Question: As shown in the figure, in $\triangle ABC$, $AC=BC$, circle O is the circumcircle of $\triangle ABC$, and the extension of $BO$ intersects side $AC$ at point D. When $\triangle ABD$ is an isosceles triangle, what is the measure of $\angle A$?

Solution:
\textbf{Solution:} Connect OC, \\
$\because$ $CA=CB$, \\
$\therefore$ $\overset{\frown}{CA}=\overset{\frown}{CB}$, $\angle CAB=\angle CBA$, \\
$\therefore$ $OC\perp AB$, \\
$\therefore$ $\angle DCO=\angle BCO$, \\
$\because$ $OC=OB$, \\
$\therefore$ $\angle OBC=\angle BCO$, \\
$\therefore$ $\angle ADB=3\angle OBC$, \\
When $BA=BD$, $\angle A=\angle BDA=3\angle OBC$, \\
Then $8\angle OBC=180^\circ$, \\
Solving for: $\angle OBC=22.5^\circ$, \\
$\therefore$ $\angle A=67.5^\circ$; \\
When $AB=AD$, $\angle ABD=\angle BDA=3\angle OBC$, \\
Then $10\angle OBC=180^\circ$, \\
Solving for: $\angle OBC=18^\circ$, \\
$\therefore$ $\angle A=\angle ABD+\angle OBC=4\angle OBC=72^\circ$, \\
The case for DA=DB does not exist, \\
In summary, when $\triangle ABD$ is an isosceles triangle, the degree of $\angle A$ is $67.5^\circ$ or $72^\circ$, \\
Hence the answer is: $\boxed{67.5^\circ}$ or $\boxed{72^\circ}$.


Question: As shown in the figure, $ \triangle ABC$ is inscribed in $ \odot O$, $ \angle BAC=60^\circ$, D is the midpoint of $ BC$, and $ \angle AOD=166^\circ$, $ AE$ and $ CF$ are the altitudes on sides $ BC$ and $ AB$, respectively. What is the measure of $ \angle BCF$ in degrees?

Solution:
\textbf{Solution:} Connect $OB, OC$,\\
since $OB=OC$ and $D$ is the midpoint of $BC$,\\
thus $\angle BOD=\frac{1}{2}\angle BOC=\angle BAC=60^\circ$\\
$\angle AOB=\angle AOD-\angle BOD=166^\circ-60^\circ=106^\circ$\\
since $OA=OB$\\
thus $\angle ABO=\frac{180^\circ-106^\circ}{2}=37^\circ$\\
$\angle OBC=90^\circ-60^\circ=30^\circ$\\
thus $\angle ABC=\angle ABO+\angle OBC=37^\circ+30^\circ=67^\circ$\\
since $CF \perp AB$\\
thus $\angle BCF=90^\circ-\angle ABC=90^\circ-67^\circ=\boxed{23^\circ}$


Question: As shown in the figure, $CD$ is the diameter of the circle $\odot O$, $AB$ is a chord of $\odot O$, $AB=10$, $AB\perp CD$ at the foot M, and $CM=2$. What is the length of the radius?

Solution:
\textbf{Solution:} Connect OA, let the radius be $r$, as shown in the following figure:\\
Then, we have $OA=r$, $OM=r-2$\\
According to the perpendicular bisector theorem, we obtain $AM=\frac{1}{2}AB=5$,\\
By the Pythagorean theorem, we find: $OM^2+AM^2=OA^2$, that is, $(r-2)^2+5^2=r^2$\\
Solving this, we find $r=\boxed{\frac{29}{4}}$


Question: As shown in the figure, the isosceles $\triangle ABC$ is inscribed in $\odot O$, with $AB=AC$, and $\angle BAC=120^\circ$. Point $D$ is on the arc $AC$, and line $BD$ is drawn. Point $E$ is on line $BD$, satisfying $\angle ABE=\angle ECB$. If $CD=2$, what is the area of $\triangle AEC$?

Solution:
\textbf{Solution:} As shown in the figure, connect $AD$,\\
since $AB=AC$ and $\angle BAC=120^\circ$,\\
thus $\angle ABC=\angle ACB=\frac{1}{2}\left(180^\circ-\angle BAC\right)=30^\circ$,\\
therefore $\angle ADB=\angle ACB=30^\circ$,\\
since $\angle ABE=\angle ECB$,\\
thus $\angle DEC=\angle DBC+\angle ECB=\angle DBC+\angle ABE=\angle ABC=30^\circ$,\\
therefore $\angle ADB=\angle DEC=30^\circ$,\\
hence $AD\parallel EC$,\\
thus the distance between $AD$ and $EC$ is equal everywhere,\\
therefore $\text{Area of }\triangle AEC=\text{Area of }\triangle DEC$,\\
since point D is a point on the arc $\overset{\frown}{AC}$,\\
thus $\angle EDC=\angle BAC=120^\circ$,\\
construct $DH\perp EC$ at point H,\\
since $\angle ABD=\angle ECB$ and $\angle ACD=\angle ABD$,\\
thus $\angle ACD=\angle ECB$,\\
since $\angle ACB=30^\circ$,\\
thus $\angle ACE+\angle ECB=30^\circ$,\\
thus $\angle ACE+\angle ACD=30^\circ$,\\
that is, $\angle DCE=30^\circ=\angle DEC$,\\
thus $\triangle DEC$ is an isosceles triangle,\\
since $CD=2$,\\
thus $DH=\frac{1}{2}CD=1$,\\
thus $CH=\sqrt{CD^{2}-DH^{2}}=\sqrt{3}$,\\
thus $EC=2\sqrt{3}$,\\
thus $\text{Area of }\triangle DEC=\frac{1}{2}EC\cdot DH=\frac{1}{2}\times 2\sqrt{3}\times 1=\boxed{\sqrt{3}}$.


Question: As shown in the figure, in $ \triangle ABC$ and $\triangle ADE$, $\angle ACB=\angle AED=90^\circ$, $\angle ABC=\angle ADE$. Connect BD and CE. If $AC:BC=3:4$, then what is the ratio of $BD:CE$?

Solution:
\textbf{Solution:} Given that $\because$ $\angle ACB=\angle AED=90^\circ$ and $\angle ABC=\angle ADE$,\\
$\therefore$ $\triangle ABC\sim \triangle ADE$,\\
$\therefore$ $\angle BAC=\angle DAE$ and $\frac{AC}{AB}=\frac{AE}{AD}$,\\
$\because$ $\angle BAC+\angle BAE=\angle DAE+\angle BAE$,\\
$\therefore$ $\angle CAE=\angle BAD$,\\
$\because$ $\frac{AC}{AB}=\frac{AE}{AD}$,\\
$\therefore$ $\triangle ACE\sim \triangle ABD$,\\
$\therefore$ $\frac{BD}{CE}=\frac{AB}{AC}$,\\
$\because$ $AC\colon BC=3\colon 4$ and $\angle ACB=\angle AED=90^\circ$,\\
let $AC=3x$, $BC=4x$,\\
In $Rt\triangle ABC$, $AB^2=AC^2+BC^2$,\\
$\therefore$ $AB=\sqrt{AC^2+BC^2}=\sqrt{(3x)^2+(4x)^2}=5x$,\\
$\therefore$ $AC\colon BC\colon AB=3\colon 4\colon 5$,\\
$\therefore$ $BD\colon CE=\boxed{5\colon 3}$,


Question: As shown in the figure, in $\triangle ABC$, points D and E are on sides $AB$ and $AC$ respectively, with $DE\parallel BC$. Given that $AE=2$ and $\frac{AD}{AB}=\frac{1}{3}$, what is the length of $EC$?

Solution:
\textbf{Solution:} Given $DE\parallel BC$, \\
hence $\angle ADE=\angle ABC$, $\angle AED=\angle ACB$, \\
therefore $\triangle ADE\sim \triangle ABC$, \\
thus $\frac{AE}{AC}=\frac{AD}{AB}$, \\
given $\frac{AD}{AB}=\frac{1}{3}$, $AE=2$, \\
thus $AC=6$, \\
hence $EC=AC-AE=6-2=\boxed{4}$.


Question: As shown in the figure, it is known that $AB\perp BD$, $ED\perp BD$, $C$ is the midpoint of line segment $BD$, and $AC\perp CE$, with $ED=1$, $BD=4$. What is the length of $AB$?

Solution:
\textbf{Solution:} Since $AB\perp BD$ and $ED\perp BD$,\\
it follows that $\angle B=\angle D=90^\circ$,\\
which means $\angle A+\angle ACB=90^\circ$,\\
Since $AC\perp CE$,\\
we have $\angle ACE=90^\circ$,\\
therefore $\angle ECD+\angle ACB=90^\circ$,\\
thus $\angle A=\angle ECD$,\\
which implies $\triangle ABC\sim \triangle CDE$,\\
hence $\frac{AB}{CD}=\frac{BC}{DE}$\\
Given $C$ is the midpoint of segment $BD$, $ED=1$, and $BD=4$,\\
we find $BC=CD=2$,\\
therefore $\frac{AB}{2}=\frac{2}{1}$\\
leading to $AB=\boxed{4}$.


Question: As shown in the figure, within square $ABCD$, point $E$ is on side $AD$. Triangle $ABE$ is folded along line $BE$ to obtain triangle $FBE$. Connect $CF$ and extend it to intersect the extension of $BE$ at point $P$. If $AB=5$ and $AE=1$, what is the measure of $\angle P$? What is the length of $PC$?

Solution:
\textbf{Solution:} Let a perpendicular be drawn from B to CP at foot N, and from C to BP at foot M, thus CM$\perp$BP.\\
$\because$ $\triangle ABE$ is folded along line BE to obtain $\triangle FBE$,\\
$\therefore$ $Rt\triangle ABE\cong Rt\triangle FEB$,\\
$\therefore$ $BF=AB=5$, $\angle ABE=\angle PBF$,\\
$\because$ Quadrilateral ABCD is a square,\\
$\therefore$ $BC=5$,\\
$\therefore$ $BC=BF$,\\
$\therefore$ $\triangle BFC$ is an isosceles triangle,\\
Also, $\because$ $BN\perp FC$,\\
$\therefore$ $BN$ bisects $\angle FBC$,\\
$\therefore$ $\angle PBN=\angle PBF+\angle FBN=\frac{1}{2}\left(\angle ABF+\angle FBC\right)=45^\circ$,\\
In $Rt\triangle BPN$, $\angle PBN=45^\circ$, $\angle BNP=90^\circ$,\\
$\therefore$ $\angle P=45^\circ$,\\
$\therefore$ $\triangle CMP$ is an isosceles right triangle,\\
$\therefore$ $CM=MP$,\\
$\because$ $\angle ABP+\angle PBC=90^\circ$, $\angle PBC+\angle MCB=90^\circ$, $\therefore$ $\angle ABP=\angle MCB$,\\
$\therefore$ $\triangle EAB\sim \triangle BMC$,\\
$\therefore$ $\frac{CM}{AB}=\frac{CB}{BE}$,\\
In $Rt\triangle ABE$, $BE=\sqrt{AB^2+AE^2}=\sqrt{26}$,\\
$\therefore$ $\frac{CM}{5}=\frac{5}{\sqrt{26}}$, $CM=\frac{25}{\sqrt{26}}$,\\
$\therefore$ In $Rt\triangle CMP$, $PM=CM=\frac{25}{\sqrt{26}}$,\\
$\therefore$ $CP=\sqrt{CM^2+PM^2}=\boxed{\frac{25\sqrt{13}}{13}}$.


Question: The shapes conforming to the golden ratio \( \left(\frac{\sqrt{5}-1}{2}\right) \) are often visually pleasing. In the pentagram shown, where \( AD=BC=\frac{\sqrt{5}+1}{2} \), and both points C and D are the golden sections of \( AB \), what is the length of \( CD \)?

Solution:
\textbf{Solution:} Given that points C and D are both golden section points of AB,\\
hence $AC=\frac{\sqrt{5}-1}{2}AB$, $BD=\frac{\sqrt{5}-1}{2}AB$,\\
thus $AC+BD=AB+CD$,\\
which means $AC+BD=\frac{\sqrt{5}-1}{2}AB+\frac{\sqrt{5}-1}{2}AB=(\sqrt{5}-1)AB=AB+CD$,\\
$AB=\frac{CD}{\sqrt{5}-2}=(\sqrt{5}+2)CD$,\\
since $AB=AD+BD+CD=\frac{\sqrt{5}+1}{2}AB+\frac{\sqrt{5}+1}{2}AB+CD=\sqrt{5}+1+CD$,\\
thus $(\sqrt{5}+2)CD=\sqrt{5}+1+CD$,\\
therefore $(\sqrt{5}+1)CD=\sqrt{5}+1$,\\
solving this gives $CD=\boxed{1}$.


Question: As shown in the figure, within a square grid with a side length of 1, where points A, B, C, and D are all located at the grid points, and line segments $AB$ and $CD$ intersect at point E, what is the value of $\frac{AE}{BE}$?

Solution:
\textbf{Solution:} We have $AD=\sqrt{1+4}=\sqrt{5}$, $BC=\sqrt{16+4}=2\sqrt{5}$,\\
since it is known from the lattice points that $AD \parallel BC$,\\
it follows that $\triangle ADE \sim \triangle BCE$,\\
therefore $\frac{AE}{BE}=\frac{AD}{BC}=\frac{\sqrt{5}}{2\sqrt{5}}=\boxed{\frac{1}{2}}$


Question: As shown in the figure, $l_{1} \parallel l_{2}$, $l_{2} \parallel l_{3}$, line $a$ and $b$ intersect $l_{1}$, $l_{2}$, $l_{3}$ at points A, B, C and points D, E, F respectively. If $AB = 3$, $DE = 2$, $BC = 6$, what is the value of $EF$?

Solution:
\textbf{Solution}: Since $l_{1} \parallel l_{2} \parallel l_{3}$,\\
it follows that $\frac{AB}{BC}=\frac{DE}{EF}$,\\
thus $\frac{3}{6}=\frac{2}{EF}$\\
Solving yields: EF=\boxed{4}


Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is a point on side $BC$, and $AE$ and $BD$ intersect at point $F$. If $\angle DAF = \angle ABD$, $\frac{AF}{FE} = \frac{3}{2}$, and $BE = \sqrt{15}$, what is the length of $DF$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a parallelogram, and $AD\parallel BC$, with $\frac{AF}{FE}=\frac{3}{2}$,\\
it follows that $\triangle ADF\sim \triangle EBF$,\\
thus $\frac{AF}{EF}=\frac{AD}{EB}=\frac{DF}{BF}=\frac{3}{2}$,\\
also, since $BE=\sqrt{15}$,\\
then $AD=\frac{3}{2}BE=\frac{3\sqrt{15}}{2}$, and $BF=\frac{2}{3}DF$\\
Given $\angle DAF=\angle ABD=\angle BEA$,\\
it implies that $\triangle ABD\sim \triangle FEB$,\\
thus $\frac{AD}{BF}=\frac{BD}{BE}$,\\
which means $\frac{AD}{\frac{2}{3}DF}=\frac{BF+DF}{BE}=\frac{\frac{2}{3}DF+DF}{BE}$,\\
hence $\frac{\frac{3\sqrt{15}}{2}}{\frac{2}{3}DF}=\frac{\frac{5}{3}DF}{\sqrt{15}}$,\\
solving gives $DF=\boxed{\frac{9}{2}}$.


Question: As shown in the figure, in the circle $\odot O$ with a radius of $6\, \text{cm}$, $C$ and $D$ are the trisection points of diameter $AB$, with points $E$ and $F$ located on the semicircles on either side of $AB$, respectively, where $\angle BCE = \angle BDF = 60^\circ$. By connecting $AE$ and $BF$, what is the area of the two shaded regions in the figure in cm$^{2}$?

