Question: As shown in the figure, consider a square $ABCD$ with side length $2$. Points $E$ and $F$ are the midpoints of $AB$ and $CD$ respectively. Connect $CE$ and $AF$, and draw $DG \perp AF$ from point $D$ with $G$ as the foot of the perpendicular. Extend $DG$ to meet $CE$ at point $H$. What is the length of $EH$?

Solution:
\textbf{Solution:} Since $DG \perp AF$ and $AF \parallel CE$, it follows that $DG \perp CE$. Therefore, $\angle DHC = \angle AGD = 90^\circ$, $\angle ADG + \angle CDH = 90^\circ$, $\angle CDH + \angle DCH = 90^\circ$. Hence, $\angle ADG = \angle DCH$. In triangles $\triangle ADG$ and $\triangle DCH$, we have
\[
\left\{\begin{array}{l}
\angle DHC = \angle AGD \\
\angle ADG = \angle DCH \\
AD = CD
\end{array}\right.
\]
Therefore, $\triangle ADG \cong \triangle DCH$ (by AAS), so $CH = DG = \frac{2\sqrt{5}}{5}$. Since parallelogram $AFCE$, we have $AF = CE = \sqrt{5}$. Therefore, $EH = CE - CH = \sqrt{5} - \frac{2\sqrt{5}}{5} = \boxed{\frac{3\sqrt{5}}{5}}$.


Question: As shown in the figure, it is known that points $E$ and $F$ are the midpoints of sides $BC$ and $AD$ of quadrilateral $ABCD$, respectively. Connect $AC$, $AE$, and $CF$. Quadrilateral $AECF$ is a parallelogram; If $BC=12$ and $\angle BAC=90^\circ$, what is the perimeter of parallelogram $AECF$?

Solution:
\textbf{Solution:} Since $\angle BAC=90^\circ$ in $\triangle ABC$, and point E is the midpoint of BC,\\
$\therefore AE=CE=\frac{1}{2}BC=6$,\\
$\therefore$ Quadrilateral AECF is a rhombus,\\
$\therefore$ The perimeter of quadrilateral AECF is $6\times 4=\boxed{24}$.


Question: As shown in the figure, BD is the diagonal of the rhombus ABCD, and $\angle CBD = 76^\circ$. The perpendicular bisector of segment AB, with the foot of the perpendicular being point E, intersects AD at point F. Connect BF. Find the degree of $\angle DBF$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\
it follows that CD=CB, $AB\parallel CD$, and $\angle A=\angle C$, \\
Because $\angle CBD=76^\circ$, \\
it follows that $\angle CDB=\angle CBD=76^\circ$, \\
hence $\angle ABD=\angle CDB=76^\circ$, \\
thus $\angle A=\angle C=180^\circ-\angle CDB-\angle CBD=28^\circ$, \\
Because FE is the perpendicular bisector of segment AB, \\
it follows that AF=BF, \\
hence $\angle FBA=\angle A=28^\circ$, \\
thus $\angle DBF=\angle ABD-\angle FBA=\boxed{48^\circ}$.


Question: As shown in the figure, quadrilateral $ABCD$ is a parallelogram, with $AD=BD$. Draw $CE\parallel BD$, intersecting the extension of $AD$ at point $E$. Quadrilateral $BDEC$ is a rhombus; draw $BE$. If $AB=4$ and $AD=7$, what is the length of $BE$?

Solution:
\textbf{Solution:} Extend BE to intersect CD at O,\\
since quadrilateral ABCD is a parallelogram,\\
thus CD = AB = 4.\\
From (1), we know quadrilateral BDEC is a rhombus,\\
thus DO = CO = $\frac{1}{2}$CD = 2, and BO = $\frac{1}{2}$BE, with CD$\perp$BE.\\
In $\triangle BDO$, BD = AD = 7, and DO = 2,\\
thus BO = $\sqrt{BD^{2}-DO^{2}}$ = $\sqrt{7^{2}-2^{2}}$ = $3\sqrt{5}$,\\
therefore, BE = 2BO = $\boxed{6\sqrt{5}}$.


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Solution:
\textbf{Solution:} As shown in Figure 2, draw $BD$ intersecting $AC$ at point $O$,\\
since quadrilateral $ABCD$ is a rhombus with $AB=2$,\\
it follows that $AB=AD$, $AO=OC$, $OB=OD=\frac{1}{2}BD$, and $AC\perp BD$,\\
given $\angle BAD=60^\circ$,\\
triangle $ABD$ is an equilateral triangle,\\
thus $BD=AB=2$, $OB=OD=1$, $OC=OA=\sqrt{AB^{2}-OB^{2}}=\sqrt{2^{2}-1^{2}}=\sqrt{3}$,\\
since $BG\perp DF$,\\
we have $OE=OB=OD=1$,\\
therefore, $CE=OC-OE=\sqrt{3}-1=\boxed{\sqrt{3}-1}$.


Question: As shown in the figure, there is a rectangular sheet of paper ABCD with point E on the side AB. Fold $\triangle BCE$ along the straight line CE, so that point B falls on the diagonal AC, denoted as point F. If $AB=4$ and $BC=3$, what is the length of $AE$?

Solution:
\textbf{Solution:} As shown in Figure 1, in the rectangular paper ABCD, $\because$AB＝4, BC＝3, thus by the Pythagorean theorem, we get AC＝5. From the fold, it follows: FC＝BC＝3, $\angle$EFC＝$\angle$B＝$90^\circ$, BE＝FE. $\therefore$ $AF=AC-FC=5-3=2$. Let AE＝x, then $BE=4-x=FE$. In $\triangle AFE$, $2^2+(4-x)^2=x^2$, solving for x, we get: $x=\frac{5}{2}$.\\
$\therefore$ $AE=\boxed{\frac{5}{2}}$.


Question: As shown in the figure, there is a rectangular piece of paper ABCD, with point E on side AB. Triangle $BCE$ is folded along line CE, such that point B falls on the diagonal AC, marked as point F. Connect DF. If points D, F, and E are on the same straight line, and $DF=2$, find the length of AE.

Solution:
\textbf{Solution:} As shown in Figure 2, in the rectangle ABCD, since $DC\parallel AB$, it follows that $\angle DCE = \angle BEC$. From the folding, we know that $\angle BEC = \angle FEC$, therefore $\angle DCE = \angle FEC$, which implies $DC = DE$. Moreover, since points D, F, and E lie on the same straight line and $\angle EFC = \angle B$, it follows that $\angle DFC = 90^\circ$, hence $\angle DFC = \angle DAE = 90^\circ$. Given $CF = CB = DA$, it follows that $\triangle CDF \cong \triangle DEA$, \\
therefore $AE = DF = \boxed{2}$.


Question: As shown in the figure, $AD$ is the median of $\triangle ABC$, and $\angle DAC=\angle B$. Point $E$ is on $AD$, with $CD=CE$. $\triangle ACE\sim \triangle BAD$: If $AB=10$, and $BC=6$, determine the length of the line segment $AD$.

Solution:
\textbf{Solution:} Given that $AD$ is the median of $\triangle ABC$,\\
thus $CD = BD = CE = \frac{1}{2} BC = 3$,\\
since $\angle DAC = \angle B$, it follows that $\angle ACD = \angle BCA$, and thus $\triangle ACD \sim \triangle BCA$,\\
so $\frac{AC}{BC} = \frac{CD}{AC}$, that is, $\frac{AC}{6} = \frac{3}{AC}$,\\
which leads to $AC = 3\sqrt{2}$. Given that $\triangle ACE \sim \triangle BAD$,\\
we have $\frac{AC}{BA} = \frac{CE}{AD}$, that is, $\frac{3\sqrt{2}}{10} = \frac{3}{AD}$,\\
therefore $AD = \boxed{5\sqrt{2}}$.


Question: As shown in the figure, within $\triangle ABC$, points D, E, and F are located on the sides AB, AC, and BC, respectively, and lines DE and EF are connected. It is known that quadrilateral BFED is a parallelogram, and $\frac{DE}{BC} = \frac{1}{4}$. If AB = 8, find the length of the line segment AD.

Solution:
\textbf{Solution}: Since quadrilateral BFED is a parallelogram,\\
it implies that DE$\parallel$BF, and hence DE$\parallel$BC, which further implies $\triangle$ADE$\sim$$\triangle$ABC,\\
thus $\frac{AD}{AB} = \frac{DE}{BC} = \frac{1}{4}$,\\
given that AB$=8$,\\
it follows that AD$=\boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, points D, E, and F are located on sides AB, AC, and BC, respectively, with DE and EF connected. It is known that quadrilateral BFED is a parallelogram, and $\frac{DE}{BC} = \frac{1}{4}$. If the area of $\triangle ADE$ is 1, what is the area of the parallelogram BFED?

Solution:
\textbf{Solution:} Given $\because \triangle ADE \sim \triangle ABC$,\\
$\therefore \frac{S_{\triangle ADE}}{S_{\triangle ABC}} = \left( \frac{DE}{BC} \right)^2 = \left( \frac{1}{4} \right)^2 = \frac{1}{16}$, since the area of $\triangle ADE$ is 1, $\therefore$ the area of $\triangle ABC$ is 16,\\
$\because$ quadrilateral BFED is a parallelogram, $\therefore EF \parallel AB$, $\therefore \triangle EFC \sim \triangle ABC$,\\
$\therefore \frac{S_{\triangle ADE}}{S_{\triangle ABC}} = \left( \frac{3}{4} \right)^2 = \frac{9}{16}$, hence the area of $\triangle EFC$ $= 9$, $\therefore$ the area of parallelogram BFED $= 16 - 9 - 1 = \boxed{6}$.


Question: As shown in the diagram, in $\triangle ABC$, $CD$ is the angle bisector, and $DE$ bisects $\angle CDB$ and intersects $BC$ at point $E$, and $DE \parallel AC$. It is given that $CD^2 = CA \cdot CE$. If $\frac{AD}{BD} = \frac{4}{3}$ and $AC = 14$, what is the length of $AD$?

Solution: \textbf{Solution:} Since $CD$ bisects $\angle ACB$, then $\angle BCD = \angle ACD$. Since $DE\parallel BC$, then $\angle EDC = \angle BCD$, which leads to $\angle ECD = \angle EDC$; hence $DE=CE$. Given $DE\parallel BC$, we have $\frac{AD}{BD}=\frac{4}{3}$, which implies $\triangle BDE \sim \triangle BAC$. Therefore, $\frac{DE}{AC}=\frac{BD}{AB}=\frac{3}{7}$. This simplifies to $\frac{CE}{14}=\frac{3}{7}$. Solving this gives $CE=DE=6$. Since $DE\parallel BC$, it follows that $\angle ACD = \angle CDE$ and $\angle A = \angle EDB$. Given $\angle CDE = \angle EDB$, then $\angle ACD = \angle A$, which implies $AD=CD$. Knowing $CD^2 = CA \cdot CE$, we find $AD^2 = 14 \times 6$. Solving this gives $AD= \boxed{2\sqrt{21}}$.


Question: As shown in the figure, $\triangle ABC$ is inscribed in $\odot O$, with $AB$ being the diameter of $\odot O$, $AD=16$, and $CE=6$. Connect $OC$, and the chord $AD$ intersects $OC$ and $BC$ at points $E$ and $F$, respectively, where point $E$ is the midpoint of $AD$. $\angle CAD=\angle CBA$. What is the length of $AB$?

Solution:
\textbf{Solution:} Given that $AB$ is the diameter,\\
then $\angle ACB=90^\circ$,\\
since $AE=DE=\frac{1}{2}AD=8$,\\
it follows that $OC\perp AD$,\\
thus $\angle AEC=90^\circ$,\\
given $CE=6$,\\
it follows that $AC=\sqrt{AE^2+CE^2}=10$,\\
since $\angle AEC=\angle ACB$ and $\angle CAD=\angle CBA$,\\
therefore $\triangle AEC\sim \triangle BCA$,\\
thus $\frac{CE}{AC}=\frac{AC}{AB}$,\\
which gives $\frac{6}{10}=\frac{10}{AB}$,\\
therefore $AB=\boxed{\frac{50}{3}}$.


Question: As shown in the figure, the angle bisector of $\angle BAC$ intersects the circumcircle of $\triangle ABC$ at point $D$ and intersects $BC$ at point $F$. The angle bisector of $\angle ABC$ meets $AD$ at point $E$. Given $DE=DB$, if $\angle BAC=90^\circ$ and $BD=4$, what is the radius of the circumcircle of $\triangle ABC$?

Solution:
\textbf{Solution:} Draw line CD, as shown,\\
$\because$ $\angle BAC=90^\circ$,\\
$\therefore$ BC is the diameter,\\
$\therefore$ $\angle BDC=90^\circ$,\\
$\because$ $\angle 1=\angle 2$,\\
$\therefore$ DB=BC,\\
$\therefore$ $\triangle DBC$ is an isosceles right triangle,\\
$\therefore$ BC=$\sqrt{2}$BD=$4\sqrt{2}$,\\
$\therefore$ the radius of the circumscribed circle of $\triangle ABC$ is $\boxed{2\sqrt{2}}$;


Question: As shown in the diagram, the bisector of $\angle BAC$ intersects the circumcircle of $\triangle ABC$ at point D and intersects BC at point E. The bisector of $\angle ABC$ intersects AD at point F. If BD$=6$ and DF$=4$, what is the length of AD?

Solution:
\textbf{Solution:} Given $\angle 5=\angle 2=\angle 1$ and $\angle FDB=\angle BDA$,\\
$\therefore \triangle DBF\sim \triangle ADB$,\\
$\therefore \frac{BD}{DA}=\frac{DF}{DB}$, thus $\frac{6}{AD}=\frac{4}{6}$,\\
$\therefore AD=\boxed{9}$.


Question: As shown in the figure, it is known that the parabola $y=ax^2+bx+3$ (where $a\neq 0$) intersects the x-axis at points $A(-4,0)$ and $B(6,0)$, and intersects the y-axis at point $C$. Suppose $G$ is a moving point between $A$ and $C$ on the parabola, and a line $GD$ is drawn through $G$ parallel to the x-axis, intersecting the parabola at point $D$. Through points $D$ and $G$, perpendiculars to the x-axis are drawn, with the feet of the perpendiculars being $E$ and $F$, respectively, forming rectangle $DEFG$. What is the equation of this parabola?

Solution:
Solution: Substituting $A(-4,0)$, $B(6,0)$ into $y=ax^2+bx+3$, we get
\[
\begin{cases}
16a-4b+3=0\\
36a+6b+3=0
\end{cases}
\]
Solving for $a$ and $b$, we find
\[
\begin{cases}
a=-\frac{1}{18}\\
b=\frac{1}{4}
\end{cases}
\]
$\therefore$ The equation of the parabola is $y=-\frac{1}{18}x^2+\frac{1}{4}x+3=\boxed{-\frac{1}{18}x^2+\frac{1}{4}x+3}$.


Question: As shown in the figure, it is known that the parabola $y=ax^2+bx+3$ ($a \neq 0$) intersects the x-axis at points $A(-4,0)$ and $B(6,0)$, and intersects the y-axis at point $C$. If $G$ is a moving point on the parabola between points $A$ and $C$, draw line $GD$ parallel to the x-axis, intersecting the parabola at point $D$. Draw vertical lines from points $D$ and $G$ to the x-axis, with the feet of the perpendiculars being $E$ and $F$, respectively, forming the rectangle $DEFG$. When point $G$ coincides with point $C$, what is the area of rectangle $DEFG$?

Solution:
\textbf{Solution:} When point $G$ coincides with $C$, the coordinates of point $G$ are $(0,3)$.\\
Substituting $y=3$ into $y=-\frac{1}{18}x^{2}+\frac{1}{4}x+3$,\\
we get $-\frac{1}{18}x^{2}+\frac{1}{4}x+3=3$.\\
Solving this equation, we find $x_{1}=0$ and $x_{2}=2$.\\
$\therefore$ The coordinates of point $D$ are $(2,3)$.\\
$\therefore GD=2$, $DE=3$.\\
$\therefore$ The area of rectangle DEFG is $S_{\text{rectangle} DEFG}=GD \cdot DE=2 \times 3=\boxed{6}$.


