{3}C_{3}$ are perfectly superimposed and arranged as depicted. Given $BC=\sqrt{10}$ and $AB=1$, what is the area of quadrilateral $CC_{1}C_{2}C_{3}$?

Solution:
\textbf{Solution:} From the given problem, we have Rt$\triangle ABC \cong$ Rt$\triangle AB_{1}C_{1} \cong$ Rt$\triangle AB_{2}C_{2} \cong$ Rt$\triangle AB_{3}C_{3}$,\\
$\therefore AC = AC_{1} = AC_{2} = AC_{3}$, and AB$ = AB_{1} = 1$.\\
In Rt$\triangle ABC$, by the Pythagorean theorem, we have $AC = \sqrt{BC^{2} - AB^{2}} = 3$,\\
$\therefore Area_{\text{quadrilateral} CC_{1}C_{2}C_{3}} = 4 \times \frac{3 \times 3}{2} = \boxed{18}$.


Question: As shown in the figure, $AD$ is the angle bisector of $\triangle ABC$, and $AE$ is the altitude of $\triangle ABC$. Given that $\angle B=40^\circ$ and $\angle DAE=15^\circ$, what is the degree measure of $\angle C$?

Solution:
\textbf{Solution:} Given the construction,\\
$\because AE$ is the altitude of $\triangle ABC$,\\
$\therefore \angle AEB=90^\circ$,\\
$\because \angle B=40^\circ$,\\
$\therefore \angle BAE=50^\circ$,\\
Also, $\because \angle DAE=15^\circ$,\\
$\therefore \angle BAD=50^\circ-15^\circ=35^\circ$,\\
$\because AD$ is the bisector of $\triangle ABC$,\\
$\therefore \angle BAD=\angle CAD=35^\circ$,\\
$\therefore \angle BAC=70^\circ$,\\
$\therefore \angle C=180^\circ-\angle B-\angle BAC=180^\circ-40^\circ-70^\circ=\boxed{70^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC = 90^\circ$, $AB = 15$, $AC = 20$, and $AD \perp BC$ with D being the foot of the perpendicular. Find the lengths of AD and BD.

Solution:
\textbf{Solution:} In the right-angled triangle $ABC$, where $\angle BAC = 90^\circ$, $AB = 15$, and $AC = 20$,\\
according to the Pythagorean theorem, we have $BC = \sqrt{AB^2 + AC^2} = \sqrt{15^2 + 20^2} = 25$,\\
since the area of $\triangle ABC$ is calculated as ${S}_{\triangle ABC} = \frac{1}{2}AB \cdot AC$ and ${S}_{\triangle ABC} = \frac{1}{2}BC \cdot AD$,\\
it follows that $\frac{1}{2}AB \cdot AC = \frac{1}{2}BC \cdot AD$.\\
Therefore, $AD = \frac{AB \cdot AC}{BC} = \frac{15 \times 20}{25} = \boxed{12}$;\\
since $AD \perp BC$,\\
it follows that $\angle ADB = 90^\circ$.\\
In the right-angled triangle $ADB$, according to the Pythagorean theorem, we have $BD = \sqrt{AB^2 - AD^2} = \sqrt{15^2 - 12^2} = 9$,\\
therefore, the lengths of $AD$ and $BD$ are $12$ and $9$ respectively.


Question: As shown in the figure, in $\triangle ABC$, $AD$ bisects $\angle BAC$. Given $AB=AC=3$ and $AD=2$, find the length of $BC$.

Solution:
\textbf{Solution:} Since $AB=AC$ and $AD$ bisects $\angle BAC$, \\
$\therefore AD \perp BC$, and $BD=DC$, \\
$\therefore BD= \sqrt{AB^{2}-AD^{2}}= \sqrt{5}$, \\
$\therefore BC=2BD=2\sqrt{5}=\boxed{2\sqrt{5}}$.


Question: In the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, $AB=AC=a$, $AD$ is the altitude of $\triangle ABC$, what is the length of $AD$?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $\angle BAC=90^\circ$ and $AB=AC=a$, \\
$\therefore$ BC=$\sqrt{AB^2+AC^2}=\sqrt{a^2+a^2}=\sqrt{2}a$,\\
since $AD$ is the height of $\triangle ABC$,\\
$\therefore$ the area $S_{\triangle ABC}=\frac{1}{2}\times AB\times AC=\frac{1}{2}\times BC\times AD$,\\
thus $\frac{1}{2}\times a\times a=\frac{1}{2}\times \sqrt{2}a\times AD$,\\
we find $AD=\boxed{\frac{\sqrt{2}}{2}a}$.


Question: As illustrated, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle A=30^\circ$, $CD\perp AB$ at point D, $DE\parallel BC$ intersects AC at point E. If $BD=2$, what is the length of $DE$?

Solution:
\textbf{Solution:}

Given $\because \angle ACB=90^\circ$ and $\angle A=30^\circ$,

$\therefore \angle B=60^\circ$ and $AB=2BC$,

Since $CD\perp AB$ at point $D$,

$\therefore \angle CDB=\angle CDA=90^\circ$,

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

Thus, $BC=2BD$,

Given $BD=2$,

$\therefore BC=4$,

Therefore, $AB=2BC=8$,

Thus, $AD=AB-BD=6$,

Since $DE\parallel BC$,

$\therefore \angle DEA=\angle ACB=90^\circ$,

Given in $\triangle ADE$, $\angle A=30^\circ$,

$\therefore DE=\frac{1}{2}AD=\boxed{3}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC = 45^\circ$. $F$ is the intersection point of altitudes $AD$ and $BE$, $AC = \sqrt{5}$, $BD = 2$. What is the length of segment $DF$?

Solution:
\textbf{Solution:} Given that $AD$ and $BE$ are altitudes of $\triangle ABC$,\\
it follows that $\angle ADB = \angle ADC = \angle BEC = 90^\circ$.\\
Thus, $\angle C + \angle DAC = 90^\circ$; $\angle C + \angle DBF = 90^\circ$.\\
This means $\angle DAC = \angle DBF$.\\
Since $\angle ABC = 45^\circ$,\\
it implies $\angle DAB = 45^\circ$.\\
Hence, $\angle ABC = \angle DAB$.\\
Therefore, $DA = DB$.\\
In $\triangle ADC$ and $\triangle BDF$,\\
\[\left\{ \begin{array}{l} \angle ADC = \angle BDF \\ DA = DB \\ \angle DAC = \angle DBF \end{array} \right.\]\\
Thus, $\triangle ADC \cong \triangle BDF$ (ASA).\\
Therefore, $AC = BF = \sqrt{5}$.\\
In right $\triangle BDF$, with $\angle BDF = 90^\circ$,\\
it follows that $BD^2 + DF^2 = BF^2$.\\
Given $BD = 2$, and $BF = \sqrt{5}$,\\
it concludes $DF = \boxed{1}$.


Question: As shown in the figure, $AD$ is the altitude of $\triangle ABC$, and $CE$ is the angle bisector of $\triangle ADC$. If $\angle BAD = \angle ECD$ and $\angle B = 70^\circ$, what is the degree measure of $\angle CAD$?

Solution:
\textbf{Solution:}

Since $AD$ is the altitude of $\triangle ABC$,
\[\therefore \angle ADB=\angle ADC=90^\circ\]

Since $\angle B=70^\circ$,
\[\therefore \angle BAD=20^\circ\]

Since CE is the bisector of $\triangle ADC$,
\[\therefore \angle ECD=\frac{1}{2}\angle ACD\]

Since $\angle BAD=\angle ECD=20^\circ$,
\[\therefore \angle ACD=40^\circ\]

Therefore, in $\triangle ACD$, $\angle CAD=90^\circ-40^\circ=\boxed{50^\circ}$.


Question: As shown in the figure, in the right-angled triangle $ABC$ with $\angle B=90^\circ$, $AB=4$, and $BC=3$, the shaded area is a rectangle, and $AE=1$. What is the area of the shaded region?

Solution:
\textbf{Solution:} In $\triangle ABC$, where $\angle B=90^\circ$, $AB=4\text{cm}$, $BC=3\text{cm}$,\\
by the Pythagorean theorem, we have $AB^2+BC^2=AC^2$,\\
thus $4^2+3^2=AC^2$\\
$\therefore AC= \sqrt{3^2+4^2}=5\text{cm}$,\\
$\because AE=1\text{cm}$,\\
$\therefore$ the area of rectangle $ACDE$ is $5\times 1=\boxed{5}\text{cm}^2$


Question: As shown in the figure, in $\triangle ABC$, point D is the midpoint of side AB, and $CD = \frac{1}{2}AB$. Prove that: $\angle ACB = 90^\circ$.

Solution:
\textbf{Solution:} Proof: Since point D is the midpoint of side $AB$,\\
$ \therefore AD=BD=\frac{1}{2}AB$,\\
Also since $CD=\frac{1}{2}AB$,\\
$ \therefore AD=BD=CD$,\\
$ \therefore \angle ACD=\angle A$, $\angle BCD=\angle B$.\\
Since $\angle A+\angle B+(\angle ACD+\angle BCD)=180^\circ$,\\
$ \therefore \angle ACD+\angle BCD+(\angle ACD+\angle BCD)=180^\circ$,\\
$ \therefore 2(\angle ACD+\angle BCD)=180^\circ$,\\
$ \therefore \angle ACD+\angle BCD=90^\circ$,\\
which means $\angle ACB=\boxed{90^\circ}$.


Question: In the figure, within $\triangle ABC$, $AD$ bisects $\angle BAC$, $AE\perp BC$. If $\angle BAD=40^\circ$ and $\angle C=70^\circ$, find the measure of $\angle DAE$?

Solution:
\textbf{Solution:} Given that $AD$ bisects $\angle BAC$,\\
thus $\angle BAC = 2\angle BAD = 80^\circ$ (definition of angle bisector),\\
Given that $\angle C = 70^\circ$,\\
in $\triangle ABC$, thus $\angle B = 180^\circ - \angle BAC - \angle C\\
= 180^\circ - 70^\circ - 80^\circ = 30^\circ$ (sum of interior angles in a triangle is $180^\circ$),\\
Given that $AE \perp BC$,\\
thus $\angle AEB = 90^\circ$ (definition of perpendicular),\\
in $\triangle ABE$, $\angle BAE = 180^\circ - \angle B - \angle AEB\\
= 180^\circ - 30^\circ - 90^\circ = 60^\circ$ (sum of interior angles in a triangle is $180^\circ$),\\
thus $\angle DAE = \angle BAE - \angle BAD = 60^\circ - 40^\circ = \boxed{20^\circ}$.


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle with a side length of 4. Point $P$ is on $AB$. Through point $P$, draw $PE\perp AC$ with the foot of the perpendicular as $E$. Extend $BC$ to point $Q$ such that $CQ=PA$. Connect $PQ$ and let it intersect $AC$ at point $D$. Find the length of $DE$.

Solution:
\textbf{Solution:} Draw $PF \parallel BC$ through point P intersecting $AC$ at F, then $\angle FPD = \angle Q$, $\angle APF = \angle B$, $\angle AFP = \angle ACB$,

$\because$ $\triangle ABC$ is an equilateral triangle,

$\therefore$ $\angle B = \angle A = \angle ACB = 60^\circ$,

$\therefore$ $\angle APF = \angle AFP = \angle A = 60^\circ$,

$\therefore$ $\triangle APF$ is an equilateral triangle, and $PE \perp AC$,

$\therefore$ AP=PF, AE=EF, and AP=CQ,

$\therefore$ PF=CQ,

In $\triangle PDF$ and $\triangle QDC$,

$\left\{\begin{array}{l}
\angle FPD = \angle Q \\
\angle PDF = \angle QDC \\
PF = CQ
\end{array}\right.$,

$\therefore$ $\triangle PDF \cong \triangle QDC$ (AAS),

$\therefore$ FD=CD,

$\therefore$ EF + FD = AE + CD,

$\therefore$ DE = $\frac{1}{2}$AC, and AC=4,

$\therefore$ DE = \boxed{2}.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=45^\circ$, $\tan C=\frac{2}{3}$, and $AC=2\sqrt{13}$. Find the length of $BC$.

Solution:
\textbf{Solution:} According to the problem, draw $AD \perp BC$ through point A, as shown:

$\therefore$ $\triangle ABD$ and $\triangle ACD$ are both right-angled triangles,

$\because$ $\tan C = \frac{AD}{CD} = \frac{2}{3}$,

Let $AD = 2x$, $CD = 3x$,

$\therefore$ $AC = \sqrt{(2x)^2 + (3x)^2} = 2\sqrt{13}$,

Solving, we get: $x = 2$ (negative value is discarded),

$\therefore$ $AD = 4$, $CD = 6$,

$\because$ $\angle B = 45^\circ$,

$\therefore$ $BD = AD = 4$,

$\therefore$ $BC = 4 + 6 = \boxed{10}$;


Question: As shown in the figure, in $ \triangle ABC$, $ \angle ACB = 90^\circ$, D is the midpoint of $ AB$, $ ED \perp AB$ intersects $ AC$ at point E, $ \tan \angle EBC = \frac{3}{4}$. What is the value of $ \tan \angle ABE$?

Solution:
\textbf{Solution:} In $\triangle EBC$, we have $\angle ECB=90^\circ$,\\
$\therefore \tan\angle EBC=\frac{CE}{BC}=\frac{3}{4}$.\\
Let $CE=3k$, $BC=4k$, so $BE=5k$.\\
Since $D$ is the midpoint of $AB$ and $ED\perp AB$,\\
$\therefore AE=BE=5k$,\\
$\therefore \angle ABE=\angle BAE$, $AC=8k$,\\
In $\triangle ABC$, $\angle ACB=90^\circ$,\\
$\therefore \tan\angle CAB=\frac{BC}{AC}=\frac{4k}{8k}=\frac{1}{2}$.\\
$\therefore$ The tangent of $\angle ABE$ is $\boxed{\frac{1}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=30^\circ$, $AB=4$, $AD\perp BC$ at point $D$ and $\tan \angle CAD=\frac{1}{2}$, what is the length of $BC$?

Solution:
\textbf{Solution:} Since $AD \perp BC$ at point D, \\
therefore, $\triangle ADB$ and $\triangle ADC$ are right-angled triangles, \\
in $\triangle ADB$, $\angle B=30^\circ$, $AB=4$, \\
therefore, $AD=\frac{1}{2}AB=2$, $BD=\sqrt{AB^{2}-AD^{2}}=\sqrt{4^{2}-2^{2}}=2\sqrt{3}$, \\
in $\triangle ADC$, $\tan\angle CAD=\frac{1}{2}$, $AD=2$, \\
therefore, $\tan\angle CAD=\frac{CD}{AD}=\frac{1}{2}$, \\
therefore, $CD=1$, \\
therefore, $BC=BD+CD=2\sqrt{3}+1=\boxed{2\sqrt{3}+1}$.


