Question: As shown in the diagram, in $\triangle ABC$ and $\triangle ADC$, $AB=AD$, $BC=DC$, $\angle B=140^\circ$, what is the measure of $\angle D$ in degrees?

Solution:
\textbf{Solution:} In $\triangle ABC$ and $\triangle ADC$, AB = AD, BC = DC, AC = AC,\\
$\therefore \triangle ABC \cong \triangle ADC$,\\
$\therefore \angle D = \angle B = \boxed{140^\circ}$.


Question: As shown in the figure, in square $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $E$ is the midpoint of $BC$, and $DE$ intersects $AC$ at $F$. If $DE=12$, what is the length of $EF$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a square, \\
hence $AD=BC$, $AD\parallel BC$, \\
since $E$ is the midpoint of $BC$, \\
hence $CE=\frac{1}{2}BC=\frac{1}{2}AD$, \\
since $AD\parallel BC$, \\
hence $\triangle ADF\sim \triangle CEF$, \\
hence $\frac{EF}{DF}=\frac{CE}{AD}=\frac{1}{2}$, \\
since $DE=12$, \\
hence $EF=\frac{1}{3}DE=\boxed{4}$.


Question: As shown in the figure, to estimate the width of the fish pond $AB$, Xiao Kai took points $C$, $D$, and $E$ on the land such that $A$, $C$, $D$ are on the same straight line and $B$, $C$, $E$ are on the same straight line. It was measured that $CD = \frac{1}{2}AC$ and $CE = \frac{1}{2}BC$. Xiao Kai measured the length of $DE$ to be 10 meters. What is the length of the fish pond width $AB$ in meters?

Solution:
\textbf{Solution:} Given that $CD=\frac{1}{2}AC, CE=\frac{1}{2}BC$,\\
it follows that $\frac{CD}{AC}=\frac{CE}{BC}=\frac{1}{2}$,\\
and since $\angle DCE=\angle ACB$,\\
it implies $\triangle DCE\sim \triangle ACB$,\\
hence $\frac{ED}{AB}=\frac{CD}{AC}=\frac{1}{2}$.\\
Given $ED=10\text{m}$,\\
therefore $AB=\boxed{20\text{m}}$.


Question: As shown in the figure, in the equilateral $\triangle ABC$, $AB=4$, and $D$ and $P$ are points on the sides $BC$ and $AC$, respectively, with $BP=1$ and $\angle APD=60^\circ$. What is the length of $AD$?

Solution:
\textbf{Solution:} Given that $\because \triangle ABC$ is an equilateral triangle,\\
$\therefore \angle B=\angle C=60^\circ$,\\
In $\triangle ABP$, $\angle B+\angle APB+\angle BAP=180^\circ$,\\
$\therefore \angle BAP+\angle APB=120^\circ$,\\
$\because \angle APB+\angle CPD=180^\circ-\angle APD=120^\circ$,\\
$\therefore \angle BAP=\angle CPD$,\\
$\therefore \triangle ABP\sim \triangle PCD$,\\
$\therefore \frac{AB}{PC}=\frac{BP}{CD}$, which gives $\frac{4}{4-1}=\frac{1}{CD}$,\\
$\therefore CD=\frac{3}{4}$,\\
$\therefore AD=AC-CD=\boxed{\frac{13}{4}}$.


Question: Given: As shown in the figure, in $\triangle ABC$, $AB=4$, $BC=8$. $D$ is a point on side $BC$, and $\angle BDA = \angle BAC$. Find the length of $BD$.

Solution:
\textbf{Solution:} In $\triangle BDA$ and $\triangle BAC$,\\
$\angle B=\angle B$, $\angle BDA=\angle BAC$,\\
$\therefore \triangle BDA\sim \triangle BAC$,\\
$\therefore \frac{AB}{BC}=\frac{BD}{AB}$,\\
$\because AB=4$, $BC=8$,\\
$\therefore \frac{4}{8}=\frac{BD}{4}$,\\
$\therefore BD=\boxed{2}$.


Question: As shown in the figure: Point D is on the side AB of $\triangle ABC$, connect CD, $\angle 1 = \angle B$, $AD = 4$, $AC = 6$, find the length of AB.

Solution:
Solution: Since $\angle 1 = \angle B$, and $\angle A = \angle A$,\\
$\therefore$ $\triangle ACD \sim \triangle ABC$,\\
$\therefore$ $\frac{AD}{AC} = \frac{AC}{AB}$,\\
$\therefore$ $\frac{4}{6} = \frac{6}{AB}$,\\
$\therefore$ AB $= \boxed{9}$


Question: As shown in the figure, $AB \parallel PQ$, $AB = 80 \, \text{m}$, $PQ = 100 \, \text{m}$, and points $P$, $A$, $C$ are on the same line, as are points $Q$, $B$, $C$. If the distance between $AB$ and $PQ$ is $20 \, \text{m}$, what is the distance from point $C$ to line $PQ$ in meters?

Solution:
\textbf{Solution:} Given that $AB\parallel PQ$ and $AB=80m$, $PQ=100m$, \\
thus $\triangle CAB\sim \triangle CPQ$, \\
hence, $\frac{AB}{PQ}=\frac{CB}{CQ}$, that is $\frac{CB}{CQ}=\frac{80}{100}=\frac{4}{5}$, \\
as shown in the figure, draw a line through point $C$ perpendicular to the extension of $PQ$, with the foot at point $D$, intersecting the extension of $AB$ at point $E$, \\
since the distance between $AB$ and $PQ$ is $20m$, that is $DE=20m$, and according to the drawing, $CE\perp AB$, $CE\perp PD$, $AB\parallel PQ$, thus $BE\parallel PD$, \\
therefore $\triangle CBE\sim \triangle CQD$, \\
therefore $\frac{CB}{CQ}=\frac{CE}{CD}=\frac{4}{5}$, \\
since $CD=CE+ED=CE+20$, \\
thus $\frac{CE}{CE+20}=\frac{4}{5}$, \\
therefore $CE=80$, then $CD=20+80=\boxed{100m}$, which means the distance from point $C$ to line $PQ$ is $100m$.


Question: As shown in the figure, $D$ is a point on side $BC$ of $\triangle ABC$, with $\angle CAD = \angle B$. Given $AD=4$, $AB=8$, $AC=6$, what is the length of $DC$?

Solution:
Solution: In $\triangle ADC$ and $\triangle BAC$, we have $\left\{\begin{array}{l}
\angle CAD=\angle B \\
\angle C=\angle C
\end{array}\right.$,\\
$\therefore \triangle ADC\sim \triangle BAC$,\\
$\therefore \frac{AD}{AB}=\frac{DC}{AC}$,\\
$\because AD=4$, $AB=8$, $AC=6$,\\
$\therefore \frac{4}{8}=\frac{DC}{6}$,\\
Solving for $DC$, we find $DC=\boxed{3}$,\\
Therefore, the length of $DC$ is 3.


Question: As shown in the figure, Xiaoxin stands under the light, casting a shadow $AB=2.4\,m$ on the ground. When she squats down, the shadow becomes $AC=1.05\,m$. Given that Xiaoming's height $AD=1.6\,m$ and the height while squatting is half the standing height, what is the height of the lamp above the ground $PH$?

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

$\because$ AD $\parallel$ PH, $\therefore \triangle ADB \sim \triangle HPB$; $\triangle AMC \sim \triangle HPC$ (M is the midpoint of AD),

$\therefore$ AB : HB = AD : PH, AC : AM = HC : PH, that is $2.4 : (2.4 + AH) = 1.6 : PH$, $1.05 : 0.8 = (1.05 + HA) : PH$,

Solving, we get: $AH = 8.4$, $PH = \boxed{7.2}$.


Question: As shown in the figure, it is known that the length of line segment $AF$ is $13cm$, and points $B$, $C$, $D$, $E$ are sequentially on $AF$, with $AB=BC=CD$. Point $E$ is the midpoint of $DF$, and $CE=5cm$. Find the length of $BE$.

Solution:
\textbf{Solution:} Let $AB=BC=CD=x\, \text{cm}$, thus $BD=2x\, \text{cm}$,\\
since $AF$ is $13\, \text{cm}$ long,\\
therefore $DF=13-3x$,\\
since $E$ is the midpoint of $DF$,\\
therefore $DE=\frac{1}{2}(13-3x)$,\\
since $CE=5$,\\
therefore $x+\frac{1}{2}(13-3x)=5$,\\
therefore $x=3$,\\
therefore $BC=3$,\\
therefore $BE=BC+CE=\boxed{8}\, (\text{cm})$.


Question: Given that $C$ is the midpoint of the segment $AB$, and $D$ is on the segment $CB$, with $DA=6$, $DB=4$, find the length of $CD$.

Solution:
\textbf{Solution:} Since $DA=6$ and $DB=4$, \\
$\therefore$ $AB=10$, \\
Since $C$ is the midpoint of line segment $AB$, \\
$\therefore$ $AC=5$, \\
Since $DA=6$, \\
$\therefore$ $CD=\boxed{1}$.


Question: As shown in the figure, it is known that $OC$ is a ray inside $\angle AOB$, $OD$ is the angle bisector of $\angle AOB$, the ratio of the degrees of $\angle AOC$ to $\angle BOC$ is $3:2$, and $\angle DOC=12^\circ$. What is the degree measure of $\angle AOB$?

Solution:
\textbf{Solution:} Solution 1,\\
$\because \angle AOC : \angle BOC = 3 : 2$,\\
$\therefore$ let $\angle AOC = (3x)^\circ$, $\angle BOC = (2x)^\circ$,\\
$\therefore \angle AOB = \angle AOC + \angle BOC = (5x)^\circ$,\\
$\therefore OD$ bisects $\angle AOB$,\\
$\therefore \angle AOD = (2.5x)^\circ$,\\
$\because \angle COD = 12^\circ$ and $\angle COD = \angle AOC - \angle AOD$,\\
$\therefore 3x - 2.5x = 12$,\\
Solving for $x$ gives $x = 24$,\\
$\therefore \angle AOB = 5 \times 24^\circ = \boxed{120^\circ}$.\\
Solution 2,\\
$\because \angle AOC : \angle BOC = 3 : 2$,\\
$\therefore \angle AOC = 2\angle BOC$,\\
$\because \angle AOB = \angle AOC + \angle BOC$,\\
$\therefore \angle AOB = \angle BOC$,\\
$\because OD$ bisects $\angle AOB$,\\
$\therefore \angle AOD = \angle AOB = \angle BOC$,\\
$\because \angle COD = \angle AOC - \angle AOD$,\\
$\therefore \angle COD = \angle BOC - \angle BOC = \angle BOC$,\\
i.e., $\angle BOC = 4\angle COD$,\\
$\because \angle COD = 12^\circ$,\\
$\therefore \angle BOC = 48^\circ$,\\
$\therefore \angle AOB = \angle BOC = \boxed{120^\circ}$.


Question: As shown in the figure, it is known that point C is a point on line segment AB, with $AC=12\, \text{cm}$ and $CB=\frac{2}{3}AC$. D is the midpoint of AB. Find the length of DC.

Solution:
Solution:\\
Since $AC=12$\\
Then $CB=\frac{2}{3}AC=\frac{2}{3}\times 12=8$\\
Then $AB=AC+CB=12+8=20$\\
Also, since $D$ is the midpoint of $AB$\\
Then $DB=\frac{1}{2}AB=\frac{1}{2}\times 20=10$\\
Therefore $DC=DB-CB=10-8=\boxed{2}$


Question: As shown in the figure: It is known that $O$ is a point on line $CD$, $OA$ bisects $\angle BOC$, $\angle AOC=40^\circ$, find the degree measure of $\angle BOD$.

Solution:
\textbf{Solution:} Given that $O$ is a point on the line $CD$, and $OA$ bisects $\angle BOC$, $\angle AOC=40^\circ$,\\
then $\angle BOC=2\angle AOC=80^\circ$,\\
thus $\angle BOD=180^\circ-\angle BOC=\boxed{100^\circ}$.


Question: Given the figure, $AB=16\text{cm}$, where $C$ is a point on $AB$ and $AC=10\text{cm}$. Point $D$ is the midpoint of $AC$, and point $E$ is the midpoint of $BC$. Find the lengths of the line segments $DE$ and $AE$.

Solution:
\textbf{Solution:} Since $D$ is the midpoint of $AC$, and $AC = 10$ cm,

$\therefore DC = \frac{1}{2}AC = 5$ cm.

Also, since $AB = 16$ cm,

$\therefore BC = AB - AC = 6$ cm.

Since $E$ is the midpoint of $BC$,

$\therefore CE = \frac{1}{2}BC = 3$ cm.

Therefore, $DE = DC + CE = \boxed{8}$ cm, $AE = AC + CE = \boxed{13}$ cm.


Question: As shown in the figure, it is known that the length of segment $AB=6\text{cm}$, $C$ is the midpoint of $AB$, point $D$ is on $AC$, and $CD=2AD$, $E$ is the midpoint of $BC$. Find the length of segment $DE$.

Solution:
\textbf{Solution:} Given that $AB=6\text{cm}$, and C is the midpoint of AB,\\
thus $AC=BC=\frac{1}{2}AB=3\text{cm}$,\\
also, since $AB=6\text{cm}$,\\
thus $AC=BC=3\text{cm}$,\\
since E is the midpoint of BC,\\
thus $CE=\frac{1}{2}BC=1.5\text{cm}$,\\
since $CD=2AD$, and $AD+DC=AC$,\\
thus $AD+2AD=AC=3AD$,\\
thus $AD=1\text{cm}$, $CD=2\text{cm}$,\\
thus $DE=CD+CE= 2+1.5=\boxed{3.5cm}$.


Question: As shown in the figure, it is known that $\angle AOE = \angle COD$, and the ray $OC$ bisects $\angle BOE$, $\angle EOD = 30^\circ$. What is the degree measure of $\angle AOD$?

Solution:
\textbf{Solution:} Since $\angle AOB=180^\circ$ and $\angle EOD=30^\circ$,\\
it follows that $\angle AOD+\angle EOC+\angle COB=150^\circ$.\\
Since $\angle AOE=\angle COD$,\\
it follows that $\angle AOD=\angle EOC$.\\
Since $OC$ bisects $\angle EOB$,\\
it follows that $\angle EOC=\angle COB$,\\
hence $\angle EOC=\angle COB=\angle AOD=\boxed{50^\circ}$.


Question: As shown in the figure, it is known that the line segment $AB=3\text{cm}$. Extend $AB$ to point $C$ such that $BC=2AB$, then extend $BA$ to point $D$ such that $AD=AC$. Find the length of the line segment $BD$.

Solution:
\textbf{Solution}: Since $AB = 3\,cm$, and $BC = 2AB$, \\
therefore $BC = 2 \times 3 = 6\,cm$, and $AC = AB + BC = 3 + 6 = 9\,cm$, \\
since $AD = AC = 9\,cm$, \\
therefore $BD = AD + AB = 9 + 3 = \boxed{12}$.


Question: As shown in the figure, six rays OA, OB, OC, OD, OE, and OF are drawn from point O, and $\angle AOB=90^\circ$, OF bisects $\angle BOC$, OE bisects $\angle AOD$. If $\angle EOF=170^\circ$, what is the degree measure of $\angle COD$?

Solution:
\textbf{Solution:} Given the problem, we have \\
\(\because \angle AOB=90^\circ\), \(\angle EOF=170^\circ\), \\
\(\therefore \angle AOE + \angle FOB = 360^\circ - 90^\circ - 170^\circ = 100^\circ\). \\
\(\because\) OF bisects \(\angle COB\), OE bisects \(\angle AOD\), \\
\(\therefore \angle COF = \angle FOB\), \(\angle AOE = \angle EOD\). \\
\(\therefore \angle EOD + \angle COF = 100^\circ\). \\
\(\therefore \angle COD = 170^\circ - 100^\circ = \boxed{70^\circ}\).


Question: As shown in the diagram, $OB$ and $OE$ are two rays within $\angle AOC$, $OD$ bisects $\angle AOB$, $\angle BOE=\frac{1}{2}\angle EOC$. Given $\angle DOE=55^\circ$ and $\angle AOC=150^\circ$, calculate the degree measure of $\angle EOC$.

Solution:
\textbf{Solution:} Let $\angle BOE = x^\circ$, then $\angle DOB = 55^\circ - x^\circ$,\\
since $\angle BOE = \frac{1}{2} \angle EOC$, we have $\angle EOC = 2x^\circ$,\\
as $OD$ bisects $\angle AOB$, we get $\angle AOB = 2\angle DOB$,\\
thus $2x + x + 2(55 - x) = 150$,\\
solving the equation gives $x = 40$,\\
hence $\angle EOC = 2x = \boxed{80^\circ}$.


Question: As shown in the figure, $O$ is a point on the line $CE$. With $O$ as the vertex, we make $ \angle AOB=90^\circ$, and $OA$, $OB$ are on opposite sides of the line $CE$, $ \angle AOC=50^\circ$, and $OB$ bisects $ \angle COD$. Find the degree measure of $ \angle DOE$?

