Question: As shown in the figure, within the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, $E$ is the midpoint of side $AD$, and $AC=6$, $BD=8$, then what is the length of the line segment $OE$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus with AC=6 and BD=8,\\
it follows that OA=OC=3 and OB=OD=4, with AO$\perp$BO,\\
In the right triangle AOD, by the Pythagorean theorem, we get: AD=$\sqrt{OA^{2}+OD^{2}}$=5,\\
Since E is the midpoint of AD,\\
it follows that OE=$\frac{1}{2}$AD=$\boxed{2.5}$


Question: As shown in the figure, in the hexagon $ABCDEF$, each interior angle is equal. If $\angle 1=58^\circ$, what is the degree measure of $\angle 2$?

Solution:
\textbf{Solution:} Since all internal angles of the hexagon ABCDEF are equal, \\
$\therefore \angle B = \angle C = \angle CDE = (6-2) \times 180^\circ \div 6 = 120^\circ$, \\
$\therefore \angle$CDA $= 360^\circ - \angle1 - \angle B - \angle C = 62^\circ$, \\
$\therefore \angle2 = \angle CDE - \angle CDA = \boxed{58^\circ}$.


Question: As shown in the diagram, in rhombus ABCD, AC intersects BD at O, and DE $\perp$ BC at E. Connect OE. If $\angle BAD=40^\circ$, then what is the degree measure of $\angle OED$?

Solution:
\textbf{Solution:}
Given that quadrilateral ABCD is a rhombus, and $\angle BAD=40^\circ$,\\
it follows that AD$\parallel$BC\\
Hence, $\angle BAD+\angle ABC=180^\circ$\\
Therefore, $\angle ABC=180^\circ-40^\circ=140^\circ$\\
Thus, $\angle ABD=\angle CBD=\frac{1}{2}\angle ABC=70^\circ$, and BO=DO,\\
Since DE$\perp$BC,\\
it follows that OE=OD=OB, and $\angle BDE=20^\circ$,\\
Therefore, $\angle ODE=\angle OED=\boxed{20^\circ}$


Question: As shown in the figure, the two diagonals $AC, BD$ of the square $ABCD$ intersect at point $O$, and point $E$ is on $BD$, with $BE=BC$. What is the degree measure of $\angle ACE$?

Solution:
\textbf{Solution:} In square $ABCD$, $\angle CBD=\angle ACB=45^\circ$,\\
$\because BE=BC$,\\
$\therefore \angle BCE=\frac{1}{2}\left(180^\circ-\angle CBD\right)=67.5^\circ$,\\
$\therefore \angle ACE=\angle BCE-\angle ACB=\boxed{22.5^\circ}$.


Question: As shown in the figure, in rectangle ABCO, point D is on side BC. Connect AD and fold $\triangle ABD$ along AD, making point B exactly coincide with point E on side OC. Given B $(10, 8)$ and that the inverse proportion function $y= \frac{k}{x}$ ($k\neq 0$) passes through point D, what is the value of $k$?

Solution:
\textbf{Solution:}

Given a rectangle ABCO with point B at (10, 8),

\(\therefore \) AB=OC=10, BC=OA=8,

\(\because \) triangle ABD is folded along AD such that point B exactly coincides with point E on side OC,

\(\therefore \) AE=AB=10, DE=BD,

\(\therefore \) OE\(=\sqrt{AE^{2}-OA^{2}}=\sqrt{10^{2}-8^{2}}=6\),

\(\therefore \) EC=OC-OE=10-6=4,

Let CD=\(b\), and BD=DE=8-\(b\),

\(\because \) \(CD^{2}+CE^{2}=DE^{2}\),

\(\therefore \) \(b^{2}+4^{2}=(8-b)^{2}\),

Solving for \(b\), we find \(b=3\),

\(\therefore \) The coordinates of point D are (10, 3),

\(\because \) the graph of the inverse proportion function passes through point D,

\(\therefore \) \(k=10\times 3=\boxed{30}\),


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

Solution:
\textbf{Solution:} Given rectangle ABCD,\\
thus OD=OB=OA=$\frac{1}{2}$BD, $\angle$ABC=$90^\circ$,\\
since points E and F are the midpoints of AO and AD respectively,\\
thus EF is the midline of $\triangle$AOD,\\
thus EF=$\frac{1}{2}$OD=4,\\
thus OD=8,\\
thus AB=OB=OA,\\
thus $\triangle$AOB is an equilateral triangle\\
thus $\angle$BAC=$60^\circ$,\\
thus $\angle$ACB=$90^\circ$−$\angle$BAC=$90^\circ$−$60^\circ$=$\boxed{30^\circ}$.


Question: As shown in the figure, this is a beautiful Pythagorean tree, in which all the quadrilaterals are squares, and all the triangles are right-angled triangles. If the side lengths of squares $A$, $B$, $C$, and $D$ are $3$, $6$, $2$, and $3$ respectively, what is the area of the largest square $E$?

Solution:
\textbf{Solution:} As shown in the figure,\\
by the Pythagorean theorem, the area of square $F$ is equal to the area of square $A$ plus the area of square $B$ = $3^{2}+6^{2}=45$,\\
Similarly, the area of square $G$ is equal to the area of square $C$ plus the area of square $D$ = $2^{2}+3^{2}=13$,\\
$\therefore$ the area of square $E$ is equal to the area of square $F$ plus the area of square $G$ = $45+13=\boxed{58}$.


Question: As shown in the figure, in rectangle $ABCD$, where $AB=4$ and $BC=5$, point $E$ is a point on $AB$. The rectangle is folded along $DE$ so that side $BC$ exactly passes through point $A$. What is the length of $BE$?

Solution:
\textbf{Solution:} In rectangle $ABCD$, fold the rectangle along $DE$ such that side $BC$ exactly passes through point $A$,\\
$\therefore \angle C'=\angle C=90^\circ$, $C'D=CD=AB=4$, $AD=BC=5$, $\angle B=\angle B'=90^\circ$,\\
In $\triangle AC'D$, we have $AC'=\sqrt{AD^2-C'D^2}=\sqrt{5^2-4^2}=3$,\\
Let $BE=x$, then $AE=AB-BE=4-x$, $AB'=B'C'-AC'=5-3=2$,\\
In $\triangle AB'E$, we have $AE^2=AB'^2+B'E^2$,\\
which gives $(4-x)^2=2^2+x^2$,\\
solving for $x$ yields $x=\boxed{\frac{3}{2}}$.


Question: As shown in the figure, quadrilateral ABCD is a rhombus. Extend BC to E so that BD = BE. Connect DE. If $\angle ABC = 80^\circ$, what is the degree measure of $\angle E$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\
it follows that $BD$ bisects $\angle ABC$,\\
and since $\angle ABC=80^\circ$,\\
it follows that $\angle DBC=\frac{1}{2}\angle ABC=40^\circ$,\\
and since $BD=BE$,\\
it follows that $\angle BDE=\angle BED=\frac{1}{2}\left(180^\circ-40^\circ\right)=\boxed{70^\circ}$.


Question: As shown in the figure, within the rhombus ABCD, point E is a point on side AB, and $DE=AD$. Connect EC. If $\angle ADE = 40^\circ$, what is the degree measure of $\angle BCE$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\
thus $AD=CD$, $\angle A=\angle BCD$, $CD\parallel AB$,\\
since $DE=AD$, $\angle ADE=40^\circ$,\\
thus $\angle DAE=\angle DEA=70^\circ$,\\
since $CD\parallel AB$,\\
thus $\angle CDE=\angle DEA=70^\circ$, and $DE=DC=DA$,\\
thus $\angle DCE=55^\circ$,\\
since $\angle DCB=\angle DAE=70^\circ$,\\
thus $\angle BCE=\angle DCB-\angle DCE=\boxed{15^\circ}$.


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

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\
it follows that OD = OB, and AB = BC, \\
since DE = EC, \\
it follows that OE is the median of $\triangle BDC$, \\
hence, AB = BC = 2OE = \boxed{8}.


Question: As shown in the figure, the perimeter of the rhombus flower bed ABCD is 80m, with $\angle ABC=120^\circ$. If two paths AC and BD are built along the diagonals of the rhombus, what is the length of path AC?

Solution:
\textbf{Solution:} As shown in the figure, let the diagonals AC and BD intersect at point O, \\
since quadrilateral ABCD is a rhombus, and $\angle ABC=120^\circ$, \\
therefore $\angle ADB=\angle CDB=60^\circ$, and AC$\perp$BD, \\
thus $\triangle AOD$ is a right triangle, with $\angle DAO=30^\circ$, \\
since the perimeter of the rhombus is 80, \\
therefore AD=80÷4=20, \\
hence OD=10, \\
using the Pythagorean theorem, we get: $OA=\sqrt{AD^{2}-OD^{2}}=10\sqrt{3}$, \\
based on the properties of a rhombus, we can derive: AC=2OA=$\boxed{20\sqrt{3}}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is a rhombus, with diagonals $AC$ and $BD$ intersecting at point $O$. Let $DH \perp AB$ at $H$ and connect $OH$. If $\angle DHO = \alpha$, what is the measure of $\angle DAB$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\
it follows that $OB=OD$, $AC\perp BD$, $AB=AD$,\\
hence $\angle ABD=\angle ADB$,\\
since $DH\perp AB$,\\
then $OH=OB=\frac{1}{2}BD$,\\
since $\angle DHO=\alpha $,\\
thus $\angle OHB=90^\circ-\angle DHO=90^\circ-\alpha $,\\
therefore $\angle ABD=\angle OHB=\angle ADB=90^\circ-\alpha $,\\
hence $\angle DAB=180^\circ-\angle ABD-\angle ADB=180^\circ-90^\circ+\alpha -90^\circ+\alpha =\boxed{2\alpha}$.


Question: As shown in the figure, the diagonals $AC$ and $BD$ of $ABCD$ intersect at point $O$. $E$ is the midpoint of $AD$. Connect $OE$. Given $AC=8$, $BC=10$, if $AC \perp CD$, then what is the length of $OE$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a parallelogram,\\
it follows that $AO=OC$, and $AD=BC=10$,\\
since $AC\perp CD$,\\
it follows that $\angle ACD=90^\circ$,\\
therefore $CD=\sqrt{AD^{2}-AC^{2}}=\sqrt{10^{2}-8^{2}}=6$,\\
since E is the midpoint of AD,\\
it follows that OE is the median of $\triangle ACD$,\\
thus $OE=\frac{1}{2}CD=\boxed{3}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $\angle C=110^\circ$, and the exterior angles adjacent to $\angle BAD$ and $\angle ABC$ are both $120^\circ$, then what is the value of $\angle \alpha$?

Solution:
\textbf{Solution:} Since in quadrilateral ABCD, $\angle C=110^\circ$,\\
the external angle adjacent to $\angle C$ is: $180^\circ-110^\circ=70^\circ$,\\
thus $\angle \alpha =360^\circ-70^\circ-120^\circ-120^\circ=\boxed{50^\circ}$.


Question: As shown in the figure, square $ABCD$ and square $CEFG$ are given, where point $D$ is on $CG$, $BC=2$, $H$ is the midpoint of $AF$, and $CH=4$. What is the length of $CE$?

Solution:
\textbf{Solution:} Connect $AC, CF$,

Since both quadrilateral $ABCD$ and quadrilateral $GCEF$ are squares,

$AC = \sqrt{AB^{2}+BC^{2}} = \sqrt{2^{2}+2^{2}} = 2\sqrt{2}$,

$\angle ACD=\angle FCD=45^\circ$,

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

Also, since point $H$ is the midpoint of $AF$,

$AF=2CH=8$．

Therefore, in $\triangle ACF$, $CF =\sqrt{AF^{2}-AC^{2}}=\sqrt{8^{2}-\left(2\sqrt{2}\right)^{2}}=2\sqrt{14}$,

Therefore, in $\triangle CEF$, $CE=EF$,

$CE^{2}+EF^{2}=CF^{2}$,

$2CE^{2}=\left(2\sqrt{14}\right)^{2}$,

therefore, $CE=\boxed{2\sqrt{7}}$.


Question: As shown in the figure, in $\triangle ABC$, $D$, $E$, and $F$ are the midpoints of the sides $AB$, $BC$, and $AC$ respectively. If the perimeter of $\triangle ABC$ is 6, what is the perimeter of $\triangle DEF$?

Solution:
\textbf{Solution:} Given $\because$ in $\triangle ABC$, D, E, F are the midpoints of the sides $AB$, $BC$, and $AC$ respectively,\\
$\therefore$DE= $\frac{1}{2}$ AC, EF= $\frac{1}{2}$ AB, DF= $\frac{1}{2}$ BC,\\
$\because$ the perimeter of $\triangle ABC$ is AB+BC+AC=6,\\
$\therefore$ the perimeter of $\triangle DEF$ is EF+DF+DE= $\frac{1}{2}$ （AB+BC+AC）= $\frac{1}{2} \times 6=\boxed{3}$


Question: As shown in the figure, in $\triangle ABC$, D, E, and F are the midpoints of BC, AC, and AB, respectively. $AH \perp BC$ at H, and $HE=8$. What is the length of DF?

Solution:
\textbf{Solution:} Given that $AH \perp BC$ and E is the midpoint of side AC, \\
$\therefore$ AC = 2HE = 16, \\
Given that D and F are the midpoints of sides BC and AB respectively, \\
$\therefore$ DF = $\frac{1}{2}$ AC = \boxed{8}


Question: As shown in the figure, the side length of square $ABCD$ is $4$, and $E$, $F$, $G$ are moving points on sides $AB$, $BC$, $AD$, respectively, with $AE=BF$. Fold $\triangle BEF$ inward along $EF$ to $\triangle B^{\prime}EF$, and connect $BB^{\prime}$, $B^{\prime}G$, and $GC$. What is the minimum value of $B^{\prime}G+GC$ when $BB^{\prime}$ is at its maximum?

Solution:
\textbf{Solution:} As shown in the figure, when the quadrilateral B$^{\prime}$EBF is a square, BB$^{\prime}$ is maximized. \\
$\therefore$ BE = BF, \\
$\because$ AE = BF, \\
$\therefore$ AE = BE, \\
$\therefore$ E, F are respectively the midpoints of sides AB, BC, \\
Draw a line through B$^{\prime}$ perpendicular to CD at point H, \\
Then AE = BE = BF = B$^{\prime}$H = CH = $\frac{1}{2}$ BC = 2, \\
Construct the symmetric point C$^{\prime}$ of C with respect to AD, and connect B$^{\prime}$C$^{\prime}$, GC$^{\prime}$, \\
$\therefore$ CG = GC$^{\prime}$, \\
$\therefore$ B$^{\prime}$G + GC = B$^{\prime}$G + GC$^{\prime}$ $\ge$ B$^{\prime}$C$^{\prime}$, that is, the minimum value of B$^{\prime}$G + GC is B$^{\prime}$C$^{\prime}$, \\
In the right triangle $\triangle$B$^{\prime}$C$^{\prime}$H, B$^{\prime}$H = 2, HC$^{\prime}$ = CC$^{\prime}$ − CH = 8 − 2 = 6, \\
By the Pythagorean theorem, B$^{\prime}$C$^{\prime}$ = $\sqrt{2^{2}+6^{2}} = 2\sqrt{10}$, \\
$\therefore$ The minimum value of B$^{\prime}$G + GC is $\boxed{2\sqrt{10}}$.


Question: As shown in the diagram, in the Cartesian coordinate system, point O is the origin, point A $(-3,0)$ and point B $(0,2)$ are both on the coordinate axes. If the graph of the inverse proportion function $y= \frac{k}{x}$ passes through the center of symmetry of rectangle AOBC, what is the value of $k$?

