Question: As shown in the figure, it is known that in $\triangle ABC$, $\angle BAC = 30^\circ$, $\angle C = 90^\circ$, the coordinate of point A is $(-1,0)$, and the coordinate of point B is $(3,0)$. The vertex of parabola $y_1$ is denoted as Q, and it passes through the three vertices A, B, and C of $\triangle ABC$ (with A to the left of B, and C below the x-axis). Parabola $y_2$ also intersects the x-axis at points A and B, with its vertex denoted as P. Suppose point M is a moving point on parabola $y_1$, located to the right of its axis of symmetry. If $\triangle PQM$ is similar to $\triangle ABC$, determine the coordinates of vertex P of parabola $y_2$.

Solution:
\textbf{Solution:} Since point M is on the right side of the symmetry axis of the parabola $y_{1}$, it follows that point Q is not the right-angle vertex. Let point M be $(a, \frac{\sqrt{3}}{3}a^{2}- \frac{2\sqrt{3}}{3}a- \sqrt{3})$, thus, the distance of point M from the symmetry axis is $a-1$, and $y_{M}-y_{Q} = \frac{\sqrt{3}}{3}a^{2}- \frac{2\sqrt{3}}{3}a- \sqrt{3} -(- \frac{4\sqrt{3}}{3}) = \frac{\sqrt{3}}{3}a^{2}- \frac{2\sqrt{3}}{3}a+ \frac{\sqrt{3}}{3}$. Since triangle $PQM$ is similar to triangle $ABC$ with $\angle ACB = 90^\circ$.

(1) When $\angle PMQ = 90^\circ$,

(i) If $\angle PQM = 30^\circ$ and $\angle QPM = 60^\circ$, then $y_{M}-y_{Q} = \sqrt{3}(a-1)$ and $y_{P}-y_{M} = \frac{\sqrt{3}}{3}(a-1)$. Therefore, $\frac{\sqrt{3}}{3}a^{2}- \frac{2\sqrt{3}}{3}a+ \frac{\sqrt{3}}{3} = \sqrt{3}(a-1)$. Solving, we find $a=1$ (discard) or $a=4$. Therefore, $M(4, \frac{5\sqrt{3}}{3})$, and $y_{P}-y_{M} = \frac{\sqrt{3}}{3}(4-1) = \sqrt{3}$. Hence, the y-coordinate of point P is $\frac{5\sqrt{3}}{3} + \sqrt{3} = \frac{8\sqrt{3}}{3}$. Therefore, $P(1, \frac{8\sqrt{3}}{3})$.

(ii) If $\angle PQM = 60^\circ$ and $\angle QPM = 30^\circ$, then $\sqrt{3}(y_{M}-y_{Q}) = a-1$, and $y_{P}-y_{M} = \sqrt{3}(a-1)$. Therefore, $\sqrt{3}(\frac{\sqrt{3}}{3}a^{2}- \frac{2\sqrt{3}}{3}a+ \frac{\sqrt{3}}{3}) = a-1$. Solving, we find $a=1$ (discard) or $a=2$. Therefore, $M(2, -\sqrt{3})$, and $y_{P}-y_{M} = \sqrt{3} \times (2-1) = \sqrt{3}$. Hence, the y-coordinate of point P is $-\sqrt{3} + \sqrt{3} = 0$. Therefore, $P(1, 0)$.

(2) When $\angle MPQ = 90^\circ$,

(i) If $\angle PQM = 30^\circ$, then $y_{M}-y_{Q} = \sqrt{3}(a-1)$. Therefore, $\frac{\sqrt{3}}{3}a^{2}- \frac{2\sqrt{3}}{3}a- \frac{\sqrt{3}}{3} = \sqrt{3}(a-1)$. Solving, we find $a=1$ (discard) or $a=4$. Therefore, $M(4, \frac{5\sqrt{3}}{3})$. Hence, $P(1, \frac{5\sqrt{3}}{3})$.

(ii) If $\angle PQM = 60^\circ$, then $\sqrt{3}(y_{M}-y_{Q}) = a-1$. Therefore, $\sqrt{3}(\frac{\sqrt{3}}{3}a^{2}- \frac{2\sqrt{3}}{3}a- \frac{\sqrt{3}}{3}) = a-1$. Solving, we find $a=1$ (discard) or $a=?$.

\boxed{P(1, \frac{8\sqrt{3}}{3})}


Question: As shown in the diagram, it is known that $AB$ is the diameter of circle $O$, and chord $CD$ intersects diameter $AB$ at point $E$, with $AC=CD$. Extend line $CO$ to intersect chord $AD$ at point $F$, and connect $AC$, $BC$, $BD$. Given $AB=10$ and $BE=2$, what is the length of $BD$?

Solution:
\textbf{Solution:} Since the diameter AB $= 10$, it follows that OA $=$ OB $=$ OC $= 5$ and $\angle ADB = \angle ACB = 90^\circ$. Since BE $= 2$, it follows that OE $=$ OB - EB $= 3$. Since AC $=$ CD, it implies $\overset{\frown}{AC} = \overset{\frown}{CD}$, thus CF $\perp$ AD, leading to AF $=$ DF. Since $\angle AFC = \angle ADB = 90^\circ$, it follows that CF $\parallel$ BD,
\[
\therefore \frac{OC}{BD}=\frac{OE}{EB},
\]
\[
\therefore \frac{5}{BD}=\frac{3}{2}, \therefore BD=\boxed{\frac{10}{3}}.


Question: As shown in the figure, it is known that $AB$ is the diameter of circle $O$, and chord $CD$ intersects with diameter $AB$ at point $E$, with $AC=CD$. Extend line $CO$ to intersect chord $AD$ at point $F$. Connect $AC$, $BC$, $BD$, with $AB=10$ and $BE=2$. What is the length of $AF$?

Solution:
\textbf{Solution:} Given that $AO=OB$ and $AF=FD$, \\
hence $OF=\frac{1}{2}BD=\frac{5}{3}$, \\
in $\triangle AOF$, $AF=\sqrt{OA^{2}-OF^{2}}=\sqrt{5^{2}-\left(\frac{5}{3}\right)^{2}}=\boxed{\frac{10\sqrt{2}}{3}}$.


Question: As shown in the figure, it is known that $AB$ is the diameter of circle $O$, and chord $CD$ intersects diameter $AB$ at point $E$. Given $AC=CD$, connect $CO$ and extend it to intersect chord $AD$ at point $F$. Connect $AC$, $BC$, $BD$, and given $AB=10$, $BE=2$. What is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} In $\triangle \mathrm{ACF}$, $\angle \mathrm{AFC} = 90^\circ$, $\mathrm{AF} = \frac{10\sqrt{2}}{3}$, $\mathrm{CF} = \mathrm{OC} + \mathrm{OF} = 5 + \frac{5}{3} = \frac{20}{3}$, \\
$\therefore \mathrm{AC} = \sqrt{\mathrm{AF}^2 + \mathrm{CF}^2} = \sqrt{\left(\frac{10\sqrt{2}}{3}\right)^2 + \left(\frac{20}{3}\right)^2} = \frac{10\sqrt{6}}{3}$, \\
In $\triangle \mathrm{ABC}$, $\mathrm{BC} = \sqrt{\mathrm{AB}^2 - \mathrm{AC}^2} = \sqrt{10^2 - \left(\frac{10\sqrt{6}}{3}\right)^2} = \frac{10\sqrt{3}}{3}$, \\
$\therefore S_{\triangle \mathrm{ABC}} = \frac{1}{2} \cdot \mathrm{AC} \cdot \mathrm{BC} = \frac{1}{2} \times \frac{10\sqrt{6}}{3} \times \frac{10\sqrt{3}}{3} = \boxed{\frac{50\sqrt{2}}{3}}.$


Question: As shown in the figure, in $\mathrm{Rt}\triangle ABC$ with $\angle A=90^\circ$, $O$ is a point on side $BC$. A semicircle centered at $O$ is tangent to side $AB$ at point $D$ and intersects sides $AC$ and $BC$ at points $E$, $F$, and $G$, respectively. Line $OD$ is drawn, given that $BD = 2$, $AE = 3$, and $\tan\angle BOD= \frac{2}{3}$. What is the radius of the circle centered at $O$, $OD$?

Solution:
\textbf{Solution:} Since AB is tangent to circle O,\\
it follows that OD$\perp$AB. In $\triangle BDO$, BD=2, $\tan\angle BOD= \frac{BD}{OD} = \frac{2}{OD}$,\\
hence, OD=\boxed{3}.


Question: As shown in the figure, in $\triangle ABC$, $\angle A = 90^\circ$, O is a point on side BC. A semicircle centered at O is tangent to side AB at point D and intersects sides AC and BC at points E, F, and G, respectively. Segment OD is drawn. It is known that BD = 2, AE = 3, and $\tan \angle BOD = \frac{2}{3}$. AE is a tangent to circle O; what is the sum of the areas of the two shaded regions?

Solution:
\textbf{Solution:} Since $OD \parallel AC$, it follows that $\frac{BD}{AB} = \frac{OD}{AC}$, that is, $\frac{2}{2+3} = \frac{3}{AC}$. Thus, $AC = 7.5$, and therefore $EC = AC - AE = 7.5 - 3 = 4.5$. Hence, the area of the shaded region $S_{\text{shaded}} = S_{\triangle BDO} + S_{\triangle OEC} - S_{\text{sector} FOD} - S_{\text{sector} EOG}$ \\
$= \frac{1}{2} \times 2 \times 3 + \frac{1}{2} \times 3 \times 4.5 - \frac{90^\circ \pi \times {3}^{2}}{360}$ \\
$= 3 + \frac{27}{4} - \frac{9\pi }{4}$ \\
$= \boxed{\frac{39-9\pi }{4}}.$


Question: As shown in the figure, $\triangle ABC$ is inscribed in $\odot O$, with $AB$ being the diameter of $\odot O$, $AB=10$, $AC=6$. Connect $OC$, and the chord $AD$ intersects $OC$, $BC$ at points $E$, $F$, where point $E$ is the midpoint of $AD$. $\angle CAD=\angle CBA$. What is the length of $OE$?

Solution:
\textbf{Solution:} Since $AB$ is a diameter, \\
$\therefore \angle ACB=90^\circ$, \\
Because $AE=DE$, \\
$\therefore OC\perp AD$, \\
$\therefore \angle AEC=90^\circ$, \\
$\therefore \angle AEC=\angle ACB$, \\
$\therefore \triangle AEC\sim \triangle BCA$, \\
$\therefore \frac{CE}{AC} = \frac{AC}{AB}$, \\
$\therefore \frac{CE}{6} = \frac{6}{10}$, \\
$\therefore CE=3.6$, \\
Since $OC= \frac{1}{2} AB=5$, \\
$\therefore OE=OC-EC=5-3.6=\boxed{1.4}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C = 90^\circ$, points D and E are located on AC and AB respectively, BD bisects $\angle ABC$, $DE \perp AB$ at point $E$, $AE = 6$, and $\cos A = \frac{3}{5}$. What is the length of CD?

Solution:
\textbf{Solution:} In right $\triangle ADE$, where $\angle AED=90^\circ$ and $AE=6$, with $\cos A=\frac{3}{5}$, we find that\\
$\therefore AD=\frac{AE}{\cos A}=10$\\
$\therefore DE=\sqrt{AD^2-AE^2}=\sqrt{10^2-6^2}=8$\\
Since $BD$ bisects $\angle ABC$, and $DE\perp AB$, $DC\perp BC$,\\
$\therefore CD=DE=\boxed{8}$


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, points D and E are on AC and AB respectively, BD bisects $\angle ABC$, $DE\perp AB$ at point $E$, $AE=6$, $\cos A=\frac{3}{5}$. What is the value of $\tan \angle DBC$?

Solution:
\textbf{Solution:} Given $AD=10$, $DC=8$,

$\therefore AC=AD+DC=18$,

In triangles $\triangle ADE$ and $\triangle ABC$,

$\because \angle A=\angle A$, $\angle AED=\angle ACB$,

$\therefore \triangle ADE \sim \triangle ABC$,

$\therefore \frac{DE}{BC}=\frac{AE}{AC}$, which means $\frac{8}{BC}=\frac{6}{18}$, thus $BC=24$,

$\therefore \tan\angle DBC=\frac{CD}{BC}=\frac{8}{24}=\boxed{\frac{1}{3}}.$


Question: As shown in Figure 1, AC is the diameter of $\odot O$, PA is the tangent to $\odot O$, with A being the point of tangency. Point B lies on $\odot O$, and PA = PB. Chord AB intersects PC at point M. PB is the tangent to $\odot O$, and segment BC is connected. If $\angle APB = 4\angle BPC$ and AP = 6, what is the length of BC?

Solution:
\textbf{Solution:} Connect PO.\\
Since PA and PB are tangents to the circle $O$,\\
it follows that PA=PB=6, and $\angle APO=\angle BPO$, that is, $\angle 1=\angle 2+\angle 3$,\\
Since $\angle APB=4\angle BPC$,\\
it follows that $\angle APM=3\angle BPM$,\\
therefore $\angle 1+\angle 2=3\angle 3$,\\
thus $2\angle 2+\angle 3=3\angle 3$, which means $\angle 2=\angle 3=\frac{1}{2}\angle 1$.\\
Let $\angle 2=\angle 3=\alpha$, then $\angle 1=2\alpha$, $\angle APB=4\alpha$,\\
thus $\angle PBA=\frac{1}{2}(180^\circ-4\alpha)=90^\circ-2\alpha$,\\
Considering that AC is a diameter,\\
thus $\angle ABC=90^\circ$,\\
therefore $\angle PBC=180^\circ-2\alpha$,\\
thus $\angle 4=180^\circ-\angle 3-\angle PBC=180^\circ-\alpha-(180^\circ-2\alpha)=\alpha$,\\
therefore $\angle 3=\angle 4$,\\
thus BC=PB=\boxed{6}.


Question: As shown in Figure 1, AC is the diameter of $\odot O$, PA is the tangent to $\odot O$ with A being the point of tangency, point B is on $\odot O$, PA = PB, and the chord AB intersects PC at point M. As shown in Figure 2, if AB = 4BM, find the value of $ \frac{MC}{MB}$.

Solution:
\textbf{Solution:} Connect PO and intersect AB at point Q, connect BO and BC.\\
$\because$ PA and PB are the tangents to $\odot$O,\\
$\therefore$ PA＝PB, $\angle$APO＝$\angle$BPO,\\
$\therefore$ AQ＝BQ＝ $ \frac{1}{2}$ AB, $\angle$PQM＝$90^\circ$,\\
$\because$ AB＝4BM,\\
$\therefore$2BQ＝4BM, that is BQ＝2BM,\\
$\therefore$ M is the midpoint of BQ.\\
Let BM＝QM＝x, then AQ＝BQ＝2x, AM＝3x,\\
$\because$ $\angle$PQM＝$\angle$CBM＝$90^\circ$, BM＝QM, $\angle$PMQ＝CMB,\\
$\therefore$ $\triangle$PQM$\cong$$\triangle$CBM (ASA),\\
$\therefore$ PM＝CM,\\
$\because$ $\angle$PAC＝$90^\circ$,\\
$\therefore$AM＝CM＝3x,\\
$\therefore$ $ \frac{MC}{MB}=\frac{3x}{x}=\boxed{3}$.


Question: As shown in the figure, $AC$ is the diameter of $\odot O$, $PA$ is the tangent of $\odot O$ with $A$ being the point of tangency, point $B$ is on $\odot O$, $PA=PB$, chord $AB$ intersects $PC$ at point $M$, if $AP=AC$, $PO$ intersects $AB$ at point $D$, $PC$ intersects $\odot O$ at point $N$, draw $DN$, then what is the value of $\frac{DP}{DN}$?

Solution: \textbf{Solution:} We have $\frac{DP}{DN}=\boxed{2\sqrt{2}}$.


Question: As shown in the figure, in $\triangle ABC$, points D and E are on sides AB and AC, respectively, with $\angle AED=\angle B$. AG bisects $\angle BAC$, and the line segment AG intersects line segments DE and BC at points F and G, respectively. If $\frac{AD}{DB}=\frac{1}{2}$, find the value of $\frac{EF}{FG}$.

Solution:
\textbf{Solution:} Since $AG$ bisects $\angle BAC$,\\
we have $\angle BAG = \angle EAF$,\\
and since $\angle AEF = \angle B$,\\
it follows that $\triangle BAG \sim \triangle EAF$,\\
thus $\frac{EF}{BG} = \frac{AF}{AG}$,\\
Since $\triangle ADF \sim \triangle ACG$,\\
we have $\frac{AD}{AC} = \frac{AF}{AG} = \frac{1}{2}$,\\
therefore, $\frac{EF}{BG} = \frac{AD}{AC} = \boxed{\frac{1}{2}}$.


