Question: As shown in the diagram, in $\triangle ABC$, $AB=AC$ and $\angle B=40^\circ$. Point D moves along the line segment $BC$ (D does not coincide with B or C). Connect $AD$ and construct $\angle ADE=40^\circ$, with $DE$ intersecting line segment $AC$ at $E$. As point D moves, the shape of $\triangle ADE$ also changes. When $\triangle ADE$ is an isosceles triangle, what is the degree measure of $\angle BDA$?

Solution:
\textbf{Solution:} Since AB = AC,\\
it follows that $\angle B = \angle C = 40^\circ$,\\
(1) When AD = AE, $\angle ADE = \angle AED = 40^\circ$,\\
since $\angle AED > \angle C$,\\
this case does not hold;\\
(2) When DA = DE, that is $\angle DAE = \angle DEA = \frac{1}{2}(180^\circ - 40^\circ) = 70^\circ$,\\
since $\angle BAC = 180^\circ - 40^\circ - 40^\circ = 100^\circ$,\\
it follows that $\angle BAD = 100^\circ - 70^\circ = 30^\circ$;\\
therefore $\angle BDA = 180^\circ - 30^\circ - 40^\circ = \boxed{110^\circ}$;\\
(3) When EA = ED, $\angle ADE = \angle DAE = 40^\circ$,\\
it follows that $\angle BAD = 100^\circ - 40^\circ = 60^\circ$,\\
therefore $\angle BDA = 180^\circ - 60^\circ - 40^\circ = 80^\circ$;\\
therefore, when $\triangle ADE$ is an isosceles triangle, the degree of $\angle BDA$ is $110^\circ$ or $80^\circ$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, and the points $D$, $E$, $F$ are on the sides $BC$, $AB$, $AC$, respectively, with $CD=BE$ and $BD=CF$. If $\angle EDF = 42^\circ$, what is the measure of $\angle BAC$?

Solution:
\textbf{Solution:} Since $AB=AC$, \\
it follows that $\angle B=\angle C$, \\
In $\triangle BDE$ and $\triangle CFD$, \\
$\begin{cases}
BD=CF\\
\angle B=\angle C\\
BE=CD
\end{cases}$, \\
hence, $\triangle BDE \cong \triangle CFD$, \\
therefore, $\angle EDB=\angle DFC$, $\angle FDC=\angle BED$, \\
since $\angle EDF+\angle BDE+\angle FDC=180^\circ$, $\angle B+\angle BED+\angle EDB=180^\circ$, \\
it follows that $\angle B=\angle EDF=42^\circ$, \\
thus, $\angle BAC=180^\circ-\angle B-\angle C=\boxed{96^\circ}$.


Question: As shown in the figure, $BD$ is the angle bisector of $\triangle ABC$, with $AB=15$, $BC=9$, and $AC=12$. What is the length of $BD$?

Solution:
\textbf{Solution:} As shown in the diagram, draw $DE\perp AB$ at point $E$,\\
because $AB=15$, $BC=9$, and $AC=12$,\\
therefore, $AB^{2}=225$, and $BC^{2}+AC^{2}=9^{2}+12^{2}=225$\\
therefore, $AB^{2}=BC^{2}+AC^{2}$\\
therefore, $\triangle ABC$ is a right-angled triangle\\
therefore, $\angle C=90^\circ$\\
therefore, $DC\perp BC$\\
because $BD$ is the angle bisector of $\triangle ABC$,\\
therefore, $DE=DC$\\
In $Rt\triangle DEB$ and $Rt\triangle DCB$\\
$\left\{\begin{array}{l}DB=DB\\ DC=DE\end{array}\right.$\\
therefore, $Rt\triangle DEB \cong Rt\triangle DCB$\\
therefore, $BE=BC=9$\\
therefore, $AE=AB-BE=15-9=6$\\
Let $DC=DE =x$, then $AD=AC-DC=12-x$\\
In $Rt\triangle ADE$, $AD^{2}=AE^{2}+DE^{2}$\\
that is $\left(12-x\right)^{2}=x^{2}+6^{2}$\\
Solving for $x$ yields $x=\frac{9}{2}$\\
In $Rt\triangle BDC$\\
$BD=\sqrt{DC^{2}+BC^{2}}=\sqrt{\left(\frac{9}{2}\right)^{2}+9^{2}}=\boxed{\frac{9}{2}\sqrt{5}}$


Question: In the figure, in equilateral triangle $ABC$, we have $AB=2$. Let $BD$ be the altitude to side $AC$. Extend $BC$ to point $E$ such that $CE=CD$. What is the length of $BE$?

Solution:
\textbf{Solution:} Since triangle $ABC$ is an equilateral triangle,\\
we have $BC=AC=2$,\\
and since $BD$ is the altitude of side $AC$,\\
we have $DC=\frac{1}{2}AC=\frac{1}{2}\times 2=1$,\\
thus $CE=CD=1$,\\
therefore $BE=BC+CE=2+1=\boxed{3}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, point D is on line AB. When $\triangle ABC$ is folded along CD, point A falls on side BC at point A. If $\angle B=35^\circ$, what is the measure of $\angle BDA$?

Solution:
\textbf{Solution:} \\
Since $\angle ACB=90^\circ$ and $\angle B=35^\circ$, \\
it follows that $\angle A=180^\circ-\angle ACB-\angle B=180^\circ-90^\circ-35^\circ=55^\circ$, \\
Since $\triangle CDA$ is obtained by folding $\triangle CDA$, \\
thus $\angle CAD=\angle A=55^\circ$, \\
Since $\angle CAD=\angle B+\angle BDA=\angle B+20^\circ$, \\
it follows that $\angle BDA=\angle CAD-\angle B=55^\circ-35^\circ=\boxed{20^\circ}$.


Question: As shown in the figure, $\angle BAC=33^\circ$, points $D$ and $E$ are on sides $AB$ and $AC$ respectively, connect $DE$. When $\angle A$ is folded along $DE$, the corresponding point of point $A$ is $A'$. If $\angle 1 + \angle 2 = 170^\circ$, what is the measure of $\angle 2$?

Solution:
\textbf{Solution:} Let point O be the intersection of AB and $AE$,\\
from the property of folding, we have: $\angle A = \angle BAC = 33^\circ$,\\
since $\angle 2 = \angle BAC + \angle AOE$, and $\angle AOE = \angle 1 + \angle A$,\\
thus $\angle 2 = \angle BAC + \angle 1 + \angle A = \angle 1 + 66^\circ$, that is $\angle 1 = \angle 2 - 66^\circ$,\\
since $\angle 1 + \angle 2 = 170^\circ$,\\
therefore $\angle 2 = \boxed{118^\circ}$.


Question: As shown in the figure, the perpendicular bisectors $l_{1}$ and $l_{2}$ of line segments AB and BC intersect at point O. If $\angle 1=37^\circ$, what is the degree measure of $\angle AOC$?

Solution:
\textbf{Solution}: Draw and extend line BO to P,\\
since the perpendicular bisectors $l_{1}$ and $l_{2}$ of line segments AB and BC intersect at point O,\\
therefore, AO = OB = OC, $\angle$BDO = $\angle$BEO = $90^\circ$,\\
therefore, $\angle$DOE + $\angle$ABC = $180^\circ$,\\
since $\angle$DOE + $\angle 1$ = $180^\circ$,\\
therefore, $\angle$ABC = $\angle 1$ = $38^\circ$,\\
since OA = OB = OC,\\
therefore, $\angle A$ = $\angle$ABO, $\angle$OBC = $\angle C$,\\
since $\angle$AOP = $\angle A$ + $\angle$ABO, $\angle$COP = $\angle C$ + $\angle$OBC,\\
therefore, $\angle$AOC = $\angle$AOP + $\angle$COP = $\angle A$ + $\angle$ABC + $\angle C$ = $2\times38^\circ$ = $\boxed{76^\circ}$;


Question: As shown in the figure, $BD$ and $CE$ are the medians of the equilateral triangle $ABC$, and $BD$ and $CE$ intersect at point $F$. What is the measure of $\angle BFC$ in degrees?

Solution:
\textbf{Solution:} Given that $\triangle ABC$ is an equilateral triangle\\
Hence, $\angle ABC=\angle ACB=60^\circ$\\
Since BD, CE are medians of the equilateral triangle ABC\\
Therefore, $\angle DBC=\frac{1}{2}\angle ABC=30^\circ$, $\angle ECB=\frac{1}{2}\angle ACB=30^\circ$\\
Additionally, $\angle BFC=180^\circ-\angle DBC-\angle ECB$\\
Therefore, $\angle BFC=180^\circ-30^\circ-30^\circ=\boxed{120^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $\angle C=37^\circ$, the perpendicular bisector of side $BC$ intersects $AC$ and $BC$ at points $D$ and $E$, respectively, and $AB=CD$, then what is the degree measure of $\angle A$?

Solution:
\textbf{Solution:} Draw BD as shown in the diagram: \\
$\because$ DE is the perpendicular bisector of BC, and AB=CD, \\
$\therefore$ BD=CD=AB, \\
$\because$ $\angle$C=37$^\circ$, \\
$\therefore$ $\angle$DBC=$\angle$C=37$^\circ$, \\
$\therefore$ $\angle$ADB=2$\angle$C=74$^\circ$, \\
$\because$ AB=BD, \\
$\therefore$ $\angle$A=$\angle$ADB=\boxed{74^\circ},


Question: As shown in the figure, $\angle AOE = \angle BOE = 15^\circ$, $EF \parallel OB$, $EC \perp OB$. If $EC = 2$, what is the length of $EF$?

Solution:
\textbf{Solution:} Construct $EG \perp OA$ at $G$, as shown in the figure:\\
Since $EF \parallel OB$, $\angle AOE = \angle BOE = 15^\circ$, and $EC \perp OB$,\\
it follows that $\angle OEF = \angle COE = 15^\circ$ and $EG = CE = 2$,\\
Since $\angle AOE = 15^\circ$,\\
it follows that $\angle EFG = 15^\circ + 15^\circ = 30^\circ$,\\
Therefore, $EF = 2EG = \boxed{4}$.


Question: As shown in the figure, $ABC$ is an acute-angled triangle with $AB=4$, $\angle BAC=60^\circ$. The bisector of $\angle BAC$ intersects $BC$ at point $D$. $M$ and $N$ are points on $AD$ and $AB$, respectively. When $BM+MN$ reaches its minimum value, what is the length of $AN$?

Solution:
\textbf{Solution:} Let $E$ be the point symmetrical to $B$ with respect to $AD$. Draw $EN \perp AB$ intersecting $AB$ at point $N$ and $AD$ at the extension of $CM$ at point $M$. Connect $BM$.

Since $\angle BAC=60^\circ$ and $AD$ bisects $\angle BAC$,

it follows that point $E$ lies on $AC$,

Since $BM + MN = EM + MN = EN$, the value of $BM + MN$ is minimized at this moment,

By symmetry, $AE = AB$,

Since $AB=4$,

Therefore, $AE=4$,

In $\triangle ABE$, $\angle EAN=60^\circ$,

Therefore, $\angle AEN=30^\circ$,

Therefore, $AN=\frac{1}{2}AE=\boxed{2}$


Question: As shown in the figure, it is known that in $\triangle ABC$, $\angle C=90^\circ$, $BC=3$, and $AC=5$. The triangle is folded along $DE$ such that point $A$ coincides with point $B$. What is the length of $AE$?

Solution:
\textbf{Solution:} By the properties of the folding, we have: $BE=AE$,\\
Let $BE=AE=x$,\\
In the right triangle $\triangle BEC$, by the Pythagorean theorem, we have: $BE^2=BC^2+EC^2$,\\
thus $x^2=3^2+(5-x)^2$,\\
thus $x=\boxed{3.4}$.


Question: As shown in the figure, $\angle A=70^\circ$, $\angle ABD=\angle BCE=30^\circ$, and $CE$ bisects $\angle ACB$. What is the measure of $\angle BEC$ in degrees?

Solution:
\textbf{Solution:} Given $CE$ bisects $\angle ACB$,\\
hence $\angle ACE = \angle BCE = 30^\circ$,\\
since $\angle A = 70^\circ$,\\
thus $\angle ABC + \angle ACB = 110^\circ$,\\
given $\angle ABE = \angle ACE = 30^\circ$,\\
thus $\angle EBC + \angle ECB = 110^\circ - 60^\circ = 50^\circ$,\\
therefore $\angle BEC = 180^\circ - 50^\circ = \boxed{130^\circ}$,


Question: As shown in the figure, the line $y=-2x+2$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively, and the ray $AP\perp AB$ at point $A$. If point $C$ is a moving point on ray $AP$, and point $D$ is a moving point on the $x$-axis, and the triangle with vertices $C$, $D$, $A$ is congruent to $\triangle AOB$, then what is the length of $OD$?

Solution:
\textbf{Solution:} In $y = -2x + 2$, \\
when $x = 0$, then $y = 2$; when $y = 0$, then $x = 1$,\\
$\therefore$ $OA = 1$, $OB = 2$, by the Pythagorean theorem we get $AB = \sqrt{OA^2 + OB^2} = \sqrt{5}$,\\
(1) When $\angle ACD = 90^\circ$, as in figure 1,\\
$\because \triangle AOB \cong \triangle DCA$,\\
$\therefore AD = AB = \sqrt{5}$,\\
$\therefore OD = 1 + \sqrt{5}$;\\
(2) When $\angle ADC = 90^\circ$, as in figure 2,\\
$\because AP \perp AB$,\\
$\therefore \angle BAP = \angle AOB = 90^\circ$,\\
$\therefore \angle ABO + \angle BAO = \angle CAD + \angle BAO = 90^\circ$,\\
$\therefore \angle ABO = \angle CAD$,\\
$\because \triangle AOB \cong \triangle CDA$,\\
$\therefore AD = OB = 2$,\\
$\therefore OA + AD = 3$,\\
In conclusion: the length of $OD$ is $1 + \sqrt{5}$ or $\boxed{3}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ bisects $\angle BAC$, $DE \perp AB$ at point $E$. If $AB=8$, $AC=6$, and $DE=4$, what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Draw $DF \perp AC$ at $F$, as shown in the figure.\\
Since $AD$ is the angle bisector of $\angle BAC$ in $\triangle ABC$, $DE \perp AB$, $DF \perp AC$,\\
therefore $DE=DF=4$,\\
Since $AB=8$, $AC=6$,\\
therefore the area of $\triangle ABC$ = area of $\triangle ADB$ + area of $\triangle ADC$\\
= $\frac{1}{2}$AB$\cdot$DE + $\frac{1}{2}$AC$\cdot$DF\\
= $\frac{1}{2}\times 8 \times 4 + \frac{1}{2}\times 6 \times 4\\
= \boxed{28}$,


Question: In $\triangle ABC$, $AC=8$, $BC=6$, and $\angle C=90^\circ$. Now, $\triangle ABC$ is folded as shown in the figure, making point A coincident with point B. The fold line is $DE$. What is the length of $DE$?

Solution:
\textbf{Solution:} Given that $AC=8$, $BC=6$, and $\angle C=90^\circ$,\\
thus $AB=\sqrt{AC^2+BC^2}=10$,\\
Given folding $\triangle ABC$ such that point A coincides with point B, with the fold line being $DE$,\\
thus $BE=AE$, $AD=BD=5$, and $DE\perp AB$,\\
In right triangle $BEC$, $BE^2=BC^2+CE^2$,\\
thus $BE^2=36+(8-BE)^2$,\\
thus $BE=\frac{25}{4}$,\\
In right triangle $BDE$, $DE=\sqrt{BE^2-BD^2}=\boxed{\frac{15}{4}}$.


Question: As shown in the figure, the quadrilateral $OABC$ is a square piece of paper placed in the Cartesian coordinate system, where point $O$ coincides with the origin, point $A$ is on the x-axis, and point $C$ is on the y-axis, with $OC=5$. Point $E$ is on side $BC$, and the coordinates of point $N$ are $\left(3,0\right)$. A line through point $N$ and parallel to the y-axis, $MN$, intersects $EB$ at point $M$. Now, the paper is folded so that vertex $C$ lands on point $G$ on $MN$, with the fold line being $OE$. On the positive x-axis, there exists a point $P$ such that the triangle formed by $P$, $O$, and $G$ is an isosceles triangle. What are the coordinates of point $P$?

Solution:
\textbf{Solution:} Given that the coordinates of point N are $(3,0)$, \\
hence $ON=3$, \\
from the folding, we have $OG=OC=5$, \\
since $\angle ONG=90^\circ$, \\
thus $NG=\sqrt{OG^{2}-ON^{2}}=\sqrt{5^{2}-3^{2}}=4$, \\
therefore, $G(3,4)$, \\
as shown in Figure 1, when $OP=OG=5$, \\
we get $P(5,0)$; \\
as shown in Figure 2, when $PG=OG$, \\
since $GN\perp PO$, \\
thus $PN=ON=3$, \\
therefore $OP=6$, \\
hence, $P(6,0)$; \\
as shown in Figure 3, when $PO=PG$, let $P(x,0)$, \\
from $PO=PG$ we have $PO^{2}=PG^{2}$, \\
thus $x^{2}=(x-3)^{2}+(0-4)^{2}$, \\
solving this gives $x=\frac{25}{6}$, \\
therefore, $P\left(\frac{25}{6},0\right)$; \\
In summary, the coordinates of point P are $(5,0)$ or $(6,0)$ or $\left(\frac{25}{6},0\right)$, \\
thus, the answer is: $\boxed{(5,0)}$ or $\boxed{(6,0)}$ or $\boxed{\left(\frac{25}{6},0\right)}$.


