Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and $P$ is a point on $\odot O$. If $\angle PAB=32^\circ$, what is the degree measure of $\angle PBA$?

Solution:
\textbf{Solution:} Since $AB$ is the diameter of $\odot O$, \\
$\therefore \angle APB=90^\circ$ \\
Since $\angle PAB=32^\circ$, \\
$\therefore \angle PBA=\boxed{58^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, a semicircle $O$ with diameter $BC$ intersects $AB$, $AC$ at points $D$, $E$, respectively. Segments $OD$, $OE$ are connected. If $\angle A=\alpha$, what is the measure of $\angle DOE$?

Solution:
\textbf{Solution}: As shown in the figure, connect CD,\\
$\because$BC is the diameter of semicircle O,\\
$\therefore \angle BDC=90^\circ$,\\
$\because \angle A=\alpha$,\\
$\therefore \angle ACD=90^\circ-\angle A=90^\circ-\alpha$,\\
$\because \angle ACD$ and $\angle DOE$ are the inscribed angle and the central angle of arc $\overset{\frown}{DE}$, respectively,\\
$\therefore \angle DOE=2\angle ACD=180^\circ-2\alpha$,\\
Therefore, $\angle DOE=\boxed{180^\circ-2\alpha}$.


Question: As shown in the figure, in circle $\odot O$, $AB$ is the diameter, and point $C$ is a point on the circle. The minor arc $AC$ is folded along chord $AC$ to intersect $AB$ at point $D$. Connect $CD$. If $\angle BAC=18^\circ$, then what is the measure of $\angle BDC$?

Solution:
\textbf{Solution:} Since $AB$ is the diameter, \\
$\therefore \angle ACB=90^\circ$, \\
Since $\angle BAC=18^\circ$, \\
$\therefore \angle B=90^\circ-\angle BAC=90^\circ-18^\circ=72^\circ$, \\
According to the property of reflection, \\
the inscribed angle corresponding to arc $\overset{\frown}{AC}$ is $\angle B$, \\
the inscribed angle corresponding to arc $\overset{\frown}{ABC}$ is $\angle ADC$, \\
$\therefore \angle ADC+\angle B=180^\circ$, \\
Since $\angle ADC+\angle BDC=180^\circ$, \\
$\therefore \angle B=\angle BDC=\boxed{72^\circ}$.


Question: As shown in the figure, for the circumcircle of $\triangle ABC$, with $AB = BC = 4$, the arc $AB$ is folded down along the chord $AB$ to intersect $BC$ at point $D$. If point $D$ is the midpoint of $BC$, what is the length of $AC$?

Solution:
\textbf{Solution:} As shown in the figure, draw AD,\\
$\because$ $AB=BC=4$,\\
$\therefore$ $\angle BAC=\angle BCA$,\\
$\because$ point D is the midpoint of BC,\\
$\therefore$ $BD=CD=2$,\\
$\because$ the arc AB folds down along the chord AB and intersects BC at point D,\\
$\therefore$ $\overset{\frown}{AB}=\overset{\frown}{ADB}$,\\
$\therefore$ $\angle BCA=\angle ABD+\angle BAD$,\\
Also, $\because$ $\angle BAC=\angle BAD+\angle CAD$,\\
$\therefore$ $\angle ABD=\angle CAD$,\\
Also, $\because$ $\angle ACB=\angle ACD$,\\
$\therefore$ $\triangle ACD\sim \triangle BCA$,\\
$\therefore$ $\frac{CD}{AC}=\frac{AC}{BC}$,\\
$\therefore$ $\frac{2}{AC}=\frac{AC}{4}$,\\
$\therefore$ $AC=\boxed{2\sqrt{2}}$.


Question: As shown in the figure, in $ \triangle ABC$, D is the midpoint of AC, and CE $\perp$ AB at point E. If DE = 3 and AE = 5, then what is the length of CE?

Solution: \textbf{Solution:} Since CE$\perp$AB, it follows that $\angle$AEC$=90^\circ$. In the right-angled triangle AEC, since point D is the midpoint of the hypotenuse AC, and DE=3, it follows that AC=2DE=6. By the Pythagorean theorem, we have $CE=\sqrt{AC^{2}-AE^{2}}=\sqrt{6^{2}-5^{2}}=\sqrt{11}=\boxed{\sqrt{11}}$.


Question: As shown in the figure, $\triangle ABC$ is folded along line MN such that point A coincides with point E on side AB. If $\angle B=54^\circ$ and $\angle C=90^\circ$, what is the measure of $\angle ENC$?

Solution:
\textbf{Solution:}
\[
\because \angle B=54^\circ, \angle C=90^\circ,
\]
\[
\therefore \angle A=90^\circ-\angle B=36^\circ,
\]
From the property of folding, we get:
\[
\angle AEN=\angle A=36^\circ,
\]
\[
\therefore \angle ENC=\angle AEN+\angle A=\boxed{72^\circ}.
\]


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, point D is on side BC, AD=BD, and DE bisects $\angle ADB$ and intersects AB at point E. If AC=12 and BC=16, what is the length of AE?

Solution:
\textbf{Solution:} As shown in the diagram, in $\triangle ABC$, $\angle C=90^\circ$, $AC=12$, $BC=16$,\\
by the Pythagorean theorem, we have $AB=\sqrt{AC^2+BC^2}=\sqrt{12^2+16^2}=20$,\\
since $AD=BD$ and $DE$ bisects $\angle ADB$ and intersects $AB$ at point $E$.\\
Therefore, $AE=BE=\frac{1}{2}AB=\boxed{10}$.


Question: As shown in the figure, in $\triangle ABC$, $AC = 10$, the perpendicular bisector of $AB$ intersects $AB$ at point $M$ and intersects $AC$ at point $D$. The perimeter of $\triangle BDC$ is $18$. What is the length of $BC$?

Solution:
\textbf{Solution:} Since $MN$ is the perpendicular bisector of $AB$,\\
then $BD = AD$,\\
thus the perimeter of $\triangle BDC = BD + CD + BC = AD + CD + BC = AC + BC = 18$,\\
since $AC = 10$,\\
therefore $BC = \boxed{8}$.


Question: As illustrated, to calculate the height of the school's flagpole, Xiao Ming pulled the flag-raising rope to the bottom of the flagpole and tied a knot at the point where the rope was as long as the flagpole. Then, he pulled the bottom of the rope to a point on the ground 5 meters away from the bottom of the flagpole, and found that the distance from the bottom end of the rope to the knot was about 1 meter. What is the height of the flagpole?

Solution:
\textbf{Solution:} Let the height of the flagpole be $x$ m, thus $AC = x$ m, $AB= (x+1)$ m, $BC=5$ m,\\
In $\triangle ABC$, by the Pythagorean theorem we have $AC^2+BC^2=AB^2$\\
$\therefore x^2+5^2=(x+1)^2$\\
Solving yields: $x=\boxed{12}$


Question: As shown in the figure, if $\triangle ABC \cong \triangle DEF$ with points $B$, $E$, $C$, and $F$ lying on the same straight line, and $BC=7$, $EC=4$, then what is the length of $CF$?

Solution:
\textbf{Solution:} Since $\triangle ABC \cong \triangle DEF$ and $BC = 7$, \\
$\therefore EF = BC = 7$, \\
$\therefore CF = EF - EC = \boxed{3}$


Question: As shown in the figure, in quadrilateral $ABCD$, $DA\perp AB$, $DA=6\,\text{cm}$, $\angle B+\angle C=150^\circ$, point $A$ is exactly the midpoint of $EB$, and $P$, $Q$ are moving points on the line segments $CE$, $BE$ respectively. What is the minimum value of $BP+PQ$?

Solution:
\textbf{Solution:} As shown in the diagram, construct point $F$ as the symmetric point of $B$ with respect to $CE$. Draw $BF$ and $EF$. Then, $EB=EF$,

since $\angle B + \angle C = 150^\circ$, it follows that $\angle BEC = 180^\circ - (\angle B + \angle C) = 30^\circ$. Since point $B$ is symmetric to point $F$ with respect to $EC$, it follows that $\angle BEC = \angle FEC = 30^\circ$, hence $\angle BEF = 60^\circ$, therefore $\triangle BEF$ is an equilateral triangle. Draw $BP$, $PF$, $PQ$, then $BP=FP$, therefore $BP + QP = FP + PQ$. Thus, when $F$, $P$, and $Q$ are on the same line and $FQ \perp EB$, then the minimum value of $BP+PQ$ is the length of $FQ$. At this moment, $Q$ coincides with the midpoint of $EB$, hence coincides with point $A$. Since $DA \perp AB$ and $DA=6$cm, it follows that $AE= 6\sqrt{3}$cm. In $\triangle QEF$, $FQ = \sqrt{3}AE = \sqrt{3} \times 6\sqrt{3} = \boxed{18}$cm. Therefore, the minimum value of $BP+PQ$ is $18$cm.


Question: As shown in the figure, it is known that $AB=AC$, $AD=AE$, $\angle BAC=\angle DAE$, $\angle 1=30^\circ$, $\angle 2=40^\circ$. What is the degree measure of $\angle 3$?

Solution:
\textbf{Solution:} Since $\angle BAC=\angle DAE$,\\
it follows that $\angle BAC-\angle DAC=\angle DAE-\angle DAC$, i.e., $\angle BAD=\angle CAE$,\\
thus $\triangle ABD\cong \triangle ACE\ (SAS)$,\\
hence $\angle ABD=\angle 2=40^\circ$,\\
since $\angle 1=30^\circ$,\\
therefore $\angle 3=\boxed{70^\circ}$.


Question: As shown in the figure, in the Cartesian coordinate system, the rectangle $AOBC$ has sides $OB$ and $OA$ on the x-axis and y-axis, respectively, with point $D$ on side $BC$. The rectangle is folded along $AD$ such that point $C$ precisely lands on point $E$ on side $OB$. If point $A(0, 8)$ and point $B(10, 0)$, what are the coordinates of point $D$?

Solution:
\textbf{Solution:}

Since rectangle AOBC,\\
thus AC=OB, $\angle$AOB=$\angle$DBE=$90^\circ$,\\
Since point A (0, 8), point B (10, 0),\\
thus OA=8, AC=OB=10,\\
Since the rectangle is folded along AD, point C perfectly falls on side OB,\\
thus AC=AE=10, DC=DE,\\
In Rt$\triangle$AOE,\\
$OE=\sqrt{AE^{2}-OA^{2}}=\sqrt{10^{2}-8^{2}}=6$,\\
thus BE=OB−OE=10−6=4,\\
Let DC=DE=x, then DB=8−x,\\
In Rt$\triangle$DEB, $DE^{2}=BE^{2}+DB^{2}$ implies $x^{2}=4^{2}+(8−x)^{2}$,\\
Solving this: $x=5$,\\
thus DB=8−5=3,\\
thus the coordinates of point D are $\boxed{(10, 3)}$.


Question: As shown in the figure, it is known that $\triangle ABC$ is an equilateral triangle with a side length of 3, $\angle BDC = 120^\circ$, $BD = CD$, points M and N are located on sides $AB$ and $AC$ respectively, and $\angle MDN = 60^\circ$. Connecting $MN$, what is the perimeter of $\triangle AMN$?

Solution:
\textbf{Solution:} Extend $AB$ to $F$ such that $BF=CN$, connect $DF$, \\
$\because$ $\triangle BDC$ is an isosceles triangle, $\angle BDC=120^\circ$, \\
$\therefore$ $\angle BCD=\angle DBC=30^\circ$, \\
$\because$ $\triangle ABC$ is an equilateral triangle with side length of 3, \\
$\therefore$ $\angle ABC=\angle BAC=\angle BCA=60^\circ$, \\
$\therefore$ $\angle DBA=\angle DCA=90^\circ$, \\
In $\triangle DBF$ and $\triangle CND$, $\left\{\begin{array}{l}BF=CN \\ \angle DBF=\angle DCN \\ DB=DC\end{array}\right.$, \\
$\therefore$ $\triangle BDF\cong \triangle CDN\left(SAS\right)$, \\
$\therefore$ $\angle BDF=\angle CDN$, $DF=DN$, \\
$\because$ $\angle MDN=60^\circ$, \\
$\therefore$ $\angle BDM+\angle CDN=60^\circ$, \\
$\therefore$ $\angle BDM+\angle BDF=60^\circ=\angle FDM=\angle MDN$, \\
In $\triangle DMN$ and $\triangle DMF$, $\left\{\begin{array}{l}DN=DF \\ \angle MDN=\angle MDF \\ DM=MD\end{array}\right.$, \\
$\therefore$ $\triangle DMN\cong \triangle DMF\left(SAS\right)$, \\
$\therefore$ $MN=MF$, \\
$\therefore$ $MF=BF+BM=BM+CN=MN$, \\
$\therefore$ The perimeter of $\triangle AMN$ $AM+AN+MN=AM+MB+BF+AN=AB+AN+CN=AB+AC=\boxed{6}$.


Question: As shown in the figure, points A, D, C, E are on the same straight line, $AB \parallel EF$, $AB=EF$, $\angle B = \angle F$, $AE=10$, $AC=6$, then what is the length of CD?

Solution:
\textbf{Solution:} Since $AB\parallel EF$,\\
$\therefore \angle A = \angle E$,\\
In $\triangle ABC$ and $\triangle EFD$,\\
$\angle A = \angle E$, $AB = EF$, $\angle B = \angle F$,\\
$\therefore \triangle ABC \cong \triangle EFD$ (ASA),\\
$\therefore AC = ED = 6$,\\
$\therefore AD = AE - ED = 10 - 6 = 4$,\\
$\therefore CD = AC - AD = 6 - 4 = \boxed{2}$.


Question: As shown in the figure, $\angle A = \angle B$, $AE = BE$, point $D$ is on side $AC$, $\angle 1 = \angle 2$, $AE$ and $BD$ intersect at point $O$, if $\angle 1 = 38^\circ$, then what is the degree measure of $\angle BDE$?

Solution:
\textbf{Solution:} Given that $AE$ and $BD$ intersect at point $O$,\\
$\therefore \angle AOD=\angle BOE$,\\
In $\triangle AOD$ and $\triangle BOE$, $\angle A=\angle B$,\\
$\therefore \angle BEO=\angle 2$,\\
$\because \angle 1=\angle 2$,\\
$\therefore \angle 1=\angle BEO$,\\
$\therefore \angle 1+\angle DEO=\angle BEO+\angle DEO$,\\
that is, $\angle AEC=\angle BED$,\\
In $\triangle AEC$ and $\triangle BED$,\\
$\left\{\begin{array}{l}
\angle A=\angle B \\
\angle AEC=\angle BED \\
AE=BE
\end{array}\right.$\\
$\therefore \triangle AEC\cong\triangle BED\left(AAS\right)$,\\
$\therefore CE=DE$,\\
$\because \angle 1=38^\circ$,\\
$\therefore \angle C=\angle EDC=\frac{1}{2}\times \left(180^\circ-\angle 1\right)=\frac{1}{2}\times \left(180^\circ-38^\circ\right)=71^\circ$,\\
$\therefore \angle BDE=\boxed{71^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, where $AB=AC=20cm$, $DE$ is perpendicular to and bisects $AB$ at point $E$, and intersects $AC$ at $D$. If $BC=15cm$, what is the perimeter of $\triangle DBC$?

Solution:
\textbf{Solution:} Since $DE$ is the perpendicular bisector of $AB$,\\
$\therefore AD=BD$,\\
$\therefore$ the perimeter of $\triangle BCD = BD + CD + BC = AD + CD + BC = AC + BC$\\
Also, $\because AB = AC = 20\,\text{cm}$, $BC = 15\,\text{cm}$,\\
$\therefore$ the perimeter of $\triangle BCD = 20 + 15 = \boxed{35}\,\text{cm}$.


