Question: As shown in the figure, in $\triangle ABC$, $\angle A=30^\circ$, $\angle B=45^\circ$, and $AC=2\sqrt{3}$. What is the length of $AB$?

Solution:
\textbf{Solution:} Draw $CD \perp AB$ at $D$, \\
$\therefore$ $\angle ADC=\angle BDC=90^\circ$.\\
$\because$ $\angle B=45^\circ$, \\
$\therefore$ $\angle BCD=\angle B=45^\circ$, \\
$\therefore$ $CD=BD$. \\
$\because$ $\angle A=30^\circ$, $AC=2\sqrt{3}$, \\
$\therefore$ $CD=\sqrt{3}$, \\
$\therefore$ $BD=CD=\sqrt{3}$. \\
By the Pythagorean theorem, $AD=\sqrt{AC^2-CD^2}=3$, \\
$\therefore$ $AB=AD+BD=3+\sqrt{3}$. \\
Hence, the answer is: $AB=\boxed{3+\sqrt{3}}$.


Question: As shown in the diagram, in square $ABCD$, $\angle B=60^\circ$, $AB=6$, $BC=4$, what is the area of square $ABCD$? Point $E$ is a moving point on $AB$, extend line $ED$ to point $F$ such that $DE=3DF$, construct square $EFGC$ with $EC$ and $EF$ as adjacent sides, connect $EG$, what is the minimum value of $EG$?

Solution:
\textbf{Solution:} Construct $CH\perp AB$ at point H, and let EG intersect CD at point O,\\
since in $\square ABCD$, $\angle B=60^\circ$ and BC=4,\\
therefore, CH=$2\sqrt{3}$,\\
thus, the area of $\square ABCD$: AB×CH=$12\sqrt{3}$;\\
since quadrilateral ECGF is a parallelogram,\\
thus, $EF\parallel CG$,\\
therefore, $\triangle EOD\sim \triangle GOC$,\\
thus, $\frac{EO}{GO}=\frac{DO}{CO}=\frac{DE}{CG}$,\\
therefore, DE＝3DF,\\
therefore, $\frac{DE}{EF}=\frac{3}{4}$,\\
thus, $\frac{ED}{CG}=\frac{3}{4}$,\\
therefore, $\frac{EO}{GO}=\frac{3}{4}$,\\
thus, when EO takes its minimum value, EG can also take its minimum value, when EO$\perp$CD, EO takes the minimum value,\\
therefore,CH=EO,\\
thus, EO=$2\sqrt{3}$,\\
therefore, GO=$\frac{8\sqrt{3}}{3}$,\\
thus, the minimum value of EG is $2\sqrt{3}+\frac{8\sqrt{3}}{3}=\boxed{\frac{14\sqrt{3}}{3}}$.


Question: As shown in the figure, within circle $\odot O$, chord $BC$ is perpendicular to radius $OA$, point $D$ is a point on the major arc $BC$, connecting $BD$, $AD$, $OC$, with $\angle ADB=30^\circ$. If chord $BC=8\sqrt{3}$ cm, what is the length of the arc corresponding to chord $BC$ in the figure in cm?

Solution: \textbf{Solution:} Draw line OB. Since OA$\perp$BC, it follows that $\overset{\frown}{AB}=\overset{\frown}{AC}$, and therefore $\angle$AOC=$\angle$AOB. According to the inscribed angle theorem, $\angle$AOB=$2\angle$ADB=$60^\circ$, hence $\angle$AOC=$\angle$AOB=$60^\circ$. Since OA$\perp$BC, it follows that BE= $\frac{1}{2}$ BC=$4\sqrt{3}$ cm. In right triangle BOE, with $\angle$AOB=$60^\circ$, it follows that $OB=\frac{BE}{\sin 60^\circ}=8$ cm. Therefore, the length of the minor arc BC is $\frac{120\pi \times 8}{180}=\boxed{\frac{16}{3}\pi}$ cm.


Question: As shown in the figure, the side length of each of the four small squares is 1. If an arc is drawn with O as the center and OG as the radius, intersecting AB and CD at points E and F, respectively, what is the length of the arc EF?

Solution:
\textbf{Solution:} From the given information, we can deduce that $OE=OG=OF=2$\\
$\therefore$ $OD=\frac{1}{2}OF$\\
$\therefore$ In $\triangle ODF$ with a right angle at $D$, $\sin\angle OFD=\frac{OD}{OF}=\frac{1}{2}$\\
$\therefore$ $\angle OFD=30^\circ$\\
Similarly, we can deduce: $\angle OEA=30^\circ$\\
Since $AB\parallel OG\parallel DC$\\
$\therefore \angle EOG=\angle OEA=30^\circ$, $\angle FOG=\angle OFD=30^\circ$\\
$\therefore$ $\angle EOF=\angle EOG+\angle FOG=60^\circ$\\
Therefore, $\overset{\frown}{EF}=\frac{n\pi r}{180}=\frac{60\pi \times 2}{180}=\boxed{\frac{2}{3}\pi}$


Question: As shown in the figure, in the semicircle \(O\) with \(AB\) as the diameter, \(C\) is the trisection point of the semicircle. Point \(P\) is a moving point on arc \(BC\), and lines \(CP\) and \(AP\) are connected. Construct \(OM\) perpendicular to \(CP\) and intersecting \(AP\) at \(N\), connect \(BN\). If \(AB=12\), what is the minimum value of \(NB\)?

Solution:
\textbf{Solution:} As shown in the figure, connect AC and OC. \\
$\because$ C is the trisecting point of the semicircle, \\
$\therefore$ $\angle$AOC = $60^\circ$, \\
$\because$ OA = OC, \\
$\therefore$ $\triangle$AOC is an equilateral triangle, \\
Construct the circumscribed circle $\odot$T of $\triangle$AOC, and connect TA=TC, TN, TB. \\
$\because$ OM$\perp$PC, \\
$\therefore$ CM = PM, \\
$\therefore$ NC = NP, \\
$\therefore$ $\angle$NPC = $\angle$NCP = $\frac{1}{2}\angle$AOC = $30^\circ$, \\
$\therefore$ $\angle$CNM = $60^\circ$, \\
$\therefore$ $\angle$CNO = $120^\circ$, \\
$\because$ $\angle$CNO + $\angle$OAC = $180^\circ$, \\
$\therefore$ point N is on $\odot$T, the trajectory is $\overset{\frown}{OC}$, \\
Draw TH$\perp$AB at H through point T. \\
In the right triangle $\triangle$ATH, AH=OH=3, $\angle$TAH=$30^\circ$, \\
$\therefore$ TH = AH$\cdot \tan30^\circ$ = $\sqrt{3}$, \\
$\therefore$ AT = TN = 2HN = 2$\sqrt{3}$, \\
In the right triangle $\triangle$BHT, BT = $\sqrt{TH^2+BH^2}=\sqrt{(\sqrt{3})^2+9^2}=2\sqrt{21}$, \\
$\because$ BN $\geq$ BT−TN, \\
$\therefore$ BN $\geq 2\sqrt{21}-2\sqrt{3}$, \\
$\therefore$ The minimum value of BN is $\boxed{2\sqrt{21}-2\sqrt{3}}$.


Question: As shown in the diagram, in rhombus $ABCD$, $\tan \angle DAB = \frac{4}{3}$, $AB=3$, and point $P$ is a moving point on side $AB$. Extend $BA$ to point $Q$ such that $AQ=2AP$, and $CQ$, $DP$ intersect at point $T$. When point $P$ moves from point $A$ to the right towards point $B$, what is the length of the path traced by point $T$?

Solution:
\textbf{Solution:} Extend $AT$ until it intersects $CD$ at $N$, as shown in the diagram: \\
Since $CD\parallel BQ$,\\
it follows that $ \frac{AP}{DN} = \frac{AT}{NT} = \frac{AQ}{CN}$,\\
therefore, $ \frac{AP}{AQ} = \frac{1}{2} = \frac{DN}{CN}$,\\
thus, point $N$ is the trisection point on $CD$ closer to $D$,\\
hence, point $T$ moves along segment $AN$,\\
when $P$ moves from point $A$ to the right until it coincides with $B$, as depicted in the diagram: \\
the path of point $T$ is along $AT$, draw $DH\perp AB$ at $H$, and draw $TM\perp AB$ at $M$,\\
in $\triangle ADH$, $\tan \angle DAB = \frac{4}{3}$,\\
let $DH = 4k$, then $AH = 3k$, $AD = 5k$,\\
since $AD = AB = 3$,\\
it follows that $5k = 3$,\\
thus, $k = \frac{3}{5}$,\\
therefore, $DH = \frac{12}{5}$, $AH = \frac{9}{5}$,\\
thus, $BH = AB - AH = \frac{6}{5}$,\\
since $ \frac{DT}{PT} = \frac{CD}{PQ} = \frac{AP}{AP+AQ} = \frac{1}{3}$,\\
it follows that $ \frac{PT}{PD} = \frac{3}{4}$,\\
since $DH\perp AB$, $TM\perp AB$,\\
it follows that $TM\parallel DH$,\\
hence, $ \frac{PT}{PD} = \frac{TM}{DH} = \frac{BM}{BH}$, namely $ \frac{3}{4} = \frac{TM}{\frac{12}{5}} = \frac{BM}{\frac{6}{5}}$,\\
thus, $TM = \frac{9}{5}$, $BM = \frac{9}{10}$,\\
therefore, $AM = AB - BM = \frac{21}{10}$,\\
in $\triangle ATM$, $AT = \sqrt{AM^{2}+TM^{2}} = \boxed{\frac{3\sqrt{85}}{10}}$.


Question: As shown in the figure, in an equilateral triangle $ABC$, $D$ is the midpoint of $AC$, and $P$ is a moving point on side $AB$. Draw $PE\perp AB$, intersecting $BC$ at point $E$, and connect $DP$, $DE$. If $AB=8$ and $\triangle PDE$ is an isosceles triangle, what is the length of $BP$?

Solution:
\textbf{Solution:} As shown in the diagram, construct DM $\perp$ AB at point M, and DN $\perp$ BC at point N.\\
$\therefore \angle AMD=\angle DNC=90^\circ$,\\
then $\triangle AMD$ and $\triangle DNC$ are both right-angled triangles.\\
$\because \triangle ABC$ is an equilateral triangle, and AB=8,\\
$\therefore \angle A=\angle B=\angle C=60^\circ$.\\
$\because D$ is the midpoint of AC,\\
$\therefore AD=CD=\frac{1}{2}AC=4$.\\
In $\triangle AMD$,\\
AM=$AD\cdot\cos\angle A=4\cdot\cos60^\circ=2$,\\
DM=$AD\cdot\sin\angle A=4\cdot\sin60^\circ=2\sqrt{3}$,\\
Similarly, we can derive CN=2, DN=$2\sqrt{3}$.\\
$\therefore BM=AB-AM=6$,\\
BN=BC-CN=6.\\
Let BP=a,\\
$\because$ EP $\perp$ AB\\
$\therefore \angle EPB=90^\circ$.\\
In $\triangle EPB$,\\
$PE=BP\cdot\tan\angle B=a\cdot\tan60^\circ=\sqrt{3}a$,\\
BE=$\frac{BP}{\cos\angle B}=\frac{a}{\cos60^\circ}=2a$.\\
$\therefore MP=BM-BP=6-a$,\\
EN=BN-BE=6-2a.\\
When $\triangle PDE$ is an isosceles triangle,\\
(1) When PE=DE,\\
in $\triangle DEN$, by the Pythagorean theorem,\\
$EN^2+DN^2=DE^2$.\\
Which gives $(6-2a)^2+(2\sqrt{3})^2=(\sqrt{3}a)^2$.\\
Solving this, we get $a_1=12-4\sqrt{6}$, $a_2=12+4\sqrt{6}$ (discard as it does not meet the requirements).\\
Hence, BP$=\boxed{12-4\sqrt{6}}$.\\
(2) When PE=PD,\\
in $\triangle DMP$, by the Pythagorean theorem,\\
$MP^2+DM^2=PD^2$.\\
Which gives $(6-a)^2+(2\sqrt{3})^2=(\sqrt{3}a)^2$.\\
Solving this, we get $a_1=\sqrt{33}-3$, $a_2=-\sqrt{33}-3$ (discard as it does not meet the requirements).\\
Hence, BP$=\boxed{\sqrt{33}-3}$.\\
(3) When PD=DE, for this to hold, point E must be to the right of point N, with EN=a-6. Since DM=DN, DP=DE, both being right-angled triangles, hence MP=EN, which gives 6-a=2a-6, solving this gets a=4.\\
In conclusion, the length of BP could be $12-4\sqrt{6}$ or $\sqrt{33}-3$ or $\boxed{4}$.


Question: As shown in the figure, the regular hexagon $ABCDEF$ inscribed in circle $\odot O$ has a side length of $2\sqrt{3}$ cm. What is the area of this regular hexagon in cm$^{2}$?

Solution:
\textbf{Solution:} Construct \(OH \perp AB\) at point H, and connect OA and OB.\\
Since the side length of the inscribed regular hexagon ABCDEF in circle \(O\) is \(2\sqrt{3}\) cm,\\
Therefore, OA = OB = AB = \(2\sqrt{3}\) cm,\\
Therefore, OH = OA\(\cdot\cos 30^\circ\) = \(2\sqrt{3} \cdot \frac{\sqrt{3}}{2}\) = 3 (cm),\\
Therefore, the area of the regular hexagon ABCDEF = 6\(\cdot\) Area of \(\triangle OAB\) = 6\(\cdot \frac{1}{2} \cdot 2\sqrt{3}\cdot 3\) = \(\boxed{18\sqrt{3}}\) (cm)\(^{2}\).


Question: As shown in the figure, points A and B are the intersections of the inverse proportion function $y_{1} = \frac{10}{x}$ and the linear function $y=2x$. Point C is on the inverse proportion function $y_{2} = \frac{k}{x}$. Connect OC, draw $AD \perp x$-axis through point A to intersect OC at point D, and connect BD. If $AD=BD$ and $OC=3OD$, then what is the value of $k$?

Solution:
\textbf{Solution:} Solve the system of equations $\begin{cases}y=2x\\ y=\frac{10}{x}\end{cases}$,\\
we get $\begin{cases}x_{1}=-\sqrt{5}\\ y_{1}=-2\sqrt{5}\end{cases}$, $\begin{cases}x_{2}=\sqrt{5}\\ y_{2}=2\sqrt{5}\end{cases}$,\\
$\therefore$ the coordinates of point A are ($-\sqrt{5}$, $-2\sqrt{5}$), and the coordinates of point B are ($\sqrt{5}$, $2\sqrt{5}$),\\
$\because$ A and B are symmetric about the origin,\\
$\therefore$ O is the midpoint of AB,\\
also, $\because$ AD = BD,\\
$\therefore$ point D lies on the perpendicular bisector of segment AB,\\
$\therefore$ CO$\perp$AB,\\
also, $\because$ AH$\perp$x-axis,\\
$\therefore$ $\angle$AOH + $\angle$OAH = $\angle$AOH + $\angle$COH = $90^\circ$,\\
$\therefore$ $\angle$OAH = $\angle$COH,\\
draw CE$\perp$x-axis at point E,\\
$\because$ OC = 3OD, and the x-coordinate of point D is $-\sqrt{5}$,\\
$\therefore$ the x-coordinate of point C is $-3\sqrt{5}$,\\
$\because  \tan \angle$OAH = $\tan \angle$COH = $\frac{OH}{AH}$ = $\frac{CE}{OE}$ = $\frac{1}{2}$,\\
$\therefore$ CE = $\frac{1}{2}$OE = $\frac{3\sqrt{5}}{2}$,\\
$\therefore$ the coordinates of point C are ($-3\sqrt{5}$, $\frac{3\sqrt{5}}{2}$),\\
$\therefore$ k = $-3\sqrt{5} \times \frac{3\sqrt{5}}{2}$ = $\boxed{-\frac{45}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, the perpendicular bisectors $DE$ and $FG$ of $AB$ and $AC$, respectively, intersect $BC$ at points $E$ and $G$. If point $G$ is to the left of point $E$ and $\angle GAE=25^\circ$, then what is the measure of $\angle BAC$ in degrees?