Question: As shown in the figure, a rectangular paper ABCD (AD > DC) is folded along a straight line passing through point D, so that point A falls on side BC at point E, and the crease intersects side AB at point F. If BE = 1, EC = 2, then what is the value of $\sin\angle EDC$?

Solution: \textbf{Solution}: We have $\sin\angle EDC = \boxed{\frac{2}{3}}$.


Question: As shown in the figure, a rectangular piece of paper ABCD (with AD > DC) is folded along a straight line passing through point D, so that point A lands on side BC, with the landing point being E, and the crease intersects side AB at point F. Given that \(BE:EC = 1:4\) and \(CD = 9\), find the length of \(BF\).

Solution: \textbf{Solution:} Let $\because BE:EC=1:4$, assume $BE=x$, $EC=4x$, by the properties of the rectangle, we have: $AD=BC=BE+EC=5x$. From the property of folding, we get: $DE=AD=5x$. In $\triangle DEC$, we have $CD^2+CE^2=DE^2$, that is $9^2+(4x)^2=(5x)^2$. Solving this gives $x=3$ (neglect the negative value), hence $BE=3$. Let $BF=a$, then $EF=AF=AB-BF=9-a$. By the Pythagorean theorem, we have $BE^2+BF^2=EF^2$, that is $3^2+a^2=(9-a)^2$. Solving this yields $a=\boxed{4}$, so $BF=4$.


Question: As shown in the figure, fold the rectangular paper ABCD (AD > DC) at point A along the line passing through point D, making point A land on the side BC, with the landing point being E, and the fold line intersecting side AB at point F. If $BE: EC = m: n$, find $AF: FB$ (expressed as an algebraic expression involving $m$ and $n$).

Solution:
\textbf{Solution:} From the given conditions, we can derive: $\angle EFD=\angle A=\angle B=\angle C=90^\circ$\\
$\therefore$ $\angle BEF+\angle DEC=90^\circ$, $\angle DEC+\angle EDC=90^\circ$\\
$\therefore$ $\angle BEF=\angle EDC$\\
$\therefore$ $\triangle BEF\sim \triangle CDE$\\
$\therefore$ $\frac{EF}{DE}=\frac{BF}{CE}$, that is $\frac{EF}{BF}=\frac{DE}{CE}$\\
Since $BE\colon EC=m\colon n$\\
Let $BE=mk$, $EC=nk$\\
Then $DE=AD=BC=BE+CE=mk+nk$\\
$\therefore$ $\frac{EF}{BF}=\frac{DE}{CE}=\frac{mk+nk}{nk}=\frac{m+n}{n}$\\
From the property of folding, we obtain: $AF=EF$\\
$\therefore$ $AF\colon FB=\boxed{\frac{m+n}{n}}$


Question: As shown in Figure 1, the quadrilateral ABCD and the quadrilateral CEFG are both rectangles, with points E and G located on the sides CD and CB respectively, and point $F$ on AC, with $AB=3$, $BC=4$. Calculate the value of $\frac{AF}{BG}$.

Solution:
\textbf{Solution:} Since quadrilateral ABCD and quadrilateral CEFG are both rectangles,\\
it follows that $\angle ABC=\angle FGC=90^\circ$\\
Thus, $AB\parallel FG, \ AC=\sqrt{AB^{2}+BC^{2}}=5$\\
Therefore, $\frac{AF}{BG}=\frac{CF}{CG}=\frac{AC}{BC}=\boxed{\frac{5}{4}}$


Question: As shown in the figure, the parabola $y=ax^2+bx+6$ intersects the x-axis at points $A(2,0)$ and $B(8,0)$, and intersects the y-axis at point $C$. Find the analytical expression of the parabola.

Solution:
\textbf{Solution:} By substituting points A(2,0) and B(8,0) according to the problem, we have:
\[
\begin{cases}
0=4a+2b+6\\
0=64a+8b+6
\end{cases},
\]
from which we derive:
\[
\begin{cases}
a=\frac{3}{8}\\
b=-\frac{15}{4}
\end{cases},
\]
Therefore, the equation of the parabola is: $\boxed{y=\frac{3}{8}x^2-\frac{15}{4}x+6}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+6$ intersects the x-axis at points $A(2,0)$ and $B(8,0)$, and intersects the y-axis at point $C$. Point $P$ is a moving point on the parabola, when $\angle PCB =\frac{1}{2}\angle BCO$, find the x-coordinate of point $P$.

Solution:
\textbf{Solution:} When point $P$ is on the parabola below $BC$, then $\angle PCB =\frac{1}{2}\angle BCO$ i.e., $CP$ bisects $\angle BCO$. As shown in the figure, draw $CP$ to bisect $\angle BCO$, intersecting the x-axis at point $D$, and draw $DE\perp BC$, with $E$ being the foot of the perpendicular,\\
$\because CP$ bisects $\angle BCO$, $DE\perp BC$,\\
$\therefore OD=DE$, $\angle DCO=\angle DCE$,\\
$\because OD=DE$, $\angle DCO=\angle DCE$, $\angle COD=\angle CED=90^\circ$,\\
$\therefore \triangle DOC\cong \triangle DEC, CO=CE=6$,\\
$\therefore BC=\sqrt{CO^2+BO^2}=\sqrt{6^2+8^2}=10, BE=BC−CE=4$,\\
Let $OD=DE=m$, $BD=8−m$,\\
By the Pythagorean theorem: $BE^2+DE^2=BD^2$, i.e., $4^2+m^2=(8−m)^2$,\\
Solving this yields: $m=3$, i.e., $OD=DE=3$, the coordinates of $D$ are $(3,0)$,\\
Let the equation of $CD$ be: $y=kx+b(k\ne 0)$, substituting points $C, D$ we get:\\
$\begin{cases} 6=b \\ 0=3k+b \end{cases}$, thus solving we get: $\begin{cases} k=-2 \\ b=6 \end{cases}$, so, the equation of $CD$ is: $y=-2x+6$,\\
$\because P$ is the intersection point of line $CD$ and the parabola,\\
$\therefore$ we solve together to get: $-2x+6=\frac{3}{8}x^2-\frac{15}{4}x+6$,\\
Solving for $x=0$ (discard) or $x=\boxed{\frac{14}{3}}$, hence the x-coordinate of $P$ is $x=\frac{14}{3}$,\\
When $P$ is on the parabola above $BC$, then $\angle PCB =\frac{1}{2}\angle BCO$. As shown in the figure, draw $\angle PCB =\frac{1}{2}\angle BCO$ intersecting the parabola at point $P$, extend $DE$ to intersect $CP$ at point $F$, draw $EH\perp$x-axis from $E$ intersecting at point $H$,\\
$\because \angle PCB =\frac{1}{2}\angle BCO$, $\angle DCB=\angle DCO$,\\
$\therefore \angle PCB=\angle DCB$,\\
$\because \angle PCB=\angle DCB, CE=CE, \angle DEC=\angle FEC$,\\
$\therefore \triangle DEC\cong \triangle FEC, DE=DF$,\\
$\because \angle CBO=\angle EBH, \angle COB=\angle EHB=90^\circ$,\\
$\therefore \triangle EHB\sim \triangle COB$,\\
$\therefore \frac{BE}{BC}=\frac{EH}{CO}=\frac{BH}{BO}, \frac{4}{10}=\frac{EH}{6}=\frac{BH}{8}$,\\
Hence $EH=\frac{12}{5}, BH=\frac{16}{5}, OH=BO−BH=\frac{24}{5}$,\\
$\therefore E(\frac{24}{5}, \frac{12}{5})$,\\
Let $F$ be $(m, n)$, by $DE=DF$ we get $\frac{m+3}{2}=\frac{24}{5}, \frac{n+0}{2}=\frac{12}{5}$, solving for: $m=\frac{33}{5}, n=\frac{24}{5}$,\\
That is, $F$ is $(\frac{33}{5}, \frac{24}{5})$,\\
Let the equation of $CF$ be: $y=kx+b(k\ne 0)$, substituting $C, F$ we find:\\
$\begin{cases} 6=b \\ \frac{24}{5}=\frac{33}{5}k+b \end{cases}$, solving, we get: $\begin{cases} k=-2 \\ b=6 \end{cases}$, therefore, the equation of $CF$ is: $y=-2x+6$,\\


Question: As shown in the figure, $AB$ is the diameter of a semicircle, and $BC$ is a chord of the semicircle. The arc $\overset{\frown}{BC}$ is folded along the chord $BC$ to intersect the diameter $AB$ at point D. When $AD=BD=5$, what is the length of $BC$?

Solution: \textbf{Solution:} We have $BC = \boxed{5\sqrt{3}}$.


Question: As shown in the figure, $AB$ is the diameter of a semicircle, and $BC$ is a chord of the semicircle. The arc $\overset{\frown}{BC}$ is folded along the chord $BC$ to intersect the diameter $AB$ at point D. When $AD=4$ and $BD=6$, what is the length of $BC$?

Solution: \textbf{Solution:} We have $BC=\boxed{4\sqrt{5}}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed within the semicircle $O$, with $AB$ being the diameter and $C$ is the midpoint of arc $\overset{\frown}{BD}$. Extending $AD$ and $BC$ to meet at point $E$, we have $CE = CD$. Given $AB = 5$ and $BC = \sqrt{5}$, find the length of $AD$.

Solution:
\textbf{Solution:} From (1) we have: $AE=AB=5$, $CD=CE=BC=\sqrt{5}$, $\triangle CED \sim \triangle AEB$,\\
$\therefore \frac{CE}{AE}=\frac{DE}{BE}$, $\therefore DE=\frac{\sqrt{5}\times 2\sqrt{5}}{5}=2.$\\
$\therefore AD=AE-DE=\boxed{3}.$


Question: As shown in the figure, point $P$ is on the positive $y$-axis, $\odot P$ intersects the $x$-axis at points $B$ and $C$. With $AC$ as one of its right-angled sides, an isosceles $\triangle ACD$ is constructed. $BD$ intersects the $y$-axis and $\odot P$ at points $E$ and $F$, respectively. $AC$ and $FC$ are connected, and $AC$ intersects $BD$ at point $G$. $\angle ACF=\angle ADB$; $CF=DF$; what is the degree measure of $\angle DBC$?

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


Question: As shown in the figure, point $P$ is located on the positive half of the y-axis, and $\odot P$ intersects the x-axis at points $B$ and $C$. An isosceles triangle $\triangle ACD$ is formed with $AC$ as the right angle side. $BD$ intersects the y-axis and $\odot P$ at points $E$ and $F$ respectively. Connect $AC$ and $FC$, with $AC$ and $BD$ intersecting at point $G$. If $OB=3$ and $OA=6$, then what is the area of $\triangle GDC$?

Solution: \textbf{Solution:} Hence, the area of $\triangle GDC$ is \boxed{15}.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AE$ bisects $\angle BAC$ and intersects $BC$ at point $E$, and point $D$ is on $AB$, with $DE\perp AE$. The circumcircle $\odot O$ of $\triangle ADE$ intersects $AC$ at point $F$. The line $BC$ is a tangent to the circle $\odot O$; if the radius of $\odot O$ is $10$ and $AC=16$, find the area of $\triangle ADE$.

Solution:
\textbf{Solution:} Given that $AD$ is the diameter of $\odot O$,\\
$\therefore \angle AED=90^\circ$,\\
$\therefore \angle C=\angle AED=90^\circ$,\\
Given that $\angle 1=\angle 2$,\\
$\therefore \triangle ACE\sim \triangle AED$,\\
$\therefore \frac{AC}{AE}=\frac{AE}{AD}$, that is $\frac{16}{AE}=\frac{AE}{20}$,\\
$\therefore AE=8\sqrt{5}$ (negative values are discarded),\\
$\therefore DE=\sqrt{AD^{2}-AE^{2}}=\sqrt{20^{2}-\left(8\sqrt{5}\right)^{2}}=4\sqrt{5}$,\\
$\therefore S_{\triangle ADE}=\frac{1}{2}AE\cdot DE=\frac{1}{2}\times 8\sqrt{5}\times 4\sqrt{5}=\boxed{80}$.


Question: As shown in the trapezoid $ABCD$, $\angle ABC = 90^\circ$, $AD \parallel BC$, $BC = 2AD$, and the diagonals $AC$ and $BD$ intersect at point $E$. Point $F$ is a point on the line segment $EC$, and $\angle BDF = \angle BAC$. If $BC = 6$ and $\sin\angle BAC = \frac{2}{3}$, what is the length of $FC$?

Solution:
\textbf{Solution:}

Since $\angle ABC=90^\circ$, $BC=6$, and $\sin\angle BAC=\frac{2}{3}$,

it follows that $\frac{BC}{AC}=\frac{2}{3}$, hence $AC=9$,

thus $AB=\sqrt{AC^2-BC^2}=3\sqrt{5}$,

since $BC=2AD$,

therefore $AD=3$,

given that $AD\parallel BC$,

it follows that $\angle BAD=90^\circ$,

hence $BD=\sqrt{AB^2+AD^2}=3\sqrt{6}$,

because $\triangle EAD\sim \triangle ECB$,

it follows that $\frac{EA}{EC}=\frac{ED}{EB}=\frac{AD}{BC}=\frac{3}{6}=\frac{1}{2}$,

thus $\frac{EA}{AC}=\frac{1}{3}$, $\frac{ED}{BD}=\frac{1}{3}$,

therefore $EA=\frac{1}{3}AC=3$, $ED=\frac{1}{3}BD=\sqrt{6}$,

hence $EC=6$, $BE=2\sqrt{6}$,

since $EB^2=EF\cdot EC$,

it follows that $(2\sqrt{6})^2=6EF$,

therefore $EF=4$,

therefore $FC=EC-EF=6-4=\boxed{2}$.


Question: As shown in the figure, in the Cartesian coordinate system \(xOy\), the parabola \(y=x^2+bx+c\) intersects with the x-axis at points A \((-1,0)\) and B \((3,0)\), and intersects with the y-axis at point C, with the vertex being point D. Find the expression of the parabola and the coordinates of point C?

Solution:
\textbf{Solution:} Substitute A ($-1$, $0$) and B ($3$, $0$) into $y=x^{2}+bx+c$, we have
\[
\left\{
\begin{array}{l}
1-b+c=0, \\
9+3b+c=0.
\end{array}
\right.
\]
Solving we get:
\[
\left\{
\begin{array}{l}
b=-2, \\
c=-3.
\end{array}
\right.
\]
Therefore, $y=x^{2}-2x-3$.

When $x=0$, $y=-3$. $\therefore$ The coordinates of point C are $\boxed{(0, -3)}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the parabola $y=x^2+bx+c$ intersects the x-axis at point A ($-1$, $0$) and point B ($3$, $0$), and intersects the y-axis at point C. The vertex is point D. Connecting BC and BD, what is the value of the tangent of $\angle CBD$?

Solution:
\textbf{Solution:} Let's connect CD, and draw DE $\perp$ y-axis at point E,\\
$\because y={x}^{2}-2x-3={(x-1)}^{2}-4$,\\
$\therefore$ the coordinates of point D are (1, -4).\\
$\because$ B(3, 0), C(0, -3), D(1, -4), E(0, -4),\\
$\therefore$ OB=OC=3, CE=DE=1,\\
$\therefore$ BC=$3\sqrt{2}$, DC=$\sqrt{2}$, BD=$2\sqrt{5}$.\\
$\therefore BC^{2}+DC^{2}=18+2=20=DB^{2}$.\\
$\therefore\angle BCD=90^\circ$.\\
$\therefore\tan\angle CBD=\frac{DC}{BC}=\frac{\sqrt{2}}{3\sqrt{2}}=\boxed{\frac{1}{3}}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the parabola $y=x^{2}+bx+c$ intersects the x-axis at points A ($-1$, $0$) and B ($3$, $0$), and intersects the y-axis at point C, with the vertex being point D. If point P is a point on the x-axis, when $\triangle BDP$ is similar to $\triangle ABC$, determine the coordinates of point P.

