Question: As shown in the figure, in $\triangle ABC$, the perpendicular bisectors of sides $AB$ and $AC$ intersect at point $P$. Connect $BP$ and $CP$. If $\angle A = 50^\circ$, what is the measure of $\angle BPC$?

Solution:
\textbf{Solution:} As the diagram shows, connect AP and extend BP to intersect AC at D,\\
$\because$ the perpendicular bisectors of AB and AC intersect at point P,\\
$\therefore$ point P is the circumcenter of $\triangle ABC$,\\
$\therefore$ $\angle BPC = 2\angle BAC$\\
$\because$ $\angle BAC = 50^\circ$,\\
$\therefore$ $\angle BPC = \boxed{100^\circ}$.


Question: As shown in the figure, in circle $\odot O$, the diameter $CD$ is perpendicular to chord $AB$ at point $E$, and $OB$ and $CB$ are connected. Given that the radius of $\odot O$ is $2$ and $AB=2\sqrt{3}$, 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$\sqrt{3}$, \\
hence, EB = $\frac{1}{2}$AB = $\sqrt{3}$, \\
since the radius of circle O is 2, \\
therefore, $\sin\angle EOB = \frac{EB}{OB} = \frac{\sqrt{3}}{2}$, \\
thus, $\angle EOB = 60^\circ$, \\
therefore, $\angle BCD = \boxed{30^\circ}$.


Question: As shown in the figure, within circle $O$, $AB$ is a diameter, and point $C$ is a point on the circle. The minor arc $AC$ is folded along the chord $AC$ to intersect $AB$ at point $D$, and $CD$ is connected. If point $D$ does not coincide with the center $O$ and $\angle BAC=24^\circ$, what is the degree measure of $\angle DCA$?

Solution:
\textbf{Solution:} As shown in the figure, connect BC,\\
$\because$ AB is the diameter of $ \odot O$\\
$\therefore$ $\angle ACB=90^\circ$\\
$\therefore$ $\angle B=180^\circ-\angle BAC-\angle B=180^\circ-90^\circ-24^\circ=66^\circ$\\
According to the reflection property, $\angle B$ is the inscribed angle corresponding to arc $\overset{\frown}{AC}$, $\angle ADC$ is the inscribed angle corresponding to arc $\overset{\frown}{ABC}$\\
$\therefore$ $\angle B+\angle ADC=180^\circ$\\
$\therefore$ $\angle ADC=114^\circ$\\
$\therefore$ $\angle DCA=180^\circ-\angle ADC-\angle BAC=180^\circ-114^\circ-24^\circ=\boxed{42^\circ}$


Question: As shown in the figure, AB is the diameter of the circle $O$, and point C is a point on circle $O$, $\angle ABC=70^\circ$. What is the measure of $\angle BAC$?

Solution:
\textbf{Solution:} Since AB is the diameter of circle O, and point C is a point on circle O,\\
it follows that $\angle ACB=90^\circ$,\\
Since $\angle ABC=70^\circ$,\\
it follows that $\angle BAC=180^\circ-\left(\angle ACB+\angle ABC\right)=180^\circ-90^\circ-70^\circ=\boxed{20^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, with points $C, D$ on $\odot O$, and $\angle ACD=35^\circ$, then what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Since $\overset{\frown}{AD}=\overset{\frown}{AD}$,\\
$\therefore \angle AOD=2\angle ACD=2\times 35^\circ=70^\circ$,\\
$\therefore \angle BOD=180^\circ-\angle AOD=180^\circ-70^\circ=\boxed{110^\circ}$.


Question: As shown in the figure, $\triangle ABC$ is inscribed in circle $\odot O$, with $\angle C=46^\circ$. Connecting $OA$, what is the measure of $\angle OAB$?

Solution:
\textbf{Solution:} As shown in the figure:\\
Connect OB,\\
$\because \angle C=46^\circ$,\\
$\therefore \angle AOB=92^\circ$,\\
Also, $\because OA=OB$,\\
$\therefore \angle OAB=\angle OBA=(180-92^\circ)÷2=\boxed{44^\circ}$.


Question: As shown in the figure, $\triangle ABC$ is inscribed in $\odot O$, with $CA=CB$. If $\angle C=40^\circ$, what is the degree measure of $\overset{\frown}{BC}$?

Solution:
\textbf{Solution:} Given $\because$CA＝CB, $\angle$C＝$40^\circ$,\\
$\therefore$$\angle$A＝$\angle$B＝$\frac{1}{2}\times(180^\circ-40^\circ)＝70^\circ$,\\
$\therefore$ the degree of arc $\overset{\frown}{BC}$ is $\boxed{140^\circ}$.


Question: As shown in the figure, in the circle $\odot O$, $AB$ is the diameter of $\odot O$, and the chord $CD\perp AB$ at point $H$. If $AH=5$ and $HB=1$, then what is the length of $CD$?

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


Question: As shown in the figure, with $O$ in the center, point $C$ is on the arc $\overset{\frown}{AB}$, and $\angle ADC$, $\angle BEC$ are the angles subtended by arcs $\overset{\frown}{AC}$, $\overset{\frown}{BC}$ on the circumference, respectively. If $\angle AOB=110^\circ$ and $\angle ADC=20^\circ$, what is the measure of $\angle BEC$?

Solution:
\textbf{Solution:} Since $\angle AOB=110^\circ$,\\
therefore $\angle ADC+\angle BEC= \frac{1}{2} \angle AOB=55^\circ$,\\
since $\angle ADC=20^\circ$,\\
therefore $\angle BEC=\boxed{35^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, point $D$ is the midpoint of the arc $AC$, draw $DE \perp AB$ at point $E$, extend $DE$ to intersect $\odot O$ at point $F$. If $AC = 12$ and $AE = 3$, what is the length of the diameter of $\odot O$?

Solution:
\textbf{Solution:} As shown in the figure, connect OF.\\
Since DE$\perp$AB, it follows that DE=EF, $\overset{\frown}{AD} = \overset{\frown}{AF}$. Since point D is the midpoint of arc AC, it follows $\overset{\frown}{AD} = \overset{\frown}{CD}$, hence $\overset{\frown}{AC} = \overset{\frown}{DF}$, therefore AC=DF=12, thus EF=$\frac{1}{2}$DF=6. Let OA=OF=x, in the right triangle $\triangle$OEF, we have $x^{2}=6^{2}+(x-3)^{2}$. Solving this, we get $x=\frac{15}{2}$, thus AB=2x=\boxed{15}.


Question: As shown in the figure, with point A of square ABCD as the center and AB as the radius, draw the arc $\overset{\frown}{BD}$. Choose a point F on arc $\overset{\frown}{BD}$ such that $DF=DC$, and let point E be another point on arc $\overset{\frown}{BD}$ (not coinciding with points D or F). What is the value of $\angle DEF$?

Solution:
\textbf{Solution:} Draw AE and AF.\\
$\because$ With A as the center of the circle and AB as the radius, construct arc $\overset{\frown}{BD}$, and choose a point F on arc $\overset{\frown}{BD}$ such that DF=DC.\\
$\therefore$AD=AE=AF=DF,\\
$\therefore$$\triangle$ADF is an equilateral triangle,\\
$\therefore$$\angle$DAF=$60^\circ$, $\angle$ADE=$\angle$DEA, $\angle$AEF=$\angle$AFE,\\
$\because$$\angle$DAE+$\angle$ADE+$\angle$DEA=$180^\circ$, then 2$\angle$ADE=$180^\circ$−$\angle$DAE,\\
$\angle$EAF+$\angle$AEF+$\angle$AFE=$180^\circ$, then 2$\angle$AEF=$180^\circ$−$\angle$EAF,\\
$\therefore$2$\angle$ADE+2$\angle$AEF=$360^\circ$−($\angle$DAE+$\angle$EAF)=$360^\circ$−$\angle$DAF=$300^\circ$,\\
$\therefore$$\angle$ADE+$\angle$AEF=$150^\circ$, thus $\angle$DEF=\boxed{150^\circ}.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=BC=6$. Points D and E are located on AC and BC, respectively, $CD=2$. If the circle $\odot O$ with diameter DE intersects the midpoint F of AB, then what is the diameter of $\odot O$?

Solution:
\textbf{Solution:} Construct $FG\perp AC$ and $FH\perp CB$, with the feet of the perpendiculars being G and H, respectively, as shown in the figure. \\
Then, quadrilateral BCGF is a rectangle, with $AC\parallel FH$ and $CB\parallel FG$, \\
since $AC=BC=6$ and point F is the midpoint of AB, \\
thus, $CG=CH=GF=HF=\frac{1}{2} \times 6=3$, \\
therefore, quadrilateral BCGF is a square, \\
therefore, $\angle GFH=90^\circ$, \\
since DE is a diameter, then $\angle DFE=90^\circ$, \\
therefore, $\angle DFG+\angle DFH=\angle DFH+\angle EFH=90^\circ$, \\
therefore, $\angle DFG=\angle EFH$, \\
since $\angle DGF=\angle EHF=90^\circ$ and $GF=HF=3$, \\
therefore, $\triangle DFG \cong \triangle EFH$, \\
therefore, DF=EF, \\
since, in the right-angled $\triangle DFG$, $DG=3-2=1$ and $GF=3$, \\
therefore, $DF=\sqrt{1^2+3^2}=\sqrt{10}=EF$, \\
In the right-angled $\triangle DEF$, \\
$DE=\sqrt{(\sqrt{10})^2+(\sqrt{10})^2}=2\sqrt{5}$; \\
Thus, the answer is: $\boxed{2\sqrt{5}}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is a rectangle with $AB=3$ and $BC=4$. Point $P$ is a moving point on line segment $BC$, and point $M$ is a point on line segment $AP$ such that $\angle ADM=\angle BAP$. What is the minimum value of $BM$?

Solution:
\textbf{Solution:} As shown in the figure, let O be the midpoint of AD and connect OB and OM.\\
$\because$ Quadrilateral ABCD is a rectangle,\\
$\therefore\angle$BAD$=90^\circ$, AD$=$BC$=4$,\\
$\therefore\angle$BAP$+\angle$DAM$=90^\circ$,\\
$\because\angle$ADM$=\angle$BAP,\\
$\therefore\angle$ADM$+\angle$DAM$=90^\circ$,\\
$\therefore\angle$AMD$=90^\circ$,\\
$\because$AO$=$OD$=2$,\\
$\therefore$OM$= \frac{1}{2}$AD$=2$,\\
$\therefore$ the trajectory of point M is a circle $\odot$O with O as the center and 2 as the radius.\\
$\because$OB$= \sqrt{AB^{2}+AO^{2}} = \sqrt{3^{2}+2^{2}} = \sqrt{13}$,\\
$\therefore$BM$\geq$OB$-$OM$=\sqrt{13}-2$,\\
$\therefore$ the minimum value of BM is $\boxed{\sqrt{13}-2}$.


Question: As shown in the figure, $\triangle ABC$ is inscribed in $\odot O$, and $AD$ is the diameter of $\odot O$. If $\angle B = 20^\circ$, then what is the degree measure of $\angle CAD$?

Solution: \textbf{Solution:} Draw line BD. Since AD is the diameter of circle O, it follows that $\angle ABD = 90^\circ$. Since $\angle ABC = 20^\circ$, it follows that $\angle CBD = \angle ABD - \angle ABC = 70^\circ$. Therefore, $\angle CAD = \angle CBD = \boxed{70^\circ}$.


Question: As shown in the figure, the parabola $y_{1} = -x^{2} + 4x$ and the line $y_{2} = 2x$. What is the range of $x$ when $y_{1} < y_{2}$?

Solution:
\textbf{Solution:} Given $-x^{2}+4x=2x$, we find $x_{1}=0$ and $x_{2}=2$. Thus, the $x$-coordinates of the intersection points of the parabola and the linear function are 0 and 2. Hence, when $x<0$ or $x>2$, $y_{1}<y_{2}$.\\
Therefore, the answer is: $\boxed{x<0$ or $x>2}$.


Question: As shown in the figure, place the right-angle vertex of the triangular cardboard on one side of the ruler. If $\angle 1=20^\circ$ and $\angle 3=30^\circ$, what is the value of $\angle 2$?

Solution:
\textbf{Solution:} As shown in the figure, \\
\(\because\) \(\angle 1=20^\circ\), \(\angle 3=30^\circ\), \\
\(\therefore\) \(\angle 4=\angle 1+\angle 3=20^\circ+30^\circ=50^\circ\), \\
\(\because\) The two sides of the straightedge are parallel to each other, \\
\(\therefore\) \(\angle 2=\angle 4= \boxed{50^\circ}\).


Question: As shown in the figure, with vertex B of $\triangle ABC$ as the center and the length of BA as the radius, an arc is drawn intersecting side BC at point D, and AD is connected. If $\angle B=50^\circ$ and $\angle C=36^\circ$, what is the degree measure of $\angle DAC$?

Solution:
\textbf{Solution:} Given that $\angle B=50^\circ$ and $\angle C=36^\circ$,\\
hence $\angle BAC=180^\circ-(\angle B+\angle C)=180^\circ-(50^\circ+36^\circ)=94^\circ$,\\
since BA=BD,\\
thus $\angle BAD=\angle BDA$,\\
therefore $\angle BAD=\frac{1}{2}(180^\circ-\angle B)=\frac{1}{2}(180^\circ-50^\circ)=65^\circ$,\\
thus $\angle DAC=\angle BAC-\angle BAD=94^\circ-65^\circ=\boxed{29^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ and $CE$ are medians. If the area of the quadrilateral $BDFE$ is $6$, what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Connect BF,\\
since AD and CE are the medians of $\triangle ABC$,\\
therefore AE=BE, BD=CD,\\
therefore $S_{\triangle CBE}=\frac{1}{2}S_{\triangle ABC}$, $S_{\triangle ACD}=\frac{1}{2}S_{\triangle ABC}$,\\
therefore $S_{\triangle CBE}=S_{\triangle ACD}$,\\
therefore $S_{\triangle ACF}=S_{\text{quadrilateral } BDFE}=6$,\\
since AE=BE, BD=CD,\\
therefore $S_{\triangle AEF}=S_{\triangle BEF}$, $S_{\triangle CDF}=S_{\triangle BDF}$,\\
therefore $S_{\triangle AEF}+S_{\triangle CDF}=S_{\text{quadrilateral } BDFE}=6$,\\
therefore $S_{\triangle ABC}=3S_{\text{quadrilateral } BDFE}=\boxed{18}$


Question: As shown in the figure, in $\triangle ABC$, $DM$ and $EN$ are the perpendicular bisectors of $AB$ and $AC$, respectively, with footpoints $M$ and $N$. They intersect $BC$ at points $D$ and $E$, respectively. If $\angle DAE=20^\circ$, what is the measure of $\angle BAC$?

Solution:
\textbf{Solution:} Since $\angle$DAE=$20^\circ$,\\
then $\angle$ADE+$\angle$DEA=$160^\circ$.\\
Since DM and EN are the perpendicular bisectors of AB and AC respectively,\\
then AD=BD, AE=EC,\\
then $\angle$B=$\angle$BAD, $\angle$C=$\angle$CAE,\\
Moreover, since $\angle$ADE, $\angle$DEA are the external angles of $\triangle$ABD and $\triangle$AEC respectively,\\
then $\angle$ADE=$\angle$B+$\angle$BAD=$2\angle$B, $\angle$DEA=$\angle$C+$\angle$CAE=$2\angle$C,\\
then $\angle$B+$\angle$C=$80^\circ$,\\
then $\angle$BAC=$180^\circ$−($\angle$B+$\angle$C)=$\boxed{100^\circ}$.


Question: As shown in the figure, in $\triangle ABC$ and $\triangle ADC$, $\angle B=\angle D=90^\circ$, $CB=CD$, $\angle 1=30^\circ$, what is the value of $\angle 2$?

