Question: As shown, $\triangle ABC$ and $\triangle A'B'C'$ are similar figures with point O as the center of similitude. If $OA: AA'=2:5$, then what is the ratio of the perimeters of $\triangle ABC$ and $\triangle A'B'C'$?

Solution:
\textbf{Solution:} Since $OA:AA'=2:5$, \\
therefore, $OA:OA'=2:3$, \\
thus, the ratio of similarity between $\triangle ABC$ and $\triangle A'B'C'$ is $2:3$, \\
hence, the ratio of the perimeters of $\triangle ABC$ and $\triangle A'B'C'$ is $\boxed{2:3}$.


Question: As shown in the figure: $\triangle AOB$ and $\triangle A_1OB_1$ are similar figures with the origin as the center of similarity, and the scale factor of similarity is $1\colon 3$. If the coordinates of point B are ($-1$, $2$), what are the coordinates of point $B_1$?

Solution:
\textbf{Solution:} Since $\triangle AOB$ and $\triangle A_1OB_1$ are homothetic figures with the origin as the center of homothety, and the ratio of homothety is $1\colon 3$, the coordinates of point $B$ are ($-1$, $2$),\\
therefore, the coordinates of point $B_1$ are ($-1\times(-3)$, $2\times(-3)$), which is $\boxed{(3,-6)}$.


Question: As shown in the figure, in the right-angled triangle $ABC$, where $\angle ABC=90^\circ$ and the side $AB$ is on the x-axis, taking $O$ as the center of dilation, the triangle $A_1B_1C_1$ is similar to the triangle $ABC$. If the corresponding point of $C(4,6)$ is $C_1(2,3)$, what are the coordinates of $B_1$?

Solution:
\textbf{Solution:} Given that in the right triangle $ABC$, $\angle ABC=90^\circ$, and $C(4,6)$,\\
thus $B(4,0)$,\\
therefore, taking $O$ as the center of similarity, construct $\triangle A_1B_1C_1$ similar to $\triangle ABC$, with the corresponding point of $C(4,6)$ being $C_1(2,3)$,\\
thus, the coordinates of ${B}_1$ are: $(4\times \frac{1}{2},0)$, that is $\boxed{(2,0)}$.


Question: As shown in the figure, within the rectangular paper $ABCD$, where $AD=9$, $E$ is a point on $CD$. Connect $AE$, and after folding $\triangle ADE$ along the straight line $AE$, point $D$ falls on point $F$. Draw $FG\perp AD$ at $G$. If $AG=2GD$, what is the value of $DE$?

Solution: \textbf{Solution:} As shown in the figure, draw line EH $\perp$ FG at point H, then $\angle$EHG = 90$^\circ$. It is easy to deduce from the folding property that $\triangle AED \cong \triangle AEF$, thus $AF = AD = 9$, $DE = EF$. Since $AG = 2GD$, it follows that $GD = 3$, $AG = 6$. Given $FG \perp AD$, it follows that $\angle FGD = \angle AGF = 90^\circ$. Therefore, in $\triangle AGF$, $FG = \sqrt{AF^2 - AG^2} = \sqrt{9^2 - 6^2} = 3\sqrt{5}$. Since ABCD is a rectangle, it follows that $\angle D = 90^\circ$, hence quadrilateral GHED is a rectangle. Therefore, $GH = DE$, $HE = GD = 3$. Let $DE = x$, then $GH = EF = x$, $HF = 3\sqrt{5} - x$. Therefore, in $\triangle HEF$, $HF^2 + HE^2 = EF^2$, which gives $(3\sqrt{5} - x)^2 + 3^2 = x^2$. Solving for $x$ gives $x = \frac{9\sqrt{5}}{5}$. Therefore, $DE = \boxed{\frac{9\sqrt{5}}{5}}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB = 5$, $BC = 6$, point $E$ is on side $BC$, and $BE = 2$. $F$ is a moving point on side $AB$. Connect $EF$, and construct an equilateral $\triangle EFG$ with $EF$ as a side, and point $G$ is inside rectangle $ABCD$. Connect $CG$. What is the minimum value of $CG$?

Solution:
\textbf{Solution:} Given the problem, we know that point F is the driving point, and point G is the driven point. Point F moves on a line segment, and point G also definitely moves on a linear trajectory.\\
Rotate $\triangle EFB$ about point E by $60^\circ$, making EF coincide with EG, and thus obtaining $\triangle EFB \cong \triangle EGH$,\\
therefore, it can be inferred that $\triangle EBH$ is an equilateral triangle, and point G is on the line HN, which is perpendicular to HE,\\
Draw CM$\perp$HN, then CM is the minimum value of CG,\\
Draw EP$\perp$CM, it can be known that quadrilateral HEPM is a rectangle,\\
then $CM=MP+CP=HE+ \frac{1}{2}EC=2+2=\boxed{4}$.


Question: As illustrated in the Cartesian coordinate system, the vertices of $\triangle ABC$ are respectively at $A(2, 2)$, $B(4, 2)$, and $C(4, 4)$. By taking the origin as the center of similitude, and drawing $\triangle DEF$ on the opposite side of the origin such that $\triangle DEF$ is similar to $\triangle ABC$ with a similarity ratio of $1:2$, what is the length of the line segment DF?

Solution:
\textbf{Solution:}

Given points A(2,2), B(4,2), and C(4,4),\\
hence, AB = 2, and BC = 2,\\
by the Pythagorean theorem, we find that AC = $\sqrt{AB^{2} + BC^{2}} = 2\sqrt{2}$,\\
since a triangle DEF is drawn on the opposite side of the origin with the origin as the center of dilation, such that triangle DEF and triangle ABC are similar with a scale factor of 1:2,\\
therefore, the length of segment DF is $\frac{1}{2}AC = \boxed{\sqrt{2}}$.


Question: As shown in the figure, $\triangle ABC$ and $\triangle A'B'C'$ are similar figures with the point $O$ as the center of similitude. If $OA:OA'=2:3$, what is the ratio of the areas of $\triangle ABC$ and $\triangle A'B'C'$?

Solution:
\textbf{Solution:} Given that $\triangle ABC$ and $\triangle A'B'C'$ are similar figures with point O as the center of similitude,\\
it follows that $\triangle ABC \sim \triangle A'B'C'$, $AC \parallel A'C'$,\\
therefore $\angle ACO = \angle A'C'O$, $\angle CAO = \angle C'A'O$,\\
hence $\triangle AOC \sim \triangle A'OC'$,\\
thus $AC : A'C' = OA : OA' = 2 : 3$,\\
therefore $\frac{S_{\triangle ABC}}{S_{\triangle A'B'C'}} = \left(\frac{AC}{A'C'}\right)^2 = \boxed{\frac{4}{9}}$.


Question: As shown in the figure, $\triangle ABC$ is rotated $40^\circ$ clockwise around point C to obtain $\triangle A'CB'$. If $AC \perp A'B'$, and $AA'$ is connected, what is the measure of $\angle AA'B'$?

Solution:
\textbf{Solution:} Let the intersection point of $AC$ and ${A}^{\prime}{B}^{\prime}$ be O, \\
from the property of rotation it follows that $AC={A}^{\prime}C$, $\angle {A}^{\prime}CA=\angle BC{B}^{\prime}=40^\circ$, \\
$\therefore$ $\angle CA{A}^{\prime}=\angle C{A}^{\prime}A=\frac{180^\circ-\angle {A}^{\prime}CA}{2}=70^\circ$, \\
$\because$ $AC\perp {A}^{\prime}{B}^{\prime}$, \\
$\therefore$ $\angle {A}^{\prime}OC=90^\circ$, \\
$\therefore$ $\angle C{A}^{\prime}O=180^\circ-\angle {A}^{\prime}CO-\angle {A}^{\prime}OC=50^\circ$, \\
$\therefore$ $\angle A{A}^{\prime}{B}^{\prime}=\angle C{A}^{\prime}A-\angle C{A}^{\prime}O=\boxed{20^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $\angle ABC=30^\circ$, $AC=\sqrt{5}$. By rotating $\triangle ABC$ counterclockwise around point A, we get $\triangle AB'C'$, making point C fall on the side AB. Connect $BB'$. What is the length of $BB'$?

Solution:
\textbf{Solution:} Given that $\angle C=90^\circ$,\\
$\angle ABC=30^\circ$,\\
$AC=\sqrt{5}$,\\
therefore $AB=2\sqrt{5}$,\\
$BC=\sqrt{15}$,\\
since right triangle $ABC$ is rotated counterclockwise around point $A$ to get $\triangle AB'C'$,\\
thus $AC'=AC=\sqrt{5}$,\\
$B'C'=BC=\sqrt{15}$,\\
therefore $BC'=\sqrt{5}$,\\
since $\angle BC'B'=90^\circ$,\\
thus $BB'=\sqrt{B'C'^2+BC'^2}=\sqrt{(\sqrt{15})^2+(\sqrt{5})^2}=\boxed{2\sqrt{5}}$


Question: As shown in the figure, in rectangle $ABCD$, $AD=2\sqrt{3}$ and $AB=5$. Let $M$ be a point on $CD$. $\triangle ADM$ is folded along the straight line $AM$ to obtain $\triangle ANM$. If $AN$ bisects $\angle MAB$, what is the length of $CN$?

Solution:
\textbf{Solution}: As shown in the diagram, draw a perpendicular line from point N to CD, intersecting $CD$ at point E. \\
From the fold, we know: $DM=MN$, \\
$DA=AN=2\sqrt{3}$, \\
$\angle DAM=\angle NAM$ \\
Because AN bisects $\angle MAB$, \\
Therefore, $\angle NAM=\angle BAN$ \\
Therefore, $\angle DAM=\angle NAM=\angle BAN$ \\
Because $\angle DAB=90^\circ$, \\
Therefore, $\angle DAM=30^\circ$, \\
$\angle DAN=60^\circ$ \\
Therefore, $DM=AD\cdot\tan30^\circ=2\sqrt{3}\times \frac{\sqrt{3}}{3}=2$ \\
Therefore, $MN=DM=2$ \\
Because $\angle DAN+\angle DMN=180^\circ$, \\
$\angle CMN+\angle DMN=180^\circ$ \\
Therefore, $\angle CMN=\angle DAN=60^\circ$ \\
Therefore, $NE=MN\cdot\sin60^\circ=2\times \frac{\sqrt{3}}{2}=\sqrt{3}$, \\
$ME=1$ \\
Therefore, $CE=5-2-1=2$ \\
Therefore, in right triangle $\triangle ECN$, by the Pythagorean theorem, we get: \\
$CN=\sqrt{{(\sqrt{3})}^{2}+{2}^{2}}=\sqrt{3+4}=\boxed{\sqrt{7}}$


Question: As shown in the diagram, $\triangle ABC$ is rotated clockwise by $50^\circ$ around point $C$ to obtain $\triangle A'B'C'$. If $\angle A=40^\circ$ and $\angle B=110^\circ$, what is the degree measure of $\angle BCA'$?

Solution:
\textbf{Solution:} Given that $\angle A=40^\circ$ and $\angle B=110^\circ$, \\
it follows that $\angle ACB=30^\circ$, \\
since $\triangle ABC$ is rotated $50^\circ$ clockwise around point C to obtain $\triangle A'B'C$, \\
thus $\angle ACA'=50^\circ$, \\
therefore, $\angle BCA'=\angle BCA+\angle ACA'=30^\circ+50^\circ=\boxed{80^\circ}$.


Question: As shown in the figure, within the right-angled triangle $ABC$ with a $30^\circ$ angle, point $M$ is the midpoint on the hypotenuse $AB$. After rotating the triangle clockwise by $90^\circ$ around the right-angle vertex $C$, we obtain $\triangle A^\prime B^\prime C$. If $AC=6$, what is the length of the path traveled by point $M$?

Solution:
\textbf{Solution:} As illustrated, connect CM and CM',\\
since in the $30^\circ$ right-angled triangle ABC, point M is the midpoint on the hypotenuse AB,\\
therefore CM = $\frac{1}{2}AB$,\\
since $\angle B = 30^\circ$,\\
therefore AB = 2AC = 12,\\
therefore CM = 6,\\
According to the problem, the path of point M is an arc with CM as the radius and point C as the center,\\
since the triangle is rotated $90^\circ$ clockwise around the right-angle vertex C to form $\triangle A^{\prime}B^{\prime}C$,\\
therefore $\angle MCM' = 90^\circ$,\\
thus, the length of the path traveled by point M is\\
$ \frac{90^\circ\pi \times 6}{180} = \boxed{3\pi} $.


Question: As shown in the figure, the pentagon ABCDE is obtained through a similarity transformation from the pentagon FGHMN, with point O as the center of similarity. Points F, G, H, M, and N are the midpoints of OA, OB, OC, OD, and OE, respectively. What is the ratio of the areas of the pentagons ABCDE and FGHMN?

Solution:
\textbf{Solution:} Since $F$ is the midpoint of $AO$,\\
it follows that $OF:OA=1:2$,\\
since pentagon $ABCDE$ is obtained from pentagon $FGHMN$ through a similarity transformation,\\
it follows that $FN \parallel AE$,\\
thus $\triangle OFN \sim \triangle OAE$,\\
hence $OF:OA=FN:AE=1:2$,\\
therefore, the ratio of the areas of pentagons $ABCDE$ and $FGHMN$ is: $4:1$,\\
so, the ratio of the areas of pentagons $ABCDE$ and $FGHMN$ is $\boxed{4:1}$.


Question: As shown in the figure, it is known that $\triangle ABC$ and $\triangle EDC$ are similar figures with point $C$ as the center of similarity, and the similarity ratio between $\triangle ABC$ and $\triangle EDC$ is $1\colon 2$. If the area of $\triangle ABC$ is $2$, what is the area of $\triangle EDC$?

Solution:
\textbf{Solution:} Since triangles $\triangle ABC$ and $\triangle EDC$ are homothetic,\\
it follows that $\triangle ABC \sim \triangle EDC$,\\
and since the homothety ratio is $1\colon 2$,\\
the similarity ratio is $1\colon 2$,\\
thus, the area ratio of $\triangle ABC$ to $\triangle EDC$ is: $1\colon 4$,\\
since the area of $\triangle ABC$ is $2$,\\
the area of $\triangle EDC$ is: $2 \times 4 = \boxed{8}$.


Question: As shown in the figure, in the Cartesian coordinate system, $A\left(1, 0\right)$, $B\left(0, 2\right)$, and $\triangle ABC$ is an isosceles right triangle. By using A as the center of dilation and enlarging $\triangle ABC$ by a similarity ratio of $1:2$, the enlarged figure is denoted as $\triangle AB'C'$. What are the coordinates of $C'$?

Solution:
\textbf{Solution:}\\
Since $A$ is the center of similitude, if we enlarge $\triangle ABC$ by the similarity ratio $1\colon 2$, the resulting figure is denoted as $\triangle AB'C'$,\\
therefore $AB=\frac{1}{2}AB'$,\\
thus point $B$ is the midpoint of segment $AB'$,\\
since $A(1,0)$ and $B(0,2)$,\\
by the Pythagorean theorem, $AB=\sqrt{4+1}=\sqrt{5}$,\\
thus $AB' =2AB=2\sqrt{5}$,\\
based on similarity, $\triangle AB'C'$ is an isosceles right triangle,\\
$AC'=\sqrt{AB'^{2}+B'C'^{2}}=2\sqrt{10}$,\\
A, when $C'(-6,-2)$, by the Pythagorean theorem, $AC'=\sqrt{(1+6)^{2}+(0+2)^{2}}=\sqrt{53}\ne 2\sqrt{10}$, the option is incorrect and does not match the condition;\\
B, when $C'(-5,2)$, by the Pythagorean theorem, $AC'=\sqrt{(1+5)^{2}+(0-2)^{2}}=\sqrt{40}=2\sqrt{10}$, the option is correct and matches the condition;\\
C, when $C'(-4,-2)$, by the Pythagorean theorem, $AC'=\sqrt{(1+4)^{2}+(0+2)^{2}}=\sqrt{29}\ne 2\sqrt{10}$, the option is incorrect and does not match the condition;\\
D, when $C'(-4,2)$, by the Pythagorean theorem, $AC'=\sqrt{(1+4)^{2}+(0-2)^{2}}=\sqrt{29}\ne 2\sqrt{10}$, the option is incorrect and does not match the condition;\\
Therefore, the answer is $C'(-5,2)$, that is, the coordinates of $C'$ are $\boxed{(-5,2)}$.