Solution:
\textbf{Solution:}

Given that, $\angle BOC = \angle AOB - \angle AOC = 90^\circ - 50^\circ = 40^\circ$,\\
since OB bisects $\angle COD$,\\
it follows that $\angle COD = 2\angle BOC = 80^\circ$,\\
thus, $\angle DOE = 180^\circ - \angle COD = \boxed{100^\circ}$.


Question: As shown in the figure, it is given that $\angle AOC = 2\angle BOC$, $OD$ bisects $\angle AOB$, and $\angle COD = 20^\circ$. Find the measure of $\angle AOB$.

Solution:
\textbf{Solution:} Given that $\angle AOC = 2\angle BOC$,\\
and $\angle AOC + \angle BOC = \angle AOB$,\\
it follows that $\angle AOB = 3\angle BOC$,\\
Since $OD$ bisects $\angle AOB$,\\
we have $\angle AOB = 2\angle BOD$,\\
thus, $\angle BOC = \frac{2}{3}\angle BOD$,\\
Given $\angle BOD = \angle BOC + \angle COD$,\\
it follows that $\angle COD = \frac{1}{3}\angle BOD$,\\
Given $\angle COD = 20^\circ$,\\
it follows that $\angle BOD = 60^\circ$,\\
therefore, $\angle AOB = \boxed{120^\circ}$.


Question: As shown in the diagram, it is known that $\angle AOB=90^\circ$, $\angle BOC=30^\circ$, $OM$ bisects $\angle AOC$, and $ON$ bisects $\angle BOC$. What is the measure of $\angle MON$ in degrees?

Solution:
\textbf{Solution:} Given that $\angle AOB = 90^\circ$ and $\angle BOC = 30^\circ$,\\
it follows that $\angle AOC = 90^\circ + 30^\circ = 120^\circ$,\\
since OM bisects $\angle AOC$,\\
we have $\angle AOM = \frac{1}{2} \angle AOC = \frac{1}{2} (\angle AOB + \angle BOC) = \frac{1}{2} \times 120^\circ = 60^\circ$,\\
since ON bisects $\angle BOC$,\\
it follows that $\angle CON = \frac{1}{2} \angle BOC = \frac{1}{2} \times 30^\circ = 15^\circ$,\\
thus, $\angle MON = \angle AOC - \angle AOM - \angle CON = 120^\circ - 60^\circ - 15^\circ = \boxed{45^\circ}$.


Question: As shown in the figure, it is known that point $M$ is the midpoint of line segment $AB$, point $N$ is on line segment $MB$, $AM=\frac{5}{3}MN$, if $MN=3\,cm$, find the lengths of line segments $AB$ and $NB$.

Solution:
\textbf{Solution:} Given that $AM= \frac{5}{3}MN$, and $MN=3\,\text{cm}$\\
$\therefore AM=5\,\text{cm}$\\
Furthermore, since point $M$ is the midpoint of line segment $AB$\\
$\therefore AM=BM= \frac{1}{2}AB$,\\
$\therefore AB=\boxed{10\,\text{cm}}$\\
Also, since $NB=AB-AM-MN$,\\
$\therefore NB=\boxed{2\,\text{cm}}$


Question: As shown in the figure, it is known that $AE$ is the bisector of $\angle BAC$, and $ED\parallel CA$. Given that $BE=2$, $EC=3$, and $AC=4$, what is the length of $AD$?

Solution:
\textbf{Solution:} Given that \\
$\because ED\parallel CA$, \\
$\therefore \triangle BED\sim \triangle BCA$, $\angle DEA=\angle EAC$, \\
$\therefore \frac{BE}{BC}=\frac{ED}{CA}$, \\
$\because BE=2$, $EC=3$, $AC=4$, \\
$\therefore BC=5$, \\
$\therefore ED=\frac{BE\cdot CA}{BC}=\frac{8}{5}$, \\
$\because AE$ bisects $\angle BAC$, \\
$\therefore \angle EAC=\angle DAE$, \\
$\therefore \angle DAE=\angle DEA$, \\
$\therefore \triangle ADE$ is an isosceles triangle, \\
$\therefore ED=AD=\boxed{\frac{8}{5}}$.


Question: As shown in the figure, in $ \triangle ABC$, $AD$ is the median to side $BC$, and point $E$ is on $AD$ with $ \frac{DE}{AE}=\frac{1}{3}$, ray $CE$ intersects $AB$ at point $F$. Find the value of $ \frac{AF}{FB}$.

Solution:
\textbf{Solution:} As shown in the figure, draw $DM\parallel AB$ through $D$, intersecting $CF$ at $M$,\\
$\therefore \triangle DME\sim \triangle AFE$, $\triangle CDM\sim \triangle CBF$,\\
$\therefore \frac{DE}{AE}=\frac{DM}{AF}=\frac{1}{3}$, $\frac{CD}{CB}=\frac{DM}{BF}$,\\
$\because D$ is the midpoint of $BC$, $\therefore \frac{CD}{CB}=\frac{DM}{BF}=\frac{1}{2}$,\\
$\therefore \frac{AF}{BF}=\boxed{\frac{3}{2}}$.


Question: As shown in the figure, in rectangle $ABCD$, $E$ is the midpoint of $BC$, and $DF\perp AE$ with the foot of the perpendicular being $F$. If $AB=12$ and $BC=10$, find the length of $DF$.

Solution:
\textbf{Solution:}

Since quadrilateral ABCD is a rectangle, \\
it follows that $\angle B = \angle BAD = 90^\circ$, $\angle BAE + \angle DAF = 90^\circ$, \\
since DF$\perp$AE, \\
it follows that $\angle AFD = 90^\circ$, $\angle FAD + \angle ADF = 90^\circ$, \\
thus, $\angle BAE = \angle FDA$, \\
therefore, $\triangle ABE \sim \triangle DFA$, \\
therefore, $\frac{DF}{AB} = \frac{DA}{AE}$, \\
Given that, AD = BC = 10, AB = 12, \\
since E is the midpoint of BC, \\
it follows that BE = 5, \\
In $\triangle ABE$, \\
$AE = \sqrt{AB^2 + BE^2} = \sqrt{12^2 + 5^2} = 13$, \\
therefore, $\frac{DF}{12} = \frac{10}{13}$, \\
therefore, $DF = \boxed{\frac{120}{13}}$


Question: As shown in the figure, points D and E are on sides AC and AB of $\triangle ABC$, respectively, with $AE=6$, $AC=10$, $BC=15$, and $\frac{AD}{AB}=\frac{3}{5}$. Find the length of DE.

Solution:
\textbf{Solution:} Since $AE:AC = 6:10 = 3:5$, and $AD:AB = 3:5$,\\
it follows that $\frac{AD}{AB} = \frac{AE}{AC}$.\\
Also, $\angle A = \angle A$,\\
therefore $\triangle ADE \sim \triangle ABC$,\\
thus $\frac{DE}{BC} = \frac{AE}{AC} = \frac{3}{5}$.\\
Given that $BC = 15$,\\
it follows that $DE = \frac{3}{5} \times 15 = \boxed{9}$.


Question: As shown in the figure, $D$ and $E$ are points on $AC$ and $AB$ respectively, $\triangle ADE \sim \triangle ABC$, and $DE=8$, $BC=24$, $CD=18$, $AD=6$. Find the lengths of $AE$ and $BE$.

Solution:
\textbf{Solution:} Since $\triangle ADE \sim \triangle ABC$,\\
we have $\frac{AE}{AC}=\frac{AD}{AB}=\frac{DE}{BC}$,\\
Since $DE=8$, $BC=24$, $CD=18$, $AD=6$,\\
thus $AC=AD+CD=24$,\\
thus $AE=\boxed{8}$, $AB=18$,\\
thus $BE=AB-AE=\boxed{10}$.


Question: In the rectangle $ABCD$, points $E$ and $F$ are located on sides $AD$ and $DC$ respectively. Given that $\triangle ABE \sim \triangle DEF$, $AB=6$, $AE=9$, and $DE=2$, find the length of $EF$.

Solution:
\textbf{Solution}: Given $\triangle ABE \sim \triangle DEF$, with AB=6, AE=9, and DE=2,\\
then $\frac{AB}{DE}=\frac{AE}{DF}$, which means $\frac{6}{2}=\frac{9}{DF}$, thus solving for DF gives DF=3,\\
since quadrilateral ABCD is a rectangle,\\
then $\angle D=90^\circ$,\\
by the Pythagorean theorem,\\
$EF=\sqrt{DE^{2}+DF^{2}}=\sqrt{2^{2}+3^{2}}=\boxed{\sqrt{13}}$


Question: As shown in the figure, $\angle AOC = 30^\circ$, $\angle BOC = 80^\circ$, and $OC$ bisects $\angle AOD$. What is the measure of $\angle BOD$?

Solution:
\textbf{Solution:} Since $\angle AOC=30^\circ$ and OC bisects $\angle AOD$,\\
it follows that $\angle COD=\angle AOC=30^\circ$,\\
Since $\angle BOC=80^\circ$,\\
it follows that $\angle BOD=\angle BOC-\angle COD=\boxed{50^\circ}$.


Question: As shown in the figure, the lines AB and CD intersect at point O, $\angle 1 = 35^\circ$, $\angle 2 = 75^\circ$. What is the measure of $\angle EOB$ in degrees?

Solution:
\textbf{Solution:} Given that $\because$ $\angle 1=35^\circ$,\\
$\therefore$ $\angle BOD=\angle 1=35^\circ$,\\
$\therefore$ $\angle EOB=\angle 2+\angle BOD=75^\circ+35^\circ=\boxed{110^\circ}$.


Question: As shown in the figure, the straight lines AB and CD intersect at point O, with $\angle 1: \angle 2 = 2:3$. Find the measure of $\angle BOD$ in degrees?

Solution:
\textbf{Solution:} Let us assume $\because$ $\angle 1\colon \angle 2=2\colon 3$,\\
$\therefore$ let $\angle 1=2x$, $\angle 2=3x$,\\
$\because$ $\angle 1+\angle 2=180^\circ$,\\
$\therefore$ $2x+3x=180^\circ$,\\
$\therefore$ $x=36^\circ$,\\
$\therefore$ $\angle 1=72^\circ$,\\
$\therefore$ $\angle BOD=\angle 1=\boxed{72^\circ}$.


Question: As shown in the figure, $C$ is a point on line segment $AB$, and $D$ and $E$ are the midpoints of line segments $AC$ and $BC$, respectively. If $AB = 10$ and $AD = 2$, find the length of $CE$.

Solution:
\textbf{Solution:} Since $D$ is the midpoint of line segment $AC$,\\
$AD=2$,\\
therefore $AC=2AD=4$,\\
since $AB=10$,\\
therefore $BC=10-4=6$,\\
since $E$ is the midpoint of line segment $BC$,\\
therefore $CE=\frac{1}{2}BC=\boxed{3}$


Question: As shown in the figure, it is known that $O$ is a point on line $AC$, $OB$ is a ray, $OD$ bisects $\angle AOB$, $OE$ is within $\angle BOC$, $2\angle BOE = \angle EOC$, $\angle DOE = 70^\circ$, find the degree measure of $\angle EOC$.

Solution:
\textbf{Solution:} Let $\angle AOB = x^\circ$, so $\angle BOC = (180-x)^\circ$.\\
Since OD bisects $\angle AOB$,\\
it follows that $\angle DOB= \frac{1}{2}\angle AOB$,\\
thus $\angle DOB= \frac{1}{2}x^\circ$\\
Since $2\angle BOE=\angle EOC$,\\
it follows that $\angle BOE= \frac{1}{3}\angle BOC= \frac{1}{3}(180-x)^\circ$.\\
Since $\angle DOE=\angle DOB+\angle BOE=70^\circ$,\\
it follows that $\frac{1}{2}x+ \frac{1}{3} (180-x)=70$,\\
Solving this, we find $x=60$,\\
Therefore $\angle BOC= 180^\circ-60^\circ= 120^\circ$,\\
Therefore $\angle EOC= \frac{2}{3}\angle BOC= \frac{2}{3}\times120^\circ= \boxed{80^\circ}$.


Question: As shown in the figure, the line segment AB is divided into three parts in the ratio 2:4:7 by points C and D. Points M and N are the midpoints of AC and DB, respectively. If MN equals 17 cm, what is the length of BD in cm?

Solution:
\textbf{Solution:} Since segment AB is divided into three parts with ratios 2:4:7 by points C and D,\\
let AC = 2x, CD = 4x, and BD = 7x.\\
Since M and N are the midpoints of segments AC and DB, respectively,\\
then CM = $\frac{1}{2}$AC = x and DN = $\frac{1}{2}$BD = $\frac{7}{2}$x.\\
Given MN = 17 cm, then x + 4x + $\frac{7}{2}$x = 17,\\
thus x = 2,\\
therefore BD = \boxed{14} cm.


Question: As shown in the figure, $BC=6\,cm$, $BD=7\,cm$, $D$ is the midpoint of $AC$, find the length of $AB$.

Solution:
\textbf{Solution:} Since $BC=6\,\text{cm}$ and $BD=7\,\text{cm}$,\\
$\therefore CD=BD-BC=1\,\text{cm}$,\\
Since point $D$ is the midpoint of $AC$,\\
$\therefore AD=CD=1\,\text{cm}$,\\
$\therefore AB=AD+BD=1+7=\boxed{8}\,\text{cm}$.


Question: Given, as shown in the figure, $AB \parallel CD$, $\angle ABE = 80^\circ$, $EF$ bisects $\angle BEC$, $EF \perp EG$, find the degree measure of $\angle DEG$.

Solution:
\textbf{Solution:} Given $AB\parallel CD$, $\angle ABE=80^\circ$,\\
thus, $\angle BEC=180^\circ-\angle ABE=100^\circ$,\\
since EF bisects $\angle BEC$,\\
thus, $\angle CEF= \frac{1}{2} \angle BEC=50^\circ$,\\
since $EF\perp EG$,\\
thus, $\angle FEG=90^\circ$,\\
hence, $\angle DEG=180^\circ-\angle CEF-\angle FEG=\boxed{40^\circ}$


Question: As shown in the figure, OD is the bisector of $\angle BOC$, OE is the bisector of $\angle AOC$, and $\angle AOB : \angle BOC = 3 : 2$. If $\angle BOE = 13^\circ$, what is the degree measure of $\angle DOE$?

Solution:
\textbf{Solution:} Let $\angle AOB=3x$ and $\angle BOC=2x$.\\
Then, $\angle AOC=\angle AOB+\angle BOC=5x$.\\
Since OE is the bisector of $\angle AOC$,\\
$\therefore \angle AOE= \frac{1}{2}\angle AOC= \frac{5}{2}x$,\\
$\therefore \angle BOE=\angle AOB-\angle AOE=3x- \frac{5}{2}x= \frac{1}{2}x$,\\
Since $\angle BOE=13^\circ$,\\
$\therefore \frac{1}{2}x=13^\circ$,\\
Solving, we find: $x=26^\circ$,\\
Since OD is the bisector of $\angle BOC$,\\
$\therefore \angle BOD= \frac{1}{2}\angle BOC=x=26^\circ$,\\
$\therefore \angle DOE=\angle DOB+\angle BOE=26^\circ+13^\circ=\boxed{39^\circ}$.


Question: As shown in the figure, the two diagonals of rectangle ABCD intersect at point O, where $ \angle AOB=60^\circ$ and $ AB=6$. What is the length of the diagonals of the rectangle?

Solution:
\textbf{Solution:} Given rectangle ABCD, \\
it follows that AC = BD, OA = OC, OD = OB, \\
therefore, OA = OB, \\
since $\angle AOB = 60^\circ$, \\
hence $\triangle AOB$ is an equilateral triangle, \\
therefore OA = OB = AB = 6cm, \\
hence AC = BD = 2$\times$6cm = \boxed{12cm}, \\


Question: As shown in the figure, in rectangle $ABCD$, where $AB=3$ and $AD=4$, two lines passing through points $A$ and $C$ intersect at point $E$. If $AE=13$ and $CE=12$, find the area of the shaded region in the figure.

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a rectangle, with $AB=3$ and $AD=4$,\\
it follows that $CD=AB=3$ and $\angle D=90^\circ$,\\
thus, in right triangle $ACD$, $AC=\sqrt{AD^2+CD^2}=\sqrt{4^2+3^2}=5$,\\
in triangle $ACE$, where $AE=13$, $CE=12$, and $AC=5$,\\
since $AC^2+CE^2=5^2+12^2=169$ and $AE^2=13^2=169$,\\
it follows that $AC^2+CE^2=AE^2$,\\
thus, $\angle ACE=90^\circ$,\\
hence, the area of the shaded region, ${S}_{\text{shaded}}={S}_{\triangle ACE}-{S}_{\triangle ACD}$\\
$=\frac{1}{2}AC\cdot CE-\frac{1}{2}AD\cdot CD$\\
$=\frac{1}{2}\times 5\times 12-\frac{1}{2}\times 4\times 3$\\
$=30-6$\\
$=\boxed{24}$.\\
Therefore, the area of the shaded region in the figure is $24$.