Question: As shown in the figure, right-angled triangles $AOB$ and $COD$ are positioned as depicted (with the right-angled vertices coinciding). Given that $\angle AOC=45^\circ$, what is the measure of $\angle DEB$?

Solution:
\textbf{Solution:} Since $\angle AOC=45^\circ$ and $\angle C=45^\circ$,\\
it follows that $\angle AFE=\angle AOC+\angle C=90^\circ$.\\
Since $\angle A=30^\circ$,\\
it follows that $\angle AEF=60^\circ$,\\
therefore, $\angle DEB=\angle AEF=\boxed{60^\circ}$.


Question: As shown in the figure, it is known that $ \triangle ABC \cong \triangle DCB$, $ \angle A = 80^\circ$, and $ \angle ACB = 40^\circ$. What is the measure of $ \angle ABD$?

Solution:
\textbf{Solution:} Given that $\because$ $\triangle ABC\cong \triangle DCB$\\
$\therefore$ $\angle DBC=\angle ACB=40^\circ$\\
In $\triangle ABC$\\
$\angle ABC=180^\circ-\angle A-\angle ACB=60^\circ$\\
$\therefore$ $\angle ABD=\angle ABC-\angle DBC=20^\circ$\\
Hence, the answer is: $\boxed{20^\circ}$.


Question: As shown in the figure, $ \triangle ABC \cong \triangle DEF$. If $AC = 5$ and $CF = 2$, what is the length of $CD$?

Solution:
\textbf{Solution:} Given that $\triangle ABC\cong \triangle DEF$, $\angle A$ and $\angle D$ are corresponding angles, and $AB$ and $DE$ are corresponding sides,\\
$\therefore AC=DF=5$,\\
Also, since $CF=2$,\\
$\therefore CD=DF-CF=5-2=\boxed{3}$.


Question: As shown in the figure, $AB=AC$, $AD=AE$, $\angle BAC=\angle DAE$, and points $B$, $D$, $E$ are on the same line. If $CE=3$, $DE=5$, then what is the length of $BE$?

Solution:
\textbf{Solution:} Given that $\angle BAC = \angle BAD + \angle DAC$ and $\angle DAE = \angle CAE + \angle DAC$, \\
Since $\angle BAC = \angle DAE$, \\
$\therefore \angle BAD = \angle CAE$, \\
In $\triangle BAD$ and $\triangle CAE$, \\
$\left\{\begin{array}{l}
AB = AC \\
\angle BAD = \angle CAE \\
AD = AE
\end{array}\right.$, \\
$\therefore \triangle BAD \cong \triangle CAE \left( SAS \right)$, \\
$\therefore BD = CE = 3$, \\
$\therefore BE = BD + DE = 3 + 5 = \boxed{8}$.


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

Solution:
\textbf{Solution:} Given that $\triangle ABC \cong \triangle DBE$ and $\angle ABD = 23^\circ$, $\angle DBE = 45^\circ$,\\
it follows that $\angle ABC = \angle DBE = 45^\circ$,\\
therefore, $\angle DBC = \angle ABC - \angle ABD = \boxed{22^\circ}$.


Question: As shown in the diagram, in $\triangle ABC$, $\angle BAC=115^\circ$, the perpendicular bisectors of $AB$ and $CD$ intersect $BC$ at points $E$ and $F$, respectively. What is the measure of $\angle EAF$?

Solution:
\textbf{Solution:} Since the perpendicular bisectors of AB and AC intersect BC at points E and F, respectively,\\
we have EA = EB, thus $\angle B = \angle EAG$,\\
Let $\angle B = \angle EAG = x$ degrees,\\
Since FA = FC, then $\angle C = \angle FAH$,\\
Let $\angle C = \angle FAH = y$,\\
Since $\angle BAC = 115^\circ$,\\
Therefore, $x + y + \angle EAF = 115^\circ$,\\
According to the triangle angle sum theorem, $x + y + x + y + \angle EAF = 180^\circ$,\\
Solving this gives $\angle EAF = \boxed{50^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, point $D$ is on the extension line of $BC$. If $\angle ACD = 108^\circ$ and $\angle B = 45^\circ$, then what is the degree measure of $\angle A$?

Solution:
\textbf{Solution:} Since $\angle ACD$ is an exterior angle of $\triangle ABC$,\\
it follows that $\angle ACD=\angle A+\angle B$,\\
Given that $\angle ACD=108^\circ$, $\angle B=45^\circ$,\\
thus $\angle A=\angle ACD−\angle B=108^\circ−45^\circ=\boxed{63^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=130^\circ$, the perpendicular bisectors of AB and AC respectively intersect BC at points E and F, and intersect AB and AC at points D and G, respectively. What is the measure of $\angle EAF$?

Solution:
\textbf{Solution:} Since $\angle$BAC=$130^\circ$,\\
it follows that $\angle$B+$\angle$C=$180^\circ-\angle$BAC=$180^\circ-130^\circ$=$50^\circ$;\\
Since DE bisects AB perpendicularly, and FG bisects AC perpendicularly,\\
it follows that BE=AE, and AF=CF,\\
which implies that $\angle$B=$\angle$BAE, $\angle$C=$\angle$CAF,\\
hence $\angle$EAF=$\angle$BAC-$\angle$BAE-$\angle$CAF=$\angle$BAC-$\angle$B-$\angle$C=$130^\circ-50^\circ=\boxed{80^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=3$, $BC=4$, $BE$ bisects $\angle ABC$, $CD\perp AB$ at $D$, and $BE$ intersects $CD$ at $F$. What is the length of $CF$?

Solution:
\textbf{Solution:} Construct line EG such that EG $\perp$ AB at point G. Since CD $\perp$ AB, it follows that EG $\parallel$ CD, thus $\angle$BEG = $\angle$CFE. Since BE bisects $\angle$ABC and EG $\perp$ AB, with $\angle$ACB = 90$^\circ$ implying AC $\perp$ BC, it follows that EG = CE. In right triangles $\triangle$BCE and $\triangle$BGE, we have
\[\left\{\begin{array}{l}
BE = BE\\
CE = EG
\end{array}\right.\]
Therefore, right $\triangle$BCE $\cong$ right $\triangle$BGE (HL), leading to BC = BG = 4, and $\angle$BEG = $\angle$BEC = $\angle$CFE, thus CE = CF = EG. In right $\triangle$ABC, we have
\[AB = \sqrt{AC^{2} + BC^{2}} = \sqrt{3^{2} + 4^{2}} = 5,\]
therefore AG = AB - BG = 5 - 1 = 4. Let CF = x, then AE = 3 - x. In right $\triangle$AEG, we have
\[AG^{2} + EG^{2} = AE^{2}\]
which means
\[1^{2} + x^{2} = (3 - x)^{2}.\]
Solving this equation, we find
\[x = \frac{4}{3},\]
therefore CF = \boxed{\frac{4}{3}}.


Question: As shown in the figure, $\angle AOP=\angle BOP=15^\circ$, $PC \parallel OA$ intersects $OB$ at point C, $PD \perp OA$ at D, if $PC=3$, then what is the value of $PD$?

Solution:
\textbf{Solution:} Draw PE $\perp$ OB at point E,\\
since PC $\parallel$ OA,\\
it follows that $\angle$CPO = $\angle$AOP,\\
since $\angle$AOP = $\angle$BOP = $15^\circ$,\\
it follows that $\angle$BOP = $\angle$CPO = $15^\circ$,\\
thus, $\angle$BCP = $\angle$BOP + $\angle$CPO = $30^\circ$,\\
therefore, PE = $\frac{1}{2}$PC = $\frac{3}{2}$,\\
since $\angle$AOP = $\angle$BOP, and PE $\perp$ OB, PD $\perp$ OA at D,\\
thus, PD = PE = $\frac{3}{2}$ = \boxed{1.5}.


Question: As shown in the figure, let there be an equilateral triangle $ABC$ with side length $a$. Let $P$ be a point on side $AB$, and draw $PE\perp AC$ at point $E$. Let $Q$ be a point on the extension of $BC$. When $AP=CQ$, line $PQ$ intersects $AC$ at point $D$. What is the length of $DE$?

Solution:
\textbf{Solution:} As shown in the figure, extend line $AD$ through point $Q$ and draw a perpendicular line to it at point $F$, \\
since $\triangle ABC$ is an equilateral triangle, \\
therefore $\angle A=\angle ACB=60^\circ$, \\
since $\angle ACB=\angle QCF$, \\
therefore $\angle QCF=\angle A=60^\circ$, \\
also, since $PE\perp AC$, $QF\perp AC$, \\
therefore $\angle AEP=\angle CFQ=90^\circ$, \\
and $AP=CQ$, \\
therefore $\triangle AEP\cong \triangle CFQ$ (AAS), \\
therefore $AE=CF$, $PE=QF$, \\
Similarly, it can be proved that $\triangle DEP\cong \triangle DFQ$, \\
therefore $DE=DF$, \\
therefore $AC=AE+DE+CD=DE+CD+CF=DE+DF=2DE$, \\
therefore $DE=\frac{1}{2}AC=\frac{1}{2}a$. \\
Hence, $DE=\boxed{\frac{1}{2}a}$.


Question: As shown in the figure, point \(O\) is a point inside \(\triangle ABC\), \(\angle ABO=30^\circ\), \(\angle ACO=15^\circ\), \(\angle BOC=100^\circ\), then what is the degree measure of \(\angle A\)?

Solution:
\textbf{Solution:} Since, in $\triangle OBC$, $\angle BOC=100^\circ$,\\
it follows that $\angle OBC+\angle OCB=180^\circ-\angle BOC=80^\circ$,\\
In $\triangle ABC$,\\
$\angle A=180^\circ-(\angle ABO+\angle ACO+\angle OBC+\angle OCB)=180^\circ-(30^\circ+15^\circ+80^\circ)=\boxed{55^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AC=BC$, point $D$ is on side $AC$, and point $E$ is on the extension line of $CB$. $DE$ intersects $AB$ at point $F$. If $\angle C=50^\circ$ and $\angle E=25^\circ$, then what is the degree measure of $\angle BFE$?

Solution:
\textbf{Solution:} Given $AC=BC$, \\
$\therefore \angle A=\angle ABC$, \\
Since $\angle C=50^\circ$, and $\angle A+\angle ABC+\angle C=180^\circ$, \\
$\therefore \angle ABC= \frac{1}{2}(180^\circ-\angle C)=65^\circ$, \\
Given $\angle E=25^\circ$, and $\angle ABC=\angle E+\angle BFE$, \\
$\therefore \angle BFE=\angle ABC-\angle E=65^\circ-25^\circ=\boxed{40^\circ}$.


Question: As shown in the figure, D and E are the midpoints of BC and AC respectively, the area of $\triangle CDE$ is 2, then what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Since $D$ and $E$ are the midpoints of $BC$ and $AC$ respectively, \\
therefore, $AD$ and $DE$ are the medians of $\triangle ABC$ and $\triangle ADC$ respectively, \\
therefore, ${S}_{\triangle ADC}=2{S}_{\triangle CDE}=2\times 2=4$, \\
therefore, ${S}_{\triangle ABC}=2{S}_{\triangle ADC}=2\times 4=\boxed{8}$.


Question: As shown in the figure, in $\triangle ABC$, $AD \perp BC$ at point $D$, $BE \perp AC$ at point $E$, and $BE$ intersects $AD$ at point $F$. If $AD=BD=5$ and $CD=3$, what is the length of $AF$?

Solution:
\textbf{Solution:} Since $BE\perp AC$ and $AD\perp BC$,\\
it follows that $\angle AEB=\angle ADC=\angle BDF=90^\circ$,\\
Given $\angle AFE=\angle BFD$ and $\angle FBD+\angle BDF+\angle BFD=180^\circ$, $\angle AEB+\angle AFE+\angle DAC=180^\circ$,\\
it implies $\angle DAC=\angle DBF$,\\
In $\triangle BDF$ and $\triangle ADC$, we have\\
$\left\{\begin{array}{l}
\angle DBF=\angle DAC\\
BD=AD\\
\angle BDF=\angle ADC
\end{array}\right.$,\\
which leads to $\triangle BDF \cong \triangle ADC\left(ASA\right)$,\\
thus $DF=CD=3$,\\
Since $AF+DF=AD=5$,\\
it results in $AF=\boxed{2}$.


Question: As shown in Figure 1, in $\triangle ABC$, $\angle B=80^\circ$, $\angle C=30^\circ$. If $\triangle ABC \cong \triangle ADE$, $\angle DAC=32^\circ$, then what is the degree measure of $\angle EAC$?

Solution:
\textbf{Solution:} Given: $\because$ $\angle B=80^\circ$ and $\angle C=30^\circ$,\\
$\therefore$ $\angle BAC=180^\circ-80^\circ-30^\circ=70^\circ$,\\
$\because$ $\triangle ABC \cong \triangle ADE$,\\
$\therefore$ $\angle BAC=\angle DAE=70^\circ$,\\
$\therefore$ $\angle EAC=\angle DAE-\angle DAC=70^\circ-32^\circ=\boxed{38^\circ}$


Question: As shown in the figure, $ \triangle ABC \cong \triangle A'B'C'$, $ \angle BCB' = 30^\circ$, what is the degree measure of $ \angle ACA'$?

Solution:
\textbf{Solution:} Since $\triangle ABC \cong \triangle A'B'C$,\\
it follows that $\angle ACB = \angle A'CB'$,\\
therefore $\angle ACB - \angle A'CB = \angle A'CB' - \angle A'CB$,\\
which means $\angle ACA' = \angle BCB' = 30^\circ$,\\
thus A meets the conditions stated in the problem.\\
Hence, the answer is: $\boxed{30^\circ}$.


Question: As shown in the figure, points B, E, C, and F lie on the same straight line. Given that triangles $ABC \cong DEF$, with $BC = 8$ and $BF = 13$, what is the length of BE?

Solution:
\textbf{Solution:} Since $\triangle ABC \cong \triangle DEF$, BC=8, BF=13,\\
therefore, EF=BC=8,\\
therefore, BE=BF-EF=13-8=\boxed{5}


Question: When a set of triangular plates are stacked together as shown in the figure, what is the degree measure of $\angle \alpha$ in the diagram?

Solution:
\textbf{Solution:} As shown in the figure,\\
according to the special nature of the angles on the drafting triangle, it is known that $\angle 1=45^\circ,\ \angle 2=60^\circ$. By applying the property of the external angle of a triangle, it follows that $\angle \alpha =\angle 2-\angle 1=60^\circ-45^\circ=\boxed{15^\circ}$.


Question: As shown in the figure, in quadrilateral ABCD, AB $\parallel$ CD, $\angle B = \angle D$. Point E is on the extension of BC. Line segment AE is drawn and intersects CD at point H. The bisector of $\angle DCE$ intersects AE at point G. If AB $= 2$AD $= 10$, point H is the midpoint of CD, and HE $= 6$, then what is the value of AC?

Solution:
\textbf{Solution:} Given that $\because$ AB＝2AD＝10,\\
$\therefore$AD=5,\\
$\because$AB$\parallel$CD,\\
$\therefore$$\angle$B=$\angle$DCE,\\
$\because$$\angle$B=$\angle$D,\\
$\therefore$$\angle$B=$\angle$DCE,\\
$\therefore$AD$\parallel$BC,\\
$\therefore$Quadrilateral ABCD is a parallelogram, $\angle$DAH=$\angle$E,\\
$\therefore$CD=AB=10,\\
$\because$ point H is the midpoint of CD,\\
$\therefore$CH=DH=5,\\
$\therefore$$\triangle $AHD$\cong $$\triangle $ECH (AAS),\\
$\therefore$CE=AD=5=CH, AH=EH=6,\\
$\because$ CG bisects $\angle$DCE,\\
$\therefore$CG$\perp$EH, HG=EG=3,\\
$\therefore$CG=$ \sqrt{CH^{2}-HG^{2}}=\sqrt{5^{2}-3^{2}}$=4, AG=AH+HG=9,\\
$\therefore$AC=$ \sqrt{AG^{2}+CG^{2}}=\sqrt{9^{2}+4^{2}}=\boxed{\sqrt{97}}$


Question: As shown in the figure, the area of $\triangle ABC$ is 24. Point $D$ is a point on side $AC$, and extending $BD$ intersects the line parallel to $BC$ through point $A$ at point $E$. Connect $EC$, and construct parallelogram $DECF$ with $DE$ and $EC$ as adjacent sides. $DF$ intersects side $BC$ at point $H$. Connect $AH$. When $AD = \frac{1}{2}CD$, what is the area of $\triangle AHC$?