Question: As shown in the figure, $ \triangle ABC$ and $\triangle A'B'C'$ are similar figures with the coordinate origin $O$ as the center of similarity. If point $A$ is the midpoint of $OA'$, and the area of $\triangle ABC$ is $6$, what is the area of $\triangle A'B'C'$?

Solution:
\textbf{Solution:} Given the problem, we know the similarity ratio between triangles $ \triangle ABC$ and $ \triangle A'B'C'$ is $ \frac{1}{2}$.\\
$\therefore$ $ \frac{{S}_{\triangle ABC}}{{S}_{\triangle A'B'C'}}={\left(\frac{1}{2}\right)}^{2}$\\
$\because$ $ {S}_{\triangle ABC}=6$\\
$\therefore$ $ {S}_{\triangle A'B'C'}=\boxed{24}$


Question: As shown in the Cartesian coordinate system, the vertices of $\triangle OAB$ are $O(0,0)$, $A(4,3)$, and $B(3,0)$. Taking point $O$ as the center of similitude, a similar figure $\triangle OCD$ is constructed in the third quadrant, with a similitude ratio of $\frac{1}{3}$ to $\triangle OAB$. What are the coordinates of point $C$?

Solution: Solution: Since $\triangle OAB$ and $\triangle OCD$ have the origin $O$ as the center of similitude, with a similitude ratio of $\frac{1}{3}$, and point $C$ is located in the third quadrant, with the coordinates of point $A$ being $(4,3)$, and given that point $A$ corresponds to point $C$, it follows that the coordinates of point $C$ are $\left(-4\times\frac{1}{3}, -3\times\frac{1}{3}\right)$, which simplifies to $\left(-\frac{4}{3}, -1\right)$. Therefore, the answer is: $C=\boxed{\left(-\frac{4}{3}, -1\right)}$.


Question: In rectangle $ABCD$, $AB = 12$, and $BC = 8$. The rectangle is folded along $MN$ so that point $C$ precisely falls on the midpoint $F$ of side $AD$. If the circle with center $O$, which is the center of symmetry of the rectangle, is tangent to $FN$ at point $G$, what is the radius of circle $\odot O$?

Solution:
\textbf{Solution:} As shown in the figure, connect OF, OG, ON, and draw OH$\perp$DC at point H through O,\\
$\because$ point O is the center of symmetry for the rectangle, and AB=12, BC=8, F is the midpoint of AD,\\
$\therefore$ OF=DH=HC=$\frac{1}{2}$DC=6, OH=FD=$\frac{1}{2}$AD=4,\\
$\because$ circle O is tangent to FN at point G,\\
$\therefore$ OG$\perp$FN,\\
By the property of folding, we get: FN=NC, let FN=NC=a, then DN=12−a\\
In the right-angled triangle FDN, FN$^{2}$=FD$^{2}$+DN$^{2}$, which gives a$^{2}$=4$^{2}$+(12−a)$^{2}$, solving for a, we get a=$\frac{20}{3}$,\\
$\therefore$ DN=$\frac{16}{3}$,\\
$\therefore$ NH=DH−DN=6−$\frac{16}{3}$=$\frac{2}{3}$,\\
In the right-angled triangle OHN, by Pythagoras' theorem, ON$^{2}$=NH$^{2}$+OH$^{2}$=$\frac{4}{9}$+16,\\
Let FG=b, then GN=$\frac{20}{3}$−b,\\
In the right-angled triangles OGF and OGN, by Pythagoras' theorem, OF$^{2}$−FG$^{2}$=OG$^{2}$=ON$^{2}$−GN$^{2}$,\\
$\therefore$ 6$^{2}$−b$^{2}$=$\frac{4}{9}$+16−($\frac{20}{3}$−b)$^{2}$,\\
Solving for b, we get b=$\frac{24}{5}$,\\
$\therefore$ OG$^{2}$=36−($\frac{24}{5}$)$^{2}$,\\
Solving for OG, we obtain OG=\boxed{3.6}, which means the radius is 3.6.


Question: As shown in the diagram, square ABCD and square EFOG are similar figures, where point A corresponds to point E. The coordinate of point A is $(-4, 2)$, and the coordinate of point E is $(-1, 1)$. What are the coordinates of the center of similitude of these two squares?

Solution:
\textbf{Solution:} As shown in the diagram, extend lines AE and DG until they intersect the x-axis at point N. \\
Assume the equation of line $AE$ is $y=kx+b$. Substituting points $A(-4,2)$ and $E(-1,1)$ into the equation, we get $\left\{\begin{array}{l}-4k+b=2 \\ -k+b=1\end{array}\right.$. \\
Solving these equations gives $\left\{\begin{array}{l}k=-\frac{1}{3} \\ b=\frac{2}{3}\end{array}\right.$. \\
Thus, $y=-\frac{1}{3}x+\frac{2}{3}$. \\
Setting $y=0$ yields $x=2$, \\
$\therefore$ the coordinates of N are $(2,0)$; \\
Hence, the answer is: the coordinates of $N$ are $\boxed{(2,0)}$.


Question: As shown in the figure, in square $ABCD$, $AB = 4$, point $M$ is on side $CD$, and $DM = 1$. Triangles $AEM$ and $ADM$ are symmetric about the line containing $AM$. Rotating $\triangle ADM$ $90^\circ$ clockwise around point $A$ results in $\triangle ABF$. When connecting $EF$, what is the length of segment $EF$?

Solution:
\textbf{Solution:} \\
Since $\triangle AEM$ and $\triangle ADM$ are symmetrical about the line containing AM,\\
therefore $\triangle AEM \cong \triangle ADM$\\
thus $AE=AD=AB=4$\\
Connect BM, as shown in the figure,\\
since $\triangle ADM$ is rotated clockwise by $90^\circ$ around point A to get $\triangle ABF$,\\
therefore $\triangle ABF \cong \triangle ADM$\\
thus $AF=AM$, $\angle FAB=\angle MAD$\\
thus $\angle FAB=\angle MAE$\\
therefore $\angle FAB+\angle BAE=\angle MAE + \angle BAE$\\
therefore $\angle FAE=\angle MAB$\\
thus $\triangle FAE\cong \triangle MAB$ (SAS)\\
thus $EF=BM$\\
In the square ABCD,\\
therefore $BC=CD=AB=4$\\
since $DM=1$\\
therefore $CM=3$\\
In $\triangle BCM$,\\
$BM=\sqrt{3^2+4^2}=5$\\
therefore $EF=\boxed{5}$


Question: In rectangle $ABCD$, $AB = 3$, $BC = 2\sqrt{10}$. Point $E$ is on side $BC$. Connect $DE$. Fold $\triangle DEC$ along $DE$ to get $\triangle DEC'$. $C'E$ intersects $AD$ at point $F$. Connect $AC'$. If point $F$ is the midpoint of $AD$, what is the length of $AC'$?

Solution:
\textbf{Solution:} As shown in the figure, draw $C'H \perp AD$ at point H,

since point F is the midpoint of AD, and AD = BC = $2\sqrt{10}$,

thus AF = DF = $\sqrt{10}$,

since $\triangle DEC$ is folded along DE,

thus CD = C'D = 3, $\angle C = \angle EC'D = 90^\circ$,

in $\triangle DC'F$, C'F = $\sqrt{DF^{2} - C'D^{2}} = \boxed{1}$,

since $S_{\triangle C'F} = \frac{1}{2} \times DF \times C'H = \frac{1}{2} \times C'F \times C'D$,

thus $\sqrt{10} \times C'H = 1 \times 3$,

thus C'H = $\frac{3\sqrt{10}}{10}$,

thus FH = $\sqrt{C'F^{2} - C'H^{2}} = \frac{\sqrt{10}}{10}$,

thus AH = AF + FH = $\frac{11\sqrt{10}}{10}$,

in $\triangle AC'H$, AC'$ = \sqrt{AH^{2} + C'H^{2}} = \boxed{\sqrt{13}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=6$, $BC=9$, and point D is the midpoint of side $BC$. The $\triangle ACD$ is folded along $AD$ so that point C falls on point $C'$ in the same plane. Connect $BC'$. What is the length of $BC'$?

Solution:
\textbf{Solution:} Suppose the intersection point of CC' and AD is E, as shown in the figure: \\
Since point D is the midpoint on side BC, and BC = 9, \\
Thus, CD = BD = 4.5, \\
Since $\triangle ACD$ is folded along AD, making point C coincide with point C' in the same plane, \\
Thus, CD = C'D, $CC'\perp AD$, CE = C'E, \\
Thus, CD = BD = C'D, \\
Thus, $\angle DCC'=\angle DC'C$, $\angle DBC'=\angle DC'B$, \\
Since $\angle DCC'+\angle DC'C+\angle DBC'+\angle DC'B=180^\circ$, \\
Thus, $\angle CC'B=\angle CC'D+\angle DC'B= \frac{1}{2} (\angle DCC'+\angle DC'C+\angle DBC'+\angle DC'B)=90^\circ$, \\
Since $\angle C=90^\circ$ and AC = 6, \\
Thus, AD = $\sqrt{AC^{2}+CD^{2}}$ = 7.5, \\
Since $S_{\triangle ACD}$ = $\frac{1}{2}$ AC$\cdot$CD = $\frac{1}{2}$ AD$\cdot$CE, \\
Thus, CE = $\frac{AC\cdot CD}{AD}$ = 3.6, \\
Thus, CC' = 7.2, \\
In Rt$\triangle BC'C$, BC' = $\sqrt{BC^{2}-CC'^{2}}= \boxed{\frac{27}{5}}$.


Question: As shown in the figure, the four vertices of rectangle $EFGH$ are respectively on the four sides of rhombus $ABCD$, with $BE=BF$. $\triangle AEH$ and $\triangle CFG$ are folded along edges $EH$ and $FG$ respectively. When the overlapping section forms a rhombus whose area is $\frac{1}{16}$ of the area of rhombus $ABCD$, then what is the value of $\frac{AE}{EB}$?

Solution:
\textbf{Solution:} Let the overlapping rhombus have a side length of $x$, and $BE=BF=y$,\\
due to the symmetry of the rectangle and the rhombus as well as the property of folding, the quadrilateral $AHME$ and the quadrilateral $BENF$ are rhombuses,\\
$\therefore AE=EM$, $EN=BE=y$, $EM=x+y$,\\
$\because$ when the overlapping part is a rhombus and its area is $\frac{1}{16}$ of the area of the rhombus $ABCD$, and the two rhombuses are similar,\\
$\therefore AB=4MN=4x$,\\
$\therefore AE=AB-BE=4x-y$,\\
$\therefore 4x-y=x+y$,\\
Solving gives: $x=\frac{2}{3}y$,\\
$\therefore AE=\frac{5}{3}y$,\\
$\therefore \frac{AE}{EB}=\frac{\frac{5}{3}y}{y}=\boxed{\frac{5}{3}}$;


Question: As shown in the figure, $A$ is a point on the graph of the inverse proportion function $y = \frac{k}{x}$, and $AB$ is perpendicular to the $x$-axis at $B$. If the area of $\triangle AOB = 2$, what is the value of $k$?

Solution:
\textbf{Solution:} Since ${S}_{\triangle AOB}=2$,\\
therefore $\frac{k}{2}=2$,\\
thus $k=\boxed{4}$;


Question: As shown in the figure, the parabola $y=ax^2+bx+c\,(a\ne 0)$ passes through the points $(-1, 0)$ and $(0, -3)$, and its vertex is in the fourth quadrant. Let $P=a+b+c$. What is the range of values for $P$?

Solution:
\textbf{Solution:} Given the parabola $y=ax^2+bx+c$ (where $a\neq 0$) passes through the points $(-1, 0)$ and $(0, -3)$,\\
thus $\because$ $0=a(-1)^2-b+c$ and $-3=c$,\\
it leads to $\therefore$ $b=a-3$,\\
Considering $\because$ when $x=1$, $y=ax^2+bx+c=a+b+c$,\\
we find that $\therefore$ $P=a+b+c=a+a-3-3=2a-6$,\\
Given that the vertex is in the fourth quadrant, it implies that $a>0$,\\
therefore $\therefore$ $b=a-3<0$,\\
which implies $\therefore$ $a<3$,\\
Hence $\therefore$ $0<a<3$,\\
resulting in $\therefore$ $-6<2a-6<0$, i.e., $-6<P<0$. \\
Therefore, the range of $P$ is $P \in \boxed{(-6, 0)}$.


Question: The graph of the linear function $y_1=mx+n$ ($m\neq 0$) and the quadratic function $y_2=ax^2+bx+c$ ($a\neq 0$) are shown in the figure. What is the solution set of the inequality $ax^2+bx+c>mx+n$?

Solution:
\textbf{Solution}: From the graph, we know that the quadratic function $y_{2}=ax^{2}+bx+c$ ($a\neq 0$) and the linear function $y_{1}=mx+n$ ($m\neq 0$) intersect at the x-coordinates $-4$ and $3$, \\
$\because$ the inequality $ax^{2}+bx+c>mx+n$ implies that the graph of the quadratic function is above that of the linear function, \\
$\therefore$ the solution set of the inequality $ax^{2}+bx+c>mx+n$ is $-4<x<3$; \\
Hence, the answer is: $\boxed{-4<x<3}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=-\frac{2}{3}x$ and the line $y=-\frac{2}{3}x+2$ intersect with the graph of the function $y=\frac{k}{x}$ $(x<0)$ at points $A$ and $B$, respectively. Connecting $AB$ and $OB$, if the area of $\triangle OAB$ is $3$, what is the value of $k$?

Solution:
\textbf{Solution:} Let the line $y=-\frac{2}{3}x+2$ intersect the $y$ axis at point $C$, and connect $AC$, \\
$\therefore$ the point $C(0,2)$, \\
$\because$ $OA\parallel CB$, \\
$\therefore$ ${S}_{\triangle AOC}={S}_{\triangle AOB}=3$, \\
$\therefore$ ${S}_{\triangle AOC}=\frac{1}{2}\times OC\times |{x}_{A}|=3$, \\
$\therefore$ $|{x}_{A}|=3$, \\
$\therefore$ ${x}_{A}=\pm 3$, \\
$\because$ point $A$ is in the second quadrant, \\
$\therefore$ ${x}_{A}=-3$, \\
$\because$ point $A$ is on the line $y=-\frac{2}{3}x$, \\
$\therefore$ $y=-\frac{2}{3}\times (-3)=2$, \\
$\therefore$ the coordinates of point $A$ are: $A(-3,2)$, \\
$\therefore$ substituting point $A(-3,2)$ into $y=\frac{k}{x}\ (x<0)$, \\
$\therefore$ $2=\frac{k}{-3}$, \\
$\therefore$ $k=\boxed{-6}$.


Question: As shown in the figure, point $A$ is on the graph of the inverse proportion function $y=\frac{a}{x}$ in the first quadrant, and point $B$ is on the positive half-axis of the $x$-axis. Given $OA=AB$ and the area of $\triangle AOB$ is 2, what is the value of $a$?

Solution:
\textbf{Solution:} Refer to the diagram, draw $AC\perp OB$ at point C through point A, \\
Let the coordinates of point A be $A(m,n)$, hence $OC=m$, and $AC=n$, \\
Since $OA=AB$, \\
It follows that $OB=2OC=2m$, \\
Given that the area of $\triangle AOB$ is 2, \\
It follows that $\frac{1}{2}OB\cdot AC=\frac{1}{2}\times 2m\cdot n=2$, \\
Upon rearranging: $mn=2$, \\
Substituting point $A(m,n)$ into the inverse proportion function $y=\frac{a}{x}$ yields：$a=mn=\boxed{2}$.


Question: As shown in the diagram, the line $y=ax+b$ intersects the x-axis at point $A(1, 0)$ and intersects the graph of the function $y=\frac{k}{x}$ at points B and C. The x-coordinate of point B is 4, and the x-coordinate of point C is $-3$. What is the solution to the system of inequalities $0<ax+b<\frac{k}{x}$?

Solution:
\textbf{Solution:} Observing the graph yields: when $-3<x<0$, the line $y=ax+b$ is above the x-axis and below the graph of the inverse proportion function $y=\frac{k}{x}$,\\
$\therefore$ the solution to the system of inequalities $0<ax+b<\frac{k}{x}$ is $-3<x<0$.\\
Thus, the answer is: $\boxed{-3<x<0}$.


