Question: As shown in the figure, point $C$ is on the circle $\odot O$, point $P$ is on the extension of the diameter $AB$ of $\odot O$. Connect $AC$, $OC$, $BC$, $CP$, and it is given that $BO=BC=BP$. $CP$ is the tangent to the circle $\odot O$; If $OP=4$, find the length of $AC$.

Solution:
\textbf{Solution:} Since $OP=4$, and $BO=BC=BP$.\\
Therefore, $OB=OC=OA=BC=BP=2$, and $AB=4$,\\
Since $AB$ is the diameter,\\
Therefore, $\angle ACB=90^\circ$,\\
Thus, $AC=\sqrt{AB^{2}-BC^{2}}=\sqrt{4^{2}-2^{2}}=\boxed{2\sqrt{3}}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, point $C$ is a point on $\odot O$, the bisector of $\angle CAB$ intersects $\overset{\frown}{BC}$ at point $D$, a line is drawn through point $D$ such that $DE\parallel BC$ and it intersects the extension of $AC$ at point $E$. $DE$ is a tangent line of $\odot O$; if $DE=2$ and $CE=1$, find the length of $BD$.

Solution:
\textbf{Solution}: Connect CD,\\
$\because$ $\angle OAD = \angle CAD$,\\
$\therefore$ $\overset{\frown}{CD} = \overset{\frown}{BD}$,\\
$\therefore$ CD = BC,\\
In $\triangle CED$, $\angle E = 90^\circ$, DE = 2, CE = 1,\\
$\therefore$ $CD = \sqrt{DE^2 + CE^2} = \sqrt{2^2 + 1^2} = \sqrt{5}$,\\
$\therefore$ BD = $\boxed{\sqrt{5}}$.


Question: As shown in the figure, AB is the diameter of semicircle O, D is the midpoint of BC, extending OD intersects the arc $BC$ at point E, and point F is a point on the extension of OD such that $\angle B = \angle F$. CF is a tangent to the circle $O$; if AB = 4, $\angle B = 30^\circ$, connect AD, find the length of AD.

Solution:
\textbf{Solution:} Join AD,\\
$\because$ AB is the diameter of semicircle $O$,\\
$\therefore \angle ACB=90^\circ$.\\
$\because$ AB$=4$, and $\angle B=30^\circ$,\\
$\therefore$ AC$=2$, and $BC=\sqrt{4^2-2^2}=2\sqrt{3}$,\\
$\therefore$ CD$=\sqrt{3}$,\\
In $\triangle ACD$, $AD=\sqrt{2^2+(\sqrt{3})^2}=\boxed{\sqrt{7}}$.


Question: As shown in the figure, with AB as the diameter, construct $\odot O$. Take a point C on $\odot O$, extend AB to point D, and connect DC, where $\angle DCB=\angle DAC$. Through point A, draw $AE\perp AD$ to meet the extension of DC at point E. CD is the tangent to $\odot O$; if $CD=4$ and $DB=2$, find the length of AE.

Solution:
\textbf{Solution:} Since $\angle DCO = 90^\circ$ and $OC = OB$, it follows that $OC^2 + CD^2 = OD^2$, which implies $OB^2 + 4^2 = (OB + 2)^2$. Hence, $OB = 3$, leading to $AB = 6$ and $AD = 8$. \\
Since $AE \perp AD$ and $AB$ is the diameter of the circle $O$, it follows that $AE$ is the tangent of circle $O$. \\
Since $CD$ is the tangent of circle $O$, it implies $AE = CE$. In $\triangle ADE$, we have $AD^2 + AE^2 = DE^2$, so $8^2 + AE^2 = (4+AE)^2$, resulting in $AE = \boxed{6}$.


Question: As shown in the figure, it is known that $AB$ is the diameter of $\odot O$, and $BD$ is the tangent of $\odot O$ with the point of tangency at $B$. Through the point $C$ on $\odot O$, draw $CD\parallel AB$, intersecting $BD$ at point $D$. Connect $AC$ and $BC$. If $DC$ is the tangent of $\odot O$ with the point of tangency at $C$. What are the measures of $\angle BCD$ and $\angle DBC$?

Solution:
\textbf{Solution:} Given that AB is the diameter of circle O, and BD is a tangent to circle O at point B,\\
\[\therefore \text{DB} \perp \text{AB},\]
\[\therefore \angle \text{DBA} = 90^\circ,\]
Given that DC is a tangent to circle O at point C,\\
\[\therefore \text{DC} = \text{DB},\]
Given that CD is parallel to AB,\\
\[\therefore \angle D + \angle \text{DBA} = 180^\circ,\]
\[\therefore \angle D = 90^\circ,\]
\[\therefore \angle \text{BCD} = \angle \text{DBC} = \boxed{45^\circ};\]


Question: Given that $AB$ is the diameter of $\odot O$, and $BD$ is a tangent to $\odot O$ at point $B$. Through point $C$ on $\odot O$, construct $CD\parallel AB$, intersecting $BD$ at point $D$. Connect $AC$ and $BC$. As shown in Figure (2), when $CD$ intersects $\odot O$ at point $E$, connect $BE$. If $\angle EBD=30^\circ$, what are the measures of $\angle BCD$ and $\angle DBC$?

Solution:
\textbf{Solution:} Given that,\\
$\because$ AB is the diameter of $\odot$O, and DB is the tangent of $\odot$O touching at B,\\
$\therefore$ DB$\perp$AB,\\
$\therefore$ $\angle$DBA$=90^\circ$,\\
$\angle$DEB$=\angle$EBA,\\
$\therefore$ $\angle$BDC$=90^\circ$,\\
$\because$ $\angle$EBD$=30^\circ$,\\
$\therefore$ $\angle$DEB$=60^\circ$,\\
$\therefore$ $\angle$EBA$=60^\circ$,\\
$\therefore$ $\angle$ACE$=120^\circ$,\\
$\because$ AB is the diameter of $\odot$O,\\
$\therefore$ $\angle$BCA$=90^\circ$,\\
$\therefore$ $\angle$BCD$=\boxed{30^\circ}$,\\
$\therefore$ $\angle$DBC$=\boxed{60^\circ}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, and AC intersects $\odot O$ at point C. The bisector of $\angle BAC$ intersects $\odot O$ at point D, DE $\perp$ AC at point E. DE is the tangent to $\odot O$; if the diameter AB $= 10$, and the chord AC $= 6$, what is the length of DE?

Solution:
\textbf{Solution:} Draw OF$\perp$AC, with F being the foot of the perpendicular.\\
$\therefore AF=\frac{1}{2}AC=3$,\\
In $\triangle AFO$, $AF^{2}+OF^{2}=AO^{2}$, $AO=\frac{1}{2}AB=5$,\\
$\therefore 3^{2}+OF^{2}=5^{2}$,\\
$\therefore OF=4$,\\
$\because \angle AED=\angle ODE=\angle OFE=90^\circ$,\\
$\therefore$ Quadrilateral ODEF is a rectangle,\\
$\therefore DE=OF=\boxed{4}$.


Question: As shown in the figure, it is known that the side length of square $ABCD$ is $6\,cm$, and $E$ is a point on side $AB$ with $AE$ being $1\,cm$ long. A moving point $P$ starts from point $B$ and moves along the direction of ray $BC$ at a speed of $1\,cm$ per second. $\triangle EBP$ is folded along $EP$, with point $B$ landing at point ${B}^{\prime}$, and the movement time is $t$ seconds. When $t$ equals what value, is $\angle {B}^{\prime}PC$ a right angle?

Solution: \textbf{Solution:} Given that the problem information is insufficient, it is impossible to provide specific solution steps. Please provide a complete problem description for an accurate mathematical solution.


Question: As shown in the figure, it is known that the side length of square $ABCD$ is $6\,cm$, and $E$ is a point on side $AB$, with the length of $AE$ being $1\,cm$. A moving point $P$ starts from point $B$ and moves in the direction of ray $BC$ at a speed of $1\,cm$ per second. Triangle $EBP$ is folded along $EP$, and point $B$ falls at point ${B}^{\prime}$. Let the moving time be $t$ seconds. If the distance from point ${B}^{\prime}$ to line $AD$ is $3\,cm$, what is the length of $BP$ in $cm$?

Solution: \textbf{Solution:} We have the length of $BP$ as $\boxed{\frac{5}{7}\sqrt{21}}$ or $\boxed{15cm}$.


Question: In right-angled triangle $ABC$, with $\angle BAC=90^\circ$, E and F are the midpoints of BC and AC, respectively. Extend BA to point D such that $AB = 2AD$. Connect DE, DF, AE, EF, and the intersection of AF and DE is point O. If $AB=8$ and $BC=12$, what is the length of DE?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $\angle BAC=90^\circ$, $AB=8$, $BC=12$,\\
therefore, by the Pythagorean theorem, $AC=\sqrt{BC^2-AB^2}=\sqrt{12^2-8^2}=4\sqrt{5}$\\
Additionally, given that $OA=OF$ and $AF=CF$,\\
thus $OA=\frac{1}{2}AF=\frac{1}{4}AC=\sqrt{5}$,\\
therefore, in $\triangle AOD$, $\angle DAO=90^\circ$, $AD=\frac{1}{2}AB=4$, $OA=\sqrt{5}$,\\
thus by the Pythagorean theorem, $DO=\sqrt{DA^2+OA^2}=\sqrt{4^2+(\sqrt{5})^2}=\sqrt{21}$,\\
hence $DE=2DO=2\sqrt{21}=\boxed{2\sqrt{21}}$.


Question: As illustrated in the diagram, in $\triangle ABC$, $\angle BAC=90^\circ$ and $\angle B=45^\circ$. A line is drawn from point A such that $AD \parallel BC$, with point D being to the right of point A. Points P and Q are respectively on rays $AD$ and $CB$. Point E is a point on segment $CQ$ and $CQ=2AP$. Let $AP=x$ and $CE=y$, it follows that $y=2x-2$. When point Q is the midpoint of $BC$, $y=3$. What are the lengths of $QE$ and $BC$?

Solution:
\textbf{Solution}:
As shown in the figure below,\\
Given that $y=3$, that is $3=2x-2$,\\
we find $x=\frac{5}{2}$, thus $AP=\frac{5}{2}$,\\
$\therefore CE=y=2x-2=2\times \frac{5}{2}-2=3$, $CQ=2AP=2\times \frac{5}{2}=5$,\\
$\therefore QE=CQ-CE=5-3=\boxed{2}$,\\
$\because$ point Q is the midpoint of BC,\\
$\therefore BC=2CQ=2\times 5=\boxed{10}$;


Question: As shown in the diagram, in $\triangle ABC$, $\angle BAC=90^\circ$ and $\angle B=45^\circ$. A line is drawn through point A such that $AD \parallel BC$, with point D to the right of point A. Points P and Q are located on ray $AD$ and ray $CB$, respectively. Point E is a point on line segment $CQ$, and $CQ=2AP$. Let $AP=x$ and $CE=y$, then $y=2x-2$. When point Q is the midpoint of $BC$, $y=3$. If $PE \perp BC$, find the length of $BQ$.

Solution:
\textbf{Solution}: As shown in the diagram, draw $AM\perp BC$ at point M, PE intersects AC at point N,\\
$\because$ $\angle BAC=90^\circ$, $\angle B=45^\circ$,\\
$\therefore$ $\angle C=180^\circ-\angle BAC-\angle B=45^\circ$,\\
$\therefore$ $\angle B=\angle C$, $BM=MC=\frac{1}{2}BC=\frac{1}{2}\times 10=5$,\\
$\because$ $AM\perp BC$, $PE\perp BC$,\\
$\therefore$ $AM\parallel PE$,\\
Also, $\because$ $AD\parallel BC$,\\
$\therefore$ Quadrilateral AMEP is a parallelogram,\\
$\therefore$ $ME=AP=x$,\\
$\because$ $CE=y=2x-2$,\\
From $MC=CE+ME$, we get, $5=2x-2+x$,\\
Solving yields $x=\frac{7}{3}$, that is $AP=\frac{7}{3}$,\\
$\therefore$ $CQ=2AP=2\times \frac{7}{3}=\frac{14}{3}$,\\
$\therefore$ $BQ=BC-CQ=10-\frac{14}{3}=\boxed{\frac{16}{3}}$;


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, $\angle B=45^\circ$. A line through point A such that $AD\parallel BC$ is drawn, and point D is to the right of point A. Points P and Q are located on the rays $AD$ and $CB$ respectively. Point E is a point on the line segment $CQ$, and $CQ=2AP$. Let $AP=x$, $CE=y$, then $y=2x-2$. When point Q is the midpoint of $BC$, $y=3$. There exists a value of x such that the quadrilateral with vertices A, B, E, and P is a parallelogram. Find the value of x.

Solution:
\textbf{Solution:} Existence is proven as follows:

(1) As shown in the figure below, when points Q, E are on the line segment BC, if the quadrilateral formed by vertices A, B, E, P is a parallelogram, then $AP=BE=x$,\\
since $CE=y=2x-2$,\\
thus $BE=BC-CE=10-(2x-2)=12-2x$,\\
therefore $x=12-2x$,\\
solving this gives $x=\boxed{4}$;\\
(2) As shown in the figure below, when points Q, E are on the extension of the line segment CB, if the quadrilateral formed by vertices A, B, E, P is a parallelogram, then $AP=BE=x$,\\
since $CE=y=2x-2$,\\
thus $BE=CE-BC=2x-2-10=2x-12$,\\
therefore $x=2x-12$,\\
solving this gives $x=\boxed{12}$.

In conclusion, when the quadrilateral formed by vertices A, B, E, P is a parallelogram, $x=\boxed{4}$ or \boxed{12}.


Question: As shown in the figure, in square $ABCD$, point $E$ is on side $BC$, and point $F$ is on side $AD$, with $AF=CE$, and $EF$ intersects the diagonal $BD$ at point $O$. $O$ is the midpoint of $BD$. If $EF\perp BD$ and the perimeter of square $ABCD$ is 24, upon connecting $BF$, what is the perimeter of $\triangle ABF$?

Solution: \textbf{Solution:} Since there's insufficient information provided by the question, specific steps and results cannot be provided. Please provide the complete question information for an accurate solution.


Question: As shown in the figure, in the square $ABCD$, $A(-10,0)$, $B(-2,0)$, $C(0,4)$. Points $P$, $Q$ are respectively on the line segments $OA$, $CD$, with $OP=2DQ$. Connect $PQ$, $PC$, and let $DQ=x$. For what value of $x$ is $PQ \parallel AD$?

Solution:
\textbf{Solution:} Given, $A(-10,0)$, $B(-2,0)$, $C(0,4)$,\\
thus $OA=10$, $OB=2$, $OC=4$,\\
thus $AB=OA-OB=10-2=8$,\\
since quadrilateral $ABCD$ is a parallelogram,\\
thus $CD=AB=8$, $AB \parallel CD$,\\
since $OP=2DQ$, $DQ=x$,\\
thus $OP=2x$,\\
thus $AP=AO-OP=10-2x$,\\
since $PQ \parallel AD$, $AP \parallel DQ$,\\
thus quadrilateral $APQD$ is a parallelogram,\\
thus $AP=DQ$,\\
thus $10-2x=x$,\\
thus $x=\boxed{\frac{10}{3}}$;


Question: As shown in the figure, in $\square ABCD$, $A(-10,0)$, $B(-2,0)$, $C(0,4)$. Points $P$ and $Q$ are on the line segments $OA$ and $CD$, respectively, with $OP=2DQ$. Connect $PQ$ and $PC$, and let $DQ=x$. There exists an $x$ such that the distance from point $Q$ to $PC$ is $4$. What is the value of $x$ obtained?