Question: As shown in the figure, it is known that the parabola $y=ax^2+bx+3$ ($a\neq 0$) intersects with the x-axis at points $A(-4,0)$ and $B(6,0)$, and intersects with the y-axis at point $C$. If $G$ is a moving point between $A$ and $C$ on the parabola, draw line $GD\parallel x$ axis through point $G$, intersecting the parabola at point $D$. Draw perpendiculars to the x-axis through points $D$ and $G$, with feet at $E$ and $F$ respectively, to form rectangle $DEFG$. If line $BC$ intersects $DG$ and $DE$ at points $M$ and $N$ respectively, what is the maximum area of $\triangle DMN$?

Solution:
\textbf{Solution:} Let the equation of line BC be $y=kx+m$ ($k\neq 0$). By substituting B(6,0) and C(0,3) into the equation, we get
\[
\left\{
\begin{array}{l}
6k+m=0\\
m=3
\end{array}
\right.
\]
From which we solve
\[
\left\{
\begin{array}{l}
k=-\frac{1}{2}\\
m=3
\end{array}
\right.
\]
Therefore, the expression for line BC is $y=-\frac{1}{2}x+3$.

Assume the x-coordinate of point D is n, due to symmetry, $2\leq n\leq 6$,

Therefore, the coordinates of points D and N are respectively D$\left(n, -\frac{1}{8}n^{2}+\frac{1}{4}n+3\right)$, N$\left(n, -\frac{1}{2}n+3\right)$.

Therefore $DN=-\frac{1}{8}n^{2}+\frac{1}{4}n+3-\left(-\frac{1}{2}n+3\right)=-\frac{1}{8}\left(n-3\right)^{2}+\frac{9}{8}$.

Therefore, when $n=3$, DN reaches its maximum value of $\frac{9}{8}$.

Since DG is parallel to the x-axis,

Therefore, $\angle DMN=\angle OBC$.

Also, since $\angle MDN=\angle BOC=90^\circ$

Therefore, $\triangle DMN\sim \triangle OBC$.

Therefore, $\frac{S_{\triangle DMN}}{S_{\triangle OBC}}=\left(\frac{DN}{OC}\right)^{2}$.

Therefore, when DN is at its maximum, the area of $\triangle DMN$ is also at its maximum.

Since $S_{\triangle OBC}=3\times 6\times \frac{1}{2}=9$,

Therefore $S_{\triangle DMN}=S_{\triangle OBC}\times \left(\frac{DN}{OC}\right)^{2}=9\times \left(\frac{9}{8}\div 3\right)^{2}=\boxed{\frac{81}{64}}$.

Therefore, the maximum area of $\triangle DMN$ is $\boxed{\frac{81}{64}}$.


Question: As shown in the figure, there is a square $ABCD$ with a side length of $10$. $E$ is a moving point on side $BC$ (not coinciding with $B$ or $C$), and $AE$ is connected. $G$ is a point on the extension of $BC$. A perpendicular line through point $E$ to $AE$ intersects the angle bisector of $\angle DCG$ at point $F$. If $FG \perp BG$, then $\triangle ABE \sim \triangle EGF$; if $EC = 2$, find the area of $\triangle CEF$.

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a square,\\
$ \therefore \angle DCG=90^\circ$,\\
Since $CF$ bisects $\angle DCG$,\\
$ \therefore \angle FCG=\frac{1}{2}\angle DCG=45^\circ$,\\
Since $\angle CGF=90^\circ$,\\
$ \therefore \angle GCF=\angle CFG=45^\circ$,\\
$ \therefore FG=CG$,\\
Since $AB=BC=10$, $CE=2$,\\
$ \therefore BE=8$,\\
$ \therefore EG=CE+CG=2+FG$,\\
From $(1)$, we know that $\triangle BAE \sim \triangle GEF$,\\
$ \therefore \frac{AB}{EG}=\frac{BE}{FG}$,\\
$ \therefore \frac{10}{2+FG}=\frac{8}{FG}$,\\
$ \therefore FG=8$,\\
$ \therefore {S}_{\triangle ECF}=\frac{1}{2}CE\cdot FG=\frac{1}{2}\times 2\times 8=\boxed{8}$;


Question: As shown in the figure, there is a square $ABCD$ with a side length of 10. Let $E$ be a moving point on side $BC$ (not coinciding with $B$ or $C$), connect $AE$. Let $G$ be a point on the extension line of $BC$. A perpendicular line is drawn from point $E$ to the angle bisector of $\angle DCG$, intersecting at point $F$. If $FG \perp BG$, please directly write down the value of $EC$ for which the area of $\triangle CEF$ is maximized.

Solution:
\textbf{Solution:} Let $CE=x$, thus $BE=10-x$, \\
$\therefore EG=CE+CG=x+FG$, \\
According to (1), $\triangle BAE \sim \triangle GEF$, \\
$\therefore \frac{AB}{EG}=\frac{BE}{FG}$, \\
$\therefore \frac{10}{x+FG}=\frac{10-x}{FG}$, \\
$\therefore FG=10-x$, \\
$\therefore \text{Area of }\triangle ECF=\frac{1}{2}\times CE\times FG=\frac{1}{2}\cdot x\cdot (10-x)=-\frac{1}{2}(x^2-10x)=-\frac{1}{2}(x-5)^2+\frac{25}{2}$, \\
When $x=\boxed{5}$, $\text{Area of }\triangle ECF$ is maximal.


Question: As shown in the figure, the parabola $y=ax^{2}+bx+c$ passes through three points $A(-1,0)$, $B(4,0)$, and $C(0,3)$. Point $D$ is a moving point on the parabola above line $BC$, and $DE\perp BC$ at $E$. Find the expression of the parabola.

Solution:
\textbf{Solution:} Given the problem, we have
\[
\begin{cases}
a-b+c=0\\
16a+4b+c=0\\
c=3
\end{cases}
\]
Solving these, we find
\[
\begin{cases}
a=-\frac{3}{4}\\
b=\frac{9}{4}\\
c=3
\end{cases}
\]
Thus, the equation of the parabola is \boxed{y=-\frac{3}{4}x^{2}+\frac{9}{4}x+3}.


Question: As shown in the figure, the parabola $y=ax^{2}+bx+c$ passes through three points $A(-1,0)$, $B(4,0)$, and $C(0,3)$. $D$ is a moving point on the parabola above the line $BC$, and $DE\perp BC$ at point $E$. What is the maximum length of the line segment $DE$?

Solution:
\textbf{Solution:} Let the equation of line $BC$ be $y=kx+b$, with the system $\left\{\begin{array}{l}4k+b=0\\ b=3\end{array}\right.$. Solving, we find $\left\{\begin{array}{l}k=-\frac{3}{4}\\ b=3\end{array}\right.$, $\therefore y=-\frac{3}{4}x+3$.

Let $D(a,-\frac{3}{4}a^2+\frac{9}{4}a+3)$ (where $0<a<4$). Construct $DM\perp x$-axis, intersecting $BC$ at point $M$, to have $M(a,-\frac{3}{4}a+3)$.

Hence, $DM= (-\frac{3}{4}a^2+\frac{9}{4}a+3)-(-\frac{3}{4}a+3)=-\frac{3}{4}a^2+3a$.

Given $\angle DME=\angle OCB$ and $\angle DEM=\angle BOC$,

it follows that $\triangle DEM \sim \triangle BOC$,

hence $\frac{DE}{DM}=\frac{OB}{BC}$.

Given that $OB=4$, $OC=3$,

it follows that $BC=5$,

hence $DE=\frac{4}{5}DM$,

thus $DE=-\frac{3}{5}a^2+\frac{12}{5}a=-\frac{3}{5}(a-2)^2+\frac{12}{5}$.

When $a=2$, $DE$ reaches its maximum value, which is $\boxed{\frac{12}{5}}$.


Question: As shown in the diagram, the parabola $y=ax^2+bx+c$ passes through points $A(-1,0)$, $B(4,0)$, and $C(0,3)$. Point $D$ lies on the parabola above line $BC$, and $DE$ is perpendicular to $BC$ at point $E$. In diagram 2, the midpoint of $AB$ is marked as $F$, and lines $CD$ and $CF$ are drawn. There exists a point $D$ such that one of the angles in $\triangle CDE$ is equal to $\angle CFO$. Find the x-coordinate of point $D$.

Solution:
\textbf{Solution:} Existence exists. Assume that there is such a point $D$, $\triangle CDE$ where there is an angle equal to $\angle CFO$,\\
$\because$ point $F$ is the midpoint of $AB$,\\
$\therefore OF=\frac{3}{2}$, $\tan\angle CFO=\frac{OC}{OF}=2$,\\
Draw $BG\perp BC$ through point $B$, intersecting the extension of $CD$ at point $G$, and draw $GH\perp x$ axis through point $G$, with the foot of the perpendicular at $H$,\\
(1) If $\angle DCE = \angle CFO$,\\
$\therefore\tan\angle DCE=\frac{GB}{BC}=2$,\\
$\therefore BG=10$,\\
$\because\triangle GBH\sim \triangle BCO$,\\
$\therefore\frac{GH}{BO}=\frac{HB}{OC}=\frac{GB}{BC}$,\\
$\therefore GH=8$, $BH=6$,\\
$\therefore G(10,8)$,\\
Let the analytical equation of line $CG$ be $y=kx+b'$,\\
$\therefore\left\{\begin{array}{l}b'=3\\ 10k+b'=8\end{array}\right.$,\\
Solving we get $\left\{\begin{array}{l}k=\frac{1}{2}\\ b'=3\end{array}\right.$,\\
$\therefore$ the analytical equation of line $CG$ is $y=\frac{1}{2}x+3$,\\
$\therefore\left\{\begin{array}{l}y=\frac{1}{2}x+3\\ y=\frac{3}{4}x^{2}+\frac{9}{4}x+3\end{array}\right.$,\\
Solving we get $x=\boxed{\frac{7}{3}}$, or $x=0$ (discard).\\
(2) If $\angle CDE = \angle CFO$,\\
Similarly, we get $BG=\frac{5}{2}$, $GH=2$, $BH=\frac{3}{2}$,\\
$\therefore G\left(\frac{11}{2},2\right)$,\\
Similarly, we can obtain that the analytical equation of line $CG$ is $y=-\frac{2}{11}x+3$,\\
$\therefore\left\{\begin{array}{l}y=-\frac{2}{11}x+3\\ y=\frac{3}{4}x^{2}+\frac{9}{4}x+3\end{array}\right.$,\\
Solving we get $x=\boxed{\frac{107}{33}}$ or $x=0$ (discard),\\
In conclusion, there exists a point $D$, such that there is an angle in $\triangle CDE$ equal to $\angle CFO$, and the $x$-coordinate of point $D$ is \boxed{\frac{7}{3}} or \boxed{\frac{107}{33}}.


Question: As shown in Figure 1, there is a lamp post $AB$ beside a straight road. Under the light, Xiao Li moves straight from the bottom of the lamp post at point $B$ for $4\,\text{m}$ to reach point $D$. At point $D$, the length of her shadow is measured to be $DE=1\,\text{m}$. Xiao Li's height is $CD=1.2\,\text{m}$. What is the height of the lamp post $AB$?

Solution:
\textbf{Solution:} As shown in figure 1, according to the given problem: $AB\parallel CD$, $BE=1+4=5$ (meters),\\
$\therefore \triangle EAB \sim \triangle ECD$,\\
$\therefore \frac{AB}{CD}=\frac{BE}{DE}$,\\
thus $\frac{AB}{1.2}=\frac{5}{1}$,\\
Solving gives: $AB=\boxed{6}$ (meters).


Question: As shown in Figure 1, there is a lamppost $AB$ beside a straight road. Under the light, Xiao Li progresses in a straight line from the bottom of the lamppost $B$ for $4m$ to reach point $D$. At point $D$, her shadow $DE$ is measured to be $1m$. The height of Xiao Li, $CD$, is $1.2m$. If Xiao Li continues to move in a straight line from point $D$ for another $4m$ to reach point $G$ (as shown in Figure 2), what is the length of her shadow $GH$ at that point?

Solution:
\textbf{Solution:} As shown in Figure 2, according to the given conditions: \(AB\parallel FG\), \(BE=5\) (meters), \\
\(\therefore \triangle HGF \sim \triangle HBA\), \\
\(\therefore \frac{AB}{FG}=\frac{BH}{GH}\), \\
that is \(\frac{6}{1.2}=\frac{8+GH}{GH}\), \\
solving this we get: \(GH=\boxed{2}\) (meters).


Question: As shown in the figure, in rectangle $ABCD$, point $E$ is the midpoint of $BC$, and $EF\perp AE$ intersects $CD$ at point $F$. $\triangle ABE\sim \triangle ECF$; if $AB=3$ and $BC=8$, find the length of $EF$.

Solution:
\textbf{Solution:} Since $E$ is the midpoint of $BC$, and $BC=8$\\
$ \therefore BE=EC=\frac{1}{2}BC=4$\\
Since $\angle B=90^\circ$, and $AB=3$\\
$ \therefore AE=5$\\
$ \therefore \triangle ABE\sim \triangle ECF$\\
$ \therefore \frac{AB}{EC}=\frac{AE}{EF}$     i.e., $ \frac{3}{4}=\frac{5}{EF}$\\
$ \therefore EF=\boxed{\frac{20}{3}}$.


Question: As shown in the figure, the graph of the quadratic function $y=ax^{2}+bx-3$ intersects the $x$-axis at points $A(-1,0)$ and $B(3,0)$, and intersects the $y$-axis at point $C$. The vertex of the parabola is point $M$. What is the analytical expression of the quadratic function?

Solution:
\textbf{Solution:} Substitute point A ($-1$, $0$) and point B ($3$, $0$) into $y=ax^{2}+bx-3$,\\
we get $\begin{cases}a-b-3=0\\ 9a+3b-3=0\end{cases}$.\\
Solving this, we find $\begin{cases}a=1\\ b=-2\end{cases}$.\\
Therefore, the equation of the quadratic function is $\boxed{y=x^{2}-2x-3}$.


Question: As shown in the figure, the graph of the quadratic function $y=ax^{2}+bx-3$ intersects the x-axis at points $A(-1,0)$ and $B(3,0)$, and intersects the y-axis at point $C$. The vertex of the parabola is point $M$. Point $P$ is a point on the parabola, with the x-coordinate of point $P$ being $m$ ($m>3$). Point $Q$ is on the axis of symmetry, and $AQ \perp PQ$. If $AQ=2PQ$, find the value of $m$.

Solution:
\textbf{Solution:} Draw a line parallel to the x-axis passing through point Q, intersecting the line parallel to the y-axis passing through point P at point N, and intersecting the line parallel to the y-axis passing through point A at point M.\\
$\therefore \angle AMQ=\angle QNP=90^\circ$\\
$\therefore \angle QPN+\angle PQN=90^\circ$\\
$\because \angle AQP=90^\circ$,\\
$\therefore \angle AQM+\angle PQN=90^\circ$,\\
$\therefore \angle AQM=\angle QPN$,\\
$\therefore \triangle AMQ\sim \triangle QNP$.\\
$\therefore \frac{AM}{QN}=\frac{MQ}{NP}=\frac{AQ}{QP}=2$.\\
The axis of symmetry of this parabola is the line $x=1$,\\
thus let the coordinates of point Q be (1, t),\\
given that the coordinates of point P are ($m$, $m^{2}-2m-3$),\\
hence $AM=t$, $QN=m-1$, $MQ=2$,\\
$NP=t-m^{2}+2m+3$.\\
$\therefore \frac{t}{m-1}=\frac{2}{t-m^{2}+2m+3}=2$.\\
Solving, we get $m=0$ (discard) or $4$.\\
Therefore, $m=\boxed{4}$.


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

Solution:
Solution: In right-angled $\triangle ADE$, given $AE=6$ and $\cos A= \frac{AE}{AD}=\frac{3}{5}$, it follows that $AD=10$,\\
by Pythagoras' theorem, $DE= \sqrt{AD^{2}-AE^{2}}=\sqrt{10^{2}-6^{2}}=\boxed{8}$\\
Since $BD$ bisects $\angle ABC$ and $DE\perp AB$, $\angle C=90^\circ$,\\
by the property of angle bisector, we have $DC=DE=\boxed{8}$.