Solution:
\[
\text{Solution:} \quad \because \text{Point A} (-3, 0) \text{ and Point B} (0, 2) \text{ are both on the coordinate axes}, \\
\therefore \text{the center of rectangle AOBC is} \left( -\frac{3}{2} ,1\right), \\
\because \text{the graph of the inverse proportion function } y= \frac{k}{x} \text{ passes through the center of symmetry of rectangle AOBC}, \\
\therefore k= -\frac{3}{2} \times 1 = \boxed{-\frac{3}{2}}
\]


Question: As shown in the figure, quadrilateral ABCD is a rhombus. Points E and F are located on sides BC and CD, respectively, with BE = DF and AB = AE. If $\angle EAF=75^\circ$, what is the degree measure of $\angle C$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rhombus, \\
it follows that AB = AD, $\angle B$ = $\angle D$, and $\angle C$ = $\angle BAD$. \\
In $\triangle ABE$ and $\triangle ADF$, \\
since $\left\{\begin{array}{l} AB = AD \\ \angle B = \angle D \\ BE = DF \end{array}\right.$, \\
it implies that $\triangle ABE \cong \triangle ADF$ (SAS), \\
thus, $\angle DAF = \angle BAE$. \\
Let $\angle BAE = \angle DAF = x$, \\
then $\angle DAE = 75^\circ + x$, \\
and since AD $\parallel$ BC, \\
$\angle AEB = 75^\circ + x$. \\
Given that AB = AE, \\
$\angle B = \angle AEB = 75^\circ + x$. \\
Since $\angle BAE + \angle ABE + \angle AEB = 180^\circ$, \\
$x + 75^\circ + x + 75^\circ + x = 180^\circ$, \\
thus $x = 10^\circ$, \\
therefore $\angle BAD = 95^\circ$, \\
which means $\angle C = \boxed{95^\circ}$.


Question: As shown in the figure, in the square ABCD, points E and F are located on BC and CD, respectively. AE intersects BF at point G, and AC intersects BF at point H. If CE = DF, BG = GH, and AB = 2, what is the area of $\triangle CFH$?

Solution:
\textbf{Solution:} As shown in the figure, draw $FM\perp CH$ at point $M$. \\
$\because$ Quadrilateral $ABCD$ is a square,\\
$\therefore \angle ABC=\angle BCD=\angle BAD=90^\circ$, $AB=BC=2$, $\angle BAC=\frac{1}{2}\angle BAD$, $\angle ACD=\frac{1}{2}\angle BCD$.\\
$\therefore AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{2^{2}+2^{2}}=2\sqrt{2}$, $\angle BAC=45^\circ$, $\angle ACD=45^\circ$.\\
Moreover, $\because CE=DF$,\\
$\therefore BC-CE=CD-DF$.\\
$\therefore BE=CF$.\\
In $\triangle ABE$ and $\triangle BCF$,\\
$\begin{cases} AB=BC, \\ \angle ABE=\angle BCF, \\ BE=CF. \end{cases}$\\
$\therefore \triangle ABE\cong \triangle BCF$ (SAS).\\
$\therefore \angle 1=\angle 2$.\\
Moreover, $\because \angle ABE=\angle ABG+\angle 2=90^\circ$,\\
$\therefore \angle ABG+\angle 1=90^\circ$.\\
$\therefore \angle AGB=180^\circ-(\angle ABG+\angle 1)=90^\circ$.\\
Moreover, $\because BG=GH$,\\
$\therefore AH=AB=2$.\\
$\therefore CH=AC-AH=2\sqrt{2}-2$.\\
In $\triangle ABG$ and $\triangle AHG$,\\
$\begin{cases} AB=AH, \\ AG=AG, \\ BG=HG. \end{cases}$\\
$\therefore \triangle ABG\cong \triangle AHG$ (SSS).\\
$\therefore \angle 1=\angle HAG$.\\
$\therefore \angle BAC=\angle 1+\angle HAG=2\angle 1=45^\circ$.\\
$\therefore \angle 1=22.5^\circ$.\\
$\therefore \angle 2=\angle 1=22.5^\circ$.\\
$\therefore \angle BFC=180^\circ-\angle 2-\angle BCF=180^\circ-22.5^\circ-90^\circ=67.5^\circ$.\\
$\therefore \angle CHF=180^\circ-\angle HCF-\angle HFC=180^\circ-45^\circ-67.5^\circ=67.5^\circ$.\\
$\therefore \angle CHF=\angle CFH$.\\
$\therefore CH=CF=2\sqrt{2}-2$.\\
$\because MF\perp HC$, $\angle ACD=45^\circ$,\\
$\therefore \angle MFC=180^\circ-\angle FMC-\angle MCF=45^\circ$\\
$\therefore FM=MC$.\\
In $\triangle FMC$, $FC^{2}=MC^{2}+MF^{2}$.\\
$\therefore {(2\sqrt{2}-2)}^{2}=2MF^{2}$.\\
$\therefore MF=2-\sqrt{2}$.\\
$\therefore S_{\triangle FHC}=\frac{1}{2}MF\cdot HC=\frac{1}{2}(2\sqrt{2}-2)(\sqrt{2}-2)=\boxed{3\sqrt{2}-4}$.


Question: As shown in the figure, take the hypotenuse $BC$ of the right triangle $\triangle ABC$ as a side to construct a square $BCEF$ on the same side of $\triangle ABC$. Let the center of the square be $O$, and connect $AO$. If $AB=2$ and $AO=3\sqrt{2}$, what is the length of $AC$?

Solution:
\textbf{Solution:} As shown in the diagram, draw line OD $\perp$ OA, intersecting AC at D, \\
$\because$ Quadrilateral BCEF is a square, \\
$\therefore$ $\angle$BOC=90$^\circ$, OB=OC, \\
$\therefore$ $\angle$AOB+$\angle$BOD=90$^\circ$, $\angle$DOC+$\angle$BOD=90$^\circ$, \\
$\therefore$ $\angle$AOB=$\angle$DOC, \\
$\because$ $\angle$ODC=$\angle$OAC+$\angle$AOD, $\angle$OAB=$\angle$OAC+$\angle$BAC, $\angle$AOD=$\angle$BAC=90$^\circ$, \\
$\therefore$ $\angle$OAB=$\angle$ODC, \\
In $\triangle$ABO and $\triangle$DCO, $ \{\begin{array}{l}\angle OAB=\angle ODC\hfill \\ \angle AOB=\angle DOC\hfill \\ OB=OC\hfill \end{array}$, \\
$\therefore$ $\triangle$ABO$\cong$$\triangle$DCO, \\
$\therefore$ OA=OD= $3\sqrt{2}$, CD=AB=2, \\
$\therefore$ $\triangle$OAD is an isosceles right triangle, \\
$\therefore$ AD= $ \sqrt{2}OA$ =6, \\
$\therefore$ AC=AD+CD=\boxed{8}.


Question: As shown in the figure, in the Cartesian coordinate system, A(8, 0), point B is a moving point on the graph of the linear function $y=x$. Taking OB as a side, a square OBCD is constructed. When AB is at its minimum length, point D just falls on the graph of the inverse proportion function $y=\frac{k}{x}$. Then, $k=$ (\quad)

Solution:
\textbf{Solution:} According to the principle that a perpendicular line segment is the shortest, when AB is perpendicular to the line $y=x$, AB is at its shortest,\\
$\because$ $\angle AOB=45^\circ$\\
$\therefore$ $\angle BAO=45^\circ$\\
$\therefore$ $\triangle AOB$ is an isosceles right triangle,\\
$\because$ the coordinates of point A are (8, 0)\\
$\therefore$ OA=8\\
$\therefore$ $BO=BA=4\sqrt{2}$\\
$\because$ quadrilateral OBCD is a square,\\
$\therefore$ $DO=BO=4\sqrt{2}$, $\angle DOB=90^\circ$\\
$\therefore$ $\angle DOC=\angle BOC=45^\circ$\\
Draw DE$\perp$x-axis through point D at point E,\\
$\therefore$ $\angle ODE=\angle DOE=45^\circ$\\
$\therefore$ $\triangle DEO$ is an isosceles right triangle,\\
$\therefore$ $DE=OE=\frac{DO}{\sqrt{2}}=\frac{4\sqrt{2}}{\sqrt{2}}=4$\\
$\because$ point D is in the second quadrant,\\
$\therefore$ D(−4, 4)\\
Also, point D is on the graph of the reciprocal function $y=\frac{k}{x}$\\
$\therefore$ $k=(−4)\times 4=\boxed{-16}$


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the vertex $O$ of the rhombus $ABOC$ is at the origin, the side $BO$ is on the negative half axis of the $x$ axis, and the coordinates of vertex $C$ are $(-3, 4)$. The graph of the inverse proportion function $y = \frac{k}{x}$ intersects the diagonal $AO$ of the rhombus at point $D$. Line $BD$ is drawn. When $BD \perp x$ axis, what is the value of $k$?

Solution:
\textbf{Solution:} Given that $C(-3,4)$,\\
hence $OC= \sqrt{3^{2}+4^{2}}=5$,\\
since the quadrilateral OBAC is a rhombus,\\
it follows that $AC=OB=OC=5$, and $AC\parallel OB$,\\
thus $B(-5,0)$, $A(-8,4)$,\\
let the equation of line OA be $y=mx$,\\
substituting $A(-8,4)$ into it gives $-8m=4$, leading to $m= -\frac{1}{2}$,\\
hence, the equation of line OA is $y= -\frac{1}{2}x$,\\
for $x=-5$, we have $y= -\frac{1}{2}x = \frac{5}{2}$, therefore $D(-5,\frac{5}{2})$,\\
substituting $D(-5,\frac{5}{2})$ into $y= \frac{k}{x}$,\\
thus $k= -\frac{5\times 5}{2} = -\frac{25}{2}$,\\
so, the answer is: $k=\boxed{-\frac{25}{2}}$.


Question: As shown in the figure, the length of line segment $AB$ is $8$, point $D$ is on $AB$, $\triangle ACD$ is an equilateral triangle with a side length of $3$. A ray $DP$ is drawn perpendicular to $CD$ through point $D$, and a rectangle $CDGH$ is constructed with a moving point $G$ (not coinciding with $D$) on $DP$. Let the intersection point of the diagonals of rectangle $CDGH$ be $O$. Connect $OB$. What is the minimum value of the length of segment $BO$?

Solution:
\textbf{Solution:} Draw line AO. Since quadrilateral CDGH is a rectangle, we have CG=DH, OC= $\frac{1}{2}$ CG, and OD= $\frac{1}{2}$ DH, which means OC=OD. Since $\triangle$ACD is an equilateral triangle, we have AC=AD and $\angle$CAD=$60^\circ$. In triangles $\triangle$ACO and $\triangle$ADO,
\[\left\{\begin{array}{l}
AC=AD\\
AO=AO\\
CO=DO
\end{array}\right.\]
, therefore, $\triangle$ACO$\cong$$\triangle$ADO (SSS), thus $\angle$OAB=$\angle$CAO=$30^\circ$. This means point O must move along the bisector of $\angle$CAB. Therefore, when OB$\perp$AO, the length of OB is minimized. Since $\angle$OAB=$30^\circ$ and $\angle$AOB=$90^\circ$, we get OB= $\frac{1}{2}$ AB= $\frac{1}{2}$ ×8=4, thus the minimum value of OB is $\boxed{4}$.


Question: As shown in the figure, in $\triangle ABC$, points $D$, $E$, and $F$ are the midpoints of sides $AB$, $BC$, and $CA$, respectively, and $AB=8\,cm$, $AC=10\,cm$. What is the perimeter of the quadrilateral $ADEF$ in $cm$?

Solution:
\textbf{Solution:} Let's consider triangle $\triangle ABC$, where points D, E, and F are the midpoints of sides AB, BC, and CA respectively, and we have AB = 8cm, AC = 10cm,\\
Hence, AD = 4cm, AF = 5cm,\\
Since EF and DE are the midlines of $\triangle ABC$,\\
it follows that $DE \parallel \frac{1}{2}AC$ and $EF \parallel \frac{1}{2}AB$\\
Also, since AB = 8cm and AC = 10cm,\\
it results in EF = 4cm and DE = 5cm,\\
Therefore, the perimeter of the quadrilateral $ADEF$ equals $AD + DE + EF + AF = \boxed{18}$cm.


Question: As shown in the figure, let's consider a square $ABCD$ with a side length of $2$, where points $E$ and $F$ are located on sides $AD$ and $BC$, respectively. Points $G$ and $H$ are located on the sides of the square $ABCD$. Connect $GH$. If the angle between line segments $GH$ and $EF$ is $45^\circ$ and $GH=\sqrt{5}$, what is the length of $EF$?

Solution:
\textbf{Solution:} As shown in the figure, draw BK $\parallel$ EF intersecting AD at K, and BM $\parallel$ GH intersecting CD at M,\\
then BK $=$ EF, BM $=$ GH $=$ $\sqrt{5}$,\\
since the angle between line segments GH and EF is $45^\circ$,\\
$\therefore \angle KBM=45^\circ$,\\
$\therefore \angle ABK+\angle CBM=90^\circ-45^\circ=45^\circ$,\\
draw $\angle KBN=45^\circ$ intersecting the extension of AD at N,\\
then $\angle ABN+\angle ABK=45^\circ$,\\
$\therefore \angle ABN=\angle CBM$,\\
in $\triangle ABN$ and $\triangle CBM$,\\
$\left\{\begin{array}{l}
\angle ABN=\angle CBM \\
AB=BC \\
\angle BAN=\angle C=90^\circ
\end{array}\right.$,\\
$\therefore \triangle ABN \cong \triangle CBM$ (ASA),\\
$\therefore BN=BM, AN=CM$,\\
in $\triangle BCM$, CM $=$ $\sqrt{BM^2-BC^2}=\sqrt{5-4}=1$,\\
draw KP$\perp$BN at point K through point P,\\
since $\angle KBN=45^\circ$,\\
$\therefore \triangle BKP$ is an isosceles right-angled triangle,\\
let EF $=$ BK $=$ x, then BP $=$ KP $=$ $\frac{\sqrt{2}}{2}x$,\\
$\therefore NP=\sqrt{5}-\frac{\sqrt{2}}{2}x$, $AK=\sqrt{x^2-4}$,\\
since $\frac{1}{2}BN \cdot KP=\frac{1}{2}KN \cdot AB$,\\
$\therefore \sqrt{5} \times \frac{\sqrt{2}}{2}x=2(1+\sqrt{x^2-4})$,\\
$\therefore 3x^2+4\sqrt{10}x-40=0$,\\
$\therefore (3x-2\sqrt{10})(x+2\sqrt{10})=0$,\\
solving, we get $x=\frac{2\sqrt{10}}{3}$, $x=-2\sqrt{10}$,\\
upon verification: $x=\frac{2\sqrt{10}}{3}$ is a root of the original equation, $x=-2\sqrt{10}$ is an extraneous root, discard it,\\
thus, EF $=$ $\boxed{\frac{2\sqrt{10}}{3}}$.


Question: As shown in the figure, in rectangle $ABCD$, where $E$ is the midpoint of $AD$, $\triangle ABE$ is folded along line $BE$ to obtain $\triangle GBE$. Extend $BG$ to intersect $CD$ at point $F$. If $AB = 3$ and $BC = 2\sqrt{6}$, what is the length of $FD$?

Solution:
\textbf{Solution:}

Given that $E$ is the midpoint of $AD$, it follows that $AE=DE$.

Since $\triangle ABE$ is folded along $BE$ to obtain $\triangle GBE$, we have $AE=EG$ and $AB=BG$.

Thus, $ED=EG$.

Considering rectangle $ABCD$ implies that $\angle A=\angle D=90^\circ$, leading to $\angle EGF=90^\circ$.

By connecting $EF$ and considering right triangles $EDF$ and $EGF$ with $ED=EG$ and $EF=EF$, it is concluded that right triangle $EDF$ is congruent to right triangle $EGF$ by HL congruence.

Hence, $DF=FG$.

Let $DF=x$, then $BF=3+x$ and $CF=3-x$.

In right triangle $BCF$, using the Pythagorean theorem $BC^{2}+CF^{2}=BF^{2}$ translates to $(2\sqrt{6})^{2}+(3-x)^{2}=(3+x)^{2}$.

Solving this equation gives $x=\boxed{2}$, which means $DF=2$.


Question: As shown in the figure, it is known that the quadrilateral $ABCD$ is a rectangle, point $M$ is on $BC$, $BM=CD$, point $N$ is on $CD$, and $DN=CM$, $DM$ intersects with $BN$ at point $P$, then what is the ratio of $DM\colon BN$?