Question: As shown in the figure, the graph of the quadratic function $y_{1}=-x^{2}+bx+c$ intersects with the graph of the linear function $y_{2}=kx+2$ at points $A(-1, 3)$ and $B(4, m)$. In order for $y_{1}<y_{2}$, what is the range of values for $x$?

Solution:
Solution: Since $y_1 < y_2$, \\
it follows that the range of $x$ for which the line is above the parabola is, \\
according to the graph, when $x < -1$ or $x > 4$, \\
therefore $x < -1$ or $x > 4$, \\
hence, the answer is:$\boxed{ x<-1 or x>4}$.


Question: As shown in the figure, the parabola $y=\frac{1}{4}x^{2}-4$ intersects the x-axis at points A and B. Let P be a moving point on the circle with center C(0,3) and radius 2, and let Q be the midpoint of line segment PA. Connect OQ. What is the maximum value of line segment OQ?

Solution:
\textbf{Solution:} Let us connect BP as shown in the diagram.\\
When $y=0$, we have $\frac{1}{4}x^{2}-4=0$. Solving this, we find $x_{1}=4$ and $x_{2}=-4$, thus $A(-4,0)$ and $B(4,0)$.\\
Since Q is the midpoint of segment PA,\\
it follows that OQ is the median of $\triangle ABP$,\\
therefore OQ$=\frac{1}{2}BP$.\\
When BP is at its maximum, OQ is also at its maximum.\\
Moreover, when BP passes through the circle's center C, PB is maximized. As shown in the diagram, when point P moves to position P', BP is maximized.\\
Since BC$=\sqrt{3^{2}+4^{2}}=5$,\\
it follows that BP'$=5+2=7$.\\
Therefore, the maximum value of segment OQ is $\boxed{\frac{7}{2}}$.


Question: As shown in the figure, the graph of the linear function $y_1=mx+n$ ($m\neq 0$) and the graph of the quadratic function $y_2=ax^2+bx+c$ ($a\neq 0$) are illustrated. What is the solution set of the inequality $ax^2+bx+c>mx+n$?

Solution:
\textbf{Solution:} Given the linear function $y_1=mx+n$ ($m\neq 0$) and the quadratic function $y_2=ax^2+bx+c$ ($a\neq 0$), by examining their graphs, it is known that:\\
The solution set of the inequality $ax^2+bx+c>mx+n$ is: $-4<x<3$,\\
Hence, the solution set is $\boxed{-4<x<3}$.


Question: As shown in the figure, during the final physical examination, the relationship between the flight height $y$ (meters) and the horizontal distance $x$ (meters) of the shot put thrown by Xiao Zhu approximately satisfies the quadratic function $y=\frac{1}{10}x^{2}+\frac{3}{5}x+\frac{8}{5}$. What was Xiao Zhu's score in this shot put event?

Solution:
\textbf{Solution:} Solve: In $y=\frac{1}{10}x^{2}+\frac{3}{5}x+\frac{8}{5}$, let $y=0$ to get:\\
$-\frac{1}{10}x^{2}-\frac{3}{5}x-\frac{8}{5}=0$,\\
we find $x=-2$ (discard) or $x=8$,\\
$\therefore$ Therefore, Xiao Zhu's shot put result is $\boxed{8}$ meters.


Question: As shown in the figure, the parabola $y=ax^2+bx+2$ is symmetric about the line $x=1$, and the point $(3, 0)$ lies on the parabola. What is the range of $x$ for which $ax^2+bx+2<0$?

Solution:
\textbf{Solution} \\
Since the parabola $y=ax^2+bx+2$ is symmetrical about the line $x=1$, and it intersects the x-axis at $\left(3,0\right)$,\\
it follows that the other intersection point with the x-axis is $\left(-1,0\right)$, \\
therefore, when $ax^2+bx+2<0$, the value range of $x$ is $x<-1$ or $x>3$, \\
hence, the solution is: $\boxed{x\in(-\infty,-1)\cup(3,\infty)}$.


Question: As shown in the figure, in quadrilateral $ABCD$, a circle $\odot O$ with diameter $AB$ exactly passes through point $C$, and $AC$ and $DO$ intersect at point $E$. It is known that $AC$ bisects $\angle BAD$, $\angle ADC=90^\circ$, and $CD:BC=2:\sqrt{5}$. What is the value of $CE:AE$?

Solution:
\textbf{Solution:} As shown in the figure, connect OC\\
$\because$ AB is the diameter of the circle,\\
$\therefore$ $\angle$ACB=$\angle$ADC=90$^\circ$,\\
$\because$ AC bisects $\angle$BAD,\\
$\therefore$ $\angle$DAC=$\angle$CAB, $\angle$DAB=2$\angle$CAB,\\
$\therefore$ $\triangle ADC\sim \triangle ACB$,\\
$\therefore$  $\frac{CD}{BC}=\frac{AC}{AB}=\frac{2}{\sqrt{5}}=\frac{AD}{AC}$,\\
$\therefore$  $AC=\frac{2\sqrt{5}}{5}AB$,\\
$\therefore$  $BC=\sqrt{AB^{2}-AC^{2}}=\frac{\sqrt{5}}{5}AB$,  $AD=\frac{2}{\sqrt{5}}AC=\frac{4}{5}AB$\\
$\therefore$  $CD=\frac{2}{\sqrt{5}}BC=\frac{2}{5}AB$,\\
Also $\because$ $\angle$BOC=2$\angle$CAB,\\
$\therefore$ $\angle$BOC=$\angle$DAB,\\
$\therefore$ AD$\parallel$OC,\\
$\therefore$ $\triangle OCE\sim \triangle DAE$,\\
$\therefore$  $\frac{CE}{AE}=\frac{OC}{DA}=\frac{\frac{1}{2}AB}{\frac{4}{5}AB}=\boxed{\frac{5}{8}}$,\\


Question: As shown in the figure, in $\triangle ABC$, points $D$ and $E$ are located on sides $AB$ and $AC$ respectively. If $\frac{AD}{AC}=\frac{AE}{AB}=\frac{2}{3}$ and $BC=2$, what is the length of $DE$?

Solution:
\textbf{Solution:} Since $\angle DAE = \angle CAB$,\\
it follows that $\triangle DAE \sim \triangle CAB$,\\
hence $\frac{DE}{BC} = \frac{AE}{AB} = \frac{2}{3}$,\\
therefore $DE = \frac{2}{3} \times BC = \frac{2}{3} \times 2 = \boxed{\frac{4}{3}}$


Question: As shown in the figure, in $\triangle ABC$, $DE \parallel BC$, $\frac{AD}{AB} = \frac{1}{3}$. If the area of $\triangle ABC$ is $9$, then what is the area of quadrilateral $BCED$?

Solution:
\textbf{Solution:} Given that DE$\parallel$ BC, it follows that $\triangle ADE \sim \triangle ABC$. Therefore, $\frac{S_{\triangle ADE}}{S_{\triangle ABC}} = \left(\frac{AD}{AB}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$.\\
Since $S_{\triangle ABC} = 9$, it implies that $S_{\triangle ADE} = 1$. Consequently, the area of quadrilateral BCDE equals $9 - 1 = \boxed{8}$.


Question: As shown in the figure, in $\triangle ABC$, points $D, E, F$ are on the sides $AB$, $AC$, and $BC$ respectively, with $DE \parallel BC$, $EF \parallel AB$, and $AD: DB = 3:5$, what is the ratio of $BF: CF$?

Solution:
\textbf{Solution:} Given that DE$\parallel$BC,\\
it follows that $\frac{AD}{BD}=\frac{AE}{CE}=\frac{3}{5}$;\\
since EF$\parallel$AB,\\
it also follows that $\frac{AE}{CE}=\frac{BF}{CF}=\boxed{\frac{3}{5}}$.


Question: As shown in the figure, $AD \parallel BE \parallel CF$, where points $B, E$ are on $AC, DF$ respectively, $\frac{AB}{BC} = \frac{2}{3}, EF = 6$. What is the length of $DE$?

Solution:
\textbf{Solution:} Since $AD\parallel BE\parallel CF$, \\
it follows that $\frac{AB}{BC}=\frac{DE}{FE}$, \\
thus $DE=\boxed{4}$


Question: According to the figure, it is a schematic diagram of the pinhole imaging principle in physics. Given that the object $AB=30$, based on the dimensions in the figure $(AB\parallel CD)$, what should be the length of $CD$?

Solution:
Solution: According to the problem, $\triangle ODC \sim \triangle OAB$\\
$\therefore \frac{AB}{CD} = \frac{36}{12}$\\
$\because AB = 30$\\
$\therefore CD = \frac{1}{3} \times 30 = \boxed{10}$


Question: As shown in the figure, within a $2\times 4$ square grid, line segments $AB$ and $CD$ intersect at point $E$. If the side length of each small square is 1, what is the length of $DE$?

Solution: \textbf{Solution:} From the figure, we know: $BC \parallel AD$, $\therefore \triangle CBE \sim \triangle DAE$, $\therefore \frac{CB}{AD} = \frac{CE}{DE}$, $\because CB = \sqrt{2^2 + 2^2} = 2\sqrt{2}$, $AD = \sqrt{1^2 + 1^2} = \sqrt{2}$, $\therefore \frac{CE}{DE} = \frac{2}{1}$, $\because CD = \sqrt{1^2 + 2^2} = \sqrt{5}$, $\therefore DE = \boxed{\frac{\sqrt{5}}{3}}$


Question: As shown in the figure, it is known that $AB\parallel CD\parallel EF$, $AD:DF=2:3$, $BE=15$, what is the length of $BC$?

Solution:
\textbf{Solution:} Since $AB \parallel CD \parallel EF$, \\
it follows that $\frac{BC}{BE} = \frac{AD}{AF}$, \\
which means $\frac{BC}{15} = \frac{2}{2+3}$, \\
hence, BC$=\boxed{6}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the isosceles right triangle $ABC$ and the isosceles right triangle $DEF$ are similar figures, with their hypotenuses perpendicular to the x-axis. Let $O$ be the center of similarity, $\angle ABC = \angle DEF = 90^\circ$, and the points $O$, $B$, $C$, $E$, $F$ are collinear. If the area ratio ${S}_{\triangle DEF} \colon {S}_{\triangle ABC} = 1 \colon 2$, and the coordinates of point $D$ are $(-1, 0)$, what are the coordinates of point $B$?

Solution: \textbf{Solution:} Since isosceles right triangle $ABC$ and isosceles right triangle $DEF$ are homothetic figures, it follows that $\triangle ABC\sim \triangle DEF$, $\frac{OD}{OA}=\frac{DE}{AB}$. Given that the area ratio $S_{\triangle DEF} : S_{\triangle ABC}=1 : 2$, then $\frac{S_{\triangle DEF}}{S_{\triangle ABC}}=\frac{1}{2}=\left(\frac{DE}{AB}\right)^2$. Hence, $\frac{OD}{OA}=\frac{DE}{AB}=\frac{1}{\sqrt{2}}$. Since point $D$ is at $(-1,0)$, OD equals 1, thus OA equals $\sqrt{2}$. Given $\triangle ABC$ is an isosceles right triangle, we have $\angle C=\angle CAB=45^\circ$, $\angle ABC=\angle OBA=90^\circ$. Therefore, AC is perpendicular to the x-axis, leading to $\angle CAB+\angle BAO=90^\circ$, which means $\angle BAO=\angle OBA=45^\circ$. Consequently, OB=AB. Draw BG perpendicular to OA, thus BG=OG=$\frac{1}{2}$OA=$\frac{\sqrt{2}}{2}$. Therefore, the coordinates of point B are $\boxed{\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the inverse proportion function $y=\frac{k}{x}$ ($x>0$) intersects with side $AB$ of rectangle $OABC$ at point $M(1,3)$, and intersects with side $BC$ at point $N$. If the symmetric point $B'$ of point $B$ with respect to line $MN$ lies exactly on the $x$-axis, what is the length of $OC$?

Solution:
Solution:It is known from the problem that the point $M(1,3)$ lies on the graph of the inverse proportion function $y=\frac{3}{x}(x>0)$,

$\therefore$ $3=\frac{k}{1}$, solving for $k$, we get $k=3$,

$\therefore$ the equation of this inverse proportion function is $y=\frac{3}{x}(x>0)$.

Let $N(a,\frac{3}{a})$, then $B(a,3)$, $C(a,0)$,

$\therefore$ $CN=\frac{3}{a}$

$\therefore$ $BN=BC-CN=3-\frac{3}{a}$, $BM=AB-AM=a-1$.

As shown in the figure, connect $MB',NB'$, and draw $MD\perp x$ axis at point D through point M.

$\therefore$ $DM=3$, $OD=1$.

Since the symmetric point $B'$ of point B about line $MN$ exactly lies on the x-axis,

$\therefore$ $B'N=BN=3-\frac{3}{a}$, $B'M=BM=a-1$.

Since $\angle DMB'+\angle DB'M=\angle CB'N+\angle DB'M=90^\circ$,

$\therefore$ $\angle DMB'=\angle CB'N$.

Also, since $\angle MDB'=\angle B'CN=90^\circ$,

$\therefore$ $\triangle DMB'\sim \triangle CB'N$,

$\therefore$ $\frac{B'M}{B'N}=\frac{DM}{B'C}=\frac{B'D}{CN}$, that is, $\frac{a-1}{3-\frac{3}{a}}=\frac{3}{B'C}=\frac{B'D}{\frac{3}{a}}$,

Solving this, we find $B'C=\frac{9}{a}$, $B'D=1$.

Since $B'M^2=MD^2+B'D^2$, i.e., $(a-1)^2=3^2+1^2$,

$\therefore$ $a_1=1+\sqrt{10}$, $a_2=1-\sqrt{10}$ (discard),

$\therefore$ $OC=1+\sqrt{10}$.

Therefore, the answer is: $OC=\boxed{1+\sqrt{10}}$.


Question: As shown in the figure, point F is on the side AD of $ \square ABCD$, and the extensions of BA and CF intersect at point E. If $AE:AB = 1:2$, what is the ratio of the area of the quadrilateral $ABCF$ to the area of $\triangle CDF$?

Solution:
Solution: Since $AE: AB = 1: 2$,\\
therefore $AE: BE=1: 3$, $AE: DC=1: 2$.\\
In parallelogram $ABCD$ where $AB \parallel CD$, $AD \parallel BC$,\\
hence $\triangle AEF \sim \triangle DCF$, $\triangle AEF \sim \triangle BEC$,\\
thus ${S}_{\triangle AEF}:{S}_{\triangle DCF}=1:4$, ${S}_{\triangle AEF}:{S}_{\triangle BEC}=1:9$,\\
therefore ${S}_{\triangle AEF}:{S}_{\text{quadrilateral }ABCF}=1:8$.\\
Hence ${S}_{\text{quadrilateral }ABCF}:{S}_{\triangle CDF}=8:4=2:1$.\\
Thus the answer is: $\boxed{2:1}$.


Question: As shown in the figure, $E$, $F$, $G$, $H$ are points on the sides of the square $ABCD$, and $AG=BE=CH=DF$, $EF$ and $GH$ cut the square into four pieces which are then reassembled into the quadrilateral $MQPN$. If the ratio of the area of the square $ABCD$ to that of the quadrilateral $MQPN$ is $9\colon 10$, then what is the ratio $AG\colon GD$?