Solution:
\textbf{Solution:} As shown in the figure, extend $HD$ to intersect $AG$ at point $Q$, \\
$\because$ quadrilateral $DECF$, \\
$\therefore$ $DF \parallel CE$, $ED \parallel CF$, \\
$\therefore$ $\angle CFH = \angle EDQ$, \\
Also, $\because$ $BC \parallel AQ$, \\
$\therefore$ quadrilateral $HQEC$ is a parallelogram, \\
$\therefore$ $EQ = CH$, \\
Also, $\angle EQD = \angle CHF$, \\
$\therefore$ $\triangle EDQ \cong \triangle CHF$ (AAS), \\
$\therefore$ $S_{\triangle EDQ} = S_{\triangle CHF}$, \\
$\therefore$ $S_{\square HQEC} = S_{\square DECF}$, \\
$\because$ the area of $\triangle ABC$ is 24, \\
$\therefore$ $S_{\triangle BEC} = 24$, \\
Also, $\because$ $AD = \frac{1}{2}CD$, \\
$\therefore$ $S_{\triangle BDC} = \frac{2}{3}S_{\triangle ABC} = 16$, \\
$\therefore$ $S_{\triangle DEC} = S_{\triangle BEC} - S_{\triangle BDC} = 24 - 16 = 8$, \\
$\therefore$ $S_{\square HQEC} = S_{\square DECF} = 2S_{\triangle DEC} = 16$, \\
$\therefore$ $S_{\triangle AHC} = \frac{1}{2}S_{\square HQEC} = \boxed{8}$.


Question: In the figure, within $\triangle ABC$, $CD\perp AB$ at point $D$, and $E$, $F$ are the midpoints of $AC$, $BC$, respectively. Given $AB=10$, $BC=8$, and $DE=4.5$, what is the perimeter of $\triangle DEF$?

Solution:
\textbf{Solution:} Since $E$ and $F$ are the midpoints of $AC$ and $BC$ respectively,\\
therefore $EF=\frac{1}{2}AB=5$,\\
since $CD\perp AB$,\\
therefore $\angle CDB=90^\circ$,\\
therefore $\triangle CDB$ is a right-angled triangle,\\
$DF=\frac{1}{2}BC=4$,\\
therefore the perimeter of $\triangle DEF$ is $DF+EF+DE=4+5+4.5=\boxed{13.5}$.


Question: As shown in the figure, if $D$, $E$, and $F$ are the midpoints of the sides of $\triangle ABC$, with $EF=6\,\text{cm}$, $DE=4\,\text{cm}$, and $DF=5\,\text{cm}$, then what is the perimeter of $\triangle ABC$?

Solution:
\textbf{Solution:} In $\triangle ABC$,\\
since points D, E, F are the midpoints of the sides of $\triangle ABC$,\\
therefore $EF = \frac{1}{2}AB$, $DE = \frac{1}{2}BC$, $DF = \frac{1}{2}AC$,\\
thus, the perimeter of $\triangle ABC$ is: $AB+BC+AC=2EF+2DE+2DF$\\
$=2(EF+DE+DF)=2\times (6+4+5)$\\
$=\boxed{30}$ (cm)


Question: As shown in the figure, diagonals $AC$ and $BD$ of rectangle $ABCD$ intersect at point $O$. If $\angle OAD = 30^\circ$, what is the measure of $\angle AOD$?

Solution:
\textbf{Solution:} Given a rectangle ABCD, where diagonals AC and BD intersect at point O,\\
it follows that $OD=OB=OA=OC$,\\
since $\angle OAD=30^\circ$,\\
it follows that $\angle ODA=\angle OAD=30^\circ$,\\
thus, $\angle AOD=180^\circ-\left(\angle ODA+\angle OAD\right)=180^\circ-\left(30^\circ+30^\circ\right)=\boxed{120^\circ}$.


Question: As shown in the figure, ABCD is a square with a side length of 4. Point E is on the diagonal BD, and $ \angle BAE=22.5^\circ$, $EF\perp AB$ with the foot of the perpendicular at F. What is the length of EF?

Solution:
\textbf{Solution:} In the square $ABCD$, $\angle ABD = \angle ADB = 45^\circ$,\\
$\because \angle BAE = 22.5^\circ$,\\
$\therefore \angle DAE = 90^\circ - \angle BAE = 67.5^\circ$.\\
In $\triangle ADE$, $\angle AED = 180^\circ - 45^\circ - 67.5^\circ = 67.5^\circ$,\\
$\therefore \angle DAE = \angle ADE$.\\
$\therefore AD = DE = 4$.\\
$\because$ The side length of the square is 4,\\
$\therefore BD = 4\sqrt{2}$.\\
$\therefore BE = BD - DE = 4\sqrt{2} - 4$.\\
$\because EF \perp AB, \angle ABD = 45^\circ$,\\
$\therefore \triangle BEF$ is an isosceles right triangle.\\
$\therefore EF = \frac{\sqrt{2}}{2}BE = \frac{\sqrt{2}}{2} \times (4\sqrt{2} - 4) = \boxed{4 - 2\sqrt{2}}$.


Question: As shown in the figure, $AB \parallel CD$, $AC$ and $BD$ intersect at point P. E and F are the midpoints of $AC$ and $BD$, respectively. If $AB = 10$ and $CD = 6$, what is the length of $EF$?

Solution:
\textbf{Solution:} Connect CF and extend it to meet AB at G, \\
$\because$AB$\parallel$DC, \\
$\therefore$$\angle$D＝$\angle$B, \\
$\because$F is the midpoint of BD, \\
$\therefore$DF＝BF, \\
In $\triangle DFC$ and $\triangle BFG$, $\left\{\begin{array}{l}\angle D=\angle B \\ DF=BF \\ \angle DFC=\angle BFG\end{array}\right.$, \\
$\therefore$$\triangle DFC\cong \triangle BFG$ by ASA, \\
$\therefore$BG＝CD＝6, CF＝FG, \\
$\therefore$AG＝AB−BG＝4, \\
$\because$CF＝FG, CE＝EA, \\
$\therefore$EF＝$\frac{1}{2}$AG＝$\frac{1}{2}\times 4＝\boxed{2}$


Question: As shown in the figure, in rectangle $ABCD$, it is known that $AE\perp BD$ at $E$, $\angle DBC=30^\circ$, and $BE=3\text{cm}$. What is the length of $AE$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\
$\therefore$ $\angle ABC = \angle BAD = 90^\circ$,\\
Since $\angle DBC = 30^\circ$,\\
$\therefore$ $\angle ABE = 60^\circ$,\\
Given that AE$\perp$BD,\\
$\therefore$ $\angle BAE = 30^\circ$,\\
$\therefore$ $AE = \sqrt{3}BE = \boxed{3\sqrt{3}}\text{cm}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is a rhombus, with points $E$ and $F$ on sides $AD$ and $CD$ respectively, and $AE=CF$, $BA=BE$. If $\angle EBF=30^\circ$, what is the measure of $\angle C$?

Solution:
\textbf{Solution}: Let the measure of $\angle C$ be $x$, thus $\angle A=\angle C=x$,\\
in $\triangle ABE$ and $\triangle CBF$,\\
since $\left\{\begin{array}{l}BA=BC\\ \angle A=\angle C\\ AE=CF\end{array}\right.$,\\
it follows that $\triangle ABE\cong \triangle CBF$ (SAS),\\
therefore $\angle ABE=\angle CBF=180^\circ-2x$,\\
thus $\angle ABC=2\left(180^\circ-2x\right)+30^\circ$,\\
since $\angle ABC=\angle D$,\\
it follows that $\angle ABC+\angle D+\angle A+\angle C=360^\circ$,\\
that is: $2\left[2\left(180^\circ-2x\right)+30^\circ\right]+2x=360^\circ$,\\
solving gives: $x=\boxed{70^\circ}$,\\
thus, the measure of $\angle C$ is: $70^\circ$.


Question: As shown in the diagram, point $E$ is a point on the diagonal $BD$ of the square $ABCD$, $EF\perp BC$, $EG\perp CD$, with the feet of the perpendiculars being $F$ and $G$, respectively. If $CG=6$ and $CF=8$, what is the length of $AE$?

Solution:
\textbf{Solution:} Connect CE, as shown in the diagram: \\
$\because$ Quadrilateral ABCD is a square, \\
$\therefore AD=CD$, $\angle ADE=\angle CDE$, $\angle C=90^\circ$, \\
In $\triangle ADE$ and $\triangle CDE$, \\
$\left\{\begin{array}{l}
AD=CD\\
\angle ADE=\angle CDE\\
DE=DE
\end{array}\right.$\\
$\therefore \triangle ADE\cong \triangle CDE$, \\
$\therefore AE=CE$, \\
$\because$ EF$\perp$BC, EG$\perp$CD, \\
$\therefore \angle EGC=\angle EFC=\angle C=90^\circ$, \\
$\therefore$ Quadrilateral EFCG is a rectangle, \\
$\therefore$ EF=CG, \\
$\because CG=6$, $CF=8$, \\
$\therefore AE=CE=\sqrt{6^2+8^2}=\boxed{10}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the vertex $C$ of the rhombus $OABC$ is on the positive half of the $x$-axis. If the coordinates of point $A$ are $(3, 4)$, what are the coordinates of point $B$?

Solution:
\textbf{Solution:} Since the coordinates of point $A$ are $(3,4)$, \\
$\therefore OA=\sqrt{(3-0)^{2}+(4-0)^{2}}=5$, \\
Since quadrilateral $OABC$ is a rhombus, \\
$\therefore AB\parallel OC$ and $AB=OA=5$, \\
$\therefore$ the x-coordinate of point $B$ is $3+5=8$, the y-coordinate of point $B$ is the same as that of point $A$, which is 4, \\
Thus, we have $B(8,4)$, \\
Hence, the answer is: $B=\boxed{(8,4)}$.


Question: As shown in the figure, on a flat ground, to measure the distance between two points A and B located beside a pond, a point O is determined first. The midpoints of OA and OB are marked as C and D respectively. It is measured that CD = 50 m. What is the distance between A and B?

Solution:
Solution: Since points C and D are the midpoints of OA and OB, respectively, and CD equals 50m, \\
it follows that CD is the midsegment of $\triangle$OAB, \\
therefore, AB equals $2 \times CD = \boxed{100}$ (m).


Question: As shown in the figure, the diagonals $AC$ and $BD$ of rectangle $ABCD$ intersect at point $O$, and $\angle AOB = 60^\circ$, $AC = 2$. What is the length of side $AB$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle and $AC=2$, \\
it follows that $AC=BD=2$, and $OA=\frac{1}{2}AC=\frac{1}{2}BD=OB=1$, \\
because $\angle AOB = 60^\circ$, \\
therefore $\triangle AOB$ is an equilateral triangle, \\
hence $AB=OA=OB=\boxed{1}$.


Question: As shown in the figure, two congruent rectangular sheets of paper, ABCD and AECF, are overlapped as illustrated, with AB=AF, AE=BC. AE and BC intersect at point G, AD and CF intersect at point H, and $\angle AGB=30^\circ$, AB=2. What is the perimeter of the quadrilateral AGCH?

Solution:
\textbf{Solution:} Given two congruent rectangular paper pieces ABCD and AECF overlapped as shown in the diagram, with AB=AF and AE=BC, and $\angle AGB=30^\circ$. \\
Therefore, $AD\parallel BC$ and $FC\parallel AE$, with $\angle B=\angle F=90^\circ$. \\
Thus, $\angle HAG=\angle AGB=30^\circ$ and $\angle FHA=\angle HAG=30^\circ$. \\
Hence, $AG=2AB$ and $AH=2AF$. \\
Given that $AB=2$, \\
it follows that $AG=AH=4$. \\
Since $AG\parallel HC$ and $AH\parallel GC$, \\
therefore, quadrilateral $AGCH$ is a parallelogram. \\
Because $AG=AH$, \\
thus, quadrilateral $AGCH$ is a rhombus. \\
Therefore, the perimeter of quadrilateral $AGCH$ is $\boxed{16}$.


Question: As shown in the diagram, in rectangle $ABCD$, $AB=3\text{cm}$, $AD=9\text{cm}$. If this rectangle is folded such that point $B$ coincides with point $D$, with the crease being $EF$, what is the area of $\triangle BEF$?

Solution:
\textbf{Solution:} As shown in the diagram, the point corresponding to point C after folding is G,\\
In the rectangle ABCD, AB=CD=3cm, AD=9cm, $\angle C=90^\circ$,\\
Based on the folding, it can be concluded that: $\angle G=\angle C=90^\circ$, BG=CD=3cm, GF=CF,\\
Let BF=xcm, then GF=CF=$(9-x)$cm,\\
In $\triangle BGF$, according to the Pythagorean theorem: \\
$3^2+(9-x)^2=x^2$, solving for x gives $x=5$,\\
$\therefore S_{\triangle BEF}=\frac{1}{2}\times BF\times AB=\frac{1}{2}\times 5\times 3=\boxed{7.5}\text{(cm}^2\text{)}$


Question: As shown in the figure, what is the area of the shaded part?

Solution:
\[ \textbf{Solution:} \text{ Given:} \text{ The length of the shaded part is: } \sqrt{13^{2}-5^{2}}=12, \]
\[ \because \text{ the width is } 4 \text{ cm}, \]
\[ \therefore \text{ the area is: } 4\times 12=\boxed{48cm^{2}} \]


Question: As shown in the figure, in rhombus $ABCD$, if $ \angle A = \angle B + 40^\circ$, then what is the measure of $\angle B$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\
therefore $\angle A+\angle B=180^\circ$.\\
Since $\angle A=\angle B+40^\circ$,\\
therefore, we find: $\angle B=\boxed{70^\circ}$.


Question: As shown in the diagram, we know that $ABCD$ is a rectangle, with $AD = 12$ and $CD = 9$. Points $R$ and $P$ are fixed on $DC$ and $BC$, respectively. Points $E$ and $F$ are the midpoints of $AP$ and $RP$, respectively. If $CR = 4$, what is the length of $EF$?

Solution:
\textbf{Solution:} Draw $AR$,\\
In $\triangle ADR$, $AD = 12$, $DR=CD-CR=9-4=5$,\\
By the Pythagorean theorem, we have:\\
$AR= \sqrt{12^2+5^2}=13$ ,\\
Since points $E$ and $F$ are the midpoints of $AP$ and $RP$ respectively,\\
Then $EF$ is the median of $\triangle APR$,\\
Therefore, $EF=\frac{1}{2}AR=\boxed{6.5}$.


Question: As shown in the figure, in $\triangle ABC$, $D$ and $E$ are the midpoints of sides $AB$ and $AC$ respectively. If $\angle A=75^\circ$ and $\angle ADE=35^\circ$, what is the degree measure of $\angle C$?

Solution:
\textbf{Solution:} Given that $\angle A=75^\circ$ and $\angle ADE=35^\circ$,\\
it follows that $\angle AED=180^\circ-\angle A-\angle ADE=70^\circ$,\\
since $D$ and $E$ are the midpoints of sides $AB$ and $AC$ respectively, implying that $DE$ is the midline of $\triangle ABC$,\\
thus $DE\parallel BC$,\\
therefore, $\angle C=\angle AED=\boxed{70^\circ}$.


