Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $y=-2x+3$ intersects the $y$-axis at point $A$ and the $x$-axis at point $B$. Point $P$ is a moving point on segment $AB$. Perpendicular lines are drawn from $P$ to the $x$-axis and the $y$-axis, and the feet of the perpendiculars are denoted as $C$ and $D$, respectively. Connect $CD$. When $CD$ is minimized, what is the value of $PA\cdot PB$?

Solution:
\textbf{Solution:} Since $PC\perp OB$, $PD\perp OA$, and $\angle DOB=90^\circ$, \\
it follows that quadrilateral $OCPD$ is a rectangle. \\
By connecting $OP$, we have $CD=OP$. \\
Therefore, when $OP\perp AB$, $OP$ is at its shortest, meaning $CD$ is also at its shortest. \\
Since $y=-2x+3$, \\
when $x=0$, $y=3$, and when $y=0$, $x=\frac{3}{2}$. \\
Thus, $A(0,3)$ and $B\left(\frac{3}{2},0\right)$, \\
which means $OA=3$ and $OB=\frac{3}{2}$, \\
and so $AB=\sqrt{OA^2+OB^2}=\sqrt{3^2+\left(\frac{3}{2}\right)^2}=\frac{3}{2}\sqrt{5}$. \\
$\tan\angle OBA=\frac{OP}{PB}=\frac{OA}{OB}=2$, \\
Letting $PB=x$, then $PO=2x$. \\
Therefore, $x^2+4x^2=\left(\frac{3}{2}\right)^2$, \\
Solving this, we find $x=\frac{3}{10}\sqrt{5}$, \\
Hence, $PB=\frac{3}{10}\sqrt{5}$ and $PA=AB-PB=\frac{3\sqrt{5}}{2}-\frac{3\sqrt{5}}{10}=\frac{6}{5}\sqrt{5}$, \\
Therefore, $PA\cdot PB=\frac{3\sqrt{5}}{10}\times \frac{6\sqrt{5}}{5}=\boxed{\frac{9}{5}}$;


Question: As shown in the figure, the roof of the workshop is an isosceles triangular (in the shape of a chevron) steel frame with the central column $AD$ (where $D$ is the midpoint of the base) measuring 10 meters in length, and $\angle B = 36^\circ$. What is the length of the span $BC$?

Solution:
\textbf{Solution:} Given that: $AB=AC$, $BD=CD$, $AD=10$ meters, \\
$\therefore$ $AD\perp BC$, \\
$\because$ $\angle B=36^\circ$, \\
$\therefore$ $BD=\frac{AD}{\tan B}=\frac{10}{\tan 36^\circ}$ meters, \\
$\therefore$ $BC=2BD=\frac{20}{\tan 36^\circ}$ meters. \\
Hence, the answer is: $BC=\boxed{\frac{20}{\tan 36^\circ}}$ meters.


Question: As shown in the figure, it is known that the three vertices of $\triangle ABC$ are all on the grid points of a square. What is the value of $\cos A$?

Solution:
\textbf{Solution:} As shown in the figure: Draw $BD\perp AC$ at D,\\
By the Pythagorean theorem,\\
$AB=\sqrt{3^2+3^2}=3\sqrt{2}$, $AC=\sqrt{3^2+1^2}=\sqrt{10}$,\\
$\because$ ${S}_{\triangle ABC}=\frac{1}{2}AB\cdot CD=\frac{1}{2}BC\times 3$, thus $CD=\frac{3\times 2}{3\sqrt{2}}=\sqrt{2}$,\\
In triangle $\triangle ABD$, $\angle ADC=90^\circ$, $AC=\sqrt{10}$, $CD=\sqrt{2}$,\\
$AD=\sqrt{AC^2-CD^2}=\sqrt{(\sqrt{10})^2-(\sqrt{2})^2}=2\sqrt{2}$,\\
$\therefore$ $\cos A=\frac{AD}{AC}=\frac{2\sqrt{2}}{\sqrt{10}}=\boxed{\frac{2\sqrt{5}}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, CD is the altitude from C to the hypotenuse AB, $BC=\sqrt{7}$, and $AC=3$. What is the value of $\sin\angle ACD$?

Solution:
\textbf{Solution:} Given $\triangle ABC$, where $\angle ACB = 90^\circ$,\\
$\therefore AB = \sqrt{3^2 + (\sqrt{7})^2} = 4$,\\
$\angle ACD + \angle BCD = 90^\circ$,\\
$\because CD$ is the height from the hypotenuse $AB$,\\
$\therefore CD \perp AB$,\\
$\therefore \angle B + \angle BCD = 90^\circ$,\\
$\therefore \angle ACD = \angle B$,\\
$\therefore \sin \angle ACD = \sin \angle B = \frac{AC}{AB} = \boxed{\frac{3}{4}}$


Question: As shown in the figure, AB represents a ski jump track. The angle of elevation of point B from point A is measured to be $35^\circ$. The distance between the bottom point C and the top point B is 50 meters. What is the length of the track AB? Meters.

Solution:
\textbf{Solution:} In the right triangle $ABC$,\\
since $\angle A=35^\circ$ and $BC=50$ meters,\\
it follows that $\sin 35^\circ = \frac{BC}{AB}$,\\
therefore, $AB = \frac{50}{\sin 35^\circ}$ (meters).\\
Thus, $AB = \boxed{\frac{50}{\sin 35^\circ}}$ (meters).


Question: Figure 1 shows a rectangular material $ABCD$, which is divided into three pieces, $\angle AEB=30^\circ$, and $GF\perp AD$. These three pieces are seamlessly and non-overlappingly assembled into the shape of Figure 2, which is exactly an axisymmetric figure. Therefore, $\frac{AB}{BC}=($     $)$

Solution:
\textbf{Solution:} As shown in the figure, $\because$ Figure 2 is an axisymmetric figure,\\
$\therefore BM=BE$, $MN=TE$,\\
Let $AB=m$, then $BE=\sqrt{3}m$, $AM=FG=DF=(\sqrt{3}-1)m$,\\
$\because AF=\sqrt{3}FG$,\\
$\therefore AF=(3-\sqrt{3})m$,\\
$\therefore AD=BC=DF+AF=2m$,\\
$\therefore \frac{AB}{BC}=\frac{m}{2m}=\boxed{\frac{1}{2}}$,


Question: In right triangle $ABC$, $\angle C = 90^\circ$, and $AB = 10$. If a circle centered at point $C$ with radius $CB$ exactly passes through the midpoint $D$ of $AB$, what is the length of $AC$?

Solution:
\textbf{Solution:} As shown in the figure, connect CD,\\
$\because$ $\angle C=90^\circ$ and D is the midpoint of AB,\\
$\therefore$ CD=DA=DB.\\
And CD=CB,\\
$\therefore$ CD=CB=DB, thus $\triangle CDB$ is an equilateral triangle.\\
$\therefore$ $\angle B=60^\circ$.\\
$\because$ AB=10,\\
$\therefore$ $AC=AB\cdot \sin\angle B=10\times \frac{\sqrt{3}}{2}=\boxed{5\sqrt{3}}$.


Question: The ancient Chinese mathematician Zhao Shuang, while annotating the "Zhou Bi Suan Jing", assembled 4 congruent right-angled triangles into a square (as illustrated), and utilized it to prove the Pythagorean theorem. This illustration is referred to as the "Gougu Diagram". If the area of the small square within the "Gougu Diagram" is 1, and the area of each right-angled triangle is $6$, and $\alpha$ is one of the acute angles in the right-angled triangle, then what is the value of $\tan \alpha$?

Solution:
\textbf{Solution:} Let the length of the shorter leg in the right-angled triangle be $x$, then the longer leg is $x+1$. $\because$ The area of each right-angled triangle is 6. $\therefore$ $\frac{1}{2}x(x+1)=6$\\
Solving gives $x_{1}=3$, $x_{2}=-4$ (discard). $\therefore$ The shorter leg is 3, the longer leg is 4. $\therefore$ $\tan\alpha = \frac{4}{3}$\\
Thus, the answer is: $\tan\alpha = \boxed{\frac{4}{3}}$


Question: As shown in the figure, in a square grid, it is known that the three vertices of $\triangle ABC$ are all on the grid points. What is the value of the tangent of $\angle ACB$?

Solution:
\textbf{Solution:} Extend CB to intersect the grid at D and connect AD as shown in the figure:\\
Then $\angle ADC = 45^\circ + 45^\circ = 90^\circ$,\\
$\because AD = \sqrt{1^2 + 1^2} = \sqrt{2}$, $CD = \sqrt{2^2 + 2^2} = 2\sqrt{2}$,\\
$\therefore$ the tangent of $\angle ACB = \frac{AD}{CD} = \frac{\sqrt{2}}{2\sqrt{2}} = \boxed{\frac{1}{2}}$;


Question: As shown in the figure, in $\triangle ABC$, $AC=4$, $BC=3$, and $\angle C=90^\circ$, then what is the value of $\sin A$?

Solution:
\textbf{Solution:} Since $AC=4$, $BC=3$, and $\angle C=90^\circ$, \\
therefore $AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{3^{2}+4^{2}}=5$, \\
hence $\sin A=\frac{BC}{AB}=\boxed{\frac{3}{5}}$;


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AB=c$, $AC=b$, $BC=a$, what is the value of $\tan A$?

Solution:
\textbf{Solution:} Since in the right triangle $ABC$, $\angle C=90^\circ$, $AC=b$, $BC=a$, \\
therefore $\tan A = \frac{a}{b}$, thus $\tan A = \boxed{\frac{a}{b}}$


Question: As shown in the figure, a ladder $AB$ rests against a wall that is perpendicular to the ground level. The distance $AC$, which is the gap between the base of the ladder $A$ and the wall, is 6 meters. If the angle between the ladder and the ground is $\alpha$, what is the length of the ladder $AB$?

Solution:
\textbf{Solution:} Given the problem, we have $\cos\alpha =\frac{AC}{AB}$, where $AC=6$ meters.\\
Therefore, $AB=\frac{6}{\cos\alpha}$ meters,\\
So, the answer is: $AB=\boxed{\frac{6}{\cos\alpha}}$ meters.


Question: As shown in the figure, with point O as the center, draw an arc with any length as the radius, which intersects the ray $OA$ at point B. Then, with point B as the center and the length of $BO$ as the radius, draw another arc. If the two arcs intersect at point C, draw ray $OC$. What is the value of $\tan\angle AOC$?

Solution:
\textbf{Solution:}
Connect $BC$, \\
given the problem, we have: $OB=OC=BC$, \\
thus, $\triangle OBC$ is an equilateral triangle, \\
hence, $\tan\angle AOC=\tan60^\circ=\boxed{\sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, $D$ is the midpoint of $BC$. If $AB=8$ and $AD=5$, what is the value of $\sin B$?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $\angle BAC=90^\circ$, and D is the midpoint of $BC$,\\
it follows that $AD$ is the median to the hypotenuse $BC$ of the right-angled triangle $ABC$,\\
hence, $BC=2AD=2\times 5=10$,\\
given $AB=8$,\\
then, $AC=\sqrt{BC^{2}-AB^{2}}=\sqrt{10^{2}-8^{2}}=6$,\\
thus, $\sin B=\frac{AC}{BC}=\frac{6}{10}=\boxed{\frac{3}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$. Extend $AB$ to point $D$ such that $BD=AB$. Connect $CD$. If $\tan\angle BCD=\frac{1}{3}$, what is the value of $\frac{BC}{AC}$?

Solution:
\textbf{Solution:} Draw $BE\parallel AC$ through point B, intersecting $CD$ at E, as shown in the diagram:\\
$\therefore$ $\angle ACB=\angle CBE=90^\circ$,\\
$\because$ $\tan\angle BCD=\frac{1}{3}$,\\
$\therefore$ in $\triangle BCE$, $\frac{BE}{BC}=\frac{1}{3}$,\\
$\therefore$ $BC=3BE$,\\
$\because$ $BE\parallel AC$,\\
$\therefore$ $\triangle DBE\sim \triangle DAC$,\\
$\therefore$ $\frac{BE}{AC}=\frac{BD}{AD}$,\\
$\because$ $BD=AB$,\\
$\therefore$ $\frac{BE}{AC}=\frac{BD}{AD}=\frac{1}{2}$,\\
$\therefore$ $AC=2BE$,\\
$\therefore$ $\frac{BC}{AC}=\frac{3BE}{2BE}=\boxed{\frac{3}{2}}$,


Question: As shown in the figure, the vertices of $\triangle ABC$ are all lattice points in a square grid. What is the value of $\cos\angle ABC$?

Solution:
\textbf{Solution:} Let $\triangle ABE$ be a right-angled triangle, where $CE=2, BE=4$, \\
$\therefore$ $BC= \sqrt{CE^2+BE^2} = \sqrt{2^2+4^2} = 2\sqrt{5}$, \\
$\therefore$ $\cos\angle ABC = \frac{BE}{BC} = \frac{4}{2\sqrt{5}} = \boxed{\frac{2\sqrt{5}}{5}}$


Question: In the diagram, it is known that in $\triangle ABC$, $\angle C=90^\circ$, $\tan A= \frac{1}{2}$, D is a point on AC, and $\angle CBD=\angle A$. Then, what is the value of $\cos \angle CDB$?

Solution:
\[
\text{Solution:} \quad \because \angle CBD=\angle A,
\]
\[
\therefore \tan\angle CBD=\tan A=\frac{1}{2},
\]
\[
\text{Let } CD=a,
\]
\[
\therefore \tan\angle CBD=\frac{CD}{BC}=\frac{1}{2},
\]
\[
\therefore BC=2a,
\]
\[
\text{In } \triangle CBD,
\]
\[
BD=\sqrt{CD^2+BC^2}=\sqrt{a^2+(2a)^2}=\sqrt{5}a,
\]
\[
\therefore \cos\angle CDB=\frac{CD}{BD}=\frac{a}{\sqrt{5}a}=\boxed{\frac{\sqrt{5}}{5}}.
\]


Question: As shown in the figure, within a $4\times4$ square grid, points A, B, and C are grid intersections, with $AD\perp BC$ and the foot of the perpendicular being D. What is the value of $\sin\angle BAD$?

Solution:
\textbf{Solution:} As shown in the figure, connect AC. \\
In $\triangle BEC$, $BC=\sqrt{BE^{2} + CE^{2}} = \sqrt{4^{2} + 3^{2}} = 5$, \\
since $AD \perp BC$, \\
thus $\frac{1}{2} \times BC \times AD = 8$, \\
which means $\frac{1}{2} \times 5 \times AD = 8$, \\
solving this we get $AD = \frac{16}{5}$, \\
in $\triangle ADB$, $BD = \sqrt{AB^{2} - AD^{2}} = \frac{12}{5}$, \\
therefore $\sin\angle BAD = \frac{BD}{AB} = \frac{\frac{12}{5}}{4} = \boxed{\frac{3}{5}}$


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, $AB=3$, $AC=4$. Point $D$ is the midpoint of $BC$. $\triangle ABD$ is folded along $AD$ to obtain $\triangle AED$. Connect $CE$. What is the distance from point $E$ to $BC$?

Solution:
\textbf{Solution:} As shown in the diagram, draw line $EB$, and construct $EH\perp BC$ at point $H$, \\
since $\angle BAC=90^\circ$, $AB=3$, $AC=4$,\\
then $BC=\sqrt{AB^2+AC^2}=\sqrt{9+16}=5$,\\
since point $D$ is the midpoint of $BC$,\\
thus $AD=BD=CD=2.5$,\\
since by folding $\triangle ABD$ along $AD$ we get $\triangle AED$,\\
thus $AE=AB=3$, $BD=DE=CD$,\\
thus $\angle CEB=90^\circ$, $AD$ perpendicularly bisects $BE$,\\
thus $EO=BO$,\\
since $S_{\triangle ADB}=\frac{1}{2}S_{\triangle ABC}=\frac{1}{2}\times \frac{1}{2}\times 3\times 4=3$,\\
thus $\frac{1}{2}\times \frac{5}{2}\times BO=3$,\\
thus $BO=\frac{12}{5}$,\\
thus $BE=\frac{24}{5}$,\\
thus $DO=\sqrt{BD^2-BO^2}=\sqrt{\frac{25}{4}-\frac{144}{25}}=\frac{7}{10}$,\\
since $\sin\angle DBO=\frac{DO}{DB}=\frac{EH}{BE}=\frac{\frac{7}{10}}{\frac{5}{2}}=\frac{EH}{\frac{24}{5}}$,\\
thus $EH=\boxed{\frac{168}{125}}$.