Solution:
\textbf{Solution:} There exists $x$ such that the distance from Q to PC is 4, for the following reasons: Draw line QE$\perp$PC at point E, then QE=4, $\angle$CEQ=$\angle$POC=$90^\circ$, \\
$\because$OC=4, \\
$\therefore$QE=OC, \\
$\because$AB$\parallel$CD, \\
$\therefore$$\angle$QCE=$\angle$CPO, \\
In $\triangle$QCE and $\triangle$CPO, \\
$\left\{\begin{array}{l}
\angle QCE=\angle CPO \\
\angle CEQ=\angle POC \\
QE=OC
\end{array}\right.$, \\
$\therefore$$\triangle$QCE$\cong$$\triangle$CPO, \\
$\therefore$CQ=PC, \\
$\therefore$CQ$^2$=PC$^2$, \\
$\because$CQ=CD−DQ=8−x, PC$^2$=OC$^2$+OP$^2$=$4^2$+$(2x)^2$, \\
$\therefore$(8−x)$^2$=$4^2$+$(2x)^2$, \\
Solving this, we find: $x=\frac{4\sqrt{13}−8}{3}$ or $x=\frac{−4\sqrt{13}−8}{3}<0$ (discard the negative solution), \\
$\therefore x=\boxed{\frac{4\sqrt{13}−8}{3}}$;


Question: As shown in the diagram, in quadrilateral $ABCD$, we have $A(-10, 0)$, $B(-2, 0)$, and $C(0, 4)$. Points $P$ and $Q$ are located on line segments $OA$ and $CD$, respectively, with $OP=2DQ$. By connecting $PQ$ and $PC$, and denoting $DQ$ as $x$, construct the symmetric point $Q'$ of point $Q$ with respect to line $PC$. When $Q'$ lies on an axis, determine the value of $x$ that satisfies this condition.

Solution:
\textbf{Solution:} When point $Q'$ is on the y-axis, as shown in the diagram:\\
\[
\because \text{Point Q and point $Q'$ are symmetric with respect to PC,}
\]
\[
\therefore QQ' \perp PC, \text{ CQ=C$Q'$},
\]
\[
\therefore \text{PC bisects } \angle DCO,
\]
\[
\therefore \angle PCO=45^\circ,
\]
\[
\because \text{PO}\perp\text{CQ},
\]
\[
\therefore \text{PO=CO},
\]
\[
\therefore 2x=4,
\]
\[
\therefore x=\boxed{2};
\]
When point $Q'$ is on the x-axis, as shown in the diagram:\\
Connect C$Q'$, C$Q'$=CQ=8−x,\\
\[
\because \text{DQ=x, CD=8, OP=2DQ},
\]
\[
\therefore \text{CQ=CD−DQ}=8−x, \text{OP=2x},
\]
\[
\because \text{Point Q's symmetric point with respect to line PC is $Q'$},
\]
\[
\therefore \text{PC}\perp QQ', \text{ QE=$Q'E$}, \text{C$Q'$=CQ=8−x},
\]
\[
\because \text{AB}\parallel\text{CD},
\]
\[
\therefore \angle QCE=\angle Q'PE,
\]
In $\triangle QCE$ and $\triangle Q'PE$,
\[
\left\{ \begin{array}{l}
\angle QCE=\angle Q'PE \\
\angle CEQ=\angle Q'EP \\
QE=Q'E
\end{array} \right.,
\]
\[
\therefore \triangle QCE \cong \triangle Q'PE,
\]
\[
\therefore \text{CQ=PQ'}=8−x,
\]
\[
\therefore \text{OQ'}=\left|OP-PQ'\right|=\left|3x−8\right|,
\]
In $\triangle OCQ'$,
\[
OQ'^{2}+OC^{2}=CQ'^{2},
\]
\[
\text{i.e., } \left(3x−8\right)^{2}+4^{2}=\left(8−x\right)^{2},
\]
Solving this, we find $x=2+\sqrt{2}$ or $x=2-\sqrt{2}$,\\
In conclusion, when $Q'$ lies on the coordinate axes, the values of $x$ that satisfy the conditions are \boxed{2} or \boxed{2+\sqrt{2}} or \boxed{2-\sqrt{2}}.


Question: As shown in the diagram, within quadrilateral $ABCD$, points $E$ and $F$ are on the diagonal $AC$ such that $AE=EF=FC$. Quadrilateral $DEBF$ is a parallelogram; if $\angle CDE=90^\circ$, $DC=8$, and $DE=6$, what is the perimeter of quadrilateral $DEBF$?

Solution:
\textbf{Solution:} Given, \\
$\because \angle CDE=90^\circ$, $DC=8$, $ED=6$, \\
$\therefore CE= \sqrt{CD^2+DE^2}=\sqrt{64+36}=10$, \\
$\because EF=CF$, \\
$\therefore DF= \frac{1}{2}CE=5$, \\
$\therefore$ the perimeter of $\square DEBF$ $=2(DF+DE)=2\times(5+6)=\boxed{22}$.


Question: As shown in the quadrilateral $ABCD$, where $AB\parallel CD$ and $\angle BCD=90^\circ$, with $AB=AD=10\text{cm}$ and $BC=8\text{cm}$. Point $P$ starts moving from point $A$, moving at a constant speed of $3\text{cm/s}$ along the polyline in the direction of $ABCD$; point $Q$ starts moving from point $D$, moving at a constant speed of $2\text{cm/s}$ along the line segment in the direction of $DC$. It is known that both points start moving at the same time, and when one point reaches its endpoint, the other point also stops moving. Let the movement time be $t\text{(s)}$. Find the length of $CD$, in cm.

Solution:
\textbf{Solution:} As shown in the figure, draw $AE\perp DC$ at E,\\
since $AB\parallel CD$ and $\angle BCD=90^\circ$, with $AB=AD=10\text{cm}$ and $BC=8\text{cm}$, and $AE\perp DC$,\\
thus, quadrilateral $ABCE$ is a rectangle, that is, $AE=BC=8$ and $AB=AD=CE=10$,\\
therefore, in $\triangle ADE$, $DE=\sqrt{AD^{2}-AE^{2}}=\sqrt{10^{2}-8^{2}}=6$,\\
thus, $CD=DE+EC=6+10=\boxed{16}$,\\
hence, the length of $CD$ is $16\text{cm}$.


Question: In the quadrilateral $ABCD$, $AB \parallel CD$, $\angle BCD = 90^\circ$, $AB = AD = 10 \text{cm}$, $BC = 8 \text{cm}$. Point $P$ starts from point $A$ and moves at a constant speed of $3\text{cm/s}$ along the polyline $ABCD$ direction; point $Q$ starts from point $D$ and moves at a constant speed of $2\text{cm/s}$ along the segment $DC$ direction. It is known that both points start simultaneously, and when one point reaches the end, the other point also stops moving, let the motion time be $t (s)$. When the quadrilateral $PBQD$ is a parallelogram, what is the perimeter of the quadrilateral $PBQD$?

Solution:
\textbf{Solution:} Draw lines $PD$ and $BP$, \\
when the quadrilateral $PBQD$ forms a parallelogram, we have $PB = DQ$, \\
given that point P starts from point A and moves in the direction of the polyline $ABCD$ at a constant speed of $3\text{cm/s}$; point Q starts from point D and moves along the line segment $DC$ at a constant speed of $2\text{cm/s}$, \\
let the time of movement for point P be t seconds, then $AP = 3t$, $DQ = 2t$, \\
$\therefore PB = 10 - 3t$, \\
$\therefore 10 - 3t = 2t$, solving this equation, we find $t = 2$, \\
$\therefore PB = DQ = 10 - 3 \times 2 = 4$, hence $CQ = 16 - 4 = 12$, \\
in $\triangle BCQ$, $BQ = \sqrt{BC^2 + CQ^2} = \sqrt{8^2 + 12^2} = 4\sqrt{13}$, \\
$\therefore$ the perimeter of the parallelogram $PBQD$ is: $2(PB + BQ) = 2 \times (4 + 4\sqrt{13}) = 8 + 8\sqrt{13}\text{cm}$, \\
$\therefore$ the perimeter of the quadrilateral $PBQD$ is $\boxed{8 + 8\sqrt{13}\text{cm}}$.


Question: In quadrilateral $ABCD$, $AB\parallel CD$, $\angle BCD=90^\circ$, $AB=AD=10\text{cm}$, $BC=8\text{cm}$. Point $P$ starts moving from point $A$ at a constant speed of $3\text{cm/s}$ along the polyline $ABCD$ direction; point $Q$ starts moving from point $D$ at a constant speed of $2\text{cm/s}$ along the line segment $DC$ direction. It is known that both points start moving at the same time, and when one point reaches its destination, the other point stops moving, let the motion time be $t(s)$. During the movement of points $P$ and $Q$, there exists a moment when the area of $\triangle BPQ$ is $16\text{cm}^2$. Find all possible values of $t$ that satisfy this condition?

Solution:
\textbf{Solution:} As shown in the figure, we connect $PQ$, $BQ$, and let the time of motion be $t(s)$,\\
$\because AB=10$, $BC=8$, the speed of point P is $3cm/s$, $DC=16$, the speed of point Q is $2cm/s$,\\
$\therefore$ the time for point P to reach point B is $\frac{10}{3}$ seconds, the time for point P to reach point C is $6$ seconds, and the time for point Q to reach point C is $8$ seconds,\\
(1) When $0<t<\frac{10}{3}$, $BP=10-3t$, the height of $\triangle BPQ$ is $BC=8$,\\
$\therefore \frac{1}{2}BP\cdot BC=\frac{1}{2}\times (10-3t)\times 8=16$, solving this equation, we get $t=\boxed{2}$,\\
Verification: When $t=2$, $BP=10-3t=10-3\times 2=4$, $BC=8$, the area of $\triangle BPQ$ is $16cm^2$, and $0<2<\frac{10}{3}$,\\
$\therefore t=2$, satisfies the condition;\\
(2) When $t=\frac{10}{3}$, point P coincides with point B, $\triangle BPQ$ does not exist, which does not meet the requirements, hence $t\ne \frac{10}{3}$;\\
(3) When $\frac{10}{3}<t\le 6$, as shown in the figure,\\
$\therefore BP=3t-10$, $DQ=2t$, $CQ=16-2t$,\\
$\therefore$ the base of $\triangle BPQ$ is $BP=3t-10$, and the height is $CQ=16-2t$,\\
$\therefore$ the area of $\triangle BPQ$ is $\frac{1}{2}BP\cdot CQ=\frac{1}{2}\times (3t-10)\times (16-2t)=16$, after simplification, $3t^2-34t+96=0$, solving this equation, we get $t_1=\frac{16}{3}$, $t_2=6$,\\
$\therefore$when $t=6$ or $t=\boxed{\frac{16}{3}}$, the condition is met;\\
④ When $6<t<8$, as shown in the figure,\\
$\therefore CP=3t-18$, $CQ=16-2t$,\\
$\therefore PQ=CQ-CP=16-2t-(3t-18)=34-5t$, the height of $\triangle BPQ$ is $BC=8$,\\
$\therefore$ the area of $\triangle BPQ$ is $\frac{1}{2}PQ\cdot BC=\frac{1}{2}\times (34-5t)\times 8=16$, solving this equation, we get $t=6$,\\
$\therefore$ when $6<t<8$, the condition is not met;\\
⑤ When $t=8$, point Q coincides with point C, motion stops, as shown in the figure,\\
$\therefore$ the position of point P is $CP=3t-18=3\times 8-18=6$, the height of $\triangle BPQ$ is $BC=8$,\\
$\therefore$ the area of $\triangle BPQ$ is $\frac{1}{2}CP\cdot BC=\frac{1}{2}\times 6\times 8=24$, which does not meet the condition.\\
In summary, when the value of $t$ is $2$ or $\frac{16}{3}$ or $6$, the area of $\triangle BPQ$ is $16cm^2$.


Question: As shown in Figure 1, in the Cartesian coordinate system, let point $A(0,6)$ and let point $B$ be a moving point on the positive half of the $x$-axis. Connect $AB$ and take the midpoint $M$ of $AB$. Rotate the line segment $MB$ around point $B$ in a clockwise direction by $90^\circ$ to obtain the line segment $BC$. Construct a perpendicular line to the $x$-axis through point $B$, intersecting line $AC$ at point $D$. Suppose the coordinates of point $B$ are $(2t,0)$. When $t=2$, what are the coordinates of point $M$?

Solution: \textbf{Solution:} When $t=2$, the coordinates of point $M$ are $\boxed{(2, 3)}$.


Question: As shown in Figure 1, in the Cartesian coordinate system, point $A(0,6)$, and point $B$ is a moving point on the positive x-axis. Connect $AB$, take the midpoint $M$ of $AB$, and rotate the segment $MB$ $90^\circ$ clockwise around point $B$ to obtain the segment $BC$. Draw a perpendicular line from point $B$ to the x-axis, intersecting line $AC$ at point $D$. Let the coordinates of point $B$ be $(2t,0)$. Find the coordinates of point $C$ (expressed in terms of $t$).

Solution:
\textbf{Solution:} As shown in Figure 1, let $ME\perp OB$ at $E$, and $CF\perp$ the x-axis at $F$.\\
Since $B(2t,0)$,\\
it follows that $OB=2t$,\\
Since $ME\parallel OA$ and $AM=BM$,\\
it follows that $OE=EB=t$ and $ME=\frac{1}{2}OA=3$,\\
Since $\angle MEB=\angle CFB=\angle CBM=90^\circ$,\\
it follows that $\angle MBE+\angle CBF=90^\circ$ and $\angle MBE+\angle BME=90^\circ$,\\
Therefore, $\angle BME=\angle CBF$,\\
Since $BM=BC$,\\
it follows that $\triangle MEB\cong \triangle BFC$ (AAS),\\
Therefore, $BF=ME=3$ and $CF=BE=t$,\\
Therefore, $OF=OB+BF=2t+3$,\\
Therefore, \boxed{C(2t+3,t)}.


Question: In the diagram, within the quadrilateral $ABCD$, points $G$ and $H$ are respectively the midpoints of $AB$ and $CD$, points $E$ and $F$ lie on the diagonal $AC$, and $AE=CF$. Quadrilateral $EGFH$ is a parallelogram; connecting $BD$ intersects with $AC$ at point $O$, if $BD=10$, $AE+CF=EF$, what is the length of $EG$?

Solution:
\textbf{Solution:} Draw line $BD$ intersecting $AC$ at point $O$, as shown in the figure:\\
Since quadrilateral $ABCD$ is a parallelogram,\\
$\therefore OA=OC$, $OB=OD$,\\
Since $BD=10$,\\
$\therefore OB=OD=5$,\\
Since $AE=CF$, $OA=OC$,\\
$\therefore OE=OF$,\\
Since $AE+CF=EF$,\\
$\therefore 2AE=EF=2OE$,\\
$\therefore AE=OE$,\\
Moreover, since point $G$ is the midpoint of $AB$,\\
$\therefore EG$ is the median of $\triangle ABO$,\\
$\therefore EG=\frac{1}{2}OB=2.5$. \\
Therefore, the length of $EG$ is $\boxed{2.5}$.


Question: As shown in the figure, within the rectangle $ABCO$, point $C$ is on the x-axis, and point $A$ is on the y-axis. The coordinates of point $B$ are $(-6, 8)$. The rectangle $ABCO$ is folded along the line $BD$ such that point $A$ falls on point $E$ which is on the diagonal $OB$. The line containing $BD$ intersects $OA$ and the x-axis at points $D$ and $F$, respectively. What is the length of the segment $BO$?

Solution: \textbf{Solution:} Since $B(-6,8)$, it follows that $OC=6$ and $BC=8$. Given that quadrilateral $ABCO$ is a rectangle, it implies $\angle BCO=90^\circ$. By the Pythagorean theorem, we have: $BO= \sqrt{BC^{2}+OC^{2}}=\sqrt{8^{2}+6^{2}}=\boxed{10}$.


Question: As shown in the figure, in rectangle $ABCO$, point $C$ is on the x-axis, point $A$ is on the y-axis, and point $B$ has coordinates $(-6, 8)$. The rectangle $ABCO$ is folded along line $BD$ such that point $A$ falls on point $E$ on the diagonal $OB$, and the line containing $BD$ intersects $OA$ and the x-axis at points $D$ and $F$, respectively. What is the equation of line $BD$?