Solution:
\textbf{Solution:} Construct the axial symmetric figure of $\triangle DBF$ as $\triangle CAG$, and draw $AM \perp CG$, $ON \perp CE$, \\
Since $\triangle DBF$ and its axial symmetric figure $\triangle CAG$, \\
Because points C and D are trisection points of the diameter AB, then point H coincides with point C \\
$\therefore \triangle ACG \cong \triangle BDF$, \\
$\therefore \angle ACG = \angle BDF = 60^\circ$, \\
Since $\angle ECB = 60^\circ$, \\
$\therefore$ Points G, C, and E are collinear, \\
Since $AM \perp CG$ and $ON \perp CE$, \\
$\therefore AM \parallel ON$, \\
$\therefore \frac{AM}{ON} = \frac{AC}{OC}$, \\
In $\triangle ONC$, $\angle OCN = 60^\circ$, \\
$\therefore ON = \sin \angle OCN \cdot OC = \frac{\sqrt{3}}{2} \cdot OC$, \\
Since $OC = OA - AC = 6 - 12 \div 3 = 2$, \\
$\therefore ON = \frac{\sqrt{3}}{2} \times 2 = \sqrt{3}$, \\
Similarly: $AM = 2\sqrt{3}$, \\
Since $ON \perp GE$, \\
$\therefore NE = GN = \frac{1}{2}GE$, \\
Connect OE, \\
In $\triangle ONE$, $NE = \sqrt{OE^2 - ON^2} = \sqrt{6^2 - (\sqrt{3})^2} = \sqrt{33}$, \\
$\therefore GE = 2NE = 2\sqrt{33}$, \\
$\therefore S_{\triangle AGE} = \frac{1}{2}GE \cdot AM = \frac{1}{2} \times 2\sqrt{3} \times 2\sqrt{33} = 6\sqrt{11}$, \\
$\therefore$ The total area of the two shaded parts in the figure is $\boxed{6\sqrt{11}}$.


Question: As shown in the figure, the edge $DE$ of the rectangle $DEFG$ lies on the edge $BC$ of $\triangle ABC$, with vertices $G$ and $F$ located on edges $AB$ and $AC$, respectively. Given that $BC=6\,cm$, $DE=3\,cm$, and $EF=2\,cm$, what is the area of $\triangle ABC$ in $cm^{2}$?

Solution:
\textbf{Solution:} Let $AH \perp BC$ at $H$, and intersect $GF$ at $M$, as shown in the figure,\\
then $MH=EF=2\,cm$,\\
since quadrilateral $DEFG$ is a rectangle,\\
thus $GF\parallel BC$, $DG=EF=2\,cm$, $GF=DE=3\,cm$,\\
since $GF\parallel BC$,\\
thus $\triangle AGF\sim \triangle ABC$,\\
therefore $\frac{AM}{AH}=\frac{GF}{BC}$,\\
that is $\frac{AM}{AM+2}=\frac{3}{6}$,\\
solving this, we get: $AM=2\,cm$,\\
thus $AH=AM+MH=4\,cm$,\\
therefore, the area of $\triangle ABC$ $=\frac{1}{2}BC \cdot AH=\frac{1}{2} \times 6 \times 4 = \boxed{12}\,\left(cm^{2}\right)$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, the net of a cube with edge length of 1 has two edges respectively on $AC$, $BC$, and two vertices on the hypotenuse $AB$. What is the area of $\triangle ABC$?

Solution:
\textbf{Solution:}\\
From the given information, we have: $DE \parallel MF$,\\
$\therefore \triangle BDE \sim \triangle BMF$,\\
$\therefore \frac{BD}{BM} = \frac{DE}{MF}$, which gives $\frac{BD}{BD+1} = \frac{2}{4}$,\\
Solving gives $BD=1$,\\
Similarly, we find that $AN=6$;\\
Furthermore, since quadrilateral $DENC$ is a rectangle,\\
$\therefore DE=CN=2$, $DC=EN=3$,\\
$\therefore BC=BD+DC=4$, $AC=CN+AN=8$,\\
$\therefore$ The area of $\triangle ABC$ $= BC \times AC \div 2 = 4 \times 8 \div 2 = \boxed{16}$.


Question: As shown in the figure, $AD \parallel EB \parallel FC$, $AB = 2BC$, $DF = 6$. What is the length of $DE$?

Solution:
\textbf{Solution}: Since $AD\parallel EB\parallel FC$ and $AB=2BC$,\\
it follows that $\frac{DE}{EF}=\frac{AB}{BC}$. Therefore, $\frac{DE}{EF}=\frac{2BC}{BC}=2$,\\
Given $DF=DE+EF=6$,\\
it results in $DE=\boxed{4}$.


Question: As shown in the figure, $BC$ bisects $\angle ABD$, $\angle A = \angle BCD$. If $AB = 4$ and $BD = 6$, what is the length of $BC$?

Solution:
\textbf{Solution:} Since $BC$ bisects $\angle ABD$, \\
it follows that $\angle ABC = \angle CBD$, \\
Moreover, since $\angle A = \angle BCD$, \\
it follows that $\triangle ABC \sim \triangle CBD$,\\
thus, $\frac{AB}{BC} = \frac{BC}{BD}$, \\
Given $AB = 4$ and $BD = 6$, \\
it follows that $BC^{2} = AB \times BD = 4 \times 6 = 24$, \\
hence, $BC = \boxed{2\sqrt{6}}$.


Question: As illustrated in the figure, within $ \triangle ABC$, D is the midpoint of AB. A line passing through D intersects AC at E and intersects the extension of BC at F. When $ BF=9$ and $ CF=4$, what is the value of $ \frac{AE}{EC}$?

Solution:
\textbf{Solution:} Construct $CM\parallel AB$ through $C$, intersecting $DF$ at point $M$,\\
$\therefore$ $\triangle CME\sim \triangle ADE$, $\triangle FMC\sim \triangle FDB$,\\
$\therefore$ $\frac{AE}{EC}=\frac{AD}{CM}$, $\frac{BD}{CM}=\frac{BF}{CF}$,\\
Moreover, $\because$ $AD=BD$,\\
$\therefore$ $\frac{AE}{EC}=\frac{BF}{CF}$,\\
$\because$ $BF=9$, $CF=4$,\\
$\therefore$ $\frac{AE}{EC}=\boxed{\frac{9}{4}}$.


Question: As shown in the figure, points $D$ and $E$ are the midpoints of $AB$ and $AC$, respectively. What is the ratio of the area of $\triangle ADE$ to the area of the quadrilateral $BCED$?

Solution:
Solution: Since points $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively, \\
$\therefore$ $DE=\frac{1}{2}BC$, i.e., $DE$ is the midline of $\triangle ABC$, \\
and $\triangle ADE \sim \triangle ABC$, \\
$\therefore$ $\frac{AD}{AB}=\frac{1}{2}$, \\
based on the fact that the area ratio of similar triangles equals the square of the similarity ratio, \\
$\therefore$ $S_{\triangle ADE} : S_{\triangle ABC} = 1 : 4$, thus $S_{\text{quadrilateral} BCED} = S_{\triangle ABC} - S_{\triangle ADE}$, \\
$\therefore$ $S_{\triangle ADE} : S_{\text{quadrilateral} BCED} = 1 : (4-1) = 1 : 3 = \boxed{\frac{1}{3}}$.


Question: As shown in the figure, it is known that in the right $\triangle ABC$, $\angle ACB=90^\circ$, $AC=6$, $BC=4$. Rotating $\triangle ABC$ clockwise by $90^\circ$ around the right angle vertex $C$ yields $\triangle DEC$. If the point $F$ is the midpoint of $DE$, and $AF$ is connected, what is the length of $AF$?

Solution:
\textbf{Solution:} Let FG$\perp$AC, \\
according to the properties of rotation, EC=BC=4, DC=AC=6, $\angle$ACD=$\angle$ACB=$90^\circ$, \\
since point F is the midpoint of DE, \\
therefore FG$\parallel$CD \\
therefore $\triangle EFG\sim \triangle EDC$ \\
therefore $\frac{GF}{CD}=\frac{EF}{ED}=\frac{EG}{EC}=\frac{1}{2}$ \\
therefore GF=$\frac{1}{2}$CD=$\frac{1}{2}$AC=3 \\
EG=$\frac{1}{2}$EC=$\frac{1}{2}$BC=2 \\
since AC=6, EC=BC=4 \\
therefore AE=2 \\
therefore AG=4 \\
According to the Pythagorean theorem, $AF=\sqrt{AG^{2}+GF^{2}}=\sqrt{4^{2}+3^{2}}=\boxed{5}$


Question: As shown in the figure, in rectangle $ABCD$, $AB=6$, $BC=2$. Point $B$ is on line $l$, and the distance from point $A$ to line $l$, denoted as $AE=4$. What is the distance from point $C$ to line $l$, denoted as $CF$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\
it follows that $\angle ABC=90^\circ$,\\
hence $\angle ABE+\angle FBC=90^\circ$,\\
since $\angle EAB+\angle ABE=90^\circ$,\\
it follows that $\angle EAB=\angle FBC$,\\
also, since $\angle AEB=\angle BFC$,\\
it follows that $\triangle AEB\sim \triangle BFC$,\\
therefore $\frac{AB}{BC}=\frac{AE}{BF}$, that is, $\frac{6}{2}=\frac{4}{BF}$,\\
hence $BF=\frac{4}{3}$,\\
thus $CF=\sqrt{BC^{2}-BF^{2}}=\boxed{\frac{2\sqrt{5}}{3}}$.


Question: As shown in the figure, on the graph of the inverse proportion function $y=\frac{2}{3x}$, there is a moving point A. Connect AO and extend it to intersect another branch of the graph at point B. In the fourth quadrant, there is a point C, satisfying $AC=BC$. As point A moves, point C always moves on the graph of the function $y=\frac{k}{x}$. If $\tan\angle CAB=3$, what is the value of $k$?

Solution:
\textbf{Solution:} Connect OC, and draw AE $\perp$ y-axis at point E, and draw CF $\perp$ x-axis at point F, as shown in the diagram.\\
Due to the symmetry of line AB and the inverse function $y=\frac{2}{3x}$, points A and B are symmetrical about point O,\\
$\therefore$ AO = BO,\\
Also, since AC = BC,\\
$\therefore$ CO $\perp$ AB.\\
Since $\angle$AOE + $\angle$AOF = $90^\circ$, and $\angle$AOF + $\angle$COF = $90^\circ$,\\
$\therefore$ $\angle$AOE = $\angle$COF,\\
Also, since $\angle$AEO = $\angle$CFO = $90^\circ$,\\
$\therefore$ $\triangle AOE \sim \triangle COF$,\\
$\therefore$ $\frac{AE}{CF} = \frac{OE}{OF} = \frac{OA}{OC}$,\\
Since $\tan \angle$CAB = $\frac{OC}{OA}$ = 3,\\
$\therefore$ CF = 3AE, OF = 3OE,\\
Also, since AE$\cdot$OE = $\left|\frac{2}{3}\right|$ = $\frac{2}{3}$, CF$\cdot$OF = $|k|$,\\
$\therefore$ |k| = 6,\\
$\therefore$ k = $\pm$6,\\
Since point C is in the fourth quadrant,\\
$\therefore$ k = \boxed{-6}.


Question: As shown in the figure, in rectangle $ABCD$, $M$ and $N$ are points on sides $AB$ and $BC$, respectively. After folding the rectangle $ABCD$ along line $MN$, point $B$ falls on point $E$ on side $AD$. If $AB=4$, $AD=6$, and $AE=2\sqrt{2}AM$, then what is the length of $CN$?

Solution:
\textbf{Solution:} Given the property of folding, we know that $BM=EM$.\\
Let $AM=x$, then $AE=2\sqrt{2}x$, $BM=EM=AB-AM=4-x$,\\
since in $\triangle AEM$, $AM^2+AE^2=EM^2$,\\
therefore $x^2+(2\sqrt{2}x)^2=(4-x)^2$,\\
solving this, we get $x_1=1$, $x_2=-2$ (discard),\\
hence $AM=1$, $AE=2\sqrt{2}$.\\
As illustrated, draw $NF\perp AD$ at point F through point N.\\
Therefore, $\angle MAE=\angle EFN=90^\circ$, $NF=AB=4$, $BN=AF$,\\
therefore, $\angle FEN+\angle FNE=90^\circ$.\\
Since $\angle FEN+\angle AEM=90^\circ$,\\
therefore, $\angle AEM=\angle FNE$,\\
therefore, $\triangle MAE\sim \triangle EFN$,\\
therefore, $\frac{AM}{EF}=\frac{AE}{NF}$, that is $\frac{1}{EF}=\frac{2\sqrt{2}}{4}$,\\
therefore, $EF=\sqrt{2}$,\\
therefore, $BN=AF=AE+EF=2\sqrt{2}+\sqrt{2}=3\sqrt{2}$,\\
therefore, $CN=BC-BN=6-3\sqrt{2}$.\\
Thus, the answer is: $CN=\boxed{6-3\sqrt{2}}$.


Question: As shown in the figure, $D, E$ are two points on sides $AB, AC$ of $\triangle ABC$, respectively, and $DE \parallel BC$, $DE:BC = 1:3$. What is the ratio of $AD:AB$?

Solution:
\textbf{Solution:} Since $DE \parallel BC$, \\
then $\triangle ADE \sim \triangle ABC$, \\
therefore $\frac{AD}{AB} = \frac{DE}{BC} = \boxed{\frac{1}{3}}$.


Question: As shown in the figure, in $\triangle ABC$, points D and E are the midpoints of AB and AC, respectively. What is the ratio of the area of $\triangle ADE$ to the area of the quadrilateral DBCE?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, points $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively,\\
thus $\frac{AD}{AB}=\frac{1}{2}$,\\
therefore $\frac{{S}_{\triangle ADE}}{{S}_{\triangle ABC}}={\left(\frac{AD}{AB}\right)}^{2}=\frac{1}{4}$,\\
thus $\frac{{S}_{\triangle ADE}}{{S}_{\text{quadrilateral }DBCE}}=\frac{1}{3}$,\\
meaning the ratio of the area of $\triangle ADE$ to that of quadrilateral DBCE is $\boxed{1\colon 3}$.


Question: As shown in the diagram, points D and E are the midpoints of AB and AC, respectively. What is the ratio of $S_{\triangle ADE}$ to $S_{\text{quadrilateral} BCED}$?

Solution:
\textbf{Solution:} Since points D and E are the midpoints of AB and AC respectively,\\
it follows that DE$\parallel$BC,\\
which implies $\triangle ADE \sim \triangle ABC$,\\
thus, DE$\colon$ BC=1$\colon$ 2,\\
leading to $S_{\triangle ADE}\colon S_{\triangle ABC}=1\colon 4$,\\
hence, $S_{\triangle ADE}\colon S_{\text{quadrilateral} BCED}=1\colon 3$.\\
Therefore, the answer is: $S_{\triangle ADE} \colon S_{\text{quadrilateral} BCED}=\boxed{1\colon 3}$.


Question: As shown in the figure, in rectangle $ABCD$ and rectangle $AEGH$, the ratios $AD:AB=AH:AE=1:2$. What are the ratios of $DH:CG:BE$?

Solution:
\textbf{Solution:} Connect AC, AG,\\
$\because$ in rectangles ABCD and AEGH,\\
$\therefore$ $\angle$DAH+$\angle$HAB=90$^\circ$, $\angle$BAE+$\angle$HAB=90$^\circ$, HG=AE,\\
$\therefore$ $\angle$DAH=$\angle$BAE,\\
$\because$ AD$\colon$AB=AH$\colon$AE=1$\colon$2,\\
$\therefore$ $\triangle$ADH$\sim$ $\triangle$ABE,\\
$\therefore$ DH$\colon$BE=AD$\colon$AB=1$\colon$2,\\
In right $\triangle$AHG, HG=2AH,\\
$\therefore$ AG=$\sqrt{AH^2+HG^2}=\sqrt{5}AH$,\\
$\therefore$ AH$\colon$AG$\colon$AE=1$\colon\sqrt{5}\colon$2,\\
When AH overlaps with DA, rotate rectangle AHGE around point A to rectangle AHGE,\\
$\therefore$ $\triangle$ACG$\sim$ $\triangle$ABE$\sim$ $\triangle$ADH,\\
$\therefore$ $\frac{HD}{CG}=\frac{AH}{AG}=\frac{1}{\sqrt{5}}$, $\frac{CG}{BE}=\frac{AG}{AE}=\frac{\sqrt{5}}{2}$\\
$\therefore$ DH$\colon$CG$\colon$BE=1$\colon\sqrt{5}\colon$2.\\
Thus, the answer is: DH$\colon$CG$\colon$BE=$\boxed{1\colon\sqrt{5}\colon2}$.


Question: As shown in the figure, it is known that in quadrilateral $ABCD$, $BD$ bisects $\angle ABC$, $AB=4$, and $BC=9$. If $\triangle ABD$ is similar to $\triangle DBC$, what is the length of $BD$?

Solution:
\textbf{Solution:} Given that $\triangle ABD \sim \triangle DBC$,\\
it follows that $\frac{BD}{BC}=\frac{AB}{BD}$,\\
which implies $BD^{2}=AB \cdot BC=4 \times 9=36$,\\
therefore, $BD=\boxed{6}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $CD$ is the median to the hypotenuse $AB$. A line through point A perpendicular to $CD$ intersects $CD$ and $CB$ at points E and F, respectively. If $\tan B = \frac{2}{3}$, what is the value of $\frac{AE}{EF}$?