Solution:
\textbf{Solution:} As shown in the diagram, connect $EH, FG, GE, HF$,
\[
\because \text{quadrilateral ABCD is a square},
\]
\[
\therefore \angle A=\angle B=\angle C=\angle D=90^\circ, AB=BC=CD=DA.,
\]
\[
\because AG=BE=CH=DF,
\]
\[
\therefore DG=AE=BH=CF,
\]
\[
\therefore \triangle AEG\cong \triangle BHE\cong \triangle CFH\cong \triangle DGF,
\]
\[
\therefore EG=FG=EH=HF,
\]
\[
\therefore \text{quadrilateral EHFG is a rhombus},
\]
\[
\because \triangle DGF\cong \triangle AEG,
\]
\[
\therefore \angle DGF=\angle AEG,
\]
\[
\because \angle AEG+\angle AGE=90^\circ,
\]
\[
\therefore \angle DGF+\angle AGE=90^\circ,
\]
\[
\therefore \angle EGF=90^\circ,
\]
\[
\therefore \text{quadrilateral EHFG is a square},
\]
\[
\therefore GH\perp EF,
\]
Combining the above, it is known that both the quadrilateral MQPN and the quadrilateral A'B'C'D' are squares, $AG={A}^{\prime}{G}^{\prime}$, $DG={D}^{\prime}{G}^{\prime}$,
\[
\therefore {A}^{\prime}{D}^{\prime}={A}^{\prime}{G}^{\prime}-{G}^{\prime}{D}^{\prime}=AG-GD.
\]
\[
\because \text{The ratio of the area of square ABCD to the area of quadrilateral MQPN is } 9\colon 10,
\]
\[
\therefore \text{The ratio of the area of square ABCD to the area of quadrilateral A'B'C'D' is } 9\colon 1,
\]
\[
\therefore {\left(AG+GD\right)}^{2}\colon {\left({A}^{\prime}{G}^{\prime}-{G}^{\prime}{D}^{\prime}\right)}^{2}=9\colon 1,
\]
\[
\therefore \left(AG+GD\right)\colon \left(AG-GD\right)=3\colon 1,
\]
\[
\therefore AG=2GD,
\]
\[
\therefore AG\colon GD=\boxed{2}.
\]


Question: As shown in the diagram, $P$ is a point inside rectangle $ABCD$. Lines are drawn connecting $P$ to each vertex of rectangle $ABCD$. The vertices of rectangle $EFGH$ lie respectively on $AP$, $BP$, $CP$, and $DP$. It is known that $AE = 2EP$ and $EF \parallel AB$. The sum of the areas of the two shaded regions in the diagram is $S$. What is the area of rectangle $ABCD$?

Solution:
\textbf{Solution:} Given that $AE=2EP$,\\
it follows that $\frac{PE}{PA}=\frac{1}{3}$,\\
since quadrilaterals $ABCD$ and $EFGH$ are rectangles,\\
we have $\angle DAB=\angle HEF=90^\circ$,\\
since $EF\parallel AB$,\\
it results in $\angle PEF=\angle PAB$,\\
which leads to $\angle PEH=\angle PAD$,\\
hence $EH\parallel AD$,\\
similarly, $FG\parallel BC$,\\
because $EH\parallel AD$,\\
it implies $\triangle PEH\sim \triangle PAD$ with a similarity ratio of $\frac{PE}{PA}=\frac{1}{3}$,\\
thus $\frac{S_{\triangle PEH}}{S_{\triangle PAD}}=\left(\frac{PE}{PA}\right)^2=\frac{1}{9}$,\\
similarly, $\frac{S_{\triangle PFG}}{S_{\triangle PBC}}=\frac{1}{9}$,\\
since $S_{\triangle PAD}+S_{\triangle PBC}=\frac{1}{2}S_{\text{rectangle}ABCD}$,\\
it follows that $S=\frac{1}{9}\left(S_{\triangle PAD}+S_{\triangle PBC}\right)=\frac{1}{9}\times \frac{1}{2}S_{\text{rectangle}ABCD}$,\\
therefore, the area of rectangle $ABCD$ is $\boxed{18S}$.


Question: In an isosceles triangle with a vertex angle of $36^\circ$, we refer to such a triangle as a "golden triangle", where the ratio of the base to the leg is the golden ratio. As shown in the figure, in $\triangle ABC$, $\angle A=36^\circ$, $AB=AC$, $BD$ bisects $\angle ABC$ and intersects $AC$ at point $D$. If $CD=1$, what is the length of $AC$?

Solution:
\textbf{Solution:} Given that $\angle A=36^\circ$ and AB＝AC, with BD bisecting $\angle ABC$ and intersecting AC at point D,\\
it follows that $\angle ABC=\angle ACB=72^\circ$, $\angle A=\angle DBC=36^\circ$, and $\angle BCD=\angle BDC=72^\circ$,\\
implying that $\angle ACB=\angle BCD$,\\
which means $\triangle BCD\sim \triangle ABC$,\\
leading to the ratio BC$\colon$ AB=DC$\colon$ BC,\\
and given BC$\colon$ AB=$\frac{\sqrt{5}-1}{2}$ and DC=1,\\
we find BC=$\frac{\sqrt{5}+1}{2}$,\\
thus $\frac{\sqrt{5}+1}{2}\colon$ AB=$\frac{\sqrt{5}-1}{2}$,\\
resulting in AB=$\frac{\sqrt{5}+3}{2}$,\\
so AC=$\boxed{\frac{\sqrt{5}+3}{2}}$.


Question: As shown in the figure, in parallelogram $ABCD$, $E$ is a point on $CD$. $DE:CE=2:3$. Connect $AE$ and $BD$ which intersect at point $F$. What is the value of $\frac{S_{\triangle DEF}}{S_{\triangle ABF}}$?

Solution:
\textbf{Solution:} In parallelogram $ABCD$, we have $DC=AB$ and $DC\parallel AB$,\\
$\therefore$ $\triangle DEF\sim \triangle BAF$,\\
$\because$ $DE\colon CE=2\colon 3$,\\
$\therefore$ $DE\colon DC=2\colon 5$,\\
$\therefore$ $\frac{DE}{AB}=\frac{2}{5}$,\\
$\therefore$ $\frac{S_{\triangle DEF}}{S_{\triangle ABF}}=\left(\frac{2}{5}\right)^2=\boxed{\frac{4}{25}}$;


Question: As shown in the figure, the edge OA of $\triangle OABC$ is on the positive half of the x-axis, point D is on edge AB, and the extension of segment CD intersects the x-axis at point E. Lines OB and OD are connected. The inverse proportion function $y= \frac{k}{x}$ ($k>0$) passes through points B and D. If the area of $\triangle OBC$ is 4 and the area of $\triangle BDE$ is 2, what is the value of $k$?

Solution: \textbf{Solution:} In parallelogram $ABCD$, $CB \parallel OA$, therefore $S_{\triangle OBC} = S_{\triangle EBC} = 4$, because $S_{\triangle BDE} = 2$, therefore $S_{\triangle BDC} = S_{\triangle BDE} = 2$ therefore $CD = DE$. Because $\triangle CBD \sim \triangle EAD$, therefore $\frac{CD}{DE} = \frac{BD}{DA}$ therefore $BD = DA$, $D$ is the midpoint of $AB$. Draw $BM \perp OA$, $DN \perp OA$ then $MN = AN$, $DN = \frac{1}{2}BM$ because the inverse proportion function $y = \frac{k}{x}$ (k>0) passes through points $B$, $D$ therefore $S_{\triangle BOM} = S_{\triangle DON}$ therefore $OM \cdot BM = ON \cdot DN$ therefore $\frac{OM}{ON} = \frac{DN}{BM} = \frac{1}{2}$. Let $OM = t$, then $ON = 2t$, $MN = t = AN$, $OA = 3t$ therefore $OM = \frac{1}{3}OA$ therefore $S_{\triangle BOM} = \frac{1}{3}S_{\triangle BOA} = \frac{1}{3}S_{\triangle BOC} = \frac{4}{3}$. Because the inverse proportion function $y = \frac{k}{x}$ (k>0) passes through point $D$ therefore $S_{\triangle DON} = \frac{1}{2}k = \frac{4}{3}$ therefore $k = \boxed{\frac{8}{3}}$.


Question: As shown in the diagram, within $\triangle ABC$, $AB=6$, $AC=5$, $BC=3$. Point $M$ is on $AB$ with $AM=2$. A line $MN$ intersects $\triangle ABC$ at point $M$, and it satisfies $\angle AMN = \angle C$. What is the length of $MN$?

Solution:
\textit{Solution}: In triangles $\triangle AMN$ and $\triangle ACB$,\\
$\because$ $\left\{\begin{array}{l}
\angle AMN=\angle C \\
\angle MAN=\angle CAB
\end{array}\right.$ \\
$\therefore$ $\triangle AMN\sim \triangle ACB$ \\
$\therefore$ $\frac{MN}{BC}=\frac{AM}{AC}$ \\
$\therefore$ $\frac{MN}{3}=\frac{2}{5}$ \\
$\therefore$ $MN=\boxed{\frac{6}{5}}$


Question: As shown in the figure, the graphs of the inverse proportion functions $y_{1}=-\frac{1}{x}$ and $y_{2}=-\frac{4}{x}$ are depicted. Points A and C are located on the x-axis and y-axis, respectively. Quadrilateral $OABC$ is a square. The sides $AB$ and $BC$ intersect the graphs of the inverse proportion functions $y_{2}$ and $y_{1}$ at points F, H and points E, G, respectively. If $OA=3$, what is the value of $\frac{EF}{GH}$?

Solution:
\textbf{Solution:} Given that quadrilateral $OABC$ is a square, and $OA=3$,\\
thus $A(-3,0)$, $B(-3,3)$, $C(0,3)$,\\
substituting $x=-3$ into ${y}_{1}=-\frac{1}{x}$ and ${y}_{2}=-\frac{4}{x}$ yields, ${y}_{1}=-\frac{1}{-3}=\frac{1}{3}$, ${y}_{2}=-\frac{4}{-3}=\frac{4}{3}$,\\
thus $H(-3,\frac{1}{3})$, $F(-3,\frac{4}{3})$,\\
therefore $BF=3-\frac{4}{3}=\frac{5}{3}$, $BH=3-\frac{1}{3}=\frac{8}{3}$,\\
substituting $y=3$ into ${y}_{1}=-\frac{1}{x}$ and ${y}_{2}=-\frac{4}{x}$ yields, ${x}_{1}=-\frac{1}{3}$, ${x}_{2}=-\frac{4}{3}$,\\
thus $E(-\frac{4}{3},3)$, $G(-\frac{1}{3},3)$,\\
therefore $BG=-\frac{1}{3}-(-3)=\frac{8}{3}$, $BE=-\frac{4}{3}-(-3)=\frac{5}{3}$,\\
therefore $\frac{BF}{BH}=\frac{BE}{BG}=\frac{\frac{5}{3}}{\frac{8}{3}}=\frac{5}{8}$,\\
therefore $EF\parallel GH$,\\
thus $\triangle BEF\sim \triangle BGH$,\\
therefore $\frac{EF}{GH}=\frac{BE}{BG}=\boxed{\frac{5}{8}}$,


Question: As shown in the figure, D is a point on side AB of the equilateral $\triangle ABC$, and $AD: DB = 2: 3$. Now, fold $\triangle ABC$ such that point C coincides with D, with the fold line being EF, where points E and F lie on AC and BC, respectively. What is the ratio $CE: CF$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is an equilateral triangle,\\
it follows that $\angle A = \angle B = \angle C = 60^\circ$.\\
From the properties of folding, we know that $\angle EDF = \angle A = 60^\circ$, $DE = CE$, and $DF = CF$,\\
hence, $\angle ADE + \angle BDF = 180^\circ - \angle EDF = 120^\circ$, and $CE:CF = DE:DF$.\\
Since $\angle AED + \angle ADE = 180^\circ - \angle A = 120^\circ$,\\
it follows that $\angle AED = \angle BDF$,\\
hence, $\triangle AED \sim \triangle BDF$,\\
thus, $\frac{DE}{DF} = \frac{7x}{8x} = \frac{7}{8}$,\\
therefore, $CE:CF = \boxed{\frac{7}{8}}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is a parallelogram, point $E$ is on side $CD$, with $DE=2CE$, connect $AE$ and intersect $BD$ at point $F$. What is the ratio of $DF:BD$?

Solution:
\textbf{Solution}: Since $DE=2CE$,\\
it follows that $\frac{DE}{DC}=\frac{2}{3}$,\\
since quadrilateral $ABCD$ is a parallelogram,\\
it follows that $AB=CD$, $AB\parallel CD$,\\
therefore, $\triangle DFE\sim \triangle BFA$,\\
thus, $\frac{DF}{BF}=\frac{DE}{AB}=\frac{DE}{CD}=\frac{2}{3}$,\\
hence, $DF\colon BD=\boxed{2\colon 5}$.


Question: As shown in the figure, in $\triangle ABC$, points $D$, $E$, $F$ are the midpoints of $AB$, $BC$, $AC$, respectively. If the area of $\triangle ABC$ is $1$, what is the area of $\triangle DEF$?

Solution:
\textbf{Solution:} Given points $D$, $E$, $F$ are the midpoints of $AB$, $BC$, $AC$ respectively,\\
thus $EF=\frac{1}{2}BA$, $DF=\frac{1}{2}BC$, $DE=\frac{1}{2}AC$,\\
thus $\frac{EF}{BA}=\frac{DF}{BC}=\frac{DE}{AC}=\frac{1}{2}$,\\
thus $\triangle ABC\sim \triangle EDF$,\\
thus $\frac{{S}_{\triangle EDF}}{{S}_{\triangle ABC}}={\left(\frac{1}{2}\right)}^{2}=\frac{1}{4}$,\\
given ${S}_{\triangle ABC}=1$,\\
thus ${S}_{\triangle EDF}=\boxed{\frac{1}{4}}$,


Question: As shown in the figure, the lines, $l_{1} \parallel l_{2} \parallel l_{3}$ and lines $AC$ and $DF$ are intersected by $l_{1}$, $l_{2}$, $l_{3}$. Given $AB = 4$, $BC = 6$, $EF = 9$, what is the length of $DE$?

Solution:
\textbf{Solution:} Since the lines, $l_1\parallel l_2\parallel l_3$\\
Therefore, $\frac{AB}{BC}=\frac{DE}{EF}$\\
Given $AB=4$, $BC=6$, $EF=9$\\
Thus $DE=\frac{AB\cdot EF}{BC}=\frac{4\times 9}{6}=\boxed{6}$


Question: As shown in the figure, within square $ABCD$, point $E$ is on side $BC$, and $AB = 3BE$. Draw $BF\perp AE$ through point $B$, intersecting with side $CD$ at point $F$. With $C$ as the center and $CF$ as the radius, draw a circle intersecting side $BC$ at point $G$. Connect $DG$, intersecting $BF$ at point $H$. What is the ratio of $DH$ to $HG$?

Solution:
\textbf{Solution:} As shown in the diagram, connect AH and CH. Let AE and BF intersect at M,\\
$\because BF \perp AE$,\\
$\therefore \angle AMB = 90^\circ$,\\
$\therefore \angle BAM + \angle ABM = 90^\circ$,\\
$\because$ Quadrilateral ABCD is a square,\\
$\therefore AB = BC = CD = AD$, $\angle ABE = \angle BCF = 90^\circ$,\\
$\therefore \angle ABM + \angle CBF = 90^\circ$,\\
$\therefore \angle BAE = \angle CBF$,\\
$\therefore \triangle ABE \cong \triangle BCF$ (ASA),\\
$\therefore BE = CF$,\\
$\therefore BF = DF$,\\
$\because CG = CF$, $\angle DCG = \angle BCF$, DC = BC,\\
$\therefore \triangle BCF \cong \triangle DCG$ (SAS),\\
$\therefore \angle CBF = \angle CDG$,\\
Also, $\because \angle BHG = \angle DHF$,\\
$\therefore \triangle BHG \cong \triangle DHF$ (AAS),\\
$\therefore HG = HF$,\\
Also, $\because HC = HC$, CG = CF,\\
$\therefore \triangle HCG \cong \triangle HCF$ (SSS),\\
$\therefore \angle HCG = \angle HCF = 45^\circ$,\\
$\therefore$ Points A, H, and C are collinear,\\
$\because AD \parallel CG$,\\
$\therefore \triangle ADH \sim \triangle CGH$,\\
$\therefore \frac{DH}{HG} = \frac{AD}{CG} = \frac{AD}{BE} = \frac{AB}{BE} = \boxed{3\colon 1}$,\\


Question: As shown in the figure, in rhombus $ABCD$, point $E$ is on side $AD$, and ray $CE$ intersects the extension of $BA$ at point $F$. If $\frac{AE}{ED}=\frac{1}{2}$ and $AB=3$, then what is the length of $AF$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rhombus, \\
$\therefore$ $AE\parallel BC$, $AD=AB=BC=3$, \\
$\therefore$ $\triangle FAE\sim \triangle FBC$, \\
$\therefore$ $\frac{AF}{BF}=\frac{AE}{BC}$, \\
Since $\frac{AE}{ED}=\frac{1}{2}$, \\
$\therefore$ $\frac{AE}{BC}=\frac{AE}{AD}=\frac{1}{3}$, \\
$\therefore$ $\frac{AF}{BF}=\frac{AE}{BC}=\frac{1}{3}$, that is: $\frac{AF}{AF+AB}=\frac{AF}{AF+3}=\frac{1}{3}$, \\
Solving gives: $AF=\boxed{\frac{3}{2}}$;


Question: As shown in the figure, it is known that line $l_{1} \parallel l_{2} \parallel l_{3}$. Line AC intersects lines $l_{1}$, $l_{2}$, and $l_{3}$ at points A, B, and C, respectively. Line DF intersects lines $l_{1}$, $l_{2}$, and $l_{3}$ at points D, E, and F, respectively. Lines AC and DF intersect at point O. If BC = 2AO = 2OB and OD = 1, then what is the length of OF?