Solution: \textbf{Solution:} Let $D(0, b)$. Then, according to the problem, we have $\angle DEO=90^\circ$, $DE=DA=8-b$, $BE=BA=6$, and $EO=4$. Applying the Pythagorean theorem to the right triangle $DEO$: $OE^2+DE^2=DO^2$, that is, $4^2+(8-b)^2=b^2$, solving this, we get $b=5$, hence $D(0, 5)$. The equation of line $BD$ is given by $y=kx+b$. Given $B(-6, 8)$ and $D(0, 5)$, we can establish: $\begin{cases}-6k+b=8\\ b=5\end{cases}$, from which we find $\begin{cases}k=-\frac{1}{2}\\ b=5\end{cases}$. Therefore, the equation of line $BD$ is: $y=-\frac{1}{2}x+5$; hence, the equation of line $BD$ is $\boxed{y=-\frac{1}{2}x+5}$.


Question: As shown in the figure, in rectangle $ABCO$, point $C$ is on the x-axis, point $A$ is on the y-axis, and the coordinates of point $B$ are ($-6$, $8$). Rectangle $ABCO$ is folded along line $BD$, such that point $A$ lands on point $E$ located on diagonal $OB$. Line $BD$ intersects $OA$ and the x-axis at points $D$ and $F$, respectively. Point $M$ is a moving point on line $BD$. Through point $M$, construct $MN\perp x$ axis, with $N$ being the foot of the perpendicular. During the movement of point $M$, is there a scalene triangle with vertices $N$, $E$, $O$? Write down the coordinates of point $N$ and find the coordinates of point $M$.

Solution:
\textbf{Solution:} Exist;\\
(1) The coordinates of point N are $(4,0)$ or $(-4,0)$, corresponding to the coordinates of point M being $(4,3)$ or $(-4,7)$;\\
(2) $\therefore$ The coordinates of point N are $\boxed{\left(-\frac{24}{5},0\right)}$, corresponding to the coordinates of point M being $\boxed{\left(-\frac{24}{5},\frac{37}{5}\right)}$;\\
(3) The coordinates of point N are $\boxed{\left(-\frac{10}{3},0\right)}$, corresponding to the coordinates of point M being $\boxed{\left(-\frac{10}{3},\frac{20}{3}\right)}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AB\parallel CD$, $AB=BC=2CD$, $E$ is the midpoint of diagonal $AC$, and $F$ is the midpoint of side $BC$, with segments $DE$ and $EF$ connected. Quadrilateral $CDEF$ is a rhombus; segment $DF$ intersects $AC$ at point $G$. Given that $DF=2$ and $CD=\frac{5}{3}$, find the area of quadrilateral $CDEF$.

Solution:
\textbf{Solution:} As shown in the figure, $DF$ intersects $EC$ at point $G$\\
Since quadrilateral $CDEF$ is a rhombus and $DF=2$,\\
it follows that $DG=1$, $DF\perp CE$, and $EG=GC$,\\
Since $CD=\frac{5}{3}$,\\
we have $CG=\sqrt{CD^{2}-DG^{2}}=\sqrt{\left(\frac{5}{3}\right)^{2}-1^{2}}=\frac{4}{3}$\\
Therefore, $CE=2CG=\frac{8}{3}$\\
Hence, the area of quadrilateral $CDEF$ is $S_{\text{quadrilateral}CDEF}=\frac{1}{2}DF\cdot CE=\frac{1}{2}\times \frac{8}{3}\times 2=\boxed{\frac{8}{3}}$


Question: As shown in the figure, in quadrilateral $ABCD$, $AD \parallel CB$, and $O$ is the midpoint of the diagonal $AC$. A line passing through point $O$ intersects the sides $AD$ and $BC$ of quadrilateral $ABCD$ at points $M$ and $N$, respectively. The segments $CM$ and $AN$ are then connected. Quadrilateral $ANCM$ is a parallelogram; when $MN$ bisects $\angle AMC$,

Solution:
\textbf{Solution:}\\
(1) Since $MN$ bisects $\angle AMC$,\\
it follows that $\angle AMN=\angle CMN$,\\
and since $AD\parallel BC$,\\
it follows that $\angle AMN=\angle CNM$,\\
therefore, $\angle CMN=\angle CNM$,\\
hence $CM=CN$,\\
thus parallelogram $ANCM$ is a rhombus;\\
(2) Since quadrilateral $ABCD$ is a rectangle,\\
it implies that $\angle ABN=90^\circ$, $BC=AD=8$, and $AM=CN$,\\
thus $AB=\sqrt{AC^{2}-BC^{2}}=\sqrt{(4\sqrt{5})^{2}-8^{2}}=4$, $AM=AN=NC=AD-DM$, $BN=DM$,\\
in $\triangle ABN$ right-angled at $B$, by the Pythagorean theorem, we have $AN^{2}=AB^{2}+BN^{2}$,\\
thus, $(8-DM)^{2}=4^{2}+DM^{2}$, solving this yields $DM=\boxed{3}$.


Question: As shown in the figure, in rectangle ABCD, the diagonals AC and BD intersect at point O. A line parallel to BD is drawn through point C, and a line parallel to AC is drawn through point D, with the two lines intersecting at point P. Quadrilateral CODP is a rhombus. Given that $AD=7$, and $AC=25$, what is the area of quadrilateral CODP?

Solution:
\textbf{Solution:} Given that $AD=7$ and $AC=25$, \\
therefore $DC=\sqrt{AC^2-AD^2}=24$, \\
since $AO=CO$, \\
thus the area of $\triangle COD$ is half of that of $\triangle ADC$, which equals $\frac{1}{2}\times \frac{1}{2}AD\cdot CD=42$, \\
since the quadrilateral CODP is a rhombus, \\
therefore the area of the rhombus CODP is twice the area of $\triangle COD$, which equals $\boxed{84}$.


Question: As shown in the figure, it is known that $E$ is the midpoint of side $BC$ of square $ABCD$. Connect $AE$ and extend it to intersect the extension line of $DC$ at point $F$. Connect $AC$, $BF$, and $\angle AEC=2\angle ABC$. Quadrilateral $ABFC$ is a rectangle; if $\triangle AFD$ is an equilateral triangle with a side length of $4$, what is the area of quadrilateral $ABFC$?

Solution:
\textbf{Solution:} Since $\triangle AFD$ is an equilateral triangle, \\
$\therefore$ $\angle AFC=60^\circ$, AF=DF=4, \\
$\therefore$ CF=CD=2, \\
Since quadrilateral ABFC is a rectangle, \\
$\therefore$ $\angle ACF=90^\circ$, $\angle AFC=60^\circ$, $\angle FAC=30^\circ$, \\
By the Pythagorean theorem, AC=$2\sqrt{3}$, \\
$\therefore$ The area of quadrilateral ABFC=AC$\cdot$CF=\boxed{4\sqrt{3}}.


Question: As shown in the figure, in the diamond $ABCD$, $AB=BD=5$, what is the degree measure of $\angle BAC$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\
$\therefore$ AB=AD, and AC bisects $\angle$BAD, \\
Since AB=BD=5, \\
$\therefore$ AB=AD=BD, \\
$\therefore$ $\triangle$ABD is an equilateral triangle, \\
$\therefore$ $\angle$BAD$=60^\circ$, \\
$\therefore$ $\angle$BAC$=\boxed{30^\circ}$;


Question: As shown in the figure, in the rhombus $ABCD$, $AB=BD=5$, find: What is the length of $AC$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\
it follows that OA = OC, BO = OD = $\frac{1}{2}$BD = $\frac{5}{2}$, and BD$\perp$AC,\\
In $\triangle AOB$, OA = $\sqrt{AB^{2}-BO^{2}}=\frac{5\sqrt{3}}{2}$,\\
therefore, AC = $2\text{OA}=\boxed{5\sqrt{3}}$.


Question: As shown in the figure, in the right triangle $ABC$, $\angle ACB=90^\circ$. A square $ACED$ is constructed on the side $AC$ to the left of $AC$. A line is drawn through point $E$ such that $EF\perp AB$, with the foot of the perpendicular at point $F$, intersecting $AC$ at point $G$. It is given that $CG=CB$; if point $G$ is the midpoint of $AC$, and the length of $EC$ is 2, find the length of $BF$.

Solution:
\textbf{Solution:} Given that point $G$ is the midpoint of $AC$ and the length of $EC$ is $2$,\\
$\therefore CG=\frac{1}{2}AC=\frac{1}{2}EC=1=AG$,\\
From (1) we know $CB=CG$,\\
$\therefore CB=1$,\\
$\therefore EG=\sqrt{CE^2+CG^2}=\sqrt{5}=AB$,\\
Let $BF=x$,\\
In $\triangle BFE$, by the Pythagorean theorem: $BF^2+EF^2=BE^2$,\\
$\therefore x^2+(\sqrt{5}+GF)^2=3^2$, rearranging: $x^2+5+2\sqrt{5}GF+GF^2=9$ (1),\\
$\therefore GF^2=4-x^2-2\sqrt{5}GF$,\\
In $\triangle AGF$, by the Pythagorean theorem: $GF^2+AF^2=AG^2$,\\
$\therefore GF^2+(\sqrt{5}-x)^2=1^2$, rearranging: $GF^2+5-2\sqrt{5}x+x^2=1$ (2),\\
$\therefore 4-x^2-2\sqrt{5}GF+5-2\sqrt{5}x+x^2=1$,\\
$\therefore GF+x=\frac{4\sqrt{5}}{5}$ (3),\\
Draw line $BG$,\\
In $\triangle BGF$, by the Pythagorean theorem: $GF^2+BF^2=BG^2$,\\
$\because BG^2=1^2+1^2=2$,\\
$\therefore GF^2+x^2=2$,\\
(1) + (2) gives: $x^2+5+2\sqrt{5}GF+GF^2+GF^2+5-2\sqrt{5}x+x^2=10$,\\
Rearranging gives: $x-GF=\frac{2\sqrt{5}}{5}$④,\\
(3) + ④ gives: $2x=\frac{6\sqrt{5}}{5}$,\\
$\therefore x=\boxed{\frac{3\sqrt{5}}{5}}$,\\
Therefore, the length of $BF$ is $\boxed{\frac{3\sqrt{5}}{5}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AB\parallel CD$, the diagonals $AC$ and $BD$ intersect at point $O$, and $AO=CO$. It is known that $\triangle ABO$ is an equilateral triangle. If quadrilateral $ABCD$ is a rectangle and $AB=2$, what is the length of $BC$?

Solution: \textbf{Solution:} Since $\triangle ABO$ is an equilateral triangle and $AB=2$, it follows that $OA=AB=2$. Given that quadrilateral $ABCD$ is a rectangle, we have $\angle ABC=90^\circ$, and thus $AC=2OA=4$. Therefore, in right triangle $\triangle ABC$, we have $BC=\sqrt{AC^{2}-AB^{2}}=\sqrt{4^{2}-2^{2}}=2\sqrt{3}$. Hence, the length of $BC$ is $\boxed{2\sqrt{3}}$.


Question: As shown in the diagram, the diagonals $AC$ and $BD$ of square $ABCD$ intersect at point $O$. Extend $BD$ in both directions to points $E$ and $F$ respectively, such that $BE=DF$. If $AB=3\sqrt{2}$ and $BE=3$, what is the perimeter of quadrilateral $AECF$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a square, \\
therefore $AO=\frac{1}{2}AC$, $BO=\frac{1}{2}BD$, $AC=BD$, and $AC \perp BD$, \\
thus $AO=BO$, $\angle AOB=90^\circ$. \\
In the right triangle $\triangle AOB$, by the Pythagorean theorem, $AB=\sqrt{AO^{2}+BO^{2}}=3\sqrt{2}$, \\
therefore $AO=BO=3$. \\
Hence $EO=OB+BE=6$. \\
In $\triangle AOE$, $\angle AOE = 90^\circ$, $AE=\sqrt{AO^{2}+EO^{2}}=\sqrt{3^{2}+6^{2}}=3\sqrt{5}$. \\
Given that quadrilateral AECF is a rhombus, \\
therefore $AE=EC=CF=AF$. \\
Thus, the perimeter of quadrilateral AECF is $4AE=12\sqrt{5}$. \\
Therefore, the perimeter of quadrilateral AECF is $\boxed{12\sqrt{5}}$.


Question: As shown in the figure, square $ABEF$ and square $ACGD$ are constructed externally on sides $AB$ and $AC$ of $\triangle ABC$, respectively. Lines $BD$, $CF$, and $DF$ are drawn, with $CF$ intersecting $BD$ at $O$ and $AD$ at $H$. If $AB=2$ and $AC=4$, what is the value of $BC^2+DF^2$?

Solution: \textbf{Solution:} We have $BC^2 + DF^2 = \boxed{40}$.


Question: As shown in the figure, in square \(ABCD\), \(E\) and \(F\) are located on \(BD\) and \(AD\) respectively. Connect \(EF\) and construct an isosceles right triangle \(EFG\) on the right side of \(EF\) such that \(\angle EFG=90^\circ\), and connect \(DG\). \(DG\perp BD\); If \(AB=8\), \(FD=2\), \(BE=2\sqrt{2}\), write down the length of segment \(DG\) directly.

Solution: \textbf{Solution:} The length of the line segment $DG$ is $\boxed{4\sqrt{2}}$.


Question: As shown in the figure, it is known that in $ \triangle ABC$, $ \angle BAC=45^\circ$, $ AD\perp BC$ with the foot of the perpendicular at D, $ BD=2$, and $ DC=3$. Determine the length of AD. Following the approach of folding the figure as Xiaoming did, explore and solve the following question: What is the length of AD?

Solution:
\textbf{Solution:} Let $AD=x$, then $EG=GF=AE=x$,\\
since $BD=2$, $DC=3$,\\
thus $BE=2$, $CF=3$,\\
therefore $BG=x-2$, $CG=x-3$,\\
since quadrilateral AEGF is a square,\\
thus $\angle EGC=90^\circ$,\\
in $\triangle ABC$, $BG^2+CG^2=BC^2$,\\
thus $(x-2)^2+(x-3)^2=5^2$,\\
after simplifying, $(x-6)(x+1)=0$, solving gives $x=6$, or $x=-1$ (discard),\\
therefore $AD=\boxed{6}$.


Question: As shown in the figure, in the Cartesian coordinate system, the quadrilateral OABC is a square. An equilateral triangle $\triangle ABD$ is constructed with AB as the side, and point D is precisely on the x-axis. The diagonal CA intersects the x-axis at point E. If the coordinate of point E is $\left(4, 0\right)$, write down directly the degree measure of $\angle BDO$.

Solution: \textbf{Solution:} We have $\angle BDO = \boxed{45^\circ}$.


Question: As shown in the figure, in the Cartesian coordinate plane, the quadrilateral OABC is a square. An equilateral $\triangle ABD$ is constructed with AB as the side, and point D is exactly on the x-axis. The diagonal CA intersects the x-axis at point E. If the coordinates of point E are $\left(4, 0\right)$, what are the coordinates of point B?

Solution:
\textbf{Solution:} In square $OABC$, $OC=BC$, $\angle OCA = \angle BCA = \angle OAC = 45^\circ$,
\[
\because EC = EC,
\]
\[
\therefore \triangle OEC \cong \triangle BEC \text{ (SAS)},
\]
\[
\therefore \angle OEC = \angle BEC, \text{ BE=OE},
\]
\[
\because \angle EOA = \angle EDA = 15^\circ,
\]
\[
\therefore \angle OEC = 60^\circ,
\]
\[
\therefore \angle BED = 180^\circ - 60^\circ - 60^\circ = 60^\circ,
\]
As shown in the figure, draw BH $\perp$ ED at point H,
\[
\therefore \angle EBH = 30^\circ,
\]
\[
\because E(4,0),
\]
\[
\therefore OE = 4 = EB,
\]
\[
\therefore EH = \frac{1}{2} EB = 2,
\]
According to the Pythagorean theorem, we get BH=$2\sqrt{3}$,
\[
\therefore \text{The coordinates of point B are} \boxed{(6，2\sqrt{3})};
\]


Question: As shown in the figure, in the Cartesian coordinate system, the quadrilateral OABC is a square. An equilateral $\triangle$ABD is formed with AB as the side, where point D is precisely on the x-axis. The diagonal CA is extended to intersect the x-axis at point E. Given that the coordinates of point E are $\left(4, 0\right)$. What is the area of the square OABC?