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

Solution:
\textbf{Solution:} Given that $AD=10$ and $DC=8$, we have $AC=AD+DC=18$.\\
In triangles $\triangle ADE$ and $\triangle ABC$, $\angle A=\angle A$, and $\angle AED=\angle ACB$,\\
therefore $\triangle ADE \sim \triangle ABC$. It follows that $\frac{DE}{BC}=\frac{AE}{AC}$, that is $\frac{8}{BC}=\frac{6}{18}$, hence $BC=24$,\\
Therefore, $\tan\angle DBC=\frac{CD}{BC}=\frac{8}{24}=\boxed{\frac{1}{3}}$.


Question: As shown in the figure, $\triangle ABC$ is an equilateral triangle, CE is an external angle bisector, and point D is on AC. Extend BD to intersect with CE at point E. $\triangle ABD\sim \triangle CED$, if $AB=6$ and $AD=2CD$, find the length of $BE$.

Solution:
\textbf{Solution:} Draw BM$\perp$AC at point M, where AC=AB=6.\\
$\therefore$AM=CM=3, BM=AB$\cdot\sin 60^\circ=3\sqrt{3}$.\\
$\because$AD=2CD,\\
$\therefore$CD=2, AD=4, MD=1.\\
In $\triangle BDM$, BD=$\sqrt{BM^2+MD^2}=2\sqrt{7}$.\\
Since $\triangle ABD\sim\triangle CED$,\\
$\frac{BD}{ED}=\frac{AD}{CD}$, $\frac{2\sqrt{7}}{ED}=2$,\\
$\therefore$ED=$\sqrt{7}$\\
$\therefore$BE=BD+ED=$3\sqrt{7}$.\\
Hence, the length of $BE$ is $\boxed{3\sqrt{7}}$.


Question: As shown in the figure, in $\triangle ABC$, $D$ is a point on $AB$, and $\angle ABC = \angle ACD$. $\triangle ABC \sim \triangle ACD$; if $AC=3$, $AD=2$, find the length of $AB$.

Solution:
\textbf{Solution:} Since $\triangle ABC \sim \triangle ACD$, \\
$\therefore \frac{AB}{AC}=\frac{AC}{AD}$, \\
which means $\frac{AB}{3}=\frac{3}{2}$, \\
$\therefore AB=\boxed{\frac{9}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, points $P$ and $D$ are on sides $BC$ and $AC$ respectively, and $\angle APD = \angle B$. $AC\cdot CD = CP\cdot BP$; If $AB=10$, $BC=12$, when $PD \parallel AB$, find the length of $BP$.

Solution:
\textbf{Solution:} Since $PD\parallel AB$,\\
$\therefore \angle APD=\angle BAP$.\\
Also, since $\angle APD=\angle C$,\\
$\therefore \angle BAP=\angle C$.\\
Also, since $\angle B=\angle B$,\\
$\therefore \triangle BAP \sim \triangle BCA$,\\
$\therefore \frac{BA}{BC}=\frac{BP}{BA}$.\\
Also, since $AB=10$, $BC=12$,\\
$\therefore \frac{10}{12}=\frac{BP}{10}$,\\
$\therefore BP=\boxed{\frac{25}{3}}$.


Question: As shown in the figure, $AB$ is the diameter of circle $\odot O$, and points $C$ and $D$ are two points on circle $\odot O$. Lines $BC$ and $DC$ are drawn, $BC=CD$, and $CE\perp DA$, intersecting the extension of $DA$ at point $E$. $CE$ is the tangent to circle $\odot O$; if $AE=2$ and $CE=4$, what is the length of $AD$?

Solution:
\textbf{Solution:} Draw line $AC$, as shown in the diagram,\\
from $\left(1\right)$ we know $\angle ECA+\angle ACO=90^\circ$,\\
since $AB$ is the diameter of $\odot O$,\\
therefore $\angle ACB=90^\circ$,\\
therefore $\angle ACO+\angle OCB=90^\circ$,\\
thus $\angle ECA=\angle OCB$.\\
Since $OC=OB$,\\
thus $\angle B=\angle OCB$,\\
therefore $\angle ECA=\angle B$,\\
since $\angle B=\angle ADC$,\\
thus $\angle ADC=\angle ECA$.\\
Since $\angle E=\angle E$,\\
thus $\triangle EAC\sim \triangle ECD$,\\
therefore $\frac{EA}{EC}=\frac{EC}{ED}$,\\
since $AE=2$, $CE=4$,\\
thus $ED=8$,\\
therefore $AD=DE−AE=\boxed{6}$.


Question: As shown in the figure, it is known that the vertex of the quadratic function $y=ax^2+bx+c$ is $M(2, 1)$, and it passes through the point $N(3, 2)$. What is the equation of this quadratic function?

Solution:
\textbf{Solution:} Let the quadratic function be defined as $y=a(x-2)^2+1$.\\
By substituting $x=3$ and $y=2$, we get $a+1=2$, solving for $a$ gives $a=\boxed{1}$.\\
Therefore, the equation of the quadratic function is $y=(x-2)^2+1$ (or written as $y=x^2-4x+5$).


Question: Given the graph of the quadratic function $y=ax^2+bx+c$ with its vertex at $M(2, 1)$, and it passes through point $N(3, 2)$. If the graph of the linear function $y=-\frac{4}{3}x-4$ intersects the x-axis at point $A$ and the y-axis at point $B$. Let $P$ be a moving point on the parabola. Draw line $PQ\parallel y$ axis intersecting line $AB$ at point $Q$. A circle with diameter $PQ$ intersects line $AB$ at point $D$. Let the x-coordinate of point $P$ be $n$. What is the value of $n$ when the length of segment $DQ$ is minimized, and what is the minimum value?

Solution:
\textbf{Solution:} Given $P\left(n, n^{2}-4n+5\right)$, $Q\left(n, -\frac{4}{3}n-4\right)$. \\
Therefore, $PQ=n^{2}-4n+5-\left(-\frac{4}{3}n-4\right)=n^{2}-\frac{8}{3}n+9=\left(n-\frac{4}{3}\right)^{2}+\frac{65}{9}$. \\
Hence, when $n=\frac{4}{3}$, $PQ$ reaches its minimum value of $\frac{65}{9}$. \\
It is easy to prove that $\triangle DPQ \sim \triangle OAB$, \\
therefore $\frac{DQ}{PQ}=\frac{OB}{AB}$, \\
resulting in $DQ=\frac{4}{5}PQ$. \\
Thus, when $n=\frac{4}{3}$, $DQ$ reaches its minimum value, which is $\boxed{\frac{52}{9}}$.


Question: As shown in the figure, $\Delta DBE$ is inscribed in $\odot O$, with $BD$ as the diameter and $DE=EB$. Point $C$ is on $\odot O$ (not coinciding with $D$, $B$, $E$), $\angle A=45^\circ$, point $A$ is on line $CD$, and line $AB$ is drawn. $\triangle ABD \sim \triangle CBE$, $DC=6$, $DB=10$. What is the length of segment $CE$?

Solution:
\textbf{Solution:} By the given problem, \\
since $\angle BCD=90^\circ$, \\
therefore $BC= \sqrt{BD^{2}-CD^{2}}=8$, \\
since $\triangle ABC$ and $\triangle BED$ are isosceles right triangles,\\
therefore $AB= \sqrt{2}BC= 8\sqrt{2}$, $BE= \frac{\sqrt{2}}{2}BD= 4\sqrt{2}$,\\
therefore $AD=AC-CD=8-6=2$,\\
by similarity $\triangle ADB\sim \triangle CEB$,\\
therefore $\frac{BE}{BC}=\frac{CE}{AD}$, that is $\frac{4\sqrt{2}}{8}=\frac{CE}{2}$,\\
solving for $CE$ yields $CE= \boxed{\sqrt{2}}$.


Question: As shown in the diagram, $\Delta DBE$ is inscribed in $\odot O$, with $BD$ as the diameter and $DE=EB$. Point $C$ is on $\odot O$ (not coincident with $D$, $B$, or $E$), $\angle A=45^\circ$, and point $A$ is on line $CD$. Connect $AB$. If line $BC$ intersects line $DE$ at point $F$, when $\frac{DC}{CB}=\frac{1}{3}$, find the value of $\frac{BF}{DF}$.

Solution:
\textbf{Solution:} Join $DG$, construct $EH \perp BC$ at point $H$,\\
since $\frac{DC}{CB} = \frac{1}{3}$, let $DC = k$, $CB = 3k$,\\
as $\triangle ABC$ is an isosceles right-angled triangle, then $BC = AC = 3k$, $AD = 2k$,\\
since $BD$ is the diameter,\\
therefore $\angle DGB = \angle DGA = 90^\circ$,\\
since $\angle A = 45^\circ$,\\
therefore $DG = AD\sin\angle A = \sqrt{2}k$,\\
since $\angle ABC = \angle DBE$, that is $\angle DBG + \angle CBD = \angle CBH + \angle CBD$,\\
therefore $\angle DBG = \angle EBH$,\\
therefore $\triangle DBG \sim \triangle EHB$,\\
therefore $\frac{DB}{EB} = \frac{DG}{EH} = \sqrt{2}$,\\
therefore $EH = k$,\\
since $DC \parallel EH$,\\
therefore $\triangle DCF \sim \triangle EHF$,\\
therefore $\frac{EF}{DF} = \frac{EH}{DC} = 1$;\\
As shown in figure 2, join $DG$, construct $EH \perp BC$ at point $H$,\\
since $\frac{DC}{CB} = \frac{1}{3}$, let $DC = k$, $CB = 3k$,\\
as $\triangle ABC$ is an isosceles right-angled triangle, then $BC = AC = 3k$, $AD = 4k$,\\
since $BD$ is the diameter,\\
therefore $\angle DGB = \angle DGA = 90^\circ$,\\
since $\angle A = 45^\circ$,\\
therefore $DG = \sqrt{2}k$,\\
since $\angle ABC = \angle DBE$,\\
therefore $\angle DBG = \angle EBH$,\\
therefore $\triangle DBG \sim \triangle EHB$,\\
therefore $\frac{DB}{EB} = \frac{DG}{EH} = \sqrt{2}$,\\
therefore $EH = 2k$,\\
since $DC \parallel EH$,\\
therefore $\triangle DCF \sim \triangle EHF$,\\
therefore $\frac{EF}{DF} = \frac{EH}{DC} = 2$;\\
In conclusion, the value of $\frac{BF}{DF}$ is $\boxed{1}$ or $\boxed{2}$.


Question: As shown in the figure, in the square $ABCD$, $M$ is a point on $BC$, $ME\perp AM$, $ME$ intersects $CD$ at $F$, and intersects the extension of $AD$ at point $E$. $\triangle ABM\sim \triangle MCF$; if $AB=4$ and $BM=2$, what is the area of $\triangle DEF$?

Solution:
\textbf{Solution:}

Since $\triangle ABM\sim \triangle MCF$,

it follows that $\frac{AB}{MC}=\frac{BM}{CF}$,

Since quadrilateral $ABCD$ is a square, $AB=4$, $BM=2$,

therefore $BC\parallel AD$, $BC=CD=4$, $MC=BC-BM=4-2=2$,

$\angle FDE=\angle CDA=90^\circ$,

thus $CF=\frac{MC\cdot BM}{AB}=\frac{2\times 2}{4}=1$,

thus $DF=DC-CF=4-1=3$,

Since $BC\parallel AD$,

it follows that $\triangle MCF\sim \triangle EDF$,

thus $\frac{MC}{DE}=\frac{CF}{DF}=\frac{1}{3}$,

thus $DE=6$,

thus $S_{\triangle DEF}=\frac{1}{2}DF\times DE=\frac{1}{2}\times 3\times 6=\boxed{9}$.


Question: As shown in the figure, in $\triangle ABC$, points D, E, and F are located on sides AB, BC, and AC, respectively, with $DE \parallel AC$ and $EF \parallel AB$. $\triangle BDE \sim \triangle EFC$. If $\frac{AF}{FC} = \frac{1}{2}$ and $AD = 6$, find the length of AB.

Solution: \textbf{Solution:} Given $\because$ EF $\parallel$ AB, $\therefore$ $\frac{BE}{EC}=\frac{AF}{FC}=\frac{1}{2}$, and $\because$ DE $\parallel$ AC, $\therefore$ $\frac{BD}{AD}=\frac{BE}{EC}=\frac{1}{2}$, thus $\frac{BD}{AD}=\frac{AF}{FC}=\frac{1}{2}$. Given AD＝6, $\therefore$ $\frac{BD}{6}=\frac{1}{2}$, thus BD=3, hence AB＝\boxed{9}.


Question: As shown in the figure, $AC$ is the diameter of $\odot O$, $BC$ is the chord of $\odot O$, point $P$ is a point outside $\odot O$, and $PB$ and $AB$ are connected, with $\angle PBA=\angle C$. $PB$ is the tangent of $\odot O$; connect $OP$, if $OP\parallel BC$ and $OP=8$, the radius of $\odot O$ is $3$, find the length of $BC$.

Solution:
Solution:\\
Since the radius of $\odot O$ is $3$,\\
hence $OB=3$, $AC=6$,\\
Since $OP \parallel BC$,\\
thus $\angle CBO = \angle BOP$,\\
Since $OC=OB$,\\
thus $\angle C = \angle CBO$,\\
thus $\angle C = \angle BOP$,\\
Also, since $\angle ABC = \angle PBO = 90^\circ$,\\
thus $\triangle ABC \sim \triangle PBO$,\\
thus $\frac{BC}{OB} = \frac{AC}{OP}$,\\
that is $\frac{BC}{3} = \frac{6}{8}$,\\
hence $BC = \boxed{\frac{9}{4}}$.


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 $F$. $\triangle ABE \sim \triangle DFA$; If $AB = 10$ and $BC = 4$, find the length of $DF$.

Solution: \textbf{Solution:} Given that $E$ is the midpoint of $BC$, and $BC=4$. Since quadrilateral $ABCD$ is a rectangle, it follows that $AD=BC=4$. Hence, $BE=EC=\frac{1}{2}BC=2$. Since $\angle ABE=90^\circ$, we have $AE=\sqrt{AB^{2}+BE^{2}}=2\sqrt{26}$. Given that $\triangle ABE\sim \triangle DFA$, we have $\frac{AE}{AD}=\frac{AB}{DF}$. Thus, $DF=\frac{AB\cdot AD}{AE}=\boxed{\frac{10\sqrt{26}}{13}}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, with $C$ and $D$ being two points on the circle. Connect $AC$ and $CD$, and let $AC=CD$. Extend $DC$ to intersect with the extension of $BA$ at point $E$. $\triangle EAC\sim \triangle ECO$; if $\tan\angle EOC=\frac{3}{4}$, find the value of $\frac{EC}{EO}$.

Solution:
\textbf{Solution:} Draw $CF\perp EO$ through point C, \\
since $\tan\angle EOC=\frac{3}{4}$, \\
therefore, $\frac{CF}{OF}=\frac{3}{4}$. Let $CF=3x$, then $OF=4x$, \\
thus, $OC= \sqrt{(3x)^{2}+(4x)^{2}}=5x=OA$, \\
therefore, $AF=5x-4x=x$, \\
thus, $AC= \sqrt{(3x)^{2}+x^{2}}=\sqrt{10}x$, \\
therefore, $\frac{AC}{OC}=\frac{\sqrt{10}x}{5x}=\frac{\sqrt{10}}{5}$. From (1), we get $\triangle EAC\sim \triangle ECO$, \\
thus, $\frac{EC}{EO}=\frac{AC}{OC}$, \\
therefore, $\frac{EC}{EO}=\boxed{\frac{\sqrt{10}}{5}}$.


Question: As shown in the figure, in rectangle $ABCD$, point $M$ is on $BC$, and $EM \perp AM$ meets the extension of $AD$ at point $E$. $\triangle ABM \sim \triangle EMA$; if $AB=4$ and $BM=3$, find the value of $AE$.

Solution:
Solution: Since $AB=4$ and $BM=3$, \\
we have $AM= \sqrt{AB^{2}+BM^{2}}=\sqrt{4^{2}+3^{2}}=5$, \\
Because $\triangle ABM \sim \triangle EMA$, \\
it follows that $\frac{AM}{AE}=\frac{AB}{AM}$, \\
Thus, $AE= \frac{AM^{2}}{AB}=\boxed{\frac{25}{4}}$.