Question: As shown in the figure, in the right-angled triangle $ABC$ where $\angle C=90^\circ$, $D$ is the midpoint of $BC$, and $\angle CAD=30^\circ$, with $BC=6$, what is the length of $AD+DB$?

Solution:
\textbf{Solution:} Since $\angle C=90^\circ$ and D is the midpoint of BC, with $\angle CAD=30^\circ$ and BC=6,\\
it follows that $AD=2CD$, BD=CD=$\frac{1}{2}$BC=3,\\
therefore $AD+BD=3CD=\boxed{9}.$


Question: As shown in the figure, within $\triangle ABC$, AD and AE are the median and altitude from vertex A to side BC, respectively, with AE$=6$ and the area of $\triangle ABD=S_{\triangle ABD}=15$. What is the length of CD?

Solution:
\textbf{Solution:} Since $S_{\triangle ABC}=15$ and AE is the altitude on side BC, we have
\[
\frac{1}{2} \times BD \times AE=15,
\]
which gives
\[
\frac{1}{2} \times 6 \times BD=15,
\]
solving this yields $BD=5$. Since AD is the median to side BC, it follows that $CD=BD=\boxed{5}$.


Question: As shown in the figure, in parallelogram $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $E$ is the midpoint of side $CD$, and the length of $OE$ is $3$ cm. What is the length of $BC$ in centimeters?

Solution:
\textbf{Solution:} Since ABCD is a parallelogram, \\
therefore $OB=OD$, \\
and since point E is the midpoint of side CD, \\
therefore $BC=2OE=\boxed{6}$.


Question: As shown in the figure, in square $ABCD$ with $AB=4$, $BE=CE$, and $F$ being a moving point on side $AB$, connect $EF$. Fold $\triangle BEF$ over to $\triangle GEF$ such that point $B$ falls on point $G$, and connect $DG$. When the perimeter of quadrilateral $AFGD$ reaches its minimum value, what is the length of $BF$?

Solution:
\textbf{Solution:} As shown in the figure, connect ED, FG, DF,\\
since square ABCD, with AB=4,\\
therefore $\angle ABC = \angle DCB = 90^\circ$, and DC=AD=BC=4,\\
since the triangle BEF is folded onto triangle GEF, we have BE=EC,\\
therefore BE=EG=EC=2,\\
therefore point G moves on a circle with E as its center and a radius of 2,\\
therefore, when points E, G, and D are collinear, the length of GD is the shortest, thus the length of FG=BF is the maximum,\\
therefore, the length of AF is the shortest,\\
therefore, the perimeter of quadrilateral AFGD is the minimum,\\
in $\triangle DCE$, ED=$\sqrt{2^2+4^2}=2\sqrt{5}$,\\
therefore GD=$2\sqrt{5}-2$,\\
let FG=BF=x, AF=4-x,\\
in triangles $\triangle FAD$ and $\triangle FGD$, we have DF$^2$=AF$^2$+AD$^2$=FG$^2$+GD$^2$,\\
therefore $(4-x)^2+16=x^2+(2\sqrt{5}-2)^2$,\\
solving this, we get $x=\sqrt{5}+1$,\\
therefore BF=$\boxed{\sqrt{5}+1}$.


Question: As shown in the figure, the diagonals AC and BD of the parallelogram ABCD intersect at point O, and E and F are the midpoints of AO and BO, respectively. If $AC+BD=24$ cm, and the perimeter of $\triangle OAB$ is 18 cm, then how long is $EF$ in cm?

Solution:
Solution: Since $ABCD$ is a parallelogram, \\
it follows that $OB=OD$ and $OA=OC$, \\
since $AC+BD=24cm$, it follows that $OA+OB=12cm$, \\
since the perimeter of $\triangle OAB$ is $18cm$, it follows that $AB=18-12=6cm$, \\
since $E$ and $F$ are the midpoints of $AO$ and $BO$ respectively, it follows that $EF$ is the midline of $\triangle OAB$, \\
therefore $EF=\frac{1}{2}AB=3cm$, \\
thus the answer is: $\boxed{3}$.


Question: In square $ABCD$, $AB=4$, point $P$ is on line $AC$, draw $PM\perp AB$ at point $M$, and $PN\perp BC$ at point $N$. What is the minimum value of $MN$?

Solution:
\textbf{Solution:} Connect PB, and through point B draw $BP' \perp AC$ at $P'$, as shown in the figure: \\
$\because$ Quadrilateral ABCD is a square,\\
$\therefore \angle DCB=\angle ABC=90^\circ$, and $DC=BC=AD=AB=4$,\\
$\therefore AC=\sqrt{AD^2+DC^2}=4\sqrt{2}$,\\
$\because PM\perp AB$ and $PN\perp BC$,\\
$\therefore$ Quadrilateral BNPM is a rectangle,\\
$\therefore PB=MN$,\\
When $BP\perp AC$, BP is minimal, that is, as shown in the figure, the length of $BP'$,\\
$\because$ The area of $\triangle ABC = \frac{1}{2}AB\times CB = \frac{1}{2}AC\times BP'$,\\
$\therefore BP'=\frac{AB\cdot BC}{AC}=\frac{4\times 4}{4\sqrt{2}}=2\sqrt{2}$,\\
$\therefore$ The minimum value of MN is $\boxed{2\sqrt{2}}$;


Question: As shown in the figure, in quadrilateral $ABCD$, $\angle A=90^\circ$, $AB=AD$, and $E$, $F$ are the midpoints of $AB$, $AD$, respectively. If $EF=\sqrt{2}$, $BC=\sqrt{11}$, and $CD=\sqrt{3}$, then what is the area of quadrilateral $ABCD$?

Solution:
\textbf{Solution:} Connect BD, \\
$\because$ E and F are the midpoints of AB and AD respectively, \\
$\therefore$ BD=2EF=2$\sqrt{2}$, \\
$\because$ $\angle$A=90$^\circ$ and AB=AD, \\
$\therefore$ 2AB$^{2}$=BD$^{2}$=8, \\
$\therefore$ AB=AD=2, \\
$\therefore$ Area of $\triangle ABD$=$\frac{1}{2}$×2×2=2, \\
$\because$ BC=$\sqrt{11}$, CD=$\sqrt{3}$, and BD=2$\sqrt{2}$, \\
$\therefore$ BC$^{2}$=CD$^{2}$+BD$^{2}$, \\
$\therefore$ $\angle$BDC=90$^\circ$, \\
$\therefore$ Area of $\triangle ABD$=$\frac{1}{2}$×$\sqrt{3}$×2$\sqrt{2}$=$\sqrt{6}$, \\
$\therefore$ Area of the quadrilateral ABCD = Area of $\triangle ABD$ + Area of $\triangle ABD$ = 2 + $\sqrt{6}$. \\
Therefore, the answer is: $S_{\text{quadrilateral} ABCD} = \boxed{2+\sqrt{6}}$.


Question: As shown in the figure, within square $ABCD$, point $E$ is on diagonal $AC$, $EF \perp AB$ at point $F$, $EG \perp BC$ at point $G$. Connect $FG$. If $AB = 8$, what is the minimum value of $FG$?

Solution:
\textbf{Solution:} As shown in the diagram, connect BE, \\
$\because$ Quadrilateral ABCD is a square, \\
$\therefore \angle B=90^\circ$, AB=BC=8, \\
$\therefore$AC=8$\sqrt{2}$, \\
$\because$EF$\perp$AB, EG$\perp$BC, \\
$\therefore \angle$BFE=$\angle$FEG=$\angle$B=$90^\circ$, \\
$\therefore$ Quadrilateral BFEG is a rectangle, \\
$\therefore$FG=BE, \\
$\therefore$ When BE is the shortest, FG is also the shortest, \\
$\therefore$ When BE$\perp$AC, BE is the shortest, \\
$\therefore$AE=CE, \\
$\therefore$BE=$\frac{1}{2}$AC=4$\sqrt{2}$, \\
$\therefore$ The minimum value of FG is $\boxed{4\sqrt{2}}$.


Question: As shown in the figure, a point E is taken outside the square ABCD, and AE, BE, and DE are connected respectively. A perpendicular line to AE is drawn from point A and intersects DE at point P. If AE = AP = 1, and PB = $ \sqrt{5}$. What is the area of the square ABCD, $S_{\text{square ABCD}}$=?

Solution:
\textbf{Solution:}

Since quadrilateral ABCD is a square,
\begin{align*}
&\therefore AB=AD, \angle BAD=90^\circ,\\
&\because AE\perp AP, AE=AP=1,\\
&\therefore \angle AEP=\angle APE=45^\circ, \angle EAP=\angle BAD=90^\circ,\\
&\therefore \angle EAB=\angle PAD,\\
&\text{In } \triangle EAB \text{ and } \triangle PAD,\\
&\because AB=AD, \angle EAB=\angle PAD, AE=AP,\\
&\therefore \triangle EAB \cong \triangle PAD \text{ (SAS)},\\
&\therefore \angle EBA=\angle ADP, BE=DP, \angle APD=\angle AEB=180^\circ-45^\circ=135^\circ,\\
&\therefore \angle PEB=135^\circ-45^\circ=90^\circ, \text{ i.e., } \triangle BEP\text{ is a right-angled triangle},\\
&\because AE=AP=1,\\
&\therefore EP=\sqrt{1^2+1^2}=\sqrt{2},\\
&\therefore BE=DP=\sqrt{BP^2-EP^2}=\sqrt{3},\\
&\text{Draw line BF }\perp\text{ AE extended to meet at point F, connect BD, then }\angle FEB=180^\circ-135^\circ=45^\circ,\\
&\therefore \angle EBF=45^\circ=\angle FEB,\\
&\therefore EF=BF,\\
&\because BE=\sqrt{3},\\
&\therefore BF=EF=\frac{\sqrt{6}}{2},\\
&\therefore {S}_{\triangle APB}+{S}_{\triangle APD}={S}_{\triangle APB}+{S}_{\triangle AEB}={S}_{\text{quadrilateral }AEBP}={S}_{\triangle AEP}+{S}_{\triangle PEB}\\
&=\frac{1}{2}\times 1\times 1+\frac{1}{2}\times \sqrt{2}\times \sqrt{3}\\
&=\frac{1}{2}+\frac{1}{2}\sqrt{6},\\
&\because {S}_{\triangle DPB}=\frac{1}{2}\times DP\times BE=\frac{1}{2}\times \sqrt{3}\times \sqrt{3}=\frac{3}{2},\\
&\therefore {S}_{\text{square }ABCD}=2{S}_{\triangle ABD}=2({S}_{\triangle BPD}+{S}_{\triangle APD}+{S}_{\triangle APB})=2\times (\frac{1}{2}+\frac{1}{2}\sqrt{6}+\frac{3}{2})=\boxed{4+\sqrt{6}}.
\end{align*}


Question: As shown in the figure, all quadrilaterals in the figure are squares, and the triangle is a right-angled triangle. The areas of the two smaller squares are respectively 1 and 2. What is the area of the largest square?

Solution:
\textbf{Solution}: Since the areas of the two small squares are respectively 1 and 2,\\
it follows that they are the squares of the two legs of a right triangle,\\
thus, according to the Pythagorean theorem, we have: $S_{\text{large square}}=S_{\text{area of the small square is 1}}+S_{\text{area of the small square is 2}}=\boxed{3}$


Question: As shown in the figure, in rhombus $ABCD$, the diagonals $AC=6$ and $BD=10$. What is the area of rhombus $ABCD$?

Solution:
\textbf{Solution:} Given that in the rhombus $ABCD$, the diagonals $AC=6$ and $BD=10$, \\
therefore, the area of rhombus $ABCD$ is $= \frac{1}{2}AC \cdot BD = \boxed{30}$.


Question: As shown in the figure, in $\text{Rt}\triangle ABC$, $\angle ACB=90^\circ$, points $D$ and $E$ are the midpoints of sides $AB$ and $AC$, respectively. Extend $BC$ to point $F$ such that $CF=\frac{1}{2}BC$. If $AB=10$, what is the length of $EF$?

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

$\therefore$ $AD=BD=\frac{1}{2}AB=5$, DE is the midline of $\triangle ACB$, AE=CE

$\therefore$ $DE\parallel BC$, $DE=\frac{1}{2}BC$,

$\because$ $\angle ACB=90^\circ$, $CF=\frac{1}{2}BC$,

$\therefore$ $\angle DEA=\angle FCE=90^\circ$, CF=ED,

$\therefore$ $\triangle AED\cong \triangle ECF$ (SAS),

$\therefore$ EF=AD=\boxed{5}.


Question: As shown in the figure, in the square $ABCD$, point $F$ is on the side $CD$, and $BF$ intersects $AC$ at point $E$. If $\angle CBF=20^\circ$, then what is the measure of $\angle AED$ in degrees?

Solution:
\textbf{Solution:} Given $\because$ quadrilateral ABCD is a square\\
$\therefore \angle ACB = \angle ACD = \angle BAC = \angle CAD = 45^\circ$, $\angle ABC = 90^\circ$, AB = AD\\
$\because \angle FBC = 20^\circ$\\
$\therefore \angle ABF = 70^\circ$\\
$\therefore$ in $\triangle ABE$, $\angle AEB = 65^\circ$\\
In $\triangle ABE$ and $\triangle ADE$\\
$ \left\{\begin{array}{l}
AB = AD \\
\angle BAE = \angle EAD = 45^\circ \\
AE = AE
\end{array}\right.$ \\
$\therefore \triangle ABE \cong \triangle ADE$\\
$\therefore \angle AED = \angle AEB = \boxed{65^\circ}$


Question: As shown in the figure, within the square $ABCD$, points $M$ and $N$ are the midpoints of $AD$ and $BC$ respectively, point $P$ is on $CD$. After folding $\triangle BCP$ along $BP$, the corresponding point of $C$, point $Q$, falls on $MN$. If $AB=6$, what is the length of $PD$?

Solution:
\textbf{Solution:} As shown in the figure, connect CQ, \\
$\because$ Quadrilateral ABCD is a square, \\
$\therefore$ AD=BC=AB, $AD\parallel BC$, $\angle D=90^\circ$, \\
$\because$ M and N are the midpoints of AD and BC respectively, \\
$\therefore$ DM=$\frac{1}{2}AD=CN=\frac{1}{2}BC=BN$, \\
$\therefore$ Quadrilateral CNMD is a rectangle, \\
$\therefore$ $\angle QNB=\angle QNC=90^\circ$, \\
In $\triangle QNB$ and $\triangle QNC$, \\
$\left\{\begin{array}{l}BN=CN\\ \angle QNB=\angle QNC\\ QN=QN\end{array}\right.$, \\
$\therefore$ $\triangle QNB\cong \triangle QNC$ (SAS), \\
$\therefore$ QB=QC, \\
By the property of folding, it is known that BC=BQ, $\angle QBP=\angle CBP$, \\
$\therefore$ BC=BQ=CQ, \\
$\therefore$ $\triangle QBC$ is an equilateral triangle, \\
$\therefore$ $\angle QBC=60^\circ$, \\
$\therefore$ $\angle PBC=30^\circ$, \\
$\therefore$ PB=2PC, \\
$\because$ $BC^2+PC^2=BP^2$, \\
$\therefore$ $4PC^2=PC^2+6^2$, \\
$\therefore$ $PC=2\sqrt{3}$, \\
$\therefore$ $PD=CD-PC=6-2\sqrt{3}$, \\
Hence, the answer is: $PD=\boxed{6-2\sqrt{3}}$.


Question: As illustrated in the diagram, two strips with a width of 2 and parallel opposite edges are placed crosswise on top of each other. By rotating one of the strips, we obtain a quadrilateral ABCD, where $\angle ABC=60^\circ$. What is the area of the quadrilateral ABCD?

Solution:
\textbf{Solution:} As shown in the figure, draw CF$\perp$AB at point F, and draw AE$\perp$BC at point E through point A.\\
According to the problem, we know: AE=CF=2,\\
$\because$ $\angle$ABC=60$^\circ$,\\
$\therefore$ $\angle$FCB=30$^\circ$,\\
$\therefore$ let BF=x, then BC=2x,\\
$\therefore$ $4x^{2}-x^{2}=4$,\\
$\therefore$ x=$\frac{2\sqrt{3}}{3}$,\\
$\therefore$ BC=$\frac{4\sqrt{3}}{3}$,\\
$\because$ AB$\parallel$CD, AD$\parallel$BC,\\
$\therefore$ Quadrilateral ABCD is a parallelogram,\\
$\therefore$ $S_{\square ABCD}$=BC·AE=$\frac{4\sqrt{3}}{3} \times 2=\boxed{\frac{8\sqrt{3}}{3}}$.


