Question: As shown in the figure, rectangle $ABCD$ is folded along $EF$, such that point $B$ exactly coincides with point ${B}^{\prime}$ on side $AD$. If $AE=5$ and $\angle EFB=60^\circ$, then what is the length of $EF$?

Solution:
\textbf{Solution:} Since it is folded,\\
$\therefore \angle EFB = \angle EFB' = 60^\circ$,\\
$\therefore \angle CFB' = 180^\circ - \angle EFB - \angle EFB' = 60^\circ$,\\
Since quadrilateral ABCD is a rectangle,\\
$\therefore AD \parallel BC$,\\
$\therefore \angle EB'F = \angle CFB' = 60^\circ$,\\
$\therefore \triangle EFB'$ is an equilateral triangle,\\
$\therefore EB' = EF$,\\
Since $\angle A'B'E = 90^\circ - \angle EB'F = 90^\circ - 60^\circ = 30^\circ$,\\
And $\angle EA'B' = 90^\circ$,\\
$\therefore EB' = 2EA' = 10$,\\
$\therefore EF = EB' = \boxed{10}$.


Question: Given the figure, in the Cartesian coordinate system, the midpoint of diagonal $AC$ of rectangle $ABCD$ coincides with the origin. Point $E$ is a point on the $x$-axis, and $AE$ is connected. If $AD$ bisects $\angle OAE$, the graph of the inverse proportion function $y=\frac{k}{x}$ ($k>0$, $x>0$) passes through two points $A$ and $F$ on $AE$, and $AF=EF$, the area of $\triangle ABE$ is $18$, then what is the value of $k$?

Solution:
\textbf{Solution:} As shown in the figure, connect BD and AC at point O. Draw AG $\perp$ x-axis at point G from point A, and draw FH $\perp$ x-axis at point H from point F. Connect OF, \\
$\therefore$ FH $\parallel$ AG, \\
$\because$ AE=EF, \\
$\therefore$ FH is the median line of $\triangle$ AGE, \\
$\therefore$ GH=HE, AG=2FH \\
$\because$ points A and F are on the graph of the inverse proportion function y=$\frac{k}{x}$ (k>0, x>0), \\
$\therefore$ S$_{\triangle AOG}$=S$_{\triangle FOH}$=$\frac{k}{2}$, \\
$\therefore$ OG·AG=OH·FH, \\
$\therefore$ OH=2OG, \\
$\therefore$ OG=GH=HE, \\
$\because$ of the rectangle ABCD, \\
$\therefore$ OA=OD, \\
$\therefore$ $\angle$OAD=$\angle$ODA, \\
Additionally, $\because$ AD bisects $\angle$OAE, \\
$\therefore$ $\angle$OAD=$\angle$EAD, \\
$\therefore$ $\angle$ODA=$\angle$EAD, \\
$\therefore$ AE $\parallel$ BD, \\
$\therefore$ S$_{\triangle AOE}$=S$_{\triangle ABE}$=18, \\
$\therefore$ S$_{\triangle AOG}$=$\frac{1}{3}$S$_{\triangle AOE}$=6, \\
$\therefore$ $\frac{k}{2}$=6, \\
$\therefore$ k=\boxed{12}.


Question: As shown in the diagram, in the rhombus $ABCD$, $\angle ABC=30^\circ$. Connect $AC$. Follow the steps below to draw: Take $C$ and $B$ as centers, and with the length of $BC$ as the radius, draw arcs. The arcs intersect at point $P$. Connect $BP$ and $CP$. What is the measure of $\angle ACP$?

Solution:
\textbf{Solution:} According to the steps given in the question, the diagram is completed as follows:\\
Since rhombus ABCD, $\angle$ABC=30$^\circ$,\\
therefore, $\angle$BCD=150$^\circ$,\\
thus, $\angle$ACB=75$^\circ$,\\
(1) When point P is above BC,\\
since BP=PC=BC,\\
thus, $\triangle BPC$ is an equilateral triangle,\\
therefore, $\angle$PCB=60$^\circ$,\\
hence, $\angle$ACP=$\angle$ACB−$\angle$PCB=75$^\circ$−60$^\circ$=\boxed{15^\circ};\\
(2) When point P' is below BC,\\
since BP'=P'C=BC,\\
thus, $\triangle BP'C$ is an equilateral triangle,\\
therefore, $\angle$P'CB=60$^\circ$,\\
hence, $\angle$ACP'=$\angle$ACB+$\angle$P'CB=75$^\circ$+60$^\circ$=135$^\circ$,\\
In conclusion, the degree of $\angle$ACP is either $\boxed{15^\circ}$ or 135$^\circ$.


Question: As shown in the figure, given a rectangle $ABCD$ with two sides $AB=6$ and $AD=8$, point $E$ is the intersection of diagonals $AC$ and $BD$, point $P$ is a moving point on side $AD$. Construct point $D'$, which is the symmetric point of $D$ with respect to line $PE$. When $ED'$ is perpendicular to a side of the rectangle, what is the length of $PD$?

Solution:
\textbf{Solution:} As shown in the figure, when $ED^{\prime} \perp AD$,

$\because$ rectangle ABCD,

$\therefore$ AE=CE, CD$\perp$AD, AB=CD=6,

$\therefore$ EF$\parallel$CD,

$\therefore$ EF is the median of $\triangle ACD$,

$\therefore$ EF=$\frac{1}{2}CD=\frac{1}{2}\times 6=3$, DF=$\frac{1}{2}AD=4$, AC=BD=2DE, in $\triangle ADC$

$AC=\sqrt{AD^{2}+CD^{2}}=\sqrt{8^{2}+6^{2}}=10$,

$\therefore$ DE=5

$\because$ point D is symmetrical to point D$^{\prime}$ regarding line PE,

$\therefore$ DE=ED$^{\prime}=5$, DP=D$^{\prime}P$,

$\therefore$ FD$^{\prime}=5-3=2$, let DP=D$^{\prime}P=x$, then PF=4-x, in $\triangle PFD^{\prime}$

$D^{\prime}F^{2}+PF^{2}=D^{\prime}P^{2}$, which implies $x^{2}=(4-x)^{2}+4$

Solving this: $x=\boxed{\frac{5}{2}}$;

When $D^{\prime}E\perp AB$

$\because$ DA$\perp$AB,

$\therefore$ AD$\parallel D^{\prime}E$,

$\therefore \angle DPE=\angle PED^{\prime}$,

$\because$ point D is symmetrical to point D$^{\prime}$ regarding line PE,

$\therefore \triangle DPE$ and $\triangle D^{\prime}PE$ are symmetrical regarding line PE,

$\therefore \angle DPE=\angle D^{\prime}PE$,

$\therefore \angle PED^{\prime}=\angle D^{\prime}PE$,

$\therefore DP^{\prime}=DP=D^{\prime}E=5$.

Hence, the length of PD is $\boxed{\frac{5}{2}}$ or 5.


Question: As shown in the diagram, it is known that in the Cartesian coordinate system $xOy$, an isosceles right-angled triangle $ABC$ has its hypotenuse $AB \perp x$-axis at point $A$. The graph of the inverse proportion function $y = \frac{k}{x}$ ($k>0$, $x>0$) passing through point $B$ intersects side $AC$ at point $D$. Connect $OB$ and $OC$. If point $D$ is the midpoint of $AC$ and the area of $\triangle OBC$ is $1$, then what is the value of $k$?

Solution:
\textbf{Solution:}

Draw line $EF \perp y$-axis at point F through point B. Draw line $CE \perp EF$ at point E through point C. Draw line $CM \perp AB$ at point M through point C.\\
Therefore, $\angle ABE = \angle E = \angle BMC = 90^\circ$,\\
Thus, quadrilateral $BMCE$ is a rectangle,\\
Since isosceles right triangle $\triangle ABC$ has hypotenuse $AB \perp x$-axis at point A,\\
Therefore, $\angle ABC = 45^\circ$,\\
Therefore, $\angle EBC = 45^\circ$,\\
Therefore, $\triangle EBC$ is an isosceles right triangle,\\
Therefore, $BE = CE$,\\
Therefore, quadrilateral $BMCE$ is a square,\\
Therefore, $BE = CE = CM = \frac{1}{2}AB$,\\
Since the graph of the inverse proportion function $y = \frac{k}{x}$ ($k>0$, $x>0$) passing through point B intersects side $AC$ at point D,\\
Let point A be $(x,0)$, then $B\left(x, \frac{k}{x}\right)$\\
Therefore, $BE = CE = CM = \frac{k}{2x}$, $AB = \frac{k}{x}$,\\
Therefore, point $C\left(x+\frac{k}{2x}, \frac{k}{2x}\right)$\\
Since point D is the midpoint of $AC$,\\
Therefore, $D\left(x+\frac{k}{4x}, \frac{k}{4x}\right)$,\\
Since $S_{\triangle OBC} = S_{\text{trapezoid} EFOC} - S_{\triangle BFO} - S_{\triangle BEC} = 1$\\
Therefore, $\frac{\left(\frac{k}{2x}+\frac{k}{x}\right)\left(x+\frac{k}{2x}\right)}{2}-\frac{1}{2}x\cdot\frac{k}{x}-\frac{1}{2}\cdot\frac{k}{2x}\cdot\frac{k}{2x}=1$\\
Rearranging gives $k^2 = (4-k)x^2$;\\
Since points B and D are on the graph of the inverse proportion function,\\
Therefore, $k = \left(x+\frac{k}{4x}\right)\cdot\frac{k}{4x}$\\
Rearranging gives $x^2 = \frac{k}{12}$\\
Substituting the above equation yields $k^2 = (4-k) \times \frac{k}{12}$\\
Since $k \neq 0$\\
Solving yields: $k = \boxed{\frac{4}{13}}$.


Question: As shown in the figure, in the rhombus $ABCD$, the perpendicular bisector of $AD$ intersects $AD$ at point $E$, and intersects $AC$ at point $F$, $\angle ABC=4\angle DFC$, what is the measure of $\angle BAD$?

Solution:
\textbf{Solution:} As shown in the diagram,\\
$\because EF$ perpendicularly bisects $AD$.\\
$\therefore AF=DF$.\\
$\therefore \angle FAD=\angle FDA$.\\
$\because \angle DFC=\angle FAD+\angle FDA$.\\
$\therefore \angle FDC=2\angle FAD$.\\
In rhombus $ABCD$, $AC$ is the diagonal,\\
$\therefore \angle BAD=2\angle FAD$.\\
$\therefore \angle BAD=\angle DFC$.\\
$\because \angle ABC=4\angle DFC$.\\
$\therefore \angle ABC=4\angle BAD$.\\
$\because \angle BAD+\angle ABC=180^\circ$.\\
$\therefore \angle BAD+4\angle BAD=180^\circ$.\\
$\therefore \angle BAD=\boxed{36^\circ}$


Question: As shown in the figure, points $E$, $F$, $G$, and $H$ are the midpoints of the sides of square $ABCD$, and lines $BE$, $DG$, $CF$, $AH$ are connected. If $AB=10$, what is the area of quadrilateral $MNPQ$?

Solution:
\textbf{Solution:} Given that points $E$, $F$, $G$, $H$ are midpoints of the sides of the square $ABCD$,\\
$\therefore DE=BG=DH=CG=\frac{1}{2}AB$, and $DE\parallel BG$,\\
$\therefore$ quadrilateral $DEBG$ is a parallelogram,\\
$\therefore BE\parallel DG$,\\
$\therefore AM=QM$,\\
Similarly, $DQ=PQ$ can be deduced,\\
In $\triangle ADH$ and $\triangle DCG$,\\
$\left\{\begin{array}{l}AD=DC\\ \angle ADH=\angle DCG\\ DH=CG\end{array}\right.$,\\
$\therefore \triangle ADH \cong \triangle DCG\ (SAS)$,\\
$\therefore \angle DAH=\angle CDG$,\\
$\because \angle CDG+\angle ADQ=90^\circ$,\\
$\therefore \angle DAH+\angle ADQ=90^\circ$,\\
$\therefore \angle AQD=90^\circ$,\\
Similarly, it can be deduced that $\angle BMA=\angle CNB=\angle DPC=90^\circ$,\\
$\therefore$ quadrilateral $MNPQ$ is a rectangle,\\
In $\triangle ADQ$ and $\triangle DCP$,\\
$\left\{\begin{array}{l}\angle AQD=\angle DPC\\ \angle DAQ=\angle CDP\\ AD=DC\end{array}\right.$,\\
$\therefore \triangle ADQ \cong \triangle DCP\ (AAS)$,\\
$\therefore AQ=DP$,\\
$\therefore AM=MQ=DQ=PQ$,\\
$\therefore$ quadrilateral $MNPQ$ is a square,\\
Let $MQ=x$, then $AQ=2x$, $DQ=x$,\\
In right triangle $\triangle ADQ$, $x^2+(2x)^2=100$,\\
$\therefore x^2=20$,\\
$\therefore$ the area of quadrilateral $MNPQ$ is $\boxed{20}$.


Question: As shown in the figure, points $E$ and $F$ are located on sides $AD$ and $CD$ of the square $ABCD$, respectively, and $OE \perp OF$. Given that $AD = 6$, what is the area of the shaded region in the figure?

Solution:
\textbf{Solution:}

Since quadrilateral ABCD is a square, \\
$\therefore \angle EDO = \angle FCO$, AC $\perp$ BD, OD = $ \frac{1}{2}$ BD, OC = $ \frac{1}{2}$ AC, and AC = BD, \\
$\therefore \angle DOC = 90^\circ$, and OD = OC, \\
Since OE $\perp$ OF, \\
$\therefore \angle EOF = 90^\circ$, \\
$\therefore \angle DOE = \angle COF$, \\
Thus, $\triangle ODE \cong \triangle OCF$ (ASA), \\
Therefore, the area of the shaded part in the figure = Area of $\triangle AOD$ = $ \frac{1}{4}$ Area of square ABCD, \\
Since AD = 6, \\
Therefore, the area of the shaded part in the figure = $ \frac{1}{4} \times 6^{2}$ = $\boxed{9}$.


Question: As shown in the figure, in rectangle $ABCD$, if $AB=7$ and $\angle DBC=30^\circ$, what is the length of $AC$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a rectangle, and $AB=7$,

hence $CD=AB=7$, $AC=BD$, and $\angle BCD=90^\circ$,

since in $\triangle BCD$, $\angle DBC=30^\circ$,

it follows that $BD=2CD=14$,

therefore $AC=BD=\boxed{14}$.


Question: As shown in the figure, the side length of square $ABCD$ is $4$, $E$ is the midpoint of side $AD$, connect $BE$, fold $\triangle ABE$ along line $BE$ to $\triangle FBE$, extend $EF$ to intersect $CD$ at point $G$, what is the length of $CG$?

Solution: \textbf{Solution:} As shown in the diagram, construct line $BG$. Because quadrilateral $ABCD$ is a square, therefore $AB=BC=4$, $\angle A=\angle ABC=\angle C=90^\circ$. From folding, $AB=BF=BC=4$, $AE=EF=\frac{1}{2}AD=2=DE$, $\angle A=\angle BFE=90^\circ=\angle C$. Because $\angle BFE+\angle BFG=180^\circ$, therefore $\angle C=\angle BFG=90^\circ$. In $\triangle BFG$ and $\triangle BCG$, we have $\left\{\begin{array}{c}BF=BC\\ BG=BG\end{array}\right.$, therefore $\triangle BFG\cong \triangle BCG$ (HL), therefore $FG=CG$. Let $CG=x$, then $DG=4-x$, $EG=2+x$. In $\triangle DEG$, by the Pythagorean theorem, $EG^{2}=DE^{2}+DG^{2}$, solving $(2+x)^{2}=2^{2}+(4-x)^{2}$ yields $x=\boxed{\frac{4}{3}}$, which means the length of $CG$ is $\frac{4}{3}$.


Question: As shown in the figure, the side length of square $ABCD$ and square $EFCG$ are respectively 3 and 1, with points $F$ and $G$ respectively on sides $BC$ and $CD$. Let $P$ be the midpoint of $AE$. When connecting $PG$, what is the length of $PG$?

Solution:
\textbf{Solution:} Extend $GE$ to meet $AB$ at point $O$, and draw $PH \perp OE$ at point $H$,\\
since square $ABCD$, and square $EFCG$,\\
therefore, $EF=1$, $AB=3$, quadrilateral $EFBO$, $ODG$ are rectangles,\\
therefore, $EF=OB=1$ therefore $AO=BF=AB-OB=3-1=2$;\\
since $PH\parallel AB$, and since $P$ is the midpoint of $AE$,\\
therefore, $PH$ is the midline of $\triangle AOE$,\\
therefore, $PH=\frac{1}{2}OA=\frac{1}{2}\times 2=1$,\\
since in $\triangle AOE$, $\angle OAE=45^\circ$,\\
therefore, $\triangle AOE$ is an isosceles right triangle,\\
therefore, $OA=OE=2$, similarly, it is known that $HE=PH=1$.\\
therefore, $HG=HE+EG=1+1=2$.\\
In $\triangle PHG$,\\
$PG=\sqrt{PH^{2}+HG^{2}}=\sqrt{1^{2}+2^{2}}=\boxed{\sqrt{5}}$.