Solution:
Solution: $\because BC = 2AO = 2OB$,\\
$\therefore OC = 3AO$,\\
$\because$ line $l_1 \parallel l_2 \parallel l_3$,\\
$\therefore \frac{AO}{OC} = \frac{OD}{OF}$,\\
$\therefore \frac{AO}{OC} = \frac{OD}{OF} = \frac{1}{3}$,\\
$\because OD = 1$,\\
$\therefore OF = \boxed{3}$


Question: As shown in the diagram, in the rhombus $ABCD$, the heights $CE$ and $CF$ are drawn from point $C$ onto sides $AB$ and $AD$ respectively. Line $BF$ is drawn and intersects $CE$ at point $G$. If point $E$ is the midpoint of $AB$, what is the value of $\frac{EG}{CG}$?

Solution:
\textbf{Solution:} As the diagram shows, construct $FH \perp CE$ intersecting $CE$ at $H$,\\
$\because \angle D = \angle EBC, \angle DFC = \angle BEC = 90^\circ, CD = BC$,\\
$\therefore \triangle DFC \cong \triangle BEC \left( AAS \right)$,\\
$\therefore BE = DF, CF = CE, \angle DCF = \angle BCE$,\\
$\because AD \parallel BC, CF \perp AD$,\\
$\therefore \angle FCB = 90^\circ$,\\
Similarly, $\angle DCE = 90^\circ$,\\
$\because \angle FCE + \angle DCF = 90^\circ, \angle D + \angle DCF = 90^\circ$,\\
$\therefore \angle D = \angle FCE$,\\
$\because FH \perp CE$,\\
$\therefore \triangle DFC \sim \triangle CHF$,\\
$\therefore \frac{CD}{CF} = \frac{CF}{FH}$,\\
Let $BE = x$, then $DF = BE = x, CD = AB = BC = 2x, CF = CE = \sqrt{3}x$,\\
$\therefore \frac{2x}{\sqrt{3}x} = \frac{\sqrt{3}x}{FH}$,\\
$\therefore FH = \frac{3}{2}x$,\\
$\therefore CH = \frac{\sqrt{3}}{2}x$,\\
$\because FH \perp CE, CE \perp AB$,\\
$\therefore \triangle FHG \sim \triangle BEG$,\\
$\therefore \frac{FH}{BE} = \frac{HG}{EG} = \frac{\frac{3}{2}x}{x}$,\\
$\because HE = CE - CH = \frac{\sqrt{3}}{2}x$,\\
$\therefore HG = \frac{3\sqrt{3}}{10}x, GE = \frac{\sqrt{3}}{5}x$,\\
$\therefore \frac{EG}{CG} = \frac{\frac{\sqrt{3}}{5}x}{\frac{8\sqrt{3}}{10}x} = \boxed{\frac{1}{4}}$,


Question: As shown in the diagram, in $\triangle ABC$, $\angle ABC=90^\circ$, $\angle A=60^\circ$, and D is a point on side $AC$ such that $\angle DBC=30^\circ$, $AC=12\text{cm}$. What is the perimeter of $\triangle ABD$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a parallelogram,\\
it follows that $DC\parallel AB$, $AD\parallel BC$, $DC=AB$, $AD=BC$,\\
therefore, $\triangle CDF\sim \triangle BEF$,\\
thus, $BE\colon DC=BF\colon CF$,\\
since $BE\colon AB=3\colon 2$, $DC=AB$,\\
it follows that $BE\colon DC=BF\colon CF=3\colon 2$,\\
therefore, $CF\colon BF=2\colon 3$,\\
thus, $CF\colon BC=2\colon 5$,\\
since $AD=BC=10$,\\
it follows that $CF\colon 10=2\colon 5$,\\
therefore, $CF=\boxed{4}$.


Question: As shown in the figure, the lines $l_1 \parallel l_2 \parallel l_3$, and the lines AC and DF are intersected by $l_1$, $l_2$, $l_3$. Given that AB = 8, BC = 12, and EF = 9, what is the length of DE?

Solution:
\textbf{Solution:} Since $l_1\parallel l_2\parallel l_3$,\\
it follows that $\frac{AB}{BC}=\frac{DE}{EF}$,\\
given $AB=8, BC=12, EF=9$,\\
thus $\frac{8}{12}=\frac{DE}{9}$, solving gives $DE=\boxed{6}$.


Question: As shown in the figure, it is known that quadrilateral $ABFE \sim$ quadrilateral $EFCD$, $AB=2$, and $EF=3$. What is the length of $DC$?

Solution:
\textbf{Solution:} Since quadrilateral $ABFE\sim$ quadrilateral $EFCD$,\\
we have $\frac{AB}{EF}=\frac{EF}{CD}$,\\
Considering $AB=2$ and $EF=3$,\\
we obtain $CD=\frac{EF^2}{AB}=\boxed{\frac{9}{2}}$


Question: As shown in the figure, in quadrilateral $ABCD$, point $E$ is on side $BC$. Connect $DE$ and extend it to intersect the extension of side $AB$ at point $F$. If $\frac{CE}{BE}=\frac{4}{3}$, what is the ratio of the perimeter of $\triangle BEF$ to the perimeter of $\triangle ADF$?

Solution:
\textbf{Solution:} Given that parallelogram ABCD,\\
thus AD=DC, BC$\parallel$AD,\\
given $\frac{CE}{BE}=\frac{4}{3}$,\\
thus $\frac{BE}{BC}=\frac{3}{7}$,\\
therefore, BE:AD=3:7,\\
given BE$\parallel$AD,\\
thus $\triangle ADF \sim \triangle BEF$,\\
therefore, the perimeter ratio of $\triangle BEF$ to $\triangle ADF$ is equal to the similarity ratio $3:7$.\\
Hence, the answer is: $\boxed{3:7}$.


Question: As shown in the figure, \(l_1, l_2, l_3\) are a set of parallel lines, and the lines AC, DF intersect this set of parallel lines successively at points A, B, C, and D, E, F, respectively. If \(\frac{AB}{BC}=\frac{2}{3}\), then what is the value of \(\frac{EF}{DF}\)?

Solution: \textbf{Solution:} Since $l_1 \parallel l_2 \parallel l_3$, we have $\frac{EF}{DF}=\frac{BC}{AC}$. Furthermore, since $\frac{AB}{BC}=\frac{2}{3}$, it follows that $\frac{BC}{AC}=\frac{3}{5}$. Hence, $\frac{EF}{DF}=\boxed{\frac{3}{5}}$.


Question: As shown in the diagram, in square $ABCD$, $E$ is a point on side $BC$. If $BE: EC=2:3$ and $AE$ intersects $BD$ at $F$, what is the ratio of $BF: FD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\
$\therefore$ AD $\parallel$ BC and AD $=$ BC, \\
Since BE: EC $=$ 2: 3, \\
$\therefore$ BE: AD $=$ 2: 5, \\
Since AD $\parallel$ BC, \\
$\therefore$ $\triangle$ADF $\sim$ $\triangle$EBF, \\
$\therefore$ BF: FD $=$ BE: AD $=$ 2: 5, \\
Hence, the answer is: \boxed{2:5}.


Question: As shown in the figure, the grid is a square grid, and A, B, C, D, E, F are the intersections of the grid lines. What is the ratio of the area of $\triangle ABC$ to the area of $\triangle DEF$?

Solution:
\textbf{Solution:} As shown in the diagram, let the side length of the small squares in the square grid be 1,\\
then we have $AB=1$, $BC=\sqrt{1^2+2^2}=\sqrt{5}$, $AC=\sqrt{1^2+1^2}=\sqrt{2}$, $DE=2$, $EF=\sqrt{2^2+2^2}=2\sqrt{2}$, $DF=\sqrt{2^2+4^2}=2\sqrt{5}$,\\
$\therefore \frac{AB}{DE}=\frac{BC}{DF}=\frac{AC}{EF}=\frac{1}{2}$,\\
$\therefore \triangle ABC\sim \triangle DEF$,\\
$\therefore S_{\triangle ABC} : S_{\triangle DEF}=\left(\frac{1}{2}\right)^2=\boxed{\frac{1}{4}}$,


Question: As shown in the figure, the right triangle has its right-angled vertex at the coordinate origin, $\angle OAB=30^\circ$. If point A lies on the graph of the inverse proportional function $y=\frac{6}{x}$ $(x>0)$, what is the value of $k$ in the inverse proportional function $y=\frac{k}{x}$ passing through point B?

Solution:
\textbf{Solution:} Draw $BC\perp x$ axis at point $C$ through point $B$, and draw $AD\perp x$ axis at point $D$ through point $A$, as shown in the figure.\\
$\because \angle BOA=90^\circ$,\\
$\therefore \angle BOC+\angle AOD=90^\circ$,\\
$\because \angle AOD+\angle OAD=90^\circ$,\\
$\therefore \angle BOC=\angle OAD$,\\
Also, $\because \angle BCO=\angle ADO=90^\circ$,\\
$\therefore \triangle BCO\sim \triangle ODA$,\\
$\therefore \frac{OB}{OA}=\tan30^\circ=\frac{\sqrt{3}}{3}$,\\
$\therefore \frac{{S}_{\triangle BCO}}{{S}_{\triangle ODA}}=\frac{1}{3}$,\\
$\because \frac{1}{2}\times AD\times DO=\frac{1}{2}xy=3$,\\
$\therefore {S}_{\triangle BCO}=\frac{1}{2}\times BC\times CO=\frac{1}{3}{S}_{\triangle ODA}=1$,\\
$\because$ The graph of the inverse proportion function that passes through point $B$ is in the second quadrant,\\
therefore, the equation of the inverse proportion function is: $y=-\frac{2}{x}$,\\
$\therefore k=\boxed{-2}$.


Question: As shown in the figure, it is known that $AB \parallel CD \parallel EF$, and they intersect lines $l_1$ and $l_2$ at points A, C, E and B, D, F respectively. If $AC: CE = 2:3$ and $BD = 4$, then what is the value of $BF$?

Solution:
\[
\textbf{Solution:} \quad \text{Given,} \quad AB \parallel CD \parallel EF,\\
\text{it follows that} \quad AC:CE = BD:DF,\\
\text{Given} \quad AC:CE = 2:3, \quad \text{and} \quad BD=4,\\
\text{then} \quad 2:3=4:DF,\\
\text{thus,} \quad DF=6,\\
\text{therefore,} \quad BF=BD+DF=4+6=\boxed{10}.
\]


Question: As shown in the figure, the lines $l_1 \parallel l_2 \parallel l_3$, and the lines $a$ and $b$ intersect with these three lines at points $A$, $B$, $C$ and $E$, $F$, $F$, respectively. If $AB:BC = 5:3$, then what is the ratio of $EF:DF$?

Solution:
\textbf{Solution:} Since ${l}_{1}\parallel {l}_{2}\parallel {l}_{3}$\\
$\therefore \frac{AB}{BC}=\frac{DE}{EF}=\frac{5}{3}$\\
$\therefore \frac{EF}{DE}=\frac{3}{5}$\\
$\therefore \frac{EF}{DF}=\boxed{\frac{3}{8}}$


Question: As shown in the figure, in $\square ABCD$, $EF\parallel AB$, $DE:EA=2:3$, and $EF=4$, what is the length of $CD$?

Solution:
\textbf{Solution:} Given $EF \parallel AB$,\\
it follows that $\angle DEF = \angle A$, $\angle DFE = \angle DBA$,\\
which means $\triangle DEF \sim \triangle DAB$.\\
Given $DE : EA = 2 : 3$,\\
it implies $EF : AB = DE : DA = 2 : 5$,\\
and since $EF = 4$,\\
we find that $AB = 10$.\\
Therefore, in quadrilateral $ABCD$, $CD = AB = \boxed{10}$.


Question: As shown in the figure, $A$, $B$, $C$, and $D$ are four points in sequence on a straight line, $M$ and $N$ are the midpoints of $AB$ and $CD$ respectively, and $MN=6\,\text{cm}$, $BC=3\,\text{cm}$. What is the length of $AD$?

Solution:
\textbf{Solution:} Let $M$ and $N$ be the midpoints of $AB$ and $CD$ respectively.\\
Therefore, $AB = 2MB$ and $CD = 2CN$.\\
Since $MN = 6$ and $BC = 3$,\\
it follows that $MB + CN = MN - BC = 6 - 3 = 3$,\\
and therefore $AB + CD = 2MB + 2CN = 2(MB + CN) = 2 \times 3 = 6$,\\
hence $AD = AB + CD + BC = 6 + 3 = \boxed{9}$.


Question: As shown in the figure, $\triangle ABC$ is rotated clockwise around point C to obtain $\triangle EDC$. If point A is exactly on the extension of line ED and $\angle ABC=110^\circ$, then what is the degree measure of $\angle ADC$?

Solution:
Solution: It is known from the rotation that $\triangle ABC \cong \triangle EDC$,\\
$\therefore \angle EDC = \angle ABC = 110^\circ$,\\
$\because$ Points A, D, E are collinear,\\
$\therefore \angle EDC + \angle ADC = 180^\circ$,\\
$\therefore \angle ADC = 180^\circ - \angle EDC = 180^\circ - 110^\circ = \boxed{70^\circ}$.


Question: As shown in the figure, given that $\angle A=n^\circ$, if point $P_{1}$ is the intersection of the angle bisector of $\angle ABC$ and the external angle bisector of $\angle ACE$, point $P_{2}$ is the intersection of the angle bisector of $\angle P_{1}BC$ and the external angle bisector of $\angle P_{1}CE$, point $P_{3}$ is the intersection of the angle bisector of $\angle P_{2}BC$ and the external angle bisector of $\angle P_{2}CE$, and so on, then what is the degree measure of $\angle P_{2023}$?

Solution:
\textbf{Solution:} Given that $\angle ACE = \angle ABC + \angle A$, $\angle P_{1}CE = \angle P_{1}BC + \angle P_{1}$, and point $P_{1}$ is the intersection of the angle bisector of $\angle ABC$ and the exterior angle $\angle ACE$,\\
$\therefore \angle ACE = 2\angle P_{1}CE$, $\angle ABC = 2\angle P_{1}BC$,\\
$\therefore \angle ABC + \angle A = 2(\angle P_{1}BC + \angle P_{1})$,\\
$\therefore \angle P_{1} = \frac{1}{2}\angle A = \frac{1}{2}n^\circ$; By the same reasoning: $\angle P_{2} = \frac{1}{2}\angle P_{1} = \frac{1}{2} \times \frac{1}{2}n^\circ = \frac{n^\circ}{{2}^{2}}$, ..., $\angle P_{m} = \frac{n^\circ}{{2}^{m}}$;\\
$\therefore \angle P_{2023} = \frac{n^\circ}{{2}^{2023}}$.\\
Hence, the answer is: $\boxed{\frac{n^\circ}{{2}^{2023}}}$.


Question: As shown in the figure, it is known that straight lines AB and CD intersect at point O. OA bisects $\angle EOC$, and $\angle EOC = 70^\circ$. What is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} Since OA bisects $\angle EOC$ and $\angle EOC=70^\circ$,
\[
\therefore \angle AOC=\frac{1}{2}\angle EOC=\frac{1}{2}\times 70^\circ=35^\circ,
\]
\[
\therefore \angle BOD=\angle AOC=\boxed{35^\circ}
\]


Question: As shown in the figure, $O$ is a point on the straight line $AB$. An arbitrary ray $OM$ passes through $O$, $OC$ bisects $\angle AOM$, and $OD$ bisects $\angle BOM$. What is the measure of $\angle COD$?