Question: As shown in the figure, in $\square ABCD$, $AC$ is the diagonal, $\angle ACD=90^\circ$, point $E$ is the midpoint of $BC$, $AF$ bisects $\angle BAC$, $CF\perp AF$ at point $F$, and $EF$ is connected. Given $AB=5$ and $BC=13$, what is the length of $EF$?

Solution:
\textbf{Solution}: Extend $AB$ and $CF$ to meet at point $H$, \\
$\because$ Quadrilateral $ABCD$ is a parallelogram, \\
$\therefore AB\parallel CD$, \\
$\therefore \angle ACD=\angle BAC=90^\circ$, \\
$\therefore AC=\sqrt{BC^{2}-AB^{2}}=\sqrt{169-25}=12$, \\
$\because AF$ bisects $\angle BAC$, \\
$\therefore \angle BAF=\angle CAF=45^\circ$, \\
In $\triangle AFH$ and $\triangle AFC$, \\
$\left\{\begin{array}{l}
\angle HAF=\angle CAF \\
AF=AF \\
\angle AFH=\angle AFC=90^\circ
\end{array}\right.$, \\
$\therefore \triangle AFH \cong \triangle AFC\left(ASA\right)$, \\
$\therefore AC=AH=12$, $HF=CF$, \\
$\therefore BH=AH-AB=7$, \\
$\because$ Point $E$ is the midpoint of $BC$, $HF=CF$, \\
$\therefore EF=\frac{1}{2}BH=\boxed{\frac{7}{2}}$,


Question: As shown in the figure, it is a centrally symmetric figure with $A$ as the center of symmetry. If $\angle C=90^\circ$, $\angle B=30^\circ$, and $AC=3$, what is the length of $BB'$?

Solution:
\textbf{Solution:} In the right triangle $ABC$, $\angle B=30^\circ$, $AC=3$,\\
$\therefore AB=2AC=6$,\\
$\because$ B and B' are centrally symmetric about A,\\
$\therefore BB'=2AB=\boxed{12}$.


Question: As shown in the figure, the diagonals $AC$ and $BD$ of the rhombus $ABCD$ intersect at point $O$, and $E$ is the midpoint of $AD$. If $OE=3$, then what is the perimeter of the rhombus $ABCD$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a rhombus,\\
it follows that $AC \perp BD$ and $AB = BC = CD = DA$,\\
which means $\triangle AOD$ is a right-angled triangle.\\
Given that $OE = 3$, and point $E$ is the midpoint of segment $AD$,\\
it follows that $AD = 2OE = 6$.\\
Therefore, the perimeter of the rhombus is: $6 \times 4 = \boxed{24}$.


Question: As shown in the figure, on the exterior of the square $ABCD$, an equilateral triangle $ADE$ is constructed. What is the measure of $\angle AEB$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a square,\\
it follows that $\angle$BAD=90$^\circ$, and AB=AD.\\
Since $\triangle$ABD is an equilateral triangle,\\
it follows that $\angle$DAE=60$^\circ$, and AD=AE.\\
Therefore, $\angle$BAE=$\angle$DAE+$\angle$BAD=90$^\circ$+60$^\circ$=150$^\circ$, and AB=AE.\\
Hence, $\triangle$BAE is an isosceles triangle,\\
which means $\angle AEB=\frac{180^\circ−\angle BAE}{2}=\frac{180^\circ−150^\circ}{2}=15^\circ$.\\
Thus, the answer is: $\boxed{15^\circ}$.


Question: As shown in the figure, the quadrilateral ABCD is a rhombus with $AC=16$ and $DB=12$. If $DH \perp AB$ at point H, what is the length of $DH$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rhombus,\\
thus $\angle AOB = 90^\circ$, OA = OC = 8, OB = OD = 6,\\
thus AB = $\sqrt{OA^{2} + OB^{2}}$ = $\sqrt{8^{2} + 6^{2}}$ = 10,\\
thus, since the area of rhombus $ABCD = AB \cdot DH = \frac{1}{2}AC \times BD$,\\
thus DH = $\frac{AC \cdot BD}{2AB}$ = $\frac{16 \times 12}{2 \times 10}$ = $\boxed{\frac{48}{5}}$.


Question: As shown in the diagram, in $\triangle ABC$, BD bisects $\angle ABC$, and $AD \perp BD$. Let E be the midpoint of AC, with $AD=6\,\text{cm}$, $BD=8\,\text{cm}$, and $BC=16\,\text{cm}$. What is the length of $DE$ in $\,\text{cm}$?

Solution:
\textbf{Solution:} As shown in the figure, extend AD to meet BC at point F,\\
$\because$AD$\perp$BD,\\
$\therefore$AB=$\sqrt{AD^{2}+BD^{2}}$=$\sqrt{6^{2}+8^{2}}=10$,\\
$\because$BD bisects $\angle$BC, and AD$\perp$BD, BD=BD,\\
$\therefore$$\triangle$ABD$\cong$$\triangle$FBD (ASA),\\
$\therefore$AD=DF, BF=AB=10,\\
Also $\because$AE=EC, FC=BC−BF=16−10=6,\\
$\therefore$DE is the median of $\triangle$ACF,\\
$\therefore$DE=$\frac{1}{2}$FC=\boxed{3}.


Question: As shown in the figure, in quadrilateral $ABCD$, points $E$ and $F$ are the midpoints of $AD$ and $DC$, respectively, $BF\perp CD$. Given that $BF=8$ and $EF=5$, what is the perimeter of quadrilateral $ABCD$?

Solution:
\textbf{Solution:}

As shown in the figure, connect BE, extend BE to intersect the extension line of CD at point H, and draw EG $\perp$ BF.

Since parallelogram ABCD,

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

$\therefore$ $\angle$EAB = $\angle$EDH, $\angle$EBA = $\angle$EHD,

Also, since E is the midpoint of AD,

$\therefore$ AE = ED,

$\therefore$ $\triangle$AEB $\cong$ $\triangle$DEH,

$\therefore$ AB = DH, BE = EH,

Since BF $\perp$ CD, EG $\perp$ BF,

$\therefore$ EG $\parallel$ HF

$\therefore$ EG is the midline of $\triangle$BHF,

$\therefore$ GF = $\frac{1}{2}$BF = $\frac{1}{2} \times 8 = 4$,

$\therefore$ EG = $\sqrt{EF^2 - GF^2} = \sqrt{5^2 - 4^2} = 3$,

$\therefore$ HF = 6,

Since F is the midpoint of CD,

$\therefore$ AB = DH = DC = 2DF = 2FC,

Let FC = x, then HF = DH + DF = 2x + x = 3x = 6,

$\therefore$ x = 2,

$\therefore$ AB = 4, BC = $\sqrt{BF^2 + FC^2} = \sqrt{8^2 + 2^2} = 2\sqrt{17}$,

$\therefore$ The perimeter of parallelogram ABCD = 2(AB + BC) = 2(4 + 2$\sqrt{17}$) = $\boxed{8 + 4\sqrt{17}}$.


Question: As shown in the figure, the lines $AB, IL, JK, DC$ are parallel to each other, and the lines $AD, IJ, LK, BC$ are also parallel to each other. The area of the quadrilateral $ABCD$ is 60, and the area of quadrilateral $EFGH$ is 35. What is the area of quadrilateral $IJKL$?

Solution:
\textbf{Solution:} Since lines $AB$, $IL$, $JK$, $DC$ are parallel to each other, and lines $AD$, $IJ$, $LK$, $BC$ are parallel to each other, \\
therefore, quadrilaterals AFJE, DGKF, CHLG, BEIH are parallelograms, \\
therefore, $S_{\triangle AEF}=S_{\triangle EFJ}$, $S_{\triangle BEH}=S_{\triangle EHI}$, $S_{\triangle CHG}=S_{\triangle HGL}$, $S_{\triangle DGF}=S_{\triangle GFK}$, \\
since the area of quadrilateral $ABCD$ is 60, and the area of quadrilateral $EFGH$ is 35, \\
therefore, $S_{\triangle AEF}+S_{\triangle BEH}+S_{\triangle CHG}+S_{\triangle DGF}=25$, \\
therefore, $S_{\triangle EFJ}+S_{\triangle EHI}+S_{\triangle HGL}+S_{\triangle GFK}=25$, \\
therefore, $S_{\text{quadrilateral }IJKL}=S_{\text{quadrilateral }EFGH}-(S_{\triangle EFJ}+S_{\triangle EHI}+S_{\triangle HGL}+S_{\triangle GFK})=35-25=\boxed{10}$.


Question: As shown in the figure, for the rhombus $ABCD$ with side length $AB=2\sqrt{3}+2$ and $\angle ABC=60^\circ$, two points $E$ and $H$ are taken on the diagonal $BD$ such that $BE=DH$ and $AE\parallel FH$. When $\angle AED=75^\circ$, what is the sum $AE+FH$?

Solution:
\textbf{Solution:} As shown in the diagram, construct FM$\perp$BD at M, EN$\perp$AB at N. \\
Because quadrilateral ABCD is a rhombus and $\angle$ABC=$60^\circ$, \\
it follows that $\angle$ABD=$\angle$ADB=$30^\circ$, \\
Since $\angle$AED=$75^\circ$=$\angle$ABE+$\angle$BAE, \\
it results in $\angle$BAE=$45^\circ$, \\
Given $\angle$ENA=$90^\circ$, \\
it leads to $\angle$NAE=$\angle$NEA=$45^\circ$, \\
hence NA=NE, \\
Let NA=NE=x, then BE=2x, and BN=$ \sqrt{3}$x, \\
Since AB=2$ \sqrt{3}$+2, \\
it implies that $ \sqrt{3}$x+x=2$ \sqrt{3}$+2, \\
thus x=2, \\
Therefore, BE=4, AE=2$ \sqrt{2}$, \\
Since $\angle$HFD=$180^\circ$−$\angle$FHD−$\angle$FDH=$75^\circ$ and $\angle$FHD=$75^\circ$, \\
it concludes that $\angle$DFH=$\angle$DHF, \\
Therefore, DF=DH=BE=2, \\
In the right triangle $\triangle$FMD, \\
since $\angle$FMD=$90^\circ$ and $\angle$FDM=$30^\circ$ with DF=4, \\
it follows that FM=2, DM=2$ \sqrt{3}$, and HM=4−2$ \sqrt{3}$, \\
therefore, FH=$ \sqrt{FM^{2}+HM^{2}}$ \\
=$ \sqrt{2^{2}+(4−2\sqrt{3})^{2}}$ \\
=2$ \sqrt{6}$−2$ \sqrt{2}$, \\
Thus, AE+FH=2$ \sqrt{6}$−2$ \sqrt{2}+2$ $\sqrt{2}$ \\
=$\boxed{2 \sqrt{6}}$.


Question: As shown in the diagram, in quadrilateral $ABCD$, with $AB=10$, $BC=18$, the angle bisectors of $\angle ABC$ and $\angle BCD$ intersect $AD$ at points E and F, respectively. If $BE=12$, then what is the length of $CF$?

Solution:
\textbf{Solution:} In parallelogram $ABCD$, $BE$ bisects $\angle ABC$, and $CF$ bisects $\angle BCD$,

$\therefore$ $\angle ABE=\angle CBE$, $\angle BCF=\angle DCF$,

$\because$ $AB\parallel CD$,

$\therefore$ $\angle ABC+\angle DCB=180^\circ$, $\angle CBE=\angle AEB$, $\angle BCF=\angle DFC$,

$\therefore$ $\angle CBE+\angle BCF=90^\circ$, $\angle ABE=\angle AEB$, $\angle DCF=\angle DFC$,

$\therefore$ $AE=AB=10$, $DF=DC=10$,

$\because$ $AD=BC=18$,

$\therefore$ $AF=AD-DF=8$,

$\therefore$ $EF=AE-AF=2$,

Extend $CG$ such that $CG=EF$,

$\therefore$ $EFCG$ is a parallelogram,

$\therefore$ $CG=EF=2$, $BG=20$, $EG=CF$,

$\therefore$ $\angle BCF=\angle G$,

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

$\because$ $BE=12$, $BG=20$,

$\therefore$ $EG=\sqrt{BG^2-BE^2}=\sqrt{20^2-12^2}=16$,

$\therefore$ $CF=EG=\boxed{16}$.


Question: As shown in the figure, seven squares are arranged such that each pair of adjacent squares shares a common vertex, and the area of each square is represented by numeric or alphabetic characters. What is the value of $S_{1} + S_{2} + S_{3} + S_{4}$?

Solution:
\textbf{Solution:} As shown in the diagram, \\
since AB=BE, $\angle$ACB=$\angle$BDE=$90^\circ$, \\
therefore, $\angle$ABC+$\angle$BAC=$90^\circ$, $\angle$ABC+$\angle$EBD=$90^\circ$, \\
thus, $\angle$BAC=$\angle$EBD, \\
hence, $\triangle ABC\cong \triangle BDE$ (AAS), \\
therefore, BC=ED, \\
since $ AB^{2}=AC^{2}+BC^{2}$, \\
thus $ AB^{2}=AC^{2}+ED^{2}=S_{1}+S_{2}$, \\
namely $ S_{1}+S_{2}=1$, \\
similarly, $ S_{3}+S_{4}=3$.\\
Then $ S_{1}+S_{2}+S_{3}+S_{4}=1+3=\boxed{4}$.


Question: As shown in the figure, rectangle $ABCD$ ($AD > AB$), equilateral triangles are constructed inwards using $AD$ and $BC$ as sides respectively (Figure 1); similarly, equilateral triangles are constructed inwards using $AB$ and $CD$ as sides respectively (Figure 2). The overlapping areas of the two equilateral triangles are shown as the shaded areas. Let the area of the shaded portion in Figure 1 be $S_{1}$, and the area of the shaded portion in Figure 2 be $S_{2}$. If $\frac{S_{1}}{S_{2}}=8$, then what is the value of $\frac{AD}{AB}$?

Solution:
\textbf{Solution:} Let us set $AD=BC=a$, $AB=CD=b$,\\
According to the problem: $\angle ADN=\angle BCH=60^\circ$,\\
$\therefore \angle NDC=\angle HCD=30^\circ$,\\
$\therefore FD=FC$,\\
$\because$ quadrilateral $ABCD$ is a rectangle,\\
$\therefore AD\parallel BC$,\\
$\therefore \angle FNC=\angle ADN=60^\circ$,\\
$\therefore \triangle FNC$ is an equilateral triangle,\\
$\therefore FN=FC$,\\
$\therefore FN=FD$,\\
$\therefore S\triangle FNC=S\triangle FDC=\frac{1}{2}S\triangle DCN$,\\
In $\triangle DNC$, $DN=2CN$,\\
By the Pythagorean theorem, we get $NC=\frac{\sqrt{3}}{3}b$,\\
$\therefore S\triangle FNC=S\triangle FDC=\frac{1}{2}S\triangle DCN=\frac{1}{2}\times\frac{1}{2}NC\cdot CD=\frac{\sqrt{3}}{12}b^2$,\\
Similarly: $S\triangle DHF=S\triangle AGE=S\triangle ABE=S\triangle BEM=\frac{\sqrt{3}}{12}b^2$,\\
$\therefore S_1=S_{\text{rectangle}}ABCD-S\triangle NFC-S\triangle DFC-S\triangle DHF-S\triangle MBE-S\triangle ABE-S\triangle AGE=ab-\frac{\sqrt{3}}{2}b^2$,\\
Draw $HM\perp AD$ at $M$ through point $H$, and $GN\perp AB$ at point $N$ through point $G$,\\
According to the problem: $\angle HEF=\angle HGF=\angle GAB=\angle EDC=60^\circ$,\\
$GA=AB=CD=ED=EC=GB$,\\
$\therefore \angle HAD=\angle HDA=30^\circ$,\\
$\therefore HA=HD$,\\
$\because HM\perp AD$,\\
$\therefore AM=\frac{1}{2}AD=\frac{1}{2}a$,\\
By the Pythagorean theorem, we get $MH=\frac{\sqrt{3}}{6}a$,\\
$\therefore S\triangle HAD=\frac{1}{2}\cdot AD\cdot MH=\frac{\sqrt{3}}{12}a^2$,\\
Similarly: $S\triangle FBC=\frac{\sqrt{3}}{12}a^2$,\\
$\because \triangle GAB$ is an equilateral triangle, $GN\perp AB$,\\
$\therefore AN=\frac{1}{2}AB=\frac{1}{2}b$,\\
$\because AG=AB=b$,\\
$\therefore GN=\sqrt{AG^2-AN^2}=\frac{\sqrt{3}}{2}b$,\\
$\therefore S\triangle ABG=\frac{1}{2}AB\cdot NG=\frac{1}{2}b\cdot\frac{\sqrt{3}}{2}b=\frac{\sqrt{3}}{4}b^2$,\\
Similarly: $S\triangle CDE=\frac{\sqrt{3}}{4}b^2$,\\
$\therefore S_2=S\triangle ABG+S\triangle CDE+S\triangle ADH+S\triangle BFC-S_{\text{rectangle}}ABCD=\frac{\sqrt{3}}{6}a^2+\frac{\sqrt{3}}{2}b^2-ab$,\\
$\because \frac{S_1}{S_2}=8$,\\
$\therefore \frac{ab-\frac{\sqrt{3}}{2}b^2}{\frac{\sqrt{3}}{6}a^2+\frac{\sqrt{3}}{2}b^2-ab}=\frac{8}{1}$,\\
Solving, we get: $a=\frac{3\sqrt{3}}{2}b$ or $a=\frac{3\sqrt{3}}{4}b$,\\
According to the problem, we know: $a<\sqrt{3}$


Question: As shown in the figure, it is known that in rectangle $ABCD$, point $E$ is on the extension line of side $BC$, and $CE=BD$. Connect $AE$ and intersect $BD$ at $F$. If $\angle E=20^\circ$, what is the degree measure of $\angle AFB$?