Solution:
\textbf{Solution:}
Given that $DE$ is the perpendicular bisector of $AB$, and $FG$ is the perpendicular bisector of $AC$,\\
thus $EA=EB$, $GA=GC$,\\
therefore, $\angle B=\angle BAE$, $\angle C=\angle GAC$,\\
since $\angle B+\angle BAE+\angle GAC+\angle C-\angle GAE=180^\circ$, $\angle GAE=25^\circ$,\\
it follows that $2\angle B+2\angle C-25^\circ=180^\circ$,\\
therefore, $\angle B+\angle C=102.5^\circ$,\\
hence, $\angle BAC=180^\circ-(\angle B+\angle C)=\boxed{77.5^\circ}$.


Question: As illustrated, in the isosceles triangle $ABC$ where $AB=AC$, $AD$ and $BE$ are the altitudes of the isosceles triangle $ABC$. When $DE$ is connected and given $AE=4$ and $CE=1$, what is the length of $DE$?

Solution:
\textbf{Solution:}
Given that $AE=4$ and $CE=1$, hence $AC=AB=AE+CE=4+1=5$. Since $AD$ and $BE$ are the altitudes of isosceles triangle $ABC$, it follows that $\angle AEB=\angle BEC=90^\circ$ and $BD=CD$. In right triangle $ABE$, we have $BE=\sqrt{AB^{2}-AE^{2}}=\sqrt{5^{2}-4^{2}}=3$; In right triangle $BEC$, $BC=\sqrt{BE^{2}+CE^{2}}=\sqrt{3^{2}+1^{2}}=\sqrt{10}$. In right triangle $BEC$, with $DE$ being the median, therefore $DE=\frac{1}{2}BC=\frac{\sqrt{10}}{2}$. Hence, the answer is: $DE=\boxed{\frac{\sqrt{10}}{2}}$.


Question: As shown in the figure, $\angle ACD$ is an exterior angle of $\triangle ABC$, $CE$ bisects $\angle ACD$. If $\angle A=60^\circ$ and $\angle B=40^\circ$, then what is the measure of $\angle DCE$ in degrees?

Solution:
\textbf{Solution}: Since $CE$ bisects $\angle ACD$, \\
it follows that $\angle ACD=2\angle ECD$, \\
since $\angle ACD$ is an exterior angle of $\triangle ABC$, \\
thus $\angle ACD=2\angle ECD=\angle A+\angle B$, \\
therefore, $2\angle DCE=60^\circ+40^\circ=100^\circ$, \\
hence, $\angle DCE=\boxed{50^\circ}$.


Question: As shown in the figure, in the right-angled triangle $\triangle ABC$ where $\angle ACB = 90^\circ$, squares are constructed externally on the three sides of the right-angled $\triangle ABC$, with their areas being $S_{1}$, $S_{2}$, and $S_{3}$, respectively, and $S_{1} = 4$, $S_{2} = 16$. What is the value of $S_{3}$?

Solution:
\textbf{Solution:} Since in the right-angled triangle $ABC$,\\
thus $AB^2=AC^2+BC^2$;\\
Since three squares are constructed externally on the three sides of the right-angled triangle $ABC$,\\
thus $S_{3}=S_{1}+S_{2}$,\\
therefore $S_{3}=4+16=\boxed{20}$.


Question: As shown in the figure, a set of triangle rulers is placed as shown in figure a, where $\angle ACB=\angle DEC=90^\circ$, $\angle A=45^\circ$, $\angle D=30^\circ$, the hypotenuse $AB=6\text{cm}$, $DC=7\text{cm}$. The triangle ruler DCE is rotated clockwise by $15^\circ$ around point C to get $\triangle D_{1}CE_{1}$ (as shown in figure b). At this time, $AB$ intersects $CD_{1}$ at point $O$, and intersects $D_{1}E_{1}$ at point $F$. What is the degree of $\angle OFE_{1}$, and what is the length of the line segment $AD_{1}$ in cm?

Solution:
\textbf{Solution:}

Since $\angle ACB = \angle DEC = 90^\circ$, $\angle A = 45^\circ$, $\angle D = 30^\circ$,\\
thus $\angle DCE = 60^\circ$, $\angle B = 45^\circ$\\
Since rotating triangle DCE around point C clockwise by $15^\circ$ results in $\triangle D_1CE_1$,\\
thus $\angle D_1CE_1 = 60^\circ$, $\angle E_1 = \angle DEC = 90^\circ$, $\angle BCE_1 = 15^\circ$\\
therefore $\angle OCB = 45^\circ$, and $\angle B = 45^\circ$,\\
thus $\angle COB = 90^\circ$,\\
since $\angle COB + \angle D_1CE_1 + \angle E_1 + \angle OFE_1 = 360^\circ$\\
therefore $\angle OFE_1 = 120^\circ$;\\
since $\triangle ACB$ is an isosceles right triangle and $\angle COB = 90^\circ$,\\
thus $AO = CO = BO = 3\text{cm}$,\\
therefore $D_1O = 4\text{cm}$,\\
thus $AD_1 = \sqrt{AO^2 + D_1O^2} = \sqrt{3^2 + 4^2} = \boxed{5}$


Question: As shown in the figure, in quadrilateral $ABCD$, points $M$ and $N$ are located on $AB$ and $BC$, respectively. Triangle $\triangle BMN$ is folded along $MN$ to obtain $\triangle FMN$. If $MF\parallel AD$ and $FN\parallel DC$, what is the measure of $\angle B$?

Solution:
\[
\text{Solution: Since } MF \parallel AD, \text{ and } FN \parallel DC,
\]
\[
\therefore \angle FMB = \angle A = 110^\circ, \angle FNB = \angle C = 90^\circ,
\]
\[
\text{Since } \triangle FMN \text{ and } \triangle BMN \text{ are symmetric about } MN,
\]
\[
\therefore \angle BMN = \angle FMN = 55^\circ, \angle BNM = \angle FNM = 45^\circ,
\]
\[
\therefore \angle B = 180^\circ - \angle BMN - \angle BNM = \boxed{80^\circ}.
\]


Question: As shown in the figure, $\triangle ABC \cong \triangle DEF$, $BC=7$, and $EC=4$, then what is the length of $CF$?

Solution:
\textbf{Solution:} Since $\triangle ABC \cong \triangle DEF$,\\
it follows that $BC = EF = 7$,\\
hence $CF = EF - EC = 7 - 4 = \boxed{3}$.


Question: As shown in the figure, in $\triangle ABC$, we have $AB=AC$, $AD=BC=8$. Points $E$ and $F$ are on the median $AD$. What is the area of the shaded region?

Solution:
\textbf{Solution:} Given $\because$ AB = AC, AD = BC = 8, AD$\perp$BC at point D,\\
$\therefore$ BD=CD=$\frac{1}{2}$BC=4,\\
$\therefore$ $\triangle ABC$ is symmetrical about AD,\\
$\therefore$ S$_{\triangle BEF}$=S$_{\triangle CEF}$,\\
$\therefore$ the area of the shaded part in the figure = $\frac{1}{2}$S$_{\triangle BEF}$=$\frac{1}{2} \times \frac{1}{2} \times 8 \times 8 = \boxed{16}$


Question: As shown in the figure, it is known that $\angle MON=40^\circ$, P is a fixed point inside $\angle MON$, point A is on OM, and point B is on ON. When the perimeter of $\triangle PAB$ is at its minimum, what is the degree measure of $\angle APB$?

Solution:
\textbf{Solution:} Construct the symmetric points $P_{1}$ and $P_{2}$ of $P$ with respect to $OM$ and $ON$, respectively. Draw the line segment $P_{1}P_{2}$, intersecting $OM$ and $ON$ at points $A$ and $B$, respectively.\\
$\therefore$ $PA=P_{1}A$, $PB=P_{2}B$, at this moment the perimeter of $\triangle PAB$ is minimal,\\
$\therefore$ $\angle P_{1}PA=\angle P_{1}$, $\angle P_{2}PB=\angle P_{2}$,\\
According to the problem, $\angle P_{1}PP_{2}=180^\circ-\angle MON=180^\circ-40^\circ=140^\circ$,\\
$\therefore$ $\angle P_{1}PA+\angle P_{2}PB=\angle P_{1}+\angle P_{2}=180^\circ-\angle P_{1}PP_{2}=40^\circ$,\\
$\therefore$ $\angle APB=140^\circ-40^\circ=\boxed{100^\circ}$.


Question: As shown in the diagram, triangle $\triangle ABC$ is folded along line $l$, positioning point $B$ on point $F$. If $\angle 1 - \angle 2 = 70^\circ$, what is the degree measure of $\angle B$?

Solution:
\textbf{Solution:} As shown in the figure, \\
due to the property of folding, we have: $\angle F=\angle B$, \\
according to the external angle property of triangles, we get: $\angle 1=\angle 3+\angle B$, $\angle 2=\angle 3-\angle F=\angle 3-\angle B$, \\
$\therefore \angle 1-\angle 2=\angle 3+\angle B-(\angle 3-\angle B)=2\angle B=70^\circ$, \\
thus $\angle B=\boxed{35^\circ}$.


Question: As shown in the figure, $\triangle ABC \cong \triangle DEF$, where $BC = 7$ and $EC = 4$. What is the length of $CF$?

Solution:
\textbf{Solution:} Since $\triangle ABC \cong \triangle DEF$,\\
it follows that $BC=EF=7$,\\
thus, $CF=EF-EC=7-4=\boxed{3}$.


Question: As shown in the figure, it is known that $\angle BDC=142^\circ$, $\angle B=34^\circ$, and $\angle C=28^\circ$. What is the degree measure of $\angle A$?

Solution:
\textbf{Solution:} Construct ray AE passing through points A and D, \\
$\because$ $\angle CDE = \angle CAD + \angle C$, $\angle BDE = \angle BAD + \angle B$, \\
$\therefore$ $\angle BDC = \angle CDE + \angle BDE = \angle CAD + \angle C + \angle BAD + \angle B = 142^\circ$, \\
$\therefore$ $\angle CAB + 34^\circ + 28^\circ = 142^\circ$, \\
$\therefore$ $\angle CAB = \boxed{80^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, points D, E, and F are the midpoints of $BC$, $AD$, and $CE$, respectively, and $S_{\triangle AEF} = 4\, \text{cm}^2$, then what is the area of $\triangle ABC$ in $\text{cm}^2$?

Solution:
\textbf{Solution:} Since point F is the midpoint of CE and the area of $\triangle AEF = 4\text{cm}^2$,\\
it follows that AF is the median of $\triangle AEC$, \\
thus the area of $\triangle AEC = 2$ times the area of $\triangle AEF = 2 \times 4 = 8$;\\
Since point E is the midpoint of AD,\\
it follows that CD is the median of $\triangle ADC$,\\
thus the area of $\triangle ADC = 2$ times the area of $\triangle AEC = 2 \times 8 = 16$;\\
Since point D is the midpoint of BC,\\
it follows that AD is the median of $\triangle ABC$,\\
thus the area of $\triangle ABC = 2$ times the area of $\triangle ADC = 2 \times 16 = \boxed{32}\text{cm}^2$;


Question: As shown in the figure, the isosceles $\triangle ABC$ has a base $BC=20$ and an area of $120$. Point $F$ is on side $BC$, and $BF=3FC$. $EG$ is the perpendicular bisector of the leg $AC$. If point $D$ moves on $EG$, what is the minimum perimeter of $\triangle CDF$?

Solution:
\textbf{Solution:} As shown in the figure,\\
$\because$ EG is the perpendicular bisector of the line segment AC,\\
$\therefore$ DA=DC,\\
$\therefore$ DF+DC=AD+DF,\\
$\therefore$ when A, D, F are collinear, DF+DC is minimized, and the minimum value is the length of the line segment AF.\\
$\because$ $ \frac{1}{2}\cdot BC\cdot AH=120$,\\
$\therefore$ AH=12,\\
$\because$ AB=AC, AH$\perp$BC,\\
$\therefore$ BH=CH=10,\\
$\because$ BF=3FC,\\
$\therefore$ CF=FH=5,\\
$\therefore$ $AF=\sqrt{AH^{2}+HF^{2}}=\sqrt{12^{2}+5^{2}}=13$,\\
$\therefore$ the minimum value of DF+DC is 13,\\
$\therefore$ the shortest perimeter of $\triangle CDF$=13+5=\boxed{18}.


Question: As shown in the diagram, in rectangle $ABCD$, it is known that $AB=8\,cm$, $BC=10\,cm$. Fold $AD$ along the straight line $AE$, making point D land on point F which is on side $BC$. What is the length of $CF$ in $cm$?

Solution:
\textbf{Solution:} In the rectangle $ABCD$, \\
$\because$ $AB=8\text{cm}$, $BC=10\text{cm}$, let $CF=x\text{cm}$, \\
$\therefore$ $AD=BC=10\text{cm}$, $BF=(10-x)\text{cm}$, $\angle B=90^\circ$, \\
$\because$ folding $AD$ along the line $AE$, such that point $D$ falls on point $F$ on side $BC$, \\
$\therefore$ $\triangle ADE \cong \triangle AFE$, \\
$\therefore$ $AD=AF=10\text{cm}$, \\
In $\triangle ABF$, by the Pythagorean theorem, $AB^2+BF^2=AF^2$, \\
$\therefore$ $8^2+(10-x)^2=10^2$, solving for $x$, we find $x=\boxed{4}$, \\
$\therefore$ $CF=4\text{cm}$.


Question: On a line, seven squares are placed in sequence (as shown). It is known that the areas of the three diagonally placed squares are 1, 2, and 3, respectively, and the areas of the four upright squares are $S_1$, $S_2$, $S_3$, and $S_4$. What is the value of $S_3 + S_4 - S_1 - S_2$?

Solution:
\textbf{Solution:} Observe that, \\
$\because AB=BE$, and $\angle ACB=\angle BDE=90^\circ$, \\
$\therefore \angle ABC+\angle BAC=90^\circ$, and $\angle ABC+\angle EBD=90^\circ$, \\
$\therefore \angle BAC=\angle EBD$, \\
$\because$ in triangles $\triangle ABC$ and $\triangle BDE$, \\
$\left\{\begin{array}{l}
\angle ACB=\angle BDE=90^\circ \\
\angle BAC=\angle EBD \\
AB=BE
\end{array}\right.$, \\
$\therefore \triangle ABC\cong \triangle BDE(\text{AAS})$, \\
$\therefore BC=ED$, \\
$\because AB^2=AC^2+BC^2$, \\
$\therefore AB^2=AC^2+ED^2={S}_{1}+{S}_{2}$, \\
i.e., ${S}_{1}+{S}_{2}=1$, \\
Similarly, ${S}_{3}+{S}_{4}=3$. \\
Hence, ${S}_{3}+{S}_{4}-{S}_{1}-{S}_{2}={S}_{3}+{S}_{4}-({S}_{1}+{S}_{2})=3-1=\boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACD=110^\circ$, $\angle B=30^\circ$, then what is the measure of $\angle A$ in degrees?

Solution:
\textbf{Solution:} Since $\angle ACD$ is an exterior angle of $\triangle ABC$,\\
therefore $\angle ACD = \angle A + \angle B$,\\
thus $\angle A = 110^\circ - 30^\circ = \boxed{80^\circ}$.


Question: As shown in the figure, a ship sails in the direction indicated by the arrow. There is a lighthouse at point C. When the ship sails from point A to point B, $\angle ACB = 90^\circ$.

Solution:
\textbf{Solution:} Since $\angle CBD = \angle A + \angle C$,\\
thus $\angle C = \angle CBD - \angle A$\\
$= 70^\circ - 30^\circ$\\
$= \boxed{40^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $\angle BAC=90^\circ$, point $D$ is on segment $BC$, $\angle EDB=\frac{1}{2}\angle ACB$, $BE\perp DE$, and $DE$ intersects $AB$ at point $F$. If $BE=4$, what is the length of $DF$?

Solution:
\textbf{Solution:} Construct $DH\parallel AC$ through $D$ and meet $AB$ at $H$. Extend $BE$ to intersect with the extension of $DH$ at point $G$, as shown in the figure:\\
In $\triangle ABC$, $AB=AC$, $\angle BAC=90^\circ$,\\
$\therefore \angle ABC=\angle C=45^\circ$,\\
$\because DH\parallel AC$,\\
$\therefore \angle BDH=\angle C=45^\circ$, $\angle BHD=\angle A=90^\circ$,\\
$\therefore \triangle HBD$ is an isosceles right triangle,\\
$\therefore HB=HD$,\\
$\because \angle EDB=\frac{1}{2}\angle ACB=22.5^\circ$,\\
$\therefore \angle GDE=\angle EDB=22.5^\circ$,\\
$\therefore DE$ bisects $\angle BDG$,\\
$\because DE\perp BG$,\\
$\therefore \angle GED=\angle BED=90^\circ$,\\
In $\triangle BDE$ and $\triangle GDE$,\\
$\left\{\begin{array}{l}
\angle GED=\angle BED=90^\circ \\
ED=ED \\
\angle GDE=\angle BDE=22.5^\circ
\end{array}\right.$,\\
$\therefore \triangle BDE\cong \triangle GDE\left(AAS\right)$,\\
$\therefore BE=GE$, that is $BG=2BE$,\\
In $\triangle BGH$ and $\triangle DFH$,\\
$\left\{\begin{array}{l}
\angle GBH=\angle FDH=22.5^\circ \\
BH=DH \\
\angle GHB=\angle FHD
\end{array}\right.$,\\
$\therefore \triangle BGH\cong \triangle DFH(ASA)$,\\
$\therefore BG=DF$,\\
$\therefore DF=2BE=\boxed{8}$.