Question: As shown in the diagram, the angle between the two legs $OA$ and $OB$ of the compass is $ \angle AOB = \alpha$, with $OA = OB = 10$. What is the distance $AB$ between the two legs?

Solution:
\textbf{Solution:} Let $OC \perp AB$ with the foot of the perpendicular at C, \\
$\because OA=OB=10$, $OC \perp AB$, \\
$\therefore \angle AOC=\frac{1}{2}\angle AOB=\frac{\alpha}{2}$, $AC=\frac{1}{2}AB$, \\
In $\triangle AOC$, $AC=AO\cdot \sin\frac{\alpha}{2}=10\sin\frac{\alpha}{2}$, \\
$\therefore AB=2AC=20\sin\frac{\alpha}{2}$, \\
Thus, $AB=\boxed{20\sin\frac{\alpha}{2}}$


Question: As shown in the figure, $O$ is the midpoint of the seesaw $AB$. The support column $OC$ is perpendicular to the ground $MN$, with the foot of the perpendicular being point $C$. When one end $B$ of the seesaw touches the ground, the angle between seesaw $AB$ and the ground $MN$ is $20^\circ$. Given that $AB=1.6m$, what is the length of $OC$?

Solution:
\textbf{Solution:} Since O is the midpoint of AB, and AB=1.6,\\
it follows that OB=$\frac{1}{2}$AB=0.8,\\
In $\triangle OCB$, $\sin\angle OBC=\frac{OC}{OB}$,\\
Therefore, OC=OB$\cdot\sin\angle OBC=0.8\sin20^\circ=\boxed{0.8\sin20^\circ}$


Question: Fold the rectangular paper ABCD as shown in the figure, and you get the rhombus AECF exactly. If \(AD = \sqrt{3}\), what is the area of the rhombus AECF?

Solution:
\textbf{Solution:} We know from the properties of folding that $\angle DAF = \angle OAF$ and $OA = AD = \sqrt{3}$,\\
In rhombus AECF, $\angle OAF = \angle OAE$,\\
$\therefore \angle OAE = \frac{1}{3} \times 90^\circ = 30^\circ$,\\
$\therefore AE = \frac{AO}{\cos 30^\circ} = \frac{\sqrt{3}}{\frac{\sqrt{3}}{2}} = 2$,\\
$\therefore$ the area of rhombus AECF $= AE \cdot AD = 2\sqrt{3}$.\\
Thus, the answer is: \boxed{2\sqrt{3}}.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $BC=\sqrt{5}$, and point D is on segment $AC$, with $BD$ connected. If $\tan A=\frac{1}{2}$ and $\tan \angle ABD=\frac{1}{3}$, what is the length of $CD$?

Solution:
\textbf{Solution:} Draw $DE \perp AB$ at $E$ from point $D$, \\
$\because \tan A = \frac{1}{2}, \tan \angle ABD = \frac{1}{3}$, \\
$\therefore \frac{DE}{AE} = \frac{1}{2}, \frac{DE}{BE} = \frac{1}{3}$, \\
$\therefore AE = 2DE, BE = 3DE$, \\
$\therefore AE + BE = 2DE + 3DE = 5DE = AB$, \\
$\because$ In $\triangle ABC$, $\angle C = 90^\circ$, $BC = \sqrt{5}$, $\tan A = \frac{1}{2}$, \\
$\therefore \frac{BC}{AC} = \frac{\sqrt{5}}{AC} = \frac{1}{2}$, \\
$\therefore AC = 2\sqrt{5}$, \\
$\therefore AB = \sqrt{AC^2 + BC^2} = 5$, \\
$\therefore DE = 1$, \\
$\therefore AE = 2$, \\
$\therefore AD = \sqrt{AE^2 + DE^2} = \sqrt{5}$, \\
$\therefore CD = AC - AD = \boxed{\sqrt{5}}$.


Question: As shown in the figure, the archaeological team measured the angle of elevation of the top $C$ of the ancient tower $BC$ at point $A$ to be $60^\circ$. The length of the slope $AD$ is 5 meters, with a gradient of $i=3\colon 4$, and the length of $BD$ is 6 meters. What is the height of the ancient tower $BC$?

Solution:
\textbf{Solution:} As shown in the diagram, draw $AE\perp BC$ with $E$ as the foot of the perpendicular, and draw $AF\perp BD$ with $F$ as the foot of the perpendicular, then quadrilateral $AEBF$ forms a rectangle, hence $AE=FB$,\\
since the slope of the ramp $AD$ has a gradient of $i=3\colon 4$, let $AF=3x$, $DF=4x$,\\
therefore $AD=5x$,\\
since $AD=5$, then $x=1$,\\
thus, $AF=3$, $DF=4$,\\
since the length of $BD$ is $6$,\\
therefore, $AE=FB=DF+BD=4+6=10$,\\
since $\angle CAE=60^\circ$,\\
therefore, $CE=AE\times \tan\angle CAE=10\sqrt{3}$,\\
thus, $BC=BE+CE=10\sqrt{3}+3$,\\
which means the height of the ancient tower $BC$ is $\boxed{3+10\sqrt{3}}$ meters.


Question: As shown in the figure, $\triangle ABC$ and $\triangle CDE$, with the right-angled vertices coinciding at point $C$, point $D$ on $AB$, $\angle BAC=\angle DEC$ and $\sin\angle BAC=\frac{1}{3}$, connect $AE$. If $BD=2$, $AD=7$, then what is the length of $AE$?

Solution:
\textbf{Solution:} Given that $BD=2$, $AD=7$,

therefore $AB=AD+BD=7+2=9$, in $\triangle ABC$, $\sin\angle BAC=\frac{BC}{AB}=\frac{1}{3}$,

hence $BC=\frac{1}{3}AB=3$,

thus $AC=\sqrt{9^2-3^2}=6\sqrt{2}$, and in $\triangle BCA$ and $\triangle DCE$, $\angle BAC=\angle DEC$,

thus $\tan\angle BAC=\tan\angle DEC$,

therefore $\frac{BC}{AC}=\frac{DC}{EC}$,

given $\angle BCA=\angle DCE=90^\circ$,

hence $\angle BCA-\angle DCA=\angle DCE-\angle DCA$, that is, $\angle BCD=\angle ACE$,

therefore $\triangle BCD \sim \triangle ACE$,

thus $\angle CAE=\angle B$, $\frac{BC}{AC}=\frac{BD}{AE}$,

hence $\angle DAE=\angle DAC+\angle CAE=90^\circ$,

thus $\frac{3}{6\sqrt{2}}=\frac{2}{AE}$, solving gives: $AE=\boxed{4\sqrt{2}}$.


Question: As shown in the figure, Alpha and Beta are two buildings, with a horizontal distance $BC = 30m$ between them. The elevation angle $\angle EAD$ measured from point A to point D is $45^\circ$, and the elevation angle $\angle CBD$ measured from point B to point D is $60^\circ$. What is the height of building Beta?

Solution:
\textbf{Solution:} As shown in the figure, draw $AF\perp CD$ at point F. In $Rt\triangle BCD$,\\
$\angle DBC=60^\circ$, $BC=30m$, $\tan\angle DBC=\frac{CD}{BC}$\\
$\therefore CD=BC\cdot \tan60^\circ=30\sqrt{3}m$,\\
In $Rt\triangle ADF$, $\angle DAF=45^\circ$,\\
$\therefore DF=AF=BC=30m$,\\
$\therefore AB=CF=CD-DF=(30\sqrt{3}-30)m$,\\
this means the height of building B is $\boxed{(30\sqrt{3}-30)m}$.


Question: As shown in the figure, the vertices of $\triangle ABC$ are lattice points of a square grid. What is the value of $\sin B$?

Solution:
\textbf{Solution:} By the Pythagorean theorem, we have $AB=\sqrt{3^2+3^2}=3\sqrt{2}$,\\
therefore, $\sin B=\frac{3}{3\sqrt{2}}=\boxed{\frac{\sqrt{2}}{2}}$.


Question: As shown in the figure, $AB$ is the diameter of the circle $\odot O$ with $AB=26$, and point $C$ is a point on the semicircle of $\odot O$. $CE \perp AB$ at point $E$, the angle bisector of $\angle OCE$ intersects $\odot O$ at point $D$, and the chord $AC=10$. What is the area of the triangle $\triangle ACD$?

Solution:
\textbf{Solution:} Connect OD and draw AF $\perp$ CD with the foot at F, \\
$\because$ AB is the diameter of $\odot$O, \\
$\therefore \angle ACB=90^\circ$, \\
$\therefore \angle ACE+\angle ECB=90^\circ$, \\
$\because CE\perp AB$, \\
$\therefore \angle CEB=90^\circ$, \\
$\therefore \angle CBE+\angle ECB=90^\circ$, \\
$\therefore \angle ACE=\angle CBE$, \\
$\because OB=OC$, \\
$\therefore \angle CBE=\angle OCB$, \\
$\therefore \angle ACE=\angle OCB$, \\
$\because CD$ bisects $\angle OCE$, \\
$\therefore \angle OCD=\angle ECD$, \\
$\therefore \angle ACE+\angle DCE=\angle OCB+\angle OCD= \frac{1}{2} \angle ACB=45^\circ$, \\
$\therefore \angle ACD=45^\circ$, \\
$\therefore \angle AOD=2\angle ACD=90^\circ$, \\
In $\triangle ACF$, AC=10, \\
$\therefore AF=AC\cdot\sin45^\circ=10\times \frac{\sqrt{2}}{2} =5\sqrt{2}$, \\
$CF=AC\cdot\cos45^\circ=10\times \frac{\sqrt{2}}{2} =5\sqrt{2}$, \\
In $\triangle AOD$, AO=OD=$\frac{1}{2}$AB=13, \\
$\therefore AD=\sqrt{2}AO=13\sqrt{2}$, \\
$\therefore DF=\sqrt{AD^2-AF^2}=\sqrt{(13\sqrt{2})^2-(5\sqrt{2})^2}=12\sqrt{2}$, \\
$\therefore CD=CF+DF=17\sqrt{2}$, \\
$\therefore$ the area of $\triangle ACD$= $\frac{1}{2}$CD$\cdot$AF\\
= $\frac{1}{2}\times17\sqrt{2}\times5\sqrt{2}$\\
= $\boxed{85}$.


Question: As shown in Figure 1, fold the square paper $ABCD$ so that $AB$ coincides with $CD$, with the crease being $EF$. As shown in Figure 2, unfold it and then fold it once more, making point C coincide with point E, with the crease being $GH$. The corresponding point of point B is point M, and $EM$ intersects $AB$ at N. What is the ratio of $AE$ to $AN$?

Solution: \textbf{Solution:} Let the side length of the square be $2a$ and $DH=b$. Therefore, $CH=2a−b$. Because when the square paper $ABCD$ is folded to make $AB$ coincide with $CD$, the crease is $EF$. As shown in Figure 2, after unfolding and then folding again to make point C coincide with point E, the crease is $GH$, and the corresponding point to point B is point M. Therefore, $EH=CH=2a−b$, $DE=AE=\frac{1}{2}\times2a=a$, $\angle BEH=\angle C=90^\circ$. In $\triangle DEH$, $DE^{2}+DH^{2}=EH^{2}$, i.e., $a^{2}+b^{2}=(2a−b)^{2}$. Solving gives $b=\frac{3}{4}a$. Because $\angle DEH+\angle AEN=90^\circ$ and $\angle AEN+\angle DEH=90^\circ$, therefore $\angle ANE=\angle DEH$. Therefore, $\tan\angle ANE=\tan\angle DEH=\frac{AE}{AN}=\frac{DH}{DE}=\frac{b}{a}=\frac{\frac{3}{4}a}{a}=\boxed{\frac{3}{4}}$.


Question: As shown in the figure, in rectangle $ABCD$, where $AB=3$ and $AD=5$, point $E$ is on $DC$. When rectangle $ABCD$ is folded along $AE$, point $D$ falls exactly on point $F$ on side $BC$. What is the value of $\cos\angle EFC$?

Solution:
\textbf{Solution:} Given the properties of reflection transformations, we know that $\angle AFE = \angle D = 90^\circ$, and AF = AD = 5,\\
$\therefore \angle EFC + \angle AFB = 90^\circ$,\\
$\because \angle B = 90^\circ$,\\
$\therefore \angle BAF + \angle AFB = 90^\circ$,\\
$\therefore \angle EFC = \angle BAF$,\\
$\cos \angle BAF = \frac{BA}{BF} = \frac{3}{5}$,\\
$\therefore \cos \angle EFC = \boxed{\frac{3}{5}}$.


Question: As shown in the figure, the vertices of $\triangle ABC$ are on the grid points. What is the value of $\cos B$?

Solution:
\textbf{Solution:} As shown in the figure, we select lattice point $D$ such that $\angle ADB=90^\circ$, and connect $AD$ and $BD$.\\
Since $AD=3$ and $BD=3$\\
Therefore, $AB=\sqrt{AD^{2}+BD^{2}}=\sqrt{3^{2}+3^{2}}=3\sqrt{2}$\\
Therefore, $\cos B=\frac{BD}{AB}=\frac{3}{3\sqrt{2}}=\boxed{\frac{\sqrt{2}}{2}}$


Question: As shown in the diagram, within a grid formed by small squares, the vertices of $\triangle ABC$ are all lattice points (intersections of the grid lines). What is the value of $\tan\angle ABC$?

Solution:
\textbf{Solution}: As shown in the figure, draw AD$\perp$BC at D. \\
In $\triangle ABD$, $\tan\angle ABC=\frac{AD}{BD}=\frac{1}{2}$, \\
thus, $\tan\angle ABC=\boxed{\frac{1}{2}}$.


Question: As shown in the cross-section of the dam, the height of the dam $BC$ is $6$ meters, and the slope of the inclined surface is $1\colon 2$. What is the length of the slope $AB$?

Solution:
\textbf{Solution:} Given that $\angle ACB = 90^\circ$,\\
since the slope of the inclined plane is $1:2$,\\
it follows that $\frac{BC}{AC} = \frac{1}{2}$,\\
thus $AC = 2BC = 12$ meters,\\
hence $AB = \sqrt{AC^2 + BC^2} = \boxed{6\sqrt{5}}$ meters.


Question: As shown in the figure, it is known in $\triangle ABC$ that $\angle C=90^\circ$, $AC=8$, and $BC=15$. Then, the value of $\tan A$ is (\quad)

Solution:
\textbf{Solution:} Given that in the right triangle $\text{Rt}\triangle ABC$, $\angle C=90^\circ$, $AC=6$, $BC=15$, \\
thus $\tan A= \frac{BC}{AC}=\boxed{\frac{15}{6}}$.


Question: As shown in the figure, AB is a generatrix of the cone, and BC is the diameter of the base. Given that BC = 6 cm and the lateral surface area of the cone is $15\pi$, what is the value of $\cos\angle ABO$?

Solution: \textbf{Solution:} From the given information, we have $\frac{1}{2}\pi \times 6\times AB=15\pi$, from which we find $AB=5\,\text{cm}$. Since $BO=\frac{1}{2}$BC$=3\,\text{cm}$, it follows that $\cos\angle ABO=\frac{BO}{AB}=\boxed{\frac{3}{5}}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ (where $k\neq 0$) is shown. Then, what is the solution set of the inequality $kx+b>3$?

Solution: \textbf{Solution:} As can be seen from the graph, the solution set of the inequality $kx+b>3$ is $\boxed{x>2}$.


Question: As shown in the figure, the line $y=x+b$ intersects with the line $y=kx+6$ at the point $P(1,3)$. What is the solution set for the inequality $x+b>kx+6$ with respect to $x$?

Solution:
\textbf{Solution:} When $x>1$, we have $x+b>kx+6$,\\
which means the solution set of the inequality $x+b>kx+6$ is $x>1$,\\
therefore, the answer is $x>1$.\\
Hence, the answer is: $\boxed{x>1}$.


Question: As shown in the figure, the graphs of the functions $y=2x$ and $y=nx+6$ intersect at point $A(m, 4)$. Then, how many integer solutions does the system of inequalities $0<nx+6<2x$ have?