Solution:
\textbf{Solution:} As shown in the diagram, connect AC, intersecting BD at point O,\\
since ABCD is a rectangle, then BD=AC,\\
CE=BD, then CE=AC,\\
$\angle E=20^\circ$, thus $\angle CAE=20^\circ$,\\
$\therefore \angle ACB=\angle CAE+\angle E=40^\circ$,\\
$\because$ OB=$\frac{1}{2}$BD, OC=$\frac{1}{2}$AC,\\
$\therefore$ OB=OC, $\angle OBC=\angle OCB=40^\circ$,\\
$\therefore \angle BOC=180^\circ-\angle OBC-\angle OCB=100^\circ$,\\
$\therefore \angle AOF=\angle BOC=100^\circ$,\\
$\because \angle OAF=20^\circ$,\\
$\therefore \angle AFO=180^\circ-\angle OAF-\angle AOF=60^\circ$,\\
$\therefore \angle AFB=\boxed{60^\circ}$,


Question: As shown in the figure, in rectangle $ABCD$, $AB=3$ and $AD=4$. Let $E$ be a moving point on line segment $BC$, and construct point $F$ which is the reflection of point B across line $AE$. Connect $EF$ and $DF$. Let $G$ be the midpoint of $DF$. When points $D$, $F$, and $E$ are collinear, what is the length of $CE$? During the entire motion of $E$, what is the minimum distance between points $C$ and $G$?

Solution:
\textbf{Solution:} As shown in the diagram, connect AF, take the midpoint H of AD, and connect GH, GC, HC,\\
Quadrilateral ABCD is a rectangle, thus AB=CD=3, AD=BC=4, $\angle$B=90$^\circ$,\\
By the property of folding, it can be obtained that AF=AB=3, BE=EF, $\angle$AFE=$\angle$B=90$^\circ$,\\
Let CE=x, then BE=EF=4−x,\\
When points E, F, and D are collinear, AF$\perp$DE,\\
In $\triangle AFD$, DF=$\sqrt{AD^{2}-AF^{2}}=\sqrt{7}$,\\
In $\triangle DCE$, DE$^{2}$=DC$^{2}$+CE$^{2}$,\\
$\therefore$ ($\sqrt{7}$+4−x)$^{2}$=x$^{2}$+9,\\
Solving yields: x=$\boxed{\sqrt{7}}$;\\
$\because$H is the midpoint of AD, G is the midpoint of DF,\\
$\therefore$HG is the midline in $\triangle DAF$,\\
$\therefore$HG=$\frac{1}{2}$AF=$\frac{3}{2}$,\\
In $\triangle DHC$, HC=$\sqrt{DH^{2}+DC^{2}}=\sqrt{13}$,\\
When point G is not on line HC,\\
In $\triangle GHC$, GC>HC−HG,\\
When point G is on line HC, GC=HC−HG,\\
$\therefore$The minimum value of CG is $\boxed{\sqrt{13}-\frac{3}{2}}$;


Question: As shown in the figure, in the rhombus $ABCD$, $AB=5$ and $AC+BD=14$. What is the area of the rhombus $ABCD$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a rhombus,\\
$\therefore OA=\frac{1}{2}AC$, $OB=\frac{1}{2}BD$, and $AC\perp BD$,\\
$\therefore \angle AOB=90^\circ$,\\
$\therefore OA^{2}+OB^{2}=AB^{2}=25$,\\
$\because AC+BD=14$,\\
$\therefore OA+OB=7$,\\
$\therefore (OA+OB)^{2}=7^{2}=49$,\\
which means $OA^{2}+2AO\cdot OB+OB^{2}=49$,\\
$\therefore 2OA\cdot OB=49-25=24$,\\
$\therefore S_{\text{rhombus} ABCD}=\frac{1}{2}AC\cdot BD=2OA\cdot OB=\boxed{24}$.


Question: As shown in the figure, it is known that $\triangle ABC$ is an equilateral triangle with a side length of 6, and point D is on side $BC$ with $BD=2$. Construct an equilateral triangle $ADE$ with $AD$ as its side. Through point E, draw $EF\parallel BC$ intersecting $AC$ at point F. Connect $BF$. What is the area $S_{\text{quadrilateral }BDEF}$?

Solution:
\textbf{Solution:} Draw $EC$, and construct $CH\perp EF$ at $H$,\\
since $\triangle ABC$ and $\triangle ADE$ are both equilateral triangles,\\
therefore $AB=AC$, $AD=AE$, $\angle BAC=\angle DAE=\angle ABC=\angle ACB=60^\circ$,\\
thus $\angle BAD=\angle CAE$,\\
in $\triangle BAD$ and $\triangle CAE$,\\
$\left\{\begin{array}{l}
AB=AC\\
\angle BAD=\angle CAE\\
AD=AE
\end{array}\right.$,\\
thus $\triangle BAD \cong \triangle CAE(SAS)$,\\
thus $BD=EC=2$, $\angle ACE=\angle ABD=60^\circ$,\\
since $EF\parallel BC$,\\
thus $\angle EFC=\angle ACB=60^\circ$,\\
thus $\triangle EFC$ is an equilateral triangle,\\
therefore $CH=\sqrt{3}$, $EF=EC=BD=2$,\\
since $EF\parallel BD$,\\
thus quadrilateral $BDEF$ is a parallelogram,\\
thus the area of quadrilateral $BDEF=BD\cdot CH=2\sqrt{3}$.\\
Hence, the answer is: $S_{\text{quadrilateral}BDEF}=\boxed{2\sqrt{3}}$.


Question:
As shown in the graph, the function

\[
y =
\begin{cases}
3-x & (0 \le x \le 2)\\
x-1 & (2 \le x \le 4)
\end{cases}
\]

is illustrated. What is the minimum value of this function?

Solution:
\textbf{Solution:} Since the minimum value of the function is $(2,1)$, \\
therefore, the minimum value of the function is $\boxed{1}$.


Question: As shown in the figure, it is known that the line $y=kx$ intersects the line $y=ax+4$ at the point $A(2,1)$. What is the solution set of the inequality $ax+4>kx$?

Solution:
\textbf{Solution:} From the graph, we can see that the coordinates of $A$ are $(2,1)$,\\
when $x<2$, the line $y=ax+4$ is above the line $y=kx$,\\
meaning the solution set for the inequality $ax+4>kx$ with respect to $x$ is: $x<2$,\\
hence, the answer is: $\boxed{x<2}$.


Question: As the figure shows, the line $y=kx-\frac{3}{2}$ intersects with $y=mx-5$ at point $A\left(1, -3\right)$. What is the solution set of the inequality $kx-\frac{3}{2}>mx-5$ with respect to $x$?

Solution:
\textbf{Solution:} By observing the graph, we know that when $x=1$, $kx-\frac{3}{2}=mx-5$.\\
Therefore, when $x<1$, $kx-\frac{3}{2}>mx-5$.\\
Thus, the answer is $\boxed{x<1}$.


Question: As shown in the diagram, the graph $l_1$ of the linear function $y=k_1x+b_1$ intersects with the graph $l_2$ of $y=k_2x+b_2$ at point P. What is the solution set for the inequality in terms of $x$, $\left(k_1-k_2\right)x+\left(b_1-b_2\right)>0$?

Solution:
\textbf{Solution:} Given that $y_1-y_2=(k_1x+b_1)-(k_2x+b_2)=(k_1-k_2)x+(b_1-b_2)$,\\
as can be seen from the graph: to the left of their intersection point, $y_1>y_2$,\\
meaning that when $x<-2$, $y_1-y_2>0$,\\
therefore, the solution set of $(k_1-k_2)x+(b_1-b_2)>0$ is $\boxed{x<-2}$.


Question: As shown in the figure, the line $y=kx-3$ intersects the x-axis and y-axis at points B and A, respectively, with $OB=\frac{1}{3}OA$. Point C is a point on line AB, located in the second quadrant. When the area of $\triangle OBC$ is 3, what are the coordinates of point C?

Solution:
\textbf{Solution:} Draw $CH\perp x$ axis at point H, as shown in the figure:\\
Since the line $y=kx-3$ intersects the x-axis and y-axis at points B and A respectively,\\
thus, by setting $x=0$, we get $y=-3$, i.e., $OA=3$,\\
since $OB=\frac{1}{3}OA$,\\
thus, $OB=1$, i.e., $B(-1,0)$, substituting into the line equation yields: $0=-k-3$, solving for $k$ yields: $k=-3$;\\
thus, the equation of line AB is $y=-3x-3$,\\
since the area of $\triangle OBC$ is 3,\\
thus, $\frac{1}{2}OB\cdot CH=3$,\\
therefore, $CH=6$, i.e., the y-coordinate of point C is 6,\\
therefore, $-3x-3=6$, solving for: $x=-3$,\\
therefore, $C\left(-3,6\right)$;\\
Thus, the answer is $\boxed{\left(-3,6\right)}$.


Question: In the figure, point $A(-1, m)$ lies on the line $y=2x+3$. When connecting $OA$ and $\angle AOB=90^\circ$, point $B$ is on the line $y=-x+b$ and $OA=OB$. What is the value of $b$?

Solution:
\textbf{Solution:} Substituting point A ($-1$, $m$) into the line equation $y=2x+3$, we get: $m=-2+3=1$,\\
since $\angle AOB=90^\circ$ and $OA=OB$,\\
therefore, rotating line segment $OA$ around point $O$ clockwise by $90^\circ$, we obtain line segment $OB$. Hence, the coordinates of point $B$ are ($1$, $1$),\\
substituting point $B$ into the line $y=-x+b$, we get: $1=-1+b$,\\
therefore $b=\boxed{2}$.


Question: As shown in the figure, the line $l_{1}: y=kx-1$ intersects with the line $l_{2}: y=-x+5$ at point $P(2, m)$. What is the solution set of the inequality $kx-1 \ge -x+5$ with respect to $x$?

Solution:
\textbf{Solution:} Since the graph of the function shows that when $x \geq 2$, the line $l_1\colon y=kx-1$ is not below the line $l_2\colon y=-x+5$, \\
therefore, the solution set for the inequality in terms of $x$, $kx-1 \geq -x+5$, is $\boxed{x \geq 2}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects the x-axis at point $A(-3,0)$. What is the solution set of the inequality $kx+b>0$ with respect to $x$?

Solution:
\textbf{Solution:} Since the solution set of the inequality $kx+b>0$ represents the range of the independent variable $x$ when $y>0$, \\
therefore, from the graph in the question, it indicates the range of the coordinate $x$ where the graph of the linear function is above the x-axis, \\
therefore, according to the graph in the question, when $x<-3$, the graph of the linear function is above the x-axis, \\
therefore, when $x<-3$, we have $kx+b>0$, \\
Hence, the answer is: $\boxed{x<-3}$.


Question: The graph of the function $y=kx+b$ is shown in the figure. What is the solution set of the inequality $kx+b<0$ with respect to $x$?

Solution:
\textbf{Solution:} From the graph, we know that the function $y=kx+b$ intersects the x-axis at the point $(2,0)$. Therefore, the intersection point of the function $y=kx+b$ and the x-axis is $(2,0)$. Moreover, $y$ decreases as $x$ increases. \\
$\therefore$ When $x>2$, we have $kx+b<0$, \\
$\therefore$ The solution set of the inequality $kx+b<0$ with respect to $x$ is $x>2$. \\
Thus, the answer is: $\boxed{x>2}$.


Question: As shown in the figure, line $l_{1}: y=kx (k\neq 0)$ and line $l_{2}: y=ax-2 (a\neq 0)$ intersect at point B. Line $l_{2}$ intersects the x-axis at point A ($-4$, $0$). If the x-coordinate of point B is $-2$, what is the value of $k$?

Solution:
\textbf{Solution:} Since line $l_{2}\colon y=ax-2\left(a\ne 0\right)$ intersects the x-axis at point A ($-4$, $0$),\\
$\therefore -4a-2=0$,\\
Solving for $a$, we get: $a=-\frac{1}{2}$,\\
$\therefore$ the equation for line $l_{2}$ is $y=-\frac{1}{2}x-2$,\\
Since line $l_{1}$ intersects line $l_{2}$ at point B, and the x-coordinate of point B is $-2$,\\
$\therefore$ for line $l_{2}\colon y=-\frac{1}{2}x-2$,\\
when $x=-2$, $y=1-2=-1$,\\
$\therefore$ the coordinates of point B are ($-2$, $-1$),\\
Substituting point ($-2$, $-1$) into $y=kx$,\\
we get: $-2k=-1$,\\
Solving for $k$, we find: $k=\boxed{\frac{1}{2}}$.


Question: Given the function $y=2x+b$ and the function $y=kx-3$ intersect at point $P(4, -6)$, what is the solution set of the inequality $kx-3>2x+b$?

Solution:
\[ \text{{\bf Solution:}} \, \text{Given that the function} \, y=2x+b \, \text{and the function} \, y=kx-3 \, \text{intersect at point} \, P(4,-6), \]
\[ \text{therefore, the solution set of the inequality} \, kx-3>2x+b \, \text{is} \, \boxed{x<4}. \]


Question: As shown in the figure, given that the graphs of the functions $y=x+b$ and $y=ax+3$ intersect at point $P$, what is the solution set of the inequality $x+b<ax+3$?

Solution:
\textbf{Solution:} Since the x-coordinate of the intersection point of the two lines is: $x=1$,\\
by examining the graph, it is evident that when $x<1$, $x+b<ax+3$;\\
therefore, the answer is $\boxed{x<1}$.


Question: As shown in the figure, the line $y=-x+m$ intersects with $y=nx+4n\;\left(n\neq 0\right)$ at a point with an abscissa of $-2$. What is the solution set of the inequality $-x+m>nx+4n>0$ with respect to $x$?

Solution:
\textbf{Solution:} Given that the line $y=-x+m$ intersects with $y=nx+4n (n\ne 0)$ at a point where the x-coordinate is $-2$,\\
it follows that the solution set of the inequality $-x+m>nx+4n$ with respect to $x$ is $x<-2$,\\
thus, when $y=nx+4n=0$, we have $x=-4$,\\
therefore, the solution set for the inequality $-x+m>nx+4n>0$ is $-4<x<-2$.\\
Hence, the answer is: $\boxed{-4<x<-2}$.


Question: As shown in the Cartesian coordinate system $xOy$, the graphs of the linear functions ${y}_{1}=kx$ and ${y}_{2}=-x+b$ intersect at point $A(1, 2)$. What is the solution set of the inequality $kx>-x+b$ with respect to $x$?

Solution:
\textbf{Solution:} Since the graph of linear functions $y_{1}=kx$ and $y_{2}=-x+b$ intersects at point $A(1,2)$,\\
it follows from the graph that the solution set of the inequality $kx>-x+b$ with respect to $x$ is $x>1$.\\
Therefore, the answer is: $\boxed{x>1}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ is depicted. What is the range of $x$ when $y<0$?

Solution:
\textbf{Solution:} From the graph, we can conclude: when $y<0$, the range of $x$ is $x>2$. \\
Therefore, the answer is $\boxed{x>2}$.


Question: As shown in the figure, in $\triangle ABC$, $AC = BC = 13$. Place $\triangle ABC$ in the Cartesian coordinate system, with the coordinates of points A and B being $(2, 0)$ and $(12, 0)$, respectively. Translate $\triangle ABC$ along the x-axis to the left such that point C lies on the line $y = -x + 8$. What is the distance of this translation of $\triangle ABC$?

Solution:
\textbf{Solution:} Construct CD $\perp$ x-axis at point D, as shown in the figure.\\
Since the coordinates of points A and B are respectively (2, 0) and (12, 0), and AC=BC=13,\\
it follows that AD=BD=$\frac{1}{2}$AB=5,\\
hence CD=$\sqrt{AC^{2}-AD^{2}}=12$.\\
Therefore, the coordinates of point C are (7, 12).\\
When y=12, we have 12=$-x+8$,\\
solving this gives: x=$-4$,\\
thus, the coordinates of point C after translation are ($-4$, 12).\\
Therefore, $\triangle ABC$ is translated 7-(-4)=\boxed{11} units length to the left along the x-axis.


Question: If the graph of the function $y=kx+b$ is shown in the figure, what is the solution set of the inequality $kx+b>0$?

Solution:
\textbf{Solution:} From the graph, we can determine that the function $y=kx+b$ intersects the x-axis at $(2,0)$. Since $y$ decreases as $x$ increases, \\
$\therefore$ the solution set of the inequality $kx+b>0$ is $\boxed{x<2}$.