Solution:
\textbf{Solution:} Since $OC$ bisects $\angle AOM$ and $OD$ bisects $\angle BOM$, \\
thus $\angle MOC= \frac{1}{2}\angle AOM$ and $\angle MOD= \frac{1}{2}\angle BOM$, \\
hence $\angle COD=\angle MOC+\angle MOD= \frac{1}{2}\angle AOM+\frac{1}{2}\angle BOM=\frac{1}{2}(\angle AOM+\angle BOM)=\frac{1}{2}\times 180^\circ=\boxed{90^\circ}$.


Question: The figure on the right is the unfolded shape of a certain solid. What is this solid?

Solution: \textbf{Solution:} According to the net of the solid, we can conclude that the solid is a \boxed{\text{pentagonal prism}}.


Question: As shown in the figure, $AB=9$, and $C$ is the midpoint of $AB$. Point $D$ is on the line segment $AC$, and $AD\colon CB=1\colon 3$. What is the length of $DC$?

Solution:
\textbf{Solution:} Since $C$ is the midpoint of $AB$ and $AB = 9$, \\
then $AC = BC = \frac{1}{2} AB = 4.5$, \\
and since the ratio $AD : CB = 1 : 3$, \\
it follows that $AD : AC = 1 : 3$, \\
hence $AD = \frac{1}{3} AC = 1.5$, \\
thus $DC = AC - AD = 4.5 - 1.5 = \boxed{3}$.


Question: As shown in the figure, the shapes of a certain geometric body viewed from three different directions are as shown. What is this geometric body?

Solution: \textbf{Solution:} From the top view, it can be concluded that the solid is a prism. Since both the front view and the left view are triangles, the solid is identified as a \boxed{triangular prism}.


Question: As shown in the figure, point O is on line DB. Given that $\angle 1=15^\circ$ and $\angle AOC=90^\circ$, what is the measure of $\angle 2$ in degrees?

Solution:
\textbf{Solution:} Given that $\angle 1=15^\circ$ and $\angle AOC=90^\circ$, \\
it follows that $\angle BOC=75^\circ$, \\
therefore, $\angle 2=180^\circ-75^\circ=\boxed{105^\circ}$.


Question: As shown in the figure, it is known that $AB=24$, $C$ is the midpoint of $AB$, point $D$ is on line segment $AC$, and $AD:CB=1:3$, then what is the length of $DB$?

Solution:
\textbf{Solution:} Since $C$ is the midpoint of $AB$ and $AB=24$,\\
$\therefore$AC=CB=12,\\
Since $AD\colon CB=1\colon 3$,\\
$\therefore$AD=$\frac{1}{3}$BC=4,\\
$\therefore$DB=AB−AD=24−4=\boxed{20}.


Question: As shown in the figure, $C$ is a point on the line segment $AB$, $M$ is the midpoint of $AC$, and $N$ is the midpoint of $BC$. If $MC$ is $2\,\text{cm}$ longer than $NC$, how much longer is $AC$ than $BC$?

Solution:
\textbf{Solution:} Given that $M$ is the midpoint of $AC$, and $N$ is the midpoint of $BC$,
\\
hence $MC=\frac{1}{2}AC$, $CN=\frac{1}{2}BC$,
\\
since $MC-NC=2cm$,
\\
it follows that $\frac{1}{2}AC-\frac{1}{2}BC=2cm$,
\\
therefore $AC-BC=\boxed{4cm}$.


Question: As shown in the figure, what will be the shape after the planar figure is pulled into a three-dimensional form?

Solution: \textbf{Solution:} From the development diagram, we know: the two bases are regular pentagons, $\therefore$ after the plane figure is extruded into a three-dimensional shape, it will form a \boxed{\text{pentagonal prism}}.


Question: As shown in the diagram, what is the solid represented by its net?

Solution:
\textbf{Solution:} According to the unfolded figure, there are a total of 5 faces, with two triangles serving as the base and three rectangles as the lateral faces,\\
$\therefore$ this solid is a \boxed{triangular prism}.


Question: As shown in the figure, lines AB and CD intersect at point E, with EF$\perp$AB at point E. If $\angle FEC - \angle AEC = 20^\circ$, what is the measure of $\angle AED$?

Solution:
\textbf{Solution:} Let $\angle AEC$ be $x$, then $\angle FEC=x+20^\circ$. \\
Since EF$\perp$AB, \\
it follows that $\angle AEF=90^\circ$, \\
which means $\angle AEC+\angle FEC= 90^\circ$, \\
hence $x+x+20^\circ=90^\circ$, \\
Solving this gives: $x=35^\circ$, \\
that is $\angle AEC=35^\circ$, \\
Therefore, $\angle AED=180^\circ−35^\circ=\boxed{145^\circ}$.


Question: As shown in the figure, lines \(m\), \(n\), and \(l\) intersect, and \(m \perp n\), \(\angle 1 = 48^\circ\). What is the degree measure of \(\angle 3\)?

Solution:
\textbf{Solution:} Since line m intersects with line n,\\
Therefore $\angle 1 = \angle 2 = 48^\circ$,\\
Since m $\perp$ n,\\
Therefore $\angle 2 + \angle 3 = 90^\circ$,\\
Therefore $\angle 3 = 90^\circ - 48^\circ = \boxed{42^\circ}$.


Question: As shown in the diagram, it is known that lines $a \parallel b \parallel c$, and lines $m$, $n$ intersect $a$, $b$, $c$ at points $A$, $C$, $E$, $B$, $D$, $F$, respectively. If $AC=8$, $CE=12$, and $BD=6$, what is the value of $DF$?

Solution:
\textbf{Solution:} Since line $a \parallel b \parallel c$,\\
thus, $\frac{AC}{CE}=\frac{BD}{DF}$,\\
thus, $\frac{8}{12}=\frac{6}{DF}$,\\
we get $DF=\boxed{9}$.


Question: As shown in the figure, it is known that lines $a\parallel b\parallel c$, and lines $m, n$ intersect with $a, b, c$ at points $A, C, E, B, D, F$, respectively. If $AC=8$, $CE=12$, and $BD=6$, what is the value of $DF$?

Solution:
\textbf{Solution:} Since $a\parallel b\parallel c$,\\
it follows that $\frac{AC}{CE}=\frac{BD}{DF}$,\\
Given $AC=8$, $CE=12$, and $BD=6$,\\
we have $\frac{8}{12}=\frac{6}{DF}$. Solving this yields: $DF=\boxed{9}$.


Question: As shown in the figure, it is known that point $E$ is a moving point on the diagonal $AC$ of rectangle $ABCD$, and the vertices $G$ and $H$ of square $EFGH$ are both on side $BC$. If $AB = 3$ and $BC = 4$, then what is the value of $\tan \angle CFE$?

Solution:
Solution: In the square EFGH, we have EF$\parallel$GH and EH$\perp$BC,\\
which means EF$\parallel$BC,\\
$\therefore \angle CFE=\angle BCF$,\\
In the rectangle ABCD, AB$\perp$BC,\\
$\therefore$EH$\parallel$AB,\\
$\triangle ABC\sim \triangle EHC$,\\
$\therefore \frac{EH}{AB}=\frac{CH}{BC}$, i.e., $\frac{AB}{BC}=\frac{EH}{CH}=\frac{3}{4}$,\\
Let $EH=3a$, $CH=4a$, then $FG=3a$, $GH=3a$,\\
$\therefore CG=7a$,\\
$\therefore \tan\angle CFE=\tan\angle BCF=\frac{FG}{CG}=\frac{3a}{7a}=\boxed{\frac{3}{7}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, point $E$ is a point on side $BC$, and $BE: EC=2:1$, $AE$ intersects $BD$ at point $F$. What is the ratio of the area of $\triangle BEF$ to the area of $\triangle ABD$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a parallelogram,\\
$\therefore$ $AD\parallel BC$ and $AD=BC$,\\
$\therefore$ $\angle ADB=\angle CBD$,\\
Given $\angle AFD=\angle BFE$,\\
$\therefore$ $\triangle AFD\sim \triangle EFB$,\\
Given $BE:EC=2:1$,\\
$\therefore$ $\frac{BE}{AD}=\frac{2}{3}$,\\
$\therefore$ $\frac{{h}_{\triangle BEF}}{{h}_{\triangle DAF}}=\frac{2}{3}$,\\
$\therefore$ $\frac{{h}_{\triangle BEF}}{{h}_{\triangle ABD}}=\frac{2}{5}$,\\
Let $BE=2x$, then $AD=3x$,\\
Let ${h}_{\triangle BEF}=2a$, then ${h}_{\triangle ABD}=5a$,\\
$\therefore$ $\frac{{S}_{\triangle BEF}}{{S}_{\triangle ABD}}=\frac{2x\cdot 2a}{3x\cdot 5a}=\boxed{\frac{4}{15}}$.


Question: As shown in the figure, $ \triangle ABO \sim \triangle CDO$, if $ BO=8$, $ DO=4$, and $ CD=3$, then what is the length of $ AB$?

Solution:
\textbf{Solution:} Since $\triangle ABO \sim \triangle CDO$, $BO=8$, $DO=4$, $CD=3$,
\[
\therefore \frac{BO}{DO}=\frac{AB}{CD}, \text{ which means } \frac{8}{4}=\frac{AB}{3},
\]
\[
\therefore AB=\boxed{6}.
\]


Question: As shown in the figure, D and E are the midpoints of sides AB and AC of $\triangle ABC$, respectively. CD and BE intersect at point O. What is the value of $\frac{S_{\triangle COE}}{S_{\triangle BOC}}$?

Solution:
\textbf{Solution:} Since $D$ and $E$ are the midpoints of sides $AB$ and $AC$ of $\triangle ABC$ respectively,\\
it follows that $DE \parallel BC$ and $DE = \frac{1}{2}BC$,\\
thereby $\triangle DOE \sim \triangle COB$,\\
which implies $\frac{DE}{BC}=\frac{OE}{OB}=\frac{1}{2}$,\\
resulting in ${S}_{\triangle COE}:{S}_{\triangle BOC}=\frac{OE}{OB}=\boxed{\frac{1}{2}}$;


Question: As shown in the figure, in parallelogram $ABCD$, point $M$ is the midpoint of $CD$, and $AM$ intersects $BD$ at point $N$. If it is known that the area of $\triangle DMN$ is 3, then what is the area of $\triangle ADN$?

Solution:
\textbf{Solution:} Since point $M$ is the midpoint of $CD$, \\
$\therefore$ DM=$\frac{1}{2}CD$, \\
In parallelogram $ABCD$ where AB$\parallel$CD, AB=CD, \\
$\therefore$ $\angle$ABN=$\angle$MDN, $\angle$BAN=$\angle$DMN, \\
$\therefore$ $\triangle ABN\sim \triangle MDN$, \\
$\therefore$ $\frac{AB}{MD}=\frac{AN}{MN}=\frac{CD}{DM}=2$, \\
$\therefore$ AN=2MN, \\
$\therefore$ $\frac{{S}_{\triangle ADN}}{{S}_{\triangle DMN}}=AN:MN=2:1$, \\
Since ${S}_{\triangle DMN}=3$, \\
$\therefore$ ${S}_{\triangle ADN}=2{S}_{\triangle DMN}=2\times 3=\boxed{6}$.


Question: As shown in the figure, in $\triangle ABC$, D and E are the midpoints of AB and AC, respectively. If $DE=4$, what is the length of $BC$?

Solution:
\textbf{Solution:} Since $D$ and $E$ are the midpoints of sides $AB$ and $AC$ respectively,\\
it follows that $DE$ is the midsegment of $\triangle ABC$.\\
Since $DE = 4$,\\
it follows that $BC = 2DE = 2 \times 4 = \boxed{8}$.


Question: As shown in the figure, in rectangle $ABCD$, $\angle BOC = 120^\circ$, $BD = 12$, where $P$ is a moving point on side $AD$. What is the minimum value of $OP$?

Solution:
\textbf{Solution:} As shown in the figure, if we draw OP$\perp$AD from point O, then the length of OP is the smallest at this time.\\
Since quadrilateral ABCD is a rectangle,\\
it follows that AC=BD,\\
Since $AO=\frac{1}{2}AC$ and $DO=\frac{1}{2}DB$,\\
it follows that AO=DO=$\frac{1}{2}DB=6$,\\
Since $\angle$AOD=$\angle$BOC=120$^\circ$,\\
it follows that $\angle$OAD=30$^\circ$,\\
Since $\angle$OPA=90$^\circ$,\\
it follows that OP=$\frac{1}{2}AO=\boxed{3}$


Question: As shown in the figure, $O$ is the midpoint of the diagonal $AC$ of rectangle $ABCD$, and $E$ is the midpoint of $AD$. If $AB=6$ and $BC=8$, then what is the perimeter of $\triangle BOE$?

Solution:
\textbf{Solution:} Since point $O$ is the midpoint of the diagonal $AC$ of rectangle $ABCD$, and point $E$ is the midpoint of $AD$, \\
$\therefore AB=CD=6$, $AD=BC=8$, $OE=\frac{1}{2}CD=3$, $AE=\frac{1}{2}AD=4$, \\
In $Rt\triangle ABE$, $BE=\sqrt{AB^{2}+AE^{2}}=\sqrt{6^{2}+4^{2}}=2\sqrt{13}$, \\
In $Rt\triangle ABC$, $AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{6^{2}+8^{2}}=10$, \\
$\therefore BO=\frac{1}{2}AC=5$, \\
Thus, the perimeter of $\triangle BOE$ is: $5+3+2\sqrt{13}=8+2\sqrt{13}$, \\
Therefore, the answer is: $\boxed{8+2\sqrt{13}}$.


Question: As shown in Figure 1, there is a right-angled triangle with one of its legs measuring 1 unit and the hypotenuse measuring 3 units. By arranging them as illustrated in Figure 2 to form a square, and ensuring that the pieces do not overlap or leave gaps at the joints, what is the area of the quadrilateral ABCD, which is the blank space in the middle of Figure 2?

Solution:
\textbf{Solution:} From Figure 2, we have $AB=BC=CD=DA$ and $\angle BAD=90^\circ$, \\
$\therefore$ quadrilateral ABCD is a square, \\
$\because$ one of the legs of the right-angled triangle paper is 1 and the hypotenuse is 3, \\
according to the Pythagorean theorem, we can find that the other leg of the right-angled triangle is $\sqrt{3^2-1^2}=2\sqrt{2}$, \\
$\therefore$ the area of square ABCD is $(2\sqrt{2}-1)^2=9-4\sqrt{2}$, \\
thus, the answer is: $\boxed{9-4\sqrt{2}}$.


Question: As shown in the diagram, in the Cartesian coordinate system, if the vertices $A$ and $B$ of the rhombus $ABCD$ are at $(-3, 0)$ and $(2, 0)$ respectively, and point $D$ is on the y-axis, what are the coordinates of point $C$?

Solution:
Solution: Since the coordinates of points A and B are respectively $(-3,0)$ and $(2,0)$, \\
it follows that $AB=5$ \\
Since quadrilateral $ABCD$ is a rhombus, \\
it follows that $AB=AD=BC=5$, and $DC\parallel AB$ \\
Since $AO=3$, \\
it follows that $DO=\sqrt{AB^{2}-AO^{2}}=\sqrt{5^{2}-3^{2}}=4$ \\
Therefore, $C(5,4)$ \\
Hence, the coordinates of point C are $\boxed{(5,4)}$


Question: As shown in the figure, in rectangle $ABCD$, $AB=8$. The rectangle is folded along the line $AC$, with point $B$ falling onto point $E$, and $AE$ intersects $CD$ at point $F$. If $AF=\frac{25}{4}$, what is the length of $AD$?

Solution:
\textbf{Solution:} Given $DC\parallel AB$,\\
$\therefore \angle FCA=\angle CAB$, also $\angle FAC=\angle CAB$,\\
$\therefore \angle FAC=\angle FCA$,\\
$\therefore FA=FC=\frac{25}{4}$,\\
$\therefore FD=FE$,\\
Given $DC=AB=8$, $AF=\frac{25}{4}$,\\
$\therefore FD=FE=8-\frac{25}{4}=\frac{7}{4}$,\\
$\therefore AD=BC=EC=\sqrt{FC^{2}-FE^{2}}=\boxed{6}$.


Question: As shown in the figure, in the Cartesian coordinate system, the quadrilateral $ABCD$ is a square. The coordinate of point $A$ is $(0,2)$, and the coordinate of point $B$ is $(4,0)$. What is the coordinate of point $C$?