Solution:
\textbf{Solution:} We have $\because \tan\angle ACO=\frac{AO}{OC}=\frac{1}{3}$,
$\therefore \angle ACO=\angle CBD$.
$\because OC=OB$,
$\therefore \angle OCB=\angle OBC=45^\circ$.
$\therefore \angle ACO+\angle OCB=\angle CBD+\angle OBC$.
That is: $\angle ACB=\angle DBO$.
$\therefore$, when $\triangle BDP$ is similar to $\triangle ABC$, point P is to the left of point B.
(i) When $\frac{AC}{CB}=\frac{DB}{BP}$,
$\therefore \frac{\sqrt{10}}{3\sqrt{2}}=\frac{2\sqrt{5}}{BP}$.
$\therefore BP=6$.
$\therefore P(-3,0)$.
(ii) When $\frac{AC}{CB}=\frac{BP}{DB}$,
$\therefore \frac{\sqrt{10}}{3\sqrt{2}}=\frac{BP}{2\sqrt{5}}$.
$\therefore BP=\frac{10}{3}$.
$\therefore P\left(-\frac{1}{3},0\right)$.
In conclusion, the coordinates of point P are $\boxed{(-3,0)}$ or $\boxed{\left(-\frac{1}{3},0\right)}$.


Question: As shown in the diagram, $\triangle ABC$ is inscribed in circle $\odot O$, with $D$ being a point on the extension of diameter $AB$ of $\odot O$, and $\angle DCB = \angle OAC$. A line parallel to $BC$ passing through the center $O$ of the circle intersects the extension of $DC$ at point $E$. $CD$ is a tangent to circle $\odot O$; if $CD = 4$ and $CE = 6$, find the radius of circle $\odot O$ and the value of $\tan \angle OCB$.

Solution:
\textbf{Solution:} Since OE $\parallel$ BC, \\
$\therefore \frac{BD}{OB}=\frac{CD}{CE}$, \\
Given that CD = 4, CE = 6, \\
$\therefore \frac{BD}{OB}=\frac{4}{6}=\frac{2}{3}$, \\
Let BD = 2x, then OB = OC = 3x, OD = OB + BD = 5x, \\
Since OC $\perp$ DC, \\
$\therefore \triangle OCD$ is a right-angled triangle, \\
In $\triangle OCD$, $OC^{2}+CD^{2}=OD^{2}$, \\
$\therefore (3x)^{2}+4^{2}=(5x)^{2}$, \\
Solving this, we find x = 1, \\
$\therefore OC=3x=3$, i.e., the radius of $\odot O$ is $\boxed{3}$, \\
Since BC $\parallel$ OE, \\
$\therefore \angle OCB=\angle EOC$, \\
In $\triangle OCE$, $\tan \angle EOC= \frac{EC}{OC}=\frac{6}{3}=2$, \\
$\therefore \tan \angle OCB=\tan \angle EOC=\boxed{2}$.


Question: As shown in the diagram, in the parallelogram $ABCD$, point $E$ is on the side $AD$, and $CE$ and $BD$ intersect at point $F$, with $BF = 3DF$. What is the value of $AE\colon ED$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\
it follows that AD$\parallel$BC and AD=BC. \\
Therefore, $\frac{BC}{ED}=\frac{BF}{DF}$. \\
Since BF=3DF, \\
it follows that $\frac{BF}{DF}=3$. \\
Therefore, $\frac{BC}{ED}=3$. \\
Therefore, $\frac{AD}{ED}=3$, which means $AE\colon ED=\boxed{2}$.


Question: As shown in the figure, in parallelogram $ABCD$, diagonal $AC$ is perpendicular to side $CD$, $\frac{AB}{AC}=\frac{3}{4}$, the perimeter of quadrilateral $ABCD$ is $16$, point $E$ is on the extension of $AD$, point $F$ is on ray $AB$, $\angle CED=\angle CDF$. If point $F$ coincides with point $B$, find the cotangent of $\angle AFD$?

Solution:
\textbf{Solution:} Given the problem, \\
since $\frac{AB}{AC}=\frac{3}{4}$, \\
let AB=3k, AC=4k, and the intersection of AC and BD be O, \\
since quadrilateral ABCD is a parallelogram, \\
thus OA=OC=$\frac{1}{2}AC=2k$, and CD$\parallel$AB, \\
since AC$\perp$CD, \\
thus AC$\perp$AB, \\
thus BC=$\sqrt{AC^2+AB^2}=5k$, \\
since the perimeter of quadrilateral ABCD is 16, \\
thus $5k+5k+3k+3k=16$, \\
solving gives $k=1$, \\
thus AB=3, AC=4, BC=5, OA=2, \\
thus $\cot \angle AFD = \frac{AB}{AO} = \boxed{\frac{3}{2}}$;


Question: In parallelogram $ABCD$, diagonal $AC$ is perpendicular to side $CD$, $\frac{AB}{AC}=\frac{3}{4}$, and the perimeter of quadrilateral $ABCD$ is $16$. Point $E$ is on the extension of line $AD$, and point $F$ is on the ray $AB$, $\angle CED=\angle CDF$. As shown in Figure 2, point $F$ is a point on side $AB$. Let $AE=x$, $BF=y$, find the functional relationship between $y$ and $x$?

Solution:
\textbf{Solution:} From the given information, \\
$\because CD \parallel AB$, \\
$\therefore \angle EDC = \angle FAD$, $\angle CDF = \angle AFD$, \\
$\because \angle CED = \angle CDF$, \\
$\therefore \angle CED = \angle AFD$, \\
$\therefore \triangle CDE \sim \triangle DAF$, \\
$\therefore \frac{DE}{AF} = \frac{DC}{AD}$, \\
$\because AE = x$, $BF = y$, $AD = BC = 5$, \\
$\therefore DE = x - 5$, $DC = AB = 3$, $AF = 3 - y$, \\
$\therefore \frac{x - 5}{3 - y} = \frac{3}{5}$, \\
$\therefore y = -\frac{5}{3}x + \frac{34}{3}$, \\
The domain is: $5 < x \le \frac{34}{5}$. \\
$\therefore \boxed{y = -\frac{5}{3}x + \frac{34}{3}}$.


Question: In parallelogram $ABCD$, the diagonal $AC$ is perpendicular to side $CD$, and $\frac{AB}{AC} = \frac{3}{4}$. The perimeter of quadrilateral $ABCD$ is $16$. Point $E$ is on the extension of side $AD$, and point $F$ is on the ray $AB$, with $\angle CED = \angle CDF$. If $BF:FA = 1:2$, what is the area of $\triangle CDE$?

Solution:
\textbf{Solution:} Given point $F$ on the ray $AB$, we can deduce: $\triangle CDE\sim \triangle DAF$, \\
$\therefore \frac{S_{\triangle CDE}}{S_{\triangle DAF}}=\left(\frac{DC}{AD}\right)^{2}$, \\
(1) When point $F$ is on side $AB$, \\
$\because BF:FA=1:2$, $AB=3$, $\therefore AF=2$, \\
According to the problem, we have $S_{\triangle DAF}=\frac{1}{2}AF\cdot AC$, \\
$\because AC=4$, \\
$\therefore S_{\triangle DAF}=\frac{1}{2}AF\cdot AC=4$, \\
$\therefore \frac{S_{\triangle CDE}}{4}=\left(\frac{3}{5}\right)^{2}$, \\
$\therefore S_{\triangle CDE}=\boxed{\frac{36}{25}}$, \\
(2) When point $F$ is on the extension of $AB$, \\
$\because BF:FA=1:2$, $AB=3$, \\
$\therefore AF=6$, \\
According to the problem, we have $S_{\triangle DAF}=\frac{1}{2}AF\cdot AC$, \\
$\therefore S_{\triangle DAF}=\frac{1}{2}AF\cdot AC=12$, \\
$\therefore \frac{S_{\triangle CDE}}{12}=\left(\frac{3}{5}\right)^{2}$, \\
$\therefore S_{\triangle CDE}=\boxed{\frac{108}{25}}$, \\
To sum up, the area of $\triangle CDE$ is either \boxed{\frac{36}{25}} or \boxed{\frac{108}{25}}.


Question: As shown in the figure, it is known that in $\triangle ABC$, $CD \perp AB$ with the foot of the perpendicular at point $D$, $AD=2$, $BD=6$, $\tan\angle B=\frac{2}{3}$. Point $E$ is the midpoint of side $BC$. Find the length of side $AC$.

Solution:
\textbf{Solution:} Since $CD \perp AB$ \\
Therefore, both $\triangle ACD$ and $\triangle BCD$ are right-angled triangles, \\
Since $\tan B = \frac{2}{3}$ \\
Therefore $\frac{CD}{BD} = \frac{2}{3}$ \\
Since $BD = 6,$ \\
Therefore $CD = \frac{2}{3}BD = \frac{2}{3} \times 6 = 4$ \\
Since $AD = 2$ \\
By the Pythagorean theorem, we have $AC = \sqrt{CD^{2}+AD^{2}} = \sqrt{4^{2}+2^{2}} = \boxed{2\sqrt{5}}$


Question: As shown in the figure, it is known that in $\triangle ABC$, $CD\perp AB$ with the foot of the perpendicular at point $D, AD=2, BD=6, \tan\angle B=\frac{2}{3}$. Point $E$ is the midpoint of side $BC$. Find the sine of $\angle EAB$.

Solution:
\textbf{Solution:} Construct $EF\perp AB$ at point F through point $E$, as shown in the figure,\\
since $CD\perp AB$,\\
therefore $EF\parallel CD$\\
therefore $\triangle BEF\sim \triangle BCD$\\
therefore $\frac{BE}{BC}=\frac{BF}{BD}=\frac{EF}{CD}$\\
since point $E$ is the midpoint of side $BC$\\
therefore $BE=CE=\frac{1}{2}BC$\\
therefore $\frac{BF}{BD}=\frac{EF}{CD}=\frac{1}{2}$\\
since $CD=4, BD=6$\\
therefore $EF=2, BF=3$\\
therefore $DF=3$\\
therefore $AF=DF+DF=2+3=5$\\
In $\triangle EAF$, since $AF^{2}+EF^{2}=AE^{2}$\\
therefore $AE=\sqrt{AF^{2}+EF^{2}}=\sqrt{5^{2}+2^{2}}=\sqrt{29}$\\
therefore $\sin\angle EAB=\frac{EF}{AE}=\frac{2}{\sqrt{29}}=\boxed{\frac{2\sqrt{29}}{29}}


Question: As shown in the figure, it is known that in the parallelogram $ABCD$, $G$ is a point on the extension line of $AB$. Connect $DG$, which intersects $AC$ and $BC$ at points $E$ and $F$ respectively, and $AE: EC = 3: 2$. If $AB = 10$, what is the length of $BG$?

Solution:
Solution: Since quadrilateral ABCD is a parallelogram, \\
$\therefore$AB$\parallel$CD, \\
$\therefore$$\angle GAE=\angle CDE$, $\angle AGE=\angle CDE$, \\
$\therefore$$\triangle AGE\sim \triangle CDE$, \\
$\therefore$$\frac{AG}{CD}=\frac{AE}{CE}=\frac{3}{2}$, \\
Furthermore, since AB=CD=10, \\
$\therefore$AG=$\frac{3}{2}$CD=$\frac{3}{2}\times 10=15$, \\
$\therefore$BG=AG−AB=15−10=\boxed{5};


Question: As shown in the figure, it is known that in the parallelogram $ABCD$, $G$ is a point on the extension line of $AB$. Connect $DG$, and let it intersect $AC$ and $BC$ at point $E$ and point $F$, respectively, and $AE: EC = 3: 2$. Find the value of $\frac{EF}{FG}$.

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\
it follows that AD $\parallel$ BC, which means AD $\parallel$ CF, \\
hence, $\angle ADE = \angle CFE$, and $\angle DAE = \angle FCE$, \\
thus, $\triangle ADE \sim \triangle CFE$, \\
therefore, $\frac{DE}{EF} = \frac{AE}{EC} = \frac{3}{2}$, \\
Furthermore, since $\triangle AGE \sim \triangle CDE$, \\
it follows that $\frac{DE}{GE} = \frac{EC}{AE} = \frac{2}{3}$, \\
therefore, $\frac{EF}{GE} = \frac{EF}{DE} \times \frac{DE}{GE} = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$, \\
hence, $\frac{EF}{FG} = \frac{EF}{GE-EF} = \boxed{\frac{4}{5}}$.


Question: In quadrilateral $ABCD$, $AD \parallel BC$, $AB= \sqrt{5}$, $AD=2$, $DC= 2\sqrt{5}$, $\tan \angle ABC=2$ (as shown in the figure). Point $E$ is on ray $AD$, and point $F$ is on side $BC$, with $BE$ and $EF$ connected, and $\angle BEF=\angle DCB$. What is the length of segment $BC$?

Solution:
\textbf{Solution:} Draw $AH\perp BC$ and $DG\perp BC$ through points A and D respectively, with H and G as the foot of the perpendiculars.\\
We can get: $AD=HG=2$, $AH=DG$,\\
$\because \tan\angle ABC=2$, $AB= \sqrt{5}$,\\
$\therefore AH=2$, $BH=1$,\\
$\therefore DG=2$,\\
$\because DC= 2\sqrt{5}$,\\
$\therefore CG= \sqrt{DC^{2}-DG^{2}}=4$,\\
$\therefore BC=BH+HG+GC=1+2+4=\boxed{7}$.


Question: In quadrilateral $ABCD$, $AD \parallel BC$, $AB= \sqrt{5}$, $AD=2$, $DC= 2\sqrt{5}$, $\tan \angle ABC=2$ (as shown in the figure). Point $E$ is on ray $AD$, and point $F$ is on side $BC$. Lines $BE$ and $EF$ are drawn such that $\angle BEF=\angle DCB$. When $FB=FE$, what is the length of segment $BF$?

Solution:
\textbf{Solution:} Construct $EM\perp BC$ at point M. It follows that $EM=2$, \\
From (1), we have $\tan\angle C= \frac{DG}{GC}=\frac{1}{2}$, \\
Since $FB=FE$, \\
it follows that $\angle FEB=\angle FBE$, \\
Since $\angle FEB=\angle C$, \\
it follows that $\angle FBE=\angle C$, \\
thus, $\tan\angle FBE= \frac{1}{2}$, \\
which leads to $\frac{EM}{BM}=\frac{1}{2}$, \\
therefore, $BM=4$, \\
Given that $FM^2+EM^2=FE^2$, \\
it follows that $(4−FB)^2+2^2=FB^2$, \\
thus, $BF=\boxed{\frac{5}{2}}$.


Question: In quadrilateral $ABCD$, $AD \parallel BC$, $AB= \sqrt{5}$, $AD=2$, $DC= 2\sqrt{5}$, and $\tan \angle ABC=2$ (as shown in the figure). Point $E$ is on the ray $AD$, and point $F$ is on side $BC$. $BE$ and $EF$ are connected, and $\angle BEF=\angle DCB$. When point $E$ is on the extension line of segment $AD$, let $DE=x$, $BF=y$. Find the analytical expression of $y$ as a function of $x$.

Solution:
\textbf{Solution:} Construct line $EN\parallel DC$ through point E, intersecting the extension of BC at point N.\\
Since $DE\parallel CN$,\\
therefore quadrilateral DCNE is a parallelogram,\\
thus $DE=CN$, $\angle DCB=\angle ENB$,\\
Since $\angle FEB=\angle DCB$,\\
thus $\angle FEB=\angle ENB$,\\
Also, since $\angle EBF=\angle NBE$,\\
thus $\triangle BEF\sim \triangle BNE$,\\
thus $\frac{BF}{BE}=\frac{BE}{BN}$,\\
thus $BE^2=BF\cdot BN$,\\
Construct $EM\perp BC$ through point E, with M being the foot of the perpendicular. We find $EM=2$, $BM=x+3$.\\
thus $BE^2=ME^2+BM^2=(x+3)^2+2^2=x^2+6x+13$,\\
thus $y(7+x)=x^2+6x+13$,\\
thus $y=\frac{x^2+6x+13}{7+x}$\\
When $y=7$, $x^2-x-36=0$,\\
Solving yields: $x=\frac{1+\sqrt{145}}{2}$, (neglect the negative root)\\
thus \boxed{y=\frac{x^2+6x+13}{7+x}} \left(0<x\le \frac{1+\sqrt{145}}{2}\right).


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=90^\circ$, $BC=3$. Point D is on the extension of AC, such that $AC = 3CD$. Through point D, draw $DH \parallel AB$, which intersects the extension of BC at point H. Find the value of $BD \cdot \cos\angle HBD$?