Solution:
\textbf{Solution}: Since $\angle B=\angle D=90^\circ$\\
Therefore, both $\triangle ABC$ and $\triangle ADC$ are right-angled triangles. \\
In $\triangle ABC$ and $\triangle ADC$: \\
Since $\left\{\begin{array}{l}CB=CD\\ AC=AC\end{array}\right.$\\
Therefore, $\triangle ABC\cong \triangle ADC$ (HL criterion)\\
Therefore, $\angle 1 = \angle CAD$\\
Since $\angle 2 + \angle CAD + \angle D = 180^\circ$\\
Therefore, $\angle 2 = 180^\circ - 90^\circ - 30^\circ = \boxed{60^\circ}$


Question: As shown in the figure, $BD$ bisects $\angle ABC$, and $CD$ bisects $\angle ACD$. If $\angle A = 80^\circ$, what is the degree measure of $\angle D$?

Solution:
\textbf{Solution:}

Given that $\angle A=80^\circ$,\\
it follows that $\angle ABC + \angle ACB = 180^\circ - \angle A = 100^\circ$,\\
Since BD bisects $\angle ABC$ and CD bisects $\angle ACB$,\\
we have $\angle CBD = \frac{1}{2} \angle ABC$ and $\angle BCD = \frac{1}{2} \angle ACB$,\\
In $\triangle BCD$,\\
$\angle D = 180^\circ - (\angle CBD + \angle BCD)$\\
$= 180^\circ - (\frac{1}{2} \angle ABC + \frac{1}{2} \angle ACB)$\\
$= 180^\circ - \frac{1}{2} (\angle ABC + \angle ACB)$\\
$= 180^\circ - \frac{1}{2} \times 100^\circ$\\
$= 180^\circ - 50^\circ$\\
$= \boxed{130^\circ}$.


Question: "The cross section of a dam in a reservoir is a trapezoid, and the slope of a ramp inside the dam is $i=1\colon \sqrt{3}$. What is the angle of this slope?

Solution:
\textbf{Solution:} As shown in the figure, \\
since the slope is $i=1\colon \sqrt{3}$, \\
therefore $\tan A= \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$, \\
since $\tan 30^\circ= \frac{\sqrt{3}}{3}$, \\
therefore, the angle of the slope is $30^\circ$, \\
hence, the answer is: $A=\boxed{30^\circ}$.


Question: As shown in the figure, in rhombus $ABCD$, $DE \perp AB$, $\cos A = \frac{3}{5}$, and $AE = 6$, then what is the length of $AB$?

Solution:
\textbf{Solution:} Since $DE\perp AB$,\\
it follows that $\triangle ADE$ is a right-angled triangle,\\
hence $DA=\frac{AE}{\cos A}=6\times \frac{5}{3}=10$,\\
since quadrilateral ABCD is a rhombus,\\
it follows that AB=AD=10,\\
thus, the answer is: AB=\boxed{10}.


Question: As shown in the figure, in the Cartesian coordinate system, point A has coordinates $ \left(3, 4\right)$. What is the value of $ \cos\alpha $?

Solution:
\textbf{Solution:} Construct $AB \perp x$-axis at $B$, as shown in the figure,\\
since the coordinates of point A are (3,4),\\
hence $OB = 3$, $AB = 4$,\\
therefore $OA = \sqrt{3^2 + 4^2} = 5$,\\
in $\triangle AOB$ right, $\cos\alpha = \frac{OB}{OA} = \frac{3}{5}$ .\\
Thus $\cos\alpha = \boxed{\frac{3}{5}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AD\parallel BC$, $\angle ABC=90^\circ$, $O$ is the midpoint of the diagonal $BD$, $OA=2$, $BC=5$, and $CD=3$. What is the value of $\tan\angle DCB$?

Solution:
\textbf{Solution:} Since AD $\parallel$ BC and $\angle$ABC$=90^\circ$, \\
it follows that $\angle$BAD$=90^\circ$, \\
Given that O is the midpoint of the diagonal BD and OA=2, \\
it follows that BD=2OA=4, \\
Given BC=5 and CD=3, \\
it follows that BD$^{2}$+CD$^{2}$=BC$^{2}$, \\
thus $\angle$BDC$=90^\circ$, \\
hence $\tan$($\angle$DCB) = $\frac{BD}{CD}$ = $\boxed{\frac{4}{3}}$.


Question: As shown in the figure, the barrier at the entrance of a parking lot rotates from the horizontal position $AB$ around point $O$ to the position of $AB$. Given that $AO=4$ meters and the rotation angle $\angle AOA=47^\circ$, what is the vertical distance $AH$ that the endpoint $A$ rises?

Solution:
\textbf{Solution:} Refer to the diagram, draw $A^{\prime}H\perp AB$ at $H$ through point $A^{\prime}$,\\
according to the problem, we get $OA^{\prime}=OA=4$ meters,\\
in $\triangle OA^{\prime}H$, $\angle A^{\prime}OH=47^\circ$, $\sin\angle AOH=\frac{AH}{OA}$,\\
$\therefore$ the vertical distance the railing's endpoint $A$ rises is $AH=\sin\angle AOH\cdot OA=4\sin47^\circ$ meters,\\
thus, $AH=\boxed{4\sin47^\circ}$ meters.


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

Solution:
\textbf{Solution:} In right triangle $ABC$, $\angle C=90^\circ$, AC=4, BC=3,\\
then AB$=\sqrt{AC^{2}+BC^{2}}=\sqrt{4^{2}+3^{2}}=5$,\\
$\therefore \sin A=\frac{BC}{AB}=\boxed{\frac{3}{5}}$


Question: As shown in the figure, in a $6\times6$ square grid, where each small square has a side length of $1$, points $A$, $B$, and $C$ are all located on the grid intersections, and $\odot O$ is the circumcircle of $\triangle ABC$, what is the sine value of $\angle BAC$?

Solution:
\textbf{Solution:} Let $E$ be the midpoint of $BC$, and connect $BO$, $CO$, and $OE$. \\
$\because$ $\angle BAC = \frac{1}{2}\angle BOC$\\
$\therefore$ $\angle BAC = \angle BOE$\\
$\because$ $BE=1$, $EO=2$\\
$\therefore$ $BO=\sqrt{BE^2+EO^2}=\sqrt{5}$\\
$\therefore$ $\sin\angle BAC=\sin\angle BOE=\frac{BE}{BO}=\frac{1}{\sqrt{5}}=\boxed{\frac{\sqrt{5}}{5}}$


Question: As shown in the figure, in a $3\times 3$ grid, A and B are both lattice points. With point A as the center and the length of AB as the radius, an arc is drawn. Point C in the figure is the intersection point of the arc and the grid lines. Then, what is the value of $\tan\angle BAC$?

Solution:
\textbf{Solution:} As shown in the figure, since $CD\parallel AB$,\\
it follows that $\angle BAC = \angle DCA$.\\
Since the radii of the same circle are equal, it follows that $AC = AB = 3$, and $AD = 2$,\\
therefore, $CD = \sqrt{AC^2 - AD^2} = \sqrt{5}$,\\
in $\triangle ACD$, $\tan \angle ACD = \frac{AD}{CD} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$.\\
Thus, $\tan \angle BAC = \tan \angle ACD = \boxed{\frac{2\sqrt{5}}{5}}$.


Question: In the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, point D is the midpoint of side AB, and CD is connected. If $BC=4$ and $CD=3$, what is the value of $\sin\angle DCB$?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $\angle ACB=90^\circ$, and point D is the midpoint of side AB,\\
$\therefore$AD=BD=CD=$\frac{1}{2}$AB,\\
$\therefore$ $\angle DCB=\angle B$,\\
Also, since CD=3,\\
$\therefore$AB=6,\\
$AC=\sqrt{AB^{2}-BC^{2}}=\sqrt{6^{2}-4^{2}}=2\sqrt{5}$,\\
$\therefore$ $\sin\angle DCB=\sin B=\frac{AC}{AB}=\frac{2\sqrt{5}}{6}=\boxed{\frac{\sqrt{5}}{3}}$.


Question: Figure 1 is the emblem of the 7th International Congress on Mathematical Education (ICME), featuring two adjacent right-angled triangles in its main pattern, which together form the quadrilateral $OABC$ as shown in Figure 2. In the figure, if $AB=BC=1$ and $\angle AOB=\alpha$, then what is the value of $\tan\angle BOC$?

Solution:
\textbf{Solution}: Since $AB=BC=1$,\\
in the right triangle $OAB$, $\sin\alpha =\frac{AB}{OB}$,\\
therefore $OB=\frac{1}{\sin\alpha}$,\\
in the right triangle $OBC$, $\tan\angle BOC=\frac{BC}{OB}=\frac{1}{\frac{1}{\sin\alpha}}=\sin\alpha$.\\
Hence, the answer is: $\boxed{\sin\alpha}$.


Question: As shown in the figure, in $\triangle ABC$, AD is the altitude from A to side BC, AE bisects $\angle BAC$, $\angle C=46^\circ$, $\angle DAE=10^\circ$, what is the measure of $\angle B$?

Solution:
\textbf{Solution:} Given that $AD$ is the perpendicular from $A$ to the side $BC$,\\
it follows that $\angle ADC=90^\circ$,\\
since $\angle C=46^\circ$,\\
we have $\angle CAD=90^\circ-\angle C=44^\circ$,\\
given $\angle DAE=10^\circ$,\\
it follows that $\angle CAE=\angle CAD-\angle DAE=34^\circ$,\\
since $AE$ bisects $\angle BAC$,\\
we have $\angle BAC=2\angle CAE=68^\circ$,\\
consequently, $\angle B=180^\circ-\angle BAC-\angle C=180^\circ-68^\circ-46^\circ=\boxed{66^\circ}$.


Question: As shown in the figure, $\triangle ABC$ is rotated about point $A$ to $\triangle ADE$. What is the angle of rotation?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is rotated about point A to $\triangle ADE$,\\
the angle of rotation is either $\angle BAD$ or $\angle CAE$.


Question: As shown in the figure, in $\triangle ABC$, $\angle CAB=65^\circ$. When $\triangle ABC$ is rotated counterclockwise about point A in the plane to the position of $\triangle AB'C'$, such that $CC'\parallel AB$, what is the degree measure of the rotation angle?

Solution:
\textbf{Solution:} Since $CC^{\prime} \parallel AB$,\\
it follows that $\angle ACC^{\prime} = \angle CAB = 65^\circ$,\\
Since $\triangle ABC$ is rotated about point A to obtain $\triangle AB^{\prime}C^{\prime}$,\\
it follows that AC = AC$^{\prime}$,\\
Therefore, $\angle CAC^{\prime} = 180^\circ - 2 \times 65^\circ = 50^\circ$,\\
Hence, $\angle CAC^{\prime} = \angle BAB^{\prime} = \boxed{50^\circ}$.


Question: As shown in the figure, $BC = \frac{1}{2}AB$, D is the midpoint of $AC$, and $DC = 3\,cm$. What is the length of $AB$?

Solution:
\textbf{Solution:} Since $D$ is the midpoint of $AC$ and $DC=3\,cm$,\\
$\therefore AC=6\,cm$,\\
and since $BC=\frac{1}{2}AB$,\\
$\therefore BC=\frac{1}{3}AC=\frac{1}{3}\times 6=2\,cm$,\\
$\therefore AB=AC-BC=6-2=\boxed{4\,cm}$.


Question: As shown in the figure, $\triangle COD$ is the shape obtained by rotating $\triangle AOB$ clockwise around point $O$ by $30^\circ$, and point $C$ is exactly on $AB$. What is the degree measure of $\angle A$?

Solution:
\textbf{Solution:} Since $\triangle COD$ is the shape obtained by rotating $\triangle AOB$ $30^\circ$ clockwise around point O,\\
thus $\angle AOC=30^\circ$, $OA=OC$,\\
therefore $\angle A=\frac{180^\circ-\angle AOC}{2}=75^\circ$,\\
hence, the answer is: $\boxed{75^\circ}$.


Question: As shown, it is known that points O and C are on the same side of line $m$. Draw $\odot O$ intersecting $m$ at points A and B. Connect AC, BC, OA, and OB. If point C is outside of $\odot O$ and $\angle AOB=110^\circ$, then what could be the possible degree of $\angle C$?

Solution:
\textbf{Solution:} As shown in the diagram, let $AC$ intersect $\odot O$ at point $D$, and connect $BD$,

$\because$ $\angle AOB=110^\circ$,

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

$\because$ $\angle ADB=\angle C+\angle CBD$,

$\therefore$ $\angle C<55^\circ$,

$\therefore$ $A$ meets the requirements, $BCD$ does not.

Thus, the answer is: $\boxed{\angle C<55^\circ}$.


Question: As shown in the figure, it is known that in circle $\odot O$, CD is the diameter of $\odot O$, and points A and B are on $\odot O$ with $AC=AB$. If $\angle BCD=26^\circ$, what is the degree measure of $\angle ABC$?

Solution:
\textbf{Solution:} Since CD is the diameter,\\
$\therefore\angle CAD=90^\circ$,\\
$\therefore\angle ACD+\angle ADC=90^\circ$,\\
Since AC$=$AB,\\
$\therefore\angle ACB=\angle B$,\\
Since $\angle D=\angle B$,\\
$\therefore\angle ACB=\angle D$,\\
$\therefore\angle ACB+26^\circ+\angle D=90^\circ$,\\
$\therefore\angle ACB=32^\circ$,\\
$\therefore\angle ABC=\angle ACB=\boxed{32^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $\angle BAC=70^\circ$, $\odot O$ is the circumcircle of $\triangle ABC$, point D is on the minor arc $\overset{\frown}{AC}$. What is the degree measure of $\angle D$?

Solution:
\textbf{Solution:} Given that AB=AC, and $\angle$BAC=$70^\circ$, \\
thus $\angle$B=$\angle$ACB=$55^\circ$, \\
since quadrilateral ABCD is inscribed in circle O, \\
therefore $\angle$D=$180^\circ$−$\angle$B=$\boxed{125^\circ}$,


Question: As shown in the figure, the diameter of $O$ is $20$, the length of chord $AB$ is $16$, $ON\perp AB$ with the foot of the perpendicular being $N$. What is the length of $ON$?

Solution:
\textbf{Solution:} From the given information, we have \\
OA=10, $\angle$ONA=$90^\circ$, AB=16, \\
$\therefore$AN=8, \\
$\therefore$ON= $ \sqrt{OA^{2}-AN^{2}}=\sqrt{10^{2}-8^{2}}=\boxed{6}$,


Question: As shown in the figure, in circle $\odot O$, CD is a tangent with the point of tangency at D, and line CO intersects $\odot O$ at B and A, where $\angle A = 35^\circ$. What is the degree measure of $\angle C$?

Solution:
Solution: Connect OD, \\
$\because$ CD is the tangent line of $\odot$O, \\
$\therefore$ $\angle$ODC = $90^\circ$, \\
Also $\because$ $\angle$COD = 2$\angle$A = $70^\circ$, \\
$\therefore$ $\angle$C = $90^\circ - 70^\circ = \boxed{20^\circ}$


Question: In the figure, within $\triangle ABC$, $\angle A=60^\circ$, $BD\perp AC$ at $D$, $CE\perp AB$ at $E$, $BD$ and $CE$ intersect at point $H$. If $CE=4$ and $BD=5$, what is the value of $\frac{DH}{HB}$?