Solution:
\textbf{Solution:} As shown in the figure, \\
$\because \angle BDO=45^\circ$, $\angle BHD=90^\circ$, \\
$\therefore \angle HBD=45^\circ$, \\
$\therefore \triangle BHD$ is an isosceles right triangle, \\
$\because BH=2\sqrt{3}$, \\
$\therefore BD=2\sqrt{6}$, \\
$\therefore AB=BD=2\sqrt{6}$, \\
$\therefore$ the area of the square OABC is: ${\left(2\sqrt{6}\right)}^{2}=\boxed{24}$.


Question: As shown in the figure, quadrilateral $ABCD$ is a rectangle, and diagonals $AC$ and $BD$ intersect at point $O$. Extend $BC$ to point $E$ so that $CE = BC$, and connect $DE$. It is given that $AC = DE$; If $F$ is the midpoint of $BC$, and connecting $OF$, with $AC = 5$ and $OF = 2$, what is the perimeter of $\triangle BDE$?

Solution:
\textbf{Solution:} \\
Since quadrilateral ABCD is a rectangle,\\
$AC=BD=5$,\\
$DE=5$,\\
$OB=OD=OC=OA=\frac{5}{2}$,\\
Since F is the midpoint of BC,\\
$\angle OFB=90^\circ$,\\
By the Pythagorean theorem: $BF=\sqrt{OB^{2}-OF^{2}}=\sqrt{\left(\frac{5}{2}\right)^{2}-2^{2}}=\frac{3}{2}$,\\
$BC=CE=2BF=3$,\\
$BE=6$,\\
The perimeter of $\triangle BDE$ = BE+BD+DE=6+5+5=\boxed{16}.


Question: As shown in the figure, the diagonals of rhombus ABCD intersect at point O. Lines are drawn through points C and D such that $CE \parallel BD$ and $DE \parallel AC$, and these parallel lines intersect at point E. Quadrilateral OCED is a rectangle; if $CE=1$ and $DE=2$, calculate the area of rhombus ABCD.

Solution:
\textbf{Solution:} From equation (1), we know that quadrilateral OCED is a rectangle, \\
$\therefore OD=CE=1, OC=DE=2$\\
$\because$ABCD is a rhombus,\\
$\therefore BD=2OD=2, AC=2OC=4$\\
$\therefore S_{ABCD}=\frac{1}{2}\times 2\times 4=\boxed{4}$


Question: As shown in the figure, in $\triangle ABC$, $\angle A=90^\circ$, $AB=3$, $AC=4$. $D$, $E$, and $F$ are the midpoints of sides $AB$, $BC$, and $AC$ respectively. Connect $DE$, $DF$, and $EF$, with the measure of $\angle ADF$ being $53^\circ$. What is the measure of $\angle C$?

Solution:
\textbf{Solution:} Since $D$ and $F$ are the midpoints of sides $AB$ and $AC$ respectively, \\
$\therefore DF \parallel BC$, \\
Since $\angle ADF = 53^\circ$, \\
$\therefore \angle B = \angle ADF = 53^\circ$, \\
Since $\angle A = 90^\circ$, \\
$\therefore \angle C = 180^\circ - 90^\circ - 53^\circ = \boxed{37^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle A=90^\circ$, $AB=3$, and $AC=4$. D, E, and F are the midpoints of sides AB, BC, and AC, respectively. Lines DE, DF, and EF are connected, and the measure of $\angle ADF$ is $53^\circ$. What is the perimeter of quadrilateral ADEF?

Solution:
\textbf{Solution:} Given that D, E, F are the midpoints of sides AB, BC, AC respectively, \\
$\therefore$ DE$\parallel$ AC and $DE=\frac{1}{2}AC$, EF$\parallel$ AB and $EF=\frac{1}{2}AB$, AF$=\frac{1}{2}$AC, and AD$=\frac{1}{2}$AB.\\
Since $AB=3$ and $AC=4$, \\
$\therefore$ AF=DE=2 and $AD=EF=\frac{3}{2}$.\\
Therefore, the perimeter of quadrilateral ADEF $=\left(2+\frac{3}{2}\right)\times 2=\boxed{7}$.


Question: Place a square piece of paper in the Cartesian coordinate system as shown in the figure. The side length of the square paper $ABCD$ is $3$, and points $E$ and $F$ are located on $BC$ and $CD$, respectively. Fold $AB$ and $AD$ along $AE$ and $AF$, respectively, so that points $B$ and $D$ exactly overlap at point $G$. It is known that $3BE=BC$. Please directly write the coordinates of points $D$ and $E$, and find the equation of line $EF$.

Solution:
\textbf{Solution:} The coordinates of point $D$ are $(3,3)$; the coordinates of point $E$ are $(1,0)$;\\
Let the equation of line $EF$ be $y=kx+b$,\\
therefore $\left\{\begin{array}{l}
k+b=0 \\
3k+b=\frac{3}{2}
\end{array}\right.$,\\
thus, $\left\{\begin{array}{l}
k=\frac{3}{4} \\
b=-\frac{3}{4}
\end{array}\right.$,\\
hence the equation of line $EF$ is $\boxed{y=\frac{3}{4}x-\frac{3}{4}}$;


Question: As shown in the diagram, place a square paper in the Cartesian coordinate system. The side length of the square paper ABCD is 3, with points E and F located on BC and CD, respectively. Fold AB and AD along AE and AF such that points B and D both precisely fall on point G. Given that $3BE=BC$, there exists a point M on line EF, such that the area of $\triangle AFM$ is half the area of $\triangle AEF$. What are the coordinates of point M?

Solution:
\textbf{Solution:} Suppose there exists a point $M$ on the line $EF$ such that the area of $\triangle AFM$ is half of the area of $\triangle AEF$, \\
$\because$ the heights of $\triangle AEF$ and $\triangle AFM$ are equal, \\
$\therefore \frac{S_{\triangle AFM}}{S_{\triangle AEF}}=\frac{FM}{EF}=\frac{1}{2}$, \\
$\therefore FM=\frac{1}{2}EF$, \\
When $M$ is on the segment $EF$, specifically, when $M$ is the midpoint of $EF$, we denote this point as $M_{1}$, \\
$\because$ the coordinate of $E$ is $(1,0)$, and the coordinate of $F$ is $(3,\frac{3}{2})$, \\
$\therefore \left\{\begin{array}{l}x_{M_{1}}=\frac{1+3}{2}=2 \\ y_{M_{1}}=\frac{0+\frac{3}{2}}{2}=\frac{3}{4}\end{array}\right.$, \\
$\therefore$ the coordinates of $M_{1}$ are $\boxed{(2,\frac{3}{4})}$; \\
When $M$ lies on the extension of $EF$, we denote this point as $M_{2}$, then $FM_{1}=FM_{2}$, which means $F$ is the midpoint of $M_{1}M_{2}$, \\
$\therefore \left\{\begin{array}{l}3=\frac{2+x_{M_{2}}}{2} \\ \frac{3}{2}=\frac{y_{M_{2}}+\frac{3}{4}}{2}\end{array}\right.$, \\
$\therefore \left\{\begin{array}{l}x_{M_{2}}=4 \\ y_{M_{2}}=\frac{9}{4}\end{array}\right.$, \\
$\therefore$ the coordinates of $M_{2}$ are $\boxed{(4,\frac{9}{4})}$; \\
$\therefore$ In summary, when the coordinates of $M$ are $\boxed{(2,\frac{3}{4})}$ or $\boxed{(4,\frac{9}{4})}$, the area of $\triangle AFM$ is half of the area of $\triangle AEF$.


Question: As shown in the figure, the side length of the square paper ABCD is 3, with points E and F located on BC and CD, respectively. AB and AD are folded along AE and AF, respectively, so that points B and D just coincide at point G. It is known that $3BE=BC$. If points P and Q are moving points on segments AG and AF, respectively, what is the minimum value of $EP+PQ$, and what are the coordinates of point Q at this time?

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

$\because EP+PQ \ge EQ$,

$\therefore$ when points Q, P, and E are collinear, $EP+PQ$ reaches its minimum value EQ,

Furthermore, since the perpendicular from a point to a line is the shortest segment, when $EQ \perp AF$, EQ reaches its minimum value, that is, $EP+PQ$ reaches its minimum value,

From (1) we have $EF=EG+GF=\frac{3}{2}+1=\frac{5}{2}$, and by the property of folding, we get AG=AB=3, $\angle AGE=\angle AGF=\angle ABC=90^\circ$,

$\therefore AF=\sqrt{AG^2+GF^2}=\frac{3\sqrt{5}}{2}$,

$\because S_{\triangle AEF}=\frac{1}{2}AG\cdot EF=\frac{1}{2}AF\cdot QE$,

$\therefore QE=\frac{AG\cdot EF}{AF}=\sqrt{5}$,

Suppose the equation of line AF is $y=mx+n$,

$\therefore \left\{\begin{array}{l}3k+b=\frac{3}{2} \\ b=3 \end{array}\right.$,

$\therefore \left\{\begin{array}{l}k=-\frac{1}{2} \\ b=3 \end{array}\right.$,

$\therefore$ the equation of line AF is $y=-\frac{1}{2}x+3$;

Suppose the coordinates of point Q are $(t, -\frac{1}{2}t+3)$,

$\therefore QE^2=(t-1)^2+\left(-\frac{1}{2}t+3\right)^2=5$,

Solving this yields $t=2$,

$\therefore$ the coordinates of point Q are $(2, 2)$,

$\therefore$ the minimum value of $EP+PQ$ is $\boxed{\sqrt{5}}$, at this moment, the coordinates of point Q are $\boxed{(2, 2)}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=3$, and $BC=4$. Squares $ABMN$, $ACDE$, and $BCFG$ are constructed on the sides of $\triangle ABC$ on the same side as $AB$. Point $M$ is on side $FG$, $MN$ intersects $CF$ at point $P$, and $AN$ intersects $CD$ at point $Q$. $\triangle ABQ \cong \triangle NAP$; what is the area of quadrilateral $CPNQ$?

Solution:
\textbf{Solution:} Since $\triangle ABQ \cong \triangle NAP$,\\
it follows that $S_{\triangle ABQ} = S_{\triangle NAP}$,\\
hence $S_{\triangle ABC} = S_{\text{quadrilateral} CPNQ}$,\\
Since $\angle ACB = 90^\circ$, $AC = 3$, and $BC = 4$,\\
it follows that $S_{\triangle ABC} = \frac{1}{2}AC \cdot BC = 6$,\\
thus $S_{\text{quadrilateral} CPNQ} = \boxed{6}$.


Question: As shown in the figure, in parallelogram $ABCD$, points $E$ and $F$ are on sides $BC$ and $AD$ respectively, with $BE=DF$, $\angle AEC=90^\circ$. Quadrilateral $AECF$ is a rectangle; $BF$ is drawn, and if $AB=4$, $\angle ABC=60^\circ$, and $BF$ bisects $\angle ABC$, what is the length of $AD$?

Solution:
\textbf{Solution:} Given the construction, $BF$ bisects $\angle ABC$,\\
$\therefore \angle ABF = \angle FBC$,\\
$\because BC \parallel AD$,\\
$\therefore \angle AFB = \angle FBC$,\\
$\therefore \angle AFB = \angle ABF$,\\
$\therefore AF = AB = 4$,\\
In $\triangle ABE$,\\
$\angle AEB = 90^\circ$, $\angle ABE = 60^\circ$, $AB = 4$,\\
$\therefore \angle BAE = 30^\circ$,\\
$\therefore BE = 2$,\\
$\therefore FD = BE = 2$,\\
$\therefore AD = AF + FD = \boxed{6}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle B=30^\circ$, point $D$ is on the extension of $BA$, $\angle CDB=45^\circ$, point $M$ is on side $BC$, and quadrilateral $ACMN$ is a square. What is the degree measure of $\angle ACD$?

Solution: \textbf{Solution:} Since the information provided by the problem is insufficient, a direct solution cannot be offered. Therefore, a specific mathematical solution process cannot be provided. \boxed{}


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle B=30^\circ$, point $D$ is on the extension of $BA$, $\angle CDB=45^\circ$, point $M$ is on side $BC$, and quadrilateral $ACMN$ is a square. What is the value of $\frac{BM}{AD}$?

Solution:
\textbf{Solution:} As shown in Figure 1, draw $CQ\perp AB$ at point $Q$ passing through point $C$,\\
then $\angle AQC=\angle BQC=90^\circ$,\\
let $AQ=x$,\\
since $\angle B=30^\circ$,\\
it follows that $BC=2CQ$,\\
in $\triangle ACQ$, $\angle CAQ=60^\circ$,\\
thus $\angle ACQ=90^\circ-\angle CAQ=30^\circ$,\\
therefore $AC=2AQ=2x$,\\
therefore $CQ=\sqrt{AC^{2}-AQ^{2}}=\sqrt{(2x)^{2}-x^{2}}=\sqrt{3}x$,\\
therefore $BC=2\sqrt{3}x$,\\
since quadrilateral $ACMN$ is a square,\\
therefore $CM=AC=2x$,\\
therefore $BM=BC-CM=(2\sqrt{3}-2)x$,\\
since $\angle DQC=90^\circ$, $\angle BDC=45^\circ$,\\
therefore $\triangle CDQ$ is an isosceles right triangle,\\
therefore $DQ=CQ=\sqrt{3}x$,\\
therefore $AD=DQ-AQ=(\sqrt{3}-1)x$,\\
therefore $\frac{BM}{AD}=\frac{(2\sqrt{3}-2)x}{(\sqrt{3}-1)x}=\boxed{2}$


Question: It is known that in the parallelogram $ABCD$, points $E$ and $F$ are respectively on sides $AB$ and $BC$, with $DE=AF$, $DE\perp AF$ at point $G$. If $\angle BAD=90^\circ$, parallelogram $ABCD$ is a square; extending $CB$ to point $H$, making $BH=AE$, determine the shape of $\triangle AHF$ and explain the reason.

Solution:
\textbf{Solution:}
To prove: $\triangle AHF$ is an isosceles triangle, for the following reasons:\\
From (1) we have: $\triangle ADE \cong \triangle BAF$,\\
$\therefore AE=BF$,\\
$\because BH=AE$,\\
$\therefore BF=BH$,\\
$\because$ quadrilateral ABCD is a square,\\
$\therefore \angle ABC=90^\circ$,\\
$\therefore AB \perp BC$ i.e., AB bisects FH perpendicularly,\\
$\therefore AH=AF$,\\
$\therefore \boxed{\triangle AHF \text{ is an isosceles triangle}}$.


Question: As shown in the figure, in parallelogram $ABCD$, points $E$ and $F$ are located on sides $AB$ and $BC$ respectively, with $DE=AF$ and $DE \perp AF$ at point $G$. If $AB=AD$, $\angle AED=60^\circ$, $AE=6$, and $BF=2$, what is the length of $DE$?

Solution:
\textbf{Solution:} Extend $CB$ to point $H$, such that $BH=AE$, and connect $AH$, as shown in the figure:\\
$\because$ Quadrilateral $ABCD$ is a parallelogram, and $AB=AD$,\\
$\therefore$ Quadrilateral $ABCD$ is a rhombus,\\
$\therefore$ $AD\parallel BC$,\\
$\therefore$ $\angle ABH=\angle BAD$.\\
$\because$ $BH=AE$,\\
$\therefore$ $\triangle DAE\cong \triangle ABH$ (SAS),\\
$\therefore$ $DE=AH$, $\angle AHB=\angle DEA=60^\circ$,\\
$\because$ $DE=AF$,\\
$\therefore$ $AH=AF$,\\
$\therefore$ $\triangle AHF$ is an equilateral triangle,\\
$\therefore$ $AH=HF=BH+BF=AE+BF=6+2=8$,\\
$\therefore$ $DE=AH=\boxed{8}$.