Solution:
\textbf{Solution:} From the given information, \\
since $DH\parallel AB$, \\
then $\angle BHD=\angle ABC =90^\circ$, \\
thus $\triangle ABC\sim \triangle DHC$, \\
resulting in $\frac{AC}{CD}=\frac{BC}{CH}$, \\
since $AC=3CD$ and $BC=3$, \\
it follows that $CH=1$, \\
$BH=BC+CH=4$, \\
In $\triangle BHD$, $\cos\angle HBD= \frac{BH}{BD}$, \\
thus $BD\cdot\cos\angle HBD=BH=\boxed{4}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC = 90^\circ$, $BC = 3$. D is a point on the extension line of $AC$ with $AC = 3CD$. A line through point D is drawn, $DH \parallel AB$, intersecting the extension line of $BC$ at point H. If $\angle CBD = \angle A$, what is the length of $AB$?

Solution:
\textbf{Solution:} From method one, we have\\
$\because \angle A=\angle CBD, \angle ABC=\angle BHD$,\\
$\therefore \triangle ABC\sim \triangle BHD$,\\
$\therefore \frac{BC}{HD}=\frac{AB}{BH}$,\\
$\because \triangle ABC\sim \triangle DHC$,\\
$\therefore \frac{DH}{AB}=\frac{DC}{AC}=\frac{1}{3}$,\\
$\therefore AB=3DH$,\\
$\therefore \frac{3}{DH}=\frac{3DH}{4}$,\\
$\therefore DH=2$,\\
$\therefore AB=\boxed{6}$.\\
From method two, we have,\\
$\because \angle CBD =\angle A, \angle ADB=\angle ADB$,\\
$\therefore \triangle CDB\sim \triangle BDA$,\\
$\therefore \frac{CD}{BD}=\frac{BD}{AD}$, hence $BD^2=CD\cdot AD$,\\
$\therefore BD^2=CD\cdot 4CD=4CD^2$,\\
$\therefore BD=2CD$,\\
$\because \triangle CDB\sim \triangle BDA$,\\
$\therefore \frac{CD}{BD}=\frac{BC}{AB}$,\\
$\therefore \frac{CD}{2CD}=\frac{3}{AB}$,\\
$\therefore AB=\boxed{6}$.


Question: As shown in the figure, the line $y=\frac{1}{2}x+2$ intersects with the $x$-axis and the $y$-axis at points $A$ and $C$, respectively. The parabola $y=-\frac{1}{2}x^2+bx+c$ passes through points $A$ and $C$, and intersects with the $x$-axis at another point $B$. Point $D$ is a moving point on the parabola. What is the analytical expression of the parabola?

Solution:
\textbf{Solution:} Let $y=0$, we get, $\frac{1}{2}x+2=0$, which gives $x=-4$,\\
Let $x=0$, we get, $y=2$,\\
$\therefore A(-4, 0)$, $C(0, 2)$\\
$\because$ the parabola $y=-\frac{1}{2}x^2+bx+c$ passes through points $A$ and $C$,\\
$\therefore \left\{\begin{array}{l}
-\frac{1}{2}\times (-4)^2-4b+c=0 \\
c=2
\end{array}\right.$,\\
Upon solving, we get $\left\{\begin{array}{l}
b=-\frac{3}{2} \\
c=2
\end{array}\right.$,\\
$\therefore y=-\frac{1}{2}x^2-\frac{3}{2}x+2=\boxed{-\frac{1}{2}x^2-\frac{3}{2}x+2}$


Question: As shown in the figure, the line $y=\frac{1}{2}x+2$ intersects the $x$-axis and $y$-axis at points $A$ and $C$, respectively. The parabola $y=-\frac{1}{2}x^2+bx+c$ passes through points $A$ and $C$, and its other intersection point with the $x$-axis is $B$. Point $D$ is a moving point on the parabola. When point $D$ is above line $AC$, by connecting $BC$, $CD$, $BD$, and with $BD$ intersecting $AC$ at point $E$, let the area of $\triangle CDE$ be $S_1$ and the area of $\triangle BCE$ be $S_2$. What is the maximum value of $\frac{S_1}{S_2}$?

Solution:
\textbf{Solution:} Let $y=0$,

$\therefore -\frac{1}{2}x^2-\frac{3}{2}x+2=0$,

$\therefore x_1=-4$, $x_2=1$,

$\therefore B(1,0)$,

Draw $DM\perp x$ axis from $D$ meeting $AC$ at point $M$, draw $BN\perp x$ axis from $B$ meeting $AC$ at $N$,

$\therefore DM\parallel BN$,

$\therefore \triangle DME\sim \triangle BNE$,

$\therefore \frac{S_1}{S_2}=\frac{DE}{BE}=\frac{DM}{BN}$,

Let $D\left(a, -\frac{1}{2}a^2-\frac{3}{2}a+2\right)$,

$\therefore M\left(a, \frac{1}{2}a+2\right)$,

$\because B(1,0)$,

$\therefore N\left(1, \frac{5}{2}\right)$,

$\therefore \frac{S_1}{S_2}=\frac{DM}{BN}=-\frac{1}{5}(a+2)^2+\frac{4}{5}$,

$\therefore$ When $a=-2$, the maximum value of $\frac{S_1}{S_2}$ is $\boxed{\frac{4}{5}}$.


Question: As illustrated, the line $y=\frac{1}{2}x+2$ intersects the $x$-axis and the $y$-axis at points $A$ and $C$, respectively. The parabola $y=-\frac{1}{2}x^2+bx+c$ passes through points $A$ and $C$, and intersects the $x$-axis at another point $B$. $D$ is a moving point on the parabola. $F$ is a moving point on the axis of symmetry of the parabola. There exists a parallelogram with vertices $B$, $C$, $D$, and $F$. Please find the coordinates of point $D$.

Solution: \textbf{Solution:} The coordinates of point $D$ are $\boxed{\left(-\frac{1}{2}, \frac{21}{8}\right)}$ or $\left(-\frac{5}{2}, \frac{21}{8}\right)$ or $\left(\frac{5}{2}, -\frac{39}{8}\right)$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=90^\circ$, $AB=12\,cm$, $BC=16\,cm$. Point $P$ starts from point $A$ and moves towards point $B$ along side $AB$ at a speed of $1\,cm/s$, and point $Q$ starts from point $B$ and moves towards point $C$ along side $BC$ at a speed of $2\,cm/s$. If $P$ and $Q$ depart from $A$ and $B$ simultaneously, when one point reaches its destination, the other point also stops moving. Let the time of movement be $t$ seconds. What value of $t$ makes the area of $\triangle PBQ$ equal to $35\,cm^{2}$?

Solution:
\textbf{Solution:} According to the problem, we know that $BQ=2t$, $AP=t$, \\
$\therefore QC=BC-BQ=16-2t$, $BP=AB-AP=12-t$. \\
Using the formula for the area of a triangle, we have $\frac{1}{2}BQ\cdot BP=35$, \\
$\frac{1}{2}\times 2t(12-t)=35$, \\
$t^{2}-12t+35=0$, \\
Solving this, we get $t_{1}=\boxed{5}$, $t_{2}=7$. \\
Therefore, when $t$ is 5 or 7, the area of $\triangle PBQ$ equals $35\,cm^{2}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=90^\circ$, $AB=12\,cm$, $BC=16\,cm$. Point $P$ starts from point $A$ and moves towards point $B$ along edge $AB$ at a speed of $1\,cm/s$, and point $Q$ starts from point $B$ and moves towards point $C$ along edge $BC$ at a speed of $2\,cm/s$. If $P$ and $Q$ start simultaneously from $A$ and $B$, respectively, and one of them stops moving when it reaches its endpoint, the other stops as well. Let the time of movement be $t$ seconds. For what value of $t$ is the length of $PQ$ equal to $8\sqrt{2}\,cm$?

Solution:
\textbf{Solution:} Let $t$ seconds later, the length of $PQ$ be $8\sqrt{2}\,cm$. According to the Pythagorean theorem, we have
\[ PQ^{2}=BP^{2}+BQ^{2}=(12-t)^{2}+(2t)^{2}=128, \]
\[ 5t^{2}-24t+144-128=0, \]
Solving this, we obtain \[ t_{1}=\frac{4}{5},\] \[t_{2}=4.\]
Therefore, when $t$ is \boxed{\frac{4}{5}} or \boxed{4}, the length of $PQ$ equals $8\sqrt{2}\,cm$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=90^\circ$, $AB=12\,cm$, and $BC=16\,cm$. Point $P$ starts from point $A$ and moves towards point $B$ along side $AB$ at a speed of $1\,cm/s$. Point $Q$ starts from point $B$ and moves towards point $C$ along side $BC$ at a speed of $2\,cm/s$. If $P$ and $Q$ start moving from points $A$ and $B$ simultaneously, and when one point reaches the end, the other point also stops moving. Let the moving time be $t$ seconds. Explore after how many seconds, the triangle with vertices $B$, $P$, $Q$ is similar to $\triangle CBA$?

Solution:
\textbf{Solution:} Let's assume after $x$ seconds, the two triangles become similar.\\
(1) When $\frac{BP}{BA}=\frac{BQ}{BC}$, it implies $\triangle BPQ\sim \triangle BAC$, which gives $\frac{12-x}{12}=\frac{2x}{16}$,\\
Solving this, we get: $x=\boxed{\frac{24}{5}}$,\\
(2) When $\frac{BP}{BC}=\frac{BQ}{BA}$, it implies $\triangle BQP\sim \triangle BAC$, which gives $\frac{12-x}{16}=\frac{2x}{12}$,\\
Solving this, we get: $x=\boxed{\frac{36}{11}}$,\\
Therefore, after $\boxed{\frac{24}{5}}$ seconds or $\boxed{\frac{36}{11}}$ seconds, the triangle formed by the points $B$, $P$, $Q$ becomes similar to $\triangle CBA$.


Question: As shown in the figure, $AD$ and $BE$ are the altitudes of $\triangle ABC$, and $DE$ is connected. $\triangle ACD \sim \triangle BCE$; If point $D$ is the midpoint of $BC$, $CE=6$, and $BE=8$, find the length of $AB$.

Solution:
\textbf{Solution:} Since point D is the midpoint of BC, and AD is perpendicular to BC,\\
it follows that AB=AC,\\
In right triangle $\triangle BEC$,\\
since CE=6 and BE=8,\\
we have BC=$\sqrt{CE^{2}+BE^{2}}=\sqrt{6^{2}+8^{2}}=10$,\\
thus CD=$\frac{1}{2}$BC=5,\\
since $\triangle ACD\sim\triangle BCE$,\\
it follows that $\frac{AD}{CD}=\frac{BE}{EC}$,\\
thus AD=$\frac{8\times 5}{6}=\frac{20}{3}$,\\
therefore, AC=$\sqrt{AD^{2}+CD^{2}}=\sqrt{\left(\frac{20}{3}\right)^{2}+5^{2}}=\frac{25}{3}$,\\
hence AB=AC=$\boxed{\frac{25}{3}}$.


Question: As shown in the diagram, with O as the center and AB as the diameter, a circle is drawn. A point on $\odot O$ is chosen, and AB is extended to point D. Line DC is drawn, and a tangent to $\odot O$ at point A intersects the extension of DC at point E, with $\angle DCB = \angle DAC$. CD is a tangent to $\odot O$; if AD = 6 and $\tan\angle DCB = \frac{2}{3}$, find the length of AE.

Solution:
\textbf{Solution:} Since $EA$ is the tangent to circle $\odot O$,\\
$\therefore EC=EA$, and $OE\perp AC$,\\
$\therefore \angle BAC=\angle OEA$ (the complementary angles of equal angles are equal),\\
$\therefore \angle CDB=\angle OEA$.\\
Since $\tan\angle DCB= \frac{2}{3}$,\\
$\therefore \tan\angle OEA= \frac{OA}{AE}=\frac{2}{3}$,\\
Since right $\triangle DCO\sim \triangle DAE$,\\
$\therefore \frac{DC}{DA}=\frac{OC}{AE}=\frac{OA}{AE}=\frac{2}{3}$,\\
$\therefore CD= \frac{2}{3}\times6=4$,\\
In the right $\triangle DAE$, let $AE=x$,\\
$\therefore (x+4)^{2}=x^{2}+6^{2}$,\\
Solving this, we get $x=\boxed{\frac{5}{2}}$.\\
That is, the length of $AE$ is $\boxed{\frac{5}{2}}$.


Question: As shown in the figure, point P is a point on the diagonal BD of the rhombus ABCD. Line CP is extended to intersect the extension of BA at point F and AD at point E. $\angle DCP=\angle DAP$; if $PE=4$ and $EF=5$, find the length of the line segment PA.

Solution:
\textbf{Solution:} Since $\triangle APD \cong \triangle CPD$,\\
$\therefore \angle DAP = \angle DCP$,\\
$\because CD \parallel AB$,\\
$\therefore \angle DCF = \angle DAP = \angle CFB$,\\
Also, $\because \angle FPA = \angle FPA$,\\
$\therefore \triangle APE \sim \triangle FPA$,\\
$\therefore \frac{AP}{PF} = \frac{PE}{PA}$,\\
$\therefore PA^{2} = PE \cdot PF$,\\
$\because PE = 4, EF = 5$,\\
$\therefore PF = PE + EF = 9$,\\
$\therefore PA^{2} = PE \cdot PF = 4 \times 9 = 36$,\\
$\therefore PA = \boxed{6}$.


Question: As shown in the figure, the linear function $y=x+b$ intersects with the inverse proportion function $y=\frac{k}{x}$ ($k\neq 0$) at points $A$ and $B$, and the coordinates of point $A$ are (1, 3). The linear function $y=x+b$ intersects with the x-axis at point $C$. Connect $OA$ and $OB$. What are the expressions for the linear function and the inverse proportion function?

Solution:
\textbf{Solution:} Given point $A(1, 3)$, substituting into the linear function $y=x+b$ yields: $3=1+b$,\\
Solving, we find: $b=2$,\\
Thus, the expression for the linear function is $y=x+2$,\\
Substituting $A(1, 3)$ into the inverse proportion function $y=\frac{k}{x}$ yields: $3=\frac{k}{1}$, implying $k=3$,\\
Therefore, the expression for the inverse proportion function is $\boxed{y=\frac{3}{x}}$;


Question: As shown in the figure, the linear function $y=x+b$ and the inverse proportion function $y=\frac{k}{x}$ (where $k\neq 0$) intersect at points $A$ and $B$. The coordinates of point $A$ are $(1,3)$. The linear function $y=x+b$ intersects the $x$-axis at point $C$. Connect $OA$ and $OB$. What are the coordinates of point $B$ and the area of $\triangle AOB$?

Solution:
\textbf{Solution:} From the system of equations
\[
\begin{cases}
y=x+2\\
y=\frac{3}{x}
\end{cases}
\]
we find:
\[
\begin{cases}
x=-3\\
y=-1
\end{cases}
\]
or
\[
\begin{cases}
x=1\\
y=3
\end{cases}
\]
$\therefore \boxed{B(-3, -1)}$.

Let $y=0$ and substitute into $y=x+2$ to get: $x=-2$,

$\therefore C(-2, 0)$,

$\therefore {S}_{\triangle AOB}={S}_{\triangle AOC}+{S}_{\triangle BOC}=\frac{1}{2}\times 2\times 3+\frac{1}{2}\times 2\times 1=\boxed{4}$.


Question: As shown in the figure, the linear function $y=x+b$ and the hyperbolic function $y=\frac{k}{x}$ (where $k \neq 0$) intersect at points $A$ and $B$. The coordinates of point $A$ are (1, 3). The linear function $y=x+b$ intersects the $x$-axis at point $C$. Lines $OA$ and $OB$ are drawn. A perpendicular line to the $y$-axis is drawn through point $A$, and its foot is point $D$. Point $M$ is a moving point on the graph of the hyperbolic function $y=\frac{k}{x}$ in the first quadrant. A perpendicular line to the $x$-axis is drawn through point $M$, intersecting the $x$-axis at point $N$, and line $CM$ is drawn. When $\triangle ADO$ is similar to $\triangle CNM$, find the coordinates of point $M$.