Solution:
\textbf{Solution:} Given $BD \perp AC$, $CE \perp AB$,\\
thus $\angle ADB = \angle AEC = 90^\circ$,\\
In $\triangle BDA$ with $\angle A = 60^\circ$, and $BD = 5$,\\
thus $\angle ABD = 30^\circ$,\\
thus $AB = 2AD$,\\
Given $AB^{2} = AD^{2} + BD^{2}$, that is $4AD^{2} = AD^{2} + 25$,\\
thus $AD = \frac{5\sqrt{3}}{3}$,\\
In $\triangle ACE$ with $\angle A = 60^\circ$, and $CE = 4$,\\
thus $\angle ACE = 30^\circ$,\\
thus $AC = 2AE$,\\
Given $AC^{2} = AE^{2} + CE^{2}$, that is $4AE^{2} = AE^{2} + 16$,\\
thus $AE = \frac{4\sqrt{3}}{3}$, and $AC = \frac{8\sqrt{3}}{3}$,\\
thus $CD = AC - AD = \frac{8\sqrt{3}}{3} - \frac{5\sqrt{3}}{3} = \sqrt{3}$,\\
In $\triangle CDH$ with $\angle DCH = 30^\circ$, and $CD = \sqrt{3}$,\\
thus $CH = 2DH$,\\
Given $CH^{2} = DH^{2} + CD^{2}$, that is $4DH^{2} = DH^{2} + 3$,\\
thus $DH = 1$,\\
thus $BH = BD - DH = 5 - 1 = 4$,\\
thus $\frac{DH}{HB} = \boxed{\frac{1}{4}}$,


Question: As shown in the figure, in $\triangle ABC$, taking points A and B as the centers respectively, arcs are drawn with radii greater than $\frac{1}{2}AB$ and intersecting at points M and N. Line $MN$ is drawn, intersecting $BC$ at point D. $AD$ is connected. If the perimeter of $\triangle ABC$ is $17$ and $AB=7$, what is the perimeter of $\triangle ADC$?

Solution:
Solution: From the method, we know: MN is the perpendicular bisector of AB,\\
$\therefore$AD=BD,\\
$\because$ The perimeter of $\triangle ABC$ is 17 and AB=7,\\
$\therefore$AC+BC=10,\\
$\therefore$ The perimeter of $\triangle ADC$ is:\\
AC+CD+AD=AC+CD+BD=AC+BC=\boxed{10}


Question: As shown in the figure, point B is the midpoint of segment AD, C is on segment BD and satisfies BD = 3CD. If the sum of the lengths of all segments in the figure is 30, what is the length of segment BC?

Solution:
\textbf{Solution:} Let $CD=x$, then $BC=2x$, $AB=BD=3x$,\\
the sum of all line segments is $AB+AC+AD+BC+BD+CD$\\
$=AB+(AB+BC)+(AB+BD)+BC+BD+CD$\\
$=3x+5x+6x+2x+3x+x$\\
$=20x$,\\
According to the problem, $20x=30$,\\
solving this gives $x=1.5$,\\
thus, $BC=2 \times 1.5=\boxed{3}$,\\
meaning the length of segment BC is 3.


Question: As shown in the figure, $\angle AOB = 120^\circ$, $\angle AOC = \frac{1}{3}\angle BOC$, $OM$ bisects $\angle BOC$. What is the degree measure of $\angle AOM$?

Solution:
\textbf{Solution:} Since $\angle AOB=120^\circ$, and $\angle AOC= \frac{1}{3} \angle BOC$,\\
thus $\angle AOC= \frac{1}{4} \angle AOB=30^\circ$, and $\angle BOC= \frac{3}{4} \angle AOB=90^\circ$,\\
given that OM bisects $\angle BOC$,\\
thus $\angle BOM=\angle COM= \frac{1}{2} \angle BOC=45^\circ$,\\
therefore $\angle AOM=\angle AOC+\angle COM=\boxed{75^\circ}$


Question: As shown in the figure, in the equilateral triangle $\triangle ABC$, the point $E$ is the midpoint of side $AC$, and point $P$ is a moving point on the median $AD$ of $\triangle ABC$. If $AD = 6$, what is the minimum value of $EP+CP$?

Solution:
\textbf{Solution:} Construct point F as the symmetric point of E with respect to AD, and connect CF, which intersects AD at point P,\\
since $\triangle ABC$ is an equilateral triangle, and AD is the median to side BC,\\
therefore $AD\perp BC$,\\
thus AD is the perpendicular bisector of BC,\\
since the corresponding point of E with respect to AD is point F,\\
therefore CF is the minimum value of EP+CP.\\
Since $\triangle ABC$ is an equilateral triangle, and E is the midpoint of side AC,\\
thus F is the midpoint of AB,\\
therefore CF is the median of $\triangle ABC$,\\
thus $CF=AD=\boxed{6}$,\\
which means the minimum value of EP+CP is 6.


Question: As shown in the figure, a right-angled triangle ruler with an angle of $45^\circ$ is placed on two parallel lines $m$ and $n$. If $\angle \alpha = 118^\circ$, then what is the degree measure of $\angle \beta$?

Solution:
\textbf{Solution:} As shown in the diagram, \\
$\because m\parallel n$, \\
$\therefore \angle 1=\angle 2$, \\
$\because \angle \alpha =\angle 2+\angle A$, \\
and $\angle A=45^\circ$, $\angle \alpha =118^\circ$, \\
$\therefore \angle 2=118^\circ-45^\circ=73^\circ$, \\
$\therefore \angle 1=73^\circ$, \\
$\therefore \angle \beta =\boxed{73^\circ}$.


Question: As shown in the figure, $\triangle A'B'C'$ is obtained from $\triangle ABC$ through a similarity transformation centered at point O. If $OA':AA = 1:2$, what is the ratio of the perimeter of $\triangle A'B'C'$ to that of $\triangle ABC$?

Solution:
\textbf{Solution:} Given that $OA'\colon AA=1\colon 2$,\\
it follows that $OA'\colon OA=1\colon 3$,\\
since $\triangle A'B'C'$ is obtained from $\triangle ABC$ through a similarity transformation centered at point O,\\
it implies $A'B' \parallel AB$ and $\triangle A'B'C' \sim \triangle ABC$,\\
thus $\frac{A'B'}{AB}=\frac{OA'}{OA}=\frac{1}{3}$,\\
therefore, the ratio of the perimeter of $\triangle A'B'C'$ to the perimeter of $\triangle ABC$ is $\boxed{1\colon 3}$.


Question: As shown in the figure, in circle $O$, $\overset{\frown}{AB} = \overset{\frown}{AC}$. If $\angle B = 70^\circ$, then what is the measure of $\angle A$?

Solution:
\textbf{Solution:} Given that in circle $\odot O$, $\widehat{AB}=\widehat{AC}$ and $\angle B=70^\circ$,\\
it follows that $\angle C=\angle B=70^\circ$,\\
thus $\angle A=180^\circ-\angle B-\angle C=180^\circ-70^\circ-70^\circ=\boxed{40^\circ}$.


Question: In $\triangle ABC$, $AD$ is the angle bisector, and points $E$ and $F$ are the midpoints of line segments $AC$ and $CD$, respectively. If the areas of $\triangle ABD$ and $\triangle EFC$ are 21 and 7, respectively, what is the value of $\frac{AB}{AC}$?

Solution:
\textbf{Solution:} Construct $DM \perp AB$ at $M$, and $DN \perp AC$ at $N$.\\
Since the point F is the midpoint of DC,\\
it follows that $S_{\triangle DEF} = S_{\triangle FCE} = 7$,\\
Since E is the midpoint of AC,\\
it follows that $S_{\triangle ADE} = S_{\triangle CDE} = 14 + 7 = 21$,\\
Therefore, $S_{\triangle ADC} = 28$,\\
Since AD is the angle bisector, $DM \perp AB$, and $DN \perp AC$,\\
it follows that $DM = DN$,\\
Since $S_{\triangle ABD} : S_{\triangle ADC} = \frac{1}{2} AB \cdot DM : (\frac{1}{2} \cdot AC \cdot DN) = 21:28$\\
Therefore, $\frac{AB}{AC} = \boxed{\frac{3}{4}}$.


Question: As shown in the figure, in a $4\times4$ square grid, each small square has a side length of $1$. Points $A$, $B$, and $C$ are all on the grid points, and $AD\perp BC$ at point $D$. What is the length of $AD$?

Solution:
\textbf{Solution:} From the Pythagorean theorem, we have: $AB =\sqrt{2^{2}+4^{2}}=2\sqrt{5}$, $AC =\sqrt{1^{2}+2^{2}}=\sqrt{5}$, and $BC =\sqrt{3^{2}+4^{2}}=5$,\\
$\because AB^{2}+AC^{2}=25$ and $BC^{2}=25$,\\
$\therefore AB^{2}+AC^{2}=BC^{2}$,\\
$\therefore \angle BAC=90^\circ$,\\
$\therefore S_{\triangle ABC} =\frac{1}{2}AC\cdot AB=\frac{1}{2}BC\cdot AD$,\\
$\therefore \sqrt{5}\times 2\sqrt{5}=5\times AD$,\\
$\therefore AD=\boxed{2}$,


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $\angle BAC=70^\circ$. The triangle $\triangle ABC$ is rotated $70^\circ$ clockwise around point A, with the corresponding points of B and C after rotation being B' and C', respectively. Connecting BB', what is the degree measure of $\angle BB'C'$?

Solution:
\textbf{Solution:}

Given the properties of rotation, we know that: $AB=AB'$, $\angle BAB'=70^\circ$, and $\angle ACB=\angle AC'B'=90^\circ$ \\
$\therefore \angle ABB'=\angle AB'B=\frac{1}{2}\times (180^\circ-70^\circ)=55^\circ$ \\
In $\triangle AB'C'$, $\angle AB'C'=90^\circ-\angle B'AB=20^\circ$ \\
$\because \angle BB'C+\angle AB'C'=\angle AB'B$ \\
$\therefore \angle BB'C'=\angle AB'B-\angle AB'C'=55^\circ-20^\circ=\boxed{35^\circ}$


Question: As shown in the figure, in the same plane, if one side of a regular hexagon and a square, both with equal side lengths, coincide, then what is the measure of $\angle 1$?

Solution:
\textbf{Solution:} To find: $\angle 1=\frac{(6-2)\times 180^\circ}{6}-\frac{(4-2)\times 180^\circ}{4}$\\
$=120^\circ-90^\circ$\\
$=\boxed{30^\circ}$.


Question: As shown in the figure, line AC intersects the graph of the inverse proportion function $y=\frac{k}{x}$ ($x>0$) at points A and C (with point A to the left of point C), and intersects the x-axis at point B. A rectangle ADMN is drawn downwards with point A as the vertex, and its diagonals intersect at point O, with AD bisecting $\angle OAB$. Given that AC=CB and line CD is drawn, if the area of $\triangle ACD$ is 6, what is the value of $k$?

Solution:
\textbf{Solution:} As shown in the figure, draw AN $\perp$ x-axis at point N through point A, and draw CM $\perp$ x-axis at point M through point C, connect CO, \\
$\because$ it's a rectangle ADMN, \\
$\therefore$ OA=OD, \\
$\therefore$ $\angle$OAD=$\angle$ODA, \\
Also, $\because$ AD bisects $\angle$OAB, \\
$\therefore$ $\angle$OAD=$\angle$DAB, \\
$\therefore$ $\angle$ODA=$\angle$DAB, \\
$\therefore$ OD$\parallel$AB, \\
$\because$ the area of $\triangle$ACD is 6, and AC=CB, \\
$\therefore$ $S_{\triangle AOC}$=$S_{\triangle OCB}$=6, CM is the median line of $\triangle$ANB, \\
$\therefore$ AN=2CM, NM=MB, \\
Let point A be (x, $\frac{k}{x}$), then ON=x, AN=$\frac{k}{x}$, \\
$\therefore$ CM=$\frac{k}{2x}$, \\
$\therefore$ C(2x, $\frac{k}{2x}$), \\
$\therefore$ MB=x, OB=3x, \\
$\therefore$ $S_{\triangle OCB}$=6=$\frac{1}{2}$OB·CM=$\frac{1}{2}$×3x·$\frac{k}{2x}$=$\frac{3k}{4}$, \\
$\therefore$ k=\boxed{8}.


Question: As shown in the diagram, in rhombus ABCD with an area of 24, and $AC+BD=14$, what is the length of AB?

Solution:
\textbf{Solution:} As shown in the diagram, AC intersects BD at point O, \\
$\because$ ABCD is a rhombus with an area of 24, \\
$\therefore$ AC $\perp$ BD, and AC$\cdot$BD = 48, \\
Also, since AC + BD = 14, \\
Let AC = a, then BD = 14 − a, \\
$\therefore$ a(14−a) = 48, \\
Upon rearrangement and solving, we get: a = 8 or a = 6 (which does not fit the diagram, hence discarded), \\
$\therefore$ AC = 8, BD = 6, \\
$\therefore$ AO = 4, BO = 3, \\
$\therefore$ AB = $\sqrt{AO^{2}+BO^{2}}$ = $\sqrt{4^{2}+3^{2}}$ = \boxed{5}.


Question: In the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, point $D$ is the midpoint of the hypotenuse $AB$, $DE$ bisects $\angle ADC$, and $BC=4$, what is the length of $DE$?

Solution:
\textbf{Solution:} Let point $D$ be the midpoint of the hypotenuse $AB$, and $\angle ACB=90^\circ$, \\
$\therefore$AD=CD=BD,\\
$\because$ $DE$ bisects $\angle ADC$,\\
$\therefore$DE$\perp$AC, and AE=CE,\\
$\therefore$DE is the midline of $\triangle ABC$,\\
$\therefore$DE= $\frac{1}{2}BC=\boxed{2}$.


Question: As shown in the figure, in the Cartesian coordinate system, rectangle ABCD is located in the first quadrant, and AB is parallel to the y-axis. The line M: $y=-x$ is translated in the positive direction of the x-axis, and the segment EF intercepted by rectangle ABCD has its length $l$ as a function of the translation distance $a$ as shown in the figure. What is the area of rectangle ABCD?

Solution:
\textbf{Solution}: From the graph, it is known that when $a=1$, the line $M$ passes through point $A$.\\
When $a=4$, line $M$ goes through point $B$.\\
When $a=6$, line $M$ goes through point $D$.\\
When $a=9$, line $M$ goes through point $C$.\\
Therefore, when $F$ moves on segment $BC$, $4\leq a \leq 9$, and $BC=9-4=5$,\\
When $F$ moves on segment $AB$, $1\leq a \leq 4$,\\
Also, at this time $AF=AE$,\\
$\therefore AB=4-1=3$.\\
Hence, the area of rectangle $ABCD$ is $5\times 3=\boxed{15}$.


Question: As shown in the figure, D and E are the midpoints on the sides AB and AC of $\triangle ABC$, respectively, and F is a point on the segment DE. If $AB=10$, $BC=12$, and $\angle AFB=90^\circ$, then what is the length of segment EF?

Solution:
\textbf{Solution:} Since $\angle AFB=90^\circ$ and D is the midpoint of AB,\\
it follows that DF=$\frac{1}{2}$AB=5,\\
Since D and E are the midpoints of AB and AC respectively,\\
DE is the midline of $\triangle ABC$,\\
thus DE=$\frac{1}{2}$BC=6,\\
therefore EF=DE−DF=6−5=\boxed{1}.


Question: As shown in the figure, D, E, and F are the midpoints of the sides of $\triangle ABC$. If the perimeter of $\triangle ABC$ is 12, what is the perimeter of $\triangle DEF$?

Solution:
\textbf{Solution:} Since D, E, F are the midpoints of the sides of $\triangle ABC$ respectively,\\
it follows that DE, DF, and EF are the midlines of $\triangle ABC$,\\
therefore, BC = 2DF, AC = 2DE, AB = 2EF,\\
thus, the perimeter of $\triangle ABC$ = AB + BC + AC = 2(DF + EF + DE) = 12,\\
hence, DF + EF + DE = $12 \times \frac{1}{2} = \boxed{6}$, which means the perimeter of $\triangle DEF$ is 6.


Question: As shown in the figure, rectangle $ABCD$ is folded along the diagonal $BD$ such that point $C$ falls on point $F$, and $BF$ intersects $AD$ at point $E$. If $\angle BDC = 62^\circ$, what is the measure of $\angle DBF$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\
it follows that $AD\parallel BC$, and $\angle ADC=90^\circ$,\\
Also, since $\angle BDC=62^\circ$,\\
it then follows that $\angle BDE=90^\circ-\angle BDC=90^\circ-62^\circ=28^\circ$,\\
thus, $\angle CBD=\angle BDE=28^\circ$,\\
Since the rectangle $ABCD$ is folded along the diagonal $BD$,\\
it then follows that $\angle FBD=\angle CBD=\boxed{28^\circ}$.