Question: As shown in the figure, in rectangle $ABCD$, point $E$ is on $AD$, and $EC$ bisects $\angle BED$. If $\angle ABE=45^\circ$ and $AB=\sqrt{5}$, find the length of $BC$.

Solution:
\textbf{Solution:} In rectangle $ABCD$, $\angle A=90^\circ$, and $\angle ABE=45^\circ$, \\
$\therefore \triangle ABE$ is an isosceles right triangle,\\
$\therefore AE=AB=\sqrt{5}$,\\
$\therefore BE=\sqrt{AE^2+AB^2}=\sqrt{10}$,\\
From (1) we know $BC=BE$,\\
$\therefore BC=\boxed{\sqrt{10}}$.


Question: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Draw a perpendicular line from point $A$ to $BC$ with the foot of the perpendicular being point $E$. Extend $BC$ to point $F$ such that $CF=BE$. Connect $DF$. Prove that the quadrilateral $AEFD$ is a rectangle.

Solution:
\textbf{Solution:} \\
Since $CF=BE$, \\
it follows that $EF=BC=AD$, \\
and since $EF\parallel AD$, \\
it implies that quadrilateral $AEFD$ is a parallelogram, \\
since $AE\perp EF$, \\
it follows that $\angle AEF=90^\circ$, \\
therefore, the quadrilateral AEFD is a \boxed{rectangle}.


Question: As shown in the figure, in the rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. A perpendicular is drawn from point $A$ to $BC$ at right foot $E$, and $BC$ is extended to point $F$ such that $CF=BE$. Segment $DF$ is drawn. Segment $OE$ is drawn. If $AB=13$ and $OE=5$, what is the length of $AE$?

Solution:
\textbf{Solution:} Since point O is the midpoint of AC, and triangle AEC is a right-angled triangle,\\
we have $AC=2OE=10$.\\
Let $EC=x$, then $BE=BC-EC=13-x$.\\
Hence, $AB^2-BE^2=AC^2-EC^2$, that is $13^2-(13-x)^2=10^2-x^2$.\\
Solving this, we find $x=\frac{50}{13}$.\\
Therefore, $AE=\sqrt{AC^2-EC^2}=\boxed{\frac{120}{13}}$.


Question: As shown in the figure, point $O$ is a point inside $\triangle ABC$. Connect $OA$, $OB$, and $OC$, and connect the midpoints $D$, $E$, $F$, $G$ of $AB$, $OB$, $OC$, $AC$ respectively, to form quadrilateral $DEFG$. Quadrilateral $DEFG$ is a parallelogram; if $BO$ is on $CO$ and $M$ is the midpoint of $EF$, and $OA=8$, $OM=3$, what is the perimeter of the parallelogram $DEFG$?

Solution:
Solution: From the given, $BO\perp CO$ and M is the midpoint of EF, $OM=3$,\\
$\therefore EF=2OM=6$. Given that quadrilateral DEFG is a parallelogram,\\
$\therefore DG=EF=6$,\\
$\because$ D is the midpoint of AB and E is the midpoint of BO,\\
$\therefore DE=\frac{1}{2}AO=4$,\\
$\therefore$ the perimeter of the quadrilateral DEFG is: $4+4+6+6=\boxed{20}$.


Question: As shown in the figure, the diagonals $AC$ and $BD$ of rhombus $ABCD$ intersect at point $O$. Through point $D$, draw $DE \parallel AC$ and $DE = \frac{1}{2}AC$. Connect $CE$ and $OE$, and connect $AE$ which intersects $OD$ at point $F$. $OE = CD$; If the side length of rhombus $ABCD$ is $2$, and $\angle ABC=60^\circ$, what is the length of $AE$?

Solution:
\textbf{Solution:} Given that in diamond ABCD, $\angle ABC=60^\circ$,\\
therefore, $\triangle ABC$ is an equilateral triangle,\\
therefore, $AC=AB=2$,\\
since $OA=\frac{1}{2}AC=1$ and $AC\perp BD$, $AD=2$,\\
thus, $OD=\sqrt{AD^{2}-OA^{2}}=\sqrt{3}$,\\
therefore, in rectangle OCED, $CE=OD=\sqrt{AD^{2}-AO^{2}}=\sqrt{3}$,\\
therefore, in $\triangle ACE$, $AE=\sqrt{AC^{2}+CE^{2}}=\boxed{\sqrt{7}}$.


Question: As shown in the diagram, point E is a point on the side CD of the rhombus ABCD. Folding $\triangle ADE$ along AE, the corresponding point of point D, which is point F, exactly falls on the side BC. Let $\frac{DE}{CE} = k$. If point F coincides with point C, what is the value of k?

Solution: The information you have provided seems to be incomplete. You have only given a number "1", without providing any LaTeX code or a specific description of the mathematical problem. In order to assist you with cleaning and correcting the LaTeX code, please provide the complete math problem and the corresponding LaTeX code.


Question: As shown in the figure, point E is a point on side CD of rhombus ABCD. Triangle ADE is folded along AE such that the corresponding point of D, point F, exactly lies on side BC. Let $ \frac{DE}{CE}=k$. If point F is the midpoint of side BC, what is the value of $k$?

Solution:
Solution: Given that point F is the midpoint of side BC in rhombus ABCD, we need to find the value of \( k \) such that \( \frac{DE}{CE} = k \).

Since F is the midpoint of BC, we know that \( BF = FC \).

When triangle ADE is folded along AE, point D maps to point F. Since F is the midpoint of BC, and D maps to F, it implies that E must be the midpoint of CD. This is because the fold symmetry ensures that DE and CE are equal when D maps to the midpoint of BC.

Therefore, \( DE = CE \), which means \( \frac{DE}{CE} = 1 \).

Thus, the value of \( k \) is \(\boxed{1}\).


Question: In the rhombus $ABCD$, $\angle ABC=120^\circ$ and $AB=10$, with points $M$ and $N$ located on sides $AD$ and $AB$, respectively. What is the length of the diagonal $AC$?

Solution: \textbf{Solution:} The length of the diagonal $AC$ is $10\sqrt{3}=\boxed{10\sqrt{3}}$.


Question: As shown in the diagram, in rhombus $ABCD$, $\angle ABC=120^\circ$, $AB=10$, and points $M$ and $N$ are on sides $AD$ and $AB$, respectively. If $\triangle AMN$ is folded along $MN$ to obtain $\triangle PMN$, with point $P$ exactly at the midpoint of side $CD$, then what is the length of $AN$?

Solution: \textbf{Solution:} The length of $AN$ is \boxed{\frac{35}{4}}.


Question: As shown in the figure, points $M$ and $N$ are respectively on the side $AD$ and diagonal $BD$ of the rectangle $ABCD$. By connecting $BM$, $MN$, we have $AB=1$, $BC=2$. What is the value of $\sin \angle ADB$?

Solution: \textbf{Solution}: Thus, $\sin\angle ADB = \boxed{\frac{\sqrt{5}}{5}}$.


Question: As shown in the figure, points \(M\) and \(N\) are on side \(AD\) and diagonal \(BD\) of rectangle \(ABCD\), respectively. Connect \(BM\), \(MN\), where \(AB=1\), \(BC=2\), what is the minimum value of \(BM+MN\)?

Solution: \textbf{Solution:} The minimum value of $BM+MN$ is $\boxed{\frac{4\sqrt{5}}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BCA=90^\circ$, $BC=8$, and $AC=6$. Point $D$ is the midpoint on side $AB$, and point $E$ is a moving point on side $BC$. Connect $DE$, and fold $\triangle BDE$ along $DE$ to get $\triangle FDE$. As shown in figure (1), segment $DF$ intersects segment $BC$ at point $G$. When $BE=2$, what is the value of $\frac{GE}{GD}$?

Solution: \textbf{Solution}: Solve: $\frac{GE}{GD}=\boxed{\frac{2}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BCA=90^\circ$, $BC=8$, $AC=6$. Point $D$ is the midpoint of side $AB$, and point $E$ is a moving point on side $BC$. Connect $DE$, and fold $\triangle BDE$ along $DE$ to get $\triangle FDE$. As shown in figure (2), when point $E$ coincides with point $C$, segment $EF$ intersects segment $AB$ at point $P$. Find the length of $DP$.

Solution:
\[
\textbf{Solution:} \text{Given that,}
\]
\[
\because \angle PCD = \angle BCD = \angle B, \\
\therefore \angle PCD = \angle B, \\
\because \angle CPD = \angle BPC, \\
\therefore \triangle CPD \sim \triangle BPC, \\
\therefore \frac{DP}{CP} = \frac{CP}{BP} = \frac{CD}{BC} = \frac{5}{8}, \\
\text{Let } DP = 5k, CP = 8k, \\
\because CP^2 = PD \cdot PB, \\
\therefore 64k^2 = 5k(5k + 8k), \\
\therefore k = \frac{25}{39}, \\
\therefore PD = 5k = \boxed{\frac{125}{39}}.
\]


Question: As shown in the figure, in $\triangle ABC$, $\angle BCA=90^\circ$, $BC=8$, and $AC=6$. Point $D$ is the midpoint on side $AB$, and point $E$ is a movable point on side $BC$. Connect $DE$, and fold $\triangle BDE$ along $DE$ to get $\triangle FDE$. As shown in figure (3), connect $CD$, and the line segment $EF$ intersects with the line segment $CD$ at point $M$. When $\triangle DFM$ is a right-angled triangle, find the length of $BE$.

Solution:
\textbf{Solution:}\\
When $\angle FMD=90^\circ$,\\
$\because \angle F=\angle B, \angle FMD=\angle ACB=90^\circ$,\\
$\therefore \triangle FDM\sim \triangle BAC$,\\
$\therefore \frac{DF}{AB}=\frac{DM}{AC}$,\\
$\therefore \frac{5}{10}=\frac{DM}{6}$,\\
$\therefore DM=3$,\\
$\therefore CM=CD-DM=2$,\\
$\because \angle ECM=\angle B$,\\
$\therefore \angle CME=\angle ACB=90^\circ$,\\
$\therefore \triangle CEM\sim \triangle BAC$,\\
$\therefore \frac{CE}{AB}=\frac{CM}{BC}$,\\
$\therefore \frac{CE}{10}=\frac{2}{8}$,\\
$\therefore CE=\frac{5}{2}$,\\
$\therefore BE=\frac{11}{2}$;\\
When $\angle FDM=90^\circ$,\\
$\because \angle F=\angle BCD, \angle FMD=\angle CME$,\\
$\therefore \angle CEM=\angle FDM=90^\circ$,\\
$\therefore \angle FED=\angle BED=45^\circ$,\\
Construct $DH\perp BC$ at $H$,\\
Then $\triangle BDH\sim \triangle BAC$,\\
$\therefore \frac{DB}{BA}=\frac{DH}{AC}$,\\
$\therefore \frac{5}{10}=\frac{DH}{6}$,\\
$\therefore DH=3, BH=4$,\\
$\therefore EH=DH=3$,\\
$\therefore BE=3+4=7$.\\
In conclusion, $BE=\boxed{\frac{11}{2}}$ or \boxed{7}.


Question: As shown in the figure, the line $y = -x + 1$ intersects with the graph of the inverse proportion function $y = \frac{k}{x}$, where the $x$-coordinate of one of the intersection points is $-2$. What is the expression of the inverse proportion function?

Solution:
\textbf{Solution:} Substituting $x=-2$ into $y=-x+1=3$ gives us one of the points of intersection as $\left(-2,3\right)$.\\
Substituting $\left(-2,3\right)$ into the expression for the inverse proportionality function and solving yields: $k=-2\times 3=-6$,\\
$\therefore$ the expression for the inverse proportionality function is: $\boxed{y=-\frac{6}{x}}$.


Question: As shown in the figure, the line $y=-x+1$ intersects with the graph of the inverse proportion function $y=\frac{k}{x}$. One of the intersection points has an abscissa of $-2$. After vertically translating the line $y=-x+1$ downwards by 2 units, find the coordinates of the intersection points between the translated line and the graph of the inverse proportion function $y=\frac{k}{x}$.

Solution:
\textbf{Solution:} The graph of the linear function $y=-x+1$ is translated downward by 2 units to obtain $y=-x-1$,\\
by solving the system $\left\{\begin{array}{l}y=-\frac{6}{x}\\ y=-x-1\end{array}\right.$, we obtain: $\left\{\begin{array}{l}x=2 \\ y=-3\end{array}\right.$ or $\left\{\begin{array}{l}x=-3 \\ y=2\end{array}\right.$,\\
$\therefore$ the coordinates of the intersections between the translated line and the graph of the hyperbolic function $y=\frac{k}{x}$ are $\boxed{\left(2，-3\right)}$ and $\boxed{\left(-3，2\right)}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the vertices A and B of the square ABCD are on the x-axis, and the parabola $y=-x^2+bx+c$ passes through point A and $C(4, -5)$, and intersects with line DC at another point E. Determine the equation of the parabola.

Solution:
\textbf{Solution}: Since quadrilateral ABCD is a square, and $C(4, -5)$,

it follows that $AD=AB=5$, and $B(4, 0)$,

thus $OA=1$,

therefore $A(-1, 0)$,

Substituting the coordinates of points A and C into $y=-{x}^{2}+bx+c$ yields: $\begin{cases}-16+4b+c=-5 \\ -1-b+c=0\end{cases}$,

Solving this, we find: $\begin{cases}b=2 \\ c=3\end{cases}$,

Therefore, the equation of the parabola is $y=-{x}^{2}+2x+3=\boxed{-x^2+2x+3}$.


Question: As shown in the diagram, within the Cartesian coordinate system, the vertices A and B of square ABCD are located on the x-axis, and the parabola $y=-x^2+bx+c$ passes through points A and $C(4, -5)$, and intersects with line DC at another point E. Let P be a point on the y-axis, and draw a perpendicular line from P to the axis of symmetry of the parabola, with Q being the foot of the perpendicular. Connect EQ and AP. Find the minimum value of $EQ+PQ+AD$.

Solution:
\textbf{Solution:} Connect OC, intersecting the line of symmetry $x=1$ at point Q.\\
Since $PQ \perp y$-axis,\\
it follows that $AO \parallel PQ$,\\
Given that $AO = PQ = 1$,\\
it follows that quadrilateral AOQP is a parallelogram,\\
Hence, $AP=OQ$,\\
Therefore, $EQ + PQ + AP = EQ + 1 + OQ$\\
To minimize the value of $EQ + PQ + AP$, the sum $EQ + OQ$ must be minimized.\\
Since E and C are symmetric about the axis $x=1$,\\
it follows that $EQ = CQ$,\\
therefore, $EQ + OQ = CQ + OQ$,\\
at this point, the value of $EQ + OQ$ is minimized, with the minimum value being the length of segment OC.\\
Given $C(4, -5)$,\\
thus $OC = \sqrt{4^{2}+5^{2}} = \sqrt{41}$,\\
So, the minimum value of $EQ + PQ + AP$ is $\sqrt{41} + 1$,\\
meaning the minimum value of $EQ + PQ + AP$ is $\boxed{\sqrt{41} + 1}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertices A and B of the square ABCD are on the x-axis, and the parabola $y=-x^2+bx+c$ passes through points A and $C(4, -5)$, and intersects with line DC at another point E. Let N be a point in the plane. Is there a point M on the axis of symmetry of the parabola such that the quadrilateral formed by points M, N, E, A is a rhombus? If such a point exists, please write down the coordinates of point M directly; if it does not exist, please explain why.

Solution: \textbf{Solution:} Exist, the coordinates of point M are $\boxed{\left(1, -3\right)}$.


Question: As shown in the figure, $\triangle ABC$ and $\triangle ADE$ are equilateral triangles. Connect $BE$, $BD$, $CD$, $EC$. As shown in Figure 1, if $\angle ADC=30^\circ$ and $AE=7$, $EC=9$, what is the length of $EB$?