Question: As shown in the diagram, a right-angled triangle is placed with point C on the extension of FD, AB $\parallel$ CF, $\angle F=\angle ACB=90^\circ$, $\angle E=45^\circ$, $\angle A=60^\circ$, and AC=5. Determine the length of CD.

Solution:
\textbf{Solution:} Draw BM$\perp$FD at point M,\\
In $\triangle ACB$, $\angle ACB=90^\circ$, $\angle A=60^\circ$, AC=5,\\
$\therefore \angle ABC=30^\circ$, BC=AC$\cdot \tan 60^\circ=5\sqrt{3}$,\\
$\because AB\parallel CF$,\\
$\therefore \angle BCM=30^\circ$,\\
$\therefore$BM=BC$\cdot \sin 30^\circ=\frac{5\sqrt{3}}{2}$, CM=BC$\cdot \cos 30^\circ=\frac{15}{2}$,\\
In $\triangle EFD$, $\angle F=90^\circ$, $\angle E=45^\circ$,\\
$\therefore \angle EDF=45^\circ$,\\
$\therefore \angle EDF=\angle DBM=45^\circ$,\\
$\therefore$MD=BM=$\frac{5\sqrt{3}}{2}$,\\
$\therefore$CD=$CM-MD=\boxed{\frac{15}{2}-\frac{5\sqrt{3}}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC = 90^\circ$, extend the hypotenuse $BC$ to point $D$ so that $CD = \frac{1}{2}BC$. Connect $AD$. If $\tan B = \frac{4}{3}$, what is the value of $\tan \angle CAD$?

Solution:
\textbf{Solution:} Construct line $CF \perp AC$ intersecting $AD$ at point $F$, \\
it follows that $\angle ACF=90^\circ$, \\
since $\angle BAC=90^\circ$, \\
it implies $AB\parallel CF$, \\
hence $\angle DCF=\angle DBA$, \\
thus $\triangle DCF\sim \triangle DBA$, \\
implying $\frac{CF}{AB}=\frac{CD}{BD}$, \\
since $CD=\frac{1}{2}BC$, \\
thus $CD=\frac{1}{3}BD$, \\
that is $\frac{CD}{BD}=\frac{1}{3}$, \\
thus $\frac{CF}{AB}=\frac{1}{3}$, \\
hence $CF=\frac{1}{3}AB$, \\
In right $\triangle ABC$, \\
since $\tan B=\frac{4}{3}$, that is $\frac{AC}{AB}=\frac{4}{3}$, \\
let $AC=4x$, then $AB=3x$, \\
thus $CF=\frac{1}{3} \times 3x=x$, \\
therefore $\tan\angle CAD=\frac{CF}{AC}=\frac{x}{4x}=\boxed{\frac{1}{4}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle A=30^\circ$, $AC=2\sqrt{3}$, $\tan B=\frac{\sqrt{3}}{2}$, find the length of $AB$.

Solution:
\textbf{Solution}: As shown in the diagram, draw $CD \perp AB$ at point D,\\
$\therefore \angle ADC=\angle BDC=90^\circ$\\
$\because \angle A=30^\circ$, $AC=2\sqrt{3}$,\\
$\therefore CD=\frac{1}{2}AC=\sqrt{3}$,\\
$\therefore AD=\sqrt{AC^{2}-CD^{2}}=3$\\
$\because \tan B=\frac{CD}{BD}=\frac{\sqrt{3}}{2}$,\\
$\therefore BD=2$,\\
$\therefore AB=AD+BD=\boxed{5}$


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $BC=1$, and $AC=\sqrt{3}$. Find the values of $\tan A$ and $\tan B$.

Solution:
\textbf{Solution:} Given in right-angled triangle $ABC$, $\angle C=90^\circ$, $BC=1$, and $AC=\sqrt{3}$,\\
thus, $\tan A=\frac{BC}{AC}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$,\\
$\tan B=\frac{AC}{BC}=\frac{\sqrt{3}}{1}=\boxed{\sqrt{3}}$.


Question: As illustrated in the diagram, a rectangular wooden box slides down an inclined plane. When the box slides to the position shown in the diagram, we have $AB=2\text{m}$. Given that the height of the box $BE=1\text{m}$, and the angle of the inclined plane $\angle BAC$ is $30^\circ$, calculate the height of the endpoint $E$ above the ground $AC$.

Solution:
\textbf{Solution:} Draw $EH\perp AC$ intersecting $AC$ at $H$ and $AB$ at $G$, as shown in the figure, \\
From the given, \\
$\angle BAC+\angle AGH=90^\circ$, $\angle BEG+\angle BGE=90^\circ$, $\angle AGH=\angle BGE$, and $\angle BAC$ is $30^\circ$, \\
$\therefore \angle BAC=\angle BEG=30^\circ$, \\
$\because \cos\angle BEG=\cos30^\circ=\frac{BE}{EG}$, $\tan\angle BEG=\tan30^\circ=\frac{BG}{BE}$, $BE=1m$, \\
$\therefore GE=\frac{BE}{\cos30^\circ}=\frac{2\sqrt{3}}{3}$, $BG=BE\times \tan30^\circ=\frac{\sqrt{3}}{3}$, \\
$\because AB=2m$, \\
$\therefore AG=AB−BG=2−\frac{\sqrt{3}}{3}$, \\
$\because \cos\angle BAC=\cos30^\circ=\frac{GH}{GA}$, \\
$\therefore GH=AG\times \cos30^\circ=(2−\frac{\sqrt{3}}{3})\times \frac{\sqrt{3}}{2}=\sqrt{3}−\frac{1}{2}$, \\
$EH=EG+GH=\frac{2\sqrt{3}}{3}+\sqrt{3}−\frac{1}{2}=\boxed{\frac{5\sqrt{3}}{3}−\frac{1}{2}}$.


Question: Given: In $\triangle ABC$, $CD \perp AB$, $\sin A = \frac{4}{5}$, $CD = 4$, $AB = 5$. Find the length of $AD$ and the value of $\tan B$.

Solution:
\textbf{Solution:}

Since $CD \perp AB$,\\
it follows that $\angle CDA = 90^\circ$\\
Since $\sin A = \frac{CD}{AC} = \frac{4}{5}$, and $CD = 4$,\\
it follows that $AC = 5$ \\
By the Pythagorean theorem,\\
$AD = \sqrt{AC^{2} - CD^{2}} = \boxed{3}$\\
$BD = AB - AD = 4$\\
Therefore, $\tan B = \frac{CD}{BD} = \boxed{2}$


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=6$, $\sin B=\frac{2}{3}$. Point D is on the extension line of side BC, and $CD=BC$. What is the value of $\tan D$?

Solution:
\textbf{Solution:} As shown in the figure, draw $AE\perp BD$ at point E through point $A$.\\
$\therefore$ $\sin B=\frac{AE}{AB}=\frac{2}{3}$.\\
$\because$ $AB=AC=6$,\\
$\therefore$ $AE=4$,\\
$\therefore$ $BE=\sqrt{AB^{2}-AE^{2}}=2\sqrt{5}$.\\
$\because$ $AE\perp BD$, $AB=AC=6$,\\
$\therefore$ $CE=BE=2\sqrt{5}$,\\
$\therefore$ $CD=BC=BE+CE=4\sqrt{5}$,\\
$\therefore$ $DE=CD+CE=6\sqrt{5}$,\\
$\therefore$ $\tan D=\frac{AE}{DE}=\frac{4}{6\sqrt{5}}=\boxed{\frac{2\sqrt{5}}{15}}$


Question: As shown in the diagram, in $ \triangle ABC$, $ BA=BC$, point E lies on $ BC$, and $ AE\perp BC$, $\cos\angle B=\frac{4}{5}$, $ EC=3$, find the length of $ BE$ and the value of $\tan\angle CAE$.

Solution:
\textbf{Solution:} Given $AE\perp BC$ and $\cos\angle B=\frac{4}{5}$,
\[\therefore \frac{BE}{AB}=\frac{4}{5},\] suppose $BE=4x$, then $AB=5x$,
\[\because BA=BC,\] thus $BC=5x$,
\[\because EC=3\text{ and }EC=BC-BE,\]
\[\therefore 5x-4x=3,\] that is $x=3$,
\[\therefore BE=4x=\boxed{12},\] $BA=BC=5x=15$,
\[\therefore AE=\sqrt{AB^{2}-BE^{2}}=\sqrt{15^{2}-12^{2}}=9,\]
\[\therefore \tan\angle CAE=\frac{CE}{AE}=\frac{3}{9}=\boxed{\frac{1}{3}}.\]


Question: As shown in the figure, in $\triangle ABC$, $\angle C=30^\circ$, $AC=12$, $\sin B=\frac{3}{5}$, find the length of $BC$.

Solution:
\textbf{Solution:} Construct $AD\perp BC$ at point A, with D as the foot\\
In $\triangle ADC$, $AC=12$, $\angle C=30^\circ$\\
$\therefore AD=\frac{1}{2}AC=6$, $CD=\sqrt{3}AD=6\sqrt{3}$\\
In $\triangle ABD$, $\sin B=\frac{3}{5}$\\
$\therefore AB=10$\\
$\therefore BD=8$\\
$\therefore BC=BD+CD=8+6\sqrt{3}$\\
$\therefore$ The length of $BC$ is $\boxed{8+6\sqrt{3}}$


Question: As shown in the figure, in $\triangle ABC$, $AC=12$, $\angle C=45^\circ$, $\angle B=120^\circ$, find the length of $BC$.

Solution:
\textbf{Solution:} Draw $AD\perp BC$ through point A, where it intersects the extension of $BC$ at point D, then $\angle ADC=90^\circ$,\\
In $\triangle ADC$, $\angle C=45^\circ$,\\
$\therefore AD=DC$,\\
According to the Pythagorean theorem, we have: $AD^2+DC^2=AC^2$,\\
that is $2AD^2=AC^2=12^2$,\\
Solving this gives $AD=DC=6\sqrt{2}$,\\
$\because \angle ABC=120^\circ$,\\
$\therefore \angle ABD=60^\circ$,\\
In $\triangle ABD$,\\
let $BD=x$, then $AB=2x$,\\
According to the Pythagorean theorem, we have: $AD^2+DB^2=AB^2$,\\
that is $(6\sqrt{2})^2+x^2=(2x)^2$,\\
Solving this gives $x=2\sqrt{6}$,\\
that is $BD=2\sqrt{6}$,\\
$\therefore BC=DC-DB=6\sqrt{2}-2\sqrt{6}=\boxed{6\sqrt{2}-2\sqrt{6}}$.


Question: As shown in the figure, in $\triangle ABC$, $AD \perp BC$ at point $D$, $\cos B = \frac{4}{5}$, $\tan C = \sqrt{3}$, $AB = 5$. Find the length of $AC$.

Solution:
\textbf{Solution:}\\
Since $AD\perp BC$,\\
it follows that $\angle ADB=\angle ADC=90^\circ$,\\
Since $\cos B=\frac{4}{5}$ and $AB=5$,\\
it follows that $BD=AB\cdot\cos B=5\times \frac{4}{5}=4$,\\
Therefore, $AD=\sqrt{AB^{2}-BD^{2}}=\sqrt{5^{2}-4^{2}}=3$,\\
Since $\tan C=\sqrt{3}$,\\
it follows that $CD=\frac{AD}{\tan C}=\frac{3}{\sqrt{3}}=\sqrt{3}$,\\
Therefore, $AC=\sqrt{AD^{2}+CD^{2}}=\sqrt{3^{2}+(\sqrt{3})^{2}}=\boxed{2\sqrt{3}}$.


Question: In $\triangle ABC$, $\angle C=90^\circ$, $\tan A=\frac{3}{4}$, $BC=9$, find the length of $AC$ and the value of $\sin A$.

Solution:
\textbf{Solution}: Given that in $\triangle ABC$,\\
$\tan A = \frac{3}{4}$ and $BC = 9$,\\
then $\frac{BC}{AC} = \frac{3}{4}$,\\
thus $AC = \boxed{12}$,\\
and $AB = \sqrt{AC^2 + BC^2} = \sqrt{9^2 + 12^2} = 15$,\\
therefore $\sin A = \frac{BC}{AB} = \frac{9}{15} = \boxed{\frac{3}{5}}$


Question: As shown in the figure, it is known that the line $y=kx+b$ passes through point $A$ and point $B$. Find the coordinates of point $C$, the intersection of this line with the $x$-axis.

Solution: \textbf{Solution:} To solve this, we substitute points $A(-3, 3)$ and $B(0, -3)$ into $y=kx+b$ to get $\begin{cases}-3k+b=3 \\ b=-3\end{cases}$. Solving this system, we obtain $\begin{cases}k=-2 \\ b=-3\end{cases}$. Therefore, the equation of line $AB$ is $y=-2x-3$. When $y=0$, solving $-2x-3=0$ yields $x=-\frac{3}{2}$. Therefore, the coordinates of point $C$ are $\left(-\frac{3}{2}, 0\right)$. Hence, the intersection point $C$ of this line with the $x$-axis is $\boxed{\left(-\frac{3}{2}, 0\right)}$.


Question: As shown in the figure, the line $l_{1}$ intersects the x-axis and y-axis at points A and B, respectively, with coordinates $(2, 0)$ and $(0, 3)$. The line $l_{2}: y=\frac{1}{2}x+3$ passing through point B intersects the x-axis at point C. Point $D(n, 6)$ is a point on the line $l_{1}$, and line segment $CD$ is drawn.

Solution:
\textbf{Solution:}

(I) Let the equation of line $l_1$ be $y=kx+b$. Substituting A(2,0) and B(0,3) into it, we get
\[
\left\{
\begin{array}{l}
2k+b=0\\
b=3
\end{array}
\right.
\]
Solving this, we find
\[
\left\{
\begin{array}{l}
k=-\frac{3}{2}\\
b=3
\end{array}
\right.
\]
Therefore, the equation of line $l_1$ is $y=-\frac{3}{2}x+3$;

(II) When $y=0$, $\frac{1}{2}x+3=0$, solving this gives $x=-6$, therefore the coordinates of point C are (-6,0). Substituting D(n,6) into $y=-\frac{3}{2}x+3$ gives $-\frac{3}{2}n+3=6$, solving this gives $n=-2$, therefore the coordinates of point D are (-2,6);

(III) $S_{\triangle BCD}=S_{\triangle DAC}-S_{\triangle BAC}=\frac{1}{2}\times(2+6)\times6-\frac{1}{2}\times(2+6)\times3=\boxed{12}$.