Question: As shown in the figure, point C is on the semicircle O with diameter AB, and $AB=2$. AD bisects $\angle BAC$ and intersects BC at point D. CP bisects $\angle BCA$ and intersects AD at point P. $PF\perp AC$, $PE\perp BC$. Quadrilateral CEPF is a square; what is the maximum value of $AC\cdot BC$?

Solution: \textbf{Solution:} Let line $CG\perp AB$ pass through point C. From the area formula of triangle $ABC$, $S_{\triangle ABC}=\frac{1}{2}AB\cdot CG=\frac{1}{2}AC\cdot BC$, we know that when $CG$ is at its maximum, $AC\cdot BC$ will also reach its maximum value, that is, $CG=\frac{1}{2}AB=\frac{1}{2}\times 2=1$. Using the area formula of a triangle, $S_{\triangle ABC}=\frac{1}{2}AB\cdot CG=\frac{1}{2}AC\cdot BC$, since $AB=2$, it follows that $\frac{1}{2}\times 2\times 1=\frac{1}{2}AC\cdot BC$, therefore, $AC\cdot BC=\boxed{2}$.


Question: As shown in the figure, point C is a point on the semicircle O with AB as its diameter, and $AB=2$. AD bisects $\angle BAC$ and intersects BC at point D. CP bisects $\angle BCA$ and intersects AD at point P. $PF\perp AC$, $PE\perp BC$. Find the minimum value of $\frac{1}{AC}+\frac{1}{DC}$.

Solution:
\textbf{Solution:} Let $PE=PF=CE=CF=x$,\\
$\because$ $PE\perp BC$, $AC\perp BC$,\\
$\therefore PE\parallel AC$,\\
$\therefore \triangle PED\sim \triangle ACD$,\\
$\therefore \frac{PE}{AC}=\frac{PD}{AD}$ (1);\\
Similarly, $PF\parallel BC$, $\triangle PAF\sim \triangle DAC$,\\
$\therefore \frac{PF}{CD}=\frac{AP}{AD}$ (2),\\
From (1) + (2), we get $\frac{PE}{AC}+\frac{PF}{CD}=\frac{PD}{AD}+\frac{AP}{AD}=\frac{AD}{AD}=1$,\\
$\therefore \frac{PE}{AC}+\frac{PF}{CD}=1$,\\
that is $\frac{x}{AC}+\frac{x}{CD}=1$,\\
$\therefore \frac{1}{AC}+\frac{1}{DC}=\frac{1}{x}$;\\
When x is at its maximum, $\frac{1}{AC}+\frac{1}{DC}$ is at its minimum;\\
$\because AD$ bisects $\angle BAC$,\\
$\therefore$ Point P is the incenter of $\triangle ACB$,\\
$\therefore PE, PF$ are the radii of the incircle;\\
Draw $PH\perp AB$, with H as the foot,\\
then it is easy to see $AF=AH$, $BE=BH$,\\
$\therefore AF+BE=AH+BH=AB$,\\
$\therefore CE=CF=\frac{AC+BC-AB}{2}$,\\
Let $AC=b$, $BC=a$, $AB=c=2$,\\
$\therefore x=\frac{a+b-c}{2}=\frac{a+b-2}{2}=\frac{a+b}{2}-1$,\\
$\because AC^{2}+BC^{2}=AB^{2}$,\\
$\therefore a^{2}+b^{2}=4$,\\
$\because (a-b)^{2}=a^{2}+b^{2}-2ab\ge 0$,\\
$\therefore 2ab\le a^{2}+b^{2}=4$,\\
$\therefore ab\le 2$,\\
$\because (a+b)^{2}=a^{2}+b^{2}+2ab=4+2ab\le 4+4=8$,\\
$\therefore a+b\le 2\sqrt{2}$,\\
$\therefore$ the maximum value of $a+b$ is $2\sqrt{2}$;\\
$\therefore x=\frac{a+b}{2}-1=\frac{2\sqrt{2}}{2}-1=\sqrt{2}-1$,\\
$\therefore$ the maximum value of $x$ is $\sqrt{2}-1$,\\
$\therefore \frac{1}{x}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$,\\
$\therefore$ the minimum value of $\frac{1}{AC}+\frac{1}{DC}$ is $\boxed{\sqrt{2}+1}$.


Question: As shown in the figure, in the Cartesian coordinate system, the sides of the rhombus OABC are such that OC is on the x-axis, and B is at (18, 6). The graph of the inverse proportion function $y=\frac{k}{x} (k \neq 0)$ passes through point A and intersects OB at point E. What is the length of the sides of the rhombus OABC?

Solution:
\textbf{Solution:} Draw line $BF \perp x$-axis at point F, according to the given information, we have $BF=6$ and $OF=18$.\\
Since quadrilateral OABC is a rhombus,\\
it follows that $OC=BC$.\\
In $\triangle BCF$, $6^{2}+(18-BC)^{2}=BC^{2}$.\\
Solving this, we find $BC=\boxed{10}$,\\
thus, the side length of rhombus OABC is $\boxed{10}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the side OC of the rhombus OABC lies on the x-axis, with B(18,6). The graph of the inverse proportion function $y=\frac{k}{x}$ ($k \ne 0$) passes through point A and intersects OB at point E. Find the value of k.

Solution: \textbf{Solution:} From equation (1), we find that the coordinate of point A is (8, 6). Substituting point A (8, 6) into $y= \frac{k}{x}$, we solve for $k=\boxed{48}$.


Question: As shown in the figure, in the Cartesian coordinate system, the sides of the rhombus OABC are located with OC on the x-axis, and point B is at (18, 6). The graph of the inverse proportion function $y=\frac{k}{x}$ ($k\neq 0$) passes through point A and intersects with OB at point E. Find the value of OE: EB.

Solution:
\textbf{Solution:} Set $E(a, \frac{48}{a})$. Draw line $EG$ perpendicular to the x-axis at point $G$. According to the problem, it is known that $OG=a$ and $EG=\frac{48}{a}$. From the drawing, we know that $EG \parallel BF$,\\
$\therefore \triangle OGE \sim \triangle OBF$,\\
$\therefore \frac{\frac{48}{a}}{6}=\frac{a}{18}$. Solving this gives $a=12$,\\
$\therefore \frac{OE}{OB}=\frac{OG}{OF}=\frac{12}{18}=\frac{2}{3}$,\\
$\therefore \frac{OE}{EB}=\boxed{2}$.


Question: As shown in the figure, inside $\triangle ABC$, points $D$, $E$, and $F$ lie on $AB$, $BC$, and $AC$ respectively, with $DE\parallel AC$, and $EF\parallel AB$. $\triangle BDE\sim \triangle EFC$. Given that $\frac{AF}{FC}=\frac{4}{3}$ and the area of $\triangle EFC$ is 9, find the area of $\triangle ABC$.

Solution:
\textbf{Solution:} Given that $FE\parallel AB$,\\
$\therefore \angle A=\angle EFC, \angle B=\angle FEC$.\\
$\therefore \triangle ABC\sim \triangle FEC$.\\
Given that $\frac{AF}{FC}=\frac{4}{3}$,\\
$\therefore AF=\frac{4}{3}FC$.\\
$\therefore \frac{FC}{AC}=\frac{FC}{AF+FC}=\frac{FC}{\frac{4}{3}FC+FC}=\frac{3}{7}$.\\
$\therefore \frac{S_{\triangle EFC}}{S_{\triangle ABC}}=\left(\frac{3}{7}\right)^2=\frac{9}{49}$.\\
Since the area of $\triangle EFC$ is 9,\\
$\therefore S_{\triangle ABC}=\boxed{49}$.


Question: As shown in the figure, in right triangle $ABC$ with $\angle ACB=90^\circ$, we have $AB=25$ and $AC=15$. Point $P$ starts from point $B$ and moves towards $A$ at a speed of $2$ units per second. Simultaneously, point $Q$ starts from $C$ and moves towards $B$ at a speed of $1$ unit per second. When one point reaches its endpoint, the other point stops moving. Connect $PQ$ and on the ray $PB$, take $PN=PQ$ to form quadrilateral $PQMN$ with $PN$ and $PQ$ as sides. Let the movement time be $t$ seconds ($t>0$). What is the length of $BC$?

Solution: \textbf{Solution:} From the given conditions, we have $BC=\boxed{20}$.


Question: As shown in the figure, in $\mathrm{Rt}\triangle ABC$ where $\angle ACB=90^\circ$, $AB=25$, and $AC=15$, point $P$ starts from point $B$ and moves toward $A$ at a speed of $2$ units per second. At the same time, point $Q$ starts from $C$ and moves toward $B$ at a speed of $1$ unit per second. When one point reaches its destination, the other point stops moving. Connect $PQ$. On ray $PB$, take $PN=PQ$ and construct $\square PQMN$ using $PN$ and $PQ$ as sides. Let the movement time be $t$ seconds ($t>0$). Find the value of $t$ when $\square PQMN$ is a square.

Solution:
\textbf{Solution:} As shown in the figure,\\
$\because$ Quadrilateral PQMN is a square,\\
$\therefore$ $\angle$BPQ$=90^\circ$,\\
$\therefore$ $\triangle BPQ \sim \triangle BCA$,\\
$\therefore$ $\frac{PB}{BQ}=\frac{BC}{AB}=\frac{20}{25}=\frac{4}{5}$,\\
$\therefore$ BP$=\frac{4}{5}$BQ,\\
$\because$ BP$=2t$, CQ$=t$,\\
$\therefore$ $2t=\frac{4}{5}(20-t)$,\\
$\therefore$ $t=\boxed{\frac{40}{7}}$;


Question: As shown in the figure, within $\triangle ABC$, $\angle ACB=90^\circ$, $AB=25$, and $AC=15$. Point $P$ starts from point $B$ and moves towards $A$ at a speed of $2$ units per second, while point $Q$ starts from $C$ and moves towards $B$ at a speed of $1$ unit per second. The movement of one point ceases immediately when the other point arrives at its destination. Connect $PQ$, and on ray $PB$, take $PN=PQ$, then construct $\square PQMN$ with $PN$ and $PQ$ as edges. Let the movement time be $t$ seconds ($t>0$). Construct the symmetric point $C'$ of point $C$ with respect to the line $PQ$. When points $C$, $Q$, and $C'$ are not collinear, and $\angle CQC'$ equals twice the internal angle of $\triangle ABC$, directly write out the value of $t$.

Solution: \textbf{Solution:} We have $t=\boxed{\frac{40}{7}}$ or $t=\boxed{\frac{100}{21}}$.


Question: As shown in the figure, $AB=4$, $CD=6$, point $F$ is on $BD$, lines $BC$ and $AD$ intersect at point $E$, and $AB \parallel CD \parallel EF$. If $AE=3$, what is the length of $ED$?

Solution:
\textbf{Solution:} Given $AB\parallel CD$,\\
$\therefore \triangle AEB\sim \triangle DEC$,\\
$\therefore \frac{AE}{DE}=\frac{AB}{CD}$,\\
Given $AB=4$, $CD=6$, $AE=3$,\\
$\therefore \frac{3}{DE}=\frac{4}{6}$,\\
Solving gives: $DE=\boxed{\frac{9}{2}}$;


Question: As shown in the figure, $AB=4$, $CD=6$, point $F$ is on $BD$, $BC$ and $AD$ intersect at point $E$, and $AB \parallel CD \parallel EF$. What is the length of $EF$?

Solution:
\textbf{Solution:} Since $CD\parallel EF$, \\
$\therefore \triangle BEF\sim \triangle BCD$, \\
$\therefore \frac{EF}{CD}=\frac{BF}{BD}$, \\
Similarly: $\frac{EF}{AB}=\frac{DF}{BD}$, \\
$\therefore \frac{EF}{CD}+\frac{EF}{AB}=\frac{BF}{BD}+\frac{DF}{BD}=1$, \\
$\therefore \frac{EF}{6}+\frac{EF}{4}=1$, \\
Solving for $EF$, we get $EF=\boxed{\frac{12}{5}}$.


Question: As shown in the figure, point B is a point on line segment AC, and $AB=21$, $BC= \frac{1}{3}AB$. What is the length of line segment AC?

Solution:
\textbf{Solution:} Given that $AB=21$ and $BC= \frac{1}{3}AB$, \\
it follows that $BC=7$ and $AC=AB+BC=21+7=\boxed{28}$.


Question: As shown in the figure, point B is a point on line segment AC, with $AB=21$ and $BC= \frac{1}{3}AB$. If point O is the midpoint of line segment AC, what is the length of line segment QB?

Solution:
\textbf{Solution:} Given that point O is the midpoint of line segment AC, \\
thus $OA = \frac{1}{2}AB = \frac{1}{2} \times 28 = 14$, \\
$OB = AB - OA = 21 - 14 = \boxed{7}$.


Question: As shown in Figure 1, it is known that point $B$ is on line segment $AC$, and point $D$ is on line segment $AB$. If $AB=10\, \text{cm}$, $BC=6\, \text{cm}$, and $D$ is the midpoint of line segment $AC$, what is the length of line segment $DB$ in $\text{cm}$?

Solution:
\textbf{Solution:} As shown in Figure 1 in the problem, \\
since AB=10cm, and BC=6cm, \\
therefore AC=AB+BC=10+6=16cm. \\
Also, since D is the midpoint of segment AC, \\
therefore DC=$\frac{1}{2}AC=\frac{1}{2}\times 16=8$cm, \\
therefore DB=DC−BC=8−6=$\boxed{2}$cm.


Question: As shown in Figure 2, it is known that point $B$ is on line segment $AC$, and point $D$ is on line segment $AB$. If $BD = \frac{1}{4}AB = \frac{1}{3}CD$, with $E$ being the midpoint of line segment $AB$ and $EC=16\, \text{cm}$, find the length of line segment $AC$.

Solution:
\textbf{Solution:} Let $BD=x$,\\
since $BD=\frac{1}{4}AB=\frac{1}{3}CD$,\\
thus $AB=4BD=4x$ and $CD=3BD=3x$,\\
therefore $BC=DC-DB=3x-x=2x$,\\
thus $AC=AB+BC=4x+2x=6x$.\\
Since $E$ is the midpoint of the segment $AB$,\\
thus $BE=\frac{1}{2}AB=\frac{1}{2}\times 4x=2x$,\\
thus $EC=BE+BC=2x+2x=4x$.\\
Furthermore, since $EC=16$,\\
thus $4x=16$,\\
solving this gives: $x=4$,\\
therefore $AC=6x=6\times 4=\boxed{24}$.


Question: As shown in the figure, \(C\) is a point on line segment \(AB\), \(M\) is the midpoint of line segment \(AC\), and \(N\) is the midpoint of line segment \(BC\). If \(AM = \frac{5}{4} BC = 5 \, \text{cm}\), find the length of \(MN\).

Solution:
\textbf{Solution:} Since $M$ is the midpoint of segment $AC$, $AM=\frac{5}{4}BC=5\text{cm}$,\\
thus $AM=CM=5\text{cm}$, and $BC=4\text{cm}$.\\
Additionally, since $N$ is the midpoint of segment $BC$,\\
thus $CN=\frac{1}{2}BC=2\text{cm}$,\\
therefore $MN=MC+CN=\boxed{7\text{cm}}$.


Question: As shown in the figure, C is a point on the line segment AB, M is the midpoint of the line segment AC, N is the midpoint of the line segment BC. If $AC=x$ cm, and $BC= (10-x)$ cm, then what is the length of $MN$ in cm?

Solution:
\textbf{Solution:} Since $M$ is the midpoint of line segment $AC$ and $N$ is the midpoint of line segment $BC$, \\
it follows that $CM=\frac{1}{2}AC=\frac{1}{2}x$ and $CN=\frac{1}{2}BC=\frac{1}{2}\left(10-x\right)$, \\
therefore, $MN=CM+CN=\boxed{5}$ cm.