Question: As shown in the figure, in rectangle ABCD, AB = 2, $\angle DAC=30^\circ$, the point M is the midpoint of side BC, point P is a moving point on the diagonal AC ($0<CP<1.5$). Triangle CPM is folded along PM, such that point C falls on point C$^{\prime}$, segment MC$^{\prime}$ intersects AC at point N. Segment AC$^{\prime}$ is drawn. When $\triangle ANC^{\prime}$ forms a right triangle, what is the length of segment AC$^{\prime}$?

Solution:
\textbf{Solution:} When $\angle ANC'$ is a right angle, $\triangle ANC'$ is a right-angled triangle, then $\angle CNM=90^\circ$,\\
$\because$ in the rectangle ABCD, AB=2, $\angle DAC=30^\circ$,\\
$\therefore \angle ACB=30^\circ$, AC=2AB=4, BC=$\sqrt{AC^2-AB^2}=2\sqrt{3}$,\\
$\because$ point M is the midpoint of side BC, $\therefore MC=\frac{1}{2}BC=\sqrt{3}$,\\
In $\triangle CNM$, $\angle CNM=90^\circ$, $\angle MCN=30^\circ$,\\
$\therefore MN=\frac{1}{2}MC=\frac{\sqrt{3}}{2}$, NC=$\sqrt{MC^2-MN^2}=\frac{3}{2}$,\\
$\therefore AN=AC-NC=\frac{5}{2}$,\\
As folded $MC'=MC=\sqrt{3}$,\\
$\therefore NC'=MC'-MN=\frac{\sqrt{3}}{2}$,\\
In $\triangle ANC'$, $\angle ANC'=90^\circ$,\\
$\therefore AC'=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=\sqrt{7}$;\\
When $\angle AC'N=90^\circ$, $\triangle ANC'$ is a right-angled triangle, then $\angle AC'M=90^\circ$,\\
Connecting AM,\\
In $\triangle ABM$, AB=2, MC=BM=$\sqrt{3}$,\\
AM=$\sqrt{2^2+(\sqrt{3})^2}=\sqrt{7}$,\\
As folded $MC'=MC=\sqrt{2}$,\\
$\therefore AC'=\sqrt{(\sqrt{7})^2-(\sqrt{3})^2}=2$;\\
In conclusion, the length of the line segment AC' is $\boxed{2}$ or $\sqrt{7}$.


Question: As shown in the figure, in $\triangle ABC$, $D$ and $E$ are the midpoints of $AB$ and $AC$, respectively. Point $F$ is on segment $DE$, and $AF\perp BF$. If $AB=4$ and $BC=7$, what is the length of $EF$?

Solution:
\textbf{Solution:} Since points D and E are the midpoints of sides AB and AC, respectively,

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

Since BC = 7,

we have DE = $\frac{1}{2}$BC = $\frac{1}{2} \times 7 = \frac{7}{2}$.

Since AF$\perp$BF and D is the midpoint of AB, with AB = 4,

it follows that DF = $\frac{1}{2}$AB = $\frac{1}{2} \times 4 = 2$.

Therefore, EF = DE - DF = $\frac{7}{2} - 2 = \boxed{\frac{3}{2}}$.


Question: In the figure, within $\triangle ABC$, $\angle ACB=90^\circ$, CD is the median. Extend CB to point E such that BE=BC; connect DE, let F be the midpoint of DE, and connect BF. If AC=8, BC=6, what is the length of BF?

Solution:
\textbf{Solution:} In right-angled triangle $\triangle ABC$, $\angle ACB=90^\circ$, $AC=8$, and $BC=6$,\\
$\therefore AB=\sqrt{AC^{2}+BC^{2}}=10$,\\
$\because CD$ is the median to the hypotenuse $AB$,\\
$\therefore CD=\frac{1}{2}AB=5$,\\
$\because BE=BC$, and $F$ is the midpoint of $DE$,\\
$\therefore BF=\frac{1}{2}CD=\boxed{\frac{5}{2}}$.


Question: As shown in the figure, with the hypotenuse $AB$ of the right-angled $\triangle ABC$ as one side, construct a square $ABED$ on the right side of $AB$, with the diagonals of the square intersecting at point $O$. Connect $CO$. If $AC=4$ and $CO=6$, what is the length of $BC$?

Solution:
\textbf{Solution:} As illustrated, extend $CB$ to the point $G$, such that $BG=AC=4$,\\
$\because$ according to the problem, quadrilateral $ABED$ is a square,\\
$\therefore \angle 4=\angle 5=45^\circ$, $\angle$EBA$=90^\circ$,\\
$\therefore \angle 1+\angle 2=90^\circ$,\\
Furthermore, $\because \triangle ABC$ is a right-angled triangle with $AB$ being the hypotenuse,\\
$\therefore \angle 2+\angle 3=90^\circ$,\\
$\therefore \angle 1=\angle 3$,\\
$\because \angle 1+\angle 5=\angle 3+\angle 4$,\\
$\therefore \angle CAO=\angle GBO$,\\
In $\triangle CAO$ and $\triangle GBO$,\\
$\left\{\begin{array}{l}CA=GB \\ \angle CAO=\angle GBO \\ AO=BO \end{array}\right.$,\\
$\therefore \triangle CAO \cong \triangle GBO$ (SAS),\\
$\therefore CO=GO=6$, $\angle 6=\angle 8$,\\
$\because \angle 7+\angle 8=90^\circ$,\\
$\therefore \angle 6+\angle 7=90^\circ$,\\
$\therefore \angle COG=90^\circ$,\\
$\therefore CG=\sqrt{6^2+6^2}=6\sqrt{2}$,\\
$\therefore BC=CG-BG=6\sqrt{2}-4$.\\
Thus, the answer is: $BC=\boxed{6\sqrt{2}-4}$.


Question: As shown in the figure, the side length of square $ABCD$ is $4$, and point $E$ is the midpoint of $AD$. When $\triangle ABE$ is folded along the line containing $BE$ to obtain $\triangle FBE$, and $BF$ is extended to intersect $CD$ at point $G$, what is the length of $DG$?

Solution:
\textbf{Solution:}
As shown in the figure, connect EG,

$\because$ $\triangle FBE$ is obtained by folding $\triangle ABE$,

$\therefore$ AE=FE, $\angle EFB=\angle EAB=90^\circ$,

$\therefore$ $\angle EFG=\angle EDG=90^\circ$.

$\because$ AE=DE,

$\therefore$ FE=DE.

$\because$ EG=EG,

$\therefore$ Rt$\triangle EFG \cong$ Rt$\triangle EDG$ (HL).

$\therefore$ DG=FG,

Let DG=FG=x,

$\because$ AB=BF=4,

According to the problem: BG=4+x, CG=4-x,

By the Pythagorean theorem: BG$^{2}$=BC$^{2}$+CG$^{2}$,

We get: $(4+x)^{2}=4^{2}+(4-x)^{2}$,

Solve to get: $x=\boxed{1}$.

$\therefore$ DG=1,


Question: As shown in the figure, the side length of the square $ABCD$ is $8$, with point $E$ on $AB$ and $BE=2$. Points $M$ and $N$ are on $AC$ such that $MN=2\sqrt{2}$. Connecting $BN$ and $EM$, what is the minimum perimeter of the quadrilateral $BEMN$?

Solution:
\textbf{Solution:} Connect BD and DN, construct point E's symmetric point F with respect to BD, thus BE=BF=2,\\
connect NF and DF,\\
$\because$ Quadrilateral ABCD is a square, $\therefore \angle ABC=\angle BCD=90^\circ$, DB$\perp$AC, BN=DN, point F is on BC,\\
$\therefore$EF$\parallel$AC, EF= $\sqrt{BE^2+BF^2}=2\sqrt{2}=MN$,\\
$\therefore$ Quadrilateral MEFN is a parallelogram,\\
$\therefore$ME=NF,\\
$\therefore$ME+BN=NF+DN$\geq$DF (equality holds when D, N, F are collinear),\\
In $\triangle DCF$, CD=8, CF=8-2=6, then DF= $\sqrt{CD^2+CF^2}=10$,\\
$\therefore$ME+BN$\geq 10$,\\
$\therefore$MN+BE+ME+BN$\geq 2\sqrt{2}+2+10=12+2\sqrt{2}$,\\
thus, the minimum perimeter of quadrilateral $BEMN$ is $\boxed{12+2\sqrt{2}}$.


Question: As shown in the diagram, in a rhombus $ABCD$ with $AB=2$ and $\angle A=120^\circ$, perpendiculars are drawn from the center of symmetry $O$ of the rhombus $ABCD$ to sides $AB$ and $BC$, intersecting the sides at points $E$, $F$, $G$, and $H$. What is the area of the rhombus $ABCD$? What is the perimeter of the quadrilateral $EFGH$?

Solution:
\textbf{Solution:} Connect $BD$ and $AC$ as shown in the diagram:\\
Since quadrilateral $ABCD$ is a rhombus with $\angle BAD=120^\circ$,\\
it follows that $AB=BC=CD=AD=2$, $\angle BAO=\angle DAO=60^\circ$, and $BD\perp AC$,\\
therefore $\angle ABO=\angle CBO=30^\circ$,\\
thus $OA=\frac{AB}{2}=1$, $OB=\sqrt{3}\times OA=\sqrt{3}$, meaning $AC=2\times OA=2$, $BD=2\times OB=2\sqrt{3}$,\\
hence the area of rhombus $ABCD$ is $\frac{1}{2}AC\cdot BD=\frac{1}{2}\times 2\times 2\sqrt{3}=\boxed{2\sqrt{3}}$;\\
Since $OE\perp AB$ and $OF\perp BC$,\\
it means $\angle BEO=\angle BFO=90^\circ$,\\
in $\triangle BEO$ and $\triangle BFO$,\\
$\left\{\begin{array}{l}
\angle BEO=\angle BFO \\
\angle EBO=\angle FBO \\
BO=BO \\
\end{array}\right.$,\\
thus $\triangle BEO\cong \triangle BFO(AAS)$,\\
meaning $OE=OF$ and $BE=BF$,\\
since $\angle EBF=60^\circ$,\\
it follows that $\triangle BEF$ is an equilateral triangle,\\
thus $EF=BE=\sqrt{3}\times \frac{\sqrt{3}}{2}=\frac{3}{2}$,\\
By the same reasoning, $\triangle DGH$, $\triangle OEH$, $\triangle OFG$ are all equilateral triangles,\\
hence $EF=GH=\frac{3}{2}$, $EH=FG=\frac{\sqrt{3}}{2}$,\\
therefore, the perimeter of quadrilateral $EFGH$ is $3+\sqrt{3}$,\\
Thus, the answer is: $\boxed{3+\sqrt{3}}$.


Question: As shown in the figure, in the rhombus $ABCD$, $AC=4$ and $BD=2$, what is the perimeter of this rhombus?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rhombus,\\
it follows that $OA=\frac{1}{2}AC=2$, $OB=\frac{1}{2}BD=1$, and $AC\perp BD$, with AB=BC=CD=DA,\\
therefore, in $\triangle OAB$, $AB=\sqrt{OA^{2}+OB^{2}}=\sqrt{2^{2}+1^{2}}=\sqrt{5}$,\\
thus, the perimeter of this rhombus is: $4\sqrt{5}$,\\
hence, the answer is: $\boxed{4\sqrt{5}}$.


Question: As shown in the figure, the diagonals $AC$ and $BD$ of the rhombus $ABCD$ intersect at point $O$. $P$ is a moving point on side $AB$ (not coinciding with points $A$ or $B$). $PE \perp OA$ at point $E$, and $PF \perp OB$ at point $F$. If $AB=8$ and $\angle BAD=60^\circ$, what is the minimum length of segment $EF$?

Solution:
\textbf{Solution}: Connect OP, \\
$\because$ Quadrilateral ABCD is a rhombus, \\
$\therefore$ AC$\perp$BD, $\angle$CAB=$\frac{1}{2}\angle$DAB=$30^\circ$, \\
$\because$ PE$\perp$OA at point E, PF$\perp$OB at point F, \\
$\therefore$ $\angle$EOF=$\angle$OEP=$\angle$OFP=$90^\circ$, \\
$\therefore$ Quadrilateral OEPF is a rectangle, \\
$\therefore$ EF=OP, \\
$\because$ When OP takes the minimum value, the value of EF is also minimum, \\
$\therefore$ When OP$\perp$AB, OP is minimized, \\
$\because$ AB=8, \\
$\therefore$ OB=$\frac{1}{2}$AB=4, OA=$\sqrt{AB^2−OB^2}$=$4\sqrt{3}$, \\
$\therefore$ Area of $\triangle ABO$=$\frac{1}{2}$OA$\cdot$OB=$\frac{1}{2}$AB$\cdot$OP, \\
$\therefore$ OP=$\frac{4\times 4\sqrt{3}}{8}=\boxed{2\sqrt{3}}$, \\
$\therefore$ The minimum value of EF is $2\sqrt{3}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, AD is the altitude from A to side BC, E and F are the midpoints of sides AB and AC, respectively. If AB = 8 and AC = 6, what is the perimeter of $\triangle DEF$?

Solution:
\textbf{Solution:} Given $AD$ is the altitude from $A$ to side $BC$,\\
thus $\angle ADB=\angle ADC=90^\circ$,\\
since $E,F$ are the midpoints of sides $AB,AC$ respectively, and $AB=8,AC=6$,\\
then $DE=\frac{1}{2}AB=4,DF=\frac{1}{2}AC=3$,\\
since in $\triangle ABC$, $\angle BAC=90^\circ,AB=8,AC=6$,\\
then $BC=\sqrt{AB^{2}+AC^{2}}=10$,\\
also since $E,F$ are the midpoints of sides $AB,AC$ respectively,\\
then $EF=\frac{1}{2}BC=5$,\\
therefore, the perimeter of $\triangle DEF$ is $DE+DF+EF=4+3+5=\boxed{12}$.


Question: As shown in the diagram, in rhombus $ABCD$, $\angle A=120^\circ$, and the length of diagonal $BD$ is 3. What is the length of the sides of the rhombus?

Solution:
\textbf{Solution:} As shown in the figure, connect AC, intersecting BD at point O.\\
$\because$ Quadrilateral ABCD is a rhombus, hence AC$\perp$BD, OB=OD=$\frac{1}{2}$BD=$\frac{1}{2}\times 3=\frac{3}{2}$,\\
$\angle$BAC=$\frac{1}{2}\angle BAD=\frac{1}{2}\times 120^\circ=60^\circ$,\\
$\therefore \angle$ABO=$90^\circ-\angle$BAO=$30^\circ$\\
$\therefore$AB=2AO\\
In $\triangle AOB$, let AO=x, then AB=2x\\
By the Pythagorean theorem, we have: $AO^{2}+BO^{2}=AB^{2}$\\
$\therefore x^{2}+\left(\frac{3}{2}\right)^{2}=(2x)^{2}$\\
Solving for x, we get $x=\frac{\sqrt{3}}{2}$\\
Hence, AB=2x=$\boxed{\sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=12$, $BC=5$. Point $D$ is outside $\triangle ABC$, and segments $AD$ and $BD$ are drawn. Point $E$ is the midpoint of segment $BD$, $AD=4$, $\angle CAD=\angle CAB$. What is the length of segment $CE$?

Solution:
\textbf{Solution:} As shown in the figure, extend AD and BC to meet at point F\\
$\because$ $\angle ACB=90^\circ$,\\
$\therefore$ $\angle ACF=90^\circ=\angle ACB$,\\
Also, $\because$ $\angle CAD=\angle CAB$ and AC=AC,\\
$\therefore$ $\triangle ABC \cong \triangle AFC$,\\
$\therefore$ AF=AB and BC=CF\\
In $\triangle ABC$, AC=12 and BC=5,\\
$\therefore$ $AB=\sqrt{AC^2+BC^2}=\sqrt{12^2+5^2}=13$\\
$\therefore$ AF=13\\
$\because$ AD=4,\\
$\therefore$ DF=AF−AD=13−4=9\\
$\because$ E is the midpoint of BD,\\
$\therefore$ CE is the midline of $\triangle BDF$\\
$\therefore$ $CE=\frac{1}{2}DF=\boxed{\frac{9}{2}}$


Question: As shown in the figure, in rhombus $ABCD$, $\angle D=120^\circ$, then what is the degree measure of $\angle 1$?

Solution:
\textbf{Solution:} Given rhombus $ABCD$,\\
it follows that $AB\parallel CD$ and $\angle 1=\frac{1}{2}\angle DAB$,\\
since $\angle D=120^\circ$,\\
it follows that $\angle DAB=60^\circ$,\\
hence $\angle 1=\frac{1}{2}\angle DAB=\boxed{30^\circ}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertices A and C of the rhombus ABCD are located at (3,3) and (-1,-1), respectively. The diagonal BD intersects AC at point M and intersects the x-axis at point N. If BN=2ND, what are the coordinates of point D?