Question: After filling a cylindrical oil tank with a diameter of 52 cm with some oil, the cross-section is shown in the figure. If the maximum depth CD of the oil is 16 cm, calculate the width AB of the oil surface.

Solution:
\textbf{Solution:} From the problem statement, we have: $OC \perp AB$ at point D,\\
According to the perpendicular diameter theorem, point D is the midpoint of AB, $AB = 2BD$,\\
$\because$ the diameter is $52\,\text{cm}$,\\
$\therefore$OB = $26\,\text{cm}$,\\
$\therefore$OD = OC - CD = $26 - 16 = 10\,\text{cm}$,\\
By the Pythagorean theorem,\\
$BD = \sqrt{BO^2 - OD^2} = 24\,\text{cm}$,\\
$\therefore$AB = \boxed{48}\,\text{cm}.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and the chord $CD \perp AB$ at point $E$. If $AB=26$ and $EB=8$, find the length of chord $CD$.

Solution:
\textbf{Solution:} Draw line $OC$ as shown in the diagram:\\
Since $AB$ is the diameter of circle $O$ and $CD \perp AB$,\\
it follows that $CE=DE=\frac{1}{2}CD$, $OC=OB=\frac{1}{2}AB=13$,\\
therefore $OE=OB-EB=13-8=5$,\\
In the right-angled triangle $\triangle OCE$, by the Pythagorean theorem, we have $CE=\sqrt{OC^{2}-OE^{2}}=12$,\\
thus, $CD=2CE=\boxed{24}$.


Question: As shown in the diagram, the square $MNPQ$ is inscribed in $\triangle ABC$, with points M, N on side $BC$, and points P, Q on sides $AC$ and $AB$ respectively. Given that the height $AD$ from side $BC$ is $6\,cm$ and $BC=12\,cm$, what is the side length of the square $MNPQ$ in centimeters?

Solution:
\textbf{Solution:} Given that: square $MNPQ$ is inscribed in $\triangle ABC$, with the height from $BC$ being $AD=6\,cm$, \\
$\therefore$ $\angle ADC=\angle ADB=\angle CNP=\angle BMQ=90^\circ$, \\
Since $\angle B=\angle B$, $\angle C=\angle C$, \\
$\therefore$ $\triangle BMQ\sim \triangle BDA$, $\triangle CNP\sim \triangle CDA$, \\
$\therefore$ $\frac{BM}{BD}=\frac{QM}{AD}$, $\frac{NP}{AD}=\frac{CN}{CD}$, \\
Since $QM=NP$, let the side length of the square be $x$, \\
$\therefore$ $BM=\frac{x}{AD}\cdot BD$, $CN=\frac{x}{AD}\cdot CD$ \\
$\therefore$ $BM+CN=\frac{x}{AD}\cdot (BD+CD)=\frac{x}{AD}\cdot BC=BC-x$ \\
Since $AD=6\,cm$, $BC=12\,cm$, \\
$\therefore$ $\frac{x}{6}\cdot 12=12-x$, \\
Solving gives: $x=\boxed{4}$ \\
$\therefore$ The side length of square $MNPQ$ is $4\,cm$.


Question: As shown in the figure, in $\triangle ABC$, where $AB = AC = 4$ and $BC = 5$, points D and E are on line segments $BC$ and $AC$ respectively, with $\angle ADE = \angle B$. When $BD = 2$, find the length of $EC$.

Solution:
\textbf{Solution:} Given: $AB=AC$,\\
thus $\angle B=\angle C$.\\
Since $\angle B+\angle BAD+\angle ADB=180^\circ$,\\
and $\angle ADE+\angle EDC+\angle ADB=180^\circ$,\\
$\angle ADE=\angle B$,\\
therefore $\angle BAD=\angle EDC$,\\
thus $\triangle ABD\sim \triangle DCE$,\\
hence $\frac{EC}{BD}=\frac{DC}{AB}$.\\
Given $AB=4, BC=5, BD=2$,\\
thus $\frac{EC}{2}=\frac{5-2}{4}$,\\
therefore $EC=\boxed{\frac{3}{2}}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=8$, $BC=4\sqrt{2}$, $DE\perp AC$ at point $F$, and intersects $AB$ at point $E$. Find the length of $DE$.

Solution:
\textbf{Solution:} In rectangle $ABCD$,\\
$\therefore$ $\angle DAB = \angle ADC = 90^\circ$, $CD = AB = 8$, $AD = BC = 4\sqrt{2}$,\\
$\therefore$ $\angle DAF + \angle ACD = 90^\circ$,\\
$\because$ $DE \perp AC$ at point F, $\therefore$ $\angle ADE + \angle DAF = 90^\circ$,\\
$\therefore$ $\angle ACD = \angle EDA$,\\
$\therefore$ $\triangle ACD \sim \triangle EDA$,\\
$\therefore$ $\frac{AD}{EA} = \frac{CD}{AD}$,\\
$\therefore$ $EA = \frac{{\left(4\sqrt{2}\right)}^2}{8} = 4$,\\
$\therefore$ $DE = \sqrt{AD^2 + EA^2} = \sqrt{{\left(4\sqrt{2}\right)}^2 + 4^2} = \boxed{4\sqrt{3}}$.


Question: As shown in the diagram, in parallelogram $ABCD$, point $E$ is on side $BC$. Connect $AE$ and extend it to intersect diagonal $BD$ at point $F$ and the extension of $DC$ at point $G$. If $\frac{CE}{BE}=\frac{2}{3}$, what is the value of $\frac{FE}{EG}$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a parallelogram,\\
thus AD=BC, and AD$\parallel$BC,\\
therefore, $\triangle ADF\sim \triangle EBF$, $\triangle GEC\sim \triangle GAD$,\\
thus, $\frac{EF}{AF}=\frac{BE}{AD}$, $\frac{EG}{AG}=\frac{EC}{AD}$,\\
given that $\frac{CE}{BE}=\frac{2}{3}$,\\
thus, $\frac{BE}{AD}=\frac{3}{5}$, $\frac{CE}{AD}=\frac{2}{5}$,\\
thus, $\frac{FE}{AF}=\frac{3}{5}$, $\frac{EG}{AG}=\frac{2}{5}$,\\
thus, $\frac{FE}{AE}=\frac{3}{8}$, $\frac{EG}{AE}=\frac{2}{3}$,\\
thus, $\frac{FE}{EG}=\boxed{\frac{9}{16}}$.


Question: As shown in the figure, in rectangle $ABCD$, points $E$ and $F$ are on sides $AD$ and $DC$, respectively, $BE\perp EF$. Given $AB=6$, $AE=9$, $DE=2$, what is the length of $EF$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a rectangle, \\
$\therefore \angle A=\angle D=90^\circ$, \\
$\therefore \angle ABE+\angle AEB=90^\circ$, \\
$\because BE\perp EF$, \\
$\therefore \angle DEF+\angle AEB=90^\circ$, \\
$\therefore \angle DEF=\angle ABE$, \\
In $\triangle DEF$ and $\triangle ABE$, $\left\{\begin{array}{l}\angle D=\angle A=90^\circ \\ \angle DEF=\angle ABE\end{array}\right.$, \\
$\therefore \triangle DEF\sim \triangle ABE$, \\
$\therefore \frac{DF}{AE}=\frac{DE}{AB}$, \\
$\because AB=6, AE=9, DE=2$, \\
$\therefore \frac{DF}{9}=\frac{2}{6}$, \\
Solving gives $DF=3$, \\
In $\triangle DEF$, $EF=\sqrt{DE^2+DF^2}=\boxed{\sqrt{13}}$.


Question: As shown, it is known that in $\triangle ABC$, $AD$ is the median to $BC$, and point $E$ is on $AD$ with $\frac{DE}{AE} = \frac{1}{3}$. The ray $CE$ intersects $AB$ at point $F$. Find the value of $\frac{AF}{FB}$.

Solution:
\textbf{Solution:} As shown in the figure, draw $DM\parallel AB$ through $D$, intersecting $CF$ at $M$, \\
$\therefore \triangle DME\sim \triangle AFE, \triangle CDM\sim \triangle CBF,$ \\
$\therefore \frac{DE}{AE}=\frac{DM}{AF}=\frac{1}{3}, \frac{CD}{CB}=\frac{DM}{BF},$ \\
$\because D$ is the midpoint of $BC$, \\
$\therefore \frac{CD}{CB}=\frac{DM}{BF}=\frac{1}{2},$ \\
$\therefore \frac{AF}{BF}=\boxed{\frac{3}{2}}.$


Question: As shown in the figure, points D and E are located on the sides $AB$ and $AC$ of $\triangle ABC$, respectively, with $DE \parallel BC$. If ${S}_{\triangle ADE} = 2$ and ${S}_{\triangle BCE} = 7.5$, what is the area of ${S}_{\triangle BDE}$?

Solution:
\textbf{Solution:} Given $DE\parallel BC$, it follows that $\frac{AD}{DB}=\frac{AE}{EC}$. \\
Also, since $\frac{S_{\triangle ADE}}{S_{\triangle BDE}}=\frac{AD}{BD}$, it implies $\frac{S_{\triangle ABE}}{S_{\triangle BCE}}=\frac{AE}{EC}$, hence $\frac{S_{\triangle ADE}}{S_{\triangle BDE}}=\frac{S_{\triangle ABE}}{S_{\triangle BCE}}$. \\
Let $S_{\triangle BDE}=x$, with $S_{\triangle ADE}=2$, and $S_{\triangle BCE}=7.5$, therefore $\frac{2}{x}=\frac{2+x}{7.5}$. Consequently, $x_1=3$, $x_2=-5$ (discard). Therefore, $S_{\triangle DBE}=\boxed{3}$.


Question: As shown in the figure, in rectangle $ABCD$, points $E$ and $F$ are respectively on sides $AD$ and $DC$, $\triangle ABE\sim \triangle DEF$, $DF>DE$, $AB=9$, $AE=12$, $DE=3$, what is the length of $FC$?

Solution:
\textbf{Solution:} Given that $\triangle ABE\sim \triangle DEF$, \\
therefore $\frac{AB}{DE}=\frac{AE}{DF}$, \\
since $AB=9$, $AE=12$, and $DE=3$, \\
then $\frac{9}{3}=\frac{12}{DF}$, \\
solving this gives $DF=4$, \\
since quadrilateral $ABCD$ is a rectangle, \\
therefore $DC=AB=9$, \\
thus $FC=DC−DF=9−4=\boxed{5}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=6$, $BC=8$. Point $D$ is on side $AB$, $DE \perp AC$, and $DF \perp BC$ with the feet of the perpendiculars being points $E$ and $F$, respectively. Join $EF$. What is the minimal value of segment $EF$?

Solution: \textbf{Solution:} Therefore, the minimum value of line segment $EF$ is $\boxed{4.8}$.


Question: As shown in the figure, in $\triangle ABC$, $M$ is the midpoint of $BC$, $AN$ bisects $\angle BAC$, $BN\perp AN$ at point $N$, and $MN$ is connected. If $AB=10$, $BC=15$, and $MN=3$, what is the perimeter of $\triangle ABC$?

Solution:
\textbf{Solution:} Extend $BN$ to meet $AC$ at point $D$,\\
$\because AN$ bisects $\angle BAC$, that is $\angle BAN = \angle DAN$,\\
$BN \perp AN$, that is $\angle BNA = \angle DNA = 90^\circ$, and $AN$ is the common side of $\triangle ABN$ and $\triangle ADN$\\
$\therefore \triangle ABN \cong \triangle ADN$, $AD = AB = 10$\\
$BN = DN$, that is, $N$ is the midpoint of $BD$\\
Also, $M$ is the midpoint of $BC$, $\therefore CD = 2MN = 6$\\
$\therefore$ The perimeter is: $AB + BC + CD + AD = 10 + 15 + 6 + 10 = \boxed{41}$


Question: As shown in the figure, given an equilateral $\triangle ABC$ with a side length of 4, where points $D$ and $E$ are the midpoints of $AC$ and $BC$ respectively. Draw $DF\perp DE$, intersecting the extension of $BC$ at point $F$. Find the length of $DF$.

Solution:
\textbf{Solution:} Given that $\triangle ABC$ is an equilateral triangle with side length 4,\\
$\therefore AC=BC=AB=4, \angle ACB=60^\circ=\angle B=\angle A,$\\
Given that points D and E are the midpoints of $AC$ and $BC$, respectively,\\
$\therefore DE\parallel AB$ and $DE=\frac{1}{2}AB=2,$\\
$\therefore \angle DEC=\angle B=60^\circ,$\\
Given that $DE\perp DF,$\\
$\therefore \angle F=30^\circ$, and $DE\colon DF\colon EF=1\colon \sqrt{3}\colon 2,$\\
$\therefore DF=\sqrt{3}DE=2\sqrt{3}=\boxed{2\sqrt{3}}.$


Question: As shown in the diagram, within the rectangle $ABCD$, point $E$ is the midpoint of $AB$, $EF\perp AB$ intersects $CD$ at point $F$, and point $M$ is located on $AD$. Connect $BM$, and fold $\triangle ABM$ along $BM$. When the corresponding point $A'$ of point $A$ exactly falls on $EF$, what is the degree measure of $\angle CBA'$?

Solution:
\textbf{Solution:} As shown in the figure, connect $AA'$.\\
Since point E is the midpoint of $AB$, $EF\perp AB$ and intersects $CD$ at point F,\\
it follows that $EF$ bisects $AB$ perpendicularly,\\
thus $A'A=A'B$,\\
due to the properties of folding, it is known that $A'B=AB$,\\
therefore $A'A=A'B=AB$,\\
hence $\triangle ABA'$ is an equilateral triangle,\\
thus $\angle ABA'=60^\circ$,\\
in rectangle $ABCD$, $\angle ABC=90^\circ$,\\
thus $\angle CBA'=\boxed{30^\circ}$.


Question: As shown in the figure, in rhombus ABCD, the diagonals AC and BD intersect at point O, with $BD = 6$ and $AC = 4$. Find the perimeter of rhombus ABCD.