Solution:
\textbf{Solution:} Since point A($m$, 4) lies on the graph of the function $y=2x$, \\
it follows that $4=2m$, solving this gives $m=2$, \\
therefore, point A is (2, 4), \\
substituting point A(2, 4) into $y=nx+6$, we get: $4=2n+6$, solving for $n$ gives: $n=-1$, \\
thus, the equation is: $y=-x+6$, \\
substituting $y=0$ into $y=-x+6$, we find: $x=6$, \\
therefore, point B is (6, 0), \\
from the graph of the function, it is known that when $2<x<6$, both functions $y=2x$ and $y=nx+6$ are above the $x$-axis, and the graph of $y=2x$ is above that of $y=nx+6$, \\
therefore, the solution set of the system of inequalities $0<nx+6<2x$ is: $2<x<6$, \\
therefore, there are 3 integer solutions: $3, 4, 5$, for a total of $\boxed{3}$ solutions.


Question: As shown in the figure, points A, B, C lie on the graph of the linear function $y= -2x+m$, with their abscissas being $-1$, $1$, and $2$ respectively. Perpendicular lines are drawn from these points to the $x$-axis and the $y$-axis. What is the sum of the areas of the shaded parts in the figure?

Solution:
\textbf{Solution:} As shown in the figure,\\
based on the problem statement, we have: the coordinates of point A are $(-1, 2+m)$, point B are $(1, -2+m)$, point C are $(2, m-4)$, point D are $(0, 2+m)$, point E are $(0, m)$, point F are $(0, -2+m)$, and point G are $(1, m-4)$.\\
Hence, $DE=EF=BG=2+m-m=m-(-2+m)=2-2+m-(m-4)=2$,\\
and since $AD=BF=GC=1$,\\
the total area of the shaded regions in the figure is equal to $S=3\times \frac{1}{2}\times 1\times 2=\boxed{3}$.


Question: As shown in the figure, the intersection of the line $y=-x+m$ with $y=nx+4n\ (n\ne 0)$ has an abscissa of $-2$. How many integer solutions are there for the inequality $-x+m>nx+4n>0$ with respect to $x$?

Solution: \textbf{Solution}: When $y=0$, for $y=nx+4n\ (n\ne 0)$, then $x=-4$. Therefore, the solution set for $nx+4n>0$ is $x>-4$. Since the $x$-coordinate of the intersection point between $y=-x+m$ and $y=nx+4n\ (n\ne 0)$ is $-2$, observing the graph shows that the solution set for $-x+m>nx+4n$ is $x<-2$. Hence, the solution set for $-x+m>nx+4n>0$ is $-4<x<-2$. Since $x$ is an integer, it follows that $x=\boxed{-3}$.


Question: As shown in the figure, the graphs of the functions $y_{1}=\frac{1}{2}x$ and $y_{2}=ax+5$ intersect at the point $A(m, 2)$. What is the solution set for the inequality $\frac{1}{2}x\le ax+5$ with respect to $x$?

Solution:
\textbf{Solution:} Substitute $A\left(m,2\right)$ into $y_{1}=\frac{1}{2}x$ to get $\frac{1}{2}m=2$\\
Solving this gives $m=4$\\
Thus, the coordinates of point A are $\left(4,2\right)$\\
Therefore, the solution set for the inequality in $x$, $\frac{1}{2}x\le ax+5$, is $\boxed{x\le 4}$


Question: As shown in the figure, the line $y=x+8$ intersects with the line $y=ax+b$ at the point $P$. Based on the graph, what is the solution set of the inequality $x+8>ax+b$?

Solution:
\textbf{Solution:} From the given information, we can determine that the line $y=x+8$ intersects with the line $y=ax+b$ at point P(20, 25).\\
By examining the graphical representation of the functions, it is evident that for $x>20$,\\
the graph of the line $y=x+8$ is located above the graph of the line $y=ax+b$,\\
meaning the solution set for the inequality $x+8>ax+b$ regarding $x$ is: $x>20$,\\
therefore, the answer is: $\boxed{x>20}$.


Question: As shown in the figure, the line $y_{1}=x+b$ intersects with the line $y_{2}=kx-1$ at point $P$, where the $x$-coordinate of point $P$ is $-1$. What is the solution set of the inequality $x+b<kx-1$ in terms of $x$?

Solution:
\textbf{Solution:} By observing the graph of the function, it is found that when $x < -1$, the graph of the function $y = x + b$ is below the graph of $y = kx - 1$,\\
therefore, the solution set of the inequality $x + b < kx - 1$ is $x < -1$.\\
Hence, the answer is: $\boxed{x < -1}$.


Question: As shown in the figure, the graph of the direct proportion function $y_1=-2x$ intersects with the graph of the linear function $y_2=ax+3$ at point $A(-1, m)$. What is the solution set of the inequality $-2x>ax+3$ with respect to $x$?

Solution:
\textbf{Solution:} From the graph, it is known that when $x<-1$, $-2x>ax+3$. \\
Therefore, the answer is $\boxed{x<-1}$.


Question: As shown in the figure, let $A$ be a point on the negative half-axis of the x-axis. A line $AB$ perpendicular to the x-axis is drawn through point $A$ and intersects the line $y=x$ at point $B$. The triangle $\triangle ABO$ is translated upwards along the line $y=x$ by $5\sqrt{2}$ units to obtain $\triangle A'B'O'$. If the coordinates of point $A$ are $(-3, 0)$, what are the coordinates of point $B'$?

Solution:
\textbf{Solution:} Because the coordinates of point A are $(-3, 0)$, and AB is perpendicular to the x-axis and intersects the line $y = x$ at point B,
$\therefore$ the coordinates of B are $(-3, -3)$,
Moving $\triangle ABO$ along the line $y = x$ upwards by $5\sqrt{2}$ units essentially translates $\triangle ABO$ 5 units to the right and 5 units upwards,
$\therefore$ the coordinates of $B'$ are $(-3+5, -3+5)$, that is $B'(2,2)$,
Thus, the answer is: the coordinates of $B'$ are $\boxed{(2,2)}$.


Question: As shown in the figure, it is known that the functional relationship between $y$ and $x$ is as illustrated. Specifically, when $x \geq 0$, $y = x$; when $x < 0$, $y = -2x + 1$. Then, what is the range of $x$ when the function value $y > 3$?

Solution:
\textbf{Solution:} Substitute $y=3$ into $y=x$, then $x=3$.\\
Substitute $y=3$ into $y=-2x+1$ to get, $3=-2x+1$, solving this gives $x=-1$.\\
Therefore, the points of intersection between the line $y=3$ and the graph of the function are $(3,3)$ and $(-1,3)$.\\
Observing the graph, when the function value $y>3$, the range of $x$ is $x<-1$ or $x>3$,\\
thus, the answer is: $\boxed{x \in (-\infty, -1) \cup (3, +\infty)}$.


Question: As shown in the figure, for rectangle $ABOC$ with side $BO$ and $CO$ on the x-axis and y-axis respectively, the coordinate of point $A$ is $(-12, 8)$. Points $D$ and $E$ are the midpoints of $AC$ and $OC$ respectively. Point $P$ is a moving point on $OB$. What are the coordinates of point $P$ when $PD+PE$ is at its minimum?

Solution:
\textbf{Solution:} Consider point E's symmetric point with respect to the x-axis, denoted as E', and connect DE' to intersect OB at point P. \\
$\because$ The sides BO and CO of rectangle ABOC are respectively on the x-axis and y-axis, and the coordinates of point A are (-12, 8),\\
$\therefore$ AC = 12, OC= 8,\\
$\because$ Points D and E are the midpoints of AC and OC, respectively,\\
$\therefore$ D (-6, 8), E (0, 4),\\
$\therefore$ E' (0, -4),\\
Let the equation of line DE' be $y= kx + b$,\\
From the problem statement, we have:\\
$\left\{\begin{array}{l}
-6k+b=8\\
b=4
\end{array}\right.$,\\
$\therefore$ $\left\{\begin{array}{l}
k=-2\\
b=-4
\end{array}\right.$,\\
$\therefore$ The equation of line DE' is $y= -2x - 4$\\
When y=0, $-2x - 4=0$,\\
$\therefore$ x = -2,\\
$\therefore$ The coordinates of P are $\boxed{(-2, 0)}$.


Question: The graph of the linear functions $y_1=4x+5$ and $y_2=3x+10$ is shown in the figure. What is the solution set of $4x+5>3x+10$?

Solution:
\textbf{Solution:} Solve: The graphs of the linear functions $y_1=4x+5$ and $y_2=3x+10$ intersect at point P(5, 25).\\
Therefore, the solution set for $4x+5>3x+10$ is: $\boxed{x>5}$.


Question: As shown in the figure, the graph of line $l_{1}: y=k_{1}x+b$ and the graph of line $l_{2}: y=k_{2}x$ are shown in the same Cartesian coordinate system. The line $l_{1}: y=k_{1}x+b$ intersects the x-axis at point $(-3,0)$. What is the solution set of the inequality $k_{2}x<k_{1}x+b<0$ with respect to $x$?

Solution:
\textbf{Solution:} From the graph, it is known that the intersection point of line $l_{1}$ and line $l_{2}$ is ($-1$, $-2$), and on line $l_{1}$, $y$ decreases as $x$ increases.\\
Since $y=k_{1}x+b$ intersects the $x$-axis at the point ($-3$, $0$), the solution set of the inequality $k_{2}x < k_{1}x+b$ with respect to $x$ is $x<-1$,\\
therefore, the solution set of the inequality $k_{2}x < k_{1}x+b < 0$ with respect to $x$ is $-3<x<-1$,\\
hence, the answer is: $\boxed{-3<x<-1}$.


Question: Given the graphs of the linear functions $y_1=kx+b$ and $y_2=x+a$ as shown in the figure, what is the solution set of the inequality $kx+b<x+a$?

Solution:
Since when $x>3$, the graph of the linear function $y_{1}=kx+b$ is below the graph of the linear function $y_{2}=x+a$, the solution set of the inequality $kx+b<x+a$ is $x>3$. \\
Therefore, the solution set is $\boxed{x>3}$.


Question: As shown in the diagram, the graphs of the functions $y=3x$ and $y=kx+4$ intersect at the point $P(m, 3)$. What is the solution set of the inequality $3x>kx+4$ with respect to $x$?

Solution:
\textbf{Solution:} Substituting $P(m, 3)$ into $y=3x$, we have $3m=3$, from which we solve $m=1$. Thus, the coordinates of point P are $(1, 3)$, \\
hence for $x>1$, we have $3x>kx+4$, \\
meaning the solution set for the inequality $3x>kx+4$ is $\boxed{x>1}$.


Question: As illustrated, the lines $l_{1}$ and $l_{2}$ are the graphs of the functions $y=k_{1}x+b_{1}$ and $y=k_{2}x+b_{2}$, respectively. Based on the graph, what is the solution set of the inequality $k_{1}x+b_{1}>k_{2}x+b_{2}$ with respect to $x$?

Solution:
\textbf{Solution:} When $x<-2$, $k_{1}x+b_{1}>k_{2}x+b_{2}$,\\
thus the solution set of the inequality $k_{1}x+b_{1}>k_{2}x+b_{2}$ is $x<-2$.\\
Therefore, the answer is: $x=\boxed{-2}$.


Question: As shown in the figure, the graphs of the functions $y=ax$ and $y=bx+5$ intersect at point A. What is the solution set of the inequality $ax\geq bx+5$?

Solution:
\textbf{Solution:} According to the graph of the function, when $x \geq 2$, $ax \geq bx + 5$.\\
Therefore, the answer is: $\boxed{x \geq 2}$.


Question: As shown in the figure, the line $AB$: $y=kx+b$ intersects with the linear function $y=2x$ at point $B$. What is the value of $k+b$?

Solution:
\textbf{Solution:} Substitute $x=1$ into $y=2x$, to get $y=2\times1=2$,\\
$\therefore B(1,2)$,\\
Substitute $x=1$, $y=2$ into $y=kx+b$, we get: $k+b=\boxed{2}$


Question: As shown in the figure, AB is the diameter of the semicircle O, and point D is the midpoint of the arc AC. If $\angle BAC = 44^\circ$, then what is the value of $\angle DAC$?

Solution:
\textbf{Solution:} Given that $AB$ is the diameter of the semicircle $O$,\\
it follows that $\angle ACB = 90^\circ$,\\
since $\angle BAC = 44^\circ$,\\
then $\angle B = 90^\circ - \angle BAC = 46^\circ$,\\
as quadrilateral $ABCD$ is inscribed in semicircle $O$,\\
then $\angle D = 180^\circ - \angle B = 134^\circ$,\\
given that point $D$ is the midpoint of arc $AC$,\\
it follows that $ \overset{\frown}{AD} =  \overset{\frown}{DC}$,\\
thus $AD = DC$,\\
therefore $\angle DAC = \angle DCA = \frac{1}{2} (180^\circ - \angle D) = \boxed{23^\circ}$,


Question: As shown in the figure, within the circle $ \odot O$, $AB = AC$. If $\angle ABC = 65^\circ$, what is the degree measure of $\angle BOC$?

Solution:
\textbf{Solution:} Given that $AB=AC$, \\
hence $\angle ABC=\angle ACB=65^\circ$, \\
therefore $\angle A=180^\circ-2\angle ABC=180^\circ-2\times65^\circ=50^\circ$, \\
since $\overset{\frown}{BC}=\overset{\frown}{BC}$, \\
thus $\angle BOC=2\angle A=2\times50^\circ=\boxed{100^\circ}$.


Question: As shown in the figure, the diameter $AB=20$ of the semicircle $O$, and the chord $AC=12$. The chord $AD$ bisects $\angle BAC$. What is the length of $AD$?

Solution:
\textbf{Solution:} Connect BD, BC, OD, intersecting BC at point F,\\
$\because$ AB is the diameter,\\
$\therefore \angle ACB = \angle ADB = 90^\circ$,\\
$\therefore BC = \sqrt{AB^2 - AC^2} = \sqrt{20^2 - 12^2} = 16$,\\
$\because$ AD bisects $\angle BAC$,\\
$\therefore \angle CAD = \angle DAB$,\\
$\therefore \overset{\frown}{CD} = \overset{\frown}{BD}$,\\
$\therefore$ OD$\perp$CB,\\
$\therefore$ BF = $\frac{1}{2}BC = 8$, $\angle BFO = 90^\circ$,\\
$\therefore OF = \sqrt{OB^2 - BF^2} = \sqrt{10^2 - 8^2} = 6$,\\
$\therefore DF = OD - OF = 10 - 6 = 4$,\\
In the right $\triangle DFB$,\\
$BD = \sqrt{DF^2 + BF^2} = \sqrt{4^2 + 8^2} = 4\sqrt{5}$,\\
In the right $\triangle ABD$,\\
$AD = \sqrt{AB^2 - BD^2} = \sqrt{20^2 - (4\sqrt{5})^2} = \boxed{8\sqrt{5}}$.


Question: As shown in the figure, place the right-angle vertex of a triangular ruler, which includes a $30^\circ$ angle, against one side of a straightedge. If $\angle 1 = 37^\circ$, then what is the degree measure of $\angle 2$?

Solution:
\textbf{Solution:}\\
Since $\angle 1=37^\circ$, the triangle is a right-angled triangle.\\
Therefore, $\angle 3=90^\circ-\angle 1=53^\circ$.\\
Since the two lines on the ruler are parallel,\\
Therefore, $\angle 4=\angle 3=53^\circ$.\\
Therefore, $\angle 2=180^\circ-\angle 4=\boxed{127^\circ}$.


Question: As shown in the figure, $OB$ is the bisector of $\angle AOC$, and $OD$ is the bisector of $\angle COE$. If $\angle AOB=30^\circ$ and $\angle COE=60^\circ$, then what is the measure of $\angle BOD$?

Solution:
\textbf{Solution:} Since OB is the bisector of $\angle AOC$, and OD is the bisector of $\angle COE$, \\
$\therefore \angle COD= \frac{1}{2} \angle COE$ and $\angle BOC=\angle AOB$, \\
Also, since $\angle AOB=30^\circ$ and $\angle COE=60^\circ$, \\
$\therefore \angle BOC=30^\circ$ and $\angle COD=30^\circ$, \\
$\therefore \angle BOD=\angle BOC+\angle COD=30^\circ+30^\circ=\boxed{60^\circ}$.


Question: As shown in the diagram, the straight line $DE$ intersects with $BC$ at point $O$, $\angle COE$ and $\angle AOE$ are complementary, and $\angle BOD=35^\circ$. What is the measure of $\angle AOE$ in degrees?

Solution:
Solution: Since $\angle COE = \angle BOD$,\\
$\angle BOD = 35^\circ$,\\
therefore $\angle COE = 35^\circ$,\\
since $\angle COE$ and $\angle AOE$ are complementary,\\
thus $\angle COE + \angle AOE = 90^\circ$,\\
therefore $\angle AOE = 90^\circ - \angle COE = \boxed{55^\circ}$.