Solution:
\textbf{Solution:} As illustrated in the figure, draw AE $\perp$ x-axis at point E, draw DF $\perp$ x-axis at point F, and draw MP $\perp$ x-axis at point P,\\
$\therefore$ MP $\parallel$ DF, \\
$\because$ Quadrilateral ABCD is a rhombus, \\
$\therefore$ MA=MC, MB=MD, AC$\perp$BD, \\
$\because$ Point A is (3, 3), and point C is (-1, -1), \\
$\therefore$ M is (1, 1), \\
$\therefore$ OP=MP=1, \\
$\therefore$ $\triangle$OPM is an isosceles right triangle, \\
$\therefore$ $\angle$MOP = 45$^\circ$, \\
$\because$ AC$\perp$BD, $\angle$OMN = 90$^\circ$, \\
$\therefore$ $\triangle$OMN is an isosceles right triangle, \\
$\therefore$ MN=MO, \\
$\therefore$ NP=MP=OP=1, \\
$\because$ BN=2ND, \\
$\therefore$ BD=3DN, \\
$\because$ BD=2DM, \\
$\therefore$ $DM=\frac{3}{2}DN$, \\
$\therefore$ DN=2MN, \\
$\because$ MP$\parallel$DF, \\
$\therefore$ $\triangle$PMN$\sim$$\triangle$FDN, \\
$\therefore$ $\frac{DF}{MP}=\frac{FN}{PN}=\frac{DN}{MN}=2$, \\
$\therefore$ DF=2, FN=2, \\
$\therefore$ OF=4, \\
$\therefore$ The coordinates of point D are \boxed{(4, -2)}.


Question: As shown in the figure, the central angle $\angle AOB$ corresponding to arc AB is $90^\circ$, and the radius is 4. Point C is the midpoint of OB, and point D is a point on arc AB. CD rotates $90^\circ$ counterclockwise around point C to get CE. What is the minimum value of AE?

Solution:
\textbf{Solution:}

Construct $MC\perp OB$ passing through point C such that $CM=OC$. Connect EM and OD. Then, $\angle OCM=90^\circ$,

Since point C is the midpoint of OB,

$\therefore OC=BC=\frac{1}{2}OB=2$,

$\therefore CM=OC=2$,

Since CD rotates $90^\circ$ counterclockwise around point C to get CE,

$\therefore CD=CE$, $\angle DCE=90^\circ$,

$\therefore \angle OCM=\angle DCE$,

$\therefore \angle OCM+\angle OCE=\angle DCE+\angle OCE$,

$\therefore \angle ECM=\angle DCO$,

In $\triangle ECM$ and $\triangle DCO$,

$\left\{\begin{array}{l}
EC=DC \\
\angle ECM=\angle DCO \\
CM=CO
\end{array}\right.$,

$\therefore \triangle ECM\cong\triangle DCO$ (SAS),

$\therefore EM=OD=4$,

$\therefore$ point E is on the circle with center M and radius 4,

$\therefore$ when points A, E, M are collinear, AE takes the minimum value,

Construct $MN\perp AO$ extending AO and intersecting at point N,

$\therefore \angle MNO=\angle MCO=\angle CON=90^\circ$,

$\therefore$ quadrilateral COMN is a rectangle,

Since $CM=OC$,

$\therefore$ quadrilateral COMN is a square,

$\therefore MN=OC=ON=2$,

$\therefore AN=AO+ON=6$,

$\therefore AM=\sqrt{AN^2+MN^2}=2\sqrt{10}$,

$\therefore$ the minimum value of AE is $AM-EM=2\sqrt{10}-4$,

Thus, the answer is: $AE=\boxed{2\sqrt{10}-4}$.


Question: As shown in the figure, quadrilateral $OABC$ is a rectangle, with points $A$ and $C$ located on the $x$-axis and $y$-axis respectively. Connect $AC$. The coordinates of point $B$ are $(12, 5)$. The bisector of $\angle CAO$ intersects the $y$-axis at point $D$. What are the coordinates of point $D$?

Solution:
\textbf{Solution:} Let line DE be perpendicular to AC at E, \\
because quadrilateral ABCO is a rectangle, with B(12,5), \\
therefore OC = AB = 5, OA = BC = 12, $\angle COA = 90^\circ$, \\
because AD bisects $\angle OAC$, \\
therefore OD = DE, \\
by the Pythagorean theorem: OA$^{2}$ = AD$^{2}$ - OD$^{2}$, AE$^{2}$ = AD$^{2}$ - DE$^{2}$, \\
therefore OA = AE = 12, \\
by the Pythagorean theorem: AC = $\sqrt{5^{2} + 12^{2}} = 13$, \\
in right triangle $\triangle DEC$, DE$^{2}$ + EC$^{2}$ = CD$^{2}$, \\
that is, OD$^{2}$ + (13 - 12)$^{2}$ = (5 - OD)$^{2}$, \\
solving this gives: OD = 2.4, \\
therefore, the coordinates of D are \boxed{(0, 2.4)}.


Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is on side $AB$. $\triangle ADE$ is folded along $DE$ to obtain $\triangle A'DE$. Given that $AB=14$, $AD=13$, $BD=15$, and let $BE=x$. When point $A'$ lies inside (including on the boundary) of $\triangle CDE$, what is the range of values for $x$?

Solution:
\textbf{Solution:} Construct $DH\perp AB$ through point D, with the foot being H, and construct $CI\perp AB$ through point C, with the foot being I.\\
Since quadrilateral ABCD is a parallelogram,\\
therefore $AB\parallel CD$, $AD=BC$, $AB=CD$,\\
Since $DH\perp AB$, $CI\perp AB$,\\
therefore $DH\parallel CI$, $\angle DHA=\angle DHB=\angle CIB=90^\circ$,\\
therefore quadrilateral DHIC is a rectangle, $\triangle ADH$, $\triangle BDH$, $\triangle BCI$ are right-angled triangles,\\
therefore $DH=CI$,\\
In $\triangle ADH$ and $\triangle BCI$,\\
$\left\{\begin{array}{l}DH=CI\\ AD=BC\end{array}\right.$\\
therefore $\triangle ADH \cong \triangle BCI\ (HL)$,\\
therefore $AH=BI$,\\
Let $AH=m$,\\
Since $AB=14$,\\
therefore $HB=14-m$,\\
In $\triangle ADH$, using Pythagoras' theorem, we get: $DH^2=AD^2-AH^2$,\\
In $\triangle BDH$, using Pythagoras' theorem, we get: $DH^2=BD^2-HB^2$,\\
therefore $AD^2-AH^2=BD^2-HB^2$,\\
Since $AD=13$, $BD=15$,\\
therefore $13^2-m^2=15^2-(14-m)^2$, simplifying gives: $28m=140$, solving gives: $m=5$,\\
therefore $AH=5$, hence $BI=5$,\\
therefore $DH=\sqrt{AD^2-AH^2}=\sqrt{13^2-5^2}=12$, hence $CI=12$,\\
(1) When $A'$ is on CE, as shown in the figure,\\
Since $\triangle ADE$ folds along DE to form $\triangle A'DE$\\
therefore $\angle A=\angle DA'E$, $AD=A'D$,\\
Since $\angle DA'E + \angle DA'C=180^\circ$,\\
therefore $\angle A + \angle DA'C =180^\circ$,\\
Since quadrilateral ABCD is a parallelogram,\\
therefore $AD\parallel BC$, $AB\parallel CD$, $AB=CD$,\\
therefore $\angle CBE + \angle A=180^\circ$, $\angle CEB=\angle DCA'$, $AD=BC$,\\
therefore $\angle DA'C=\angle CBE$, $DA'=BC$,\\
In $\triangle BCE$ and $\triangle A'DC$,\\
$\left\{\begin{array}{l}\angle CEB=\angle DCA'\\ \angle CBE=\angle DA'C\\ BC=A'D\end{array}\right.$\\
$\triangle BCE \cong \triangle A'DC\ (AAS)$,\\
therefore $EC=CD$,\\
Since $CD=AB=14$,\\
therefore $EC=14$\\
Since $BE=x$,\\
therefore $EI=BE+BI=x+5$,\\
In $\triangle ECI$, using Pythagoras' theorem, we get: $EC^2=EI^2+CI^2$,\\
therefore $14^2=(x+5)^2+12^2$, simplifying gives: $x^2+10x-27=0$, solving gives: $x_1=-5+2\sqrt{13}$, $x_2=-5-2\sqrt{13}$ (discard),\\
therefore $x=\boxed{-5+2\sqrt{13}}$,\\


Question: As shown in the figure, in square $ABCD$, $AB=3$, $E$ is a point on $AD$, $AE=2$. Connect $BE$ and draw $AF\perp BE$ to meet $DC$ at $F$. What is the length of $EF$?

Solution:
\textbf{Solution:}

Since quadrilateral ABCD is a square, \\
$\therefore AB=3=AD$, $\angle BAE=90^\circ=\angle ADF$,\\
$\therefore \angle DAF+\angle AFD=90^\circ$,\\
Since $AF\perp BE$,\\
$\therefore \angle AEB+\angle DAF=90^\circ$,\\
$\therefore \angle AFD=\angle AEB$,\\
In $\triangle ABE$ and $\triangle DAF$,\\
$\left\{\begin{array}{l}
\angle AEB=\angle AFD \\
\angle BAE=\angle ADF \\
AB=AD
\end{array}\right.$,\\
$\therefore \triangle ABE \cong \triangle DAF \left(AAS\right)$,\\
$\therefore AE=DF=2$,\\
Since $AD=3$, $AE=2$,\\
$\therefore DE=AD-AE=3-2=1$,\\
In $Rt\triangle DEF$, $EF=\sqrt{DE^2+DF^2}=\sqrt{1^2+2^2}=\boxed{\sqrt{5}}$.


Question: As shown in the diagram, within the square $ABCD$, $BD$ is the diagonal. Point $E$ is the midpoint of $CD$, and through point $E$ a line $EF\perp BD$ is drawn, where the foot of the perpendicular is point $F$. If $BC= 2\sqrt{2}$, what is the length of $EF$?

Solution:
\textbf{Solution:} In the square $ABCD$, we have $CD=BC=2\sqrt{2}$ and $\angle C=90^\circ$,\\
$\therefore$ $BD=\sqrt{BC^2+CD^2}=4$,\\
$\because$ point $E$ is the midpoint of $CD$,\\
$\therefore$ $CE=DE=\sqrt{2}$,\\
$\therefore$ $BE=\sqrt{BC^2+CE^2}=\sqrt{10}$,\\
Let $DF=x$, then $BF=4-x$,\\
$\because$ $EF\perp BD$,\\
$\therefore$ $\angle DFE=\angle BFE=90^\circ$,\\
$\therefore$ $DE^2-DF^2=EF^2=BE^2-BF^2$,\\
$\therefore$ $\left(\sqrt{10}\right)^2-\left(4-x\right)^2=\left(\sqrt{2}\right)^2-x^2$,\\
Solving gives: $x=1$,\\
i.e., $DF=1$,\\
$\therefore$ $EF=\sqrt{DE^2-DF^2}=\boxed{1}$.


Question: As shown in the figure, in $ \square ABCD$, $ \angle B=45^\circ$, $AB=2$, and points H and G are moving points on sides $DC$ and $BC$ respectively (point G does not coincide with B or C). Connect $AH$ and $HG$. If point E is the midpoint of $AH$ and point F is the midpoint of $GH$, what is the minimum value of $EF$?

Solution:
\textbf{Solution}: As shown in the figure, draw $AN\perp BC$ at point N, \\
$\because \angle B=45^\circ$, $AB=2$, \\
In $\triangle ABN$, \\
$\therefore AN=\sqrt{2}$, \\
In $\triangle AHG$, \\
Point E is the midpoint of AH, and point F is the midpoint of GH, \\
$\therefore EF=\frac{1}{2}AG$, \\
$\because$ Point G is a moving point on side BC (G does not coincide with B or C), \\
$\therefore$ when G coincides with N, AG is minimal, \\
$\therefore$ The minimum value of AG is $\sqrt{2}$, the minimum value of EF is $\frac{\sqrt{2}}{2}$. \\
Hence, the answer is: $\boxed{\frac{\sqrt{2}}{2}}$.


Question: As shown in the rectangle $ABCD$, point $E$ is on side $BC$, and $AE$ is connected. Folding $\triangle ABE$ along line $AE$ results in $\triangle AFE$, with $EF$ intersecting $AC$ at point $M$. Given $AB=8$, $BC=10$, and $BE= \frac{3}{5}BC$, what is the distance from point $F$ to line $AD$?

Solution:
\textbf{Solution:} Draw line $MN \perp BC$ through point $F$,\\
since $BE= \frac{3}{5}BC$ and $BC=10$,\\
therefore, $BE=6$,\\
since a fold is made,\\
therefore, $\triangle ABE \cong \triangle AFE$,\\
therefore, $EF=BE=6$, $\angle AFE=\angle B= 90^\circ$, $AF= AB=8$,\\
therefore, $\angle AFN+\angle EFM=90^\circ$,\\
since $\angle AFN+\angle FAN=90^\circ$,\\
therefore, $\angle FAN=\angle EFM$,\\
therefore, $\triangle AFN \sim \triangle FEM$,\\
therefore, $\frac{AF}{EF}=\frac{AN}{FM}=\frac{FN}{EM}=\frac{4}{3}$,\\
Let $AN=4x$, $FM=3x$, $FN=8−3x$, $EM=4x−6$,\\
therefore, $FN=8−3x$, $EM=4x−6$,\\
therefore, $\frac{8−3x}{4x−6}=\frac{4}{3}$,\\
therefore, $x=\frac{48}{25}$,\\
Upon verification: $x=\frac{48}{25}$ is the root of the original equation,\\
therefore, $FN=\boxed{\frac{56}{25}}$.


Question: In rhombus $ABCD$, if $AC=6$ and $BD=8$, then what is the perimeter of rhombus $ABCD$?

Solution:
\textbf{Solution:} Given that the quadrilateral $ABCD$ is a rhombus,\\
$\therefore$ $AO=\frac{1}{2}AC=\frac{1}{2}\times 6=3$,\\
$BO=\frac{1}{2}BD=\frac{1}{2}\times 8=4$,\\
$AC\perp BD$,\\
$\therefore$ $\triangle ABO$ is a right triangle,\\
By Pythagorean theorem, we have $AB=\sqrt{3^2+4^2}=5$,\\
$\therefore$ The perimeter of the rhombus $ABCD$ is: $4\times 5=\boxed{20}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertices $A$ and $B$ of the isosceles $\triangle ABC$ are located on the $x$-axis and $y$-axis, respectively. The graph of the inverse proportion function $y= \frac{k}{x}$ ($k>0$, $x>0$) passes through the right-angle vertex $C$. If $OA=7$ and $OB=3$, what is the value of $k$?

Solution: \textbf{Solution:} As shown in the figure, construct $CE\perp y$-axis, $CD\perp x$-axis. Since $\angle O=90^\circ$, it follows that quadrilateral $ODCE$ is a rectangle. Therefore, $\angle ECD=90^\circ$. Since $\angle BCA=90^\circ$, it follows that $\angle ECB=\angle DCA$. Therefore, in $\triangle CEB$ and $\triangle CDA$, we have $\left\{\begin{array}{l} \angle CEB=\angle CDA \\ \angle ECB=\angle DCA \\ CB=CA \end{array}\right.$ Therefore, $\triangle CEB\cong \triangle CDA\left(AAS\right)$, which implies $CE=CD$, $EB=DA$. Consequently, quadrilateral $ODCE$ is a square. Let $EB=DA=x$, so $OD=OA-DA=7-x$, $OE=OB+BE=3+x$. Since $OE=OD$, we get $7-x=3+x$, which is solved to find $x=2$. Therefore, $OD=7-2=5$, $OE=3+2=5$. Hence, the coordinates of point C are $(5,5)$. Substituting $C(5,5)$ into $y=\frac{k}{x}$ gives $5=\frac{k}{5}$. Solving this yields $k=\boxed{25}$.


Question: As shown in the diagram, in rectangle $ABCD$, $AB=12$, and point $E$ is on side $BC$. Connect $DE$ and $AE$, $\angle DEC=30^\circ$, and $AE\perp DE$. Fold $\triangle CDE$ along $DE$ to obtain $\triangle DEF$. Connect $AF$. What is the distance from $E$ to $AF$?