Question: As shown in the figure, $AC$ is the diagonal of rhombus $ABCD$. If $\angle ABC = 60^\circ$, what is the degree measure of $\angle CAD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\
$AB=BC=CD=AD$, and $AD\parallel BC$,\\
since $\angle ABC=60^\circ$,\\
then $\triangle ABC$ is an equilateral triangle, i.e., $\angle ACB = \angle CAB = 60^\circ$,\\
since $AD\parallel BC$,\\
then $\angle CAD=\angle ACB=60^\circ$.\\
Therefore, the answer is: $\boxed{60^\circ}$.


Question: As shown in the figure, in $ \triangle ABC$, D, E, and F are the midpoints of the three sides respectively. Given $ AB=6$ and $AC=4$, what is the perimeter of the quadrilateral AEDF?

Solution:
\textbf{Solution:}
Given that in $\triangle ABC$, E, D, and F are the midpoints of AB, BC, and CA, respectively,
$\therefore$ DE=AF=$\frac{1}{2}$AC=2, DF=AE=$\frac{1}{2}$AB=3,
$\therefore$ the perimeter of quadrilateral AEDF is $(2+3)\times 2=\boxed{10}$.


Question: As shown in the figure, the diagonals of the rhombus $ABCD$ are $AC=6$ and $BD=8$. What is the perimeter of the rhombus $ABCD$?

Solution:
\textbf{Solution:} As shown in the figure:\\
Let the intersection point of AC and BD be O,\\
since in rhombus ABCD, AC=6, BD=8,\\
thus OA= $\frac{1}{2}$ AC=3, OB= $\frac{1}{2}$ BD=4, and AC$\perp$BD,\\
therefore, $AB=\sqrt{OA^{2}+OB^{2}}=5$,\\
thus, the perimeter of rhombus ABCD=4$\times$5=\boxed{20},


Question: As shown in the figure, in $\triangle ABC$, $AB=14$, $BC=8$, D and E are the midpoints of side AC and BC, respectively, point F is on DE, and $\angle BFC=90^\circ$, then what is the length of DF?

Solution:
\textbf{Solution:} Since points D and E are the midpoints of sides AB and AC, respectively,\\
thus, DE is the midline of $\triangle ABC$,\\
since AB=14,\\
thus, DE= $\frac{1}{2}AB = 7$,\\
since $\angle$BFC=$90^\circ$ and BC=8,\\
thus, EF= $\frac{1}{2}BC = 4$,\\
therefore, DF=DE−EF=7−4=\boxed{3}


Question: As shown in the figure, in the square $ABCD$, $AB=4$. Let $E$ be a point on side $AB$, and let $F$ be a point on side $BC$ such that $BF=1$. Rotate point $E$ clockwise around point $F$ by $90^\circ$ to obtain point $G$. Connect $DG$. What is the minimum length of $DG$?

Solution:
\textbf{Solution:} Draw line $GP\perp BC$ at point P, extend PG to intersect AD at point H,\\
then $\angle GPF=90^\circ$,\\
since quadrilateral ABCD is a square,\\
$\therefore \angle ADC=\angle C=\angle B=90^\circ$,\\
$\therefore$ quadrilateral CDHP is a rectangle,\\
$\therefore CD=PH=AB=4$, $PC=DH$,\\
since $\angle EFG=90^\circ$,\\
$\therefore \angle BFE+\angle PFG=90^\circ$,\\
also, $\angle BFE+\angle BEF=90^\circ$,\\
$\therefore \angle PFG=\angle BEF$,\\
since $FE=FG$, $\angle B=\angle GPF=90^\circ$,\\
$\therefore \triangle BEF\cong \triangle PFG\left(AAS\right)$,\\
$\therefore BE=PF$, $PG=BF=1$,\\
$\therefore GH=PH-PG=4-1=3$,\\
let $BE=PF=x$, then $PC=DH=4-1-x=3-x$,\\
in $\triangle DGH$, by the Pythagorean theorem,\\
$DG^2=DH^2+GH^2=(3-x)^2+3^2=(3-x)^2+9$,\\
when $x=3$, $DG^2$ has its minimum value of $9$,\\
$\therefore$ the minimum value of $DG$ is $\boxed{3}$.


Question: As shown in the figure, point $O$ is the midpoint of the diagonal $BD$ of rectangle $ABCD$, point $E$ is on $CD$, and $OE \parallel AD$. If $AB = 8$ and $OE = 3$, what is the length of $OC$?

Solution:
\textbf{Solution:} Given that point O is the midpoint of the diagonal BD of rectangle ABCD, and OE $\parallel$ AD,\\
it follows that OE $\parallel$ CB, and OE is the midline of $\triangle$BCD;\\
Given that OE = 3,\\
it follows that AD = BC = 2OE = 6,\\
Given that AB = 8, and $\angle$BAD = 90$^\circ$,\\
it follows that BD = 10,\\
In $\triangle$BCD, OC = $\frac{1}{2}$BD = \boxed{5};


Question: As shown in the figure, in the rhombus $ABCD$, $AC$ and $BD$ intersect at point $O$, $E$ is the midpoint of $AB$, and $DE \perp AB$. If $BD = 2\sqrt{3}$, what is the length of $DE$?

Solution:
\textbf{Solution:} Let $E$ be the midpoint of $AB$ and $DE\perp AB$,\\
hence $AD=DB$,\\
since quadrilateral $ABCD$ is a rhombus,\\
thus $AB=AD$ and $AC\perp BD$,\\
hence $AD=DB=AB$ and $BO=\frac{1}{2}BD=\sqrt{3}$,\\
thus $\triangle ABD$ is an equilateral triangle,\\
hence $\angle DAB=60^\circ$,\\
thus $\angle OAB=30^\circ$,\\
hence $AB=2OB=2\sqrt{3}$,\\
thus $AO=\sqrt{AB^{2}-OB^{2}}=3$,\\
since $AO$ and $DE$ are both heights of the equilateral $\triangle ABD$,\\
hence $DE=AO=\boxed{3}$.


Question: As shown in the figure, in rectangle $ABCD$, the perpendicular bisector of the diagonal $AC$, $EF$, intersects $BC$ and $AD$ at points $E$ and $F$ respectively. If $BE=3$ and $AF=5$, what is the perimeter of rectangle $ABCD$?

Solution:
\textbf{Solution:} Connect AE, \\
$\because$ EF is the perpendicular bisector of AC, \\
$\therefore$ $\angle$AOF = $\angle$COE = 90$^\circ$, AO = CO, AE = CE, \\
$\because$ Quadrilateral ABCD is a rectangle, \\
$\therefore$ AD$\parallel$BC, $\angle$B = 90$^\circ$, AD = BC, \\
$\therefore$ $\angle$FAO = $\angle$ECO, \\
In $\triangle$AOF and $\triangle$COE, \\
$\left\{\begin{array}{l} \angle FAO = \angle ECO\hfill \\ AO = CO\hfill \\ \angle AOF = \angle COE\hfill \end{array}\right.$ , \\
$\therefore$ $\triangle$AOF $\cong$ $\triangle$COE (ASA), \\
$\therefore$ AF = CE, \\
$\therefore$ AE = CE, \\
$\because$ BE = 3, AF = 5, \\
$\therefore$ CE = 5, \\
$\therefore$ AE = 5, BC = BE + CE = 8, \\
$\therefore$ AB = $ \sqrt{AE^{2}-BE^{2}}$ = 4, \\
$\therefore$ The perimeter of rectangle ABCD is 2(4 + 8) = $\boxed{24}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $\angle B=90^\circ$, $AB=8$, $BC=6$. Point $M$ is the midpoint of diagonal $AC$, and point $N$ is the midpoint of side $AD$. Connecting $BM$, $MN$, if $BM = 3MN$, then what is the length of the line segment $CD$?

Solution:
\textbf{Solution:} Given $\angle B=90^\circ$ and lengths AB=8, BC=6, \\
$\therefore$ the length of AC is given by $\sqrt{8^2+6^2}=10$, \\
Since M is the midpoint of AC, \\
$\therefore$ BM= $\frac{1}{2}$AC=5, \\
Given that BM is three times MN, \\
$\therefore$ MN=$\frac{5}{3}$, \\
Also, since N is the midpoint of AD, \\
$\therefore$ MN is the midsegment of $\triangle ACD$, \\
$\therefore$ CD is twice MN, which is $2 \times \frac{5}{3} = \boxed{\frac{10}{3}}$.


Question: As shown in the figure, rhombus \(ABCD\) has a side length of 10, \(E\) is the midpoint of \(AD\), \(O\) is the intersection point of the diagonals, and rectangular \(OEFG\) has one side on \(AB\), with \(EF = 4\). What is the length of \(BG\)?

Solution:
\textbf{Solution:} Since $ABCD$ is a rhombus, and $E$ is the midpoint of side $AD$,\\
therefore $AD=AB=10$, $AE=\frac{1}{2}AD=5$, $AC\perp BD$,\\
thus $EO=\frac{1}{2}AD=5$. Since $OEFG$ is a rectangle,\\
therefore $FG=EO=5$, $\angle EFA=90^\circ$,\\
in $\triangle EFA$, where $AE=5$, $EF=4$,\\
thus $AF=\sqrt{5^2-4^2}=3$,\\
therefore $BG=AB-AF-FG=10-3-5=\boxed{2}$.


Question: As shown in the figure, there are two congruent rectangles AEFG and ABCD placed as shown. The line containing CD intersects AE and GF respectively at points H and M. If AB = 3, BC = $ \sqrt{3} $, and CH = MH, what is the length of the segment MH?

Solution:
\textbf{Solution:} As shown in the figure, draw $HQ \perp GF$, \\
$\therefore \angle HQM = 90^\circ$, \\
$\because$ Rectangle $AEFG$ and rectangle $ABCD$ are congruent, \\
$\therefore AD = EF = HQ$, $\angle ADH = \angle MQH = \angle QHA = 90^\circ$, \\
$\therefore \angle QHM + \angle AHD = \angle QHM + \angle QMH$, \\
$\therefore \angle AHD = \angle QHM$, \\
$\therefore \triangle ADH \cong \triangle MQH$ (AAS), \\
$\therefore AH = MH$, \\
$\because AB = 3$, \\
$\therefore$ Let $DH = x$, then $CH = MH = AH = 3 - x$, \\
$\because BC = \sqrt{3}$, \\
$\therefore AD = \sqrt{3}$, \\
In $\triangle ADH$, $AD^2 + DH^2 = AH^2$, \\
$\therefore 3 + x^2 = (3 - x)^2$, \\
Upon simplification, we find: $x = 1$, \\
$\therefore MH = 3 - 1 = \boxed{2}$.


Question: As shown in the figure, point $O$ is the midpoint of the diagonal $AC$ of rectangle $ABCD$, $E$ is the midpoint of side $BC$, $AD=8$, $OE=3$. What is the length of the line segment $OD$?

Solution:
\textbf{Solution:} Given that in the rectangle ABCD, $AD=8$ and $OE=3$, where O is the midpoint of the diagonal AC, and E is the midpoint of the side BC, \\
it follows that $BC=AD=8$ and $AB=2OE=6$, with $\angle B=90^\circ$, \\
hence $AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{6^{2}+8^{2}}=10$, \\
since the point O is the midpoint of AC and $\angle ADC=90^\circ$, \\
it follows that $OD=\frac{1}{2}AC=\boxed{5}$.


Question: As shown in the figure, in rectangle ABCD, point E is the midpoint of CD, point F is a point on BC, and FC = 2BF. Connect AE, EF, and AF. If AB = 2 and AD = 3, what is the measure of $\angle AEF$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle, and AD = 3, AB = 2, \\
thus $\angle B = \angle C = 90^\circ$, CD = AB = 2, BC = AD = 3.\\
Since point E is the midpoint of CD, and FC = 2BF, \\
thus CE = DE = 1, BF = 1, CF = 2.\\
Thus AB = CF = 2, CE = BF = 1.\\
In $\triangle ABF$ and $\triangle FCE$,\\
$\begin{cases} AB=CF\\ \angle B = \angle C\\ BF=CE \end{cases}$, \\
thus $\triangle ABF \cong \triangle FCE$ (by SAS).\\
Thus AF = EF, $\angle BAF = \angle CFE$.\\
Since $\angle B = 90^\circ$, \\
thus $\angle BAF + \angle AFB = 90^\circ$.\\
Thus $\angle CFE + \angle AFB = 90^\circ$.\\
Thus $\angle AFE = 180^\circ - (\angle CFE + \angle AFB) = 180^\circ - 90^\circ = 90^\circ$.\\
Thus $\triangle AFE$ is an isosceles right triangle.\\
Thus $\angle AEF = \boxed{45^\circ}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB = 8$, and $BC = 4$. The rectangle is folded along $AC$, and point $D$ falls on point $D'$. What is the area of the overlapping part, $\triangle AFC$?

Solution:
\textbf{Solution:} Given that rectangle $ABCD$ is folded,\\
it follows that $AD=BC=AD^{\prime}, \angle D=\angle B=90^\circ=\angle D^{\prime},$ \\
since $\angle AFD^{\prime}=\angle CFB,$ \\
it can be concluded that $\triangle AFD^{\prime} \cong \triangle CFB,$ \\
hence, $D^{\prime}F=BF,$ \\
let $D^{\prime}F=x$, then $AF=8-x$,\\
in $\triangle AFD^{\prime}$, $(8-x)^{2}=x^{2}+4^{2}$,\\
solving this yields: $x=3$,\\
thus, $AF=AB-FB=8-3=5$,\\
therefore, the area of $\triangle AFC$ is $\frac{1}{2} \cdot AF \cdot BC = \boxed{10}$.


Question: As shown in the figure, the diagonals of square $ABCD$ intersect at point $O$, and point $O$ is also a vertex of square $A_1B_1C_1O$, with both squares having equal side lengths. No matter how square $A_1B_1C_1O$ is rotated around point $O$, the area of the overlapping part of the two squares is always equal to what fraction of the area of a square?

Solution:
\textbf{Solution}: As shown in the figure, construct OG$\perp$AB at G, and OH$\perp$BC at H. Assume AB intersects A$_{1}$O at point M.\\
Since quadrilateral ABCD is a square,\\
$\therefore \angle$OBH=$\angle$OBG=$45^\circ$,\\
$\therefore$ quadrilateral OGBH is a square,\\
$\therefore$OG=OH,\\
Since $\angle$GOM+$\angle$FOG=$\angle$FOH+$\angle$FOG=$90^\circ$,\\
$\therefore \angle$GOM=$\angle$FOH,\\
$\therefore \triangle FOH \cong \triangle MOG$ (ASA),\\
$\therefore$Area of quadrilateral OFBM=Area of quadrilateral OGBH=$\boxed{\frac{1}{4}}$ Area of square ABCD.


Question: As shown in the figure, point $P$ is a point inside $\triangle ABC$, with $AP \perp BP$, $BP=12$, and $CP=15$. Points $D$, $E$, $F$, and $G$ are the midpoints of $AP$, $BP$, $BC$, and $AC$, respectively. If the perimeter of quadrilateral $DEFG$ is 28, what is the length of $AP$?

Solution:
\textbf{Solution:} By the problem, $DE$, $EF$, $FG$, and $GD$ are the medians of $\triangle PAB$, $\triangle BPC$, $\triangle CAB$, and $\triangle ACP$, respectively, \\
$\therefore$DE=GF=$\frac{1}{2}$AB, EF=GD=$\frac{1}{2}$PC, \\
$\because$ The perimeter of the quadrilateral $DEFG$ is 28, \\
$\therefore$AB+PC=28, \\
$\because$PC=15, \\
$\therefore$AB=13, \\
In $\triangle PAB$, $\angle APB=90^\circ$, BP=12, \\
$\therefore$AP=$\sqrt{AB^{2}-BP^{2}}=\sqrt{169-144}=\boxed{5}$


Question: As shown in the figure, the four vertices of the square paper $ABCD$ lie on four parallel lines $a$, $b$, $c$, $d$, respectively. The distances between two adjacent lines among these four lines are ${h}_{1}$, ${h}_{2}$, ${h}_{3}$ (${h}_{1}>0$, ${h}_{2}>0$, ${h}_{3}>0$) respectively. If ${h}_{1}=6$ and ${h}_{2}=3$, what is the area $S$ of the square $ABCD$?