Question: As shown in the figure, one corner of a rectangular piece of paper is folded so that the vertex \(A\) falls on \(A'\), with \(EF\) as the crease. If \(EA'\) exactly bisects \(\angle FEB\), what is the degree measure of \(\angle FEB\)?

Solution:
\textbf{Solution:}\\
Since $EA'$ exactly bisects $\angle FEB$,\\
it follows that $\angle FEA'=\angle A'EB$,\\
From the folding, we get: $\angle FEA=\angle FEA'$,\\
thus $\angle FEA'=\angle A'EB=\angle FEA$,\\
Since $\angle FEA'+\angle A'EB+\angle FEA=180^\circ$,\\
it follows that $\angle FEA'=\angle A'EB=\angle FEA=60^\circ$,\\
Therefore, $\angle FEB=\angle FEA'+\angle A'EB=\boxed{120^\circ}$.


Question: As shown in the figure, points C and D are on line segment AB, where D is the midpoint of line segment AB, $AC = \frac{1}{3}AD$, and $CD = 4\, \text{cm}$. What is the length of line segment AB?

Solution:
\textbf{Solution:} Since $AC = \frac{1}{3}AD$ and $CD = 4cm$,\\
it follows that $CD = AD - AC = AD - \frac{1}{3}AD = \frac{2}{3}AD = 4$,\\
hence $AD = 6$,\\
given that $D$ is the midpoint of segment $AB$,\\
it follows that $AB = 2AD = 12$,\\
thus $AB = \boxed{12}$cm.


Question: Place a set of set squares as shown in the figure, what is the degree measure of $\angle AOB$?

Solution:
\textbf{Solution:} As shown in the figure below: \\
From the diagram, we can see that $\angle 1=30^\circ$, $\angle 2=45^\circ$, \\
$\therefore$ $\angle 3=\angle 1+\angle 2=75^\circ$, \\
$\therefore$ $\angle AOB=180^\circ-\angle 3=\boxed{105^\circ}$,


Question: As shown in the figure, $AB \parallel CD$, $\angle FGB = 155^\circ$, $FG$ bisects $\angle EFD$, what is the measure of $\angle AEF$?

Solution:
\textbf{Solution:} Given $\because$ $\angle FGB=155^\circ$, \\
$\therefore$ $\angle EGF=25^\circ$, \\
$\because$ $AB\parallel CD$, \\
$\therefore$ $\angle EGF=\angle GFD=25^\circ$, \\
$\because$ FG bisects $\angle EFD$, \\
$\therefore$ $\angle EFD=2\angle GFD=50^\circ$, \\
$\therefore$ $\angle AEF=\angle EFD=\boxed{50^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, points D and E are located on sides $AB$ and $AC$ respectively, with $DE\parallel BC$. If $DE:BC = 2:3$, then what is the ratio of ${S}_{\triangle ADE}$ to ${S}_{\triangle ABC}$?

Solution:
\textbf{Solution:} Since $DE\parallel BC$,\\
it follows that $\triangle ADE\sim \triangle ABC$,\\
therefore $\frac{{S}_{\triangle ADE}}{{S}_{\triangle ABC}}=\left(\frac{DE}{BC}\right)^{2}=\left(\frac{2}{3}\right)^{2}=\boxed{\frac{4}{9}}$.


Question: In $\triangle ABC$, points D and E are located on sides AB and AC respectively, such that $DE\parallel BC$ and $AD: BD = 3: 2$. What is the ratio of the area of $\triangle ADE$ to that of quadrilateral $BCED$?

Solution:
\textbf{Solution:} Given that $\because \frac{AD}{BD}=\frac{3}{2}$,\\
$\therefore \frac{AD}{AB}=\frac{3}{5}$,\\
$\because DE \parallel BC$,\\
$\therefore \triangle ADE \sim \triangle ABC$,\\
$\therefore \frac{S_{\triangle ADE}}{S_{\triangle ABC}}=\left(\frac{AD}{AB}\right)^2=\left(\frac{3}{5}\right)^2=\frac{9}{25}$,\\
$\therefore \frac{S_{\triangle ADE}}{S_{\text{quadrilateral} BCED}}=\boxed{\frac{9}{16}}$.


Question: As shown in the figure, the line $y=kx+b$ (where $k$ and $b$ are constants) intersects the $x$ axis and $y$ axis at points $A(-4,0)$ and $B(0,3)$, respectively. The parabola $y=-x^2+2x+1$ intersects the $y$ axis at point $C$. Point $E$ moves on the axis of symmetry of the parabola $y=-x^2+2x+1$, and point $F$ moves on the line $AB$. What is the minimum value of $CE+EF$?

Solution:
\textbf{Solution:} As shown in the figure, let the point $C^{\prime}$ be the symmetric point of C with respect to the axis of symmetry of the parabola, by the property of symmetry we have $CE = C^{\prime}E$, \\
$\therefore CE + EF = C^{\prime}E + EF$, \\
$\therefore$ when points F, E, $C^{\prime}$ are collinear and $C^{\prime}F$ is perpendicular to AB, $CE + EF$ is minimized, \\
According to the problem we have $\left\{\begin{array}{l}-4k+b=0\\ b=3\end{array}\right.$, solving this gives $\left\{\begin{array}{l}k=\frac{3}{4}\\ b=3\end{array}\right.$, \\
$\therefore$ the equation of the line is $y= \frac{3}{4}x+3$; \\
$\because C(0,1)$, $\therefore C^{\prime}(2,1)$, \\
$\therefore$ the equation of line $C^{\prime}F$ is $y=-\frac{3}{4}x+\frac{11}{3}$, \\
From $\left\{\begin{array}{l}y=\frac{3}{4}x+3\\ y=−\frac{3}{4}x+\frac{11}{3}\end{array}\right.$ we solve $\left\{\begin{array}{l}x=\frac{8}{25}\\ y=\frac{81}{25}\end{array}\right.$, \\
$\therefore F\left(\frac{8}{25}, \frac{81}{25}\right)$, \\
$\therefore CF = \sqrt{{\left(\frac{8}{25}−2\right)}^{2}+{\left(\frac{81}{25}−1\right)}^{2}} = \frac{14}{5} = 2.8$, \\
hence, the minimum of $CE+EF$ is $\boxed{2.8}$.


Question: As shown in the figure, AC intersects BD at point E, with AD $\parallel$ BC. If $AE : AC = 1 : 3$, what is the ratio ${S}_{\triangle AED} : {S}_{\triangle CEB}$?

Solution:
\textbf{Solution:} Given $\because AE: AC = 1: 3$,\\
$\therefore AE: EC = 1: 2$,\\
$\because AD \parallel BC$,\\
$\therefore \triangle ADE \sim \triangle CBE$,\\
$\therefore \frac{S_{\triangle ADE}}{S_{\triangle CBE}} = \left(\frac{AE}{EC}\right)^2 = \left(\frac{1}{2}\right)^2 = \boxed{\frac{1}{4}}$


Question: As shown in the figure, $O$ is the centroid of $\triangle ABC$. A straight line passing through $O$ intersects $AB$ and $AC$ at $G$ and $H$ (neither coinciding with the vertices of $\triangle ABC$) respectively. Let ${S}_{\mathrm{quadrilateral\ BCHG}}$ and ${S}_{\triangle AGH}$ represent the areas of the quadrilateral $BCHG$ and the triangle $AGH$, respectively. What is the maximum value of $\frac{{S}_{\mathrm{quadrilateral\ BCHG}}}{{S}_{\triangle AGH}}$?

Solution:
\textbf{Solution}: Draw MN$\parallel$BC through O intersecting AB at N and AC at M. Draw ME$\parallel$AB through M intersecting GH at E. \\
Since O is the centroid of $\triangle ABC$, \\
therefore $\frac{AO}{OD}=2$, where D is the midpoint of BC \\
thus, BD=CD, $\frac{AO}{AD}=\frac{2}{3}$ \\
Since MN$\parallel$BC \\
therefore $\triangle AMN\sim \triangle ACB$ \\
therefore $\frac{NO}{BD}=\frac{MO}{CD}=\frac{AO}{AD}=\frac{2}{3}$, $\frac{{S}_{\triangle AMN}}{{S}_{\triangle ABC}}=\left(\frac{AO}{AD}\right)^{2}=\frac{4}{9}$ \\
thus, MO=NO \\
Since ME$\parallel$AB \\
therefore $\angle MOH=\angle NOG$ \\
therefore $\triangle MOE\cong \triangle NOG$ (AAS) \\
therefore ${S}_{\triangle MOE}={S}_{\triangle NOG}$ \\
Let ${S}_{\triangle MEH}=x$, ${S}_{\triangle ABC}=9y$ \\
therefore ${S}_{\triangle AMN}=4y$, ${S}_{\text{quadrangle}BCMN}=5y$, \\
therefore ${S}_{\text{quadrangle}BCHG}={S}_{\text{quadrangle}BCMN}-{S}_{\triangle HOM}+{S}_{\triangle NOG}$ \\
$={S}_{\text{quadrangle}BCMN}-({S}_{\triangle MOE}+{S}_{\triangle MEH})+{S}_{\triangle NOG}$ \\
$=5y-x$ \\
${S}_{\triangle AHG}={S}_{\triangle AMN}-{S}_{\triangle NOG}+{S}_{\triangle HOM}$ \\
$={S}_{\triangle AMN}-{S}_{\triangle NOG}+({S}_{\triangle MOE}+{S}_{\triangle MEH})$ \\
$=4y+x$ \\
therefore $\frac{{S}_{\text{quadrangle}BCHG}}{{S}_{\triangle AGH}}=\frac{5y-x}{4y+x}$ \\
Since $x$ is a constant \\
therefore, as $y$ becomes smaller, the value of $\frac{{S}_{\text{quadrangle}BCHG}}{{S}_{\triangle AGH}}=\frac{5y-x}{4y+x}$ becomes larger \\
therefore, when $y=0$, $\frac{{S}_{\text{quadrangle}BCHG}}{{S}_{\triangle AGH}}=\boxed{\frac{5}{4}}$ is maximized, at this point GH$\parallel$BC


Question: As shown in the figure, in $\triangle ABC$, points D and E are the midpoints of AB and AC, respectively. If the area of $\triangle ADE$ is 1, what is the area of quadrilateral BDEC?

Solution:
Solution: Since points D and E are the midpoints of AB and AC respectively,

it follows that DE is the midline of $\triangle ABC$,

therefore DE$\parallel$BC, and DE$=\frac{1}{2}$BC,

thus, $\triangle ADE$ is similar to $\triangle ABC$,

which implies $\frac{S_{\triangle ADE}}{S_{\triangle ABC}}=\left(\frac{DE}{BC}\right)^{2}=\frac{1}{4}$,

given that the area of $\triangle ADE$ is 1,

it follows that $S_{\triangle ABC}=4$,

hence, the area of the quadrilateral BCED equals $S_{\triangle ABC}-S_{\triangle ADE}=3$,

Therefore, the answer is: $S_{\text{quadrilateral } BCED}=\boxed{3}$.


Question: As shown in the figure, it is known that $ \triangle ADE \sim \triangle ACB$. If $AD=6$, $DE=4$, and $BC=8$, then what is the length of AC?

Solution:
\textbf{Solution:} Since $\triangle ADE \sim \triangle ACB$,\\
we have $\frac{AD}{AC}=\frac{DE}{CB}$,\\
thus $AC=\frac{AD \cdot CB}{DE}=\boxed{12}$.


Question: As shown in the figure, in right-angled triangle $ABC$, with $\angle ACB = 90^\circ$, squares $ADEB$, $CBFG$, and $ACHI$ are constructed externally on sides $AB$, $BC$, and $AC$ of $\triangle ABC$, respectively. The square $ABED$ is folded along the line $AB$ to obtain square $ABE'D'$, where $AD'$ intersects $CH$ at point $N$, point $E'$ lies on side $FG$, and $D'E'$ intersects $CG$ at point $M$. Let the area of $\triangle ANC$ be $S_1$ and the area of quadrilateral $BCME'$ be $S_2$. Given that $CN = 2NH$ and $S_1 + S_2 = 14$, what is the area of square $ABED$?

Solution:
\textbf{Solution:} Let $NH=x$, then $CN=2x$,\\
According to the problem, we have $CA=CH=3x$,\\
In $\triangle ACN$ and $\triangle BCA$ right triangles,\\
$\angle ACN=\angle BCA=90^\circ$,\\
Since $\angle CAN+\angle CNA=\angle CAN+\angle CAB=90^\circ$,\\
Therefore, $\angle CNA=\angle CAB$,\\
Thus, $\triangle ACN \sim \triangle BCA$ right triangles,\\
Hence, $\frac{CN}{AC}=\frac{AC}{BC}=\frac{2x}{3x}=\frac{2}{3}$,\\
So, $BC=\frac{9}{2}x$,\\
In the $\triangle ABC$ right triangle, by the Pythagorean theorem, we find:\\
$AB^{2}=AC^{2}+BC^{2}=9x^{2}+\frac{81}{4}x^{2}=\frac{117}{4}x^{2}$,\\
Since $S_{ABED}=AB^{2}$,\\
Therefore, $S_{ABED}=\frac{117}{4}x^{2}$,\\
In $\triangle ABN$ and $\triangle D'AM$ right triangles, $\angle ABN=\angle D'AM$, AB=AD', $\angle BAN=\angle D'$,\\
Therefore, $\triangle ABN \cong \triangle D'AM$ (ASA),\\
Hence, $S_{CND'M}=S_{\triangle ABC}$,\\
So, $S_{1}+S_{2}=S_{ABE'D'}-2S_{\triangle ABC}=14$,\\
Thus, $\frac{117}{4}x^{2}-2\times \frac{1}{2}\cdot 3x\cdot \frac{9}{2}x=14$,\\
Solving yields: $x^{2}=\frac{56}{63}$,\\
Therefore, $S_{ABED}=\frac{117}{4}x^{2}=\frac{117}{4}\times \frac{56}{63}=\boxed{26}$,\\


Question: As shown in the figure, it is known that point $M$ is the centroid of $\triangle ABC$, and $AB=18$. If $MN\parallel AB$, what is the value of $MN$?

Solution:
\textbf{Solution:} Given that point $M$ is the centroid of $\triangle ABC$ and $AB=18$,\\
it follows that $AD=DB=\frac{1}{2}AB=9$, and $\frac{CM}{CD}=\frac{2}{3}$.\\
Since $MN\parallel AB$,\\
it implies that $\triangle CMN\sim \triangle CDB$,\\
therefore, $\frac{MN}{BD}=\frac{CM}{CD}=\frac{2}{3}$, which means $\frac{MN}{9}=\frac{2}{3}$,\\
Solving this gives: $MN=\boxed{6}$.


Question: In the figure, in right-angled $\triangle ABC$, $AB=9$, $BC=6$, $\angle B=90^\circ$. Fold $\triangle ABC$ such that point $A$ coincides with the midpoint $D$ of $BC$. If the fold line is $MN$, then the length of segment $AN$ is (　)

Solution:
\textbf{Solution:} Let $AN=x$. By the property of folding, we know $DN=AN=x$, so $BN=9-x$.\\
Since $D$ is the midpoint of $BC$,\\
it follows that $BD=\frac{1}{2}\times 6=3$.\\
In the right triangle $\triangle BDN$, by the Pythagorean theorem, we have: $ND^{2}=NB^{2}+BD^{2}$, that is $x^{2}=(9-x)^{2}+3^{2}$,\\
Solving for $x$ yields: $x=\boxed{5}$.


Question: As shown in the figure, in $ \triangle ABC$, $ \angle ABC = 90^\circ$, $ \angle A = 60^\circ$, and $D$ is a point on side $AC$, with $ \angle DBC = 30^\circ$ and $ AC = 12\,cm$. What is the perimeter of $ \triangle ABD$?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $\angle ABC=90^\circ$, $\angle A=60^\circ$,\\
thus $\angle C=30^\circ$,\\
and since $\angle DBC=30^\circ$,\\
it follows that $\angle ABD=\angle ABC-\angle DBC=60^\circ$, and $BD=DC$,\\
thus $\angle ABD=\angle A=60^\circ$,\\
therefore $\triangle ABD$ is an equilateral triangle,\\
thus $BD=AD=AB$,\\
since $BD=DC$ and $AC=12\text{cm}$,\\
it follows that $AD=DC=\frac{1}{2}AC=6\text{cm}$,\\
thus $BD=AD=AB=6\text{cm}$,\\
therefore, the perimeter of $\triangle ABD$ is: $BD+AD+AB=6\times 3=\boxed{18\text{cm}}$.