Solution: \textbf{Solution}: Since quadrilateral ABCD is a rectangle, we have $\angle ABC=\angle BAC=\angle ADC=\angle BCD=90^\circ$, and CD=AB=12. Draw FM$\perp$BC at M, intersecting AD at N. Draw AH$\perp$EF at H through point A, and draw EG$\perp$AF at G through point E. Then, quadrilaterals ABMN and CDNM are rectangles, thus $AB=MN=CD=12$, AN=BM, and by the folding property, CE=EF. Also, $\angle CED=\angle DEF=30^\circ$, thus DE=2CD=24, and $\angle CEF=60^\circ$. By the Pythagorean theorem, CE=EF=$\sqrt{DE^2-CD^2}=\sqrt{24^2-12^2}=12\sqrt{3}$, which makes $\triangle CEF$ an equilateral triangle. Thus, ME=CM=$\frac{1}{2}$CE=$6\sqrt{3}$. By the Pythagorean theorem again, FM=$\sqrt{EF^2-ME^2}=\sqrt{(12\sqrt{3})^2-(6\sqrt{3})^2}=18$. Since AE$\perp$DE and $\angle CDE=\angle DEF=30^\circ$, it follows that $\angle AEH=\angle AEB=60^\circ$, and $\angle ABE=\angle AHE=90^\circ$, making AH=AB=12 and $\angle HAE=\angle BAE=30^\circ$. Hence, AE=2BE. By applying the Pythagorean theorem: $AB^2+BE^2=AE^2$ gives $12^2+BE^2=4BE^2$, from which we get BE=$4\sqrt{3}$. Consequently, AN=BE+ME=$4\sqrt{3}+6\sqrt{3}=10\sqrt{3}$, and FN=FM−MN=18−12=6. Therefore, AF=$\sqrt{AN^2+FN^2}=\sqrt{(10\sqrt{3})^2+6^2}=4\sqrt{21}$. From $S_{\triangle AEF}=\frac{1}{2}AF\cdot EG=\frac{1}{2}EF\cdot AH$, we derive $\frac{1}{2}\times 4\sqrt{21}\cdot EG=\frac{1}{2}\times 12\sqrt{3}\times 12$, solving for EG yields $\boxed{\frac{36\sqrt{7}}{7}}$.


Question: As shown in the figure, the side length of the square paper $ABCD$ is 4, and point $F$ is on side $AD$. Connect $BF$, fold the paper along the line $BF$, and the corresponding point of point $A$ is point $G$. Extend line $AG$ to intersect $CD$ at point $E$. If $DE=3$, what is the length of $GE$?

Solution:
\textbf{Solution:}
Since quadrilateral $ABCD$ is a square,
$\therefore \angle D=\angle DAB=90^\circ$, $AD=BA$,
Since it is folded,
$\therefore AH\perp BF$,
$\therefore \angle AHB=90^\circ$,
$\therefore \angle ABF+\angle BAH=90^\circ$,
Since $\angle BAH+\angle DAE=90^\circ$,
$\therefore \angle DAE=\angle ABF$,
$\therefore \triangle DAE\cong \triangle ABF$ (ASA),
$\therefore AE=BF$, $AF=DE$,
Since $AB=4$, $DE=4$,
$\therefore$ $AF=3$, $BF=\sqrt{AB^2+AF^2}=\sqrt{3^2+4^2}=5$,
$\therefore$ $AH=\frac{AF\cdot AB}{BF}=\frac{3\times 4}{5}=\frac{12}{5}$,
$\therefore AG=2AH=\frac{24}{5}$,
$\therefore$ $GE=AE-AG=5-\frac{24}{5}=\boxed{\frac{1}{5}}$.


Question: As shown in the figure, point $E$ is a point outside the square $ABCD$, and $ED=CD$, connect $AE$ and intersect $BD$ at point $F$. If $\angle CDE=30^\circ$, then what is the measure of $\angle DFC$?

Solution:
\textbf{Solution}:
Since quadrilateral $ABCD$ is a square,\\
thus $AD=DC, \angle ADC=90^\circ$,\\
therefore $\angle ADB=\angle BDC=45^\circ$,\\
since $DC=DE$,\\
thus $AD=DE$,\\
therefore $\angle DAE=\angle DEA$,\\
since $\angle ADE=90^\circ+30^\circ=120^\circ$,\\
thus $\angle DAE=\frac{180^\circ−120^\circ}{2}=30^\circ$,\\
therefore $\angle AFD=180^\circ−30^\circ−45^\circ=105^\circ$,\\
In $\triangle ADF$ and $\triangle CDF$,\\
since $\left\{\begin{array}{c}AD=DC\\ \angle ADB=\angle BDC\\ DF=DF\end{array}\right.$,\\
thus $\triangle ADF\cong \triangle CDF$,\\
therefore $\angle DFC=\angle AFD=\boxed{105^\circ}$.


Question: As shown in the figure, in rectangle $ABCD$, $E$ and $F$ are the midpoints of $AB$ and $AD$ respectively. If $AC = 4$, what is the length of $EF$?

Solution:
\textbf{Solution:} As shown in the figure, connect BD,\\
since quadrilateral ABCD is a rectangle,\\
therefore AC=BD=4,\\
since E, F are the midpoints of AD, AB respectively,\\
therefore EF= $\frac{1}{2}$ BD= \boxed{2}


Question: As shown in the figure, the four corners of the rectangular paper ABCD are folded inward to perfectly form a quadrilateral EFGH with no gaps and no overlaps. If $BC=8$ and $GH=7$, then what is the length of $EH$?

Solution:
\textbf{Solution:} Given that folding the four corners of the rectangle ABCD inward perfectly forms a quadrilateral EFGH without any gaps or overlaps,\\
it follows that $\angle CFE = \angle EFH = \frac{1}{2} \angle CFH$, $\angle BFG = \angle HFG = \frac{1}{2} \angle BFH$, FM = CF, FN = BF, $\angle GNH = \angle EMH = 90^\circ$,\\
hence $\angle EFH + \angle HFG = \frac{1}{2} (\angle CFH + \angle BFH) = 90^\circ$, meaning $\angle EFG = 90^\circ$,\\
Similarly, it can be derived that $\angle FEH = \angle GHE = 90^\circ$,\\
thus quadrilateral EFGH is a rectangle,\\
hence $\angle GHF = \angle EFH$, EF = GH = 7,\\
In $\triangle GHN$ and $\triangle EFM$,  $\left\{\begin{array}{l} \angle GNH = \angle EMF\hfill \\ \angle GHN = \angle EFM\hfill \\ GH = EF\hfill \end{array} \right.$,\\
thus $\triangle GHN \cong \triangle EFM$,\\
hence HN = FM,\\
thus CF = HN,\\
hence FH = NH + FN = CF + BF = BC = 8,\\
thus EH = $\sqrt{FH^2 - EF^2}$ = $\sqrt{FH^2 - GH^2}$ = $\sqrt{8^2 - 7^2}$ = $\boxed{\sqrt{15}}$.


Question: As shown in the diagram, square $ABCD$ has a side length of 1, and point $E$ is on the extension line of $BC$. If $BE=BD$, then what is the length of $CE$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a square, \\
it follows that BC=CD=1, \\
By the Pythagorean theorem, we have $BD=\sqrt{BC^{2}+CD^{2}}=\sqrt{2}$, \\
therefore, $CE=BE-BC=BD-BC= \boxed{\sqrt{2}-1}$.


Question: As illustrated, in $\triangle ABC$, $\angle ACB=90^\circ$, $AB=10\,\text{cm}$, $BC=8\,\text{cm}$. Point $D$ is on side $AB$ with $AD=AC$, $AE\perp CD$ at foot $E$. Point $F$ is the midpoint of $BC$. Then, what is the length of $EF$ in $\text{cm}$?

Solution:
\textbf{Solution}: In $\triangle ABC$, where $\angle ACB=90^\circ$, $AB=10\text{cm}$, $BC=8\text{cm}$,\\
we have $AC=\sqrt{AB^{2}-BC^{2}}=\sqrt{10^{2}-8^{2}}=6\text{cm}$.\\
Thus, $AD=AC=6\text{cm}$,\\
and $BD=AB-AD=10-6=4\text{cm}$.\\
Since $AD=AC$ and $AE\perp CD$,\\
it follows that $CE=DE$, meaning point $E$ is the midpoint of $CD$,\\
and given that point $F$ is the midpoint of $BC$,\\
$EF$ serves as the midsegment of $\triangle CBD$,\\
thus $EF=\frac{1}{2}BD=\boxed{2}\text{cm}$.


Question: As shown in the figure, in the Cartesian coordinate system, the square ABCD with a side length of 2 has its side AB on the x-axis, and the midpoint of AB is the origin O. Fix points A and B, and push the square in the direction of the arrow until point D falls on the positive half-axis of the y-axis at point $D'$. Point E is a moving point on the x-axis. When $ED'+EC'$ takes the minimum value, what are the coordinates of point E?

Solution:
\textbf{Solution:} Construct point $F$ as the reflection of point $D'$ across the $x$ axis, and connect $C'F$. The intersection of $C'F$ and the $x$ axis is the desired point $E$.\\
Since the $x$ axis bisects $D'F$ perpendicularly,\\
it follows that $ED'=EF$\\
Therefore, $ED'+EC'=EF+EC'=FC'$\\
This implies that at point $E$, the sum $ED'+EC'$ is minimized.\\
Since $AD'=2$, $AO=1$, and $\angle AOD'=90^\circ$,\\
it follows that $OD'=\sqrt{AD{'}^{2}-AO^{2}}=\sqrt{2^{2}-1^{2}}=\sqrt{3}$,\\
Therefore, $F(0,-\sqrt{3})$, $C'(2,\sqrt{3})$,\\
Thus, the equation of line $FC'$ is: $y=\sqrt{3}x-\sqrt{3}$,\\
Hence, the coordinates of point $E$ are $\boxed{(1,0)}$.


Question: As the diagram shows, points $A$ and $B$ are separated by a pond and their distance cannot be measured directly. Therefore, Xiao Ming selects a point $C$ on the shore, connects $CA$ and $CB$, and extends them to points $M$ and $N$ respectively, such that $AM=AC$ and $BN=BC$. Given that the length of $MN$ is measured to be $20m$, what is the distance between points $A$ and $B$ in meters?

Solution:
\textbf{Solution:} Since $AM=AC$, and $BN=BC$\\
therefore, A and B are respectively the midpoints of CM and CN,\\
therefore, AB is the mid-segment of $\triangle CMN$,\\
therefore, $MN=2AB$,\\
therefore, AB= $\frac{1}{2}MN=\boxed{10}$ m


Question: In the parallelogram $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, $AC=AB$, $E$ is the midpoint of side $AB$, $G$ and $F$ are points on side $BC$, and $OG$ and $EF$ are connected. If $AB=25$, $BC=14$, and $GF=7$, what is the area of the shaded region in the figure?

Solution:
\textbf{Solution}: Connect OE, let OG intersect EF at point H, and draw AP$\perp$BC at point P, intersecting OE at point Q, as shown in the figure.\\
$\because$AB=AC, AP$\perp$BC,\\
$\therefore$ $BP=\frac{1}{2}BC=7.$\\
$\therefore$ $AP=\sqrt{AB^{2}-BP^{2}}=\sqrt{25^{2}-7^{2}}=24.$\\
$\because$Quadrilateral ABCD is a parallelogram,\\
$\therefore$OA=OC.\\
$\because$EA=EB,\\
$\therefore$OE is the median line of $\triangle ABC$.\\
$\therefore$OE$\parallel$BC and $OE=\frac{1}{2}BC=7.$\\
$\therefore \angle AEO=\angle ABC,$ $\angle AOE=\angle ACB.$\\
$\because AB=AC,$\\
$\therefore \angle ABC=\angle ACB.$\\
$\therefore \angle AEO=\angle AOE.$\\
$\therefore AO=AE.$ That is, $\triangle AEO$ is an isosceles triangle.\\
$\because$OE$\parallel$BC,\\
$\therefore$$\angle$AQE=$\angle$APB=90$^\circ$.\\
$\therefore$AP$\perp$OE at point Q.\\
$\therefore EQ=OQ=\frac{1}{2}OE=\frac{7}{2}.$\\
$\therefore$ $AQ=\sqrt{AE^{2}-EQ^{2}}=\sqrt{\left(\frac{25}{2}\right)^{2}-\left(\frac{7}{2}\right)^{2}}=12.$\\
$\therefore$ ${S}_{\triangle AEO}=\frac{1}{2}OE\cdot AQ=\frac{1}{2}\times 7\times 12=42.$\\
Connect EG, OF,\\
$\because$OE$\parallel$GF, OE=GF=7,\\
$\therefore$Quadrilateral OEGF is a parallelogram.\\
$\therefore$HE=HF, HO=HG.\\
$\because$PQ=AP−AQ=24−12=12,\\
$\therefore$ ${S}_{\text{parallelogram} OEGF}=GF\cdot PQ=7\times 12=84.$\\
$\therefore$ ${S}_{\triangle OEH}={S}_{\triangle GFH}=\frac{1}{4}{S}_{\text{parallelogram} OEGF}=\frac{1}{4}\times 84=21.$\\
$\therefore$ ${S}_{\text{shaded}}={S}_{\triangle AEO}+{S}_{\triangle OEH}+{S}_{\triangle GFH}=42+21+21=\boxed{84}.$


Question: As illustrated in the diagram, the sides of the rhombus $ABCD$ are of length $\sqrt{3}$, with $\angle ABC=60^\circ$. Let $M$ be any point on line segment $CD$ (which can coincide with points $C$ or $D$). Perpendicular lines to ray $BM$ are drawn from points $A$, $C$, and $D$, meeting $BM$ at points $E$, $F$, and $G$, respectively. Let $AE+CF+DG=m$. What is the range of possible values for $m$?

Solution:
\textbf{Solution:} As shown in the diagram, connect AC and BD, which intersect at point O, and also connect AM,\\
since quadrilateral ABCD is a rhombus and $\angle ABC=60^\circ$,\\
therefore, AC$\perp$BD, AO＝CO, BO＝DO, AB＝BC＝$\sqrt{3}$, $\angle ABO=30^\circ$,\\
therefore, AO＝$\frac{1}{2}$AB＝$\frac{\sqrt{3}}{2}$,\\
thus, AC＝$\sqrt{3}$,\\
since BO＝$\sqrt{AB^2-AO^2}=\sqrt{3-\frac{3}{4}}=\frac{3}{2}$,\\
thus, BD＝3,\\
thus, the area of rhombus ABCD ＝$\frac{AC\times BD}{2}=\frac{3\sqrt{3}}{2}$,\\
since $S_{\triangle ABM}=\frac{1}{2}\times$BM$\times$AE, $S_{\triangle BCM}=\frac{1}{2}\times$BM$\times$CF, $S_{\triangle BMD}=\frac{1}{2}\times$BM$\times$DG,\\
thus, $S_{\triangle ABM}$+$S_{\triangle BCM}$+$S_{\triangle BMD}=\frac{1}{2}S_{\text{rhombus ABCD}}+\frac{1}{2}S_{\text{rhombus ABCD}}=\frac{1}{2}\times$BM$\times$(AE+CF+DG),\\
thus, $\frac{3\sqrt{3}}{2}＝\frac{1}{2}\times$BM$\times$m,\\
since $\sqrt{3}\leq$BM$\leq$3,\\
thus, $\sqrt{3}\leq$m$\leq$3,\\
therefore, the answer is: $\boxed{\sqrt{3}\leq m\leq 3}$.


Question: As shown in the diagram, rhombus $ABCD$ has a side length of $2\sqrt{3}$, and $\angle ABC=60^\circ$. The diagonals $AC$ and $BD$ intersect at point $O$. Point $E$ is a moving point on line $AD$, and $CE$ is connected. After rotating the line segment $EC$ clockwise around point $C$ by the angle of $\angle BCD$, we get the corresponding line segment $CF$ (i.e., $\angle ECF=\angle BCD$). What is the minimum length of $DF$?

Solution:
\textbf{Solution:} Let's connect BE and construct BH $\perp$ AD, intersecting the extension line of DA at H. In diamond ABCD, $\angle ABC=60^\circ$, thus $\angle BCD=120^\circ$. Since $\angle ECF=120^\circ$, it follows that $\angle BCD=\angle ECF$, and thus $\angle BCE=\angle DCF$. By rotation, we get: EC=FC. In $\triangle BEC$ and $\triangle DFC$, we have:
\[
\left\{
\begin{array}{l}
BC=DC \\
\angle BCE=\angle DCF \\
EC=FC
\end{array}
\right.
\]
Therefore, $\triangle DCF \cong \triangle BCE$ (SAS), hence $DF=BE$. Thus, finding the minimum value of DF is equivalent to finding the minimum value of BE. Since in $\triangle AHB$, $\angle BAH=60^\circ$ and AB= $2\sqrt{3}$, it follows that BH= $3$. When E coincides with H, the minimum value of BE is 3. Therefore, the minimum value of DF is $\boxed{3}$.


Question: In $\triangle ABC$, $\angle C=90^\circ$, $AC=3$, $BC=4$. Point $N$ is on side $BC$, and point $M$ is a moving point on side $AB$. Points $D$ and $E$ are the midpoints of $CN$ and $MN$, respectively. What is the minimum value of $DE$?