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\
it follows that $\angle AOB = 90^\circ$,\\
$BO = \frac{1}{2}BD = \frac{1}{2} \times 6 = 3$,\\
$AO = \frac{1}{2}AC = \frac{1}{2} \times 4 = 2$,\\
$AB = BC = CD = AD$.\\
In $\triangle AOB$, by the Pythagorean theorem, we have\\
$AB = \sqrt{AO^{2} + BO^{2}} = \sqrt{13}$,\\
thus, the perimeter ${C}_{\text{rhombus} ABCD} = 4AB = 4\sqrt{13} = \boxed{4\sqrt{13}}$.


Question: As shown in the figure, point $E$ is a point inside the square $ABCD$. Connect $AE$, $BE$, $DE$, where $\angle AEB=90^\circ$, and the area of $\triangle ADE$ is 4. What is the length of $AE$?

Solution:
\textbf{Solution:} Extend AE, and draw DF perpendicular to AE from point D, where F is the foot of the perpendicular,\\
$\therefore$ $\angle$DFA=$90^\circ$,\\
$\therefore$ $\angle$DAF+$\angle$ADF=$90^\circ$,\\
$\because$ In the square ABCD, $\angle$DAB=$90^\circ$, and AD=AB,\\
$\therefore$ $\angle$DAF+$\angle$BAE=$90^\circ$,\\
$\therefore$ $\angle$ADF=$\angle$BAE,\\
$\because$ $\angle$AEB=$90^\circ$,\\
$\therefore$ $\angle$AEB=$\angle$DFA,\\
$\therefore$ $\triangle$ABE$\cong$ $\triangle$DAF (by AAS),\\
$\therefore$ AE=DF,\\
Given $S_{\triangle ADE}=4$, we have\\
$\frac{1}{2}AE\cdot DF=\frac{1}{2}AE^2=4$,\\
Solving this gives AE=$\boxed{2\sqrt{2}}$,\\
$\therefore$ the length of AE is $2\sqrt{2}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$. Point $D$ is the midpoint of hypotenuse $AB$, $DE\perp AC$ at foot $E$. If $DE=2$ and $CD=2\sqrt{5}$, what is the length of $BE$?

Solution: \textbf{Solution:} Given that in the right-angled triangle $ABC$, angle $\angle ACB=90^\circ$ and $DE \perp AC$, we have $DE \parallel BC$. Since point $D$ is the midpoint of $AB$, $DE$ is the median of triangle $ABC$. Also, given $DE=2$, we infer $BC=4$. In the right-angled triangle $DCE$, with $DE=2$ and $CD=2\sqrt{5}$, by Pythagoras' theorem, we have $CE=\sqrt{CD^2-DE^2}=\sqrt{(2\sqrt{5})^2-2^2}=4$. Therefore, in the right-angled triangle $BCE$, applying Pythagoras' theorem again, we find $BE=\sqrt{BC^2+CE^2}=\sqrt{4^2+4^2}=4\sqrt{2}$. Hence, the length of $BE$ is $\boxed{4\sqrt{2}}$.


Question: As shown in the figure, quadrilateral $ABCD$ is a rectangle. Fold $\triangle ACD$ along $AC$ to $\triangle ACD^{\prime}$. $AD^{\prime}$ intersects $BC$ at point $E$. Given $AD=4$ and $DC=3$, find the length of $BE$.

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle, \\
it follows that AB=DC=3, BC=AD=4, AD$\parallel$BC, and $\angle$B=90$^\circ$, \\
Since $\triangle$ACD is folded along AC to $\triangle$ACD$^{\prime}$, and AD$^{\prime}$ intersects BC at point E, \\
it follows that $\angle$DAC=$\angle$D$^{\prime}$AC, \\
Since AD$\parallel$BC, it follows that $\angle$DAC=$\angle$ACB, \\
hence $\angle$D$^{\prime}$AC=$\angle$ACB, therefore AE=EC, \\
Let BE=x, then EC=4−x, and AE=4−x, \\
In $\triangle$ABE, since AB$^{2}$+BE$^{2}$=AE$^{2}$, \\
it follows that 3$^{2}$+x$^{2}$=$(4−x)^{2}$, solving this gives x= $\boxed{\frac{7}{8}}$, \\
thus, the length of BE is $\frac{7}{8}$.


Question: As shown in the figure, $\angle AOB$ is a straight angle, $\angle BOC=36^\circ$, $OD$ bisects $\angle AOC$, $\angle DOE=90^\circ$. Find the measure of $\angle AOE$.

Solution:
\textbf{Solution:} Since $\angle AOB$ is a straight angle and $\angle BOC=36^\circ$,\\
it follows that $\angle AOC=180^\circ-\angle BOC=180^\circ-36^\circ=144^\circ$.\\
Since $OD$ bisects $\angle AOC$,\\
then $\angle AOD= \frac{1}{2}\angle AOC=\frac{1}{2}\times144^\circ=72^\circ$.\\
Since $\angle DOE=90^\circ$,\\
thus $\angle AOE=\angle DOE-\angle AOD=90^\circ-72^\circ=\boxed{18^\circ}$.


Question: As shown in the figure, in triangle \(ABC\), \(AB=10\text{cm}\), \(AC=6\text{cm}\), \(D\) is the midpoint of \(BC\), and \(E\) is a point on side \(AB\). If the perimeter of triangle \(BDE\) is equal to the perimeter of quadrilateral \(ACDE\), find the length of segment \(AE\).

Solution:
\textbf{Solution:} Given the diagram, we have:\\
The perimeter of triangle $BDE = BE + BD + DE$, the perimeter of quadrilateral $ACDE = AE + AC + DC + DE$,\\
And since the perimeter of triangle $BDE$ equals the perimeter of quadrilateral $ACDE$, and $D$ is the midpoint of $BC$,\\
Therefore, $BD = DC$, $BE + BD + DE = AE + AC + DC + DE$,\\
Therefore, $BE = AE + AC$,\\
And since $AB= 10\text{cm}$, $AC = 6\text{cm}$, $BE = AB - AE$,\\
Therefore, $AE + AC = AB - AE$,\\
Therefore, $10 - AE = AE + 6$,\\
Therefore, $AE = \boxed{2\text{cm}}$.


Question: In $\triangle ABC$, $\angle BAC=50^\circ$, $\angle B=48^\circ$, $AD$ is the angle bisector of $\triangle ABC$. What is the measure of $\angle ADC$?

Solution:
\textbf{Solution:} Given that $\angle BAC=50^\circ$ and $\angle B=48^\circ$, \\
it follows that $\angle C=180^\circ-\angle BAC-\angle B=82^\circ$, \\
since $AD$ is the bisector of $\triangle ABC$, \\
it implies that $\angle CAD=\frac{1}{2}\angle BAC=25^\circ$, \\
hence, $\angle ADC=180^\circ-\angle CAD-\angle C=\boxed{73^\circ}$.


Question: As shown in the figure, it is known that the lines $AB$ and $CD$ intersect at point $O$, $\angle COE$ is a right angle, $OF$ bisects $\angle AOE$, $\angle COF = 28^\circ$. Find the measure of $\angle BOE$.

Solution:
\textbf{Solution:} Since $\angle COE$ is a right angle and $\angle COF=28^\circ$,\\
it follows that $\angle EOF=90^\circ-28^\circ=62^\circ$,\\
since $OF$ bisects $\angle AOE$,\\
it follows that $\angle AOF=\angle EOF=62^\circ$,\\
therefore, $\angle EOB=180^\circ-62^\circ-62^\circ=\boxed{56^\circ}$.


Question: As shown in the figure, $OB$ is the angle bisector of $\angle AOC$, and $OD$ is the angle bisector of $\angle COE$. If $\angle AOB=40^\circ$ and $\angle COE=60^\circ$, what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Given that OB is the angle bisector of $\angle AOC$, and OD is the angle bisector of $\angle COE$, with $\angle AOB=40^\circ$ and $\angle COE=60^\circ$,\\
thus, $\angle BOC=\angle AOB=40^\circ$ and $\angle COD= \frac{1}{2}\angle COE=\frac{1}{2}\times 60^\circ=30^\circ$,\\
hence, $\angle BOD=\angle BOC+\angle COD=40^\circ+30^\circ=\boxed{70^\circ}$.


Question: As shown in the figure, point O is on line AB, $\angle COE$ is a right angle, OF bisects $\angle AOE$, $\angle COF=30^\circ$, what is the measure of $\angle EOB$?

Solution:
\textbf{Solution:} Given, $\because$ $\angle COE$ is a right angle,\\
$\therefore$ $\angle COE=90^\circ$,\\
$\because$ $\angle COE=\angle COF+\angle FOE$, $\angle COF=30^\circ$,\\
$\therefore$ $\angle FOE=\angle COE−\angle COF=90^\circ−30^\circ=\boxed{60^\circ}$,\\
$\because$ OF bisects $\angle AOE$,\\
$\therefore$ $\angle FOE=\angle AOF=\frac{1}{2}\angle AOE$,\\
$\therefore$ $\angle AOE=2\angle FOE=120^\circ$,\\
$\because$ $\angle AOE+\angle BOE=180^\circ$,\\
$\therefore$ $\angle BOE=180^\circ−120^\circ=60^\circ$


Question: As shown in the diagram, $\angle AOB$ is a straight angle, $\angle AOC=80^\circ$, $\angle BOD=30^\circ$, $OM$ and $ON$ are the angle bisectors of $\angle AOC$ and $\angle BOD$ respectively. What is the degree measure of $\angle MON$?

Solution:
\textbf{Solution:} Given that $\angle AOB$ is a straight angle,\\
$\therefore \angle AOB=180^\circ$,\\
since OM and ON are the bisectors of $\angle AOC$ and $\angle BOD$, respectively, and $\angle AOC=80^\circ$, $\angle BOD=30^\circ$,\\
$\therefore \angle AOM=\frac{1}{2}\angle AOC=40^\circ$, $\angle BON=\frac{1}{2}\angle BOD=15^\circ$,\\
$\therefore \angle MON=\angle AOB-\angle AOM-\angle BON=180^\circ-40^\circ-15^\circ=\boxed{125^\circ}$.


Question: As the figure shows, $\angle AOB$ is a right angle, $\angle AOC=50^\circ$, $ON$ is the bisector of $\angle AOC$, and $OM$ is the bisector of $\angle BOC$. What is the measure of $\angle MON$?

Solution:
\textbf{Solution:} Since $\angle AOB$ is a right angle,\\
it follows that $\angle AOB = 90^\circ$.\\
Therefore, $\angle BOC = \angle AOB + \angle AOC = 90^\circ + 50^\circ = 140^\circ$.\\
Since OM bisects $\angle BOC$,\\
it follows that $\angle COM = \frac{1}{2} \angle BOC = 70^\circ$.\\
Since ON bisects $\angle AOC$,\\
it follows that $\angle CON = \frac{1}{2} \angle AOC = 25^\circ$.\\
Therefore, $\angle MON = \angle COM - \angle CON = 70^\circ - 25^\circ = \boxed{45^\circ}$.


Question: As shown in the figure, $OC$ is the angle bisector of $\angle AOB$, $\angle AOD = 2\angle BOD$, and $\angle COD = 18^\circ$. Please find the measure of $\angle BOD$ in degrees.

Solution:
\textbf{Solution:} From the construction we get \\
Since $OC$ is the bisector of $\angle AOB$, \\
Therefore $\angle BOC=\frac{1}{2}\angle AOB$; \\
Since $\angle AOD=2\angle BOD$, \\
Therefore $\angle AOB=3\angle BOD$, i.e., $\angle BOD=\frac{1}{3}\angle AOB$; \\
Therefore $\angle COD=\frac{1}{2}\angle AOB-\frac{1}{3}\angle AOB=\frac{1}{6}\angle AOB$, \\
Therefore $\angle BOD=2\angle COD$, \\
Since $\angle COD=18^\circ$, \\
Therefore $\angle BOD=\boxed{36^\circ}$.


Question: As shown in the figure, the lines AB and CD intersect at point O, $\angle EOC = 90^\circ$, OF is the angle bisector of $\angle AOE$, $\angle COF = 34^\circ$. What is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Given that $\angle EOC=90^\circ$ and $\angle COF=34^\circ$, \\
it follows that $\angle EOF=56^\circ$,\\
since OF is the angle bisector of $\angle AOE$,\\
it follows that $\angle AOF=\angle EOF=56^\circ$,\\
thus $\angle AOC=\angle AOF-\angle COF=22^\circ$,\\
hence $\angle BOD=\angle AOC=\boxed{22^\circ}$.


Question: As shown in the figure, $\angle ACD = \angle BCE = 90^\circ$. If $\angle ACB = 150^\circ$, can you calculate the measure of $\angle DCE$? Please write down the solution process.

Solution:
\textbf{Solution:} Since $\angle ACD = \angle BCE = 90^\circ$,\\
it follows that $\angle ACD + \angle ECB = 180^\circ$\\
Also, since $\angle ACB + \angle ECD = \angle ACD + \angle ECB = 180^\circ$, and $\angle ACB = 150^\circ$,\\
it follows that $\angle ECD = 180^\circ - \angle ACB = \boxed{30^\circ}$


Question: A rectangular piece of iron, with a length of 25 cm and a width of 15 cm, has small squares with a side length of 2 cm cut from each of its four corners. Then, by folding up the four sides, a lidless iron box is made. What is the volume of the iron box in liters?

Solution:
\textbf{Solution:} By the problem statement, we have: the length of the iron box $= 25 - 2 \times 2 = 21\,\text{cm}$, the width of the iron box $= 15 - 2 \times 2 = 11\,\text{cm}$, the height of the iron box $= 2\,\text{cm}$,\\
$\therefore$ the volume of the iron box $= 21 \times 11 \times 2 = 462\,\text{cm}^3 = 0.462$ litres.\\
Thus, the volume of the iron box is $\boxed{0.462}$ litres.


Question: As shown in the figure, $\angle AOB$ and $\angle BOC$ are adjacent supplementary angles. $OD$ is the angle bisector of $\angle AOB$, and $OE$ is within $\angle BOC$ such that $\angle BOE=\frac{1}{2}\angle EOC$, and $\angle DOE={72}^\circ$. Find the degree measure of $\angle EOC$?