Question: As shown in the figure, $\triangle ABD \cong \triangle EBC$, $AB=4\text{cm}$, $BC=7\text{cm}$, then what is the length of $DE$ in cm?

Solution:
\textbf{Solution:}

Since $\triangle ABD \cong \triangle EBC$, and given $AB=4\, \text{cm}$, $BC=7\, \text{cm}$,

it follows that $BE=AB=4\, \text{cm}$, $BD=BC=7\, \text{cm}$,

thus $DE=BD-BE=\boxed{3}\, \text{cm}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=BC$, the coordinates of point $A$ are $(-7,3)$, and the coordinates of point $C$ are $(-2,0)$. What are the coordinates of point $B$?

Solution:
\textbf{Solution:} Construct $AD \perp x$-axis at point D, and $BE \perp x$-axis at point E, as shown in the figure,\\
then $\angle ADC=\angle CEB=90^\circ$,\\
$\therefore \angle ACD+\angle CAD=90^\circ$,\\
$\therefore \angle ACB=90^\circ$,\\
$\therefore \angle ACD+\angle BCE=90^\circ$,\\
$\therefore \angle CAD=\angle BCE$,\\
In $\triangle ACD$ and $\triangle CBE$,\\
$\left\{\begin{array}{l} \angle ADC=\angle CEB \\ \angle CAD=\angle BCE \\ AC=CB \end{array}\right.$,\\
$\therefore \triangle ACD \cong \triangle CBE$ (AAS),\\
$\therefore AD=CE$, $DC=EB$,\\
$\because$ the coordinates of point A are $(-7, 3)$, and the coordinates of point C are $(-2, 0)$,\\
$\therefore OD=7$, $AD=3$, $OC=2$,\\
$\therefore CE=3$, $BE=OD-OC=7-2=5$,\\
$\therefore OE=CE-OC=3-2=1$,\\
$\therefore$ the coordinates of point B are $\boxed{(1, 5)}$.


Question: In the quadrilateral $ABDC$, the diagonal $AD$ bisects $\angle BAC$, $\angle ACD=136^\circ$, $\angle BCD=44^\circ$, what is the measure of $\angle ADB$?

Solution:
\textbf{Solution:} Draw line DE$\perp$AB through point D intersecting the extension of AB at E, and draw DF$\perp$AC through point D intersecting the extension of AC at F, and draw DG$\perp$BC through point D at point G, as shown in the figure: \\
$\because$ $AD$ bisects $\angle BAC$, $DE\perp AB$, $DF\perp AC$\\
$\therefore$ $\angle BAD=\frac{1}{2}\angle BAC$, $DE=DF$\\
$\because$ $\angle ACD=136^\circ$\\
$\therefore$ $\angle DCF=180^\circ-\angle ACD=44^\circ$\\
$\because$ $\angle BCD=44^\circ$, $\angle ACB=\angle ACD-\angle BCD=92^\circ$\\
$\therefore$ $CD$ bisects $\angle BCF$\\
$\because$ $DF\perp AC$, $DG\perp BC$\\
$\therefore$ $DF=DG$\\
$\therefore$ $DE=DG$\\
$\because$ $DE\perp AB$, $DG\perp BC$\\
$\therefore$ $BD$ bisects $\angle CBE$\\
$\therefore$ $\angle DBE=\frac{1}{2}\angle CBE$\\
$\therefore$ $\angle ADB=\angle DBE-\angle BAD$\\
$=\frac{1}{2}\angle CBE-\frac{1}{2}\angle BAC$\\
$=\frac{1}{2}\left(\angle CBE-\angle BAC\right)$\\
$=\frac{1}{2}\angle BCA$\\
$=\boxed{46^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle BAC=30^\circ$, and $BC=1$. Equilateral triangles $ABD$ and $ACE$ are constructed on sides $AB$ and $AC$, respectively. Line $DE$ is drawn and intersects $AB$ at point $F$. What is the length of $DF$?

Solution:
\textbf{Solution:} As shown in the diagram, \\
draw DG$\perp$AB at G, and EH$\perp$DA intersecting the extension of DA at H, \\
since $\angle$EAC＝$60^\circ$ and $\angle$BAC＝$30^\circ$, \\
thus $\angle$EAG＝$\angle$AGD＝$90^\circ$, \\
since BC＝1, \\
thus in $\triangle ABC$, AC＝$\sqrt{3}$ and AB＝2, \\
also since $\triangle ABD$ and $\triangle ACE$ are equilateral triangles, \\
thus AE＝$\sqrt{3}$ and DG＝$\sqrt{3}$, \\
thus DG＝AE, \\
also since $\angle$DFG＝$\angle$EFA, \\
thus $\triangle AEF \cong \triangle GDF$ (AAS), \\
thus DF＝$\frac{1}{2}$DE, \\
also since in $\triangle AEH$, $\angle$EAH＝$30^\circ$, \\
thus HE＝$\frac{1}{2}$AE＝$\frac{1}{2}\sqrt{3}$ and AH＝$\frac{3}{2}$, \\
thus DH＝DA+AH＝2+$\frac{3}{2}$＝$\frac{7}{2}$, \\
thus in $\triangle DEH$, DE＝$\sqrt{HE^2+DH^2}$＝$\sqrt{\left(\frac{1}{2}\sqrt{3}\right)^2+\left(\frac{7}{2}\right)^2}$＝$\sqrt{13}$, \\
thus the length of DF is $\boxed{\frac{\sqrt{13}}{2}}$.


Question: As shown in the figure, point $M$ is on the side $BC$ of the equilateral triangle $\triangle ABC$, with $BM = 8$. The ray $CD$ is perpendicular to $BC$ at point $C$. Point $P$ is a moving point on ray $CD$, and point $N$ is a moving point on line segment $AB$. When the value of $MP+NP$ is minimized, $BN = 9$. What is the length of $AC$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is an equilateral triangle, \\
thus $AC=BC$ and $\angle B=60^\circ$, \\
Let point $M$ be the symmetric point with respect to line $CD$ and construct $GN\perp AB$ at $N$ intersecting $CD$ at $P$, \\
then, at this time, the value of $MP+PN$ is minimized, \\
Since $\angle B=60^\circ$ and $\angle BNG=90^\circ$, \\
thus $\angle G=30^\circ$, \\
Since $BN=9$, \\
thus $BG=2BN=18$, \\
thus $MG=BG-BM=18-8=10$, \\
thus $CM=CG=5$, \\
thus $AC=BC=\boxed{13}$.


Question: As shown in the figure, within $\triangle ABC$, $\angle ACB=90^\circ$, $AC=BC$, points $D$ and $E$ are on $AB$ and $BC$ respectively, connecting $CD$ and $DE$. If $BC=BD$, $AC=2$, and $\angle CDE=45^\circ$, what is the length of $BE$?

Solution:
\textbf{Solution:} Given $\because \angle ACB=90^\circ$ and AC=BC,\\
$\therefore \angle A=\angle B=45^\circ$,\\
$\because \angle BDC=\angle CDE+\angle BDE=\angle A+\angle ACD$ and $\angle CDE=45^\circ$,\\
$\therefore \angle BDE=\angle ACD$,\\
$\because BC=BD$,\\
$\therefore BD=AC$,\\
In $\triangle BDE$ and $\triangle ACD$,\\
$\left\{\begin{array}{l}
\angle B=\angle A \\
BD=AC \\
\angle BDE=\angle ACD
\end{array}\right.$,\\
$\therefore \triangle BDE \cong \triangle ACD$ (ASA),\\
$\therefore BE=AD$,\\
$\because BC=AC=2$ and $\angle ACB=90^\circ$,\\
$\therefore AB =\sqrt{AC^2+BC^2}=\sqrt{2^2+2^2}=2\sqrt{2}$,\\
$\because BD=BC=2$,\\
$\therefore AD=AB-BD=2\sqrt{2}-2$,\\
$\therefore BE=\boxed{2\sqrt{2}-2}$.


Question: As shown in the figure, in $\triangle ABC$, $CE$ bisects $\angle ACB$, and $CF$ bisects the supplementary angle of $\angle ACD$. If $EF \parallel BC$ and intersects $AC$ at $M$, with $CE=8\, cm$ and $CF=6\, cm$, what is the length of $EM$ in $cm$?

Solution:
\textbf{Solution:} Since CE bisects $\angle ACB$, and CF bisects the supplementary angle $\angle ACD$, \\
it follows that $\angle MCE = \angle BCE = \frac{1}{2}\angle ACB$, $\angle FCM = \angle FCD = \frac{1}{2}\angle ACD$, \\
hence $\angle ECF = \frac{1}{2}\angle ACB + \frac{1}{2}\angle ACD = \frac{1}{2}(\angle ACD + \angle ACD) = 90^\circ$, \\
according to the Pythagorean theorem, it can be found that $EF = \sqrt{EC^{2} + CF^{2}} = 10$cm, \\
since $EF\parallel BC$, \\
it is inferred that $\angle MEC = \angle BCE$, $\angle MFC = \angle FCD$, \\
which implies $MEC = \angle MCE$, $\angle MFC = \angle MCF$, \\
thus $EM = CM = FM$, \\
hence $EM = \frac{1}{2}EF = \boxed{5}$cm,


Question: As shown in the diagram, in $\triangle ABC$, $\angle ACB=90^\circ$. Squares are constructed externally on the three sides of $\triangle ABC$, with their areas being $S_1$, $S_2$, and $S_3$, respectively, and it is given that $S_1=9$ and $S_3=16$. What is the value of $S_2$?

Solution:
\textbf{Solution:} Given that ${S}_{1}=9$ and ${S}_{3}=16$,\\
it follows that $BC^{2}=9$ and $AB^{2}=16$,\\
since $\angle ACB=90^\circ$,\\
in $\triangle ABC$,\\
it implies that $BC^{2}+AC^{2}=AB^{2}$,\\
therefore $9+AC^{2}=16$,\\
hence $AC^{2}=7$,\\
thus, ${S}_{2}=\boxed{7}$.


Question: As shown in the figure, in $\triangle ABC$, $AD \perp BC$, $BE \perp AC$, with the feet of the perpendiculars being points D and E, respectively, $AD$ and $BE$ intersect at point F, $BF=AC$, $\angle ABE=20^\circ$, then what is the degree measure of $\angle CAD$?

Solution:
\textbf{Solution:} Given that $AD\perp BC$, $BE\perp AC$,\\
thus $\angle BDF = \angle ADC = 90^\circ$, $\angle BEC = \angle ADC = 90^\circ$,\\
therefore $\angle DAC + \angle C = 90^\circ$, $\angle DBF + \angle C = 90^\circ$,\\
therefore $\angle DBF = \angle DAC$,\\
in $\triangle DBF$ and $\triangle DAC$,\\
$\left\{\begin{array}{l} \angle BDF = \angle ADC \hfill \\ \angle DBF = \angle DAC \hfill \\ BF = AC \hfill \end{array}\right.$, \\
thus $\triangle DBF \cong \triangle DAC$ (AAS),\\
thus $AD = BD$,\\
given that $\angle ADB = 90^\circ$,\\
therefore $\angle ABD = \angle DAB = 45^\circ$,\\
given that $\angle ABE = 20^\circ$,\\
therefore $\angle CAD = \angle DBF = \angle ABD - \angle ABE = 45^\circ - 20^\circ = \boxed{25^\circ}$.


Question: As shown in the figure, two identically sized set squares, ACF and CFB, containing $45^\circ$ angles, are placed as depicted. Point D is on side AC, and point E is on side BC, with $\angle CFE=13^\circ$, and $\angle CFD=32^\circ$. What is the measure of $\angle DEC$?

Solution:
\textbf{Solution:} Construct $FH$ perpendicular to $FE$, intersecting $AC$ at point $H$,
\[
\because \angle AFC=\angle EFH=90^\circ
\]
Also,
\[
\because \angle AFC=\angle AFH+\angle CFH,\ \angle HFE=\angle CFE+\angle CFH
\]
\[
\therefore \angle AFH=\angle CFE=13^\circ
\]
\[
\because \angle A=\angle FCE=45^\circ,\ FA=CF
\]
\[
\therefore \triangle FAH\cong \triangle FCE\ (\text{ASA})
\]
\[
\therefore FH=FE
\]
\[
\because \angle DFE=\angle DFC+\angle EFC=32^\circ+13^\circ=45^\circ
\]
\[
\because \angle DFH=\angle HFE−\angle DFE=90^\circ−45^\circ=45^\circ
\]
\[
\therefore \angle DFE=\angle DFH
\]
Also,
\[
\because DF=DF
\]
\[
\therefore \triangle HDF\cong \triangle EDF\ (\text{SAS})
\]
\[
\therefore \angle DHF=\angle DEF
\]
\[
\because \angle DHF=\angle A+\angle HFA=45^\circ+13^\circ=58^\circ
\]
\[
\therefore \angle DEF=58^\circ
\]
\[
\because \angle CFE+\angle CEF+\angle FCE=180^\circ
\]
\[
\therefore \angle CEF=180^\circ−\angle CFE−\angle FCE=180^\circ−13^\circ−45^\circ=122^\circ
\]
\[
\therefore \angle DEC=\angle CEF−\angle DEF=122^\circ−58^\circ=\boxed{64^\circ}
\]


Question: As shown in the figure, $\triangle ADB \cong \triangle ECB$, and the corresponding point of A is E, the corresponding point of D is C. If $\angle CBD = 40^\circ$ and $BD \perp EC$, then what is the degree measure of $\angle D$?

Solution:
\textbf{Solution:} Let BD and EC intersect at point G,\\
since $\triangle ADB \cong \triangle ECB$,\\
it follows that $\angle C = \angle D$,\\
since BD$\perp$EC,\\
it follows that $\angle BGC = 90^\circ$,\\
therefore, $\angle C = \angle D = 90^\circ - \angle CBD = 90^\circ - 40^\circ = \boxed{50^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, where $AB=5$, $AC=7$, and $AB\perp AC$, $EF$ is the perpendicular bisector of $BC$, and point $P$ is a moving point on line $EF$. What is the minimum perimeter of $\triangle ABP$?

Solution:
\textbf{Solution:} As shown in the diagram, connect PC,

$\because$ AB=5,

$\therefore$ the perimeter of $\triangle ABP$ is AB+PA+PB=5+PA+PB,

To minimize the perimeter of $\triangle ABP$, it is required to minimize the value of PA+PB,

$\because$ EF is a perpendicular bisector of BC,

$\therefore$ PC=PB,

$\therefore$ PA+PB=PA+PC,

According to the principle that the shortest distance between two points is a straight line, when points A, P, and C are collinear,

that is, when point P is on the side AC, PA+PC reaches its minimum value, which is AC,

$\therefore$ the minimum value of PA+PB is AC=7,

$\therefore$ the minimum perimeter of $\triangle ABP$ is 5+7=\boxed{12},


Question: As shown in the figure, in $\triangle CEF$, $\angle E=80^\circ$, $\angle F=55^\circ$, $AB\parallel CF$, $AD\parallel CE$, connect BC and CD, then what is the degree measure of $\angle A$?

Solution:
\textbf{Solution}: Connect AC and extend it to intersect EF at point M,\\
since AB$\parallel$CF, and AD$\parallel$CE,\\
therefore, $\angle$3=$\angle$1, $\angle$2=$\angle$4,\\
therefore, $\angle$BAD=$\angle$3+$\angle$4=$\angle$1+$\angle$2=$\angle$FCE,\\
since $\angle$FCE=$180^\circ$−$\angle$E−$\angle$F=$180^\circ−80^\circ−55^\circ=45^\circ$,\\
therefore, $\angle$BAD=\boxed{45^\circ}.