Question: As shown in the figure, it is known that $P$ is the centroid of $\triangle ABC$. Connect $AP$ and extend it to intersect $BC$ at point $D$. If the area of $\triangle ABC$ is $20$, what is the area of $\triangle ADC$?

Solution:
\textbf{Solution:} Since $P$ is the centroid of $\triangle ABC$,

it follows that $AD$ is the median of $\triangle ABC$,

hence, the area of $\triangle ADC$ is equal to half the area of $\triangle ABC$,

and since the area of $\triangle ABC$ is 20,

thus, the area of $\triangle ADC$ is $\boxed{10}$.


Question: As shown in the figure, it is known that $ \triangle ABC \cong \triangle DEF$. If $ DF = 6$, $ AB = 3$, $ EF = 5$, and $ DC = 4$, what is the length of $ AD$?

Solution:
\textbf{Solution:} Since $\triangle ABC \cong \triangle DEF$,\\
therefore $AC = DF = 6$,\\
thus $AD = AC - DC = 6 - 4 = \boxed{2}$


Question: As shown in the diagram, in $\triangle ABC$, it is known that $\angle C=90^\circ$, $AC=BC=1$, $AB=\sqrt{2}$, the bisector of $\angle BAC$ intersects side $BC$ at point $D$, and $DE\perp AB$ at point $E$. What is the perimeter of $\triangle DBE$?

Solution:
\textbf{Solution:} Because $DE\perp AB$ and $\angle C=90^\circ$, and $AD$ bisects $\angle BAC$,\\
hence $DC=DE$,\\
thus $DE+BD=DC+BD=BC=1$,\\
in right triangles $\triangle ADC$ and $\triangle ADE$,\\
$\left\{\begin{array}{l}
AD=AD\\
DC=DE
\end{array}\right.$,\\
therefore, by HL (Hypotenuse-Leg), $\triangle ADC \cong \triangle ADE$,\\
so, $AE=AC=1$,\\
therefore $BE=AB-AE=\sqrt{2}-1$,\\
hence the perimeter of $\triangle DBE$ is $BE+DE+BD=\sqrt{2}-1+1=\boxed{\sqrt{2}}$.


Question: As shown in the figure, in $\triangle ABC$, points $D$, $E$, and $F$ are located on the three sides respectively, $E$ is the midpoint of $AC$, $AD$, $BE$, $CF$ intersect at point $G$, $BD=2DC$, ${S}_{\triangle GEC}=3$, ${S}_{\triangle GDC}=4$, then what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} $BD=2DC$,\\
$\therefore {S}_{\triangle ABD}=2{S}_{\triangle ACD}$,\\
$\therefore {S}_{\triangle ABC}=3{S}_{\triangle ACD}$,\\
$\because E$ is the midpoint of $AC$,\\
$\therefore {S}_{\triangle AGE}={S}_{\triangle CGE}$,\\
Also, $\because {S}_{\triangle GEC}=3$, ${S}_{\triangle GDC}=4$,\\
$\therefore {S}_{\triangle ACD}={S}_{\triangle AGE}+{S}_{\triangle CGE}+{S}_{\triangle CGD}=3+3+4=10$,\\
$\therefore {S}_{\triangle ABC}=3{S}_{\triangle ACD}=3\times 10=\boxed{30}$.


Question: As shown in the figure, $P$ is the midpoint of side $AC$ of the equilateral triangle $\triangle ABC$, $E$ is a point on the extension line of side $BC$, and $PE = PB$, what is the degree measure of $\angle CPE$?

Solution:
\textbf{Solution:} Since $P$ is the midpoint of side $AC$ of the equilateral triangle $\triangle ABC$,\\
it follows that $BP$ bisects $\angle ABC$, and $\angle ABC=60^\circ=\angle ACB$,\\
thus, $\angle PBC=30^\circ$,\\
since $PE=PB$,\\
it follows that $\angle PBC=\angle E=30^\circ$,\\
hence, $\angle CPE=\angle ACB-\angle E=\boxed{30^\circ}$.


Question: As shown in the figure, it is known that lines AB $\parallel$ CD $\parallel$ EF, BD $= 2$, and DF $= 4$. What is the value of $ \frac{AC}{AE}$?

Solution:
\textbf{Solution:} Given that $BD=2$ and $DF=4$,\\
thus $BF=BD+DF=2+4=6$;\\
since $AB\parallel CD\parallel EF$,\\
it follows that $\frac{AC}{AE}=\frac{BD}{BF}$,\\
hence $\frac{AC}{CE}=\frac{2}{6}=\boxed{\frac{1}{3}}$.


Question: In the given figure, in $\triangle ABC$, $AH \perp BC$ at $H$, with $BC=12$ and $AH=8$. Points $D$ and $E$ are located on $AB$ and $AC$ respectively, while points $G$ and $F$ are on $BC$. The quadrilateral $DEFG$ is a square. What is the length of the side $DE$ of the square?

Solution:
Solution:As shown in the figure,

$\because$ square DEFG,

$\therefore$ DE=DG, $\angle$NDG=$\angle$DGH=90$^\circ$, DE$\parallel$BC,

$\because$ AH$\perp$BC,

$\therefore$ AH$\perp$DE,

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

$\therefore$ quadrilateral DNHG is a rectangle,

$\therefore$ DG=NH=DE=x,

$\therefore$ AN=AH−NH=8−x,

$\because$ DE$\parallel$BC,

$\therefore$ $\triangle$ADE$\sim$$\triangle$ABC,

$\therefore$ $\frac{DE}{BC}=\frac{AN}{AH}$, that is, $\frac{x}{12}=\frac{8−x}{8}$

Solve it: x=4.8,

Upon verification, x=4.8 is a solution of the equation,

$\therefore$ DE=\boxed{4.8}.


Question: As shown in the figure, in $\triangle ABC$, point D is on side AB, and $\angle A = \angle BCD$, the area ratio $S_{\triangle ADC} : S_{\triangle BDC} = 5 : 4$, $CD = 4$, what is the length of AC?

Solution:
\textbf{Solution}: Given that $S_{\triangle ADC} : S_{\triangle BDC} = 5 : 4$,\\
$\therefore S_{\triangle BDC} : S_{\triangle ABC} = 4 : 9$,\\
$\because \angle A = \angle BCD, \angle B = \angle B$,\\
$\therefore \triangle BCD \sim \triangle BAC$,\\
$\therefore \frac{S_{\triangle BAC}}{S_{\triangle BCD}} = \left(\frac{AC}{CD}\right)^2$,\\
$\therefore \left(\frac{AC}{4}\right)^2 = \frac{9}{4}$,\\
Solving for it, we have: $AC = \boxed{6}$ (taking the positive value).


Question: The large square ABCD is composed of four congruent right-angled triangles and one small square, as shown in the diagram. Draw a perpendicular line from point D to F, intersecting the extension of the diagonal EF of the small square at point G. Connect CG and extend BE to intersect CG at point H. If $AE=2BE$, what is the value of $\frac{CG}{BH}$?

Solution:
\textbf{Solution:} As shown in the figure, draw a line GM$\perp$DG at point G, intersecting the extended line of CF at point M,\\
$\because$DF$\perp$DG,\\
$\therefore$$\angle$GDF=$\angle$DFM=$\angle$MGD=$90^\circ$,\\
$\therefore$Quadrilateral DFMG is a rectangle,\\
$\therefore$DF=MG;\\
$\because$The large square ABCD is composed of four congruent right-angled triangles and a small square,\\
$\therefore$let BE=AT=CN=x,\\
$\because$AE=2BE,\\
$\therefore$BH=CF=2x, TF=DF=x,\\
$\because$Square ETFH,\\
$\therefore$$\angle$TFE=$\angle$DFG=$45^\circ$,\\
In $\triangle$DFG, $\angle$GDF=$90^\circ$,\\
$\therefore$$\triangle$DFG is an isosceles right-angled triangle,\\
$\therefore$DF=DG,\\
$\therefore$Quadrilateral DFMG is a square,\\
$\therefore$MF=DF=x,\\
$\therefore$CM=2x+x=3x\\
In Rt$\triangle$CMG,\\
$CG=\sqrt{M{G}^{2}+C{M}^{2}}=\sqrt{{x}^{2}+{\left(3x\right)}^{2}}=\sqrt{10}x$;\\
$\because$NH$\parallel$MG,\\
$\therefore$$\triangle$CHN$\sim$$\triangle$CMG\\
$\therefore$$\frac{NH}{MG}=\frac{CN}{CM}$, thus $\frac{NH}{x}=\frac{x}{3x}$\\
$\therefore$$NH=\frac{1}{3}x$,\\
$\therefore$$BH=BN+NH=2x+\frac{1}{3}x=\frac{7}{3}x$\\
$\therefore$$\frac{CG}{BH}=\frac{\sqrt{10}x}{\frac{7}{3}x}=\boxed{\frac{3\sqrt{10}}{7}}$.


Question: As shown in the figure, the lines $l_{1} \parallel l_{2} \parallel l_{3}$, and the lines $a$ and $b$ intersect $l_{1}$, $l_{2}$, and $l_{3}$ at points A, B, C and D, E, F, respectively. If $AB: BC=1: 2$ and $DE=2$, then what is the length of $EF$?

Solution:
\textbf{Solution:} Given $\because$ lines $l_{1}\parallel l_{2}\parallel l_{3}$,\\
$\therefore$ $\frac{AB}{BC}=\frac{DE}{EF}=\frac{1}{2}$,\\
$\therefore$ EF=2DE=2$\times$2=$\boxed{4}$.


Question: As shown in the figure, the large square ABCD is composed of four congruent right-angled triangles and a small square. A perpendicular line from point $G$ to $D$ intersects $AB$ at point $I$. If $GI = \frac{4}{3}GD$, what is the value of $\frac{EH}{AD}$?

Solution:
Solution: Draw $IM \perp BG$ at point $M$,\\
since the large square $ABCD$ is composed of four congruent right-angled triangles and a small square,\\
therefore, let $AF=BG=CH=DE=a$, $BF=CG=DH=AE=b$,\\
thus, $EH=HG=DE-DH=a-b$,\\
since $DG \perp IG$,\\
thus, $\angle HGD+\angle HGI=90^\circ$,\\
since $\angle HGI+\angle IGM=90^\circ$,\\
thus, $\angle IGM=\angle HGD$,\\
since $\angle IMG=\angle DHG=90^\circ$,\\
thus, $\triangle IMG \sim \triangle DHG$,\\
thus, $\frac{IG}{DG}=\frac{IM}{DH}=\frac{GM}{GH}$\\
since $GI=\frac{4}{3}GD$\\
thus, $IM=\frac{4}{3}DH=\frac{4}{3}b$, $GM=\frac{4}{3}GH=\frac{4}{3}(a-b)$,\\
thus, $BM=BG-MG=a-\frac{4}{3}(a-b)=\frac{4}{3}b-\frac{1}{3}a$,\\
since $IM \parallel AF$,\\
thus, $\triangle BMI \sim \triangle BFA$,\\
thus, $\frac{IM}{AF}=\frac{BM}{BF}$, solving this gives $a=2b$;\\
thus, $AB^{2}=AD^{2}=AF^{2}+BF^{2}=a^{2}+b^{2}=4b^{2}+b^{2}=5b^{2}$,\\
thus, $AD=\sqrt{5}b$,\\
$EH=a-b=2b-b=b$\\
thus, $\frac{EH}{AD}=\frac{b}{\sqrt{5}b}=\boxed{\frac{\sqrt{5}}{5}}$.


Question: As shown in the figure, within $\triangle ABC$, $D$ and $E$ are the midpoints of sides $AB$ and $AC$, respectively. If the area of $\triangle ADE$ is $\frac{1}{2}$, what is the area of quadrilateral $DBCE$?

Solution:
\textbf{Solution:} Since D and E are the midpoints of AB and AC, respectively,\\
it follows that DE$\parallel$BC, and AE = CE = $\frac{1}{2}$AB,\\
therefore, $\frac{AE}{AC}$ = $\frac{1}{2}$, and $\triangle$ADE$\sim$$\triangle$ABC,\\
thus, $\frac{S_{\triangle ADE}}{S_{\triangle ABC}}$ = $\left(\frac{AE}{AC}\right)^2$ = $\left(\frac{1}{2}\right)^2$ = $\frac{1}{4}$,\\
hence, S$_{\triangle ABC}$ = 4S$_{\triangle ADE}$ = 4$\times$$\frac{1}{2}$ = 2,\\
therefore, S$_{\text{quadrilateral DBCE}}$ = S$_{\triangle ABC}$ - S$_{\triangle ADE}$ = 2 - $\frac{1}{2}$ = $\boxed{\frac{3}{2}}$,


Question: In the Cartesian coordinate system, the position of the square $ABCD$ is shown in the diagram, with the coordinates of point $A$ being $\left(1, 0\right)$ and the coordinates of point $D$ being $\left(0, 2\right)$. Extend $CB$ to meet the $x$-axis at point ${A}_{1}$, and construct the square ${A}_{1}{B}_{1}{C}_{1}C$; extend ${C}_{1}{B}_{1}$ to meet the $x$-axis at point ${A}_{2}$, and construct the square ${A}_{2}{B}_{2}{C}_{2}{C}_{1}$, and so on according to this pattern, what is the area of the 2021st square?

Solution:
\textbf{Solution:} \\
Let point A be (1,0), and point D be (0,2),\\
therefore, OA=1, OD=2, BC=AB=AD=$\sqrt{5}$,\\
considering squares ABCD and A$_{1}$B$_{1}$C$_{1}$C,\\
therefore, $\angle$OAD+$\angle$A$_{1}$AB=$90^\circ$, $\angle$ADO+$\angle$OAD=$90^\circ$,\\
thus, $\angle$A$_{1}$AB=$\angle$ADO,\\
since $\angle$AOD=$\angle$A$_{1}$BA=$90^\circ$,\\
it follows that $\triangle$AOD $\sim$ $\triangle$A$_{1}$BA,\\
thus, $\frac{AO}{{A}_{1}B}=\frac{OD}{AB}$,\\
thus, $\frac{1}{{A}_{1}B}=\frac{2}{\sqrt{5}}$,\\
therefore, ${A}_{1}B=\frac{\sqrt{5}}{2}$,\\
hence, ${A}_{1}{B}_{1}={A}_{1}C={A}_{1}B+BC=\frac{3}{2}\sqrt{5}$,\\
Similarly, it can be derived that, ${A}_{2}{B}_{2}=\frac{9}{4}\sqrt{5}=\left(\frac{3}{2}\right)^{2}\sqrt{5}$,\\
Similarly, it can be derived that, ${A}_{3}{B}_{3}=\left(\frac{3}{2}\right)^{3}\sqrt{5}$,\\
Similarly, it can be derived that, ${A}_{2020}{B}_{2020}=\left(\frac{3}{2}\right)^{2020}\sqrt{5}$,\\
Therefore, the area of the 2021st square is $\left[\left(\frac{3}{2}\right)^{2021−1}\times\sqrt{5}\right]^{2}=5\times\left(\frac{3}{2}\right)^{4040}$.\\
Hence, the answer is: $\boxed{5\times\left(\frac{3}{2}\right)^{4040}}$.


Question: As shown in the figure, line AB $ \parallel $ CD $ \parallel $ EF. If AC = 3 and CE = 4, what is the value of $\frac{BD}{BF}$?

Solution:
\textbf{Solution:} Given that $\because$ AB $\parallel$ CD $\parallel$ EF\\
$\therefore$ $\frac{BD}{BF}=\frac{AC}{AE}$\\
$\because$ AC=3, CE=4\\
$\therefore$ $\frac{BD}{BF}=\boxed{\frac{3}{7}}$


Question: As shown in the figure, in $\triangle ABC$, $AB=6$, $BC=5$, points $D$ and $E$ are on $BC$ and $AC$ respectively, $CD=2BD$, $CE=2AE$, $BE$ intersects $AD$ at point $F$, what is the maximum area of $\triangle AFE$?