Solution:
\textbf{Solution:} Let $BM=CD=a$, $DN=CM=b$,\\
$\therefore BC=a+b$, $NC=a-b$,\\
$\because$ Quadrilateral $ABCD$ is a rectangle,\\
$\therefore \angle DCB=90^\circ$,\\
In right triangles $\triangle DCM$ and $\triangle BCN$, by the Pythagorean theorem, we have\\
$DM=\sqrt{DC^2+MC^2}=\sqrt{a^2+b^2}$,\\
$BN=\sqrt{BC^2+CN^2}=\sqrt{(a+b)^2+(a-b)^2}=\sqrt{2}\cdot\sqrt{a^2+b^2}$,\\
$\therefore DM\colon BN=1\colon \sqrt{2}$,\\
Therefore, the answer is: $DM\colon BN=\boxed{1\colon \sqrt{2}}$.


Question: As illustrated, with point C of the regular pentagon ABCDE fixed, and rotating in the clockwise direction, to have the vertex \(D^{\prime}\) of the new pentagon \(A^{\prime}B^{\prime}CD^{\prime}E^{\prime}\) fall onto the line BC, what is the angle of rotation?

Solution:
\textbf{Solution:} Since the external angle of a regular pentagon is $360^\circ \div 5 = 72^\circ$,\\
by fixing point C of the regular pentagon ABCDE and rotating it clockwise, a rotation of $72^\circ$ will result in the vertices of the new pentagon A$^{\prime}$B$^{\prime}$C$^{\prime}$D$^{\prime}$E$^{\prime}$ with vertex D$^{\prime}$ falling on the line BC.\\
Therefore, the answer is: \boxed{72^\circ}.


Question: As shown in the figure, in rectangle $ABCD$, $E$ and $F$ are two points on $CD$, $EG \perp AC$, and $FH \perp AC$, with the feet of the perpendiculars being $G$ and $H$ respectively. If $AD=2$, $DE=1$, $CF=2$, and $AG=CH$, what is the sum of $EG+FH$?

Solution:
\textbf{Solution:} Draw line $EM \perp AB$ at $M$, and extend $EG$ to intersect $AB$ at $Q$, then $\triangle EQM$ is a right-angled triangle.\\
Since $EG \perp AC$ and $FH \perp AC$,\\
$\therefore \angle CHF=\angle AGQ=90^\circ$,\\
Since in rectangle $ABCD$, $CD\parallel AB$,\\
$\therefore \angle FCH=\angle QAG$,\\
In $\triangle FCH$ and $\triangle QAG$,\\
$\left\{\begin{array}{l}
\angle CHF=\angle AGQ\\
CH=AG\\
\angle FCH=\angle QAG
\end{array}\right.$,\\
$\therefore \triangle FCH\cong \triangle QAG$ (ASA),\\
$\therefore AQ=CF=2$, FH=QG,\\
Since $\angle D=\angle DAM=\angle AME=90^\circ$,\\
$\therefore$ quadrilateral $ADEM$ is a rectangle,\\
$\therefore AM=DE=1$, EM=AD=2,\\
$\therefore MQ=2-1=1$,\\
$\therefore$ in $\triangle EMQ$,\\
$EQ= \sqrt{EM^2+QM^2}=\sqrt{5}$,\\
i.e., $EG+QG=EG+FH= \boxed{\sqrt{5}}$.


Question: As shown in the figure, in square $ABCD$, points $E$ and $F$ are on sides $AD$ and $CD$ respectively, and $AE=DF$. Lines $AF$ and $BE$ intersect at point $G$. If $\frac{S_{\triangle AEG}}{S_{\text{quadrilateral} DEGF}}=\frac{1}{4}$, what is the value of $\frac{S_{\triangle AEG}}{S_{\triangle ABE}}$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a square,\\
it follows that AD=AB, and $\angle EAB=\angle FDA=90^\circ$.\\
Given AE=DF,\\
it then follows that $\triangle EAB \cong \triangle FDA$,\\
thus $S_{\triangle EAB}=S_{\triangle FDA}$.\\
Since $\frac{S_{\triangle AEG}}{S_{\text{quadrilateral} DEGF}}=\frac{1}{4}$,\\
it then follows that $\frac{S_{\triangle AEG}}{S_{\triangle ADF}}=\frac{1}{5}$,\\
thus $\frac{S_{\triangle AEG}}{S_{\triangle ABE}}= \boxed{\frac{1}{5}}$.


Question: As shown in the figure, a "clover" pattern is composed of two congruent rhombuses (rhombus ABCD and rhombus EFGH), where the overlapping part is a regular octagon (shaded area). Points B and D are on EG, and points F and H are on AC. If CF equals 2, what is the length of BD?

Solution:
\textbf{Solution:} As shown in the diagram: connect DF, BD, CF, and extend lines BD and CF to meet at point O,\\
since it's a regular octagon,\\
therefore $\angle CDA=(8-2)\times 180^\circ \div 8=135^\circ$,\\
therefore $\angle DCB=180^\circ - \angle CDA=45^\circ$,\\
therefore $\angle DCO= \frac{1}{2} \angle DCB=22.5^\circ$.\\
Since $MF=MD$ and $\angle DMF=135^\circ$,\\
therefore $\angle MDF= \frac{1}{2}(180^\circ - \angle DMF)=22.5^\circ$,\\
therefore $\angle MDF=\angle DCO$,\\
therefore $DF=FC=2$,\\
since $\angle DOF=360^\circ \div 8 \times 2=90^\circ$,\\
therefore $DO^2+OF^2=DO^2+DO^2=DF^2=4$,\\
therefore $DO= \sqrt{2}$,\\
therefore $BD=2DO=2\sqrt{2}=\boxed{2\sqrt{2}}$.


Question: As shown in the figure, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, and $AE \perp BD$ at point $E$. If $\angle BAE = 2\angle OAE$, what is the measure of $\angle AOB$?

Solution:
\textbf{Solution}: Given that quadrilateral ABCD is a rectangle, \\
thus AO=BO, \\
thus $\angle$OAB=$\angle$OBA.\\
Given that $\angle$BAE=2$\angle$OAE, \\
thus let $\angle$OAE=x, then $\angle$BAE=2x, $\angle$ABO=$\angle$OAB=x+2x=3x, $\angle$AOE=$90^\circ-$.\\
Given that $\angle$AOB+$\angle$OAB+$\angle$OBA=$180^\circ$, \\
thus 3x+3x+$90^\circ-x=180^\circ$, \\
solving for x gives x=$18^\circ$, \\
thus $\angle$AOB=$90^\circ-18^\circ=\boxed{72^\circ}$.


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

Solution:
\textbf{Solution:} Since $AC= \sqrt{AB^{2}+BC^{2}}=\sqrt{8^{2}+6^{2}}=10$,\\
and in $\triangle ABC$ with right angle at $C$, $M$ is the midpoint of $AC$,\\
thus $BM= \frac{1}{2}AC=5$,\\
and $MN= \frac{1}{3}BM=\frac{5}{3}$,\\
since $M$ and $N$ are midpoints of $AC$ and $AD$ respectively,\\
it follows that $CD=2MN=2\times \frac{5}{3}=\boxed{\frac{10}{3}}$.


Question: As shown in the figure, ABCD is a square with a side length of \(a\). Point \(P\) is on the diagonal \(AC\), and line \(PB\) is drawn. Line \(PQ\) is perpendicular to \(BP\) and intersects segment \(CD\) at point \(Q\). When \(DQ = 2CQ\), what is the length of \(BP\)?

Solution:
\textbf{Solution:} As shown in the figure, draw PF$\perp$CD at point F, and PE$\perp$BC at point E.\\
Since square ABCD has a side length of $a$,\\
it follows that CP bisects $\angle$BCD, and $\angle$PEC=$\angle$ECF=90$^\circ$,\\
therefore PE=PF,\\
therefore quadrilateral PECF is a square,\\
hence EC=CF.\\
Since $\angle$BPE+$\angle$EPQ=90$^\circ$ and $\angle$EPQ+$\angle$QPF=90$^\circ$,\\
it follows that $\angle$BPE=$\angle$QPF.\\
Since $\angle$BPE=$\angle$QPF and $\angle$PEB=$\angle$PFQ=90$^\circ$ and PE=PF,\\
it follows that $\triangle$PEB$\cong$$\triangle$PFQ,\\
therefore PB=PQ, and BE=QF.\\
Since DQ=2CQ and CD=$a$,\\
it follows that QD=$\frac{2}{3}a$ and CQ=$\frac{1}{3}a$.\\
Let BE=FQ=$m$, then CE=$a-m$, and CF=QC+FQ=$\frac{1}{3}a+m$,\\
thus $a-m=\frac{1}{3}a+m$,\\
therefore $m=\frac{1}{3}a$,\\
hence BE=$\frac{1}{3}a$ and PE=CF=$\frac{2}{3}a$,\\
therefore BP=$\sqrt{B{E}^{2}+P{E}^{2}}$=$\sqrt{(\frac{1}{3}a{)}^{2}+(\frac{2}{3}a{)}^{2}}$=$\boxed{\frac{\sqrt{5}}{3}a}$.


Question: As shown in the figure, within rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, $AC=10$, and $\angle ACB = 30^\circ$. What is the length of $CD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
it follows that $\angle ABC=90^\circ$.\\
Since AC=10 and $\angle ACB=30^\circ$,\\
it follows that AB=5.\\
Since quadrilateral ABCD is a rectangle,\\
it follows that CD=AB=5.\\
Therefore, the length of CD is $\boxed{5}$.


Question: As shown in the figure, it is known that the diameter of $\odot O$ is 4, and $\angle ACB = 45^\circ$. What is the length of AB?

Solution:
\textbf{Solution:} Connect OA and OB, as shown in the figure,\\
$\because$ $\angle AOB = 2\angle ACB = 2\times 45^\circ = 90^\circ$,\\
$\therefore$ $\triangle AOB$ is an isosceles right triangle,\\
$\therefore$ AB = $\sqrt{2}$OA = $2\sqrt{2}$.\\
Therefore, AB = \boxed{2\sqrt{2}}.


Question: As shown in the figure, the vertex $P$ of a $30^\circ$ angle of a right-angled triangle board falls on the circle $\odot O$, and the two sides intersect the circle $\odot O$ at points $A$ and $B$, respectively. If $AO$ and $BO$ are connected, what is the degree measure of $\angle AOB$?

Solution:
\textbf{Solution}: Since $\angle P=30^\circ$, $\angle AOB=2\angle P$\\
Therefore, $\angle AOB=\boxed{60^\circ}$


Question: As shown in the figure, within the circle $\odot O$, the radius $OC$ is perpendicular to the chord $AB$ at point $D$, and given $AB=8$ and $OC=5$, what is the length of $OD$?

Solution:
\textbf{Solution:} Given that radius $OC$ is perpendicular to chord $AB$ at point $D$, and $AB=8$,\\
hence, $AD=\frac{1}{2}AB=4$, and $\angle ADO=90^\circ$,\\
since $OA=OC=5$,\\
thus, $OD=\sqrt{OA^{2}-AD^{2}}=\boxed{3}$.


Question: As shown in the figure, in the circle $\odot O$, $CD$ is a diameter, $AB$ is a chord, and $AB\perp CD$ at $M$, with $AB=8$ and $OC=5$, what is the length of $MD$?

Solution:
\textbf{Solution:} Draw connection OA, \\
since $CD$ is the diameter, $AB$ is a chord, and $AB\perp CD$ at $M$, with $AB=8$, \\
thus $AM=BM=4$, \\
since $OC=5$, \\
thus $OA=OD=5$, \\
thus $OM=\sqrt{OA^{2}-AM^{2}}=\sqrt{5^{2}-4^{2}}=3$. \\
Therefore, $DM=OD-OM=5-3=\boxed{2}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, CD is a chord of $\odot O$, and $\angle ACD = 40^\circ$, what is the measure of $\angle BAD$?

Solution:
\textbf{Solution:} Connect BD,\\
since the angles subtended by the same arc in a circle are equal,\\
therefore, $\angle B = \angle C = 40^\circ$,\\
since AB is a diameter,\\
therefore, $\angle ADB = 90^\circ$,\\
thus, $\angle DAB + \angle B = 90^\circ$,\\
therefore, $\angle DAB = 90^\circ - 40^\circ = \boxed{50^\circ}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is an inscribed quadrilateral of the circle $\odot O$. If $\angle BOD = 70^\circ$, what is the measure of $\angle C$ in degrees?

Solution:
\textbf{Solution:} Since $\angle BOD=70^\circ$,\\
it follows that $\angle A=\frac{1}{2}\angle BOD=35^\circ$,\\
since quadrilateral $ABCD$ is inscribed in circle $\odot O$,\\
it follows that $\angle C+\angle A=180^\circ$,\\
therefore, $\angle C=\boxed{145^\circ}$.


Question: As shown in the figure, circle O is the circumcircle of right $\triangle ABC$, with $\angle ACB=90^\circ$ and $\angle A=25^\circ$. A tangent to circle O passes through point C and intersects the extension of AB at point D. What is the measure of $\angle D$?

Solution:
\textbf{Solution:} Connect OC, \\
$\because$ Circle O is the circumcircle of the Rt$\triangle ABC$, $\angle ACB=90^\circ$, \\
$\therefore$ AB is the diameter, \\
$\because$ $\angle A=25^\circ$, \\
$\therefore$ $\angle BOC=2\angle A=50^\circ$, \\
$\because$ CD is a tangent to circle O, \\
$\therefore$ OC$\perp$CD, \\
$\therefore$ $\angle D=90^\circ-\angle BOC=\boxed{40^\circ}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\odot O$. If $\angle BOD = 100^\circ$, then what is the measure of $\angle DCB$?

Solution:
\textbf{Solution:} Given that $\because \angle BOD=100^\circ$,\\
$\therefore \angle BAD=\frac{1}{2}\angle BOD=50^\circ$;\\
Since $\because \angle BAD+\angle BCD=180^\circ$,\\
$\therefore \angle BCD=180^\circ-\angle BAD=180^\circ-50^\circ=\boxed{130^\circ}$.


Question: As shown in the figure, $PA$ and $PB$ are tangent to the circle $\odot O$ at points $A$ and $B$, respectively. Point $C$ is a point on the circle $\odot O$. Connect $AC$ and $BC$. If $\angle P = 80^\circ$, what is the degree measure of $\angle ACB$?

Solution:
\textbf{Solution:} As shown in the figure, draw $OA$ and $OB$.\\
Since $PA$ and $PB$ are tangent to the circle $\odot O$ at points A and B, respectively,\\
it follows that $OA\perp PA$ and $OB\perp PB$,\\
which means $\angle OAP=\angle OBP=90^\circ$,\\
thus $\angle AOB=180^\circ-\angle P=180^\circ-80^\circ=100^\circ$,\\
therefore $\angle ACB=\frac{1}{2}\angle AOB=\frac{1}{2}\times 100^\circ=\boxed{50^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and point C is the midpoint of the arc $BE$. Draw $CD\perp AB$ at point G, intersecting $\odot O$ at point D. If $BE=8, BG=2$, what is the radius of $\odot O$?

Solution:
\textbf{Solution:} Let us connect $OD$, as shown in the diagram. Assume the radius of $\odot O$ to be $r$,\\
since $CD\perp AB$,\\
it follows that $\overset{\frown}{BC}=\overset{\frown}{BD}$, and $CG=DG$,\\
since point C is the midpoint of arc BE,\\
it follows that $\overset{\frown}{CE}=\overset{\frown}{CB}$,\\
therefore $\overset{\frown}{BE}=\overset{\frown}{CD}$,\\
therefore $CD=BE=8$,\\
therefore $DG=\frac{1}{2}CD=4$,\\
In $\triangle ODG$ with right angle at $D$,\\
since $OG=r-2$ and $OD=r$,\\
it follows that $4^2+(r-2)^2=r^2$, solving this gives $r=\boxed{5}$,\\
which means the radius of $\odot O$ is $5$.


Question: As shown in the figure, $AB$ is the diameter of the circle $O$. The chord $CD \perp AB$ at point $E$, $OC = 5 \text{cm}$, $CD = 6 \text{cm}$, then what is the value of $BE$?

Solution:
\textbf{Solution:} Given that AB is the diameter and CD$\perp$AB,\\
it follows that CE=$\frac{1}{2}$CD=$\frac{1}{2} \times 6=3$, and $\angle$OEC=$90^\circ$,\\
thus, OE=$\sqrt{OC^{2}-CE^{2}}=\sqrt{5^{2}-3^{2}}=4$,\\
which means BE=OB−OE=5−4=\boxed{1}cm.