Question: As shown in the figure, the midpoints of the sides of the quadrilateral $ABCD$ are $E$, $F$, $G$, and $H$ respectively. The diagonals $AC$ and $BD$ intersect at point $O$. If the perimeter of the quadrilateral $EFGH$ is $3$, what is the length of $AC + BD$?

Solution:
\textbf{Solution:} As shown in the figure,\\
since E, F, G, H are the midpoints of line segments AB, BC, CD, AD, respectively,\\
therefore, EH and FG are the midlines of $\triangle ABD$ and $\triangle BCD$, respectively, EF and HG are the midlines of $\triangle ACD$ and $\triangle ABC$, respectively,\\
based on the properties of a triangle's midline, we know:\\
EF$\parallel$AC, GH$\parallel$AC and EF=GH=$\frac{1}{2}$AC, EH=FG=$\frac{1}{2}$BD,\\
therefore, quadrilateral EFGH is a parallelogram,\\
since the perimeter of quadrilateral EFGH is 3, i.e., EF+GH+EH+FG=3,\\
therefore, AC+BD=\boxed{3},


Question: As shown in the figure, the side length of square $ABCD$ is $2$. Diagonals $AC$ and $BD$ intersect at point $O$, and $E$ is a point on the extension line of $AC$, with $OE = 2CO$. What is the length of $BE$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a square, \\
$\therefore AC\perp BD$, $AO=BO=CO=DO$, \\
Since the side length of square $ABCD$ is 4, \\
$\therefore BC=2$, \\
In $\triangle BOC$ (right-angled triangle), \\
$BO^{2}+CO^{2}=BC^{2}$, \\
which means $2CO^{2}=2^{2}$, \\
solving this gives $CO=\sqrt{2}$, \\
Since $OE=2CO$, \\
$\therefore OE=2\sqrt{2}$, \\
In $\triangle BOE$ (right-angled triangle), \\
$BE=\sqrt{OE^{2}+BO^{2}}=\sqrt{(2\sqrt{2})^{2}+(\sqrt{2})^{2}}=\boxed{\sqrt{10}}$.


Question: As shown in the figure, $D$ and $E$ are the midpoints of sides $AB$ and $AC$ of $\triangle ABC$, respectively. If $DE = 12$, what is the length of $BC$?

Solution:
Solution: Since $D$ and $E$ are the midpoints of sides $AB$ and $AC$ of $\triangle ABC$ respectively, \\
it follows that $DE$ is the midline of $\triangle ABC$, \\
hence $BC = 2DE$, \\
since $DE = 12$, \\
it follows that $BC = \boxed{24}$.


Question: As shown in the figure, point O is the midpoint of the diagonal AC of rectangle ABCD, and $OM \parallel AB$ intersects AD at point M. If $OM = 3$ and $BC = 8$, what is the length of OB?

Solution:
\textbf{Solution:}
Given that quadrilateral ABCD is a rectangle,\\
it follows that AB$\parallel$CD, and AD=BC=8,\\
since $OM\parallel AB$,\\
it also follows that OM$\parallel$CD, and since point O is the midpoint of the diagonal AC of rectangle ABCD, OM=3,\\
thus, CD=6,\\
in $\triangle ADC$, $AC=\sqrt{AD^{2}+DC^{2}}=10$,\\
since in $\triangle ABC$, point O is the midpoint of hypotenuse AC,\\
it follows that BO=$\frac{1}{2}AC= \boxed{5}$,


Question: As shown in the figure, within rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, $AE$ perpendicular bisects $BO$, with $AE = \sqrt{3}\,\text{cm}$, what is the length of $OD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
it follows that OB=OD, OA=OC, AC=BD,\\
which implies OA=OB,\\
since AE perpendicularly bisects OB,\\
we have AB=AO,\\
therefore, OA=AB=OB,\\
given that AE=$\sqrt{3}$cm,\\
it results in OB=OD=\boxed{2};


Question: In rhombus $ABCD$, $\angle BAD=60^\circ$, $AB=3$, point $E$ is on side $AB$, and point $F$ is a moving point on diagonal $AC$. Lines $BF$ and $EF$ are connected. What is the minimum value of $BF+EF$?

Solution:
Solution: Draw lines BD and DF, and construct DH$\perp$AB at H.\\
Since quadrilateral ABCD is a rhombus,\\
it follows that $AD=AB$,\\
Since $\angle BAD=60^\circ$,\\
$\triangle ADB$ is an equilateral triangle,\\
Since B and D are symmetric about AC,\\
it follows that $BF=DF$,\\
Thus, $BF+FE=DF+FE$,\\
According to the principle that vertical segments are the shortest, when D, F, and E are collinear and coincide with DH, the value of $BF+EF$ is minimized, and the minimum value is the length of DH,\\
In $\triangle ADH$, $\angle BAD=60^\circ$, $AD=AB=3$,\\
thus $\angle ADH=30^\circ$\\
therefore $AH=\frac{1}{2}AD=\frac{3}{2}$,\\
therefore $DH=\sqrt{AD^{2}-AH^{2}}=\frac{3\sqrt{3}}{2}$,\\
Hence, the minimum value of $BF+EF$ is $\boxed{\frac{3\sqrt{3}}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, $AB=3$, $AC=4$. Let P be a moving point on side $BC$, with $PE \perp AB$ at E and $PF \perp AC$ at F. What is the minimum value of $EF$?

Solution:
\textbf{Solution:} Connect AP, as shown in the diagram: \\
Since $PE\perp AB$ and $PF\perp AC$, \\
Therefore, $\angle AEP=\angle AFP=90^\circ$, \\
Since $\angle BAC=90^\circ$, \\
Therefore, the quadrilateral AFPE is a rectangle, \\
Therefore, $EF=AP$, \\
To minimize EF, we only need to minimize AP, \\
When $AP\perp BC$, AP is the shortest, \\
Since $\angle BAC=90^\circ$ and $AB=3$ and $AC=4$, \\
Therefore, $BC=\sqrt{AB^{2}+AC^{2}}=\sqrt{3^{2}+4^{2}}=5$, \\
Since the area of $\triangle ABC$ $=\frac{1}{2}\times 4\times 3=\frac{1}{2}\times 5\times AP$, \\
Therefore, $AP=2.4$, \\
That is, $EF=\boxed{2.4}$.


Question: As shown in the figure, given a rectangular piece of paper $ABCD$, with $AB=4$ and $BC=3$, point $P$ is on side $BC$. Triangle $CDP$ is folded along $DP$ such that point $C$ falls on point $E$, and $PE$, $DE$ intersect $AB$ at points $O$, $F$ respectively, and $OP=OF$. What is the length of $DF$?

Solution:
Solution:By the properties of the rectangle, we have: $DC=AB=4$, $AD=BC=3$, $\angle A=\angle B=\angle C=90^\circ$,\\
From the folding properties, we get: $DC=DE=4$, $CP=EP$, $\angle E=\angle C=90^\circ$,\\
In $\triangle OEF$ and $\triangle OBP$,\\
$\left\{\begin{array}{l}
\angle EOF=\angle BOP \\
\angle E=\angle B \\
OF=OP
\end{array}\right.$,\\
$\therefore \triangle OEF\cong \triangle OBP$,\\
$\therefore OE=OB$, $EF=BP$,\\
$\therefore BF=EP=CP$,\\
Let $BF=EP=CP=x$,\\
Then $AF=4-x$, $BP=EF=3-x$, $DF=DE-EF=4-(3-x)=x+1$,\\
In $\triangle ADF$, $AF^2+AD^2=DF^2$,\\
That is, $(4-x)^2+9=(x+1)^2$,\\
$\therefore x=\frac{12}{5}$,\\
$\therefore DF=x+1=\boxed{\frac{17}{5}}$


Question: As shown in the figure, within rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Points $E$ and $F$ are respectively the midpoints of $AO$ and $AD$. Connecting $EF$, if $AB = 6\,cm$ and $BC = 8\,cm$, then what is the length of $EF$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
it follows that $\angle$ABC$=90^\circ$, BD = AC, and BO = OD,\\
given that AB = 6 cm, and BC = 8 cm,\\
by the Pythagorean theorem, we find: AC$= \sqrt{AB^{2}+BC^{2}} = \sqrt{6^{2}+8^{2}} = 10$ (cm),\\
hence, BD = 10 cm, and DO = 5 cm,\\
given that points E and F are the midpoints of AO and AD respectively,\\
it follows that EF is the midline of $\triangle AOD$,\\
therefore, EF$= \frac{1}{2} \times$OD = \boxed{2.5} cm.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, the quadrilaterals with sides AB, BC, AC in the diagram are all squares, and through measurement, the areas of the three squares are 225, 400, and \(S\) respectively. What is the value of \(S\)?

Solution:
\textbf{Solution:} Since $\angle ACB=90^\circ$,\\
it follows that $AB^{2}=BC^{2}+AC^{2}=400+225=625$,\\
and the value of S is exactly $AB^{2}$,\\
hence, the value of S is $\boxed{625}$.


Question: As shown in the diagram, in the right-angled $\triangle ABC$, $\angle BAC=90^\circ$, AB=3, AC=4, P is a moving point on side BC, PE$\perp$AB at E, PF$\perp$AC at F, M is the midpoint of EF, what is the minimum value of AM?

Solution:
\textbf{Solution:} Connect AP, \\
$\because$ PE $\perp$ AB, PF $\perp$ AC, \\
$\therefore \angle$AEP = $\angle$AFP = $90^\circ$, \\
Also, $\because \angle$BAC = $90^\circ$, \\
$\therefore$ Quadrilateral AEPF is a rectangle, \\
$\therefore$ AP = EF, \\
$\because \angle$BAC = $90^\circ$, M is the midpoint of EF, \\
$\therefore$ AM = $\frac{1}{2}$EF = $\frac{1}{2}$AP, \\
$\because$ in $\triangle$ABC, $\angle$BAC = $90^\circ$, AB = 3, AC = 4, \\
$\therefore$ BC = $\sqrt{AB^2 + AC^2}$ = 5, \\
When AP $\perp$ BC, the value of AP is minimal, \\
At this moment, $S_{\triangle ABC}$ = $\frac{1}{2} \times 3 \times 4$ = $\frac{1}{2} \times 5 \times$ AP, \\
Solving this, we get AP = $\frac{12}{5}$, \\
$\therefore$ The minimum value of AP is $\frac{12}{5}$, \\
$\therefore$ The minimum value of AM is $\boxed{\frac{6}{5}}$,


Question: As shown in the figure, the pentagon ABCDE is a regular pentagon, and F, G are points on the sides CD, DE, respectively, with BF$\parallel$AG. If $\angle$CFB$=57^\circ$, what is the value of $\angle$AGD?

Solution:
\textbf{Solution:} Since pentagon ABCDE is a regular pentagon, \\
$\therefore$ $\angle$D=$180^\circ \times 3 \div 5=108^\circ$, \\
Since $\angle$CFB=$57^\circ$, \\
$\therefore$ $\angle$BFD=$180^\circ - \angle$CFB=$180^\circ - 57^\circ =123^\circ$, \\
Since BF$\parallel$AG, \\
$\therefore$ $\angle$BAG+$\angle$FBA=$180^\circ$, \\
In pentagon ABFDG, \\
$\therefore$ $540^\circ=\angle$BAG+$\angle$FBA+$\angle$BFD+$\angle$D+$\angle$AGD=$180^\circ+123^\circ+108^\circ+\angle$AGD, \\
$\therefore$ $\angle$AGD=\boxed{129^\circ}.


Question: As shown in the figure, a rectangular field $ABCD$ has the ratio of its length $AB$ to its width $BC$ as $\sqrt{2} : 1$. $DE \perp AC$, and $BF \perp AC$, with the perpendicular feet being points $E$ and $F$, respectively. Now, there is a plan to plant flowers and grass in the quadrilateral region $DEBF$. What is the ratio of the area of quadrilateral $DEBF$ to the area of the rectangle $ABCD$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\
$ \therefore AD\parallel BC$, $AD=BC$, $\angle ABC=90^\circ$,\\
$ \therefore \angle DAE=\angle BCF$.\\
Since $BF\perp AC$, $DE\perp AC$,\\
$ \therefore \angle AED=\angle CFB=90^\circ$, $BF\parallel DE$.\\
In $\triangle ADE$ and $\triangle CBF$,\\
$\left\{\begin{array}{l}
\angle DAE=\angle BCF\\
\angle AED=\angle CFB\\
AD=CB
\end{array}\right.$,\\
$ \therefore \triangle ADE \cong \triangle CBF$ (AAS),\\
$ \therefore DE=BF$, $AE=CF$,\\
Since $BF\parallel DE$,\\
$ \therefore $ Quadrilateral $DEBF$ is a parallelogram,\\
Let $AD=BC=x$, then $CD=AB=\sqrt{2}x$,\\
$ \therefore AC=\sqrt{AB^2+BC^2}=\sqrt{(\sqrt{2}x)^2+x^2}=\sqrt{3}x$,\\
Since $DE\perp AC$ at point $E$,\\
$ \therefore S_{\triangle ADC}=\frac{1}{2}AD\cdot CD=\frac{1}{2}AC\cdot DE$,\\
$ \therefore x\cdot \sqrt{2}x=\sqrt{3}x\cdot DE$,\\
$ \therefore DE=\frac{\sqrt{6}}{3}x$,\\
In $\triangle ADE$, $AE=\sqrt{x^2-\left(\frac{\sqrt{6}}{3}x\right)^2}=\frac{\sqrt{3}}{3}x$,\\
$CF=\frac{\sqrt{3}}{3}x$,\\
$ \therefore EF=AC-AE-CF=\frac{\sqrt{3}}{3}x$,\\
$ \therefore S_{\text{quadrilateral }DEBF}=EF\cdot DE=\frac{\sqrt{3}}{3}x\cdot \frac{\sqrt{6}}{3}x=\frac{\sqrt{2}}{3}x^2$,\\
Since $S_{\text{rectangle }ABCD}=x\cdot \sqrt{2}x=\sqrt{2}x^2$,\\
$ \therefore $ The ratio of the area of quadrilateral $DEBF$ to rectangle $ABCD$ is $\boxed{1\colon 3}$.


Question: As shown in the figure, in parallelogram $ABCD$, $\angle ABC = 45^\circ$. Points $E$ and $F$ are located on the extensions of $CD$ and $BC$ respectively, with $AE \parallel BD$ and $\angle EFC = 30^\circ$. Given $AB = 1$, what is the length of $CF$?

Solution:
\textbf{Solution:} As shown in the figure, draw line EH $\perp$ BF at point H,\\
$\because$ Quadrilateral ABCD is a parallelogram,\\
$\therefore AB \parallel CD$, AB=DC,\\
$\because AE \parallel DB$,\\
$\therefore$ Quadrilateral ABDE is a parallelogram,\\
$\therefore AB=DE=CD$, thus D is the midpoint of CE.\\
$\because AB=1$,\\
$\therefore CE=2$,\\
$\therefore AB \parallel CD$,\\
$\therefore \angle ECF=\angle ABC=45^\circ$,\\
$\because CE=8$, $\angle ECF=45^\circ$,\\
$\therefore EH=CH=\sqrt{2}$,\\
$\because \angle EFC=30^\circ$,\\
$\therefore FH=\sqrt{3}EH=\sqrt{6}$,\\
$\therefore CF=\sqrt{2}+\sqrt{6}=\boxed{\sqrt{2}+\sqrt{6}}$.


Question: As shown in the figure, given that $\square ABCD$ has three vertices at $A(-1, 0)$, $B(-2, -3)$, and $C(2, -1)$, what are the coordinates of the fourth vertex $D$?

