Question: As shown in the figure, there are four congruent squares placed inside the parallelogram ABCD, all arranged parallely. The top-left vertices of each square, B, E, F, and G, all lie on the line AB, with $BE = EF = FG$. If the line PQ passes exactly through the point D, and $AB = 14$, $CH = 5$, with $\angle A = 60^\circ$, then what is the area of each square?

Solution:
\textbf{Solution:} As shown in the diagram, extend EP to meet CD at K,\\
given $BH\parallel EP$ and ABCD is a parallelogram,\\
$\therefore AB\parallel CD$, $AB=CD=14$, and $\angle A=\angle C=60^\circ$,\\
$\therefore$ quadrilateral BCKE is a parallelogram, and $\angle EKD=\angle C=60^\circ$,\\
$\therefore BC=EK$, $BE=CK$,\\
by the property of congruent squares we get EP=BH, $\angle KPD=90^\circ$,\\
$\therefore BC-BH=EK-EP$, that is $PK=CH=5$, $\angle PDK=30^\circ$,\\
$\therefore DK=2PK=10$,\\
$\therefore CK=CD-DK=4$,\\
$\therefore BE=4$,\\
Also, since $BE=EF=FG$,\\
$\therefore AG=AB-BE-EF-FG=2$,\\
Similarly, it is found $AG=2AM$,\\
$\therefore AM=1$,\\
$\therefore GM=\sqrt{AG^2-AM^2}=\sqrt{3}$,\\
$\therefore$ the area of each small square is $\boxed{3}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $E$, $F$, $G$, and $H$ are respectively the midpoints of $AB$, $BD$, $CD$, and $AC$. Given that $AD=4$ and $BC=5$, what is the perimeter of quadrilateral $EFGH$?

Solution:
\textbf{Solution:} Given that E, F, G, and H are the midpoints of AB, BD, CD, and AC, respectively,

$\therefore EF = \frac{1}{2}AD = 2, FG = \frac{1}{2}BC = 2.5$, $GH = \frac{1}{2}AD = 2, EH = \frac{1}{2}BC = 2.5$,

$\therefore$ the perimeter of the quadrilateral EFGH = EF + FG + GH + HE = \boxed{9}


Question: As shown in the figure, in a parallelogram $ABCD$ made of paper, $\angle BAD=45^\circ$, with $AB=10$. The paper is folded such that the corresponding point $A'$ of point $A$ lies on the side $BC$, and the fold line $EF$ intersects $AB$, $AD$, and $AA'$ at points $E$, $F$, and $G$, respectively. Continue to fold the paper so that the corresponding point $C'$ of point $C$ falls on $A'F$, and connect $GC'$. What is the distance from point $G$ to $AD$? What is the minimum value of $GC'$?

Solution:
\textbf{Solution:} Construct $BH \perp AD$ at $H$, and $GP \perp AD$ at $P$, with $GQ \perp A'F$ at $Q$, and draw $A'R \perp AD$ at $R$, as shown in the diagram: \\
Since $\angle BAD = 45^\circ$ and $AB = 10$, \\
it follows that $BH = \frac{\sqrt{2}}{2}AB = 5\sqrt{2}$, \\
Since quadrilateral $ABCD$ is a parallelogram, and $BH \perp AD$, $A'R \perp AD$, \\
it follows that quadrilateral $BHR{A'}$ is a rectangle, \\
thus, $A'R = BH = 5\sqrt{2}$, \\
Since $GP \perp AD$ and $A'R \perp AD$, \\
it follows that $GP \parallel A'R$, \\
Since the paper is folded so that point $A$ maps to $A'$ on side $BC$, with fold line $EF$, \\
it follows that $AG = A'G$ and $\angle AFE = \angle A'FE$, \\
therefore, $GP$ is the median of $\triangle A'AR$, \\
thus, $GP = \frac{1}{2}A'R = \frac{5\sqrt{2}}{2}$, \\
Since $GP \perp AD$ and $GQ \perp A'F$, \\
it follows that $GP = GQ = \frac{5\sqrt{2}}{2}$, \\
Since the paper is folded so that point $C$ maps to $C'$ on $A'F$, \\
it follows that when $C'$ coincides with $Q$, $GC'$ is minimized, with the minimum value being the length of $GQ$, \\
thus, the minimum value of $GC'$ is $\boxed{\frac{5\sqrt{2}}{2}}$.


Question: As shown in the figure, the perimeter of the rhombus ABCD is 16, and $\angle DAB=60^\circ$. The point E is the midpoint of the side BC, and the point P is a moving point on the diagonal AC. When connecting BP and EP, what is the minimum value of $PB+PE$?

Solution:
\textbf{Solution:} Let's connect BD, which intersects AC at O, and connect DE which intersects AC at P, as shown in the diagram.\\
Since the diagonals of a rhombus are perpendicular bisectors of each other, it can be inferred that B and D are symmetrical about the line AC, hence PD=PB.\\
Considering that the shortest distance between two points is a straight line,\\
$\therefore PE+PB=PE+PD=DE$\\
meaning DE is the minimum value of $PB+PE$.\\
$\because$ Quadrilateral ABCD is a rhombus with a perimeter of 16, and $\angle DAB=60^\circ$,\\
$\therefore \angle DCB=\angle DAB=60^\circ$, and $BC=DC=4$.\\
$\therefore \triangle BDC$ is an equilateral triangle.\\
$\because$ Point E is the midpoint of side BC,\\
$\therefore CE=\frac{1}{2}BC=2$, and $DE\perp BC$.\\
$\therefore \angle DEC=90^\circ$.\\
In $\triangle DCE$, by the Pythagorean theorem, we find that $DE=\sqrt{BD^2-BE^2}=2\sqrt{3}$.\\
Thus, the minimum value of $PB+PE$ is $\boxed{2\sqrt{3}}$.