Solution:
\textbf{Solution:} Let's join CM,\\
since points D and E are the midpoints of CN and MN respectively,\\
therefore DE = $\frac{1}{2}$ CM,\\
when CM $\perp$ AB, the value of CM is minimal, and at this point, the value of DE is also minimal.\\
According to the Pythagorean theorem: $AB = \sqrt{AC^{2} + BC^{2}} = \sqrt{3^{2} + 4^{2}} = 5$,\\
since $S_{\triangle ABC} = \frac{1}{2} \times AB \times CM = \frac{1}{2} \times AC \times BC$,\\
therefore $CM = \frac{12}{5}$,\\
therefore $DE = \frac{1}{2}CM = \boxed{\frac{6}{5}}$.


Question: As shown in the figure, the side length of the square $ABCD$ is $8$, $E$ is a point on side $BC$, and $BE=2.5$. $F$ is a moving point on side $AB$ connected with $EF$. An equilateral triangle $\triangle EFG$ is constructed to the right side with $EF$ as one of its sides. Connect $CG$. What is the minimum value of $CG$?

Solution:
\textbf{Solution:} It is known from the problem that point $F$ is the driving point, and point $G$ is the driven point. Point $F$ moves on a line segment, thus point $G$ must also move on a straight line trajectory.\\
Rotate $\triangle EFB$ about point $E$ by $60^\circ$ so that $EF$ coincides with $EG$, obtaining $\triangle EFB\cong \triangle EHG$,\\
Therefore, it can be inferred that $\triangle EBH$ is an equilateral triangle, and point $G$ is on line $HN$, which is perpendicular to $HE$,\\
Draw $CM\perp HN$ through point $C$, then $CM$ is the minimum value of $CG$,\\
Draw $EP\perp CM$ through point $E$, it can be seen that quadrilateral $HEPM$ forms a rectangle,\\
Thus $CM=MP+CP=HE+\frac{1}{2}EC=2.5+\frac{11}{4}=\boxed{\frac{21}{4}}$,


Question: As shown in the figure, it is known that point $E$ is located on the side $AB$ of the square $ABCD$. A square $BEFG$ is constructed outside square $ABCD$ with $E$ as one of its sides. Line segments $DF$ and $MN$ are drawn, where $M$ and $N$ are the midpoints of $DC$ and $DF$, respectively. If $AB=7$ and $BE=5$, what is the length of $MN$?

Solution:
\textbf{Solution:} Draw line CF, \\
$\because$ In square ABCD and square BEFG, we have AB=7, BE=5, \\
$\therefore$ GF=GB=5, BC=7, \\
$\therefore$ GC=GB+BC=5+7=12, \\
$\therefore$ CF= $\sqrt{GF^{2}+GC^{2}}=\sqrt{5^{2}+12^{2}}=13$, \\
$\because$ M and N are the midpoints of DC and DF respectively, \\
$\therefore$ MN= $\frac{1}{2}$ CF= $\boxed{\frac{13}{2}}$;


Question: As shown in the diagram, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, with $AC=16$ and $BD=12$. Let $E$ be a point on side $AD$, and line $OE$ intersects $BC$ at point $F$. Fold the rhombus along line $EF$, making the corresponding point of $B$ as $B'$ and the corresponding point of $A$ as $A'$. If $AE=4$, what is the length of $B'F$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a rhombus, $AC=16$, and $BD=12$,\\
it follows that $OA=OC=\frac{1}{2}AC=8$, $OB=\frac{1}{2}BD=6$, $AC\perp BD$, $AD\parallel BC$,\\
hence $BC=\sqrt{OB^2+OC^2}=10$, $\angle OCF=\angle OAE$, $\angle OFC=\angle OEA$,\\
in $\triangle COF$ and $\triangle AOE$, we have $\left\{\begin{array}{l}\angle OFC=\angle OEA \\ \angle OCF=\angle OAE \\ OC=OA\end{array}\right.$,\\
therefore, $\triangle COF\cong \triangle AOE$ (AAS),\\
thus, $CF=AE=4$,\\
hence $BF=BC-CF=10-4=6$,\\
by the property of folding, we get: ${B'}^F=BF=\boxed{6}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is a rhombus with a perimeter of 24, and the coordinates of point $A$ are $\left(4, 0\right)$. What are the coordinates of point $D$?

Solution:
\textbf{Solution}: Since quadrilateral $ABCD$ is a rhombus with a perimeter of 24,\\
it follows that AB$=24\div4=6$,\\
since the coordinates of point A are $\left(4, 0\right)$,\\
it follows that OA$=4$,\\
since the diagonals of a rhombus are perpendicular bisectors of each other,\\
it follows that in right-angled triangle AOB, by Pythagoras' theorem, OB$=\sqrt{6^2-4^2}=2\sqrt{6}$,\\
hence OD=OB$=2\sqrt{6}$,\\
therefore, the coordinates of D are $\boxed{\left(0, -2\sqrt{6}\right)}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $CD$ bisects the angle between the extensions of sides $AC$ and $BC$, $AD \perp DC$, and point $E$ is the midpoint of $AB$. If $AC=4$ and $BC=6$, what is the length of segment $DE$?

Solution:
\textbf{Solution:} Extend $BC$ and $AD$ respectively to meet at point $G$, \\
$\because$ $CD$ bisects the angle between the extended lines of $AC$ and $BC$\\
$\therefore \angle 1=\angle 2$\\
$\because$ $AD\perp DC$\\
$\therefore \angle CDA=\angle CDG=90^\circ$\\
$\because CD=CD$\\
$\therefore \triangle CDG\cong \triangle CDA$ (ASA)\\
$\therefore CG=AC$, $AD=GD$\\
$\because$ $AC=4$, $BC=6$\\
$\therefore BG=BC+CG=6+4=10$\\
Also, $\because$ point $E$ is the midpoint of $AB$\\
$\therefore DE=\frac{1}{2}BG=\boxed{5}$.


Question: As shown in the figure, in rectangle $ABCD$, where $AB=4$ and $AD=2$, $E$ is the midpoint of $AB$, $F$ is a moving point on $EC$, and $P$ is the midpoint of $DF$. Connect $PB$. What is the minimum value of $PB$?

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

When point $F$ coincides with point $C$, point $P$ is at ${P}_{1}$, and $C{P}_{1}=D{P}_{1}$,

When point $F$ coincides with point $E$, point $P$ is at ${P}_{2}$, and $E{P}_{2}=D{P}_{2}$,

$\therefore {P}_{1}{P}_{2}\parallel CE$ and ${P}_{1}{P}_{2}=\frac{1}{2}CE$.

When point $F$ is on segment $EC$ except at points $C$ and $E$, $DP=FP$.

According to the Midline Theorem: ${P}_{1}P\parallel CE$ and ${P}_{1}P=\frac{1}{2}CF$.

$\therefore$ The trajectory of point $P$ is the line segment ${P}_{1}{P}_{2}$,

$\therefore$ When $BP\perp {P}_{1}{P}_{2}$, $PB$ reaches its minimum value.

$\because$ In rectangle $ABCD$, $AB=4$, $AD=2$, $E$ is the midpoint of $AB$,

$\therefore \triangle CBE$, $\triangle ADE$, $\triangle BC{P}_{1}$ are isosceles right triangles, and $C{P}_{1}=2$.

$\therefore \angle ADE=\angle CDE=\angle C{P}_{1}B=45^\circ$, $\angle DEC=90^\circ$.

$\therefore \angle D{P}_{2}{P}_{1}=90^\circ$.

$\therefore \angle D{P}_{1}{P}_{2}=45^\circ$.

$\therefore \angle {P}_{2}{P}_{1}B=90^\circ$, i.e., $B{P}_{1}\perp {P}_{1}{P}_{2}$,

$\therefore$ The minimum value of $BP$ is the length of $B{P}_{1}$.

In the isosceles right triangle $BC{P}_{1}$, $C{P}_{1}=BC=2$,

$\therefore B{P}_{1}=2\sqrt{2}$

$\therefore$ The minimum value of $PB$ is $\boxed{2\sqrt{2}}$.


Question: As shown in the figure, given point $E$ is on the edge $AB$ of the rectangular sheet $ABCD$. The sheet is folded along $DE$ such that the corresponding point of $A$, denoted as $A'$, falls exactly on the line segment $CE$. If $AD=3$ and $AE=1$, then what is the length of $CD$?

Solution:
\textbf{Solution:} Given that the quadrilateral ABCD is a rectangle,\\
hence, $AB\parallel CD$ ,\\
therefore, $\angle CDE=\angle AED$,\\
based on the property of folding, it is known that,\\
$A'D=AD=3$, $AE=A'E=1$, $\angle AED=\angle A'ED$\\
therefore, $\angle CDE=\angle CED$\\
thus, $CD=CE$, $CA'=BE$,\\
since $A'$ lies on $CE$,\\
therefore, $\angle DA'C=\angle B=90^\circ$\\
Let $CD=x$, then $AB=x$,\\
thus $BE=x-1$,\\
therefore $x^2=3^2+(x-1)^2$,\\
Solving for $x$ gives: $x=5$\\
i.e., $CD=\boxed{5}$


Question: As shown in the figure, in rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. If $AB=3$ and $BD=4$, what is the area of rhombus $ABCD$?

Solution:
\textbf{Solution:}\\
Since quadrilateral $ABCD$ is a rhombus and $BD=4$,\\
it follows that $OB=2$,\\
Since $AB=3$,\\
we have $OA=\sqrt{AB^{2}-OB^{2}}=\sqrt{3^{2}-2^{2}}=\sqrt{5}$,\\
Therefore, $AC=2OA=2\sqrt{5}$,\\
So, the area of rhombus $ABCD$ is $S_{\text{rhombus}ABCD}=\frac{1}{2}AC\cdot BD=\frac{1}{2}\times 2\sqrt{5}\times 4=4\sqrt{5}$,\\
Hence, the answer is: $S_{\text{rhombus}ABCD}=\boxed{4\sqrt{5}}$.


Question: In square $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Point $E$ is on $DO$ with $DE = 2EO$. Connect $AE$ and fold $\triangle ADE$ along $AD$ to get $\triangle ADE'$. Point $F$ is the midpoint of $AE$, and connect $E'F$. If $DE = 2\sqrt{2}$, what is the area of $\triangle AFE'$?

Solution:
\textbf{Solution:} Connect $EE'$,\\
since quadrilateral $ABCD$ is a square,\\
it follows that $OA=OD$, $\angle ADO=45^\circ$, $\angle AOD=90^\circ$\\
Given $DE=2\sqrt{2}$, $DE=2EO$\\
it follows that $OD=OA=3\sqrt{2}$\\
thus, ${S}_{\triangle ADE}=\frac{1}{2}DE\times OA=6$\\
According to the problem, $\angle ADE'=\angle ADO=45^\circ$, $DE=DE'=2\sqrt{2}$\\
thus, ${S}_{\triangle DEE'}=\frac{1}{2}DE\times DE=4$, ${S}_{\text{quadrilateral} AED{E}'}=2{S}_{\triangle ADE}=12$\\
therefore, ${S}_{\triangle AEE'}={S}_{\text{quadrilateral}AED{E}'}-{S}_{\triangle DEE'}=8$\\
Also, since $F$ is the midpoint of $AE$\\
it follows that ${S}_{\triangle AEE'F}=\frac{1}{2}{S}_{\triangle AEE'}=\boxed{4}$


Question: Given: As shown in the figure, in rectangle $ABCD$, $AB=3$, $AD=4$. Point $P$ is a point inside rectangle $ABCD$, and satisfies ${S}_{\triangle PBC}=\frac{1}{3}{S}_{\text{rectangle} ABCD}$, what is the minimum perimeter of $\triangle ADP$?

Solution:
\textbf{Solution:} Draw $MN\perp AD$ through point $P$, intersecting $AD$ at point $M$ and $BC$ at point $N$, \\
since $S_{\triangle PBC}=\frac{1}{3}S_{\text{rectangle }ABCD}$, \\
therefore $\frac{1}{2}\times BC\times PN=\frac{1}{3}\times BC\times MN$, \\
hence $PN=\frac{2}{3}MN$, \\
since $AB=3$, \\
thus $MP=1$, \\
Draw $GH\parallel AD$ through point $P$, intersecting $AB$ at point $G$ and $CD$ at point $H$, construct the symmetric point $A'$ of point $A$ about $GH$, and the intersection of $A'D$ with $GH$ will be the desired point $P$, \\
since $AP=A'P$, \\
thus $AP+PD=A'D$, \\
since $AG=1$, \\
thus $AA'=2$, \\
In $\triangle AA'D$, with $AD=4$ and $AA'=2$, \\
thus $A'D=2\sqrt{5}$, \\
therefore, the minimum perimeter of $\triangle ADP$ is $2\sqrt{5}+4$, \\
hence the answer is $\boxed{4+2\sqrt{5}}$.


Question: In the rhombus $ABCD$, $E$ and $F$ are respectively on $BC$ and $DC$, with $BE=DF$ and $AE=AB$. If $\angle EAF=30^\circ$, what is the degree measure of $\angle D$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rhombus,\\
it follows that $AB=AD$, $\angle B=\angle D$,\\
Given that $BE=DF$,\\
it follows that $\triangle ABE\cong \triangle ADF$ (SAS),\\
therefore, $\angle BAE=\angle DAF$,\\
Let $\angle BAE=\angle DAF=x$,\\
Since $\angle EAF=30^\circ$,\\
it follows that $\angle BAD=2x+30^\circ$,\\
Given that $AE=AB$,\\
it follows that $\angle B=\angle AEB=\frac{180^\circ-x}{2}$,\\
Since $\angle B+\angle BAD=180^\circ$,\\
it follows that $\frac{180^\circ-x}{2}+30^\circ+2x=180^\circ$,\\
Solving gives: $x=40^\circ$,\\
therefore, $\angle D=\angle B=\frac{180^\circ-40^\circ}{2}=\boxed{70^\circ}$;


Question: As shown in the figure, in $\triangle ABC$, points D and E are the midpoints of $AB$ and $AC$ respectively. Connect $BE$. If $AE=6$, $DE=5$, and $\angle BEC=90^\circ$, what is the length of $BE$?

Solution:
Given that points D and E are the midpoints of AB and AC respectively, with AE=6 and DE=5,

it follows that EC=AE=6 and BC=$2\cdot DE$=10,

in the right triangle $\triangle BEC$, we have $BE= \sqrt{BC^{2}-EC^{2}} = \boxed{8}$.


Question: As shown in the figure, rectangle $ABCD$ is folded along diagonal $AC$, making point $B$ coincide with point $E$, and $CE$ intersects $AD$ at point $F$. If $AF=5$ and $DF=4$, then what is the length of $AC$?

Solution: \textbf{Solution:} From the property of folding, we have $\angle ACF = \angle ACB$. Since $AD \parallel BC$, it follows that $\angle ACB = \angle FAC$, which implies $\angle FAC = \angle ACF$, hence $AF = CF = 5$. Given that $CF = 5$, $DF = 4$, and $\angle D = 90^\circ$, it follows $CD = \sqrt{CF^2 - DF^2} = 3$. Since $AD = AD + DF = 5 + 4 = 9$, it follows that $AC = \sqrt{AD^2 + CD^2} = \sqrt{9^2 + 3^2} = \boxed{3\sqrt{10}}$.


Question: As shown in the figure, in $\triangle CDE$, $CD=1$ and $\angle CDE=45^\circ$, respectively construct squares $ABCD$ and $CEFG$ externally on sides $CD$ and $CE$. If $AE=BD$, what is the value of $EF^2$?

Solution:
\textbf{Solution:} Draw $EN \perp AD$ and $EM \perp CD$ at $E$,\\
since $\angle CDN=45^\circ$,\\
thus, quadrilateral $DMEN$ is a square,\\
let $DN=x$, thus $EN=DM=EM=x$,\\
since $AE=BD=\sqrt{2}CD=\sqrt{2}$,\\
in $\triangle AEN$, since $AN^{2}+EN^{2}=AE^{2}$,\\
i.e., $(1+x)^{2}+x^{2}=2$,\\
we get $x=\frac{\sqrt{3}-1}{2}$ or $x=\frac{-\sqrt{3}-1}{2}$ (discard),\\
since in $\triangle CME$, since $CM^{2}+EM^{2}=CE^{2}$,\\
i.e., $CE^{2}=\left(\frac{\sqrt{3}-1}{2}+1\right)^{2}+\left(\frac{\sqrt{3}-1}{2}\right)^{2}=4-2\sqrt{3}$,\\
since quadrilateral $CEFG$ is a square,\\
thus $EF^{2}=\boxed{4-2\sqrt{3}}$.