Solution:
\textbf{Solution:} Draw $BE\perp x$-axis at $E$, $DM\perp x$-axis at $M$, and $CF\perp BE$ at $F$, where $DM$ and $CF$ intersect at $N$,\\
then quadrilateral $EFNM$ is a rectangle,\\
thus $EF=MN$, $EM=FN$, $FN\parallel EM$,\\
$\therefore \angle EAB = \angle AQC$,\\
$\because$ Quadrilateral $ABCD$ is a parallelogram,\\
$\therefore AB=DC$, $AB\parallel DC$,\\
$\therefore \angle AQC = \angle DCN$,\\
$\therefore \angle DCN = \angle EAB$,\\
In $\triangle DCN$ and $\triangle BAE$\\
$\lbrace \begin{array}{l} \angle N=\angle BEA=90^\circ\hfill \\ \angle DCN=\angle EAB\hfill \\ CD=AB\hfill \end{array}$, \\
$\therefore \triangle DCN \cong \triangle BAE$ (AAS),\\
$\therefore BE=DN$, $AE=CN$,\\
$\because A(-1,0)$, $B(-2,-3)$, $C(2,-1)$,\\
$\therefore CN=AE=2-1=1$, $DN=BE=3$,\\
$\therefore DM=3-1=2$, $OM=2+1=3$,\\
$\therefore$ The coordinates of $D$ are $\boxed{(3,2)}$.


Question: As shown in the figure, point A is on the graph of the inverse proportion function $y = \frac{k}{x}$ ($x < 0$), AB is perpendicular to the x-axis at point B, and C is the midpoint of OB. Connect AO and AC. If the area of $\triangle ABC$ is 4, what is the value of $k$?

Solution:
\textbf{Solution:} Since $C$ is the midpoint of $OB$, and the area of $\triangle ABC$ is 4, \\
it follows that the area of $\triangle AOB$ is 8, \\
Let $A(a, b)$, \\
Since $AB$ is perpendicular to the x-axis at point $B$, \\
then $-ab = 16$, \\
Since point $A$ lies on the graph of the inverse proportion function $y = \frac{k}{x}$ (x < 0), \\
thus $k = \boxed{-16}$.


Question: As shown in the figure, a parabola (with a fixed shape) intersects the x-axis at points E and F (point E is to the left of point F), and its vertex P moves along the segment $AB$. If the coordinates of points A and B are respectively $(-2, -3)$ and $(4, -3)$, and the minimum value of the x-coordinate of point E is $-5$, what is the maximum value of the x-coordinate of point F?

Solution:
\textbf{Solution:} Let the axis of symmetry be the line $x=a$, and E, F are symmetric about $x=a$,\\
then $a-x_{E}=x_{F}-a$, $2a=x_{F}+x_{E}$,\\
since the shape of the parabola is fixed,\\
thus, the opening size of the parabola remains unchanged,\\
by translation, it is known that, when P is at point A, the x-coordinate of E is the minimum,\\
when P is at point A, we have $2 \times (-2) = -5 + x_{F}$, solving gives $x_{F}=1$,\\
thus $x_{F}-x_{E}$ is a constant value $1-(-5)=6$,\\
when P moves to B, the maximum of $x_{F}$ is $4+\frac{6}{2}=\boxed{7}$,


Question: As shown in the figure, point A is on the hyperbola $y=\frac{4}{x}$, point B is on the hyperbola $y=\frac{k}{x}$, and $AB\parallel x$-axis, intersecting the y-axis at point C. If $AB=2AC$, what is the value of $k$?

Solution:
\textbf{Solution:} Let us draw AD $\perp$ x-axis at D through point A, and draw BE $\perp$ x-axis at E through point B,\\
$\because$ AB $\parallel$ x-axis,\\
$\therefore$ the quadrilateral ACOD is a rectangle, and the quadrilateral BCOE is a rectangle,\\
$\because$ AB = 2AC,\\
$\therefore$ BC = 3AC,\\
$\because$ point A is on the hyperbola $y = \frac{4}{x}$,\\
$\therefore$ the area of the rectangle ACOD = 4,\\
Similarly, the area of the rectangle BCOE = k,\\
$\therefore$ the area of the rectangle BCOE = $3 \times$ the area of the rectangle ACOD = 12,\\
$\therefore$ k = \boxed{12}


Question: As shown in the figure, the graphs of the hyperbolas $C_1\colon y=\frac{k}{x} (k\neq 0)$ and $C_2\colon y=-\frac{1}{x}$ are illustrated. Point A lies on $C_1$. Perpendiculars are drawn from point A to the x-axis and the y-axis, meeting them at points B and C, respectively. The line segment $AB$ intersects $C_2$ at point D. If the area of the triangle $\triangle AOD$ is 2, what is the value of $k$?

Solution:
Solution: Since $S_{\triangle AOD} = S_{\triangle AOB} - S_{\triangle DOB}$,
\[
\therefore \frac{|k|}{2}-\frac{1}{2}=2,
\]
\[
\therefore |k|=5,
\]
Since the inverse proportion function is located in the third quadrant,
\[
\therefore k=\boxed{-5}.
\]


Question: As illustrated, the parabola $y = -(x+m)^2 + 5$ intersects the x-axis at points $A$ and $B$. After shifting this parabola 3 units to the right, it intersects the original parabola at point $C$. What is the y-coordinate of point $C$?

Solution:
\textbf{Solution:} To solve, we move the parabola $y=-(x+m)^2+5$ 3 units to the right to obtain $y=-(x+m-3)^2+5$, \\
According to the problem: $\left\{\begin{array}{l}y=-(x+m)^2+5 \\ y=-(x+m-3)^2+5\end{array}\right.$, \\
Solving we get: $\left\{\begin{array}{l}x=\frac{3}{2}-m \\ y=\frac{11}{4}\end{array}\right.$, \\
$\therefore$ The coordinates of point $C$ are $\left(\frac{3}{2}-m, \frac{11}{4}\right)$, \\
Hence, the y-coordinate of point $C$ is $\boxed{\frac{11}{4}}$.


Question: As shown in the figure, the vertex B of the rhombus OABC is on the y-axis, and the coordinates of vertex C are $(-3, 2)$. If the graph of the inverse proportion function $y = \frac{k}{x}$ $(x>0)$ passes through point A, what is the value of $k$?

Solution:
\textbf{Solution:} Since diamond OABC is axially symmetrical, \\
$\therefore$ A and C are symmetrical with respect to the y-axis, \\
then A(3, 2), \\
$\because$ A lies on the graph of y=$\frac{k}{x}$, \\
$\therefore k=3\times 2=\boxed{6}$.


Question: As shown in the figure, in the Cartesian coordinate system $xoy$, the line $y=k_1x+4$ intersects the $y$-axis at point C and intersects the graph of the inverse proportion function $y=\frac{k_2}{x}$ at point B in the first quadrant. Connect BO. If the area of $\triangle OBC$ is $2$ and $\tan\angle BOC=\frac{1}{5}$, what is the value of $k_2$?

Solution:
\textbf{Solution:} Given that the line $y=k_1x+4$ intersects with the x-axis at point A, and with the y-axis at point C,\\
thus, the coordinates of point C are (0, 4),\\
hence, OC=4,\\
draw BD $\perp$ y-axis at D,\\
given that $S_{\triangle OBC}=2$,\\
thus, $\frac{1}{2}OC\cdot BD=\frac{1}{2}\times 4\cdot BD=2$,\\
hence, BD=1,\\
given that $\tan \angle BOC=\frac{1}{5}$,\\
thus, $\frac{BD}{OD}=\frac{1}{5}$,\\
hence, OD=5,\\
thus, the coordinates of point B are (1, 5),\\
given that the graph of the inverse proportion function $y=\frac{k_2}{x}$ intersects at point B within the first quadrant,\\
thus, $k_{2}=1\times 5=\boxed{5}$.


Question: As shown in the figure, there are points A and B on the graphs of the inverse proportion functions $y=-\frac{a}{x}$ and $y=\frac{6}{x}$, respectively, and $AB\parallel x$-axis, $AD\perp x$-axis at D, $BC\perp x$-axis at C. If the area of rectangle ABCD is 8, what is the value of $a$?

Solution:
\textbf{Solution:} According to the problem, the area of rectangle ABCD is $a+6=8$,
\[\therefore a=\boxed{2}.\]


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

Solution: \textbf{Solution:} The information provided by the \LaTeX\ code of this problem is insufficient to address the core issue, thus preventing the provision of specific solution steps and results.


Question: How many of the following plane figures, when rotated around the dashed line for one complete turn, can form a geometric solid in the shape of a vase as shown in the figure?

Solution:
\textbf{Solution:}\\
A. The shape after rotation does not match the required 3D figure, thus it is not applicable;\\
B. The shape after rotation is a cylinder, which is not the required 3D figure, thus it is not applicable;\\
C. The shape after rotation matches the required 3D figure, thus it is applicable;\\
D. The shape after rotation does not match the required 3D figure, thus it is not applicable;\\
Hence, the answer is: \boxed{C}


Question: Intersecting a cylinder as shown in the figure with a plane, it is impossible for the cross-section to be what?

Solution: \textbf{Solution:} In this problem, using a plane to cut through a cylinder, a horizontal cut results in a circle, a vertical cut results in a rectangle, and an oblique cut results in an \boxed{ellipse}. The only impossibility is a trapezoid.


Question: As shown in the figure is the scene of drumming at the moment of a festive gathering and the three-dimensional shape of the drum. What is the shape of the drum when viewed from the front?

Solution: \textbf{Solution:} From the front view, the shape of the \boxed{drum} looks like:


Question: As shown in the figure, a cut is made on a cubic box, with the intersections of the cut surface and the edges being points $A$, $B$, and $C$ respectively. After cutting off the corner and unfolding the box into a plane, what are the impossible configurations for the unfolded figure?

Solution:
\textbf{Solution:} As shown in the figure, the square's net is as follows:

\begin{itemize}
  \item "141" type:
  \item "132" type:
  \item "222" type:
  \item "33" type:
\end{itemize}

Considering the characteristics of the square's net, after cutting off corners and unfolding the box into a plane, we know that the faces with corners cut off are three adjacent faces; it cannot be that the faces with corners cut off are opposite faces.

Hence, the answer is: \boxed{B}.


Question: As shown in the figure, the solid shape is composed of a rectangular prism and a cylinder. What shape is obtained when viewing this figure from above?

Solution: \textbf{Solution:} From the given problem, we can deduce that the top view is a rectangle with a circle in the middle. Thus, the answer is \boxed{C}.


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

Solution:
\textbf{Solution:} Draw line $AD \perp BC$ at $D$,\\
$\therefore$ $DC = 1$, $AD = 3$,\\
$\therefore$ $AC =\sqrt{AD^{2}+DC^{2}}=\sqrt{10}$,\\
$\therefore$ $\cos\angle ACB =\frac{DC}{AC}=\frac{1}{\sqrt{10}}=\frac{\sqrt{10}}{10}=\boxed{\frac{\sqrt{10}}{10}}$


Question: As shown in the figure, the vertices of $\triangle ABC$ are all located on the grid points of a square grid. Then, what is the value of $\sin\angle ABC$?

Solution:
\textbf{Solution:} As shown in the figure, connect CD\\
$\because$ $BC=\sqrt{1+3^2}=\sqrt{10}$, $CD=\sqrt{1+1}=\sqrt{2}$, $BD=\sqrt{2^2+2^2}=2\sqrt{2}$\\
$\therefore$ $BC^2=CD^2+BD^2$\\
$\therefore$ $\angle BDC=90^\circ$\\
$\therefore$ $\sin\angle ABC=\frac{CD}{BC}=\frac{\sqrt{2}}{\sqrt{10}}=\boxed{\frac{\sqrt{5}}{5}}$


Question: As shown in the figure, in a grid composed of small squares with side length of 1, points A, B, and C are all on the grid points. If a circle with AB as its diameter passes through points C and D, then what is the value of $\tan\angle ADC$?

Solution:
\textbf{Solution:} Since $AB$ is the diameter,\\
$\therefore\angle ACB=90^\circ$,\\
In $\triangle ABC$, $\tan\angle ABC= \frac{AC}{BC}=\frac{3}{2}$,\\
Since $\angle ADC=\angle ABC$,\\
$\therefore\tan\angle ADC= \frac{3}{2}$,\\
Therefore $\tan\angle ADC = \boxed{\frac{3}{2}}$.


Question: As shown in the figure, in a square grid with a side length of 1, points A, O, and B are all on the grid points. What is the value of $\tan\angle AOB$?

Solution:
\textbf{Solution:} As shown in the figure, connect AB.\\
In the right-angled $\triangle AOB$, since $\angle OBA=90^\circ$, AB=2, OB=4,\\
therefore $\tan\angle AOB= \frac{AO}{BO}=\frac{2}{4}=\boxed{\frac{1}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C = 90^\circ$, if $\sin B = \frac{4}{5}$, what is the value of $\sin A$?

Solution:
\textbf{Solution:} In $\triangle ABC$, where $\angle C=90^\circ$ and $\sin B=\frac{AC}{AB}=\frac{4}{5}$,\\
let $AC=4a$ and $AB=5a$,\\
thus $BC=\sqrt{AB^{2}-AC^{2}}=3a$,\\
therefore $\sin A=\frac{BC}{AB}=\frac{3a}{5a}=\boxed{\frac{3}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=4$, and $BC=3$, with $CD$ being the altitude of $\triangle ABC$. What is the value of $\tan\angle BCD$?

Solution:
\textbf{Solution:} In $\triangle ABC$ and $\triangle BCD$, $\angle A + \angle B = 90^\circ$, $\angle BCD + \angle B = 90^\circ$.\\
$\therefore \angle A = \angle BCD$.\\
$\therefore \tan \angle BCD = \tan A = \frac{BC}{AC} = \boxed{\frac{3}{4}}$


Question: As shown in the figure, points A, B, and C are all on the lattice points of a $4 \times 4$ square grid. What is the value of $\tan\angle BAC$?

Solution:
Solution: As shown in the figure, take lattice point D and connect BD.\\
From the lattice graph, it can be inferred that $BD\perp AC$,\\
Using the properties of lattice triangles, we find: $AD=\sqrt{3^2+3^2}=3\sqrt{2}$, $BD=\sqrt{1^2+1^2}=\sqrt{2}$,\\
$\therefore$ $\tan\angle BAC=\frac{BD}{AD}=\frac{\sqrt{2}}{3\sqrt{2}}=\boxed{\frac{1}{3}}$


Question: As shown in the figure, in rectangle $ABCD$, $AB=5$, $AD=3$, point $E$ is on $AB$, and point $F$ is on $BC$. If $AE=2$, $CF=1$, what is the value of $\sin(\angle 1+\angle 2)$?

Solution:
\textbf{Solution:} Connect EF, \\
$\because$ Quadrilateral ABCD is a rectangle, \\
$\therefore$ $\angle A=\angle B=\angle C=\angle ADC=90^\circ$, BC=AD=3, CD=AB=5, \\
In $\triangle ADE$, AD=3, AE=2, \\
$\therefore$ $DE^2=AD^2+AE^2=3^2+2^2=13$, \\
$\because$ AB=5, \\
$\therefore$ BE=AB−AE=3, \\
$\because$ CF=1, \\
$\therefore$ BF=BC−CF=2, \\
In $\triangle EBF$, \\
$\therefore$ $EF^2=BE^2+BF^2=3^2+2^2=13$, \\
$\therefore$ EF=DE \\
In $\triangle CDF$, \\
$\therefore$ $DF^2=DC^2+CF^2=5^2+1^2=26$, \\
$\because$ 26=13+13, i.e., $DF^2=DE^2+EF^2$, \\
$\therefore$ $\angle DEF=90^\circ$, \\
$\therefore$ $\angle EDF=\angle DFE=45^\circ$, \\
$\therefore$ $\angle 1+\angle 2=\angle ADC-\angle EDF=45^\circ$, \\
$\therefore$ $\sin(\angle 1+\angle 2)=\sin45^\circ=\boxed{\frac{\sqrt{2}}{2}}$.