Solution:
\textbf{Solution:}

(1) When $\frac{MN}{OD}=\frac{CN}{AD}$, $\triangle ADO\sim \triangle CNM$,\\
$MN=\frac{3}{x}$, $OD=3$, $AD=1$, $CN=2+x$,\\
$\therefore \frac{\frac{3}{x}}{3}=\frac{2+x}{1}$, i.e., $x^{2}+2x-1=0$,\\
Solving for $x$, we get: $x_{1}=\sqrt{2}-1$, $x_{2}=-\sqrt{2}-1$,\\
Since $M$ is in the first quadrant,\\
$\therefore x=\sqrt{2}-1$, $y=\frac{3}{\sqrt{2}-1}=3\sqrt{2}+3$,\\
$\therefore M(\sqrt{2}-1, 3\sqrt{2}+3)$,\\
(2) When $\frac{MN}{AD}=\frac{CN}{OD}$, $\triangle ADO\sim \triangle CNM$,\\
$\therefore \frac{\frac{3}{x}}{1}=\frac{2+x}{3}$, i.e., $x^{2}+2x-9=0$,\\
Solving for $x$, we get: $x_{1}=\sqrt{10}-1$, $x_{2}=-\sqrt{10}-1$,\\
Since $M$ is in the first quadrant,\\
$\therefore x=\sqrt{10}-1$, $y=\frac{3}{\sqrt{10}-1}=\frac{\sqrt{10}+1}{3}$,\\
$\therefore M\left(\sqrt{10}-1, \frac{\sqrt{10}+1}{3}\right)$,\\
In summary, when $\triangle ADO$ and $\triangle CNM$ are similar, the coordinates of point $M$ are $\boxed{(\sqrt{2}-1, 3\sqrt{2}+3)}$ or $\boxed{\left(\sqrt{10}-1, \frac{\sqrt{10}+1}{3}\right)}$.


Question: As shown in Figure 1, in the Cartesian coordinate system, $\odot P$ is the inscribed circle of $\triangle AOC$, with points D, E, and F being the points of tangency. Line CD intersects $\odot P$ at point G. The radius of $\odot P$ is 2, $EG \parallel$ to the x-axis, and $AB=AC$. A parabola passes through points A, B, and C. $\triangle ADP\cong \triangle COD$; what is the equation of the parabola?

Solution: \textbf{Solution:} From (1), we know $\triangle APD\cong \triangle CDO$, $\therefore OD=PD=2$, $AD=OC=AE$, $\because$ circle $P$ is the inscribed circle of $\triangle AOC$, $\therefore CE=CF$, let $CE=CF=n$, then $AE=AD=OC=n+2$, $OA=n+4$, $AC=2n+2$. In right triangle $\triangle ACO$, $OA^2+OC^2=AC^2$, $\therefore (n+4)^2+(n+2)^2=(2n+2)^2$, solving this gives: $n_1=4$, $n_2=-2$ (discard), $\therefore OA=8$, $OC=6$, $AC=10$, $\because AB=AC$, $\therefore AB=10$, $\therefore OB=AB-OA=2$, $\therefore A(-8, 0)$, $B(2, 0)$, $C(0, 6)$. The equation of the parabola is $y=a(x+8)(x-2)$. Substituting $C(0, 6)$ into it gives: $6=a(0+8)(0-2)$, solving for $a$ gives: $a=-\frac{3}{8}$, $\therefore y=-\frac{3}{8}(x+8)(x-2)=-\frac{3}{8}x^2-\frac{9}{4}x+6$, $\therefore$ the equation of the parabola is $y=-\frac{3}{8}x^2-\frac{9}{4}x+6$; thus, the equation of the parabola is $\boxed{y=-\frac{3}{8}x^2-\frac{9}{4}x+6}$.


Question: As shown in Figure 1, in the Cartesian coordinate system, $\odot P$ is the inscribed circle of $\triangle AOC$, with points D, E, and F as the tangency points. The line CD intersects $\odot P$ at point G. The radius of $\odot P$ is 2, $EG$ is parallel to the x-axis, $AB=AC$, and the parabola passes through points A, B, and C. As shown in Figure 2, point M is on the parabola and above the line AC, $MN$ is parallel to the y-axis and intersects AC at point N. A perpendicular line from point N to BC is drawn, forming $NK\perp BC$ at the foot K. Let $t=MN+\frac{2}{5}\sqrt{10}NK$, find the maximum value of $t$ and the coordinates of point M at that time.

Solution: \textbf{Solution:} In the right-angled triangle $Rt\triangle BCO$, we have $BC=\sqrt{OB^{2}+OC^{2}}=\sqrt{2^{2}+6^{2}}=2\sqrt{10}$. Let the equation of the line $AC$ be $y=kx+b$. Then, we have the system of equations $\left\{\begin{array}{l}-8k+b=0 \\ b=6\end{array}\right.$. Solving this, we get $\left\{\begin{array}{l}k=\frac{3}{4} \\ b=6\end{array}\right.$, therefore, the equation of the line $AC$ is $y=\frac{3}{4}x+6$. Let $M(m, -\frac{3}{8}m^{2}-\frac{9}{4}m+6)$, since $MN$ is parallel to the $y$ axis and intersects $AC$ at $N$, we have $N(m, \frac{3}{4}m+6)$, therefore $MN=-\frac{3}{8}m^{2}-\frac{9}{4}m+6-(\frac{3}{4}m+6)=-\frac{3}{8}m^{2}-3m$, $CN=-\frac{5}{4}m$. Since $AB=AC$, we have $\angle CBO=\angle NCK$, and since $NK\perp BC$ with the foot at $K$, we have $\angle NKC=\angle COB=90^\circ$, therefore, $\triangle NCK\sim \triangle CBO$, which implies $\frac{NK}{CN}=\frac{OC}{BC}$, hence $\frac{NK}{-\frac{5}{4}m}=\frac{6}{2\sqrt{10}}$, leading to $NK=-\frac{3\sqrt{10}}{8}m$. Therefore, $t=MN+\frac{2}{5}\sqrt{10}NK=-\frac{3}{8}m^{2}-3m+\frac{2\sqrt{10}}{5}\times (-\frac{3\sqrt{10}}{8}m)=-\frac{3}{8}m^{2}-\frac{9}{2}m=-\frac{3}{8}(m+6)^{2}+\frac{27}{2}$. Therefore, when $m=-6$, $t$ achieves its maximum value $\boxed{\frac{27}{2}}$, and at this point the coordinates of $M$ are $\boxed{(-6, 6)}$.


Question: As shown in the figure, in $\triangle ABC$, $D$ and $E$ are the midpoints of sides $AB$ and $AC$, respectively, and $F$ is a point on the extension of side $CA$, with $\angle F = \angle C$. If $BC = 8$, what is the length of $FD$?

Solution:
\textbf{Solution:} Since $D$ and $E$ are the midpoints of the sides $AB$ and $AC$ respectively,
\newline
it follows that $DE=\frac{1}{2}BC$ and $DE \parallel BC$,
\newline
Given that $BC=8$,
\newline
it follows that $DE=4$,
\newline
Since $DE \parallel BC$,
\newline
it follows that $\angle AED=\angle C$,
\newline
Since $\angle F=\angle C$,
\newline
it follows that $\angle AED=\angle F$,
\newline
Therefore, $DF=DE=\boxed{4}$.


Question: As shown in the figure, it is known that quadrilateral $ABCD$ is inscribed in circle $\odot O$, $AB\parallel CD$, $BC\parallel AD$, $AB=6$, and $BC=8$. Point $E$ is a point on circle $\odot O$ (point $E$ is not coincident with $A$ or $C$). Connect $EA$, $EC$, and $OE$. Point $I$ is the incenter of $\triangle AEC$, and point $M$ is on line segment $OE$ such that $ME=2MO$. What is the minimum value of the length of segment $MI$?

Solution: \textbf{Solution:} The minimum value of segment $MI$ is $\boxed{\frac{5\sqrt{2}-20}{3}}$.


Question: As shown in Figure 1, in $\triangle ABC$, the endpoints $D$ and $E$ of the line segment $DE$ are located on the sides $AB$ and $AC$, respectively. If the product of the two line segments $AD$ and $AE$ above $DE$ is equal to the product of the two line segments $BD$ and $CE$ below $DE$, that is, $AD \times AE = BD \times CE$, then $DE$ is called a "friendly partition" segment of $\triangle ABC$. As shown in Figure 1, if $DE$ is a "friendly partition" segment of $\triangle ABC$, what is the length of $AC$?

Solution:
\textbf{Solution:} Let $AE = x$, \\
since $DE$ is a "friendly split" line segment of $\triangle ABC$, \\
it follows that $AD \cdot AE = BD \cdot EC$. \\
Since $AD = 2CE$ and $AB = 8$, \\
it follows that $2EC \cdot AE = (8 - AD) \cdot EC$. \\
Therefore, $2x = 8 - 2EC$. \\
Therefore, $x = 4 - EC$, \\
which means $AE = 4 - EC$. \\
Thus, $AC = AE + EC = \boxed{4}$.


Question: As shown in the figure, $D$ and $E$ are points on $AB$ and $AC$, respectively, in $\triangle ABC$, with $\angle AED = \angle B$, $AD = 2$, $AC = 3$. The bisector of $\angle A$ in $\triangle ABC$, $AF$, intersects $DE$ at point $G$ and $BC$ at point $F$. Given that $\triangle ADE \sim \triangle ACB$, find the value of $\frac{AG}{GF}$.

Solution:
Solution: Since $\triangle ADE \sim \triangle ACB$, it follows that $\angle ADE = \angle C$. Since AF bisects $\angle BAC$, we have $\angle DAG = \angle CAF$. Consequently, $\triangle ADG \sim \triangle ACF$, which means $\frac{AG}{AF} = \frac{AD}{AC}$. Given that $AD = 2$ and $AC = 3$, we find $\frac{AG}{AF} = \frac{2}{3}$. Letting $AG = 2a$ and $AF = 3a$, we get $FG = a$.

Therefore, $\frac{AG}{GF} = \boxed{2}


Question: As shown in the diagram, in the Cartesian coordinate system, the coordinates of point B are $(6, 0)$, and $AO=AB=5$, $AH \perp x$-axis at point H. A reciprocal function $y=\frac{k}{x}\ (x>0)$ passes through point A and intersects the $x$-axis at point B, creating line $BC \perp x$-axis. Line $OC$ intersects line $AB$ at point D and line $AH$ at point M. What is the expression of this reciprocal function?

Solution:
Since point B has the coordinates $(6, 0)$, and $AO=AB=5$, $AH \perp x$-axis, \\
thus $OB=6$, $OH=HB=3$,\\
thus $AH= \sqrt{AO^{2}-OH^{2}}=\sqrt{5^{2}-3^{2}}=4$,\\
thus $A(3,4)$,\\
thus $k=3\times 4=12$,\\
thus, the expression of the inverse proportion function is: \boxed{y=\frac{12}{x} \left(x>0\right)}.


Question: As shown in the figure, in the Cartesian coordinate system, the coordinates of point B are $(6, 0)$, and $AO=AB=5$, $AH \perp x$-axis at point H. A hyperbola $y = \frac{k}{x} (x>0)$ passing through point A intersects the $x$-axis at B and intersecting it at point C when a line perpendicular to the $x$-axis is drawn through B. Connect $OC$, which intersects $AB$ at point D and $AH$ at point M. Find the value of $\frac{AD}{DB}$.

Solution:
\textbf{Solution:}

Given that point B has coordinates $(6, 0)$, and $y=\frac{12}{x}$ $(x>0)$, with $BC\perp x$ axis intersecting the reciprocal function $y=\frac{k}{x}$ $(x>0)$ at point C,\\
thus $y= \frac{12}{6} =2$,\\
hence point C is $(6,2)$,\\
thus $BC=2$,\\
Let the equation of OC be $y=kx$,\\
thus $2=6k$,\\
hence $k= \frac{1}{3}$,\\
therefore, the equation of OC is $y= \frac{1}{3} x$,\\
when $x=3$, then $y= \frac{1}{3} x=1$,\\
therefore, point M is $(3,1)$,\\
thus $MH=1$,\\
therefore $AM=3$,\\
since $AH\perp x$ axis and $BC\perp x$ axis,\\
thus $AM\parallel BC$,\\
therefore $\triangle ADM\sim \triangle BDC$,\\
hence $AD: DB=AM: BC=3:2$.\\
Thus, $\frac{AD}{DB}=\boxed{\frac{3}{2}}$.


Question: As shown in the diagram, point $P$ is a point on the diagonal $BD$ of the rhombus $ABCD$. Line $CP$ is extended to meet $AD$ at point $E$ and the extended line of $BA$ at point $F$. It is given that $\triangle APD \cong \triangle CPD$; $\triangle APE \sim \triangle FPA$; If $PE = 4$ and $PF = 12$, what is the length of $PC$?

Solution:
\textbf{Solution:} Given that $\triangle APE \sim \triangle FPA$,\\
$\therefore \frac{AP}{PF} = \frac{PE}{PA}$,\\
$\therefore PA^{2} = PE \cdot PF$,\\
Since $\triangle APD \cong \triangle CPD$,\\
$\therefore PA = PC$,\\
$\therefore PC^{2} = PE \cdot PF$,\\
Given that $PE = 4$, $PF = 12$,\\
$\therefore PC = 4\sqrt{3}$.\\
Thus, the length of $PC$ is $\boxed{4\sqrt{3}}$.


Question: As shown in the figure, in the Cartesian coordinate system, the quadrilateral $ABCD$ is a parallelogram with $AD=6$. If the lengths of $OA$ and $OB$ are the two roots of the quadratic equation in $x$, ${x}^{2}-7x+12=0$, and $OA>OB$, find the lengths of $OA$ and $OB$.

Solution:
\textbf{Solution:} To solve the equation $x^2-7x+12=0$, factorize it to get: $\left(x-3\right)\left(x-4\right)=0$. So, we have: $x-3=0$ and $x-4=0$. Therefore, the solutions are: $x_1=3$ and $x_2=4$. Since $OA>OB$, \\
it follows that $OA=\boxed{4}$ and $OB=\boxed{3}$.


Question: As shown in the figure, in the Cartesian coordinate system, quadrilateral $ABCD$ is a parallelogram with $AD=6$. If the lengths of $OA$ and $OB$ are the two roots of the quadratic equation in $x$, ${x}^{2}-7x+12=0$, and $OA>OB$, and if point M is within the Cartesian coordinate system, then is there a point F on line $AB$ such that the quadrilateral with vertices A, C, F, and M, with $AC$ and $AF$ as adjacent sides, forms a rhombus? If such a point exists, provide the coordinates of point F. If it does not exist, explain why.

Solution:
\textbf{Solution:} According to the calculated data, $OB=OC=3$, $\because$ $AO\perp BC$, $\therefore$ $AO$ bisects $\angle BAC$. Consider two scenarios:
- (1) If $AC$ and $AF$ are adjacent sides, and point F is on the ray $AB$, then $AF=AC=5$, $\therefore$ point F coincides with B, that is $F\left(-3, 0\right)$;
- (2) If $AC$ and $AF$ are adjacent sides, and point F is on the ray $BA$, then M should be on line $AD$, and $FC$ perpendicularly bisects $AM$. Given $A\left(0, 4\right)$, $D\left(6, 4\right)$, $C\left(3, 0\right)$, $FC=8$, ${x}_{C}={x}_{F}=3$, $\therefore$ in this case, the coordinates of point F are $\left(3, 8\right)$.

In conclusion, there are two points that meet the conditions: \boxed{{F}_{1}\left(-3, 0\right)}; \boxed{{F}_{2}\left(3, 8\right)}.


Question: As shown in the figure, in the square $ABCD$, $E$ is the midpoint of $BC$, $F$ is a point on side $CD$, and $CD=4CF$. Extend line $EF$ to intersect the extension of line $AD$ at point $G$. $\triangle ABE\sim \triangle ECF$. If the side length of the square $ABCD$ is 8, find the length of $AG$.

Solution: \textbf{Solution:} Given that the side length of square ABCD is 8, it follows that BC=CD=AD=8, and BC is parallel to AD, leading to $\angle CEF=\angle G$. Since $\angle CFE=\angle DFG$, triangles CEF and DGF are similar, implying $\frac{CF}{DF}=\frac{CE}{DG}$. Given that E is the midpoint of BC and CD equals 4 times CF, we find CF=2, DF=6, and CE=4. Thus, $\frac{2}{6}=\frac{4}{DG}$, which gives DG=12. Therefore, AG=DG+AD=\boxed{20}.


Question: As shown in Figure 1, the parabola $y=-x^{2}+bx+c$ passes through points $A(-1,0)$ and $B(4,0)$ and intersects the $y$-axis at point $C$. Connect points $B$ and $C$. Point $P$ is a moving point on the parabola. A perpendicular line $l$ to the $x$-axis is drawn from point $P$, intersects line $BC$ at point $G$, and intersects the $x$-axis at point $E$. What is the expression of the parabola?