Solution:
\textbf{Solution:} Connect DE, and draw DH $\perp$ AB at H, as shown in the figure:\\
$\because$ CD = 2BD, CE = 2AE,\\
$\therefore \frac{CD}{BD}=\frac{CE}{AE}=2$,\\
$\therefore$ DE $\parallel$ AB,\\
$\therefore \triangle CDE \sim \triangle CBA, \triangle DEF \sim \triangle ABF$,\\
$\therefore \frac{DE}{BA}=\frac{CD}{CB}=\frac{2}{3}$, $\frac{DF}{AF}=\frac{DE}{AB}$,\\
$\therefore \frac{DF}{AF}=\frac{DE}{AB}=\frac{2}{3}$,\\
$\because$ DE $\parallel$ AB,\\
$\therefore S_{\triangle ABE}=S_{\triangle ABD}$,\\
$\therefore S_{\triangle ABE} - S_{\triangle ABF}=S_{\triangle ABD} - S_{\triangle ABF}$,\\
$\therefore S_{\triangle AEF}=S_{\triangle BDF}$,\\
$\therefore S_{\triangle AEF}=\frac{2}{5}S_{\triangle ABD}$,\\
$\because BD = \frac{1}{3}BC = \frac{5}{3}$, and $AB = 6$,\\
$\therefore$ when DH is maximal, the area of $\triangle ABD$ is maximal,\\
$\because$ DH $\leq$ BD,\\
$\therefore$ when DH = BD = $\frac{5}{3}$, the area of $\triangle ABD$ is maximal, and the maximum value is $\frac{1}{2} \times \frac{5}{3} \times 6 = 5$,\\
$\therefore$ the maximum area of $\triangle AFE$ is $\frac{2}{5} \times 5 = \boxed{2}$,


Question: As shown in the figure, in $ \triangle ABC$, $ \angle C=90^\circ$, O is a point on side AB, circle $O$ is tangent to both AC and BC. If $ BC=3$ and $ AC=4$, what is the radius of circle $O$?

Solution:
\textbf{Solution:} Construct OD$\perp$AC at D, OE$\perp$BC at E, as shown in the figure, let the radius of $\odot O$ be $r$,\\
since $\odot O$ is tangent to both AC and BC,\\
thus OD=OE=r,\\
and $\angle C=90^\circ$,\\
thus quadrilateral ODCE is a square,\\
thus CD=OD=r,\\
since OD$\parallel$BC,\\
thus $\triangle ADO\sim \triangle ACB$,\\
thus $\frac{AF}{AC}=\frac{OF}{BC}$ \\
since AF=AC−r, BC=3, AC=4,\\
substituting in, we get $\frac{4−r}{4}=\frac{r}{3}$\\
thus $r=\boxed{\frac{12}{7}}$.


Question: As shown in the diagram, in parallelogram $ABCD$, $AC$ and $BD$ intersect at point $O$. $E$ is the midpoint of $OD$. When $AE$ is extended to intersect $DC$ at point $F$, what is the ratio of the area of $\triangle DEF$ to the area of $\triangle OAB$?

Solution:
\textbf{Solution:} Let $O$ be the intersection point of the diagonals of the parallelogram $ABCD$,

$\therefore$ $DO=BO$, $AB\parallel DC$,

Also, since $E$ is the midpoint of $OD$,

$\therefore$ $DE=\frac{1}{2}OD=\frac{1}{4}DB$,

$\therefore$ $DE:EB=1:3$,

Also, since $AB\parallel DC$,

$\therefore$ $\triangle DEF\sim \triangle BAE$,

$\therefore$ $\frac{S_{\triangle DEF}}{S_{\triangle BEA}}=\left(\frac{1}{3}\right)^2=\frac{1}{9}$,

$\therefore$ $S_{\triangle DEF}=\frac{1}{9}S_{\triangle BEA}$,

Since $\frac{S_{\triangle AOB}}{S_{\triangle BEA}}=\frac{OB}{EB}=\frac{2}{3}$,

$\therefore$ $S_{\triangle AOB}=\frac{2}{3}S_{\triangle BEA}$,

$\therefore$ $S_{\triangle DEF}:S_{\triangle OAB}=\frac{\frac{1}{9}S_{\triangle BEA}}{\frac{2}{3}S_{\triangle BEA}}=\boxed{1:6}$


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, a point light source is located at $P(2, 2)$. The coordinates of the ends of the rod $AB$ are respectively $(0, 1)$ and $(3, 1)$. What is the length of the shadow $CD$ of the rod $AB$ on the $x$-axis?

Solution:
\textbf{Solution:} Construct PE$\perp$x-axis at E, intersecting AB at M, as shown in the figure,\\
since P(2, 2), A(0, 1), B(3, 1).\\
Therefore, PM = 1, PE = 2, AB = 3,\\
since AB$\parallel$CD\\
Therefore $\frac{AB}{CD}=\frac{PM}{PE}$ \\
Therefore $\frac{3}{CD}=\frac{1}{2}$ \\
Therefore CD= \boxed{6},


Question: As shown in the figure, $l_{1}\parallel l_{2}\parallel l_{3}$, and lines a, b intersect $l_{1}$, $l_{2}$, $l_{3}$ at points A, B, C, and D, E, F, respectively. If $\frac{AB}{BC}=\frac{2}{3}$ and $DE=4$, what is the length of $DF$?

Solution:
\textbf{Solution:} Since $l_1 \parallel l_2 \parallel l_3$,\\
it follows that $\frac{DE}{EF} = \frac{AB}{BC} = \frac{2}{3}$. Given that $DE=4$,\\
it implies that $EF=6$,\\
hence $DF=DE+EF=\boxed{10}$.


Question: As shown in the figure, on a billiard table, a ball is at point A and is to be shot from point A, reflected off the side cushion CD to hit ball B. Draw AC$\perp$ CD at point C, and BD$\perp$ CD at point D. Given that $\angle AEC=\angle BED$, with AC = 10cm, BD = 15cm, CD = 20cm, if the player is able to hit ball B, what is the length of DE?

Solution:
\textbf{Solution}: Since $AC\perp CD$ and $BD\perp CD$,\\
it follows that $\angle ACE=\angle BDE=90^\circ$,\\
Since $\angle AEC=\angle BED$,\\
it implies that $\triangle AEC\sim\triangle BED$,\\
Therefore, $\frac{AC}{BD}=\frac{CE}{DE}$,\\
Given that $AC=10\text{cm}$, $BD=15\text{cm}$,\\
it leads to $\frac{10}{15}=\frac{20-DE}{DE}$,\\
which results in $DE=\boxed{12}\text{cm}$,\\
Hence, the length of $DE$ is $12\text{cm}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=a$, $BC=b$ ($a>b$). Within $\triangle ABC$, construct $\angle CBD=\angle A$, $\angle DCE=\angle CBD$, $\angle EDF=\angle DCE$, then what is the length of $EF$?

Solution:
\textbf{Solution:} \\
\[
\because \angle CBD = \angle A, \angle BCD = \angle ACB,
\]
\[
\therefore \triangle BCD \sim \triangle ABC,
\]
\[
\therefore \frac{BC}{CD} = \frac{AB}{BC},
\]
\[
\therefore \frac{b}{CD} = \frac{a}{b},
\]
\[
\therefore CD = \frac{b^2}{a},
\]
\[
\because \triangle BCD \sim \triangle ABC,
\]
\[
\therefore \frac{AB}{AC} = \frac{BC}{BD},
\]
\[
\because AB = AC,
\]
\[
\therefore BC = BD = b,
\]
\[
\because \angle DCE = \angle CBD, \angle CDE = \angle BDC,
\]
\[
\therefore \triangle CDE \sim \triangle BDC,
\]
\[
\therefore \frac{DE}{CD} = \frac{CD}{BC},
\]
\[
\therefore DE = \frac{CD^2}{BC} = \frac{b^4}{a^2}{b} = \frac{b^3}{a^2},
\]
\[
\because \triangle CDE \sim \triangle BDC,
\]
\[
\therefore \frac{BD}{BC} = \frac{CD}{CE},
\]
\[
\because BD = BC,
\]
\[
\therefore CD = CE = \frac{b^2}{a},
\]
\[
\because \angle DEC = \angle CED, \angle EDF = \angle DCE,
\]
\[
\therefore \triangle DEF \sim \triangle CDE,
\]
\[
\therefore \frac{DE}{EF} = \frac{CD}{DE},
\]
\[
\therefore EF = \frac{DE^2}{CD} = \frac{\left(\frac{b^6}{a^4}\right)}{\left(\frac{b^2}{a}\right)} = \boxed{\frac{b^4}{a^3}}.
\]


Question: As shown in the figure, it is known that $AB \parallel CD \parallel EF$. If $AC = 6$, $CE = 3$, and $DF = 2$, then what is the length of $BD$?

Solution:
\textbf{Solution:} Given that $AB\parallel CD\parallel EF$,\\
it follows that $\frac{AC}{EC}=\frac{BD}{FD}$,\\
since $AC=6$, $CE=3$, $DF=2$,\\
we have $\frac{6}{3}=\frac{BD}{2}$,\\
thus, $BD=\boxed{4}$.


Question: As shown in the diagram, in parallelogram $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $M$ is the midpoint of $AD$, and line $CM$ intersects $BD$ at point $N$. What is the value of $\frac{S_{\triangle MDN}}{S_{\triangle BCD}}$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a parallelogram, \\
$\therefore AD\parallel BC$ and $AD=BC$, \\
Given that point M is the midpoint of AD, \\
$\therefore MD = \frac{1}{2}AD = \frac{1}{2}BC$, \\
Since $AD\parallel BC$, \\
$\therefore \triangle MDN \sim \triangle CBN$, \\
$\therefore \frac{MD}{BC} = \frac{DN}{BN} = \frac{MN}{NC} = \frac{1}{2}$, \\
$\therefore BN = 2DN$, $CN = 2MN$, \\
$\therefore S_{\triangle CDN} = 2S_{\triangle MDN}$, $S_{\triangle BNC} = 2S_{\triangle CDN}$, \\
$\therefore S_{\triangle BCD} = 6S_{\triangle MDN}$, \\
$\therefore \frac{S_{\triangle MDN}}{S_{\triangle BCD}} = \boxed{\frac{1}{6}}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=2BC$, point $M$ is the midpoint of side $CD$, points $E$ and $F$ are on sides $AB$ and $BC$ respectively, and $AF \perp ME$, with $G$ being the foot of the perpendicular. If $EB=2$ and $BF=1$, what is the area of quadrilateral $BFGE$?

Solution:
\textbf{Solution:} Set $BC=a$, then $AB=2a$, $DM=MC=a$.\\
Construct $MH\perp AB$ at $H$,\\
then $\angle EMH=90^\circ-\angle MEA=\angle FAB$.\\
Hence $Rt\triangle EMH\backsim Rt\triangle FAB$.\\
Therefore, $\frac{MH}{HE}=\frac{AB}{BF}$,\\
that is, $\frac{a}{a-2}=\frac{2a}{1}$,\\
solving this gives $a=\frac{5}{2}$.\\
Thus, $BC=\frac{5}{2}$, $AB=5$.\\
Therefore, $AF=\sqrt{AB^2+BF^2}=\sqrt{5^2+1^2}=\sqrt{26}$,\\
$S_{\triangle ABF}=\frac{1}{2}AB\times BF=\frac{1}{2}\times 5\times 1=\frac{5}{2}$.\\
Also, $Rt\triangle AEG\backsim Rt\triangle AFB$,\\
Therefore, $\frac{S_{\triangle AEG}}{S_{\triangle AFB}}=\left(\frac{AE}{AF}\right)^2=\left(\frac{5-2}{\sqrt{26}}\right)^2=\frac{9}{26}$.\\
Thus, $S_{\triangle AEG}=\frac{9}{26}S_{\triangle AFB}=\frac{9}{26}\times \frac{5}{2}=\frac{45}{52}$.\\
Thus, $S_{\text{quadrilateral} BFGE}=S_{\triangle AFB}-S_{\triangle AEC}=\frac{5}{2}-\frac{45}{52}=\boxed{\frac{85}{52}}$.


Question: As shown in the figure, five congruent isosceles triangles are pieced together to form two regular pentagons of different sizes, inside and outside. If the side length of the smaller regular pentagon is 1, what is the side length of the larger regular pentagon?

Solution:
\textbf{Solution:} Extend CF to intersect BA at point H, \\
since assembling 5 congruent isosceles triangles forms two differently sized regular pentagon patterns, \\
therefore $\angle$BAE=$\angle$CBA=108$^\circ$, \\
therefore $\angle$ABG+$\angle$BAG=108$^\circ$, \\
$\angle$ABG+$\angle$BAG+$\angle$AGB=180$^\circ$, \\
therefore $\angle$AGB=$\angle$BAG=180$^\circ$−108$^\circ$=72$^\circ$, \\
therefore $\angle$ABG=180$^\circ$−72$^\circ \times 2$=$36^\circ$, \\
since AB=BC, \\
therefore $\angle$BHC=(180$^\circ$−108$^\circ$)÷2=$36^\circ$, \\
therefore $\angle$CBF=$\angle$CFB=108$^\circ$−36$^\circ$=72$^\circ$=$\angle$AFG, \\
therefore $\angle$FHG=180$^\circ$−72$^\circ \times 2$=36$^\circ$=$\angle$ABG, \\
therefore points A and H coincide, \\
therefore $\angle$FAG=72$^\circ$−36$^\circ$=36$^\circ$=$\angle$ABG, \\
therefore $\triangle$BAG$\sim$$\triangle$FGA, \\
therefore $\frac{BA}{AG}=\frac{AG}{FG}$ \\
Let BF=AF=AG=x, then BG=AB=x+1, \\
therefore $\frac{1+x}{x}=\frac{x}{1}$ \\
Solving it: $x_{1}=\frac{1+\sqrt{5}}{2}$, $x_{2}=\frac{1−\sqrt{5}}{2}$ (discard since it does not meet the conditions), \\
therefore AB=1+$\frac{1+\sqrt{5}}{2}$=\boxed{\frac{3+\sqrt{5}}{2}}.


Question: As shown in the figure, $A$ and $B$ are located on the horizontal lines of the circular grid. Points $C$ and $D$ are the intersections of the diameter $AB$ with the grid lines. What is the ratio of $BC:CD:DA$?

Solution:
\textbf{Solution:} Given the figure, draw CE$\perp$BE at point E, BF$\perp$CF at point F, and AG$\perp$DG at point G,\\
thus $\angle$BEC=$\angle$CFD=$\angle$AGD=$90^\circ$,\\
thus BE$\parallel$CF$\parallel$DG,\\
thus $\angle$B=$\angle$FCD=$\angle$ADG,\\
thus $\triangle$BEC$\sim$ $\triangle$CFD$\sim$ $\triangle$DGA,\\
thus BC$\colon$ CD$\colon$ AD=EC$\colon$ FD$\colon$ GA=1$\colon$ 3$\colon$ 2.\\
Therefore, the answer is: BC$\colon$ CD$\colon$ AD=\boxed{1\colon 3\colon 2}.


Question: As shown in the figure, in $ \triangle ABC$, $E$ and $F$ are located on sides $AB$ and $AC$ respectively, with $EF\parallel BC$. Given that $AB=3$, $AE=2$, and $EF=4$, what is the length of $BC$?

Solution:
\textbf{Solution:} Since $EF\parallel BC$,\\
thus $\frac{EF}{BC} = \frac{AE}{AB}$, so $\frac{4}{BC} = \frac{2}{3}$\\
Solving this yields: $BC=\boxed{6}$.


Question: As shown in the figure, $l_1\parallel l_2\parallel l_3$, the lines $a, b$ intersect $l_1, l_2, l_3$ at points $A, B, C$ and $D, E, F$ respectively. If $AB:BC=2:3$ and $EF=6$, what is the length of $DE$?

Solution:
\textbf{Solution:} From the theorem that parallel lines divide segments proportionally, we have: $\frac{DE}{EF} = \frac{AB}{BC}$, that is, $\frac{DE}{6} = \frac{2}{3}$, \\
Thus, we find $DE = \boxed{4}$.


Question: As shown in the figure, in $\triangle ABC$, points D and E are respectively the midpoints of AB and AC. If the area of $\triangle ABC$ is 16, what is the area of quadrilateral BCED?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, points D and E are the midpoints of AB and AC, respectively,\\
then $DE \parallel BC$ and $DE= \frac{1}{2}BC$,\\
thus $\triangle ADE \sim \triangle ABC$,\\
since $\frac{DE}{BC}=\frac{1}{2}$,\\
it follows that $\frac{S_{\triangle ADE}}{S_{\triangle ABC}}=\frac{1}{4}$,\\
given that $S_{\triangle ABC}=16$,\\
then $S_{\triangle ADE}=\frac{1}{4}S_{\triangle ABC}=\frac{1}{4}\times 16=4$,\\
therefore the area of quadrilateral $BCED= S_{\triangle ABC}-S_{\triangle ADE}=16-4=\boxed{12}$.