Question: As shown in the figure, it is known that line segment $AB=m$ (where $m$ is a constant), point $C$ is a point on the line $AB$ (not coinciding with $A$ or $B$), and points $P$, $Q$ are respectively on segments $BC$, $AC$, and satisfy $CQ=2AQ$, $CP=2BP$. As shown in Figure 1, with point $C$ on segment $AB$, what is the length of $PQ$ (expressed in terms of $m$)?

Solution:
\textbf{Solution:} Let $AQ=a$ and $BP=b$,\\
then $CQ=2a$ and $CP=2b$,\\
since $AB=AQ+CQ+CP+BP=a+2a+2b+b=3a+3b=m$,\\
it follows that $a+b=\frac{m}{3}$,\\
thus $PQ=CQ+CP=2a+2b=2(a+b)=\boxed{\frac{2m}{3}}$.


Question: As shown in the figure, it is known that the length of segment $AB=m$ (where $m$ is a constant), point $C$ is on the line segment $AB$ (not coinciding with $A$ or $B$), points $P$ and $Q$ are respectively on the segments $BC$ and $AC$, and it satisfies $CQ=2AQ$, $CP=2BP$. As shown in figure 2, if point $C$ is to the left of point $A$, while point $P$ is on the segment $AB$ (not coinciding with the endpoints), find the value of $2AP+CQ-2PQ$.

Solution:
\textbf{Solution:} Let $AQ=x$, and $BP=y$,

then we have $CQ= 2x$, and $CP= 2y$,

thus $AP=CP-CA=2y- 3x$, $PQ=CP-CQ=2y- 2x$,

hence $2AP+CQ- 2PQ=2(2y- 3x) + 2x - 2(2y-2x)=4y - 6x+2x-4y+4x=0$.

$\therefore$ The value of $2AP+CQ- 2PQ$ is $\boxed{0}$.


Question: As shown in the figure, $C$ is the midpoint of the line segment $AB$, and $D$ is on the line segment $CB$, and $DA=6$, $DB=4$. Find the length of $AB$.

Solution:
\textbf{Solution:} Since $DA=6, DB=4$,\\
it follows that $AB=6+4=\boxed{10}$.


Question: As shown in the figure, $C$ is the midpoint of segment $AB$, and $D$ is on segment $CB$, with $DA=6, DB=4$. Find the length of $CD$.

Solution:
\textbf{Solution:} Since point C is the midpoint of the line segment $AB$,\\
$\therefore AC=\frac{1}{2}AB=5$.\\
$\therefore CD=AD-AC=6-5=\boxed{1}$.


Question: As shown in the figure, $C$ is a point on line segment $AD$, $B$ is the midpoint of $CD$, and $AD=13\, \text{cm}$, $BC=3\, \text{cm}$. What is the length of $AC$ in centimeters?

Solution:
\textbf{Solution:} Since $B$ is the midpoint of $CD$, and $BC=3\text{cm}$, \\
then $CD=2BC=6\text{cm}$. \\
Since $AD=13\text{cm}$, \\
then $AC=AD-CD=13-6=\boxed{7}\text{cm}$.


Question: As shown in the diagram, point \(C\) lies on segment \(AD\), and \(B\) is the midpoint of \(CD\), with \(AD=13 \, \text{cm}\), and \(BC=3 \, \text{cm}\). If point \(E\) is on line \(AD\), and \(EA=4 \, \text{cm}\), what is the length of \(BE\) in cm?

Solution:
\textbf{Solution:} As shown in Figure 1, when point E is on segment AC, \\
$\because AB=AC+BC=10\text{cm}$, $EA=4\text{cm}$, \\
$\therefore BE=AB-AE=10-4=\boxed{6}\text{cm}$. \\
As shown in Figure 2, when point E is on the extension line of segment CA, \\
$\because AB=10\text{cm}$, $AE=4\text{cm}$, \\
$\therefore BE=AE+AB=14\text{cm}$. \\
In summary, the length of BE is either \boxed{6\text{cm}} or \boxed{14\text{cm}}.


Question: As shown in the figure, point $C$ is a point on line segment $AB$, and points $M$, $N$, $P$ are the midpoints of line segments $AC$, $BC$, $AB$, respectively. If $AB = 10\, \text{cm}$, what is the length of line segment $MN$?

Solution:
\textbf{Solution:} Since $M$ and $N$ are the midpoints of $AC$ and $BC$ respectively,

$\therefore$ $MC = \frac{1}{2}AC$, $CN = \frac{1}{2}BC$,

$\therefore$ $MN = MC + CN = \frac{1}{2}AC + \frac{1}{2}BC = \frac{1}{2}(AC + BC) = \frac{1}{2}AB = \frac{1}{2} \times 10 = \boxed{5}$ (cm).


Question: As shown in the figure, point \(C\) is a point on line segment \(AB\), and points \(M\), \(N\), and \(P\) are the midpoints of line segments \(AC\), \(BC\), and \(AB\), respectively. If \(AC = 3\, \text{cm}\) and \(CP = 1\, \text{cm}\), what is the length of line segment \(PN\)?

Solution:
\textbf{Solution:} \\
Since $AC=3$ and $CP=1$, \\
thus $AP=AC+CP=4$, \\
Since point P is the midpoint of segment AB, \\
thus $AB=2AP=8$, $CB=AB-AC=5$, \\
Since point N is the midpoint of segment CB, \\
thus $CN=\frac{1}{2}CB=\frac{5}{2}$, \\
therefore $PN=CN-CP=\frac{5}{2}-1=\boxed{\frac{3}{2}}$ (cm).


Question: As shown in the figure, it is known that point C is on line segment AB, and points M and N are on line segments AC and BC, respectively, with $AM=2MC$, $BN=2NC$. If $AC=9$ and $BC=6$, what is the length of line segment MN?

Solution:
\textbf{Solution:} Given that $AC=9$ and $BC=6$, and also that $AM=2MC$ and $BN=2NC$.\\
Therefore, $MC= \frac{1}{3} AC=3$ and $NC= \frac{1}{3} BC=2$,\\
Therefore, $MN=MC+NC=3+2=\boxed{5}$.


Question: As shown in the figure, it is known that point C is on line segment AB, and points M and N are on line segments AC and BC, respectively, with $AM=2MC$ and $BN=2NC$. If $MN=5$, what is the length of line segment AB?

Solution:
\textbf{Solution:} Since $AM=2MC$ and $BN=2NC$,\\
it follows that $MC= \frac{1}{3} AC$ and $NC= \frac{1}{3} BC$,\\
Thus $MN=MC+NC= \frac{1}{3} AC+ \frac{1}{3} BC= \frac{1}{3} AB$,\\
If $MN=5$, then $AB=3MN=15$,\\
Answer: The length of $AB$ is $\boxed{15}$.


Question: As shown in the figure, point $C$ is the midpoint of line segment $AB$, point $D$ is the midpoint of line segment $AC$, and $BE=\frac{1}{3}BC$. If $BE=2\,\text{cm}$, what is the length of $AB$ in cm?

Solution:
\textbf{Solution:} Since $BE=\frac{1}{3}BC$ and $BE=2\,\text{cm}$, \\
then $BC=3BE=6\,\text{cm}$. \\
Given that point $C$ is the midpoint of line segment $AB$, \\
it follows that $AB=2BC=\boxed{12}\,\text{cm}$.


Question: As shown in the figure, point $C$ is the midpoint of line segment $AB$, point $D$ is the midpoint of line segment $AC$, and $BE=\frac{1}{3}BC$. If $DE=14\,\text{cm}$, find the length of $AB$.

Solution:
\textbf{Solution:} Let $BE = x\,cm$, hence $BC = 3x\,cm$,\\
since point $C$ is the midpoint of segment $AB$,\\
therefore $AB = 2BC = 6x\,cm$, $AC = 3x\,cm$,\\
since point $D$ is the midpoint of segment $AC$,\\
therefore $AD = \frac{1}{2}AC = \frac{3}{2}x\,cm$,\\
therefore $DE = AB - AD - BE = 6x - \frac{3}{2}x - x = \frac{7}{2}x\,cm$,\\
since $DE = 14\,cm$,\\
therefore $\frac{7}{2}x = 14$, solving for $x$ yields: $x = 4$,\\
therefore $AB = 6x = \boxed{24\,cm}$.


Question: As shown in the figure, it is known that the length of segment AB is $a$, and segment AB is extended to point C so that $BC = \frac{1}{2}AB$. Find the length of segment AC (expressed as an algebraic expression involving $a$).

Solution:
\textbf{Solution:} Given that $\because AB=a$,\\
$\therefore BC=\frac{1}{2}a$,\\
$\therefore AC=AB+BC=a+\frac{1}{2}a=\boxed{\frac{3}{2}a}$.


Question: As shown in the figure, it is known that the length of line segment $AB$ is $a$. Extend line segment $AB$ to point $C$, making $BC=\frac{1}{2}AB$. Take the midpoint $D$ of line segment $AC$. If $DB=2$, find the value of $a$.

Solution:
Solution: Since $AC = \frac{3}{2}a$ and D is the midpoint of AC,\\
it follows that $CD = \frac{1}{2}AC = \frac{3}{4}a$,\\
Since $BC = \frac{3}{2}a$ and $DB = CD - BC = 2$,\\
it follows that $\frac{3}{4}a - \frac{3}{2}a = 2$,\\
Solving this, we find: $a = \boxed{8}$.


Question: As shown in the figure, point $C$ is a point on the line segment $AB$ ($AC>BC$), $D$ is on the line segment $BC$, with $BD=2CD$, and point $E$ is the midpoint of $AB$. If $AD=10$ and $EC=3CD$, what is the length of line segment $CD$?

Solution:
\textbf{Solution:} Let $CD=x$, then $BD=2CD=2x$,\\
since $AD=10$,\\
thus $AB=10+2x$,\\
since point E is the midpoint of $AB$,\\
thus $AE=\frac{1}{2}AB=5+x=BE$,\\
thus $EC=EB-CB=5+x-3x=5-2x$,\\
thus $5-2x=3x$,\\
solving gives $x=\boxed{1}$,\\
hence $CD=\boxed{1}$.


Question: As shown in the figure, point C is a point on segment $AB$ ($AC>BC$), D is on segment $BC$, $BD=2CD$, and point E is the midpoint of $AB$. If $AC=2BC$, find the value of $\frac{EC}{BD}$.

Solution:
\textbf{Solution:} From the given information, we have $BC=CD+BD=3x$, $AC=2BC=6x$,
\[\therefore AB=3x+6x=9x,\]
\[\because \text{E is the midpoint of } AB,\]
\[\therefore EB=\frac{1}{2}AB=4.5x,\]
\[\therefore EC=EB-BC=1.5x,\]
\[\therefore \frac{EC}{BD}=\frac{1.5x}{2x}=\boxed{\frac{3}{4}}.\]


Question: As shown in the figure, it is known that the length of line segment AB is 24 cm. AB is extended to C such that BC is equal to $\frac{1}{2}$ of AB. What is the length of AC?

Solution:
\textbf{Solution:} Given that $BC = \frac{1}{2}AB$ and $AB = 24$ cm, \\
then $BC = \frac{1}{2} \times 24 = 12\, \text{cm}$, \\
therefore $AC = AB + BC = \boxed{36}\, \text{cm}$.


Question: As shown in the figure, it is known that the segment $AB=24\,cm$, extending $AB$ to $C$ such that $BC=\frac{1}{2}AB$. If $D$ is the midpoint of $AB$, and $E$ is the midpoint of $AC$, what is the length of $DE$ in $cm$?

Solution:
\textbf{Solution:} Since $D$ is the midpoint of $AB$, and $E$ is the midpoint of $AC$,\\
$\therefore$ $AD = \frac{1}{2}AB = 12$cm, $AE = \frac{1}{2}AC = 18$cm,\\
$\therefore$ $DE = 18 - 12 = \boxed{6}$cm.


Question: As shown in the diagram, extend the line segment $AB$ to $C$ so that $BC=4AB$. Point $D$ is the midpoint of the line segment $BC$. If $CD=4\,cm$, find the length of $AC$.

Solution:
\textbf{Solution:} Let $AB=x$,\\
$\therefore BC=4AB=4x$, $AC=AB+BC=5x$\\
$\because$ point $D$ is the midpoint of line segment $BC$,\\
$\therefore CD=BD=\frac{1}{2}BC=2x$,\\
then $2x=4$, solving this yields: $x=2$,\\
$CD=BD=2x=4\text{cm}$,\\
$\therefore AC=5x=\boxed{10}\text{cm}$.


Question: As shown in the figure, extend the line segment $AB$ to $C$ so that $BC=4AB$. Let $D$ be the midpoint of line segment $BC$. If $CD=4\text{cm}$, and let $E$ be the midpoint of line segment $AC$, what is the length of $ED$ in cm?

Solution:
\textbf{Solution:} Given that point $E$ is the midpoint of segment $AC$,\\
it follows that $AE=\frac{1}{2}AC=5\text{cm}$,\\
since $BD=4\text{cm}$ and $AB=2\text{cm}$,\\
we have $AD=AB+BD=6\text{cm}$,\\
therefore $ED=AD-AE=6-5=\boxed{1}\text{cm}$.


Question: As shown in the figure, segment $AB=12$, point $C$ is the midpoint of segment $AB$, and point $D$ is the midpoint of segment $BC$. What is the length of segment $AD$?

Solution:
\textbf{Solution:} Given $\because AB=12$ and C is the midpoint of AB,\\
$\therefore AC=BC=6$,\\
$\because D$ is the midpoint of $BC$,\\
$\therefore CD=\frac{1}{2}BC=3$,\\
$\therefore AD=AC+CD=\boxed{9}$.


Question: As shown in the figure, the length of segment AB is 12, and point C is the midpoint of segment AB, while point D is the midpoint of segment BC. If there is a point E on segment AB such that $CE = \frac{1}{3}BC$, calculate the length of AE.

Solution:
\textbf{Solution:} Given that $BC=6$, and $CE=\frac{1}{3}BC$, \\
thus $CE=\frac{1}{3}\times 6=2$, \\
when E is to the left of C, $AE=AC-CE=6-2=4$; \\
when E is to the right of C, $AE=AC+CE=6+2=8$. \\
Therefore, the length of $AE$ is $\boxed{4}$ or $\boxed{8}$.


Question: As shown in the figure, $B$ and $C$ are two points on the line segment $AD$, and $AB:BC:CD=2:3:4$, $M$ is the midpoint of $AD$, if $MC=1$. What is the length of the line segment $AD$?

Solution:
\textbf{Solution:} Let $AB=2x$, hence $BC=3x, CD=4x$.\\
$AD=2x+3x+4x=9x.$\\
Since $M$ is the midpoint of $AD$,\\
it follows that $MD=\frac{1}{2}AD=\frac{9}{2}x.$\\
$MC=MD-CD=\frac{9}{2}x-4x=\frac{1}{2}x$,\\
According to the problem, $\frac{1}{2}x=1$,\\
Solving this gives $x=2$,\\
Therefore, $AD=9x=\boxed{18}$.


Question: As shown in the figure, $B$ and $C$ are two points on the line segment $AD$, and $AB:BC:CD = 2:3:4$. Let $M$ be the midpoint of $AD$, with $MC = 1$. If $N$ is a point on the line segment $AD$ such that $CN = \frac{1}{7}DB$, find the length of the line segment $DN$.

Solution:
\textbf{Solution:} From (1), we know that $DB=18\times \frac{3+4}{2+3+4}=14, CN=\frac{1}{7}DB$,\\
$\therefore CN=\frac{1}{7}\times 14=2$.\\
Consider the following two cases:\\
(1) When point $N$ is on segment $CD$, $DN=CD-CN=8-2=6$;\\
(2) When point $N$ is on segment $CB$, $DN=CD+CN=8+2=10$.\\
In conclusion, the length of segment $DN$ is either $6$ or $\boxed{10}$.


Question: As shown in the figure, the length of line segment $AB=8$, point $C$ is the midpoint of line segment $AB$, and point $D$ is the midpoint of line segment $BC$. What is the length of line segment $AD$?

Solution: \textbf{Solution:} Since $AB=8$ and C is the midpoint of AB, it follows that $AC=BC=4$. Since D is the midpoint of BC, it follows $CD=DB=\frac{1}{2}BC=2$. Therefore, $AD=AC+CD=4+2=\boxed{6};$


Question: As shown in the figure, the line segment $AB=8$, point $C$ is the midpoint of line segment $AB$, and point $D$ is the midpoint of line segment $BC$. There is a point $E$ on line segment $AC$ such that $CE=BC$. What is the length of $AE$?