Question: As shown in the diagram, in rhombus ABCD, E and F are on BC and DC, respectively, with BE=DF and AE=AB. If $\angle EAF=30^\circ$, what is the degree measure of $\angle D$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rhombus, \\
$\therefore$AB=AD, $\angle B=\angle D$, \\
$\because$BE=FD, \\
$\therefore \triangle ABE \cong \triangle ADF$ (SAS), \\
$\therefore$AE=AF, $\angle BAE=\angle DAF$, \\
Let $\angle B=x$, $\angle BAE=y$, \\
$\because \angle BAE+\angle B+\angle AEB=2x+y=180^\circ$, $\angle BAD+\angle B=2y+x+30^\circ=180^\circ$, \\
$\therefore 4x+2y-2y-x-30^\circ=180^\circ$, \\
Simplifying, we get $3x=210^\circ$, \\
$\therefore x=\boxed{70^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, and D, E, and F are the midpoints of sides AB, AC, and BC, respectively. If the length of CD is 3, what is the length of EF?

Solution:
Solution: Since $\angle ACB=90^\circ$ and point D is the midpoint of side AB, with CD=3, it follows that AB=2CD=6.
Since E and F are the midpoints of sides AC and BC, respectively, it follows that EF is the median of $\triangle ABC$,
which implies EF=$\frac{1}{2}$AB=$\boxed{3}$.


Question: As shown in the figure, $O$ is any point inside the equilateral triangle $ABC$. Lines $OD$, $OE$, and $OF$ are drawn through point $O$ parallel to $AB$, $AC$, and $BC$, respectively, intersecting $AC$, $BC$, and $AB$ at points $G$, $H$, and $I$. Given that the perimeter of the equilateral triangle $ABC$ is 18, what is the sum of $OD + OE + OF$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is an equilateral triangle with a perimeter of 18, \\
$\therefore$ $\angle B=\angle C=\angle A=60^\circ$, AC=6\\
Since OD$\parallel$AB, OE$\parallel$AC, and OF$\parallel$BC, \\
$\therefore$ Quadrilateral OHCF is a parallelogram, Quadrilateral AEOG is a parallelogram, $\angle C=\angle OHD=\angle GFO=60^\circ$, $\angle ODH=\angle B=60^\circ$, $\angle OGF=\angle A=60^\circ$\\
$\therefore$ $\triangle ODH$ and $\triangle OGF$ are equilateral triangles, OH=CF\\
$\therefore$ OD=OH, OF=FG, OE=AG\\
$\therefore$ OD=CF,\\
Since AC=AG+GF+CF\\
$\therefore$ AC=OE+OF+OE=6.\\
Thus, the answer is: $\boxed{6}$.


Question: As shown in the figure, it is known that point $P$ is a point on diagonal $BD$ of square $ABCD$, and $AP=3$, $PF \perp CD$ at point $F$, $PE \perp BC$ at point $E$. When connecting $EF$, what is the length of $EF$?

Solution:
\textbf{Solution}: Let's connect CP.\\
Since square ABCD,\\
thus $\angle C=90^\circ$, AB=BC, and $\angle ABP=\angle CBP=45^\circ$. Let's connect CP again,\\
In $\triangle ABP$ and $\triangle CBP$\\
\[\left\{\begin{array}{l}
AB=BC\\
\angle ABP=\angle CBP\\
BP=BP
\end{array}\right.\]\\
thus $\triangle ABP\cong \triangle CBP$ (SAS)\\
thus AP=CP=3\\
Since PF$\perp$CD, and PE$\perp$BC,\\
thus $\angle PFC=\angle PEC=90^\circ$,\\
thus quadrilateral CEPF is a rectangle,\\
thus EF=PC=\boxed{3}.


Question: As shown in the pentagon $ABCDE$, $\angle D=120^\circ$, the exterior angle adjacent to $\angle EAB$ is $80^\circ$, and the exterior angles adjacent to $\angle DEA$ and $\angle ABC$ are both $60^\circ$. What is the measure of $\angle C$ in degrees?

Solution:
\textbf{Solution:} Since the external angle adjacent to $\angle$EAB is $80^\circ$, and the external angles adjacent to both $\angle$DEA and $\angle$ABC are $60^\circ$,\\
it follows that $\angle$DEA=$180^\circ-60^\circ=120^\circ$, $\angle$ABC=$180^\circ-60^\circ=120^\circ$, $\angle$EAB=$180^\circ-80^\circ=100^\circ$;\\
The sum of the interior angles of a pentagon is $(5-2)\times180^\circ=540^\circ$;\\
Therefore, $\angle$C=$540^\circ-120^\circ-120^\circ-120^\circ-100^\circ=\boxed{80^\circ}$.


Question: As shown in the diagram, in square $ABCD$, $AB=6$, $E$ is a point on the diagonal $AC$, and line $BE$ is drawn. Through point $E$, line $EF\perp BE$ intersects $AD$ at point $F$. The areas of $\triangle BCE$ and $\triangle AEF$ are ${S}_{1}$ and ${S}_{2}$, respectively. If $2{S}_{1}=3{S}_{2}$, then what is the length of $CE$?

Solution:
\textbf{Solution:} Connect ED, and through E draw $MN \perp BC$ at N intersecting AD at M,\\
since quadrilateral ABCD is a square,\\
therefore $AB=BC=CD=AD=6$, $\angle ABC=\angle BCD=\angle ADC=\angle DAB=90^\circ$,\\
therefore $\angle 1=\angle 2=45^\circ$,\\
since $MN \perp BC$,\\
therefore $\angle ENC=\angle ENB=90^\circ$,\\
therefore quadrilateral MNCD is a rectangle,\\
therefore $MN=CD=6$, $DM=CN$, $\angle DME=90^\circ$,\\
in $\triangle CDE$ and $\triangle CBE$,\\
$\left\{ \begin{array}{l} CD=CB \\ \angle 2=\angle 1 \\ CE=CE \end{array} \right.$,\\
therefore $\triangle CDE \cong \triangle CBE$ (SAS),\\
therefore $ED=EB$, $\angle EDC=\angle EBC$,\\
since $\angle CDA=\angle CBA=90^\circ$,\\
therefore $\angle CDA-\angle EDC=\angle CBA-\angle EBC$,\\
that is $\angle ADE=\angle ABE$,\\
since $EF \perp BE$,\\
therefore $\angle FEB=90^\circ$,\\
since $\angle FEB+\angle DAB+\angle AFE+\angle ABE=360^\circ$,\\
therefore $\angle AFE+\angle ABE=360^\circ-\angle FEB-\angle DAB=180^\circ$,\\
since $\angle AFE+\angle EFD=180^\circ$,\\
therefore $\angle ABE=\angle EFD$,\\
therefore $\angle ADE=\angle EFD$,\\
therefore $ED=EF$,\\
since $\angle DME=90^\circ$,\\
therefore $EM \perp DF$,\\
therefore $DM=MF$,\\
in $\triangle NEC$, $\angle 1=45^\circ$,\\
therefore $\triangle NEC$ is an isosceles right triangle,\\
let $NE=NC=x$,\\
then $CE= \sqrt{2}x$, $DM=MF=CN=x$,\\
therefore $AF=AD-DM-MF=6-2x$,\\
$ME=MN-EN=6-x$,\\
therefore $S_{1}=\frac{1}{2}BC\cdot EN=\frac{1}{2}\times 6\times x=3x$, $S_{2}=\frac{1}{2}(6-2x)(6-x)=x^{2}-9x+18$,\\
since $2S_{1}=3S_{2}$,\\
therefore $2\cdot 3x=3(x^{2}-9x+18)$,\\
solving, we find: $x_{1}=2$, $x_{2}=9$ (discard),\\
therefore $CE= \sqrt{2}x=2\sqrt{2}$,\\
hence the answer is: $CE = \boxed{2\sqrt{2}}$.


Question: As shown in the figure, the diagonals of rectangle $ABCD$ intersect at point $O$. If $\angle ACB = 30^\circ$ and $BO = 2$, then what is the length of $BC$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
$\therefore \angle ABC=90^\circ$, AC=BD, AO=CO, BO=DO,\\
$\therefore$ AO=OC=OB=OD,\\
Since BO=2,\\
$\therefore$ AO=OC=BO=2,\\
$\therefore$ AC=2+2=4,\\
Since $\angle BCA=30^\circ$,\\
$\therefore$ AB= $\frac{1}{2}$ AC=2,\\
$\therefore$ BC= $\sqrt{AC^2-AB^2}=2\sqrt{3}$,\\
Therefore, the answer is: $BC=\boxed{2\sqrt{3}}$.


Question: As shown in the figure, points $D$ and $E$ are on the sides of $\triangle ABC$. The angle bisector of $\angle ABC$ is perpendicular to $AE$ with the foot of the perpendicular being $Q$, and the angle bisector of $\angle ACB$ is perpendicular to $AD$ with the foot of the perpendicular being $P$. If $BC=12$ and $PQ=3$, what is the perimeter of $\triangle ABC$?

Solution:
Solution: Since line BQ bisects $\angle ABC$ and BQ $\perp$ AE,

it follows that $\angle ABQ = \angle EBQ$,

Since $\angle ABQ + \angle BAQ = 90^\circ$ and $\angle EBQ + \angle BEQ = 90^\circ$,

it follows that $\angle BAQ = \angle BEQ$,

Therefore, $AB = BE$, similarly, $CA = CD$,

Thus, point Q is the midpoint of AE, and point P is the midpoint of AD (triangle median theorem),

Therefore, PQ is the midline of $\triangle ADE$,

Since $PQ = 3$,

it follows that $DE = 2 \times PQ = 6$,

The perimeter of $\triangle ABC$ is: $AB + AC + BC = BE + CD + BC = DE + BC + BC = 6 + 12 + 12 = \boxed{30}$.


Question: As shown in the rectangle $ABCD$, where $BC=4$, the diagonals $AC$ and $BD$ intersect at point $O$, $AN \perp BD$ with the foot being $N$, $BN = 3DN$, what is the length of $AN$?

Solution:
\textbf{Solution:} Since $BN=3DN$,\\
it follows that $BD=4DN$,\\
Since quadrilateral $ABCD$ is a rectangle and $BC=4$,\\
it follows that $AD=BC=4$ and $BO=DO$,\\
thus $DO=\frac{1}{2}BD=2DN$,\\
hence $DN=ON$,\\
Since $AN\perp BD$,\\
it follows that $\angle ANO=\angle AND=90^\circ$,\\
therefore, $\triangle ANO$ and $\triangle AND$ are right-angled triangles,\\
In $\triangle ANO$ and $\triangle AND$,\\
$\left\{\begin{array}{l}
AN=AN\\
\angle ANO=\angle AND\\
NO=ND
\end{array}\right.$\\
hence $\triangle ANO\cong \triangle AND$ (SAS),\\
thus $AO=AD$, $\angle NAO=\angle NAD$,\\
therefore, $AO=AD=OD$,\\
hence $\triangle AOD$ is an equilateral triangle,\\
thus $\angle OAD=60^\circ$,\\
therefore, $\angle NAD=\frac{1}{2}\angle OAD=\frac{1}{2}\times 60^\circ=30^\circ$,\\
thus $ND=\frac{1}{2}AD=\frac{1}{2}\times 4=2$,\\
thus $AN=\sqrt{AD^2-ND^2}=\sqrt{4^2-2^2}=\boxed{2\sqrt{3}}$


Question: As shown in the figure, in $ \triangle ABC$, we have $AB=AC$ and $AD$ is the median to side $BC$. Point $E$ is on side $AB$, and $DE \parallel AC$. If $DE=3$, what is the value of $AB$?

Solution:
\textbf{Solution:} Given that $AB=AC$, and $AD$ is the median to side $BC$,\\
hence $\angle EAD=\angle CAD$, $\angle B=\angle C$,\\
Given $DE\parallel AC$,\\
hence $\angle EDA=\angle CAD$, $\angle EDB=\angle C$,\\
thus $\angle EAD=\angle EDA$, $\angle EDB=\angle B$,\\
hence $DE=AE$, $DE=BE$,\\
thus $AB=2DE=\boxed{6}$.


Question: As shown in the figure, in the equilateral $\triangle ABC$, D is the midpoint of $AB$, $DE \perp AC$ at point E, and $EF \perp BC$ at point F. Given $AB=8$, what is the length of $BF$?

Solution:
\textbf{Solution:} Given that $\triangle ABC$ is an equilateral triangle and $D$ is the midpoint of $AB$,\\
thus $\angle A=\angle C=60^\circ$, and $AC=AB=BC=8$, with $AD=\frac{1}{2}AB=4$,\\
since $DE\perp AC$ and $EF\perp BC$,\\
thus $\angle AED=\angle CFE=90^\circ$,\\
therefore $\angle EDA=\angle FEC=30^\circ$,\\
hence $AE=\frac{1}{2}AD=2$,\\
therefore $CE=AC-AE=8-2=6$, and $CF=\frac{1}{2}CE=3$,\\
thus $BF=BC-CF=8-3=\boxed{5}$;


Question: As shown in the figure, $CE$ is the bisector of the exterior angle $\angle ACD$ of $\triangle ABC$. If $\angle B = 45^\circ$ and $\angle ACE = 65^\circ$, then what is the measure of $\angle A$?

Solution:
\textbf{Solution:} Since $CE$ is the bisector of $\angle ACD$ and $\angle ACE=65^\circ$,\\
it follows that $\angle ACD=130^\circ$.\\
Since $\angle ACD$ is an exterior angle of $\triangle ABC$,\\
we have $\angle ACD=\angle A + \angle B$.\\
Given that $\angle B=45^\circ$,\\
it follows that $\angle A=\boxed{85^\circ}$.


Question: As shown in the diagram, in $\triangle ABC$, $AB=AC$, $\angle B=25^\circ$, and point D is on side $BC$. Fold $\triangle ACD$ along $AD$ to get $\triangle ADE$. If $AD=DE$, then what is the measure of $\angle AFB$?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $AB=AC$, and $\angle B=25^\circ$, \\
thus, $\angle C=\angle B=25^\circ$, \\
by folding, we know $\angle E=\angle C=25^\circ$, \\
since $AD=DE$, \\
hence, $\angle DAE=\angle E=25^\circ$, \\
since $\angle DAE=\angle DAC=25^\circ$, \\
thus, $\angle CAF=50^\circ$, \\
therefore, $\angle AFB=\angle C+\angle CAF=\boxed{75^\circ}$.


Question: As shown in the figure, $AB=AC$, point D is a point inside $\triangle ABC$, $\angle D=110^\circ$, $\angle 1=\angle 2$, then what is the measure of $\angle A$ in degrees?

Solution:
\textbf{Solution:}\\
Since $\angle D=110^\circ$ and $\angle 1=\angle 2$,\\
it follows that $\angle D=180^\circ-\angle 1-\angle DCB=110^\circ$,\\
hence $\angle 1+\angle DCB=70^\circ$,\\
Since $AB=AC$,\\
it follows that $\angle ABC=\angle ACB$,\\
thus $\angle ABD=\angle BCD$,\\
Since $\angle 1+\angle DCB=70^\circ$ and $\angle 1=\angle 2$,\\
it follows that $\angle ACB=\angle 2+\angle DCB=70^\circ$,\\
hence $\angle ABC+\angle ACB=140^\circ$,\\
thus $\angle A=180^\circ-140^\circ=\boxed{40^\circ}$.


Question: As shown in the figure, quadrilateral $OABC$ is a rectangle with $OA=1$. What is the value represented by point $P$?

Solution:
\textbf{Solution:} Given that $\because$OA=1, OC=3,\\
$\therefore$OB=$\sqrt{3^2+1^2}$=$\sqrt{10}$,\\
thus, the number represented by point P is $\boxed{\sqrt{10}}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=8\,\text{cm}$, $BC=10\,\text{cm}$. A point $E$ is taken on side $CD$, and $\triangle ADE$ is folded such that point $D$ aligns exactly with point $F$ on side $BC$. What is the length of $CE$ in $\text{cm}$?

Solution:
\textbf{Solution:}
Given: rectangle ABCD, \\
hence AD=BC=10cm, and CD=AB=8cm, \\
By the problem statement: Rt$\triangle$ADE$\cong$Rt$\triangle$AFE,\\
thus $\angle$AFE=90$^\circ$, AF=10cm, and EF=DE,\\
Let CE=xcm, then DE=EF=CD−CE=$(8−x)$cm,\\
In Rt$\triangle$ABF, by the Pythagorean theorem: AB$^{2}$+BF$^{2}$=AF$^{2}$,\\
that is $8^{2}$+BF$^{2}$=$10^{2}$,\\
thus BF=6cm,\\
thus CF=BC−BF=10−6=4(cm),\\
In Rt$\triangle$ECF, by the Pythagorean theorem: EF$^{2}$=CE$^{2}$+CF$^{2}$,\\
that is $(8−x)^{2}$=x$^{2}$+4$^{2}$,\\
thus 64−16x+x$^{2}$=x$^{2}$+16,\\
thus x=\boxed{3},\\
which means CE=3cm.


Question: As shown in the figure, in rectangle $ABCD$, $AD=3$, $AB=5$, point $E$ is a moving point on the ray $CD$ (not coincident with $D$). $\triangle ADE$ is folded along $AE$ to obtain $\triangle D'AE$. Connect $D'B$. If $\triangle ABD'$ is a right triangle, then what is the length of $AE$?