Question: As shown in the figure, in the rectangular paper ABCD, AB = 4, BC = 8. Now fold the rectangular paper ABCD along the diagonal BD, so that point C coincides with point \(C'\). Find the length of AF.

Solution:
\textbf{Solution:} Given that ABCD is a rectangle, \\
hence AB-CD = 4, BC = AD = 8, $\angle A = \angle ABC = \angle C = \angle CDA = 90^\circ$, \\
after folding: CD = C$^{\prime}$D = 4, BC = BC$^{\prime}$ = 8, $\angle CBD = \angle C^{\prime}BD$, \\
since $\angle CBD = \angle ADB$, \\
therefore $\angle ADB = \angle C^{\prime}BD$, \\
resulting in FB = FD, \\
let AF = x, then FC$^{\prime}$ = x, and FB = FD = 8-x, \\
in the right triangle $\triangle ABF$, by Pythagoras theorem, \\
$4^2 + x^2 = (8-x)^2$, \\
solving this, x = $\boxed{3}$, which means AF = 3.


Question: As shown in the figure, if two small squares with areas of $15\text{cm}^2$ and $24\text{cm}^2$ are cut from a large square, find the area of the remaining part.

Solution:
\textbf{Solution:} Because the areas of the two small squares are $15\text{cm}^2$ and $24\text{cm}^2$ respectively,\\
hence their side lengths are $\sqrt{15}$cm and $\sqrt{24}=2\sqrt{6}$cm respectively,\\
therefore, the side length of the large square is $\sqrt{15}+2\sqrt{6}$,\\
therefore, the area of the remaining part (i.e., the shaded area) is $(\sqrt{15}+2\sqrt{6})^2−15−24$\\
$=15+2\times \sqrt{15}\times 2\sqrt{6}+24−15−24$\\
$=4\sqrt{9\times 10}$\\
$=12\sqrt{10}\text{cm}^2$,\\
thus, the area of the remaining part is $\boxed{12\sqrt{10}\text{cm}^2}$.


Question: As shown in the figure, the square $ABCD$ and the equilateral triangle $\triangle CDE$ share the common side $CD$. Connect $AE$ and $AC$. What is the measure of $\angle EAC$ in degrees?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a square, \\
it follows that $DA=DC$, $\angle ADC=90^\circ$, and $\angle CAD=45^\circ$, \\
since $\triangle CDE$ is an equilateral triangle, \\
it follows that $DE=DC$, $\angle CDE=60^\circ$, \\
therefore, $DA=DE$, $\angle ADE=90^\circ+60^\circ=150^\circ$, \\
thus, $\triangle ADE$ is an isosceles triangle, \\
therefore, $\angle DAE=\angle DEA$, \\
hence, $\angle DAE = \frac{1}{2}(180^\circ-150^\circ)=15^\circ$, \\
therefore, $\angle EAC=\angle CAD - \angle DAE = 45^\circ - 15^\circ = \boxed{30^\circ}$.


Question: As shown in the figure, rectangle $ABCD$ has side $AB$ on the x-axis, with points $A$ and $B$ being symmetrically positioned about the origin. The coordinates of point $D$ are $(-2, 3)$. Find the analytical expression of line $AC$.

Solution:
\textbf{Solution:} Since the side $AB$ of rectangle $ABCD$ lies on the x-axis, and points A, B are symmetrical about the origin, with point D's coordinates being $\left(-2, 3\right)$, \\
$\therefore A\left(-2, 0\right), B\left(2, 0\right), AB=CD, AB\parallel CD, \angle ABC=\angle BCD=\angle DAB=90^\circ,$ \\
$\therefore C\left(2, 3\right),$ \\
Let $AC$ be defined by $y=kx+b,$ \\
$\therefore \begin{cases}-2k+b=0\\ 2k+b=3\end{cases},$ \\
Solving, we get: $\begin{cases}k=\frac{3}{4}\\ b=\frac{3}{2}\end{cases},$ \\
$\therefore$The equation of AC is: $\boxed{y=\frac{3}{4}x+\frac{3}{2}}.$


Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=kx+b(k\neq 0)$ intersects the line $y=x$ at point $A(2, a)$ and intersects the y-axis at point $B(0, 6)$. Find the analytical expression of the function corresponding to the line $y=kx+b$.

Solution:
\textbf{Solution:} Substituting point A (2, a) into $y=x$, \\
we get $a=2$, \\
$\therefore$ A (2, 2), \\
Substituting points A (2, 2), B (0, 6) into $y=kx+b$, \\
we obtain $\begin{cases}2k+b=2\\ b=6\end{cases}$, \\
Solving this, we find $\begin{cases}k=-2\\ b=6\end{cases}$, \\
$\therefore$ the equation of the line is $\boxed{y=-2x+6}$.


Question: As shown in the figure, the three vertices of $\triangle ABC$ are all on the coordinate axes of the Cartesian coordinate system, $BC=6$, and the expression for the line containing side $AB$ is $y=x+2$. What is $\sin\angle ACB$?

Solution:
\textbf{Solution:} Given that the equation of line AB is $y=x+2$,\\
it follows that when $y=0$, $x=-2$, and when $x=0$, $y=2$,\\
therefore, point A is $(0,2)$ and point B is $(-2,0)$,\\
thus, OA$=2$ and OB$=2$,\\
since BC$=6$,\\
it follows that OC$=BC-OB=6-2=4$,\\
therefore, AC$=\sqrt{OA^2+OC^2}=\sqrt{2^2+4^2}=2\sqrt{5}$,\\
hence, $\sin\angle ACB=\frac{OA}{AC}=\frac{2}{2\sqrt{5}}=\boxed{\frac{\sqrt{5}}{5}}$.


Question: As shown in the figure, it is known that in rectangle $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Point $E$ is a moving point on side $AB$, and $CE$ intersects $BD$ at point $F$. Connect $OE$. If point $E$ is the midpoint of $AB$, what is the value of $\frac{OF}{FB}$?

Solution: Solution: $\frac{OF}{FB} = \boxed{\frac{1}{2}}$.


Question: As shown in the figure, it is known that in rectangle $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, and point $E$ is a moving point on side $AB$. The line $CE$ intersects $BD$ at point $F$. Connect $OE$. If point $F$ is the midpoint of $OB$, then what is the value of $\frac{AE}{BE}$?

Solution: \textbf{Solution:} Given the information, it is insufficient to provide a detailed mathematical solution process. Therefore, it is impossible to directly determine the specific value of \boxed{\frac{AE}{BE}}.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, point $D$ is on line $AB$, $DE\parallel BC$, $BE\perp AB$. $\triangle DEB\sim \triangle BAC$; If $BE=2$, $AC=3$, and the area of $\triangle BDE$ is 1, find the area of $\triangle ABC$.

Solution:
\textbf{Solution:} From (1), we can obtain $ \triangle DEB\sim \triangle BAC$,\\
$\therefore \frac{S_{\triangle BDE}}{S_{\triangle ABC}}=\left(\frac{BE}{AC}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}$,\\
$\because S_{\triangle BDE}=1$,\\
$\therefore \frac{1}{S_{\triangle ABC}}=\frac{4}{9}$, solving this we get: $S_{\triangle ABC}=\boxed{\frac{9}{4}}$.


Question: Given $AC$ and $BD$ are diameters of $\odot O$, connect $AB$, $BC$, and let point $F$ be a point on $OC$ with $CF=2OF$. As shown in the figure, if $BC=6$ and $\angle BAC=30^\circ$, find the length of $OF$.

Solution:
\textbf{Solution:} It follows from the construction that $AC$ is the diameter of $\odot O$,\\
$\therefore \angle ABC=90^\circ$,\\
$\because \angle BAC=30^\circ$,\\
$\therefore \angle C=60^\circ$,\\
$\because OB=OC$,\\
$\therefore \triangle BOC$ is an equilateral triangle,\\
$\therefore OC=BC=6$,\\
$\because CF=2OF$,\\
$\therefore OF=\frac{1}{3}OC=2$;\\
Therefore, the length of $OF$ is $\boxed{2}$.


Question: As shown in the figure, in rectangle $ABCD$, $P$ is a point on $AB$, and $PB>PC$. Connect $PC$ and fold rectangle $ABCD$ along $PC$, such that point $B$ falls onto ${B}^{\prime}$. Extend ${B}^{\prime}C$ to intersect the extension of $AB$ at point $Q$. Given that $AB=10, BC=4$, and $PB=6$, find the length of $BQ$?

Solution:
\textbf{Solution:} Let $BQ=x$, $CQ=y$,

Since $ABCD$ is folded along $PC$, and point $B$ falls onto $B'$, with $BC=4$, $PB=6$,

Therefore, $PB'=PB=6$, $\angle PB'Q=\angle PBC=90^\circ$,

Since $\angle Q=\angle Q$,

Therefore, $\triangle PB'Q \sim \triangle CBQ$,

Thus, $\frac{BC}{CQ} = \frac{PB'}{PQ}$,

Therefore, $\frac{4}{y}=\frac{6}{6+x}$,

Solving gives $y=\frac{2x+12}{3}$,

By the Pythagorean theorem, we get ${x}^{2}+16={y}^{2}$,

Therefore, ${x}^{2}+16=\left(\frac{2x+12}{3}\right)^{2}$,

Solving gives $x=\boxed{\frac{48}{5}}$, $x=0$ (discard),

Therefore, $BQ=\boxed{\frac{48}{5}}$.


Question: As shown in the figure, in rectangle $ABCD$, $P$ is a point on side $AB$, and $PB>PC$. Connect $PC$, and fold rectangle $ABCD$ along $PC$, such that point $B$ falls onto $B'$. Extend $B'C$ to intersect the extended line of $AB$ at $Q$, given that $AB=10, BC=4$. If $DP \perp PC$, find the length of $BQ$.

Solution:
\textbf{Solution:} Let $PB = m$, $PA = n$, then $m + n = 10$,\\
since $DP \perp PC$, and rectangle $ABCD$,\\
therefore $\angle A = \angle B = \angle DPC = 90^\circ$,\\
since $\angle BPC = 90^\circ - \angle APD$, $\angle ADP = 90^\circ - \angle APD$,\\
therefore $\angle BPC = \angle ADP$,\\
therefore $\triangle BPC \sim \triangle ADP$,\\
therefore $\frac{BP}{DA} = \frac{BC}{AP}$,\\
therefore $\frac{m}{4} = \frac{4}{n}$, that is, $mn = 16$,\\
therefore $(m - n)^{2} = (m + n)^{2} - 4mn = 100 - 64 = 36$,\\
therefore $m - n = 6$,\\
since $m + n = 10$,\\
therefore $m = 8 = PB$.\\
Let $BQ = x$, $CQ = y$,\\
since $ABCD$ is folded along $PC$, point B falls on ${B}^{\prime}$, $BC = 4$, $PB = 6$.\\
therefore $P{B}^{\prime} = PB = 6$, $\angle P{B}^{\prime}Q = \angle PBC = 90^\circ$,\\
since $\angle Q = \angle Q$,\\
therefore $\triangle P{B}^{\prime}Q \sim \triangle CBQ$,\\
therefore $\frac{BC}{CQ} = \frac{P{B}^{\prime}}{PQ}$,\\
therefore $\frac{4}{y} = \frac{8}{8 + x}$,\\
solving gives $y = \frac{x + 8}{2}$,\\
by Pythagorean theorem, ${x}^{2} + 16 = {y}^{2}$,\\
therefore ${x}^{2} + 16 = \left(\frac{x + 8}{2}\right)^{2}$,\\
solving gives $x = \boxed{\frac{16}{3}}, x = 0$ (discard $x = 0$).


Question: As shown in the figure, in quadrilateral $ABCD$, point $E$ is on the extension of line $BA$, lines $EC$ and $BD$ intersect at point $G$, and line $EC$ intersects $AD$ at point $F$. It is known that $EA: AB = 1:2$. What is $EF: EC$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a parallelogram, \\
$\therefore$ $AD \parallel BC$, \\
$\therefore$ $\triangle EAF\sim \triangle EBC$, \\
$\therefore$ $\frac{EA}{EB}=\frac{EF}{EC}$, \\
Let $EA=x$, \\
$\because$ $EA:AB=1:2$, \\
$\therefore$ $AB=2x$, \\
$\therefore$ $EB=3x$, \\
$\therefore$ $\frac{x}{3x}=\frac{EF}{EC}$, \\
$\therefore$ $\frac{EF}{EC}=\frac{1}{3}$, \\
$\therefore$ $EF:EC=1:3$, \\
$\therefore$ $EF:EC=\boxed{1:3}$.


Question: As shown in the figure, inside the quadrilateral $ABCD$, point $E$ lies on the extension of side $BA$. Lines $EC$ and $BD$ are drawn and they intersect at point $G$. Line $EC$ intersects with line $AD$ at point $F$. It is known that $EA:AB=1:2$. What is the ratio of $FG:GC$?

Solution:
\textbf{Solution:} Given that $AD\parallel BC$,\\
$\therefore \triangle FGD\sim \triangle CGB$,\\
$\therefore \frac{FG}{GC}=\frac{FD}{BC}$,\\
From $\triangle EAF\sim \triangle EBC$,\\
Based on the conclusion from (1), $\frac{AF}{BC}=\frac{1}{3}$,\\
$\therefore AF=\frac{BC}{3}$,\\
$\therefore FD=AD-AF=BC-\frac{BC}{3}=\frac{2BC}{3}$,\\
$\therefore \frac{FG}{GC}=\frac{\frac{2BC}{3}}{BC}=\boxed{\frac{2}{3}}$,\\
$\therefore FG\colon GC=2\colon 3$.


Question: In the figure, $AB$ is the diameter of the circle $\odot O$, and $C$, $D$ are two points on the circle. $OC\parallel BD$, and the chord $AD$ intersects $BC$ and $OC$ at $E$ and $F$, respectively. $\overset{\frown}{AC}=\overset{\frown}{CD}$; connect $AC$. If $CE=1$, and $EB=3$, find the length of $AC$.

Solution:
\textbf{Solution:} As shown in the figure, \\
$\because \overset{\frown}{AC}=\overset{\frown}{CD}$, \\
$\therefore \angle CAD=\angle ABC$, \\
$\because \angle ECA=\angle ACB$, \\
$\therefore \triangle ACE\sim \triangle BCA$, \\
$\therefore \frac{AC}{CE}=\frac{BC}{AC}$, that is $AC^{2}=CE\cdot CB$, \\
$\because CE=1$, $EB=3$, \\
$\therefore CB=CE+BE=4$, \\
thus $AC^{2}=1\times (1+3)$, \\
$\therefore AC=\boxed{2}$ (negative values are discarded).


Question: As shown in the figure, it is known that the parabola $y = ax^2 + bx + c$ ($a \neq 0$) passes through points $A(4, 0)$, $B(-1, 0)$, and $C(0, -4)$. Find the equation of this parabola.

Solution:
\textbf{Solution:} Substitute the coordinates of A, B, and C into the equation of the parabola to get
\[
\begin{cases}
16a+4b+c=0 \\
a-b+c=0 \\
c=-4
\end{cases}
\]
Solving this system gives
\[
\begin{cases}
a=1 \\
b=-3 \\
c=-4
\end{cases}
\]
Thus, the equation of the parabola is $y=x^2-3x-4$. Therefore, the answer is $\boxed{y=x^2-3x-4}$.


Question: As shown in the graph, it is known that the parabola $y=ax^2+bx+c$ ($a\neq 0$) passes through points $A(4, 0)$, $B(-1, 0)$, and $C(0, -4)$. If the circle centered at point $A$ is tangent to line $BC$ at point $M$, find the coordinates of the tangent point $M$.

Solution:
\textbf{Solution:} Draw $AM\perp BC$ through point A, with M as the foot,

thus, $\angle AMB = \angle COB = 90^\circ$,

since $\angle ABM = \angle CBO$,

therefore, $\triangle BOC \sim \triangle BMA$,

therefore, $\frac{OB}{BM}=\frac{BC}{AB}$,

since $BC = \sqrt{OC^2 + OB^2} = \sqrt{17}$,

therefore, $\frac{1}{BM} = \frac{\sqrt{17}}{5}$,

therefore, $BM = \frac{5}{\sqrt{17}}$,

Draw $MN \perp AB$ through M, with N as the foot,

thus, $MN \parallel OC$,

therefore, $\triangle BOC \sim \triangle BNM$,

therefore, $\frac{OC}{MN} = \frac{OB}{NB} = \frac{\sqrt{17}}{\frac{5}{\sqrt{17}}}$,

therefore, $MN = \frac{4}{MN} = \frac{1}{NB} = \frac{20}{17}$, $NB = \frac{5}{17}$,

therefore, $ON = \frac{12}{17}$,

therefore, $\boxed{M\left(-\frac{12}{17}, -\frac{20}{17}\right)}$.