Solution:
\textbf{Solution:} Given that $\angle BOE = \frac{1}{2}\angle EOC$, \\
then $\angle BOC = \angle BOE + \angle EOC = 3\angle BOE$, \\
since $\angle DOE = 72^\circ$, \\
thus $\angle BOD = \angle DOE - \angle BOE = 72^\circ - \angle BOE$, \\
since $OD$ bisects $\angle AOB$, \\
thus $\angle AOB = 2\angle BOD = 144^\circ - 2\angle BOE$, \\
since $\angle AOB$ and $\angle BOC$ are supplementary, \\
thus $\angle AOB + \angle BOC = 180^\circ$, \\
therefore $144^\circ - 2\angle BOE + 3\angle BOE = 180^\circ$, \\
therefore $\angle BOE = 36^\circ$, \\
therefore $\angle EOC = 2\angle BOE = \boxed{72^\circ}$.


Question: As shown in the diagram, $\angle AOB$ is a right angle, $OB$ bisects $\angle COD$, $\angle AOC=62^\circ$, find the degree measure of $\angle AOD$.

Solution:
\textbf{Solution:} Since $\angle AOB$ is a right angle,\\
it follows that $\angle AOB=90^\circ$,\\
therefore, $\angle BOC=\angle AOB-\angle AOC=90^\circ-62^\circ=28^\circ$.\\
Since $OB$ bisects $\angle COD$,\\
it follows that $\angle BOD=\angle BOC=28^\circ$,\\
therefore, $\angle AOD=\angle AOB+\angle BOD=90^\circ+28^\circ=\boxed{118^\circ}$.


Question: As shown in the figure, it is known that $ \angle AOD=60^\circ$, $ \angle BOC$ is a right angle, and OC bisects $ \angle AOD$. What is the measure of $ \angle BOD$ in degrees?

Solution:
\textbf{Solution:} \\
$\because$ OC bisects $\angle AOD$,\\
$\therefore \angle COD=\frac{1}{2}\angle AOD$,\\
$\because \angle AOD=60^\circ$,\\
$\therefore \angle COD=30^\circ$,\\
$\because \angle BOC$ is a right angle,\\
$\therefore \angle BOC=90^\circ$,\\
$\therefore \angle BOD=\angle BOC-\angle COD=90^\circ-30^\circ=\boxed{60^\circ}$.


Question: As shown in the figure, it is known that $ \angle AOB = 120^\circ$ and $ \angle BOC = 30^\circ$. If $OD$ is the angle bisector of $ \angle AOC$, find the measure of $ \angle BOD$.

Solution:
\textbf{Solution:} Given that $\because \angle AOB=120^\circ$ and $\angle BOC=30^\circ$,\\
$\therefore \angle AOC=\angle AOB - \angle BOC=90^\circ$,\\
Also, $\because$ OD is the bisector of $\angle AOC$,\\
$\therefore \angle COD=\frac{1}{2}\angle AOC=45^\circ$,\\
$\therefore \angle BOD=\angle COD+\angle BOC =45^\circ+30^\circ=\boxed{75^\circ}$.


Question: As shown in the figure, it is known that $OD$ bisects $\angle AOB$, ray $OC$ is inside $\angle AOD$, $\angle BOC = 2\angle AOC$, $\angle AOB = 120^\circ$, find the complement of $\angle COD$.

Solution:
\textbf{Solution:} Since $\angle AOB = \angle BOC + \angle AOC$ and $\angle BOC = 2\angle AOC$,\\
it follows that $\angle AOB = 2\angle AOC + \angle AOC$, that is $3\angle AOC = 120^\circ$,\\
hence $\angle AOC = 40^\circ$.\\
Since OD bisects $\angle AOB$,\\
it follows that $\angle AOD = \angle BOD = \frac{1}{2}\angle AOB = \frac{1}{2} \times 120^\circ = 60^\circ$.\\
Since $\angle COD = \angle AOD - \angle AOC$,\\
it follows that $\angle COD = 60^\circ - 40^\circ = 20^\circ$.\\
Therefore, the supplement of $\angle COD$ is $180^\circ - \angle COD = 180^\circ - 20^\circ = \boxed{160^\circ}$.


Question: As shown in the figure, $\angle AOB$ is a straight angle, $\angle AOC = 28^\circ$, $\angle BOD = 58^\circ$, $OM$ and $ON$ are the angle bisectors of $\angle AOC$ and $\angle BOD$, respectively. Find the measure of $\angle MON$.

Solution:
\textbf{Solution:} From the given, $OM$ and $ON$ are the angle bisectors of $\angle AOC$ and $\angle BOD$, respectively, with $\angle AOC=28^\circ$ and $\angle BOD=58^\circ$.\\
Therefore, $\angle AOM=\frac{1}{2}\angle AOC=14^\circ$ and $\angle BON=\frac{1}{2}\angle BOD=29^\circ$.\\
Therefore, $\angle MON=\angle AOB-\angle AOM-\angle BON=180^\circ-14^\circ-29^\circ=\boxed{137^\circ}$.


Question: As shown in the figure, point $O$ lies on line $DE$, $\angle COD = 60^\circ$, $\angle BOD = \frac{1}{2}\angle COB$, and $OC$ is the bisector of $\angle AOB$. Find the degree measure of $\angle AOE$.

Solution:
\[ \text{Solution:} \quad \because \angle COD = 60^{\circ} = \angle DOB + \angle BOC, \]
\[ \therefore \angle DOB = 20^{\circ}, \, \angle BOC = 40^{\circ}, \]
\[ \because OC \text{ is the bisector of } \angle AOB, \]
\[ \therefore \angle AOB = 2\angle BOC = 80^{\circ}, \]
\[ \therefore \angle AOD = \angle DOB + \angle AOB = 100^{\circ}, \]
\[ \because D, O, E \text{ are collinear}, \]
\[ \therefore \angle DOE = 180^{\circ}, \]
\[ \therefore \angle AOE = \angle DOE - \angle AOD = 180^{\circ} - 100^{\circ} = \boxed{80^{\circ}}. \]


Question: As shown in the figure, $\angle AOB$ is a right angle, $\angle AOC = 30^\circ$, $ON$ is the bisector of $\angle AOC$, and $OM$ is the bisector of $\angle BOC$. What is the measure of $\angle MON$ in degrees?

Solution:
\textbf{Solution:} Given that $ON$ is the bisector of $\angle AOC$ and $OM$ is the bisector of $\angle BOC$, \\
it follows that $\angle NOC= \frac{1}{2} \angle AOC$ and $\angle MOC= \frac{1}{2} \angle BOC$\\
Therefore, $\angle MON=\angle MOC-\angle NOC$\\
$= \frac{1}{2} \angle BOC- \frac{1}{2} \angle AOC$\\
$= \frac{1}{2} (\angle BOC-\angle AOC)$\\
$= \frac{1}{2} \angle AOB$\\
$= \frac{1}{2} \times 90^\circ$\\
$= \boxed{45^\circ}$


Question: As shown in the figure, $O$ is a point on the line $MN$, $\angle AOB$ is a right angle, and $OC$ bisects $\angle MOB$. If $OD$ bisects $\angle CON$, and $\angle DON - \angle AOM = 21^\circ$, what is the measure of $\angle BON$ in degrees?

Solution:
\textbf{Solution:} Since $OC$ bisects $\angle MOB$,\\
$\therefore \angle BOC = \angle MOC$,\\
Since $OD$ bisects $\angle CON$ and $\angle AOB$ is a right angle,\\
$\therefore \angle COD = \angle NOD = \frac{1}{2}\left(180^\circ - \angle BOC\right)$,\\
$\angle AOB = 90^\circ$,\\
Since $\angle AOM = 90^\circ - 2\angle BOC$,\\
Since $\angle DON - \angle AOM = 21^\circ$,\\
$\therefore \left\{\begin{array}{l}
\frac{1}{2}\angle BOC + \angle AOM = 69^\circ \\
\angle AOM + 2\angle BOC = 90^\circ
\end{array}\right.$,\\
Solving yields $\left\{\begin{array}{l}
\angle AOM = 62^\circ \\
\angle BOC = 14^\circ
\end{array}\right.$,\\
$\therefore \angle BON = 180^\circ - 2\angle BOC = 180^\circ - 28^\circ = \boxed{152^\circ}$.


Question: As shown in the figure, it is known that $\angle AOB = 50^\circ$, $\angle BOC = 90^\circ$, and $OM$ and $ON$ are the angle bisectors of $\angle AOB$ and $\angle BOC$ respectively. What is the degree measure of $\angle MON$?

Solution:
\textbf{Solution:} Since $\angle AOB=50^\circ$, and $OM$ is the angle bisector of $\angle AOB$,\\
$\therefore \angle BOM=25^\circ$.\\
Also, since $\angle BOC=90^\circ$, and $ON$ is the angle bisector of $\angle BOC$,\\
$\therefore \angle BON=45^\circ$.\\
$\therefore \angle MON=25^\circ+45^\circ=\boxed{70^\circ}$.


Question: As shown in the figure, it is known that $\angle AOE$ is a straight angle, $OD$ bisects $\angle COE$, $OB$ bisects $\angle AOC$, and the ratio $\angle COD : \angle BOC = 2 : 3$. Find the measures of $\angle COD$ and $\angle BOC$ in degrees.

Solution:
\textbf{Solution:} Since OD bisects $\angle COE$, \\
then $\angle COD= \frac{1}{2} \angle COE$, \\
Since OB bisects $\angle AOC$, \\
then $\angle BOC= \frac{1}{2} \angle AOC$, \\
therefore $\angle BOD=\angle COD+\angle BOC= \frac{1}{2} (\angle COE+\angle AOC)=90^\circ$, \\
Since the ratio $\angle COD\colon \angle BOC=2\colon 3$, \\
thus $\angle COD= \frac{2}{5}\times 90^\circ =\boxed{36^\circ}$, $\angle BOC= \frac{3}{5}\times 90^\circ =\boxed{54^\circ}$.


Question: As shown in the figure, O is a point on line AB, and $\angle COD$ is a right angle. OE bisects $\angle BOD$, and OF bisects $\angle AOE$. If $\angle BOC=54^\circ$, what are the degrees of $\angle COE$ and $\angle DOF$?

Solution:
\textbf{Solution:} Since $\angle COD=90^\circ$ and $\angle BOC=54^\circ$,\\
it follows that $\angle BOD=90^\circ-54^\circ=36^\circ$,\\
Since OE bisects $\angle BOD$,\\
it follows that $\angle DOE=\angle BOE=18^\circ$,\\
thus $\angle COE=\angle BOC+\angle BOE=54^\circ+18^\circ=\boxed{72^\circ}$, $\angle AOE=180^\circ-\angle BOE=180^\circ-18^\circ=162^\circ$.\\
Since OF bisects $\angle AOE$,\\
it follows that $\angle EOF=\frac{1}{2}\angle AOE=81^\circ$\\
thus $\angle DOF=\angle EOF-\angle DOE=81^\circ-18^\circ=\boxed{63^\circ}$


Question: As shown in the figure, $\angle AOB$ is a right angle, $OP$ bisects $\angle AOB$, $OQ$ bisects $\angle AOC$, $\angle POQ=70^\circ$, find the degree of $\angle AOC$.

Solution:
\textbf{Solution:}\\
Since $\angle AOB=90^\circ$ and OP bisects $\angle AOB$,\\
it follows that $\angle POA=45^\circ$,\\
Since $\angle POQ=70^\circ$,\\
it results in $\angle AOQ=\angle POQ-\angle POA=25^\circ$,\\
Since OQ bisects $\angle AOC$,\\
it concludes that $\angle AOC=2\angle AOQ=\boxed{50^\circ}$.


Question: As shown in the figure, it is known that the straight lines AB and CD intersect at point O, $\angle COE$ is a right angle, $OF$ bisects $\angle AOE$, $\angle COF=34^\circ$, find the degree measure of $\angle BOD$.

Solution:
\textbf{Solution:} Given that $\angle COE$ is a right angle, and $\angle COF=34^\circ$,\\
$\therefore \angle EOF=90^\circ-34^\circ=56^\circ$,\\
Since $OF$ bisects $\angle AOE$,\\
$\therefore \angle AOF=\angle EOF=56^\circ$,\\
Given that $\angle COF=34^\circ$,\\
$\therefore \angle AOC=56^\circ-34^\circ=22^\circ$,\\
Since $\angle AOC$ and $\angle BOD$ are vertically opposite angles,\\
$\therefore \angle BOD=\angle AOC=\boxed{22^\circ}$.


Question: As shown in the figure, lines AB and CD intersect at point O, $\angle COE$ is a right angle, $OF$ bisects $\angle AOE$, $\angle COF=34^\circ$, what are the measures of $\angle AOC$ and $\angle BOD$?

Solution:
\textbf{Solution:} Since $\angle COE=90^\circ$ and $\angle COF=34^\circ$,\\
it follows that $\angle EOF=\angle COE-\angle COF=56^\circ$,\\
Given that OF is the bisector of $\angle AOE$,\\
it implies $\angle AOE=2\angle EOF=112^\circ$,\\
thus, $\angle AOC=112^\circ-90^\circ=\boxed{22^\circ}$,\\
Since $\angle BOD$ and $\angle AOC$ are vertical angles,\\
we conclude that $\angle BOD=\boxed{22^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ and $AE$ are respectively the altitude and the angle bisector, $\angle B=33^\circ$, $\angle C=67^\circ$. What is the measure of $\angle EAD$ in degrees?

Solution:
\textbf{Solution:} Given $\angle B=33^\circ$ and $\angle C=67^\circ$,\\
it follows that $\angle BAC=80^\circ$. Since AE is the bisector of $\angle BAC$,\\
it follows that $\angle BAE=40^\circ$ and thus $\angle AED=\angle BAE+\angle B=33^\circ+40^\circ=73^\circ$.\\
Since AD is the altitude of the triangle,\\
it follows that $\angle EAD=90^\circ-73^\circ=\boxed{17^\circ}$.


Question: As shown in the figure, it is known that $\angle AOD$ and $\angle BOC$ are both right angles, $\angle AOC = 38^\circ$, and $OE$ bisects $\angle BOD$. What is the measure of $\angle COE$ in degrees?

Solution:
\textbf{Solution}: Since $\angle AOD$ is a right angle and $\angle AOC=38^\circ$,\\
it follows that $\angle COD=\angle AOD-\angle AOC=90^\circ-38^\circ=52^\circ$.\\
Since $\angle BOC$ is also a right angle,\\
we have $\angle BOD=\angle BOC-\angle COD=90^\circ-52^\circ=38^\circ$.\\
Given that OE bisects $\angle BOD$,\\
it follows that $\angle DOE= \frac{1}{2} \angle BOD= \frac{1}{2} \times 38^\circ=19^\circ$.\\
Therefore, $\angle COE=\angle COD+\angle DOE=52^\circ+19^\circ=\boxed{71^\circ}$.