Question: As shown in the diagram, within a $5\times 4$ grid of small squares, points A, B, and C all lie on the lattice points (lattice points refer to the vertices of the small squares). What is the value of $\tan\angle ABC$?

Solution:
\textbf{Solution:} As shown in the figure, take lattice point D, connect AD and BD, then $\angle BDA=90^\circ$, BD=5, AD=4,\\
$\therefore$ $\tan\angle ABC=\tan\angle ABD=\frac{AD}{BD}=\frac{4}{5}=\boxed{\frac{4}{5}}$


Question: As shown in the figure, $AB$ is the slope of the water-facing side of the river embankment, with a height of $AC=1$ and a horizontal distance $BC=\sqrt{3}$. What is the slope of the slope $AB$?

Solution:
\textbf{Solution}: Let $\because$ the height of the slope $AC=1$, the horizontal distance $BC=\sqrt{3}$,\\
$\therefore$ the gradient of the slope $AB$ is $\tan B=\frac{AC}{BC}=\frac{1}{\sqrt{3}}=\boxed{\frac{\sqrt{3}}{3}}$.


Question: As shown in the figure, in $\triangle ABC$, if $\tan A = \frac{1}{2}$ and $AB = 10$, then what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} We have $\because \tan A=\frac{1}{2}=\frac{BC}{AC}$\\
$\therefore AC=2BC$\\
By the Pythagorean theorem, we have $AB^{2}=AC^{2}+BC^{2}$, which gives $100=(2BC)^{2}+BC^{2}$\\
Solving this, we find $BC^{2}=20$\\
$\therefore {S}_{\triangle ABC}=\frac{1}{2}AC\times BC=BC^{2}=\boxed{20}$


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

Solution:
\textbf{Solution:} Since $\angle C=90^\circ$, AC=4, BC=3, \\
therefore AB=5, \\
therefore $\sin B= \frac{AC}{AB}=\boxed{\frac{4}{5}}$


Question: As shown in the figure, composed of small squares with a side length of 1, points A, B, and C are all located at the intersections of the grid lines. Circle $O$ with AB as its diameter passes through point C. If point D is on circle $O$, then what is the value of $\tan \angle ADC$?

Solution:
\textbf{Solution:} Construct $AC$ and $BC$, \\
since $AB$ is the diameter of the circle $O$, \\
it follows that $\angle ACB = 90^\circ$, \\
since $\angle ABC = \angle ADC$, \\
it follows that $\tan\angle ADC = \tan\angle ABC = \frac{AC}{BC} = \boxed{\frac{4}{3}}$


Question: As shown in the figure, a protractor is combined with a triangular plate with an angle of $30^\circ$ ($\angle CAB=30^\circ$). The hypotenuse AB of the triangular plate coincides with the diameter MN of the circle on which the protractor lies. Now, point P is exactly the midpoint of the semicircular arc of the protractor, and we connect CP. If BC = 4, what is the length of CP?

Solution:
\textbf{Solution:} As shown in the figure, let the intersection of CP and AB be H, construct $HI\perp BC$ at I, and $HJ\perp AP$ at J,\\
$\because$ AB is the diameter of the semicircle, $\angle ACB=90^\circ$,\\
$\therefore$ C lies on the semicircle Q where the protractor is located, and P is the midpoint of $\overset{\frown}{APB}$,\\
$\therefore \angle BCP=\angle ACP=45^\circ$,\\
$\therefore$ $CI=HI$,\\
$\because$ $\angle CAB=30^\circ$, $\angle ACB=90^\circ$,\\
$\therefore$ $\angle IBH=60^\circ$, $\angle IHB=30^\circ$,\\
Let $BI=x$, then $BH=2x$, $HI=\sqrt{3}x$, $CI=4−x$,\\
$\therefore$ $4−x=\sqrt{3}x$,\\
Solving yields: $x=2\sqrt{3}−2$,\\
$\therefore$ $BH=2(2\sqrt{3}-2)=4\sqrt{3}-4$, $CH=\sqrt{2} \times \sqrt{3}(2\sqrt{3}-2)=6\sqrt{2}-2\sqrt{6}$,\\
Similarly, it can be obtained that: $AB=2BC=8$,\\
$\therefore$ $AH=8−BH=12-4\sqrt{3}$,\\
Similarly, it can be obtained that: $\angle JAH=45^\circ$, $HJ=AJ=\frac{\sqrt{2}}{2}AH=6\sqrt{2}-2\sqrt{6}$,\\
And $\angle HPJ=\angle ABC=60^\circ$,\\
$\therefore$ $HP=\frac{HJ}{\sin 60^\circ}=4\sqrt{6}-4\sqrt{2}$,\\
$\therefore$ $CP=4\sqrt{6}-4\sqrt{2}+6\sqrt{2}-2\sqrt{6}=\boxed{2\sqrt{2}+2\sqrt{6}}$.


Question: As shown in Figure 1, in $\triangle ABC$, $\angle ACB = 90^\circ$, $\angle CAB= 30^\circ$, and $\triangle ABD$ is an equilateral triangle. As shown in Figure 2, when quadrilateral $ACBD$ is folded to make $D$ coincide with $C$, and $EF$ is the crease, what is the sine value of $\angle ACE$?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle BAC=30^\circ$, let AB=2a,\\
then it follows AC= $\sqrt{3}a$, BC=a;\\
Given that $\triangle ABD$ is an equilateral triangle,\\
thus AD=AB=2a;\\
Let DE=EC=x, then AE=2a−x;\\
In $\triangle AEC$, by the Pythagorean theorem, we get: $(2a−x)^{2}+3a^{2}=x^{2}$,\\
solving for x yields $x= \frac{7}{4}a$;\\
thus AE= $\frac{1}{4}a$, EC= $\frac{7}{4}a$,\\
thus $\sin\angle ACE= \frac{AE}{CE}=\boxed{\frac{1}{7}}$.


Question: As shown in the figure, point E is on the side BC of the square ABCD. If $CE=1$ and $DE=2$, what is the length of AD?

Solution:
\textbf{Solution}: Since quadrilateral $ABCD$ is a square,\\
it follows that $AD=CD$ and $\angle C=90^\circ$,\\
since $CD=\sqrt{DE^{2}-CE^{2}}=\sqrt{4-1}=\sqrt{3}$,\\
therefore $AD=CD=\boxed{\sqrt{3}}$


Question: In the figure, within $\triangle ABC$, $\angle C=90^\circ$, $AC=4$, $BC=3$. Let point P be a moving point on the hypotenuse $AB$. Draw $PE\perp AC$ at point E and $PF\perp BC$ at point F. Connect $EF$. Then, the minimum value of the line segment $EF$ is

Solution:
\textbf{Solution:} Draw $PC$,\\
since $PE\perp AC$ and $PF\perp BC$,\\
therefore $\angle PEC = \angle PFC = 90^\circ$,\\
thus quadrilateral $ECFP$ is a rectangle,\\
hence $EF=PC$,\\
thus, when $PC$ is at its minimum, $EF$ is also at its minimum,\\
which is when $CP\perp AB$, $PC$ is at its minimum,\\
since $AC=4$ and $BC=3$,\\
therefore $AB=5$,\\
thus, the minimum value of $PC$ is: $ \frac{AC\cdot BC}{AB}=\frac{4\times 3}{5}=\boxed{2.4}$\\
therefore, the minimum length of segment $EF$ is 2.4.


Question: As shown in the figure, on two sides of $\triangle MON$, segments $OA$ and $OB$ are intercepted such that $OA = OB$; arcs are drawn with points A and B as centers and the length of $OA$ as the radius, intersecting at point C; segments $AC$, $BC$, $AB$, and $OC$ are connected. If $AB = 2\,\text{cm}$, and the area of quadrilateral $OACB$ is $4\,\text{cm}^{2}$, what is the length of $OC$?

Solution:
Solution: According to the drawing method, we have $AC=BC=OA$,\\
$\because$ $OA=OB$,\\
$\therefore$ $OA=OB=BC=AC$,\\
$\therefore$ Quadrilateral $OACB$ is a rhombus.\\
$\because$ $AB=2\text{cm}$, and the area of quadrilateral $OACB$ is $4\text{cm}^2$,\\
$\therefore$ $\frac{1}{2}AB\times OC=\frac{1}{2}\times 2\times OC=4$,\\
Solving yields $OC=\boxed{4}$.


Question: As shown in the diagram, the quadrilateral $ABCD$ is a rhombus. The coordinates of vertices $A$ and $C$ are $(0, 2)$ and $(8, 2)$ respectively. Given that point $D$ is on the x-axis, what are the coordinates of vertex $B$?

Solution:
\textbf{Solution:} Connect AC and BD, where AC and BD intersect at point N, as shown in the figure,\\
$\because$A(0,2), C(8,2),\\
$\therefore$Points A and C lie on the line $y=2$, and AC=8, OA=2,\\
$\therefore$AC$\perp$y-axis,\\
$\because$In the rhombus ABCD, AC$\perp$BD, and AC, BD bisect each other,\\
$\therefore$BD$\parallel$y-axis, BN=ND=$\frac{1}{2}BD$, AN=NC=$\frac{1}{2}AC=4$,\\
$\therefore$BD$\perp$x-axis,\\
$\because$BD$\perp$x-axis, AC$\perp$y-axis,\\
$\therefore$$\angle$AOD=$\angle$AND=$\angle$NAO=$\angle$ODN=$90^\circ$,\\
$\therefore$Quadrilateral ANDO is a rectangle,\\
$\therefore$OA=DN=2, OD=AN=4,\\
$\therefore$BD=2DN=4,\\
$\therefore$The coordinates of point B are $\boxed{(4,4)}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=90^\circ$, $\angle A=60^\circ$. Point D lies on side $AC$, $\angle DBC=30^\circ$, $AC=12\text{cm}$. What is the perimeter of $\triangle ABD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\
it follows that OA=OC, \\
and since E is the midpoint of AB, \\
it follows that OE is the median of $\triangle ABC$, \\
thus OE$=\frac{1}{2}$BC=5.


Question: As shown in the diagram, in rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Given $AB=5$ and $AC=8$, what is the area of rhombus $ABCD$?

Solution:
\textbf{Solution:} Given that rhombus $ABCD$,\\
therefore $AC\perp BD$, $AC=2OA$, $BD=2BO$,\\
thus $OA=4$, $\angle AOB=90^\circ$,\\
therefore $BO=\sqrt{AB^{2}-OA^{2}}=\sqrt{5^{2}-4^{2}}=3$,\\
thus $BD=2\times 3=6$,\\
therefore the area of the rhombus $ABCD$ is $\frac{1}{2}AC\cdot BD=\frac{1}{2}\times 8\times 6=\boxed{24}$.


Question: As shown in Figure 1, point $P$ is a moving point on the edge of rectangle $ABCD$. Point $P$ starts from $A$ and moves along the four edges of the rectangle, finally returning to $A$. Suppose the distance traveled by point $P$ is $x$, and the area of $\triangle ABP$ is $y$. Figure 2 is the graph of function $y$ as $x$ varies. Then, what is the length of the diagonal $BD$ of rectangle $ABCD$?

Solution:
\textbf{Solution:} According to Figure 2, it is known that $AB=5$,\\
when $P$ moves to point $C$,\\
$y=\frac{1}{2}AB\cdot BC=10$,\\
$\therefore \frac{1}{2}\times 5\cdot BC=10$,\\
$\therefore BC=4$,\\
$\because$ the diagonals of a rectangle are equal,\\
$\therefore BD=AC=\sqrt{5^{2}+4^{2}}=\sqrt{41}$.\\
Hence, the answer is: $BD=\boxed{\sqrt{41}}$.


Question: As shown in the figure, in rectangle $ABCD$, where $AB = 3$ and $BC = 4$, connect $BD$ and draw the bisector of $\angle CBD$ intersecting $CD$ at point $E$. What is the length of $CE$?

Solution:
\textbf{Solution:} Construct EH$\perp$BD at H, as shown in the figure,\\
since quadrilateral ABCD is a rectangle,\\
therefore AB＝CD＝3, BC＝AD＝4, $\angle$C＝$90^\circ$,\\
therefore BD＝$\sqrt{BC^{2}+CD^{2}}$＝5,\\
since BE bisects $\angle$CBD,\\
therefore $\angle$EBC＝$\angle$EBH,\\
In $\triangle$EBH and $\triangle$EBC,\\
$\left\{\begin{array}{l}
\angle EHB=\angle C=90^\circ\\
\angle EBH=\angle EBC\\
BE=BE
\end{array}\right.$,\\
therefore $\triangle$EBH$\cong$$\triangle$EBC,\\
therefore BC＝BH＝4, EC＝EH, let EC＝EH＝x,\\
In $\triangle$DEH,\\
since DE$^{2}$＝DH$^{2}$+EH$^{2}$,\\
therefore $(3−x)^{2}$＝$1^{2}$+$x^{2}$,\\
therefore x＝$\boxed{\frac{4}{3}}$,\\
therefore CE＝$\frac{4}{3}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ bisects $\angle BAC$, $E$ is the midpoint of $BC$, $AD \perp BD$, $AC = 7$, $AB = 3$, what is the value of $DE$?

Solution:
\textbf{Solution:} As shown in the diagram, extend $BD$ to intersect $AC$ at point $F$, \\
$\because AD$ bisects $\angle BAC$, \\
it follows that $\angle BAD=\angle FAD$, \\
$\because AD \perp BC$, \\
it follows that $\angle ADB=\angle ADF=90^\circ$, \\
In $\triangle ABD$ and $\triangle AFD$, \\
$\therefore \left\{ \begin{array}{l} \angle BAD=\angle FAD \\ AD=AD \\ \angle ADB=\angle ADF \end{array} \right.$, \\
$\therefore \triangle ABD \cong \triangle AFD \text{ (ASA) }$, \\
$\therefore AF=AB=3$, $BD=DF$, \\
$\therefore FC=AC-AF=7-3=4$, \\
$\because BD=DF$, $BE=EC$, \\
$\therefore DE$ is the median of $\triangle BFC$, \\
$\therefore DE=\frac{1}{2}FC=\boxed{2}$.


Question: As shown in the figure, quadrilateral $ABCD$ is a kite, with diagonals $AC$ and $BD$ intersecting at point $O$, and $DH \perp AB$ at point $H$. Connect $OH$, and $\angle CAD=25^\circ$. What is the measure of $\angle DHO$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\
it follows that AD=AB, BO=OD, $\angle$DAO=$\angle$BAO=25$^\circ$, AC$\perp$BD, \\
therefore $\angle$ABD=65$^\circ$, \\
since DH$\perp$AB, BO=DO, \\
it follows that HO=DO, \\
thus $\angle$DHO=$\angle$BDH=90$^\circ$-$\angle$ABD=\boxed{25^\circ}.


Question: As shown in the figure, point O is the midpoint of the diagonal $AC$ of rectangle $ABCD$, and point M is the midpoint of $AD$. If $AB=3$ and $BC=4$, what is the perimeter of quadrilateral $ABOM$?

Solution:
\textbf{Solution:} Since $AB=3$ and $BC=4$,\\
it follows that $AC=\sqrt{3^2+4^2}=5$. Since point O is the midpoint of AC, then $BO=\frac{1}{2}AC=2.5$.\\
Also, M is the midpoint of AD, thus MO is the midline of $\triangle ACD$, hence $OM=\frac{1}{2}CD=1.5$.\\
Therefore, the perimeter of the quadrilateral ABOM is $AB+BO+MO+AM=3+2.5+2+1.5=\boxed{9}$.


Question: As shown in the figure, in square $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Let $M$ be a point on side $AD$, and draw $OM$. Construct $ON \perp OM$ intersecting $CD$ at point $N$. If the area of quadrilateral $MOND$ is $1$, what is the length of $AB$?