Solution:
\textbf{Solution:} To solve, substitute the coordinates of points A (-1,0) and B (4,0) into the expression of the function, we obtain:
\[
\begin{cases}
-1-b+c=0, \\
-16+4b+c=0.
\end{cases}
\]
Rearranging gives:
\[
\begin{cases}
-b+c=1, \\
4b+c=16,
\end{cases}
\]
Solving for the variables, we get:
\[
\begin{cases}
b=3, \\
c=4,
\end{cases}
\]
$\therefore$The equation of the parabola is $y=-x^{2}+3x+4$; thus, the answer is $\boxed{y=-x^{2}+3x+4}$.


Question: As shown in Figure 1, the parabola $y=-x^2+bx+c$ passes through points $A(-1,0)$ and $B(4,0)$, and intersects with the y-axis at point $C$. Connect $BC$, and let point $P$ be a moving point on the parabola. Perpendicular line $l$ is drawn from point $P$ to the x-axis, intersecting line $BC$ at point $G$ and the x-axis at point $E$. When $P$ moves on the section of the parabola to the right of the y-axis, draw $CF\perp l$ through point $C$, where $F$ is the foot of the perpendicular. At what position of $P$ will the triangle formed by $P$, $C$, and $F$ be similar to $\triangle OBC$? Also, find the coordinates of point $P$ at this time.

Solution:
\textbf{Solution:} Set $x=0$, then we get $y=4$, and the point $C(0,4)$,\\
\(\therefore OC=4\).\\
\(\therefore OC=OB\).\\
\(\therefore \triangle OBC\) is an isosceles right triangle,\\
\(\because \angle CFP=\angle COB=90^\circ\),\\
\(\therefore\) when $FC=PF$, \(\triangle PFC\sim \triangle OBC\).\\
Let the coordinates of point $P$ be \((a, -a^{2}+3a+4)\) (\(a>0\)).\\
Then \(CF=a\), \(PF= |-a^{2}+3a+4-4|=|a^{2}-3a|\).\\
\(\therefore |a^{2}-3a| =a\).\\
\(\therefore a^{2}-3a=a\),\\
Solving this gives $a=4$, $a=0$ (discard the latter),\\
When $a=4$, $y=-4^{2}+3\times 4+4=0$,\\
\(\therefore\) Point $P(4,0)$,\\
or $a^{2}-3a=-a$,\\
Solving this gives $a=2$, $a=0$ (discard the latter),\\
When $a=2$ then, $y=-2^{2}+3\times 2+4=6$.\\
\(\therefore\) Point $P(2,6)$,\\
\(\therefore\) In synthesis, the coordinates of point $P$ can be \boxed{(2,6)} or \boxed{(4,0)}.


Question: As shown in the figure, quadrilateral ABCD is a parallelogram. The circle $O$ with diameter AD intersects AB at point E. Connect DE. It is given that DA $= 2\sqrt{2}$, DE $= \sqrt{7}$, and DC $= 5$. A line $l$ is drawn through point E. Through point C, draw CH $\perp$ $l$, with H being the foot of the perpendicular. If $l \parallel$ AD, and $l$ intersects the circle $O$ at another point F, connect DF. What is the length of DF?

Solution:
\textbf{Solution:} As shown in the figure, connect DF, \\
$\because$AD$\parallel$l, \\
$\therefore$ $\angle$ADE=$\angle$DEF, \\
$\therefore$ AE=DF, \\
$\because$AD is the diameter of circle O, \\
$\therefore$ $\angle$AED=$90^\circ$, \\
$\therefore$ DF=AE=$\sqrt{AD^{2}-DE^{2}}=\boxed{1}$;


Question: As shown in the figure, quadrilateral ABCD is a parallelogram. The circle $\odot O$ with diameter AD intersects AB at point E. Line DE is drawn, DA $= 2\sqrt{2}$, DE $= \sqrt{7}$, and DC $= 5$. A line $l$ is drawn through point E. Through point C, draw CH $\perp l$, with the foot of the perpendicular being H. Line BH is connected. As line $l$ rotates around point E, what is the maximum value of BH?

Solution:
\textbf{Solution:} As shown in the figure, connect CE and take the midpoint K of CE. Draw KM $\perp$ BE at M,\\
$\because$ CH $\perp$ EH,\\
$\therefore \angle CHE = 90^\circ$,\\
$\therefore$ H lies on the circle centered at K with CE as diameter,\\
$\because BH \le HK + BK$,\\
$\therefore$ as shown in the figure, when H moves to the position of H, i.e., H, B, K are collinear at this moment, BH reaches its maximum value $BH$,\\
$\because$ quadrilateral ABCD is a parallelogram,\\
$\therefore$ AB = CD = 5, AB $\parallel$ CD,\\
$\therefore$ BE = AB - AE = 4, $\angle CDE = \angle AED = 90^\circ$, $\angle DCE = \angle MEK$,\\
$\therefore$ CE = KE = $\sqrt{DE^2 + CD^2} = 4\sqrt{2}$,\\
$\therefore$ KH = $\frac{1}{2}CE = 2\sqrt{2}$,\\
$\because \angle CDE = \angle EMK = 90^\circ$,\\
$\therefore \triangle CDE \sim \triangle EMK$,\\
$\therefore \frac{KM}{DE} = \frac{EK}{CE} = \frac{EM}{CD} = \frac{1}{2}$,\\
$\therefore$ KM = $\frac{1}{2}DE = \frac{\sqrt{7}}{2}$, $EM = \frac{1}{2}CD = \frac{5}{2}$,\\
$\therefore$ BM = AB - AE - EM = $\frac{3}{2}$,\\
$\therefore$ BK = $\sqrt{KM^2 + BM^2} = 2$,\\
$\therefore$ BH = $2 + 2\sqrt{2}$,\\
$\therefore$ The maximum value of BH is $\boxed{2 + 2\sqrt{2}}$;


Question: As shown in the figure, the quadrilateral ABCD is a parallelogram, with circle $\odot O$ centered at O and passing through AB at point E. Line segments DE and DA are connected, with $DA = 2\sqrt{2}$, $DE = \sqrt{7}$, and $DC = 5$. A straight line $l$ is drawn through point E. Through point C, draw $CH \perp l$ with the foot of the perpendicular at H. Through point A, draw $AM \perp l$ with the foot of the perpendicular at M. As the straight line $l$ rotates about point E, what is the maximum value of $CH - 4AM$?

Solution:
\textbf{Solution:} Draw BN $\perp l$ at N through point B and draw BT $\parallel l$ intersecting CH at T, \\
$\because$BN $\perp l$, CH $\perp l$, \\
$\therefore$BN $\parallel$ CH, \\
$\therefore$ Quadrilateral BCHN is a parallelogram, \\
$\therefore$HT=BN, \\
Similarly, it can be proved that AM $\parallel$ BN, \\
$\therefore$ $\triangle AME \sim \triangle BNE$, \\
$\therefore$ $\frac{BN}{AM}=\frac{BE}{AE}=4$, \\
$\therefore$BN=4AM, \\
$\therefore$HT=4AM, \\
$\therefore$CH-4AM=CH-HT=CT, \\
Also, $\because$ $CT \le BC$ \\
$\therefore$ when the line $l$ is perpendicular to the line BC, $CT=BC$, i.e., at this time, the maximum value of CH-4AM is BC, \\
$\because$ Quadrilateral ABCD is a parallelogram, \\
$\therefore$ $BC=AD=2\sqrt{2}$, \\
$\therefore$ The maximum value of CH-4AM is $\boxed{2\sqrt{2}}$.


Question: As shown in the figure, in $\triangle ABC$ and $\triangle ADE$, $\angle BAC = \angle DAE = 90^\circ$, $\angle B = \angle ADE = 30^\circ$, AC and DE intersect at point F, connect CE, point D lies on side BC. $\triangle ABD \sim \triangle ACE$; If $\frac{AD}{BD} = \sqrt{3}$, find the value of $\frac{DF}{CF}$.

Solution:
\textbf{Solution:} Given that $\triangle ABD \sim \triangle ACE$,\\
$\therefore \frac{AD}{AE}=\frac{BD}{EC}$, which means $\frac{AD}{BD}=\frac{AE}{EC}=\sqrt{3}$,\\
In $Rt\triangle ADE$, $\angle ADE=30^\circ$,\\
$\therefore \tan 30^\circ =\frac{AE}{AD}=\frac{\sqrt{3}}{3}$,\\
$\therefore \frac{AD}{CE}=\frac{AE}{EC} \div \frac{AE}{AD}=\sqrt{3} \times \sqrt{3}=3$,\\
$\because \triangle BAD \sim \triangle CAE$,\\
$\therefore \angle B=\angle ACE=\angle ADE$,\\
$\because \angle AFD=\angle EFC$,\\
$\therefore \triangle ADF \sim \triangle ECF$,\\
$\therefore \frac{DF}{CF}=\frac{AD}{EC}=\boxed{3}$.


Question: Given: As shown in the figure, in trapezoid $ABCD$, $AD\parallel BC$, $AB=DC=6$, $E$ is a point on the diagonal $BD$, $DE=4$, and $\angle BCE=\angle ABD$. $\triangle ABD\sim \triangle ECB$; if the ratio $AD:BC=3:5$, what is the length of $AD$?

Solution:
\textbf{Solution:} In trapezoid ABCD, we have $AD\parallel BC$ and $AB=DC=6$, \\
$\therefore \angle ABC=\angle BCD$, \\
Also, since $\angle BCE=\angle ABD$, \\
$\therefore \angle DBC=\angle DCE$, \\
Since $\angle BDC=\angle CDE$, \\
$\therefore \triangle BDC\sim \triangle CDE$, \\
$\therefore \frac{CD}{DE}=\frac{BD}{CD}$, \\
Since $DC=6$ and $DE=4$, \\
$\therefore BD=9$, \\
$\therefore BE=BD-DE=5$, \\
Since $\triangle ABD\sim \triangle ECB$, \\
$\therefore \frac{AD}{BE}=\frac{BD}{BC}$, \\
Given $AD\colon BC=3\colon 5$, \\
Let $AD=3x$ and $BC=5x$, \\
$\therefore \frac{3x}{5}=\frac{9}{5x}$, \\
Solving, we get $x=\sqrt{3}$ (neglecting the negative value), \\
$\therefore x=\sqrt{3}$, hence $AD=3\sqrt{3}$. Therefore, the length of $AD$ is $\boxed{3\sqrt{3}}$.


Question: As shown in Figure 1, the line AB intersects the x-axis and the y-axis at points A and B, respectively, and point C is on the negative half-axis of the x-axis. The coordinates of these three points are A(4,0), B(0,4), and C(-1,0), respectively. Find the equation of line AB.

Solution:
\textbf{Solution:} Since line AB passes through points A(4,0) and B(0,4), \\
let the equation of line AB be $y=kx+4$, \\
substituting A(4,0) into it gives: $4k+4=0$, \\
solving for $k$ gives: $k=-1$, \\
therefore, the equation of line AB is $\boxed{y=-x+4}$.


Question: As shown in Figure 1, line AB intersects the x-axis and y-axis at points A and B, respectively. Point C is located on the negative half-axis of the x-axis. The coordinates of these three points are \(A(4,0)\), \(B(0,4)\), and \(C(-1,0)\) respectively. Connect BC. Suppose point E is a movable point on segment AC (not coinciding with A or C). Draw \(EF \parallel BC\) intersecting AB at point F. When the area of \(\triangle BEF\) is \(\frac{5}{2}\), find the coordinates of point E.

Solution:
\textbf{Solution:} Let the coordinates of point E be $(m,0)$,\\
the equation of line BC be $y=4x+4$,\\
since EF$\parallel$BC,\\
thus, let the equation of line EF be: $y=4x+n$, substituting the coordinates of point E into the above equation yields: $0=4m+n$,\\
therefore $n=-4m$,\\
thus, the equation of line EF is: $y=4x-4m$,\\
hence $-x+4=4x-4m$,\\
solving this gives: $x=\frac{4}{5}(m+1)$,\\
substituting the value of $x$ into $y=-x+4$, we get $y=\frac{16-4m}{5}$,\\
therefore, the coordinates of point F are $\left(\frac{4m+4}{5},\frac{16-4m}{5}\right)$,\\
${S}_{\triangle BEF}={S}_{\triangle OAB}-{S}_{\triangle OBE}-{S}_{\triangle AEF}=\frac{1}{2}\times 4\times 4-\frac{1}{2}\times 4m-\frac{1}{2}\left(4-m\right)\times \frac{16-4m}{5}=\frac{5}{2}$,\\
solving this yields: $m=\frac{3}{2}$,\\
thus, the coordinates of point E are $\boxed{\left(\frac{3}{2},0\right)}$;


Question: As shown in Figure 1, the straight line AB intersects the x-axis and y-axis at points A and B respectively, with point C on the negative half of the x-axis. The coordinates of these three points are A(4,0), B(0,4), and C(-1,0). As shown in Figure 2, point B is translated 1 unit to the right to obtain point D. There exists a moving point P on the x-axis. If $\angle DCO+\angle DPO=\alpha$ and $\tan\alpha=4$, please directly write down the coordinate of point P.

Solution: \textbf{Solution:} The coordinates of point $P$ are $(19,0)$ or $(-17,0)$, therefore the answer is \boxed{(19,0)} or \boxed{(-17,0)}.


Question: As shown in the figure, with point $O$ as the center and the length of $AB$ as the diameter, a circle is drawn. A point $C$ is taken on $\odot O$, and $AB$ is extended to point $D$. Line $DC$ is drawn, with $\angle DCB = \angle DAC$. A line is drawn through point $A$ such that $AE \perp AD$ and it intersects the extension of $DC$ at point $E$. $CD$ is the tangent line of $\odot O$. Given that $CD = 4$ and $DB = 2$, find the length of $AE$.

Solution:
\textbf{Solution:} Let the radius of the circle $\odot O$ be $r$, thus $OA = OB = OC = r$,\\
$\because BD = 2$,\\
$\therefore OD = OB + BD = r + 2, AD = OA + OD = 2r + 2$,\\
In right triangle $\triangle OCD$, $CD^{2} + OC^{2} = OD^{2}$, that is $4^{2} + r^{2} = (r + 2)^{2}$,\\
Solving for $r$ yields $r = 3$,\\
$\therefore OC = 3, AD = 8$,\\
In triangles $\triangle ADE$ and $\triangle CDO$, $\left\{\begin{array}{l} \angle DAE = \angle DCO = 90^\circ \\ \angle D = \angle D \end{array}\right.$,\\
$\therefore \triangle ADE \sim \triangle CDO$,\\
$\therefore \frac{AE}{OC} = \frac{AD}{CD}$, that is $\frac{AE}{3} = \frac{8}{4}$,\\
Solving yields $AE = \boxed{6}$.


Question: As shown in the diagram, within the square $ABCD$, $E$ is a point on the side $BC$. Connect $AE$ and draw $DF\parallel AE$ through point $D$ to intersect the extension of $BC$ at point $F$. $AG$ bisects $\angle DAE$ and intersects $DF$ at point $G$, and intersects $DC$ at point $M$. If $\angle DAM=30^\circ$ and $CF=1$, what is the length of $AB$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a square,\\
it follows that AD $ \parallel $ EF, AB=DC, $\angle$ABE=$\angle$DCB=$\angle$DCF=$\angle$DAB=$90^\circ$,\\
Since AE $ \parallel $ DF,\\
it implies that quadrilateral AEFD is a parallelogram,\\
hence, AE=DF,\\
In right triangles $\triangle$ABE and $\triangle$DCF, $\left\{\begin{array}{l}AE=DF\\ AB=DC\end{array}\right.$,\\
it follows that right triangle $\triangle$ABE$\cong$$\triangle$DCF (HL),\\
therefore $\angle$BAE=$\angle$CDF,\\
Since AM bisects $\angle$DAE, and $\angle$DAM=$30^\circ$,\\
it follows that $\angle$EAM=$\angle$DAM=$30^\circ$,\\
thus $\angle$BAE=$90^\circ-30^\circ-30^\circ=30^\circ$,\\
therefore $\angle$CDF=$\angle$BAE=$30^\circ$,\\
hence, DF=2CF,\\
Since CF=1,\\
it follows that DF=2,\\
thus, DC=$\sqrt{DF^2-CF^2}$=$\sqrt{2^2-1^2}=\sqrt{3}$,\\
therefore AB=DC=$\boxed{\sqrt{3}}$;


Question: In square $ABCD$ and square $AEFG$, connect $AC$, $AF$, and $CF$ respectively, with point $H$ being the midpoint of $CF$. As shown in Figure 1, when points $F$ and $G$ are respectively on line segments $AB$ and $AC$, what is the measure of $\angle BHG$?