Solution:
\textbf{Solution:} Given $\because$ $\angle ACB=90^\circ$ and CD is the median to the hypotenuse AB,\\
$\therefore$ CD=BD, $\angle ACD+\angle BCD=90^\circ$,\\
$\therefore$ $\angle BCD=\angle B$,\\
$\therefore$ $\tan\angle B=\tan\angle BCD=\frac{EF}{CE}=\frac{2}{3}$,\\
$\therefore$ let EF=2x, CE=3x,\\
$\because$ AE$\perp$CD,\\
$\therefore$ $\angle AEC=\angle CEF=90^\circ$,\\
$\therefore$ $\angle CAF+\angle ACD=90^\circ$,\\
$\therefore$ $\angle CAF=\angle BCD$,\\
$\therefore$ $\triangle AEC\sim \triangle CEF$,\\
$\therefore$ $\frac{AE}{CE}=\frac{CE}{EF}$,\\
$\therefore$ AE=$\frac{CE^2}{EF}=\frac{9}{2}x$,\\
$\therefore$ $\frac{AE}{EF}=\frac{\frac{9}{2}x}{2x}=\boxed{\frac{9}{4}}$.


Question: As shown in the figure, it is known that trapezoid $ABCD$ satisfies $AD \parallel BC$. The diagonal $AC$ intersects the midline $EF$ at point $G$. If $AD = 3$ and $EF = 5$, what is the length of $EG$?

Solution:
\textbf{Solution:} Since $EF$ is the midline of trapezoid $ABCD$,\\
it follows that $CF=DF$,\\
and since $AD\parallel BC$,\\
it also follows that $\triangle CFG\sim \triangle CDA$,\\
thus $\frac{GF}{AD}=\frac{CF}{CD}=\frac{1}{2}$,\\
therefore $GF=\frac{1}{2}AD=\frac{3}{2}$,\\
hence $EG=EF-GF=5-\frac{3}{2}=\boxed{\frac{7}{2}}$.


Question: As shown in the figure, D is a point on the extension of side BC of $\triangle ABC$, and $CD=BC$, with line DE intersecting AB and AC at points E and F, respectively. If $EA = EB$, then $\frac{AF}{AC}=?$.

Solution:
\textbf{Solution:} Construct line $EG \parallel BD$ through point $E$, intersecting $AC$ at point $G$,\\
since $AE=BE$ and $BC=CD$,\\
therefore $AB=2AE$,\\
therefore $\triangle AEG \sim \triangle ABC$,\\
therefore $\frac{AE}{AB}=\frac{AG}{AC}=\frac{EG}{BC}=\frac{EG}{CD}=\frac{1}{2}$,\\
since $EG\parallel BD$,\\
therefore $\triangle EFG \sim \triangle DFC$,\\
therefore $\frac{EG}{CD}=\frac{GF}{CF}=\frac{1}{2}$,\\
therefore $CF=2GF$,\\
therefore $AG=CG=3GF$,\\
therefore $AF=AG+CF=4GF$,\\
therefore $AC=2CG=6GF$,\\
therefore $\frac{AF}{AC}=\frac{4GF}{6GF}=\boxed{\frac{2}{3}}$.


Question: As shown in the diagram, it demonstrates the principle of pinhole imaging, where $OA=30\,cm$, $OC=10\,cm$, and $AB\parallel CD$. If the height of the object $AB$ is $15\,cm$, what is the height of the image $CD$ in $cm$?

Solution:
\textbf{Solution:} Since AB $\parallel$ CD,\\
therefore $\frac{OA}{OC}=\frac{AB}{CD}$,\\
thus $\frac{30}{10}=\frac{15}{CD}$,\\
therefore CD = \boxed{5} cm.


Question: As shown in the figure, the sides of square $ABCD$ and square $DEFG$ are respectively $3$ and $2$, with points $E$ and $G$ located on sides $AD$ and $CD$, respectively. Point $H$ is the midpoint of $BF$. Connect $HG$. What is the length of $HG$?

Solution:
\textbf{Solution:} Extend $GF$ to intersect $AB$ at $M$, and draw $HN \perp GM$ at $N$,\\
since squares $ABCD$ and $DEFG$,\\
thus $GM \perp AB$, $FM=3-2=1$, $BM=3-2=1$,\\
thus $FM=BM$, $NH\parallel BM$,\\
since $H$ is the midpoint of $BF$,\\
thus $NH=FN=\frac{1}{2}BM=\frac{1}{2}$,\\
thus $GN=GF+FN=\frac{5}{2}$,\\
thus $GH=\sqrt{GN^{2}+NH^{2}}=\sqrt{\left(\frac{5}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}=\boxed{\frac{\sqrt{26}}{2}}$.


Question: As shown in the figure, in the hexagon $ABCDEF$, if $\angle 1 + \angle 2 + \angle 3 = 140^\circ$, then $\angle 4 + \angle 5 + \angle 6 = ?$ degrees.

Solution:
\textbf{Solution:} Given that the sum of the exterior angles of any polygon equals $360^\circ$, we have $\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6=360^\circ$,\\
since $\angle 1+\angle 2+\angle 3=140^\circ$,\\
therefore $\angle 4+\angle 5+\angle 6=360^\circ-140^\circ=\boxed{220^\circ}$.


Question: As shown in the figure, in the rectangle $ABCD$, $AB = 8$, $AD = 10$. Point $E$ is on the side $DC$. The triangle $ADE$ is folded along the straight line $AE$, and point $E$ precisely falls on point $F$ on the side $BC$. What is the length of $CE$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
it follows that CD = AB = 8, BC = AD = 10, and $\angle B = \angle D = \angle C = 90^\circ$\\
Since $\triangle ADE$ is folded along the straight line AE, with point D exactly landing on point F on side BC,\\
it follows that AF = AD = 10, and EF = DE\\
In $\triangle ABF$, BF = $\sqrt{AF^2 - AB^2} = \sqrt{10^2 - 8^2} = 6$\\
Therefore, CF = BC - BF = 10 - 6 = 4,\\
Let CE = x, then DE = EF = 8 - x\\
In $\triangle CEF$,\\
since $CF^2 + CE^2 = EF^2$,\\
it follows that $4^2 + x^2 = (8 - x)^2$, solving for x gives $x = \boxed{3}$,\\
Therefore, the length of CE is 3.


Question: As shown in the figure, the diagonals $AC$ and $BD$ of rectangle $ABCD$ intersect at point $O$. $AE$ bisects $\angle BAD$ and intersects $BC$ at point $E$. Connect $OE$. If $OE \perp BC$ and $OE = 1$, what is the length of $AC$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle, \\
$\therefore \angle ABC=\angle BAD=90^\circ$, $OA=OC$, $OB=OD$, $AC=BD$, \\
$\therefore OB=OC$, \\
Since $OE\perp BC$, \\
$\therefore BE=CE$, \\
$\therefore OE$ is the median of $\triangle ABC$, \\
$\therefore AB=2OE=2$, \\
Since $AE$ bisects $\angle BAD$, \\
$\therefore \angle BAE=45^\circ$, \\
$\therefore \triangle ABE$ is an isosceles right triangle, \\
$\therefore BE=AB=2$, \\
$\therefore BC=2BE=4$, \\
$\therefore AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{2^{2}+4^{2}}=\boxed{2\sqrt{5}}$;


Question: In the rhombus $ABCD$, the diagonals are $AC=6$ and $BD=8$. Point $E$ is the midpoint of side $AB$, and points $F$ and $P$ are moving points on $BC$ and $AC$, respectively. What is the minimum value of $PE+PF$?

Solution:
\textbf{Solution:} Let AC and BD intersect at point H. Construct the symmetric point $E'$ of E with respect to AC. Draw a perpendicular from $E'$ to AD, intersecting BC at point $F'$. Connect $E'F'$ and $E'P$ as shown in the figure.\\
By the property of symmetry, we know that $E'P=PE$, thus PE+PF=$E'P+PF$.\\
Then, when $E'$, P, and F are collinear and $E'P\perp BC$, $E'P+PF$ is minimized,\\
Therefore, the length of $E'F'$ is the minimum value of PE+PF,\\
$\because$ Quadrilateral ABCD is a rhombus with diagonals AC=6 and BD=8,\\
$\therefore$ AC$\perp$BD, AH=HC=3, BH=HD=4,\\
$\therefore AD=\sqrt{AH^2+DH^2}=\sqrt{3^2+4^2}=5$,\\
$\because$ The area of the rhombus $S_{\text{rhombus }ABCD}=\frac{1}{2}\times AC\times BD=AD\times E'F'$,\\
$\therefore E'F'=\frac{AC\times BD}{2AD}=\frac{6\times 8}{2\times 5}=\frac{24}{5}$,\\
Therefore, the answer is: $\boxed{\frac{24}{5}}$.


Question: As shown in the figure, in square $ABCD$, $G$ is a point on diagonal $BD$, $GE \perp CD$, $GF \perp BC$, with $E$ and $F$ being the respective foot of the perpendiculars. Connect $EF$. Let $M$ and $N$ be the midpoints of $AB$ and $BG$ respectively. If $EF=5$, what is the length of $MN$?

Solution:
\textbf{Solution:} Connect AG and CG, \\
$\because$ In the square ABCD, $\angle$BCD$=90^\circ$, \\
$\because$ GE$\perp$CD, GF$\perp$BC, \\
$\therefore$ Quadrilateral CFGE is a rectangle, \\
$\therefore$ CG$=EF=5$, \\
$\because$ AB$=BC$, $\angle$ABD$=\angle$CBD$=45^\circ$, \\
$\because$ BG$=BG$, \\
$\therefore$ $\triangle ABG\cong \triangle CBG$ (SAS), \\
$\therefore$ AG$=CG=5$, \\
$\because$ M and N are respectively the midpoints of AB and BG, \\
$\therefore$ MN$=\frac{1}{2}AG=\boxed{2.5}$, \\


Question: As shown in the figure, fold the square paper $ABCD$ along the line that passes through the midpoint of the opposite sides and then unfold it; the crease is marked as $MN$. Fold the paper again through point $B$ making point $A$ land on point $F$ on $MN$, with the crease marked as $BE$. If $FN=3$, what is the side length of the square paper?

Solution:
\textbf{Solution:} Let the side length of the square be $a$. From the folding, it is known that $BN=\frac{1}{2}a$ and $BF=AB=a$. \\
In $\triangle BFN$, according to the Pythagorean theorem, we have $BF^2=FN^2+BN^2$, \\
which implies: $a^2=3^2+\left(\frac{a}{2}\right)^2$, \\
Solving this gives: $a=2\sqrt{3}$ or $a=-2\sqrt{3}$ (discard this solution). \\
Therefore, the answer is: $a=\boxed{2\sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, points D and E are the midpoints of AB and BC, respectively. If AC=6, what is the length of DE?

Solution:
\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AB$ and $BC$, respectively,\\
therefore $DE$ is the midline of $\triangle ABC$,\\
and since $AC=6$,\\
therefore $DE=\frac{1}{2}BC=\boxed{3}$.


Question: As shown in the figure, it is known that the area of the rhombus $ABCD$ is 24 and the area of the square $AECF$ is 18. What is the side length of the rhombus?

Solution:
\textbf{Solution:} As shown in the diagram, connect AC and BD, intersecting at point O, \\
$\because$ the area of square AECF is 18, \\
$\therefore$ AC $= \sqrt{18} \times \sqrt{2} = 6$, \\
$\therefore$ AO $= 3$, \\
$\because$ the area of diamond ABCD is 24, \\
$\therefore$ BD $= \frac{24 \times 2}{6} = 8$, \\
$\therefore$ BO $= 4$, \\
$\therefore$ in $\triangle$AOB, $AB = \sqrt{AO^{2} + BO^{2}} = \sqrt{3^{2} + 4^{2}} = \boxed{5}$.


Question: As shown in the figure, the distance from the midpoint \(M\) of one side of the rhombus \(ABCD\) to the intersection point \(O\) of the diagonals is \(1.5\, \text{cm}\). What is the perimeter of the rhombus \(ABCD\) in centimeters?

Solution:
\textbf{Solution:} Let point $M$ be the midpoint of side $AB$ of rhombus $ABCD$, and let point $O$ be the midpoint of diagonal $BD$, with $OM=1.5\,cm$, \\
$\therefore$ in $\triangle ABD$, we have $OM=\frac{1}{2}AD=1.5$, \\
$\therefore$ $AB=BC=CD=DA=2OM=3$, \\
thus, the perimeter of the rhombus $ABCD$ is $\boxed{12}$.


Question: As shown in the figure, in square $ABCD$, point $E$ is on $AD$, and point $F$ is on $AC$, with $\angle BFE=90^\circ$. Line $BE$ intersects $AC$ at point $G$. If $AG=24$ and $CG=32$, what is the length of $GF$?

Solution:
\textbf{Solution:} As shown in the figure, connect $DF$, and rotate $\triangle BCF$ $90^\circ$ counterclockwise about point $B$ to get $\triangle BAM$, connect $MG$,

$\because$ Quadrilateral $ABCD$ is a square,

$\therefore \angle BAC = \angle DAC = \angle ACB = 45^\circ$, $AB=AD$, $\angle BAD = 90^\circ$,

In $\triangle ABF$ and $\triangle ADF$,

$\begin{cases}
AB=AD\\
\angle BAC=\angle DAC\\
AF=AF
\end{cases}$,

$\therefore \triangle ABF \cong \triangle ADF$ (SAS),

$\therefore BF=DF$, $\angle ABF = \angle ADF$;

$\because \angle BFE = 90^\circ$,

$\therefore \angle BFE = \angle BAD = 90^\circ$,

$\because \angle ABF + \angle AEF + \angle BFE + \angle BAD = 360^\circ$,

$\therefore \angle ABF + \angle AEF = 180^\circ$,

Also, $\because \angle DEF + \angle AEF = 180^\circ$,

$\therefore \angle DEF = \angle ABF$,

$\therefore \angle DEF = \angle ADF$,

$\therefore EF = DF$,

$\therefore EF = BF$,

$\because \angle BFE = 90^\circ$,

$\therefore \angle EBF = \angle BEF = 45^\circ$;

From the rotation, we know: $\angle BAM = \angle BCF = 45^\circ$, $AM=CF$, $BM=BF$, $\angle MBF = \angle ABC = 90^\circ$,

$\therefore \angle MAG = \angle BAM + \angle BAC = 45^\circ + 45^\circ = 90^\circ$,

In $\triangle AMG$, according to the Pythagorean theorem: $AM^2 + AG^2 = MG^2$;

$\because AM = CF = CG - GF = 32 - GF$, $AG=24$,

$\therefore (32 - GF)^2 + 24^2 = MG^2$;

$\because \angle MBF = 90^\circ$, $\angle EBF = 45^\circ$,

$\therefore \angle MBG = 90^\circ - \angle EBF = 45^\circ$,

$\therefore \angle MBG = \angle FBG = 45^\circ$,

In $\triangle MBG$ and $\triangle FBG$,

$\begin{cases}
BM=BF\\
\angle MBG=\angle FBG\\
BG=BG
\end{cases}$,

$\therefore \triangle MBG \cong \triangle FBG$,

$\therefore MG = FG$,

$\therefore (32 - GF)^2 + 24^2 = GF^2$,

Rearranging gives: $64GF = 1600$,

Solving gives $GF = \boxed{25}$.


Question: As shown in the figure, in parallelogram $ABCD$, $AE \perp BC$, $AF \perp CD$ with perpendicular feet at $E$ and $F$ respectively, $\angle EAF = 60^\circ$, and $DF = 3\,\text{cm}$. What is the length of $AD$ in cm?

Solution:
\textbf{Solution:} Since $AE \perp BC$ and $AF \perp CD$, and $\angle EAF = 60^\circ$, \\
it follows that $\angle AEC = \angle AFC = \angle AFD = 90^\circ$,\\
Moreover, since the sum of interior angles of quadrilateral AECF is $360^\circ$, we get\\
$\angle C + \angle AEC + \angle AFC + \angle EAF = 360^\circ$,\\
thus $\angle C + 90^\circ + 90^\circ + 60^\circ = 360^\circ$,\\
therefore $\angle C = 120^\circ$,\\
Since quadrilateral ABCD is a parallelogram,\\
it follows that $AD \parallel BC$,\\
thus $\angle D = 180^\circ - \angle C = 60^\circ$,\\
therefore $\angle DAF = 180^\circ - \angle D - \angle AFD = 30^\circ$.\\
thus $AD = 2DF = \boxed{6} \text{cm}$.