Solution:
\textbf{Solution:} According to the problem, quadrilateral ABCD is a rectangle, with $AD=3$ and $AB=5$. By folding $\triangle ADE$ along AE, we obtain $\triangle D^{\prime}AE$, then $\angle D=\angle ED^{\prime}A=90^\circ$, $AD=BC=AD^{\prime}=3$, $AB=CD=5$,\\
(1) As shown in Figure 1, when point E is on segment CD,\\
since $\angle ED^{\prime}A=\angle D=\angle AD^{\prime}B=90^\circ$,\\
then points $B$, $D^{\prime}$, and $E$ are collinear,\\
since $S_{\triangle ABE}=\frac{1}{2}AB\cdot AD=\frac{1}{2}BE\cdot AD^{\prime}$,\\
thus $BE=AB=5$,\\
since $BD^{\prime}=\sqrt{AB^{2}-AD^{\prime 2}}=\sqrt{5^{2}-3^{2}}=4$,\\
thus $DE=D^{\prime}E=BE-BD^{\prime}=5-4=1$;\\
therefore, in $Rt\triangle ADE$, $AE=\sqrt{AD^{2}+DE^{2}}=\sqrt{3^{2}+1^{2}}=\sqrt{10}$;\\
(2) As shown in Figure 2, when point E is on the ray extending from CD,\\
since $\angle AD^{\prime}B=\angle BCE=90^\circ$, $AD=BC=AD^{\prime}=3$, $AB=CD=5$,\\
thus $BD^{\prime}=\sqrt{AB^{2}-AD^{\prime 2}}=4$,\\
let $CE=x$, then $D^{\prime}E=DE=x+5$,\\
thus $BE=D^{\prime}E-BD^{\prime}=x+1$,\\
since $CE^{2}+BC^{2}=BE^{2}$, that is, $x^{2}+3^{2}=(x+1)^{2}$,\\
solving for $x$ yields $x=4$,\\
thus $DE=CD+CE=5+4=9$,\\
therefore, in $Rt\triangle ADE$, $AE=\sqrt{AD^{2}+DE^{2}}=\sqrt{3^{2}+9^{2}}=3\sqrt{10}$.\\
In summary, the value of AE is either $\boxed{\sqrt{10}}$ or $\boxed{3\sqrt{10}}$.


Question: As shown in the figure, it is known that the angle bisectors of the three interior angles of \( \triangle ABC \) intersect at point D, and the perpendicular bisectors of the three sides intersect at point E. If \( \angle BDC = 130^\circ \), what is the measure of \( \angle BEC \) in degrees?

Solution:
\textbf{Solution:} As shown in Figure 1, connect $AD$,\\
since in $\triangle BDC$, $\angle BDC=130^\circ$,\\
therefore, $\angle DBC+\angle DCB=180^\circ-\angle BDC=50^\circ$.\\
Since the angle bisectors of the interior angles of $\triangle ABC$ meet at point D,\\
therefore, $AD$ bisects $\angle BAC$,\\
thus, $\angle BAD=\angle DAC$,\\
Similarly, it can be concluded that $\angle ABD=\angle CBD$, $\angle BCD=\angle ACD$,\\
Since in $\triangle ABC$,\\
$\angle BAD+\angle DAC+\angle ABD+\angle CBD+\angle BCD+\angle ACD=180^\circ$,\\
and since $\angle BAD=\angle DAC$, $\angle ABD=\angle CBD$, $\angle BCD=\angle ACD$,\\
therefore, $\angle DBC+\angle DCB+\angle DAC=90^\circ$,\\
since $\angle DBC+\angle DCB=50^\circ$,\\
thus, $\angle DAC=90^\circ-\left(\angle DBC+\angle DCB\right)=40^\circ$.\\
As shown in Figure 2, connect $AE$,\\
since the perpendicular bisectors of the sides of $\triangle ABC$ meet at point E,\\
thus, $EA=EB=EC$,\\
therefore, $\angle EAB=\angle EBA$, $\angle EBC=\angle ECB$, $\angle EAC=\angle ECA$,\\
since $\angle DAC=40^\circ$, $\angle DAC=\angle DAB$,\\
therefore, $\angle BAC=\angle DAC+\angle DAB=80^\circ$,\\
since $\angle BAC=\angle BAE+\angle CAE$,\\
thus, $\angle BAE+\angle CAE=80^\circ$,\\
since $\angle EAB=\angle EBA$, $\angle EAC=\angle ECA$,\\
therefore, $\angle ABE+\angle ACE=80^\circ$,\\
that is, $\angle ABD+\angle DBE+\angle ACD+\angle DCE=80^\circ$,\\
since $\angle ABD=\angle CBD$, $\angle BCD=\angle ACD$,\\
therefore, $\angle CBD+\angle DBE+\angle BCD+\angle DCE=80^\circ$,\\
since $\angle DBC+\angle DCB=50^\circ$,\\
thus, $\angle DBE+\angle DCE=80^\circ-50^\circ=30^\circ$,\\
since $\angle DBC+\angle DCB=50^\circ$\\
therefore, $\angle DBE+\angle EBC+\angle ECB+\angle DCE=50^\circ$,\\
since $\angle EBC=\angle ECB$, $\angle DBE+\angle DCE=30^\circ$,\\
therefore, $2\angle ECB=50^\circ-\left(\angle DBE+\angle DCE\right)=20^\circ$,\\
thus, $\angle ECB=\angle EBC=10^\circ$,\\
therefore, in $\triangle BEC$,\\
$\angle BEC=180^\circ-\angle ECB-\angle EBC=\boxed{160^\circ}$.


Question: As shown in the figure, in right-angled triangle $ABC$, where $\angle C=90^\circ$, $AC=3$, and $BC=4$, the line $AD$ bisects $\angle CAB$ and intersects $BC$ at $D$, what is the length of $BD$?

Solution:
\textbf{Solution:} Construct $DE \perp AB$ at $E$, \\
since $AD$ is the bisector of $\angle CAB$, $\angle C=90^\circ$, and $DE \perp AB$ at $E$, \\
it follows that $CD=DE$, \\
In the right triangles $\triangle ACD$ and $\triangle AED$, \\
$\left\{\begin{array}{l} AD=AD\\ DC=DE \end{array}\right.$, \\
hence, right $\triangle ACD \cong$ right $\triangle AED$ (HL), \\
thus $AC=AE=3$, \\
by the Pythagorean theorem, $AB=\sqrt{AC^2+BC^2}=\sqrt{3^2+4^2}=5$, \\
thus $BE=AB-AE=5-3=2$, \\
Let $BD=x$, then $CD=DE=4-x$ \\
In the right $\triangle BDE$, $DE^2+BE^2=BD^2$ \\
thus $(4-x)^2+2^2=x^2$ \\
Solving gives $x=\frac{5}{2}$ \\
That is, $BD=\boxed{\frac{5}{2}}$


Question: As shown in the figure, $OA$ bisects $\angle BOD$, $AC \perp OB$ at point $C$, and $AC=3$. Given that the distance from point $A$ to the $y$-axis is 4, what are the coordinates of point $A$?

Solution:
\textbf{Solution:} As shown in the figure, construct $AE\perp OD$ at point E, \\
since $OA$ bisects $\angle BOD$, $AE\perp OD$, and $AC\perp OB$, \\
thus, $AE=AC=3$. \\
Since the distance from point A to the y-axis is 4, \\
thus, $OE=4$. \\
Since point A is in the second quadrant, \\
thus, the coordinates of point A are $\boxed{(-4,3)}$.


Question: As can be seen from the diagram, in the Cartesian coordinate system, an isosceles right-angled triangle board is placed as shown. If $A(3, 0), B(0, 2)$, then what are the coordinates of point $C$?

Solution:
\textbf{Solution:} As shown in the figure, draw $CH\perp x$-axis at point H.\\
Since $\angle AHC=\angle CAB=\angle AOB=90^\circ$,\\
it follows that $\angle BAO+\angle CAH=90^\circ$, $\angle CAH+\angle ACH=90^\circ$,\\
thus, $\angle ACH=\angle BAO$,\\
in $\triangle AHC$ and $\triangle BOA$,\\
$\left\{\begin{array}{l}
\angle AHC=\angle AOB \\
\angle ACH=\angle OAB \\
AC=AB
\end{array}\right.$,\\
therefore, $\triangle AHC\cong \triangle BOA\left(AAS\right)$,\\
hence, $AH=OB$, $CH=OA$,\\
since $A(3,0)$, $B(0,2)$,\\
it follows that $OA=CH=3$, $OB=AH=2$,\\
therefore, $OH=OA+AH=5$,\\
thus, $C\left(5,3\right)$.\\
Therefore, the answer is $C=\boxed{\left(5,3\right)}$.


Question: As shown in the figure, $AD$ is the median of $\triangle ABC$. If $AB=AC=13$ and $BC=10$, then what is the length of $AD$?

Solution:
\textbf{Solution:} Since $AD$ is the median of $\triangle ABC$, if $AB=AC=13$, $BC=10$, \\
then $AD\perp BC$, $BD=\frac{1}{2}BC=5$, \\
In $Rt\triangle ABD$, by the Pythagorean theorem, we have: $AD=\sqrt{AB^{2}-BD^{2}}=\boxed{12}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle A=90^\circ$, $\angle B=38^\circ$, points E and F are located on sides BC and AC respectively. Fold $\triangle CEF$ along the line containing EF, so that the corresponding point $C'$ of C falls on AB, and $CE=BC$, what is the measurement of $\angle AFC'$?

Solution:
\textbf{Solution:} Given $\because$ $\angle A=90^\circ$,\\
$\therefore$ $\angle B+\angle C=90^\circ$,\\
$\therefore$ $\angle C=90^\circ-38^\circ=52^\circ$,\\
$\because$ folding $\triangle CEF$ along the line containing $EF$, such that the corresponding point of $C$, denoted $C'$, falls on $AB$,\\
$\therefore$ $C'E=CE=BC'$\\
$\therefore$ $\angle CEF=\angle C'EF=\frac{1}{2}(180^\circ-38^\circ)=71^\circ$, $\angle CFE=\angle C'FE$,\\
$\therefore$ $\angle CFE=\angle C'FE=180^\circ-\angle C-\angle CEF=180^\circ-52^\circ-71^\circ=57^\circ$,\\
$\therefore$ $\angle AFC' = 180^\circ-\angle CFE-\angle C'FE=180^\circ-57^\circ-57^\circ=\boxed{66^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=4$, $BC=5$, $CA=6$, and the three angle bisectors intersect at point $O$. The area of $\triangle CAO$ equals $9$. What is the area of $\triangle ABO$?

Solution:
\textbf{Solution:} Draw $OM\perp AB$ at point M, and $OD\perp AC$ at point D, \\
since the three angle bisectors intersect at point O, \\
hence $OM=OD$, \\
since the area of $\triangle CAO$ is 9, \\
thus ${S}_{\triangle CAO}=\frac{1}{2}AC\cdot OD=\frac{1}{2}\times 6\cdot OD=9$, \\
Solving this, we get $OD=3$, \\
thus $OM=3$, \\
therefore ${S}_{\triangle ABO}=\frac{1}{2}AB\cdot OM=\frac{1}{2}\times 4\times 3=\boxed{6}$.


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

Solution:
\textbf{Solution:} Construct $CF \perp AB$ at point $F$,\\
Since $BE \perp CD$,\\
$\therefore \angle E=\angle DFC=90^\circ$,\\
Since point $D$ is the midpoint of $AB$,\\
$\therefore CD$ is the median of right triangle $ABC$,\\
$\therefore CD=AD=BD$,\\
In $\triangle CDF$ and $\triangle BDE$, $\left\{\begin{array}{l}\angle CFD=\angle E\\ \angle CDF=\angle BDE\\ CD=BD\end{array}\right.$\\
$\therefore \triangle CDF \cong \triangle BDE$ (AAS)\\
$\therefore CF=BE=4$,\\
In right triangle $ACF$, $AF=\sqrt{AC^{2}-CF^{2}}=\sqrt{5^{2}-4^{2}}=3$;\\
Let $AB=x$, then $BF=x-3$,\\
In right triangle $ABC$, $BC^{2}=x^{2}-9$,\\
In right triangle $BCF$, $BC^{2}=16+(x-3)^{2}$\\
$\therefore 16+(x-3)^{2}=x^{2}-9$,\\
Solving for $x$: $x=\frac{25}{3}$.\\
$\therefore AB=\boxed{\frac{25}{3}}$.


Question: As shown in the figure, $AB \perp BC$ at point $B$, $AB \perp AD$ at point $A$, and point $E$ is the midpoint of $CD$. Given that $BC=5$, $AD=12$, and $BE=12.5$, what is the length of $AB$?

Solution:
\textbf{Solution:} Extend BE to meet AD at point G, as shown in the diagram: \\
$\because$ AB$\perp$BC at point B, AB$\perp$AD at point A,\\
$\therefore$ BC$\parallel$AD, $\angle$A=90$^\circ$,\\
$\therefore$ $\angle$C=$\angle$D, $\angle$CBE=$\angle$DGE\\
$\because$ point E is the midpoint of CD,\\
$\therefore$ CE=DE,\\
In $\triangle$BCE and $\triangle$GDE,\\
$ \left\{\begin{array}{l}\begin{array}{c}\angle C=\angle D\\ \angle CBE=\angle DGE\\ CE=DE\end{array}\\ \end{array}\right.$\\
$\therefore$ $\triangle$BCD$\cong$ $\triangle$GDE (AAS),\\
$\therefore$ BE=GE=12.5, BC=DG=5,\\
$\therefore$ BG=BE+GE=25,\\
$\because$ AD=12,\\
$\therefore$ AG=AD−DG=7,\\
$\therefore$ $AB=\sqrt{BG^{2}−AG^{2}}=\sqrt{25^{2}−7^{2}}=\boxed{24}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$ and $AD\perp BC$ at point $D$. If $AB=6$ and $CD=4$, what is the perimeter of $\triangle ABC$?

Solution:
\textbf{Solution:}

In triangle $\triangle ABC$, we have $AB=AC$, and $AD \perp BC$ at point D,

$\therefore BC=2CD=8$,

$\therefore$ the perimeter of $\triangle ABC$ is: $AB+AC+BC=6+6+8=\boxed{20}$.


Question: As shown in the figure, in $\triangle ABC$, we have $AB=AC$, and $AD$ is the altitude on $BC$. $E$ is a point on side $AB$, and $EF\perp BC$ at point $F$, intersecting the extension of $CA$ at point $G$. It is given that $EF=2$ and $EG=3$. What is the length of $AD$?

Solution:
\textbf{Solution:} Draw $AH \perp FG$ at point H,\\
since AB=AC, and AD is an altitude,  \\
therefore, $\angle DAC = \angle DAB$,\\
since $EF \perp BC$,\\
therefore, $\angle ADC = \angle GFC = 90^\circ$,\\
therefore, $AD \parallel GF$,\\
therefore, $\angle G = \angle DAC$, $\angle AEG = \angle BAD$,\\
therefore, $\angle G = \angle GEA$,\\
therefore, $AG = AE$,\\
since $AH \perp EG$,\\
therefore, $EH= \frac{1}{2}EG=1.5$;\\
since $\angle AHF = \angle ADF = \angle HFD = 90^\circ$,\\
therefore, quadrilateral ADFH is a rectangle,\\
therefore, $AD = HF = HE + EF = 1.5 + 2 = \boxed{3.5}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, CE is the angle bisector of $\triangle ABC$, and $\angle AEC = 105^\circ$. What is the measure of $\angle B$?

Solution:
\textbf{Solution:} Given that $\angle ACB=90^\circ$ and CE is the angle bisector of $\triangle ABC$, \\
it follows that $\angle ACE=\angle BCE=45^\circ$,\\
Furthermore, since $\angle AEC=105^\circ$,\\
it results in $\angle B=\angle AEC-\angle BCE=105^\circ-45^\circ=\boxed{60^\circ}$.


Question: As shown in the figure, $\angle C=90^\circ$, $AD$ bisects $\angle BAC$ and intersects $BC$ at $D$. If $BC=7\text{cm}$ and $BD=4\text{cm}$, what is the distance from point $D$ to $AB$ in $\text{cm}$?

Solution:
\textbf{Solution:} As shown in the figure, draw DH $\perp$ AB at point H, \\
$\because$ BC = 7cm, BD = 4cm, \\
$\therefore$ CD = 7 - 4 = 3cm, \\
Also, $\because$ AD is the bisector of $\angle BAC$, $\angle C=90^\circ$, and DH $\perp$ AB, \\
$\therefore$ DH = CD = 3cm, \\
$\therefore$ the distance from point D to AB is \boxed{3} cm.


Question: As shown in the figure, in $\triangle ABC$, where $AB=AC=10$ and $BC=8$, the line $AD$ bisects $\angle BAC$ and intersects $BC$ at point $D$. Point $E$ is the midpoint of $AC$. After connecting $DE$, what is the perimeter of $\triangle CDE$?