Question: As shown in the figure, in $\triangle ABD$, C is a point on $AD$, and $AD = 4CD$, $\angle CBD = \angle A$. A line is drawn through point D such that $DH \parallel AB$, intersecting the extension of $BC$ at point H. We have $\triangle HCD \sim \triangle HDB$. If $BC = 6$, find the length of $DH$.

Solution:
\textbf{Solution:} Since $AD=4CD$,\\
it follows that $\frac{CD}{AC}=\frac{1}{3}$,\\
Given $DH\parallel AB$,\\
it leads to $\triangle HCD\sim \triangle BCA$,\\
which implies $\frac{HC}{BC}=\frac{CD}{AC}=\frac{1}{3}$,\\
Since $BC=6$,\\
we find that $HC=2$ and $HB=HC+BC=8$,\\
Because $\triangle HCD\sim \triangle HDB$,\\
it can be deduced that $\frac{HC}{HD}=\frac{HD}{HB}$, thus $HD^2=HC\cdot HB=2\times 8=16$,\\
Therefore, $HD=\boxed{4}$ (the negative value has been discarded).


Question: As shown in the figure, points C and D lie on a semicircle with AB as the diameter. AD bisects $\angle BAC$, and the line segment OD intersects BC at point E. OD $\parallel$ AC. If AB $=10$ and BC $=8$, and we connect BD, find the length of BD.

Solution:
\textbf{Solution:} Let AD intersect BC at M,\\
because AB is the diameter of circle O,\\
therefore, $\angle C=\angle ADB=90^\circ$,\\
because OD$\parallel$AC,\\
therefore, $\angle DEM=90^\circ$,\\
by the Pythagorean theorem, we have $AC=\sqrt{AB^{2}-BC^{2}}=\sqrt{10^{2}-8^{2}}=6$,\\
because AO=BO and OD$\parallel$AC,\\
therefore, CE=BE=$\frac{1}{2}\times 8=4$,\\
because AC=6,\\
therefore, OE=$\frac{1}{2}$AC=3,\\
because OD=$\frac{1}{2}$AB=$\frac{1}{2}\times 10=5$,\\
therefore, DE=5-3=2,\\
because OD$\parallel$AC,\\
therefore, $\triangle DEM\sim \triangle ACM$,\\
therefore, $\frac{AC}{DE}=\frac{CM}{EM}$,\\
therefore, $\frac{6}{2}=\frac{4-EM}{EM}$,\\
solving for EM gives: EM=1, thus, BM=4+1=5,\\
in $\triangle DEM$, by the Pythagorean theorem, we have $DM=\sqrt{DE^{2}+ME^{2}}=\sqrt{2^{2}+1^{2}}=\sqrt{5}$,\\
in $\triangle MDB$, by the Pythagorean theorem, we have $BD=\sqrt{BM^{2}-DM^{2}}=\sqrt{5^{2}-(\sqrt{5})^{2}}=\boxed{2\sqrt{5}}$.


Question: In the Cartesian coordinate system, the parabola $y=x^2+(k-1)x-k$ intersects the line $y=kx+1$ at points $A$ and $B$, where point $A$ is to the left of point $B$. As shown in Fig. 1, when $k=1$, directly write down the coordinates of points $A$ and $B$.

Solution: \textbf{Solution}: Let $\boxed{A(-1,0)}$, $\boxed{B(2,3)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=x^{2}+(k-1)x-k$ intersects with the line $y=kx+1$ at points $A$ and $B$, where point $A$ is to the left of point $B$. Given $A(-1, 0)$ and $B(2, 3)$, let point $P$ be a moving point on the parabola, situated below the line $AB$. Determine the maximum area of $\triangle ABP$ and the coordinates of point $P$ under this condition.

Solution:
\textbf{Solution}: Let $P(x, x^{2}-1)$. Draw $PF\parallel y$-axis through point P, intersecting line $AB$ at point $F$, hence $F(x, x+1)$.\\
$\therefore PF=y_{F}-y_{P}=(x+1)-(x^{2}-1)=-x^{2}+x+2$.\\
$S_{\triangle ABP}=\frac{1}{2}PF(x_{F}-x_{A})+\frac{1}{2}PF(x_{B}-x_{F})=\frac{1}{2}PF(x_{B}-x_{A})=\frac{3}{2}PF$\\
$\therefore S_{\triangle ABP}=\frac{3}{2}(-x^{2}+x+2)=-\frac{3}{2}(x-\frac{1}{2})^{2}+\frac{27}{8}$\\
When $x=\frac{1}{2}$, $y_{P}=x^{2}-1=-\frac{3}{4}$.\\
$\therefore$ The maximum area of $\triangle ABP$ is $\boxed{\frac{27}{8}}$, with point $P$ coordinates at $\boxed{\left(\frac{1}{2}, -\frac{3}{4}\right)}$.


Question: In the Cartesian coordinate system, the parabola $y=x^{2}+(k-1)x-k$ intersects with the line $y=kx+1$ at points $A$ and $B$, where point $A$ is to the left of point $B$. As shown in the figure, the parabola $y=x^{2}+(k-1)x-k$ ($k>0$) intersects with the x-axis at points $C$ and $D$, with point $C$ being to the left of point $D$. There exists a unique point $Q$ on the line $y=kx+1$ such that $\angle OQC=90^\circ$. What is the value of $k$ in this case?

Solution:
\textbf{Solution:} Method 1:\\
Let line AB: $y=kx+1$ intersect the x-axis and y-axis at point E and F, respectively,\\
then E$\left(-\frac{1}{k},0\right)$, F$\left(0,1\right)$, OE$=\frac{1}{k}$, OF$=1$.\\
In $\triangle EOF$, by the Pythagorean theorem, we have: $EF=\sqrt{\left(\frac{1}{k}\right)^2+1}=\frac{\sqrt{1+k^2}}{k}$\\
Let $y=x^2+(k-1)x-k=0$, that is, $(x+k)(x-1)=0$, solving this yields: $x=-k$ or $x=1$.\\
$\therefore$C$\left(-k,0\right)$, OC$=k$.\\
I. Assume there exists a unique point Q such that $\angle OQC=90^\circ$,\\
then the circle with OC as its diameter touches line AB at point Q, according to the inscribed angle theorem, this implies $\angle OQC=90^\circ$.\\
Let point N be the midpoint of OC, and connect NQ, then NQ$\perp$EF, NQ$=CN=ON=\frac{k}{2}$.\\
$\therefore$EN$=OE-ON=\frac{1}{k}-\frac{k}{2}$.\\
$\because \angle NEQ=\angle FEO$, $\angle EQN=\angle EOF=90^\circ$,\\
$\therefore \triangle EQN\sim \triangle EOF$,\\
$\therefore \frac{NQ}{OF}=\frac{EN}{EF}$, that is: $\frac{\frac{k}{2}}{1}=\frac{\frac{1}{k}-\frac{k}{2}}{\frac{\sqrt{1+k^2}}{k}}$\\
Solving this yields: $k=\pm\frac{2\sqrt{5}}{5}$,\\
$\because k>0$,\\
$\therefore k=\boxed{\frac{2\sqrt{5}}{5}}$.\\
II. If line AB passes through point C, at that point the line and circle have only another point Q as their intersection, hence there exists a unique point Q such that $\angle OQC=90^\circ$,\\
Substituting C$\left(-k,0\right)$ into $y=kx+1$,\\
we get $k=1$, $k=-1$ (discard),\\
thus, there exists a unique point Q such that $\angle OQC=90^\circ$, at this time $k=1$.\\
To sum up, when $k=\frac{2\sqrt{5}}{5}$ or $1$, there exists a unique point Q such that $\angle OQC=90^\circ$.


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle, with points $D$ and $E$ located on sides $BC$ and $AC$ respectively, and $\angle ADE = 60^\circ$. $\triangle ABD \sim \triangle DCE$; if $AB=3$ and $EC=\frac{2}{3}$, what is the length of $DC$?

Solution:
\textbf{Solution:} From equation (1), we have proved that $\triangle ABD \sim \triangle DCE$,\\
$\therefore \frac{BD}{AB}=\frac{CE}{DC}$\\
Let $CD = x$, then $BD = 3 - x$,\\
$\therefore \frac{3-x}{3}=\frac{\frac{2}{3}}{x}$\\
$\therefore x = 1$ or $x = 2$,\\
$\therefore DC = \boxed{1}$ or $DC = \boxed{2}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and chord $CD\perp AB$. Point $E$ is on the extension of $CA$. Connect $DE$ to intersect $\odot O$ at point $F$, and connect $AF, CF$. If the degree of $\overset{\frown}{BD}$ is $40^\circ$, what is the degree of $\angle AFC$?

Solution:
\textbf{Solution:} As shown in Figure 1, connect $OD, AD$, and let $AB$ intersect $CD$ at $H$. \\
Since the degree of $\overset{\frown}{BD}$ is $40^\circ$, \\
therefore $\angle BOD=40^\circ$, \\
therefore $\angle DAB=\frac{1}{2}\angle DOB=20^\circ$, \\
since $AB\perp CD$, \\
therefore $\angle AHD=90^\circ$, \\
therefore $\angle ADH=90^\circ-\angle DAB=70^\circ$, \\
therefore $\angle AFC=\angle ADH=\boxed{70^\circ}$.


Question: As shown in the diagram, $AB$ is the diameter of the circle $\odot O$, with chord $CD\perp AB$. Point E is located on the extension line of $CA$, and line $DE$ intersects $\odot O$ at point F. Lines $AF$ and $CF$ are drawn. Given that $AB=5$ and $CD=4$, and that $CF$ passes through the center of the circle, calculate the length of $CE$.

Solution:
\textbf{Solution:} As shown in Figure 2, suppose $AB$ intersects $CD$ at $H$. \\
Since $AB$ is a diameter and $AB\perp CD$, $CD=4$ \\
Thus, $CH=DH=2$, \\
Since $OC=\frac{1}{2}CF=\frac{1}{2}AB=\frac{5}{2}$, $\angle OHC=90^\circ$, \\
Thus, $OH=\sqrt{OC^{2}-CH^{2}}=\sqrt{\left(\frac{5}{2}\right)^{2}-2^{2}}=\frac{3}{2}$, \\
Therefore, $HA=OH+OA=4$, \\
As a result, $AC=\sqrt{CH^{2}+AH^{2}}=\sqrt{2^{2}+4^{2}}=2\sqrt{5}$, \\
Since $CF$ is a diameter, \\
Thus, $\angle CDF=\angle AHC=90^\circ$, \\
Therefore, $AH\parallel DE$, \\
Therefore, $\triangle AHC\sim \triangle EDC$, \\
Thus, $\frac{CH}{CD}=\frac{AC}{CE}$ \\
Since $CH=HD$, \\
Thus, $CD=2CH$ \\
Therefore, $CE=2AC=4\sqrt{5}$, \\
Therefore, $CE=\boxed{4\sqrt{5}}$.


Question: In the figure, $AB$ is the diameter of $\odot O$, and the chord $CD\perp AB$. Point $E$ is on the extension of $CA$. Connect $DE$ intersecting $\odot O$ at point $F$, and connect $AF, CF$. Given the conditions in (3), connect $AD$, $EH$, intersecting at point $G$. What is the length of $GH$?

Solution:
\textbf{Solution:} As shown in the diagram below, connect $AD$ and $EH$, intersecting at point G,\\
$\because DE^2+DC^2=CE^2$,\\
$\therefore DE=\sqrt{CE^2-DC^2}=8$,\\
$\because DE^2+DH^2=HE^2$,\\
$\therefore HE=2\sqrt{17}$,\\
$\because AH\parallel DE$,\\
$\therefore \triangle AHG\sim \triangle DGE$,\\
$\therefore \frac{AH}{DE}=\frac{HG}{GE}$,\\
$\therefore \frac{AH}{DE}=\frac{HG}{HE-HG}$,\\
$\therefore \frac{HG}{2\sqrt{17}-HG}=\frac{4}{8}$,\\
$\therefore HG=\boxed{\frac{3}{4}\sqrt{17}}$.


Question: As shown in Figure 1, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=8$, and $BC=6$. Points D and E are movable points on sides $BC$ and $AB$, respectively, with $BE=BD$. Connect $DE$ and fold $\triangle BDE$ along $DE$ such that point B falls on point F. Connect $AF$. As shown in Figure 2, when point F is on side $AC$, what is the length of $BE$?

Solution:
\textbf{Solution:} Solving: By the property of folding, we get $BE=BD=FE=FD$, \\
$\therefore$ quadrilateral $BEFD$ is a rhombus, \\
$\therefore$ $EF\parallel BC$. \\
$\because$ point F is on side $AC$, \\
$\therefore$ $\triangle AEF\sim \triangle ABC$, \\
$\therefore$ $\frac{EF}{BC}=\frac{AE}{AB}=\frac{AF}{AC}$. \\
$\because$ $AC=8, BC=6, \angle ACB=90^\circ$, \\
$\therefore$ in $Rt\triangle ABC$, $AB=\sqrt{6^2+8^2}=10$. \\
Let $BE=EF=x$, \\
then $\frac{x}{6}=\frac{10-x}{10}$, \\
$\therefore$ $x=\frac{15}{4}$, \\
$\therefore$ $BE=\boxed{\frac{15}{4}}$.


Question: As shown in Figure 1, in right triangle $ABC$, where $\angle ACB=90^\circ, AC=8, BC=6$, $D$ and $E$ are moving points located on sides $BC$ and $AB$ respectively, and $BE=BD$. Connect $DE$. Triangle $BDE$ is folded along $DE$ such that point $B$ coincides with point $F$. Connect $AF$. As shown in Figure 3, during the movement of points $D$ and $E$, when $AF \parallel DE$, find the length of $AF$.

Solution:
\textbf{Solution:} As shown in the figure, join $BF$ intersecting $ED$ at point O, and let $EF$ intersect $AC$ at point G.\\
Since quadrilateral $BEFD$ is a rhombus,\\
it follows that $BF\perp ED, OB=OF, EF\parallel BC$.\\
Since $AF\parallel ED$,\\
it implies $FA\perp BF, \frac{BE}{BA}=\frac{BO}{BF}=\frac{1}{2}$,\\
thus, $\angle AFB=90^\circ, BE=\frac{1}{2}AB=5$,\\
Since E is the midpoint of $AB$,\\
it follows that $EF=AE=BE=5$.\\
Since $EF\parallel BC$,\\
it implies $\frac{EG}{BC}=\frac{AE}{AB}=\frac{AG}{AC}=\frac{1}{2}$,\\
thus, $EG=3, GF=2, AG=4$.\\
In $\triangle AGF$ with a right angle at $F$, $AF=\sqrt{AG^{2}+GF^{2}}=\sqrt{4^{2}+2^{2}}=\boxed{2\sqrt{5}}$.


Question: As shown in Figure 1, in $\triangle ACB$, $\angle ACB=90^\circ, AC=4, BC=3$. Point $D$ is a moving point on the hypotenuse $AB$ ($0<AD<3.2$). Taking point $A$ as the center and $AD$ as the radius, draw circle $A$ intersecting $AC$ at point $F$. Extend $CD$ to intersect circle $A$ at point $E$, and connect $AE$ and $DF$. $\angle FAE=2\angle FDC$. As shown in Figure 2, if $AE\parallel CB$, find the length of $EC$.

Solution:
\textbf{Solution:} Given that $AE\parallel CB$,\\
$\therefore \angle EAC+\angle ACB=180^\circ$, $\angle E=\angle BCD$,\\
Since $\angle ACB=90^\circ$, $AE=AD$,\\
$\therefore \angle EAC=90^\circ$, $\angle E=\angle ADE$,\\
Since $\angle ADE=\angle CDB$,\\
$\therefore \angle BCD=\angle CDB$,\\
$\therefore BD=BC=3$,\\
Since $\angle ACB=90^\circ, AC=4, BC=3$,\\
$\therefore AB=\sqrt{AC^{2}+BC^{2}}=5$,\\
$\therefore AE=AD=2$,\\
$\therefore CE=\sqrt{AC^{2}+AE^{2}}=\boxed{2\sqrt{5}}$;


Question: Definition: If a moving point $P$ satisfies the condition $PA=4PB$ with respect to the two endpoints of a line segment $AB$, then $P$ is called the $KZ$ point of the line segment $AB$, but point $P$ is not the $KZ$ point of line segment $BA$. As shown in Figure 1, in $\triangle ABC$, $\angle C=90^\circ$, $AB=17$. If point $C$ is the $KZ$ point of the line segment $AB$, what is the length of $AC$?