Solution:
\textbf{Solution:} As illustrated, draw CE$\perp$x-axis at point C, with E being the foot of the perpendicular.\\
Since quadrilateral ABCD is a square, and points A(0, 2), B(4, 0),\\
then AB=BC, $\angle$ABC=90$^\circ$, AO=2, OB=4,\\
thus $\angle$AOB=$\angle$BEC= 90$^\circ$,  $\angle$ABO=$\angle$BCE=90$^\circ$−$\angle$CBE,\\
therefore $\triangle$AOB$\cong$$\triangle$BEC,\\
thus BE=AO=2, EC=OB=4,\\
therefore OE=OB+BE=2+4=6,\\
hence, the coordinates of point C are $\boxed{(6, 4)}$.


Question: As shown in the diagram, the diagonals $AC$ and $BD$ of the rhombus $ABCD$ intersect at point $O$, with $BD = 6$ and $AC = 8$. If $E$ is the midpoint of $CD$, then what is the length of $OE$?

Solution:
Solution: Since quadrilateral ABCD is a rhombus,\\
it follows that OC=OA=$\frac{1}{2}AC=4$, OB=OD=$\frac{1}{2}BD=3$, and $\angle$DOC=$90^\circ$,\\
thus CD=$\sqrt{OC^2+OA^2}=5$,\\
given that E is the midpoint of AB, and O is the midpoint of AC,\\
thus OE is the median to the hypotenuse of $\triangle COD$,\\
therefore OE=$\frac{1}{2}CD=\boxed{2.5}$.


Question: In rectangle $ABCD$, where $AB=6, BC=8$, the diagonals intersect at point $O$. What is the length of $BO$?

Solution:
\textbf{Solution:}
Since quadrilateral ABCD is a rectangle, and $AB=6$, $BC=8$, \\
it follows that $\angle ABC=90^\circ$, \\
therefore $BD=AC=\sqrt{AB^{2}+BC^{2}}=10$, \\
since $OB=\frac{1}{2}BD$, \\
it follows that $OB=\boxed{5}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=10$, $AC=7$, $BC=9$, and points $D$, $E$, and $F$ are the midpoints of $AB$, $AC$, and $BC$, respectively. What is the perimeter of the quadrilateral $DBFE$?

Solution:
\textbf{Solution:} Since points $D$, $E$, and $F$ are the midpoints of $AB$, $AC$, and $BC$, respectively,\\
$\therefore DE$ is a midline of $\triangle ABC$, $EF$ is also a midline of $\triangle ABC$,\\
$\therefore DE=BF=\frac{1}{2}BC=\frac{1}{2}\times 9=\frac{9}{2}$, $EF=BD=\frac{1}{2}AB=\frac{1}{2}\times 10=5$,\\
$\therefore$ the perimeter of the quadrilateral $DBFE$ is $DE+BF+EF+BD=9+10=\boxed{19}$.


Question: As shown in the figure, three squares with the same side length overlap each other. $O_1$ and $O_2$ are the centers of two of the squares. If the sum of the areas of the shaded parts is 4, what is the side length of the squares?

Solution:
\textbf{Solution:} Connect $O_1B$ and $O_1C$, as shown in the figure: \\
$\because \angle BO_1F + \angle FO_1C = 90^\circ$ and $\angle FO_1C + \angle CO_1G = 90^\circ$, \\
$\therefore \angle BO_1F = \angle CO_1G$, \\
$\because$ quadrilateral $ABCD$ is a square, \\
$\therefore \angle O_1BF = \angle O_1CG = 45^\circ$, \\
In $\triangle O_1BF$ and $\triangle O_1CG$, \\
$\left\{\begin{array}{l}
\angle FO_1B = \angle CO_1G \\
BO_1 = CO_1 \\
\angle FBO_1 = \angle GCO_1
\end{array}\right.$, \\
$\therefore \triangle O_1BF \cong \triangle O_1CG$ (by ASA), \\
$\therefore S_{\triangle O_1BF} = S_{\triangle O_1CG}$, \\
$\therefore$ The area of the shaded parts in squares $O_1$ and $O_2$ is $\frac{1}{4}S_{\text{square}} ABCD$, \\
Similarly, the area of the shaded parts in the other two squares is also $\frac{1}{4}S_{\text{square}} ABCD$, \\
$\therefore$ The total area of the shaded parts $= 4 = \frac{1}{2}S_{\text{square}} ABCD$, \\
$\therefore S_{\text{square}} ABCD = 8 = AD^2$, \\
$\therefore AD = \boxed{2\sqrt{2}}$.


Question: As shown in the diagram, in rhombus $ABCD$ with a perimeter of 24, $\angle BAD=120^\circ$, points E and F are moving points on sides $AB, AD$ respectively, and point P is a moving point on diagonal $BD$. What is the minimum value of the segment $PE+PF$?

Solution:
\textbf{Solution:} As shown in the figure, connect $AC$, intersecting $BD$ at point $O$. Draw a perpendicular line from $AE'$ to $BC$ at foot $E'$, intersecting $BD$ at point $P$.\\
When point $F$ coincides with point $A$, draw a perpendicular line from $E'$ to $BD$, extending it to intersect $AB$ at point $E$. The sum of $PE+PF$ achieves its minimum value, which is $AE'$.\\
Since the perimeter of the rhombus $ABCD$ is 24 and $\angle BAD=120^\circ$\\
Therefore, $AB=BC=6$ and $\angle BAC=\frac{1}{2}\angle BAD=60^\circ$,\\
Thus, $\triangle ABC$ is an equilateral triangle,\\
Therefore, $AB=AC$,\\
Since $AE'\perp BC$ at foot $E'$,\\
Therefore, $BE'=CE'=\frac{1}{2}BC=3$,\\
In $\triangle ABC$, by Pythagoras' theorem,\\
$AE'=\sqrt{AB^2-BE'^2}=\sqrt{6^2-3^2}=3\sqrt{3}$.\\
Hence, the minimum value of $PE+PF$ is $\boxed{3\sqrt{3}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, point $P$ is the midpoint of diagonal $BD$, points $E$ and $F$ are the midpoints of $AB$ and $CD$, respectively, $AD=BC$, $\angle CBD=30^\circ$, and $\angle ADB=100^\circ$, then what is the measure of $\angle PFE$?

Solution:
\textbf{Solution:} Since point P is the midpoint of BD, and point E is the midpoint of AB,\\
it follows that PE is the midline of $\triangle ABD$,\\
therefore PE=$\frac{1}{2}$AD, PE$\parallel$AD,\\
thus $\angle$EPD=$180^\circ-\angle$ADB=$80^\circ$,\\
Similarly, PF=$\frac{1}{2}$BC, PF$\parallel$BC,\\
thus $\angle$FPD=$\angle$CBD=$30^\circ$,\\
since AD=BC,\\
it follows that PE=PF,\\
thus $\angle$PFE=$\frac{1}{2}\times(180^\circ-110^\circ)=\boxed{35^\circ}$.


Question: As shown in the figure, the large square $ABCD$ is composed of four congruent right-angled triangles and one small square. If $\angle ADE = \angle AED$ and $AD = 4\sqrt{5}$, what is the area of $\triangle ADE$?

Solution:
\textbf{Solution:} As shown in the figure:\\
$\because \angle ADE = \angle AED$,\\
$\therefore AD = AE = AB$,\\
$\therefore \angle AEF = \angle ABF$,\\
$\because AF \perp BE$,\\
$\therefore EF = BF = \frac{1}{2}BE$,\\
$\therefore GE = AH$,\\
$\because \angle GEM = \angle HAM, \angle MGE = \angle MHA$,\\
$\therefore \triangle GEM \cong \triangle HAM$ (ASA),\\
$\therefore S_{\triangle HAM} = S_{\triangle GEM}$,\\
$\therefore S_{\triangle ADE} = S_{\triangle ADH} + S_{\triangle DGE}$,\\
$\because AD = 4\sqrt{5}$, DH = 2AH, $AD^2 = DH^2 + AH^2$,\\
$\therefore AH = 4, DH = 8$,\\
$\therefore DG = GE = 4$,\\
$\therefore S_{\triangle ADE} = \frac{1}{2} \times 4 \times 8 + \frac{1}{2} \times 4 \times 4 = \boxed{24}$.


Question: As shown in the figure, the distance from point $M$, the midpoint of one side of rhombus $ABCD$, to point $O$, the intersection of the diagonals, is $5\text{cm}$. What is the perimeter of rhombus $ABCD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\
it follows that $\angle$AOB=$90^\circ$,\\
since M is the midpoint of side AB,\\
it follows that AB=2OM=10,\\
thus, the perimeter of the rhombus ABCD is $10\times 4=\boxed{40}$.


Question: As shown in the figure, in $\triangle ABC$, $AM$ is the bisector of $\angle CAB$, and $CN$ is the bisector of the exterior angle $\angle GCB$. $BE \perp AM$ at point E, and $BD \perp CN$ at point D, with $DE$ connected. If $AB=4$, $BC=5$, and $AC=6$, what is the length of $DE$?

Solution:
\textbf{Solution:} As shown in the figure, \\
extend $BE$ to meet $AC$ at point $F$, and extend $AC$ and $BD$ to meet at point $G$,\\
since $AM$ bisects $\angle CAB$ and $BE\perp AM$,\\
it follows that $\angle CAM=\angle BAM$, $\angle AEF=\angle AEB=90^\circ$,\\
thus $\angle AFE=90^\circ-\angle CAM$, $\angle ABE=90^\circ-\angle BAM$,\\
hence $\angle AFE=\angle ABE$,\\
which implies $AF=AB$,\\
given that $AB=4$, $AC=6$, and $BC=5$,\\
it follows that $FC=AC-AF=AC-AB=6-4=2$,\\
since $CN$ bisects $\angle GCB$ and $BD\perp CN$,\\
it follows that $\angle GCN=\angle BCN$, $\angle CDB=\angle CDG=90^\circ$,\\
thus $\angle CGD=90^\circ-\angle GCN$, $\angle CBD=90^\circ-\angle BCN$,\\
hence $\angle CGD=\angle CBD$,\\
which implies $CG=CB=5$,\\
therefore $FG=FC+CG=2+5=7$,\\
since $AF=AB$ and $BE\perp AM$,\\
it follows that $AE$ is the median to side $BF$, that is, point $E$ is the midpoint of $BF$,\\
since $CG=CB$ and $BD\perp CN$,\\
it follows that $CD$ is the median to side $BG$, that is, point $D$ is the midpoint of $BG$,\\
thus $DE$ is the midsegment of $\triangle BFG$,\\
therefore $DE=\frac{1}{2}FG=\boxed{\frac{7}{2}}$.


Question: As shown in the figure, in diamond ABCD, M and N are on AB and CD, respectively, and $AM = CN$, MN intersects AC at point O, connect $BO$, if $\angle DAC = 31^\circ$, what is the degree measure of $\angle OBC$?

Solution:
\textbf{Solution}: Since quadrilateral ABCD is a rhombus,\\
it follows that $AB\parallel CD$ and $AB=BC$,\\
therefore $\angle MAO=\angle NCO$ and $\angle AMO=\angle CNO$,\\
in $\triangle AMO$ and $\triangle CNO$,\\
since $\left\{\begin{array}{l}\angle MAO=\angle NCO\hfill \\ AM=CN\hfill \\ \angle AMO=\angle CNO\hfill \end{array}\right.$,\\
it follows that $\triangle AMO \cong \triangle CNO\left(ASA\right)$,\\
therefore $AO=CO$,\\
since $AB=BC$,\\
it follows that $BO\perp AC$,\\
therefore $\angle BOC=90^\circ$,\\
since $\angle DAC=31^\circ$,\\
it follows that $\angle BCA=\angle DAC=31^\circ$,\\
therefore $\angle OBC=90^\circ-31^\circ=\boxed{59^\circ}$.


Question: As shown in the figure, it is known that the diagonals \(AC\) and \(BD\) of rhombus \(ABCD\) intersect at point \(O\). \(E\) is the midpoint of \(CD\). If \(OE = 6\), what is the perimeter of the rhombus?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\
$\therefore$ AC $\perp$ BD, AB = BC = CD = DA, \\
$\therefore$ $\triangle$COD is a right-angled triangle, \\
Since OE = 6, and point E is the midpoint of the segment CD, \\
$\therefore$ CD = 2OE = 12, \\
$\therefore$ the perimeter of the rhombus ABCD = 4CD = $4\times12 = \boxed{48}$


Question: As shown in the figure, AC is the diagonal of rectangle ABCD. Fold side AB along AE so that point B falls on point M on AC, and fold side CD along CF so that point D falls on point N on AC. It is easy to prove that quadrilateral AECF is a parallelogram. When $\angle BAE$ is how many degrees, the quadrilateral AECF becomes a rhombus.

Solution:
\textbf{Solution}: When $\angle BAE = 30^\circ$, the quadrilateral AECF is a rhombus,\\
Reason: It is known from folding that $\angle BAE = \angle CAE = 30^\circ$,\\
$\because \angle B = 90^\circ$,\\
$\therefore \angle ACE = 90^\circ - 30^\circ - 30^\circ = 30^\circ$,\\
which means $\angle CAE = \angle ACE$,\\
$\therefore EA = EC$,\\
$\because$ the quadrilateral AECF is a parallelogram,\\
$\therefore$ the quadrilateral AECF is a rhombus,\\
Therefore, the answer is: $\boxed{30^\circ}$.


Question: As shown in the figure, in parallelogram $ABCD$, $O$ is the midpoint of the diagonal $AC$, $AC\perp AB$, point $E$ is the midpoint of $AD$, and $OF\perp BC$, $\angle D=53^\circ$, then what is the degree measure of $\angle FOE$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\
$\therefore$ AB $\parallel$ CD, AD $\parallel$ BC, \\
Since AC $\perp$ AB, \\
$\therefore$ $\angle$BAC = $\angle$DCA = 90$^\circ$, \\
Since point O is the midpoint of AC, and point E is the midpoint of AD, \\
$\therefore$ OE $\parallel$ CD, \\
$\therefore$ $\angle$COE + $\angle$ACD = 180$^\circ$, \\
$\therefore$ $\angle$COE = 90$^\circ$ \\
Since $\angle$D = $\angle$B = 53$^\circ$, and OF $\perp$ BC, \\
$\therefore$ $\angle$FOC = $\angle$B = 53$^\circ$, \\
$\therefore$ $\angle$EOF = $\angle$EOC + $\angle$FOC = 90$^\circ$ + 53$^\circ$ = \boxed{143^\circ}, \\


Question: As shown in the figure, the side length of square $ABCD$ is 8, and point $E$ is on $AB$ with $BE=2$. Point $F$ is a moving point on the diagonal $AC$. What is the minimum perimeter of $\triangle BFE$?

Solution:
\textbf{Solution:} As shown in the figure, connect $DB$, $DE$, $DF$, \\
since the quadrilateral $ABCD$ is a square with side length equal to 8, \\
therefore $AB=AD=8, \angle BAD=90^\circ$, and $AC$ is the perpendicular bisector of $BD$, \\
thus $BF=DF$, \\
since $BE=2$, \\
therefore $AE=AB-BE=6$, the perimeter of $\triangle BFE$ is $BE+BF+EF=2+DF+EF$, \\
according to the principle that the length of the segment between two points is shortest, when points $D$, $F$, $E$ are collinear, the sum of $DF+EF$ is minimal, with the minimum being the length of $DE$, \\
in $\triangle ADE$, $DE=\sqrt{AD^2+AE^2}=10$, \\
thus, the minimum value of the perimeter of $\triangle BFE$ is $2+DE=2+10=\boxed{12}$.


Question: As shown in the figure, in rhombus ABCD, the diagonals AC and BD intersect at point O, BE $\perp$ AD at point E, and $OA = 4$, $OB = 3$. What is the length of BE?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\
$\therefore$ $AC\perp BD$, AC=2OA, BD=2OB, \\
Since $OA=4$, $OB=3$, \\
$\therefore$ AD=$AB=\sqrt{OA^2+OB^2}=5$, AC=8, BD=6, \\
$\therefore$ $S_{\text{rhombus} ABCD}=\frac{1}{2}AC\cdot BD=AD\cdot BE$, \\
Hence $24=5BE$, \\
$\therefore$ $BE=\boxed{4.8}$


Question: As shown in the figure, the side length of rhombus ABCD is 6, and $\angle D = 120^\circ$. Point E is the midpoint of AB, and point P is a moving point on diagonal AC. What is the minimum value of $BP+EP$?