Question: As shown in the figure, it is known that circle $O$ is the circumscribed circle of $\triangle ABD$, where $AB$ is the diameter of circle $O$, and $CD$ is a chord of circle $O$. Given that $\angle ABD=58^\circ$, what is the measure of $\angle BCD$?

Solution:
\textbf{Solution:} Given that AB is the diameter of circle O, \\
thus, $\angle ADB=90^\circ$, \\
which means $\angle A=90^\circ-\angle ABD=90^\circ-58^\circ=32^\circ$, \\
since $\overset{\frown}{BD}=\overset{\frown}{BD}$, \\
therefore, $\angle BCD=\angle A=\boxed{32^\circ}$.


Question: As shown in the figure, it is known that point $O$ is the circumcenter of $\triangle ABC$, with $\angle A=35^\circ$. By connecting $BO$ and $CO$, what is the measure of $\angle OBC$?

Solution:
\textbf{Solution:} Since point O is the circumcenter of $\triangle ABC$ and $\angle A=35^\circ$,

$\therefore \angle BOC=2\angle A=70^\circ$, and BO=OC,

$\therefore \angle OBC=(180^\circ-\angle BOC)\div 2$,

$\therefore \angle OBC=\boxed{55^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of the semicircle $O$, and $\angle BAC=28^\circ$. What is the measure of $\angle D$?

Solution:
\textbf{Solution:} As shown in the figure, connect CB, \\
$\because$ AB is the diameter of the semicircle O, and $\angle BAC=28^\circ$, \\
$\therefore \angle B=\angle ACB - \angle BAC=90^\circ - 28^\circ = 62^\circ$, \\
Also, $\because$ quadrilateral ABCD is inscribed in the circle, \\
$\therefore \angle D + \angle ABC= 180^\circ$, \\
$\therefore \angle D=180^\circ - 62^\circ = \boxed{118^\circ}$.


Question: As shown in the figure, the chord $CD \perp AB$ at point E, $AB$ passes through the center O, $BD=5$, $BE=3$, then what is the length of $CD$?

Solution:
\textbf{Solution:} Since $CD \perp AB$ and $AB$ passes through the center $O$,\\
it follows that $DE = \frac{1}{2}CD$,\\
In the right triangle $\triangle DEB$, with $BD=5$ and $BE=3$,\\
by the Pythagorean theorem, we have: $DE = \sqrt{BD^{2}-BE^{2}} = 4$,\\
therefore, $CD = \boxed{8}$.


Question: As shown in the figure, $AB$ is a chord of $\odot O$, diameter $CD \perp AB$, intersecting $AB$ at point $H$, and $OA$ is connected. If $\angle A=45^\circ$ and $AB=2$, what is the length of $DH$?

Solution:
\textbf{Solution:}

Since $AB$ is a chord of the circle $O$, and the diameter $CD \perp AB$,

it follows that $H$ is the midpoint of $AB$, meaning $AH=\frac{1}{2}AB=1$,

Moreover, since $\angle A=45^\circ$ and $CD \perp AB$,

it follows that $\triangle AHO$ is an isosceles right triangle,

Therefore, $OH=AH=1$, $OA=\sqrt{OH^2+AH^2}=\sqrt{2}$,

Therefore, $DH=OD+OH=OA+OH=\sqrt{2}+1=\boxed{\sqrt{2}+1}$


Question: As shown in the figure, $CD$ is the diameter of $\odot O$, $AB$ is a chord, $CD\perp AB$ at $E$, $DE=2$, and $AB=8$. What is the length of $AC$?

Solution:
\textbf{Solution:} Draw line OA as shown in the diagram:\\
Since $CD \perp AB$,\\
thus $AE=BE=\frac{1}{2}AB=4$,\\
Let $OA=r$, then $OE=OD-DE=r-2$,\\
In $\triangle OAE$, $OA^2=OE^2+AE^2$,\\
which gives $r^2=(r-2)^2+4^2$,\\
Solving this, we find: $r=5$,\\
thus $CE=2r-DE=10-2=8$,\\
thus $AC=\sqrt{CE^2+AE^2}=\sqrt{8^2+4^2}=\boxed{4\sqrt{5}}$.


Question: As shown in the figure, it is known that the radius of circle $O$ is 6, and $AB$, $BC$ are chords of circle $O$. If $\angle ABC=50^\circ$, then what is the length of arc $\overset{\frown}{AC}$?

Solution:
\textbf{Solution:} Since $AB$ and $BC$ are chords of the circle $\odot O$, and $\angle ABC=50^\circ$,\\
it follows that $\angle AOC=2\angle ABC=100^\circ$,\\
Given that the radius of $\odot O$ is 6,\\
it follows that $\overset{\frown}{AC}=\frac{100}{360}\times 2\pi \times 6=\boxed{\frac{10}{3}\pi}$


Question: As shown in the figure, points A, B, and C are three points on the circle $ \odot O$. If $ \angle BAC=40^\circ$, what is the degree measure of $ \angle BOC$?

Solution:
\textbf{Solution:} Given that $\because \angle BOC=2\angle BAC$ and $\angle BAC=40^\circ$, \\
$\therefore \angle BOC=2\times 40^\circ=\boxed{80^\circ}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\odot O$, where $\angle A=100^\circ$. What is the degree measure of $\angle C$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is inscribed in a circle,\\
it follows that $\angle A+\angle C=180^\circ$,\\
since $\angle A=100^\circ$,\\
thus $\angle C=180^\circ-\angle A=180^\circ-100^\circ=\boxed{80^\circ}$.


Question: As shown in the figure, $ \triangle ABC$ is an inscribed triangle of $ \odot O$ with $ \angle C=30^\circ$ and the radius of $ \odot O$ is $ 2\,cm$. If point P is a point on $ \odot O$ and $ PB=AB$, what is the length of $ PA$?

Solution:
\textbf{Solution:} Connect $OA$ and $OP$, extend $OB$ to intersect $AP$ at $H$,\\
by the Inscribed Angle Theorem, $\angle AOB=2\angle C=60^\circ$,\\
since $PB=AB$,\\
thus $\angle POB=\angle AOB=60^\circ$, and $OB\perp AP$, hence $\angle OAP=30^\circ$, therefore $OH=\frac{1}{2}AO=1$,\\
then $AH=PH=\sqrt{AO^2-AH^2}=\sqrt{3}\ (\text{cm})$,\\
thus $PA=2AH=2\sqrt{3}\ (\text{cm})$,\\
therefore, the answer is: $PA=\boxed{2\sqrt{3}}\ (\text{cm})$.


Question: As shown in the figure, a regular hexagon $ABCDEF$ is inscribed in circle $\odot O$. Connecting $BD$, what is the degree measure of $\angle CDB$?

Solution:
\textbf{Solution}: As shown in the figure, connect $OB$, $OC$,\\
since regular hexagon $ABCDEF$ is inscribed in $\odot O$,\\
therefore $\angle BOC=\frac{360^\circ}{6}=60^\circ$,\\
therefore $\angle CBD=\frac{1}{2}\angle COD=\boxed{30^\circ}$.


Question: As shown in the figure, points A, B, and C are all on the circle $ \odot O$. When $ \angle OAC=50^\circ$, what is the degree measure of $ \angle B$?

Solution:
\textbf{Solution:} Given that points A, B, and C are all on the circle $\odot O$, $\angle OAC=50^\circ$,\\
it follows that $OA=OC$,\\
thus $\angle OAC=\angle OCA=50^\circ$,\\
therefore $\angle AOC=180^\circ-2\times 50^\circ=80^\circ$,\\
since $\overset{\frown}{AC}=\overset{\frown}{AC}$,\\
it implies that $\angle B=\frac{1}{2}\angle AOC=\frac{1}{2}\times 80^\circ=\boxed{40^\circ}$;


Question: As shown in the diagram, $AB$ is the diameter of $\odot O$, and $C$ is a point on $\odot O$. If $\angle BCO=35^\circ$ and $AO=2$, then what is the length of $\overset{\frown}{AC}$?

Solution:
\textbf{Solution:} Given that $\angle BCO=35^\circ$,\\
it follows that $\angle AOC=2\angle BCO=70^\circ$,\\
and given that $AO=2$,\\
thus, the arc $\overset{\frown}{AC}=\frac{n\pi r}{180}=\frac{70\pi \times 2}{180}=\frac{140\pi }{180}=\boxed{\frac{7}{9}\pi}$.


Question: As shown in the figure, AB is the diameter of the circle $\odot O$, and the quadrilateral ABCD is inscribed in $\odot O$. If $BC=CD=DA=4\,cm$, then what is the circumference of $\odot O$?

Solution:
\textbf{Solution:} As shown in the figure, connect OD and OC.\\
Since AB is the diameter of $\odot O$, and quadrilateral ABCD is inscribed in $\odot O$, if BC = CD = DA = 4cm,\\
then $\overset{\frown}{AD}$ = $\overset{\frown}{CD}$ = $\overset{\frown}{BC}$,\\
thus $\angle AOD = \angle DOC = \angle BOC = 60^\circ$.\\
Also, since OA = OD,\\
thus $\triangle AOD$ is an equilateral triangle,\\
therefore OA = AD = 4cm,\\
thus the circumference of $\odot O$ = 2×4$\pi$ = $\boxed{8\pi}$ (cm).


Question: As shown in the figure, within circle $O$, chord $AB=4$, point $C$ moves along $AB$. Connect $OC$, and draw $CD\perp OC$ intersecting circle $O$ at point $D$. What is the maximum value of $CD$?

Solution:
\textbf{Solution:} As shown in the diagram, connect $OD$,\\
since $CD \perp OC$,\\
it follows that $\angle DCO = 90^\circ$,\\
thus $CD = \sqrt{OD^{2} - OC^{2}} = \sqrt{r^{2} - OC^{2}}$,\\
when the value of $OC$ is minimal, the value of $CD$ is maximal,\\
$OC \perp AB$ when $OC$ is minimal, at this time points $D$ and $B$ coincide,\\
thus $CD = CB = \frac{1}{2}AB = 2$,\\
that is, the maximal value of $CD$ is $\boxed{2}$,


Question: As shown in the figure, AB is a chord of $\odot O$, OC$\perp$AB at C, AB=8, OC=3, what is the radius length of $\odot O$?

Solution:
\textbf{Solution:} Since OC$\perp$AB at C,\\
it follows that AC=CB,\\
Given AB=8,\\
thus AC=CB=4,\\
In $\triangle AOC$, OC=3,\\
By Pythagoras' theorem,\\
OA=$\sqrt{3^2+4^2}=5$.\\
Hence, the answer is: $OA=\boxed{5}$.


Question: As shown in the figure, points $A$, $B$, and $C$ are on circle $\odot O$, with $OB\parallel AC$. Line $BC$ intersects $OA$ at point $D$. If $\angle ADB=60^\circ$, what is the degree measure of $\angle AOB$?

Solution:
\textbf{Solution:} Let $ \angle ACB=x^\circ$,\\
then $ \angle AOB=2\angle ACB=2x^\circ$,\\
since $OB\parallel AC$,\\
thus $ \angle OBD=\angle ACB=x^\circ$,\\
since $ \angle ADB=60^\circ$,\\
thus $ \angle AOB+\angle OBD=\angle ADB=60^\circ$, that is $ 2x+x=60$,\\
solving for $x$ gives $ x=20$,\\
then $ \angle AOB=2x^\circ=\boxed{40^\circ}$.


Question: As shown in the diagram, a circle $\odot O$ with radius 8 is folded along line $AB$, such that the arc $\overset{\frown}{AB}$ exactly passes through point $D$, which is the midpoint of the radius $OC$ that is perpendicular to $AB$. What is the length of the crease $AB$?

Solution:
\textbf{Solution:} Extend $CO$ to meet $AB$ at point $E$, and connect $OB$,\\
since $CE\perp AB$,\\
thus, $E$ is the midpoint of $AB$,\\
according to the problem, we have $CD=4$, $OD=4$, and $OB=8$,\\
$DE=\frac{1}{2}\left(8\times 2-4\right)=\frac{1}{2}\times 12=6$,\\
$OE=6-4=2$,\\
in $\triangle OEB$, by the Pythagorean theorem, we have: $OE^{2}+BE^{2}=OB^{2}$,\\
substituting the values, we find $BE=2\sqrt{15}$,\\
thus, $AB=4\sqrt{15}$.\\
So, the answer is: $AB=\boxed{4\sqrt{15}}$.


Question: As shown in the figure, within circle $O$ with a radius of $5$, and chord $AB=8$, $P$ is a moving point on chord $AB$. What is the minimum value of $OP$?

Solution:
\textbf{Solution:} As shown in the diagram, connect $OA$ and draw $OC\perp AB$ through point $O$ with the foot at $C$. Hence, $AC=BC=\frac{1}{2}AB=4$,\\
In $\triangle AOC$,\\
$OC=\sqrt{OA^2-AC^2}=3$,\\
which means the minimum value for point $P$ moving on $AB$ is $\boxed{3}$.


Question: As shown in the figure, point A is a trisection point on the semicircle, point B is the midpoint of arc AN, and point P is a moving point on the diameter MN. Given that the radius $O$ is 1, what is the minimum value of $AP+BP$?

Solution:
\textbf{Solution:} Let $A'$ be the symmetrical point of A with respect to line $MN$, and draw line $A'B$, which intersects $MN$ at point $P$. Then, $PA+PB=PA'+PB=A'B$ is minimized.\\
Draw lines $OA'$ and $AA'$.\\
Since point $A$ and $A'$ are symmetrical with respect to $MN$, and point $A$ is a trisection point on the semicircle,\\
hence, $\angle A'ON=\angle AON=60^\circ$, and $PA=PA'$,\\
Since point $B$ is the midpoint of arc $AN$,\\
therefore, $\angle BON=30^\circ$,\\
thus, $\angle A'OB=\angle A'ON + \angle BON=90^\circ$,\\
Also, since $OA=OA'=1$,\\
therefore, $A'B= \sqrt{2}$\\
hence, $PA+PB=PA'+PB=A'B= \boxed{\sqrt{2}}$


Question: As shown in the figure, the three vertices of $\triangle ABC$ are on the circle $\odot O$, $D$ is a point on $\overset{\frown}{AB}$, and $E$ is a point on $\overset{\frown}{AC}$. If $\angle BAC=50^\circ$, what is the sum of $\angle D+\angle E$?

Solution:
\textbf{Solution:} As shown in the figure, connect OA, OB, and OC.\\
According to the problem: $\angle BOC=2\angle BAC=100^\circ$,\\
$\therefore \angle AOB+\angle AOC=260^\circ$,\\
Also, $\because \angle D= \frac{1}{2}(\angle BOC+\angle AOC)$, $\angle E= \frac{1}{2}(\angle BOC+\angle AOB)$,\\
$\therefore \angle D+\angle E= \frac{1}{2}(\angle BOC+\angle AOC+\angle BOC+\angle AOB)= \frac{1}{2}(260^\circ+100^\circ+100^\circ)=\boxed{230^\circ}$.


Question: As shown in the figure: AB is the diameter of $\odot O$, chord CD $\perp$ AB at E. If AB $= 20$ and CD $= 16$, what is the length of the line segment BE?

Solution:
\textbf{Solution:} Draw line OC, \\
$\because$AB=20, \\
$\therefore$OC=OA=OB=10, \\
$\because$AB$\perp$CD and AB passes through O, \\
$\therefore$CE=DE=$\frac{1}{2}$CD=8, \\
In $\triangle OCE$, by the Pythagorean theorem, we have: OE=$\sqrt{10^{2}-8^{2}}$=6, \\
$\therefore$BE=10-6=\boxed{4}.


Question: As shown in the figure, the diameter $CD$ of the circle $\odot O$ is perpendicular to the chord $AB$ at point $E$, and $BE=4$, $DE=8$, then what is the length of $CE$?

Solution:
\textbf{Solution:} Let the radius of the circle $\odot O$ be $r$. Thus, $OB=r$ and $OE = DE - OD = 8 - r$,\\
since $AB \perp CD$,\\
in $\triangle OEB$, we have $OB^{2} = OE^{2} + BE^{2}$,\\
which leads to $r^{2} = (8 - r)^{2} + 4^{2}$,\\
solving this gives $r = 5$,\\
thus, $CD = 2r = 10$,\\
therefore, $CE = CD - DE = 10 - 8 = \boxed{2}$.