Question: As shown in the figure, in $ \triangle ABC$, $DE$ is the median line of $ \triangle ABC$. $DC$ and $BE$ intersect at point $F$. If $ {S}_{\triangle DEF} = 1$, then what is $ {S}_{\triangle ADE}$?

Solution:
\textbf{Solution:} Since $DE$ is the median of $\triangle ABC$, \\
it follows that $DE \parallel BC$ and $BC = 2DE$, \\
hence $\triangle DEF \sim \triangle CBF$, \\
thus $\frac{{S}_{\triangle CBF}}{{S}_{\triangle DEF}} = \left(\frac{BC}{DE}\right)^2 = 2^2$, \\
therefore ${S}_{\triangle CBF} = 4{S}_{\triangle DEF}$, \\
since ${S}_{\triangle DEF} = 1$, \\
then ${S}_{\triangle CBF} = 4$, \\
since $BE$ is the median, \\
it follows that ${S}_{\triangle ABE} = {S}_{\triangle CBE}$, \\
since $DE$ is the median of $\triangle ABC$, \\
it follows that $DE \parallel BC$, \\
thus ${S}_{\triangle BDE} = {S}_{\triangle CDE}$, \\
hence ${S}_{\triangle BDF} = {S}_{\triangle CFE}$, \\
therefore ${S}_{\triangle BDF} + {S}_{\triangle ADE} + {S}_{\triangle DEF} = {S}_{\triangle CFE} + {S}_{\triangle CBF}$, \\
thus ${S}_{\triangle ADE} + {S}_{\triangle DEF} = {S}_{\triangle CBF}$, \\
therefore ${S}_{\triangle ADE} = \boxed{3}$.


Question: As shown in the figure, rhombus $ABCD$ has a side length of $4$, and points $E$ and $F$ are located on $AB$ and $AD$, respectively. $AC$ intersects $EF$ at point $G$. If $BE = AF = 1$ and $\angle BAD = 120^\circ$, what is the length of $FG$?

Solution:
\textbf{Solution:} Construct $EM \parallel BC$ intersecting $AC$ at $M$, and $EN \perp BC$ at $N$, as shown in the diagram:\\
Since the side length of rhombus $ABCD$ is 4, and $\angle BAD = 120^\circ$,\\
therefore, $AB = BC = 4$, $\angle BAC = \angle FAC = \frac{1}{2}\angle BAD = 60^\circ$, $AD \parallel BC$,\\
therefore, $\triangle ABC$ is an equilateral triangle,\\
therefore, $\angle B = \angle ACB = 60^\circ$, $BC = AC$,\\
since $EM \parallel BC$,\\
therefore, $EM \parallel AD$, $\angle AEM = \angle B = 60^\circ = \angle BAC$,\\
therefore, $\triangle AEM$ is an equilateral triangle,\\
therefore, $AM = AE = AB - BE = 4 - 1 = 3$,\\
since $EM \parallel AD$,\\
therefore, $\triangle AGF \sim \triangle MGE$,\\
therefore, $\frac{FG}{EG} = \frac{AF}{EM} = \frac{1}{3}$,\\
therefore, $FG = \frac{1}{4}EF$,\\
In $\triangle BCE$ and $\triangle ACF$,\\
$\begin{cases} BC = AC \\ \angle B = \angle FAC \\ BE = AF \end{cases}$,\\
therefore, $\triangle BCE \cong \triangle ACF$ (SAS),\\
therefore, $CE = CF$, $\angle BCE = \angle ACF$,\\
therefore, $\angle ACF + \angle ACE = \angle ACF + \angle ACE = \angle ACB = 60^\circ$,\\
therefore, $\triangle CEF$ is an equilateral triangle,\\
therefore, $EF = CE$,\\
since $EN \perp BC$, $\angle B = 60^\circ$,\\
therefore, $\angle BEN = 30^\circ$,\\
therefore, $BN = \frac{1}{2}BE = \frac{1}{2}$,\\
therefore, $EN = \sqrt{3}BN = \frac{\sqrt{3}}{2}$, $CN = BC - BN = 4 - \frac{1}{2} = \frac{7}{2}$,\\
therefore, $EF = CE = \sqrt{EN^2 + CN^2} = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{7}{2}\right)^2} = \sqrt{13}$,\\
therefore, $FG = \frac{1}{4}EF = \boxed{\frac{\sqrt{13}}{4}}$.


Question: As shown in the diagram, in rectangle $ABCD$, with $AB=2$ and $CB=4$, connect $AC$. Form rectangle $ACC_1B_1$ with diagonal $AC$ as a side, rotating in a counterclockwise direction, such that rectangle $ACC_1B_1\sim$ rectangle $ADCB$. Then, connect $AC_1$ and form rectangle $AC_1C_2B_2$ with diagonal $AC_1$ as a side, rotating in a counterclockwise direction, so that rectangle $AC_1C_2B_2\sim$ rectangle $ACC_1B_1$, and so on, following this pattern. Let the area of rectangle $ABCD$ be denoted as $S_1$, the area of rectangle $ACC_1B_1$ as $S_2$, the area of rectangle $AC_1C_2B_2$ as $S_3$, ..., then what is the value of $S_{2021}$?

Solution:
\textbf{Solution:}\\
$\because$ Quadrilateral $ABCD$ is a rectangle,\\
$\therefore AB\perp BC$,\\
$\therefore AC=\sqrt{AB^{2}+CB^{2}}=\sqrt{2^{2}+4^{2}}=2\sqrt{5}$,\\
$\because$ Constructing a similar rectangle $AB_{1}C_{1}C$ to rectangle $ABCD$ in a counterclockwise direction,\\
$\therefore$ The ratio of the side lengths of rectangle $AB_{1}C_{1}C$ to rectangle $ABCD$ is $\sqrt{5}:2$,\\
$\therefore$ The ratio of the area of rectangle $AB_{1}C_{1}C$ to the area of rectangle $ABCD$ is $5:4$,\\
$\because S_{1}=2\times 4=8$,\\
$S_{2}=8\times \frac{5}{4}$,\\
$S_{3}=8\times \left(\frac{5}{4}\right)^{2}$, $\dots$\\
$\therefore S_{2020}=8\times \left(\frac{5}{4}\right)^{2019}$,\\
$S_{2021}=8\times \left(\frac{5}{4}\right)^{2020}=\boxed{8\times \left(\frac{5}{4}\right)^{2020}}$.


Question: As shown in the figure, in rectangle $ABCD$, $E$, $F$, and $G$ are on $AB$, $BC$, and $CD$ respectively, $DE \perp EF$, $EF \perp FG$, $BE=3$, $BF=2$, and $FC=6$. What is the length of $DG$?

Solution:
\textbf{Solution:} Let us consider right-angled triangle $BEF$, where $BF=2$ and $BE=3$\\
Thus, $EF= \sqrt{BE^{2}+BF^{2}}=\sqrt{3^{2}+2^{2}}=\sqrt{13}$\\
Draw a line from $G$ such that $GH \perp DE$, with $H$ as the foot of the perpendicular,\\
Since $DE \perp EF$ and $EF \perp FG$\\
Hence, quadrilateral $EFGH$ is a rectangle\\
Therefore, $HG=EF= \sqrt{13}$\\
Given rectangle $ABCD$\\
Therefore, $\angle A=\angle B=90^\circ$\\
Thus, $\angle AED+\angle ADE=90^\circ$\\
Since $DE \perp EF$\\
Therefore, $\angle AED+\angle BEF=90^\circ$\\
Thus, $\angle BEF=\angle ADE$\\
Again, since $\angle A=\angle B=90^\circ$\\
Therefore, $\triangle EBF\sim \triangle DAE$\\
Similarly, $\triangle DAE\sim \triangle GHD$\\
Thus, $\triangle EBF\sim \triangle GHD$\\
Therefore, $\frac{DG}{EF}=\frac{HG}{BE}$, that is $\frac{DG}{\sqrt{13}}=\frac{\sqrt{13}}{3}$, solving this gives $DG= \boxed{\frac{13}{3}}$.


Question: As shown in the figure, lines $l_1 \parallel l_2 \parallel l_3$, and lines AC, DF intersect $l_1$, $l_2$, $l_3$ at points A, B, C and D, E, F, respectively. Connect AF. Let $BG \parallel AF$. If $\frac{DE}{EF} = \frac{2}{3}$ and $BG = 6$, then what is the length of AF?

Solution:
\textbf{Solution:} Since $l_1 \parallel l_2 \parallel l_3$, $\frac{DE}{EF}=\frac{2}{3}$\\
Therefore, $\frac{AB}{BC}=\frac{DE}{EF}=\frac{2}{3}$\\
Therefore, $\frac{BC}{AC}=\frac{3}{5}$\\
Since $BG \parallel AF$,\\
Therefore, $\triangle CBG \sim \triangle CAF$,\\
Therefore, $\frac{CB}{AC}=\frac{BG}{AF}$\\
Therefore, $\frac{3}{5}=\frac{6}{AF}$\\
Therefore, $AF=\boxed{10}$


Question: As shown in the figure, the medians $AD$ and $BE$ of $\triangle ABC$ intersect each other at point $F$. Through point $E$, draw $EG\parallel AD$ where it intersects $BC$ at point $G$. What is the value of the ratio $EG\colon AF$?

Solution:
\textbf{Solution:} Connect DE.\\
AD, BE are the medians of the triangle\\
$\therefore$DE$\parallel$AB, DE=$\frac{1}{2}$AB\\
$\therefore$$\triangle$DEF$\sim$$\triangle$ABF\\
$\therefore$$\frac{DF}{AF}=\frac{DE}{AB}=\frac{1}{2}$\\
$\therefore$AF=$\frac{2}{3}AD$\\
$\because$ED$\parallel$AD\\
$\therefore$$\triangle$EGC$\sim$$\triangle$ADC\\
$\therefore$$\frac{EG}{AD}=\frac{CE}{AC}=\frac{1}{2}$\\
$\therefore$EG=$\frac{1}{2}AD$\\
$\therefore$EG$\colon$AF=$\boxed{\frac{3}{4}}$


Question: In parallelogram ABCD shown in the figure, F is the midpoint of BC. Extend AD to E such that $AD: AE = 3: 4$. Connect EF and intersect DC at point G. What is the ratio ${S}_{\triangle DEG} : {S}_{\triangle CFG}$?

Solution:
\textbf{Solution:} Given that $AD: AE = 3:4$,\\
let $AD=3x$ and $AE=4x$,\\
hence $DE=AE-AD=4x-3x=x$,\\
since quadrilateral ABCD is a parallelogram,\\
it follows that $AD\parallel BC$ and $BC=AD=3x$,\\
given that point F is the midpoint of BC,\\
thus $CF=\frac{1}{2}BC=\frac{3}{2}x$,\\
since $AD\parallel BC$,\\
it results in $\angle DEG=\angle CFG$ and $\angle EDG=\angle FCG$,\\
hence $\triangle DEG\sim \triangle CFG$,\\
thus $\frac{{S}_{\triangle DEG}}{{S}_{\triangle CFG}}=\left(\frac{DE}{CF}\right)^{2}=\left(\frac{x}{\frac{3}{2}x}\right)^{2}=\boxed{\frac{4}{9}}$


Question: As shown in the figure, $AD\parallel BE\parallel CF$, the lines $l_1, l_2$ intersect these three parallel lines at points A, B, C, and points D, E, F, respectively. If $AB=9, BC=6, EF=4$, what is the length of DE?

Solution:
\textbf{Solution:} From the property that parallel lines divide line segments proportionally, we know that $\frac{AB}{BC}=\frac{DE}{EF}$\\
$\therefore$ $\frac{9}{6}=\frac{DE}{4}$\\
Solving this, we find $DE=\boxed{6}$


Question: As shown in the figure, the diameter \(CD=30\) of the circle \(\odot O\), and \(AB\) is a chord of \(\odot O\) such that \(AB \perp CD\) with the foot of the perpendicular being \(M\), and \(OM:OC=3:5\). What is the length of \(AB\)?

Solution:
\textbf{Solution:} Since the diameter $CD$ of the circle $\odot O$ is 30, and $AB\perp CD$, with $OM\colon OC=3\colon 5$,\\
it follows that $AO=OC=15$ and $OM=\frac{3}{5}OC=\frac{3}{5}\times 15=9$. Given that $AM=MB$,\\
then $AM=\sqrt{AO^{2}-MO^{2}}=\sqrt{15^{2}-9^{2}}=12$,\\
thus, $AB=2AM=\boxed{24}$.


Question: As shown in the figure, it is known that A, B, and C are three points on the circle $\odot$O. If $\angle AOB=70^\circ$, what is the degree measure of $\angle ACB$?

Solution:
\textbf{Solution:} Since $\angle AOB=70^\circ$,\\
then $\angle ACB=\frac{1}{2}\angle AOB=35^\circ$,\\
therefore, the answer is: $\boxed{35^\circ}$.


Question: In the figure, within circle $O$, $AB$ is the diameter of circle $O$, and chord $CD$ is perpendicular to $AB$ at point H. If $AH=5$ and $HB=1$, then what is the length of $CD$?

Solution:
\textbf{Solution:} Link $OD$, \\
$\because$ $AB$ is the diameter of the circle $\odot O$, and the chord $CD\perp AB$, \\
$\therefore$ $CD=2HD$, \\
$\because$ $AH=5$, $HB=1$, \\
$\therefore$ $AB=AH+HB=6$, \\
$\therefore$ $OD=OA=3$, \\
$\therefore$ $OH=AH−OA=2$, \\
In $\triangle ODH$, $DH=\sqrt{OD^{2}−OH^{2}}=\sqrt{5}$, \\
$\therefore$ $CD=2DH=2\sqrt{5}$, \\
Thus, the answer is: $CD=\boxed{2\sqrt{5}}$.


Question: As shown in the figure, points A, B, and C are on the circle $\odot O$, with $OA\perp OB$. What is the degree measure of $\angle ACB$?

Solution:
\textbf{Solution:} Since $OA \perp OB$, \\
$\therefore$ $\angle AOB=90^\circ$,\\
$\therefore$ $\angle ACB=\frac{1}{2}\angle AOB=45^\circ$,\\
Thus, the answer is: $\boxed{45^\circ}$.


Question: As shown in the diagram, $AB$ is the diameter of $\odot O$, $AC$ is a chord, and $\angle BAC=25^\circ$. A point D is arbitrarily chosen on $\odot O$, and point D and point C are located on opposite sides of diameter $AB$. Connecting $AD$ and $DC$, what is the measure of $\angle D$?

Solution:
\textbf{Solution:} Draw line $BC$, as shown in the diagram,\\
since $AB$ is the diameter of $\odot O$,\\
thus $\angle ACB=90^\circ$,\\
since $\angle BAC=25^\circ$,\\
thus $\angle B=65^\circ$,\\
since $\overset{\frown}{AC}=\overset{\frown}{AC}$,\\
therefore $\angle D=\angle B=\boxed{65^\circ}$;


Question: As shown in the figure, $BD$ is the diameter of $\odot O$, with $A$ and $C$ on the circle. If $\angle A=50^\circ$, what is the measure of $\angle DBC$?

Solution:
\textbf{Solution:} Given that BD is the diameter of circle $O$,\\
it follows that $\angle BCD = 90^\circ$,\\
which implies $\angle D = \angle A = 50^\circ$,\\
thus, $\angle DBC = 90^\circ - \angle D = \boxed{40^\circ}$,


Question: As shown in the figure, $AB$ is the diameter of the circle $\odot O$, with chord $CD \perp AB$. Connecting $OD$ and $BC$, if $\angle CBA = 70^\circ$, what is the measure of $\angle BOD$?

Solution:
\textbf{Solution:} Since $CD\perp AB$ and $\angle CBA=70^\circ$, \\
$\therefore \angle BCD=20^\circ$, \\
$\therefore \angle BOD=2\angle BCD=40^\circ$, \\
Hence, the answer is: $\boxed{40^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and chord $CD$ is perpendicular to $AB$ at point $E$, $OC=5\,cm$, $CD=8\,cm$. What is the length of $OE$?