Solution:
Solution: Since squares $ABCD$ and $AEFG$, where $AC$ and $AF$ are diagonals,\\
$\therefore$ $\angle AFG=\angle ACB=45^\circ$, $\angle AGF=\angle ABC=90^\circ$,\\
Since point $F$ is on segment $AB$,\\
$\therefore$ $\angle GFB=180^\circ-\angle ACB=135^\circ$.\\
Since in $\triangle CGF$ and $\triangle CBF$, point $H$ is the midpoint of hypotenuse $CF$,\\
$\therefore$ $GH=\frac{1}{2}CF=HF$, $HB=\frac{1}{2}CF=HF$,\\
$\therefore$ $\angle HGF=\angle HFG$, $\angle HFB=\angle HBF$,\\
$\therefore$ $\angle HGF+\angle HBF=\angle HFG+\angle HFB=\angle GFB=135^\circ$,\\
$\therefore$ $\angle BHG=360^\circ-\angle HGF-\angle HBF-\angle GFB=360^\circ-135^\circ-135^\circ=\boxed{90^\circ}$;


Question: In squares $ABCD$ and $AEFG$, connect $AC$, $AF$, and $CF$ respectively. Let point $H$ be the midpoint of $CF$. As shown in Figure 3, when point $G$ is on segment $AB$ and $AB=2AE=2$, what is the area of $\triangle BGH$?

Solution:
\textbf{Solution:} Given point $G$ on segment $AB$, if $AB=2AE=2$, then $BC=AB=2$, $AG=FG=AE=1$, \\
$\therefore BG=AB-AG=1$, \\
Extend $GH$ to intersect $BC$ at point $Q$, as shown in the diagram: \\
$\because$ Point $H$ is the midpoint of $CF$, \\
$\therefore HC=HF$, \\
In squares $ABCD$ and $AEFG$, $\angle ABC=\angle AGF=90^\circ$, \\
$\therefore \angle BGF=90^\circ=\angle ABC$, \\
$\therefore FG\parallel BC$, \\
$\therefore \angle GFH=\angle QCH$, \\
In $\triangle GFH$ and $\triangle QCH$, \\
$\left\{\begin{array}{l}
\angle GFH=\angle QCH \\
FH=CH \\
\angle GHF=\angle QHC
\end{array}\right.$, \\
$\therefore \triangle GFH \cong \triangle QCH\left(ASA\right)$, \\
$\therefore CQ=FG=1$, $QH=GH$, \\
$\therefore BQ=BC-CQ=1$, $S_{\triangle BGH}=S_{\triangle BQH}=\frac{1}{2}S_{\triangle BGQ}$, \\
$\because S_{\triangle BGQ}=\frac{1}{2}BG\cdot BQ=\frac{1}{2}\times 1\times 1=\frac{1}{2}$, \\
$\therefore S_{\triangle BGH}=\frac{1}{2}S_{\triangle BGQ}=\frac{1}{2}\times \frac{1}{2}=\boxed{\frac{1}{4}}$, \\
i.e., the area of $\triangle BGH$ is $\frac{1}{4}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=\frac{4}{3}x+8$ intersects the x-axis and y-axis at points A and B, respectively. Point C is on the positive half of the x-axis, and $AB=AC$. Draw the line $BC$. What is the length of $AB$?

Solution:
\textbf{Solution:} Given $y=\frac{4}{3}x+8$,\\
thus, when $x=0$, $y=8$,\\
therefore, $B(0, 8)$,\\
therefore, $OB=8$,\\
when $y=0$, we have $\frac{4}{3}x+8=0$, solving for $x$ gives $x=-6$,\\
therefore, $A(-6, 0)$,\\
therefore, $OA=6$,\\
therefore, $AB=\sqrt{OA^2+OB^2}=\boxed{10}$,


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=\frac{4}{3}x+8$ intersects the x-axis and the y-axis at points A and B, respectively. Point C is on the positive half of the x-axis, with $AB=AC$. Construct line $BC$. $AD$ is the median of $\triangle ABC$, intersecting the y-axis at point F. What is the equation of line $AD$?

Solution:
\textbf{Solution:} Given that $AB=AC=10$, $OA=6$,\\
hence $OC=AC-OA=4$,\\
since $AD$ is the median of $\triangle ABC$,\\
it follows that $BD=CD$, and $AD\perp BC$,\\
Let the midpoint of $OC$ be $Q$, and connect $DQ$, hence $DQ$ is the median of $\triangle OBC$,\\
thus $DQ=\frac{1}{2}OB=4$, $OQ=\frac{1}{2}OC=2$,\\
hence $D(2, 4)$, $Q(2, 0)$,\\
Let the equation of line $AD$ be $y=kx+b$, substituting $A(-6, 0), D(2, 4)$ into it yields:\\
$\begin{cases}
-6k+b=0\\
2k+b=4
\end{cases}$, solving gives $\begin{cases}
k=\frac{1}{2}\\
b=3
\end{cases}$,\\
thus $\boxed{y=\frac{1}{2}x+3}$


Question: As shown in the figure, in the Cartesian coordinate system, the line \(y=\frac{4}{3}x+8\) intersects the x-axis and y-axis at points A and B, respectively. Point C is on the positive half of the x-axis, and \(AB=AC\). Line \(BC\) is drawn. Given condition (2), point E is on \(AF\), point G is on \(CD\), and \(CG=AE\). Line \(EG\) is drawn. Triangle \(\triangle EGM\) is an isosceles right triangle with \(EG\) as the hypotenuse. Point M is located above line \(AD\). The symmetric point of M with respect to line \(EG\) is marked as N. Lines \(NC\), \(NE\), and \(NG\) are drawn. \(MG\) intersects \(AD\) at point H. When \(\angle NCD=90^\circ\), find the coordinates of point H.

Solution:
Solution: Construct $MP\perp MD$ through point M, meeting $AD$ at point P,\\
since $\angle PMD=\angle EMG=90^\circ$, $\angle EMP+\angle PMG=\angle PMG+\angle DMG$,\\
thus, $\angle EMP=\angle DMG$.\\
Since $\angle MHE=\angle DHG$, $\angle EGM=\angle ADC=90^\circ$,\\
thus, $\angle MEP=\angle MGD$,\\
since $ME=MG$,\\
thus $\triangle MPE\cong \triangle MDG$,\\
thus $DG=PE$.\\
Since $D\left(2, 4\right)$,\\
thus $DQ=4, AQ=OA+OQ=8, CQ=2$,\\
thus $AD=\sqrt{DQ^{2}+AQ^{2}}=4\sqrt{5}$, $CD=\sqrt{CQ^{2}+DQ^{2}}=2\sqrt{5}$.\\
Since $AE=CG$, $PE=DG$,\\
thus $DG=PE$,\\
thus $AP=AE+PE=CG+DG=CD=2\sqrt{5}$,\\
thus $PD=AD−AP=2\sqrt{5}$.\\
Thus $PD=AP$,\\
Take the midpoint V of $AQ$, connect $PV$,\\
thus $PV$ is the median of $\triangle ADQ$,\\
since $DQ=4$,\\
thus $PV=\frac{1}{2}DQ=2$,\\
since $A(-6, 0)$, $Q\left(2, 0\right)$,\\
thus $V\left(-2, 0\right)$\\
thus $P\left(-2, 2\right)$.\\
$\triangle PMD$ is an isosceles right triangle, construct $MS\perp PV$ at point S through M, construct $DT\perp MS$ at point T through D,\\
since $\angle MPS=\angle DMT, \angle MSP=\angle MTD, PM=MD$,\\
thus $\triangle MSP\cong \triangle DTM$,\\
since $D\left(2, 4\right)$, $P\left(-2, 2\right)$,\\
thus $MS+MT=4, PS-DT=2$,\\
thus $MT=3, DT=1$,\\
thus $M\left(-1, 5\right)$.\\
Construct $MR\perp DP$ at R through M, construct $MK\perp BD$ at point K through M,\\
since $MP=MD, \angle PMD=90^\circ$\\
thus $\triangle DMP$ is an isosceles right triangle.\\
thus $MR=RD=RP=\sqrt{5}$,\\
since $\angle MDP=\angle MDK=90^\circ$,\\
thus $MK=MR=\sqrt{5}$.\\
Points M, N are symmetric with respect to line $EG$,\\
thus quadrilateral $MENG$ is a rhombus,\\
since $\angle EMG=90^\circ$,\\
thus rhombus $MENG$ is a square.\\
Since $\angle MGK+\angle KMG=\angle MGK+\angle NGC=90^\circ$,\\
thus $\angle KMG=\angle NGC$.\\
Since $\angle MKG=\angle NCG=90^\circ, GM=GN$,\\
thus $\triangle MKG\cong \triangle GCN$,\\
thus $CG=MK=\sqrt{5}$.\\
Since $CD=2\sqrt{5}$,\\
thus $CD=DG$, take the midpoint W of $CQ$, connect $GW$,\\
thus $GW$ intersects $AD$ at point H,\\
thus the coordinates of point H are \boxed{(-2, 2)}.\\


Question: In the Cartesian coordinate system, the sides $OA$ and $OC$ of rectangle $OABC$ are located on the positive $x$-axis and the positive $y$-axis, respectively. The lengths of $OA$ and $OC$ are the two roots of the equation ${x}^{2}-20x+100=0$. Find the coordinates of point $C$.

Solution:
\textbf{Solution:} From the given, we get $(x-10)^{2}=0$, \\
$\therefore x_{1}=x_{2}=10$, \\
$\therefore OA=OC=10$, \\
$\therefore \boxed{C(10, 0)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the sides $OA$ and $OC$ of rectangle $OABC$ lie on the positive x-axis and the positive y-axis, respectively, and the lengths of $OA$ and $OC$ are the two roots of the equation ${x}^{2}-20x+100=0$. In figure 2, point D is on $CB$, point E is on the extension line of $BA$, and $CD=AE$. Connect $AD$ and $CE$. Draw $OF\perp AD$ through point O and intersect $CE$ at point G, with the foot of the perpendicular being point F. Let the length of $CD$ be $m$, and let the x-coordinate of point G be $n$. What is the functional relationship between $n$ and $m$?

Solution:
\textbf{Solution:} Extend $OF$ to intersect $AB$ at $H$,\\
\[
\because \angle AFO = \angle B = \angle OAB = 90^\circ,
\]
\[
\therefore \angle OAF + \angle BAD = \angle AOH + \angle OAF = 90^\circ,
\]
\[
\therefore \angle AOB = \angle BAD,
\]
\[
\because OA = AB,
\]
\[
\therefore \triangle OAH \cong \triangle ABD,
\]
\[
\therefore AH = BD = 10 - m, \text{ and } AE = CD = m,
\]
\[
\therefore EH = 10 = OC,
\]
\[
\therefore \triangle EGH \cong \triangle CGO,
\]
\[
\therefore G \text{ is the midpoint of } EC,
\]
Draw $EM \perp OC$ intersecting the extended line of $CO$ at $M$, take the midpoint $N$ of $MC$, connect $GN$,\\
\[
\therefore GN \text{ is the midline of } \triangle EMC,
\]
\[
\therefore GN \parallel EM, \text{ i.e., } GN \perp OC,
\]
\[
\because ON = OC - NC = 10 - \frac{10 + m}{2} = -\frac{m}{2} + 5,
\]
\[
\therefore n = \boxed{-\frac{1}{2}m + 5}.
\]


Question: In the Cartesian coordinate system, the sides $OA$ and $OC$ of the rectangle $OABC$ lie on the positive half of the x-axis and the positive half of the y-axis, respectively, and the lengths of $OA$ and $OC$ are the two roots of the equation ${x}^{2}-20x+100=0$. As shown in the figure, under the condition when $FG: GO = 3:5$, find the equation of line $AD$.

Solution:
\textbf{Solution:}

Let $\because FG:GO=3:5$, \\
let $FG=3a(a>0)$, \\
$\therefore GO=5a$, \\
It is known from (2) that $\triangle EGH\cong \triangle CGO$, \\
$\therefore HG=GO=5a$, \\
$\therefore FH=2a$, \\
$\therefore AD=OH=10a$, \\
Connect $OD$, \\
$S_{\triangle AOD}=\frac{1}{2}AO\cdot |x_D|=\frac{1}{2}AD\cdot OF$, \\
$10a\cdot 8a=10\times 10$, \\
$\therefore$ solving gives $a=\pm \frac{\sqrt{5}}{2}$, \\
$\because a>0$, \\
$\therefore a=\frac{\sqrt{5}}{2}$, \\
$\therefore AD=10a=5\sqrt{5}$, \\
In $Rt\triangle ABD$, $BD=\sqrt{AD^2-AB^2}=5$, \\
$\therefore CD=BC-BD=5$, \\
$\therefore D(10, 5)$; \\
Let the equation of line $AD$ be $y=kx+b(k\ne 0)$, \\
$\because A(0, 10), D(10, 5)$, \\
$\therefore \left\{\begin{array}{l}b=10 \\ 10k+b=5\end{array}\right.$, \\
$\therefore \left\{\begin{array}{l}k=-\frac{1}{2} \\ b=10\end{array}\right.$, \\
$\therefore$ the equation of line $AD$ is: \boxed{y=-\frac{1}{2}x+10}.


Question: As shown in the diagram, in parallelogram $ABCD$, $\angle ACB=90^\circ$, and through point $D$, draw $DE\perp BC$ to meet the extension of $BC$ at point $E$. Quadrilateral $ACED$ is a rectangle; connect $AE$ and intersect $CD$ at point $F$, then connect $BF$. If $\angle ABC=60^\circ$ and $CE=2$, find the length of $BF$.

Solution:
\textbf{Solution:} As shown in the diagram:\\
Since quadrilateral $ACED$ is a rectangle, it follows that $AD=CE=2, AF=EF, AE=CD$,\\
Since quadrilateral $ABCD$ is a parallelogram, it follows that $BC=AD=2$, $AB=CD$,\\
Therefore, $AB=AE$, and since $\angle ABC=60^\circ$, it follows that $\triangle ABE$ is an equilateral triangle,\\
Therefore, $\angle BFE=90^\circ$, $\angle FBE=\frac{1}{2}\angle ABE=30^\circ$, $BE=2BC=4$,\\
Therefore, $EF=\frac{1}{2}BE=2$, in $\triangle BFE$, $BF=\sqrt{BE^{2}-EF^{2}}=\boxed{2\sqrt{3}}$


Question: In rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Rectangle $AEFO$ is constructed with $OA$ as its side, where point $E$ lies exactly on the perpendicular bisector of segment $AB$, and point $F$ is on $OB$. Let $G$ be a point on $OD$ such that $OG=BF$. Connect $CG$ and $BF$. As illustrated, given that $BO=2AO=8$ and $\angle AMB=45^\circ+\frac{1}{2}\angle MAC$, connect $MG$. Determine the length of segment $MG$.

Solution:
\textbf{Solution:} As shown in figure 3, draw $GH\perp MC$ through point $G$ meeting the extended line of $MC$ at point $H$.\\
From (1) and (2), it is known that $AC\perp BD$, $\angle AMB=BAC$, $\angle AMC=\angle EBC+\angle ABO$, $\angle EBA=\angle ABO$,\\
$\therefore OB\perp AC$, $\angle AOB=\angle COG=90^\circ$,\\
$\therefore \angle BAO+\angle ABO=90^\circ$, $\angle AMC=2\angle ABO$,\\
$\therefore \angle AMB+\frac{1}{2}\angle AMC=90^\circ$,\\
$\because \angle AMB=45^\circ+\frac{1}{2}\angle MAC$,\\
$\therefore 45^\circ+\frac{1}{2}\angle MAC+\frac{1}{2}\angle AMC=90^\circ$,\\
$\therefore \angle MAC+\angle AMC=90^\circ$,\\
$\therefore \angle ACM=90^\circ$,\\
$\because \angle COG=\angle OCH=\angle H=90^\circ$,\\
$\therefore$ quadrilateral $OCGH$ is a rectangle,\\
$\because \angle BAM+\angle MAC=\angle BAC=\angle AMB=45^\circ+\frac{1}{2}\angle MAC$,\\
$\therefore \angle BAM=45^\circ-\frac{1}{2}\angle MAC$,\\
$\because \angle MAC+\angle AMC=90^\circ$, $\angle AMC=2\angle ABO$,\\
$\therefore \angle MAC+2\angle ABO=90^\circ$,\\
$\therefore \angle ABO=45^\circ-\frac{1}{2}\angle MAC$,\\
$\therefore \angle BAM=\angle ABO$,\\
$\therefore BN=AN$,\\
$\therefore \angle EBA=\angle EAB=\angle ABN=\angle BAN$,\\
$\because AB=AB$,\\
$\therefore \triangle ABN\cong \triangle ABE(ASA)$,\\
$\therefore BE=BN$,\\
$\therefore AE=BE=BN=AN$,\\
$\because BO=2AO=8$,\\
$\therefore OA=4$,\\
$\therefore OC=OA=4$, $NO=BO-BN=BO-AN=8-AN$,\\
$\because AN^2=NO^2+OA^2$,\\
$\therefore AN^2=(8-AN)^2+4^2$,\\
$\therefore AN=5$, $ON=8-5=3$,\\
$\therefore AN=AE=BE=5$,\\
$\therefore CG=BE=5$, $OF=5$, $BF=BO-OF=3$,\\
$\therefore OG=BG=3$,\\
$\therefore OG=CH=3$, $GH=OC=5$,\\
$\because MC\parallel OB$,\\
$\therefore \triangle ANO\sim \triangle AMC$,\\
$\therefore \frac{NO}{MC}=\frac{AO}{AC}$, which means $MC=\frac{NO\cdot (AO+OC)}{AO}=\frac{3\times (4+4)}{4}=6$,\\
$\therefore MG=\sqrt{MH^2+GH^2}=\sqrt{9^2+4^2}=\boxed{\sqrt{97}}$.