Question: As shown in the figure, to measure the distance between points $A$ and $B$ on the side of a pond, Xiao Jun selects a point $P$ on one side of the pond. The midpoints of $PA$ and $PB$ are $D$ and $E$ respectively, and the length of $DE$ is $16$ meters. What is the distance between points $A$ and $B$ in meters?

Solution:
\textbf{Solution:} Since D and E are the midpoints of PA and PB, respectively, \\
it follows that DE is the midline of $\triangle PAB$, \\
therefore, AB $=$ 2DE, \\
since DE $=$ 16 meters, \\
it follows that AB $=$ \boxed{32} meters.


Question: As shown in the figure, in square $ABCD$, points $E$ and $F$ are the midpoints of $AD$ and $DC$, respectively. Through point $C$, construct $CM\perp AB$ to meet the extension of $AB$ at $M$. Connect $EF$. If $CD=4, BM=2, CM=6$, then what is the length of $EF$?

Solution:
\textbf{Solution:} As illustrated in the diagram, connect AC,

$\because$ Quadrilateral ABCD is a parallelogram,

$\therefore$ AB=CD=4,

$\therefore$ AM=AB+BM=4+2=6,

$\because$ CM$\perp$AB, CM=6,

$\therefore$ AC=$\sqrt{AM^{2}+CM^{2}}=\sqrt{6^{2}+6^{2}}=6\sqrt{2}$,

$\because$ Points E and F are the midpoints of AD and CD, respectively,

$\therefore$ EF is the midline of $\triangle ADC$,

$\therefore$ EF=$\frac{1}{2}$AC=$\boxed{3\sqrt{2}}$.


Question: As shown in the figure, in rhombus $ABCD$, $\angle ABC=60^\circ$. Quadrilateral $EFGH$ is a rectangle. If $FA=FB=2\sqrt{2}$, what is the area of rectangle $EFGH$?

Solution:
\textbf{Solution:} Draw $AM\perp BC$ at M and $GN\perp BC$ at N through point A and G respectively, and connect GM,\\
$\because$ Quadrilateral EFGH is a rectangle,\\
$\therefore \angle AFB=\angle AED=\angle BGC=\angle CHD=90^\circ$,\\
$\because FA=FB=2\sqrt{2}$,\\
$\therefore AB=\sqrt{AF^{2}+BF^{2}}=4$, $\angle ABF=\angle BAF=45^\circ$,\\
$\because$ Quadrilateral ABCD is a rhombus, $\angle ABC=60^\circ$,\\
$\therefore AB=BC=CD=AD$, $\angle ABC=\angle ADC=60^\circ$, $\angle BAD=\angle BCD=120^\circ$,\\
$\therefore \angle CBG=15^\circ$, $\angle DAF=75^\circ$,\\
$\therefore \angle CDH=\angle DCH=45^\circ$, $\angle ADE=15^\circ$, $\angle BCG=75^\circ$,\\
$\therefore \angle BAF=\angle DCH=\angle ABF=\angle CDH$, $\angle ADE=\angle CBG$, $\angle DAE=\angle BCG$,\\
In $\triangle ABF$ and $\triangle CDH$,\\
$\left\{\begin{array}{l}
\angle BAF=\angle DCH\\
AB=CD\\
\angle ABF=\angle CDH
\end{array}\right.$,\\
$\therefore \triangle ABF \cong \triangle CDH(ASA)$,\\
Similarly, $\triangle BCG \cong \triangle DAE(ASA)$,\\
$\because AM\perp BC$, $\angle ABC=60^\circ$,\\
$\therefore \angle BAM=30^\circ$,\\
$\therefore BM=\frac{1}{2}AB=2$,\\
$\therefore AM=\sqrt{3}BM=2\sqrt{3}$, $BC=2BM$,\\
$\because \angle BGC=90^\circ$,\\
$\therefore BM=CM=GM=2$,\\
$\therefore \angle CMG=2\angle CBG=30^\circ$,\\
$\because GN\perp BC$,\\
$\therefore GN=\frac{1}{2}GM=1$,\\
$\therefore S_{\text{rhombus}ABCD}=BC\cdot AM=4\times 2\sqrt{3}=8\sqrt{3}$,\\
$S_{\triangle ABF}=\frac{1}{2}AF\cdot BF=\frac{1}{2}\times 2\sqrt{2}\times 2\sqrt{2}=4$,\\
$S_{\triangle BCG}=\frac{1}{2}BC\cdot GN=\frac{1}{2}\times 4\times 1=2$,\\
$\therefore S_{\text{rectangle}EFGH}=S_{\text{rhombus}ABCD}-2S_{\triangle ABF}-2S_{\triangle BCG}=8\sqrt{3}-12$.\\
Therefore, the answer is: $S_{\text{rectangle}EFGH}=\boxed{8\sqrt{3}-12}$.


Question: As shown in the diagram, in rectangle $ABCD$, $AC$ intersects $BD$ at point $O$. If $\angle COB=120^\circ$ and $AB=6$, what is the length of the diagonal $BD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
$ \therefore BD = 2OB$, $AC = 2OA$, $AC = BD$,\\
$ \therefore OA = OB$,\\
Since $\angle BOC = 120^\circ$,\\
$ \therefore \angle AOB = 60^\circ$,\\
$ \therefore \triangle AOB$ is an equilateral triangle,\\
$ \therefore OB = AB = 6$,\\
$ \therefore BD = 2OB = \boxed{12}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AB \parallel CD$, $AB \perp BD$, $AB = 5$, $BD = 4$, $CD = 3$. If point $E$ is the midpoint of $AC$, what is the length of $BE$?

Solution:
\textbf{Solution:}\\
Since $AB \parallel CD$ and $AB \perp BD$,\\
it follows that $CD \perp BD$,\\
thus $\angle ABD = \angle CDB = 90^\circ$.\\
Extend $CD$ to $F$, and construct $AF \perp CF$ at point $F$.\\
Then, $\angle BDF= 90^\circ$, $\angle F= 90^\circ$.\\
Therefore, quadrilateral $ABDF$ is a rectangle.\\
Therefore, $AF=BD=4$, and $DF=AB=5$.\\
Since $CD=3$,\\
thus $CF=5+3=8$.\\
Since $AC= \sqrt{8^2+4^2}=\sqrt{80}=4\sqrt{5}$,\\
in right triangle $BCD$ where $CD=3$ and $BD=4$,\\
hence $BC=5$,\\
thus $AB=BC$.\\
Since $\triangle ABC$ is an isosceles triangle,\\
and point $E$ is the midpoint of $AC$,\\
hence $AE=\frac{1}{2}AC=2\sqrt{5}$, and $BE \perp AC$.\\
Since $BE^2=AB^2−AE^2 = 5^2−(2\sqrt{5})^2 =25-20=5$,\\
thus $BE=\sqrt{5}$.\\
Therefore, the answer is: $BE=\boxed{\sqrt{5}}$.


Question: As shown in the figure, an equilateral triangle $\triangle DCE$ is constructed on the exterior of the rhombus $ABCD$. The segments $AE$ and $DE$ are connected. If the diagonal $AC$ is equal to the side $AB$, then what is the measure of $\angle DEA$ in degrees?

Solution:
\textbf{Solution:} Draw AC, \\
since quadrilateral ABCD is a rhombus, AC=AB,\\
therefore, AD=CD=AB=AC,\\
since $\triangle DCE$ is an equilateral triangle,\\
thus DE=CD=CE, $\angle$CED=60$^\circ$,\\
therefore, AD=AC=CE=DE,\\
thus quadrilateral ACED is a rhombus,\\
hence $\angle$DEA= $\frac{1}{2}$ $\angle$CED=30$^\circ$.\\
Therefore, the answer is: $\boxed{30^\circ}$.


Question: As shown in the figure, the side length of rhombus ABCD is 6, with diagonals AC and BD intersecting at point O, and $ \angle ABC = 120^\circ$. Point P is a moving point on AC, and point E is the midpoint of AB. What is the minimum value of $PB + PE$?

Solution:
\textbf{Solution:} Construct the line segment DE, intersecting AC at point P, and draw BP as shown in the figure: \\
Then $PB+PE=DE$, at this point $PB+PE$ has its minimum value;\\
Since in the rhombus ABCD, $\angle ABC=120^\circ$,\\
Hence, AD=BD, $\angle DAB=60^\circ$,\\
Thus, $\triangle ABD$ is an equilateral triangle,\\
Since point E is the midpoint of AB,\\
Hence, DE $\perp$ AB,\\
Since $AB=BD=6$, $BE=\frac{1}{2}AB=3$, \\
Hence, $DE=\sqrt{6^2-3^2}=\boxed{3\sqrt{3}}$;


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, two squares are constructed externally on the sides AC and AB of $\triangle ABC$, denoted as $S_1$ and $S_2$ representing the areas of these two squares respectively. If $S_1=8$ and $S_2=17$, what is the length of BC?

Solution: \textbf{Solution:} We have $BC = \sqrt{S_2 - S_1} = \sqrt{17 - 8} = \boxed{3}$


Question: As shown in the figure, in rectangle $ABCD$, $M$ is a point on side $BC$. Connect $AM$ and draw $DE \perp AM$ with $E$ as the foot of the perpendicular. If $DE=DC=1$ and $AE=2EM$, what is the length of $BM$?

Solution:
\textbf{Solution}: Since rectangle ABCD, with DE$\perp$AM, \\
it follows that $\angle$C=$\angle$AED=$\angle$B=$90^\circ$, with AB=DE=CD=1, and AD$\parallel$BC,\\
hence, $\angle$DAE=$\angle$AEM. In $\triangle$ADE and $\triangle$MAB
\[ \left\{\begin{array}{l}
\angle B=\angle AED\\
\angle DAE=\angle AMB\\
DE=AB
\end{array}\right. \]
it follows that $\triangle$ADE$\cong$$\triangle$MAB (by AAS),\\
thus, AE=BM=2EM;\\
In the right triangle ABM, AB$^{2}$+BM$^{2}$=AM$^{2}$, that is, 1$^{2}$+($2EM$)$^{2}$=($3EM$)$^{2}$,\\
Solving for this yields: $EM=\frac{\sqrt{5}}{5}$,\\
therefore, BM=\boxed{\frac{2\sqrt{5}}{5}}.


Question: In the rhombus $ABCD$, diagonal $AC=8$, $BD=10$, then what is the area of $\triangle ADO$?

Solution:
\textbf{Solution:} In diamond $ABCD$, the diagonals have lengths $AC=8$ and $BD=10$,\\
$\therefore$OA=4, OD=5, and $AC\perp BD$,\\
$\therefore$the area of $\triangle ADO$ is $\frac{1}{2}OA\cdot OD=\frac{1}{2}\times 4\times 5=\boxed{10}$.


Question: As shown in the figure, in rectangle $ABCD$, it is given that $AB=8$, $BC=12$. Points $O$ and $P$ are the midpoints of sides $AB$ and $AD$, respectively. Point $H$ is a moving point on side $CD$. Connect $OH$. Quadrilateral $OBCH$ is folded along $OH$ to obtain quadrilateral $OFEH$. Connect $PE$. What is the minimum value of the length of $PE$?

Solution:
\textbf{Solution:} As shown in the figure, connect EO, PO, OC, \\
$\because$ Quadrilateral ABCD is a rectangle, \\
$\therefore \angle B=\angle OAP=90^\circ$, \\
$\because$ Points O, P are the midpoints of sides AB, AD respectively, \\
$\therefore$ OB=AO=4, AP=DP=6, \\
In $\triangle OBC$, BC=12, OB=4, \\
$\therefore$ OC=$\sqrt{OB^2+BC^2}=\sqrt{4^2+12^2}=4\sqrt{10}$, \\
In $\triangle AOP$, OA=4, PA=6, \\
$\therefore$ OP=$\sqrt{OA^2+PA^2}=\sqrt{4^2+6^2}=2\sqrt{13}$, \\
Due to folding, OE=OC=$4\sqrt{10}$, \\
$\because$ PE$\geq$ OE−OP, \\
$\therefore$ The minimum value of PE=OE−OP=$4\sqrt{10}-2\sqrt{13}$, \\
Thus, the answer is: $PE_{\min} = \boxed{4\sqrt{10}-2\sqrt{13}}$.


Question: As shown in the figure, the side length of the square $ABCD$ is $8$. On each side, we sequentially take $AE=BF=CG=DH=5$. What is the area of the quadrilateral $EFGH$?

Solution:
\textbf{Solution:}\\
Since quadrilateral ABCD is a square,\\
$\therefore$ $\angle A=\angle B=\angle C=\angle D=90^\circ$, AB=BC=CD=DA,\\
Since AE=BF=CG=DH,\\
$\therefore$ AH=BE=CF=DG.\\
In $\triangle AEH$, $\triangle BFE$, $\triangle CGF$, and $\triangle DHG$,\\
$\left\{\begin{array}{l}
AE=BF=CG=DH \\
\angle A=\angle B=\angle C=\angle D \\
AH=BE=CF=DG
\end{array}\right.$\\
$\therefore$ $\triangle AEH \cong \triangle BFE \cong \triangle CGF \cong \triangle DHG$ (SAS),\\
$\therefore$ EH=FE=GF=GH, $\angle AEH=\angle BFE$,\\
$\therefore$ Quadrilateral EFGH is a rhombus,\\
Since $\angle BEF+\angle BFE=90^\circ$,\\
$\therefore$ $\angle BEF+\angle AEH=90^\circ$,\\
$\therefore$ $\angle HEF=90^\circ$,\\
$\therefore$ Quadrilateral EFGH is a square,\\
Since AB=BC=CD=DA=8, AE=BF=CG=DH=5,\\
$\therefore$ EH=FE=GF=GH=$\sqrt{5^2+3^2}=\sqrt{34}$\\
Thus, the area of the square EFGH is $\boxed{\sqrt{34}}$.


Question: As shown in the diagram, the squares $A_1B_1C_1O$, $A_2B_2C_2C_1$, $A_3B_3C_3C_2\ldots$ are placed as illustrated. Points $A_1$, $A_2$, $A_3\ldots$ and points $C_1$, $C_2$, $C_3\ldots$ are respectively on the line $y=kx+b$ (where $k>0$) and the x-axis, given that points $B_1(1, 1)$, $B_2(3, 2)$, then what are the coordinates of $B_{12}$?

Solution:
\textbf{Solution:} \\
Let $B_{1}$ have coordinates (1, 1), and point $B_{2}$ have coordinates (3, 2), \\
therefore, the side length of square $A_{1}B_{1}C_{1}O_{1}$ = 1, and the side length of square $A_{2}B_{2}C_{2}C_{1}$ = 2, \\
hence, the coordinates of $A_{1}$ are (0, 1), and the coordinates of $A_{2}$ are: (1, 2), \\
plugging into $y=kx+b$, we get: $\left\{\begin{array}{l}b=1 \\ k+b=2\end{array}\right.$, solving gives: $\left\{\begin{array}{l}k=1 \\ b=1\end{array}\right.$, \\
therefore, the equation of the line is: $y=x+1$.\\
Since $A_{1}B_{1}=1$, and the coordinates of point $B_{2}$ are (3, 2), \\
thus, the coordinates of point $A_{3}$ are (3,4), \\
hence, $A_{3}C_{2}=A_{3}B_{3}=B_{3}C_{3}=4$, \\
thus, the coordinates of point $B_{3}$ are (7, 4), \\
therefore, the vertical coordinate of $B_{1}$ is: 1 = $2^{0}$, and the horizontal coordinate of $B_{1}$ is: 1 = $2^{1}$ - 1, \\
therefore, the vertical coordinate of $B_{2}$ is: 2 = $2^{1}$, and the horizontal coordinate of $B_{2}$ is: 3 = $2^{2}$ - 1, \\
therefore, the vertical coordinate of $B_{3}$ is: 4 = $2^{2}$, and the horizontal coordinate of $B_{3}$ is: 7 = $2^{3}$ - 1, \\
therefore, the vertical coordinate of $B_{n}$ is: $2^{n-1}$, and the horizontal coordinate is: $2^{n}$ - 1, \\
thus $B_{n}(2^{n}-1, 2^{n-1})$. \\
Therefore, the coordinates of $B_{12}$ are: $\boxed{(2^{12}-1, 2^{11})}$.


Question: As shown in the figure, fold the rectangular paper ABCD, first creating a crease through BD, then fold it again so that AD falls on the diagonal BD, with A corresponding to A$^{\prime}$, creating the fold crease DG. If AB = 2 and BC = 1, what is the length of AG?