Solution:
\textbf{Solution:} As shown in the diagram:\\
Since $AB$ is the diameter of the circle, and $CD \perp AB$,\\
Therefore, $\angle OEC = 90^\circ$, $CE = \frac{1}{2}CD = 4\,\text{cm}$,\\
In $\triangle OCE$ right-angled at $E$, $OE = \sqrt{OC^{2} - CE^{2}} = \sqrt{5^{2} - 4^{2}} = \boxed{3}\,\text{cm}$


Question: As shown in the diagram, the quadrilateral ABCD is inscribed in circle $\odot O$, with point C being the midpoint of arc $\overset{\frown}{BD}$. If $\angle A=50^\circ$, what is the measure of $\angle CBD$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is inscribed in circle O, $\angle A=50^\circ$, \\
$\therefore \angle C=180^\circ-\angle A=180^\circ-50^\circ=130^\circ$, \\
since point C is the midpoint of arc BD, \\
$\therefore CD=CB$, \\
$\therefore \angle CBD=\angle CDB$, \\
$\therefore \angle CBD=\frac{1}{2}\times(180^\circ-130^\circ)=\boxed{25^\circ}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $O$. Point $I$ is the intersection point of the angle bisectors of $\triangle ABC$, and $\angle AIC = 124^\circ$. Point $E$ is on the extension line of side $AD$. What is the measure of $\angle CDE$?

Solution:
\textbf{Solution:} Given that point I is the intersection of the angle bisectors of $\triangle ABC$, \\
$\therefore \angle BAC = 2\angle IAC$, $\angle ACB = 2\angle ICA$, \\
Given $\angle AIC = 124^\circ$, \\
$\therefore \angle IAC + \angle ICA = 180^\circ - \angle AIC = 56^\circ$, \\
$\therefore \angle B = 180^\circ - (\angle BAC + \angle ACB) = 180^\circ - 2(\angle IAC + \angle ICA) = 180^\circ - 2 \times 56^\circ = 68^\circ$, \\
Furthermore, since quadrilateral ABCD is inscribed in circle O, \\
$\therefore \angle CDE = \angle B = \boxed{68^\circ}$.


Question: As shown in the figure, it is known that $AB$ is the diameter of the circumcircle of $\triangle ABC$, and $\angle A=25^\circ$, then what is the degree measure of $\angle B$?

Solution:
\textbf{Solution:} Since $AB$ is the diameter of the circumcircle of $\triangle ABC$, \\
then $\angle ACB = 90^\circ$, \\
thus $\angle A + \angle B = 90^\circ$, \\
moreover, since $\angle A = 25^\circ$, \\
therefore $\angle B = 90^\circ - \angle A = 90^\circ - 25^\circ = \boxed{65^\circ}$.


Question: As shown in the figure, AB is a chord of circle $O$, radius OC is perpendicular to AB at point D, and AB = 6, OD = 4. What is the length of DC?

Solution:
\textbf{Solution:} Extend line OB. \\
Since AB is a chord of circle O, and radius OC is perpendicular to AB at point D, and AB=6, \\
therefore BD=3, let CD=x, then OC=OB=x+4, \\
in $\triangle BDO$, $OD^{2}+BD^{2}=OB^{2}$, \\
thus $4^{2}+3^{2}=(x+4)^{2}$, \\
solving this, we find $x=\boxed{1}$, \\
therefore, CD=1.


Question: As shown in the figure, in the circle $\odot O$, the diameter $CD$ is perpendicular to the chord $AB$ at point $E$, and $OB$ and $BC$ are connected. Given that the radius of $\odot O$ is $2$ and $AB=2$, what is the measure of $\angle BCD$?

Solution:
\textbf{Solution:} Given that diameter $CD$ is perpendicular to chord $AB$ at point $E$, and $AB=2$,\\
it follows that $\angle OEB=90^\circ$, and $BE=1$,\\
since $OB=2$,\\
thus $\angle BOE=30^\circ$,\\
therefore $\angle BCD=\boxed{15^\circ}$.


Question: As shown in the figure, for the square $ABCD$ with a side length of 4, and a semicircle $O$ with diameter $BC$ intersecting the diagonal $AC$ at point $E$, what is the area of the shaded region?

Solution:
\textbf{Solution:} As shown in the figure,\\
$\because$ $BC$ is the diameter of the circle.\\
$\therefore$ $BE\perp AC$\\
Also, $\because$ $BC=AB$\\
$\therefore$ $BE=EA=EC$,\\
$\because$ $O$ is the midpoint of $BC$\\
$\therefore$ $EO$ is the median of $\triangle ABC$,\\
$\therefore$ $EO\parallel AB$,\\
$\therefore$ $\angle BOE=90^\circ$,\\
$\therefore$ the area of the shaded part $=S_{\triangle ABC}-S_{\text{sector} BOE}$\\
$=\frac{1}{2}AB\times BC-\frac{90^\circ\times \pi \times OB^{2}}{360}$\\
$=\frac{1}{2}\times 4\times 4-\frac{90^\circ\times \pi \times 2^{2}}{360}$\\
$=8-\pi$.\\
Therefore, the answer is: \boxed{8-\pi}.


Question: As shown in the figure, the cross-section of a horizontally placed cylindrical drainage pipe is $\odot O$, with the height of the water-containing segment being 2, and the chord $AB=4\sqrt{3}$. What is the radius of the cross-section?

Solution:
\textbf{Solution:} Draw OD$\perp$AB through point O, with the foot at point C, intersecting $\odot O$ at point D,\\
since $AB=4\sqrt{3}$ and $OD\perp AB$,\\
therefore $AC=\frac{1}{2}AB=2\sqrt{3}$,\\
Let the radius be $r$,\\
since $CD=2$,\\
therefore $OC=r-2$,\\
In $\triangle OAC$, by the Pythagorean theorem, we obtain:\\
$OC^2+AC^2=OA^2$, which means ${(r-2)}^2+{(2\sqrt{3})}^2=r^2$,\\
Solving yields: $r=\boxed{4}$.


Question: As shown in the figure, $C$ is a point on the semicircle $O$ with $AB$ as its diameter. Connect $AC$ and $BC$. Construct squares $ACDE$ and $BCFG$ externally on sides $AC$ and $BC$, respectively. Let $M$, $N$, $P$, and $Q$ be the midpoints of $DE$, $FG$, arc $AC$, and arc $BC$, respectively. If $MP + NQ = 14$ and $AC + BC = 18$, what is the length of $AB$?

Solution:
\textbf{Solution:} As shown in the figure, connect OP and OQ which intersect AC and BC at points I and H respectively,\\
$\because$ The midpoints of DE, FG, $\overset{\frown}{AC}$, and $\overset{\frown}{BC}$ are M, N, P, and Q respectively,\\
$\therefore$ OP $\perp$ AC, OQ $\perp$ BC,\\
$\therefore$ H and I are the midpoints of AC and BD,\\
$\therefore$ OH+OI=$\frac{1}{2}$(AC+BC)=9,\\
$\therefore$ MH+NI=AC+BC=18, MP+NQ=14,\\
$\therefore$ PH+QI=18−14=4,\\
$\therefore$ AB=OP+OQ=OH+OI+PH+QI=9+4=\boxed{13},


Question: As shown in the figure, quadrilateral $ABCD$ is an inscribed quadrilateral of circle $\odot O$, the radius of $\odot O$ is 2, and $\angle B=135^\circ$. What is the measure of $\angle ADC$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in circle $O$, and $\angle B=135^\circ$,

$\therefore$ $\angle ADC=180^\circ-\angle B=45^\circ$,

thus $\angle ADC=\boxed{45^\circ}$


Question: As shown in the figure, $AB$ is the diameter of $\odot O$. Points $D$ and $C$ are on $\odot O$. Connect $AD$, $DC$, $AC$. If $\angle C=65^\circ$, then what is the degree measure of $\angle BAD$?

Solution:
\textbf{Solution:} As shown in the figure, connect BD.\\
Since $AB$ is the diameter of $\odot O$,\\
$\therefore$ $\angle ADB=90^\circ$.\\
Since $\angle C=65^\circ$,\\
$\therefore$ $\angle ABD=65^\circ$,\\
$\therefore$ $\angle BAD=90^\circ-65^\circ=\boxed{25^\circ}$.


Question: As shown in the figure, within circle $O$, there is a broken line $OABC$, where $OA=8$, $AB=12$, and $\angle A=\angle B=60^\circ$. What is the length of $BC$?

Solution:
\textbf{Solution:} Extend $AO$ to meet $BC$ at point $D$, and draw $OE\perp BC$ at point $E$.\\
$\because \angle A=\angle B=60^\circ$,\\
$\therefore \angle ADB=60^\circ$,\\
$\therefore \triangle ADB$ is an equilateral triangle,\\
$\therefore BD=AD=AB=12$,\\
$\therefore OD=4$,\\
Furthermore, $\because \angle ADB=60^\circ$,\\
$\therefore DE=OD\cdot \cos 60^\circ=\frac{1}{2}OD=2$,\\
$\therefore BE=10$,\\
$\because OE\perp BC$,\\
$\therefore BC=2BE=\boxed{20}$.


Question: As shown in the figure, the cross-section of a drainpipe is a circle with a radius of $5$. If the width of the water surface $AB = 8$, then what is the depth of the water $CD$?

Solution:
\textbf{Solution:} Connect OB. \\
According to the problem, OD $\perp$ AB, intersecting AB at point C, \\
since AB = 8, \\
thus BC = $\frac{1}{2}$ AB = $\frac{1}{2} \times 8$ = 4, \\
In $\triangle OBC$, \\
since OB = 5, BC = 4, \\
thus OC = $\sqrt{OB^{2} - BC^{2}} = \sqrt{5^{2} - 4^{2}}$ = 3, \\
therefore CD = OD - OC = 5 - 3 = \boxed{2}.


Question: As shown in the figure, point D is on the circumcircle of the equilateral triangle $\triangle ABC$, and $\angle DAC = 20^\circ$. What is the measure of $\angle ACD$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is an equilateral triangle,\\
it follows that $\angle BAC = \angle ACB = 60^\circ$,\\
thus $\angle BAD = \angle BAC + \angle DAC = 60^\circ + 20^\circ = 80^\circ$,\\
since quadrilateral ABCD is a cyclic quadrilateral,\\
it follows that $\angle BCD = 180^\circ - \angle BAD = 180^\circ - 80^\circ = 100^\circ$,\\
therefore $\angle ACD = \angle BCD - \angle ACB = 100^\circ - 60^\circ = \boxed{40^\circ}$.


Question: As shown in the figure, in $ \triangle ABC$, $ \angle BAC=60^\circ$. If the radius $ OC$ of circle $ O$ is 2, what is the length of chord $ BC$?

Solution:
\textbf{Solution:} Construct line OD $\perp$ BC at point D, \\
$\therefore \angle$ODC$=90^\circ$, BC$=2$CD,\\
$\because$ arc BC$=$ arc BC,\\
$\therefore \angle$BOC$=2\angle$BAC$=2\times 60^\circ=120^\circ$,\\
$\therefore \angle C= \frac{1}{2}(180^\circ-120^\circ)=30^\circ$,\\
$\therefore$OD$=\frac{1}{2}$CO$=1$ in $\triangle$COD,\\
$CD=\sqrt{OC^{2}-OD^{2}}=\sqrt{2^{2}-1^{2}}=\sqrt{3}$,\\
$\therefore$BC$=2\sqrt{3}$. Hence, the answer is: \boxed{2\sqrt{3}}


Question: As shown in the figure, the quadrilateral $ABCD$ is an inscribed quadrilateral of the circle, with $AB$ as the diameter. It is given that $\overset{\frown}{DC} = \overset{\frown}{CB}$, and if $\angle DCB = 110^\circ$, what is the degree measure of $\angle ABC$?

Solution:
\textbf{Solution:} Connect AC, \\
$\because$ Quadrilateral ABCD is inscribed in circle O, \\
$\therefore$ $\angle$DAB + $\angle$DCB = 180$^\circ$, \\
$\therefore$ $\angle$DAB = 180$^\circ$ − 110$^\circ$ = 70$^\circ$, \\
$\because$ $ \overset{\frown}{DC} = \overset{\frown}{CB}$, \\
$\therefore$ $\angle$DAC = $\angle$CAB = $\frac{1}{2}$ $\angle$DAB = $\frac{1}{2}$ × 70$^\circ$ = 35$^\circ$; \\
$\because$ AB is the diameter, \\
$\therefore$ $\angle$ACB = 90$^\circ$, \\
$\therefore$ $\angle$ABC = 90$^\circ$ − $\angle$CAB = 90$^\circ$ − 35$^\circ$ = \boxed{55^\circ}.


Question: In the figure, A, B, and C are three points on the circle $\odot O$, with $\angle AOB = 72^\circ$. What is the measure of $\angle ACB$?

Solution: \textbf{Solution:} We have $\angle ACB=\frac{1}{2}\angle AOC=\frac{1}{2}\times 72^\circ=\boxed{36^\circ}$.


Question: As shown in the diagram, quadrilateral ABCD is inscribed in circle $O$. If $\angle C = 100^\circ$, what is the measure of $\angle BOD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is inscribed in circle $O$, and $\angle BCD=100^\circ$, \\
it follows that $\angle A=180^\circ-\angle BCD=80^\circ$, \\
By the inscribed angle theorem, we have $\angle BOD=2\angle A= \boxed{160^\circ}$.


Question: As shown in the figure, it is known that points A, B, and C are sequentially on the circle $\odot o$. Given that $\angle B - \angle A = 40^\circ$, what is the degree measure of $\angle AOB$?

Solution:
\textbf{Solution:} As shown in the figure, $AC$ intersects $OB$ at point $D$,\\
$\because$ $\angle A + \angle O + \angle ADO = 180^\circ$, $\angle B + \angle C + \angle CDB = 180^\circ$, $\angle ADO = \angle CDB$,\\
$\therefore$ $\angle A + \angle O = \angle B + \angle C$,\\
$\therefore$ $\angle B - \angle A = \angle O - \angle C = 40^\circ$,\\
$\because$ $\overset{\frown}{AB} = \overset{\frown}{AB}$,\\
$\therefore$ $\angle O = 2\angle C$,\\
$\therefore$ $2\angle C - \angle C = 40^\circ$,\\
Solving this: $\angle C = 40^\circ$,\\
$\therefore$ $\angle AOB = 2 \times 40^\circ = \boxed{80^\circ}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\odot O$, $\angle BAD=70^\circ$, then what is the degree measure of $\angle BCD$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is inscribed in circle O, $\angle$BAD=$70^\circ$,\\
therefore, $\angle$BAD+$\angle$BCD=$180^\circ$,\\
therefore, $\angle$BCD=$\boxed{110^\circ}$.


Question: As shown in the figure, OA and OB are two radii of the circle $O$, and point C is on the circle $O$. If $\angle AOB = 80^\circ$, what is the measure of $\angle ACB$?

Solution:
\textbf{Solution:} Since OA and OB are two radii of the circle O, with point C on the circle O, and $\angle$AOB=80°, \\
it follows that $\angle$ACB=$\frac{1}{2}\angle$AOB=$\boxed{40^\circ}$


Question: As shown in the figure, there is a broken line OABC inside circle O, where OA = 8, AB = 12, and $\angle A = \angle B = 60^\circ$. What is the length of BC?

Solution:
\textbf{Solution:} Extend AO and intersect BC at D, and draw OE$\perp$BC at E.\\
Since $\angle A = \angle B = 60^\circ$,\\
then $\angle ADB = 60^\circ$,\\
thus $\triangle ADB$ is an equilateral triangle,\\
thus BD = AD = AB = 12,\\
thus OD = 4,\\
Additionally, since $\angle ADB = 60^\circ$,\\
then DE = OD $\cdot \cos 60^\circ = \frac{1}{2}$ OD = 2,\\
thus BE = 10,\\
Since OE $\perp$ BC,\\
thus BC = 2BE = \boxed{20}.


Question: As shown in the figure, C is a point on the semicircle O with AB as the diameter. Lines AC and BC are connected, and squares ACDE and BCFG are constructed externally on sides AC and BC, respectively. The midpoints of DE, FG, $\overset{\frown}{AC}$, and $\overset{\frown}{BC}$ are M, N, P, Q, respectively. If $MP+NQ=14$ and $AC+BC=18$, what is the length of AB?

Solution:
\textbf{Solution:} \\
Let's connect OP and OQ, \\
since the midpoints of DE, FG, $\overset{\frown}{AC}$, and $\overset{\frown}{BC}$ are respectively M, N, P, and Q, \\
therefore OP $\perp$ AC, and OQ $\perp$ BC, \\
thus H and I are the midpoints of AC and BC, respectively, \\
hence OH + OI = $\frac{1}{2}$(AC + BC) = 9, \\
since MH + NI = AC + BC = 18, and MP + NQ = 14, \\
therefore PH + QI = 18 - 14 = 4, \\
thus AB = OP + OQ = OH + OI + PH + QI = 9 + 4 = \boxed{13},