Solution:
\textbf{Solution}: Draw $BM \perp a$ at point M through point B, and draw $CN \perp a$ at point N through point C, as shown in the figure:

Thus, we have $\angle BMA=\angle ANC=90^\circ$,

$\therefore \angle MBA+\angle MAB=90^\circ$,

In square $ABCD$, $AB=AC$, $\angle BAC=90^\circ$,

$\therefore \angle MAB+\angle CAN=90^\circ$,

$\therefore \angle CAN=\angle MBA$,

$\therefore \triangle MAB \cong \triangle NCA\left(AAS\right)$,

$\therefore AN=MB$,

Since $h_{1}=6$, $h_{2}=3$,

$\therefore AN=MB=6$, $NC=6+3=9$,

By the Pythagorean theorem, we get $AC=\sqrt{6^2+9^2}=\sqrt{117}$,

$\therefore$ the area $S$ of square $ABCD$ is $\boxed{117}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AB \parallel CD$, $AB = 5$, $DC = 11$, and the sum of $AD$ and $BC$ is $12$. Points $E$, $F$, and $G$ are the midpoints of $BD$, $AC$, and $DC$ respectively. What is the perimeter of $\triangle EFG$?

Solution:
\textbf{Solution:}

As shown in the figure, extend AE and continue it to intersect CD at point K,\\
since AB$\parallel$CD,\\
it follows that $\angle$BAE$=\angle$DKE and $\angle$ABD$=\angle$EDK,\\
since points E, F, G are the midpoints of BD, AC, DC respectively.\\
It follows that BE$=$DE,\\
so, triangle AEB is congruent to triangle KED (AAS),\\
thus, DK$=$AB, AE$=$EK,\\
therefore, EF is the midline of triangle ACK,\\
therefore, EF$=\frac{1}{2}$CK$=\frac{1}{2}$(DC$-$AB),\\
since EG is the midline of triangle BCD,\\
it follows that EG$=$BC,\\
also, since FG is the midline of triangle ACD,\\
it follows that FG$=$AD,\\
hence, EG+GF$=\frac{1}{2}$(AD+BC),\\
since AD+BC$=12$, AB$=5$, DC$=11$,\\
thus, EG+GF$=6$, FE$=3$,\\
therefore, the perimeter of triangle EFG is $6+3=\boxed{9}$.


Question: As shown in the figure, the line $y_1=x+3$ intersects with the $x$-axis and the $y$-axis at points $A$ and $C$ respectively, and the line $y_2=-x+3$ intersects with the $x$-axis and the $y$-axis at points $B$ and $C$ respectively. If point $P(m, 2)$ is a point inside (including on the edges of) $\triangle ABC$, then what is the difference between the maximum and minimum values of $m$?

Solution:
\textbf{Solution:} Let point P$(m,2)$ be a point inside (including on the edge of) $\triangle ABC$, \\
$\therefore$ point P lies on the line $y=2$, as shown in the figure, \\
When P is the intersection of line $y=2$ and line $y_{2}$, $m$ takes its maximum value, \\
When P is the intersection of line $y=2$ and line $y_{1}$, $m$ takes its minimum value, \\
$\because$ in $y_{1}=-x+3$, letting $y=2$, we get $x=1$, \\
in $y_{2}=x+3$, letting $y=2$, we get $x=-1$, \\
$\therefore$ the maximum value of $m$ is $1$, and the minimum value of $m$ is $-1$. \\
Thus, the difference between the maximum and minimum values of $m$ is: $1-(-1)=\boxed{2}$


Question: As shown in the figure, the graphs of the functions $y=2x$ and $y=ax+4$ intersect at point $A(m, 3)$. What is the solution set of the inequality $2x < ax+4$?

Solution:
\textbf{Solution:} Since the graphs of the functions $y=2x$ and $y=ax+4$ intersect at point A$(m,3)$,\\
it follows that $3=2m$,\\
yielding $m= \frac{3}{2}$,\\
thus, the coordinates of point A are $\left( \frac{3}{2}, 3\right)$,\\
therefore, the solution set for the inequality $2x < ax+4$ is $\boxed{x < \frac{3}{2}}$;


Question: As shown in the figure, a Cartesian coordinate system is established with the center of the square $ABCD$ as the origin. The coordinates of point $A$ are $(2,2)$, what are the coordinates of point $D$?

Solution:
\textbf{Solution:} As shown in the figure:

Since the coordinate system is established with the center O of the square ABCD as the origin, and the coordinates of point A are (2, 2), and point D is symmetrical to point A with respect to the y-axis,

Therefore, the coordinates of point D are: $\boxed{(-2, 2)}$.


Question: As shown in the figure, the line ${l}_{1}: y={k}_{1}x+b$ and the line ${l}_{2}: y={k}_{2}x$, with their graphs depicted in the same Cartesian coordinate system as illustrated. What is the solution set for the inequality regarding $x$, ${k}_{1}x+b>{k}_{2}x>0$?

Solution:
\textbf{Solution:} Given the graph, when $-1<x<0$, it follows that $k_1x+b>k_2x>0$,\\
hence the solution set for $k_1x+b>k_2x>0$ is $-1<x<0$.\\
Therefore, the answer is: \boxed{-1<x<0}.


Question: As shown in the graph, in the same Cartesian coordinate system, the graph of the linear function $y=-2x+4$ intersects with the graph of the direct proportion function $y=kx\ (k>0)$ at point A, and the y-coordinate of point A is 2. Then, what is the solution set for the inequality $kx>-2x+4$?

Solution:
\textbf{Solution:} Since the graph of the linear function $y=-2x+4$ intersects with the graph of the directly proportional function $y=kx\ (k>0)$ at point $A$, and the ordinate of point $A$ is 2,\\
we substitute ${y}_{A}=2$ into $y=-2x+4$ to get: ${x}_{A}=1$,\\
Since $kx>-2x+4$,\\
it follows that $x>1$,\\
Therefore, the answer is: $\boxed{x>1}$.


Question: In the Cartesian coordinate system xOy, the graph of the line $l_1: y_1 = k_1x + 5$ and the graph of the line $l_2: y_2 = k_2x$ are shown as in the figure. What is the solution set of the inequality $k_2x < k_1x + 5$ with respect to $x$?

Solution:
\textbf{Solution:} According to the graph, the coordinates of the intersection point of the two lines are $(-2, 3)$,

$\therefore$ the solution set of the inequality ${k}_{2}x<{k}_{1}x+5$ is $\boxed{x>-2}$.


Question: As shown, the moving point P starts from vertex A of the rectangle ABCD, and moves uniformly along the direction A→B→C on the sides AB, BC at a speed of $1\, \text{cm/s}$ to point C. The area $S$ ($\text{cm}^{2}$) of $\triangle APC$ as a function of the movement time $t$ (s) is illustrated in the figure. What is the length of AB?

Solution:
\textbf{Solution:} Given that point P moves from point A to B, the distance S increases with time t. When point P moves from point B to C, S decreases as t increases.\\
Therefore, based on Graph 2, the duration of point P's movement from B to C is 4s,\\
Since the speed of point P is 1cm/s,\\
Thus, $BC=1\times4=4$ (cm),\\
Given that when point P moves along the line AB reaching point B, the area of $\triangle APC$ is at its maximum,\\
Hence, from Graph 2: the area of $\triangle APC$ is $6\text{cm}^{2}$,\\
Therefore, the area of $\triangle ABC$, ${S}_{\triangle ABC}=\frac{1}{2}AB\cdot BC=6$,\\
Thus, $AB=\boxed{3}$cm.


Question: As shown in the figure, the line $y=kx+b(k\ne 0)$ passes through point $A(-3, 2)$. What is the solution set of the inequality $kx+b<2$ with respect to $x$?

Solution:
\textbf{Solution:} From the diagram, it can be observed that when $x<-3$, $kx+b<2$, \\
thus, the solution set is $\boxed{x<-3}$.


Question: The graph of the linear function $y=kx+b$ is shown in the figure. What is the solution to the inequality $kx+b\leq 0$?

Solution:
\textbf{Solution:} From the graph, we can see that\\
the graph of the linear function $y=kx+b$ intersects the $x$-axis at the point $(-2,0)$,\\
the line rises from left to right, meaning $y$ increases as $x$ increases,\\
$\therefore$ the solution set of the inequality $kx+b\leq 0$ is $x\leq -2$,\\
hence, the answer is: $\boxed{x\leq -2}$.


Question: As shown in the figure, the line $y=x+2$ intersects with the line $y=ax+4$ at point $P(m, 3)$. What is the solution set of the inequality $x+2<ax+4$ with respect to $x$?

Solution:
\textbf{Solution:} Given the problem statement, we know that point $P$ is also on $y=x+2$. Thus, when $y=3$, $x=1$\\
$\therefore$ when $x<1$, $x+2<ax+4$\\
Hence, the answer is: $\boxed{x<1}$.


Question: Given the graph, the linear function $y_{1}=x+b$ and the linear function $y_{2}=kx+5$ intersect at point $P(1, 4)$. What is the solution set of the inequality $x+b>kx+5$ in terms of $x$?

Solution:
\textbf{Solution:} From the graph, we can see that $P(1,4)$, \\
hence when $x>1$, $x+b>kx+5$, \\
therefore, the solution set is $\boxed{x>1}$.


Question: As shown in the figure, the line $y_1=kx+b$ passes through points A and B, and the line $y_2=2x$ passes through point A. What is the solution set of the inequality $2x<kx+b$?

Solution:
\textbf{Solution:} Since $A(-1, -2)$,\\
by observing the graph, the solution set of the inequality $2x<kx+b$ is $x<-1$.\\
Therefore, the answer is: \boxed{x<-1}.


Question: As shown in the figure, the line $y=ax+b\ (a\neq 0)$ passes through points A and B. What is the solution set of the inequality $ax+b>0$?

Solution:
\textbf{Solution:} According to the function graph, it is known that when $x > -3$, $y > 0$, \\
therefore, the solution set of the inequality $ax+b>0$ is $x > -3$. \\
Hence, the answer is: $\boxed{x > -3}$.


Question: As shown in the figure is the graph of the function $y_{1}=|x|$. It is known that the graph of the function $y_{2}=\frac{1}{3}x+\frac{4}{3}$ intersects with the graph of $y_{1}=|x|$ at points A and B, and $A(-1, 1)$, then what is the range of values of $x$ for which $y_{2}>y_{1}$?

Solution:
\textbf{Solution:} As shown in the figure,\\
by solving the system of equations $\begin{cases}y=x \\ y=\frac{1}{3}x+\frac{4}{3}\end{cases}$, we get: $\begin{cases}x=2 \\ y=2\end{cases}$,\\
$\therefore$ Point B is (2, 2),\\
From the graph, it is found that when $y_{2}>y_{1}$, $-1<x<2$,\\
thus, the answer is: $x \in \boxed{(-1, 2)}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is a rhombus, where the coordinates of points $A$ and $B$ are $\left(0,3\right)$ and $\left(4,0\right)$, respectively, and point $D$ is on the $y$-axis. What are the coordinates of point $C$?

Solution:
\textbf{Solution:} Given that the coordinates of points A and B are respectively $(0,3)$ and $(4,0)$,\\
thus OA=3, OB=4,\\
hence AB$=\sqrt{AO^{2}+BO^{2}}=5$,\\
since quadrilateral ABCD is a rhombus,\\
thus BC=AB=5, and AD$\parallel$BC,\\
since A and D are on the y-axis,\\
thus BC$\parallel$y-axis,\\
hence the coordinates of C are $\boxed{(4,-5)}$.


Question: As shown in the figure, the line $y=x+2$ intersects with the line $y=ax+c$ at point $P(m,3)$. What is the solution to the inequality $x+2\le ax+c$ with respect to $x$?

Solution:
\textbf{Solution:} It is known from the problem that point P also lies on $y=x+2$. Hence, when $y=3$, it follows that $x=1$.

$\therefore$ When $x\leq 1$, $x+2\le ax+c$.

Therefore, the answer is $\boxed{x\leq 1}$.


Question: As shown in the diagram, the line $y=kx+b \left(k \ne 0\right)$ passes through the point $\left(-2, 0\right)$. What is the solution set of the inequality $kx+b>0$?

Solution:
\textbf{Solution:} From the graphic, we have: when $x>-2$, $kx+b>0$, \\
Therefore, the answer is: $\boxed{x>-2}$


Question: As shown in the figure, the line $y=kx+b\ (k<0)$ intersects the y-axis at point A and intersects the x-axis at point B, and $(AB+OA)(AB-OA)=\frac{9}{4}$. What is the solution set of the inequality $kx+b>0$?

Solution:
\textbf{Solution:}\\
Since $(AB+OA)(AB−OA)=\frac{9}{4}$,\\
it follows that $AB^{2}−OA^{2}=\frac{9}{4}$,\\
therefore $OB^{2}=\frac{9}{4}$,\\
hence $OB=\frac{3}{2}$,\\
therefore, the solution set of the inequality $kx+b>0$ is $\boxed{x<\frac{3}{2}}$.


Question: As shown in the figure, points B and C are on $y=2x$ and $y=kx-2a$, respectively. Points A and D are two points on the x-axis, and the y-coordinate of point B is $a$. If quadrilateral ABCD is a rectangle and $AB=\frac{1}{2}AD$, then what is the value of $k$?

Solution:
\textbf{Solution:}

Given that point B is on the line $y = 2x$ and the y-coordinate of point B is $a$,

it follows that $2x = a$ and $AB = a$,

from which we find $x = \frac{a}{2}$,

and since $AB = \frac{1}{2}AD$,

it follows that $AD = 2a$,

thus $OD = OA + AD = \frac{a}{2} + 2a = \frac{5a}{2}$,

therefore $C\left(\frac{5a}{2}, a\right)$,

hence $\frac{5a}{2}k - 2a = a$,

solving this gives $k = \boxed{\frac{6}{5}}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ is presented. What is the range of $x$ when $kx+b>3$?

Solution:
\textbf{Solution:} Given that $\because kx + b > 3$ and $y = kx + b$,\\
$\therefore y > 3$,\\
$\because$ when $y > 3$, by analyzing the graph, it is determined that the graph meeting the requirements intersects the y-axis above the given point,\\
$\therefore x < 0$, thus, the answer is: $\boxed{x < 0}$.


Question: As shown in the figure, line $l_1\colon y=2x-1$ intersects with line $l_2\colon y=kx+b\ (k\neq 0)$ at point $P(2, 3)$. What is the solution set of the inequality $2x-1>kx+b$ with respect to $x$?

Solution: \textbf{Solution:} According to the graph, the solution set for the inequality $2x-1>kx+b$ is: $\boxed{x>2}$.


Question: As shown in the figure, the line $y=2x$ intersects the line $y=kx+b$ at point $P(m, 2)$. What is the solution of the equation $kx+b=2$ with respect to $x$?

Solution:
\textbf{Solution:} Since the line $y=2x$ intersects with $y=kx+b$ at point $P(1,2)$, \\
therefore, when $x=1$, $y=kx+b=2$, \\
thus, the solution to the equation in terms of $x$, $kx+b=2$, is $x=\boxed{1}$.


Question: As shown in the figure, the line $y=-x$ intersects with the line $y=x+b$ at point A. If the ordinate of point A is 1, what is the solution set of the inequality $x+b<-x$?