Solution:
\textbf{Solution:} Given the conditions, we find that $AB=2$ and $AD=BC=1$.\\
In $\triangle ABD$, according to the Pythagorean theorem, $BD=\sqrt{AB^2+AD^2}=\sqrt{5}$.\\
By the property of folding, $AG=A'G$, $DA=DA'=1$. Let $AG=x$, $A'G=AG=x$, $BG=2-x$, $BA'=\sqrt{5}-1$,\\
In $\triangle BGA'$, by the Pythagorean theorem,\\
$BG^2=BA'^2+A'G^2$,\\
$(2-x)^2=(\sqrt{5}-1)^2+x^2$,\\
Solving, we find $x=\frac{\sqrt{5}-1}{2}$,\\
Hence, the length of $AG$ is: $\boxed{\frac{\sqrt{5}-1}{2}}$.


Question: As shown in the figure, DE is the median line of $\triangle ABC$, the bisector of $\angle ABC$ intersects DE at point F. If $\angle DFB=32^\circ$ and $\angle A=75^\circ$, then what is the measure of $\angle AED$ in degrees?

Solution:
\textbf{Solution}: Since $DE$ is the median line of $\triangle ABC$,\\
thus $DE\parallel BC$,\\
therefore $\angle ADE=\angle ABC$, $\angle DFB=\angle FBC$.\\
Since $\angle DFB=32^\circ$,\\
thus $\angle DFB=\angle FBC=32^\circ$.\\
Since the bisector of $\angle ABC$ intersects $DE$ at point F,\\
thus $\angle DBF=\angle FBC=\frac{1}{2}\angle ABC=32^\circ$,\\
therefore $\angle ADE=\angle ABC=2\times 32^\circ=64^\circ$.\\
Since $\angle A=75^\circ$,\\
thus $\angle AED=180^\circ−\angle A−\angle ADE=180^\circ−75^\circ−64^\circ=\boxed{41^\circ}$.


Question: As shown in the diagram, in parallelogram $ABCD$, $AC$ and $BD$ intersect at point $O$, and $E$ is the midpoint of $AB$. If $OE = 6$, what is the length of $AD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\
it follows that $BO=DO$, \\
since point E is the midpoint of AB, \\
it follows that $OE$ is the median line of $\triangle ABD$, \\
hence $AD=2OE$, \\
since $OE=6$, \\
it follows that $AD=\boxed{12}$


Question: As illustrated, in square $ABCD$ with side length $AB=2$, point $P$ is located on side $AB$ (not coinciding with $A$ or $B$). A square $PBEF$ is constructed inside square $ABCD$ with $PB$ as a side, intersecting side $BC$ at point $E$. Lines $DF$ and $CF$ are drawn. When $\triangle CDF$ is an isosceles triangle with $D$ as the apex, what is the length of $PB$?

Solution:
\textbf{Solution:} Connect BF,\\
Quadrilaterals ABCD, PBEF are squares, AB=2,\\
$\therefore$ Points B, F, D are collinear, CD=AB=2, $\angle A=90^\circ$\\
$\therefore$ $BD=\sqrt{AD^{2}+AB^{2}}=\sqrt{2^{2}+2^{2}}=2\sqrt{2}$\\
$\because$ $\triangle CDF$ is an isosceles triangle with D as the vertex,\\
$\therefore$FD=CD=2, BF=BD-FC=$2\sqrt{2}-2$,\\
$\because$2PB$^{2}$=BF$^{2}$\\
$\therefore$ $PB=\frac{\sqrt{2}}{2}BF=\frac{\sqrt{2}}{2}\left(2\sqrt{2}-2\right)=\boxed{2-\sqrt{2}}$;


Question: As shown in the figure, if quadrilateral $ABCD$ is a rhombus with $AC=24$ and $BD=10$, what is the side length of the rhombus $ABCD$?

Solution:
\textbf{Solution:} Let AC and BD intersect at point O, \\
since quadrilateral ABCD is a rhombus, \\
therefore AC$\perp$BD, $OC=\frac{1}{2}AC=12$, $OD=\frac{1}{2}BD=5$, \\
therefore $\angle$COD$=90^\circ$, \\
therefore $CD=\sqrt{OC^{2}+OD^{2}}=\boxed{13}$


Question: As shown in the figure, square $ABCD$ has a side length of $3$, and points $E$ and $F$ are two moving points on the diagonal $AC$ (with point $E$ to the left of point $F$), and $EF=1$. What is the minimum value of $DE+BF$?

Solution:
\textbf{Solution:} Let us solve for when $DE=BF$. The minimum value of $DE+BF$ is obtained when $DE=BF$. Connect $BD$ and $AC$, intersecting at point $O$, and connect $BE$,\\
$\because$ $ABCD$ is a square,\\
$\therefore$ $AD=AB$, $\angle DAE=\angle BAE=45^\circ$, $AC\perp BD$, $BD=\sqrt{3^2+3^2}=3\sqrt{2}$,\\
In $\triangle ADE$ and $\triangle ABE$ we have\\
$\left\{\begin{array}{l}
AD=AB \\
\angle DAE=\angle BAE=45^\circ \\
AE=AE,
\end{array}\right.$\\
$\therefore$ $\triangle ADE \cong \triangle ABE$,\\
$\therefore$ $DE=BE=BF$,\\
$BO\perp EF$,\\
$OE=OF=\frac{1}{2}EF=\frac{1}{2}$,\\
$OB=\frac{1}{2}BD=\frac{3\sqrt{2}}{2}$,\\
$\therefore DE=BE=BF=\sqrt{\left(\frac{3\sqrt{2}}{2}\right)^2+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{19}}{2}.$\\
$DE+BF =\frac{\sqrt{19}}{2}+\frac{\sqrt{19}}{2}=\boxed{\sqrt{19}}.$


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, and $D$, $E$, and $F$ are the midpoints of $AB$, $BC$, and $CA$ respectively. If $CD = 4$, what is the length of $EF$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is a right-angled triangle, and $CD$ is the median to the hypotenuse,\\
therefore $CD=\frac{1}{2}AB$,\\
thus $AB=2CD=8$,\\
since $EF$ is the median of $\triangle ABC$,\\
therefore $EF=\frac{1}{2}AB= \boxed{4}$.


Question: It is known that the side length of square $ABCD$ is 4. Square $OMNE$ (with $OM > \frac{1}{2}AC$) rotates around the center of symmetry $O$ of square $ABCD$. What is the area of the overlapping region of the two squares?

Solution:
\textbf{Solution}: As shown in the figure, connect AC and BD. Let OE intersect CD at point Q, and OM intersect BC at point P,

$\because$ Point O is the center of the square ABCD,

$\therefore$ The diagonals of square ABCD intersect at point O,

$\therefore \angle$ODQ=$\angle$OCP=$45^\circ$, OD=OC, $\angle$COD=$90^\circ$,

$\because$ Quadrilateral OMNE is a square,

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

$\because \angle$DOC=$\angle$DOQ+$\angle$COQ=$90^\circ$, $\angle$EOM=$\angle$COQ+$\angle$COP=$90^\circ$,

$\therefore \angle$DOQ=$\angle$COP,

In $\triangle DOE$ and $\triangle AOF$,

$\left\{ \begin{array}{l} \angle DOQ=\angle COP \\ OD=OC \\ \angle ODQ=\angle OCP \end{array} \right.$,

$\therefore \triangle DOQ \cong \triangle COP$ (ASA),

$\therefore S_{\triangle DOQ}=S_{\triangle COP}$,

$\therefore$ The area of quadrilateral OPCQ=$S_{\triangle OCP}+S_{\triangle QOC}=S_{\triangle ODQ}+S_{\triangle QOC}=S_{\triangle COD}$,

$\because S_{\triangle COD}=\frac{1}{4}S_{\text{square ABCD}}=\frac{1}{4}\times 4\times 4=4$,

$\therefore$ The area of the overlapping part of the two squares is $\boxed{4}$.


Question: As shown, for the square $ABCD$ with side length $2$, where point $E$ is the midpoint of side $BC$, and point $P$ moves along the diagonal $AC$, what is the minimum value of $PE+PB$?

Solution:
\textbf{Solution:} Draw line DE, intersecting AC at point P, and draw BD. \\
$\because$ Point B and point D are symmetric about AC, \\
$\therefore$ DP=PB, \\
$\therefore$ PB+PE=PD+PE=DE \\
$\therefore$ The length of DE is the minimum value of PE+PB, \\
$\because$ In the square ABCD, AB=2, and E is the midpoint of BC, \\
$\therefore$ CE=1, \\
In $\triangle CDE$, \\
DE=$\sqrt{CD^{2}+CE^{2}}$ \\
=$\sqrt{2^{2}+1^{2}}$ \\
=$\sqrt{5}$ \\
Hence, the answer is: $\boxed{\sqrt{5}}$.


Question: Given a rectangle $ABCD$ with a point $P$ satisfying $PA=1$, $PB=2$, $PC=3$, what is the value of $PD$?

Solution:
\textbf{Solution:} As shown in the figure, draw GH$\parallel$BC through point P intersecting AB and CD at G and H, respectively; draw EF$\parallel$AB through point P intersecting AD and BC at E and F, respectively. Let AG=DH=x, GB=HC=y, AE=BF=m, and ED=FC=n.\\
Therefore, AP$^{2}$=x$^{2}$+m$^{2}$, CP$^{2}$=y$^{2}$+n$^{2}$, BP$^{2}$=y$^{2}$+m$^{2}$, DP$^{2}$=n$^{2}$+x$^{2}$,\\
Thus, AP$^{2}$+CP$^{2}$=BP$^{2}$+DP$^{2}$=x$^{2}$+y$^{2}$+m$^{2}$+n$^{2}$,\\
Given that, PA=1, PB=2, PC=3,\\
This implies $1+9=4+DP^{2}$,\\
Solving yields: $DP=\boxed{\sqrt{6}}$ or $-\sqrt{6}$ (discard the latter).


Question: As shown in the figure, in $\triangle ABC$, $DE\parallel BC$, intersecting $AB$ and $AC$ at $D$ and $E$ respectively, and $CD\perp BE$, with $CD=3$ and $BE=5$. Calculate the value of $BC+DE$.

Solution:
\textbf{Solution:} Construct EF$\parallel$DC intersecting the extension of BC at F,\\
since DE$\parallel$BC, EF$\parallel$DC,\\
thus quadrilateral DCFE is a parallelogram,\\
thus EF=CD=3, CF=DE,\\
since CD$\perp$BE,\\
thus EF$\perp$BE,\\
thus BC+DE=BC+CF=BF=$\sqrt{BE^{2}+EF^{2}}$=$\sqrt{5^{2}+3^{2}}$=$\boxed{\sqrt{34}}$.


Question: Given a square $ABCD$ with side length 1, let $E$ and $F$ be two movable points on sides $BC$ and $CD$, respectively, satisfying $BE=CF$. Connect $AE$ and $AF$. What is the minimum value of $AE+AF$?

Solution:
\textbf{Solution:} Draw line DE, \\
since $BE=CF$ and the quadrilateral ABCD is a square, \\
thus $CD-CF=BC-BE$, i.e., $DF=CE$, in $\triangle ADF$ and $\triangle DCE$, \\
\[\begin{cases}
AD=DC, \\
\angle ADF=\angle DCE, \\
DF=CE,
\end{cases}\]
thus $\triangle ADF \cong \triangle DCE$, \\
therefore, $AF=DE$; $AE+AF=AE+DE$, \\
by taking BC as the axis of symmetry, construct the symmetrical point $A'$ of point A with respect to BC and connect $DA'$, intersecting BC at point E, \\
since point A and point $A'$ are symmetric with respect to BC, \\
thus AE=$A'E$, \\
$AE+DE=A'E+DE=A'D$, \\
by the Pythagorean theorem, we have: $A'D=\sqrt{AD^{2}+AA'^{2}}=\sqrt{2^{2}+1^{2}}=\sqrt{5}$, \\
therefore, the minimum value of $AE+AF$ is $\boxed{\sqrt{5}}$.


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

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


Question: As shown in the diagram, in rhombus $ABCD$, the midpoint of side $AB$ is $M$, and the diagonals intersect at point $O$. If $OM = 3\,\text{cm}$, what is the perimeter of the rhombus $ABCD$ in centimeters?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rhombus, \\
it follows that AO=CO, and AB=BC=CD=AD, \\
since point M is the midpoint of side $AB$ and $OM=3$, \\
it follows that AD=2OM=6, \\
therefore, the perimeter of rhombus $ABCD$ is $4AD=\boxed{24}$.


Question: Sorry, it seems like you didn't provide any specific text to translate. Could you please share the text you'd like to be translated?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\
it follows that $\angle BAD=90^\circ$, $OD= \frac{1}{2}BD$, and $AD=BC=8$,\\
thus $BD=\sqrt{AB^{2}+AD^{2}}=\sqrt{6^{2}+8^{2}}=10\text{cm}$,\\
which means $OD=5\text{cm}$,\\
since points $E$ and $F$ are the midpoints of $AO$ and $AD$ respectively,\\
it follows that $EF$ is the midline of $\triangle AOD$,\\
therefore $EF= \frac{1}{2}OD=\boxed{2.5}\text{cm}$;


Question: As shown in the figure, in rectangle $ABCD$, points $E$ and $F$ are the midpoints of sides $AB$ and $BC$, respectively. Lines $EC$ and $FD$ are connected, and points $G$ and $H$ are the midpoints of $EC$ and $FD$, respectively. Line $GH$ is then connected. If $AB=6$ and $BC=10$, what is the length of $GH$?

Solution:
\textbf{Solution:} Draw line CH and extend it to intersect AD at P, and connect PE, as shown in the figure: \\
$\because$ Quadrilateral ABCD is a rectangle, \\
$\therefore$ $\angle$A＝$90^\circ$, $ AD\parallel BC$, $ AD=BC=10$, \\
$\because$ E, F are the midpoints of side AB, BC respectively, AB＝6, BC＝10, \\
$\therefore$ AE＝$ \frac{1}{2}$AB＝$ \frac{1}{2}\times $6＝3, CF＝$ \frac{1}{2}$BC＝$ \frac{1}{2}\times $10＝5, \\
$\because$ $ AD\parallel BC$, \\
$\therefore$ $\angle$DPH＝$\angle$FCH, \\
$\because$ H is the midpoint of DF, \\
$\therefore$ DH=FH \\
$\because$ In $\triangle $PDH and $\triangle $CFH $\left\{\begin{array}{l}\angle DPH=\angle FCH\hfill \\ \angle DHP=\angle FHC\hfill \\ DH=FH\hfill \end{array} \right.$, \\
$\therefore$ $\triangle $PDH$\cong $$\triangle $CFH (AAS), \\
$\therefore$ PD＝CF＝5, CH＝PH, \\
$\therefore$ AP＝AD−PD＝5, \\
$\therefore$ PE＝$ \sqrt{AP^{2}+AE^{2}}$＝$ \sqrt{5^{2}+3^{2}}$＝$ \sqrt{34}$, \\
$\because$ Point G is the midpoint of EC, CH＝PH, \\
$\therefore$ GH is the midline of $\triangle $CEP, \\
$\therefore$ GH＝$ \frac{1}{2}$EP＝$ \frac{\sqrt{34}}{2}$, \\
Hence, the answer is: $GH = \boxed{\frac{\sqrt{34}}{2}}$.


Question: As shown in the diagram, in rectangle $ABCD$, $AC$ and $BD$ intersect at point $O$, with $M$ and $N$ being the midpoints of $BC$ and $OC$, respectively. If $MN = 4$, what is the length of $AC$? If the area of $\triangle BOC$ is 5, what is the area of rectangle $ABCD$?

Solution:
\textbf{Solution:} Since M and N are the midpoints of BC and OC, respectively,

it follows that $BO = 2MN = 8$.

Since quadrilateral ABCD is a rectangle,

it follows that $AC = BD = 2BO = \boxed{16}$.

Since the area of $\triangle BOC$ is 5,

the area of rectangle ABCD is $5 \times 4 = \boxed{20}$.


Question: As shown in the figure, the side length of the square $ABCD$ is $4$, and the point $P$ is on the diagonal $AC$ (not including endpoints). With $PA$ and $PB$ as adjacent sides, a square $PAQB$ is formed. What is the minimum value of the diagonal length $PQ$?