Solution:
\textbf{Solution:} Given that point $C$ is on the line segment $AB$ as the $KZ$ point, \\
$\therefore$ $AC=4BC$, \\
Let $BC=m$, then $AC=4m$, \\
Given that $\angle C=90^\circ$, \\
$\therefore$ $AC^{2}+BC^{2}=AB^{2}$, \\
$\therefore$ $m^{2}+(4m)^{2}=17^{2}$, \\
$\therefore$ $m^{2}=1$, \\
Since $m>0$, \\
$\therefore$ $m=1$, \\
$\therefore$ $AC=\boxed{4}$;


Question: Definition: If a moving point P satisfies the condition that the distances from P to the two endpoints of a line segment $AB$ are in the ratio $PA = 4PB$, then P is called the $KZ$ point of the line segment $AB$. However, point P is not considered a $KZ$ point of the line segment $BA$. As illustrated in the diagram, within a rhombus $ABCD$ where $AB=8$ and $\angle B=120^\circ$, points E and F are located on $BC$ and $CD$ respectively, satisfying $\angle AEF=120^\circ$. By connecting $AF$, assuming point E is the $KZ$ point of line segment $AF$, what is the length of $DF$?

Solution:
\textbf{Solution:} As shown in Figure 3, take $CJ$ on $CB$ such that $CJ=CF$.\\
Since quadrilateral $ABCD$ is a rhombus,\\
it follows that $AB=BC=CD=AD=8$, $AB\parallel CD$,\\
hence $\angle B+\angle C=180^\circ$,\\
since $\angle B=120^\circ$,\\
it follows that $\angle C=60^\circ$,\\
since $CJ=CF$,\\
it follows that $\triangle CFJ$ is an equilateral triangle,\\
thus $FJ=CF=CJ$, $\angle CJF=60^\circ$,\\
therefore $\angle FJE=120^\circ=\angle B$,\\
since point E is the $KZ$ point on the line segment $AF$,\\
it follows that $AE=4EF$,\\
since $\angle AEC=\angle AEF+\angle FEJ=\angle B+\angle BAE$, $\angle AEF=\angle B=120^\circ$,\\
it follows that $\angle FEJ=\angle BAE$,\\
therefore $\triangle ABE\sim \triangle EJF$,\\
thus $\frac{AB}{EJ}=\frac{BE}{FJ}=\frac{AE}{EF}=4$,\\
thus $EJ=2$, $BE=4FJ=4CJ$,\\
thus $BC=BE+EJ+CJ=4CJ+2+CJ=8$,\\
i.e., $CJ=\frac{6}{5}$,\\
therefore $CF=CJ=\frac{6}{5}$,\\
thus $DF=CD-CF=8-\frac{6}{5}=\boxed{\frac{34}{5}}$.


Question: As shown in the diagram, $O$ is a point on line segment $PB$. Taking $O$ as the center and the length of $OB$ as the radius, circle $\odot O$ intersects $PB$ at point A. Point C is on circle $\odot O$. Line $PC$ is drawn, satisfying $PC^{2}=PA\cdot PB$. $\triangle PAC\sim \triangle PCB$. If $AB=3PA$, find the value of $\frac{AC}{BC}$.

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

$\because \triangle PAC\sim \triangle PCB$,

$\therefore \frac{PA}{PC}=\frac{AC}{BC}$, $ \angle PCA=\angle ABC$

$\because OC=OA$, $OC=OB$

$\therefore \angle OCA=\angle CAO$, $\angle OCB=\angle OBC$

$\because \angle PCA=\angle ABC$, $\angle OCB=\angle OBC$

$\therefore \angle PCA=\angle OCB$

$\because AB$ is the diameter,

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

$\therefore \angle ACO+\angle OCB=90^\circ$,

$\therefore \angle ACO+\angle PCA=90^\circ$,

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

$\because AB=3PA$, $PB=PA+AB$,

$\therefore PB=4PA$,

$\because AB=2OC=2OA$,

$\therefore OA=OC=1.5PA$,

$\therefore PO=2.5PA$,

$\because PC^2+OC^2=PO^2$

$\therefore PC=\sqrt{PO^2−OC^2}=2PA$,

$\therefore \frac{AC}{BC}=\frac{PA}{PC}=\boxed{\frac{1}{2}}$


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\odot O$, $BD$ bisects $\angle ABC$, through point $D$ draw $DE\parallel AB$, intersecting $BC$ at point $E$, connect $AE$ and intersect $BD$ at point $F$. Given that $\angle AFD = \angle ADB + \angle CDE$, if $AB^2 = BF \cdot BD$ and $AD = 2$, find the length of $CB$.

Solution:
\textbf{Solution}: Given from equation (1), we have  $\angle AFD=180^\circ-3\alpha$,\\
$\therefore$ $\angle FAB=180^\circ-4\alpha$,\\
$\because$ it is about an inscribed quadrilateral,\\
$\therefore$ $\angle CDE+2\alpha +\alpha +2\alpha =180^\circ$, that is $\angle CDE=180^\circ-5\alpha$,\\
$\therefore$ $\angle CDF=\angle CDE+\angle EDB=\alpha +180^\circ-5\alpha =180^\circ-4\alpha$,\\
$\therefore$ $\angle BAF=\angle BDC=180^\circ-4\alpha$, and $\angle ABF=\angle FBC=\alpha$,\\
$\therefore$ $\triangle ABF\sim \triangle DBC$,\\
$\therefore$ $\frac{AB}{DC}=\frac{FB}{BC}$, thus $BC=\frac{FB\cdot DC}{AB}$,\\
$\therefore$ $\angle AFB=\angle DCB=180^\circ-(180^\circ-3\alpha )=3\alpha$,\\
$\because$ $\angle DAF+\angle FAB+\angle DCB=180^\circ$, that is $\angle DAF+(180^\circ-4\alpha )+3\alpha =180^\circ$,\\
$\therefore$ $\angle DAF=\angle FDE=\angle DBE=\alpha$,\\
$\therefore$ $\triangle ADE\sim \triangle BCD$,\\
$\therefore$ $\frac{AD}{BC}=\frac{AE}{BD}$, thus $BC=\frac{AD\cdot BD}{AE}$,\\
$\therefore$ $BC=\frac{FB\cdot DC}{AB}=\frac{AD\cdot BD}{AE}$,\\
and $\triangle ABE\cong \triangle BAD$ (by SAS),\\
$\because$ $AB^{2}=BF\cdot BD$, $AD=2$, and $AB=AE$, $AD=EC=DC=2$\\
$\therefore$ $CB=\boxed{4}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\odot O$, with $BD$ bisecting $\angle ABC$. Through point $D$, draw $DE\parallel AB$, intersecting $BC$ at point $E$. Connect $AE$ and intersect $BD$ at point $F$. Given $\angle AFD = \angle ADB + \angle CDE$, and if $CE = 2$, $AB = 8$, what is the length of $DE$?

Solution:
\textbf{Solution:} From equation (2), it is known that $\triangle ABE \cong \triangle BAD$ (SAS), $CE=2$, $AB=8$, \\
$\therefore BE=BC-CE=4-2=2$, \\
$\therefore DE=BE=\boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=\sqrt{5}, BC=2$, and a line is drawn from point $B$ perpendicular to $AC$, intersecting $AC$ at point $D$. Find the value of $\cot\angle ACB$.

Solution:
\textbf{Solution:} As shown in the figure, draw $AG \perp BC$ at point $G$,\\
since $AB=AC$,\\
then $CG= \frac{1}{2}BC=1$, and $AG \perp BC$,\\
in $\triangle AGC$, by the Pythagorean theorem, we have $AG=\sqrt{AC^{2}-CG^{2}}=\sqrt{5-1}=2$,\\
thus, $\cot\angle ACB=\frac{CG}{AG}=\boxed{\frac{1}{2}}$,


Question: As shown in the figure, in $\triangle ABC$, we have $AB=AC=\sqrt{5}$, and $BC=2$. A line is drawn from point $B$ such that $BD \perp AC$ with $D$ being the foot of the perpendicular. Point $E$ is a point on the extension of $BD$. Connect $CE$. When $\angle E = \angle A$, find the length of the segment $CE$.

Solution:
\textbf{Solution:} Since $S_{\triangle ABC}=\frac{1}{2}AG\cdot BC=\frac{1}{2}BD\cdot AC$,\\
it follows that $BD= \frac{4\sqrt{5}}{5}$,\\
and since $\cot\angle ACB=\frac{1}{2}$,\\
it further follows that $\frac{CD}{BD}=\frac{1}{2}$,\\
thus, $CD= \frac{2\sqrt{5}}{5}$,\\
and since $\angle BAC=\angle E$,\\
it can be concluded that $\triangle ADB\sim \triangle EDC$,\\
therefore, $\frac{EC}{AB}=\frac{CD}{BD}=\frac{1}{2}$,\\
hence, $EC=\frac{1}{2}AB=\boxed{\frac{\sqrt{5}}{2}}$.


Question: As shown in the figure, points $D$ and $E$ are located on sides $AB$ and $AC$ of $\triangle ABC$, respectively, with $AD=2$, $AC=6$, and $\frac{AE}{AB}=\frac{1}{3}$, $BC=\frac{15}{4}$. Given that $\triangle ADE\sim \triangle ACB$; what is the length of $DE$?

Solution:
\textbf{Solution:} Given that $\because \triangle ADE \sim \triangle ACB$,\\
$\therefore \frac{DE}{BC}=\frac{AD}{AC}=\frac{1}{3}$,\\
$\because BC = \frac{15}{4}$,\\
$\therefore DE = \frac{1}{3} \times \frac{15}{4} = \boxed{\frac{5}{4}}$.


Question: As shown in the diagram, in $\triangle ABC$, $AB=AC$, $AD \perp BC$ at $D$, draw $DE \perp AC$ at $E$, and let $F$ be the midpoint of $AB$. Connect $EF$ and intersect $AD$ at point $G$. It is given that $AD^{2}=AB\cdot AE$; if $AB=5$ and $AE=4$, find the value of $DG$.

Solution:
\textbf{Solution:} As shown in the figure, connect DF.\\
$\because$AB $= 5$, $\angle$ADB $= 90^\circ$, BF $=$ AF,\\
$\therefore$DF $= \frac{1}{2}$AB $= \frac{5}{2}$,\\
$\because$AB $=$ AC, AD $\perp$ BC,\\
$\therefore$BD $=$ DC,\\
$\therefore$DF $\parallel$ AC,\\
$\therefore$ $ \frac{DF}{AE} = \frac{DG}{AG} = \frac{\frac{5}{2}}{4} = \frac{5}{8}$,\\
$\therefore$ $ \frac{DG}{AD} = \frac{5}{13}$\\
$\because$AD$^{2}$ $=$ AB $\cdot$ AE.\\
$\therefore$ $ AD = \sqrt{AB \cdot AE} = \sqrt{5 \times 4} = 2\sqrt{5}$ \\
$\therefore$ $ DG = \frac{5}{13} \times 2\sqrt{5} = \boxed{\frac{10\sqrt{5}}{13}}$.


Question: As shown in the figure, in $ \triangle ABC$, $ AB=AC$, points E, F are on side BC, and $ \angle EAF=\angle C$. $ \triangle ABF\sim \triangle ECA$; if $ BF=8$ and $ CE=6$, find the length of AB.

Solution:
\textbf{Solution:} Since $\triangle ABF\sim \triangle ECA$, \\
we have $\frac{AB}{EC}=\frac{BF}{AC}$, \\
Since $BF=8$, $CE=6$, $AB=AC$, \\
it follows that $\frac{AB}{6}=\frac{8}{AB}$, that is $AB^{2}=6\times 8=48$, \\
Hence, $AB=\boxed{4\sqrt{3}}$ (negative value discarded).


Question: In rectangle $ABCD$, $AB=4$, $BC=5$, P is a moving point on ray $BC$, draw $PE\perp AP$, $PE$ intersects ray $DC$ at point E, ray $AE$ intersects ray $BC$ at point F, let $BP=x$, $CF=y$. When $\sin\angle APB=\frac{4}{5}$, what is the length of $CE$?

Solution:
\textbf{Solution:} In the right triangle $Rt\triangle APB$, where $AB=4$ and $\sin\angle APB=\frac{4}{5}$,
\[
\therefore \sin\angle APB=\frac{AB}{AP}=\frac{4}{5}, \text{ thus } \frac{4}{AP}=\frac{4}{5},
\]
\[
\therefore AP=5,
\]
\[
\therefore PB=\sqrt{AP^2-AB^2}=3,
\]
\[
\therefore PC=BC-PB=5-3=2,
\]
\[
\because PE\perp AP,
\]
\[
\therefore \angle CPE+\angle APB=90^\circ,
\]
\[
\because \angle BAP+\angle APB=90^\circ,
\]
\[
\therefore \angle CPE=\angle BAP,
\]
\[
\because \angle ECP=\angle B=90^\circ,
\]
\[
\therefore \triangle PCE\sim \triangle ABP,
\]
\[
\therefore \frac{CE}{BP}=\frac{PC}{AB},
\]
\[
\therefore CE=\frac{BP\cdot PC}{AB}=\frac{3\times 2}{4}=\boxed{\frac{3}{2}}
\]


Question: In rectangle $ABCD$, $AB=4$, $BC=5$, $P$ is a moving point on ray $BC$, draw $PE\perp AP$, $PE$ intersects ray $DC$ at point $E$, ray $AE$ intersects ray $BC$ at point $F$. Let $BP=x$, $CF=y$. As shown in the figure, when point $P$ is on side $BC$ (point $P$ does not coincide with points $B$ or $C$), what is the function relationship between $y$ and $x$?

Solution:
\textbf{Solution:} Given the construction, \\
$\because EC\parallel AB$, \\
$\therefore \triangle ECF\sim \triangle ABF$, \\
$\therefore \frac{FC}{FB}=\frac{EC}{AB}$, \\
then $CE=\frac{FC\cdot AB}{FB}=\frac{4y}{5+y}$, \\
$\because \triangle PCE\sim \triangle ABP$, \\
$\therefore \frac{CE}{BP}=\frac{PC}{AB}$, \\
then $CE=\frac{BP\cdot PC}{AB}=\frac{x(5-x)}{4}$, \\
$\therefore \frac{4y}{5+y}=\frac{x(5-x)}{4}$, \\
Upon rearrangement, we get: $y=\frac{25x-5x^2}{16+x^2-5x} \left(0<x<5\right)$, \\
Hence, the function of $y$ in terms of $x$ is $\boxed{y=\frac{25x-5x^2}{16+x^2-5x}}\left(0<x<5\right)$.


Question: As shown in the figure, in rectangle $ABCD$, where $AB=4$ and $BC=5$, $P$ is a moving point on ray $BC$. Construct $PE\perp AP$, with $PE$ intersecting ray $DC$ at point $E$, and ray $AE$ intersecting ray $BC$ at point $F$. Let $BP=x$ and $CF=y$. When $\frac{PE}{AP}=\frac{1}{2}$, what is the length of $CF$?

Solution:
\textbf{Solution:}

(1) When point $P$ is on segment $BC$, and $E$ is on segment $CD$,

$\because$ $\triangle PCE\sim \triangle ABP$,

$\therefore$ $\frac{PC}{AB}=\frac{CE}{BP}=\frac{PE}{AP}=\frac{1}{2}$,

$\therefore$ $PC=2$, $PB=3$,

$\therefore$ $CE=\frac{3}{2}$,

$\because$ $\triangle ECF\sim \triangle ABF$,

$\therefore$ $\frac{CF}{BF}=\frac{CE}{AB}$, that is $\frac{CF}{5+CF}=\frac{\frac{3}{2}}{4}$,

Solving, we get $CF=\boxed{3}$;

(2) When $P$ is to the right of $C$, and $E$ is on the extension of segment $DC$,

$\because$ $\angle EPC+\angle BPA=90^\circ$, $\angle BPA+\angle BAP=90^\circ$,

$\therefore$ $\angle EPC=\angle BAP$,

$\because$ $\angle B=\angle PCE$,

$\therefore$ $\triangle ABP\sim \triangle PCE$,

$\therefore$ $\frac{PE}{AP}=\frac{CE}{BP}=\frac{PC}{AB}=\frac{1}{2}$,

$\therefore$ $CP=2$,

$\therefore$ $CE=\frac{7}{2}$,

$\because$ $\angle AFB=\angle CFE$, $\angle B=\angle FCE$,

$\therefore$ $\triangle ABF\sim \triangle ECF$,

$\therefore$ $\frac{AB}{EC}=\frac{BF}{CF}$, that is $\frac{4}{\frac{7}{2}}=\frac{5-CF}{CF}$,

$\therefore$ $CF=\boxed{\frac{7}{3}}$,

In summary, the length of $CF$ could be \boxed{3} or \boxed{\frac{7}{3}}.


Question: As shown in the figure, in $\triangle ABC$, $\angle B = 90^\circ$, point $D$ is on line segment $AC$, and circle $\odot O$ with diameter $CD$ intersects $AB$ at point $E$. Line segment $CE$ is drawn, and $CE$ bisects $\angle ACB$. $AB$ is a tangent to circle $\odot O$; line segment $DE$ is drawn. If $BE=2$ and $BC=4$, what is the radius of circle $\odot O$?