Solution:
\textbf{Solution:} Given that $AB=AC$ and $AD$ bisects $\angle BAC$, with $BC=8$, \\
it follows that $AD\perp BC$, and $CD=BD= \frac{1}{2} BC=4$, \\
since point E is the midpoint of AC, \\
then $DE=CE= \frac{1}{2} AC=5$, \\
hence, the perimeter of $\triangle CDE$ is $CD+DE+CE=4+5+5=\boxed{14}$.


Question: In right triangle $ABC$, where $AC=8$, $BC=6$, and $\angle C=90^\circ$, the triangle $ABC$ is folded such that point $A$ coincides with point $B$, with $DE$ being the crease. What is the length of $DE$?

Solution:
\textbf{Solution:} Since $\angle C=90^\circ$, AC=8, and BC=6,\\
then AB=$\sqrt{AC^{2}+BC^{2}}$=10,\\
Since folding $\triangle ABC$ as shown in the figure, making point A coincide with point B, where the crease is DE,\\
then BE=AE, AD=BD=5, DE$\perp$AB,\\
In $\triangle BEC$, BE$^{2}$=BC$^{2}$+CE$^{2}$,\\
then BE$^{2}$=36+(8-BE)$^{2}$,\\
thus BE=$\frac{25}{4}$,\\
In $\triangle BDE$, DE=$\sqrt{BE^{2}-BD^{2}}$=$\boxed{\frac{15}{4}}$.


Question: As shown in the figure, $D$ is a point inside $\angle MAN$, $DE\perp AM$ at $E$, $DF\perp AN$ at $F$, and $DE=DF$. Point $B$ is on ray $AM$ with $AB=6$, $BE=2$. A point $C$ is taken on ray $AN$ such that $DC=DB$. What is the length of $AC$?

Solution:
Solution:
(1) As shown in Figure 1, when point C is on line segment $AF$, connect $AD$, \\
$\because$ $DE\perp AM$ at E, $DF\perp AN$ at F, \\
$\therefore$ $\angle DEB=\angle DFC=90^\circ$, \\
In $Rt\triangle DEB$ and $Rt\triangle DFC$, $\left\{\begin{array}{l}DC=DB\\ DF=DE\end{array}\right.$, \\
$\therefore$ $Rt\triangle DEB\cong Rt\triangle DFC\left(HL\right)$, \\
$\therefore$ $CF=BE=2$, \\
Also, $\because$ in $Rt\triangle DEA$ and $Rt\triangle DFA$, $\left\{\begin{array}{l}DA=DA\\ DF=DE\end{array}\right.$, \\
$\therefore$ $Rt\triangle DEA\cong Rt\triangle DFA\left(HL\right)$, \\
$\therefore$ $AF=AE=AB+BE=6+2=8$, \\
$\therefore$ $AC=AF−CF=8−2=\boxed{6}$; \\
(2) As shown in Figure 2, when point C is on the extension of line segment $AF$, \\
Similarly, it is obtained that $AF=AE=8$, $CF=BE=2$, \\
$\therefore$ $AC=AF+CF=8+2=10$.


Question: As shown in the figure, in trapezoid $ABCD$, where $AD \parallel BC$, and the diagonals $AC$ and $BD$ intersect at point $O$, the area of $\triangle AOD$ is 9, and the area of $\triangle BOC$ is 16. What is the area of $\triangle AOB$?

Solution:
\textbf{Solution:} Since $AD \parallel BC$,

it follows that $\triangle AOD \sim \triangle COB$,

and given $S_{\triangle AOD} = 9$, $S_{\triangle BOC} = 16$,

then $\frac{AO}{OC} = \sqrt{\frac{S_{\triangle AOD}}{S_{\triangle BOC}}} = \frac{3}{4}$.

Since $\triangle AOB$ and $\triangle BOC$ have the same height,

it follows that $S_{\triangle AOB} = \boxed{12}$.


Question: As shown in the figure, it is known that point $D$ is the midpoint of side $AC$ of $\triangle ABC$, $AE \parallel BC$, line $ED$ intersects $AB$ at point $G$ and extends to intersect $BC$ at point $F$. If $\frac{BG}{GA} = 3$ and $BC = 8$, what is the length of $AE$?

Solution:
\textbf{Solution:} Since $AE \parallel BC$, \\
it follows that $\triangle AEG \sim \triangle BFG$, and $\triangle AED \sim \triangle CFD$, \\
thus, $\frac{AE}{BF}=\frac{AG}{BG}=\frac{1}{3}$, and $\frac{AE}{CF}=\frac{AD}{CD}=1$, \\
which means $AE=CF$, \\
Also, given $BC=8$, \\
therefore $\frac{AE}{8+AE}=\frac{1}{3}$\\
$AE=\boxed{4}$.


Question: As shown in the figure, square $EFGH$ is inscribed in right-angled triangle $ABC$, with $\angle A=90^\circ$ and $BC=12$. If the area of $\triangle ABC$ is 36, what is the length of $EH$?

Solution:
\textbf{Solution:} As shown in the figure, draw $AD\perp BC$ at $D$, intersecting $EH$ at $M$, \\
$\because$ The area of $\triangle ABC$ is $36$, $BC=12$, \\
$\therefore$ $\frac{1}{2}BC\times AD=36$, \\
$\therefore$ $\frac{1}{2}\times 12\times AD=36$, \\
$\therefore$ $AD=6$, \\
$\because$ Square $EFGH$ is inscribed in $\angle ABC$, \\
$\therefore$ $EH\parallel FG$, let $EH=EF=FG=HG=x$, \\
$\therefore$ $\angle AEH=\angle B$, $\angle AHE=\angle C$, \\
$\therefore$ $\triangle AEH\sim \triangle ABC$, \\
$\therefore$ $\frac{AE}{AB}=\frac{EH}{BC}$, \\
$\because$ $AD\perp BC$, \\
$\therefore$ $\angle ADG=90^\circ$, \\
$\because$ $EH\parallel FG$, \\
$\therefore$ $\angle ADG=\angle AMH=90^\circ$, \\
$\therefore$ $AM\perp EH$, \\
Again, $\because$ $EH\parallel FG$, $AD\perp BC$, \\
$\therefore$ $DM=EF=EH=x$, $\frac{AE}{AB}=\frac{AM}{AD}$, \\
$\therefore$ $AM=6-x$, \\
$\because$ $\frac{AE}{AB}=\frac{EH}{BC}$, \\
$\therefore$ $\frac{EH}{BC}=\frac{AM}{AD}$, \\
$\therefore$ $\frac{x}{12}=\frac{6-x}{6}$, \\
$\therefore$ $x=\boxed{4}$, \\
$\therefore$ $EH=4$.


Question: As shown in the diagram, in the parallelogram $ABCD$, $AB=10$, $AD=15$, and the bisector of $\angle BAD$ intersects $BC$ at point $E$ and the extension of $DC$ at point $F$. $BG\perp AE$ at point $G$. If $BG=8$, then what is the perimeter of $\triangle CEF$?

Solution:
\textbf{Solution:} Let us consider quadrilateral $ABCD$, where $CD=AB=10$, $BC=AD=15$. The bisector of $\angle BAD$ meets $BC$ at point $E$,\\
therefore, $AB\parallel DC$, $\angle BAF=\angle DAF$,\\
thus, $\angle BAF=\angle F$,\\
and hence, $\angle DAF=\angle F$,\\
leading to $DF=AD=15$, similarly $BE=AB=10$,\\
hence, $CF=DF-CD=15-10=5$;\\
since $BG\perp AE$,\\
it follows that $AE=2AG$,\\
within $\triangle ABG$, where $BG\perp AE$, and $AB=10$, $BG=8$,\\
in right triangle $\triangle ABG$, $AG=\sqrt{AB^{2}-BG^{2}}=\sqrt{100-64}=6$,\\
thus, $AE=2AG=12$,\\
hence, the perimeter of $\triangle ABE$ equals $10+10+12=32$,\\
since quadrilateral ABCD is a parallelogram,\\
it follows that $AB\parallel CF$,\\
hence, $\triangle CEF\sim \triangle BEA$,\\
thus, $\frac{CF}{AB}=\frac{5}{10}=\frac{1}{2}$, i.e., the ratio of similarity is $1\colon 2$,\\
hence, the perimeter of $\triangle CEF$ is $\boxed{16}$.


Question: As shown in the figure, it is known that $AD$ is the angle bisector of $\triangle ABC$, and $DE \parallel AB$. If $\frac{AE}{EC} = \frac{3}{4}$, then what is the value of $\frac{AE}{AB}$?

Solution:
\textbf{Solution:} Since $DE \parallel AB$, \\
it follows that $\triangle CDE \sim \triangle CBA$, \\
hence $\frac{DE}{AB}=\frac{CE}{AC}$, \\
Given $\frac{AE}{EC} = \frac{3}{4}$, \\
it then follows $\frac{DE}{AB}=\frac{CE}{AC} = \frac{4}{7}$, \\
Since $AD$ is the angle bisector of $\triangle ABC$ and $DE \parallel AB$, \\
it follows that $\angle ADE = \angle BAD = \angle DAE$, \\
consequently, $AE = DE$, \\
thus, $\frac{AE}{AB}=\frac{DE}{AB}=\frac{CE}{AC} = \boxed{\frac{4}{7}}$.


Question: As shown in the figure, in the equilateral triangle $\triangle CDE$ with the vertex $D$ sliding along the side $AB$ of the equilateral triangle $\triangle ABC$, $DE$ intersects $BC$ at point $F$. When $AD:DB=3:4$, what is the value of $\frac{BF}{CF}$?

Solution:
\textbf{Solution:} Given that $AD:DB=3:4$,\\
we can assume $AD=3k$, $BD=4k$, hence $AB=AC=BC=7k$,\\
since $\triangle ABC$ and $\triangle CDE$ are equilateral,\\
$\angle A=\angle B=\angle CDE=60^\circ$,\\
and since $\angle A+\angle ACD=\angle CDF+\angle BDF$,\\
it follows that $\angle ACD=\angle BDF$,\\
thus $\triangle ACD \sim \triangle BDF$,\\
implying that $\frac{AC}{BD}=\frac{AD}{BF}$,\\
thus $\frac{7k}{4k}=\frac{3k}{BF}$,\\
hence $BF=\frac{12}{7}k$,\\
therefore $CF=BC−BF=7k−\frac{12}{7}k=\frac{37}{7}k$,\\
hence $\frac{BF}{CF}=\frac{\frac{12}{7}k}{\frac{37}{7}k}=\boxed{\frac{12}{37}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $CD$ is the altitude on the hypotenuse $AB$. If $AD=3$ and $BD=2$, what is the length of $CD$?

Solution:
\textbf{Solution:} Since $\angle ACB=90^\circ$,\\
it follows that $\angle ACD+\angle BCD=90^\circ$.\\
Since $CD$ is the altitude from the hypotenuse $AB$,\\
it follows that $\angle ADC=\angle CDB=90^\circ$, $\angle ACD+\angle CAD=90^\circ$,\\
therefore $\angle CAD=\angle BCD$,\\
therefore $\triangle CAD\sim \triangle BCD$,\\
therefore $\frac{AD}{CD}=\frac{CD}{BD}$,\\
therefore $CD^2=3\times 2=6$,\\
therefore $CD=\sqrt{6}$ (negative values discarded)\\
Thus, the answer is: $CD=\boxed{\sqrt{6}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=90^\circ$, $AC=13$, and $\tan A=\frac{5}{12}$. $CD$ is the median to side $AB$. $\triangle ABC$ is rotated about point D, and points A, B, C correspond to points $A'$, $B'$, $C'$, respectively. $C'D$ intersects side $AC$ at point E. During the rotation, if $\frac{AD}{AB}=\frac{DE}{BC}$, then what is the value of $\frac{AE}{AC}$?

Solution:
\textbf{Solution:} Since $\angle B=90^\circ$,\\
it follows that $\tan A=\frac{BC}{AB}=\frac{5}{12}$.\\
Let $BC=5x$, then $AB=12x$,\\
since $AC^2=AB^2+BC^2$,\\
we find that $13^2=(12x)^2+(5x)^2$,\\
resulting in: $x=1$ (negative values discarded),\\
thus $BC=5$, and therefore $AB=12$.\\
Since $CD$ is the midline on side $AB$,\\
it follows that $AD=BD=\frac{1}{2}AB=6$,\\
thus $\frac{AD}{AB}=\frac{6}{12}=\frac{1}{2}=\frac{DE}{BC}$,\\
only when $\triangle ADE\sim \triangle ABC$ does it satisfy $\frac{AD}{AB}=\frac{DE}{BC}=\frac{1}{2}$, as shown in the figure,\\
thus $\frac{AE}{AC}=\boxed{\frac{1}{2}}$.


Question: As shown in the figure, $ \triangle ABC$ is an isosceles right triangle, with $ \angle C=90^\circ$. Point D is on side $BC$, and segment $AD$ is drawn. Through point B, draw $ BE\perp AD$, intersecting the extension of $AD$ at point E. If $ \frac{CD}{BD}=\frac{1}{2}$, what is the value of $ \frac{BE}{AD}$?

Solution:
\textbf{Solution:} Since $\frac{CD}{BD}=\frac{1}{2}$,

we can assume $CD=k$, $BD=2k$, hence $CB=CA=3k$,

since $\angle C=90^\circ$,

then $AD=\sqrt{AC^2+CD^2}=\sqrt{(3k)^2+k^2}=\sqrt{10}k$,

since $BE\perp AD$,

then $\angle E=\angle C=90^\circ$,

since $\angle CDA=\angle BDE$,

then $\triangle ACD\sim \triangle BED$,

hence $\frac{AC}{BE}=\frac{AD}{BD}$,

thus $\frac{3k}{BE}=\frac{\sqrt{10}k}{2k}$,

therefore $BE=\frac{3\sqrt{10}}{5}k$,

finally, $\frac{BE}{AD}=\frac{\frac{3\sqrt{10}}{5}k}{\sqrt{10}k}=\boxed{\frac{3}{5}}$


Question: As shown in the figure, $DE$ is the median of $\triangle ABC$, $M$ is the midpoint of $DE$, and the extension line of $CM$ intersects $AB$ at point $N$. What is the value of $\frac{{S}_{\triangle DMN}}{{S}_{\text{quadrilateral}DBCM}}$?

Solution:
\textbf{Solution:} Given that $DE$ is the midline of $\triangle ABC$,\\
thus $\frac{DE}{BC}=\frac{1}{2}$, $DE\parallel BC$\\
Furthermore, since M is the midpoint of $DE$,\\
thus $\frac{DM}{DE}=\frac{1}{2}$,\\
therefore $\frac{DM}{BC}=\frac{1}{4}$,\\
since $DE\parallel BC$,\\
thus $\triangle DMN\sim \triangle BCN$,\\
therefore $\frac{{S}_{\triangle DMN}}{{S}_{\triangle BCN}}=\left(\frac{DM}{BC}\right)^2=\frac{1}{16}$,\\
therefore ${S}_{\triangle DMN}\colon {S}_{\text{quadrilateral} DBCM}=1\colon 15$.\\
Hence the answer is: $\boxed{1\colon 15}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $DE\parallel BC$ intersects sides $AB$ and $AC$ at points D and E, respectively, and $DE$ divides $\triangle ABC$ into two parts of equal area. Fold $\triangle ADE$ along line $DE$ so that point A falls onto point F, $DF$ intersects $BC$ at point G, connect $CF$. If $CF\parallel AB$, what is the value of the ratio $CF\colon AB$?

Solution:
\textbf{Solution:} Draw line $AF$, intersecting $DE$ at $M$, and $BC$ at $N$,\\
$\because$ folding triangle $ADE$ along the line $DE$, point $A$ lands on point $F$,\\
$\therefore$ $AF\perp BC$, $AM=FM$,\\
$\because$ $DE\parallel BC$,\\
$\therefore$ $\triangle ADE\sim \triangle ABC$,\\
$\because$ $DE$ divides $\triangle ABC$ into two parts of equal area,\\
$\therefore$ $\frac{S_{\triangle ADE}}{S_{\triangle ABC}}=\frac{1}{2}$,\\
$\therefore$ $\frac{AM}{AN}=\frac{1}{\sqrt{2}}$,\\
$\therefore$ $\frac{AM}{MN}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$,\\
$\therefore$ $AM=(\sqrt{2}+1)MN$,\\
$\therefore$ $MF=AM=(\sqrt{2}+1)MN$, $AN=AM+MN=(\sqrt{2}+1)MN+MN=(\sqrt{2}+2)MN$,\\
$\therefore$ $NF=MF-MN=(\sqrt{2}+1)MN-MN=\sqrt{2}MN$,\\
$\because$ $CF\parallel AB$,\\
$\therefore$ $\frac{CF}{AB}=\frac{NF}{AN}=\frac{\sqrt{2}MN}{(\sqrt{2}+2)MN}=\boxed{\sqrt{2}-1}$,