Question: As shown in the figure, there is a cross-section of a tunnel. If its shape is part of a circle with center $O$, and the road surface $AB=12$ meters, the clear height $CD=9$ meters, then what is the radius $OA$ of this circle?

Solution:
\textbf{Solution:} Let the radius of the circle be $r$, thus we have $OA=r$, and $OD=9-r$,\\
since $AB=12$ meters, and $CD\perp AB$,\\
therefore $AD=\frac{1}{2}AB=\frac{1}{2}\times 12=6$ meters,\\
in $\triangle AOD$,\\
since $OA=r$, $OD=9-r$, and $AD=6$ meters,\\
therefore $OA^{2}=OD^{2}+AD^{2}$, which means $r^{2}=(9-r)^{2}+6^{2}$,\\
solving for $r$ yields $r=\boxed{\frac{13}{2}}$ meters.


Question: As shown in the figure, $\odot O$ is the circumcircle of the right-angled triangle $ABC$, with $OE \perp AB$ intersecting $\odot O$ at point $E$, and the foot of the perpendicular is point $D$. The extensions of $AE$ and $CB$ intersect at point $F$. If $ED=4$ and $AB=16$, what is the length of $FC$?

Solution:
\textbf{Solution:} \\
Since $OE\perp AB$,\\
it follows that $\angle ADO=90^\circ$,\\
also, since $\angle ABC=90^\circ$,\\
it follows that $OD\parallel BC$,\\
since the extensions of $AE$ and $CB$ meet at point F,\\
it follows that $OE\parallel CF$,\\
since $OA=OC$,\\
it follows that $\frac{AO}{AC}=\frac{AD}{AB}=\frac{AE}{EF}=1$,\\
therefore, $DE$ is the midline of $\triangle ABF$,\\
thus $AD=BD=\frac{1}{2}AB=8$, $FB=2DE=8$,\\
In the right triangle $ADO$,\\
since $AD=8$, $OD=OE-DE=OA-DE=OA-4$,\\
it follows that $OD^2+AD^2=OA^2$,\\
hence $(OA-4)^2+8^2=OA^2$,\\
which gives $OA=10$,\\
therefore, $OD=6$,\\
hence $CB=2OD=12$,\\
thus $CF=FB+BC=8+12=\boxed{20}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, point $F$ is the centroid of the triangle. Line $DE \parallel BC$ passes through point $F$ and intersects $AB$ at point $D$ and $AC$ at point $E$. If $DE=7$, what is the value of $BC$?

Solution:
\textbf{Solution:} Draw line $CF$ and extend it to intersect $AB$ at point H, and connect $GH$, as shown in the figure,\\
Since point $F$ is the centroid of $\triangle ABC$,\\
Therefore, $AG$ and $CH$ are the medians of $\triangle ABC$,\\
Therefore, points $G$ and $H$ are respectively the midpoints of $BC$ and $AB$,\\
Therefore, $GH=\frac{1}{2}AC$ and $GH\parallel AC$,\\
Therefore, $\frac{AF}{FG}=\frac{AC}{HG}=2$,\\
Therefore, $\frac{AF}{AG}=\frac{2}{3}$,\\
Since $DE\parallel BC$,\\
Therefore, $\frac{DE}{BC}=\frac{AF}{AG}=\frac{2}{3}$,\\
Since $DE=7$,\\
Therefore, $BC=\frac{3DE}{2}=\frac{3\times 7}{2}=\boxed{\frac{21}{2}}$,


Question: As shown in the figure, $DE \parallel BC$, and $AE:AC = 2:3$, $AD=6$. What is the length of $AB$?

Solution:
\textbf{Solution:} Since $DE\parallel BC$,\\
it follows that $\frac{AE}{AC}=\frac{AD}{AB}$,\\
Given $AE\colon AC=2\colon 3$ and $AD=6$,\\
it follows that $\frac{2}{3}=\frac{6}{AB}$,\\
Solving yields: $AB=\boxed{9}$.


Question: As shown in the figure, two straight lines are intersected by three parallel lines, where $AB=4$, $BC=6$, and $EF=5.4$. What is the length of $DE$?

Solution:
\textbf{Solution:} Given $\because AD\parallel BE\parallel CF$,\\
$\therefore \frac{AB}{AC}=\frac{DE}{DF}$,\\
$\because AB=4, BC=6, EF=5.4$,\\
$\therefore AC=AB+BC=4+6=10, DF=5.4+DE$,\\
$\therefore \frac{4}{10}=\frac{DE}{5.4+DE}$,\\
$\therefore DE=\boxed{3.6}$.


Question: As shown in the figure, in $\triangle ABC$, point E is on BC, and $BE=3EC$. D is the midpoint of AC, AE and BD intersect at point F, what is the value of $AF: EF$?

Solution:
\textbf{Solution:} Construct $EH\parallel AC$ through point E and intersect BD at point H, as shown in the figure:\\
Since $EH\parallel AC$,\\
thus $\triangle BEH\sim \triangle BCD$\\
therefore $\frac{EH}{CD}=\frac{BE}{BC}$,\\
since $BE=3EC$,\\
therefore $\frac{EH}{CD}=\frac{3CE}{4CE}=\frac{3}{4}$,\\
since $D$ is the midpoint of $AC$,\\
therefore $AD=CD$,\\
therefore $\frac{EH}{CD}=\frac{EH}{AD}=\frac{3}{4}$,\\
since $EH\parallel AC$,\\
therefore $\frac{AF}{EF}=\frac{AD}{EH}=\frac{4}{3}$,\\
therefore $AF\colon EF=\boxed{4\colon 3}$.


Question: As shown in the figure, in $\triangle ABC$, D and E are the midpoints of AB and AC, respectively. If $S_{\triangle ADE}=2$, what is the area of quadrilateral DBCE?

Solution:
\textbf{Solution:}\\
Since points D and E are the midpoints of AB and AC, respectively,\\
therefore, DE is the midline of $\triangle ABC$,\\
therefore, DE$\parallel$BC, and DE=$\frac{1}{2}$BC,\\
therefore, $\triangle ADE\sim \triangle ABC$,\\
therefore, $\frac{S_{\triangle ADE}}{S_{\triangle ABC}}=\left(\frac{DE}{BC}\right)^2=\frac{1}{4}$\\
therefore, $S_{\triangle ABC}=4S_{\triangle ADE}=8$\\
therefore, $S_{\text{quadrilateral DBCE}}=S_{\triangle ABC}-S_{\triangle ADE}=8-2=\boxed{6}$.


Question: As shown in the figure, given that $AB\parallel CD\parallel EF$, if $\frac{AC}{AE}=\frac{2}{3}$, what is the value of $\frac{BD}{BF}$?

Solution:
\textbf{Solution:} Since $AB \parallel CD \parallel EF$,\\
it follows that $\frac{AC}{AE}=\frac{BD}{BF}=\boxed{\frac{2}{3}}$.


Question: As shown in the figure, quadrilateral $ABCD$ is a diamond, $AC$ and $BD$ intersect at point $O$, $E$ is the midpoint of $OA$, connect $BE$ and extend it to intersect $AD$ at point $F$. Given $BE=4$, what is the length of $EF$?

Solution:
\textbf{Solution:} Given that point $E$ is the midpoint of $OA$,\\
thus $AO=2AE$,\\
since $ABCD$ is a rhombus,\\
thus $AD \parallel BC$, and $AO=OC$,\\
therefore, $\frac{AE}{EC}=\frac{EF}{BE}$, and $EC=3AE$,\\
thus, $\frac{AE}{3AE}=\frac{EF}{4}$\\
Solving yields: $EF=\boxed{\frac{4}{3}}$.


Question: As shown in the figure, points A, B, C, and D are four points on the circle $\odot O$. AC bisects $\angle BAD$, AC intersects BD at point E, $CE=4$, $CD=6$. What is the length of AC?

Solution:
\textbf{Solution:} Given that $AC$ bisects $\angle BAD$, \\
it follows that $\overset{\frown}{BC}=\overset{\frown}{CD}$, \\
hence $\angle BDC=\angle CAD$, \\
given that $\angle ACD=\angle DCE$, \\
it follows that $\triangle CDE\sim \triangle CAD$, \\
therefore, $CD: AC=CE: CD$, \\
which implies $CD^{2}=AC\cdot CE$, \\
hence $6^{2}=4(4+AE)$, \\
thus $AE=5$, \\
therefore, $AC=AE+CE=9$, \\
thus, the length of $AC$ is $\boxed{9}$.


Question: As shown in the figure, $AB$ is the diameter of a semicircle. The bisector of $\angle ABC$ intersects the chord $AC$ and the semicircle at points $E$ and $D$, respectively. If $BE=2DE$ and $AB=4$, what is the length of $AE$?

Solution:
\textbf{Solution:}

Since $\angle CDB = \angle CAB$ and $\angle DCA = \angle DBA$,\\
it follows that $\triangle CDE \sim \triangle ABE$,\\
hence $\frac{CD}{AB} = \frac{DE}{BE} = \frac{1}{2}$,\\
thus $CD = \frac{1}{2}AB = 2$.\\
Since $BD$ bisects $\angle ABC$,\\
it implies that $\angle CBD = \angle ABD$,\\
and so $\angle DAC = \angle CBD = \angle ABD = \angle ACD$,\\
therefore $AD = CD = 2$.\\
Since $AB$ is the diameter of a semicircle,\\
it results in $\angle ADB = 90^\circ$,\\
thus $\sin\angle ABD = \frac{AD}{AB} = \frac{2}{4} = \frac{1}{2}$,\\
therefore $\angle ABD = 30^\circ$,\\
hence $\angle DAE = 30^\circ$,\\
therefore $AE = \frac{AD}{\cos30^\circ} = \boxed{\frac{4\sqrt{3}}{3}}$.


Question: As shown in the figure, in $ \triangle ABC$, where $ \angle BAC=90^\circ$, the medians $ AD$ and $ BE$ intersect at point F, and $ EG \parallel BC$ intersects $ AD$ at point G. Given that $ GF=2$, what is the length of $ BC$?

Solution:
\textbf{Solution:} Let $AD$ and $BE$ be the medians intersecting at point $F$, \\
$\therefore$ Point $F$ is the centroid of $\triangle ABC$, and $BD=CD$, \\
$\therefore$ $AF=2DF$, \\
$\because$ $EG\parallel BC$, \\
$\therefore$ $\triangle AGE\sim \triangle ADC$ \\
$\therefore$ $\frac{GE}{DC}=\frac{AE}{AC}=\frac{1}{2}$, \\
$\therefore$ $\frac{GE}{BD}=\frac{1}{2}$, \\
$\because$ $EG\parallel BC$, \\
$\therefore$ $\triangle GEF\sim \triangle DBF$, \\
$\therefore$ $\frac{DF}{GF}=\frac{BD}{EG}=2$, \\
$\therefore$ $DF=2GF=4$,\\
$\therefore$ $AF=2DF=8$, \\
$\therefore$ $AD=DF+AF=4+8=12$, \\
$\because$ $\angle BAC=90^\circ$, and $AD$ is the median of $\triangle ABC$, \\
$\therefore$ $BC=2AD=\boxed{24}$,


Question: As shown in the figure, $DE \parallel BC$, and $EC:BD = 2:3$, $AD=6$. What is the length of $AE$?

Solution:
\textbf{Solution:} Since DE$\parallel$BC, and EC$\colon$ BD=2$\colon$3, AD=6, \\
$\therefore \frac{AE}{EC}=\frac{AD}{BD}$,\\
$\therefore \frac{AE}{2}=\frac{6}{3}$,\\
$\therefore$ AE=\boxed{4}


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

Solution:
\textbf{Solution:} Given that $AD=2BD$,\\
it follows that $BD: AB=1: 3$,\\
Since $DE \parallel BC$,\\
it follows that $CE: AC=BD: AB=1: 3$,\\
Since $EF \parallel AB$,\\
it follows that $CF: CB=CE: AC=1: 3$.\\
Therefore, $\frac{CF}{CB} = \boxed{\frac{1}{3}}$


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=8$, $BC=4\sqrt{7}$. A circle with center O on side BC is drawn to be tangent to AB and AC at D and E, respectively. What is the length of AD?

Solution:
\textbf{Solution:} Join $AO$, $OD$, and $OB$.\\
Since $AB$ and $AC$ are tangent to point $D$ and $E$, respectively,\\
it follows that $OD\perp AB$ and $OE\perp AC$,\\
and since $AO=AO$,\\
it results in $\triangle ADO \cong \triangle AEO$ (by Hypotenuse-Leg),\\
thus $\angle DAO = \angle EAO$,\\
and since $AB=AC$,\\
it follows that $AO \perp BC$ and $BO=OC=2\sqrt{7}$,\\
by applying the Pythagorean theorem we get\\
$AO=\sqrt{AB^{2}-BO^{2}}=6$,\\
since $\angle DAO = \angle BAO$ and $\angle ADO = \angle AOB$,\\
it results in $\triangle ADO \sim \triangle AOB$,\\
thus $\frac{AD}{AO} = \frac{AO}{AB}$,\\
therefore, $AD=\boxed{\frac{9}{2}}$.


Question: As shown in the figure, in $ \triangle ABC$, $ AC=2$, $ BC=4$, and D is a point on BC, $ \angle CAD=\angle B$, the area of $ \triangle ADC$ is $a$, then what is the area of $ \triangle ABD$?

Solution:
\textbf{Solution:} Since $\angle CAD=\angle B$ and $\angle ACD=\angle BCA$,\\
it follows that $\triangle ACD\sim \triangle BCA$,\\
hence $\frac{{S}_{\triangle ACD}}{{S}_{\triangle BCA}}=\left(\frac{AC}{BC}\right)^{2}$,\\
therefore $\frac{a}{{S}_{\triangle BCA}}=\frac{1}{4}$,\\
solving this, the area of $\triangle BCA$ is $4a$,\\
thus, the area of $\triangle ABD$ is: $4a-a=\boxed{3a}$.


Question: As shown in the figure, in quadrilateral $ABCD$, point $E$ is the midpoint of side $AD$, and $EC$ intersects the diagonal $BD$ at point $F$. If the area of $\triangle DEF$ is $3$, what is the area of quadrilateral $ABCD$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a parallelogram,

$\therefore$ $AD\parallel BC$.

$\therefore$ $\angle EDB=\angle DBC$, $\angle DEC=\angle BCE$

$\therefore$ $\triangle EFD\sim \triangle CFB$.

Additionally, since $E$ is the midpoint of $AD$,

$\therefore$ $DE=\frac{1}{2}CB$.

$\therefore$ ${S}_{\triangle BCF}=4{S}_{\triangle EDF}=12$.

Let ${S}_{\triangle DFC}=x$, then $3+x=\frac{1}{2}(12+x)$,

Solving for $x$ gives: $x=6$.

$\therefore$ ${S}_{\triangle BCD}=12+6=18$.

$\therefore$ ${S}_{\square ABCD}=2{S}_{\triangle BCD}=\boxed{36}$.


Question: As shown in the figure, there is an isosceles right-angled triangle board, its hypotenuse $BC=8\, \text{cm}$. Inside it, $\triangle DEF$ has its sides parallel to the sides of $OABC$ respectively, and its hypotenuse $EF=4\, \text{cm}$. What is the ratio of the area of $\triangle DEF$ to the area of the shaded part?

Solution:
\textbf{Solution.} Given by the problem, we have: $\frac{S_{\triangle DEF}}{S_{\triangle ABC}}=\left(\frac{EF}{BC}\right)^2=\left(\frac{4}{8}\right)^2=\frac{1}{4}$,\\
$\therefore$ $\frac{S_{\triangle DEF}}{S_{\text{shaded area}}}=\frac{1}{3}$,\\
that is, the ratio of the area of $\triangle DEF$ to the area of the shaded portion is $\boxed{1\colon 3}$.


Question: As shown in the figure, in $\triangle ABC$, $D$, $E$, $F$ are points on $AB$, $AC$, and $BC$, respectively, and $DE \parallel BC$, $EF \parallel AB$. If $BF:FC=2:3$ and $AB=15$, then what is the value of $BD$?