Question: As shown in the figure, the graph of the function $y=kx+b$ ($k\neq 0$) passes through point $B(2,0)$ and intersects with the graph of the function $y=2x$ at point $A$. Then, what is the solution set of the inequality $0<kx+b<2x$?

Solution:
\textbf{Solution:} When $y=2$, solving $2=2x$ gives $x=1$, \\
$\therefore$ point A is (1, 2), \\
$\therefore$ for $x>1$, $2x>kx+b$, \\
$\because$ the graph of the function $y=kx+b$ ($k \ne 0$) passes through point B (2, 0), \\
$\therefore$ for $x<2$, $kx+b>0$, \\
$\therefore$ the solution set for the inequality $0<kx+b<2x$ is $1<x<2$.\\
Thus, the answer is: $\boxed{1<x<2}$.


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ passes through the vertices A and C of the square OABC. Given that the coordinates of point A are $\left(1, -3\right)$, what is the value of $b$?

Solution:
\textbf{Solution:} Construct $CM \perp x$-axis at point M, and construct $AN \perp y$-axis at point N. \\
Hence, $\angle OMC = \angle ONA$, \\
Since quadrilateral OABC is a square, \\
Thus, $OA=OC$, $\angle AOC=90^\circ$, \\
Since $\angle COM + \angle MOA = \angle MOA + \angle NOA = 90^\circ$, \\
Therefore, $\angle NOA = \angle COM$, \\
Also, since $OA=OC$, \\
Therefore, $Rt\triangle OCM \cong Rt\triangle OAN$ (AAS), \\
Thus, $OM=ON$, $CM=AN$, \\
Since the coordinate of point A is $(1, -3)$, \\
Therefore, $AN=1$, $ON=3$, \\
Thus, $OM=3$, $CM=1$, \\
Hence, the coordinate of point C is $(3, 1)$, \\
Substituting the coordinates of points A and C into $y=kx+b$, \\
We obtain $\begin{cases}k+b=-3 \\ 3k+b=1\end{cases}$, \\
Solving gives: $\begin{cases}k=2 \\ b=\boxed{-5}\end{cases}$.


Question: As shown in the figure, the graph of the linear function $y=x+2$ intersects with the graph of $y=kx+b$ $(k\ne 0)$ at point $P$. What is the solution to the system of linear equations in two variables $\begin{cases}y=x+2\\ y=kx+b\end{cases}$ for $x$ and $y$?

Solution:
\textbf{Solution:} Observing the graph, we find: The graph of the linear function $y=x+2$ intersects with the graph of $y=kx+b \ (k \neq 0)$ at point P(1,3), \\
$\therefore$ the solution to the system of linear equations $\begin{cases}y=x+2\\ y=kx+b\end{cases}$ is $\boxed{\begin{cases}x=1\\ y=3\end{cases}}$.


Question: As shown in the figure, it is known that the graphs of the linear functions $y=kx+3$ and $y=-x+b$ intersect at point $P(2,4)$. What is the solution set of the linear inequality in $x$, $kx+3>-x+b$?

Solution:
\textbf{Solution:} Solve: When $x>2$, we have $kx+3>-x+b$,\\
which means the solution set of the inequality $kx+3>-x+b$ is $x>2$.\\
Therefore, the answer is: $\boxed{x>2}$.


Question: As shown in the figure, the abscissa of the intersection point of the linear functions $y_1=kx+b$ and $y_2=x+a$ is $-2$. What is the solution set of $kx+b<x+a$?

Solution:
\textbf{Solution:} From the graph, we have $kx+b<x+a$ when $x>-2$, \\
thus the answer is $\boxed{x>-2}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertices \(A\), \(B\), and \(C\) of the rhombus \(ABCD\) are on the coordinate axes. If the coordinates of point \(B\) are \((-1, 0)\) and \(\angle BCD = 120^\circ\), what are the coordinates of point \(D\)?

Solution:
\textbf{Solution:} Given that point B has the coordinates $(-1,0)$,\\
thus OB=1,\\
in diamond $ABCD$, $\angle BCD=120^\circ$, $\angle ABC+\angle BCD=180^\circ$,\\
thus $\angle ABC=60^\circ$, $\angle BAD=120^\circ$,\\
thus $\angle BAC=30^\circ$, $\angle DAO=90^\circ$,\\
thus AB=2OB=2,\\
thus OA$=\sqrt{AB^{2}-OB^{2}}=\sqrt{2^{2}-1^{2}}=\sqrt{3}$,\\
since AD=AB=2,\\
thus the coordinates of point D are $\boxed{(2,\sqrt{3})}$.


Question: As shown in the diagram, the line $y=-\frac{1}{2}x+2$ intersects the $x$ axis at point $A$ and the $y$ axis at point $B$. Point $P$ lies on the line $y=\frac{1}{3}x+b$ that passes through point $B$. What are the coordinates of point $P$ when $\triangle PAB$ is an isosceles right triangle?

Solution:
\textbf{Solution:} Solve: For $y=-\frac{1}{2}x+2$, let $x=0$, then $y=2$,\\
let $y=0$, then $-\frac{1}{2}x+2=0$, solve to get: $x=4$,\\
$\therefore$ point A $(4,0)$, B $(0,2)$,\\
$\therefore$ $OB=2$, $OA=4$,\\
Substituting point B $(0,2)$ into $y=\frac{1}{3}x+b$, we get: $b=2$,\\
$\therefore$ the equation of line PB is $y=\frac{1}{3}x+2$,\\
According to the question, $\angle PBA \neq 90^\circ$,\\
(1) When $\angle BAP'=90^\circ$ and $AB=AP'$, draw $AP'\perp AB$ at A, draw $P'H'\perp$ axis through $P'$,\\
$\therefore$ $\angle AOB=\angle P'H'A=90^\circ$, $\angle OAB+\angle P'AH'=90^\circ$,\\
$\because$ $\angle OAB+\angle OBA=90^\circ$,\\
$\therefore$ $\angle OBA=\angle P'AH'$,\\
And $AB=AP'$,\\
$\therefore$ $\triangle AOB \cong \triangle P'AH'$ (AAS),\\
$\therefore$ $AH'=OB=2$, $P'H'=OA=4$,\\
$\therefore$ $OH'=OA+AH'=6$,\\
$\therefore$ $P'(6,4)$,\\
Substitute $x=6$ into $y=\frac{1}{3}x+2$, get $y=4$,\\
$\therefore$ point $P'$ is on the line $y=\frac{1}{3}x+2$, which is consistent with the question.\\
(2) When $\angle BPA=90^\circ$ and $BP=AP$, draw $AP\perp BP$ at point $P$, draw $PH\perp y$ axis through $P$, and draw $PQ\perp x$ axis through $P$,\\
$\therefore$ $\angle PHO=\angle PQO=\angle HOQ=90^\circ$,\\
$\therefore$ quadrilateral $OHPQ$ is a rectangle,\\
$\therefore$ $PH=OQ$, $PQ=OH$, $\angle HPB+\angle BPQ=90^\circ$,\\
$\because$ $\angle APQ+\angle BPQ=90^\circ$,\\
$\therefore$ $\angle HPB=\angle APQ$,\\
And $\because$ $BP=AP$,\\
$\therefore$ $\triangle HBP \cong \triangle QAP$ (AAS),\\
$\therefore$ $HP=PQ$, $HB=QA$,\\
$\therefore$ quadrilateral $OHPQ$ is a square,\\
$\because$ $OH+OQ=(OB+HB)+OQ=OB+AQ+OQ=OB+(AQ+OQ)=OB+OA=4+2=6$,\\
$\therefore$ $PH=PQ=3$,\\
$\therefore$ $P(3,3)$,\\
Substitute $x=3$ into $y=\frac{1}{3}x+2$ to get: $y=3$,\\
$\therefore$ point $P$ is on the line $y=\frac{1}{3}x+2$, which is consistent with the question.\\
In summary, the coordinates of point $P$ are $(6,4)$ or $(3,3)$.\\
Therefore, the answer is: $\boxed{(6,4) \text{ or } (3,3)}$.


Question: As shown in the figure, rectangle $ABCD \sim $ rectangle $BCEF$. If $AB=8$ and $BC=6$, what is the value of $CE$?

Solution:
\textbf{Solution:} Since rectangle $ABCD \sim$ rectangle $BCEF$, \\
therefore $\frac{AB}{BC} = \frac{BC}{CE}$, \\
thus $CE = \frac{BC^2}{AB} = \frac{6^2}{8} = \boxed{\frac{9}{2}}$.


Question: As shown in the figure, in the parallelogram $ABCD$, $AC$ and $BD$ intersect at point $O$, $E$ is the midpoint of $OA$. Connect $BE$ and extend it to intersect $AD$ at point $F$. What is the ratio of $FD:AF$?

Solution:
\textbf{Solution:} Given quadrilateral ABCD,\\
it follows that $AD=BC$, $AD\parallel BC$, and OA=OC,\\
furthermore, since point E is the midpoint of OA,\\
it follows that $AE=OE=\frac{1}{2}OA=\frac{1}{4}AC$, and $CE=\frac{3}{4}AC$\\
Since $AD\parallel BC$\\
it follows that $\triangle AEF\sim \triangle CEB$\\
Therefore, $\frac{AF}{BC}=\frac{AE}{CE}=\frac{\frac{1}{4}AC}{\frac{3}{4}AC}=\frac{1}{3}$\\
Thus, $AF=\frac{1}{3}BC$\\
Since $AF+FD=AD=BC$\\
it follows $FD=\frac{2}{3}BC$\\
Therefore, $FD\colon AF=2\colon 1$\\
Hence, the answer is: $FD\colon AF=\boxed{2\colon 1}$.


Question: As shown in the figure, it is known that the ratio of similarity between $\triangle ADE$ and $\triangle ABC$ is $1:2$, and the area of $\triangle ADE$ is $1$. What is the area of the quadrilateral $DBCE$?

Solution:
\textbf{Solution:} Since the similarity ratio between $\triangle ADE$ and $\triangle ABC$ is $1\colon 2$,\\
it follows that $\frac{S_{\triangle ADE}}{S_{\triangle ABC}}=\left(\frac{1}{2}\right)^2=\frac{1}{4}$,\\
thus $S_{\triangle ABC}=4S_{\triangle ADE}=4$,\\
therefore, the area of the quadrilateral $DBCE=4-1=\boxed{3}$.


Question: As shown in the figure, the line $y=-\frac{1}{2}x+5$ intersects with the coordinate axes at points A and B, and intersects with the graph of the inverse proportion function $y=\frac{k}{x}\ (x>0)$ at points C and D, with $CD=3AC$. Point E is a point on the line AB. Connect OE and construct a right triangle OEF to the right of OE, where $\angle OEF=90^\circ$, $\angle OFE=\angle ABO$. If side OF intersects the graph of the inverse proportion function at point G, and $OG=GF$, what is the value of k? What are the coordinates of point E?

Solution:
\textbf{Solution:} Let the line $y=-\frac{1}{2}x+5$ intersect the coordinate axes at points A and B,\\
$\therefore A(0, 5), B(10, 0)$. Let the coordinates of point C be $(a, b)$. Draw CM$\perp y$-axis at point M, and through point D draw DN$\perp y$-axis at point N. Then $b=-\frac{1}{2}a+5$, CM=a, AM=5−b,\\
$\triangle AMC\sim \triangle AND$\\
$\therefore \frac{AM}{AN}=\frac{AC}{AD}$. Since $CD=3AC$, AD=AC+CD\\
$\therefore \frac{5−b}{AN}=\frac{1}{4}$\\
$\therefore AN=4(5−b)$,\\
$\therefore ON=OA−AN=5−4(5−b)=4b−15$,\\
$\therefore$ the coordinates of point D are $(4a, 4b−15)$,\\
Since points C, D lie on the graph of the inverse proportion $y=\frac{k}{x}(x>0)$,\\
$\therefore k=ab=4a(4b−15)$. Solving for b, we get $b=4$, substituting $b=4$ into,\\
$b=-\frac{1}{2}a+5$ gives $a=2$. Solving for $k$, we get $k=\boxed{8}$. Draw BF. Since $\angle OFE=\angle ABO$, the points O, B, F, E lie on the same circle. Since $\angle OEF=90^\circ$ and OG=GF, point G is the center of the circle, and OF is the diameter, hence $\angle OBF=90^\circ$. Since B(10, 0), the x-coordinate of point G is 5. When x=5, \\
$y=\frac{8}{x}=\frac{8}{5}$, hence the coordinates of point G are\\
$(5, \frac{8}{5})$. Since OG=GF, the coordinates of point F are\\
$(10, \frac{16}{5})$. Let the coordinates of point E be\\
$(x, -\frac{1}{2}x+5)$. By the Pythagorean theorem, OE$^2$+EF$^2$=OF$^2$, so\\
$x^2+(-\frac{1}{2}x+5)^2+(10-x)^2+(-\frac{1}{2}x+5-\frac{16}{5})^2=10^2+(\frac{16}{5})^2$ Solving for x,\\
$x=\frac{18}{25}$ or $10$ (discard), hence the coordinates of point E are\\
$(\frac{18}{25}, \frac{116}{25})$, therefore the answer is:\\
The coordinates of point E are $\boxed{(\frac{18}{25}, \frac{116}{25})}$.


Question: As shown in the figure, in $\triangle ABC$, with $D$ being a point on side $AC$ and $\angle ABC = \angle ADB = 45^\circ$, if $AD=2$ and $CD=3$, then what is the length of $AB$?

Solution:
\textbf{Solution:} Given that, $\angle ABC = \angle ADB = 45^\circ$, $\angle BAD = \angle CAB$,\\
$\therefore \triangle ABD \sim \triangle ACB$,\\
$\therefore \frac{AD}{AB} = \frac{AB}{AC}$,\\
$\therefore AB^2 = AD \cdot AC$,\\
$\because AD = 2$, CD = 3,\\
$\therefore AC = 5$,\\
$\therefore AB^2 = 2 \times 5 = 10$, that is: $AB = \boxed{\sqrt{10}}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB = 6$, and $E$ is the midpoint of $BC$. $AE$ and $BD$ intersect at point $F$. Connect $CF$. If $AE \perp BD$, what is the length of $CF$?

Solution:
\textbf{Solution:} As shown in the diagram, draw $FM\perp BC$ through point F, with FM intersecting BC at point M\\
$\because$ rectangle ABCD\\
$\therefore$ $\angle ABE=\angle BAD=\angle ACD=90^\circ$, $BC=AD$\\
$\because$AE$\perp$BD\\
$\therefore$ $\angle BAF+\angle ABF=\angle ADB+\angle ABF=90^\circ$\\
$\therefore$ $\angle BAF=\angle ADB$\\
$\therefore$ $\triangle BAD\sim \triangle EBA$\\
$\therefore$ $\frac{AD}{AB}=\frac{AB}{BE}$\\
$\because$E is the midpoint of BC\\
$\therefore$ $AD=BC=2BE$\\
$\therefore$ $\frac{2BE}{6}=\frac{6}{BE}$\\
$\therefore$ $2BE^2=36$\\
$\therefore$ $BE=\sqrt{18}=3\sqrt{2}$\\
Upon verification, $BE\ne 0$, it is a solution to the original equation $\frac{2BE}{6}=\frac{6}{BE}$\\
$\therefore$ $AD=BC=2BE=6\sqrt{2}$\\
$\therefore$ $BD=\sqrt{AB^2+AD^2}=6\sqrt{3}$, $AE=\sqrt{AB^2+BE^2}=3\sqrt{6}$\\
$\therefore$ $BF=\frac{AB\times BE}{AE}=\frac{6\times 3\sqrt{2}}{3\sqrt{6}}=2\sqrt{3}$\\
$\because$ $\angle FBE=\angle CBD$, $\angle BFE=\angle ACD=90^\circ$\\
$\therefore$ $\triangle BFE\sim \triangle BCD$\\
$\therefore$ $\frac{FM}{CD}=\frac{BF}{BD}$\\
$\therefore$ $FM=\frac{CD\times BF}{BD}=\frac{6\times 2\sqrt{3}}{6\sqrt{3}}=2$\\
$\therefore$ $BM=\sqrt{BF^2−FM^2}=\sqrt{12−4}=2\sqrt{2}$\\
$\therefore$ $CM=BC−BM=6\sqrt{2}−2\sqrt{2}=4\sqrt{2}$\\
$\therefore$ $FC=\sqrt{FM^2+CM^2}=\sqrt{4+32}=\boxed{6}$


Question: As shown in the figure, point $E$ is on the side $BC$ of the square $ABCD$, where $BE=2CE$, and point $G$ is the foot of the perpendicular. If $FG=2$, what is the value of $DG$?