Solution:
\textbf{Solution:} Since $CE=BC$ and $BC=4$, \\
it follows that $CE=4$, \\
thus, $AE=AC-CE=4-4=\boxed{0}$.


Question: As shown in the figure, the length of the segment $AB$ is $20$, point $C$ is the midpoint of segment $AB$, point $D$ is the midpoint of segment $BC$, and point $E$ is on the segment $AC$ with $CE = \frac{2}{5}AC$. Find the length of the segment $DB$.

Solution:
\textbf{Solution:} Since point C is the midpoint of line segment AB,\\
it follows that $AC=BC=\frac{1}{2}AB=\frac{1}{2}\times 20=10$,\\
Furthermore, since point D is the midpoint of line segment BC,\\
it follows that $BD=CD=\frac{1}{2}BC=\frac{1}{2}\times 10=\boxed{5}$.


Question: As shown in the figure, segment \(AB=20\), point \(C\) is the midpoint of segment \(AB\), point \(D\) is the midpoint of segment \(BC\), and point \(E\) is on segment \(AC\) such that \(CE=\frac{2}{5}AC\). Find the length of segment \(ED\).

Solution:
\textbf{Solution:} Since $CE=\frac{2}{5}AC$,\\
then $CE=\frac{2}{5}\times 10=4$,\\
thus $ED=CE+CD=4+5=\boxed{9}$.


Question: As illustrated in the figure, point $C$ is a point on line segment $AB$, point $M$ is the midpoint of line segment $AC$, and point $N$ is the midpoint of line segment $BC$. If $AB = 12\,cm$ and $AM = 5\,cm$, what is the length of $BC$ in centimeters?

Solution:
\textbf{Solution:} Since point M is the midpoint of line segment AC,\\
it follows that AC $= 2$AM,\\
since AM $= 5$cm,\\
it follows that AC $= 10$cm,\\
since AB $= 12$cm,\\
it follows that BC $=$ AB $-$ AC $= \boxed{2}$cm;


Question: As shown in the figure, point \(C\) is a point on segment \(AB\), point \(M\) is the midpoint of segment \(AC\), and point \(N\) is the midpoint of segment \(BC\). If \(MN = 8 \, \text{cm}\), what is the length of \(AB\) in centimeters?

Solution:
\textbf{Solution:} Since point M is the midpoint of line segment AC, and point N is the midpoint of line segment BC, \\
$\therefore$ BC$=2$NC, AC$=2$MC, \\
Since MN$=$NC$+$MC$=8$cm, \\
$\therefore$ AB$=$BC$+$AC$=2$MN$=2\times 8=\boxed{16}$cm.


Question: As shown in the figure, $C$ is a point on line segment $AB$, and $M$, $N$ are the midpoints of $AC$, $BC$, respectively. If $AC=8\,cm$ and $BC=6\,cm$, what is the length of line segment $MN$ in $cm$?

Solution:
\textbf{Solution:} Given that $M$ and $N$ are the midpoints of $AC$ and $BC$ respectively, \\
$\therefore CM=\frac{1}{2}AC=4\,cm$, $CN=\frac{1}{2}BC=3\,cm$, \\
$\therefore MN=CM+CN=4+3=\boxed{7}\,cm$.


Question: As shown in the figure, $C$ is a point on line segment $AB$, and $M$ and $N$ are the midpoints of $AC$ and $BC$, respectively. If the ratio of the lengths of line segments $CM$ and $CN$ is $2:1$, and the length of line segment $CN$ is $4\,\text{cm}$, find the length of line segment $AB$.

Solution:
Solution: Since the ratio of the lengths of segments $CM$ and $CN$ is $2\colon 1$ and $CN=4\,\text{cm}$,\\
$\therefore$ segment $CM=8\,\text{cm}$.\\
Since $M$ and $N$ are the midpoints of $AC$ and $BC$ respectively,\\
$\therefore$ $AC=2CM=16\,\text{cm}$, $BC=2CN=8\,\text{cm}$,\\
$\therefore$ $AB=AC+BC=16+8=\boxed{24}\,\text{cm}$.


Question: As shown in the figure, it is known that the length of line segment $AB=23$, $BC=15$, and $M$ is the midpoint of $AC$. What is the length of line segment $CM$?

Solution:
\textbf{Solution:} Since $AB=23$, $BC=15$,\\
thus $AC=AB-BC=23-15=8$.\\
Also, since $M$ is the midpoint of $AC$,\\
thus $CM=\frac{1}{2}AC=\frac{1}{2}\times 8=\boxed{4}$.


Question: As shown in the figure, it is known that the length of segment $AB=23$, $BC=15$, and $M$ is the midpoint of $AC$. Take a point $N$ on $CB$ such that the ratio $CN: NB=1:2$. Find the length of segment $MN$.

Solution:
\textbf{Solution:} Given that $BC=15$ and $CN:NB=1:2$,\\
it follows that $CN=\frac{1}{3}BC=\frac{1}{3}\times 15=5$.\\
Since $M$ is the midpoint of $AC$,\\
then $CM=\frac{1}{2}AC=4$.\\
Therefore, $MN=CM+CN=4+5=\boxed{9}$.


Question: As shown in the figure, the line segment $AB = 21, BC = 15$, and point $M$ is the midpoint of $AC$. What is the length of the line segment $AM$?

Solution:
\textbf{Solution:} Given that $AB=21$ and $BC=15$,\\
thus $AC=AB-BC=21-15=6$.\\
Since point M is the midpoint of AC,\\
it follows that $AM=\frac{1}{2}AC=\frac{1}{2}\times 6=\boxed{3}$.


Question: As shown in the figure, for the line segments $AB=21, BC=15$, point $M$ is the midpoint of $AC$. A point $N$ is chosen on $CB$ such that $CN:NB=1:2$. What is the length of $MN$?

Solution:
\textbf{Solution:} Since $BC=15$ and $CN:NB=1:2$,\\
it follows that $CN=\frac{1}{3}BC=\frac{1}{3}\times 15=5$.\\
Additionally, since point M is the midpoint of AC and $AC=6$,\\
it follows that $MC=\frac{1}{2}AC=3$,\\
therefore $MN=MC+NC=3+5=\boxed{8}$.


Question: As shown in Figure 1, it is known that point $B$ is on line segment $AC$, and point $D$ is on line segment $AB$. If $AB=6\,\text{cm}$, $BC=4\,\text{cm}$, and $D$ is the midpoint of line segment $AC$, what is the length of line segment $DB$ in centimeters?

Solution:
\textbf{Solution:}\\
Since $AC=AB+BC$, $AB=6\text{cm}$, $BC=4\text{cm}$\\
Therefore $AC=6+4=10\text{cm}$\\
Also, since $D$ is the midpoint of line segment $AC$\\
Therefore $DC= \frac{1}{2}AC= \frac{1}{2}\times10=5\text{cm}$\\
Therefore $DB=DC-BC=6-5=\boxed{1}\text{cm}$


Question: Given point B on line segment AC and point D on line segment AB. As shown in Figure 2, if $BD=\frac{1}{4}AB=\frac{1}{3}CD$, and E is the midpoint of line segment AB, with $EC=12\,\text{cm}$, find the length of line segment AC.

Solution:
\textbf{Solution:} Let $BD=x$ cm,\\
since $BD= \frac{1}{4}AB= \frac{1}{3}CD$,\\
thus $AB=4BD=4x$ cm, $CD=3BD=3x$ cm,\\
moreover, since $DC=DB+BC$,\\
thus $BC=3x-x=2x$,\\
and since $AC=AB+BC$,\\
thus $AC=4x+2x=6x$ cm,\\
since $E$ is the midpoint of segment $AB$,\\
thus $BE= \frac{1}{2}AB= \frac{1}{2} \times 4x=2x$ cm,\\
moreover, since $EC=BE+BC$,\\
thus $EC=2x+2x=4x$ cm,\\
and since $EC=12$ cm,\\
thus $4x=12$,\\
solving gives: $x=3$,\\
thus $AC=6x=6 \times 3=\boxed{18}$ cm.


Question: As shown in the figure, it is known that the length of line segment $AB=23$, $BC=15$, and point $M$ is the midpoint of $AC$. What is the length of line segment $AM$?

Solution:
\textbf{Solution:} Given that line segment $AB=23$, and $BC=15$, \\
$\therefore AC=AB-BC=23-15=8$, \\
also, since point M is the midpoint of AC, \\
$\therefore AM=\frac{1}{2}AC=\frac{1}{2}\times 8=\boxed{4}$, meaning the length of segment AM is 4.


Question: As shown in the figure, it is known that line segment $AB = 23$, $BC = 15$, and point $M$ is the midpoint of $AC$. Take a point $N$ on $CB$ such that the ratio $CN:NB = 1:2$. What is the length of line segment $MN$?

Solution:
Solution: Since $BC=15$ and $CN:NB=1:2$, \\
it follows that $CN=\frac{1}{3}BC=\frac{1}{3}\times 15=5$, \\
and since point M is the midpoint of AC, and $AC=8$, \\
it follows that $MC=\frac{1}{2}AC=4$, \\
therefore $MN=MC+NC=4+5=\boxed{9}$, meaning the length of MN is 9.


Question: As illustrated, point C is the midpoint of segment AB, point D is a point on segment BC, and point E is the midpoint of segment CD. Given $AB=20$ and $BD=4$, find the length of segment CD.

Solution:
\textbf{Solution:} Because $AB=20$ and $BD=4$,\\
it follows that $AD=AB-BD=20-4=16$,\\
since point C is the midpoint of line segment AB,\\
it follows that $AC=BC=10$,\\
therefore $CD=AD-AC=16-10=\boxed{6};$


Question: As shown in the figure, point C is the midpoint of line segment AB, point D is a point on line segment BC, and point E is the midpoint of line segment CD. Given that $AB=20$ and $BD=4$, find the length of line segment BE.

Solution:
\textbf{Solution:} Since $E$ is the midpoint of $CD$,\\
thus, $ED = \frac{1}{2}CD = 3$,\\
therefore, $BE = BD + ED = 4 + 3 = \boxed{7}$.


Question: As shown in the figure, C is a point on line segment AB, and M, N are the midpoints of AC and BC, respectively. If AC = 8 cm and BC = 6 cm, what is the length of line segment MN?

Solution:
\textbf{Solution}: Since $M$ and $N$ are the midpoints of $AC$ and $BC$ respectively,\\
$\therefore CM=\frac{1}{2}AC=4\text{cm}$, $CN=\frac{1}{2}BC=6\text{cm}$,\\
$\therefore MN=CM+CN=4+6=\boxed{10}\text{cm}$.


Question: As shown in the figure, C is a point on the line segment AB, with M and N being the midpoints of AC and BC, respectively. If the ratio of the lengths of segments CM to CN is 2:1, and the length of segment CN is 4 cm, find the length of segment AB.

Solution:
\textbf{Solution:} Given that the ratio of the lengths of segments CM to CN is $2\colon 1$ and $CN=4\,\text{cm}$, \\
$\therefore$ the length of segment CM is $8\,\text{cm}$.\\
Since M and N are the midpoints of AC and BC, respectively, \\
$\therefore$ AC$=2\cdot CM=16\,\text{cm}$ and BC$=2\cdot CN=8\,\text{cm}$,\\
$\therefore$ AB$=AC+BC=16+8=\boxed{24}\,\text{cm}$.


Question: As shown in the figure, points A, C, E, B, D are on the same straight line, and $AB=CD$, with point E being the midpoint of line segment AD. If $AB=11$ and $CE=3$, find the length of the line segment AD.

Solution:
\textbf{Solution:} Given that $CE=3, CE=BE$,\\
hence $BC=CE+BE=3+3=6$,\\
since $AB=11$,\\
thus $AC=AB-BC=11-6=5$,\\
since $CD=AB=11$,\\
therefore $AD=AC+CD=5+11=\boxed{16}$.


Question: As shown in the figure, it is known that the length of line segment $AD=30\,cm$, points C and B are both on line segment $AD$, and point E is the midpoint of $AB$. If $BD=6\,cm$, what is the length of line segment $AE$ in $cm$?

Solution:
\textbf{Solution:} Given $\because AD=30\,cm$, $BD=6\,cm$,\\
$\therefore AB=AD-BD=24\,cm$,\\
$\because E$ is the midpoint of $AB$,\\
$\therefore AE= \boxed{12\,cm}$.


Question: As shown in the figure, it is known that the line segment $AD=30\,cm$, with points C and B both on the line segment $AD$ and point E being the midpoint of $AB$. If $AC=\frac{1}{3}AD$, and point F is the midpoint of line segment $CD$, what is the length of line segment $EF$ in $cm$?

Solution:
\textbf{Solution:} Given that $\because AC= \frac{1}{3} AD=10\,cm$,\\
$\therefore CD=AD-AC=20\,cm$,\\
since $F$ is the midpoint of $CD$,\\
$\therefore DF= \frac{1}{2} CD=10\,cm$,\\
$\therefore BF=DF-BD=4\,cm$,\\
$\therefore EF=BE-BF=AE-BF=\boxed{8}\,cm$.


Question: As shown in the figure, it is known that the length of segment $AD=30\text{cm}$, points $C$ and $B$ are both on segment $AD$, and point $E$ is the midpoint of segment $AB$. If $BD=6\text{cm}$, what is the length of segment $AE$ in $\text{cm}$?

Solution:
\textbf{Solution:} Given that $AD=30cm$, $BD=6cm$,\\
hence $AB=AD-BD=30-6=24cm$,\\
since point E is the midpoint of AB,\\
thus $AE= \frac{1}{2}AB= \frac{1}{2} \times 24=\boxed{12}cm$.


Question: As shown in the figure, it is known that the length of line segment $AD$ is 30 cm, and points $C$ and $B$ are both on line segment $AD$. Point $E$ is the midpoint of $AB$. The length of $AE$ is 12 cm. If $AC = \frac{1}{3} AD$, and point $F$ is the midpoint of line segment $CD$, find the length of line segment $EF$.

Solution:
\textbf{Solution:} As shown in the figure,\\
$\because AC= \frac{1}{3} AD= \frac{1}{3}\times30=10\,\text{cm}$,\\
$\therefore CD=AD-AC=30-10=20\,\text{cm}$,\\
$\because$ Point F is the midpoint of line segment CD,\\
$\therefore DF=\frac{1}{2}CD=10\,\text{cm}$,\\
$\therefore EF=AD-AE-DF=30-12-10=\boxed{8}\,\text{cm}$.


Question: The given line segment $AB=a$ (as shown in the figure), where $C$ is a point on the extension of $AB$ in the opposite direction, and $AC=\frac{1}{3}AB$, $D$ is the midpoint of line segment $BC$. What is the length of $CD$ expressed as an algebraic expression involving $a$?

Solution: \textbf{Solution:} We have $CD=\boxed{\frac{2}{3}a}$.


Question: Given that the length of segment \(AB=a\) (as shown in the diagram), where \(C\) is a point on the extension of \(AB\) in the opposite direction, and \(AC=\frac{1}{3}AB\), \(D\) is the midpoint of segment \(BC\). If \(AD=3\text{cm}\), find the value of \(a\).

Solution:
\textbf{Solution}: Since $AC= \frac{1}{3}a$ and $AD=3\text{cm}$,\\
it follows that $CD= \frac{1}{3}a+3$,\\
which implies $\frac{1}{3}a+3= \frac{2}{3}a$.\\
Solving for $a$ yields: $a=\boxed{9}$.


Question: As shown in the figure: Given the line segment $AB=16\,\text{cm}$, point $N$ is on the line segment $AB$ with $NB=3\,\text{cm}$, and $M$ is the midpoint of $AB$. What is the length of the line segment $MN$ in centimeters?

Solution:
\textbf{Solution:}

Given that $AB=16\,\text{cm}$, and $M$ is the midpoint of $AB$,

$\therefore AM=MB=\frac{1}{2}AB=8\,\text{cm}$,

Given that $NB=3\,\text{cm}$,

$\therefore MN=MB-NB=\boxed{5}\,\text{cm}$.