Solution:
\[
\text{{Solution:}} \quad \because \text{{ DE }} \parallel \text{{ BC}}, \text{{ EF }} \parallel \text{{ AB}}, \\
\therefore \text{{ quadrilateral BFED is a parallelogram, }} \triangle \text{{ EFC }} \sim \triangle \text{{ ABC}}, \\
\therefore \frac{{EF}}{{AB}} = \frac{{CF}}{{CB}}, \text{{ EF }} = \text{{ BD }}, \\
\because \text{{ BF:FC }} = 2:3, \text{{ AB }} = 15, \\
\therefore \text{{ EF }} = \frac{3}{5}\text{{ AB }} = 9, \\
\therefore \text{{ BD }} = \boxed{9};
\]


Question: As illustrated in the figure, $AB \parallel CD \parallel EP$, given that $\frac{AC}{CE} = \frac{3}{2}$ and $BD = 12$, what is the length of $DF$?

Solution:
\textbf{Solution:} Given $\because AB\parallel CD\parallel EP$, $\frac{AC}{CE}=\frac{3}{2}$, and $BD=12$, \\
$\therefore \frac{AC}{CE}=\frac{BD}{DF}=\frac{3}{2}$, \\
$\therefore \frac{12}{DF}=\frac{3}{2}$, \\
$\therefore DF=\boxed{8}$


Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is on $AD$. Connect $CE$ and extend it to intersect the extension of $BA$ at point $F$. If $AE: AD = 2:3$ and $CD = 3$ cm, what is the length of $AF$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a parallelogram,\\
it follows that AB $\parallel$ CD,\\
which implies $\triangle AFE \sim \triangle DEC$,\\
hence AE : DE = AF : CD,\\
given AE : AD = 2 : 3,\\
it follows that AE : DE = 2 : 1,\\
since CD = 3 cm,\\
we have AF = 2CD = \boxed{6cm}.


Question: As shown in the figure, in rectangle $ABCD$, where $AD=9$, $E$ is a point on $CD$. Connect $AE$. Triangle $ADE$ is folded along the straight line $AE$ such that point $D$ falls onto point $F$. Draw $FG\perp AD$ at $F$, with the foot being $G$. If $AG=2GD$, what is the value of $DE$?

Solution:
\textbf{Solution:} As shown in the figure, \\
from the properties of the folding, we have $AF=AD=9$, $DE=EF$, and $\angle AED=\angle AEF$, \\
$\because$ FG$\perp$AD, \\
$\therefore$ FG$\parallel$CD, \\
$\therefore$ $\angle AED=\angle AMG=\angle EMF$, \\
$\therefore$ $\angle EMF=\angle AEF$, \\
$\therefore$ MF=EF=DE, \\
$\because$ AD=9, AG=2GD, \\
$\therefore$ AG=6, \\
$\therefore$ FG=$3\sqrt{5}$, \\
$\because$ $\frac{GM}{DE}=\frac{AG}{AD}=\frac{6}{9}=\frac{2}{3}$, \\
$\therefore$ GM=$\frac{2}{3}DE$, \\
$\therefore$ $\frac{2}{3}DE+DE=3\sqrt{5}$, \\
$\therefore$ DE=$\boxed{\frac{9\sqrt{5}}{5}}$.


Question: As shown in the figure, exactly six congruent right-angled triangles are used to form two regular hexagons, one larger than the other. What is the ratio of the perimeter of the larger regular hexagon to that of the smaller one?

Solution:
\textbf{Solution:} As shown in the diagram, \\
$\because$ $\triangle ABG \cong \triangle BCH$, \\
$\therefore$ AG＝BH, \\
$\because$ $\angle ABC=(6-2) \times 180^\circ \div 6=120^\circ$, $\angle CBH=90^\circ$, \\
$\therefore$ $\angle ABG=30^\circ$, \\
$\therefore$ AB＝$\sqrt{3}$AG, BG＝2AG, \\
thus BH＋HG＝2AG, \\
$\therefore$ HG＝AG, \\
$\because$ regular hexagon ABCDEF$\sim$regular hexagon HMNPQG, \\
$\therefore$ the perimeter of regular hexagon ABCDEF : the perimeter of regular hexagon HMNPQG = AB$\colon$HG = $\sqrt{3}$AG$\colon$HG = $\sqrt{3}$HG$\colon$HG = $\boxed{\sqrt{3}}$.


Question: As shown in the figure, a large square ABCD is composed of four congruent right-angled triangles and a small square. Points P and Q are the midpoints of AB and GH, respectively. If PQ happens to pass through point F, what is the value of $\frac{AB}{EF}$?

Solution:
\textbf{Solution:} As shown in the figure, draw line PM$\perp$BE at point M,\\
Let BE=a, GH=b,\\
since $\triangle AHD \cong \triangle BEA$,\\
therefore AH=BE=a,\\
since quadrilateral EFGH is a square,\\
therefore FG=GH=EH=EF=b,\\
since $\angle AEB = \angle PMB = 90^\circ$,\\
therefore PM$\parallel$AE,\\
therefore BP:AP=BM:ME,\\
since point P is the midpoint of AB,\\
therefore AP=BP,\\
therefore BM=EM=$\frac{1}{2}$a,\\
therefore PM is the median of $\triangle ABE$,\\
therefore PM=$\frac{1}{2}$AE=$\frac{1}{2}$(a−b),\\
furthermore, since point Q is the midpoint of GH,\\
therefore GQ=$\frac{1}{2}$GH=$\frac{1}{2}$b,\\
since $\angle PMF = \angle BFC = 90^\circ$,\\
therefore $\triangle PMF \sim \triangle FGQ$,\\
therefore PM:FG=MF:GQ,\\
therefore PM$\times$GQ=FG$\times$MF,\\
since MF=EM−EF=$\frac{1}{2}$a−b,\\
therefore $\frac{1}{2}$(a−b)$\times\frac{1}{2}$b=b($\frac{1}{2}$a−b),\\
Upon simplification, we get $3b^2−ab=0$, i.e., $b(3b−a)=0$,\\
since b$\neq$0,\\
therefore 3b−a=0,\\
therefore a=3b,\\
therefore AE=AH−EH=a−b=2b,\\
therefore $AB=\sqrt{AE^2+BE^2}=\sqrt{(2b)^2+(3b)^2}=\sqrt{13}b$,\\
therefore $AB:EF = \boxed{\sqrt{13}}$.


Question: As shown in the figure, $l_{1}$, $l_{2}$, $l_{3}$, $l_{4}$ are a set of parallel lines, $l_{5}$, $l_{6}$ intersect this set of parallel lines successively at points A, B, C, D, and E, F, G, H, respectively. If the ratio of $AB:BC:CD$ is $2:3:4$, and $EG=10$, then what is the length of $EH$?

Solution:
\textbf{Solution:}\\
Since $AB:BC:CD=2:3:4$,\\
it follows that $AC:CD=5:4$,\\
Since $l_{1}\parallel l_{3}\parallel l_{4}$,\\
it results in $AC:CD=EG:GH$, i.e., $5:4=10:GH$,\\
thus $GH=8$,\\
Upon verification, $GH=8$ is a solution to the given equation and is consistent with the problem statement,\\
therefore $EH=EG+GH=10+8=\boxed{18}$.


Question: As shown in the figure, in $ \triangle ABC$, $ \angle ACB=90^\circ$, $AC=\frac{1}{2}BC=2$. Point D is a moving point on $ AB$. Connect $ CD$, rotate the line segment $ CD$ counterclockwise by $90^\circ$ around point C to obtain line segment $ CE$, and connect $ DE$ and $ BE$. When the area of $ \triangle BED$ is maximized, what is the length of $ AD$?

Solution:
\textbf{Solution:} Let's connect point A and the midpoint F of $BC$, draw $FH\perp AB$ at point H, and connect $EF$,\\
$\because$ point F is the midpoint of $BC$,\\
$\therefore$ $CF=\frac{1}{2}BC=1$,\\
$\because$ $\angle ACB=\angle DCE=90^\circ$,\\
$\therefore$ $\angle ACB-\angle DCF=\angle DCE-\angle DCF$, which means $\angle ACD=\angle FCE$,\\
In $\triangle ACD$ and $\triangle FCE$,\\
$\left\{\begin{array}{l}
AC=CF\\
\angle ACD=\angle FCE\\
DC=CE
\end{array}\right.$,\\
$\therefore$ $\triangle ACD\cong \triangle FCE(SAS)$,\\
$\therefore$ $\angle CFE=\angle CAD$,\\
$\because$ $FH\perp AB$,\\
$\therefore$ $\angle ABC+\angle BFH=\angle ABC+\angle CAD=90^\circ$,\\
$\therefore$ $\angle BFH=\angle CAD$,\\
$\therefore$ $\angle BFH=\angle CFE$,\\
$\because$ $\angle CFE+\angle BFE=180^\circ$,\\
$\therefore$ $\angle BFH+\angle BFE=180^\circ$, thus points H, F, and E are collinear,\\
Therefore, $EH$ is the height of $\triangle BED$,\\
Let $AD=x$, then $EF=x$,\\
According to the Pythagorean theorem: $AB=\sqrt{AC^2+BC^2}=\sqrt{5}$,\\
$\therefore$ $BD=\sqrt{5}-x$,\\
$\because$ $\angle BFH=\angle CAD$,\\
$\therefore$ $\triangle BFH\sim \triangle ABC$,\\
$\therefore$ $\frac{BF}{AB}=\frac{FH}{AC}$, that is $\frac{1}{\sqrt{5}}=\frac{FH}{1}$, solving gives: $FH=\frac{\sqrt{5}}{5}$,\\
$\therefore$ $EH=FH+EF=\frac{\sqrt{5}}{5}+x$,\\
$S_{\triangle BDE}=\frac{1}{2}BD\cdot EH=\frac{1}{2}\left(\sqrt{5}-x\right)\left(\frac{\sqrt{5}}{5}+x\right)=-\frac{1}{2}\left(x-\sqrt{5}\right)\left(x+\frac{\sqrt{5}}{5}\right)$,\\
$\therefore$ when the area is maximized, $x=\frac{\sqrt{5}-\frac{\sqrt{5}}{5}}{2}=\boxed{\frac{2\sqrt{5}}{5}}$.


Question: As shown in the rectangle $ABCD$, where $AB=4$. $M$ is a point on $AB$ with $AM=1$, and $N$ is a moving point on $BC$. When $\triangle BMN$ is folded along $MN$ such that point $B$ coincides with point $B'$, if the distance from point $B'$ to $AD$ is 2, then what is the length of $BN$?

Solution:
\textbf{Solution:} Construct $EF\parallel AB$ through $B'$ intersecting $AD$ at $E$ and $BC$ at $F$. Draw $MG \perp EF$ at $G$ as shown in the figure: \\
$\therefore$ Quadrilateral AMGE and quadrilateral MBFG are rectangles, \\
$\therefore$ $EG=AM=1$, $BM=GF=AB-AM=3$, \\
$\because$ The distance from point $B'$ to $AD$ is 2, \\
$\therefore$  $B'G=BE-EG=1$, \\
$\because$ Fold $\triangle BMN$ along $MN$ making point $B$ coincide with point $B'$, \\
$\therefore$ $B'M=BM=3$, \\
In $\triangle B'GM$ right-angled at $G$, $MG=\sqrt{B'M^{2}-B'G^{2}}=\sqrt{3^{2}-1^{2}}=2\sqrt{2}=BF$,\\
Let $BN=x$, then $B'N=x$,\\
$NF=2\sqrt{2}-x$ and $B'F=EF-B'E=2$,\\
In $\triangle B'NF$ right-angled at $F$, $NF^{2}+B'F^{2}=B'N^{2}$,\\
$\therefore$ $(2\sqrt{2}-x)^{2}+2^{2}=x^{2}$, solving for $x$ we get $x=\boxed{\frac{3\sqrt{2}}{2}}$.


Question: As shown in the figure, in rhombus ABCD, $ \angle BAD=120^\circ$, $AB=4$, M and N are points on sides BC and CD, respectively, and point P is a point on diagonal BD. What is the minimum value of $PM+PN$?

Solution:
\textbf{Solution:} Let point $M'$ be the reflection of point $M$ across $BD$, and connect $PM'$. Construct $CH \perp AB$ at $H$,\\
since quadrilateral $ABCD$ is a rhombus,\\
thus point $M'$ lies on $AB$\\
therefore $PM = PM'$,\\
therefore, when points $M'$, $P$, and $N$ are collinear, and $M'N \perp AB$, $PM + PN$ is minimized, and the minimum value is the length of $CH$,\\
since $\angle BAD = 120^\circ$, and $AB = 4$,\\
thus $\angle ABC = 60^\circ$, and $BC = 4$,\\
thus $BH = 2$,\\
by the Pythagorean theorem, we find $CH = 2\sqrt{3}$,\\
therefore, the minimum value of $PM + PN$ is $\boxed{2\sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B= 90^\circ$, $AB=2$, and $BC=4$, the quadrilateral $ADEC$ is a square. What is the area of the square $ADEC$?

Solution:
\textbf{Solution:} Since $\angle B= 90^\circ$, AB=2, BC=4,\\
thus $AC=\sqrt{AB^{2}+BC^{2}}$\\
$=\sqrt{2^{2}+4^{2}}$\\
$=2\sqrt{5}$\\
Therefore, the area of the square ADEC is $\boxed{20}$


Question: As shown in the figure, in the right-angled triangle $ABC$ with $ \angle ACB=90^\circ$, $AC=3$, and $BC=4$, point $M$ is a point on side $AB$ (not coinciding with points $A$ or $B$). Draw $ME\perp AC$ at point $E$ and $MF\perp BC$ at point $F$. If point $P$ is the midpoint of $EF$, what is the minimum value of $CP$?

Solution:
\textbf{Solution:} Connect $CM$, as shown in the diagram:\\
Since $\angle ACB=90^\circ$, $AC=3$, $BC=4$,\\
then $AB=\sqrt{AC^2+BC^2}=\sqrt{3^2+4^2}=5$,\\
Since $ME \perp AC$ at point $E$, $MF \perp BC$ at point $F$, $\angle ACB=90^\circ$,\\
then quadrilateral $CEMF$ is a rectangle,\\
thus $EF=CM$,\\
Since point $P$ is the midpoint of $EF$,\\
then $CP=\frac{1}{2}EF$,\\
When $CM \perp AB$, $CM$ is the shortest,\\
At this time, $EF$ is also the smallest, thus $CP$ is the smallest,\\
Since the area of $\triangle ABC =\frac{1}{2}AB\cdot CM=\frac{1}{2}AC\cdot BC$,\\
thus $CM=\frac{AC\cdot BC}{AB}=\frac{3\times 4}{5}=2.4$,\\
thus $CP=\frac{1}{2}EF=\frac{1}{2}CM=\boxed{1.2}$,


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$. If $AB=15$, what is the sum of the areas of square $ADEC$ and square $BCFG$?

Solution:
\textbf{Solution:} Let the area of square $ADEC$ be: $AC^{2}$, and the area of square $BCFG$ be: $BC^{2}$; \\
$\because$ $\angle ACB=90^\circ$, \\
$\therefore$ $\triangle ABC$ is a right triangle, \\
That is, in $\text{Rt}\triangle ABC$, $AB^{2}=AC^{2}+BC^{2}$, with $AB=15$, \\
Then $15^{2}=AC^{2}+BC^{2}=225$. \\
Therefore, the combined area of square $ADEC$ and square $BCFG$ is $\boxed{225}$.


Question: As shown in the figure, within square $ABCD$, point $E$ is located on diagonal $BD$, and $BE=AB$. Connect $CE$ and $AE$. Then, what is the measure of $\angle DAE$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a square,\\
it follows that $\angle ABD=45^\circ$, $\angle BAD=90^\circ$,\\
since BE=AB,\\
it follows that $\angle BAE=\angle BEA= \frac{1}{2} \times (180^\circ-45^\circ)=67.5^\circ$,\\
therefore, $\angle DAE=\angle BAD-\angle BAE=90^\circ-67.5^\circ=\boxed{22.5^\circ}$.


Question: As shown in the figure, within rectangle $ABCD$, points $E$ and $F$ are the midpoints of sides $AB$ and $BC$ respectively. Lines $EC$ and $FD$ are connected, and points $H$ and $G$ are the midpoints of $EC$ and $FD$ respectively. Line $GH$ is connected. If $AB = 6$ and $BC = 8$, then what is the length of $GH$?