Question: As shown in the figure, the lines AB and CD intersect at O, and the ray OE divides $\angle BOD$ into two angles. Given that $\angle BOE = \frac{1}{3} \angle AOC$ and $\angle EOD = 36^\circ$, what is the measure of $\angle AOC$ in degrees?

Solution:
\textbf{Solution:} Because $\angle AOC$ and $\angle BOD$ are vertically opposite angles, \\
therefore $\angle BOD = \angle AOC$, \\
because $\angle BOE = \frac{1}{3} \angle AOC$, $\angle EOD = 36^\circ$, \\
therefore $\angle EOD = 2\angle BOE = 36^\circ$, \\
therefore $\angle EOD = 18^\circ$, \\
therefore $\angle AOC = \angle BOE + \angle EOD = 18^\circ + 36^\circ = \boxed{54^\circ}$.


Question: As shown in the figure, it is known that: $OA\perp OB$, $\angle BOC=50^\circ$, and $\angle AOD : \angle COD=4 : 7$. Draw the angle bisector OE of $\angle BOC$, and find the degree measure of $\angle DOE$.

Solution:
\textbf{Solution:} As shown in the figure:\\
Since OA$\perp$OB,\\
It follows that $\angle$AOB=$90^\circ$,\\
Since $\angle$AOD$\colon$ $\angle$COD=4$\colon$ 7,\\
Let $\angle$AOD=$4x^\circ$ and $\angle$COD=$7x^\circ$,\\
Since $\angle$AOB+$\angle$AOD+$\angle$COD+$\angle$BOC=$360^\circ$, and $\angle$BOC=$50^\circ$,\\
It follows that $90+7x+4x+50=360$,\\
Thus, $x=20$,\\
Therefore, $\angle$COD=$140^\circ$.\\
Since OE is the bisector of $\angle$BOC,\\
It follows that $\angle COE=\frac{1}{2}\angle$BOC=$25^\circ$,\\
Therefore, $\angle$DOE=$\angle$COD+$\angle$COE=\boxed{165^\circ}.


Question: As shown in the figure, it is known that $\angle BAE=\angle CAF=110^\circ$, $\angle CAE=60^\circ$, and $AD$ is the angle bisector of $\angle BAF$. What is the measure of $\angle BAD$?

Solution:
\textbf{Solution:} Since $\angle$BAE$=110^\circ$, $\angle$CAE$=60^\circ$,

therefore $\angle$BAC$=110^\circ-60^\circ=50^\circ$,

and since $\angle$CAF$=110^\circ$,

therefore $\angle$BAF$=110^\circ+50^\circ=160^\circ$,

and since AD is the angle bisector of $\angle$BAF,

therefore $\angle$BAD= $\frac{1}{2}$ $\angle$BAF= $\frac{1}{2}$ $\times160^\circ=\boxed{80^\circ}$.


Question: As shown in the figure, circle $O$ is the incircle of $\triangle ABC$, where $AB=7, BC=5$, $AC=8$. Determine the radius of the incircle.

Solution:
\textbf{Solution:} Construct BD $\perp$ AC at D, with tangency points E, F, and G, respectively. Connect OE, OF, OG,\\
Let AD = x, CD = 8−x, and the radius of the inscribed circle be r,\\
According to the Pythagorean theorem, $AB^{2}-AD^{2}=BC^{2}-CD^{2}$, that is $7^{2}-x^{2}=5^{2}-(8-x)^{2}$,\\
Solving the equation, we get $x=\frac{11}{2}$,\\
Therefore, BD = $\sqrt{AB^{2}-AD^{2}}=\sqrt{7^{2}-\left(\frac{11}{2}\right)^{2}}=\frac{5}{2}\sqrt{3}$,\\
Because circle O is the inscribed circle of $\triangle ABC$,\\
Therefore, OE $\perp$ AC, OF $\perp$ AB, OG $\perp$ BC, OE = OF = OG = r,\\
Therefore, $S_{\triangle ABC}=\frac{1}{2}AC\cdot BD=\frac{1}{2}AB\cdot OF+\frac{1}{2}BC\cdot OG+\frac{1}{2}AC\cdot OE=\frac{1}{2}(AB+BC+AC)\cdot r$,\\
Therefore, AC$\cdot$BD = $(AB+BC+AC)\cdot r$,\\
Therefore, $r=\frac{AC\cdot BD}{AB+BC+AC}=\frac{8\times \frac{5}{2}\sqrt{3}}{20}=\boxed{\sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=48^\circ$, the angle bisectors of the exterior angles $\angle DAC$ and $\angle ACF$ intersect at point E. What is the measure of $\angle AEC$ in degrees?

Solution:
Solution: Since the bisectors of exterior angles $\angle DAC$ and $\angle ACF$ of the triangle intersect at point E, \\
it follows that $\angle CAE + \angle ACE = \frac{1}{2}(\angle B + \angle ACB) + \frac{1}{2}(\angle B + \angle BAC)\\
= \frac{1}{2}(\angle BAC + \angle B + \angle ACB + \angle B)\\
= \frac{1}{2}(180^\circ + 48^\circ)\\
= 114^\circ$\\
In $\triangle ACE$, $\angle AEC = 180^\circ - (\angle CAE + \angle ACE)\\
= 180^\circ - 114^\circ\\
= \boxed{66^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $\angle B=42^\circ$, $\angle C=68^\circ$, $AD$ is the altitude of $\triangle ABC$, $AE$ is the angle bisector of $\triangle ABC$. Find the degree measure of $\angle EAD$.

Solution:
\textbf{Solution:} Given that $\angle B=42^\circ$ and $\angle C=68^\circ$, \\
it follows that $\angle BAC=180^\circ-68^\circ-42^\circ=70^\circ$\\
Since AE is the angle bisector of $\triangle ABC$,\\
it follows that $\angle CAE=70^\circ \div 2=35^\circ$\\
Since AD is the altitude of $\triangle ABC$,\\
it follows that $\angle ADB=90^\circ$\\
Therefore, $\angle DAC=90^\circ-\angle C=90^\circ-68^\circ=22^\circ$\\
Therefore, $\angle EAD=\angle EAC-\angle DAC=35^\circ-22^\circ=\boxed{13^\circ}$,\\
that is, the measure of $\angle EAD$ is $13^\circ$


Question: As shown in the figure, \(AD\) is the angle bisector of \( \triangle ABC\), \(CE\) is the height of \( \triangle ABC\), \(\angle BAC=50^\circ\), \(\angle BCE=25^\circ\), what is the degree measure of \( \angle ADB\)?

Solution:
\textbf{Solution:} Given that $AD$ is the angle bisector of $\triangle ABC$, and $\angle BAC=50^\circ$,\\
therefore $\angle BAD=25^\circ$,\\
also given that $CE$ is the altitude of $\triangle ABC$, and $\angle BCE=25^\circ$,\\
therefore $\angle BEC=90^\circ$,\\
thus we get $\angle B=65^\circ$,\\
hence, $\angle ADB=180^\circ-\angle B-\angle BAD=180^\circ-65^\circ-25^\circ=\boxed{90^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $CA=CB$, $D$ is the midpoint of $AB$, $\angle B=42^\circ$, what is the measure of $\angle ACD$ in degrees?

Solution:
\textbf{Solution:} Given that $CA=CB$, \\
it follows that $\triangle ABC$ is an isosceles triangle, \\
since $\angle B=42^\circ$, \\
then $\angle A=\angle B=42^\circ$, \\
thus $\angle ACB=96^\circ$, \\
furthermore, since D is the midpoint of AB, meaning CD bisects the base AB, \\
it implies that CD bisects $\angle ACB$, \\
therefore $\angle ACD= \frac{1}{2}\angle ACB=\boxed{48^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, we have $AB=AC$, and the points $D$, $A$, and $E$ all lie on the straight line $m$. Moreover, it is given that $\angle BDA=\angle AEC=\angle BAC=\alpha$. If $DE=10$ and $BD=3$, find the length of $CE$.

Solution:
\textbf{Solution:} Since $\angle AEC = \angle BAC = \alpha$,\\
it follows that $\angle ECA + \angle CAE = 180^\circ - \alpha$,\\
and $\angle BAD + \angle CAE = 180^\circ - \alpha$,\\
hence $\angle ECA = \angle BAD$,\\
In $\triangle BAD$ and $\triangle ACE$,\\
we have $\left\{ \begin{array}{l} \angle BDA = \angle AEC \\ \angle BAD = \angle ACE \\ AB = AC \end{array} \right.$\\
Therefore, $\triangle BAD \cong \triangle ACE$ (AAS),\\
which implies CE = AD, and AE = BD = 3,\\
Since DE = AD + AE = 10,\\
it follows that AD = DE - AE = DE - BD = 10 - 3 = 7.\\
Therefore, CE = \boxed{7}.


Question: As shown in the figure, $CD$ is the angle bisector of $\triangle ABC$, $\angle 1 = \angle 2$, $\angle B = 30^\circ$. What is the measure of $\angle ACB$?

Solution:
\textbf{Solution:} Let $\angle ACB=2\alpha$,\\
since CD is the angle bisector of $\triangle ABC$,\\
then $\angle BCD=\angle ACD= \frac{1}{2} \angle ACB=\alpha$,\\
thus $\angle 1=\angle B+\angle BCD=30^\circ+\alpha$,\\
since $\angle 1=\angle 2$,\\
thus $\angle 2=30^\circ+\alpha$,\\
since $\angle 2+\angle B+\angle ACB=180^\circ$,\\
thus $30^\circ+\alpha +30^\circ+2\alpha =180^\circ$,\\
thus $\alpha =40^\circ$,\\
thus $\angle ACB=\boxed{80^\circ}$.


Question: As shown in the figure, ship A leaves the port at a speed of 16 nautical miles per hour, heading southeast. At the same time and from the same location, ship B heads southwest. It is known that one and a half hours after leaving the port, they reach points B and A respectively. Given that the distance between points A and B is 30 nautical miles, how many nautical miles does ship B travel per hour?

Solution:
\textbf{Solution:} As shown in the diagram: $OA = 16 \times 1.5 = 24$ (nautical miles),\\
$AB = 30$ (nautical miles),\\
By the Pythagorean theorem, we have: $OB = \sqrt{30^2 - 24^2} = 18$ (nautical miles),\\
$18 \div 1.5 = \boxed{12}$ (nautical miles/hour),\\
Thus, the ship B travels $12$ nautical miles per hour.


Question: As shown in the figure, $AD$ is the altitude of $\triangle ABC$, $BE$ bisects $\angle ABC$ and intersects $AD$ at $E$. If $\angle C=70^\circ$ and $\angle BED=68^\circ$, what is the measure of $\angle BAC$?

Solution:
\textbf{Solution:} Given that $AD$ is the height of $\triangle ABC$, and $\angle BED=68^\circ$,
\[
\therefore \angle EBD=90^\circ-\angle BED=90^\circ-68^\circ=22^\circ,
\]
Since $BE$ bisects $\angle ABC$ and intersects $AD$ at E,
\[
\therefore \angle ABC=2\angle EBD=2\times 22^\circ=44^\circ,
\]
Given that $\angle C=70^\circ$,
\[
\therefore \angle BAC=180^\circ-\angle ABC-\angle C=180^\circ-44^\circ-70^\circ=\boxed{66^\circ}.
\]


Question: As shown in the figure, points A, D, B, E are on the same line, $AC=EF$, $AD=BE$, $\angle C=\angle F=90^\circ$, $\angle ABC=65^\circ$, find the degree of $\angle ADF$.

Solution:
\textbf{Solution:} Given that $AD = BE$,\\
it follows that $AD + BD = BE + BD$, which means $AB = DE$,\\
furthermore, since $AC = EF$,\\
we have Rt$\triangle ABC \cong$ Rt$\triangle EDF$ (by HL criterion),\\
which implies $\angle FDE = \angle ABC = 65^\circ$,\\
therefore, $\angle ADF = 180^\circ - \angle FDE = 180^\circ - 65^\circ = \boxed{115^\circ}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $\angle B=90^\circ$, $\angle BCA=60^\circ$, $AC=2\sqrt{2}$, $DA=1$, $CD=3$. Find the area of quadrilateral $ABCD$.

Solution:
\textbf{Solution:}

Given $\because$ $\angle B=90^\circ$, $\angle BCA=60^\circ$, $\therefore$ $\angle A=30^\circ$, $\therefore$ BC=$\frac{1}{2}$AC=$\sqrt{2}$, AB=$\sqrt{6}$, $\because$ AC=2$\sqrt{2}$, DA=1, CD=3, $\therefore$ AC$^{2}$+DA$^{2}$=9=CD$^{2}$, $\therefore$ $\angle CAD=90^\circ$, $\therefore$ $S_{\text{quadrilateral} ABCD}=S_{\triangle ABC}+S_{\triangle ACD}$, =$\frac{1}{2}\cdot$AB$\cdot$BC+$\frac{1}{2}\cdot$AC$\cdot$DA, =$\frac{1}{2}\times\sqrt{6}\times\sqrt{2}+\frac{1}{2}\times2\sqrt{2}\times1$, =$\boxed{\sqrt{2}+\sqrt{3}}$.


Question: As shown in the figure, in the right-angled triangle $ABC$, $\angle BAC=90^\circ$, $AB=AC$, perpendicular lines $BD$ and $CE$ are drawn from points $B$ and $C$ respectively, through point $A$, with the feet of the perpendiculars being $D$ and $E$. If $BD=4\,\text{cm}$ and $CE=3\,\text{cm}$, what is the length of $DE$ in centimeters?

Solution:
\textbf{Solution:} Since $BD \perp AD$ and $CE \perp AD$,\\
it follows that $\angle D = \angle E = 90^\circ$,\\
hence $\angle DAB + \angle ABD = 90^\circ$,\\
Since $\angle BAC = 90^\circ$,\\
it follows that $\angle DAB + \angle CAE = 90^\circ$,\\
hence $\angle ABD = \angle CAE$,\\
Since $AB = AC$,\\
it follows that $\triangle DAB \cong \triangle ECA$ (AAS),\\
hence $BD = AE = 4\,\text{cm}, AD = CE = 3\,\text{cm}$,\\
therefore $DE = AE + AD = \boxed{7\,\text{cm}}$.