Solution:
\textbf{Solution:} In square ABCD, diagonal BD $\perp$ AC, \\
$\therefore \angle AOD=90^\circ$ \\
$\because ON\perp OM$ \\
$\therefore \angle MON=90^\circ$ \\
$\therefore \angle AOM=\angle DON$ \\
Also, $\because \angle MAO=\angle NDO=45^\circ$, $AO=DO$ \\
$\therefore \triangle MAO \cong \triangle NDO$ (ASA) \\
$\therefore S_{\triangle MAO}=S_{\triangle NDO}$ \\
$\because$ the area of quadrilateral MOND is 1, \\
$\therefore S_{\triangle DAO}=1$ \\
$\therefore$ the area of square ABCD is 4, \\
$\therefore AB^2=4$ \\
$\therefore AB=\boxed{2}$


Question: As shown in the figure, in rectangle $ABCD$, $AC$ and $BD$ intersect at point $O$, and $E$ and $F$ are the midpoints of $AO$ and $AD$ respectively. If $EF=4$ and $AB=8$, what is the measure of $\angle ACB$ in degrees?

Solution:
\textbf{Solution:} Given that E and F are the midpoints of AO and AD respectively,\\
it follows that EF is the midline of $\triangle AOD$,\\
hence OD = 2EF = $2\times 4$ = 8,\\
since quadrilateral ABCD is a rectangle,\\
it implies OB = OD = OA = OC = 8,\\
which means AC = 16,\\
given AB = 8,\\
it follows AC = 2AB,\\
since $\angle$ABC = 90$^\circ$,\\
it results in $\angle$ACB = \boxed{30^\circ}.


Question: As shown in the figure, in rhombus $ABCD$, $AC=6$, $DB=8$, $AH \perp BC$. What is the length of $AH$?

Solution:
\textbf{Solution}: As shown in the figure, the diagonals $AC$ and $BD$ intersect at point O,\\
$\because$ Quadrilateral $ABCD$ is a rhombus, $AC=6$, $BD=8$\\
$\therefore AC\perp BD$, $OA=OC=3$, $OB=OD=4$,\\
$\therefore BC=\sqrt{OB^{2}+OC^{2}}=\sqrt{4^{2}+3^{2}}=5$,\\
$\because AH\perp BC$,\\
$\therefore$ The area of rhombus $ABCD$ $=\frac{1}{2}AC\cdot BD=BC\cdot AH$,\\
that is, $\frac{1}{2}\times 6\times 8=5AH$,\\
$\therefore AH=\boxed{\frac{24}{5}}$


Question: As shown in the figure, the diagonals AC and BD of the rhombus ABCD intersect at point O. Through point D, draw DH $\perp$ AB at point H, and connect OH. If OA = 6 and OH = 4, what is the area of the rhombus ABCD?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\
it follows that OA=OC=6, OB=OD, and AC$\perp$BD,\\
thus AC=12,\\
since DH$\perp$AB,\\
it follows that $\angle$BHD=90$^\circ$,\\
thus BD=2OH=$2\times$ 4=8,\\
therefore, the area of rhombus ABCD = $\frac{1}{2}AC\cdot BD=\frac{1}{2}\times 12\times 8=\boxed{48}$


Question: As shown in the figure, points A and B are located at the two ends of a pond. Xiao Cong wants to measure the distance between A and B with a rope, but the rope is not long enough. A classmate helps him come up with an idea: first, find a point C on the ground that can be directly reached from both A and B, locate the midpoints D and E of AC and BC, respectively, and measure the length of DE as $10m$. What is the distance between A and B?

Solution:
\textbf{Solution:} Since D and E are respectively the midpoints of AC and BC, \\
therefore DE is the midline of $\triangle ABC$, \\
therefore AB=2DE, \\
since DE=10m, \\
therefore AB=\boxed{20m}


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $M$, $N$ are the midpoints of $AB$, $AC$ respectively, and $D$, $E$ are points on $BC$, connecting $DN$, $EM$. If $AB=13$ cm, $BC=10$ cm, and $DE=5$ cm, what is the area of the shaded region in the figure?

Solution:
\textbf{Solution:} As shown in the diagram, \\
connect MN, then MN is the median of $\triangle ABC$, \\
$\therefore$ MN=$\frac{1}{2}$BC=5cm, \\
Draw AF$\perp$BC at point F through point A, \\
$\because$ AB=AC, \\
$\therefore$ BF=$\frac{1}{2}$BC=5cm, \\
$\therefore$ AF=$\sqrt{AB^{2}-BF^{2}}=\sqrt{13^{2}-5^{2}}=12cm$, \\
$\because$ the shaded part of the diagram consists of three triangles each with a base of 5cm and the sum of their heights is 12cm, \\
$\therefore$ $S_{\text{shaded}}=\frac{1}{2}\times 5\times 12=\boxed{30cm^{2}}$.


Question: As shown in the figure, quadrilateral ABCD is a rhombus with diagonal AC = 8 and DB = 6. If DH $\perp$ AB at point H, what is the length of DH?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\
it follows that AC$\perp$BD, and OA＝OC＝ $\frac{1}{2}$ AC＝4, OB＝OD＝3,\\
thus $AB=\sqrt{OA^{2}+OB^{2}}=\sqrt{3^{2}+4^{2}}=5$,\\
thus $S_{\text{rhombus} ABCD}$＝ $\frac{1}{2}$ AC$\cdot$BD＝AB$\cdot$DH,\\
thus DH＝ $\frac{AC\cdot BD}{2AB}$ ＝$\boxed{4.8}$.


Question: As shown in the figure, the vertices B and E of squares ABCO and CDEF, respectively, lie on the hyperbola $y= \frac{4}{x}$ ($x>0$). When OB, OE, and BE are connected, what is the value of $S_{\triangle OBE}$?

Solution:
\textbf{Solution:} Connect CE. \\
Since quadrilateral ABCO and quadrilateral DEFC are both squares, \\
thus, $\angle$ECF=$\angle$BOC=$45^\circ$, \\
hence, CE$\parallel$OB, \\
therefore, $S_{\triangle OBE}$=$S_{\triangle OBC}$, \\
since $S_{\triangle OBC}$=$\frac{1}{2} \times 4=2$, \\
thus, $S_{\triangle OBE}$= $\frac{1}{2} \times 2 \times 2=\boxed{2}$,


Question: As shown in the figure, in rectangle $ABCD$, $AD = 3$, $AB = 4$. Let $M$ be a moving point on line segment $BD$. Let $MP \perp CD$ at point $P$, and $MQ \perp BC$ at point $Q$. What is the minimum value of $PQ$?

Solution:
\textbf{Solution:} Connect CM as shown in the diagram:
\[
\because \text{MP} \perp \text{CD at point P, MQ} \perp \text{BC at point Q,}
\]
\[
\therefore \angle \text{CPM} = \angle \text{CQM} = 90^\circ,
\]
\[
\because \text{Quadrilateral ABCD is a rectangle,}
\]
\[
\therefore \text{BC} = \text{AD} = 3, \text{CD} = \text{AB} = 4, \angle \text{BCD} = 90^\circ,
\]
\[
\therefore \angle \text{CPM} = \angle \text{CQM} = \angle \text{BCD} = 90^\circ,
\]
\[
\therefore \text{Quadrilateral PCQM is a rectangle,}
\]
\[
\therefore \text{PQ} = \text{CM},
\]
\[
\therefore \text{When CM is at its minimum, PQ is also at its minimum,}
\]
\[
\because \text{Point M moves along BD,}
\]
\[
\therefore \text{When CM} \perp \text{BD, CM is at its minimum, then PQ is at its minimum,}
\]
Using the Pythagorean theorem, we get:
\[
\text{BD} = \sqrt{\text{DC}^2 + \text{BC}^2} = \sqrt{4^2 + 3^2} = 5,
\]
\[
\because {S}_{\triangle \text{BCD}} = \frac{1}{2} \text{BC} \times \text{CD} = \frac{1}{2} \text{BD} \times \text{CM},
\]
\[
\therefore \text{At this moment, CM} = \frac{\text{BC} \times \text{CD}}{\text{BD}} = \frac{3 \times 4}{5} = \boxed{\frac{12}{5}},
\]
\[
\therefore \text{The minimum value of PQ is} \frac{12}{5}.
\]


Question: As shown in the figure, the diagonals AC and BD of the rhombus $ABCD$ intersect at point $O$, and $E$ is the midpoint of $AB$. Line $OE$ is drawn. If $OE = 4$, what is the perimeter of the rhombus $ABCD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\
it follows that OA=OC, and AB=BC=CD=AD,\\
Since point E is the midpoint of AB,\\
it follows that OE is the median of $\triangle ABC$,\\
hence BC=2OE=8,\\
Therefore, the perimeter of the rhombus ABCD is $8\times 4=\boxed{32}$


Question: As illustrated in the diagram, within rectangle ABCD, point E is the midpoint of BC. Line AE intersects BD at point F. If $BF = 2$, what is the length of BD?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\
it follows that $AD \parallel BC$,\\
which implies $\triangle ADF \sim \triangle EBF$,\\
therefore, $\frac{BF}{DF} = \frac{BE}{DA}$,\\
since $BF = 2$ and point E is the midpoint of BC,\\
it leads to $\frac{2}{DF} = \frac{1}{2}$,\\
thus, $DF = 4$,\\
hence, $BD = BF + DF = 2 + 4 = \boxed{6}$,\\


Question: As shown in the figure, in $\triangle ABC$, $AC=3$, $AB=4$, and $BC=5$. $P$ is a moving point on $BC$, $PG\perp AC$ at point $G$, $PH\perp AB$ at point $H$, and $M$ is the midpoint of $GH$. What is the minimum value of $PM$ during the motion of $P$?

Solution:
\textbf{Solution}: Given that $AC=3$, $AB=4$, and $BC=5$,\\
thus we have $AC^{2}=9$, $AB^{2}=16$, $BC^{2}=25$,\\
which leads us to $AC^{2}+AB^{2}=BC^{2}$,\\
therefore $\angle A=90^\circ$.\\
Since $PG\perp AC$ and $PH\perp AB$,\\
we get $\angle AGP=\angle AHP=90^\circ$,\\
hence quadrilateral AGPH is a rectangle.\\
By connecting AP,\\
thus $GH=AP$\\
Given that when $AP\perp BC$, AP is at its minimum,\\
thus $3 \times 4 = 5AP$,\\
therefore $AP= \frac{12}{5}$, \\
thus the minimum value of PM is $\boxed{1.2}$.


Question: As shown in the figure, in the rectangle $AOBD$, the coordinate of point $D$ is $(1,3)$. What is the length of $AB$?

Solution:
\textbf{Solution:} Since the quadrilateral AOBD is a rectangle,\\
it follows that AB=OD,\\
and since the coordinates of point D are $(1,3)$,\\
it then follows that $OD=\sqrt{1^2+3^2}=\sqrt{10}$,\\
therefore, $AB=OD=\boxed{\sqrt{10}}$


Question: As shown in the figure, in squares $ABCD$ and $CEFG$, where point $D$ is on $CG$, given that $BC=1$ and $CE=7$, and point $H$ is the midpoint of $AF$, what is the length of $CH$?

Solution:
\textbf{Solution:} As shown in the figure, extend $AD$ to meet $EF$ at $M$, and draw lines $AC$, $CF$.

Since in squares $ABCD$ and $CEFG$, point $D$ lies on $CG$, given that $BC=1$, $CE=7$,

Hence, $AB=AD=CD=BC=1$, $EF=CE=7$,

$\angle GCE=\angle E=90^\circ$, $\angle ADC=90^\circ$,

Therefore, $\angle CDM=180^\circ-\angle ADC=90^\circ$,

Therefore, quadrilateral $DCEM$ is a rectangle,

Therefore, $DM=CE=7$, $ME=CD=1$, $\angle DME=90^\circ$,

Therefore, $AM=AD+DM=1+7=8$, $FM=EF-ME=7-1=6$,

$\angle AMF=180^\circ-\angle DME=180^\circ-90^\circ=90^\circ$,

Therefore, in $\triangle AMF$ right-angled at $M$, $AF=\sqrt{AM^2+FM^2}=\sqrt{8^2+6^2}=10$,

Since squares $ABCD$ and $CEFG$ are squares,

Therefore, $\angle ACD=\angle GCF=45^\circ$,

Therefore, $\angle ACF=90^\circ$,

Since point $H$ is the midpoint of $AF$,

Therefore, $CH=\frac{1}{2}AF=\boxed{5}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAD = \angle CAD$, $BE = CE$, $AD \perp BD$, $DE = \frac{3}{2}$, $AB = 4$, what is the value of $AC$?

Solution:
\textbf{Solution:} Extend $BD$ to intersect $AC$ at $F$, as shown in the diagram:\\
$\because$ $\angle BAD=\angle CAD$ and $AD \perp BD$,\\
$\therefore$ $AF=AB=4$, point $D$ is the midpoint of segment $BF$,\\
Further, $\because$ $BE=CE$,\\
$\therefore$ point $E$ is the midpoint of segment $BC$,\\
$\therefore$ $DE$ is the midline of $\triangle BCF$,\\
$\therefore$ $FC=2DE=3$,\\
$\therefore$ $AC=AF+FC=4+3=\boxed{7}$.


Question: As shown in the figure, quadrilateral ABCD is a rhombus with $AC=8\,\text{cm}$ and $DB=6\,\text{cm}$. If $DH \perp AB$ at H, then what is the length of DH?

Solution:
\textbf{Solution:} As shown in the diagram, let AC and BD intersect at point O, \\
$\because$ Quadrilateral ABCD is a rhombus, AC$=8\text{cm}$, DB$=6\text{cm}$,\\
$\therefore$AC$\perp$BD, OA$=$OC$=4\text{cm}$, OB$=$OD$=3\text{cm}$,\\
$\therefore$ $AB=\sqrt{OA^{2}+OB^{2}}=\sqrt{4^{2}+3^{2}}=5\text{cm}$,\\
$\because$DH$\perp$AB,\\
$\therefore$the area of the rhombus $=AB\cdot DH=\frac{1}{2}AC\cdot BD=\frac{1}{2}\times 8\times 6=24\text{cm}^{2}$,\\
$\therefore$ $DH=\boxed{4.8}\text{cm}$.


Question: As shown in the figure, within a square $ABCD$ with a side length of $4\text{cm}$, point $P$ starts moving from point $A$, travels uniformly along the path $AB \rightarrow BC$, and stops at point $C$. Through point $P$, line $PQ$ is drawn parallel to $BD$, where $PQ$ intersects with side $AD$ (or side $CD$) at point $Q$. The length of $PQ$, $y$ (in centimeters), varies as point $P$ moves, as illustrated in the figure. What is the length of $PQ$ when point $P$ has moved for $2.5s$?

Solution:
\textbf{Solution:} Given from (2), when point P moves for 2s, PQ overlaps with BD, and the side length of quadrilateral ABCD is 4cm.\\
$\therefore$ The speed of point P is 2cm/s.\\
$\therefore$ When P moves for 2.5s, the distance covered by point P is 5cm, and at this moment, P is on side BC.\\
$\therefore$ CP$_{1}$=3cm.\\
Also, since PQ$\parallel$BD,\\
$\therefore$CQ$_{1}$=3cm.\\
$\therefore$${P}_{1}{Q}_{1}=\sqrt{{3}^{2}+{3}^{2}}=\boxed{3\sqrt{2}}$cm.


Question: As shown in Figure 1, respectively along the diagonals AC of the rectangular paper piece ABCD and EG of the square paper piece EFGH, if cut and assembled as shown in Figure 2 into $\square KLMN$, and if the quadrilateral OPQR in the middle is precisely a square, and the area of $\square KLMN$ is 50, then what is the area of the square EFGH?