Solution:
\textbf{Solution:} Join CH and extend it to intersect AD at P, connect PE,

$\because$ Quadrilateral ABCD is a rectangle,

$\therefore \angle A=90^\circ$, AD$\parallel$BC,

$\because$ E and F are the midpoints of AB and BC, respectively, AB$=6$, BC$=8$,

$\therefore$ AE$=\frac{1}{2}$AB$=3$, CF$=\frac{1}{2}$BC$=4$,

$\because$ AD$\parallel$BC,

$\therefore \angle DPH=\angle FCH$,

In $\triangle PDH$ and $\triangle CFH$, $\left\{\begin{array}{l}\angle DPH=\angle FCH \\ \angle DHP=\angle FHC \\ DH=FH \end{array}\right.$,

$\therefore \triangle PDH \cong \triangle CFH$ (AAS),

$\therefore$ PD$=$CF$=4$, CH$=$PH,

$\therefore$ AP$=$AD$-$PD$=4$,

$\therefore$ PE$=\sqrt{AP^{2}+AE^{2}}=\sqrt{4^{2}+3^{2}}=5$,

$\because$ Point G is the midpoint of EC,

$\therefore$ GH$=\frac{1}{2}$EP$=\boxed{\frac{5}{2}}$.


Question: As shown in the figure, point E is a point on the extension line of side BC of square ABCD, and $CE=BD$. AE intersects DC at point F. What is the measure of $\angle AFC$?

Solution:
\textbf{Solution:} Construct line AC,\\
since quadrilateral ABCD is a square,\\
thus AC=BD,\\
since CE=BD,\\
thus CE=AC,\\
thus $\angle$E=$\angle$CAE,\\
since AC is the diagonal of square ABCD,\\
thus $\angle$ACB=$45^\circ$,\\
thus $\angle$E+$\angle$CAE=$45^\circ$,\\
thus $\angle$E=$ \frac{1}{2} \times 45^\circ=22.5^\circ$,\\
In $\triangle$CEF, $\angle$AFC=$\angle$E+$\angle$ECF=$22.5^\circ+90^\circ=\boxed{112.5^\circ}$.


Question: As shown in the figure, in rectangle $ABCD$, $AD=1, AB=2\sqrt{2}$, $M$ is a moving point on line segment $BD$, $MP\perp CD$ at point $P$, $MQ\perp BC$ at point $Q$, what is the minimum value of $PQ$?

Solution:
\textbf{Solution:} Connect CM as shown in the figure,\\
$ \because MP\perp CD$ at point P, $MQ\perp BC$ at point Q,\\
$ \therefore \angle CPM=\angle CQM=90^\circ$,\\
$ \because $ quadrilateral ABCD is a rectangle,\\
$ \therefore BC=AD=1, CD=AB=2\sqrt{2}$, $ \angle BCD=90^\circ$,\\
$ \therefore $ quadrilateral PCQM is a rectangle,\\
$ \therefore PQ=CM$,\\
In $ \triangle BCD$, $BD=\sqrt{BC^{2}+CD^{2}}=\sqrt{1^{2}+(2\sqrt{2})^{2}}=3$,\\
When CM $ \perp $BD, CM is minimized, then PQ is minimized,\\
At this time, $ S_{\triangle BCD}=\frac{1}{2}BD\cdot CM=\frac{1}{2}BC\cdot CD$,\\
$ \therefore CM=\frac{BC\cdot CD}{BD}=\frac{1\times 2\sqrt{2}}{3}=\frac{2\sqrt{2}}{3}$,\\
$ \therefore PQ$'s minimum value is $ \boxed{\frac{2\sqrt{2}}{3}}$.


Question: As shown in the figure, the diagonals $AC$ and $BD$ of the quadrilateral $ABCD$ intersect at point $O$, with $AC \perp BD$. Let $E$ and $F$ be the midpoints of $AB$ and $CD$, respectively. If $AC = BD = 2$, what is the length of $EF$?

Solution:
\textbf{Solution}: Let G be the midpoint of BC, and H be the midpoint of AD. Connect EG, GF, FH, and HE. \\
$\because$ E and G are the midpoints of AB and BC respectively, $AC=2$ \\
$\therefore EG=\frac{1}{2}AC=1$, $EG\parallel AC$,\\
Similarly: $FH=\frac{1}{2}AC$, $FH\parallel AC$, $GF\parallel BD$, $GF=\frac{1}{2}BD=1$,\\
$\therefore$ GE$\parallel$FH, GF$\parallel$EH\\
$\therefore$ Quadrilateral EGFH is a parallelogram,\\
$\because AC=BD$,\\
$\therefore GE=GF$,\\
$\therefore$ parallelogram EGFH is a rhombus,\\
$\because AC\perp BD$, $EG\parallel AC$, $GF\parallel BD$,\\
$\therefore EG\perp GF$,\\
$\therefore$ rhombus EGFH is a square,\\
$\therefore EF=\sqrt{2}EG=\boxed{\sqrt{2}}$


Question: As shown in the figure, points E, F are the midpoints of the sides AB and BC of the rhombus ABCD, respectively, with $AB=2$ and $EF=\sqrt{3}$. What is the area of the rhombus ABCD?

Solution: \textbf{Solution:} Let the intersection of AC and BD be point P. Since points E and F are the midpoints of AB and BC respectively, it follows that $AC=2\cdot EF=2\sqrt{3}$, and $AP=\sqrt{3}$. In $\triangle ABP$, we have $BP=\sqrt{AB^{2}-AP^{2}}=\sqrt{2^{2}-(\sqrt{3})^{2}}=1$, and $BD=2\cdot BP=2$. Therefore, the area of rhombus ABCD is $\frac{1}{2}\cdot AC\cdot BD=\frac{1}{2}\times 2\sqrt{3}\times 2=\boxed{2\sqrt{3}}$.


Question: In the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, $AC=8$, $AB=10$. Let $D$ and $E$ be the midpoints of $AC$ and $AB$ respectively. Connect $DE$. What is the length of $DE$?

Solution:
\textbf{Solution:} Given $\triangle ABC$, $BC=\sqrt{AB^{2}-AC^{2}}=\sqrt{10^{2}-8^{2}}=6$, \\
since D and E are the midpoints of AC and AB respectively, \\
therefore $DE=\frac{1}{2}BC=\frac{1}{2}\times 6=\boxed{3}$.


Question: As shown in the figure, in square $ABCD$, $AB = 8\,cm$. Extend $BC$ to point $E$ so that $CE = 2\,cm$. Connect $DE$. A moving point $P$ starts from point $A$ and moves to the endpoint $A$ at a velocity of $2\,cm/s$, along the path $AB \rightarrow BC \rightarrow CD \rightarrow DA$. Let the movement time of point $P$ be $t$ seconds. When $\triangle PBC$ and $\triangle DCE$ are congruent, what is the value of $t$?

Solution:
\textbf{Solution:} As shown in Figure A, when $\triangle CBP_{1} \cong \triangle DCE$, we have $BP_{1} = CE = 2\text{cm}$, \\
thus $8-2t=2$. Solving for $t$ gives $t=\boxed{3}$, \\
As shown in Figure B, when $\triangle BCP_{2} \cong \triangle DCE$, we have $CP_{2} = CE=2\text{cm}$\\
thus $2t-2\times 8=2$. Solving for $t$ gives $t=9$.


Question: As shown in the diagram, it is known that points $D$ and $E$ are the midpoints of sides $AB$ and $CB$ of $\triangle ABC$, respectively. If $AB = 8$, $CE = 6$, and $AC = 10$, what is the perimeter of $\triangle BDE$?

Solution:
\textbf{Solution:} Given that points D and E are the midpoints of sides AB and CB of $\triangle ABC$ respectively,

$\therefore$ DE= $\frac{1}{2}$ AC=5, BE=CE=6, BD= $\frac{1}{2}$ AB=4,

$\therefore$ the perimeter of $\triangle BDE$=BD+BE+DE=4+6+5=\boxed{15}.


Question: As shown in the figure, D and E are respectively the midpoints of sides AB and AC of $\triangle ABC$. If $DE = \frac{3}{2}$, then what is the length of BC?

Solution:
\textbf{Solution:} Since D and E are the midpoints of sides AB and AC of $\triangle ABC$, \\
then DE is the midline of $\triangle ABC$, \\
hence, BC = 2DE = 2 $\times$ $\frac{3}{2} = \boxed{3}$.


Question: As shown in the figure, in the rhombus $ABCD$, the diagonal $AC=5$, and $BD=12$. What is the height of the rhombus?

Solution:
\textbf{Solution:} Let AC and BD intersect at point O. Through point A, construct AH $\perp$ BC at point H,\\
Since rhombus ABCD,\\
Therefore, AC $\perp$ BD, OB=$\frac{1}{2}$BD=6, OC=$\frac{1}{2}$AC=$\frac{5}{2}$,\\
Therefore, $BC=\sqrt{OB^{2}+OC^{2}}=\sqrt{6^{2}+\left(\frac{5}{2}\right)^{2}}=\frac{13}{2}$\\
Therefore, the area of rhombus ABCD, ${S}_{\mathrm{rhombus} ABCD}=\frac{1}{2}AC\cdot BD=BC\cdot AH$\\
Therefore, $\frac{1}{2}\times 5\times 12=\frac{13}{2}AH$\\
Solving gives: $AH=\boxed{\frac{60}{13}}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is a square, and the lines $a$, $b$, $c$ pass through points $A$, $D$, and $C$ respectively, with $a \parallel b \parallel c$. If the distance between $a$ and $b$ is $3$, and the distance between $b$ and $c$ is $5$, what is the area of square $ABCD$?

Solution:
\textbf{Solution:} Draw AE$\perp$b at point E through point A, and draw CF$\perp$b at point F through point C, \\
$\therefore \angle AED=\angle CFD=90^\circ$, $\angle CDF+\angle DCF=90^\circ$, $\because$ square ABCD, $\therefore \angle ADC=\angle ADE+\angle CDF=90^\circ$, AD=CD, $\therefore \angle DCF=\angle ADE$;\\
$\because a\parallel b\parallel c$. If the distance between a and b is 3, and the distance between b and c is 5, $\therefore AE=3, CF=5$;\\
In $\triangle ADE$ and $\triangle DCF$
$\left\{\begin{array}{l}
\angle AED=\angle CFD\\
\angle ADE=\angle DCF\\
AD=DC
\end{array}\right.$\\
$\therefore \triangle ADE\cong \triangle DCF$ (AAS), thus $AE=DF=3$,\\
In $\triangle DCF$, $DC^{2}=DF^{2}+CF^{2}=3^{2}+5^{2}=34$.\\
$\therefore$ The area of the square ABCD is $\boxed{34}$.


Question: In the Cartesian coordinate system, the position of square $ABCD$ is shown in the diagram, with point $A$ having coordinates $(0, 2)$ and point $B$ having coordinates $(-3, 0)$. What is the distance from point $C$ to the $y$-axis?

Solution:
\textbf{Solution:} Construct line CE perpendicular to the x-axis at point E, as shown in the diagram,\\
hence, the distance from point C to the y-axis is OE.\\
$\because$ The coordinates of point A are $(0,2)$, and the coordinates of point B are $(-3,0)$,\\
$\therefore OA=2$, $OB=3$.\\
$\because CE \perp x$-axis,\\
$\therefore \angle CEB = 90^\circ$.\\
$\therefore \angle ECB + \angle EBC = 90^\circ$.\\
$\because$ Quadrilateral ABCD is a square,\\
$\therefore BC=AB$, $\angle CBA=90^\circ$.\\
$\therefore \angle EBC + \angle ABO = 90^\circ$.\\
$\therefore \angle ECB = \angle ABO$.\\
In $\triangle CBE$ and $\triangle BAO$,\\
$\left\{\begin{array}{l}
\angle ECB=\angle OBA\\
\angle CEB=\angle BOA=90^\circ\\
CB=BA
\end{array}\right.$,\\
$\therefore \triangle CBE \cong \triangle BAO$ (AAS).\\
$\therefore EB=OA=2$.\\
$\therefore OE=OB+BE=3+2=\boxed{5}$.\\
$\therefore$ The distance from point C to the y-axis is 5.


Question: As shown in the figure, in $\triangle ABC$, $AB = 8$, $BC = 14$, points $D$ and $E$ are the midpoints of sides $AB$ and $AC$ respectively, point $F$ is on $DE$, and $\angle AFB = 90^\circ$. What is the length of $EF$?

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


Question: As shown in the figure, $O$ is the midpoint of the diagonal $BD$ of rectangle $ABCD$, and $E$ is the midpoint of side $AB$. Given that $AB = 12$ and $OE = \frac{5}{2}$, what is the length of segment $OC$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle, \\
$\therefore$$\angle$A=$\angle$BCD=$90^\circ$, DO=OB,\\
since E is the midpoint of side AB,\\
$\therefore$OE is the median of $\triangle$ADB, OE=$\frac{5}{2}$,\\
$\therefore$AD=$2$OE=$5$,\\
since AB=$12$,\\
$\therefore$BD=$\sqrt{AD^{2}+AB^{2}}=\sqrt{12^{2}+5^{2}}=13$,\\
since O is the midpoint of the diagonal BD of rectangle ABCD,\\
$\therefore$OC=$\frac{1}{2}$BD=$\boxed{\frac{13}{2}}$.


Question: As shown in the figure, the perimeter of the rhombus $ABCD$ is 20, and $ \angle A=60^\circ$. What is the length of the diagonal $BD$?

Solution:
\textbf{Solution:}

Since quadrilateral ABCD is a rhombus and its perimeter is 20,\\
it follows that AB=AD=5.\\
Since $\angle$A=60$^\circ$,\\
it follows that $\triangle$ABD is an equilateral triangle,\\
therefore, AB=AD=BD=5.\\
Hence, the answer is: BD=\boxed{5}.


Question: As shown in the diagram, in rhombus $ABCD$, point $P$ is a moving point on the diagonal $AC$, through point $P$, draw $PE \perp AB$ at point $E$, and $PF \perp BC$ at point $F$. If the perimeter of the rhombus $ABCD$ is 48 and its area is 48, what is the value of $PE+PF$?

Solution:
\textbf{Solution:} Extend FP to meet AD at point G, as shown in the figure:\\
In rhombus ABCD, AD$\parallel$BC, $\angle$DAC=$\angle$BAC,\\
Also, $\because$PF$\perp$BC,\\
$\therefore$PF$\perp$AD,\\
$\therefore$$\angle$AGP=$90^\circ$,\\
$\because$PE$\perp$AB,\\
$\therefore$$\angle$AEP=$90^\circ$,\\
$\therefore$$\angle$AEP=$\angle$AGP,\\
Also, $\because$$\angle$EAP=$\angle$GAP, AP=AP,\\
$\therefore$$\triangle $EAP$\cong $$\triangle $GAP (AAS),\\
$\therefore$GP=EP,\\
$\because$The perimeter of rhombus ABCD is 48,\\
$\therefore$BC=12,\\
$\because$The area of rhombus ABCD is 48, i.e., $BC\cdot FG$=48,\\
$\therefore$FG=48÷12=4,\\
$\therefore$PE+PF=GF=\boxed{4},


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, points $D$ and $E$ are the midpoints of $AB$ and $BC$ respectively, point $F$ is on the extension of $CA$, $\angle FDA=\angle B$, $AC=6$, $AB=8$. What is the perimeter of quadrilateral $AEDF$?

Solution:
Solution: In right triangle $ABC$, \\
because $AC=6$ and $AB=8$, \\
therefore $BC=10$, \\
because $E$ is the midpoint of $BC$, \\
therefore $AE=BE=5$, \\
therefore $\angle BAE=\angle B$, \\
because $\angle FDA=\angle B$, \\
therefore $\angle FDA=\angle BAE$, \\
therefore $DF\parallel AE$, \\
because $D$ and $E$ are the midpoints of $AB$ and $BC$ respectively, \\
therefore $DE\parallel AC$, and $DE= \frac{1}{2}AC=3$ \\
therefore the quadrilateral $AEDF$ is a parallelogram \\
therefore the perimeter of the quadrilateral $AEDF$ is $2\times(3+5)=\boxed{16}$.


Question: As shown in the diagram, in $\square ABCD$, $AB=13$, $AD=5$, and $AC\perp BC$, what is the length of $BD$?