Question: As shown in the figure, in the pentagon $ABCDE$, $AB \perp BC$, $AE \perp DE$, $AB=1$, $BC=2$, $CD=2\sqrt{5}$, $DE=3$, $AE=4$. Find the area of the pentagon $ABCDE$.

Solution:
\textbf{Solution:} Draw $AC$ and $AD$,\\
since $AB\perp BC$ and $AE\perp DE$,\\
thus $\angle B=90^\circ$ and $\angle E=90^\circ$,\\
hence $AB^2+BC^2=AC^2$ and $AE^2+DE^2=AD^2$,\\
since $AB=1$, $BC=2$, $DE=3$, $AE=4$,\\
thus $1^2+2^2=AC^2$ and $4^2+3^2=AD^2$,\\
therefore $AC=\sqrt{5}$ and $AD=5$,\\
since $CD=2\sqrt{5}$,\\
thus $AC^2+CD^2=AD^2$,\\
thus $\angle ACD=90^\circ$,\\
thus the area of pentagon $ABCDE$ $=S_{\triangle ABC}+S_{\triangle ADE}+S_{\triangle ACD}=\frac{1}{2}\cdot AB\cdot BC+\frac{1}{2}\cdot AE\cdot DE+\frac{1}{2}\cdot AC\cdot CD$\\
$=\frac{1}{2}\times 1\times 2+\frac{1}{2}\times 4\times 3+\frac{1}{2}\times \sqrt{5}\times 2\sqrt{5}=1+6+5=\boxed{12}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle B=50^\circ$. CD is the altitude of $\triangle ABC$, and CE is the angle bisector of $\triangle ABC$. Find the measure of $\angle DCE$.

Solution:
Solution: Since $CD$ is the height of $\triangle ABC$,\\
hence $\angle CDB = 90^\circ$,\\
thus $\angle DCB = 90^\circ - \angle B = 90^\circ - 50^\circ = 40^\circ$;\\
Since $CE$ bisects $\angle ACB$,\\
thus $\angle ECB = \frac{1}{2} \angle ACB = \frac{1}{2} \times 90^\circ = 45^\circ$,\\
therefore $\angle DCE = \angle ECB - \angle DCB = 45^\circ - 40^\circ = \boxed{5^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=82^\circ$, $\angle C=58^\circ$, $BD\perp AC$ at $D$, $AE$ bisects $\angle CAB$, and $BD$ intersects $AE$ at point $F$. What is $\angle AFB$?

Solution:
\textbf{Solution:} Given: $\because \angle CAB = 180^\circ - \angle ABC - \angle C$,\\
and $\angle ABC = 82^\circ, \angle C = 58^\circ$,\\
$\therefore \angle CAB = 40^\circ$,\\
$\because$ AE bisects $\angle CAB$,\\
$\therefore \angle DAF = 20^\circ$,\\
$\because$ BD $\perp$ AC at D,\\
$\therefore \angle ADB = 90^\circ$,\\
$\therefore \angle AFB = \angle ADB + \angle DAF = 90^\circ + 20^\circ = \boxed{110^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ is the altitude on side $BC$, and $AE$ bisects $\angle BAC$. If $\angle BAC : \angle B : \angle C = 4 : 3 : 2$, find the measure of $\angle DAE$?

Solution:
\textbf{Solution:} In $\triangle ABC$,

since $\angle BAC : \angle B : \angle C = 4 : 3 : 2$,

thus $\angle BAC = 180^\circ \times \frac{4}{9} = 80^\circ$, $\angle B = 180^\circ \times \frac{3}{9} = 60^\circ$,

since $AD$ is the altitude to side $BC$,

thus $\angle ADB = 90^\circ$,

thus $\angle BAD = 90^\circ - \angle B = 30^\circ$,

since $AE$ bisects $\angle BAC$,

thus $\angle BAE = \frac{1}{2} \angle BAC = 40^\circ$,

thus $\angle DAE = \angle BAE - \angle BAD = \boxed{10^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $CD$ is the angle bisector of $\angle ACB$, $DE \parallel BC$, $\angle A=65^\circ$, $\angle B=35^\circ$, what is the measure of $\angle EDC$?

Solution: \textbf{Solution:} In $\triangle ABC$, since $\angle A=65^\circ$ and $\angle B=35^\circ$, it follows that $\angle ACB=180^\circ-\angle A-\angle B=180^\circ-65^\circ-35^\circ=80^\circ$. Since $CD$ is the angle bisector of $\angle ACB$, it follows that $\angle BCD=\frac{1}{2}\angle ACB=\frac{1}{2}\times 80^\circ=40^\circ$. Since $DE\parallel BC$, it follows that $\angle EDC=\angle BCD=\boxed{40^\circ}$.


Question: As shown in the figure, $AB=10$, point C is the midpoint of segment AB, and point D is the midpoint of segment CB. Find the length of AD.

Solution:
\textbf{Solution:} Since point C is the midpoint of the line segment AB, \\
it follows that $AC=BC=\frac{1}{2}AB$, \\
since $AB=10$, \\
it follows that $AC=BC=\frac{1}{2}\times 10=5$. \\
Since point D is the midpoint of the line segment BC, \\
it follows that $CD=\frac{1}{2}BC$, \\
therefore $CD=\frac{1}{2}\times 5=\frac{5}{2}$. \\
Thus, $AD=AC+CD=5+\frac{5}{2}=\boxed{\frac{15}{2}}$.


Question: As shown in the figure, the length of line segment $AB=10\, \text{cm}$. Extend $AB$ to point $C$ such that $BC=6\, \text{cm}$. Let points $M$ and $N$ be the midpoints of $AC$ and $BC$, respectively. Find the lengths of the line segments $BM$ and $MN$.

Solution:
\textbf{Solution:} Given $\because$ AB=10, BC=6\\
$\therefore$ AC=16\\
Also, $\because$ M is the midpoint of AC, $\therefore$ MC=AM=8\\
$\because$ N is the midpoint of BC, $\therefore$ BN=NC=3\\
BM=AB−AM=10−8=\boxed{2}\\
MN=BM+BN=2+3=\boxed{5}


Question: As shown in the figure, a rectangle $ABCD$ with a length of $5\,cm$ and a width of $3\,cm$ is first translated $2\,cm$ to the right and then $1\,cm$ downwards to obtain rectangle $A'B'C'D'$. What is the area of the shaded part in $cm^{2}$?

Solution:
\textbf{Solution:} It can be inferred from the problem that the blank part is a rectangle, with a length of $ \left(5-2\right)cm=3cm$ and a width of $ \left(3-1\right)cm=2cm$, \\
$\therefore$ the area of the shaded part $ =5\times 3\times 2-2\times 3\times 2=\boxed{18cm^{2}}$;


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=5$, and $BC=12$. If $\triangle ABC$ is folded along $AD$ such that point $C$ coincides with point $E$ on $AB$, what is the length of $DB$?

Solution:
\textbf{Solution}: By the property of folding, we have: $AE=AC=5$, $DC=DE$, and $\angle AED=\angle BED=\angle C=90^\circ$.\\
Since $\angle C=90^\circ$ and $AC=5$, $BC=12$,\\
thus $AB=\sqrt{AC^2+BC^2}=13$,\\
and hence $BE=AB-AE=8$.\\
Let $DC=x$, then $DE=x$, and $BD=BC-DC=12-x$.\\
In $\triangle BDE$, by the Pythagorean theorem, we have $DE^2+BE^2=BD^2$,\\
thus $x^2+8^2=(12-x)^2$,\\
solving this, we find $x=\frac{10}{3}$.\\
Therefore, $BD=12-x=\boxed{\frac{26}{3}}$.


Question: As shown in the figure, if $CB=4\,\text{cm}$, $DB=7\,\text{cm}$, and $D$ is the midpoint of $AC$, find the length of $AB$.

Solution:
\textbf{Solution:} Since $CB = 4$cm, $DB = 7$cm,\\
thus $DC = DB - CB = 3$cm.\\
Also, since $D$ is the midpoint of $AC$,\\
thus $AD = DC = 3$cm,\\
therefore $AB = AD + DB = \boxed{10cm}$.


Question: As shown in the figure, the diagonals $AC, BD$ of the rhombus $ABCD$ intersect at point $O$. If $AC=4, BD=6$, what is the length of $AB$?

Solution:
\textbf{Solution:} Since rhombus $ABCD$ has diagonals $AC, BD$ intersecting at point $O$, and $AC=4, BD=6$,\\
it follows that $AC\perp BD$, $AO=\frac{1}{2}AC=2$, $BO=\frac{1}{2}BD=3$,\\
In $\triangle AOB$, $AB^2=AO^2+BO^2$,\\
thus $AB=\sqrt{AO^2+BO^2}=\sqrt{2^2+3^2}=\boxed{\sqrt{13}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB = AC = 10$. Point $P$ is on $BC$. Parallel lines to $AC$ and $AB$ are drawn through point $P$, intersecting $AB$ and $AC$ at points $M$ and $N$ respectively. What is the perimeter of the quadrilateral $AMPN$?

Solution:
\textbf{Solution:} Since $AB=AC$, \\
it follows that $\angle B=\angle C$, \\
Since $PM\parallel AC$ and $PN\parallel AB$, \\
it implies that quadrilateral AMPN is a parallelogram, and $\angle MPB=\angle C$, \\
Hence, $AM=PN$, $MP=AN$, and $\angle B=\angle MPB$, \\
Therefore, $MB=MP$, \\
Consequently, the perimeter of quadrilateral AMPN is $AM+MP+PN+AN=2(AM+MP)=2(AM+MB)=2AB=\boxed{20}$


Question: In the figure, within $\triangle ABC$, points $D$ and $E$ lie on sides $AB$ and $AC$ respectively, with $DE\parallel BC$. Given that $AB=3$, $AC=4$, and $EC=1$, determine the length of $AD$.

Solution:
\textbf{Solution:} \\
Since $DE\parallel BC$, \\
it follows that $\frac{DB}{AB}=\frac{EC}{AC}$, that is $\frac{DB}{3}=\frac{1}{4}$, \\
therefore, $DB=\frac{3}{4}$, \\
thus, $AD=AB−DB=3−\frac{3}{4}=\boxed{\frac{9}{4}}$.


Question: As illustrated in the diagram, a street lamp (point O) is 6.4m above the ground. Xiao Ming, who is 1.6m tall, walks from point A, which is 9m away from the base of the street lamp (point P), along a straight line in the direction of AP to point D. When he stops, the length of his shadow under the street lamp has reduced by 1.8m. Calculate the distance Xiao Ming has walked.

Solution:
\textbf{Solution:} From the given information, we have $OP=64m$, $AB=DE=1.6m$, $AP=9m$, $DF=AC-1.8$,\\
$\because$ $\angle OPC=\angle BAC$, $\angle BCA=\angle OCP$,\\
$\therefore$ $\triangle BCA\sim\triangle OCP$,\\
$\therefore$ $\frac{BA}{OP}=\frac{AC}{PC}$, that is $\frac{1.6}{6.4}=\frac{AC}{AC+9}$,\\
$\therefore$ $AC=3$,\\
$\therefore$ $DF=AC-1.8=3-1.8=1.2\text{(m)}$,\\
$\because$ $\angle OPC=\angle EDF=90^\circ$, $\angle EFD=\angle OFP$,\\
$\therefore$ $\triangle EFD\sim\triangle OFP$,\\
$\therefore$ $\frac{DE}{OP}=\frac{DF}{PF}$, that is $\frac{1.6}{6.4}=\frac{1.2}{PF}$,\\
$\therefore$ $PF=4.8$,\\
$\therefore$ $AD=AP-PF+DF=9-4.8+1.2=\boxed{5.4}\text{(m)}$,\\


Question: As shown in the figure, in parallelogram $ABCD$, $AB=4$, $BC=2$, point $F$ is on the extension of $BC$, $AF$ intersects $CD$ at point $E$, and $\angle 1=\angle F$, find the lengths of $CF$ and $DE$.

Solution:
\textbf{Solution:} Since $\angle B = \angle B$ and $\angle 1 = \angle F$,\\
it follows that $\triangle ABC \sim \triangle FBA$,\\
therefore, $\frac{AB}{BF} = \frac{BC}{AB}$,\\
thus, BF = $\frac{4 \times 4}{2} = 8$,\\
hence, CF = BF - BC = 8 - 2 = \boxed{6},\\
In the parallelogram ABCD,\\
thus, AD $\parallel$ BF, CD = AB = 4, AD = BC = 2,\\
therefore, $\triangle ADE \sim \triangle FCE$,\\
thus, $\frac{DE}{CE} = \frac{AD}{CF}$,\\
therefore, $\frac{DE}{4 - DE} = \frac{2}{6}$,\\
thus, DE = \boxed{1}.


Question: As shown in the figure, in rectangle $ABCD$, $E$ is the midpoint of $BC$, $DF \perp AE$ with the foot of the perpendicular being $F$. If $AB = 4$ and $BC = 6$, find the length of $DF$.

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle, \\
$\therefore AD=BC$, $AD\parallel BC$, $\angle B=90^\circ$, \\
Since $BC=AD=6$, $E$ is the midpoint of $BC$, \\
$\therefore BE=3$, \\
Since $\angle B=90^\circ$, \\
$\therefore AE=\sqrt{BE^2+AB^2}=\sqrt{3^2+4^2}=5$, \\
Since $AD\parallel BC$, \\
$\therefore \angle DAF=\angle AEB$, \\
Since $DF\perp AE$, \\
$\therefore \angle AFD=90^\circ=\angle B$, \\
$\therefore \triangle ADF\sim \triangle EAB$. \\
$\therefore \frac{DF}{AB}=\frac{AD}{AE}$, \\
That is $\frac{DF}{4}=\frac{6}{5}$, \\
We find that: $DF=\boxed{\frac{24}{5}}$.


Question: Given the diagram, in $\triangle ABC$, $AB=AC$. Point $D$ is on $AC$, and satisfies $AD^{2}=CD\cdot AC$. If $AD=BC$, by connecting $BD$, what is the measure of $\angle ABD$ in degrees?