Solution:
In the square $ABCD$, we have: $AD=AB=BC=CD$\\
$\angle DAB=\angle B=\angle C=\angle ADC=90^\circ$\\
$\therefore \angle 1+\angle 3=90^\circ$\\
$\because DF\perp AE$\\
$\therefore \angle DGE=\angle DAF=90^\circ$\\
$\therefore \angle 2+\angle 3=90^\circ$\\
Also, $\angle 1+\angle 3=90^\circ$\\
$\therefore \angle 2=\angle 1$\\
In $\triangle ADF$ and $\triangle BAE$, we have\\
$\left\{\begin{array}{l}
\angle 2=\angle 1\\
AD=AB\\
\angle DAB=\angle B=90^\circ
\end{array}\right.$\\
$\therefore \triangle ADF\cong \triangle BAE$ (ASA)\\
$\therefore AF=BE$\\
$\because BE=2CE$\\
Let $CE=x$, then $BE=2x$, $BC=3x$, $AF=2x$, $BF=x$\\
$\because \angle 1=\angle 2$, $\angle AGF=\angle B=90^\circ$\\
$\therefore \triangle AGF\sim \triangle ABE$\\
$\therefore \frac{FG}{BE}=\frac{AF}{AE}$\\
$\therefore \frac{2}{2x}=\frac{2x}{AE}$\\
$\therefore AE=2x^2$\\
In $\triangle ABE$, we have $AB^2+BE^2=AE^2$\\
$\therefore (3x)^2+(2x)^2=(2x^2)^2$\\
$\therefore 9x^2+4x^2=4x^4$\\
$\therefore 13x^2=4x^4$\\
$\because x\neq 0$\\
$\therefore 13=4x^2$\\
$\therefore x=\pm \frac{\sqrt{13}}{2}$ (discard the negative solution)\\
$\therefore x=\frac{\sqrt{13}}{2}$\\
$\therefore AE=\frac{13}{2}$\\
$AF=\sqrt{13}$\\
In $\triangle AGF\sim \triangle ABE$,\\
$\frac{AF}{AE}=\frac{FG}{BE}$\\
$\therefore \frac{\sqrt{13}}{\frac{13}{2}}=\frac{FG}{\sqrt{13}}$\\
$\therefore FG=2$\\
$\therefore DG=DF-GF=\boxed{\frac{9}{2}}$


Question: As shown in the diagram, this is a schematic of a rifle being aimed. The distance from the eye to the sight, OE, is 80 cm, the width of the sight on the rifle, AB, is 0.2 cm, and the frontal width of the target, CD, is 50 cm. What is the distance from the eye to the target, OF, in meters?

Solution:
\textbf{Solution:} Because $AB \parallel CD$, \\
therefore $\triangle OAB \sim \triangle OCD$, \\
therefore $\frac{AB}{CD} = \frac{OE}{OF}$, that is $\frac{0.2}{50} = \frac{80}{OF}$, \\
therefore $OF = \frac{80 \times 50}{0.2} = 20000\,cm = 200\,m$. \\
Hence, the answer is: $OF = \boxed{200m}$.


Question: As shown in the figure, line $AD\parallel BE\parallel CF$, if $\frac{AB}{BC}=\frac{1}{3}$, $AD=2$, and $CF=6$, then what is the length of the line segment $BE$?

Solution:
\textbf{Solution:} As shown in the diagram, draw DG $\parallel$ AC through point D, intersecting CF at point G, and BE at point H, \\
$\because$ $AD\parallel BE\parallel CF$, \\
$\therefore$ $\frac{DE}{EF}=\frac{AB}{BC}=\frac{1}{3}$, quadrilaterals ABHD and ACGD are parallelograms, \\
$\therefore$ BH=AD=CG=2, $\frac{DE}{DF}=\frac{1}{4}$, \\
$\because$ $CF=6$, \\
$\therefore$ FG=4, \\
$\because$ BE$\parallel$CF, \\
$\therefore$ $\triangle DEH\sim \triangle DFG$, \\
$\therefore$ $\frac{HE}{FG}=\frac{DE}{DF}=\frac{1}{4}$, \\
$\therefore$ HE=1, \\
$\therefore$ BE=BH+HE=$\boxed{3}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=6$, $BC=4$, point $D$ is on side $AC$, $BD=BC$, then what is the length of $AD$?

Solution:
\textbf{Solution}: Given that $AB=AC$, and $BD=BC$,\\
it follows that $\angle ABC = \angle C$, and $\angle C = \angle BDC$,\\
therefore $\triangle ABC \sim \triangle BDC$,\\
which implies $ \frac{AB}{BD}=\frac{BC}{CD}$, \\
given $AB=AC=6$, $BC=4$, and $BD=BC$,\\
it follows $ \frac{6}{4}=\frac{4}{CD}$, \\
thus $ CD=\frac{8}{3}$, \\
therefore $AD=AC-CD=6- \frac{8}{3}=\boxed{\frac{10}{3}}$,


Question: As shown in the figure, it is known that $l_{1}\parallel l_{2}\parallel l_{3}$. The line AB intersects $l_{1}$, $l_{2}$, and $l_{3}$ at A, M, and B, respectively. Similarly, the line CD intersects $l_{1}$, $l_{2}$, and $l_{3}$ at C, N, and D, respectively. Given that AM = 4, MB = 6, and CD = 9, what is the length of ND?

Solution:
\textbf{Solution:} Since $l_1 \parallel l_2 \parallel l_3$,\\
it follows that $\frac{AM}{AB} = \frac{CN}{CD}$,\\
thus $\frac{4}{4+6} = \frac{CN}{9}$,\\
solving this gives $CN = 3.6$,\\
hence $ND = CD - CN = 9 - 3.6 = \boxed{5.4}$.


Question: As shown in the figure, for square $ABCD$ with a side length of 1, $O$ is the midpoint of diagonal $BD$. Point $M$ is on side $AB$, and $BM=2AM$. Point $N$ is on side $BC$, and $BN=AM$. Lines $AN$ and $MD$ intersect at point $P$. Connect $OP$. What is the length of $OP$?

Solution:
\textbf{Solution:} Let AN and BD intersect at point Q,\\
since the side length of square ABCD is 1,\\
therefore $BD=\sqrt{AB^2+AD^2}=\sqrt{2}$,\\
since $BM=2AM$,\\
therefore $BM+AM=AB=1$,\\
therefore $AM=BN=\frac{1}{3}$, $BM=\frac{2}{3}$,\\
since $AD\parallel BN$,\\
therefore $\triangle NQB\sim \triangle AQD$,\\
therefore $\frac{BN}{AD}=\frac{BQ}{QD}=\frac{NQ}{AQ}=\frac{1}{3}$,\\
therefore $DQ=3BQ$,\\
therefore $DQ=\frac{3\sqrt{2}}{4}$, $BQ=\frac{\sqrt{2}}{4}$,\\
since O is the midpoint of BD,\\
therefore $OD=\frac{\sqrt{2}}{2}$,\\
therefore $OQ=DQ-OD=\frac{\sqrt{2}}{4}$,\\
In $\triangle ADM$ and $\triangle BAN$,\\
$\begin{cases}AD=AB\\ \angle DAM=\angle ABN\\ AM=BN\end{cases}$,\\
therefore $\triangle ADM\cong \triangle BAN$ (SAS),\\
therefore $\angle ADM=\angle BAN$,\\
since $\angle PAD+\angle BAN=90^\circ$,\\
therefore $\angle PAD+\angle ADM=90^\circ$,\\
therefore $\angle APD=90^\circ$,\\
since $\angle DAM=90^\circ$, $AM=\frac{1}{3}$, $AD=1$,\\
therefore $DM= \sqrt{AM^2+AD^2}=\frac{\sqrt{10}}{3}$,\\
since $S_{\triangle ADM}=\frac{1}{2}\times AD\cdot AM=\frac{1}{2}\times DM\cdot AP$,\\
therefore $AP=\frac{1\times \frac{1}{3}}{\frac{\sqrt{10}}{3}}=\frac{\sqrt{10}}{10}$,\\
therefore $PM=\sqrt{AM^2-AP^2}=\frac{\sqrt{10}}{30}$,\\
therefore $PD=DM-PM=\frac{3\sqrt{10}}{10}$,\\
since $\triangle ADM\cong \triangle BAN$,\\
therefore $AN=DM= \frac{\sqrt{10}}{3}$,\\
since $\frac{NQ}{AQ}=\frac{1}{3}$,\\
therefore $NQ=\frac{\sqrt{10}}{12}$, $AQ=\frac{\sqrt{10}}{4}$,\\
therefore $PQ=AQ-AP=\frac{\sqrt{10}}{4}-\frac{\sqrt{10}}{10}=\frac{3\sqrt{10}}{20}$,\\
As shown in the diagram, draw OG$\perp$DM at point G through point O,\\
since $OG\parallel PQ$,\\
therefore $\triangle OGD\sim \triangle QPD$,\\
therefore $\frac{DO}{DQ}=\frac{DG}{DP}=\frac{OG}{QP}=\frac{2}{3}$,\\
therefore $DG=\frac{2}{3}DP=\frac{2}{3}\times \frac{3\sqrt{10}}{10}=\frac{\sqrt{10}}{5}$, $OG=\frac{2}{3}QP=\frac{2}{3}\times \frac{3\sqrt{10}}{20}=\frac{\sqrt{10}}{10}$,\\
therefore $PG=DP-DG=\frac{3\sqrt{10}}{10}-\frac{\sqrt{10}}{5}=\frac{\sqrt{10}}{10}$,\\
therefore $OP=\sqrt{OG^2+PG^2}=\sqrt{\frac{1}{10}+\frac{1}{10}}=\boxed{\frac{\sqrt{5}}{5}}$.


Question: As shown, it is known that $\triangle ABC$ and $\triangle ADE$ are both isosceles right-angled triangles, with $\angle BAC=\angle ADE=90^\circ$, and $AB=AC=1$, $AD=DE=\sqrt{5}$. Point D lies on line BC, and the extension of line segment $EA$ intersects line BC at point F. What is the length of $FB$?

Solution:
\textbf{Solution:} As shown in the figure, draw AH $\perp$ BC at point H,\\
$\because$ $\angle$BAC=$90^\circ$, AB=AC=1,\\
$\therefore$ BC=$\sqrt{2}$,\\
$\because$ AH$\perp$BC,\\
$\therefore$ BH=CH=$\frac{\sqrt{2}}{2}$,\\
$\therefore$ AH=$\frac{\sqrt{2}}{2}$,\\
$\because$ AD=DE=$\sqrt{5}$,\\
$\therefore$ DH=$\sqrt{AD^{2}-AH^{2}}=\sqrt{5-\frac{1}{2}}=\frac{3\sqrt{2}}{2}$,\\
$\therefore$ CD=DH−CH=$\sqrt{2}$,\\
$\because$ $\angle$ABC=$\angle$ACB=$45^\circ$,\\
$\therefore$ $\angle$ABF=$\angle$ACD=$135^\circ$,\\
$\because$ $\angle$DAE=$45^\circ$,\\
$\therefore$ $\angle$DAF=$135^\circ$,\\
$\because$ $\angle$BAC=$90^\circ$,\\
$\therefore$ $\angle$BAF+$\angle$DAC=$45^\circ$,\\
$\because$ $\angle$BAF+$\angle$F=$45^\circ$,\\
$\therefore$ $\angle$F=$\angle$DAC,\\
$\therefore$ $\triangle$ABF$\sim$ $\triangle$DCA,\\
$\therefore$ $\frac{AB}{CD}=\frac{BF}{AC}$,\\
$\therefore$ $\frac{1}{\sqrt{2}}=\frac{BF}{1}$,\\
$\therefore$ BF=$\boxed{\frac{\sqrt{2}}{2}}$.


Question: As shown in the figure, the side length of square $ABCD$ is $4$. Diagonal $AC$, points $E$, $F$ are respectively on the extensions of $BC$, $CD$, with $CE=2$, $DF=1$. Point $G$ is the midpoint of $EF$. Connect $OE$ and intersect $CD$ at point $H$. What is the length of $GH$?

Solution:
\textbf{Solution:} Draw OM$\perp$BC at M through point O, and draw GN$\perp$CF at N through point G,\\
since square ABCD,\\
thus CO=OA, and $\angle$ABC$=90^\circ$,\\
thus OM$\parallel$AB,\\
$\triangle$COM$\sim$$\triangle$CAB,\\
thus $\frac{CO}{AC}=\frac{OM}{AB}=\frac{CM}{BC}$ that is $\frac{CO}{2OC}=\frac{OM}{4}=\frac{CM}{BC}=\frac{1}{2}$,\\
we find OM=2 and CM=$\frac{1}{2}BC=\frac{1}{2}\times 4=2$,\\
since CE=2,\\
thus MC=CE=2, ME=MC+CE=4,\\
since CH$\parallel$AB,\\
thus CH$\parallel$OM,\\
thus $\frac{CH}{OM}=\frac{CE}{ME}$ that means $\frac{CH}{2}=\frac{2}{4}=\frac{1}{2}$,\\
thus CH=1,\\
since GN$\perp$CF, and $\angle$FCE$=90^\circ$,\\
thus GN$\parallel$CE and point G is the midpoint of EF,\\
thus $\triangle$FGN$\sim$$\triangle$FEC,\\
thus $\frac{FN}{FC}=\frac{NG}{CE}=\frac{FG}{FE}$ that means $\frac{FN}{FC}=\frac{NG}{CE}=\frac{FG}{2FG}=\frac{1}{2}$,\\
since DF=1,\\
thus CF=CD+DF=4+1=5,\\
thus $\frac{FN}{5}=\frac{NG}{2}=\frac{1}{2}$,\\
we find FN=2.5, NG=1,\\
thus NH=CF−FN−CH=5−2.5−1=1.5,\\
In right $\triangle$NHG, GH=$\sqrt{NG^{2}+NH^{2}}=\sqrt{1^{2}+1.5^{2}}=\boxed{\frac{1}{2}\sqrt{13}}$.


Question: As shown in the figure, in $\square ABCD$, $AB=8$, $BC=12$. $\square ABCD$ is rotated counterclockwise about point A to obtain $\square AB'C'D'$. Point $B'$ is on side BC, $BB'=4$, and $B'C'$ intersects side CD at point E. Side $C'D'$ passes through point D. What is the length of $CE$?

Solution:
\textbf{Solution:} Given the properties of a parallelogram and rotation, it is known that $C'D'=CD=AB'=AB=8$, $AD'=AD=BC'=BC=12$, $\angle BAD=\angle B'AD'$, and $\angle B'CE=\angle DC'E$. \\
$\therefore \angle B=\angle AB'B=\angle ADD'=\angle D'$, hence $\angle BAD-\angle B'AD=\angle B'AD'-\angle B'AD$. \\
$\therefore \angle BAB'=\angle DAD'$. \\
$\therefore \triangle BAB' \sim \triangle DAD'$, \\
which yields $\frac{AB}{AD}=\frac{BB'}{DD'}$, i.e., $\frac{8}{12}=\frac{4}{DD'}$. \\
Solving for $DD'=6$. \\
$\therefore DC'=D'C'-DD'=2$. \\
Since $\angle B'CE=\angle DC'E$ and $\angle B'EC=\angle DEC'$, \\
$\therefore \triangle B'EC \sim \triangle DEC'$, \\
thus, $\frac{B'E}{DE}=\frac{CE}{C'E}=\frac{BC}{DC'}=\frac{12-4}{2}$. \\
$\therefore B'E=4DE$, and $CE=4C'E$. \\
Given $B'E=12-C'E=12-\frac{1}{4}CE$, and $DE=8-CE$. \\
$\therefore 12-\frac{1}{4}CE=4\times (8-CE)$. \\
Solving yields $CE=\boxed{\frac{16}{3}}$.


Question: In the equilateral triangle $\triangle ABC$, $P$ is a point on $BC$, and $D$ is a point on $AC$, with $\angle APD=60^\circ$, $BP=4$, and $CD=2$. What is the side length of $\triangle ABC$?

Solution:
\textbf{Solution:} Given that $\triangle ABC$ is an equilateral triangle,\\
it follows that AB = BC = AC, and $\angle B = \angle C = 60^\circ$,\\
thus $\angle BAP + \angle APB = 180^\circ - 60^\circ = 120^\circ$,\\
since $\angle APD = 60^\circ$,\\
it follows that $\angle APB + \angle DPC = 180^\circ - 60^\circ = 120^\circ$,\\
therefore, $\angle BAP = \angle DPC$,\\
which means $\angle B = \angle C$ and $\angle BAP = \angle DPC$,\\
hence $\triangle ABP \sim \triangle PCD$;\\
thus $\frac{AB}{PC} = \frac{BP}{CD}$,\\
given BP = 4 and CD = 2,\\
it follows that $\frac{AB}{AB - 4} = \frac{4}{2}$, solving gives AB = \boxed{8},\\
therefore, the side length of $\triangle ABC$ is 8.


Question: As shown, it is known that the line $y=-\frac{1}{2}x+2$ intersects the x-axis at point B and the y-axis at point A. Point P is a moving point on line AB (not coinciding with A or B). Line $PQ\perp x$ axis at point Q. If the triangle with vertices P, O, Q is similar to $\triangle AOB$, then what are the coordinates of point P?