Solution:
Solution: As shown in the diagram, connect $DE$, then $\angle DEC=90^\circ$,\\
$\therefore \angle DEC=\angle B$,\\
$\because \angle DCE=\angle ECB$,\\
$\therefore \triangle DCE \sim \triangle ECB$,\\
$\therefore \frac{CD}{CE}=\frac{CE}{BC}$,\\
$\because \angle B=90^\circ$, $BE=2$, $BC=4$,\\
in $Rt\triangle BCE$, $CE=\sqrt{BC^2+BE^2}=\sqrt{4^2+2^2}=2\sqrt{5}$,\\
$\therefore CD=\frac{(CE)^2}{BC}=\frac{(2\sqrt{5})^2}{4}=5$,\\
$\therefore$ The radius of $\odot O$ is $\boxed{\frac{5}{2}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $BD$ bisects $\angle ABC$, $\angle BDC = \angle A = 90^\circ$, $\cos\angle ABD = \frac{4}{5}$. What is the value of $\frac{AD}{CD}$?

Solution:
\textbf{Solution:}

Given that $BD$ bisects $\angle ABC$, \\
it follows that $\angle ABD = \angle CBD$, \\
since $\angle BDC = \angle A = 90^\circ$, \\
it implies that $\triangle ABD \sim \triangle DBC$. \\
Therefore, $\frac{AD}{CD} = \frac{AB}{BD}$. \\
In $\triangle ABD$, \\
since $\cos\angle ABD = \frac{AB}{BD} = \frac{4}{5}$, \\
it results in $\frac{AD}{CD} = \boxed{\frac{4}{5}}$;


Question: As shown in the figure, in quadrilateral $ABCD$, $BD$ bisects $\angle ABC$, $\angle BDC = \angle A = 90^\circ$, $\cos\angle ABD = \frac{4}{5}$. If $BC=25$, find the area of the quadrilateral $ABCD$.

Solution:
\textbf{Solution:} Since $\angle ABD=\angle CBD$,\\
we have $\cos\angle CBD=\frac{BD}{BC}=\frac{4}{5}$,\\
Since $BC=25$,\\
we get $BD=20$.\\
Thus, $CD=\sqrt{BC^{2}-BD^{2}}=15$.\\
Since $\cos\angle ABD=\frac{AB}{BD}=\frac{4}{5}$,\\
we get $AB=16$,\\
and thus $AD=\sqrt{BD^{2}-AB^{2}}=12$.\\
Therefore, the area of quadrilateral $ABCD$, $S_{\text{quadrilateral}ABCD}=S_{\triangle ABD}+S_{\triangle BCD}$\\
$=\frac{1}{2}AB\cdot AD+\frac{1}{2}BD\cdot CD$\\
$=\frac{1}{2}\times 16\times 12+\frac{1}{2}\times 20\times 15$\\
$=96+150$\\
$=\boxed{246}$.


Question: Given: As shown in the figure, in the right trapezoid $ABCD$, $AB\parallel CD$, $AB\perp BC$, $AD=CD$, $E$ is the midpoint of $AC$, extend line $BE$ to intersect line segment $AD$ at point $F$. $\triangle AEF\sim \triangle BAF$, if $CD=3$, $AB=5$, find the length of $DF$.

Solution:
\textbf{Solution:} Let $CD=3$ and $AB=5$,\\
Hence $AD=CD=3$,\\
Extend $BF$ to meet the extension of $CD$ at point G, as shown in the diagram:\\
In $\triangle ECG$ and $\triangle EAB$, we have $\left\{\begin{array}{l}
\angle ECG=\angle EAB \\
CE=AE \\
\angle CEG=\angle AEB
\end{array}\right.$,\\
Therefore, $\triangle ECG\cong \triangle EAB$ (ASA),\\
Hence $CG=AB=5$,\\
Thus $DG=CG-CD=2$,\\
Since $AB\parallel CD$,\\
$\triangle DFG\sim \triangle AFB$,\\
Hence $\frac{DF}{AF}=\frac{DG}{AB}$,\\
Thus $AB\cdot DF=DG\cdot AF$,\\
Hence $5DF=2\times (3-DF)$,\\
Therefore $DF=\boxed{\frac{6}{7}}$.


Question: As shown in the figure, in square $ABCD$, where $AB=6$, $E$ is a point on side $BC$ (point $E$ does not coincide with points $B$, $C$), and point $F$ is on the extension of $CD$, and $BE=DF$. Connect $EF$, intersecting $AD$ and $AC$ at points $M$ and $N$, respectively. Given that $MD=1$, what is the length of $BE$?

Solution:
\textbf{Solution:} Given that the quadrilateral $ABCD$ is a square, \\
$\therefore CD=BC=AB=6$, $AD\parallel BC$, \\
$\therefore \triangle FDM\sim \triangle FCE$, \\
$\therefore \frac{DM}{CE}=\frac{DF}{CF}$, \\
$\because BE=DF$, \\
$\therefore CE=BC-BE=6-BE$, $CF=CD+DF=6+BE$, \\
$\therefore \frac{1}{6-BE}=\frac{BE}{6+BE}$, \\
$\therefore BE=\boxed{2}$ or $BE=\boxed{3}$;


Question: As shown in the figure, in square $ABCD$, where $AB=6$, $E$ is a point on side $BC$ (point $E$ does not coincide with points $B$, $C$), and point $F$ lies on the extension of $CD$, with $BE=DF$. Connect $EF$, intersecting $AD$ and $AC$ at points $M$ and $N$, respectively. It's given that $EF^2=2EM\cdot FN$; when $\triangle AMN$ is an isosceles triangle, find the value of $S_{\triangle AMN}$.

Solution: \textbf{Solution:} As shown in Figure 2, connect $BD$, then $BD\perp AC$. When $AN=MN$, $\angle AMN=\angle CAD=45^\circ$, at this time $\angle ANNM=90^\circ$, i.e., $MN\perp AC$, $\therefore BD\parallel MN$, this situation does not exist. When $AM=MN$, $EF\perp AD$, this situation also does not exist. As shown in Figure 3, when $AM=AN$, construct $EG\perp AD$ at $G$, $\therefore \angle ANM=\angle AMN=\frac{180^\circ-\angle MAN}{2}=\frac{180^\circ-45^\circ}{2}=67.5^\circ$. $\because AD\parallel BC$, $\therefore \angle CEF=\angle AMN=67.5^\circ$. $\because \angle AEF=45^\circ$, $\therefore \angle AEB=180^\circ-45^\circ-67.5^\circ=67.5^\circ$, $\therefore \angle AEB=\angle CEF$. $\because \angle ABC=\angle ADF=90^\circ$, $\therefore \triangle ABE\sim \triangle FCE$, $\therefore \frac{AB}{CF}=\frac{BE}{CE}$. Let $BE=CF=a$, $\therefore CE=6-a$, $CF=6+a$, $\therefore \frac{6}{6+a}=\frac{a}{6-a}$, $\therefore a_1=6\sqrt{2}-6$, $a_2=-6\sqrt{2}-6$ (discard). $\because \angle AMN=\angle AME$, $\angle MAN=\angle AEM=45^\circ$, $\therefore \triangle AMN\sim \triangle AMA$, $\therefore \frac{AM}{AN}=\frac{EM}{AE}$, $\frac{S_{\triangle AMN}}{S_{\triangle AEM}}=\left(\frac{AM}{AE}\right)^2$. $\because AM=AN$, $\therefore AE=EM$, $\therefore GM=AG=BE$. $\because \left(\frac{BE}{AE}\right)^2=\frac{BE^2}{BE^2+AB^2}=\frac{3-2\sqrt{2}}{4-2\sqrt{2}}$, $\therefore S_{\triangle EAM}=\frac{1}{2}AM\cdot AB=\frac{1}{2}\times 6\times (12\sqrt{2}-12)=36\sqrt{2}-36$, $\therefore \frac{S_{\triangle AMN}}{36\sqrt{2}-36}=\frac{3-2\sqrt{2}}{4-2\sqrt{2}}$, $\therefore S_{\triangle AMN}=\boxed{27\sqrt{2}-36}$.


Question: As shown in the figure, within $\triangle ABC$, point D and point E are located on sides $AC$ and $AB$ respectively, $EC$ and $BD$ intersect at point O, and $\angle ABD = \angle ACE$, connect $DE$. $\triangle EOD \sim \triangle BOC$; If $\frac{DE}{BC} = \frac{2}{3}$, find the value of $\frac{AE}{AC}$.

Solution:
\textbf{Solution:} From (1), we have $ \triangle ADE\sim \triangle ABC$, \\
$\because \frac{DE}{BC}=\frac{2}{3}$, \\
$\therefore \frac{AE}{AC}=\frac{DE}{BC}=\boxed{\frac{2}{3}}$.


Question: As shown in the figure, it is known that in rhombus $ABCD$, $AB=5$, $\cos B=\frac{3}{5}$. Points E and F are on sides $BC$ and $CD$, respectively. The extension of line $AF$ intersects the extension of line $BC$ at point G, and $\angle EAF=\frac{1}{2}\angle BAD$. It is given that $AE^{2}=EC\cdot EG$; if point F is the midpoint of side $CD$, what is the value of $S_{\triangle ABE}$?

Solution:
\textbf{Solution:} Draw $AH\perp BC$ through point A, with H as the foot of the perpendicular.\\
Since quadrilateral $ABCD$ is a rhombus, $\therefore$ $AD=AB=BC=CD=5$. $\therefore$ $\frac{AD}{CG}=\frac{DF}{CF}$.\\
Since point F is the midpoint of side $CD$, $\therefore$ $DF=CF$. $\therefore$ $CG=AD=5$.\\
Let $BE=x$, then $EC=5-x$.\\
$\therefore$ $EG=EC+CG=5-x+5=10-x$.\\
Since $AE^2=EC\cdot EG$, $AE^2=(5-x)(10-x)$.\\
In $\triangle ABH$ with a right angle, $\cos B=\frac{BH}{AB}=\frac{3}{5}$,\\
and since $AB=5$, $\therefore$ $BH=3$. $\therefore$ $AH=4$, $EH=3-x$.\\
In $\triangle AEH$ with a right angle, $AE^2=AH^2+EH^2$, $\therefore$ $AE^2=(3-x)^2+4^2$.\\
$\therefore$ $(5-x)(10-x)=(3-x)^2+4^2$. Solving this, we get $x=\frac{25}{9}$. $\therefore$ the area of $\triangle ABE$ is $\frac{1}{2}BE\cdot AH=\frac{1}{2}\times \frac{25}{9}\times 4=\boxed{\frac{50}{9}}$.


Question: As shown in the figure, it is known that in rhombus $ABCD$, $AB=5$ and $\cos B=\frac{3}{5}$. Points E and F are located on sides $BC$ and $CD$ respectively, and the extension line of $AF$ intersects the extension line of $BC$ at point G, with $\angle EAF=\frac{1}{2}\angle BAD$. The extensions of $AE$ and $DC$ intersect at point H. Lines $GH$ and $AC$ are connected. If $\triangle AGH$ is similar to $\triangle ABC$, what is the length of segment $BE$?

Solution:
\textbf{Solution:} Let us denote:

Since $\angle BAC = \angle EAF$,

Therefore, when $\frac{AH}{AG}=\frac{AB}{AC}$ or $\frac{AH}{AG}=\frac{AC}{AB}$, $\triangle AGH$ is similar to $\triangle ABC$.

(i) When $\frac{AH}{AG}=\frac{AB}{AC}$,

Since $\angle BAC=\frac{1}{2}\angle BAD$, and $\angle EAF=\frac{1}{2}\angle BAD$,

Therefore, $\angle BAC=\angle EAF$. Therefore, $\angle BAE=\angle CAG$.

Since $AB\parallel CD$, therefore $\angle BAE=\angle AHC$. Therefore, $\angle CAG=\angle AHC$.

Also, since $\angle EAC=\angle AGC$, therefore, $\triangle AHC \sim \triangle GAC$. Therefore, $\frac{AH}{AG}=\frac{CH}{AC}$.

Also, $\frac{AH}{AG}=\frac{AB}{AC}$, therefore, $CH=AB$.

Since $AB\parallel CD$, therefore $\frac{BE}{EC}=\frac{AB}{CH}$. Therefore, $BE=EC=\frac{5}{2}$.

(ii) When $\frac{AH}{AG}=\frac{AC}{AB}$, similarly, it can be derived that $\triangle AHC \sim \triangle GAC$.

Therefore, $\frac{AH}{AG}=\frac{AC}{CG}$. Also, since $\frac{AH}{AG}=\frac{AC}{AB}$, therefore, $CG=AB$. Therefore, $BE=\boxed{\frac{25}{9}}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $\angle ACB=90^\circ$, point D is on side $AC$, connecting $BD$, and constructing an isosceles right triangle $BDE$ (point E is on the right side of line $BD$) with $BD$ as the hypotenuse, and connecting $CE$. $\triangle ABD \sim \triangle CBE$; If $BC=12$ and $CD=5$, find the length of segment $CE$.

Solution:
\textbf{Solution:} Let us cut $CH=CB$ on the side $CA$, and connect $BH$.\\
Similarly, it can be proved: $\triangle HBD \sim \triangle CBE$.\\
$\therefore \frac{HD}{CE}=\frac{BD}{BE}$.\\
$\because BC=12$, $\therefore CH=12$.\\
Given that $CD=5$, $\therefore HD=7$. Since $\triangle BDE$ is an isosceles triangle, $\therefore \frac{BD}{BE}=\sqrt{2}$.\\
$\therefore CE=\boxed{\frac{7}{2}\sqrt{2}}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, point C is on the extension line of AB, AD bisects $\angle CAE$ and intersects $\odot O$ at point D, and AE $\perp$ CD at point E. Line CE is a tangent to $\odot O$. If BC = 3 and CD = 3$\sqrt{2}$, find the length of radius OB and line segment AE.

Solution:
\textbf{Solution:} Let $OD=OB=x$. In $\triangle COD$, $OD^2+CD^2=OC^2$, \\
$\therefore x^2+(3\sqrt{2})^2=(x+3)^2$, \\
$\therefore x=\boxed{1.5}$, \\
$\therefore OB=1.5$, \\
$\because OD\parallel AE$, \\
$\therefore \frac{OD}{AE}=\frac{OC}{AC}$, \\
$\therefore \frac{1.5}{AE}=\frac{4.5}{6}$, \\
$\therefore AE=\boxed{2}$.


Question: As shown in the diagram, within the Cartesian coordinate system, the line AB passing through point A (4, 0) intersects with the y-axis at point B (0, 4). The parabola $y=-x^2+bx+c$ that passes through the origin O intersects the line AB at points A and C. The vertex of the parabola is D. What is the expression of the parabola $y=-x^2+bx+c$?

Solution:
Solution: From the problem statement, we have
\[
\left\{
\begin{array}{l}
c=0\\
-16+4b+c=0
\end{array}
\right.
\]
Solving this, we get
\[
\left\{
\begin{array}{l}
b=4\\
c=0
\end{array}
\right.
\]
$\therefore$ The quadratic function is: \boxed{y=-x^{2}+4x}.


Question: As shown in the figure, in the Cartesian coordinate system, the line $AB$ passing through point $A(4,0)$ intersects the y-axis at point $B(0,4)$. The parabola $y=-x^2+bx+c$ passing through the origin $O$ intersects the line $AB$ at points $A$ and $C$, with the vertex of the parabola being $D$. $M$ is a point on segment $AB$, and $N$ is a point on the parabola. When $MN$ is parallel to the y-axis and $MN=2$, what are the coordinates of point $M$?

Solution:
\textbf{Solution:} As shown in the diagram,\\
$\because$ point A (4, 0), point B (0, 4),\\
$\therefore$ let the equation of line AB be $y=kx+4$,\\
$\therefore$ $4k+4=0$, solving it: $k=-1$,\\
$\therefore$ the equation of line AB is $y=-x+4$;\\
let point N ($m$, $-m^{2}+4m$), point M ($m$, $-m+4$),\\
$\because$ point M is on the line segment AB,\\
$\therefore$ $0<m<4$,\\
$MN=|-m^{2}+4m-(-m+4)|=|-m^{2}+5m-4|=2$,\\
$\therefore$ $-m^{2}+5m-4=\pm2$,\\
when $-m^{2}+5m-4=2$, solving it: $m_{1}=3$, $m_{2}=2$,\\
when $m=3$ then $-m+4=1$;\\
when $m=2$ then $-m+4=2$;\\
$\therefore$ the coordinates of point M are $\boxed{(2,2)}$ or $\boxed{(3,1)}$;\\
when $-m^{2}+5m-4=-2$, solving it:\\
$m_{3}=\frac{5-\sqrt{17}}{2}$, $m_{4}=\frac{5+\sqrt{17}}{2}>4$ (discard),\\
$\therefore$ $-m+4=-\frac{5-\sqrt{17}}{2}+4=\frac{3+\sqrt{17}}{2}$,\\
$\therefore$ the coordinates of point M are $\boxed{\left(\frac{5-\sqrt{17}}{2}, \frac{3+\sqrt{17}}{2}\right)}$.\\
There are three possible coordinates for point M that satisfy the conditions: $\boxed{\left(\frac{5-\sqrt{17}}{2}, \frac{3+\sqrt{17}}{2}\right)}$ or $\boxed{(2,2)}$ or $\boxed{(3,1)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $AB$ passing through point $A(4,0)$ intersects the $y$-axis at point $B(0,4)$. The parabola $y=-x^2+bx+c$ that goes through the origin $O$ intersects line $AB$ at points $A$ and $C$, with the vertex of the parabola being $D$. Let $P$ be a moving point on the parabola and $Q$ a point in the Cartesian coordinate system. Is there a quadrilateral with vertices $A$, $C$, $P$, $Q$ that is a rectangle? If it exists, write down the coordinates of point $Q$ directly; if not, please explain why.