Solution:
\textbf{Solution:} Given that the line $y=-x$ intersects with the line $y=x+b$ at point A, and the ordinate of point A is 1,\\
$\therefore$ $-x=1$,\\
Solving this, we get $x=-1$,\\
which means the abscissa of point A is $-1$,\\
$\therefore$ the solution set of the inequality $x+b<-x$ is $\boxed{x<-1}$.


Question: As shown in the diagram, within the Cartesian coordinate system, the coordinates of point $B$ are $(0, 5)$. After $\triangle OAB$ is translated to the right along the $x$-axis, it becomes $\triangle O^{\prime}A^{\prime}B^{\prime}$. The corresponding point of $B$, point $B^{\prime}$, is on the line $y=\frac{5}{6}x$. What is the distance between point $A$ and its corresponding point $A^{\prime}$?

Solution:
\textbf{Solution:} Given that point B has coordinates (0,5), and triangle $OAB$ is translated to the right along the x-axis to form triangle $O'A'B'$, where the corresponding point of B, denoted as B', lies on the line $y=\frac{5}{6}x$,\\
$\therefore$ the y-coordinate of point B' is: 5,\\
So, we have $5=\frac{5}{6}x$,\\
Solving this gives: $x=6$,\\
implying the distance from B to B' is 6,\\
thus, the distance between point A and its corresponding point A' is $\boxed{6}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ (where $k$ and $b$ are constants) is depicted as illustrated. What is the solution set of the inequality $kx+b<1$?

Solution: \textbf{Solution:} From the graph, we know that the linear function $y=kx+b$ ($k,b$ are constants, $k\neq 0$) passes through the point $(0,1)$, and the value of $y$ increases as $x$ increases. Therefore, the solution set of the inequality $kx+b<1$ is $\boxed{x<0}$.


Question: As shown in the figure, the graphs of the function $y=2x$ and $y=ax+4$ intersect at point $A(m, 3)$. What is the solution set of the inequality $2x<ax+4$?

Solution:
\textbf{Solution:} Because the graphs of functions $y=2x$ and $y=ax+4$ intersect at point $A\left(\frac{3}{2}, 3\right)$,\\
therefore $3=2m$,\\
therefore $m=\frac{3}{2}$,\\
therefore the coordinates of point A are $\left(\frac{3}{2}, 3\right)$,\\
from the graph, it is known that the solution set of $2x<ax+4$ is $\boxed{x<\frac{3}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, points $D$ and $E$ are on sides $AB$ and $AC$ respectively, with $DE \parallel BC$. If $AD = 2EC$, $BD = 3$, and $AE = 4$, then what is the value of $CE$?

Solution:
\textbf{Solution:} Since $DE\parallel BC$,\\
it follows that $\frac{AD}{BD}=\frac{AE}{EC}$, which means $AD\cdot EC=AE\cdot BD$,\\
given $AD=2EC, BD=3, AE=4$,\\
we have $2EC^{2}=4\times 3=12$,\\
therefore, $EC=\boxed{\sqrt{6}}$.


Question: As shown in the diagram, in $\triangle ABC$, $D$ and $E$ are points on sides $AB$ and $AC$ respectively, with $DE \parallel BC$. If $\frac{S_{\triangle ADE}}{S_{\text{quadrilateral } DBCE}} = 1:8$, what is the value of $\frac{DE}{BC}$?

Solution:
\textbf{Solution:} Since the ratio of the area of $\triangle ADE$ to the area of quadrilateral DBCE is $1:8$, \\
it follows that the area of $\triangle ADE$ is $\frac{1}{9}$ the area of $\triangle ABC$, \\
thus $\frac{{S}_{\triangle ADE}}{{S}_{\triangle ABC}}=\frac{1}{9}$. \\
Since $DE \parallel BC$, \\
then $\triangle ADE \sim \triangle ABC$, \\
thus $\frac{{S}_{\triangle ADE}}{{S}_{\triangle ABC}}=\left(\frac{DE}{BC}\right)^{2}$, \\
hence $\left(\frac{DE}{BC}\right)^{2}=\frac{1}{9}$, \\
therefore $\frac{DE}{BC}=\boxed{\frac{1}{3}}$.


Question: As shown in the figure, point $I$ is the centroid of $\triangle ABC$, with $S_{\triangle ABC}=9$. If $\angle BAC$ is translated so that its vertex coincides with point $I$, what is the area of the shaded region in the diagram?

Solution:
\textbf{Solution:} Extend AI to meet BC at D, as shown in the figure, \\
$\because$ point I is the centroid of $\triangle ABC$, \\
$\therefore$ AI=2ID, \\
$\therefore$ ID=$\frac{1}{3}AD$, \\
$\because$ angle BAC is translated so that its vertex coincides with point I, \\
$\therefore$ $\angle$EIF=$\angle$BAC, $\angle$IEF=$\angle$B, \\
$\therefore$ $\triangle IEF\sim \triangle ABC$, \\
$\therefore$ $\frac{S_{\triangle IEF}}{S_{\triangle ABC}}=\left(\frac{DI}{DA}\right)^2=\left(\frac{1}{3}\right)^2=\frac{1}{9}$, \\
$\therefore$ $S_{\triangle IEF}=\frac{1}{9}S_{\triangle ABC}=1$, \\
$\therefore$ the area of the shaded part in the figure $=9-1=\boxed{8}$.


Question: As shown in the figure, $AB\parallel CD\parallel EF$, if $BF=3DF$, then what is the value of $\frac{AC}{CE}$?

Solution:
\textbf{Solution:} Since $BF=3DF$,\\
it follows that $BD=2DF$,\\
due to $AB\parallel CD\parallel EF$,\\
it follows that $\frac{AC}{CE}=\frac{BD}{DF}$,\\
thus, $\frac{AC}{CE}=\frac{2DF}{DF}=\boxed{2}$


Question: As shown in the figure, the side DC of the rectangle ABCD is on the x-axis, and point B is on the graph of the hyperbola \(y=\frac{3}{x}\). Point E is the trisection point of side AD closer to point A. Connect CE and extend it to intersect the y-axis at point F. What is the area of \(\triangle CDF\)?

Solution:
\textbf{Solution:} Let the lengths of the rectangle's sides be AB=CD=a, AD=BC=b,\\
since point B is on the graph of the inverse proportion function $y=\frac{3}{x}$,\\
therefore B$\left(\frac{3}{b}, b\right)$,\\
therefore OC=$\frac{3}{b}$,\\
since point E is the trisection point on side AD closer to point A,\\
therefore DE=$\frac{2}{3}b$,\\
since AD$\parallel$y-axis,\\
therefore $\triangle FOC\sim \triangle EDC$,\\
therefore $\frac{OF}{ED}=\frac{OC}{CD}$, that is, $\frac{OF}{\frac{2}{3}b}=\frac{\frac{3}{b}}{a}$,\\
therefore OF=$\frac{2}{a}$,\\
therefore the area $S_{\triangle CDF}=\frac{1}{2}CD\cdot OF=\frac{1}{2}a\cdot\frac{2}{a}=\boxed{1}$


Question: As shown in the figure, it is known that $\triangle ABO \sim \triangle DEO$, and $BO: EO = 1: 3$. What is the ratio of the areas of $\triangle ABO$ to $\triangle DEO$?

Solution:
\textbf{Solution:} Given $\because \triangle ABO \sim \triangle DEO$ and $BO:EO=1:3$, \\
$\therefore$ the ratio of the areas of $\triangle ABO$ and $\triangle DEO$ is $1:9$, \\
thus the answer is: $\boxed{1:9}$.


Question: As shown in the Cartesian coordinate system, the side OB of $\triangle OAB$ lies on the y-axis with $\angle ABO=90^\circ$, $AB=4$. Point C is on line AB with $BC=\frac{1}{4}AB$, and $\angle BOC=\angle A$. If the hyperbola $y=\frac{k}{x}$ passes through point C, what is the value of $k$?

Solution:
\textbf{Solution:} Given BC $= \frac{1}{4}$ and AB = 4, the x-coordinate of point C is 1.\\
Since in the right triangles $\triangle ABO$ and $\triangle OBC$, $\angle ABO = \angle OBC$ and $\angle BOC = \angle A$,\\
it follows that $\triangle ABO \sim \triangle OBC$,\\
which implies that $ \frac{AB}{OB} = \frac{OB}{BC}$,\\
leading to $OB^2 = AB \cdot BC = 4 \cdot 1 = 4$,\\
thus, OB = 2.\\
The coordinates of point C are (1, 2). Substituting into $y = \frac{k}{x}$, we get: $k = \boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, points $D$, $E$, $F$ are on sides $AB$, $AC$, $BC$ respectively, and $DE \parallel BC$, $EF \parallel AB$. If $AD = 2BD$, then what is the value of $\frac{CF}{BF}$?

Solution:
\textbf{Solution:} Since $DE\parallel BC$, $EF\parallel AB$, and $AD=2BD$,
\[
\therefore \frac{AD}{BD}=\frac{AE}{EC}=2, \quad \frac{AE}{EC}=\frac{BF}{CF}=2,
\]
\[
\therefore \frac{CF}{BF}=\boxed{\frac{1}{2}}.
\]


Question: As shown in the diagram, in $\triangle ABC$, $\angle ACB = 90^\circ$, point O is the midpoint of AB, CD bisects $\angle BCO$ and intersects AB at point D. Draw AE $\perp$ CD, intersecting CO and BC at points G and E, respectively. Let the area of $\triangle AGO$ be $S_{1}$, and the area of $\triangle AEB$ be $S_{2}$. When $\frac{S_{1}}{S_{2}} = \frac{2}{5}$, what is the value of $\frac{OG}{BC}$?

Solution:
\textbf{Solution:} Construct line OF through point O parallel to AE and intersecting CB at point F,\\
since CD bisects $\angle$BCO and AE$\perp$CD,\\
it is easy to prove that $\triangle$CEG and $\triangle$COF are isosceles triangles,\\
therefore CG=CE, CO=CF,\\
therefore OG=EF,\\
since $\frac{{S}_{1}}{{S}_{2}} = \frac{2}{5}$,\\
let the area of $\triangle$ABE be 20a, then S$_{1}$=8a, S$_{2}$=20a,\\
since OF$\parallel$AE,\\
therefore $\triangle$OBF $\sim$ $\triangle$AEB,\\
therefore $\frac{{S}_{\triangle OBF}}{{S}_{\triangle AEB}}={\left(\frac{OB}{AB}\right)}^{2}=\frac{1}{4}=\frac{{S}_{\triangle OBF}}{20a}$,\\
therefore S$_{\triangle OBF}$=5a,\\
$\frac{{S}_{\triangle OBF}}{{S}_{\triangle AGO}}=\frac{5}{8}=\frac{OF}{AG}$,\\
since OF:AE=1:2,\\
therefore AE=2OF,\\
therefore $\frac{5}{8}=\frac{\frac{1}{2}AE}{AG}$,\\
therefore $\frac{AE}{AG}=\frac{5}{2}$,\\
therefore $\frac{GE}{OF}=\frac{CE}{CF}=\frac{2}{5}$,\\
let CE=2x, then CF=5x, EF=OG=3x,\\
therefore BF=3x,\\
therefore BC=CE+EF+BF=2x+3x+3x=8x,\\
therefore $\frac{OG}{BC}=\frac{3x}{8x}=\boxed{\frac{3}{8}}$.


Question: As shown in the diagram, in rectangle $ABCD$, where $AB=4$ and $AD=2$, a circle is drawn with center B and radius $BC$, intersecting side $AB$ at point E. Point P is a moving point on the arc $CE$. Connecting $PD$ and $PA$, what is the minimum value of $\frac{1}{2}AP+DP$?

Solution:
\textbf{Solution:} As shown in the figure, connect BP and take the midpoint G of BE, then connect PG,\\
$\because AD=BC=BP=2$, $AB=4$,\\
$\therefore \frac{BP}{BA}=\frac{2}{4}=\frac{1}{2}$,\\
$\because$ G is the midpoint of BE,\\
$\therefore \frac{BG}{BP}=\frac{1}{2}$,\\
$\therefore \frac{BP}{BA}=\frac{BG}{BP}$,\\
$\because \angle PBG=\angle ABP$,\\
$\therefore \triangle BPG\sim \triangle BAP$,\\
$\therefore \frac{PG}{AP}=\frac{BP}{BA}=\frac{1}{2}$,\\
$\therefore PG=\frac{1}{2}AP$,\\
thus $\frac{1}{2}AP+DP=PG+DP$, when points P, D, G are collinear, it takes the minimum value, that is, the length of DG,\\
$DG=\sqrt{AD^2+AG^2}=\sqrt{4+9}=\boxed{\sqrt{13}}$.


Question: As shown in the figure, within the $10\times 6$ grid paper, each small square has a side length of 1. We refer to the vertices of each small square as lattice points, and figures with vertices at the lattice points as lattice figures. Point E is a moving point on the side AB of the lattice quadrilateral ABCD. Connect ED and EC. If the lattice $\triangle DAE$ is similar to $\triangle EBC$, then what is the length of $DE$?

Solution:
\textbf{Solution}: From the figure, we know that $DA=3$, $AB=8$, $BC=4$, $AE=8-EB$, and $\angle A=\angle B=90^\circ$.\\
If $\triangle DAE \sim \triangle EBC$,\\
then $\frac{DA}{EB}=\frac{AE}{BC}$, that is $\frac{3}{EB}=\frac{8-EB}{4}$,\\
solving this gives $EB=2$ or $EB=6$,\\
when $EB=2$,\\
$EC=\sqrt{2^2+4^2}=2\sqrt{5}$,\\
$DE=\sqrt{3^2+6^2}=3\sqrt{5}$,\\
$DE+EC=5\sqrt{5}$,\\
when $EB=6$,\\
$EC=\sqrt{6^2+4^2}=2\sqrt{13}$,\\
$DE=\sqrt{3^2+2^2}=\sqrt{13}$,\\
$DE+EC=3\sqrt{13}$,\\
If $\triangle DAE \sim \triangle CBE$,\\
then $\frac{DA}{BC}=\frac{AE}{BE}$, that is $\frac{3}{4}=\frac{8-BE}{BE}$, solving gives\\
$BE=\frac{32}{7}$ (which is not consistent with the problem statement, thus discarded),\\
Therefore, $DE+EC=5\sqrt{5}$ or $3\sqrt{13}$,\\
Hence, the answer is: $DE+EC=\boxed{5\sqrt{5}}$ or $\boxed{3\sqrt{13}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=6$, and $AD$ is the median to side $BC$. If $\angle BAC=120^\circ$, what is the length of $AD$?

Solution:
\textbf{Solution:} Given that $AB=AC$ and $AD$ is the median of side $BC$,\\
thus $AD\perp BC$ and $\angle BAD=\angle CAD$,\\
since $\angle BAC=120^\circ$,\\
therefore, $\angle BAD=60^\circ$,\\
In right triangle $ABD$, $\angle B=90^\circ-60^\circ=30^\circ$,\\
thus, $AD=\frac{1}{2}AB=\boxed{3}$


Question: As shown in the figure, $AC=DB$, $AO=DO$, and $CD=300\,m$. What is the distance in meters between points $A$ and $B$?

Solution:
\textbf{Solution:} Since $AC=DB$ and $AO=DO$,

it follows that $BO=CO$.

In triangles $AOB$ and $DOC$,

\[
\left\{\begin{array}{l}
AO=DO\\
\angle AOB=\angle DOC\\
BO=CO
\end{array}\right.,
\]

thus $\triangle AOB \cong \triangle DOC$ (SAS),

therefore $AB=DC$.

Since $CD=300\,\text{m}$,

it follows that $AB=\boxed{300\,\text{m}}$,

which means the distance between points $A$ and $B$ is $300\,\text{m}$.


Question: As shown in the diagram, the traces of compass and straightedge constructions intersect sides $BC$ and $AB$ of $\triangle ABC$ at points D and E, respectively. Construct $DF \perp AB$ at point F. On $AC$, pick a point G such that $DE = DG$. If the area of $\triangle ADG$ is $52$ and the area of $\triangle AED$ is $38$, what is the area of $\triangle DEF$?