Solution:
\textbf{Solution:}

Since point $P$ lies on the line $AB$ defined by $y=-\frac{1}{2}x+2$,

let $P\left(m, -\frac{1}{2}m+2\right)$. Therefore, $PQ=\left|-\frac{1}{2}m+2\right|$ and $OQ=\left|m\right|$

Given $\angle PQO=90^\circ$ and $\angle AOB=90^\circ$,

if the triangle formed by points $P$, $O$, and $Q$ is similar to $\triangle AOB$,

then $\triangle POQ\sim \triangle AOB$ or $\triangle PQO\sim \triangle BOA$

Thus, $\frac{OQ}{OB}=\frac{QP}{OA}$ or $\frac{OQ}{OA}=\frac{QP}{OB}$

Given $y=-\frac{1}{2}x+2$,

by setting $x=0$, we get $y=2$, and by setting $y=0$, we get $x=4$

Hence, $OA=2$ and $BO=4$

Therefore, $\frac{\left|m\right|}{4}=\frac{\left|-\frac{1}{2}m+2\right|}{2}$ or $\frac{\left|m\right|}{2}=\frac{\left|-\frac{1}{2}m+2\right|}{4}$

Solving gives $m=2$, or $m=\frac{4}{5}$, or $m=-\frac{4}{3}$

Thus, the coordinates of $P$ are $\boxed{(2,1)}$, $\left(\frac{4}{5}, \frac{8}{5}\right)$, or $\left(-\frac{4}{3}, \frac{8}{3}\right)$


Question: As shown in the figure, line segments AB and CD intersect at point E, with AC $\parallel$ BD. If AC = 3, BD = 1.5, and BE = 2, what is the length of AE?

Solution:
\textbf{Solution:} Given that $AC \parallel BD$,\\
it follows that $\angle A = \angle B$ and $\angle C = \angle D$,\\
therefore $\triangle AEC \sim \triangle BED$,\\
which leads to $\frac{AC}{BD} = \frac{AE}{BE}$,\\
given that $AC = 3$, $BD = 1.5$, and $BE = 2$,\\
it follows that $\frac{3}{1.5} = \frac{AE}{2}$,\\
solving this, we find $AE = \boxed{4}$.


Question: As shown in the figure, there is a square $ABCD$ with side length $6\sqrt{2}$. $E$ is the midpoint of side $CD$. There is a moving point $F$ on the diagonal $BD$. When $\triangle ABF$ is similar to $\triangle DEF$, what is the value of $BF$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a square, \\
it follows that BC=CD=$6\sqrt{2}$, and $\angle C=90^\circ$.\\
Since E is the midpoint of side CD, \\
it follows that DE=CE=$\frac{1}{2}$CD=$3\sqrt{2}$.\\
In $\triangle BCD$, by the Pythagorean theorem, we have \\
BD=$\sqrt{BC^{2}+CD^{2}}=\sqrt{(6\sqrt{2})^{2}+(6\sqrt{2})^{2}}=12$.\\
Let BF=x, then DF=12-x.\\
(1) When $\triangle ABF\sim \triangle FDE$,\\
from $\frac{DF}{BA}=\frac{DE}{BF}$, we get $\frac{12-x}{6\sqrt{2}}=\frac{3\sqrt{2}}{x}$,\\
which gives $x=\boxed{6}$.\\
(2) When $\triangle ABF\sim \triangle EDF$,\\
from $\frac{DF}{BF}=\frac{DE}{BA}$, we get $\frac{12-x}{x}=\frac{3\sqrt{2}}{6\sqrt{2}}$,\\
which gives $x=8$.\\
In conclusion, the value of BF can be either 6 or 8.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $O$ is the midpoint of $AB$, $OD$ bisects $\angle AOC$ and intersects $AC$ at point $G$, $OD = OA$, $BD$ intersects $AC$ and $OC$ at points $E$ and $F$ respectively, connect $AD$ and $CD$. What is the value of $OG\colon BC$? If $CE = CF$, what is the value of $CF\colon OF$?

Solution:
\textbf{Solution:}

(1) Since $O$ is the midpoint of $AB$ and $\angle ACB=90^\circ$, \\
$\therefore$ $OB=OA=OC$, \\
Since $OD$ bisects $\angle AOC$ and intersects $AC$ at point $G$, \\
$\therefore$ point $G$ is the midpoint of $AC$, \\
$\therefore$ $OG$ is the median of $\triangle ABC$, \\
$\therefore$ $OG\parallel BC$ and $OG=\frac{1}{2}BC$, that is, $\frac{OG}{BC}=\boxed{\frac{1}{2}}$;

(2) \\
Since $OD=OA$, \\
$\therefore$ $OD=OB$, \\
$\therefore$ $\angle ODB=\angle OBD$, \\
Since $OG\perp AC$, \\
$\therefore$ $\angle DGE=90^\circ$, \\
$\therefore$ $\angle GDE+\angle DEG=90^\circ$, \\
Since $CE=CF$, \\
$\therefore$ $\angle CEF=\angle CFE$, \\
Since $\angle CEF=\angle DEG$, $\angle CFE=\angle OFB$, and $\angle ODB=\angle OBD$, \\
$\therefore$ $\angle OFB+\angle OBD=90^\circ$, \\
$\therefore$ $\angle FOB=90^\circ$, that is, $CO\perp AB$, \\
$\therefore$ $\triangle OBC$ is an isosceles right triangle, \\
$\therefore$ $BC: OB=\sqrt{2}: 1$;

Knowing from (1) that $OG\parallel BC$, \\
$\therefore$ $\triangle BCF\sim \triangle DOF$, \\
$\therefore$ $\frac{CF}{OF}=\frac{BC}{OD}=\frac{BC}{OB}=\boxed{\sqrt{2}}$.


Question: As shown in the figure, the straight lines $l_{1} \parallel l_{2} \parallel l_{3}$. If $AB = 6$, $BC = 10$, and $EF = 9$, then what is the length of $DE$?

Solution:
\textbf{Solution:} Since $l_{1} \parallel l_{2} \parallel l_{3}$, and $AB = 6$, $BC = 10$, \\
it follows that $\frac{DE}{EF} = \frac{AB}{BC} = \frac{6}{10} = \frac{3}{5}$, \\
Since $EF = 9$, \\
then $DE = \boxed{\frac{27}{5}}$.


Question: As shown in the figure, in right-angled triangle $ABC$ with $\angle ACB=90^\circ$, $CD\perp AB$, $AC=5$, and $BC=12$, point $P$ is a moving point on line segment $CD$. When a circle $\odot P$ with radius $4$ is tangent to one side of $\triangle ABC$, what is the length of $CP$?

Solution:
\textbf{Solution:}

Given that in right triangle $ABC$, $AC=5$, $BC=12$,\\
therefore $AB=\sqrt{AC^2+BC^2}=\sqrt{5^2+12^2}=13$,\\
since $CD\perp AB$,\\
the area of $\triangle ABC = \frac{1}{2}AB\cdot CD=\frac{1}{2}AC\cdot BC$,\\
thus, $13CD=5\times 12$,\\
therefore, $CD= \frac{60}{13}$,\\
considering three cases:\\
(1) When circle $P$ is tangent to side $BC$, as shown in the figure:\\
Draw $PE\perp BC$ at point E,\\
since $PE \perp BC$,\\
thus $\angle PEC=90^\circ$,\\
hence $\angle PCE+\angle CPE=90^\circ$, and $\angle PCE+\angle B=90^\circ$,\\
therefore, $\angle B=\angle CPE$,\\
since $\angle CEP=\angle ACB=90^\circ$,\\
thus $\triangle BCA\sim \triangle PEC$,\\
hence $\frac{BA}{PC}=\frac{BC}{PE}$,\\
therefore, $\frac{13}{PC}=\frac{12}{4}$,\\
thus, $PC=\frac{13}{3}$,\\
(2) When circle $P$ is tangent to side $AB$, as shown in the figure:\\
since $PD\perp AB$,\\
therefore, $CP=CD-PD=\frac{60}{13}-4=\frac{8}{13}$,\\
(3) When circle $P$ is tangent to side $AC$, as shown in the figure:\\
Draw $PF\perp AC$ at point F,\\
since $PF\perp AC$,\\
thus $\angle PFC=90^\circ$,\\
hence $\angle CPF+\angle PCF=90^\circ$, and $\angle A+\angle PCF=90^\circ$\\
therefore, $\angle A=\angle CPF$,\\
since $\angle CFP=\angle ACB=90^\circ$,\\
thus $\triangle BCA\sim \triangle CFP$,\\
hence $\frac{BA}{PC}=\frac{CA}{FP}$,\\
therefore, $\frac{13}{PC}=\frac{5}{4}$,\\
thus, $PC=\frac{52}{5}$,\\
since $\frac{52}{5}>\frac{60}{13}$\\
therefore, $PC=\frac{52}{5}$ (discarded)\\
In conclusion, when circle $P$ with radius 4 is tangent to one side of $\triangle ABC$, the length of $CP$ could be: $\boxed{\frac{13}{3}}$ or $\frac{8}{13}$.


Question: As shown in the figure, in $\triangle ABC$, points D and E are on AB and AC respectively, $\angle AED=\angle B$, $AD=\frac{3}{4}AC$. If the area of the quadrilateral BCED is 7, what is the area of $\triangle ADE$?

Solution:
\textbf{Solution:} Since $\angle A = \angle A$ and $\angle AED = \angle B$,

it follows that $\triangle ADE \sim \triangle ACB$.

Thus, $\frac{AD}{AC} = \frac{\frac{3}{4}AC}{AC} = \frac{3}{4}$,

and $\frac{S_{\triangle ADE}}{S_{\triangle ABC}} = \left(\frac{AD}{AC}\right)^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$.

Let $S_{\triangle ADE} = 9x$, then $S_{\triangle ABC} = 16x$.

Therefore, $S_{BCED} = S_{\triangle ABC} - S_{\triangle ADE} = 16x - 9x = 7x = 7$,

implying $x = 1$,

and therefore $S_{\triangle ADE} = 9x = \boxed{9}$.


Question: As shown in the figure, in a grid of small squares with equal side lengths, points A, B, C, and D are all on the vertices of these small squares. Lines AB and CD intersect at point P. What is the value of $\frac{AP}{PB}$? What is the value of $\tan\angle APD$?

Solution:
\textbf{Solution}:

Since quadrilateral BCED is a square, \\
it follows that DB$\parallel$AC, \\
which means $\triangle DBP\sim \triangle CAP$, \\
thus, $\frac{AP}{PB} = \frac{AC}{DB} = \boxed{3}$, \\
Connect BE, intersecting CD at point F, \\
Since quadrilateral BCED is a square, \\
it follows that DF=CF= $ \frac{1}{2}$ CD, BF= $ \frac{1}{2}$ BE, and CD=BE, BE$\perp$CD, \\
thus, BF=CF. According to the problem, AC$\parallel$BD, \\
it follows that $\triangle ACP\sim \triangle BDP$, \\
thus, DP$\colon$ CP=BD$\colon$ AC=1$\colon$ 3, \\
which means DP$\colon$ DF=1$\colon$ 2, therefore, DP=PF= $ \frac{1}{2}$ CF= $ \frac{1}{2}$ BF, \\
In $\triangle PBF$ right-angled at B, $\tan\angle BPF= \frac{BF}{PF} = \boxed{2}$, \\
Since $\angle APD=\angle BPF$, \\
it follows that $\tan\angle APD=2$.


Question: As shown in the figure, in $\triangle ABC$, $AC>AB$, $AD$ is the angle bisector, $AE$ is the median, $BF\perp AD$ at point $G$, intersects $AE$ at point $F$, and intersects $AC$ at point $M$, the extension line of $EG$ intersects $AB$ at point $H$. If $\angle BAC=60^\circ$, what is the value of $\frac{FG}{DG}$?

Solution:
\textbf{Solution:} Extend EH to P such that GH=HP, and connect AP and BP,\\
since BF$\perp$AD,\\
it follows that $\angle$ABG+$\angle$BAG=$90^\circ$,\\
$\angle$AMG+$\angle$MAG=$90^\circ$,\\
since AD is the angle bisector,\\
it follows that $\angle$BAG=$\angle$MAG,\\
thus, $\angle$ABG=$\angle$AMG,\\
thus, AB=AM,\\
thus, BG=MG,\\
since BE=EC,\\
it follows that GE$\parallel$AC,\\
thus, AH=BH,\\
thus, Quadrilateral APBG is a parallelogram,\\
thus, AP=BG, and AP$\parallel$BG,\\
thus, $\frac{GF}{PA}=\frac{GE}{PE}$,\\
thus, $\frac{GF}{BG}=\frac{GE}{PE}$,\\
Similarly, $\frac{GD}{AG}=\frac{GE}{PE}$,\\
thus, $\frac{GF}{BG}=\frac{GD}{AG}$,\\
thus, $\frac{GF}{GD}=\frac{BG}{AG}$,\\
since $\angle$BAC=$60^\circ$ and AD is the angle bisector,\\
it follows that $\angle$BAG=$30^\circ$,\\
In $\triangle$ABG, $\frac{BG}{AG}=\tan 30^\circ=\frac{\sqrt{3}}{3}$,\\
thus, $\frac{FG}{DG}=\boxed{\frac{\sqrt{3}}{3}}$.


Question: As shown in the figure, semicircles $O_1$, $O_2$, ..., $O_n$ with their centers on the positive x-axis are tangent to the line $l$. Suppose the radii of semicircles $O_1$, $O_2$, ..., $O_n$ are $r_1$, $r_2$, ..., $r_n$ respectively. Given that the acute angle between line $l$ and the x-axis is $30^\circ$, and $r_1=2$, what is the value of $r_{2021}$?

Solution:
\textbf{Solution:} Let the points of tangency be denoted as $A_1$, $A_2$, $A_3,\dots, A_{2021}$, and connect $O_1A_1$, $O_2A_2$, $O_3A_3$, $\dots$

Since $\sin 30^\circ =\frac{1}{2}=\frac{r_1}{OO_1}=\frac{r_2}{OO_2}=\frac{r_3}{OO_3}=\cdots=\frac{r_{2021}}{OO_{2021}}$,

and $OO_1=2r_1$, $OO_2=OO_1+O_1O_2=2r_1+(r_1+r_2)=3r_1+r_2$,

therefore, $\frac{r_2}{3r_1+r_2}=\frac{1}{2}$,

and since $r_1=2$,

it follows that $r_2=3r_1=2\times 3=6$,

Similarly, we can find $r_3=2\times 3^2=18$, $r_4=2\times 3^3=54$, $r_5=2\times 3^4=162$, $\dots$

Thus, $r_{2021}=2\times 3^{2020}$,

Therefore, the answer is $2\times 3^{2020}=\boxed{2\times 3^{2020}}$.


Question: As shown in the figure, in the Cartesian coordinate system, the coordinates of points $A$ and $B$ are (3, 0) and (0, 4), respectively. Point $C$ is a point on the positive $x$-axis. Line $BC$ is drawn. The line passing through point $A$ and perpendicular to $AB$ intersects with the line passing through point $C$ and perpendicular to $BC$ at point $D$. Line $BD$ is drawn. What is the value of $\sin\angle BDC$?

Solution:
\textbf{Solution:} Since $BA \perp AD$ and $BC \perp CD$,\\
it follows that $\angle BAD = \angle BCD = 90^\circ$.\\
Hence, points A, B, C, D are concyclic.\\
Therefore, $\angle BDA = \angle BCA$.\\
Since $\angle BDA + \angle DBA = \angle BCA + \angle CBO = 90^\circ$,\\
it implies that $\angle DBA = \angle CBO$.\\
Consequently, $\angle DBA - \angle CBA = \angle CBO - \angle CBA$,\\
which means $\angle DBC = \angle ABO$.\\
Moreover, $\angle DBC + \angle BDC = \angle ABO + \angle BAO = 90^\circ$,\\
thus $\angle BDC = \angle BAO$.\\
Given that the coordinates of points A and B are (3, 0) and (0, 4) respectively,\\
it follows that $BO = 4$, $OA = 3$, and $AB = \sqrt{4^2+3^2} = 5$.\\
Hence, $\sin \angle BAO = \frac{BO}{AB} = \frac{4}{5}$.\\
Therefore, $\sin \angle BDC = \boxed{\frac{4}{5}}$.


Question: As shown in the figure, in $ \triangle ABC$, $ \angle C=90^\circ$ , $AC=12$, and the perpendicular bisector $MN$ of $AB$ intersects $AC$ at $D$, and $BD$ is connected. If $\cos\angle CDB=\frac{3}{5}$, then how much is $BC$?

Solution:
\textbf{Solution:} Given $\cos\angle CDB=\frac{3}{5}$,\\
$\therefore$ $\frac{CD}{BD}=\frac{3}{5}$,\\
Let $CD=3x$ and $BD=5x$,\\
$\therefore$ $BC=\sqrt{{(5x)}^{2}-{(3x)}^{2}}=4x$,\\
Since MN is the perpendicular bisector of AB,\\
$\therefore$ AD=BD=5x,\\
$\therefore$ AC=CD+AD=3x+5x=8x,\\
Given that AC=12,\\
$\therefore$ 8x=12,\\
$\therefore$ x= $\frac{3}{2}$,\\
$\therefore$ BC=4x=\boxed{6}.


Question: As shown in the figure, the vertices of $\triangle ABC$ are all at the vertices of a small square with equal sides. What is the value of $\cos\angle BAC$?