Question: In $\triangle ABC$, the perpendicular bisector of $AB$ intersects $AB$ and $BC$ at points $D$ and $E$, respectively, and the perpendicular bisector of $AC$ intersects $AC$ and $BC$ at points $F$ and $G$, respectively. Given that $BG=9$, $CE=11$, and the perimeter of $\triangle AEG$ is 16, what is the length of $EG$?

Solution:
Solution: Since $DE$ bisects $AB$ perpendicularly, and $GF$ bisects $AC$ perpendicularly, \\
therefore $AE=BE$, and $AG=CG$, \\
since the perimeter of $\triangle AEG$ is 16, \\
therefore $AE+EG+AG=16$, \\
hence $BE+EG+CG=16$, \\
thus $9-EG+EG+11-EG=16$, \\
hence $EG=\boxed{4}$.


Question: As shown in the figure, AD is the median of $\triangle ABC$, E is a point on AD, and BE intersects AC at F. If $BE=AC$, $BF=9$, and $CF=6$, then what is the length of $AF$?

Solution:
\textbf{Solution:} Extend AD to G such that DG = AD, and connect BG,\\
$\because$ AD is the median of $\triangle ABC$,\\
$\therefore$ CD = BD,\\
In $\triangle ACD$ and $\triangle GBD$,\\
$ \left\{\begin{array}{l}CD=BD \\ \angle ADC= \angle BDG \\ AD=DG \end{array}\right.$\\
$\therefore$ $\triangle ACD \cong \triangle GBD$ (SAS),\\
$\therefore$ $\angle CAD = \angle G$, AC = BG,\\
$\because$ BE = AC,\\
$\therefore$ BE = BG,\\
$\therefore$ $\angle G = \angle BEG$,\\
$\because$ $\angle BEG = \angle AEF$,\\
$\therefore$ $\angle AEF = \angle EAF$.\\
$\therefore$ EF = AF,\\
$\therefore$ AF + CF = BF - AF,\\
i.e., AF + 6 = 9 - AF,\\
$\therefore$ AF = $\boxed{\frac{3}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, DE is the perpendicular bisector of AC, and AE $= 3\,cm$. If the perimeter of $\triangle ABD$ is $13\,cm$, what is the perimeter of $\triangle ABC$ in centimeters?

Solution:
\textbf{Solution:} Since DE is the perpendicular bisector of AC,\\
it follows that AD = CD, and AC = 2AE = 6cm,\\
Furthermore, since the perimeter of $\triangle ABD$ equals AB + BD + AD = 13cm,\\
it implies that AB + BD + CD = 13cm,\\
which means AB + BC = 13cm,\\
Therefore, the perimeter of $\triangle ABC$ equals AB + BC + AC = 13 + 6 = \boxed{19}cm.


Question: As shown in the figure, in $\triangle ABC$, $D$ and $E$ are points on $AC$ and $BC$, respectively. If $\triangle ADB \cong \triangle EDB \cong \triangle EDC$, what is the measure of $\angle C$?

Solution:
\textbf{Solution:} Given that $\triangle ADB \cong \triangle EDB \cong \triangle EDC$,\\
it follows that $\angle A = \angle BED = \angle CED$, and $\angle ABD = \angle EBD = \angle C$,\\
since $\angle BED + \angle CED = 180^\circ$,\\
it implies that $\angle A = \angle BED = \angle CED = 90^\circ$,\\
thus $\angle ABD + \angle EBD + \angle C = 90^\circ$,\\
hence $\angle C = \boxed{30^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $\angle BAC=90^\circ$. Line $l$ passes through point $A$, and from point $B$, $BE\perp l$ at point $E$, and from point $C$, $CF\perp l$ at point $F$. If $BE=2$, $CF=5$, what is the length of $EF$?

Solution:
\textbf{Solution:}\\
Since $BE\perp l$ and $CF\perp l$,\\
it follows that $\angle AEB=\angle CFA=90^\circ$,\\
thus, $\angle EAB+\angle ABE=90^\circ$,\\
since $\angle BAC=90^\circ$,\\
it follows that $\angle EAB+\angle FAC=90^\circ$,\\
thus, $\angle ABE=\angle FAC$,\\
since $AB=AC$,\\
it follows that $\triangle AEB\cong \triangle CFA$ (AAS),\\
therefore, $AF=BE=2$, $AE=CF=5$,\\
thus, $EF=AE+AF=5+2=\boxed{7}$.


Question: As shown in the figure, in $\triangle PAB$, $PA=PB$, $\angle A=\angle B$, points $M$, $N$, and $K$ are located on $PA$, $PB$, and $AB$, respectively, and $AM=BK$, $BN=AK$. If $\angle MKN=44^\circ$, what is the degree measure of $\angle P$?

Solution:
\textbf{Solution:} Given $\because$PA=PB,\\
$\therefore \angle A=\angle B$;\\
In $\triangle AMK$ and $\triangle BKN$
\[
\left\{
\begin{array}{l}
AM=BK\\
\angle A=\angle B\\
BN=AK
\end{array}
\right.
\]
$\therefore \triangle AMK \cong \triangle BKN$ (SAS),\\
$\therefore \angle AKM=\angle BNK$;\\
$\because \angle AKN=\angle B+\angle BNK=\angle AKM+\angle MKN$,\\
$\therefore \angle MKN=\angle B=\angle A=44^\circ$,\\
$\therefore \angle P=180^\circ-\angle A-\angle B=180^\circ-2\times 44^\circ=\boxed{92^\circ}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, points C and D are on $\odot O$, $CE \perp AB$ and $OF \perp AD$, with the feet of the perpendiculars being points E and F, respectively, and $\angle COF = 90^\circ$. If $BE=3$ and $CE=6$, what is the length of $AD$?

Solution: \textbf{Solution:} Since $CE \perp AB$ and $OF \perp AD$, it follows that $\angle OEC = \angle AFO = 90^\circ$. Hence, $\angle COE + \angle C = 90^\circ$. Let the radius of the circle be $r$, then $OE = r - 3$. Therefore, $CE^{2} + OE^{2} = CO^{2}$, i.e., $6^{2} + (r - 3)^{2} = r^{2}$. Solving this yields $r = 7.5$, therefore $OE = 7.5 - 3 = 4.5$. Since $\angle COF = 90^\circ$, it follows that $\angle COE + \angle AOF = 90^\circ$. Therefore, $\angle C = \angle AOF$, which means $\triangle AOF \cong \triangle OCE$ (AAS). Hence, $OE = AF = 4.5$. Since $OF \perp AD$, it follows that $AD = 2AF = 2 \times 4.5 = \boxed{9}$.


Question: As shown in the figure, the line $y = x + 1$ intersects with the parabola $y = x^2 - 4x + 5$ at points $A$ and $B$. Point $P$ is a moving point on the y-axis. When the perimeter of $\triangle PAB$ is minimized, what are the coordinates of point $P$?

Solution:
\textbf{Solution:}\\
From $\begin{cases}y=x+1\\ y=x^2-4x+5\end{cases}$, we can obtain $\begin{cases}x=1\\ y=2\end{cases}$ or $\begin{cases}x=4\\ y=5\end{cases}$,\\
$\therefore$ the coordinates of point A are (1, 2) and the coordinates of point B are (4, 5),\\
$\therefore$ AB = $\sqrt{(5-2)^2+(4-1)^2}=3\sqrt{2}$,\\
Construct the reflection of point A across the y-axis to get point A$^{\prime}$, and connect A$^{\prime}$B to intersect the y-axis at point P, then in this case, the perimeter of $\triangle PAB$ is minimized,\\
$\because$ the coordinates of point A$^{\prime}$ are ($-1$, 2) and the coordinates of point B are (4, 5),\\
$\therefore$, let the equation of line A$^{\prime}$B be $y=kx+b$,\\
then from $\begin{cases}-k+b=2\\ 4k+b=5\end{cases}$, we can obtain $\begin{cases}k=\frac{3}{5}\\ b=\frac{13}{5}\end{cases}$,\\
$\therefore$ the equation of line A$^{\prime}$B is $y= \frac{3}{5}x+\frac{13}{5}$,\\
$\therefore$ when $x=0$, $y= \frac{13}{5}$,\\
thus, the coordinates of point P are $\boxed{(0, \frac{13}{5})}$,


Question: Given: As shown in the figure, $AB$ is the diameter of $\odot O$, the chord $CD$ intersects $AB$ at point E, $BE=1$, $AE=5$, and $\angle AEC=30^\circ$. What is the length of $CD$?

Solution:
\textbf{Solution:} As shown in the figure, draw $OF\perp CD$ at F, and connect OD, \\
$\therefore$ $\angle OFE=90^\circ$, $CD=2FD$, \\
$\because$ $\angle AEC=30^\circ$, \\
$\therefore$ $OF=\frac{1}{2}OE$, \\
$\because$ BE=1, AE=5, \\
$\therefore$ AB=BE+AE=6, \\
$\because$ AB is the diameter, \\
$\therefore$ $OB=OD=\frac{1}{2}AB=3$, \\
$\therefore$ $OE=OB-BE=2$, \\
$\therefore$ $OF=\frac{1}{2}OE=1$, \\
$\therefore$ $DF=\sqrt{OD^2-OF^2}=2\sqrt{2}$, \\
$\therefore$ $CD=2DF=\boxed{4\sqrt{2}}$.


Question: As shown in the figure, $ \triangle ABC$ is an inscribed triangle of $ \odot O$, and $ AB=2\sqrt{2}$. What is the area of the shaded region in the figure?

Solution:
\textbf{Solution:} Connect OA and OB.\\
In $\triangle OAB$, OA=OB=2, AB=2$\sqrt{2}$,\\
$\therefore$ OA$^{2}$+OB$^{2}$=2$^{2}$+2$^{2}$=8, AB$^{2}$=(2$\sqrt{2}$)$^{2}$=8,\\
$\therefore$ OA$^{2}$+OB$^{2}$=AB$^{2}$,\\
$\therefore$ $\angle$AOB=$90^\circ$,\\
$\therefore$ Area of the sector $OAB$= $\frac{90}{360}\times \pi \times 4=\pi$, Area of $\triangle OAB$= $\frac{1}{2}$ ×2×2=2,\\
$\therefore$ Area of the shaded region=Area of the sector $OAB$−Area of $\triangle OAB$=$\pi$−2.\\
Therefore, the area of the shaded region in the diagram is $\boxed{\pi-2}$.


Question: As shown in the figure, we know for the rectangle $ABCD$, where $AD=10$ and $AB=6$. Now, the side $AD$ is rotated about one of its endpoints, and when the other endpoint falls exactly at point E on the line where side $BC$ is located, what is the length of the segment $DE$?

Solution:
\textbf{Solution:} As shown in the figure,\\
$\because$ The quadrilateral ABCD is a rectangle,\\
$\therefore$ AB=CD=6, AD=BC=10, $\angle$ABC=$\angle$DCB=$90^\circ$,\\
When AD=AE=10, BE= $\sqrt{AE^{2}-AB^{2}}=\sqrt{10^{2}-6^{2}}=8$,\\
$\therefore$ DE$_{1}$= $\sqrt{CD^{2}+EC^{2}}=\sqrt{6^{2}+2^{2}}=2\sqrt{10}$, DE$_{2}$= $\sqrt{CD^{2}+EC^{2}}=\sqrt{6^{2}+18^{2}}=6\sqrt{10}$,\\
When DE=DA=10, DE=10,\\
In summary, the values of DE that meet the conditions are $\boxed{2\sqrt{10}}$, $\boxed{6\sqrt{10}}$, or $\boxed{10}$.


Question: As shown in the figure, $AB=AC=AD=2$, $\angle BCD=120^\circ$, then what is the value of $BD$?

Solution:
\textbf{Solution:} As shown in the figure, with A as the center and AB as the radius, draw $\odot A$, extend BA to intersect $\odot A$ at T, and connect DT, \\
$\because$ $AB=AC=AD$,\\
$\therefore$ point A is the center of the circumscribed circle of $\triangle BCD$,\\
$\because$ $\angle BCD+\angle T=180^\circ$, $\angle BCD=120^\circ$,\\
$\therefore$ $\angle T=60^\circ$,\\
$\because$ BT is the diameter,\\
$\therefore$ $BT=2AB=4$, $\angle BDT=90^\circ$,\\
$\therefore$ $\angle DBT=90^\circ-60^\circ=30^\circ$,\\
$\therefore$ $DT=\frac{1}{2}BT=2$,\\
$\therefore$ $BD=\sqrt{BT^2-TD^2}=\sqrt{4^2-2^2}=\boxed{2\sqrt{3}}$,\\


Question: As shown in the figure, quadrilateral $ABCD$ is a circumscribed quadrilateral of circle $\odot O$, and $AB=10$, $CD=15$. What is the perimeter of quadrilateral $ABCD$?

Solution:
\textbf{Solution:}\\
$\because$ Quadrilateral ABCD is a circumscribed quadrilateral of $\odot$O,\\
$\therefore$AE = AH, BE = BF, CF = CG, DH = DG,\\
$\therefore$AD + BC = AB + CD = 25,\\
$\therefore$The perimeter of quadrilateral ABCD = AD + BC + AB + CD = 25 + 25 = \boxed{50}


Question: As shown in the figure, in $\triangle ABC$, $AD\perp BC$, $BE\perp AC$, with the feet of the perpendiculars being points $D$ and $E$ respectively, and $AD$ and $BE$ intersecting at point $F$. If $BF=AC$, then what is the measure of $\angle ABC$ in degrees?

Solution:
\textbf{Solution:} Since $AD \perp BC$ at $D$, and $BE \perp AC$ at $E$,\\
therefore, $\angle EAF + \angle AFE = 90^\circ$, and $\angle DBF + \angle BFD = 90^\circ$,\\
Also, since $\angle BFD = \angle AFE$ (vertical angles are equal),\\
therefore, $\angle EAF = \angle DBF$,\\
In right triangles $\triangle ADC$ and $\triangle BDF$,\\
$\left\{\begin{array}{l}
\angle CAD = \angle FBD \\
\angle BDF = \angle ADC \\
BF = AC
\end{array}\right.$,\\
therefore, $\triangle ADC \cong \triangle BDF$ (by AAS),\\
therefore, $BD = AD$,\\
since $\angle ADB = 90^\circ$.\\
Therefore, $\angle ABC = \angle BAD = \boxed{45^\circ}$.


Question: As shown in the figure, in $ \triangle ABC$, the perpendicular bisector of side AB intersects BC and AB at points D and E, respectively, with AE = 3cm and the perimeter of $ \triangle ADC$ being 9cm. What is the perimeter of $ \triangle ABC$ in cm?

Solution:
\textbf{Solution:} Given that in triangle $ABC$, the perpendicular bisector of side $AB$ intersects $BC$ and $AB$ at points $D$ and $E$, respectively, with $AE=3cm$,\\
it follows that $BD=AD$ and $AB=2AE=6cm$,\\
given that the perimeter of $\triangle ADC$ is $9cm$,\\
then, $AC+AD+CD=AC+BD+CD=AC+BC=9cm$,\\
thus, the perimeter of $\triangle ABC$ is given by: AB+AC+BC=\boxed{15}cm.


Question: As shown in the figure, given three equilateral triangles arranged from left to right with side lengths of 1, 2, and 3 respectively, what is the area of the shaded region?

Solution:
\textbf{Solution:} Let points M, N, H be the intersections of AC with BE, BF, and CF respectively,

$\because$ AB=1, BC=2, CD=3,

$\therefore$ AC=CG,

$\therefore\angle CAG=\angle CGA=30^\circ$,

$\therefore$ AG $\perp$ BE,

Also $\because\angle EBA=\angle FCA=60^\circ$,

$\therefore$ EB $\parallel$ CF, AG $\perp$ CF,

$\therefore$ AH=HG, AM=MN,

$\therefore$ Area of $\triangle CHG$ = Area of $\triangle CHA$,

Similarly, Area of $\triangle BMN$ = Area of $\triangle BMA$,

In $\triangle ABM$, AB=1, BM= $\frac{1}{2}$ AB= $\frac{1}{2}$, AM= $\frac{\sqrt{3}}{2}$,

$\therefore$ Area of $\triangle ABM$= $\frac{1}{2} \times BM \times AM= \frac{1}{2}\times \frac{1}{2}\times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}$,

Similarly, it can be shown that Area of $\triangle ACH$= $\frac{1}{2} \times CH \times AH= \frac{1}{2}\times \frac{3}{2}\times \frac{3\sqrt{3}}{2} = \frac{9\sqrt{3}}{8}$,

$\therefore$ The area of the shaded part= Area of $\triangle ABM$ + Area of $\triangle ACH$= $\frac{\sqrt{3}}{8}+\frac{9\sqrt{3}}{8} = \boxed{\frac{5\sqrt{3}}{4}}$


Question: As shown in the figure, $n$ squares, each with a side length of 1, are arranged as depicted. The points $A_{1}$, $A_{2}$, ..., $A_{n}$ represent the centers of these squares, respectively. What is the total area of the overlapping parts formed by these $n$ squares?