Solution:
\textbf{Solution:} Given that, $\triangle ABC$ and $\triangle ADE$ are equilateral triangles,\\
$\therefore AC=AB, AE=AD$, and $\angle ADE=60^\circ$, $\angle EAD=\angle CAB=60^\circ$\\
$\therefore \angle EDC=90^\circ$\\
Since $\angle ADC=30^\circ$, $AE=7$, and $EC=9$,\\
$\therefore ED=AE=7$\\
In $\triangle CDE$, $CD=\sqrt{EC^{2}-ED^{2}}=\sqrt{9^{2}-7^{2}}=4\sqrt{2}$,\\
Since $\angle EAD=\angle CAB=60^\circ$\\
$\therefore \angle EAD+\angle DAB=\angle DAB+\angle AC$\\
Therefore, $\angle CAD=\angle BAE$\\
$\therefore \triangle EAB\cong \triangle DAC$\\
$\therefore EB=CD=4\sqrt{2}$, hence the length of $EB$ is $\boxed{4\sqrt{2}}$.


Question: As shown in the figure, $\triangle ABC$ and $\triangle ADE$ are equilateral triangles, with the segments BE, BD, CD, EC connected. As in figure 3, the side $BC=6$ of $\triangle ABC$, and through point D, $EC=2\sqrt{19}$. Let N be a moving point on line AB, connect DN, and fold $\triangle DBN$ along DN to get $\triangle DHN$. When AH is maximized, a perpendicular line is drawn from H through point AH. Let M be a moving point on the perpendicular line, connect MA, and rotate the segment MA $60^\circ$ counterclockwise around point M to obtain the segment MP. Connect PH, and directly write out the minimum value of PH.

Solution:
textbf{Solution}: According to the problem, $AC=AB=6$, $\triangle ABC$ and $\triangle ADE$ are equilateral triangles,\\
$\therefore AC=AB, AE=AD$, $\angle ADE=60^\circ$, $\angle EAD=\angle CAB=60^\circ$, $\angle ACB=60^\circ$\\
$\because \angle EAD=\angle CAB=60^\circ$\\
$\therefore \angle EAD+\angle DAB=\angle DAB+\angle AC$\\
That is, $\angle CAD=\angle BAE$\\
$\therefore \triangle EAB \cong \triangle DAC$\\
$\because \angle ACB=60^\circ$\\
$\therefore EB=DC, \angle ABE=\angle ACD$\\
Let $BD=2x$, extend $BE$ to $K$, such that $EK=BD=2x$, connect $KA, CK$, and draw $ES\perp KC$ at point $S$ with $AB$ intersecting $KC$ at point $O$,\\
$\therefore KB=BC=AB, \angle ABK=\angle ACB=60^\circ$\\
$\therefore \triangle ABK$ is an equilateral triangle\\
$\therefore AK=KB=AB=BC=AC$, $ \angle AKB=60^\circ$\\
$\therefore$ Quadrilateral $AKBC$ is a rhombus\\
$\therefore \angle SKE=\frac{1}{2}\angle AKB=30^\circ$\\
$\therefore SE=\frac{1}{2}KE=\frac{1}{2}BD=x$, $KS=\sqrt{3}x$\\
$\because AC=AB=6$\\
$\therefore KC=2OC=2AC\times \sin\angle BAC=6\sqrt{3}$\\
$\therefore CS=6\sqrt{3}-\sqrt{3}x$\\
In $\triangle CSE$, $CS=6\sqrt{3}-\sqrt{3}x$, $SE=x$, $CE=2\sqrt{19}$\\
$\therefore SC^2+SE^2=EC^2$\\
That is, ${\left(6\sqrt{3}-\sqrt{3}x\right)}^2+{x}^2={\left(2\sqrt{19}\right)}^2$\\
Solve to get ${x}_1=1, {x}_2=-1$ (discard)\\
$\therefore BD=2$\\
$\because$ N is a moving point on line AB, connect DN, fold $\triangle DBN$ along DN to obtain $\triangle DHN$,\\
$\therefore DB=DH=2$\\
$\because AH\leq AD+DH$\\
$\therefore$ When points $A, D, H$ are collinear, and $D$ is on segment $AH$, $AH$ is maximum, the maximum value is $AH$,\\
Draw $AJ\perp BC$ through point $A$,\\
$\because \triangle ABC$ is an equilateral triangle, $AJ\perp BC$\\
$\therefore BJ=\frac{1}{2}BC=3$, $AJ=AB\cdot \sin60^\circ=3\sqrt{3}$,\\
$\because BD=2$,\\
$\therefore JD=1$,\\
In $\triangle ADJ$, $AD=\sqrt{AJ^2+JD^2}=\sqrt{{(3\sqrt{3})}^2+1}=2\sqrt{7}$\\
$\therefore AH=AD+DH=2+2\sqrt{7}$\\
Connect $AP$, extend $AE$ to $T$, such that $ET=DH=2$, connect $TP$, let the perpendicular from $AH$ to line $TP$ at $HM$ intersect at point $Q$, connect $AQ$,\\
$\because MA=MP, \angle AMP=60^\circ$\\
$\therefore \triangle AMP$ is an equilateral triangle\\
$\therefore AM=AP$, $ \angle PAM=60^\circ$\\
$\because \angle TAP=\angle TAH$\\
$\therefore \boxed{PH=2}$


Question: As shown in the figure, in parallelogram $ABCD$, the parallelogram is folded such that point $D$ falls on point $F$ on side $BC$, with the crease being $CE$. Connect $EF$, $CE$, $DF$, where $CE$ and $DF$ intersect at point $G$, and connect $BG$. Quadrilateral $DEFC$ is a rhombus. If $AB=10$, $AD=20$, and $\angle ABC=60^\circ$, find the length of $BG$.

Solution:
\textbf{Solution:} As shown in the diagram, draw $GH\perp BC$ at point $H$ passing through $G$,\\
since quadrilateral $ABCD$ is a parallelogram and $\angle ABC=60^\circ$,\\
therefore, $\angle ADC=60^\circ$.\\
Because quadrilateral $DEFC$ is a rhombus and $AB=10$,\\
thus $AB=CD=10$, $DF\perp CE$, $\angle FDC=30^\circ$, and $\angle ECF=60^\circ$,\\
hence $CG=\frac{1}{2}CD=5$, $GH=CG\cdot \sin 60^\circ=\frac{\sqrt{3}}{2}\times 5=\frac{5\sqrt{3}}{2}$,\\
therefore, $CH=\frac{1}{2}CG=\frac{5}{2}$.\\
Because quadrilateral $ABCD$ is a parallelogram,\\
therefore $BC=AD=20$,\\
thus $BH=BC-CH=20-\frac{5}{2}=\frac{35}{2}$,\\
therefore, $BG=\sqrt{GH^2+BH^2}=\sqrt{\left(\frac{5\sqrt{3}}{2}\right)^2+\left(\frac{35}{2}\right)^2}=\boxed{5\sqrt{13}}$


Question: As shown in Figure 1, it is known that the parabola $y= -\frac{\sqrt{3}}{3}x^{2}+\frac{2\sqrt{3}}{3}x+\sqrt{3}$ intersects the $x$-axis at points $A$ and $B$ (with point $A$ to the left of point $B$) and intersects the $y$-axis at point $C$. Point $D$ is the symmetric point of point $C$ with respect to the axis of symmetry of the parabola. Connect $CD$, and draw $DH \perp x$-axis at point $H$. Draw $AE \perp AC$ from point $A$, intersecting the extension line of $DH$ at point $E$. Find the length of the line segment $DE$.

Solution: \textbf{Solution:} Let $x=0$, we get $y=\sqrt{3}$, thus $C(0,\sqrt{3})$ and $D(2,\sqrt{3})$, therefore $DH = \sqrt{3}$. Let $y=0$, that is $-\frac{\sqrt{3}}{3}x^2 + 2\frac{\sqrt{3}}{3}x + \sqrt{3} = 0$, we find $x_{1}=-1,x_{2}=3$, hence $A(-1,0)$ and $B(3,0)$, therefore $\triangle ACO \sim \triangle AEH$, hence $\frac{OC}{AH} = \frac{OA}{EH}$, that is $\frac{\sqrt{3}}{3} = \frac{1}{EH}$, therefore $EH = \sqrt{3}$, hence $DE = \boxed{2\sqrt{3}}$.


Question: As shown in Figure 1, it is known that the parabola $y= -\frac{\sqrt{3}}{3}x^{2}+\frac{2\sqrt{3}}{3}x+\sqrt{3}$ intersects the $x$-axis at points $A$ and $B$ (with point $A$ to the left of point $B$), and intersects the $y$-axis at point $C$. Point $D$ is the symmetrical point of point $C$ with respect to the axis of symmetry of the parabola. Connect $CD$, and draw $DH \perp x$-axis at point $H$ through point $D$. Draw $AE \perp AC$ through point $A$, and let it intersect the extension of $DH$ at point $E$. As shown in Figure 2, try to find a point $F$ on segment $AE$ and a point $P$ on the parabola above the straight line $PF$. Calculate the maximum value of the area of $\triangle MPF$ when the perimeter of $\triangle CPF$ is at its minimum.

Solution:
\textbf{Solution:} As shown in Fig. 2, find the symmetric point N (4, $\sqrt{3}$) of point C with respect to DE, and the symmetric point G ($-$2, $-\sqrt{3}$) of point C with respect to AE. Connect GN, intersect AE at point F and DE at point P. Connect CP and CF. When points G, F, P, and N are collinear, the perimeter of $\triangle CPF = CF + PF + CP = GF + PF + PN$ is minimized,\\
The equation of line GN: $y = \frac{\sqrt{3}}{3}x - \frac{\sqrt{3}}{3}$, and the equation of line AE: $y = -\frac{\sqrt{3}}{3}x - \frac{\sqrt{3}}{3}$,\\
By solving these equations simultaneously, we get: F (0, $-\frac{\sqrt{3}}{3}$), P (2, $\frac{\sqrt{3}}{3}$). A line parallel to the y-axis through point M intersects FP at point Q. Let point M (m, $-\frac{\sqrt{3}}{3}m^{2}+\frac{2\sqrt{3}}{3}m+\sqrt{3}$), thus Q (m, $\frac{\sqrt{3}}{3}m-\frac{\sqrt{3}}{3}$), (0 < m < 2),\\
$\therefore \text{Area of }\triangle MFP = \text{Area of }\triangle MQF + \text{Area of }\triangle MQP = \frac{1}{2} \times 2 \cdot MQ = MQ = -\frac{\sqrt{3}}{3}m^{2}+\frac{\sqrt{3}}{3}m+\frac{4\sqrt{3}}{3}$,\\
$\therefore \text{Area of }\triangle MFP = -\frac{\sqrt{3}}{3}(m-\frac{1}{2})^{2}+\frac{17}{12}\sqrt{3}$,\\
$\because$ The axis of symmetry is: $m=\frac{1}{2}<2$, and the parabola opens downwards,\\
$\therefore$ When $m=\frac{1}{2}$, $\text{Area of }\triangle MFP$ is maximized, with the maximum value being $\boxed{\frac{17}{12}\sqrt{3}}$;


Question: As shown in Figure 1, it is known that the parabola $y= -\frac{\sqrt{3}}{3}x^{2}+\frac{2\sqrt{3}}{3}x+\sqrt{3}$ intersects with the $x$-axis at points $A$ and $B$ (with point $A$ on the left side of point $B$), and intersects with the $y$-axis at point $C$. Point $D$ is the symmetric point of point $C$ with respect to the axis of symmetry of the parabola. Connect $CD$, draw $DH \perp x$-axis at point $H$ through point $D$, and draw $AE \perp AC$ through point $A$, intersecting the extension line of $DH$ at point $E$. Under the conditions of question (2), the obtained $\triangle CFP$ is translated along the line $AE$ to get $\triangle C'F'P'$, and $\triangle C'F'P'$ is folded along $C'P'$ to get $\triangle C'P'F"$, denoting that during the translation process, the line $F'P'$ intersects the $x$-axis at point $K$. When $\triangle NF'F"K$ is an isosceles triangle, directly write out the value of $OK$.

Solution: \textbf{Solution:} We have $OK = \boxed{3}$.


Question: In $\triangle ABC$, $AB = AC$, $\angle ABC = 30^\circ$, $\triangle BDE$ is an equilateral triangle, and $CD$ and $AE$ are connected. As shown in Figure 1, when points $A$, $B$, and $D$ are on the same line, $AE$ and $BC$ intersect at point $P$, and $AE \perp AC$. If $PC = 4$, what is the length of $PE$?

Solution:
\textbf{Solution:}\\
Since $AB=AC$, and $\angle ABC=30^\circ$\\
Then, $\angle ACP=30^\circ$\\
Since $AE\perp AC$\\
Then, $\angle PAC=90^\circ$\\
Hence, $AP= \frac{1}{2}PC=2$\\
Since $\angle APC=\angle ABC+\angle BAE=60^\circ$\\
Then, $\angle ABC=\angle BAE=30^\circ$\\
Hence, $BP=AP=2$\\
Since $\angle ABE=60^\circ$\\
Then, $\angle PBE=30^\circ$, $\angle PEB=90^\circ$\\
Hence, $PE= \boxed{1}$


Question: As shown in the figure, in $\triangle ABC$, the line segment $AC$ is rotated $60^\circ$ counterclockwise around point $C$ to the line segment $CD$, where point $D$ falls exactly on side $AB$. Point $E$ is a point on side $BC$, and $AE$ and $DE$ are connected, with $\angle AEC=\angle BED=60^\circ$. Given that $AC=\sqrt{6}$ and $\angle BCD=15^\circ$, what is the length of $AE$?

Solution:
\textbf{Solution:} Construct $CH\perp AE$ at point H,

$\because$ rotating line segment AC about point C by $60^\circ$ counterclockwise to line segment CD,

$\therefore AC= CD$, $\angle ACD=60^\circ$,

$\therefore \triangle ACD$ is an equilateral triangle, thus $\angle CAD=\angle CDA=60^\circ$, $AD=AC=CD$,

$\because \angle BCD=15^\circ$, $\angle AEC=\angle BED=60^\circ$,

$\therefore \angle EAD=180^\circ-\angle CDA-\angle AQD=180^\circ-\angle CEA-\angle CQE=\angle ECQ=15^\circ$,

$\therefore \angle CAH=\angle CAD-\angle EAD =45^\circ$,

$\because AC= \sqrt{6}$, $\angle CAH=45^\circ$, $CH\perp AE$,

$\therefore CH=AH= \frac{\sqrt{2}}{2}AC= \sqrt{3}$,

$\because \angle CEH=60^\circ$,

$\therefore EH= \frac{CH}{\tan 60^\circ}=1$,

$\therefore AE= AH+ EH= \sqrt{3}+1=\boxed{\sqrt{3}+1}$;


Question: As shown in the figure, in rectangle $ABCD$, $BC = 1$, $AB = 2$. A perpendicular line is drawn from a point $P$ on the diagonal $BD$ to intersect $AB$ at point $F$ and $CD$ at point $E$. Line $EG$ is drawn parallel to $BD$ from point $E$ and intersects $BC$ at point $G$. Lines $FG$ and $DF$ are drawn to intersect $BD$ at point $H$. Calculate the value of $ \frac{DE}{BG}$.

Solution:
\textbf{Solution:} As shown in Figure 1, since quadrilateral ABCD is a rectangle,\\
it follows that $\angle ABC=90^\circ$,\\
since EF$\perp$AB,\\
it follows that $\angle AFE=90^\circ$,\\
thus $\angle ABC=\angle AFE$,\\
thus EF$\parallel$BC,\\
since EG$\parallel$BD,\\
it follows that quadrilateral EPBG is a parallelogram,\\
thus EP=BG,\\
thus $\tan \angle EDP= \frac{EP}{DE} = \frac{BC}{CD} = \frac{1}{2}$,\\
thus $\frac{DE}{BG} = \boxed{2}$;


Question: As shown in the figure, in rectangle $ABCD$, $BC=1$, $AB=2$. A perpendicular line is drawn from a point $P$ on the diagonal $BD$ to intersect $AB$ at point $F$ and $CD$ at point $E$. A line is drawn through point $E$ such that $EG \parallel BD$ and intersects $BC$ at point $G$. The line $FG$ intersects $BD$ at point $H$, and $DF$ is connected. When the quadrilateral $DFGE$ has one pair of adjacent sides equal, what is the length of $BG$?