Solution:
\textbf{Solution:} Since $AB=AC$,\\
it follows that $\angle ABC=\angle ACB$,\\
Since $AD^{2}=CD\cdot AC$, $AD=BC$,\\
it follows that $BC^{2}=CD\cdot AC$,\\
which means $\frac{CB}{CD}=\frac{AC}{CB}$,\\
Also, since $\angle C=\angle C$,\\
it follows that $\triangle CBA\sim \triangle CDB$,\\
thus, $\angle BDC=\angle C=\angle ABC$, $\angle A=\angle DBC$,\\
it follows that $BD=CB$,\\
thus, $DB=DA$,\\
it follows that $\angle A=\angle ABD$,\\
Let $\angle A=\angle ABD =x^\circ$,\\
it follows that $\angle BDC=\angle A+\angle ABD=2x^\circ$,\\
it follows that $\angle ABC=\angle C=2\angle ABD=2x^\circ$,\\
In $\triangle ABC$, by the triangle angle sum theorem,\\
$x+2x+2x=180$,\\
Solving, we find $x=\boxed{36}$,\\
thus, $\angle ABD=36^\circ.$


Question: As shown in the figure, in $\triangle ABC$, $D$, $E$, and $F$ are points on $AB$, $BC$ respectively, and $DE\parallel AC$, $AE\parallel DF$. Given that $BD: AD=3:2$ and $BF=6$, find the lengths of $EF$ and $FC$.

Solution:
\textbf{Solution:} Since $AE\parallel DF$, \\
then $\frac{BD}{AD}=\frac{BF}{EF}$, that is $\frac{6}{EF}=\frac{3}{2}$, \\
therefore $EF=\boxed{4}$, \\
thus $BE=BF+EF=6+4=10$, \\
since $DE\parallel AC$, \\
then $\frac{BD}{AD}=\frac{BE}{CE}$, that is $\frac{10}{CE}=\frac{3}{2}$, \\
therefore $CE=\frac{20}{3}$, \\
thus $CF=CE+EF=\boxed{\frac{32}{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $\sin B= \frac{5}{13}$, D is on side $BC$, and $\angle ADC=45^\circ$, $AC=5$. Find the tangent value of $\angle BAD$.

Solution:
\textbf{Solution:} Since $\angle C=90^\circ$ and $\angle ADC=45^\circ$ with $AC=5$,\\
it follows that $AC=CD=5$ and $AD=5\sqrt{2}$,\\
Since $\sin B= \frac{5}{13}$,\\
then $AB=\frac{AC}{\sin B}=13$,\\
Since $\angle C=90^\circ$ and $CD=5$,\\
then $BC=12$ and $BD=7$,\\
Draw $BE\perp AD$ to extend $AD$ at $E$,\\
Since $\angle BDE=\angle ADC=45^\circ$,\\
it follows that $BE=DE=\frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2}$,\\
Therefore, $AE=AD+DE=5\sqrt{2}+\frac{7\sqrt{2}}{2}=\frac{17\sqrt{2}}{2}$,\\
Thus, $\tan \angle BAD = \boxed{\frac{7}{17}}$.


Question: As shown in the figure, in $ \triangle ABC$, $ AB=AC=5$, $ BC=6$, find the value of $\cos A$.

Solution:
\textbf{Solution:} Construct $AE\perp BC$ at $E$, and construct $BD\perp AC$ at $D$,\\
$\therefore$ $AB=AC=5$, $BC=6$,\\
$\therefore$ $BE=\frac{1}{2}BC=3$,\\
$\therefore$ $AE=\sqrt{AB^{2}-BE^{2}}=4$,\\
$\therefore$ the area of triangle $ABC =\frac{1}{2}AC\cdot BD=\frac{1}{2}BC\cdot AE$,\\
$\therefore$ $BD=\frac{BC\cdot AE}{AC}=\frac{6\times 4}{5}=\frac{24}{5}$,\\
$\therefore$ $AD=\sqrt{AB^{2}-BD^{2}}=\sqrt{5^{2}-\left(\frac{24}{5}\right)^{2}}=1.4$,\\
$\therefore$ $\cos A=\frac{AD}{AB}=\frac{1.4}{5}=\boxed{\frac{7}{25}}$.


Question: As shown in the figure, the slope of a river embankment $BC=12\,m$. In order to reinforce the embankment, it is necessary to thicken the dam. After completion, the slope ratio changes from the original $1\colon 2$ to $1\colon 3$. What is the length of the reinforced slope $AD$ in meters?

Solution:
\textbf{Solution:} Construct $CE\perp AB$ through point C, and $DF\perp AB$ through point D, with the respective perpendicular feet being E and F.\\
Let $CE=x$, then $BE=2x$, $DF=CE=x$, $AF=3x$,\\
since in $\triangle CEB$, $\angle BEC=90^\circ$, and $BC=12$,\\
thus $x^2+(2x)^2=12^2$, yielding: $x=\frac{12\sqrt{5}}{5}$.\\
Since in $\triangle ADF$, $\angle AFD=90^\circ$,\\
thus $AD=\sqrt{AF^2+DF^2}=\sqrt{10}x=\sqrt{10}\times \frac{12\sqrt{5}}{5}=12\sqrt{2}$ m.\\
Answer: The reinforced slope $AD$ is $\boxed{12\sqrt{2}}$ meters.


Question: In the figure, in $ \triangle ABC$, $ \angle BAC=90^\circ$, AD is the altitude to side BC, if $\sin\angle CAD=\frac{3}{5}$ and $ BC=25$, find the length of AC.

Solution:
\textbf{Solution:} According to the problem, $\because$ $\angle BAC=90^\circ$ and AD is the height from A to side BC, $\therefore$ $\angle BAD+\angle CAD=90^\circ$ and $\angle BAD+\angle B=90^\circ$, $\therefore$ $\angle CAD=\angle B$, $\therefore$ $\sin B=\sin CAD=\frac{3}{5}$. In $Rt\triangle ABC$ with $\angle BAC=90^\circ$, $\therefore$ $\sin B=\frac{AC}{BC}=\frac{AC}{25}=\frac{3}{5}$,\\
$\therefore$ $AC=\boxed{15}$.


Question: As shown in the figure, there is a billboard AB on top of the building. From point D, which is 12 meters away from the building BC, the angle of elevation to the top A of the billboard is $37^\circ$, and the angle of elevation to the bottom B of the billboard is $30^\circ$. Calculate the height of the billboard AB. (Leave the result in square root form, reference data: $\sin 37^\circ \approx 0.60$, $\cos 37^\circ \approx 0.80$, $\tan 37^\circ \approx 0.75$)

Solution:
\textbf{Solution:} According to the problem, we know\\
$\angle ADC=37^\circ$, $\angle BDC=30^\circ$, $CD=12$ meters, $AC\perp DC$.\\
In $\triangle DCA$, $\tan\angle ADC=\frac{AC}{CD}$, thus\\
$AC=CD\cdot \tan\angle ADC=12\times \tan37^\circ\approx 12\times 0.75=9$ meters.\\
In $\triangle DBC$, $\tan\angle BDC=\frac{BC}{CD}$, thus\\
$BC=CD\cdot \tan\angle BDC=12\times \tan30^\circ=12\times \frac{\sqrt{3}}{3}=4\sqrt{3}$ meters.\\
Therefore, $AB=AC−BC=9−4\sqrt{3}$.\\
Answer: The height of the billboard $AB$ is $\boxed{9−4\sqrt{3}}$ meters.


Question: As shown in the diagram, there is a slope behind a teaching building in the mountainous area, where the slope surface BC is parallel to the ground AD. The slope ratio of the slope AB is $i=1\colon \frac{5}{12}$, and AB=26 meters. To prevent landslides and ensure safety, the school decides to modify this slope. According to geological surveys, when the slope angle does not exceed $53^\circ$, the stability of the slope against landslides can be ensured. To eliminate safety hazards, the school plans to modify the slope AB to AF (as shown in the diagram). What is the minimum length of BF? (The result should be accurate to 1 meter, reference data: $\sin{53^\circ} \approx 0.8$, $\cos{53^\circ} \approx 0.6$, $\tan{53^\circ} \approx 1.33$)

Solution:
\textbf{Solution:} Construct FH $\perp$ AD at H,\\
then quadrilateral FHEB is a rectangle,\\
$\therefore$ FH = BE, BF = HE,\\
$\because$ the slope ratio of slope AB is $i=1\colon \frac{5}{12}$,\\
$\therefore$ BE : AE = 12 : 5,\\
Let BE = 12x meters, then AE = 5x meters,\\
In $\triangle ABE$, $AB^{2}$ = $AE^{2}$ + $BE^{2}$, that is $26^{2}$ = $(12x)^{2}$ + $(5x)^{2}$,\\
Solving gives: $x_{1}$ = 2, $x_{2}$ = -2 (discard),\\
Then AE = 10 meters, BE = FH = 24 meters,\\
In $\triangle FAH$, $\tan\angle FAH$ = $\frac{FH}{AH}$,\\
$\therefore$ AH = $\frac{FH}{\tan\angle FAH}$ $\approx$ $\frac{24}{1.33} \approx 18$ (meters),\\
$\therefore$ BF = HE = AH - AE = 18 - 10 = \boxed{8} (meters),\\
Answer: BF is at least 8 meters.


Question: As shown in the figure, AB is the diameter of $\odot O$, and chord CD $\perp$ AB at point E. If AB $=$ 10, BE $=$ 2, what is the length of chord CD?

Solution:
\textbf{Solution:} Draw line $OC$ as shown:\\
Since $AB$ is the diameter of $\odot O$ and $CD\perp AB$,\\
it follows that $CE=DE=\frac{1}{2}CD$, and $OC=OA=OB=5$,\\
thus $OE=OB-EB=5-2=3$,\\
in the right triangle $\triangle OCE$, by the Pythagorean theorem, we have: $CE=\sqrt{OC^{2}-OE^{2}}=\sqrt{5^{2}-3^{2}}=4$,\\
therefore, $CD=2CE=\boxed{8}$.


Question: In the diagram, within circle $\odot O$, radius $OC$ is perpendicular to chord $AB$ at $D$. Point $E$ is on circle $\odot O$, with $\angle E=22.5^\circ$, and $AB=2$. What is the length of the radius $OB$?

Solution:
\textbf{Solution:} Given that $\because$ radius $OC\perp$ chord AB at point $D$,\\
$\therefore \overset{\frown}{AC}=\overset{\frown}{BC}$,\\
$\therefore \angle E=\frac{1}{2}\angle BOC=22.5^\circ$,\\
$\therefore \angle BOD=45^\circ$,\\
$\therefore \triangle ODB$ is an isosceles right-angled triangle,\\
$\because AB=2$,\\
$\therefore DB=OD=1$,\\
$\therefore OB=\sqrt{1^2+1^2}=\boxed{\sqrt{2}}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in $\odot O$, with $E$ being a point on the extension of $BC$, and point $C$ being the midpoint of the arc $\overset{\frown}{BD}$. If $\angle DCE=110^\circ$, what is the degree measure of $\angle BAC$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is inscribed in circle O,\\
it follows that $\angle DCE=\angle BAD$ \\
Since $\angle DCE=110^\circ$ \\
then $\angle BAD=110^\circ$ \\
Given that point C is the midpoint of the arc $BD$\\
it follows that arc $BC=$ arc $DC$ \\
Therefore, $\angle BAC=\angle DAC=\frac{1}{2}\angle BAD=\frac{1}{2}\times 110^\circ=55^\circ$ \\
Hence, the measure of $\angle BAC$ is $\boxed{55^\circ}$


Question: Given that \(P(2, n)\) is a point on the graph of the inverse proportion function \(y= \frac{4}{x}\) (\(x>0\)). When the line \(y=-2x\) is translated to the right along the \(x\)-axis passing through point P, it intersects the \(x\)-axis at point Q. If point M is a moving point on the \(y\)-axis, find the minimum value of \(PM+QM\).

Solution:
\textbf{Solution:} Let us consider the diagram as referred,\\
since $P(2, n)$ is a point on $y= \frac{4}{x}$,\\
then $n= \frac{4}{2} = 2$, which means $P(2, 2)$.\\
Also, by translating the line $y= -2x$, the translated line becomes $y= -2x + b$,\\
thus $2 = -2 \times 2 + b$ gives $b = 6$, which means the translated line is $y= -2x + 6$.\\
Therefore, the reflection of $Q(3, 0)$ across the $y$-axis is $Q'(-3, 0)$,\\
hence $PQ'= \sqrt{5^2 + 2^2} = \sqrt{29}$,\\
therefore, the minimum value of $PM + QM$ is $\boxed{\sqrt{29}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, the perpendicular bisector $MN$ of $AB$ intersects $AC$ at point $D$ and intersects $AB$ at point $E$, $\angle A=36^\circ$, $AD=5$, find the degree measure of $\angle DBC$ and the length of side $BC$.

Solution:
\textbf{Solution:} Since in $\triangle ABC$, $AB=AC$ and $\angle A=36^\circ$,\\
it follows that $\angle ABC=\angle C=\frac{180^\circ-\angle A}{2}=72^\circ$,\\
Since the perpendicular bisector of $AB$, $MN$, intersects $AC$ at point $D$,\\
it follows that $AD=BD$,\\
Therefore, $\angle ABD=\angle A=36^\circ$,\\
Therefore, $\angle DBC=\angle ABC-\angle ABD=36^\circ$,\\
Since $\angle BDC=\angle ABD+\angle A=72^\circ$,\\
it follows that $\angle BDC=\angle C=72^\circ$,\\
Therefore, $BC=BD$,\\
Also, since $AD=BD$,\\
it follows that $BC=AD=\boxed{5}$.


Question: As shown in the diagram, in $\triangle ABC$, $\angle BAC=95^\circ$, $\angle B=25^\circ$, $\angle CAD=75^\circ$, determine the measure of $\angle ADC$.

Solution:
\textbf{Solution:} Since $\angle BAC=95^\circ$ and $\angle CAD=75^\circ$,\\
it follows that $\angle BAD=\angle BAC-\angle CAD=20^\circ$,\\
Since $\angle B=25^\circ$,\\
it follows that $\angle ADC=\angle ABC+\angle BAD=25^\circ+20^\circ=\boxed{45^\circ}$.


Question: As shown in the figure, $\angle AOP = \angle BOP = 15^\circ$, $PC \parallel OA$, $PD \perp OA$, $PE \perp OB$. If $PC = 4$, what is the length of $PD$?

Solution:
\textbf{Solution:} Since $PC \parallel OA$,\\
it follows that $\angle AOP = \angle CPO = 15^\circ$,\\
thus $\angle PCE = \angle BOP + \angle CPO = 15^\circ + 15^\circ = 30^\circ$,\\
and since $PC = 4$,\\
it follows that $PE = \frac{1}{2}PC = \frac{1}{2} \times 4 = 2$,\\
since $\angle AOP = \angle BOP$, $PD \perp OA$, $PE \perp OB$,\\
it follows that $PD = PE = \boxed{2}$.