Solution:
\textbf{Solution:} Given the problem statement, the area of one shaded part is equal to $\frac{1}{4}$ of the area of a square, which is $\frac{1}{4}$.\\
The total area of the shaded parts (overlap area) for $n$ such squares is: $\frac{1}{4}\times (n-1) = \boxed{\frac{n-1}{4}}$.


Question: As shown in the figure, the straight lines $a$ and $b$ intersect perpendicularly at point $O$. The curve $C$ is centrally symmetric with respect to point $O$. The symmetric point of point $A$ is $A'$, $AB \perp a$ at point $B$, and $A'D \perp b$ at point $D$. If $OB=4$ and $OD=3$, what is the total area of the shaded regions?

Solution:
\textbf{Solution:} As shown in the figure, \\
since line a and b are perpendicular and intersect at point O, and curve C is centrally symmetric about point O, with point A's symmetric point being point A', AB $\perp$ a at point B, and A'D $\perp$ b at point D, OB = 4, OD = 3, \\
therefore, AB = 3, \\
thus, the areas of shape (1) and shape (2) are equal, \\
hence, the sum of the areas of the shaded parts equals the area of rectangle ABOE, which is $4 \times 3 = \boxed{12}$.


Question: As shown in the figure, the vertices of the equilateral triangle \( ABC \) are on the coordinate axes, and the side length is \( 4 \). What are the coordinates of point \( A \)?

Solution:
\textbf{Solution:} Given that it’s an equilateral triangle, it follows from the rule that all medians intersect at one point that: $OC = \frac{1}{2}BC = 2$,\\
Applying Pythagoras’ theorem, we find: $OA = \sqrt{AC^2 - CO^2} = 2\sqrt{3}$,\\
$\therefore$The coordinates of point A are $\boxed{(0, 2\sqrt{3})}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=50^\circ$, $\angle DAC$ and $\angle ACF$ are two exterior angles of this triangle, then what is the sum of $\angle DAC + \angle ACF$ in degrees?

Solution:
\textbf{Solution:} Given: $ \angle DAC=\angle B+\angle ACB$, $ \angle FCA=\angle B+\angle BAC$,\\
$\therefore \angle DAC+\angle ACF=\angle B+\angle B+\angle BAC+\angle BCA$,\\
$\because \angle B=50^\circ$, $\angle B+\angle BAC+\angle BCA=180^\circ$,\\
$\therefore \angle DAC+\angle ACF=180^\circ+50^\circ=\boxed{230^\circ}$.


Question: As shown in the figure, two congruent right triangles overlap each other. One of the triangles is translated along the direction from point B to C to the position of $\triangle DEF$, where $AB=10$, $DO=4$, and the translation distance is $6$. What is the area of the shaded region?

Solution:
\textbf{Solution:} Given the property of translation, we have $BE=6$, $DE=AB=10$,\\
thus $OE=DE-DO=10-4=6$,\\
therefore, the area of quadrilateral $ODFC$ is equal to the area of trapezoid $ABEO$, which is $\frac{1}{2}\times(AB+OE)\times BE =\frac{1}{2}\times(10+6)\times 6=\boxed{48}$.


Question: As shown in the figure, $a \parallel b$, if $\angle 1=50^\circ$, then what is the measure of $\angle 2$ in degrees?

Solution:
\textbf{Solution:} As shown in the figure, \\
$\because a \parallel b$, \\
$\therefore \angle 3 = \angle 1$, \\
$\because \angle 1 = 50^\circ$, \\
$\therefore \angle 3 = 50^\circ$ \\
$\because \angle 2 + \angle 3 = 180^\circ$ \\
$\therefore \angle 2 = 180^\circ - \angle 3 = 180^\circ - 50^\circ = \boxed{130^\circ}$


Question: As shown in the figure, two flat strips of paper, each with a width of 1, are overlaid in a cross fashion. The angle between the edges of the two strips is $\alpha =30^\circ$. What is the area of their overlapping part (the shaded area in the figure)?

Solution:
\textbf{Solution:} As shown in the figure: Draw AE$\perp$BC and AF$\perp$CD at F, with the feet of the perpendiculars being E and F,\\
$\therefore$ $\angle$AEB=$\angle$AFD=$90^\circ$,\\
$\because$AD$\parallel$CB and AB$\parallel$CD,\\
$\therefore$quadrilateral ABCD is a parallelogram,\\
$\because$the width of the strip is 1,\\
$\therefore$AE=AF=1,\\
In $\triangle$ABE and $\triangle$ADF\\
$\left\{\begin{array}{l}
\angle ABE=\angle ADF=30^\circ \\
\angle AEB=\angle AFD=90^\circ \\
AE=AF
\end{array}\right.$,\\
$\therefore$ $\triangle$ABE$\cong$ $\triangle$ADF (AAS),\\
$\therefore$AB=AD,\\
$\therefore$quadrilateral ABCD is a rhombus.\\
$\therefore$BC=AB,\\
$\because$ $\frac{AE}{AB}=\sin 30^\circ$,\\
$\therefore$BC=AB=$\frac{1}{\sin 30^\circ}=2$,\\
$\therefore$the area of the overlapping part (the shaded area in the figure) is: BC×AE=2×1=\boxed{2},


Question: As shown in the figure, it is known that point $E$ is a moving point on the diagonal $AC$ of rectangle $ABCD$, and the vertices $G$ and $H$ of square $EFGH$ are both on side $AD$. If $AB=3$ and $BC=4$, then what is the value of $\tan\angle AFE$?

Solution:
\textbf{Solution:} Given that $EH\parallel CD$,\\
$\triangle AEH \sim \triangle ACD$,\\
hence $\frac{EH}{AH} = \frac{CD}{AD} = \frac{3}{4}$,\\
let $EH = 3a$, then $AH = 4a$,\\
thus, $HG = GF = EH = 3a$,\\
since $EF\parallel AD$,\\
thus $\angle AFE = \angle FAG$,\\
hence $\tan\angle AFE = \tan\angle FAG = \frac{GF}{AG} = \frac{3a}{4a+3a} = \boxed{\frac{3}{7}}$.


Question: As shown in the diagram, within triangle paper $ABC$, $\angle C=90^\circ$, $\angle A=30^\circ$, and $AC=6$. When folding the paper so that point $C$ lands on point $D$ on side $AB$, and the fold line $BE$ intersects $AC$ at point $E$, what is the length of the fold line $BE$?

Solution:
\textbf{Solution:} Given, in $\triangle ABC$, $\angle C=90^\circ$, $\angle A=30^\circ$,\\
hence $\angle ABC=60^\circ$,\\
by the property of folding, we get: $\angle ABE=\frac{1}{2}\angle ABC=30^\circ$, $\angle BDE=\angle C=90^\circ$, $DE=CE$,\\
thus $\angle ABE=\angle A$,\\
thus $BE=AE$,\\
given, in $Rt\triangle ADE$, $\angle ADE=90^\circ$, $\angle A=30^\circ$,\\
thus $DE=\frac{1}{2}AE$,\\
since $DE+AE=CE+AE=AC=6$,\\
therefore $\frac{1}{2}AE+AE=6$, solving gives $AE=4$,\\
hence $BE=\boxed{4}$.


Question: As shown in the figure, AB is the diameter of circle $O$, and chord CD $\perp$ AB at point H. If AB $= 10$ and CD $= 8$, what is the length of BH?

Solution:
\textbf{Solution:} Draw line OC.

From the given: In right triangle $OCH$, we have $OC=\frac{1}{2}AB=5$, and $CH=\frac{1}{2}CD=4$;

By the Pythagorean theorem, we find: $OH=\sqrt{OC^{2}-CH^{2}}=\sqrt{5^{2}-4^{2}}=3$;

Therefore, $BH=OB-OH=\boxed{2}$.


Question: As shown in the figure, it is known that $\triangle ABC$ with $AB=AC$ and $\angle A=70^\circ$. Points $O$ and $D$ are the midpoints of $BC$ and $AB$, respectively. A circle is drawn with $O$ as the center and $OD$ as the radius, intersecting $AB$ at another point $E$ and intersecting $AC$ at points $G$ and $F$. What is the degree measure of $\angle DOE + \angle FOG$?

Solution:
\textbf{Solution:} Since $O$ and $D$ are the midpoints of $BC$ and $AB$ respectively, and $AB=AC$,\\
it follows that $OD\parallel AC$, and $OG=OD= \frac{1}{2}AC= \frac{1}{2}AB$,\\
therefore, $\angle ODE=\angle A=70^\circ$, and $OG\parallel AB$,\\
thus, $\angle OGF=\angle A=70^\circ$,\\
since $OE=OD=OG=OF$,\\
it implies that $\angle ODE=\angle OED=70^\circ$, $\angle OFG=\angle OGF=70^\circ$,\\
thus, $\angle DOD=\angle GOF=40^\circ$,\\
hence, $\angle DOE+\angle FOG=\boxed{80^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, D is the midpoint on side AC. Connect BD and fold $\triangle BDC$ along BD to obtain $\triangle BDC'$, where DC' intersects AB at point E. Connect AC'. If AD = AC' = 2 and BD = 3, what is the distance from point D to line BC?

Solution:
\textbf{Solution:} As shown in the diagram, connect $CC'$ and let it intersect $BD$ at point $M$. Draw $DH \perp BC'$ at point $H$, \\
since $AD=AC'=2$ and $D$ is the midpoint of side $AC$, \\
therefore $DC=AD=2$. By folding, $\triangle BDC\cong \triangle BDC'$, and $BD$ perpendicularly bisects $CC'$, \\
therefore $DC=DC'=2$, $BC=BC'$, $CM=C'M$, \\
therefore $AD=AC'=DC'=2$, \\
thus $\triangle ADC'$ is an equilateral triangle, \\
therefore $\angle ADC'=\angle AC'D=\angle C'AC=60^\circ$, \\
since $DC=DC'$, \\
therefore $\angle DCC'=\angle DC'C= \frac{1}{2} \times 60^\circ=30^\circ$, \\
In $\triangle C'DM$, with $\angle DC'C=30^\circ$ and $DC'=2$, \\
thus $DM=1$, $C'M= \sqrt{3}DM= \sqrt{3}$, \\
therefore $BM=BD−DM=3−1=2$, \\
In $\triangle BMC'$, $BC'= \sqrt{BM^2+C'M^2}=\sqrt{2^2+(\sqrt{3})^2}=\sqrt{7}$, \\
since the area of $\triangle BDC'$ is $\frac{1}{2} BC'\cdot DH= \frac{1}{2} BD\cdot CM$, \\
thus $\sqrt{7}DH=3\times \sqrt{3}$, \\
therefore $DH= \frac{3\sqrt{21}}{7}$, \\
since $\angle DBC=\angle DBC'$, \\
therefore the distance from point $D$ to $BC$ is $\boxed{\frac{3\sqrt{21}}{7}}$.


Question: As shown in the figure, a sector BAC with a radius of 2 and a central angle of $90^\circ$ is rotated counterclockwise around point A. During the rotation, point B falls on point B' on the arc of sector BAC, and the corresponding point for point C is point C'. What is the area of the shaded region?

Solution:
\textbf{Solution:} Connect $BB'$, and draw $AF \perp BB'$ at $F$, then $\angle AFB = 90^\circ$, as shown in the figure,\\
$\because$ when the sector $BAC$ with radius $2$ and a central angle of $90^\circ$ is rotated counterclockwise around point $A$, during the rotation, point $B$ falls on the point $B''$ on the arc of sector $BAC$, and the corresponding point for point $C$ is point $C'$,\\
$\therefore$ the area of sector $ABC$ is equal to the area of sector $AB'C'$, and $AB = AB' = BC = BB' = 2$,\\
$\therefore$ $\triangle ABB'$ is an equilateral triangle,\\
$\therefore$ $\angle ABF = 60^\circ$,\\
$\therefore$ $\angle BAF = 30^\circ$,\\
$\therefore$ $BF = \frac{1}{2} AB = \frac{1}{2} \times 2 = 1$, by Pythagoras' theorem, we have: $AF = \sqrt{2^2 - 1^2} = \sqrt{3}$,\\
$\therefore$ the area of the shaded part $S = S_{\text{sector } ABC} - (S_{\text{sector } ABB'} - S_{\triangle ABB'})$\\
$= \frac{90^\circ\pi \times 2^2}{360} - \left(\frac{60^\circ\pi \times 2^2}{360} - \frac{1}{2} \times 2 \times \sqrt{3}\right)$\\
$= \boxed{\sqrt{3} + \frac{1}{3}\pi}$.


Question: In the equilateral triangle $\triangle ABC$, where $AB=6$ and $BD=4$, point $E$ is a moving point on side $AC$. Connect $DE$ and fold $\triangle CDE$ along $DE$ to obtain $\triangle FDE$. What is the minimum distance from point $F$ to line $AB$?

Solution:
\textbf{Solution:} As shown in the diagram, draw DT$\perp$AB at T from point D.\\
$\because \triangle ABC$ is an equilateral triangle,\\
$\therefore \angle B=60^\circ$, $BC=AB=6$,\\
$\because \angle DTB=90^\circ$, $BD=4$,\\
$\therefore CD=DF=2$, $DT=BD\cdot\sin60^\circ=2\sqrt{3}$,\\
By observing the diagram, it is known that when point F is on DT, the distance from point F to AB is the smallest, with the minimum value being $2\sqrt{3}-2$,\\
Therefore, the answer is: $2\sqrt{3}-2=\boxed{2\sqrt{3}-2}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=10$, and $BC=5$. If points $M$ and $N$ are two movable points on line segments $AC$ and $AB$ respectively, what is the perimeter of $\triangle BMN$ when $BM+MN$ is minimized?

Solution:
\textbf{Solution:}

As shown in the figure, construct the point $B'$ as the symmetric point of $B$ with respect to $AC$, and draw line $B'A$ intersecting $DC$ at point $E$. Therefore, the minimum value of $BM+MN$ is equal to the minimum value of $B'M+MN$,\\
$\therefore$ when $B'N \perp AB$, the minimum value of $BM+MN$ is achieved,\\
$\therefore$ construct $B'N' \perp AB$ intersecting $AC$ at $M'$, then $B'N'$ is the minimum value of $BM+MN$;\\
$\because$ Quadrilateral $ABCD$ is a rectangle,\\
$\therefore$ $\angle D=90^\circ$, $DC \parallel AB$,\\
$\therefore$ $\angle DCA=\angle BAC$,\\
Also, $\because$ $\angle B'AC=\angle BAC$,\\
$\therefore$ $\angle B'AC=\angle DCA$,\\
$\therefore$ $AE=CE$, let $EC=AE=x$,\\
$\therefore$ in $\triangle AED$, $DE^2+AD^2=AE^2$, i.e., $(10-x)^2+5^2=x^2$, solving we get: $x=\frac{25}{4}$,\\
$\therefore B'E=AB'−AE=10−\frac{25}{4}=\frac{15}{4}$,\\
Let the height from $E$ to side $EC$ in $\triangle B'EC$ be $h$, by symmetry we have $B'C=BC=5$,\\
$\angle AB'C=\angle ABC=90^\circ$,\\
$\therefore$ $S_{\triangle B'CE}=\frac{1}{2}\times \frac{15}{4}\times 5=\frac{1}{2}\cdot \frac{25}{4}\cdot h$, solving we get: $h=3$,\\
$B'N'=h+5=8$, that is, the minimum value of $BM+MN$ is 8,\\
$\therefore$ in $\triangle AB'N'$, $AN'=\sqrt{AB'^{2}-B'N'^{2}}=\sqrt{10^{2}-8^{2}}=6$,\\
$\therefore$ $BN'=AB-AN'=10-6=4$,\\
$\therefore$ the perimeter of $\triangle BMN=BN'+B'M'+M'N'=BN'+B'N'=4+8=\boxed{12}$.


Question: In the Cartesian coordinate system $xOy$, the coordinates of point $P$ are $(m-1, -m+5)$. The radius of the circle $\odot O$ is 1. Point $Q$ lies on $\odot O$, and line segment $PQ$ is drawn. If $PQ$ is tangent to $\odot O$, what is the minimum value of line segment $PQ$?