Question: As shown in the figure: It is known that the length of line segment $AB=16\,\text{cm}$, point $N$ is on the line segment $AB$, $NB=3\,\text{cm}$, $M$ is the midpoint of $AB$. If there is a point $C$ on the line segment $AB$ satisfying $BC=10\,\text{cm}$, find the length of line segment $MC$.

Solution:
\textbf{Solution:} Since there exists a point $C$ on line segment $AB$, and $BC=10\,\text{cm}$, \\
it follows that $MC=BC-MB=10-8=\boxed{2}\,\text{cm}$.


Question: As shown in the figure, it is known that the length of segment $AB=4$. The segment $AB$ is extended to point C such that $BC=\frac{1}{2}AB$. The segment $AB$ is also extended in the opposite direction to point $D$ such that $AC=2AD$. Find the length of segment $CD$.

Solution:
\textbf{Solution:} Given that $AB=4\text{cm}$, $BC=\frac{1}{2}AB$,\\
it follows that $BC=2\text{cm}$, $AC=AB+BC=4+2=6\text{cm}$,\\
since $AC=2AD$,\\
it implies that $AD=3\text{cm}$,\\
therefore $CD=AD+AC=3+6=\boxed{9}\text{cm}$.


Question: As shown in the figure, it is known that the length of segment $AB=4$. Extend $AB$ to point C such that $BC=\frac{1}{2}AB$, and extend $AB$ in the opposite direction to point $D$ so that $AC=2AD$. If $Q$ is the midpoint of $AB$, and $P$ is a point on the segment $CD$; and $BP=\frac{1}{2}BC$, calculate the length of the segment $PQ$.

Solution:
\textbf{Solution:} As shown in the figure, when $P$ is on segment $AB$, we have $BP_1=\frac{1}{2}BC$\\
Since $Q$ is the midpoint of $AB$,\\
it follows that $BQ= \frac{1}{2}AB=2\text{cm}$,\\
Since $BP_1=\frac{1}{2}BC$,\\
it follows that $BP_1=1\text{cm}$,\\
Therefore, $P_1Q=BQ-BP_1=2-1=1\text{cm}$.\\
When $P$ is on segment $BC$, $BP_2=\frac{1}{2}BC=1$,\\
$P_2Q=BQ+BP_2=2+1=3\text{cm}$.\\
Therefore, the length of segment $PQ$ is $\boxed{1}$ or $\boxed{3\text{cm}}$.


Question: As shown in the figure, point C is a point on line segment AB, point D is the midpoint of line segment BC, and $AB=16\,cm$, $CD=3\,cm$. Find the length of line segment AC.

Solution:
\textbf{Solution:} Since point D is the midpoint of line segment BC, and $CD=3\,\text{cm}$,\\
it follows that $BC=2CD=6\,\text{cm}$,\\
therefore, $AC=AB-BC=16-6=\boxed{10}\,\text{cm}$.


Question: As shown in the figure, point C is a point on line segment AB, point D is the midpoint of line segment BC, and $AB=16\,cm$, $CD=3\,cm$. If point E is on line AB, and $AE=2\,cm$, what is the length of line segment DE in $cm$?

Solution:
\textbf{Solution:} Given that $AC=10\,cm$, $CD=3\,cm$, \\
it follows that $AD=AC+CD=10+3=13\,cm$. \\
When point E is to the left of point A, denoted as $E_1$, $DE_1=AD+AE_1=13+2=15\,cm$. \\
When point E is to the right of point A, denoted as $E_2$, $DE_2=AD-AE_2=13-2=11\,cm$. \\
Therefore, the length of segment DE is either \boxed{11\,cm} or \boxed{15\,cm}.


Question: As shown in the figure, points D, B, E are three points on the line segment AC, with D being the midpoint of the line segment AB. If point E is the midpoint of BC and $BE= \frac{1}{5} AC=2cm$, find the length of the line segment DE.

Solution:
\textbf{Solution:}

Since point $E$ is the midpoint of $BC$, \\
thus $BC=2BE=2\times \frac{1}{5}=\frac{2}{5}$cm \\
Therefore $AB=AC−BC=2−\frac{2}{5}=\frac{8}{5}$cm \\
Since $D$ is the midpoint of line segment $AB$, \\
therefore $BD=\frac{1}{2}AB=\frac{1}{2}\times \frac{8}{5}=\frac{4}{5}$ \\
Therefore $DE=BD+BE=\frac{4}{5}+\frac{1}{5}=\boxed{1}$cm


Question: As shown in the figure, points D, B, and E are three points on line segment AC, with D being the midpoint of line segment AB. If $AC = 2DE = 20$ and $AD: EC = 3:2$, find the length of line segment EC.

Solution:
Solution: Since point E is the midpoint of BC, and D is the midpoint of segment AB,\\
it follows that $BD=AD=\frac{1}{2}AB$, and $BE=EC=\frac{1}{2}BC$,\\
Since $AC=AD+DC=AD+BD+BE+EC=2(AD+EC)=20$,\\
it follows that $AD+EC=10$,\\
Since $AD\colon EC=3\colon 2$,\\
it follows that $EC=\boxed{4}$.


Question: As shown in the figure, point B is a point on line segment AC, and \(AB = 27cm\), \(BC = \frac{1}{3}AB\). What is the length of line segment \(AC\)?

Solution:
\textbf{Solution:} Solution: According to the given problem, $AB=27\text{cm}$, $BC=\frac{1}{3}AB$,\\
$\therefore BC=9\text{cm}$,\\
$\therefore AC=AB+BC=27+9=\boxed{36}\text{cm}$.


Question: As shown in the figure, point $B$ is on line segment $AC$, and $AB=27\text{cm}$, $BC=\frac{1}{3}AB$. If point $O$ is the midpoint of line segment $AC$, find the length of line segment $OB$.

Solution:
\textbf{Solution:} Since point O is the midpoint of AC, \\
$\therefore$ OC=$\frac{1}{2}$AC=$\frac{1}{2} \times 36=18$, \\
$\therefore$ OB=OC−BC=18−9=\boxed{9}.


Question: As shown in the figure, there are three points C, D, and E sequentially on the line segment AB, dividing the line segment AB into four parts with the ratios $2: 3: 4: 5$. Given that $AB=28$, what is the length of the line segment AE?

Solution:
Solution: Let $AC=2x$, then $CD$, $DE$, and $EB$ are $3x$, $4x$, and $5x$ respectively,\\
Given that, $2x+3x+4x+5x=28$\\
Solving for $x$, we get $x=2$\\
Then, $AC$, $CD$, $DE$, and $EB$ are $4\text{cm}$, $6\text{cm}$, $8\text{cm}$, and $10\text{cm}$ respectively,\\
Therefore, $AE=AC+CD+DE=\boxed{18\text{cm}}$


Question: As shown in the figure, there are three points $C$, $D$, $E$ sequentially placed on the segment $AB$, which divide the segment $AB$ into four parts with ratios $2:3:4:5$. Given that $AB=28$, and if $M$, $N$ are the midpoints of segments $DE$ and $EB$ respectively, what is the length of segment $MN$?

Solution:
Solution: From the given problem,\\
since $M$ is the midpoint of $DE$,\\
therefore $ME = \frac{1}{2}DE = 4\,\text{cm}$,\\
since $N$ is the midpoint of $EB$,\\
therefore $EN = \frac{1}{2}EB = 5\,\text{cm}$,\\
therefore $MN = ME + EN = \boxed{9\,\text{cm}}$.


Question: As shown in the diagram, point $D$ is the midpoint of segment $AC$, and point $E$ is the midpoint of segment $BC$, with $BC=\frac{4}{7}AC$. If $BC=8$, what is the length of $DC$?

Solution:
\textbf{Solution:} Since $BC=8$ and $BC=\frac{4}{7}AC$,\\
it follows that $AC=8\div\frac{4}{7}=14$,\\
Given that point $D$ is the midpoint of line segment $AC$,\\
it follows that $DC=\frac{1}{2}AC=\frac{1}{2}\times 14=\boxed{7}$.


Question: As shown in the figure, point $D$ is the midpoint of line segment $AC$, and point $E$ is the midpoint of line segment $BC$, with $BC=\frac{4}{7}AC$. If $DE=6$, what is the length of $AC$?

Solution:
\textbf{Solution:} Given from the problem, $BC=\frac{4}{7}AC$,

$\therefore AC=AB+BC=\frac{7}{4}BC$,

$\because$ point $D$ is the midpoint of line segment $AC$, and point $E$ is the midpoint of line segment $BC$,

$\therefore DC=DA=\frac{1}{2}AC=\frac{7}{8}BC$, $EC=BE=\frac{1}{2}BC$,

$\therefore DE=DC-EC=\frac{7}{8}BC-\frac{1}{2}BC=6$,

Solving this, we find $BC=16$,

$\therefore AC=\frac{7}{4}\times 16=\boxed{28}$.


Question: As shown in the figure, there are three points $D$, $B$, and $E$ successively on the line segment $AC$, with $AD=\frac{1}{2}DB$, $E$ is the midpoint of $BC$, and $BE=\frac{1}{5}AC=2$. Find the length of the line segment $AB$.

Solution:
\textbf{Solution:} Since $BE=\frac{1}{5}AC=2$, and $E$ is the midpoint of $BC$, \\
$\therefore BC=2BE=4$, $AC=10$, \\
$\therefore AB=\frac{3}{5}AC=\boxed{6}$,


Question: As shown in the figure, there are three points $D$, $B$, and $E$ successively located on the line segment $AC$, with $AD = \frac{1}{2}DB$, $E$ being the midpoint of $BC$, and $BE = \frac{1}{5}AC = 2$. Calculate the length of line segment $DE$.

Solution:
\textbf{Solution:} Since $AD=\frac{1}{2}DB$,\\
it follows that $AB=AD+DB=\frac{3}{2}DB$,\\
Since $AB=6$,\\
it follows that $DB=4$,\\
Also, since $BE=2$,\\
it follows that $DE=DB+BE$\\
$=4+2=\boxed{6}$.


Question: The line segments $AB$ and $CD$ are on the same straight line, with $M$ and $N$ being the midpoints of line segments $AB$ and $CD$, respectively. It is known that $AB=6\,cm$ and $CD=8\,cm$. When the points $A$ and $C$ coincide, what is the length of $MN$ in $cm$?

Solution:
Solution: Since $M$ and $N$ are the midpoints of line segments $AB$ and $CD$ respectively, and $AB=6\,cm$, $CD=8\,cm$, with points $A$ and $C$ coinciding,\\
therefore, $AM=3\,cm$, $AN=4\,cm$,\\
hence $MN=AN-AM=\boxed{1\,cm}$;


Question: As shown in the figure, the line segments $AB$ and $CD$ are on the same straight line, where $M$ and $N$ are the midpoints of $AB$ and $CD$, respectively. Given that $AB = 6\,\text{cm}$ and $CD = 8\,\text{cm}$. When point $C$ is on line segment $AB$, if the common part $BC$ of the line segments $AB$ and $CD$ is $2\,\text{cm}$, then what is the length of $MN$ in cm?

Solution:
\textbf{Solution:} Since M and N are the midpoints of the line segments AB and CD, respectively, and AB=6cm, CD=8cm,\\
therefore, AM=3cm, DN=4cm.\\
Since the common part of the line segments AB and CD is BC=2cm,\\
therefore, the length of AD is AB+CD−BC=6+8−2=12cm,\\
therefore, the length of MN is AD−AM−DN=12−3−4=\boxed{5cm};


Question: It is known that the length of segment $AB=20$. $M$ is the midpoint of segment $AB$, and $P$ is any point on segment $AB$. $N$ is the midpoint of segment $PB$. When $P$ is the midpoint of segment $AM$, what is the length of segment $NB$?

Solution:
\textbf{Solution}: As shown in the figure, since $M$ is the midpoint of the segment $AB$, $AB=20$\\
therefore $MA=\frac{1}{2}AB=10$\\
since $P$ is the midpoint of the segment $AM$,\\
therefore $AP=\frac{1}{2}AM=5$\\
therefore $PB=AB-AP=20-5=15$\\
since $N$ is the midpoint of the segment $PB$\\
therefore $NB=\frac{1}{2}PB=\boxed{7.5}$


Question: As shown in the figure, it is known that the length of line segment $AB=20$, $M$ is the midpoint of line segment $AB$, $P$ is any point on line segment $AB$, and $N$ is the midpoint of line segment $PB$. When the length of line segment $MP=1$, what is the length of line segment $NB$?

Solution:
\textbf{Solution:} Since $MP=1$,\\
Therefore, when P is on the left side of M, $BP=MB+MP=11$,\\
Since N is the midpoint of the line segment $PB$,\\
Therefore, $NB=\frac{1}{2}PB=5.5$,\\
When P is on the right side of M, $BP=MB-MP=9$,\\
Since N is the midpoint of the line segment $PB$,\\
Therefore, $NB=\frac{1}{2}PB=\boxed{4.5}$.


Question: As shown in the figure, the line segment $AB=8$, point C is the midpoint of line segment AB, and point D is the midpoint of line segment BC. Find the length of line segment AD.

Solution:
\textbf{Solution:} Since $AB=8$ and C is the midpoint of AB,\\
it follows that $AC=BC=4$.\\
Since D is the midpoint of BC,\\
it follows that $CD=DB=\frac{1}{2}BC=2$.\\
Therefore, $AD=AC+CD=4+2=\boxed{6}$.


Question: As shown in the figure, the length of the line segment $AB=8$, point $C$ is the midpoint of line segment $AB$, and point $D$ is the midpoint of line segment $BC$. There is a point $E$ on the line segment $AC$, where $CE=\frac{1}{3}BC$. Find the length of $AE$.

Solution:
\textbf{Solution:} \\
Since $CE = \frac{1}{3}BC$ and $BC = 4$, \\
it follows that $CE = \frac{4}{3}$, \\
therefore $AE = AC - CE = 4 - \frac{4}{3} = \boxed{\frac{8}{3}}$.


Question: Given that point $C$ is a point on segment $AB$, point $M$ is the midpoint of segment $AC$, and point $N$ is the midpoint of segment $BC$, if $AB=10\,\text{cm}$, what is the length of $MN$?

Solution:
\textbf{Solution:} We have $MN=CM+CN=\frac{1}{2}AC+\frac{1}{2}BC=\frac{1}{2}AB=5\,\text{cm}$,\\
therefore $MN=\boxed{5\,\text{cm}}$.


Question: It is known that point $C$ is a point on segment $AB$, point $M$ is the midpoint of segment $AC$, and point $N$ is the midpoint of segment $BC$. If $AC:CB=3:2$ and $NB=3.5\,cm$, then how many $cm$ does $AB$ equal?

Solution:
\textbf{Solution:} Given that $\because NB=3.5\,cm$,\\
$\therefore BC=7\,cm$,\\
$\therefore AB=7\div \frac{2}{5}=17.5\,cm$.\\
Hence, $AB=\boxed{17.5}\,cm$.


Question: As shown in Figure 1, it is known that point $D$ is the midpoint of line segment $AB$, and point $C$ is on line segment $AB$. If $AC=8\,\text{cm}$ and $BC=6\,\text{cm}$, what is the length of line segment $CD$ in cm?

Solution:
\textbf{Solution:} Given that $AC=8\text{cm}, BC=6\text{cm}$,\\
it follows that $AB=AC+BC=8+6=14\text{cm}$,\\
since point D is the midpoint of the line segment $AB$,\\
we have $BD=\frac{1}{2}AB=\frac{1}{2}\times 14=7\text{cm}$.\\
Since $CD=BD-BC$,\\
we conclude that $CD=7-6=\boxed{1}\text{cm}$.


Question: Given that point \(D\) is the midpoint of line segment \(AB\), and point \(C\) lies on line segment \(AB\). As shown in Figure 2, if \(BC = 2CD\), and point \(E\) is the midpoint of \(BD\), with \(AE = 18cm\), find the length of line segment \(AB\).