Solution:
\textbf{Solution:} Connect $DP$, $BD$, \\
since quadrilateral $ABCD$ is a rhombus, \\
thus point $B$ and point $D$ are symmetric with respect to $AC$, \\
thus $DP=PB$, \\
thus $PB+PE=DP+PE \geq DE$, \\
when $DE \perp AB$, the value of $PB+PE$ is minimal, \\
since $\angle D=120^\circ$, \\
thus $\angle DAB=60^\circ$, \\
since $AD=AB$, \\
thus $\triangle ABD$ is an equilateral triangle, \\
since $E$ is the midpoint of $AB$, \\
thus $DE \perp AB$, \\
since $AB=6$, \\
thus $DE=3\sqrt{3}$, \\
thus the minimum value of $PB+PE$ is $\boxed{3\sqrt{3}}$.


Question: In rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $E$ is the midpoint of $AB$, $AC=8\,\text{cm}$, and $BD=6\,\text{cm}$. What is the length of $OE$?

Solution:
\textbf{Solution:} Given that in the rhombus ABCD, AC = 6cm, BD = 8cm,\\
hence AC $\perp$ BD, OB = $\frac{1}{2}$BD = 4cm, OC = $\frac{1}{2}$AC = 3cm,\\
therefore BC = $\sqrt{OC^{2}+OB^{2}}$ = 5cm,\\
since E is the midpoint of BC,\\
thus OE = $\frac{1}{2}$BC = $\boxed{2.5}$cm.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, point $D$ is the midpoint of the hypotenuse $AB$, and $DE$ bisects $\angle ADC$, $BC=4$. What is the length of $DE$?

Solution:
Solution: Since $\triangle ABC$ is a right-angled triangle and point D is the midpoint of the hypotenuse AB, \\
$\therefore CD=AD$, meaning $\triangle ACD$ is an isosceles triangle, \\
Since $DE$ bisects $\angle ADC$ and $\triangle ACD$ is an isosceles triangle, \\
$\therefore$ point E is the midpoint of AC, \\
Since $BC=4$, \\
$\therefore DE=\frac{1}{2}BC=\boxed{2}$.


Question: As shown in the figure, a point $E$ is taken outside the square $ABCD$, and lines $AE$, $BE$, $DE$ are drawn. A perpendicular line is drawn from point $A$ to line $AE$, intersecting $DE$ at point $P$. If $AE=AP=1$ and $PB= \sqrt{10}$, what is the distance from point $B$ to line $AE$?

Solution:
\textbf{Solution:} Given that $\angle EAP=\angle BAD=90^\circ$, \\
it follows that $\angle EAB+\angle BAP=90^\circ$, and $\angle PAD+\angle BAP=90^\circ$, \\
thus, $\angle EAB=\angle PAD$, \\
moreover, since AE=AP and AB=AD, \\
in triangles $\triangle APD$ and $\triangle AEB$, we have \\
$\begin{cases}
AE=AP \\
\angle EAB=\angle PAD \\
AB=AD
\end{cases}$, \\
therefore, $\triangle APD \cong \triangle AEB$ (SAS), \\
extend a line from B to F such that BF$\perp$AE and BF intersects the extended line of AE at F, \\
given AE=AP and $\angle EAP=90^\circ$, \\
it implies that $\angle AEP=\angle APE=45^\circ$, \\
thus, $\angle AEB=\angle APD=180^\circ-45^\circ=135^\circ$, \\
therefore, $\angle BEP=135^\circ-45^\circ=90^\circ$, \\
which implies EB$\perp$ED, \\
since BF$\perp$AF, \\
it follows that $\angle FEB=\angle FBE=45^\circ$, \\
given PE=$\sqrt{1^2+1^2}=\sqrt{2}$, \\
thus, BE=$\sqrt{BP^2-PE^2}=2\sqrt{2}$, \\
hence, BF=EF=$\frac{2\sqrt{2}}{\sqrt{2}}=\boxed{2}$, \\
therefore, the distance from point B to line AE is $\boxed{2}$.


Question: As shown in the figure, diagonals AC and BD of rhombus ABCD intersect at point O. A perpendicular line is drawn from point C to the extended line of AB, intersecting it at point E. Line OE is then drawn. If AB = $2\sqrt{5}$ and BD = 4, what is the length of OE?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a rhombus,\\
it follows that $OA=OC$, $BD\perp AC$,\\
Since $CE\perp AB$,\\
it follows that $OE=OA=OC$,\\
Since $BD=4$,\\
it follows that $OB=\frac{1}{2}BD=2$,\\
In $\triangle AOB$, $AB=2\sqrt{5}$, $OB=2$,\\
it follows that $OA=\sqrt{AB^{2}-OB^{2}}=4$,\\
therefore, $OE=OA=\boxed{4}$.


Question: As shown in the figure, extend the side $BA$ of the square $ABCD$ to point $E$ such that $AE=BD$, then what is the measure of $\angle E$?

Solution:
\textbf{Solution:} Draw line AC, \\
$\because$ Quadrilateral ABCD is a square, \\
$\therefore$ AC=BD, and $\angle$CAB=$45^\circ$, \\
Furthermore, $\because$BD=AE, \\
$\therefore$ AE=CA, \\
$\therefore$ $\angle$E=$\angle$ACE, \\
$\because$ $\angle$CAB=$\angle$ACE+$\angle$E=2$\angle$E=$45^\circ$, \\
$\therefore$ $\angle$E=$\boxed{22.5^\circ}$.


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

Solution:
\textbf{Solution}: Let an equilateral triangle $\triangle ADE$ be constructed on the exterior of square ABCD.\\
Therefore, $AB=AD$, $AD=AE$, $\angle BAD=90^\circ$, $\angle DAE=60^\circ$,\\
Thus, $AB=AE$, $\angle BAE=150^\circ$,\\
Hence, $\angle ABE=\angle AEB=\boxed{15^\circ}$.


Question: As shown in the figure, quadrilateral ABCD is a square, with diagonals AC and BD intersecting at point O. Point E is any point on AD other than the endpoints. A line is drawn from point O such that $OF \perp OE$ and intersects CD at point F. If $AB=6$, what is the area of quadrilateral EOFD?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a square, \\
it follows that OA=OB=OC=OD, AC$\perp$BD, $\angle$ODA=$\angle$OCD=45$^\circ$,\\
Since OE$\perp$OF,\\
it follows that $\angle$EOF=$\angle$COD=90$^\circ$,\\
therefore $\angle$EOD=$\angle$FOC,\\
thus $\triangle$EOD$\cong$$\triangle$FOC,\\
therefore the area of $\triangle$EOD = the area of $\triangle$FOC,\\
Since OA=OB and $\angle$AOB=90$^\circ$,\\
it follows that 2OA$^{2}$=AB$^{2}$=6$^{2}$,\\
Solving for OA$^{2}$ gives 18,\\
Since the area of quadrilateral EOFD = area of $\triangle$EOD + area of $\triangle$FOD = area of $\triangle$FOC + area of $\triangle$FOD = area of $\triangle$COD,\\
Therefore, the area of quadrilateral EOFD = $\frac{1}{2}OD^{2}=\frac{1}{2}OA^{2}=\boxed{9}$.


Question: As shown in the figure, the perimeter of $\triangle ABC$ is 18. D and E are the midpoints of sides AB and BC, respectively. What is the perimeter of $\triangle BDE$?

Solution:
\textbf{Solution:} Since the perimeter of $\triangle ABC$ is 18,\\
we have AB+AC+BC=18,\\
Since points D and E are the midpoints of sides AB and BC respectively,\\
then DE is the midsegment of $\triangle ABC$, BD=$\frac{1}{2}$AB, BE=$\frac{1}{2}$BC,\\
thus, DE=$\frac{1}{2}$AC,\\
therefore, the perimeter of $\triangle DBE$=BD+BE+DE=$\frac{1}{2}$(AB+AC+BC)=$\boxed{9}$,


Question: As shown in the diagram, rectangle ABCD is folded along the straight line DE, with vertex A landing on side BC at point F. Given that $BE=3$ and $CD=8$, what is the length of BF?

Solution:
\textbf{Solution:} From the property of the fold, we know that $AE=EF$, $AE=AB-BE=8-3=5$, \\
in $\triangle BEF$, $EF=AE=5$, $BE=3$, \\
by the Pythagorean theorem, we have $BF=\sqrt{EF^2-BE^2}=\sqrt{5^2-3^2}=\boxed{4}$.


Question: As shown in the figure, the diagonals of rhombus ABCD intersect at point O, E is the midpoint of CD, and OE = 3. What is the perimeter of the rhombus?

Solution:
\textbf{Solution:} Since the diagonals of the rhombus ABCD intersect at point O,\\
it follows that OB=OD,\\
given that point E is the midpoint of CD,\\
it then follows that DE=CE,\\
thus, OE is the median of $\triangle BCD$,\\
hence, BC=2OE=6,\\
therefore, the perimeter of the rhombus is: $4BC=4\times 6=\boxed{24}$.


Question: As shown in the diagram, in quadrilateral \(ABCD\), \(\angle ABC=90^\circ\), points \(E\) and \(F\) are the midpoints of \(AC\) and \(AD\) respectively, and \(BE=EF\). If \(AB=8\) and \(BC=4\), what is the length of \(CD\)?

Solution:
\textbf{Solution:} Let $\angle ABC=90^\circ$ and $AB=8$, $BC=4$,\\
therefore $AC=\sqrt{AB^2+BC^2}=\sqrt{8^2+4^2}=4\sqrt{5}$.\\
Since point E is the midpoint of AC,\\
therefore $BE=\frac{1}{2}AC=2\sqrt{5}$,\\
and since $BE=EF$,\\
thus $EF=2\sqrt{5}$.\\
Since points E and F are respectively the midpoints of AC and AD,\\
therefore $EF=\frac{1}{2}DC$.\\
Thus $DC=2EF=4\sqrt{5}$.\\
Hence, the length of $DC$ is $\boxed{4\sqrt{5}}$.


Question: As shown in the figure, within the rectangular paper ABCD, where E is a point on AD, fold $ \triangle CDE$ along CE to overlap with $ \triangle CFE$. If point F exactly falls on AB, and $ AF=4$, $ BC=10$, then what is the length of $ DE$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\
it follows that AD = BC = 10,\\
let AE = x, then DE = AD - AE = 10 - x,\\
since $\triangle CDE$ is folded along CE to $\triangle CFE$,\\
it follows that EF = DE = 10 - x,\\
in $\triangle AEF$, $AF^{2}$ + $AE^{2}$ = $EF^{2}$,\\
thus $4^{2}$ + $x^{2}$ = $(10 - x)^{2}$,\\
solving for x, we get x = 4.2,\\
thus AE = 4.2,\\
hence DE = AD - AE = \boxed{5.8},


Question: As shown in the figure, in rectangle ABCD, $AB=\sqrt{3}$, $\angle BAC=60^\circ$, point E is on the extension of AB, and $BE=AC$. If DE is connected, what is the length of DE?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\
it follows that AD=BC, $\angle$DAB=$\angle$ABC=$90^\circ$,\\
since $\angle BAC=60^\circ$,\\
it follows that $\angle$ACB=$30^\circ$,\\
since AB=$\sqrt{3}$,\\
it follows that AC=2AB=$2\sqrt{3}$,\\
in $\triangle ABC$, $BC=\sqrt{AC^2-AB^2}=3$,\\
thus AD=BC=3,\\
since AC=BE=$2\sqrt{3}$ and AB=$\sqrt{3}$,\\
thus AE=AB+BE=$3\sqrt{3}$,\\
in $\triangle DAE$, DE=$\sqrt{AD^2+AE^2}=\sqrt{3^2+(3\sqrt{3})^2}=\boxed{6}$


Question: As shown in the diagram, in parallelogram $ABCD$, $AB = 3.5\,cm$, $BC = 5\,cm$, $AE$ bisects $\angle BAD$, $CF\parallel AE$, what is the length of $AF$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\
$\therefore$ AD $\parallel$ BC, \\
$\therefore$ $\angle$DAE = $\angle$AEB, \\
Since AE $\parallel$ CF, \\
$\therefore$ quadrilateral AECF is a parallelogram, \\
$\therefore$ AF = CE, \\
Since AE bisects $\angle$BAD, \\
$\therefore$ $\angle$BAE = $\angle$EAD, \\
$\therefore$ $\angle$BAE = $\angle$AEB, \\
$\therefore$ AB = BE = 3.5\text{cm}, \\
$\therefore$ EC = BC - BE = 5 - 3.5 = 1.5\text{cm}, \\
$\therefore$ AF = \boxed{1.5}\text{cm}


Question: As shown in the figure, point $O$ is the midpoint of the diagonal $AC$ of the rectangle $ABCD$, and point $E$ is the midpoint of $AD$. If $AB=6, \angle ACB=30^\circ$, then what is the perimeter of $\triangle BOE$?

Solution:
\textbf{Solution:} Given in rectangle $ABCD$, $AB=6$,

thus $CD=AB=6$, $AD=BC$, $\angle ABC=\angle BAD=90^\circ$,

since $\angle ACB=30^\circ$,

thus $AC=2AB=12$, $AD=BC=\sqrt{AC^{2}-AB^{2}}=6\sqrt{3}$,

since point $O$ is the midpoint of the diagonal $AC$ of rectangle $ABCD$,

thus $OB=\frac{1}{2}AC=6$,

since point $E$ is the midpoint of $AD$,

thus $OE=\frac{1}{2}CD=3$, $AE=\frac{1}{2}AD=3\sqrt{3}$,

thus $BE=\sqrt{AB^{2}+AE^{2}}=3\sqrt{7}$,

then the perimeter of $\triangle BOE$ is $OB+OE+BE=6+3+3\sqrt{7}=\boxed{9+3\sqrt{7}}$.


Question: As shown in the figure, the diagonals $AC$ and $BD$ of rhombus $ABCD$ intersect at point $O$, and $E$ is the midpoint of $DC$. If the perimeter of the rhombus is 16, what is the length of $OE$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\
it follows that AB=BC=CD=AD, and BO=DO.\\
Since the perimeter of rhombus ABCD is 16, \\
it follows that BC=4.\\
Since point E is the midpoint of CD, \\
it follows that OE is the midsegment of $\triangle BCD$, \\
hence OE=$\frac{1}{2}$BC= \boxed{2}.


Question: As shown, \(O\) is the intersection point of the diagonals of rhombus \(ABCD\), \(DE \parallel AC\), \(CE \parallel BD\). If \(AC = 6\), \(BD = 8\), then what is the length of line segment \(OE\)?

Solution:
\textbf{Solution:} Given $\because$DE$\parallel$AC, CE$\parallel$BD,\\
$\therefore$ quadrilateral OCED is a parallelogram,\\
$\because$ quadrilateral ABCD is a rhombus,\\
$\therefore$ $\angle$COD$=90^\circ$,\\
$\therefore$ quadrilateral OCED is a rectangle,\\
Further $\because$AC$=6$, BD$=8$,\\
$\therefore$OC$=3$, OD$=4$,\\
$\therefore$CD$=\sqrt{OC^{2}+OD^{2}}=\sqrt{3^{2}+4^{2}}=\boxed{5}$,\\
In rectangle OCED, OE$=CD=5$.


Question: As shown in the figure, in $\triangle ABCD$, point E is the midpoint of side BC, and the diagonals AC and BD intersect at point O. The perimeter of $\triangle ABC$ is 16. If we connect OE, what is the perimeter of $\triangle OEC$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a parallelogram, \\
thus $OA=OC=\frac{1}{2}AC$, \\
since point $E$ is the midpoint of side $BC$, \\
thus $BE=CE=\frac{1}{2}BC$, and $OE$ is the median of $\triangle ABC$, \\
thus $OE=\frac{1}{2}AB$, \\
since the perimeter of $\triangle ABC$ is 16, \\
thus $AB+BC+AC=16$, \\
thus the perimeter of $\triangle OEC$ is $OE+CE+OC=\frac{1}{2}AB+\frac{1}{2}BC+\frac{1}{2}AC=\frac{1}{2}\times 16=\boxed{8}$.