Solution:
\textbf{Solution:} As shown in the diagram, draw tangent PQ from point P to the circle, and connect OQ;\\
$\because$ The coordinates of point P are $(m−1, −m+5)$\\
$\therefore$ $O{P}^{2}={\left(−m+5\right)}^{2}+{\left(m−1\right)}^{2}=2{m}^{2}−12m+26$\\
$\because$ The radius of circle $O$ is 1, that is, OQ=1\\
$\therefore$ PQ$^{2}$= OP$^{2}$−OQ$^{2}$= $2{m}^{2}−12m+25=2{\left(m−3\right)}^{2}+7$\\
Hence, when m=3, PQ$^{2}$ attains its minimum value\\
$\therefore$ When m=3, PQ attains its minimum value $\boxed{\sqrt{7}}$.


Question: As shown in the figure, it is known that the area of rectangle $OABC$ is $18$, and its diagonal $OB$ intersects with the hyperbola $y= \frac{k}{x}$ at point $D$, where $OD: DB=2:1$. What is the value of $k$?

Solution:
\textbf{Solution:} Given the problem statement, let the coordinates of point D be $(x_D, y_D)$. Therefore, the coordinates of point B are $\left(\frac{3}{2}x_D, \frac{3}{2}y_D\right)$,\\
$\therefore$ the area of rectangle OABC $=\left|\frac{3}{2}x_D \times \frac{3}{2}y_D\right|=18$,\\
$\because$ the figure lies in the first quadrant,\\
$\therefore k=x_D \cdot y_D=\boxed{8}$.


Question: As shown in the figure, in $\triangle ABC$, we have $AB=4$, $AC=6$, and $AD$ is the bisector of $\angle BAC$. $M$ is the midpoint of $BC$, and $ME\parallel AD$ intersects $AC$ at $F$ and intersects the extension of $BA$ at $E$. What is the value of $BE$?

Solution:
\textbf{Solution:} Proof:

Since $AD$ bisects $\angle BAC$,

then $\angle BAD=\angle DAC$,

Since $MF \parallel AD$,

then $\angle DAC=\angle AFE$, $\angle BAD=\angle E$,

thus $\angle E=\angle AFE$,

hence $AE=AF$;

Extend $FM$ to point $N$ such that $MN=FM$, $\angle BMN=\angle CMF$, $MB=CM$,

therefore $\triangle BMN \cong \triangle CMF$ (SAS),

therefore $CF=BN$, $\angle N=\angle MFC$,

Since $\angle EFA=\angle MFC$,

thus $\angle N=\angle EFA$,

thus $\angle N=\angle E$,

thus $BN=BE$,

thus $AB+AC=AB+AF+FC=AB+AE+FC=BE+FC$,

Since $BE=FC$,

therefore $2BE=10$,

therefore $BE=\boxed{5}$.


Question: As shown in the figure, let P be any point on side BC of the equilateral $\triangle ABC$. Point D lies on the extension of side BA. Rotate line segment PD $60^\circ$ counterclockwise around point P to obtain line segment PE. Connect BE. What is the measure of $\angle CBE$ in degrees?

Solution:
\textbf{Solution:} Draw line $DF \parallel AC$ passing through point $D$ and intersecting the extension of $BC$ at point $F$, as shown in the figure: \\
$\because$ $\triangle ABC$ is an equilateral triangle,\\
$\therefore$ $\angle ABC = \angle BAC = \angle BCA = 60^\circ$,\\
$\therefore$ $\angle BAC = \angle BDF = \angle BCA = \angle F = 60^\circ$,\\
$\therefore$ $\triangle DBF$ is an equilateral triangle,\\
$\therefore$ $DB = DF$,\\
$\because$ rotating segment $PD$ $60^\circ$ counterclockwise around point $P$ results in segment $PE$,\\
$\therefore$ $\triangle PDE$ is an equilateral triangle,\\
$\therefore$ $DE = DP$, $\angle EDP = 60^\circ$,\\
$\therefore$ $\angle EDB + \angle BDP = \angle PDF + \angle BDP = 60^\circ$,\\
$\therefore$ $\angle EDB = \angle PDF$,\\
$\therefore$ $\triangle EDB \cong \triangle PDF$ (SAS),\\
$\therefore$ $\angle EBD = \angle F = 60^\circ$,\\
$\therefore$ $\angle CBE = \angle ABC + \angle EBD = \boxed{120^\circ}$


Question: As shown in the figure, in $ \triangle ABC$, $AB=AC$, the line EF is the perpendicular bisector of AB, D is the midpoint of BC, M is a moving point on EF, the area of $ \triangle ABC$ is 12, and $BC=4$, then what is the minimum value of the perimeter of $ \triangle BDM$?

Solution:
\textbf{Solution:} As shown in the diagram, we connect AD, and AM,\\
\(\because\) EF is the perpendicular bisector of line segment AB,\\
\(\therefore\) AM=BM,\\
\(\therefore\) the perimeter of \(\triangle BDM\) = BD + BM + DM = AM + DM + BD,\\
\(\therefore\) to minimize the perimeter of \(\triangle BDM\), we need to minimize the value of AM + DM,\\
\(\therefore\) when points A, M, and D are collinear, AM + DM is minimized, which equals to AD,\\
\(\because\) AB=AC, and D is the midpoint of BC,\\
\(\therefore\) AD\(\perp\)BC, \(BD=\frac{1}{2}BC=2\),\\
\(\therefore\) \(S_{\triangle ABC}=\frac{1}{2}AD\cdot BC=12\),\\
\(\therefore\) AD=6,\\
\(\therefore\) The minimum perimeter of \(\triangle BDM\) = AD + BD = \boxed{8},


Question: As shown in the figure, $\angle B = \angle D = 90^\circ$, $BC = CD$, $\angle 1 = 40^\circ$, then what is the value of $\angle 2$?

Solution:
\textbf{Solution:} Given $\because$ $\angle B = \angle D = 90^\circ$,\\
$\therefore$ $\triangle ABC$ and $\triangle ADC$ are both right-angled triangles,\\
In $\triangle ABC$ and $\triangle ADC$,\\
$\left\{\begin{array}{c} BC = DC \\ AC = AC \end{array}\right.$,\\
$\therefore$ $\triangle ABC \cong \triangle ADC$ (HL),\\
$\therefore$ $\angle 1 = \angle CAD = 40^\circ$,\\
$\therefore$ $\angle 2 = 90^\circ - \angle CAD = \boxed{50^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, isosceles right triangles $S_1$ and $S_2$ are constructed on hypotenuses $AC$ and $BC$, respectively, and a square $S$ is constructed on side $AB$. If the sum of the areas of $S_1$ and $S_2$ equals $9$, what is the length of the side of square $S$?

Solution:
\textbf{Solution:} As shown in the figure: draw DE $\perp$ AC at E. In the isosceles right triangle $\triangle$ACD, we have AD=CD, and $\angle$ADC=$90^\circ$,\\
thus, DE=AE=CE=$\frac{1}{2}$AC,\\
therefore, ${S}_{1}=\frac{1}{2}AC\cdot \frac{1}{2}AC=\frac{1}{4}AC^{2}$,\\
Similarly, ${S}_{2}=\frac{1}{4}BC^{2}$,\\
therefore, ${S}_{1}+{S}_{2}=\frac{1}{4}AC^{2}+\frac{1}{4}BC^{2}=9$,\\
therefore, $AC^{2}+BC^{2}=36$,\\
In $\triangle$ABC, $\angle$ACB=$90^\circ$, $AC^{2}+BC^{2}=AB^{2}$,\\
therefore, S=$AB^{2}=36$,\\
therefore, the side length of the square S equals $\boxed{6}$.


Question: In the grid shown, each small square has a side length of 1, and each vertex of the small squares is called a grid point. Points $O$, $A$, and $B$ are all grid points. If the sector $AOB$ in the figure is the lateral surface development of a cone, what is the radius of the base circle of the cone?

Solution:
\textbf{Solution:} Let the radius of the circular base of the cone be $r$. \\
Since each small square has a side length of 1, \\
thus $OA=OB=\sqrt{3^2+4^2}=5$, $AB=\sqrt{1^2+7^2}=5\sqrt{2}$ \\
therefore $OA^2+OB^2=AB^2$ \\
thus $\angle AOB=90^\circ$, \\
since the sector AOB is a lateral surface of the cone unfolded, \\
thus the arc $AB$ is equal to the circumference of the base circle. \\
Let the radius of the base circle be $r$ \\
thus $\frac{90^\circ \pi \times 5}{180}=2\pi r$, \\
therefore $r=\boxed{\frac{5}{4}}$.


Question: As shown in the figure, within circle $O$, $AB$ is the diameter, and chord $CD \perp AB$ at $E$. Connect $OC$ and $OD$. If the diameter is $10$ and $CD=8$, then what is the length of $BE$?

Solution:
Solution: Since $CD \perp AB$ at $E$, and $CD=8$, with $AB$ being the diameter, \\
therefore $CE=\frac{1}{2}CD=\frac{1}{2}\times 8=4$, and $\angle OEC=90^\circ$\\
therefore $OE=\sqrt{OC^{2}-CE^{2}}=\sqrt{5^{2}-4^{2}}=3$\\
therefore $BE=OB-OE=5-3=\boxed{2}$


Question: Given the figure, $\angle BAC=30^\circ$, P is a point on the bisector of $\angle BAC$, PM$\parallel$AC, PD$\perp$AC, and PD$=6$. What is the area of $\triangle AMP$?

Solution:
\textbf{Solution:} As shown in the figure, draw PE$\perp$AB at point E through point P,\\
since P is a point on the bisector of $\angle$BAC, and PD$\perp$AC, PE$\perp$AB,\\
it follows that PE=PD=6,\\
since $\angle$BAC=30$^\circ$, and PM$\parallel$AC,\\
it follows that $\angle$PME=$\angle$BAC=30$^\circ$, and $\angle$APM=$\angle$PAD,\\
therefore PM=2PE=12,\\
since $\angle$BAP=$\angle$PAD,\\
it follows that $\angle$BAP=$\angle$APM,\\
thus AM=PM=12.\\
Therefore, the area of $\triangle AMP$=  $\frac{1}{2}\times AM\times PE=\frac{1}{2}\times 12\times 6= \boxed{36}$.


Question: In $\triangle ABC$, $AB=AC$, and $\angle B=30^\circ$. The perpendicular bisector of side $AC$ intersects $AC$ and $BC$ at points $E$ and $D$ respectively. If $DE=2$, what is the length of $BC$?

Solution:
\textbf{Solution:} Draw AD, \\
$\because$AB = AC, $\angle B = 30^\circ$, \\
$\therefore$ $\angle C = 30^\circ$, $\angle BAC = 120^\circ$, \\
$\because$ DE is the perpendicular bisector of AC, \\
$\therefore$ DC = DA, $\angle DEC = 90^\circ$, \\
$\therefore$ $\angle DAC = \angle DCA = 30^\circ$, \\
$\because$ in $\triangle DEC$, $\angle C = 30^\circ$, DE = 2, \\
$\therefore$ CD = AD = 2DE = 4, \\
$\because$ $\angle BAC = 120^\circ$, $\angle DAC = 30^\circ$, \\
$\therefore$ $\angle BAD = 90^\circ$, \\
$\therefore$ BD = 2AD = 8, \\
$\therefore$ BC = BD + CD = 8 + 4 = \boxed{12}, \\


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, point $P$ is a moving point on the bisector $CD$ of $\angle ACB$, $CE=\frac{1}{2}BC=8$, and the area of $\triangle ABC$ is $48$. What is the minimum value of $PA+PE$?

Solution:
\textbf{Solution:} Since CD is the bisector of $\angle ACB$,\\
we construct the symmetric point $E'$ of point E with respect to line CD on BC, and draw line AE', where its intersection with line CD is point P.\\
Since CD perpendicularly bisects EF,\\
we have PE=PE', and CE=CE',\\
thus PA+PE=PA+PE'=AE'.\\
Since $CE=\frac{1}{2}BC=8$,\\
we get $BC=16$,\\
which means point E' is the midpoint of BC.\\
Moreover, since AB=AC,\\
we have AE'$\perp$BC.\\
Given that the area of $\triangle ABC$ is 48,\\
we deduce ${S}_{\triangle ABC}=\frac{1}{2}BC\cdot AE'=\frac{1}{2}\times 16\times AE'=48$,\\
thus AE'=6,\\
therefore, the minimum value of PA+PE is $\boxed{6}$.


Question: As shown in the Cartesian coordinate system, $OA=OB=\sqrt{5}$, and $AB=\sqrt{10}$. If the coordinates of point $A$ are (1, 2), what are the coordinates of point $B$?

Solution:
\textbf{Solution:} Construct line BN$\perp$x-axis, AM$\perp$x-axis,\\
$\because$OA=OB=$\sqrt{5}$, AB=$\sqrt{10}$,\\
$\therefore$AO$^{2}$+OB$^{2}$=AB$^{2}$,\\
$\therefore$$\angle$BOA=$90^\circ$,\\
$\therefore$$\angle$BON+$\angle$AOM=$90^\circ$,\\
$\because$$\angle$BON+$\angle$NBO=$90^\circ$,\\
$\therefore$$\angle$AOM=$\angle$NBO,\\
$\because$$\angle$AOM=$\angle$NBO, $\angle$BNO=$\angle$AMO, BO=OA,\\
$\therefore$$\triangle$BNO$\cong$$\triangle$OMA,\\
$\therefore$NB=OM, NO=AM,\\
$\because$the coordinates of point A are (1,2),\\
$\therefore$the coordinates of point B are $\boxed{(-2,1)}$.


Question: As shown in the figure, $\angle MON=30^\circ$, 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, with the side length of the first equilateral triangle from the left being $a_1$, the second being $a_2$, and so on. If $OA_1=1$, what is the value of $a_{2020}$?

Solution:
\textbf{Solution:} As shown in the figure,\\
$\because$ $\triangle A_1B_1A_2$ is an equilateral triangle,\\
$\therefore$ $A_1B_1=A_2B_1$, $\angle 3=\angle 4=\angle 1+\angle O=60^\circ$,\\
$\therefore$ $\angle 2=120^\circ$,\\
$\because$ $\angle MON=30^\circ$,\\
$\therefore$ $\angle 1=180^\circ-120^\circ-30^\circ=30^\circ$,\\
Furthermore, $\because$ $\angle 3=60^\circ$,\\
$\therefore$ $\angle 5=180^\circ-60^\circ-30^\circ=90^\circ$,\\
$\because$ $\angle MON=\angle 1=30^\circ$,\\
$\therefore$ $OA_1=A_1B_1=1$,\\
$\therefore$ $A_2B_1=1$,\\
$\because$ $\triangle A_2B_2A_3$, $\triangle A_3B_3A_4$ are equilateral triangles,\\
$\therefore$ $\angle 11=\angle 10=60^\circ$, $\angle 13=60^\circ$,\\
$\because$ $\angle 4=\angle 12=60^\circ$,\\
$\therefore$ $A_1B_1 \parallel A_2B_2 \parallel A_3B_3$, $B_1A_2 \parallel B_2A_3$,\\
$\therefore$ $\angle 1=\angle 6=\angle 7=30^\circ$, $\angle 5=\angle 8=90^\circ$,\\
$\therefore$ $a_2=2a_1$, $a_3=4a_1=4$, $a_4=8a_1=8$, $a_5=16a_1$,\\
By analogy: $a_{2020}=2^{2019}$.\\
Therefore, the answer is: $a_{2020}=\boxed{2^{2019}}$


Question: As shown in the figure, the right-angled triangle $\triangle OAB$ has its right-angle vertex $A$ on the $y$-axis. The graph of the inverse proportional function $y=\frac{k}{x}$ (where $k>0$, $x>0$) passes through the midpoint $D$ of line segment $OB$ and intersects line segment $AB$ at point $C$. Connect $CD$. If the area of $\triangle BCD$ is $3$, what is the value of $k$?