Solution:
\textbf{Solution:}

(1) As shown in Figure 1, when $DE=EG$, let $BG=x$, then $DE=EG=2x$, $CE=2-2x$, $CG=1-x$. In $\triangle CEG$ with a right angle, $CE^{2}+CG^{2}=EG^{2}$,
\[
\therefore (2-2x)^{2}+(1-x)^{2}=(2x)^{2},
\]
solving this yields $x=5-2\sqrt{5}$,
\[
\therefore \boxed{BG=5-2\sqrt{5}};
\]

(2) As shown in Figure 1, when $EG=GF$,
\[
\because CE=BF, \angle C=\angle ABC=90^\circ,
\]
\[
\therefore \text{Rt}\triangle ECG\cong \text{Rt}\triangle FBG \text{ (by the HL congruence theorem)},
\]
\[
\therefore \boxed{BG=CG=\frac{1}{2}CB=\frac{1}{2}},
\]

(3) As shown in Figure 1, when $DF=GF$, let $BG=x$, then $AF=DE=2x$,
\[
\because DF^{2}=GF^{2},
\]
\[
\therefore DA^{2}+AF^{2}=BG^{2}+BF^{2},
\]
\[
\therefore 1^{2}+(2x)^{2}=(2-2x)^{2}+x^{2},
\]
solving this yields $x=4\pm\sqrt{13}$,
\[
\because BG<1,
\]
\[
\therefore \boxed{BG=4-\sqrt{13}};
\]

(4) $\because \angle DEF=90^\circ$,
\[
\therefore DF>DE, \text{ i.e., } DF=DE \text{ does not exist};
\]

In summary, the length of $BG$ can be \boxed{5-2\sqrt{5}}, \boxed{\frac{1}{2}}, or \boxed{4-\sqrt{13}}.


Question: As shown in the figure, in rectangle $ABCD$, $BC=1$, $AB=2$. A perpendicular line to $AB$ through a point $P$ on the diagonal $BD$ intersects $AB$ at point $F$ and $CD$ at point $E$. A line through $E$ such that $EG \parallel BD$ intersects $BC$ at point $G$. Connect $FG$ and $BD$ at point $H$, and connect $DF$. The symmetrical point of $B$ with respect to $FG$ is denoted as $B'$. If $B'$ lies inside $\triangle EFG$ (excluding the boundary), find the range of values for the length of $DP$.

Solution:
\textbf{Solution:}

When point $B^{\prime}$ is on the side $EG$, let $BG = x$ and $B^{\prime}F = BF = CE = 2-2x$,\\
$\because$ $\angle FB^{\prime}G = \angle FBG = 90^\circ$,\\
$\therefore$ $\angle EFB^{\prime} = \angle CEG = \angle CDB$, $\angle C = \angle EB^{\prime}F = 90^\circ$,\\
$\therefore$ $\triangle EFB^{\prime} \sim \triangle BDC$,\\
$\therefore$ $\frac{B^{\prime}F}{EF} = \frac{DC}{BD}$,\\
$\therefore$ $\frac{2-2x}{1} = \frac{2}{\sqrt{5}}$,\\
Solving this, we get $x = 1-\frac{\sqrt{5}}{5}$,\\
$\therefore$ $DP = \sqrt{5}$ $EP = \sqrt{5}-1$\\
When point $B^{\prime}$ is on the side $EF$,\\
$\because$ $BG = B^{\prime}G = CE$,\\
$\therefore$ $x = 2-2x$,\\
Solving this, we get $x = \frac{2}{3}$,\\
$\therefore$ $DP = \sqrt{5}x = \frac{2\sqrt{5}}{3}$,\\
In summary, $\boxed{\sqrt{5}-1 < DP < \frac{2\sqrt{5}}{3}}$.


Question: As shown in the figure, within rectangle $ABCD$, point $E$ is on side $CD$. When $\triangle BCE$ is folded along $BE$ such that point $C$ coincides with point $F$ on side $AD$, and a line is drawn from point $F$ such that $FG \parallel CD$ and intersects $BE$ at point $G$, and $CG$ is connected. Quadrilateral $CEFG$ is a rhombus. Given that $AB=6$ and $AD=10$, what is the area of quadrilateral $CEFG$?

Solution:
\textbf{Solution:} Given rectangle ABCD, where AD=10,\\
it follows that BC=10.\\
Given that $\triangle$BCE is folded along BE, such that point C lands on point F on side AD,\\
it then follows that BF=BC=10.\\
In $\triangle$ABF, where AB=6, AF=$\sqrt{BF^{2}-AB^{2}}$=8,\\
thus, DF=AD−AF=2.\\
Let EF=x, then CE=x, and DE=6−x.\\
In $\triangle$DEF, we have $DF^{2}+DE^{2}=EF^{2}$,\\
thus, $2^{2}+(6-x)^{2}=x^{2}$. Solving this, we find $x=\frac{10}{3}$,\\
therefore, CE=$\frac{10}{3}$.\\
Hence, the area of quadrilateral CEFG is: CE$\cdot$DF=$\frac{10}{3}\times 2=\boxed{\frac{20}{3}}$.


Question: As shown in the figure, it is known that the parabola $y= \frac{1}{4}x^{2}+bx+c$ passes through point $A(0, 2)$ and point $C(4, 0)$, and intersects the x-axis at another point $B$. What is the analytic equation of the parabola?

Solution:
\textbf{Solution:} Substituting points A and C into $y= -\frac{1}{4}x^2+bx+c$ yields,
\[
\left\{\begin{array}{l}
c=2\\
4+4b+c=0
\end{array}\right.
\] Solving this gives \[
\left\{\begin{array}{l}
b=\frac{3}{2}\\
c=2
\end{array}\right.
\]
$\therefore$ The equation of the parabola is $\boxed{y= -\frac{1}{4}x^{2}+ \frac{3}{2}x+2}$


Question: Given a parabola $y= \frac{1}{4}x^{2}+bx+c$ passing through the points $A(0, 2)$ and $C(4, 0)$, and intersecting the $x$-axis at another point $B$. There is a point $M$ on the parabola above the line $AC$. What is the maximum area of the triangle $ACM$ and the coordinates of the point $M$ at that moment?

Solution:
\textbf{Solution:} Through point $M$, construct $MN\perp x$ axis, intersecting $AC$ at point $N$, as shown in Figure 1,\\
Let $M(a, -\frac{1}{4}a^2+ \frac{1}{2}a+2)$, then $N(a, -\frac{1}{2}a+2)$,\\
$S_{\triangle ACM} = \frac{1}{2}MN \cdot OC = \frac{1}{2}(-\frac{1}{4}a^2+a) \times 4 = -\frac{1}{2}a^2+2a$,\\
$\therefore S_{\triangle ACM} = -\frac{1}{2}a^2+2a$,\\
$\therefore$ when $a=2$, the area of triangle $ACM$ is maximized, with its maximum value being $\boxed{2}$,\\
at this moment, the coordinates of $M$ are $\boxed{(2, 2)}$;


Question: As shown in the figure, it is known that the parabola $y= \frac{1}{4}x^{2}+bx+c$ passes through the points $A(0,2)$ and $C(4,0)$, and intersects the $x$-axis at another point $B$. When the segment $OA$ is rotated clockwise by $90^\circ$ around a moving point $P(m,0)$ on the $x$-axis to obtain the segment $O'A'$, if the segment $O'A'$ has only one intersection point with the parabola, please combine the function graph to determine the range of values for $m$.

Solution:
\textbf{Solution:}\\
Since segment 0A is rotated clockwise by $90^\circ$ around the moving point P(m, 0) on the x-axis to obtain segment O'A’,\\
thus PO'=P0=m, O'A'=OA=2,\\
thus O'(m, m), A'(m+2, m),\\
When A'(m+2, m) is on the parabola, it satisfies $-\frac{1}{4}(m+2)^{2}$+ $\frac{1}{2}(m+2)+2=m$,\\
Solving this gives, $m=-3\pm \sqrt{17}$,\\
When point O'(m, m) is on the parabola, it satisfies $-\frac{1}{4}m^{2}$+ $\frac{1}{2}m+2=m$,\\
Solving this gives, $m=-4$ or $2$,\\
therefore when $-3-\sqrt{17}\leq m\leq -4$ or $-3+\sqrt{17}\leq m\leq 2$, segment O'A' and the parabola have only one common point.\\
Hence, the range of $m$ is $m=\boxed{-4}$ or $\boxed{2}$.


Question: As shown in Fig. 1, $Rt\triangle BAC$ and $Rt\triangle DAE$ are both isosceles right-angled triangles, and $AB=AC=2\sqrt{5}$, $AD=AE=2\sqrt{2}$, $\angle BAC=\angle DAE=90^\circ$. $\triangle DAE$ rotates counterclockwise around point A; connect CD. When $\tan\angle EAC=\frac{1}{3}$, find the length of CD.

Solution:
\textbf{Solution:} Draw $DW \perp AC$ at $W$,\\
since $\angle DAE=90^\circ$ and $\tan\angle EAC=\frac{1}{3}$,\\
thus $\angle ADW+\angle DAW=\angle DAW+\angle EAC$,\\
hence $\angle EAC=\angle ADW$,\\
in right triangle $\triangle AWD$ with $\angle AWD=90^\circ$,\\
thus $\tan\angle ADW=\frac{AW}{DW}=\frac{1}{3}$,\\
assume $DW=3k$ and $AW=k$, then $(3k)^2+k^2=(2\sqrt{2})^2$,\\
hence $AW=\frac{2\sqrt{5}}{5}$ and $DW=\frac{6\sqrt{5}}{5}$,\\
thus $WC=AC-AW=\frac{8\sqrt{5}}{5}$,\\
in right triangle $\triangle DWC$ with $\angle DWC=90^\circ$,\\
thus $DW^2+WC^2=DC^2$,\\
therefore, $DC=\boxed{2\sqrt{5}}$.


Question: As shown in the figure, it is known that the graph of the inverse proportion function $y_{1}=\frac{k}{x}$ intersects with the graph of the linear function $y_{2}=ax+b$ at points $A(1, 4)$ and $B(m, -2)$. Find the value of $m$ and the equation of the linear function.

Solution:
\textbf{Solution:} Since the graph of the inverse proportion function $y_1 = \frac{k}{x}$ intersects with the graph of the linear function $y_2 = ax + b$ at points $A(1, 4)$ and $B(m, -2)$,\\
we have $k = 1 \times 4 = 4$,\\
thus, the equation of the inverse proportion function is: $y = \frac{4}{x}$,\\
hence, $-2 = \frac{4}{m}$, solving this gives $m = \boxed{-2}$, that is point $B(-2, -2)$,\\
therefore $\left\{\begin{array}{l} 4 = a + b \\ -2 = -2a + b \end{array}\right.$, solving this set gives $\left\{\begin{array}{l} a = 2 \\ b = 2 \end{array}\right.$,\\
thus, the equation of the linear function is: $\boxed{y = 2x + 2}$.


Question: As shown in the figure, the graph of the inverse proportion function $y_{1}=\frac{k}{x}$ intersects with the graph of the linear function $y_{2}=ax+b$ at points $A(1, 4)$ and $B(m, -2)$. If point $C$ is symmetrical to point $A$ with respect to the $x$-axis, what is the area of $\triangle ABC$?

Solution: \textbf{Solution:} Since point C is symmetric to point A with respect to the x-axis, and given A(1, 4), it follows that point C is at (1, -4). By plotting and connecting points AC and BC in the Cartesian coordinate system, we find that AC = 8 and BC = 3. Therefore, the area of triangle $ABC$ is given by $S_{\triangle ABC} = \frac{1}{2} \cdot AC \cdot BC = \frac{1}{2} \times 8 \times 3 = \boxed{12}$.


Question: As shown in the figure, rotate rectangle $ABCD$ clockwise about point $A$, causing point $E$ to fall on $BD$, thus forming rectangle $AEFG$. $EF$ intersects $AD$ at point $H$. Connect $AF$. $BD \parallel AF$; If $AB=1$ and $BC=2$, find the length of $AH$.

Solution: \textbf{Solution:} Given that $BD\parallel AF$, $\therefore \angle 3=\angle EFA$. Since $\angle 4=\angle EFA$, $\therefore \angle 4=\angle 3$. Hence, $EH=DH$. Given $AB=1$, $BC=2$, we have $AE=1$, $AD=2$. Let $EH=DH=x$, then $AH=2-x$. In $\triangle AEH$, by Pythagoras' theorem, we have $AE^2+EH^2=AH^2$, which implies $1^2+x^2=(2-x)^2$. Solving this, we find $x=\frac{3}{4}$, hence $AH=2-\frac{3}{4}=\boxed{\frac{5}{4}}$.


Question: As shown in the figure, AB is the diameter of circle $O$, with C and D being two points on the circle. $OC \parallel BD$, and $OC$ intersects $AD$ at point E. $AC = CD$; if $OE = 2$ and $AD = 8$, find the radius of circle $O$.

Solution:
\textbf{Solution:} From (1) we know that $OC\perp AD$,\\
$\because AD=8$,\\
$\therefore AE=\frac{1}{2}AD=4$,\\
$\because OE=2$,\\
$\therefore OA=\sqrt{OE^{2}+AE^{2}}=2\sqrt{5}$,\\
$\therefore$ the radius of circle $O$ is $\boxed{2\sqrt{5}}$.


Question: As shown in the figure, AD is the bisector of the exterior angle $\angle EAC$ of $\triangle ABC$, and AD intersects the circumcircle $O$ of $\triangle ABC$ at point D. $DB = DC$; if $\angle CAB=30^\circ$, and $BC=4$, what is the length of the minor arc $\overset{\frown}{DC}$?

Solution:
\textbf{Solution:} As shown in the diagram, connect OB, OC, and OD.\\
By the Circumference Angle Theorem, $\angle COB=2\angle CAB=60^\circ$, $\angle CDB=\angle CAB=30^\circ$,\\
$\therefore \triangle COB$ is an equilateral triangle,\\
$\therefore OC=BC=4$,\\
$\because DC=DB$, $\angle CDB=30^\circ$,\\
$\therefore \angle DCB=75^\circ$,\\
$\therefore \angle DCO=15^\circ$,\\
$\therefore \angle COD=150^\circ$,\\
Therefore, the length of the minor arc $\overset{\frown}{DC}$ is $= \frac{150\pi \times 4}{180}=\boxed{\frac{10}{3}\pi}$.


Question: As shown in the figure, it is known that $AB$ is the diameter of the circle $\odot O$, and $\odot O$ intersects side $AE$ of $\triangle ABE$ at point $D$. Connect $OD$, and it satisfies $OD\parallel BE$. Point $P$ is on the extension line of $BA$, and $PD$ intersects $BE$ at point $C$. $AB=BE$; If $PA=2$, $\angle B=60^\circ$, $PC\perp BE$, find the length of diameter $AB$.

Solution: \textbf{Solution:} Since $\angle PCB=90^\circ$ and $\angle B=60^\circ$, it follows that $\angle P=30^\circ$. From this, we have $OD\parallel BE$, hence $\angle B=\angle DOA=60^\circ$. Also, since $OA=OD$, it follows that $\triangle AOD$ is an equilateral triangle, thus $AD=OA$, and $\angle DAO=60^\circ$. Therefore, $\angle ADP=\angle P=30^\circ$, which implies that $PA=AD=OA=2$, hence $AB=2OA=\boxed{4}$.


Question: As illustrated, quadrilateral $ABCD$ is inscribed in circle $\odot O$, and diagonal $BD$ is the diameter. A tangent line $AE$ from point $A$ to circle $\odot O$ intersects the extension of $CD$ at point $E$. It is given that $DA$ bisects $\angle BDE$ and $AE \perp DE$; if the radius of $\odot O$ is $5$ and $CD=6$, what is the length of $AD$?