Solution:
\textbf{Solution:} Let $CD=x\, \text{cm}$.\\
Since $BC=2CD$,\\
it follows that $BC=2x\, \text{cm}$.\\
Since $BD=CD+BC$,\\
it follows that $BD=x+2x=3x\, \text{cm}$.\\
Since $E$ is the midpoint of $BD$,\\
it follows that $DE=\frac{1}{2}BD=\frac{3}{2}x\, \text{cm}$.\\
Moreover, since $D$ is the midpoint of $AB$,\\
it follows that $AD=BD=3x\, \text{cm}$.\\
Since $AE=AD+DE$,\\
it follows that $AE=3x+\frac{3}{2}x=\frac{9}{2}x\, \text{cm}$.\\
Since $AE=18\, \text{cm}$,\\
it follows that $\frac{9}{2}x=18$, $x=4$,\\
thus $AB=2BD=6x=24\, \text{cm}$,\\
therefore, the length of segment $AB$ is $\boxed{24\, \text{cm}}$.


Question: Given points $C$ and $D$ are on the line segment $AB$, where $AB=22$ and $CD=8$. If point $C$ is the midpoint of line segment $AB$, find the length of $BD$.

Solution:
\textbf{Solution:} Since $AB=22$ and $C$ is the midpoint of $AB$,\\
it follows that $BC=11$,\\
Since $CD=8$,\\
it follows that $BD=BC-CD=\boxed{3}$.


Question: Given points $C$ and $D$ are on segment $AB$, and $AB=22$, $CD=8$. If points $M$ and $N$ are respectively the midpoints of $AC$ and $BD$, find the length of $MN$.

Solution:
\textbf{Solution:} Given $\because AB=22$, $CD=8$.\\
$\therefore AC+BD=AB-CD=14$.\\
Also, $\because M$, $N$ are the midpoints of $AC$, $BD$ respectively,\\
$\therefore CM+DN=7$.\\
$\therefore MN=CM+DN+CD=\boxed{15}$.


Question: As shown in the figure, points $B$ and $C$ are on the line segment $AD$, and $AB:BC:CD=2:3:4$. Point $M$ is the midpoint of the line segment $AC$, and point $N$ is a point on the line segment $CD$, with $MN=9$. If point $N$ is the midpoint of line segment $CD$, what is the length of $BD$?

Solution:
\textbf{Solution}: As shown in the figure,\\
since point M is the midpoint of line segment AC, and point N is the midpoint of line segment CD,\\
it follows that $CM= \frac{1}{2} AC$, $CN= \frac{1}{2} CD$,\\
thus $MN=CM+CN= \frac{1}{2} (AC+CD)= \frac{1}{2} AD=9$,\\
therefore $AD=18$,\\
since $AB:BC:CD=2:3:4$,\\
it follows that $AB= \frac{2}{9} \times AD=4$,\\
therefore $BD=AD-AB=18-4=\boxed{14}$;


Question: As shown in the figure, points B and C are on line segment $AD$, and $AB:BC:CD=2:3:4$. Point M is the midpoint of line segment $AC$, and point N is a point on line segment $CD$, with $MN=9$. When $CN=\frac{1}{3}CD$, find the length of $BD$.

Solution:
\textbf{Solution:}\\
Since CN = $\frac{1}{3}$CD,\\
and AB:BC:CD = 2:3:4,\\
let AB = 2x, BC = 3x, CD = 4x.\\
Thus, AC = 5x,\\
and since point M is the midpoint of segment AC,\\
CM = $\frac{1}{2}$AC = 2.5x,\\
and since CN = $\frac{1}{3}$CD = $\frac{4}{3}$x,\\
CM+CN = $\frac{5}{2}$x+$\frac{4}{3}$x = MN = 9,\\
thus, x = $\frac{54}{23}$,\\
Hence, BD = 7x = $\boxed{\frac{378}{23}}$;


Question: As shown in the figure, points $C$ and $E$ are two points on segment $AB$, point $D$ is the midpoint of segment $AB$, and $AB=6$, $CD=1$. What is the length of $BC$?

Solution:
\textbf{Solution:} Since point D is the midpoint of line segment AB, \\
$\therefore$ AD=DB=$\frac{1}{2}$AB=3, \\
$\therefore$ BC=DB−DC=3−1=\boxed{2};


Question: As shown in the figure, points $C$ and $E$ are two points on line segment $AB$, point $D$ is the midpoint of line segment $AB$, and $AB=6$, $CD=1$. If $AE:EC=1:3$, what is the length of $EC$?

Solution:
\textbf{Solution:} Let $AE=x$, then $EC=3x$. \\
Since $AC=AD+DC=AE+EC$, \\
it follows that $3+1=x+3x$, solving this gives: $x=1$, \\
therefore, $EC=3x=\boxed{3}$.


Question: As shown in the figure, point $C$ is the midpoint of line segment $AB$, and point $D$ is on $AB$, with $AD=\frac{1}{3}AB$. If $AD=6$, find the length of line segment $CD$.

Solution:
\textbf{Solution:} Given $\because AD= \frac{1}{3} AB$ and $AD=6$,\\
$\therefore AB=18$,\\
$\because C$ is the midpoint of $AB$,\\
$\therefore AC= \frac{1}{2} AB=9$,\\
$\therefore CD=AC-AD=9-6=\boxed{3}$.


Question: As shown in the figure, point $C$ is the midpoint of line segment $AB$, and point $D$ is on $AB$ with $AD=\frac{1}{3}AB$. If $CD=2$, what is the length of line segment $AB$?

Solution:
\textbf{Solution:} Given that $C$ is the midpoint of $AB$, \\
we have $AC= \frac{1}{2}AB$, \\
since $AD= \frac{1}{3}AB$, \\
it follows that $CD=AC-AD= \frac{1}{6}AB$, \\
given that $CD=2$, \\
thus $AB=6CD=\boxed{12}$.


Question: As shown in the figure, points $C$, $E$ are two points on line segment $AB$, point $D$ is the midpoint of line segment $AB$, and $AB=6$, $CD=1$. What is the length of $BC$?

Solution:
\textbf{Solution:} Since point D is the midpoint of line segment AB, and AB$=6$, \\
it follows that BD$=\frac{1}{2}$AB$=3$, \\
Since CD$=1$, \\
it follows that BC$=BD-CD=3-1=\boxed{2}$.


Question: As shown in the figure, points C and E are two points on the line segment AB, point D is the midpoint of line segment AB, and AB = 6, CD = 1. If AE : EC = 1 : 3, find the length of EC.

Solution:
\textbf{Solution:} Since point D is the midpoint of line segment AB, and AB$=6$,\\
thus AD$=\frac{1}{2}$AB$=3$,\\
since CD$=1$, thus AC$=$AD$+$CD$=4$,\\
since AE:EC$=1:3$, thus EC$=\frac{3}{1+3}\times 4=\boxed{3}$.


Question: Given point $B$ is on line segment $AC$, and point $D$ is on line segment $AB$. If $AB=8\text{cm}$, $BC=6\text{cm}$, and $D$ is the midpoint of line segment $AC$, what is the length of line segment $DB$ in $\text{cm}$?

Solution:
\textbf{Solution:} As shown in Figure 1:\\
$\because AB=8\text{cm}$, $BC=6\text{cm}$\\
$\therefore AC=AB+BC=8+6=14\text{cm}$\\
Furthermore, $\because D$ is the midpoint of segment $AC$\\
$\therefore DC=\frac{1}{2}AC=\frac{1}{2}\times14=7\text{cm}$\\
$\therefore DB=DC-BC=7-6=\boxed{1\text{cm}}$;


Question: Given point $B$ is on line segment $AC$, and point $D$ is on line segment $AB$. As shown in the figure, if $BD = \frac{1}{4}AB = \frac{1}{3}CD$, $E$ is the midpoint of line segment $AB$, $EC=20\,cm$, find the length of line segment $AC$.

Solution:
\textbf{Solution:} Let $BD=x$ cm, \\
since $BD= \frac{1}{4}AB= \frac{1}{3}CD$, \\
thus $AB=4BD=4x$ cm, $CD=3BD=3x$ cm, \\
hence $BC=DC-DB=3x-x=2x$, \\
thus $AC=AB+BC=4x+2x=6x$ cm, \\
since $E$ is the midpoint of segment $AB$, \\
thus $BE= \frac{1}{2}AB= \frac{1}{2}\times4x=2x$ cm, \\
thus $EC=BE+BC=2x+2x=4x$ cm, \\
also, since $EC=20$ cm, \\
thus $4x=20$, \\
solving gives: $x=5$, \\
hence $AC=6x=6\times5=\boxed{30}$ cm.


Question: As shown in the figure, point C is the midpoint of line segment AB, point D is a point on line segment CB, and point E is the midpoint of line segment DB. Given $AB=20$ and $EB=3$, find the length of line segment DB.

Solution:
\textbf{Solution:} Since point E is the midpoint of segment DB, \\
therefore $DB = 2EB = 2 \times 3 = \boxed{6}$;


Question: As shown in the figure, point $C$ is the midpoint of segment $AB$, point $D$ is a point on segment $CB$, and point $E$ is the midpoint of segment $DB$. Given that $AB=20$ and $EB=3$, what is the length of segment $CD$?

Solution:
\textbf{Solution:} Since point C is the midpoint of line segment AB,\\
it follows that $CB=\frac{1}{2}AB=\frac{1}{2}\times 20=10$\\
Given that $CD=CB-DB$,\\
it then results in $CD=10-6=\boxed{4}$


Question: Point \(C\) lies on segment \(AB\), and point \(M\) is the midpoint of \(AC\), with \(AM=1\), \(BC=4\). If point \(N\) is the midpoint of \(BC\), what is the length of \(MN\)?

Solution:
\textbf{Solution:} Given that, \\
$ \because $ point $ M$ is the midpoint of $ AC$, and $ AM=1$, \\
$ \therefore CM=AM=1$, \\
$ \because $ point $ N$ is the midpoint of $ BC$, and $ BC=4$, \\
$ \therefore CN=\frac{1}{2}BC=2$, \\
$ \therefore MN=CM+CN=1+2=\boxed{3}$.


Question: As shown in the figure, point B is a point on line segment AC, and $AB=21$, $BC=\frac{1}{3}AB$. Calculate the length of line segment AC.

Solution:
\textbf{Solution:} We have $BC = \frac{1}{3}AB = 7$, \\
$AC = AB + BC = AB + \frac{1}{3}AB = 21 + 7 = \boxed{28}$.


Question: As shown in the figure, point B is a point on line segment AC, and $AB=21$, $BC=\frac{1}{3}AB$. If point O is the midpoint of line segment AC, calculate the length of line segment OB.

Solution:
\textbf{Solution:} Since $O$ is the midpoint of $AC$,\\
therefore, $OC=\frac{1}{2}AC=14$,\\
therefore, $OB=OC-BC=14-7=\boxed{7}$.


Question: As shown in the figure, it is known that the line segment $AB = 26$, $BC = 18$, and the point $M$ is the midpoint of $AC$. What is the length of the line segment $AC$?

Solution:
\textbf{Solution:} Because $AB=26$ and $BC=18$, \\
therefore $AC=AB-BC=\boxed{8}$.


Question: Given the diagram, we know that the length of segment $AB=26$ and $BC=18$. Point $M$ is the midpoint of $AC$. A point $N$ is chosen on $CB$ such that the ratio of $CN: NB=1:2$. What is the length of segment $MN$?

Solution:
\textbf{Solution:} Since point $M$ is the midpoint of line segment $AC$,\\
it follows that $MC= \frac{1}{2}AC$,\\
Since $AC=8$,\\
it follows that $MC=4$. Additionally, since $BC=18$ and $CN:NB=1:2$,\\
it follows that $CN= \frac{1}{3}BC=6$,\\
therefore, $MN=MC+CN=6+4=\boxed{10}$.


Question: As shown in the diagram, it is known that point $D$ is on segment $AB$, and point $C$ is the midpoint of segment $AB$. If $AB=8\,\text{cm}$ and $BD=3\,\text{cm}$, what is the length of segment $CD$ in cm?

Solution:
\textbf{Solution:} Given that point $C$ is the midpoint of line segment $AB$,\\
hence, $CB = \frac{1}{2}AB = 4\,\text{cm}$,\\
thus, $CD = CB - BD = 4 - 3 = 1\,\text{cm}$,\\
meaning the length of line segment $CD$ is \boxed{1}\,\text{cm}.


Question: As shown in the figure, it is known that point $D$ is a point on line segment $AB$, point $C$ is the midpoint of line segment $AB$, if $AB=8\,\text{cm}, BD=3\,\text{cm}$. If point $E$ is a point on line $AB$, and $BE=\frac{1}{3}BD$, point $F$ is the midpoint of $BE$, what is the length of line segment $CF$ in $\,\text{cm}$?

Solution:
\textbf{Solution}:

(1) When point E is on the left side of point B\\
$\because BE=\frac{1}{3}BD$\\
$\therefore BE=\frac{1}{3}\times 3=1\,\text{cm}$\\
$\because F$ is the midpoint of $BE$\\
$\therefore BF=\frac{1}{2}BE=0.5\,\text{cm}$\\
$\therefore CF=CB−BF=4−0.5=\boxed{3.5}\,\text{cm}$\\
(2) When point E is on the right side of point B\\
$\because BE=\frac{1}{3}BD$\\
$\therefore BE=\frac{1}{3}\times 3=1\,\text{cm}$\\
$\because F$ is the midpoint of $BE$\\
$\therefore BF=\frac{1}{2}BE=0.5\,\text{cm}$\\
$\therefore CF=CB+BF=4+0.5=\boxed{4.5}\,\text{cm}$\\
In summary, the length of segment $CF$ can be \boxed{3.5}\,\text{cm} or \boxed{4.5}\,\text{cm}.


Question: Point $C$ is on line segment $AB$, point $M$ is the midpoint of $AC$, $AM=1$, $BC=4$. If point $N$ is the midpoint of $BC$, what is the length of $MN$?

Solution:
\textbf{Solution:} Since point $M$ is the midpoint of $AC$, $AM=1$,\\
$\therefore CM=AM=1$,\\
Since point $N$ is the midpoint of $BC$, $BC=4$,\\
$\therefore CN=\frac{1}{2}BC=2$,\\
$\therefore MN=CM+CN=1+2=\boxed{3}$;


Question: Point C lies on segment AB, and point M is the midpoint of AC, where $AM=1$ and $BC=4$. If point N lies on ray AB with $AN=7$, please complete the figure and directly state the length of $BN$.

Solution: \textbf{Solution:} Given the information in the problem, it is not sufficient to complete the figure, thus we cannot directly determine the length of $\boxed{BN}$.


Question: As shown in the figure, $AB=6$. Extend the line segment $AB$ to point $C$, and $D$ is the midpoint of $AC$. If $BC=2AB$, what is the length of $BD$?

Solution:
\textbf{Solution:} Since $BC = 2AB$ and $AB = 6$,\\
it follows that $BC = 12$,\\
hence $AC = AB + BC = 18$,\\
Given that $D$ is the midpoint of $AC$,\\
it means $AD = \frac{1}{2}AC = 9$,\\
therefore $BD = AD - AB = \boxed{3};$


Question: As shown in the figure, points C and D are two points on line segment AB, with AC:BC = 3:2, and point D being the midpoint of AB. If AB = 40, find the length of line segment CD.

Solution:
\textbf{Solution}: Since $AC:BC=3:2$ and $AB=40$, \\
it follows that $AC=40\times \frac{3}{3+2}=24$, \\
since point D is the midpoint of AB, \\
it follows that $AD=\frac{1}{2}AB=20$, \\
therefore $CD=AC-AD=\boxed{4}$.


Question: As shown in the figure, points C and D are two points on line segment AB, with $AC:BC = 3:2$, and point D is the midpoint of AB. As shown in Figure 2, if E is the midpoint of AC and $ED = 7$, find the length of line segment AB.

Solution:
\textbf{Solution:} Let $AC=3x, BC=2x$, thus $AB=5x$, \\
since point D is the midpoint of AB.\\
Therefore, $AD=\frac{1}{2}AB=\frac{5}{2}x$, \\
since E is the midpoint of AC, \\
Therefore, $AE=\frac{1}{2}AC=\frac{3}{2}x$, \\
Therefore, $DE=AD−AE=\frac{5}{2}x−\frac{3}{2}x=x$, \\
since $ED=7$, \\
Therefore, $x=7$, \\
Therefore, $AB=5x=\boxed{35}$.