Question: As shown in the figure, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Draw a line through point $B$ parallel to $AC$ and a line through point $C$ parallel to $BD$, and these two parallel lines intersect at point $E$. Given that $AB=2$ and $AD=2\sqrt{3}$, and quadrilateral $OBEC$ is a rhombus; calculate the area of rhombus $OBEC$.

Solution:
\textbf{Solution:} Let quadrilateral $ABCD$ be a rectangle,\\
$\therefore \angle ABC=90^\circ$, $BC=AD=2\sqrt{3}$, $OA=OC$,\\
$\therefore S_{\triangle ABC}=2S_{\triangle OBC}$,\\
$\because S_{\text{rhombus }OBEC}=2S_{\triangle OBC}$,\\
$\therefore S_{\text{rhombus }OBEC}=S_{\triangle ABC}=\frac{1}{2}AB\cdot BC=\frac{1}{2}\times 2\times 2\sqrt{3}=\boxed{2\sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, points $D$ and $E$ are the midpoints of sides $AB$ and $AC$, respectively. Line $CF$ is drawn parallel to $AB$ through point $C$ and intersects the extension of $DE$ at point $F$. Line $BE$ is drawn. Quadrilateral $BCFD$ is a parallelogram. When $AB=BC$, if $BD=2$ and $BE=3$, find the length of $AC$.

Solution:
\textbf{Solution:} Since $AB=BC$ and $E$ is the midpoint of $AC$,
thus $BE\perp AC$.\\
Since $AB=2DB=4$ and $BE=3$,\\
thus $AE=\sqrt{4^2-3^2}=\sqrt{7}$,\\
therefore $AC=2AE=2\sqrt{7}=\boxed{2\sqrt{7}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, the bisector of $\angle ABC$ intersects $AD$ at point $F$, and $EF\parallel AB$ intersects $BC$ at point $E$. The quadrilateral $ABEF$ is a rhombus; if $AB=5$, $BF=8$, and $CE=\frac{1}{2}AE$, what is the area of quadrilateral $ABCD$?

Solution: \textbf{Solution:} Given that quadrilateral ABEF is a rhombus, it follows that $AE \perp BF$, $BO=OF=\frac{1}{2}BF=4$, and $AO=OE$. Given that $AB=BE=5$, it can be deduced that $EO=\sqrt{BE^{2}-BO^{2}}=\sqrt{25-16}=3$. Therefore, $AE=6$. Given that $CE=\frac{1}{2}AE$, it follows that $CE=3$ and therefore $BC=8$. Let the distance between AD and BC be $h$. Since the area of the rhombus ABEF is equal to $BE\cdot h$ and the area of the parallelogram ABCD is equal to $BC\cdot h$, the ratio of the area of the rhombus ABEF to the area of the parallelogram ABCD is $BE:BC=5:8$. Since the area of the rhombus ABEF is $\frac{1}{2}\times 6\times 8=24$, the area of the quadrilateral ABCD is $\frac{8}{5}\times 24=\boxed{\frac{192}{5}}$.


Question: As shown in the figure, it is known that $ABCD$ is a parallelogram. Extend $AB$ to $E$ such that $BE=AB$, and connect $BD$, $ED$, $EC$. If $ED=AD$, $CD=BE$; quadrilateral $BECD$ is a rectangle; connect $AC$, if $AD=\sqrt{7}$, $CD=2$, find the length of $AC$.

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

since quadrilateral $ABCD$ is a parallelogram,

therefore $CB=AD=\sqrt{7}$, $AB=CD=2$,

since quadrilateral $BECD$ is a rectangle,

therefore $BE=CD=2$, $\angle BEC=90^\circ$,

therefore, in $\triangle BEC$,

$CE=\sqrt{BC^2-BE^2}=\sqrt{(\sqrt{7})^2-2^2}=\sqrt{3}$,

in $\triangle AEC$,

$AC=\sqrt{AE^2+CE^2}=\sqrt{(AB+BE)^2+CE^2}=\sqrt{4^2+(\sqrt{3})^2}=\boxed{\sqrt{19}}$.


Question: As illustrated, the diagonals $AC$ and $BD$ of quadrilateral $\square ABCD$ intersect at point O, with $AC \perp BD$, $BD = 12\,cm$, and $AC = 6\,cm$. Point E moves from point B towards point O on segment $BO$ at a speed of $1\,cm/s$, and point F moves from point O towards point D on segment $OD$ at a speed of $2\,cm/s$. Assuming points E and F start moving at the same time, and defining the movement duration as t seconds, what are the lengths of $EO$ and $OF$ in cm, respectively?

Solution:
\textbf{Solution:} Given the problem, the specific values of $EO$ and $OF$ cannot be directly obtained; more information is needed.\\
Therefore, a concrete answer cannot be provided.


Question: As shown in the figure, the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at point $O$. It is given that $AC\perp BD$, $BD=12\,cm$, and $AC=6\,cm$. Point $E$ moves towards point $O$ at a speed of $1\,cm/s$ from point $B$ along the segment $BO$, and point $F$ moves towards point $D$ at a speed of $2\,cm/s$ from point $O$ along the segment $OD$. If points $E$ and $F$ start moving at the same time and the moving time is $t$ seconds, for what value of $t$ is the quadrilateral $AECF$ a parallelogram?

Solution:
\textbf{Solution:} Since quadrilateral $AECF$ is a parallelogram,\\
it follows that $AO=OC$ and $EO=OF$,\\
where $EO=6-t$ and $OF=2t$,\\
thus $6-t=2t$,\\
therefore $t=\boxed{2}$,\\
hence, when $t=2$, quadrilateral $AECF$ is a parallelogram.


Question: As shown in the figure, the diagonals $AC$ and $BD$ of $\square ABCD$ intersect at point $O$, and $AC \perp BD$, $BD = 12\,cm$, $AC = 6\,cm$. Point $E$ moves from point $B$ towards point $O$ along the segment $BO$ at a speed of $1\,cm/s$, and point $F$ moves from point $O$ towards point $D$ along the segment $OD$ at a speed of $2\,cm/s$. If points $E$ and $F$ start moving at the same time, and the movement time is t seconds, is there a value of t such that $\triangle AEF$ is an isosceles triangle with $AF$ as its base? Find the value of t.

Solution:
\textbf{Solution:} Existence is shown as follows:\\
$\because \triangle AEF$ is an isosceles triangle, with $AF$ as the base,\\
$\therefore AE=EF,$\\
$\because AC\perp BD, OE=6-t, OA=3,$\\
$\therefore AE^{2}=3^{2}+(6-t)^{2},$\\
while $EF^{2}=(6-t+2t)^{2}=(6+t)^{2},$\\
$\therefore 9+(6-t)^{2}=(6+t)^{2},$\\
Solving gives: $t=\boxed{\frac{3}{8}}.$


Question: As shown in the figure, in rhombus $ABCD$, $AD\parallel x$-axis. The coordinate of point $A$ is $\left(0, 4\right)$, and the coordinate of point $B$ is $\left(3, 0\right)$. The line containing side $CD$, described by $y_{1}=mx+n$, intersects the $x$-axis at point $C$ and intersects the hyperbola $y_{2}=\frac{k}{x}$ $(x<0)$ at point $D$. Find the equation of line $CD$ and the value of $k$?

Solution:
\textbf{Solution:} Since the coordinates of point A are $(0,4)$ and the coordinates of point B are $(3,0)$,
\[AB= \sqrt{3^2+4^2}=5,\]
since quadrilateral ABCD is a rhombus,
\[AD=BC=AB=5,\]
thus $D(-5,4)$ and $C(-2,0)$. Plugging the coordinates of points C and D into the equation of the line yields
\[\begin{cases}
-5m+n=4 \\
-2m+n=0
\end{cases},\]
from which we find
\[\begin{cases}
m=-\frac{4}{3} \\
n=-\frac{8}{3}
\end{cases},\]
therefore, the equation of line CD is $y_1=-\frac{4}{3}x-\frac{8}{3}$. Given that point D lies on the graph of an inverse ratio function,
\[4=\frac{k}{-5},\]
therefore $k=\boxed{-20}$.


Question: As shown in the figure, in the rhombus $ABCD$, $AD\parallel x$-axis, the coordinates of point $A$ are $(0, 4)$, and the coordinates of point $B$ are $(3, 0)$. The line $y_{1}=mx+n$ containing side $CD$ intersects the $x$-axis at point $C$ and intersects the hyperbola $y_{2}=\frac{k}{x} (x<0)$ at point $D$. How many units should the rhombus $ABCD$ be translated in the positive direction of the $y$-axis so that point $C$ lies on the hyperbola $y_{2}=\frac{k}{x} (x<0)$?

Solution:
\textbf{Solution:} Given $C(-2,0)$,\\
substitute $x=-2$ into ${y}_{2}=-\frac{20}{x}$ $(x<0)$ to find, $y=-\frac{20}{-2}=10$,\\
therefore, by translating the rhombus $ABCD$ $10$ units in the positive direction of the $y$ axis, point $C$ will lie on the hyperbola $y_{2}=\frac{k}{x}\left(x<0\right)$.


Question: As shown in the figure, in the diamond $ABCD$, $AD\parallel x$ axis, the coordinate of point $A$ is $\left(0, 4\right)$, and the coordinate of point $B$ is $\left(3, 0\right)$. The line $y_1=mx+n$ on which side $CD$ lies intersects the $x$ axis at point $C$, and intersects the hyperbola $y_2=\frac{k}{x}\left(x<0\right)$ at point $D$. Directly write down the range of the independent variable $x$ for which $y_1\ge y_2$.

Solution: \textbf{Solution:} The range of the independent variable $x$, which satisfies $y_1 \geq y_2$, is $x \leq -5$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=2$, $BC=4$, point $D$ is the midpoint of side $AB$, and the graph of the inverse proportion function $y_{1}=\frac{k}{x}$ $(x>0)$ passes through point $D$ and intersects $BC$ at point $E$. Find the value of $k$ and the equation of line $DE$?

Solution:
\textbf{Solution:} Given that $AB=2$ and $BC=4$,\\
thus $B(4, 2)$.\\
Since point $D$ is the midpoint of side $AB$,\\
thus $D(4, 1)$, hence $k=4$.\\
The inverse proportion function equation is $y_{1}=\frac{4}{x}(x>0)$,\\
since $E$ is a point on $BC$, we have $y_{E}=2$.\\
Thus $x_{E}=\frac{4}{2}=2$, hence $E(2, 2)$.\\
Assuming the equation of line $DE$ is $y_{2}=kx+b$, we get\\
$\begin{cases}2=2k+b\\ 1=4k+b\end{cases}$, which gives $\begin{cases}k=-\frac{1}{2}\\ b=3\end{cases}$,\\
hence the equation of line $DE$ is $y_{2}=-\frac{1}{2}x+3$,\\
therefore $k=\boxed{-\frac{1}{2}}$ and the equation of line $DE$ is $y_{2}=\boxed{-\frac{1}{2}x+3}$.


Question: As shown in the figure, in rectangle $ABCD$, where $AB=2$ and $BC=4$, point $D$ is the midpoint of side $AB$, and the graph of the inverse proportion function $y_1=\frac{k}{x}(x>0)$ passes through point $D$ and intersects $BC$ at point $E$. Find a point $P$ on the $x$-axis such that the perimeter of $\triangle PDE$ is minimized. What are the coordinates of point $P$ in this case?

Solution:
\textbf{Solution:} Let $D(4, 1)$ be the point symmetrical about the $x$-axis as $D^{\prime}(4, -1)$. Let the equation of line $D^{\prime}E$ be $y=kx+b$. We have
\[
\left\{
\begin{array}{l}
2=2k+b \\
-1=4k+b
\end{array}
\right.
\]
Solving this, we get
\[
\left\{
\begin{array}{l}
k=-\frac{3}{2} \\
b=5
\end{array}
\right.
\]
Therefore, the equation of line $D^{\prime}E$ is $y=-\frac{3}{2}x+5$ \\
Therefore, the intersection of line $D^{\prime}E$ with the $x$-axis is $P\left(\frac{10}{3}, 0\right)$, \\
Therefore, when the perimeter of $\triangle PDE$ is minimized, the coordinates of $P$ are $\boxed{\left(\frac{10}{3}, 0\right)}$.


Question: As shown in the figure, within rectangle $ABCD$, $AB=2$, $BC=4$, point $D$ is the midpoint of side $AB$, the graph of the inverse proportional function $y_1=\frac{k}{x}$ ($x>0$) passes through point $D$ and intersects $BC$ at point $E$. What is the area of $\triangle PDE$?

Solution:
\textbf{Solution:} From (1), we know the equation of line $DE$ is $y_2=-\frac{1}{2}x+3$. Let its intersection with the $x$-axis be $Q$,\\
then it is known that the coordinates of $Q$ are $Q(6, 0)$,\\
$\because P\left(\frac{10}{3}, 0\right)$, then $PQ=6-\frac{10}{3}=\frac{8}{3}$,\\
and $D(4, 1)$, $y_E=2$,\\
$\therefore {S}_{\triangle PQE}=\frac{1}{2}PQ\cdot y_E=\frac{1}{2}\times \frac{8}{3}\times 2=\frac{8}{3}$\\
$\therefore {S}_{\triangle PQD}=\frac{1}{2}PQ\cdot y_D=\frac{1}{2}\times \frac{8}{3}\times 1=\frac{4}{3}$\\
$\therefore {S}_{\triangle PDE}={S}_{\triangle PQE}-{S}_{\triangle PQD}=\frac{8}{3}-\frac{4}{3}=\boxed{\frac{4}{3}}$


Question: As shown in the diagram, in quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at point $E$. Extend $CD$ to point $F$ such that $DF=CD$, and connect $AF$. Quadrilateral $ABDF$ is a parallelogram; if $AB\perp AC$ and $AB=2$, $AF=5$, what is the area of quadrilateral $ABCF$?

Solution: \textbf{Solution:} Since quadrilateral $ABCD$ is a parallelogram and $AB\perp AC$, it follows that $CD=AB=2$ and $\angle ACD=\angle BAC=90^\circ$. Since $CD=DF$, we have $CF=2CD=4$. Given that $AF=5$, in $\triangle ACF$, we have $AC=\sqrt{AF^{2}-CF^{2}}=\sqrt{5^{2}-4^{2}}=3$. Therefore, the area of quadrilateral $ABCF$ is $S_{\text{quadrilateral } ABCF}=S_{\triangle BAC}+S_{\triangle ACF}=\frac{1}{2}AB\cdot AC+\frac{1}{2}AC\cdot CF=\frac{1}{2}\times 2\times 3+\frac{1}{2}\times 3\times 4=\boxed{9}$.


Question: In the rectangle $ABCD$, as shown in the figure, a perpendicular line $EF$ is drawn through the midpoint $O$ of diagonal $AC$, intersecting sides $BC$ and $AD$ at points $E$ and $F$, respectively. $AE$ and $CF$ are then connected. Quadrilateral $AECF$ is a rhombus. If $AD = 8$ and $AB = 4$, what is the length of $CF$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
it follows that $\angle B=90^\circ$.\\
From (1) we know that quadrilateral AECF is a rhombus,\\
thus, let AE=CE=CF=x, then BE=8-x.\\
In $\triangle ABE$, we have $AB^2+BE^2=AE^2$, which means $4^2+(8-x)^2=x^2$,\\
solving this, we find $x=\boxed{5}$,\\
therefore, CF=5.