Solution:
\textbf{Solution:} As shown in the figure, let AB and PQ intersect at point O, and draw $OE\perp AC$ through point $O$, \\
since quadrilateral $AQBP$ is a parallelogram, \\
therefore $AO=OB=\frac{1}{2}AB=2$, \\
since quadrilateral $ABCD$ is a square, \\
therefore $\angle AOE=\angle OAE=45^\circ$, \\
thus, OE=AE, \\
therefore $OE^{2}+AE^{2}=AO^{2}$, \\
thus, OE=$\frac{\sqrt{2}}{2}AO$, \\
according to the problem, $P$ is a moving point on $AC$, when point P coincides with point E, PO reaches its minimum value. \\
At this time, $PQ=2OE=2\times \frac{\sqrt{2}}{2}AO=\boxed{2\sqrt{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=20$, $AC=9$. The point $M$ is the midpoint of $BC$. The line $AD$ bisects the exterior angle $\angle CAE$ of $\triangle ABC$, intersecting the extended line of $BC$ at point $D$. Through point $M$, draw $MN \parallel to AD$, intersecting $AB$ at point $N$. What is the length of $AN$?

Solution:
\textbf{Solution:} Let NA intersect NF at BN, and connect CF, as shown in the diagram.\\
$\because$BM=MC, NF=BN,\\
$\therefore$MN$\parallel$CF,\\
$\because$MN$\parallel$AD,\\
$\therefore$CF$\parallel$AD,\\
Thus, $\angle$AFC=$\angle$EAD, $\angle$ACF=$\angle$DAC,\\
$\because$AD bisects $\angle$CAE,\\
$\therefore \angle$DAC=$\angle$EAD,\\
$\therefore \angle$ACF=$\angle$AFC,\\
$\therefore$AF=AC=9,\\
$\therefore$BF=AB−AF=11,\\
$\because$MN is the midline of $\triangle$BCF,\\
$\therefore$BN=NF=$\frac{11}{2}$,\\
$\therefore$AN=NF+AF=$\boxed{\frac{29}{2}}$.


Question: As shown in the figure, the two diagonals of rectangle ABCD intersect at point O. If $\angle AOD = 60^\circ$ and AD = 2, then what is the length of CD?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
it follows that OA=OC, OB=OD, and AC=BD,\\
which means OA=OB=OD,\\
Since $\angle AOD = 60^\circ$,\\
it follows that $\triangle AOD$ is an equilateral triangle,\\
thus, OD=AD=2,\\
which implies BD=2OD=4,\\
therefore, $DC = \sqrt{AC^{2} - AD^{2}} = \sqrt{16 - 4} = 2\sqrt{3}$,\\
Hence, the answer is: $DC = \boxed{2\sqrt{3}}$.


Question: As shown in the figure, in quadrilateral ABCD, $\angle A=90^\circ$, AB = 4, and AD = 3. Points M and N are moving points on line segments BC and AB, respectively (point M does not coincide with point B). Points E and F are the midpoints of DM and MN, respectively. What is the maximum length of EF?

Solution:
\textbf{Solution:} Connect DN and DB as shown in the diagram: \\
In $\triangle DAB$, where $\angle A=90^\circ$, $AB=4$, and $AD=3$, \\
$\therefore BD=\sqrt{AD^2+AB^2}=\sqrt{3^2+4^2}=5$, \\
$\because$ Points E and F are the midpoints of DM and MN respectively, \\
$\therefore EF=\frac{1}{2}DN$, \\
According to the problem, when point N coincides with point B, DN is at its maximum, which is 5, \\
$\therefore$ The maximum length of $EF$ is $\boxed{2.5}$.


Question: As shown in the figure, the quadrilateral $OABC$ is a rhombus with $AC=6$ and $OB=8$. What are the coordinates of vertex $C$?

Solution:
\textbf{Solution:} As shown in the figure, let $AC$ and $OB$ intersect at point D,\\
since quadrilateral $OABC$ is a rhombus, with $AC=6$, and $OB=8$,\\
therefore $OD=\frac{1}{2}OB=4$, $CD=\frac{1}{2}AC=3$, and $AC\perp OB$,\\
thus $OC=\sqrt{OD^{2}+CD^{2}}=5$,\\
hence $C(5,0)$,\\
so the answer is: $C=\boxed{(5,0)}$.


Question: If point $E$ is the midpoint of $BC$, and $CD=4$, by folding the square $ABCD$ along $FH$ such that point $D$ precisely falls on the midpoint $E$ of side $BC$, and the corresponding point of point $A$ is point $P$, then what is the length of the crease $HF$?

Solution:
\textbf{Solution:}

Draw $HG \perp CD$ at point G, and connect $DE$, as shown in the diagram:\\
Then, $\angle DGH=\angle FGH=90^\circ$,\\
In the square $ABCD$, $\angle A=\angle D=\angle C=90^\circ$, $AD=CD=BC=4$,\\
$\therefore$ Quadrilateral $ADGH$ is a rectangle,\\
$\therefore GH=AD=CD$,\\
According to the folding, $FH$ is a perpendicular bisector of $DE$,\\
$\therefore \angle FDE+\angle DFH=90^\circ$,\\
$\because \angle FDE+\angle DEC=90^\circ$,\\
$\therefore \angle DFH=\angle DEC$,\\
In $\triangle HGF$ and $\triangle DCE$,\\
$\left\{\begin{array}{l}
\angle DFH=\angle DEC\\
\angle FGH=\angle ECD\\
HG=CD
\end{array}\right.$\\
$\therefore \triangle HGF \cong \triangle DCE \left(AAS\right)$,\\
$\therefore FH=DE$,\\
$\because E$ is the midpoint of $BC$,\\
$\therefore EC=2$,\\
In $Rt\triangle CDE$, according to the Pythagorean theorem, $DE=\sqrt{4^2+2^2}=2\sqrt{5}$,\\
$\therefore FH=\boxed{2\sqrt{5}}$.


Question: In the rhombus $ABCD$, as shown in the figure, $\angle ADC = 120^\circ$. Point $E$ is on the diagonal $AC$, and $AE = AB$. When connecting $DE$, if $CE = 4$, what is the length of $DE$?

Solution:
\textbf{Solution:} As shown in the diagram, draw $EF \perp CD$ at F,\\
then $\angle EFC = \angle EFD = 90^\circ$,\\
since quadrilateral ABCD is a rhombus,\\
therefore $AB = AD = CD$,\\
therefore $\angle DCA = \angle DAC$,\\
since $\angle ADC = 120^\circ$,\\
therefore $\angle DCA = \angle DAC = 30^\circ$,\\
therefore $EF = \frac{1}{2}CE = 2$,\\
since $AE = AB$,\\
therefore $AE = AE$,\\
therefore $\angle ADE = \angle AED = \frac{1}{2}(180^\circ - \angle DAC) = 75^\circ$,\\
therefore $\angle EDF = \angle ADC - \angle ADE = 120^\circ - 75^\circ = 45^\circ$,\\
therefore $\triangle DEF$ is an isosceles right-angled triangle,\\
therefore $DE = \sqrt{2}EF = 2\sqrt{2}$,\\
thus the answer is: $DE = \boxed{2\sqrt{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=90^\circ$, $AB=4$, $BC>AB$, and point $D$ is on side $BC$. Among all the parallelograms $ADCE$ with $AC$ as the diagonal, what is the minimum value of $DE$?

Solution:
Solution: The intersection point of the diagonals of parallelogram ADCE is the midpoint O of AC. When $OD \perp BC$, OD is minimal, which means DE is minimal.\\
Since $AB \perp BC$,\\
Therefore, $OD \parallel AB$,\\
Therefore, at this time, OD is the median of $\triangle ABC$,\\
Therefore, $OD = \frac{1}{2}AB = 2$,\\
Since parallelogram ADCE is a parallelogram,\\
Therefore, $DE = 2OD = \boxed{4}$.


Question: As shown in the figure, a hexagon $ABCDEF$ has equal interior angles, and $AD\parallel BC$. What is the measure of $\angle DAB$ in degrees?

Solution:
\textbf{Solution}: Since the internal angles of the hexagon ABCDEF are all equal,\\
each internal angle is: $(6-2) \times 180^\circ \div 6 = 120^\circ$,\\
that is, $\angle ABC = 120^\circ$,\\
since BC$\parallel$AD,\\
thus $\angle DAB = 180^\circ - \angle ABC = \boxed{60^\circ}$.


Question: As shown in the figure, point $P$ is a moving point on the hypotenuse $AC$ (not coinciding with $A$ or $C$) of $\triangle ABC$. Draw $PM\perp AB$ intersecting $AB$ at point $M$, and draw $PN\perp BC$ intersecting $BC$ at point $N$. Connect $MN$. If $AB=6$ and $BC=8$, then what is the minimum value of $MN$ when point $P$ moves along $AC$?

Solution:
\textbf{Solution:} Draw a line PB, \\
$\because$ $\angle ABC=90^\circ$, AB = 6, BC = 8, \\
$\therefore$ $AC=\sqrt{AB^2+BC^2}=\sqrt{6^2+8^2}=10$, \\
$\because$ PM $\perp$ AB, PN $\perp$ BC, $\angle C=90^\circ$, \\
$\therefore$ $\angle PMB=\angle PNB=\angle B=90^\circ$, \\
$\therefore$ Quadrilateral BNPM is a rectangle, \\
$\therefore$ MN = BP, \\
$\because$ The perpendicular segment is the shortest, \\
$\therefore$ When BP $\perp$ AC, the length of MN is minimized, \\
$\therefore$ ${S}_{\triangle ABC}=\frac{1}{2}BC\cdot AB=\frac{1}{2}AC\cdot BP$, \\
$\therefore$ $\frac{1}{2}\times 8\times 6=\frac{1}{2}\times 10\cdot BP$, \\
Solve for it: BP = 4.8. \\
$\therefore$ The minimum value of MN is $\boxed{4.8}$.


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

Solution:
\textbf{Solution:} Given that $BE\parallel AC$ and $AE\parallel BD$,\\
it follows that quadrilateral AEBO is a parallelogram,\\
and since the diagonals of rhombus ABCD intersect at point O,\\
then $OA=\frac{1}{2}AC=4$, $OB=OD$, and $AC\perp BD$,\\
which means $\angle AOB=90^\circ$.\\
Thus, parallelogram AOBE is a rectangle,\\
and so $AB=OE=5$,\\
which leads to $OB=\sqrt{AB^{2}-OA^{2}}=\sqrt{5^{2}-4^{2}}=3$,\\
and therefore $BD=2OB=6$.\\
Let the height of rhombus ABCD be $h$,\\
then the area $S_{\text{rhombus }ABCD}=\frac{1}{2}AC\cdot BD=AB\cdot h$,\\
yielding $h=\frac{\frac{1}{2}\times 8\times 6}{5}=\boxed{\frac{24}{5}}$,\\
meaning the height of rhombus ABCD is $\frac{24}{5}$.


Question: As shown in the figure, within the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, and their lengths are $6\text{cm}$ and $8\text{cm}$, respectively. A line passing through point $O$ intersects $AD$ and $BC$ at points $E$ and $F$, respectively. What is the sum of the areas of the shaded regions in $\text{cm}^2$?

Solution:
\textbf{Solution:} Since $AC$ and $BD$ are the diagonals of the rhombus $ABCD$,\\
it follows that $AC \perp BD$, and $OA=OC$, $AD\parallel BC$,\\
therefore, $\angle OAE=\angle OCF$, $\angle AEO=\angle CFO$,\\
thus, $\triangle AOE\cong \triangle COF$,\\
hence, the areas are $S_{\triangle AOE}=S_{\triangle COF}$,\\
therefore, $S_{\text{shaded}}=S_{\triangle BCD}=\frac{1}{2}S_{\text{rhombus} ABCD}$,\\
Given the lengths of the diagonals of the rhombus are 6cm and 8cm,\\
which means: $S_{\text{shaded}}=S_{\triangle BCD}=\frac{1}{2}S_{\text{rhombus} ABCD}=\frac{1}{2}\times (\frac{1}{2}\times 6\times 8)=\boxed{12}\text{cm}^2$


Question: As shown in the figure, the diagonals of square $ABCD$ intersect at point $O$. An arbitrary line passing through point $O$ intersects $AD$ and $BC$ at points $E$ and $F$, respectively. If the length of the diagonal of the square is $\sqrt{2}$, what is the area of the shaded region in the figure?

Solution:
\textbf{Solution:} In the square $ABCD$,

$AO=CO=DO$, $\angle OAE=\angle OCF=45^\circ$, $\angle AOD=90^\circ$,

Also, since $\angle AOE=\angle COF$,

$\therefore$ $\triangle AOE \cong \triangle COF$,

$\therefore$ $S_{\triangle COF}=S_{\triangle AOE}$,

$\therefore$ the area of the shaded region is equal to $S_{\triangle COF}+S_{\triangle DOE}=S_{\triangle AOE}+S_{\triangle DOE}=S_{\triangle AOD}$,

Since the diagonal of a square is $\sqrt{2}$,

$\therefore$ $AO=DO=\frac{\sqrt{2}}{2}$,

$S_{\triangle AOD}=\frac{1}{2}AO\cdot DO=\frac{1}{2}\times \frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{2}=\frac{1}{4}$,

$\therefore$ the area of the shaded region is equal to $\boxed{\frac{1}{4}}$.


Question: As shown in the figure, point E is on the diagonal BD of the rhombus ABCD, and DE = DA. Line AE is drawn. If $\angle CAE = 15^\circ$, what is the degree measure of $\angle ABC$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\
it follows that AC $\perp$ BD, \\
since $\angle$CAE = 15$^\circ$, \\
thus, $\angle$AED = 90$^\circ$ - $\angle$EAC = 90$^\circ$ - 15$^\circ$ = 75$^\circ$, \\
since DE = DA, \\
it follows that $\angle$EAD = $\angle$AED = 75$^\circ$, \\
therefore, $\angle$ADB = $\angle$ABD = 30$^\circ$, \\
therefore, $\angle$ABC = 2$\angle$ADB = $2 \times 30^\circ$ = \boxed{60^\circ},


Question: As shown in the figure, within $ \square ABCD$, points E and F are both on side AD. BE bisects $\angle ABC$ and CF bisects $\angle BCD$. If $BE=8$, $CF=6$, and $EF=2$, what is the perimeter of $\square ABCD$?

Solution:
\textbf{Solution:} Extend BC to point P such that CP = EF = 2,\\
since $\square ABCD$,\\
thus EF $\parallel$ CP, $\angle ABC + \angle DCB = 180^\circ$,\\
thus quadrilateral EFCP is a parallelogram,\\
thus EP = CF = 6, EP $\parallel$ CF,\\
since BE bisects $\angle ABC$, CF bisects $\angle BCD$,\\
thus $\angle EBC + \angle FCB = \frac{1}{2}\angle ABC + \frac{1}{2}\angle DCB = 90^\circ$,\\
thus CF $\perp$ BE,\\
thus PE $\perp$ BE,\\
thus BP = $\sqrt{BE^2 + PE^2}$ = 10,\\
thus BC = 8,\\
since $\angle ABE = \angle EBC$, $\angle EBC = \angle AEB$,\\
thus $\angle ABE = \angle AEB$,\\
thus AB = AE,\\
Similarly, it is derived that: CD = DF,\\
thus AE = DF,\\
since AD + EF = AE + DF,\\
thus AE = DF = 5,\\
thus AB = CD = 5,\\
thus the perimeter of $\square ABCD$ is equal to 2 $\times$ (5 + 8) = \boxed{26}.


Question: As illustrated in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, $AC = 5$, $BC = 12$. Let $D$ be a moving point on $AB$. Construct $DE \perp AC$ at point $E$ and $DF \perp BC$ at point $F$. Connect $EF$. What is the minimum value of the line segment $EF$?

Solution:
\textbf{Solution:} As shown in the diagram, connect CD.\\
$\because$ $\angle ACB = 90^\circ$, AC = 5, BC = 12,\\
$\therefore$ AB = $\sqrt{AC^2 + BC^2}$ = $\sqrt{5^2 + 12^2}$ = 13,\\
$\because$ DE $\perp$ AC, DF $\perp$ BC, $\angle C = 90^\circ$,\\
$\therefore$ Quadrilateral CFDE is a rectangle,\\
$\therefore$ EF = CD,\\
By the principle that the perpendicular segment is the shortest, when CD $\perp$ AB, the value of segment EF is the smallest,\\
At this moment, $S_{\triangle ABC} = \frac{1}{2} \text{BC} \cdot \text{AC} = \frac{1}{2} \text{AB} \cdot \text{CD}$,\\
which means $\frac{1}{2} \times 12 \times 5 = \frac{1}{2} \times 13 \cdot \text{CD}$,\\
Solving yields: CD = $\frac{60}{13}$,\\
$\therefore$ EF = $\boxed{\frac{60}{13}}$.