Solution:
\textbf{Solution:} Join OC, and draw a perpendicular line from D to the x-axis, intersecting AB at point E and the x-axis at point F,\\
\(\because S_{\triangle CAO} = S_{\triangle DOF} = \frac{1}{2}|k|\),\\
\(\because\) D is the midpoint of the line segment OB,\\
\(\therefore S_{\triangle BDE} = S_{\triangle DOF} = \frac{1}{2}|k|\),\\
\(\because\) DE is the median of \(\triangle AOB\),\\
\(\therefore DE : OA = 1 : 2\),\\
\(\therefore S_{\triangle BDE} = \frac{1}{4}S_{\triangle AOB} = S_{\triangle CAO}\),\\
\(\therefore S_{\triangle OBC} = S_{\triangle AOB} - S_{\triangle CAO} = \frac{3}{4}S_{\triangle AOB}\),\\
\(\because\) D is the midpoint of the line segment OB,\\
\(\therefore S_{\triangle OBC} = 2S_{\triangle BCD} = 6 = \frac{3}{4}S_{\triangle AOB}\),\\
\(\therefore S_{\triangle AOB} = 8\),\\
\(\therefore S_{\triangle AOC} = \frac{1}{4}S_{\triangle AOB} = 2 = \frac{1}{2}|k|\),\\
\(\because k > 0\),\\
\(\therefore k = \boxed{4}\).


Question: As shown in the diagram, $O$ is the intersection point of the diagonals of square $ABCD$, and $E$ is a point on line segment $AO$. If $AB = 1$, and $\angle BED = 135^\circ$, what is the length of $AE$?

Solution:
\textbf{Solution:} Since it is a square ABCD,\\
it follows that $\angle BCA=\angle BAC=45^\circ$, $\angle ABC=90^\circ$, and BC=AB=1,\\
thus, AC=$\sqrt{2}$AB=$\sqrt{2}$,\\
since E is a point on the diagonal AC of the square, and $\angle DEB=135^\circ$,\\
it is easy to derive that $\angle CEB=\angle CED=67.5^\circ$,\\
thus, $\angle EBC=180^\circ-67.5^\circ-45^\circ=67.5^\circ$,\\
therefore, $\angle EBC=CEB=67.5^\circ$,\\
hence, BC=EC=1,\\
thus, AE=AC−EC=$\sqrt{2}-1$.\\
Therefore, the answer is: $AE=\boxed{\sqrt{2}-1}$.


Question: As shown in the figure, in rhombus $ABCD$, $AC=6$, and $BD=4$. What is the length of $AB$?

Solution:
\textbf{Solution:} Since ABCD is a rhombus,\\
it follows that AC $\perp$ BD, AO = OC, BO = DO,\\
since AC = 6, BD = 4,\\
it follows that AO = 3, BO = 2,\\
therefore AB = $\sqrt{AO^{2}+BO^{2}}$ = $\sqrt{3^{2}+2^{2}}$ = $\boxed{\sqrt{13}}$.


Question: As shown in the figure, square $ABCD$ has a side length of $2$, and points $E$, $F$ are movable points on diagonal $AC$, with the length of $EF$ being $1$. Connecting $BE$ and $BF$, what is the minimum perimeter of $\triangle BEF$?

Solution:
\textbf{Solution:} As shown in the diagram, draw $BH \parallel AC$ such that $BH = EF = 1$, and connect $DH$ intersecting $AC$ at $E$, thus the perimeter of $\triangle BEF$ is minimized,\\
$\because BH = EF$, $BH \parallel EF$,\\
$\therefore$ the quadrilateral $EFBH$ is a parallelogram,\\
$\therefore$ $BF = EH$,\\
$\because$ the quadrilateral $ABCD$ is a square,\\
$\therefore$ $AC$ bisects $BD$ perpendicularly,\\
$\therefore$ $EB = ED$,\\
$\therefore$ $BE + BF = EH + ED = DH$,\\
$\because$ the quadrilateral $ABCD$ is a square,\\
$\therefore$ $AC \perp BD$,\\
$\because$ $BH \parallel AC$,\\
$\therefore$ $BD \perp BH$, that is, $\angle DBH = 90^\circ$,\\
In $\triangle DBH$,\\
$DH = \sqrt{BD^2 + BH^2} = \sqrt{(2\sqrt{2})^2 + 1^2} = 3$,\\
$\therefore$ the minimum possible perimeter of $\triangle BEF$ is $3 + 1 = \boxed{4}$.


Question: As illustrated in the diagram, within the rectangle $ABCD$, the ratio of $AB$ to $BC$ is $3:4$, and $BP=3$. When $OABP$ is folded along $AP$ such that point $B$ precisely aligns with point $E$ on the diagonal $AC$, what is the length of $BC$?

Solution:
Solution: Let $AB=3x$ and $BC=4x$,\\
since quadrilateral $ABCD$ is a rectangle,\\
thus $AC= \sqrt{AB^{2}+BC^{2}}=5x$,\\
since it is folded,\\
thus $BP=PE=3$, $AE=AB=3x$, $\angle AEP=\angle B=90^\circ$,\\
thus $PC=4x-3$, $EC=AC-AE=2x$,\\
thus $PC^{2}=PE^{2}+EC^{2}$, i.e., $(4x-3)^{2}=3^{2}+(2x)^{2}$,\\
Solving gives: $x=2$ or $0$ (discard $0$),\\
thus $BC=4x=\boxed{8}$.


Question: As shown in the diagram, in rectangle $ABCD$, rectangle $ABCD$ is folded along $EF$ with point $A$ landing on point $A'$, and point $B$ landing on point $B'$ on side $CD$. Line $BB'$ intersects $EF$ at point $G$, and point $M$ is on $A'B'$ with $A'M=2B'M$. Given that $CD=3$ and $AD=6$, during the folding process, when point $B'$ is at different positions on side $CD$, what is the minimum value of $MG+ B'G$?

Solution:
\textbf{Solution:} As shown in the figure, connect GC, and choose a point Q on the segment AB such that AQ=2BQ. Connect QG and QC,\\
$\because$AB=CD=3,\\
$\therefore$BQ=1,\\
$\because$rectangle ABCD is folded along EF, point A falls on point A', and point B falls on point B' on side CD, with A'M=2B'M,\\
$\therefore$Q and M are symmetrical about EF,\\
$\therefore$QG=MG,\\
Also, $\because$G is the midpoint of BB', and $\angle$BCB'=$90^\circ$,\\
$\therefore$GC=B'G,\\
$\therefore$MG+ B'G=QG+GC, $\therefore$when points Q, G, and C are collinear, CQ=QG+GC, at this time, QG+GC is the shortest,\\
$\therefore$The minimum value of MG+ B'G is the length of CQ,\\
In right $\triangle$BCQ, BQ=1, BC=6,\\
$\therefore$CQ=$\sqrt{BQ^2+BC^2}$=$\sqrt{1^2+6^2}$=$\sqrt{37}$.\\
Therefore, the answer is: $\boxed{\sqrt{37}}$.


Question: As shown in the figure, the graph of the function $y= \frac{k}{x}$ ($x>0$) passes through the midpoint of one side of the rectangle $OBCD$ and also passes through the vertex $P$ of the rectangle $OAPE$. If the area of the shaded part is $6$, what is the value of $k$?

Solution:
\textbf{Solution:} As shown in the figure, let $M$ be the midpoint of $CD$, and draw $EF \perp x$-axis at $N$, \\
$\therefore$ Rectangle $DONM$ is equal to half of the area of rectangle $OBCD$, \\
$\because$ Points $M$ and $P$ both lie on the graph of the inverse proportion function, \\
$\therefore$ The area of rectangle $DOMN$ is equal to the area of rectangle $EOAP$, \\
$\therefore$ The area of rectangle $EOAP$ is equal to half the area of rectangle $OBCD$, \\
$\therefore$ The area of rectangle $EOAP$ = area of the shaded part = 6, \\
$\therefore k=x\boxed{y=6}$.


Question: As shown in the figure, one of the vertices $A$ of a triangular plate containing a $45^\circ$ angle coincides with the vertex of the rectangle $ABCD$, the right-angled vertex $E$ is located on side $BC$, and another vertex $F$ falls exactly at the midpoint of side $CD$. If $BC = 12$, what is the length of $AB$?

Solution:
\textbf{Solution:} Given that isosceles right triangle AEF,
\[
\therefore AE=EF, \angle AEF=90^\circ,
\]
\[
\therefore \angle AEB+\angle CEF=90^\circ,
\]
Given rectangle ABCD,
\[
\therefore \angle B=\angle C=90^\circ,
\]
\[
\therefore \angle AEB+\angle BAE=90^\circ,
\]
\[
\therefore \angle BAE=\angle CEF,
\]
\[
\therefore \triangle ABE \cong \triangle CEF \text{ (AAS)},
\]
\[
\therefore AB=EC, BE=CF,
\]
Given F is the midpoint of CD and AB=CD,
\[
\therefore BE=\frac{1}{2}AB,
\]
Given BC=12,
\[
\therefore \frac{1}{2}AB + AB = 12,
\]
\[
\therefore AB=\boxed{8}.
\]


Question: As shown in the figure, in $ \triangle ABC$, $ \angle ACB=90^\circ$, and $ \square AEDB$ is constructed outside the triangle with AB as a side, whose diagonals intersect at point F. Given that AE=2 and AB=5, what is the maximum value of CF?

Solution:
\textbf{Solution:} As depicted, let O be the midpoint of AB, and draw lines FO and CO.\\
$\because$ Quadrilateral $AEDB$, AE=2, AB=5,\\
$\therefore$ BD=2, AO=BO=2.5, AF=DF,\\
$\therefore$ OF is the median line of $\triangle ABD$,\\
$\therefore$ FO=1,\\
Also, since $\angle ACB=90^\circ$,\\
$\therefore$ OC=2.5,\\
In $\triangle FOC$, CF $<$ FO+OC,\\
$\therefore$ when points F, O, and C are collinear, CF is maximized,\\
$\therefore$ CFmax=FO+OC=1+2.5=\boxed{3.5}.


Question: As shown in the figure, in square $ABCD$, $E$ lies on the extension of $BC$, $AE$ and $BD$ intersect at point $F$. Connect $FC$. Then, what is the degree measure of $\angle BCF$?

Solution:
\textbf{Solution:} Given square ABCD,\\
it follows that AD$\parallel$BC, $\angle$DCB=$90^\circ$, AD=DC, $\angle$ADF=$\angle$CDF=$45^\circ$,\\
thus $\triangle$ADF$\cong$ $\triangle$CDF (by SAS),\\
which means $\angle$DAF=$\angle$DCF,\\
also, since AD$\parallel$BC, $\angle$E=$32^\circ$,\\
it follows that $\angle$DAF=$32^\circ$,\\
therefore $\angle$DCF=$32^\circ$,\\
thus $\angle$BCF=$\angle$DCB-$\angle$DCF=$90^\circ-32^\circ=\boxed{58^\circ}$.


Question: As shown in the figure, in the right triangle $ABC$, both quadrilaterals $DBFE$ and $GFIH$ are squares. Given that $AD=1\text{cm}$ and $DB=3\text{cm}$, what is the area of the shaded part in $\text{cm}^2$?

Solution:
\textbf{Solution:} As shown in the figure, connect $BE$, $FH$,\\
$\because$ $S_{\triangle ABC}=\frac{1}{2}AB\cdot BC=\frac{1}{2}AB\cdot DE+\frac{1}{2}BC\cdot EF$,\\
$\therefore$ $4BC=12+3BC$,\\
$\therefore$ $BC=12$,\\
$\therefore$ $FC=BC-BF=12-3=9$,\\
$\because$ $S_{\triangle EFC}=\frac{1}{2}EF\cdot FC=\frac{1}{2}EF\cdot GH+\frac{1}{2}FC\cdot HI$,\\
$\therefore$ $27=3GH+9HI$,\\
$\therefore$ $GH=\frac{9}{4}$,\\
$\therefore$ the area of square GFIH is: $\boxed{\frac{81}{16}}$.


Question: In Figure 1, the hypotenuse of the right-angled triangle is 4. Four of these right-angled triangles from Figure 1 are assembled into a square as shown in Figure 2. As illustrated, the four right-angled triangles are assembled into a square, where the areas of the shaded parts are denoted as $S_1$ and $S_2$, respectively. What is the value of $S_1+S_2$?

Solution:
\textbf{Solution}: Let the length of the longer leg of the right triangle be $a$, and the length of the shorter leg be $b$,\\
$\therefore$ $a^{2}+b^{2}=4^{2}$\\
$\because$ $S_{1}=a^{2}$, $S_{2}=b^{2}$\\
$\therefore$ $S_{1}+S_{2}=16$\\
Therefore, the answer is: $\boxed{16}$.


Question: As shown in the figure, $E$ and $F$ are the midpoints of sides $CD$ and $BC$ of rectangle $ABCD$ respectively, with segments $AC$, $AE$, $EF$ connected. Given that $AE \perp EF$ and $AC = 6$, what is the length of $AB$?

Solution:
\textbf{Solution:} Join BD and AF as shown in the diagram: \\
Since quadrilateral ABCD is a rectangle, and $AC=6$, \\
therefore $BD=AC=6$, $\angle ADC=\angle ECF=90^\circ$, $CD=AB$, \\
Because E and F are respectively the midpoints of sides CD and BC of rectangle ABCD, \\
thus $EF=\frac{1}{2}BD=3$, \\
Let $DE=CE=a$, and $CF=BF=b$, then $AB=CD=2a$, $AD=BC=2b$, \\
Because $AE\perp EF$, \\
therefore, in $\triangle ADE$, $\triangle CEF$, $\triangle AEF$, $\triangle ABF$, by applying the Pythagorean theorem, we get: \\
$a^2+4b^2+9=4a^2+b^2$, $a^2+b^2=9$, \\
thus, from the above equations, we get: $a^2=6$, \\
Solving gives: $a=\sqrt{6}$ (negative root is discarded), \\
therefore, $AB=2\sqrt{6}$; \\
Hence, the answer is: $AB=\boxed{2\sqrt{6}}$.


Question: As shown in the diagram, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. If $\angle AOD = 60^\circ$ and $AD = 1$, what is the length of $AB$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\
it follows that $AO=OC$, $BO=DO$, and $AC=BD$,\\
therefore $AO=DO$,\\
since $\angle AOD=60^\circ$,\\
it follows that $\triangle AOD$ is an equilateral triangle,\\
thus $AO=OD=OB=AD=1$,\\
therefore $BD=2$,\\
In the right triangle $ABD$, by the Pythagorean theorem, we can find: $AB=\sqrt{BD^{2}-AD^{2}}=\boxed{\sqrt{3}}$.


Question: As shown in the figure, rectangle OABC is divided into six smaller rectangles by three lines. If D and E are the trisection points on the side CO, and the inverse proportion function $y=\frac{k}{x}\left(k\ne 0\right)$ just passes through the vertices F, G of the smaller rectangles. If the area of the shaded rectangles in the figure $S_{1}+S_{2}=5$, then what is the value of the inverse proportionality constant $k$?

Solution:
\textit{Solution}: Since D and E are the trisection points on side CO, \\
it follows that $S_{1}+S_{2}=\frac{1}{3}S_{\text{rectangle} OABC}$, \\
hence $S_{\text{rectangle} OAGD}=\frac{2}{3}S_{\text{rectangle} OABC}$, \\
therefore $S_{\text{rectangle} OAGD}=2(S_{1}+S_{2})=10$, \\
thus, $k=\boxed{10}$.


Question: As shown in the diagram, the perimeter of rhombus $ABCD$ is 40, and the diagonal $BD=12$. What is the area of the rhombus $ABCD$?

Solution:
\textbf{Solution:} Given that the perimeter of rhombus $ABCD$ is 40,\\
thus, the length of side $BC=10$,\\
since $BD=12$,\\
then $OB=\frac{1}{2}BD=6$,\\
thus $OC=\sqrt{BC^{2}-OB^{2}}=\sqrt{10^{2}-6^{2}}=8$,\\
therefore $AC=2OC=16$,\\
hence the area $S_{\text{rhombus}ABCD}=\frac{1}{2}AC\cdot BD=\frac{1}{2}\times 12\times 16=\boxed{96}$.


Question: As shown in the figure, the diagonals $AC$ and $BD$ of the rhombus $ABCD$ intersect at point $O$. If $AC=8$ and $BD=6$, what is the perimeter of the rhombus?

Solution:
\textbf{Solution:} It is known from the problem that $BD\perp AC$, $AO=4$, and $BO=3$, \\
$\therefore AB= \sqrt{3^2+4^2}=5$, \\
$\therefore$ the perimeter of the rhombus is $4\times 5=\boxed{20}$


Question: As shown in the figure, DE is the median line of $\triangle ABC$, with point F on DE, and $\angle AFB=90^\circ$. If $AB=8$ and $BC=10$, what is the length of EF?