Solution:
\textbf{Solution:} As shown in the diagram, construct $DH\perp AC$ at H,\\
from the construction traces, $AD$ bisects $\angle BAC$,\\
$\because$ $AD$ bisects $\angle BAC$, and $DF\perp AB$, $DH\perp AC$,\\
$\therefore$ $DF=DH$,\\
In right triangles $\triangle DEF$ and $\triangle DGH$,\\
$\left\{\begin{array}{l}DE=DG\\ DF=DH\end{array}\right.$,\\
$\therefore$ right $\triangle DEF\cong$ right $\triangle DGH$ (HL),\\
In right triangles $\triangle DAF$ and $\triangle DAH$,\\
$\left\{\begin{array}{l}DF=DM\\ AD=AD\end{array}\right.$,\\
$\therefore$ right $\triangle DAF\cong$ right $\triangle DAH$ (HL),\\
$\therefore$ area of $\triangle DEF$ = area of $\triangle DGH$, area of $\triangle DAF$ = area of $\triangle DAH$,\\
$\because$ area of $\triangle ADG$ = area of $\triangle AHD$ + area of $\triangle DGH$, area of $\triangle AHD$ = area of $\triangle ADE$ + area of $\triangle DEF$,\\
$\therefore$ area of $\triangle ADE$ + area of $\triangle DEF$ = area of $\triangle ADG$ - area of $\triangle DGH$,\\
i.e., $38$ + area of $\triangle DEF$ = $52$ - area of $\triangle DEF$,\\
$\therefore$ area of $\triangle DEF$ = $\boxed{7}$,\\


Question: In the Cartesian coordinate system, let points $A(1, 0)$ and $B(0, 2)$. Construct $\triangle BOC$ such that $\triangle BOC$ is congruent to $\triangle ABO$ (with point $C$ not coinciding with point $A$). What are the coordinates of point $C$?

Solution:
\textbf{Solution:} According to the problem, we draw $\triangle BOC$ as shown in the figure,\\
then $OC_1=BC_2=BC_3=OA=1$, $\angle OBC_2=\angle OBC_3=90^\circ$,\\
thus, the coordinates of point C are $\boxed{\left(-1, 0\right)}$ or $\left(-1, 2\right)$ or $\left(1, 2\right)$.


Question: As shown in the figure, points E and F are the midpoints of $BD$ and $CE$ respectively. If the area of $\triangle ABC$ is 32, then what is the area of the shaded part $\triangle AEF$?

Solution:
\textbf{Solution:} \\
Since point $E$ is the midpoint of $BD$, \\
it follows that $AE$ is the median of $\triangle ABD$, and $CE$ is the median of $\triangle CBD$, \\
thus, ${S}_{\triangle ADE}=\frac{1}{2}{S}_{\triangle ABD}$ and ${S}_{\triangle CDE}=\frac{1}{2}{S}_{\triangle CBD}$, \\
hence, ${S}_{\triangle ACE}={S}_{\triangle ADE}+{S}_{\triangle CDE}=\frac{1}{2}{S}_{\triangle ABD}+\frac{1}{2}{S}_{\triangle CBD}=\frac{1}{2}{S}_{\triangle ABC}=\frac{1}{2}\times 32=16$, \\
Since point $F$ is the midpoint of $CE$, \\
it follows that $AF$ is the median of $\triangle ACE$, \\
thus, ${S}_{\triangle AEF}=\frac{1}{2}{S}_{\triangle ACE}=\frac{1}{2}\times 16=\boxed{8}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=3$, $BC=4$. Point $D$ is on side $AB$, with $AD=AC$, and $AE \perp CD$ at foot $F$, intersecting $BC$ at point $E$. What is the length of $BE$?

Solution:
\textbf{Solution}: As shown in the diagram, connect DE, \\
$\because$AD＝AC=3, AE$\perp$CD, \\
$\therefore$AE is the perpendicular bisector of CD, \\
$\therefore$CE＝DE, \\
$\therefore$$\angle$ADE＝$\angle$ACB＝$90^\circ$, \\
In $\triangle ABC$, where AC=3, BC=4, \\
$\therefore$AB＝$\sqrt{AC^{2}+BC^{2}}$=5, \\
$\therefore$BD＝AB−AD＝2, \\
$\therefore$Area of $\triangle ABC$＝Area of $\triangle ACE$+Area of $\triangle ABE$, \\
$\therefore$AC·BC＝AC·CE+AB·DE, \\
$\therefore$3×4＝3CE+5DE=8DE, \\
$\therefore$DE＝$\frac{3}{2}$, \\
In $\triangle BDE$, BE＝$\sqrt{DE^{2}+BD^{2}}$＝$\sqrt{\left(\frac{3}{2}\right)^{2}+2^{2}}$=$\boxed{\frac{5}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=5$, $D$ is the midpoint of $AB$, $BE\perp CD$ with foot $E$. If $BE=4$, then what is the length of $BC$?

Solution:
\textbf{Solution:} Let CF$\perp$AB at F, \\
since $\angle$ACB=$90^\circ$ and D is the midpoint of AB, \\
therefore CD=$\frac{1}{2}$AB=BD, \\
in $\triangle CDF$ and $\triangle BDE$, \\
$\left\{\begin{array}{l}
\angle CDF=\angle BDE \\
\angle CFD=\angle BED=90^\circ \\
CD=BD
\end{array}\right.$, \\
therefore $\triangle CDF \cong \triangle BDE$ (AAS), \\
therefore CF=BE=4, \\
in $\triangle ACF$, AF=$\sqrt{AC^{2}-CF^{2}}=\sqrt{5^{2}-4^{2}}=3$, \\
let BF=x, then AB=x+3, \\
by Pythagoras' theorem, BC$^{2}$=BF$^{2}$+CF$^{2}$, \\
thus, x$^{2}$+4$^{2}$=$(x+3)^{2}-5^{2}$, \\
solving this, x=$\frac{16}{3}$, \\
therefore BC$=\sqrt{CF^{2}+BF^{2}}=\boxed{\frac{20}{3}}$.


Question: As illustrated, $\triangle PAC \cong \triangle PBD$, given that $\angle A=40^\circ$ and $\angle BPD=20^\circ$, what is the measure of $\angle PCD$?

Solution:
Solution: Since $\triangle PAC \cong \triangle PBD$,\\
it follows that $\angle A=\angle B, \angle APC=\angle BPD, \angle PCA=\angle PDB$\\
Since $\angle A=40^\circ, \angle BPD=20^\circ$\\
it follows that $\angle APC=20^\circ$\\
Therefore, $\angle PCD=\angle A+\angle APC=40^\circ+20^\circ=\boxed{60^\circ}$.


Question: As shown in the figure, in the right-angled triangle ABC, $\angle ACB=90^\circ$, $AB=7$, the point D is the midpoint of $AB$, the point P is a moving point on the hypotenuse $AB$, $FG$ is the perpendicular bisector of the line segment $CP$, Q is a moving point on $FG$. What is the minimum value of $PQ+QD$?

Solution:
\textbf{Solution:} Let's connect $CQ$, $CD$ as shown in the figure,\\
since point P is a moving point on the hypotenuse $AB$, $FG$ is the perpendicular bisector of the line segment $CP$, Q is a moving point on $FG$,\\
therefore $CQ=PQ$,\\
thus $PQ+QD=CQ+DQ\ge CD$,\\
since $\angle ACB=90^\circ$, $AB=7$, and point D is the midpoint of $AB$,\\
therefore $CD=\frac{7}{2}$.\\
Hence, the minimum value of $PQ+QD$ is $\boxed{\frac{7}{2}}$.


Question: As shown in the figure, D is a point outside the equilateral triangle $\triangle ABC$, with $AD=5$, $CD=3$. When the length of $BD$ is maximized, what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Construct an equilateral triangle $DCE$ with $CD$ as a side. Join $AE$.\\
$\because$ $BC=AC$, $CD=CE$, $\angle BCA=\angle DCE=60^\circ$,\\
$\therefore$ $\angle BCD=\angle ACE$,\\
In $\triangle BCD$ and $\triangle ACE$,\\
$\left\{\begin{array}{l}
BC=AC\\
\angle BCD=\angle ACE\\
CD=CE
\end{array}\right.$,\\
$\therefore$ $\triangle BCD \cong \triangle ACE$ (SAS),\\
$\therefore$ $BD=AE$,\\
In $\triangle ADE$, $\because$ $AD=5$, $DE=CD=3$,\\
$\therefore$ $AE \leq AD + DE$,\\
$\therefore$ The maximum value of $AE$ is 8,\\
$\therefore$ When points A, D, E are collinear, the value of $BD$ is maximum, and it is 8,\\
As shown in the diagram, draw $CF\perp AD$ with the foot at F, and draw $BM\perp AC$ intersecting AC at point M,\\
$\because$ $\angle CDE=\angle BDC=\angle E=60^\circ$,\\
$\therefore$ $\angle DCF=30^\circ$,\\
$\therefore$ $DF=\frac{1}{2}CD=\frac{3}{2}$,\\
$\therefore$ $CF=\sqrt{3^2-\left(\frac{3}{2}\right)^2}=\frac{3\sqrt{3}}{2}$, $AF=AD+DF=5+\frac{3}{2}=\frac{13}{2}$,\\
$\therefore$ $AC=\sqrt{AF^2+CF^2}=7$,\\
According to the properties of an equilateral triangle, $AM=CM=\frac{1}{2}AC=\frac{7}{2}$,\\
$\therefore$ The height $BM$ on the side $AC$ of $\triangle ABC$ is $\sqrt{7^2-\left(\frac{7}{2}\right)^2}=\frac{7\sqrt{3}}{2}$,\\
$\therefore$ The area of $\triangle ABC$ is $\frac{1}{2} \times \frac{7\sqrt{3}}{2} \times 7=\boxed{\frac{49\sqrt{3}}{4}}$.


Question: As shown in the figure, $AB\perp CD$, and $AB=CD$. $E$ and $F$ are two points on $AD$, with $CE\perp AD$, $BF\perp AD$. If $CE=6$, $BF=3$, $EF=2$, what is the length of $AD$?

Solution:
\textbf{Solution:} Given that $AB \perp CD$, $CE \perp AD$, and $BF \perp AD$,\\
it follows that $\angle AFB = \angle CED = 90^\circ$, $\angle A + \angle D = 90^\circ$, $\angle C + \angle D = 90^\circ$,\\
thus, $\angle A = \angle C$,\\
since $AB = CD$,\\
it results in $\triangle ABF \cong \triangle CDE$,\\
hence $AF = CE = 6$, $BF = DE = 3$,\\
given $EF = 2$,\\
consequently, $AD = AF + DF = \boxed{7}$.


Question: As shown in the figure, AD is the angle bisector of the apex of the isosceles triangle $\triangle ABC$, and $BD=5$, then what is the value of $CD$?

Solution:
\textbf{Solution:} \\
Since $AD$ is the angle bisector of the vertex of the isosceles triangle $ABC$, \\
then $AD$ is the median to the base $BC$ of the isosceles triangle $ABC$, \\
hence $CD = BD$, \\
since $BD = 5$, \\
thus $CD = \boxed{5}$


Question: A right-angled triangle board is placed as shown in the figure, with $D$, $F$ on side $AB$, point $C$ on side $DE$, and $EF$, $AC$ intersecting at point $G$, $\angle A=30^\circ$, $\angle E=45^\circ$, then what is the degree measure of $\angle FGA$?

Solution:
\textbf{Solution:} Since $\triangle DEF$ is a right-angled triangle, and $\angle E=45^\circ$, $\angle A=30^\circ$,\\
then, $\angle DFE=45^\circ$.\\
Since $\angle DFE=\angle A+\angle FGA$,\\
it follows that $\angle FGA=\angle DFE-\angle A=45^\circ-30^\circ=\boxed{15^\circ}$.


Question: As shown in the figure, in $ \triangle ABC$, $ AD$ bisects $ \angle BAC$ and intersects $ BC$ at point $ D$, $ DE \perp AB$ with the foot of the perpendicular being $ E$. If $ DE=1$ and $ AC=4$, then what is the area of $ \triangle ADC$?

Solution:
\textbf{Solution:} Construct $DF \perp AC$ at $F$, as shown in the figure,\\
since $AD$ bisects $\angle BAC$, $DE\perp AB$, $DF\perp AC$,\\
therefore $DF=DE=1$,\\
therefore the area of $\triangle ADC = \frac{1}{2} \times 4 \times 1 = \boxed{2}$.


Question: As shown in the figure, the equilateral triangle $\triangle ABC$ has a side length of 8. $P$ and $Q$ are points on sides $AC$ and $BC$, respectively. Line segments $AQ$ and $BP$ intersect at point $O$, with $AP = CQ$. What is the measure of $\angle AOB$ in degrees? If $BQ = 5$, what is the length of $AQ$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is an equilateral triangle, \\
we have $AB=AC$, $\angle BAC=\angle C=60^\circ$, \\
In triangles $\triangle ABP$ and $\triangle ACQ$, \\
$\left\{\begin{array}{l}
AB=AC\\
\angle BAP=\angle CAQ\\
AP=CQ
\end{array}\right.$, \\
thus $\triangle ABP\cong \triangle ACQ$ (SAS), \\
therefore, $\angle ABP=\angle CAQ$, \\
thus $\angle BOQ=\angle ABO+\angle BAQ=\angle CAQ+\angle BAQ=\angle BAC=60^\circ$, \\
hence, $\angle AOB=120^\circ$; \\
Draw line AD $\perp$ BC at D through point A, \\
Since BC=8 and $\triangle ABC$ is an equilateral triangle, \\
then BD=CD=4, $\angle ABD=60^\circ$, \\
thus AD=$\sqrt{AB^2-BD^2}=\sqrt{8^2-4^2}=4\sqrt{3}$ \\
Given BQ=5, \\
thus DQ=1, \\
therefore, AQ=$\sqrt{AD^2+DQ^2}=\sqrt{(4\sqrt{3})^2+1^2}=\boxed{7}$.


Question: As shown in the figure, it is known that $AD$ and $CE$ are two altitudes of $\triangle ABC$, with $AD=CE$ and $\angle CAD=25^\circ$. What is the measure of $\angle OCD$ in degrees?

Solution:
\textbf{Solution:} Since AD and CE are the two altitudes of $\triangle ABC$,\\
it follows that AD$\perp$CB and CE$\perp$AD,\\
therefore $\angle ADC=\angle CEA=90^\circ$,\\
In Rt$\triangle ADC$ and Rt$\triangle CEA$,\\
$\left\{\begin{array}{l}AC=CA\\ AD=CE\end{array}\right.$,\\
thus, Rt$\triangle ADC\cong$Rt$\triangle CEA$ (HL),\\
therefore, $\angle CAD=\angle ACE=25^\circ$,\\
since $\angle ACD=90^\circ-\angle CAD=90^\circ-25^\circ=65^\circ$,\\
thus, $\angle OCD=\angle ACD-\angle ACE=65^\circ-25^\circ=\boxed{40^\circ}$.


Question: As shown in the figure above, in $\triangle ABC$, $\angle A=90^\circ$, $AB=AC$, $BD$ bisects $\angle ABC$, $CE \perp BD$ at $E$. If $BD=12$, what is the length of $CE$?