Solution:
\textbf{Solution:}

Given: When AC is the side of a rectangle, \\
$\because -x^2+4x=-x+4$ Solve for: $x_1=4$, $x_2=1$, \\
When $x=1$, we have $y=-1+4=3$, \\
$\therefore$ Point C is at $(1,3)$, \\
$\because y=-x^2+4x=-(x-2)^2+4$, \\
$\therefore$ Point D is at $(2,4)$ when $x=2$ and $y=-2+4=2$, \\
$\therefore$ Point R is at $(2,2)$. Lines perpendicular to AB through points C and A intersect the parabola at points $P_1$ and $P_2$, respectively, \\
$\because$C is at $(1,3)$ and D is at $(2,4)$, \\
$\therefore CD^2=(1-2)^2+(4-3)^2=2$, $CR^2=1$, and $RD=2$, \\
$\therefore CD^2+CR^2=DR^2$, \\
$\therefore \angle RCD=90^\circ$, \\
$\therefore$ Point $P_1$ coincides with point D, \\
When $CP_1\parallel AQ_1$ and $CP_1=AQ_1$, the quadrilateral $ACP_1Q_1$ is a rectangle, \\
$\because$ Moving C$(1,3)$ 1 unit to the right and 1 unit up yields $P_1(2,4)$, \\
$\therefore$ Moving A$(4,0)$ 1 unit to the right and 1 unit up yields $Q_1(5,1)$, \\
Now, the equation of line $P_1C$ is: $y=x+2$, \\
$\because$ Line $P_2A$ is parallel to $P_1C$ and passes through A$(4,0)$, \\
$\therefore$ The equation of line $P_2A$ is: $y=x-4$, \\
$\because$ Point $P_2$ is at the intersection of the line $y=x-4$ and the parabola $y=-x^2+4x$, \\
$\therefore -x^2+4x=x-4$, solving for: $x_1=-1$, $x_2=4$ (discard), \\
$\therefore P_2$ lies at $(-1,-5)$, \\
When $AC\parallel P_2Q_2$, the quadrilateral $ACQ_2P_2$ is a rectangle, \\
$\because$ Moving A$(4,0)$ 3 units to the left and 3 units up yields C$(1,3)$, \\
$\therefore$ Moving $P_2(-1,-5)$ 3 units to the left and 3 units up yields $Q_2(-4,-2)$; \\
When AC is the diagonal of the rectangle, \\
Let $P_3(m, -m^2+4m)$ \\
When $\angle AP_3C=90^\circ$, draw $P_3H\perp x$-axis through $P_3$, and $CK\perp P_3H$ through C, \\
$\therefore \angle P_3KC=\angle AHP_3=90^\circ$, $\angle P_3CK=\angle AP_3H$, \\
$\therefore \triangle P_3CK\sim \triangle AP_3H$, \\
$\therefore \frac{P_3K}{CK}=\frac{AH}{P_3H}$ \\
$\therefore \frac{-m^2+4m-3}{m-1}=\frac{4-m}{-m^2+4m}$, \\
$\because$ Point P does not coincide with points A or C, \\
$\therefore m\neq 1$ and $m\neq 4$, \\
$\therefore -m^2-3m+1=0$, \\
$\therefore m=\frac{3\pm \sqrt{5}}{2}$, \\
$\therefore$ As illustrated, the conditions are satisfied. \boxed{Q_1(5,1)}, \boxed{Q_2(-4,-2)}.


Question: As shown in the figure, in rectangle $ABCD$, $E$ is a point on $AD$, connecting $AC$ and $BE$ intersecting at point $F$, and $FG \perp CD$ at $G$. If $AE=DE=3$, find the length of $FG$.

Solution: \textbf{Solution:} Given that quadrilateral ABCD is a rectangle, with AE=DE=3, it follows that AD$\parallel$BC, and BC=AD=6. Hence $\triangle$AEF$\sim$$\triangle$CBF, implying that $\frac{CF}{AF}=\frac{BC}{AE}=2$. Since FG$\parallel$AD, $\triangle$CFG$\sim$$\triangle$CAD, leading to $\frac{FG}{AD}=\frac{CF}{AC}=\frac{2}{3}$, thus FG=\boxed{4}.


Question: As illustrated, quadrilateral $ABCD$ is inscribed in circle $\odot O$, with $AB$ being the diameter of circle $\odot O$. Let $DE \perp AB$ at point $E$, intersecting $AC$ at point $F$. It is given that arc $\overset{\frown}{BC} = \overset{\frown}{AD}$. We have $\triangle ABC \sim \triangle DAE$. When $AD = \sqrt{5}$ and $BE = 4$, what is the length of $CD$?

Solution:
\textbf{Solution:} Draw $CH\perp AB$ at point H.\\
Since $\triangle ABC\sim \triangle DAE$,\\
it follows that $\frac{AE}{BC}=\frac{AD}{AB}$, $\angle BAC=\angle ADE$,\\
hence $\frac{AE}{\sqrt{5}}=\frac{\sqrt{5}}{4+AE}$,\\
thus $AE=1$ (negative values are discarded),\\
therefore $AB=AE+BE=5$,\\
$DE=\sqrt{AD^{2}-AE^{2}}=\sqrt{5-1}=2$,\\
therefore $AC=\sqrt{AB^{2}-BC^{2}}=\sqrt{25-5}=2\sqrt{5}$,\\
since $\overset{\frown}{BC}=\overset{\frown}{AD}$,\\
it follows that $\angle ACD=\angle BAC$, hence $\angle ADE=\angle ACD$,\\
also, since $\angle DAF=\angle DAC$,\\
it follows that $\triangle DAF\sim \triangle CAD$,\\
thus $\frac{AD}{AC}=\frac{AF}{AD}=\frac{DF}{DC}$,\\
hence $AF=\frac{\sqrt{5}}{2}$, $DC=2DF$,\\
thus $EF=\sqrt{AF^{2}-AE^{2}}=\frac{1}{2}$,\\
therefore $DF=\frac{3}{2}$,\\
therefore $CD=2DF=\boxed{3}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, $CD \perp AB$ with the foot of the perpendicular being $D$. $\triangle ABC \sim \triangle ACD$. Given that $AC = 4$ and $BC = 3$, what is the length of $AD$?

Solution: \textbf{Solution:} Given that $\angle ACB=90^\circ$, $AC=4$, and $BC=3$, it follows that $AB=5$. Since $\triangle ABC \sim \triangle ACD$, we have $\frac{AC}{AD}=\frac{AB}{AC}$. Therefore, $\frac{4}{AD}=\frac{5}{4}$, which implies that $AD=\boxed{\frac{16}{5}}$.


Question: As shown in the figure, point $E$ is a point on the extension of side $BA$ of rectangle $ABCD$. Lines $ED$ and $EC$ are drawn, and $EC$ intersects $AD$ at point $G$. Construct $CF\parallel ED$ intersecting $AB$ at point $F$, and $DC = DE$. Quadrilateral $CDEF$ is a rhombus; if $BC = 3$ and $CD = 5$, what is the length of $AE$?

Solution: \textbf{Solution:} Given that $BC=3$, $CD=5$, it follows that $AD=3$, $ED=5$. Therefore, $AE= \sqrt{DE^{2}-AD^{2}}=\sqrt{5^{2}-3^{2}}=\boxed{4}$.


Question: As shown in the figure, point $E$ is a point on the extension line of side $BA$ of rectangle $ABCD$. Connect $ED$ and $EC$, where $EC$ intersects $AD$ at point $G$. Construct $CF \parallel ED$ intersecting $AB$ at point $F$, with $DC = DE$. Under the condition of (2), what is the length of $AG$?

Solution:
\textbf{Solution:} Since $AD \parallel BC$,\\
$\therefore \frac{AG}{BC} = \frac{AE}{EB}$,\\
$\therefore \frac{AG}{3} = \frac{4}{4+5}$,\\
$\therefore AG = \boxed{\frac{4}{3}}$.


Question: As shown in Figure 1, it is known that the radius of circle $O$ is 5, and the chord $AB=4\sqrt{5}$. Points C and D are on the major arc $\overset{\frown}{BCA}$ (points B, C, and D are arranged clockwise), $\overset{\frown}{CD}=\overset{\frown}{AB}$. Point F is a moving point on $\overset{\frown}{CD}$, connect CF and extend it to intersect the extension of chord AD at point E. $BC\parallel AD$. As shown in Figure 2, connect $AC$, $BF$, $AF$. When $\angle BFA$ is equal to one of the interior angles of $\triangle ACE$, and the length of $BF$ can be determined, find all possible lengths of $BF$ that satisfy the condition.

Solution:
\textbf{Solution}: In figure (2), \\
$\because \overset{\frown}{CD}=\overset{\frown}{AB}$, \\
$\therefore \angle BFA=\angle CAE$, $AB=CD$, \\
Thus, regardless of the value of $BF$, we always have $\angle BFA=\angle CAE$, which is inconsistent with the problem statement; \\
When $\angle BFA=\angle ACE$, $\because \angle ACE=\angle ABF$ \\
$\therefore \angle BFA=\angle ABF$, \\
$\therefore AB=AF$, thus $\triangle BAF$ is an isosceles triangle, \\
As shown in the figure, draw $AH\perp BF$ at H through A, then $BH=HF=\frac{1}{2}BF$, \\
Extend $AH$ to meet $\odot O$ at P, connect $BP$, then $AP$ is the diameter, \\
$\therefore \angle ABP=90^\circ$, and since $AB=4\sqrt{5}$, $AP=2AO=10$, \\
$\therefore BP=\sqrt{AP^{2}-AB^{2}}=2\sqrt{5}$, \\
From $S_{\triangle ABP}=\frac{1}{2}AP\cdot BH=\frac{1}{2}AB\cdot BP$, we get: $BH=\frac{AB\cdot BP}{AP}=\frac{4\sqrt{5}\times 2\sqrt{5}}{10}=4$, \\
$\therefore BF=2BH=\boxed{8}$; \\
When $\angle BFA=\angle E$, $\because \angle AFC=\angle BFA+\angle BFC=\angle E+\angle FAE$, \\
$\therefore \angle BFC=\angle FAE$ \\
$\therefore \overset{\frown}{BC}=\overset{\frown}{DF}$, thus $\overset{\frown}{BF}=\overset{\frown}{CD}$, \\
$\therefore BF=CD=AB=4\sqrt{5}$, \\
To sum up, the length of $BF$ that satisfies the condition is either $8$ or $\boxed{4\sqrt{5}}$;


Question: As shown in Figure 1, it is known that the radius of $\odot O$ is 5, and the chord $AB = 4\sqrt{5}$. Points C and D are on the major arc $\overset{\frown}{BCA}$ (points B, C, D are arranged clockwise), $\overset{\frown}{CD} = \overset{\frown}{AB}$. Point F is a moving point on $\overset{\frown}{CD}$. Connect CF and extend it to intersect the extension line of chord AD at point E. $BC \parallel AD$. As shown in Figure 3, when $\angle B = 90^\circ$ and $CE = 5\sqrt{5}$, connect AC, OE, and let OE intersect $\odot O$ at point K. Calculate the ratio of the area of $\triangle KDE$ to the area of $\triangle EOC$.

Solution:
\textbf{Solution:} In Figure 3, connect $CD$,\\
$\because$ $\angle B=90^\circ$,\\
$\therefore$ $AC$ is the diameter of $\odot O$,\\
$\therefore$ $\angle ADC=90^\circ$, $AC=2AO=10$,\\
In $Rt\triangle CDE$, $CE=5\sqrt{5}$, $CD=AB=4\sqrt{5}$,\\
$\therefore$ $DE=\sqrt{CE^{2}-CD^{2}}=3\sqrt{5}$,\\
In $Rt\triangle ADC$, $AD=\sqrt{AC^{2}-CD^{2}}=2\sqrt{5}$,\\
$\therefore$ $AE=AD+DE=5\sqrt{5}$,\\
$\therefore$ $AE=CE$, thus $\triangle AEC$ is an isosceles triangle,\\
$\because$ $AO=OC$,\\
$\therefore$ $OE\perp AC$,\\
In $Rt\triangle AOE$, $OE=\sqrt{AE^{2}-OA^{2}}=\sqrt{(5\sqrt{5})^{2}-5^{2}}=10$,\\
$\therefore$ $KE=OE-OK=5$,\\
Draw $KM\perp DE$ at M through point K, then $\angle KME=\angle AOE=90^\circ$, and also $\angle KEM=\angle AEO$,\\
$\therefore$ $\triangle KME\sim \triangle AOE$,\\
$\therefore$ $\frac{KM}{AO}=\frac{KE}{AE}$, that is $\frac{KM}{5}=\frac{5}{5\sqrt{5}}$,\\
Solving for $KM$, we get: $KM=\sqrt{5}$,\\
$\therefore$ $\frac{S_{\triangle KDE}}{S_{\triangle EOC}}=\frac{\frac{1}{2}DE\cdot KM}{\frac{1}{2}OC\cdot OE}=\frac{3\sqrt{5}\times \sqrt{5}}{5\times 10}=\boxed{\frac{3}{10}}$.


Question: As shown in the figure, the parabola $y = ax^{2} + bx + c \left(a \ne 0\right)$ intersects the y-axis at point $C(0, 4)$ and intersects the x-axis at points A and B, where the coordinates of point A are $(-2, 0)$. The axis of symmetry of the parabola, $x = 1$, intersects the parabola at point D and intersects line $BC$ at point E. What is the equation of the parabola?

Solution:
\textbf{Solution:} Given the problem, we know the parabola $y=ax^2+bx+c$ passes through points $C(0, 4)$ and $A(-2, 0)$, and its axis of symmetry is the line $x=1$,
\[
\therefore \left\{\begin{array}{l}
c=4 \\
4a-2b+c=0 \\
-\frac{b}{2a}=1
\end{array}\right.,
\]
solving this system yields
\[
\left\{\begin{array}{l}
a=-\frac{1}{2} \\
b=1 \\
c=4
\end{array}\right.,
\]
\[
\therefore \text{the equation of the parabola is } y=-\frac{1}{2}x^2+x+4=\boxed{-\frac{1}{2}x^2+x+4}.
\]


Question: As shown in the figure, the parabola $y=ax^2+bx+c\ (a\ne 0)$ intersects the y-axis at point $C(0, 4)$ and intersects the $x$-axis at points A and B, where the coordinates of point A are $(-2, 0)$. The axis of symmetry of the parabola $x=1$ intersects the parabola at point D and intersects the line $BC$ at point E. Suppose point F is a moving point on the parabola above the line $BC$. Determine whether there exists a point F that maximizes the area of the quadrilateral $ABFC$. If such a point exists, find the coordinates of point F and the maximum area; if not, please explain why.

Solution:
\textbf{Solution:} Existence is proven as follows: As shown in Figure 1, draw $FH \perp x$-axis at point $H$ intersecting $BC$ at point $G$, and let $F(x, -\frac{1}{2}x^2+x+4)$,\\
$\because$ Point $B$ is symmetric to point $A(-2, 0)$ with respect to the line $x=1$,\\
$\therefore B(4, 0)$, and $AB=6$,\\
Let the equation of line $BC$ be $y=kx+4$, then $4k+4=0$,\\
Solving gives $k=-1$,\\
$\therefore y=-x+4$,\\
$\therefore G(x, -x+4)$,\\
$\therefore FG=-\frac{1}{2}x^2+2x$,\\
$\therefore S_{\triangle FBC}=\frac{1}{2} \times 4(-\frac{1}{2}x^2+2x)=-x^2+4x$,\\
$\therefore S_{\text{quadrilateral} ABFC}=-x^2+4x+\frac{1}{2} \times 6 \times 4=-(x-2)^2+16$,\\
$\therefore$ When $x=2$, the maximum area of $\text{quadrilateral } ABFC=\boxed{16}$, and $F(2, 4)$,\\
$\therefore$ The coordinates of point $F$ are $(2, 4)$, and the maximum area of quadrilateral $ABFC$ is $\boxed{16}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+c\ (a\neq 0)$ intersects the y-axis at point $C(0, 4)$ and intersects the x-axis at points A and B, with the coordinates of point A being $(-2, 0)$. The axis of symmetry of the parabola, $x=1$, intersects the parabola at point D and intersects the line $BC$ at point E. Investigate whether there exists a point P on the axis of symmetry such that the triangle formed by points P, C, and A is an isosceles triangle. If such a point P exists, find the coordinates of all such points P that meet the condition. If it does not exist, please explain why.