Solution:
\textbf{Solution:} Given that DE is the median of $\triangle ABC$, with BC=10, \\
thus DE=$\frac{1}{2}$BC=5, \\
Given that $\angle AFB=90^\circ$, and D is the midpoint of AB, with AB=8, \\
thus DF=$\frac{1}{2}$AB=4, \\
therefore EF=DE−DF=5−4=\boxed{1}.


Question: As shown in the figure, in rectangle $ABCD$, $E$ is a point on the extension of $BC$. Connect $AC$, $DE$, and let $BE=AC$. If $\angle ACB = 40^\circ$, what is the degree measure of $\angle E$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
it follows that AC=BD, and AC, BD bisect each other,\\
therefore BO=CO,\\
thus $\angle$DBE=$\angle$ACB=40$^\circ$.\\
Since BE=AC,\\
it follows that BD=BE,\\
thus $\angle$E=$ \frac{180^\circ−\angle DBE}{2}=\frac{180^\circ−40^\circ}{2}=\boxed{70^\circ}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AD \parallel BC$, and $\angle B + \angle C = 90^\circ$. Squares $ABEF$, $ADHG$, and $DCJI$ are constructed externally on sides $AB$, $AD$, and $DC$, respectively, and their areas are denoted as $S_{1}$, $S_{2}$, and $S_{3}$ in sequence. If $S_{1} + S_{3} = 4S_{2}$, then what is the value of $\frac{BC}{AD}$?

Solution:
\textbf{Solution:} As shown in the diagram, draw $AQ\parallel CD$ through point A, and let it intersect $BC$ at point Q,\\
$\therefore$ $\angle AQB=\angle QCD$,\\
$\because$ $AD\parallel BC$,\\
$\therefore$ Quadrilateral $ADCQ$ is a parallelogram,\\
$\therefore$ $AD=QC$, $AQ=DC$, $\angle BCD=\angle AEB$,\\
$\because$ $\angle ABC+\angle BCD=90^\circ$,\\
$\therefore$ $\angle ABC+\angle QCD=90^\circ$,\\
$\therefore$ $\angle BAQ=90^\circ$,\\
$\therefore$ $AB^{2}+AQ^{2}=BQ^{2}$, that is, $AB^{2}+DC^{2}=BQ^{2}$,\\
$\therefore$ $S_{1}+S_{3}=BQ^{2}$,\\
Furthermore, $\because$ $S_{1}+S_{3}=4S_{2}$,\\
$\therefore$ $4S_{2}=BQ^{2}=4AD^{2}$,\\
$\therefore$ $BQ=2AD$,\\
$\therefore$ $BC=BQ+QC=2AD+AD=3AD$,\\
$\therefore$ $\frac{BC}{AD}=\frac{3AD}{AD}=\boxed{3}$.


Question: As shown in the figure, locations A and B are separated by a pond. In order to measure the distance between A and B, an appropriate point C outside of AB is chosen. The lines AC and BC are connected, and the midpoints E and F of the line segments AC and BC, respectively, are identified. It is measured that $EF = 6m$. What is the distance between A and B in meters?

Solution:
\textbf{Solution:} Since $E$ and $F$ are the midpoints of $AC$ and $BC$ respectively,\\
therefore, $EF$ is a midsegment of $\triangle ABC$,\\
therefore, $AB = 2EF = \boxed{12}$m.


Question: As shown in the figure, the quadrilateral $ABCO$ is a rectangle. The coordinates of point $A$ are $(4, 0)$, and the coordinates of point $C$ are $(0, 8)$. When the rectangle $ABCO$ is folded along $OB$, point $A$ falls on point $D$. What are the coordinates of point $D$?

Solution:
\textbf{Solution:} Draw $DE \perp x$-axis at $E$, as shown in the diagram:\\
$\because$ Quadrilateral $ABCO$ is a rectangle, where the coordinates of point $A$ are $(4, 0)$, and the coordinates of point $C$ are $(0, 8)$,\\
$\therefore OA = BC = 4, AB = OC = 8$,\\
$\because$ Folding rectangle $ABCO$ along $OB$,\\
$\therefore \triangle ABO \cong \triangle DBO$,\\
$\therefore \angle FDO = 90^\circ, OD = OA = 4$,\\
In $\triangle DFO$ and $\triangle CFB$,\\
$\left\{\begin{array}{l}
\angle BFC = \angle OFD\\
\angle FDO = \angle BCF = 90^\circ\\
DO = 4 = BC
\end{array}\right.$,\\
$\therefore \triangle DFO \cong \triangle CFB (AAS)$,\\
$\therefore FD = FC$,\\
Let $FD = FC = x$, then $FO = OC - FC = 8 - x$,\\
In $\triangle DFO$, where $\angle FDO = 90^\circ, DF = x, OF = 8 - x, DO = 4$, according to the Pythagorean theorem $OF^{2} = DF^{2} + DO^{2}$, we get $(8-x)^{2} = x^{2} + 4^{2}$, solving for $x$ yields $x = 3$,\\
$\therefore S_{\triangle DFO} = \frac{1}{2}DF \cdot DO = \frac{1}{2}FO \cdot OE$, substituting values we get $OE = \frac{12}{5}$,\\
In $\triangle DEO$, where $\angle DEO = 90^\circ, OE = \frac{12}{5}, DO = 4$, according to the Pythagorean theorem, we get $DE = \sqrt{DO^{2} - OE^{2}} = \sqrt{4^{2} - \left(\frac{12}{5}\right)^{2}} = \frac{16}{5}$,\\
$\because$ Point $D$ is in the second quadrant,\\
$\therefore D\left(-\frac{12}{5}, \frac{16}{5}\right)$,\\
Hence, the answer is: $\boxed{\left(-\frac{12}{5}, \frac{16}{5}\right)}$.


Question: As shown in the diagram, within the rectangle $ABCD$, point $E$ lies on the extension of $AB$. If $BE=AC=6$ and $\angle E=15^\circ$, then what is the length of $AB$?

Solution:
\textbf{Solution:} As shown in the figure, connect BD, \\
since quadrilateral ABCD is a rectangle, \\
therefore BD=AC, \\
also since BE=AC, \\
therefore BD=AC=BE=6, \\
therefore $\angle E=\angle BDE$, \\
therefore $\angle ABD=\angle E+\angle BDE=2\angle E=30^\circ$, \\
therefore AB=$BD\cos\angle ABD$=6×$\frac{\sqrt{3}}{2}$=$\boxed{3\sqrt{3}}$.


Question: As shown in the figure, squares are constructed externally on each side of the right-angled triangle $ABC$. Line segment $CG$ is drawn, where square $ACDE$ and square $ABGF$ have areas of 1 and 5, respectively. What is the length of $CG$?

Solution:
\textbf{Solution:} Draw line AH, \\
since the areas of square ACDE and square ABGF are 1 and 5, respectively, \\
therefore, AC$^{2}=1$, AB$^{2}=5$, GB=AB, \\
thus, AC=1, \\
since $\angle$ACB$=90^\circ$, \\
therefore, BC$^{2}=AB^{2}-AC^{2}=5-1=4$, \\
hence, BC=2, \\
since quadrilateral BDIH is a square, \\
therefore, CI=HI=BC=2, $\angle$ICB$=90^\circ$, CB=HB, \\
thus, $\angle$ACB+$\angle$ICB$=180^\circ$, \\
hence, points A, C, and I are on the same line, \\
thus, AI=AC+CI=1+2=3, \\
therefore, HA=$\sqrt{HI^{2}+AI^{2}}=\sqrt{2^{2}+3^{2}}=\sqrt{13}$, \\
since $\angle$ABG=$\angle$CBH$=90^\circ$, \\
therefore, $\angle$GBC=$\angle$ABH$=90^\circ+\angle$ABC, \\
in $\triangle GBC$ and $\triangle ABH$, \\
$\left\{\begin{array}{l}GB=AB \\ \angle GBC=\angle ABH \\ CB=HB \end{array}\right.$, \\
thus, $\triangle GBC\cong \triangle ABH$ (SAS), \\
therefore, CG=HA=$\sqrt{13}$, \\
hence, the length of CG is $\boxed{\sqrt{13}}$.


Question: In the figure, inside $\triangle ABC$, the median from $AB$ intersects side $AC$ at point $E$, with $BE=5$ and $CE=3$. What is the length of $AC$?

Solution:
Solution: Since the perpendicular bisector of AB intersects side AC at point E, and BE = 5, \\
it follows that AE = BE = 5. \\
Since CE = 3 and $AC = AE + CE$, \\
it follows that $AC = \boxed{8}$.


Question: As shown in the figure, in $\triangle ABC$, the altitude $AE$ intersects $BC$ at point $E$. If $\frac{1}{2}\angle ABE + \angle C = 45^\circ$, $CE = 5\sqrt{2}$, and the area of $\triangle ABC$ is 20, what is the length of $AB$?

Solution:
\textbf{Solution:} Draw CD $\perp$ BA from point C, intersecting the extension line of BA at point D,\\
thus $\angle ADC = 90^\circ$,\\
since $\frac{1}{2}\angle ABE + \angle ACB = 45^\circ$,\\
thus $\angle ABE + 2\angle ACB = 90^\circ$,\\
since $\angle ABE + \angle ACB + \angle ACD = 90^\circ$,\\
thus $\angle ACB = \angle ACD$,\\
since AE is the altitude of $\triangle ABC$,\\
thus $\angle AEC = \angle ADC = 90^\circ$,\\
in $\triangle AEC$ and $\triangle ADC$,\\
$\left\{\begin{array}{l}
\angle AEC=\angle ADC \\
\angle ACE=\angle ACD \\
AC=AC
\end{array}\right.$\\
thus $\triangle AEC\cong \triangle ADC$ (AAS),\\
thus CE$=CD=5\sqrt{2}$,\\
since $S_{\triangle ABC}=\frac{1}{2}\text{AB}\cdot\text{CD}$,\\
thus $20=\frac{1}{2}\times\text{AB}\times 5\sqrt{2}$,\\
thus AB$=\boxed{4\sqrt{2}}$.


Question: As shown in the figure, in $\triangle ABC$, it is known that points D, E, and F are the midpoints of sides BC, AD, and CE, respectively, and the area of $\triangle ABC$ is $8\,\text{cm}^2$. What is the area of the shaded region $\triangle BEF$ in $\text{cm}^2$?

Solution:
\textbf{Solution:}

Since point D is the midpoint of side BC,\\
it follows that AD is the median of $\triangle ABC$,\\
hence, the area of $\triangle ABD$ equals the area of $\triangle ADC$ equals $\frac{1}{2}$ the area of $\triangle ABC$ equals $\frac{1}{2} \times 8 = 4$;\\
Since E is the midpoint of side AD,\\
then BE and CE are the medians,\\
hence, the area of $\triangle EBD$ equals the area of $\triangle EDC$ equals $\frac{1}{2}$ the area of $\triangle ABD$ equals $\frac{1}{2} \times 4 = 2$;\\
therefore, the area of $\triangle BEC$ equals $2 + 2 = 4$,\\
Similarly, it is known that BF is the median of $\triangle BEC$,\\
therefore, the area of $\triangle BEF$ equals $\frac{1}{2} \times 4 = \boxed{2}$.


Question: As shown in the figure, what is the sum of $\angle A+\angle B+\angle C+\angle D$?

Solution:
\textbf{Solution}: Draw lines AD and BC.\\
In quadrilateral ABCD, we have $\angle DAE + \angle EAB + \angle ABF + \angle FBC + \angle DCF + \angle BCF + \angle CDE + \angle ADE = 360^\circ$,\\
Since $\angle DEA = 105^\circ$ and $\angle BFC = 120^\circ$,\\
It follows that $\angle DAE + \angle ADE = 180^\circ - 105^\circ = 75^\circ$ and $\angle FBC + \angle BCF = 180^\circ - 120^\circ = 60^\circ$,\\
Therefore, $\angle EAB + \angle ABF + \angle DCF + \angle CDE = 360^\circ - 75^\circ - 60^\circ = 225^\circ$.\\
Hence, $\angle A + \angle B + \angle C + \angle D = \boxed{225^\circ}$.


Question: As shown in the figure, in the isosceles $\triangle ABC$, $\angle ACB=90^\circ$, point D is a moving point on the ray CB, AE=AD, and AE$\perp$AD. BE intersects the line containing AC at point P. If AC=5PC, then what is the value of $ \frac{BD}{CD}$?

Solution:
\textbf{Solution:} When point D is on the extension of CB, draw EM $\perp$ AC at point M on the extension of AC,\\
$\therefore \angle$AME=$\angle$ACB=$90^\circ$,\\
$\because$AE$\perp$AD,\\
$\therefore \angle$DAC+$\angle$EAM=$90^\circ$, $\angle$EAM+$\angle$AEM=$90^\circ$,\\
$\therefore \angle$DAC=$\angle$AEM,\\
In $\triangle ADC$ and $\triangle AEM$\\
$ \left\{\begin{array}{l}
\angle DAC=\angle AEM\\
\angle AME=\angle ACB\\
AE=AD
\end{array}\right.$,\\
$\therefore \triangle ADC \cong \triangle AEM$ (AAS),\\
$\therefore AC=ME$,\\
$\because \triangle ABC$ is an isosceles right triangle,\\
$\therefore AC=BC$,\\
$\therefore BC=ME$,\\
In $\triangle PME$ and $\triangle PBC$\\
$ \left\{\begin{array}{l}
\angle PCB=\angle PME\\
\angle BPC=\angle EPM\\
BC=ME
\end{array}\right.$,\\
$\therefore \triangle PME \cong \triangle PBC$ (AAS),\\
$\therefore PC=PM,\\
Let PC=PM=x, then AC=BC=5x, AM=DC=AC+PC+PM=5x+x+x=7x$,\\
$\therefore DB=DC−BC=7x−5x=2x$,\\
$\therefore \frac{BD}{CD}=\frac{2x}{7x}=\boxed{\frac{2}{7}}$;\\
When point D is on the line segment CD, draw ME $\perp$ AC at point M,\\
Similarly, it can be proven that $\triangle ADC \cong \triangle AEM$, $\triangle PME \cong \triangle PBC$,\\
$\therefore AC=ME, PC=PM,\\
Let PC=PM=x, then AC=BC=5x, AM=DC=AC−PC−PM=5x−x−x=3x$,\\
$\therefore DB=BC−DC=5x−3x=2x$,\\
$\therefore \frac{BD}{CD}=\frac{2x}{3x}=\boxed{\frac{2}{3}}$;


Question: As shown in the figure, in $\triangle ABC$, DE is the perpendicular bisector of AC, AB=5, BC=7, then what is the perimeter of $\triangle ABD$?

Solution:
Solution: Since $DE$ is the perpendicular bisector of $AC$, \\
$\therefore$ $AD=DC$, \\
$\therefore$ the perimeter of $\triangle ABD$ is $AD+BD+AB=DC+BD+AB=BC+AB=5+7=\boxed{12}$


Question: As shown in the diagram, D is a point on the bisector of $\angle MAN$. Point B lies on ray AM, with $DE \perp AM$ at point E, and $DF \perp AN$ at point F. Connect AD. Take a point C on ray AN such that $DC=DB$. If $AB=7$ and $BE=2$, what is the length of $AC$?

Solution: \textbf{Solution:} As shown in the figure, since $D$ is a point on the bisector of $\angle MAN$, and $DE \perp AM$ at point $E$, $DF \perp AN$ at point $F$, therefore $\angle DEB = \angle DFC = 90^\circ$, $DE=DF$, and $\angle BAD=\angle CAD$. In right triangles $\triangle DEB$ and $\triangle DFC$, we have $\left\{\begin{array}{l}BD=DC\\ DE=DF\end{array}\right.$, therefore right $\triangle DEB \cong$ right $\triangle DFC$ (HL), hence $\angle DBE=\angle DCF$, and therefore $\angle ABD=\angle ACD$. In $\triangle ABD$ and $\triangle ACD$, we have $\left\{\begin{array}{l}\angle BAD=\angle CAD\\ \angle ABD=\angle ACD\\ DB=DC\end{array}\right.$, therefore $\triangle ABD \cong \triangle ACD$ (AAS), hence $AC=AB=7$. In right triangles $\triangle DEA$ and $\triangle DFA$, we have $\left\{\begin{array}{l}AD=DA\\ DE=DF\end{array}\right.$, therefore right $\triangle DEA \cong$ right $\triangle DFA$ (HL), hence $AF=AE=2+7=9$, therefore $AC=AF+CF=9+2=11$. In summary, the length of $AC$ is $\boxed{7}$ or $\boxed{11}$.