Solution:
\textbf{Solution:} As shown in the diagram, extend BA and CE to intersect at point F, \\
$\because$ $\angle$ABD + $\angle$ADB = $90^\circ$, $\angle$CDE + $\angle$ACF = $90^\circ$, \\
$\therefore$ $\angle$ABD = $\angle$ACF, \\
$\because$ $\angle$BAC = $90^\circ$, \\
$\therefore$ $\angle$FAC = $\angle$BAC = $90^\circ$, \\
In $\triangle ABD$ and $\triangle ACF$, \\
$\left\{\begin{array}{l}
\angle ABD = \angle ACF\\
AB = AC\\
\angle BAC = \angle FAC
\end{array}\right.$, \\
$\therefore$ $\triangle ABD \cong \triangle ACF$ (ASA), \\
$\therefore$ BD = CF, \\
$\because$ BD bisects $\angle$ABC, \\
$\therefore$ $\angle$ABE = $\angle$CBE, \\
$\because$ CE $\perp$ BD, \\
$\therefore$ $\angle$BEF = $\angle$BEC = $90^\circ$, \\
In $\triangle BEF$ and $\triangle BEC$, \\
$\left\{\begin{array}{l}
\angle ABE = \angle CBE\\
BE = BE\\
\angle BEF = \angle BEC
\end{array}\right.$, \\
$\therefore$ $\triangle BEF \cong \triangle BEC$, \\
$\therefore$ EF = EC, \\
$\therefore$ EC = $\frac{1}{2}$CF, \\
$\therefore$ CE = $\frac{1}{2}$BD, \\
$\because$ BD = 12, \\
$\therefore$ CE = \boxed{6}.


Question: As shown in the figure, lines a and b are intersected by lines c and d. If $a\parallel b$ and $\angle 1=120^\circ$, $\angle 2=30^\circ$, then what is the measure of $\angle 3$ in degrees?

Solution:
\textbf{Solution:} As shown in the figure,\\
$\because a \parallel b$,\\
$\therefore \angle 4 = \angle 1 = 120^\circ$,\\
$\because \angle 4 = \angle 2 + \angle 3$,\\
$\therefore \angle 3 = \angle 4 - \angle 2 = 120^\circ - 30^\circ = \boxed{90^\circ}$.


Question: As shown in the diagram, it is known that $\angle MON=30^\circ$, points $A_1$, $A_2$, $A_3$, ... are on ray $ON$, and points $B_1$, $B_2$, $B_3$, ... are on ray $OM$. The triangles $\triangle A_1B_1A_2$, $\triangle A_2B_2A_3$, $\triangle A_3B_3A_4$, ... are all equilateral triangles. If $OA_1=3$, what is the side length of $\triangle A_{10}B_{10}A_{11}$?

Solution:
\textbf{Solution:} Since $\angle MON=30^\circ$ and $\angle B_{1}A_{1}A_{2}=60^\circ$,\\
it follows that $\angle OB_{1}A_{1}=\angle B_{1}A_{1}A_{2}-\angle MON=30^\circ$.\\
Therefore, $\angle MON=\angle OB_{1}A_{1}$.\\
Thus, we have $A_{1}O=B_{1}A_{1}=3$.\\
Consequently, $A_{2}O=A_{1}O+A_{1}A_{2}=3\times 2$.\\
Similarly, we can deduce $AO_{3}=A_{3}B_{3}=3\times 2^{2}$.\\
Following the same pattern $A_{4}O=A_{4}B_{4}=3\times 2^{3}$.\\
$A_{5}O=A_{5}B_{5}=3\times 2^{4}$.\\
$\cdots$\\
$A_{n}O=A_{n}B_{n}=3\times 2^{n-1}$.\\
Therefore, the side length of $\triangle A_{10}B_{10}A_{11}$ is $3\times 2^{9}=\boxed{1536}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $D$ and $E$ are two points inside $\triangle ABC$, $AE$ bisects $\angle BAC$, $\angle D=\angle DBC=60^\circ$. If $BD=5\text{cm}$ and $DE=3\text{cm}$, what is the length of $BC$ in $\text{cm}$?

Solution:
\textbf{Solution:} Extend DE to meet BC at M, extend AE to meet BC at N, construct EF $\parallel$ BC at F, \\
$\because$ AB=AC, AE bisects $\angle$BAC, \\
$\therefore$ AN$\perp$BC, BN=CN, \\
$\because$ $\angle$DBC=$\angle$D=$60^\circ$, \\
$\therefore$ $\triangle$BDM is an equilateral triangle, \\
$\therefore$ $\triangle$EFD is an equilateral triangle, \\
$\because$ BD=5, DE=3, \\
$\therefore$ EM=2, \\
$\because$ $\triangle$BDM is an equilateral triangle, \\
$\therefore$ $\angle$DMB=$60^\circ$, \\
$\because$ AN$\perp$BC, \\
$\therefore$ $\angle$ENM=$90^\circ$, \\
$\therefore$ $\angle$NEM=$30^\circ$, \\
$\therefore$ NM=1, \\
$\therefore$ BN=4, \\
$\therefore$ BC=2BN=\boxed{8} (cm),


Question: As shown in the figure, $\triangle ABC \cong \triangle EDC$, where point D is on $AB$, and $\angle B=60^\circ$, then what is the degree measure of $\angle EDA$?

Solution:
\textbf{Solution:} Since $\triangle ABC \cong \triangle EDC$,\\
it follows that $\angle B = \angle EDC = 60^\circ$, $CD = CB$,\\
hence $\angle CDB = \angle B = 60^\circ$,\\
therefore $\angle EDA = 180^\circ - \angle EDC - \angle CDB = \boxed{60^\circ}$.


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle, and $BD=CE$, $\angle 1=15^\circ$. What is the degree measure of $\angle 2$?

Solution:
\textbf{Solution:} Given that $\triangle ABC$ is an equilateral triangle,\\
hence, $AB=BC$ and $\angle ABC=\angle ACB$,\\
in $\triangle ABD$ and $\triangle BCE$,\\
we have
\[
\left\{\begin{array}{l}
AB=BC\\
\angle ABC=\angle ACB\\
BD=CE
\end{array}\right.,
\]
therefore, $\triangle ABD \cong \triangle BCE$ (SAS),\\
thus, $\angle 1 = \angle CBE$,\\
since $\angle 2 = \angle 1 + \angle ABE$,\\
hence $\angle 2 = \angle CBE + \angle ABE = \angle ABC = 60^\circ$.\\
Therefore, the answer is: $\boxed{60^\circ}$.


Question: As shown in the figure, $CA$ bisects $\angle DCB$, $CB = CD$, and the extension line of $DA$ intersects $BC$ at point $E$. If $\angle EAC = 46^\circ$, what is the degree measure of $\angle BAE$?

Solution:
\textbf{Solution}: Since $CA$ bisects $\angle DCB$,\\
it follows that $\angle BCA=\angle DCA$,\\
and since $CB=CD$, $AC=AC$,\\
it follows that $\triangle ABC\cong \triangle ADC\ (SAS)$,\\
therefore $\angle B=\angle D$,\\
hence $\angle D+\angle ACD=\angle B+\angle ACB=\angle CAE=46^\circ$,\\
therefore $\angle BAE=180^\circ-\angle B-\angle ACB-\angle CAE=\boxed{88^\circ}$


Question: As shown in the figure, it is assembled from a pair of set squares. What is the degree measure of $\angle ABC$ in the figure?

Solution:
\textbf{Solution:} Given $\because$ $\angle F=30^\circ$ and $\angle EAC=45^\circ$, \\
$\therefore$ $\angle ABF=\angle EAC-\angle F=45^\circ-30^\circ=15^\circ$, \\
$\because$ $\angle FBC=90^\circ$, \\
$\therefore$ $\angle ABC=\angle FBC-\angle ABF=90^\circ-15^\circ=\boxed{75^\circ}$.


Question: As shown in the figure, $AC$ bisects $\angle BAD$, $\angle B+\angle D=180^\circ$, $CE\perp AD$ at point $E$, $AE=10\,\text{cm}$, $AB=8\,\text{cm}$, what is the length of $DE$ in $\text{cm}$?

Solution:
\textbf{Solution:} As shown in the figure, take $EF=DE$ on $AD$,\\
$\because CE\perp AD$,\\
$\therefore$ the line $CE$ is the perpendicular bisector of the segment $FD$,\\
$\therefore CF=CD$,\\
$\therefore \angle D=\angle CFD$, $\angle AFC+\angle CFD=180^\circ$,\\
$\because \angle B+\angle D=180^\circ$,\\
$\therefore \angle AFC=\angle B$,\\
$\because \left\{\begin{array}{l}
\angle B=\angle AFC \\
\angle BAC=\angle FAC \\
AC=AC
\end{array}\right.$,\\
$\therefore \triangle BAC\cong \triangle FAC$,\\
$\therefore AF=AB=8\text{cm}$, $EF=DE=AE-AF=AE-AB=10\text{cm}-8\text{cm}=2\text{cm}$.\\
Thus, the answer is: $DE=\boxed{2\text{cm}}$.


Question: As shown in the figure: rectangle $ABCD$ is folded along $AE$ such that point $D$ falls on point $F$ on $BC$. Given $AB=6$ and the area of $\triangle ABF$ is 24, what is the value of $FC$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle, \\
it follows that $\angle B=90^\circ$ and $AD=BC$, \\
since $AB=6$, \\
it follows that ${S}_{\triangle ABF}=\frac{1}{2}AB\cdot BF=\frac{1}{2}\times 6\times BF=24$, \\
thus, $BF=8$, \\
hence, $AF=\sqrt{AB^{2}+BF^{2}}=\sqrt{6^{2}+8^{2}}=10$, \\
By the property of folding: $AD=AF=10$, \\
thus, $BC=AD=10$, \\
hence, $FC=BC-BF=10-8=\boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, point $D$ is on side $AB$, $E$ is the midpoint of side $AC$, $CF \parallel AB$, and $CF$ intersects the extension line of $DE$ at point $F$. If $AB=4$ and $CF=3$, what is the length of $BD$?

Solution:
\textbf{Solution:} Given that $CF\parallel AB$, \\
thus $\angle A=\angle ACF$, $\angle F=\angle ADE$, \\
since E is the midpoint of side $AC$, \\
then $AE=CE$, \\
in $\triangle ADE$ and $\triangle CFE$, \\
$\left\{\begin{array}{l}
\angle A=\angle ACF \\
\angle ADE=\angle F \\
AE=CE
\end{array}\right.$, \\
therefore $\triangle ADE\cong \triangle CFE$ (AAS), \\
thus $AD=CF=3$, \\
hence $BD=AB-AD=4-3=\boxed{1}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $\angle A=65^\circ$. When $\angle B$ and $\angle C$ are folded as shown, if $\angle ADB'=35^\circ$, then what is the sum of $\angle 1+\angle 2+\angle 3$?

Solution:
\textbf{Solution:}\\
From the folding, we know: $\angle B=\angle B'$, $\angle C=\angle C'$.\\
$\because \angle 3=\angle B+\angle 4$, $\angle 4=\angle ADB'+\angle B'$,\\
$\therefore \angle 3=\angle B+\angle ADB'+\angle B'$\\
$=2\angle B+35^\circ$.\\
$\because \angle 1+\angle 2=180^\circ-\angle C'GC+180^\circ-\angle C'FC$\\
$=360^\circ-(\angle C'FC+\angle C'GC)$,\\
$\angle C'FC+\angle C'GC=360^\circ-\angle C-\angle C'$\\
$=360^\circ-2\angle C$,\\
$\therefore \angle 1+\angle 2=360^\circ-(\angle C'FC+\angle C'GC)$\\
$=360^\circ-(360^\circ-2\angle C)$\\
$=2\angle C$.\\
$\therefore \angle 1+\angle 2+\angle 3$\\
$=2\angle C+2\angle B+35^\circ$\\
$=2(\angle C+\angle B)+35^\circ$\\
$=2(180^\circ-\angle A)+35^\circ$\\
$=2(180^\circ-65^\circ)+35^\circ$\\
$=\boxed{265^\circ}$.


Question: In the figure, within $ \triangle ABC$, where $ \angle ABC=90^\circ$, and the median $CD$ on the hypotenuse $AB$ measures $5$cm, what is the length of $AB$ in cm?

Solution:
\textbf{Solution:} \\
Given that in the right-angled triangle $ABC$, with the median $CD$ on the hypotenuse $AB$ equals $5$,\\
it follows that $AB = 2CD = \boxed{10}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C = 90^\circ$, $\angle A = 30^\circ$, and $BD = 2CD$, what is the measure of $\angle ADB$ in degrees?

Solution:
\textbf{Solution:} Since $\angle C=90^\circ$ and BD=2CD,\\
it follows that $\angle DBC=30^\circ$,\\
hence $\angle ADB=\angle C+\angle DBC=90^\circ+30^\circ=\boxed{120^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, we have $AB=AC$. Point $D$ is inside $\triangle ABC$, and $AD$ bisects $\angle BAC$. Connect $CD$, and fold $\triangle ADC$ along $CD$ such that $AC$ lands on $CE$ and intersects $AB$ at $F$, where $CE\perp AB$. If $BC=10$ and $AD=7$, what is the length of $EF$?

Solution:
\textbf{Solution:} As shown in the diagram, extend AD to intersect BC at point G, \\
\[
\because AB = AC, \text{ and } AD \text{ bisects } \angle BAC,
\]
\[
\therefore \angle B = \angle ACB, \text{ and } AG \perp BC, \text{ thus } BG = CG,
\]
\[
\because BC = 10,
\]
\[
\therefore BG = CG = \frac{1}{2}BC = 5,
\]
\[
\because CE \perp AB,
\]
\[
\therefore \angle BFC = 90^\circ,
\]
\[
\therefore \angle B + \angle BCF = 90^\circ,
\]
\[
\because \angle B = \angle C = \angle BCF + \angle ACD + \angle ECD,
\]
\[
\therefore 2\angle BCF + \angle ACD + \angle ECD = 90^\circ,
\]
According to the property of folding, we know that \(\angle ACD = \angle ECD\), and \(AC = EC\),
\[
\therefore \angle BCF + \angle ECD = 45^\circ,
\]
\[
\therefore \triangle CDG is an isosceles right triangle,
\]
\[
\therefore AG = AD + DG = 7 + 5 = 12,
\]
In the right triangle \(\triangle ACG\), \(AC = \sqrt{CG^{2}+AG^{2}} = \sqrt{5^{2}+12^{2}} = 13\)
\[
\therefore CE = AB = AC = 13,
\]
\[
\because S_{\triangle ABC} = \frac{1}{2}BC \cdot AG = \frac{1}{2}AB \cdot CF,
\]
\[
\therefore \frac{1}{2} \times 10 \times 12 = \frac{1}{2} \times 13 \times CF,
\]
\[
\therefore CF = \frac{120}{13},
\]
\[
\therefore EF = CE - CF = 13 - \frac{120}{13} = \boxed{\frac{49}{13}}.
\]


Question: As shown in the diagram, $AD$ is the angle bisector of $\triangle ABC$, $DE\perp AB$ with the foot at $E$, the area of $\triangle ABC$ is $8$, $DE=2$, and $AB=5$. What is the length of $AC$?

Solution:
\textbf{Solution:} Construct $DF\perp AC$ at $F$, \\
since $AD$ is the angle bisector of $\triangle ABC$, $DE\perp AB$, and $DF\perp AC$, \\
thus $DE=DF=2$, \\
therefore the area of $\triangle ADB=\frac{1}{2}AB\cdot DE=\frac{1}{2}\times 5\times 2=5$, \\
since the area of $\triangle ABC$ is 8, \\
thus the area of $\triangle ADC$ is $8-5=3$, \\
thus $\frac{1}{2}AC\cdot DF=3$, \\
thus $\frac{1}{2}AC\times 2=3$, \\
thus $AC=\boxed{3}$,


Question: As shown in the figure, CE is the bisector of the exterior angle $\angle ACD$ of $\triangle ABC$. If $\angle B = 30^\circ$ and $\angle ACE = 60^\circ$, what is the measure of $\angle A$ in degrees?

Solution:
\textbf{Solution:} Since $CE$ bisects $\angle ACD$ and $\angle ACE=60^\circ$,

it follows that $\angle ACD=2\angle ACE=120^\circ$.

Since $\angle ACD$ is an exterior angle of $\triangle ABC$ and $\angle B=30^\circ$,

it follows that $\angle ACD=\angle B+\angle A$,

thus $\angle A=\angle ACD-\angle B$

$=120^\circ-30^\circ$

$=\boxed{90^\circ}$.