Solution:
\textbf{Solution:} As shown in the diagram, draw DE $\perp$ BC, intersecting the extended line of BC at E,\\
since quadrilateral ABCD is a parallelogram,\\
therefore BC=AD=5, AD$\parallel$BC,\\
since $AC\perp BC$,\\
therefore $\angle ACB=90^\circ$,\\
by the Pythagorean theorem, we have $AC= \sqrt{AB^{2}-BC^{2}}=\sqrt{13^{2}-5^{2}}=12$,\\
since DE$\perp$BC,\\
therefore AC$\parallel$DE,\\
therefore quadrilateral ACED is a parallelogram,\\
therefore CE=AD=5, DE=AC=12,\\
therefore BE=BC+CE=10,\\
since DE$\perp$BC,\\
therefore, by the Pythagorean theorem, we have $BD= \sqrt{BE^{2}+DE^{2}}=\sqrt{12^{2}+10^{2}}=\boxed{2\sqrt{61}}$.


Question: As shown in the diagram, the perimeter of parallelogram $ABCD$ is $20$. The diagonals $AC$ and $BD$ intersect at point $O$. Point $E$ is the midpoint of $CD$, and $BD = 6$. What is the perimeter of $\triangle DOE$?

Solution:
\textbf{Solution:} Given the perimeter of quadrilateral ABCD is 20,\\
then $2(BC+CD)=20$,\\
thus $BC+CD=10$.\\
Given ABCD is a parallelogram, and the diagonals AC, BD intersect at point O, with BD = 6,\\
thus $OD=OB=\frac{1}{2}BD=3$.\\
Given point E is the midpoint of CD, and point O is the midpoint of BD,\\
thus OE is the midsegment of $\triangle BCD$ and $DE=\frac{1}{2}CD$,\\
hence $OE=\frac{1}{2}BC$,\\
therefore, the perimeter of $\triangle DOE = OD + OE + DE = \frac{1}{2}BD + \frac{1}{2}(BC + CD) = 5 + 3 = \boxed{8}$.


Question: As shown in the figure, a straight line intersects the positive semi-axes of two coordinate axes at points $A$ and $B$, respectively. Let $P$ be any point on the line segment $AB$ (excluding the endpoints). Perpendicular lines are drawn from point $P$ to the two coordinate axes, forming a rectangle with a perimeter of 4. What is the minimum value of $OP$?

Solution:
\textbf{Solution:} Let the coordinates of point P be $(x,y)$,\\
then: $2(x+y)=4$,\\
after rearranging: $y=-x+2$,\\
since P is any point on the line segment AB,\\
therefore the equation of the line containing AB is: $y=-x+2$,\\
when $x=0$, $y=2$,\\
when $y=0$, $x=2$,\\
therefore, OA=OB=2,\\
thus, in the $\triangle OAB$, according to the Pythagorean theorem: $AB=\sqrt{OA^2+OB^2}=\sqrt{2^2+2^2}=2\sqrt{2}$,\\
therefore, when OP $\perp$ AB, OP is the shortest,\\
at this time, $S_{\triangle OAB}=\frac{1}{2}OA\cdot OB=\frac{1}{2}AB\cdot OP$,\\
which gives: $4=2\sqrt{2}OP$,\\
solving this gives: $OP=\boxed{\sqrt{2}}$,\\
therefore, the minimum value of the line segment $OP$ is: $\sqrt{2}$.


Question: As shown in the figure, in the Cartesian coordinate system $xoy$, if the coordinates of points A and B of the rhombus $ABCD$ are $(-3, 0)$ and $(2, 0)$ respectively, and point D is on the positive direction of the y-axis, then what are the coordinates of point C?

Solution:
\textbf{Solution:} Given that in the rhombus ABCD, the coordinates of vertices A and B are respectively (-3, 0) and (2, 0), and point D is on the y-axis,\\
$\therefore$AB=AO+OB=5,\\
$\therefore$AD=AB=CD=5,\\
$\therefore$DO=$\sqrt{AD^{2}-AO^{2}}=\sqrt{5^{2}-3^{2}}=4$,\\
$\therefore$the coordinates of point C are: $\boxed{(5, 4)}$.


Question: As shown in the diagram, within the square $ABCD$, where $AB = 8$, the diagonals $AC$ and $BD$ intersect at point $O$, and $N$ is the midpoint of $AO$. The point $M$ is located on side $BC$, with $BM = 6$. Let $P$ be a point on the diagonal $BD$. What is the maximum value of $PM - PN$?

Solution:
\textbf{Solution:} As shown in the diagram, take point N' as the symmetric point of N with BD as the axis of symmetry. Extend and connect MN' to intersect BD at P, and connect NP.\\
Based on the property of axis symmetry, it is known that $PN = PN'$,\\
$\therefore PM - PN = PM - PN' \leq MN'$,\\
When points P, M, and N' are collinear, take “=” in the equation,\\
$\because$ The side length of the square is 8,\\
$\therefore AC = \sqrt{2}AB = 8\sqrt{2}$,\\
$\because$ O is the midpoint of AC,\\
$\therefore AO = OC = 4\sqrt{2}$,\\
$\because$ N is the midpoint of OA,\\
$\therefore ON = 2\sqrt{2}$,\\
$\therefore O{N'} = C{N'} = 2\sqrt{2}$,\\
$\therefore A{N'} = 6\sqrt{2}$,\\
$\because BM = 6$,\\
$\therefore CM = AB - BM = 8 - 6 = 2$,\\
$\therefore \frac{CM}{BM} = \frac{C{N}^{'}}{A{N}^{'}} = \frac{1}{3}$,\\
$\therefore PM\parallel AB\parallel CD,\ \angle CMN' = 90^\circ$,\\
$\because \angle N'CM = 45^\circ$,\\
$\therefore \triangle N'CM$ is an isosceles right triangle,\\
$\therefore CM = MN' = 2$,\\
thus, the maximum value of $PM - PN$ is $\boxed{2}$,\\


Question: As shown in the figure, in square ABCD, AB=2, E is the midpoint of CD, and $\triangle BCE$ is folded along BE to $\triangle BFE$. G is the midpoint of BE. Connect FG. What is the length of FG?

Solution:
\textbf{Solution:}
Since $\triangle BCE$ is folded along $BE$ to form $\triangle BFE$,

thus $\angle C=\angle BFE=90^\circ$,

therefore $\triangle BFE$ is a right-angled triangle.

Hence, $FG=\frac{1}{2}BE=\frac{1}{2}\sqrt{1^2+2^2}=\boxed{\frac{\sqrt{5}}{2}}$.


Question: As shown in the figure, the two diagonals of rectangle $ABCD$ intersect at point $O$. If $AD=\sqrt{3}$ and $CD=1$, then what is the degree measure of $\angle ADB$?

Solution:
\textbf{Solution:} Given a rectangle ABCD,\\
it follows that $\angle BCD=90^\circ$, AD$\parallel$BC, and AD=BC.\\
Hence, $\angle ADB=\angle DBC$.\\
Thus, $BD=\sqrt{BC^{2}+CD^{2}}=\sqrt{(\sqrt{3})^{2}+1^{2}}=2$.\\
This means BD=2CD,\\
and so $\angle ADB=\angle DBC=30^\circ$.\\
Therefore, the answer is: $\boxed{30^\circ}$.


Question: As shown in the figure, the side length of rhombus $ABCD$ is 2, and $\angle ADC=120^\circ$. Point $E$ is the midpoint of side $AB$, and point $P$ is a moving point on the diagonal $AC$. What is the minimum value of $PB+PE$?

Solution:
\textbf{Solution:} Connect BD, DE, and DE intersects AC at point P$^{\prime}$. Connect BP$^{\prime}$,

$\because$ rhombus ABCD,

$\therefore$ point B and point D are symmetrical about AC,

$\therefore$ DP$^{\prime}$=BP$^{\prime}$,

$\therefore$ P$^{\prime}$B + P$^{\prime}$E = DP$^{\prime}$ + P$^{\prime}$E = DE,

$\because$ the line segment between two points is the shortest,

$\therefore$ the minimum value of P$^{\prime}$B + P$^{\prime}$E is equal to the length of DE;

$\because$ rhombus ABCD,

$\therefore$ DC$\parallel$AB, AD=AB,

$\therefore$ $\angle$ADC + $\angle$DAB = 180$^\circ$,

$\therefore$ $\angle$DAB = 60$^\circ$,

$\therefore$ $\triangle$ADB is an equilateral triangle,

$\because$ point E is the midpoint of AB,

$\therefore$ DE $\perp$ AB, BE=$\frac{1}{2}$AB=1, BD=AB=2,

In $\triangle$BDE, $BE=\sqrt{BD^{2}-BE^{2}}=\sqrt{2^{2}-1^{2}}=\sqrt{3}$,

$\therefore$ the minimum value of PB+PE is $\boxed{\sqrt{3}}$.


Question: As shown in the figure, in the rhombus $ABCD$, point $B$ is on the $x$-axis, and the coordinates of point $C$ are $(6, 2)$, and the coordinates of point $A$ are $(0, 2)$. What are the coordinates of point $D$?

Solution:
\textbf{Solution:} Connect AC, BD, with the intersection point as E, \\
$\because$ Quadrilateral ABCD is a rhombus, \\
$\therefore$ AC$\perp$BD, 2AE=AC, 2BE=BD, quadrilateral AEBO is a rectangle, \\
$\therefore$ OA=BE, AE=OB, \\
$\because$ the coordinates of point C are (6, 2), and the coordinates of point A are (0, 2), \\
$\therefore$ AC$\parallel$x-axis, \\
$\therefore$ OA=2, AC=6, \\
$\therefore$ BE=DE=OA=2, OB=3, \\
$\therefore$ BD=2DE=4, \\
$\therefore$ the coordinates of point D are: $\boxed{(3, 4)}$


Question: As shown in the figure, fold the rectangular paper ABCD so that edge DC falls on the diagonal line AC, with the fold line being CE, and point D falling on point F on the diagonal line. If AB=6 and AD=8, what is the length of ED?

Solution:
Solution: In rectangle $ABCD$, $AB=6$, $AD=8$, \\
$\therefore$ $DC=6$, \\
$\therefore$ $AC=\sqrt{AD^{2}+DC^{2}}=10$, \\
From the folding, it follows that $\triangle DEC \cong \triangle FEC$, \\
$\therefore$ $FC=DC=6$, $DE=FE$, \\
Let $ED=x$, then $FE=x$, $AF=AC−CF=4$, $AE=8−x$, \\
In $\triangle AEF$: $AF^{2}+EF^{2}=AE^{2}$, \\
$4^{2}+x^{2}=(8−x)^{2}$, \\
Solving this yields: $x=\boxed{3}$.


Question: The squares $A_1B_1C_1O$, $A_2B_2C_2C_1$, $A_3B_3C_3C_2$, … are placed as shown in the figure, where points $A_1, A_2, A_3, $… and points $C_1, C_2, C_3, $… are on the line $y=x+1$ and the x-axis, respectively. What is the y-coordinate of point $B_{2022}$?

Solution:
\textbf{Solution:} Since point $A_{1}$ is the intersection point of the line $y=x+1$ and the y-axis,\\
$\therefore A_{1}(0,1)$,\\
because the quadrilateral $A_{1}B_{1}C_{1}O$ is a square,\\
$\therefore B_{1}(1,1)$,\\
because point $A_{2}$ is on the line $y=x+1$,\\
$\therefore A_{2}(1,2)$,\\
Similarly, we can derive that $A_{3}(3,4)$, $B_{2}(3,2)$, $B_{3}(7,4)$,\\
$\therefore$ the length of the sides of the first three squares $=1+2+4=7$,\\
$\therefore A_{4}(7,8)$,\\
because $B_{1}(1,1)$, $B_{2}(3,2)$, $B_{3}(7,4)$,\\
$\therefore$ the y-coordinate of $B_{n}$ is $2^{n-1}$,\\
when $n=2022$, $2^{n-1}=2^{2022-1}=2^{2021}$.\\
Thus, the answer is: the y-coordinate of $B_{2022}$ is $\boxed{2^{2021}}$.


Question: As shown in the figure, in the square $ABCD$, $AB=6$, point $Q$ is a moving point on side $AB$ (point $Q$ does not coincide with point $B$), and points $M$, $N$ are the midpoints of $DQ$ and $BQ$, respectively. What is the length of segment $MN$?

Solution:
\textbf{Solution:} As shown in the diagram, connect BD,\\
$\because$ M and N are the midpoints of BQ and DQ, respectively,\\
$\therefore$ MN is the midline of $\triangle QBD$,\\
$\because$ MN=$\frac{1}{2}$BD,\\
$\because$ Quadrilateral ABCD is a square,\\
$\therefore$ $\angle$A=$90^\circ$, AB=AD=6,\\
$\therefore$ BD=$\sqrt{AB^{2}+AD^{2}}=6\sqrt{2}$,\\
$\therefore$ MN= $\boxed{3\sqrt{2}}$.


Question: As shown in the figure, which consists of two right-angled triangles and three squares, what is the area of the shaded part?

Solution:
\textbf{Solution:} Since $\angle ACB = 90^\circ$, \\
it follows that $BC^{2} = AB^{2} - AC^{2} = 13^{2} - 12^{2} = 25$, \\
therefore, the area of the shaded region $S_{\text{shaded}} = BC^{2} = \boxed{25}$.


Question: In $\triangle ABC$, it is given that $AC=BC=4$ and $\angle ACB=120^\circ$. Point $M$ lies on side $BC$ with $BM=1$. Point $N$ is a moving point on line $AC$, and point $P$ is a moving point on side $AB$. What is the minimum value of $PM+PN$?

Solution:
\textbf{Solution:} Construct the symmetric point $C'$ of $C$ with respect to $AB$, and connect $AC'$ and $BC'$. Let $AN'=AN$ and connect $PN'$,
\[
\therefore CA=C'A=CB=BC',
\]
\[
\therefore \text{Quadrilateral } ACBC' \text{ is a rhombus},
\]
\[
\therefore PN=PN',
\]
\[
\therefore PM+PN=PM+PN',
\]
\[
\therefore \text{When } M, P, \text{ and } N' \text{ are collinear, and } MN' \perp AC', \text{ then } PM+PN \text{ is minimized},
\]
Draw line $C'H \perp BC$ at $H$ through point $C'$,
\[
\because \angle ACB=120^\circ,
\]
\[
\therefore \angle C'BH=60^\circ,
\]
\[
\therefore C'H=\frac{\sqrt{3}}{2}BC'=2\sqrt{3},
\]
\[
\therefore \text{The minimum value of } PM+PN \text{ is the distance between } BC \text{ and } AC', \text{ which is } C'H=\boxed{2\sqrt{3}}\].


Question: As shown in the figure, in rectangle $ABCD$, the length of side $AB$ is 3. Points $E$ and $F$ are located on $AD$ and $BC$ respectively. Connect $BE$, $DF$, $EF$, and $BD$. If quadrilateral $BEDF$ is a rhombus, and $EF=AE+FC$, then what is the length of side $BC$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\
$\therefore$ $\angle A=90^\circ$, AD＝BC,\\
$\therefore$ BA$\perp$BF,\\
Given that quadrilateral BEDF is a rhombus,\\
$\therefore$ EF$\perp$BD, BE＝BF＝DE＝DF,\\
$\therefore$ $\triangle$ BEF is an isosceles triangle, AE＝CF,\\
$\therefore$ $\angle$ EBO＝$\angle$ DBF,\\
Given that EF＝AE+FC, AE＝CF, EO＝FO,\\
$\therefore$ EF＝2AE＝2EO,\\
$\therefore$ AE＝EO＝CF＝FO,\\
$\therefore$ From knowledge that angles form an axisymmetric figure, AB＝BO＝3, $\angle$ ABE＝$\angle$ EBO,\\
$\therefore$ $\angle$ ABE＝$\angle$ EBD＝$\angle$ DBC＝$30^\circ$,\\
Let OE＝x, then BE＝2x,\\
According to the Pythagorean theorem, $BE^2-OE^2=OB^2$,\\
$4x^2-x^2=3^2$,\\
Solving, we get x＝$\sqrt{3}$,\\
$\therefore$ BE＝2$\sqrt{3}$,\\
$\therefore$ BF＝BE＝2$\sqrt{3}$,\\
$\therefore$ CF＝AE＝$\sqrt{3}$,\\
$\therefore$ BC＝BF+CF＝$3\sqrt{3}$,\\
Therefore, the answer is: BC＝\boxed{3\sqrt{3}}.