Solution:
\textbf{Solution:} Let the side length of the small square be 1,\\
Draw CD $\perp$ AB at D,\\
$S_{\triangle ABC} = \frac{1}{2} \times 2 \times 2 = 2$,\\
By Pythagoras theorem, we have: $AB = \sqrt{2^2 + 4^2} = 2\sqrt{5}$, $AC = \sqrt{2^2 + 2^2} = 2\sqrt{2}$,\\
$\therefore 2 = \frac{1}{2} \times 2\sqrt{5} \times CD$,\\
Solving for this, we get: $CD = \frac{2\sqrt{5}}{5}$,\\
Using Pythagoras theorem again, we find: $AD = \sqrt{AC^2 - CD^2} = \sqrt{(2\sqrt{2})^2 - \left(\frac{2\sqrt{5}}{5}\right)^2} = \frac{6\sqrt{5}}{5}$,\\
$\therefore \cos\angle BAC = \frac{AD}{AC} = \frac{\frac{6\sqrt{5}}{5}}{2\sqrt{2}} = \boxed{\frac{3\sqrt{10}}{10}}$.


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

Solution:
\textbf{Solution:} Draw BD $\perp$ AC, \\
since BC $= \sqrt{1^{2}+3^{2}} = \sqrt{10}$, and BD $= \sqrt{1^{2}+1^{2}} = \sqrt{2}$, \\
therefore, $\sin\angle ACB = \frac{\sqrt{2}}{\sqrt{10}} = \boxed{\frac{\sqrt{5}}{5}}$.


Question: As shown in the figure, it is known that $\triangle OAC$ and $\triangle ABC$ are both isosceles right triangles, with $\angle ACO=\angle ABC=90^\circ$. The graph of the inverse proportion function $y=\frac{k}{x}$ in the first quadrant passes through point $B$. Connect $OB$, and the area of $\triangle OAB$ is $4$. What is the value of $k$?

Solution:
\textbf{Solution:} Let $OC=AC=a$\\
Since $\triangle OAC$ is an isosceles right triangle,\\
we have $\sin 45^\circ=\frac{OC}{AO}=\frac{\sqrt{2}}{2}$,\\
thus $OA=\sqrt{2}a$.\\
Since $\triangle ABC$ is also an isosceles right triangle,\\
we have $\sin 45^\circ=\frac{AB}{AC}=\frac{\sqrt{2}}{2}$,\\
thus $AB=BC=\frac{\sqrt{2}}{2}a$.\\
Since both $\triangle OAC$ and $\triangle ABC$ are isosceles right triangles,\\
we have $\angle OAC=\angle BAC=45^\circ$,\\
hence $\angle OAB=90^\circ$,\\
therefore, ${S}_{\triangle OAB}=\frac{1}{2}OA\cdot AB=\frac{1}{2}\cdot \sqrt{2}a \cdot \frac{\sqrt{2}}{2}a=4$.\\
Solving, we find $a=2\sqrt{2}$,\\
thus $OC=2\sqrt{2}$, $BC=2$.\\
Draw $BE\perp x$ axis through point $B$,\\
Since $\angle ACB=45^\circ$,\\
we have $\angle BCE=45^\circ$,\\
hence $\triangle BCE$ is an isosceles right triangle,\\
therefore $\sin 45^\circ=\frac{CE}{CB}=\frac{\sqrt{2}}{2}$,\\
thus $CE=BE=\sqrt{2}$,\\
hence $OE=3\sqrt{2}$,\\
therefore $B(3\sqrt{2},\sqrt{2})$.\\
Since the inverse proportional function $y=\frac{k}{x}$ passes through point $B$ in the first quadrant,\\
we have $k=\boxed{6}$.


Question: As shown in the figure, the graph of the inverse proportion function $y = \frac{k}{x}$ ($k \neq 0$) passes through the vertices $A$ and $B$ of an equilateral triangle $\triangle ABC$, and the origin $O$ is exactly on $AB$. Given that the coordinates of point $C$ are (3, 3), a line passing through point $C$ and parallel to $AB$ intersects the graph of the inverse proportion function $y = \frac{k}{x}$ ($k \neq 0$) at points $M$ and $N$. What is the area of $\triangle MON$?

Solution:
Solution: Given point C $(3,3)$, \\
therefore the equation for line OC is $y=kx$, substituting the coordinates of point C, \\
thus, $3=3k$, \\
therefore, $k=1$, \\
thus, the equation for OC becomes $y=x$, \\
since points A and B lie on the graph of an inverse proportional function, \\
thus, points A and B are symmetric about the origin O, with O being the midpoint of AB, \\
since $\triangle ABC$ is an equilateral triangle, \\
thus, OC $\perp$ AB, \\
thus, $\angle ACO = \frac{1}{2}\angle ACB=\frac{1}{2}\times 60^\circ=30^\circ$, \\
therefore, according to the Pythagorean theorem, OC $=\sqrt{3^2+3^2}=3\sqrt{2}$, \\
thus, AO $=OC\tan 30^\circ=3\sqrt{2}\times \frac{\sqrt{3}}{3}=\sqrt{6}$, \\
since AB $\perp$ OC, with the equation for OC being $y=x$, \\
thus, $\angle COy=45^\circ$, \\
thus, $\angle AOy=45^\circ$, \\
thus, the y-coordinate of point A is $\sqrt{6}\sin 45^\circ=\sqrt{6}\times \frac{\sqrt{2}}{2}=\sqrt{3}$, the x-coordinate of point A is $-\sqrt{6}\cos 45^\circ=-\sqrt{6}\times \frac{\sqrt{2}}{2}=-\sqrt{3}$, \\
thus, point A $(-\sqrt{3}, \sqrt{3})$, \\
since point A lies on the graph of the inverse proportion function $y=\frac{k}{x}$, \\
thus, $k=-\sqrt{3}\times \sqrt{3}=-3$, \\
thus, the inverse proportion function is $y=-\frac{3}{x}$, \\
Let the equation for OA be $y=k_1x$, by substituting the coordinates of point A, \\
$\sqrt{3}=-\sqrt{3}k_1$, \\
solving yields: $k_1=-1$, \\
thus, the equation for OA is $y=-x$, \\
since MN $\parallel$ AB, \\
let the equation for MN be $y=-x+b$, passing through point C, \\
thus, $3=-3+b$, \\
solving gives $b=6$, \\
thus, the equation for MN is $y=-x+6$, \\
points M, N also lie on the graph of the inverse proportion function $y=-\frac{3}{x}$, \\
thus, $\left\{\begin{array}{l}y=-x+6 \\ y=-\frac{3}{x}\end{array}\right.$, \\
eliminating $y$ yields $x^2-6x-3=0$, \\
solving gives $x=3\pm 2\sqrt{3}$, \\
thus, $x_1=3+2\sqrt{3}, x_2=3-2\sqrt{3}$, \\
thus, $y_1=3-2\sqrt{3}, y_2=3+2\sqrt{3}$, \\
Point M $(3-2\sqrt{3}, 3+2\sqrt{3})$, point N $(3+2\sqrt{3}, 3-2\sqrt{3})$, \\
therefore, NM $=\sqrt{((3-2\sqrt{3}-3-2\sqrt{3})^2+((3+2\sqrt{3}-3+2\sqrt{3})^2}=4\sqrt{6}$, \\
thus, $S_{\triangle MON}=\frac{1}{2}OC\cdot MN=\frac{1}{2}\times 3\sqrt{2}\times 4\sqrt{6}=\boxed{12\sqrt{3}}$.


Question: As shown in the figure, $\triangle ABC$ is inscribed in $\odot O$. If $\angle A=60^\circ$ and $\angle C=45^\circ$, what is the length of $AC$?

Solution:
\textbf{Solution:} As shown in the diagram, connect BO and extend it to intersect the circle at point D. Draw AD, BD, CD, and construct DH $\perp$ AC, \\
$\because$ BD is the diameter, \\
$\therefore$ $\angle$BAD=$\angle$BCD=$90^\circ$, \\
$\because$ $\angle$ADB=$\angle$ACB=$45^\circ$, $\angle$BDC=$\angle$BAC=$60^\circ$, \\
$\therefore$ AD=$\frac{\sqrt{2}}{2}$BD=$2\sqrt{2}$, CD=$\frac{1}{2}$BD=2, \\
$\because$ $\angle$DAH=$\angle$CBD=$30^\circ$, $\angle$DCH=$\angle$ABD=$45^\circ$, \\
CH=$\frac{\sqrt{2}}{2}$CD=$\sqrt{2}$, AH=$\frac{\sqrt{3}}{2}$AD=$\sqrt{6}$, \\
$\therefore$ AC=AH+HC=$\sqrt{6}+\sqrt{2}$. \\
Thus, the answer is: $AC=\boxed{\sqrt{6}+\sqrt{2}}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $\angle ABC=90^\circ$, point D is the midpoint of side AC, and BD is connected. $\triangle ABC$ is rotated counterclockwise around point A, such that point B precisely falls on point E on ray BD, and point C falls on point F, and then FD and FC are connected. If $AB=1$ and $BC=2$, what is the value of the tangent of $\angle CFD$?

Solution:
\textbf{Solution:} After triangle $\triangle ABC$ is rotated as shown in the figure, draw $AG\perp BE$ at $G$ through $A$, draw $CN\perp AF$ at $N$ through $C$, draw $FH\perp BE$ through $F$ intersecting the extended line of $BE$ at $H$, draw $DM\perp CF$ at $M$ through $D$,\\
since $\angle ABC=90^\circ$, $D$ is the midpoint of $AC$, $AB=1$, $BC=2$,\\
therefore $DB=DA=DC$, $AC=\sqrt{5}$,\\
thus $\angle DAB=\angle ABD$,\\
By rotation, it follows that $AE=AB=1$, $\angle FAE=\angle BAD$, $EF=BC=2$, $AC=AF=\sqrt{5}$,\\
thus $\angle ABE=\angle AEB=\angle FAE$,\\
therefore $AF\parallel BD$, $\angle GAF=90^\circ$,\\
thus, quadrilateral $AGHF$ is a rectangle,\\
since $\angle AEF=90^\circ$,\\
thus $\angle AEG+\angle FEH=90^\circ=\angle AEG+\angle GAE$,\\
therefore $\angle GAE=\angle FEH$, similarly $\angle AFE=\angle HEF$,\\
thus $\angle GAE=\angle FEH=\angle AFE$,\\
thus $\tan\angle GAE=\tan\angle FEH=\tan\angle AFE$,\\
thus $\frac{EG}{AG}=\frac{FH}{EH}=\frac{AE}{EF}=\frac{1}{2}$,\\
Let $GE=x$, then $FH=AG=2x$, $EH=4x$,\\
thus $GH=AF=AC=5x$, then $5x=\sqrt{5}$, so $x=\frac{\sqrt{5}}{5}$,\\
thus $AG=\frac{2\sqrt{5}}{5}=FH$, $EH=\frac{4\sqrt{5}}{5}$,\\
thus $BG=\sqrt{AB^{2}-AG^{2}}=\frac{\sqrt{5}}{5}$, and $BD=AD=CD=\frac{\sqrt{5}}{2}$,\\
thus $GD=\sqrt{AD^{2}-AG^{2}}=\frac{3\sqrt{5}}{10}$,\\
since $AF\parallel BD$, thus $\angle NAC=\angle ADG$,\\
thus $\cos\angle ADG=\frac{DG}{AD}=\frac{3}{5}=\cos\angle NAC=\frac{AN}{AC}$,\\
thus $AN=\frac{3\sqrt{5}}{5}$, $NF=\frac{2\sqrt{5}}{5}$,\\
thus $NC=\sqrt{AC^{2}-AN^{2}}=\frac{4\sqrt{5}}{5}$, $CF=\sqrt{NF^{2}+CN^{2}}=2$,\\
while $GH=AF=\sqrt{5}$, $DG=\frac{3\sqrt{5}}{10}$,\\
thus $DH=GH-DG=\frac{7\sqrt{5}}{10}$,\\
Connect $FD$,\\
thus $DF^{2}=FH^{2}+DH^{2}=\frac{13}{4}$, $DC^{2}=\frac{5}{4}$,\\
Let $CM=y$, then $FM=2-y$,\\
from $DC^{2}-CM^{2}=DF^{2}-FM^{2}$,\\
solving for $y$, yields $y=\frac{1}{2}$, then $FM=2-\frac{1}{2}=\frac{3}{2}$,\\
thus $DM=\sqrt{DC^{2}-CM^{2}}=1$,\\
thus $\tan\angle CFD=\frac{DM}{FM}=\boxed{\frac{2}{3}}$.


Question: As shown in the figure, in $ABCD$, $AE \perp BD$ at $E$, $\angle EAC = 30^\circ$, and $AE = 3$. What is the length of $AC$?

Solution:
\textbf{Solution:} As shown in the figure, in the right-angled $\triangle AOE$,\\
$\cos\angle EAO = \frac{AE}{OA}$,\\
$\therefore OA = \frac{AE}{\cos\angle EAO} = \frac{3}{\frac{\sqrt{3}}{2}} = 2\sqrt{3}$.\\
Furthermore, $\because$ quadrilateral ABCD is a parallelogram,\\
$\therefore AC = 2OA = 4\sqrt{3}$.\\
Therefore, the length of $AC$ is $\boxed{4\sqrt{3}}$.


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

Solution:
\textbf{Solution:} To find point $B$, move point $C$ down by 1 unit and then left by 3 units. Applying the same procedure to point $A$ gives us point $D$. Connect $AD$, $BD$, $CD$, $AB$ with $AB$ and $CD$ intersecting at point $O$, as shown in the figure: \\
Therefore, quadrilateral $ACBD$ is a parallelogram. \\
Since $AC=BC=\sqrt{3^2+1^2}=\sqrt{10}$ and $AB=\sqrt{4^2+4^2}=4\sqrt{2}$, $CD=\sqrt{2^2+2^2}=2\sqrt{2}$, \\
thus, parallelogram $ACBD$ is a rhombus. \\
Therefore, $AB\perp CD$, $OB=\frac{1}{2}AB=2\sqrt{2}$, $OC=\frac{1}{2}CD=\sqrt{2}$, \\
in $\triangle BOC$, $\tan\angle ABC=\frac{OC}{OB}=\frac{\sqrt{2}}{2\sqrt{2}}=\boxed{\frac{1}{2}}$.


Question: As shown in the figure, in the right triangle $\triangle ABC$ where $\angle ACB=90^\circ$ and $AC=BC=8$. $F$ is the midpoint of $AC$, and $D$ is a moving point on segment $AB$. Connect $CD$ and rotate segment $CD$ $90^\circ$ counterclockwise around point $C$ to obtain segment $CE$. Connect $EF$. During the movement of point $D$, what are the maximum and minimum values of $EF$ respectively?

Solution:
\textbf{Solution:} As shown in the diagram, let G be the midpoint of BC, and connect DG,\\
by rotation, we have DC=EC, $\angle$DCE$=90^\circ$,\\
also, since $\angle$ACB$=90^\circ$, and AC=BC=8, with F being the midpoint of AC,\\
therefore, CG=CF, $\angle$DCG+$\angle$ACD$=\angle$ECF+$\angle$ACD$=90^\circ$,\\
therefore, $\angle$DCG=$\angle$ECF,\\
therefore, $\triangle DCG \cong \triangle ECF$ by SAS,\\
therefore, EF=DG,\\
(1) As shown in Figure 1,\\
when GD$\perp$AB, DG is the shortest, at this time $\triangle BDG$ is an isosceles right triangle,\\
therefore, DG=BG$\times \sin 45^\circ =4\times \frac{\sqrt{2}}{2}=2\sqrt{2}$,\\
thus, the minimum value of EF is $2\sqrt{2}$;\\
(2) When D coincides with B, DG=BG=4;\\
(3) As shown in Figure 2,\\
When D coincides with A, DG=$\sqrt{CG^{2}+AC^{2}}=\sqrt{4^{2}+8^{2}}=4\sqrt{5}>4$,\\
thus, the maximum value of EF is $\boxed{4\sqrt{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $\cos B= \frac{\sqrt{2}}{2}$, $\sin C= \frac{3}{5}$, and $AC= \frac{5}{3}$. What is the area of $\triangle ABC$?

Solution:
\textbf{Solution}: Draw AD$\perp$BC at point D through point A,\\
$\because$ $\sin C=\frac{3}{5}$,\\
$\therefore$ $\frac{AD}{AC}=\frac{3}{5}$,\\
$\therefore$ AD=$\frac{3}{5}AC=\frac{3}{5}\times\frac{5}{3}=1$,\\
$\therefore$ CD=$\sqrt{AC^{2}-AD^{2}}=\sqrt{\left(\frac{5}{3}\right)^{2}-1^{2}}=\frac{4}{3}$,\\
$\because$ $\cos B=\frac{\sqrt{2}}{2}$,\\
$\therefore$ $\angle B=45^\circ$,\\
$\therefore$ BD=AD=1,\\
$\therefore$ BC=BD+CD=1+$\frac{4}{3}=\frac{7}{3}$,\\
$\therefore$ The area of $\triangle ABC$=$\frac{1}{2}\times\frac{7}{3}\times1=\boxed{\frac{7}{6}}$.