Solution:
\textbf{Solution:} As shown in the diagram, let $PM = PL = NR = KR = a$, and the side length of square $ORQP$ be $b$.\\
According to the problem: $a^{2} + b^{2} + (a + b)(a - b) = 50$,\\
$\therefore a^{2} = 25$,\\
$\therefore$ the area of square $EFGH$ = $a^{2} = \boxed{25}$.


Question: As shown in the figure, the side length of rhombus $ABCD$ is $2$, $\angle A=60^\circ$, point $G$ is the midpoint of $AB$, a rhombus $BEFG$ is constructed with $BG$ as its side, where point $E$ is on the extension line of $CB$, and point $P$ is the midpoint of $FD$. What is the length of $PB$?

Solution:
\textbf{Solution:} As shown in the figure, connect BF and BD, \\
$\because$ the side length of the rhombus ABCD is 2,\\
$\therefore$ AB = BC = CD = 2,\\
$\because$ $\angle$A = $60^\circ$,\\
$\therefore$ $\triangle$BCD is an equilateral triangle,\\
$\therefore$ BD = BC = 2, $\angle$DBC = $60^\circ$,\\
$\therefore$ $\angle$DBA = $60^\circ$,\\
$\because$ point G is the midpoint of AB,\\
$\therefore$ the side length of rhombus BEFG is 1,\\
that is, BE = EF = BG = 1,\\
$\because$ point E is on the extension line of CB, $\angle$GBE = $60^\circ$,\\
$\therefore$ $\angle$FBG = $30^\circ$,\\
connect EG,\\
$\therefore$ EG $\perp$ FB at point O,\\
$\therefore$ OB = $\frac{\sqrt{3}}{2}$,\\
$\therefore$ FB = $\sqrt{3}$,\\
$\because$ $\angle$DBF = $\angle$DBA + $\angle$FBG = $90^\circ$,\\
according to the Pythagorean theorem, we get\\
DF = $\sqrt{DB^2 + BF^2} = \sqrt{7}$,\\
$\because$ point P is the midpoint of FD,\\
$\therefore$ PB = $\frac{1}{2}$DF = $\boxed{\frac{\sqrt{7}}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $D$, $E$, and $F$ are the midpoints of sides $BC$, $CA$, and $AB$, respectively. What is the perimeter of the quadrilateral $AFDE$?

Solution:
Solution: Since D, E, and F are the midpoints of sides $BC$, $CA$, and $AB$ respectively, \\
it follows that $DF \parallel AC$, $DE \parallel AB$, and $DF = \frac{1}{2} AC$, $DE = \frac{1}{2} AB$, \\
therefore, quadrilateral AFDE is a parallelogram, \\
hence, $DF = AE = \frac{1}{2} AC$, $DE = AF = \frac{1}{2} AB$, \\
therefore, the perimeter of parallelogram AFDE is equal to $\boxed{AB + AC}$


Question: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, with $BD=8$ and $BC=5$. If $AE \perp BC$ at point $E$, what is the length of $AE$?

Solution:
\textbf{Solution:} Given that $ABCD$ is a rhombus, with $BC=5$ and $BD=8$, \\
then $OC=\sqrt{5^2-4^2}=3$, \\
hence $AC=2OC=6$, \\
since $AE \perp BC$, \\
then $BC \cdot AE = \frac{1}{2} AC \cdot BD$, \\
thus $5AE = \frac{1}{2} \times 6 \times 8$, \\
therefore $AE = \boxed{\frac{24}{5}}$.


Question: As shown in the figure, point A is an arbitrary point on the graph of the inverse proportion function $y= \frac{2}{x}$ ($x>0$). Line segment $AB$ is parallel to the $x$-axis and intersects the graph of the inverse proportion function $y= -\frac{3}{x}$ at point B. A square $\square ABCD$ is formed with $AB$ as a side, where points $C$ and $D$ lie on the $x$-axis. What is the area $S_{\square ABCD}$?

Solution:
\textbf{Solution:} As shown in the figure, draw $AF \perp x$-axis at point $F$ and $BE \perp x$-axis at point $E$ from points $A$ and $B$, respectively.

Since $\square ABCD$,

it follows that $AB \parallel x$-axis,

hence quadrilateral $ABEF$ is a rectangle,

thus $S_{\square ABCD} = S_{\text{rectangle } ABEF}$,

since $A$ and $B$ are located on the graphs of inverse proportion functions $y=\frac{2}{x}$ and $y=-\frac{3}{x}$, respectively,

it follows that $S_{\text{rectangle } ABEF} = 2 + 3 = 5$,

therefore, $S_{\square ABCD} = \boxed{5}$.


Question: As shown in the figure, point $E$ is on the side $CD$ of rectangle $ABCD$. The rectangle $ABCD$ is folded along $AE$ so that point $D$ falls on point $F$ on side $BC$. If $AB=6$ and $AD=10$, what is the length of $DE$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a rectangle, \\
$\therefore AD=BC=10$, $AB=CD=6$,\\
Folding rectangle $ABCD$ along line $AE$, vertex $D$ precisely falls on point $F$ on side $BC$,\\
$\therefore AF=AD=10$, $EF=DE$,\\
In $\triangle ABF$, $BF=\sqrt{AF^{2}-AB^{2}}=\sqrt{10^{2}-6^{2}}=8$,\\
$\therefore CF=BC-BF=10-8=2$,\\
Let $CE=x$, then $DE=EF=6-x$,\\
In $\triangle ECF$, $CE^{2}+FC^{2}=EF^{2}$,\\
$\therefore x^{2}+2^{2}=(6-x)^{2}$,\\
Solving for $x$ gives $x=\frac{8}{3}$,\\
$\therefore DE=6-x=\boxed{\frac{10}{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $BC=20$, and $D$, $E$ are the midpoints of $AB$, $AC$, respectively. What is the length of $DE$?

Solution:
\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively,\\
therefore, $DE$ is the mid-segment of $\triangle ABC$,\\
therefore, $DE = \frac{1}{2}BC$,\\
since $BC = 20$,\\
therefore, $DE = \frac{1}{2} \times 20 = \boxed{10}$;


Question: In the rhombus $ABCD$ shown in the figure, we have $AB=4$ and $\angle ABC=60^\circ$. Points $P$ and $M$ are moving points on $BD$ and $BC$, respectively, with $M$ being distinct from points $B$ and $C$. What is the minimum value of $PM+PC$?

Solution:
Solution: As shown in Figure 1, connect PA,

$\because$ rhombus ABCD,

$\therefore$ AD=DC, $\angle ADB=\angle BDC$,

In $\triangle ADP$ and $\triangle CDP$,

$\because$ $\left\{\begin{array}{l}AD=DC\\ \angle ADP=\angle PDC\\ PD=PD\end{array}\right.$,

$\therefore$ $\triangle ADP\cong \triangle CDP\left(SAS\right)$,

$\therefore$ $PA=PC$, that is $PM+PC=PM+PA\ge AM$.

$\because$ M is a moving point on BC,

$\therefore$ when AM$\perp$BC, AM has the smallest value.

As shown in Figure 2, draw AM$\perp$BC at point M through A,

$\because$ $\angle ABC=60^\circ$, $\angle AMB=90^\circ$, $AB=4$,

$\therefore$ $AM=2\sqrt{3}$,

So the minimum value of $PM+PC$ is $\boxed{2\sqrt{3}}$,


Question: As shown in the figure, the diagonals AC and BD of the rhombus ABCD intersect at point O, and E is the midpoint of AB. Connect OE. If the perimeter of the rhombus ABCD is 32, what is the length of OE?

Solution:
\textbf{Solution:} Given that the perimeter of rhombus ABCD is 32,\\
it follows that BA=8, and AC$\perp$BD,\\
which implies $\angle$AOB=90$^\circ$,\\
since point E is the midpoint of AB,\\
it follows that OE=$ \frac{1}{2}$AB=$\boxed{4}$.


Question: As shown in the figure, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $E$. $BF \parallel AC$ and $CF \parallel BD$. If the area of quadrilateral $BECF$ is 2, what is the area of the rectangle $ABCD$?

Solution:
\textbf{Solution:} Given that $BF\parallel AC$ and $CF\parallel BD$, \\
it follows that quadrilateral BECF is a parallelogram, \\
since rectangle ABCD, \\
we have BE=CE, \\
thus, quadrilateral BECF is a rhombus; \\
hence, the area of rhombus BECF, $S_{\text{rhombus} BECF}=2S_{\triangle BEC}=2$, \\
thus, the area of $\triangle BEC=1$; \\
since rectangle ABCD, \\
it follows that the area of rectangle ABCD, $S_{\text{rectangle} ABCD}=4S_{\triangle BEC}=\boxed{4}$.


Question: Four congruent right-angled triangles are assembled as shown in the figure to form two differently sized squares, $ABCD$ and $EFGH$, where the ratio of the lengths of the two perpendicular sides of each right-angled triangle is $1:2$. If the area of square $EFGH$ is $5$, what is the area of square $ABCD$?

Solution:
\textbf{Solution:} From the given problem, we have $Rt\triangle AEH \cong Rt\triangle BFE$,\\
therefore $AH=BE$,\\
suppose $AH<AE$, and let $AH=BE=x$, then $AE=2x$, $AB=AE+BE=3x$,\\
in $Rt\triangle AEH$, by the Pythagorean theorem, we get: $EH^2=AE^2+AH^2=5x^2$,\\
since the area of square $EFGH$ is 5,\\
therefore $EH^2=5x^2=5$,\\
solving this yields $x^2=1$,\\
thus, the area of square $ABCD$ is $AB^2=9x^2=\boxed{9}$.


Question: Given the rhombus $ABCD$ as shown in the figure, where the diagonals $AC$ and $BD$ intersect at point $O$, with $BD=8$ and $BC=5$, and $AE \perp BC$ at point $E$, what is the length of $AE$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a rhombus with $BD=8$, \\
$\therefore BO=DO=4$, $\angle BOC=90^\circ$, \\
In $\triangle OBC$, $OC=\sqrt{BC^2-OB^2}=\sqrt{5^2-4^2}=3$, \\
$\therefore AC=2OC=6$, \\
Thus, $S_{\triangle ABC}=\frac{1}{2}\times AE\times BC=\frac{1}{2}\times BO\times AC$, \\
Hence, $5AE=24$, \\
Solving gives: $AE=\boxed{\frac{24}{5}}$.


Question: As shown in the figure, the diagonals AC and BD of the rhombus ABCD intersect at point O, with BE $\parallel$ AC and AE $\parallel$ BD. OE intersects AB at point F. If OE $= 5$ and AC $= 8$, what is the area of the rhombus ABCD.

Solution:
\textbf{Solution:}

Since $BE\parallel AC$ and $AE\parallel BD$, \\
it follows that quadrilateral AOBE is a parallelogram, \\
Since quadrilateral ABCD is a rhombus, \\
it follows that $AC\perp BD$, \\
therefore, parallelogram AOBE is a rectangle, \\
therefore, $AB=OE=5$, \\
Since $AC=8$, \\
it follows that $AO=\frac{1}{2}AC=4$, \\
thus, $OB=\sqrt{AB^{2}-AO^{2}}=\sqrt{5^{2}-4^{2}}=3$, \\
therefore, the area of ABCD, $S_{ABCD}=OB\cdot AC=3\times 8=\boxed{24}$.


Question: As shown in the figure, construct squares outwardly using the two right-angle sides of the right-angled triangle $ABC$, with their areas being $S_1$ and $S_2$ respectively. If $S_1=8\text{cm}^2$ and $S_2=17\text{cm}^2$, what is the length of the hypotenuse $AB$?

Solution:
\textbf{Solution:} Let $S_{1}=8\text{cm}^{2}$, and $S_{2}=17\text{cm}^{2}$,\\
Therefore, $BC^{2}=8$, and $AC^{2}=17$,\\
Because $\angle ACB=90^\circ$,\\
Thus $AB^{2}=BC^{2}+AC^{2}$, that is, $AB^{2}=8+17=25$,\\
Therefore, $AB=\boxed{5\text{cm}}$.


Question: As shown in the figure, on two sides of $\triangle MON$, segments OA and OB are intercepted respectively, such that $OA=OB$; arcs are drawn with point A and B as centers and the length of OA as the radius, intersecting at point C; lines AC, BC, AB, and OC are drawn, intersecting at point D. If $AB=4\,cm$ and the area of quadrilateral OACB is $16\,cm^2$. Point E is the midpoint of CB, and line DE is drawn. What is the length of segment DE?

Solution:
\textbf{Solution}: Based on the given information, we know that $AC=BC=OA$,\\
$\because$ $OA=OB$,\\
$\therefore$ $AC=BC=OA=OB$,\\
$\therefore$ the quadrilateral OACB is a rhombus,\\
$\therefore$ $OD=\frac{1}{2}OC$, $BD=\frac{1}{2}AB=2\text{cm}$, $AB\perp OC$,\\
$\because$ $AB=4\text{cm}$, the area of the quadrilateral OACB is $16\text{cm}^2$,\\
$\therefore$ $\frac{1}{2}AB\times OC=16$,\\
$\therefore$ $OC=8\text{cm}$,\\
$\therefore$ $OD=\frac{1}{2}OC=4\text{cm}$,\\
$\because$ in $\triangle ODB$, by the Pythagorean theorem, we have $OB^2=OD^2+BD^2$,\\
$\therefore$ $OB=\sqrt{OD^2+BD^2}=\sqrt{4^2+2^2}=2\sqrt{5}\text{cm}$,\\
$\because$ point E is the midpoint of CB, point D is the midpoint of OC,\\
$\therefore$ DE is the midline of $\triangle OBC$,\\
$\therefore$ $DE=\frac{1}{2}OB=\boxed{\sqrt{5}}\text{cm}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertex $D$ of the rhombus $ABCD$ is on the x-axis, and the side $BC$ is on the y-axis. If the coordinates of point $B$ are $(0,8)$ and the coordinates of point $C$ are $(0,-5)$, what are the coordinates of point $D$?

Solution:
\textbf{Solution:}

Given $B(0, 8)$, and $C(0, -5)$,\\
thus $OB=8$, $OC=5$,\\
therefore, $BC=8+5=13$,\\
since quadrilateral ABCD is a rhombus,\\
thus $CD=BC=13$,\\
in right $\triangle ODC$, $OD=\sqrt{CD^{2}-OC^{2}}=\sqrt{13^{2}-5^{2}}=12$\\
therefore, $D(12, 0)$.\\
Hence, the answer is: $D(12, 0)$ is $\boxed{(12, 0)}$.


Question: As shown in the figure, a rectangular piece of paper $ABCD$ with sides of length 4 and 8 is folded such that point $C$ coincides with point $A$. What is the length of $BF$?

Solution:
\textbf{Solution:} Given the property of the fold, we know: $AF=CF$.\\
Let $BF=m$, then $AF=CF=8-m$,\\
In $\triangle ABF$, $\angle ABF=90^\circ$, $AB=4$, $BF=m$, $AF=8-m$,\\
$\therefore AF^2=AB^2+BF^2$, meaning $(8-m)^2=4^2+m^2$,\\
$\therefore m=\boxed{3}$.


Question: As shown in the figure, in rhombus ABCD, $AC=48$, $BD=14$, and $DH \perp AB$ at point H, what is the length of line segment BH?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, and AC=48, BD=14, \\
therefore, AO=24, OB=7, \\
since AC $\perp$ BD, \\
therefore, AB=$\sqrt{AO^{2}+OB^{2}}=\sqrt{24^{2}+7^{2}}=25$, \\
since DH $\perp$ AB, \\
since the area of rhombus ABCD = $\frac{1}{2}AC\cdot BD=AB\cdot DH$, \\
therefore $\frac{1}{2} \times 48 \times 14 = 25DH$, \\
therefore $DH=\frac{336}{25}$, \\
therefore $BH=\sqrt{BD^{2}-DH^{2}}=\sqrt{14^{2}-\left(\frac{336}{25}\right)^{2}}=\boxed{\frac{98}{25}}$.