Question: As shown in the figure, within rectangle $ABCD$, $EF$ is the perpendicular bisector of the diagonal $BD$, intersecting $AD$ and $BC$ at points $E$ and $F$, respectively. Connect $AO$. If $ AO=4$ and $EF=6$, then what is the length of $AB$?

Solution:
\textbf{Solution:} As shown in the figure, connect BE, \\
since EF is the perpendicular bisector of diagonal BD of rectangle ABCD, and AO = 4, \\
thus BD = 2DO = 2AO = 8, BE = DE, $\angle$DOE = $90^\circ$, \\
therefore, DO = 4, \\
since quadrilateral ABCD is a rectangle, \\
thus AD $\parallel$ BC, \\
thus $\angle$EDO = $\angle$FBO, \\
since OB = OD, $\angle$EOD = $\angle$FOB, \\
therefore, $\triangle EOD \cong \triangle FOB$ (ASA), \\
thus EO = OF, \\
since EF = 6, \\
thus EO = 3, \\
let AE = x, \\
in $Rt\triangle OBE$, $BE = DE = \sqrt{3^2 + 4^2} = 5$, \\
in $Rt\triangle ABD$ and $Rt\triangle ABE$, $AB^2 = BD^2 - AD^2 = BE^2 - AE^2$, \\
thus $8^2 - (5 + x)^2 = 5^2 - x^2$, \\
solving this, we get: $x = \frac{7}{5}$, \\
therefore, $AB = \sqrt{5^2 - \left(\frac{7}{5}\right)^2} = \boxed{4.8}$.


Question: In parallelogram $ABCD$, an arc is drawn with A as the center and the length of $AB$ as the radius, intersecting $AD$ at F. Then, with B and F as centers, arcs longer than $\frac{1}{2}BF$ are drawn, intersecting at point G. If $BF=6$ and $AB=5$, what is the length of $AE$?

Solution:
\textbf{Solution:} Connect $EF$, as shown in the diagram:\\
Based on the problem statement, we have: $AE$ bisects $\angle DAB$, $AF=AB=5$,\\
$\therefore \angle AOF=90^\circ$, $OB=OF=3$, $\angle BAE=\angle FAE$,\\
$\therefore OA=\sqrt{AF^{2}-OF^{2}}=4$,\\
$\because$ Quadrilateral $ABCD$ is a parallelogram,\\
$\therefore AD\parallel BC$,\\
$\therefore \angle FAE=\angle AEB$,\\
$\therefore \angle BAE=\angle AEB$,\\
$\therefore BE=AB=AF$,\\
$\therefore$ Quadrilateral $ABEF$ is a parallelogram,\\
$\therefore OA=OE=\frac{1}{2}AE$,\\
$\therefore AE=2OA=\boxed{8}$;


Question: As shown in the diagram, in rectangle $ABCD$, with $AB=4$ and $AD=3$, point $E$ is the midpoint of side $BC$. Line $AE$ is drawn, and $\triangle ABE$ is folded along $AE$ to obtain $\triangle AFE$. When $AF$ is extended to intersect $CD$ at point $G$, what is the length of $DG$?

Solution:
\textbf{Solution:} Based on the properties of folding, we know that: $EB=EF$, $AB=AF$, and $\angle EBA=\angle EFA=90^\circ$,\\
since point $E$ is the midpoint of side $BC$,\\
thus $CE=BE$,\\
therefore $CE=BE=EF$,\\
As shown in the figure: connect $EG$,\\
since $DC\perp CE$, $EF\perp AC$,\\
therefore $\angle ECG=\angle EFG=90^\circ$,\\
In $\triangle ECG$ and $\triangle EFG$,\\
$\left\{\begin{array}{l}EC=EF\\ EG=EG\end{array}\right.$\\
therefore $\triangle ECG\cong \triangle EFG\left(\text{HL}\right)$,\\
thus $GC=GF$,\\
Let $GC=GF=x$, then $DG=4-x$, $AG=4+x$,\\
In $\triangle DAG$, $AD^{2}+DG^{2}=AG^{2}$,\\
which means $3^{2}+(4-x)^{2}=(4+x)^{2}$,\\
Solving this, we get: $x=\frac{9}{16}$,\\
Thus, the answer is: $DG=4-x=\boxed{\frac{55}{16}}$.


Question: As shown in the figure, fold a rectangular piece of paper \textit{ABCD} with sides of lengths 8 and 16 respectively, such that point \textit{C} coincides with point \textit{A}. What is the length of the crease \textit{EF}?

Solution:
\textbf{Solution:} Given the properties of folding, we know that $EC=AE$, and $\angle AEF=\angle CEF$, \\
since $AD\parallel BC$, \\
it follows that $\angle AFE=\angle CEF$, \\
thus, $\angle AEF=\angle AFE$, \\
which means $AE=AF=CE$, \\
using the Pythagorean theorem, we find $AB^2+BE^2=AE^2$ that is $8^2+(16-AE)^2=AE^2$, \\
solving this gives us $AE=AF=10$, and $BE=6$, \\
let $EG\perp AF$ at point G, \\
then the quadrilateral $AGEB$ is a rectangle, with $AG=6$, $GF=4$, $GE=AB=8$, and by the Pythagorean theorem, $EF=\boxed{4\sqrt{5}}$.


Question: As shown in the figure, in rectangle $ABCD$, it is known that: $AB=3$, $AD=5$, and point $P$ is a point on $BC$, and $\triangle PAD$ is an isosceles triangle. What is the length of $BP$?

Solution:
\textbf{Solution:}
(1) When $DP=AD$,\\
since rectangle $ABCD$,\\
thus $DC=AB=3$, $AD=BC=5$,\\
since $\triangle PAD$ is an isosceles triangle,\\
thus $DP=AD=5$,\\
in right $\triangle PCD$,\\
$PC=\sqrt{PD^2-DC^2}=4$,\\
thus $BP=BC-CP=5-4=1$.\\
(2) When $AD=AP$,\\
thus $AP=AD=5$,\\
in right $\triangle ABP$,\\
by the Pythagorean theorem,\\
$BP=\sqrt{PA^2-AB^2}=4$,\\
(3) When $AP=DP$,\\
draw $PE\perp AD$ at point $E$,\\
thus $AE=\frac{1}{2}AD=2.5$,\\
since $\angle B=\angle BAE=\angle AEP=90^\circ$,\\
thus quadrilateral $ABPE$ is a rectangle,\\
thus $BP=AE=2.5$.\\
In summary, $BP=1$ or $4$ or $2.5$.\\
Therefore, the answer is: \boxed{1 \text{ or } 4 \text{ or } 2.5}.


Question: As shown in the figure, diagonals AC and BD of square $ABCD$ intersect at point O, and point E is the midpoint of AD. Line OE is connected. If $OA = 1$ and the perimeter of $\triangle AOE$ is 5, then what is the perimeter of square $ABCD$?

Solution:
\textbf{Solution:} Given that $OA=1$ and the perimeter of $\triangle AOE$ is 5,\\
it follows that $OE+AE=5-1=4$,\\
Since point E is the midpoint of AD,\\
then $AD=2AE$, and OE is the median of $\triangle ACD$,\\
hence $CD=2OE$,\\
therefore, $AD+CD=2AE+2OE=2(AE+OE)=8$,\\
thus, the perimeter of $\square ABCD$ is: $2(AD+CD)=2 \times 8=\boxed{16}$.


Question: As shown in the figure, in rectangle $ABCD$, the two diagonals $AC$ and $BD$ intersect at point $O$. If $AB = OB = 6$, what is the area of the rectangle?

Solution:
\textbf{Solution:} Because quadrilateral ABCD is a rectangle, \\
$\therefore$ $\angle ABC = 90^\circ$, OA = $\frac{1}{2}$AC, OB = $\frac{1}{2}$BD, AC = BD, \\
$\therefore$ OA = OB, \\
Also because AB = OB = 6, \\
$\therefore$ OA = OB = AB = 6, \\
$\therefore$ AC = 2OA = 12, \\
$\therefore$ BC = $\sqrt{AC^2 - AB^2}$ = $6\sqrt{3}$, \\
$\therefore$ the area of rectangle ABCD = BC $\cdot$ AB = $6\sqrt{3} \times 6 = \boxed{36\sqrt{3}}$;


Question: As shown in the figure, in rectangle $ABCD$, $BC=10$, and $\angle BAC=30^\circ$. If points $M$ and $N$ are chosen on $AC$ and $AB$, respectively, such that the value of $BM+MN$ is minimized, what is this minimum value?

Solution:
\textbf{Solution:} Construct the symmetric point $B^{\prime}$ of point $B$ with respect to $AC$. Draw $B^{\prime}N \perp AB$, intersecting $AC$ at point $M$.

Since rectangle $ABCD$,

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

Since in $\triangle ABC$, $\angle BAC = 30^\circ$,

Therefore, $\angle ACB = 90^\circ - 30^\circ = 60^\circ$,

Therefore, $AB = BC \tan \angle ACB = 10 \tan 60^\circ = 10\sqrt{3}$;

Since points $B$ and $B^{\prime}$ are symmetric about $AC$,

Therefore, $AB = AB^{\prime}$, $\angle BAC = \angle B^{\prime}AC = 30^\circ$,

Therefore, $\angle BAB^{\prime} = 60^\circ$,

Therefore, $\triangle BAB^{\prime}$ is an equilateral triangle;

Since $B^{\prime}N \perp AB$,

Therefore, $B^{\prime}N = BB^{\prime} \sin \angle B^{\prime}BN = 10\sqrt{3} \sin 60^\circ = 10\sqrt{3} \times \frac{\sqrt{3}}{2} = 15$

Since points $B$ and $B^{\prime}$ are symmetric about $AC$,

Therefore, $BM = B^{\prime}M$,

Therefore, $BM + MN = B^{\prime}M + MN = B^{\prime}N = \boxed{15}$


Question: As shown in the figure, it is known that in $\triangle ABC$ ($AB > AC$), $\angle BAC = 60^\circ$, $AC = 4$, $D$ is the midpoint of side $BC$, and the line $DF$ passing through point $D$ bisects the perimeter of $\triangle ABC$ and intersects $AB$ at point $F$. What is the length of $DF$?

Solution:
\textbf{Solution:} Solution as shown in the figure:\\
Extend BA to E, such that AE=AC=4, take the midpoint F of BE, join DF, join CE, and draw a line AG$\perp$CE at point G,\\
$\because$D is the midpoint of side BC,\\
$\therefore$BD=CD,\\
$\because$EF=BF,\\
$\therefore$BD+BF=CD+AE+FA=CD+EF,\\
$\therefore$The line DF bisects the perimeter of $\triangle ABC$,\\
$\because$AE=AC=4, $\angle BAC=60^\circ$,\\
$\therefore$$\angle ACE=\angle E=30^\circ$,\\
$\therefore$AG=$\frac{1}{2}$AE=2,\\
$\therefore$EG=$\sqrt{AE^2-AG^2}=\sqrt{4^2-2^2}=2\sqrt{3}$,\\
$\because$AE=AC, AG$\perp$CE,\\
$\therefore$GE=$\frac{1}{2}$CE,\\
$\because$D is the midpoint of BC, F is the midpoint of BE,\\
$\therefore$FD is the mid-segment of $\triangle BCE$,\\
$\therefore$DF=$\frac{1}{2}$CE=EG=$\boxed{2\sqrt{3}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $M$ is on side $AD$, $E$ and $F$ are the midpoints of $BM$ and $CM$ respectively. If $AM = 2DM$, then what is the value of $\frac{EF}{DM}$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\
we have BC=AD, \\
and since AM=2DM, \\
it follows that AD=3DM, that is BC=3DM, \\
since E and F are the midpoints of BM and CM respectively, \\
then EF is the midsegment of $\triangle$BCM, \\
therefore EF=$\frac{1}{2}$BC=$\frac{3}{2}$DM, \\
thus $\frac{EF}{DM}=\boxed{\frac{3}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC$ is an obtuse angle, and $AF$, $CE$ are the altitudes of this triangle, with $P$ being the midpoint of $AC$. If $\angle B=35^\circ$, what is the degree measure of $\angle EPF$?

Solution:
\textbf{Solution:} Given that $AF \perp BC$,\\
it follows that $\angle AFB = \angle AFC = 90^\circ$,\\
since $\angle B = 35^\circ$,\\
thus $\angle EAF = \angle B + \angle AFB = 125^\circ$,\\
given $CE \perp BE$,\\
it follows that $\angle AEC = \angle AFC = 90^\circ$,\\
hence $PE = AP = PC = PF$,\\
thus $\angle AEP = \angle EAP$, $\angle AFP = \angle FAP$,\\
therefore $\angle AEP + \angle EAF + \angle AFP = 250^\circ$,\\
hence $\angle EPF = 360^\circ - 250^\circ = \boxed{110^\circ}$,\\


Question: As shown in the figure, in the rhombus ABCD, $AC=8$, $BD=6$. Let $E$ be a moving point on side $CD$. Draw $EF\perp OC$ at point $F$ and $EG\perp OD$ at point $G$, and connect $FG$. What is the minimum value of $FG$?

Solution:
\textbf{Solution:}\\
Since quadrilateral ABCD is a rhombus,\\
it follows that AC$\perp$BD and AD=CD,\\
Given that EF$\perp$OC at point F, and EG$\perp$OD at point G,\\
it follows that quadrilateral OGEF is a rectangle,\\
By connecting OE, then OE=GF,\\
When OE$\perp$DC, the value of GF is minimal,\\
Since BD=6 and AC=8,\\
it follows that OD=3 and OC=4,\\
Therefore, CD= $\sqrt{OC^{2}+OD^{2}}=5$,\\
Since the area of $\triangle$COD is $\frac{1}{2}OC\cdot OD=\frac{1}{2}CD\cdot OE$,\\
it follows that $OE=\frac{OC\cdot OD}{CD}=\frac{4\times 3}{5}=\frac{12}{5}$,\\
Therefore, the minimum value of FG is $\boxed{\frac{12}{5}}$.


Question: As shown in the figure, points $A$ and $B$ are located at the ends of a pond. Xiaocong wants to measure the distance between $A$ and $B$ with a rope, but the rope is not long enough. A classmate helped him with an idea: firstly, select a point $C$ on the ground that can be directly reached from both $A$ and $B$. Find the midpoints $D$ and $E$ of $AC$ and $BC$, respectively, and measure the length of $DE$ to be $13\,m$. Then, what is the distance between $A$ and $B$ in meters?

Solution:
\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AC$ and $BC$, respectively, \\
therefore, $AB=2DE=\boxed{26}m$.


Question: In the rhombus $ABCD$, we have $AB=6$ and $\angle CAB=30^\circ$. If we extend $AB$ and $CD$ to form the rectangle $AECF$, what is the length of the side $CE$ of the rectangle?

Solution: Solution: Since in rhombus ABCD, AB=6, and quadrilateral AECF is a rectangle, it follows that $CB=AB=6$, $\angle E=90^\circ$. Since $\angle CAB=30^\circ$, it follows that $\angle ACB=\angle CAB=30^\circ$, $\angle ACE=60^\circ$, hence $\angle BCE=30^\circ$. Therefore, $BE=\frac{1}{2}CB=3$. In right triangle $BCE$, $CE=\sqrt{BC^{2}-BE^{2}}=\sqrt{6^{2}-3^{2}}=\boxed{3\sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $AC=BC$, a square DEFG is constructed within this triangle, where the side GF is part of side AB, with points D and E respectively on sides AC and BC. When $AC=\frac{3}{2}\sqrt{2}$, what is the length of the side of square DEFG?

Solution: \textbf{Solution:} Given that in right triangle $ABC$, $AC=BC$ and $AC=\frac{3}{2}\sqrt{2}$, it follows that $AB=\sqrt{2}AC=3$ and $\angle A=\angle B=45^\circ$. Since quadrilateral $DEFG$ is a square, it implies $\angle AGD=\angle BFE=90^\circ$ and $GF=GD=EF$. Therefore, triangles $AGD$ and $BFE$ are isosceles right triangles, which means $AG=DG$ and $BF=EF$. Consequently, $AG=FG=FB=1$. Therefore, the side length of square $DEFG$ is $\boxed{1}$.


Question: As shown in the figure, the diagonals $AC$ and $BD$ of rectangle $ABCD$ intersect at point $O$, with $\angle BOC=120^\circ$ and $AC=6$. What is the length of $AB$?