Solution:
\textbf{Solution:} Construct $OF \perp DC$ through point $O$, with the perpendicular foot at $F$,\\
$\therefore \angle OFD=90^\circ$,\\
$\because \angle OAE=\angle E=90^\circ$,\\
$\therefore$ Quadrilateral $OAEF$ is a rectangle,\\
$\therefore OA=EF=5$, $AE=OF$,\\
$\because OF\perp CD$,\\
$\therefore DF=\frac{1}{2}CD=3$,\\
$\therefore DE=EF-DF=5-3=2$,\\
$\therefore OF=\sqrt{OD^{2}-DF^{2}}=\sqrt{5^{2}-3^{2}}=4$,\\
$\because AE=OF=4$,\\
$\therefore AD=\sqrt{AE^{2}+DE^{2}}=\sqrt{4^{2}+2^{2}}=\boxed{2\sqrt{5}}$,\\
$\therefore$ The length of $AD$ is $2\sqrt{5}$.


Question: As shown in the diagram, $AB$ is the diameter of circle $O$, $AC$ is a chord of circle $O$, $D$ is the midpoint of arc $BC$, and $DE$ is drawn perpendicular to $AC$ at point $E$, intersecting the extension of $AB$ at point $F$, and $DA$ is connected. If $AB=90\text{cm}$, what is the distance from the center $O$ to the "lever $EF$"?

Solution:
\textbf{Solution:} Connect $OD$,\\
since $D$ is the midpoint of the arc $BC$,\\
it follows that $\angle CAD=\angle BAD$,\\
since $OA=OD$,\\
it follows that $\angle BAD=\angle ADO$,\\
hence $\angle CAD=\angle ADO$,\\
since $DE\perp AC$,\\
it follows that $\angle E=90^\circ$,\\
hence $\angle CAD+\angle EDA=90^\circ$, which means $\angle ADO+\angle EDA=90^\circ$,\\
it follows that $OD\perp EF$,\\
hence the length of $OD$ is the distance from the center $O$ to the "lever $EF$",\\
since $AB=90\text{cm}$,\\
it follows that $OD=OA=\boxed{45\text{cm}}$


Question: As shown in the figure, $AB$ is the diameter of circle $O$, $AC$ is a chord of $\odot O$, $D$ is the midpoint of arc $BC$, draw $DE\perp AC$ at point $E$, and intersect the extension line of $AB$ at point $F$, connect $DA$. If $DA=DF=6\sqrt{3}$, calculate the area of the shaded region. (Keep the result in terms of $\pi$)

Solution:
\textbf{Solution:} Given $DA=DF$, \\
it follows that $\angle F=\angle BAD$, \\
thus, we have $\angle CAD=\angle BAD$, \\
hence, $\angle F=\angle BAD=\angle CAD$, \\
since $\angle F+\angle BAD+\angle CAD=90^\circ$, \\
it implies $\angle F=\angle BAD=\angle CAD=30^\circ$, \\
therefore, $\angle BOD=2\angle BAD=60^\circ$, and $OF=2OD$, \\
given $DF=6\sqrt{3}$, \\
it leads to $(2OD)^{2}-OD^{2}=(6\sqrt{3})^{2}$, \\
solving yields: $OD=6$, \\
thus, $S_{\text{shaded}}=S_{\text{sector} BOD}+S_{\triangle AOD}=\frac{60\pi \times 6^{2}}{360}+\frac{1}{2}\times 6\times \frac{\sqrt{3}}{2}\times 6=6\pi +9\sqrt{3}=\boxed{6\pi +9\sqrt{3}}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\odot O$, with $AC$ being the diameter, and point $C$ as the midpoint of the minor arc $BD$. $AB=AD$, if $\angle ACD=60^\circ$, and $AD=\sqrt{3}$, find the length of $BD$.

Solution:
\textbf{Solution:} Given $\because$ $\overset{\frown}{AD}=\overset{\frown}{AD}$, $\angle ACD=60^\circ$,\\
$\therefore$ $\angle ABD=\angle ACD=60^\circ$,\\
$\because$ $AB=AD$,\\
$\therefore$ $\triangle ABD$ is an equilateral triangle,\\
$\because$ $AD=\sqrt{3}$,\\
$\therefore$ $BD=AD=\boxed{\sqrt{3}}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, with the chord $CD\perp AB$ at $E$, and $F$ is a point on the extension of $AB$. Connect $CF$ and $DF$. Given that $OE = 3$, $BE = 2$, find the length of $CD$.

Solution:
\textbf{Solution:} Connect OC, \\
$\because$ CD$\perp$AB, \\
$\therefore$ CE = DE, \\
$\therefore$ OC = OB = OE + BE = 3 + 2 = 5, \\
In $\triangle OCE$, $\angle OEC = 90^\circ$, by the Pythagorean theorem, we have: \\
CE$^{2}$ = OC$^{2}$ - OE$^{2}$, \\
$\therefore$ CE$^{2}$ = 5$^{2}$ - 3$^{2}$, \\
$\therefore$ CE = 4, \\
$\therefore$ CD = 2CE = \boxed{8}.


Question: As shown in the figure, within the circle $\odot O$, chord $AC$ intersects chord $BD$ at point $P$, with $AC=BD$ and $AP=BP$. Connect $AB$. If $AB=8$, $BP=5$, and $DP=3$, what is the radius of the circle $\odot O$?

Solution:
\textbf{Solution:} Extend line $PO$ and let it intersect $AB$ at point $E$, connect $OA$, $OB$, and draw $OF\perp AC$ at point $F$, \\
$\therefore AF=\frac{1}{2}AC$, \\
$\because AP=BP, OA=OB$, \\
$\therefore PE$ is the perpendicular bisector of $AB$, \\
$\therefore PE\perp AB, AE=\frac{1}{2}AB=4$, \\
$\because AB=8, BP=5, DP=3, AC=BD$, \\
$\therefore AC=BD=AB=8, AP=5$, \\
$\therefore AF=4=AE, PF=AP-AF=1, PE=\sqrt{AP^{2}-AE^{2}}=3$, \\
In right triangles $\triangle AOE$ and $\triangle AOF$, \\
$\left\{\begin{array}{l}AE=AF \\ OA=OA\end{array}\right.$, \\
$\therefore \triangle AOE\cong \triangle AOF(HL)$, \\
$\therefore OE=OF$, let $OE=OF=x(x>0)$, then $OP=PE-OE=3-x$, \\
In $\triangle POF$, $OF^{2}+PF^{2}=OP^{2}$, solving gives $x^{2}+1^{2}=(3-x)^{2}$, we find $x=\frac{4}{3}$, \\
In $\triangle AOE$, $OA=\sqrt{AE^{2}+OE^{2}}=\sqrt{4^{2}+{(\frac{4}{3})}^{2}}=\frac{4\sqrt{10}}{3}$, \\
thus, the radius of $\odot O$ is $\boxed{\frac{4\sqrt{10}}{3}}$.


Question: As shown in the diagram, it is known that $AB$ is the diameter of $\odot O$ and $AB=8C$. Point $D$ is on $\odot O$, $OC \parallel BD$, intersecting $AD$ at point $E$. Connect $BC$, where $\angle CBD=30^\circ$. What is the degree measure of $\angle COA$?

Solution:
\textbf{Solution}: Since $OC \parallel BD$,\\
it follows that $\angle OCB = \angle CBD = 30^\circ$,\\
Since $OC = OB$,\\
it follows that $\angle OCB = \angle OBC = 30^\circ$,\\
Therefore, $\angle COA = \angle OCB + \angle OBC = \boxed{60^\circ}$;


Question: As shown in the figure, it is known that $AB$ is the diameter of $\odot O$, and $AB=8C$. Point $D$ is on $\odot O$, $OC\parallel BD$, intersecting $AD$ at point $E$. Connect $BC$, with $\angle CBD=30^\circ$. What is the length of $CE$?

Solution:
\textbf{Solution:} Since AB is the diameter of $\odot O$,\\
$\therefore \angle ADB=90^\circ$,\\
Since OC $\parallel$ BD,\\
$\therefore \angle AEO=\angle ADB=90^\circ$,\\
Since $\angle AOC=60^\circ$,\\
$\therefore \angle OAE=30^\circ$,\\
$\therefore$ OE=$\frac{1}{2}$ OA,\\
$\therefore$ CE=$\frac{1}{2}$ OC=$\frac{1}{2} \times 4=\boxed{2}$;


Question: As shown in the diagram, it is known that $AB$ is the diameter of $\odot O$, and $AB=8C$, $D$ is a point on $\odot O$, $OC\parallel BD$, intersecting $AD$ at point $E$, and $BC$ is connected, $\angle CBD=30^\circ$. Calculate the area of the shaded region in the diagram (the result should retain $\pi$).

Solution:
\[ \text{Solution: Connect } OD, \]
\[ \because \angle CBD = \angle OBC = 30^\circ, \]
\[ \therefore \angle BOD = 60^\circ, \]
\[ \because OB = OD, \]
\[ \therefore \triangle BOD \text{ is an equilateral triangle}, \]
\[ \therefore \text{Area of the shaded region} = \text{Area of sector } BOD - \text{Area of } \triangle BOD = \frac{60\pi \times 4^2}{360} - \frac{1}{2} \times 4 \times \frac{\sqrt{3}}{2} \times 4 = \boxed{\frac{8}{3}\pi -4\sqrt{3}}. \]


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, the angle bisectors of $\angle BAC$ and $\angle ABC$ intersect at point $E$, and the extended line of $AE$ intersects the circumcircle of $\triangle ABC$ at point $D$, connect $BD$. If $AB=5$, $BC=8$, and $\angle BAD=\angle DBC$; points $B$, $E$, $C$ are on the same circle centered at point $D$; what is the distance between the incenter and the circumcenter of $\triangle ABC$?

Solution:
\textbf{Solution:} As shown in the diagram:\\
$\because BD=DC, \angle ABD=\angle ACD=90^\circ, AD=AD$,\\
$\therefore Rt\triangle ABD\cong Rt\triangle ACD(HL)$,\\
$\therefore AB=AC$,\\
$\because AH=AH, \angle BAH=\angle CAH$,\\
$\therefore \triangle ABH\cong \triangle ACH(SAS)$,\\
$\therefore BH=CH$,\\
$\therefore BH=\frac{1}{2}BC=4$,\\
$\therefore \angle AHB=\angle AHC=90^\circ$,\\
$\therefore AD\perp BC$,\\
In $Rt\triangle ABH$, $AH=3$,\\
In $Rt\triangle BHO$, let $BO=x$, $OH=x-3$,\\
Then $BO^2=BH^2+OH^2$,\\
i.e., $x^2=16+(x-3)^2$,\\
Solving yields: $x=\frac{25}{6}$,\\
i.e., $BO=\frac{25}{6}$,\\
$\because AD$ is the diameter,\\
$\therefore \angle ABD=90^\circ$,\\
In $Rt\triangle ABD$,\\
$BD=\sqrt{AD^2-AB^2}=\frac{20}{3}$,\\
$\therefore DE=\frac{20}{3}$,\\
$\therefore OE=\frac{20}{3}-\frac{25}{6}=\frac{5}{2}$,\\
$\because E$ is the intersection of the angle bisectors of $\triangle ABC$,\\
$\therefore E$ is the incenter,\\
$\therefore OE$ is the distance between the incenter and circumcenter of $\triangle ABC$,\\
$\therefore The distance between the incenter and circumcenter of \triangle ABC$ is $\boxed{\frac{5}{2}}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, PD is tangent to $\odot O$ at point C, and intersects the extension line of AB at point D, and $\angle D = 2\angle CAD$. What is the degree measure of $\angle D$?

Solution:
\textbf{Solution:} Since $OA=OC$,\\
it follows that $\angle A=\angle ACO$,\\
thus $\angle COD=\angle A+\angle ACO=2\angle A$,\\
given $\angle D=2\angle A$,\\
it implies $\angle D=\angle COD$,\\
since line $PD$ is tangent to circle $O$ at point $C$,\\
therefore $\angle OCD=90^\circ$,\\
hence $\angle D=\angle COD=\boxed{45^\circ}$;


Question: As shown in the figure, AB is the diameter of the circle $\odot O$, PD is tangent to $\odot O$ at point C, and intersects the extension of AB at point D, and $\angle D=2\angle CAD$. If CD=1, what is the length of BD?

Solution:
\textbf{Solution}: Since $\angle D = \angle COD$ and $CD=1$,\\
it follows that $OC=OB=CD=1$,\\
In $\triangle OCD$, by the Pythagorean theorem, we have $OD=\sqrt{1^2+1^2}=\sqrt{2}$,\\
therefore, $BD=OD-OB=\boxed{\sqrt{2}-1}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=45^\circ$, OC $\parallel$ AD, and AD intersects the extension line of BC at D, AB intersects OC at E. AD is a tangent to the circle $O$; If AE=$2\sqrt{5}$ and CE=2, what are the radius of circle $O$ and the length of the line segment BC?

Solution:
\textbf{Solution:} Let the radius of $\odot O$ be $R$, hence $OA=R$, $OE=R-2$.\\
In $\triangle OAE$, $AO^2+OE^2=AE^2$,\\
$\therefore R^2+(R-2)^2=(2\sqrt{5})^2$,\\
we get $R_1=\boxed{4}$ or $R_2=-2$ (which is irrelevant to the problem, so discard),\\
Extend $CO$ to intersect $\odot O$ at $F$, and draw $AF$,\\
$\because \angle AEF=\angle CEB$, $\angle B=\angle AFE$,\\
$\therefore \triangle CEB\sim \triangle AEF$,\\
$\therefore \frac{AE}{CE}=\frac{AF}{BC}$,\\
$\because CF$ is the diameter,\\
$\therefore CF=8$, $\angle CAF=90^\circ$,\\
Also $\because \angle F=\angle ABC=45^\circ$,\\
$\therefore \angle F=\angle ACF=45^\circ$,\\
$\therefore AF=4\sqrt{2}$,\\
$\therefore \frac{2\sqrt{5}}{2}=\frac{4\sqrt{2}}{BC}$,\\
$\therefore BC=\boxed{\frac{4\sqrt{10}}{5}}$.


Question: As shown in the figure, it is known that $AB$ is the diameter of $\odot O$, and the chord $AC$ is parallel to the radius $OD$. Point $D$ is the midpoint of the arc $\overset{\frown}{BC}$. If $AC=OD=6$, find the area of the shaded part (the segment $AC$).

Solution:
\textbf{Solution:} Let $CE\perp AB$ at E,\\
as shown in Figure 2, $\because AC=OD=OC=OA$,\\
$\therefore \triangle AOC$ is an equilateral triangle,\\
$\therefore \angle AOC=60^\circ$,\\
$\therefore {S}_{\text{sector} AOC}=\frac{60\pi \times {6}^{2}}{360}=6\pi$,\\
$\therefore \angle OCE=30^\circ$,\\
$\therefore OE=\frac{1}{2}OC=\frac{1}{2}\times 6=3$,\\
$\therefore CE=\sqrt{3}OE=3\sqrt{3}$,\\
$\therefore {S}_{\triangle AOC}=\frac{1}{2}AO\cdot CE=\frac{1}{2}\times 6\times 3\sqrt{3}=9\sqrt{3}$,\\
$\therefore {S}_{\text{shaded}}={S}_{\text{sector} AOC}-{S}_{\triangle AOC}=6\pi -9\sqrt{3}=\boxed{6\pi -9\sqrt{3}}$.


Question: As shown in the figure, assume $\odot O$ is the cross-section of a water pipe, with the width of the water surface $AB=48\,cm$, and the maximum depth of the water being $36\,cm$. What is the radius of $\odot O$?

Solution:
\textbf{Solution:} As shown in the figure, draw $OD\perp AB$ at D, extend DO to intersect circle O at C, and connect AO,\\
According to the problem, $AB=48\text{cm}$, $CD=36\text{cm}$,\\
$\therefore$ by the Perpendicular Diameter Theorem, $AD=\frac{1}{2}AB=24$,\\
Let $OA=OC=x\text{cm}$, then $OD=CD-OD=(36-x)$,\\
$\because$ $AD^2+OD^2=AO^2$,\\
$\therefore$ $24^2+(36-x)^2=x^2$,\\
Solving for $x$, we get $x=26$,\\
$\therefore$ the radius of circle O is $\boxed{26\text{cm}}$.