Solution:
\textbf{Solution:} Existence proved, let $P(1, m)$,

$\because A(-2, 0)$, $C(0, 4)$,

$\therefore AC^2=2^2+4^2=20$, $PA^2=m^2+(1+2)^2=m^2+9$, $PC^2=(m-4)^2+1^2=m^2-8m+17$,

When $PA=PC$, then $m^2+9=m^2-8m+17$,

Solving gives $m=1$,

$\therefore P_1(1, 1)$;

When $PC=AC$, then $m^2-8m+17=20$,

Solving gives $m_1=4-\sqrt{19}$, $m_2=4+\sqrt{19}$,

$\therefore P_2(1, 4-\sqrt{19})$, $P_3(1, 4+\sqrt{19})$;

When $PA=AC$, then $m^2+9=20$,

Solving gives $m_1=\sqrt{11}$, $m_2=-\sqrt{11}$,

$\therefore P_4(1, \sqrt{11})$, $P_5(1, -\sqrt{11})$,

Therefore, summarizing the above, the coordinates of point $P$ are \boxed{(1, 1)}, \boxed{(1, 4-\sqrt{19})}, \boxed{(1, 4+\sqrt{19})}, \boxed{(1, \sqrt{11})}, or \boxed{(1, -\sqrt{11})}.


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=-\frac{1}{4}x^{2}+bx+c$ intersects with the line at points $B(2, 0)$ and $K(-12, -14)$, and intersects with the y-axis at point C. Find the equation of the parabola and the coordinates of its vertex.

Solution:
\textbf{Solution:} Given that the parabola $y=-\frac{1}{4}x^2+bx+c$ passes through points $B(2, 0)$ and $K(-12, -14)$,\\
then $\begin{cases}-\frac{1}{4}\cdot 4+2b+c=0\\ -\frac{1}{4}\cdot 144-12b+c=-14\end{cases}$,\\
Solving this, we get $\begin{cases}b=-\frac{3}{2}\\ c=4\end{cases}$,\\
Therefore, the equation of the parabola is $y=-\frac{1}{4}x^2-\frac{3}{2}x+4$,\\
Since $y=-\frac{1}{4}x^2-\frac{3}{2}x+4=-\frac{1}{4}(x+3)^2+\frac{25}{4}$,\\
Therefore, the coordinates of the vertex are $\boxed{\left(-3, \frac{25}{4}\right)}$;


Question: As shown in the diagram, in the Cartesian coordinate system, the parabola $y = -\frac{1}{4}x^2 + bx + c$ intersects the line $KB$ at point $B(2, 0)$ and point $K(-12, -14)$, and intersects the y-axis at point $C$. Connect $BC$. Let point $P$ be a moving point on the parabola above line segment $BK$. Draw $PZ \parallel x$-axis intersecting $CB$ at point $Z$, and draw $PQ \parallel CB$ intersecting line $KB$ at point $Q$. Find the maximum value of $\frac{6}{5}\sqrt{5}PQ + PZ$ and the coordinates of point $P$ at this time?

Solution:
textbf{Solution:} Let $P\left(t, -\frac{1}{4}t^{2}-\frac{3}{2}t+4\right)\left(-12<t<2\right)$,

draw $PM\parallel y$ axis crossing $BK$ at $M$, and through $Q$ draw $QN\parallel x$ axis crossing $PM$ at $N$,

since $B(2, 0)$, $K(-12, -14)$,

therefore, the equation of line $BK$ is $y=x-2$,

since $PM\parallel y$ axis,

therefore, $P$ and $M$ have the same $x$-coordinate,

therefore, $M\left(t, t-2\right)$,

thus, $PM=-\frac{1}{4}t^{2}-\frac{3}{2}t+4-(t-2)=-\frac{1}{4}t^{2}-\frac{5}{2}t+6$,

it's easy to find $C\left(0, 4\right)$,

since $B(2, 0)$,

therefore, the equation of line $BC$ is $y=-2x+4$,

since $QN\parallel x$ axis,

therefore, $P$ and $Z$ have the same $y$-coordinate,

therefore, the $y$-coordinate of $Z$ is $-\frac{1}{4}t^{2}-\frac{3}{2}t+4$,

thus, $-\frac{1}{4}t^{2}-\frac{3}{2}t+4=-2x+4$,

solving for $x$ yields $x=\frac{1}{8}t^{2}+\frac{3}{4}t$,

thus, $Z\left(\frac{1}{8}t^{2}+\frac{3}{4}t, -\frac{1}{4}t^{2}-\frac{3}{2}t+4\right)$,

therefore, $PZ=\frac{1}{8}t^{2}+\frac{3}{4}t-t=\frac{1}{8}t^{2}-\frac{1}{4}t$,

because $PZ\parallel x$ axis, $BC\parallel PQ$, $PZ\parallel x$ axis $\parallel NQ$,

therefore, $\angle OBC=\angle PZC$, $\angle PZC=\angle ZPQ$, $\angle ZPQ=\angle PQN$,

therefore, $\angle OBC=\angle PQN$,

also $\angle BOC=\angle QNP=90^\circ$,

therefore, $\triangle BOC\sim \triangle QNP$,

therefore, $\frac{BO}{NQ}=\frac{BC}{PQ}=\frac{OC}{PN}$,

since $BO=2$, $CO=4$,

therefore, $BC=\sqrt{BO^{2}+CO^{2}}=2\sqrt{5}$,

therefore, $\frac{2}{NQ}=\frac{2\sqrt{5}}{PQ}=\frac{4}{PN}$,

therefore, $NQ=\frac{\sqrt{5}}{5}PQ$, $PN=2QN$,

since $QN\parallel x$ axis, the acute angle formed by $y=x-2$ and the $x$ axis is $45^\circ$

therefore, $\angle NQM=45^\circ$,

therefore, $\triangle MNQ$ is an isosceles right triangle,

therefore, $MN=NQ$,

therefore, $NQ=\frac{1}{3}PM$,

therefore, $\frac{1}{3}PM=\frac{\sqrt{5}}{5}PQ$,

therefore, $PQ=\frac{\sqrt{5}}{3}PM$,

therefore, $\frac{6}{5}\sqrt{5}PQ+PZ=\frac{6}{5}\sqrt{5} \times \boxed{\frac{\sqrt{5}}{3}PM} + \boxed{\frac{1}{8}t^{2}-\frac{1}{4}t}$,

therefore, $\frac{6}{5}\sqrt{5}PQ+PZ=\boxed{2PM} + \boxed{\frac{1}{8}t^{2}-\frac{1}{4}t}$,

therefore, $\frac{6}{5}\sqrt{5}PQ+PZ=\boxed{2\left(-\frac{1}{4}t^{2}-\frac{5}{2}t+6\right)} + \boxed{\frac{1}{8}t^{2}-\frac{1}{4}t}$,

therefore, $\frac{6}{5}\sqrt{5}PQ+PZ=\boxed{-\frac{1}{8}t^{2}-\frac{1}{4}t+12}$,

therefore, $\frac{6}{5}\sqrt{5}PQ+PZ$ reaches its maximum value when $t=-2$,

therefore, the maximum value of $\frac{6}{5}\sqrt{5}PQ+PZ$ is $\boxed{13}$,

therefore, the coordinates of point $P$ at this time are $\boxed{(-2, 6)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=-\frac{1}{4}x^2+bx+c$ intersects the line $KB$ at point $B(2, 0)$ and $K(-12, -14)$, and intersects the y-axis at point C. Under the given conditions, the parabola is translated downwards by $\frac{3}{8}$ units and to the right by 3 units, so that point P corresponds to point $P^{\prime}$. Point S is a point on the axis of symmetry of the new parabola. Is there a point N in the plane such that the quadrilateral with vertices $P^{\prime}$, S, A, N is a rhombus? If it exists, please write down the coordinates of point N directly; if it does not exist, please explain why.

Solution: \textbf{Solution:} Exist, $\boxed{N\left(-12, \frac{949}{80}\right)}$ or $\boxed{N\left(-4, -\frac{15}{8}\right)}$.


Question: As shown in the figure, it is known that the line $y=-x+m+1$ and the hyperbolic function $y=\frac{m}{x}$ ($x>0$, $m>0$) intersect at points A and B respectively, and intersect with the x-axis at point C and with the y-axis at point D. The coordinates of point A are $(1, 3)$. What is the equation of line AB and the hyperbolic function $y=\frac{m}{x}$?

Solution:
\textbf{Solution:} Substitute $ \left(1, 3\right)$ to get $ 3 = m$,\\
$\therefore$ the equation of line $AB$ is $y = -x + 4$, $y = \boxed{\frac{3}{x}}$.


Question: As shown in the figure, it is known that the line $y=-x+m+1$ and the inverse proportional function $y=\frac{m}{x}$ (where $x>0$, $m>0$) intersect at points A and B, respectively, and intersect the $x$-axis at point C and the $y$-axis at point D, and the coordinate of point A is $(1, 3)$. If point $P$ is a point on the $y$-axis, when the area of $\triangle ABP$ is 6, find the coordinates of point $P$.

Solution:
\textbf{Solution:} Solve the system
\[
\begin{cases}
y=\frac{3}{x}\\
y=-x+4
\end{cases},
\]
to obtain
\[
\begin{cases}
x_1=1\\
y_1=3
\end{cases},
\begin{cases}
x_2=3\\
y_2=1
\end{cases},
\]
\(\therefore\) the coordinates of point \(B\) are \((3, 1)\),

\(\therefore AB=\sqrt{(3-1)^2+(1-3)^2}=2\sqrt{2}\),

substitute \(x=0\) into \(y=-x+4\) to get \(y=4\),

substitute \(y=0\) into \(y=-x+4\) to get \(x=4\),

\(\therefore\) the coordinates of point \(D\) are \((0, 4)\), and the coordinates of point \(C\) are \((4, 0)\), \(\triangle DOC\) is an isosceles right triangle, construct a line parallel to \(AB\) through point \(P\) intersecting the \(x\)-axis at point \(M\), construct \(CF\perp PM\) at point \(F\),

then \({S}_{\triangle ABP}=\frac{1}{2}AB\cdot CF=2\),

\(\therefore CF=\sqrt{2}\)

\(\because \angle DCO=45^\circ, \angle BCF=90^\circ\), \(PM\parallel AB\),

\(\therefore \angle FCM=\angle FMC=45^\circ\)

\(\therefore \triangle CFM\) is an isosceles right triangle,

\(\therefore CM=\sqrt{2}CF=2\),

\(\therefore\) the coordinates of point \(M\) are \((6, 0)\),

let the equation of \(PM\) be \(y=-x+b\),

substitute \((6, 0)\) into \(y=-x+b\): \(0=-6+b\),

solve to get \(b=6\),

the equation of line \(PM\) is \(y=-x+6\),

let \(\frac{3}{x}=-x+6\)

solve to get \({x}_{1}=3-\sqrt{6}, {x}_{2}=3+\sqrt{6}\),

substitute \(x=3-\sqrt{6}\) into \(y=-x+6\) to get \(y=3+\sqrt{6}\),

substitute \(x=3+\sqrt{6}\) into \(y=-x+6\) to get \(y=3-\sqrt{6}\),

\(\therefore\) the coordinates of point \(P\) are \(\boxed{\left(3-\sqrt{6}, 3+\sqrt{6}\right)}\) or \(\boxed{\left(3+\sqrt{6}, 3-\sqrt{6}\right)}\).


Question: As shown in the figure, it is known that the line $y=-x+m+1$ intersects with the reciprocal function $y=\frac{m}{x}$ ($x>0$, $m>0$) at points A and B, intersects the $x$-axis at point C, and intersects the $y$-axis at point D. The coordinates of point A are given as $(1, 3)$. If the line CD is translated 2 units to the right to obtain line EF, and the part of the hyperbola below line CD is folded along line CD, and if the folded image (the dashed part in the figure) has only one common point with line EF, then find the value of $m$.

Solution:
\textbf{Solution:} Consider the line $y=-x+m+1$. After translating it 2 units to the right, its equation becomes
\[y=-(x-2)+m+1=-x+m+3,\]
For the lines $y=-x+m+1, y=-x+m+3$,
and the inverse proportional function $y=\frac{m}{x}$ symmetric about the line $y=x$,
as illustrated, draw the line $y=x$, which intersects the hyperbola at point $M$, intersects the line $y=-x+m+1$ at point $N$,
and intersects the line $y=-x+m+3$ at point $P$,
Set $x=-x+m+1$,
we get $x=\frac{m+1}{2}$,
thus, the coordinates of point $N$ are $\left(\frac{m+1}{2}, \frac{m+1}{2}\right)$,
Set $x=-x+m+3$, we obtain $x=\frac{m+3}{2}$,
thus, the coordinates of point $P$ are $\left(\frac{m+3}{2}, \frac{m+3}{2}\right)$,
Set $x=\frac{m}{x}$, solving gives $x=-\sqrt{m}$ (discard) or $x=\sqrt{m}$,
thus, the coordinates of point $M$ are $\left(\sqrt{m}, \sqrt{m}\right)$,
Based on the problem statement, points $M$, $P$ are symmetric about point $N$, meaning $N$ is the midpoint of $M$, $P$,
therefore $\sqrt{m}+\frac{m+3}{2}=2\times \frac{m+1}{2}$,
solving gives $m_1=3-2\sqrt{2}$ (discard), $m_2=\boxed{3+2\sqrt{2}}$.


Question: As shown in the figure, the parabola $y=ax^2+3ax+c$ ($a>0$) intersects the y-axis at point C and the x-axis at points A and B, with point A to the left of point B. The coordinate of point B is $(1, 0)$, and $OC=3OB$. Find the equation of the parabola.

Solution:
\textbf{Solution:} Because the coordinates of B are $(1,0)$,\\
therefore $OB = 1$.\\
Because $OC = 3OB = 3$, and point C is below the x-axis,\\
therefore C is $(0, -3)$.\\
Because substituting B $(1,0)$, C $(0, -3)$ into the quadratic equation yields $\begin{cases}4a+c=0\\ c=-3\end{cases}$, solving gives: $\begin{cases}a=\frac{3}{4}\\ c=-3\end{cases}$,\\
therefore the equation of the parabola is $\boxed{y=\frac{3}{4}x^{2}+\frac{9}{4}x-3}$.


Question: Given, as shown in the graph, the parabola $y=ax^{2}+3ax+c$ ($a>0$) intersects with the y-axis at point C, and intersects with the x-axis at points A and B, where point A is to the left of point B. The coordinates of point B are $(1, 0)$, and $OC=3OB$. If point D is a moving point on the parabola below line segment AC, what is the maximum area of quadrilateral AOCD?

Solution:
\textbf{Solution:} As shown in Figure 1: draw line DE through point D parallel to y-axis, intersecting AC at point E.

Since $x=-\frac{b}{2a}=\frac{-\frac{9}{4}}{2\times \frac{3}{4}}=-\frac{3}{2}$, B(1,0),

therefore, A(-4,0).

Thus, AB=5.

Therefore, $S_{\triangle ABC}=\frac{1}{2}$AB$\cdot$OC$=\frac{1}{2}\times 5\times 3=7.5$.

Suppose the equation of AC is $y=kx+b$.

Since substituting A(-4,0) and C(0,-3) gives $\begin{cases}-4k+b=0\\ b=-3\end{cases}$, we solve to get $\begin{cases}k=-\frac{3}{4}\\ b=-3\end{cases}$,

Therefore, the equation of line AC is $y=-\frac{3}{4}x-3$.

Let D(a, $\frac{3}{4}a^{2}+\frac{9}{4}a-3$), then E(a, $-\frac{3}{4}a-3$).

Since DE$=-\frac{3}{4}a-3-\left(\frac{3}{4}a^{2}+\frac{9}{4}a-3\right)=-\frac{3}{4}(a+2)^{2}+3$,

Therefore, when $a=-2$, DE reaches its maximum value of 3.

Therefore, the maximum area of $\triangle ADC =\frac{1}{2}$DE$\cdot$AO$=\frac{1}{2}\times 3\times 4=6$.

Therefore, $S_{\text{quadrilateral }ABCD}$=$S_{\triangle ABC}$+$S_{\triangle ACD}$=7.5+6=\boxed{13.5},

Thus, the maximum area of quadrilateral ABCD is \boxed{13.5}.