Solution:
\textbf{Solution:} Given: Quadrilateral ABCD is a rectangle,\\
therefore $OA=OC=\frac{1}{2}AC=3$, $OB=OD=\frac{1}{2}BD$, $AC=BD$,\\
therefore $OA=OB=3$,\\
since $\angle BOC=120^\circ$,\\
therefore $\angle AOB=60^\circ$,\\
thus, triangle AOB is an equilateral triangle,\\
therefore $AB=OA=OB=\boxed{3}$.


Question: As shown in the figure, in the rhombus $ABCD$, $\angle A=108^\circ$, the perpendicular bisector of $AD$ intersects diagonal $BD$ at point $P$, with the foot of the perpendicular being $N$. Connecting $CP$, what is the measure of $\angle BPC$ in degrees?

Solution:
\textbf{Solution:} As shown in the diagram, connect AP, \\
in the rhombus ABCD, where $\angle ADC=72^\circ$, \\
$\therefore \angle ADP = \frac{1}{2} \angle ADC = \frac{1}{2} \times 72^\circ = 36^\circ$, \\
$\because$ NP is the perpendicular bisector of AB, \\
$\therefore AP = CP$, \\
$\therefore \angle ADP = \angle DAP = 36^\circ$, \\
$\therefore \angle APB = \angle ADP + \angle DAP = 36^\circ + 36^\circ = 72^\circ$, \\
Due to the symmetry of the rhombus, $\angle CPB = \angle APB = 72^\circ$. \\
Therefore, the answer is: $\boxed{72^\circ}$.


Question: As shown in the figure, two strips of paper, each with a width of 2, are overlapped at an angle of $30^\circ$. What is the area of the quadrilateral $ABCD$ formed by the overlapping part?

Solution:
\textbf{Solution:} Since $AB \parallel CD$, \\
it follows that $\angle ABC = 30^\circ$, \\
Also, since the widths of both strips are 2, \\
it follows that $AB = 4$, \\
Similarly, it can be concluded that $AD = 4$, \\
Again, because $AB \parallel CD$ and $AD \parallel BC$, \\
it follows that quadrilateral $ABCD$ is a parallelogram, \\
Also, since $AB = AD = 4$, \\
it follows that quadrilateral $ABCD$ is a rhombus, \\
Therefore, $S_{\text{rhombus}}ABCD = 4 \times 2 = \boxed{8}$.


Question: As shown in the figure, points E and F are inside the square ABCD, and AE $\perp$ EF, CF $\perp$ EF. Given AE = 9, EF = 5, FC = 3, what is the side length of the square ABCD?

Solution:
\textbf{Solution:} As shown in the figure, connect AC, and draw CG $\perp$ AE extended to intersect at point G. \\
Since AE $\perp$ EF and CF $\perp$ EF, \\
then the quadrilateral EFCG is a rectangle. \\
In $\triangle ACG$, $AG=9+3=12$, $CG=EF=5$, \\
thus, $AC=\sqrt{AG^{2}+GC^{2}}=13$, \\
therefore, $AD=\frac{13\sqrt{2}}{2}$. \\
Hence, the answer is: $AD=\boxed{\frac{13\sqrt{2}}{2}}$.


Question: In rhombus $ABCD$, $AB=AC=10$, diagonals $AC$ and $BD$ intersect at point $O$. Point $M$ is on line segment $AC$, and $AM=3$. Point $P$ is a moving point on line segment $BD$. What is the minimum value of $MP+\frac{1}{2}PB$?

Solution:
\textbf{Solution:} Solve: Draw $PE\perp BC$ at E through P, and draw $MF\perp BC$ at F through M, \\
since quadrilateral ABCD is a rhombus, \\
therefore AB=BC, \\
since AB=AC=10, \\
therefore AB=BC=AB=AD=CD=10, \\
therefore $\triangle ABC$ and $\triangle ACD$ are both equilateral triangles, \\
therefore $\angle ABC=\angle ACB=\angle ADC=60^\circ$, \\
since BD is the diagonal of the rhombus ABCD, \\
therefore BD bisects $\angle ABC$, \\
therefore $\angle CBO= \frac{1}{2}\angle ABC=30^\circ$, \\
therefore PE=$ \frac{1}{2}$BP, \\
therefore PM+$ \frac{1}{2}$BP=PM+PE$\leq$ MF, \\
since point P is on BD, \\
therefore the shortest distance is when points M, P, E are collinear, $MP+\frac{1}{2}PB_{\text{min}}=MF$, \\
since AM=3, \\
therefore CM=AC−AM=10−3=7, \\
in $\triangle MFC$, $\angle MCF=60^\circ$, \\
therefore $\sin\angle MCF= \frac{MF}{MC}$, \\
therefore MF=$MC\sin60^\circ= 7\times \frac{\sqrt{3}}{2}=\boxed{\frac{7\sqrt{3}}{2}}$.


Question: As shown in the figure, in the rectangle $ABCD$, where $AB=6$, $BC=8$, and $AE\perp BD$ at point $E$, what is the length of $AE$?

Solution:
\textbf{Solution:}\\
Since quadrilateral ABCD is a rectangle,\\
thus AO$=\frac{1}{2}$AC.\\
Therefore, S$\triangle$AOB$=\frac{1}{2}$S$\triangle$ABC.\\
Since in right-angled triangle ABC, AB=6, BC=8.\\
Therefore, S$\triangle$ABC$=\frac{1}{2}$AB$\cdot$BC=24\\
Therefore, S$\triangle$AOB=12.\\
Since in right-angled triangle ABC,\\
AC$=\sqrt{AB^{2}+BC^{2}}=\sqrt{6^{2}+8^{2}}=10$. \\
Therefore, BO$=\frac{1}{2}$AC=5.\\
Since AE$\perp$BD.\\
Therefore, S$\triangle$AOB$=\frac{1}{2}$OB$\cdot$AE=12\\
Therefore, AE$=\frac{24}{OB}=\frac{24}{5}=\boxed{4.8}$.


Question: In the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AD$ is the median, $E$ is the midpoint of $AD$, and $F$ is the midpoint of $BE$. Connect $DF$. If $AC=4\sqrt{3}$ and $DF\perp BE$, then what is the length of $DF$?

Solution:
\textbf{Solution:} Connect CE. \\
Since $AD$ is the median of side $BC$, and $F$ is the midpoint of $BE$, \\
therefore $DF$ is the median of $\triangle BCE$, \\
thus $CE = 2DF$, and $DF \parallel CE$, \\
therefore $\angle BDF = \angle DCE$, $\angle EDF = \angle DEC$, \\
also since $DF \perp BE$, \\
therefore $\angle EDF = \angle BDF$, \\
therefore $\angle DEC = \angle DCE$, \\
therefore $CD = ED$, \\
since $E$ is the midpoint of $AD$, and $\angle ACB = 90^\circ$, \\
therefore $CE = ED = CD = \frac{1}{2}AD$, \\
therefore $AD = 4DF$, \\
since $AC = 4\sqrt{3}$, \\
therefore $AD^{2} - CD^{2} = AD^{2} - (\frac{1}{2}AD)^{2} = AC^{2} = 48$, \\
therefore $AD = 8$, \\
therefore $DF = \boxed{2}$.


Question: As shown in the diagram, the quadrilateral $ABCD$ is a rhombus, $\angle DAB=120^\circ$, $M$ is a moving point on side $BC$, and $AM$ intersects side $BD$ at point $N$. When the length of segment $AM$ is minimal, $NM=1$. At this point, what is the distance from point $N$ to line $CD$?

Solution:
\textbf{Solution}: Since quadrilateral ABCD is a rhombus and $\angle DAB=120^\circ$,
it follows that $AD\parallel BC$, and BD bisects $\angle ADC$ and $\angle ABC$.
Therefore, $\angle ABC=60^\circ$,
which means $\angle DBC=30^\circ$.
When AM$\perp BC$ and AM is shortest, then $\angle BAM=30^\circ$.
Given $NM=1$,
we find $BN=2$ and $BM=\sqrt{2^2-1^2}=\sqrt{3}$.
Thus, $AB=2\sqrt{3}$ and $AM=\sqrt{AB^2-BM^2}=3$,
leading to $AN=AM-NM=3-1=2$.
Since $AD\parallel BC$ and AM$\perp BC$,
it follows that AM$\perp AD$.
Because BD bisects $\angle ADC$,
the distance from point N to line CD is equal to its distance to line AD, with NA=2.
Therefore, at this point, the distance from point N to the line CD is $\boxed{2}$.


Question: As shown in the figure, extend the side BC of rectangle ABCD to point E so that $CE=BD$. Connect AE, intersecting the diagonal BD at point F, and intersecting side CD at point G. If $\angle ADB = 38^\circ$, what is the measure of $\angle E$?

Solution:
\textbf{Solution:} As shown in the figure: Connect AC. Let the intersection point of AC and BD be O. \\
Since ABCD is a rectangle, \\
it follows that AC=BD, \\
which means OA=OC=OB=OD, \\
Since $\angle ADB=38^\circ$, \\
it implies $\angle BCA=\angle DBC=38^\circ$, \\
Since CE=BD, \\
it leads to AC=CE, \\
Therefore, $\angle E=\boxed{19^\circ}$.


Question: As shown in the diagram, in rectangle $ABCD$, $AB=2$, $AD=4$, and $E$ is the midpoint of $AD$, $F$ is a moving point on the line segment $EC$, $P$ is the midpoint of $BF$. When the line segment $PD$ is drawn, what is the range of possible values for $PD$?

Solution:
\textbf{Solution:} As shown in the diagram:\\
When point F coincides with point C, point P is at ${P}_{1}$, where $C{P}_{1}=B{P}_{1}$,\\
When point F coincides with point E, point P is at ${P}_{2}$, where $E{P}_{2}=B{P}_{2}$,\\
$\therefore$ ${P}_{1}{P}_{2}\parallel EC$ and ${P}_{1}{P}_{2}=\frac{1}{2}CE$,\\
When point F lies on EC excluding points C and E, we have BP=FP,\\
By the Midline Theorem, it is known that: ${P}_{1}P\parallel CF$ and ${P}_{1}P=\frac{1}{2}CF$,\\
$\therefore$ The trajectory of point P is the line segment ${P}_{1}{P}_{2}$,\\
$\because$ In rectangle ABCD, AB=2, AD=4, E is the midpoint of AD,\\
$\therefore$ $\triangle ABE$, $\triangle BEC$, $\triangle DCP_{1}$ are isosceles right triangles,\\
$\therefore$ $\angle ECB=45^\circ$, $\angle DP_{1}C=45^\circ$,\\
$\because$ ${P}_{1}{P}_{2}\parallel EC$,\\
$\therefore$ $\angle P_{2}P_{1}B=\angle ECB=45^\circ$,\\
$\therefore$ $\angle P_{2}P_{1}D=90^\circ$,\\
$\therefore$ The length of DP is smallest at $D{P}_{1}$ and largest at $D{P}_{2}$,\\
$\because$ $CD=C{P}_{1}=DE=2$,\\
$\therefore$ $D{P}_{1}=2\sqrt{2}$, $CE=2\sqrt{2}$,\\
$\therefore$ ${P}_{1}{P}_{2}=\sqrt{2}$,\\
$\therefore$ $D{P}_{2}=\sqrt{(2\sqrt{2})^2+(\sqrt{2})^2}=\sqrt{10}$,\\
Therefore, the answer is: $\boxed{2\sqrt{2}\le PD\le \sqrt{10}}$.


Question: As shown in the figure, for the rhombus $ABCD$, if $DB = 10$ and $AB = 13$, then what is the area of the rhombus $ABCD$?

Solution:
\textbf{Solution:} As shown in the figure, connect $AC$, intersecting $DB$ at point O, \\
since quadrilateral $ABCD$ is a rhombus and $DB=10$,\\
therefore $AC=2OA$, $OB=\frac{1}{2}DB=5$, $AC\perp BD$,\\
since $AB=13$,\\
therefore $OA=\sqrt{AB^{2}-OB^{2}}=12$,\\
thus $AC=24$,\\
then the area of the rhombus $ABCD$ is $\frac{1}{2}AC\cdot DB=\frac{1}{2}\times 24\times 10=\boxed{120}$.


Question: As shown in the figure, the diagonals AC and BD of the rhombus ABCD intersect at point O. Point E is a point on side AD, and $OD = OE$. If $AC = 8$ and $BD = 6$, then what is the length of AE?

Solution:
\textbf{Solution:} As shown in the figure, draw ON $\perp$ AD at point N,\\
$\because$ Quadrilateral ABCD is a rhombus,\\
$\therefore$ OD=$\frac{1}{2}$BD=3, OA=$\frac{1}{2}$AC=4, AC$\perp$BD,\\
$\therefore$ AD=$\sqrt{OA^{2}+OD^{2}}$=5,\\
$\because$ OD=OE, ON$\perp$AD at point N,\\
$\therefore$ $\angle$OND=90$^\circ$, DN=EN,\\
$\therefore$ S$_{\triangle AOD}$=$\frac{1}{2}$AD·ON=$\frac{1}{2}$OD·OA,\\
$\therefore$ 5ON=12,\\
$\therefore$ ON=$\frac{12}{5}$,\\
$\therefore$ DN=$\sqrt{OD^{2}-ON^{2}}=\sqrt{3^{2}-\left(\frac{12}{5}\right)^{2}}=\frac{9}{5}$,\\
$\therefore$ DE=2DN=$\frac{18}{5}$,\\
$\therefore$ AE=AD-DE=5-$\frac{18}{5}=\boxed{\frac{7}{5}}$.


Question: As shown in the figure, within a square $ABCD$ with a side length of 2, points $E$ and $F$ are located on sides $BC$ and $AD$, respectively. By connecting $EF$ and folding quadrilateral $ABEF$ along $EF$, the corresponding point of $B$ precisely falls on side $CD$, and the corresponding point of $A$ is $H$. After connecting $BH$, what is the minimum value of $BH+EF$?

Solution:
\textbf{Solution:} Construct $FK \perp BC$ at point $K$, extend $BC$ to point $M$ such that $CM = BC$, connect $AM$ and intersect $CD$ at point $N$, connect $MG$, $GA$, $BG$,

$\because$ Quadrilateral $ABCD$ is a square,

$\therefore$ $\angle BAD = \angle ABC = \angle BCD = 90^\circ$, $AB = BC$,

$\therefore$ $CD \perp BM$,

$\therefore$ $CD$ is the perpendicular bisector of $BM$,

$\therefore$ $MG = BG$,

$\because$ Folding quadrilateral $ABEF$ along $EF$, the corresponding point of $B$ on side $CD$ is exactly $G$, the corresponding point of $A$ is $H$,

$\therefore$ $AB = HG$, $BE = EG$, $\angle ABE = \angle HGE$, $EF \perp BG$,

$\therefore$ $\angle GBE = \angle BGE$,

$\therefore$ $\angle ABG = \angle HGB$,

In $\triangle ABG$ and $\triangle HGB$

$\left\{\begin{array}{l} BG = GB\\ \angle ABG = \angle HGB\\ AB = HG \end{array}\right.$

$\therefore$ $\triangle ABG \cong \triangle HGB$ (SAS),

$\therefore$ $GA = BH$,

$\because$ $EF \perp BG$,

$\because$ $FK \perp BC$,

$\therefore$ $\angle FKE = \angle BCG = 90^\circ$,

$\therefore$ $\angle EFK + \angle FEK = \angle GBC + \angle FEK = 90^\circ$,

$\therefore$ $\angle EFK = \angle GBC$,

$\because$ $\angle BAD = \angle ABC = \angle BKF = 90^\circ$,

$\therefore$ Quadrilateral $ABKF$ is a rectangle,

$\therefore$ $AB = FK$,

$\therefore$ $FK = BC$,

In $\triangle FEK$ and $\triangle BGC$

$\left\{\begin{array}{l} \angle FKE = \angle BCG\\ FK = BC\\ \angle EFK = \angle GBC \end{array}\right.$

$\therefore$ $\triangle FEK \cong \triangle BGC$ (ASA),

$\therefore$ $EF = BG$,

$\therefore$ $EF = MG$,

$\therefore$ $BH + EF = AG + MG$,

$\because$ $AG + MG \geq AM$,

$\therefore$ $BH + EF \geq AM$,

$\therefore$ When point $G$ coincides with point $N$, $AG + MA = AM$, at this time $AG + MA$ has the minimum value,

$\therefore$ The value of $BH + EF = AM$ is also the minimum,

$\because$ $\angle ABM = 90^\circ$, $AB = 2$, $BM = 2BC = 4$,

$\therefore$ $AM = \sqrt{AB^{2} + BM^{2}} = \sqrt{2^{2} + 4^{2}} = \boxed{2\sqrt{5}}$,

$\therefore$ The minimum value of $BH + EF$ is $\boxed{2\sqrt{5}}$.