Question: As shown in the figure, $\angle ACB = 90^\circ$, D is the midpoint of AB, and point E is the midpoint of AF, such that E, C, D are collinear, and $CE = \frac{1}{4}CD$. If $BF = 10$, what is the length of AB?

Solution:
\textbf{Solution:} Since point D is the midpoint of AB and point E is the midpoint of AF, and BF = 10,\\
it follows that $DE = \frac{1}{2}BF = 5$,\\
Since $CE = \frac{1}{4}CD$,\\
it then follows that $CD = \frac{4}{5}DE = 4$,\\
Since $\angle ACB = 90^\circ$ and point D is the midpoint of AB,\\
thus $AB = 2CD = \boxed{8}$.


Question: As shown in the figure, in the square $ABCD$, $E$ is a point on the diagonal $BD$, and $BE=BC$, then what is the value of $\angle DCE$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a square, \\
$\therefore \angle CBE=45^\circ$, $\angle BCD=90^\circ$, \\
Since BE=BC, \\
$\therefore \angle BCE= \frac{1}{2} \times (180^\circ-45^\circ)=67.5^\circ$, \\
Therefore, $\angle DCE=\angle BCD-\angle BCE=90^\circ-67.5^\circ=\boxed{22.5^\circ}$.


Question: As shown in the figure, in rectangle $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, and $AE$ bisects $\angle BAD$ and intersects $BC$ at point $E$. Connect $OE$. If $\angle EAO = 15^\circ$, what is the degree measure of $\angle COE$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle, \\
$\therefore$ AD $\parallel$ BC, AC = BD, OA = OC, OB = OD, $\angle$BAD = $90^\circ$, \\
$\therefore$ OA = OB, $\angle$DAE = $\angle$AEB, \\
Since AE bisects $\angle$BAD, \\
$\therefore$ $\angle$BAE = $\angle$DAE = $45^\circ$ = $\angle$AEB, \\
$\therefore$ AB = BE, \\
Since $\angle$EAO = $15^\circ$, \\
$\therefore$ $\angle$BAC = $45^\circ + 15^\circ$ = $60^\circ$, \\
$\therefore$ $\triangle$BAO is an equilateral triangle, \\
$\therefore$ AB = OB, $\angle$ABO = $\angle$AOB = $60^\circ$, \\
$\therefore$ $\angle$OBC = $90^\circ - 60^\circ$ = $30^\circ$, $\angle$BOC = $180^\circ - 60^\circ$ = $120^\circ$, \\
Since AB = OB = BE, \\
$\therefore$ $\angle$BOE = $\angle$BEO = $\frac{1}{2}$($180^\circ - 30^\circ$) = $75^\circ$, \\
$\therefore$ $\angle$COE = $120^\circ - 75^\circ$ = $\boxed{45^\circ}$, \\


Question: As shown in the figure, in $ABCD$, $E$ and $F$ are the midpoints of $AB$ and $BC$ respectively, $EH \perp AC$ with the foot at $H$, and it intersects $AF$ at point $G$. If $AC = 24$ and $GF = 6\sqrt{5}$, then what is the length of $EG$?

Solution:
\textbf{Solution:} As shown in the diagram, connect EF, \\
because E and F are the midpoints of AB and BC respectively, \\
therefore EF is the midline of $\triangle$BAC, \\
therefore $EF=\frac{1}{2}AC=12$, $EF\parallel AC$, \\
because EH$\perp$AC, \\
therefore EH$\perp$EF, that is $\angle$FEG$=90^\circ$, \\
therefore $EG=\sqrt{GF^{2}-EF^{2}}=\boxed{6}$.


Question: \textit{When $n$ squares, each with a side length of 1, are arranged as shown in the figure, with $A_1$, $A_2$, ..., $A_n$ being the intersection points of the diagonals of the squares, what is the total area of the overlapping regions formed by overlapping 2022 such squares in this manner?}

Solution:
\textbf{Solution:}

As shown in the figure, the center of square ABCD is $A_{1}$, and BC, CD intersect with the square where $A_{2}$ is located at points E, and F, respectively. Connect $A_{1}C$, $A_{1}D$,

In square ABCD, $\angle A_{1}CB = \angle A_{1}DA_{2} = 45^\circ$, $A_{1}C = A_{1}D$, $\angle CA_{1}D = 90^\circ$,

Also, because $\angle EA_{1}F = 90^\circ$,

Therefore, $\angle EA_{1}C = \angle FA_{1}D$,

In $\triangle EA_{1}C$ and $\triangle FA_{1}D$, $\left\{\begin{array}{l}\angle E{A}_{1}C=\angle F{A}_{1}D \\ \angle {A}_{1}CE=\angle {A}_{1}DF \\ {A}_{1}C={A}_{1}D\end{array}\right.$,

Therefore, $\triangle EA_{1}C \cong \triangle FA_{1}D$ (AAS),

Therefore, $S_{\triangle EA_{1}C} = S_{\triangle FA_{1}D}$,

Therefore, $S_{\text{Quadrilateral} EA_{1}FC} = S_{\triangle A_{1}DC} = \frac{1}{4}S_{\text{Square} ABCD} = \frac{1}{4} \times 1 \times 1 = \frac{1}{4}$,

Similarly, the area of each shaded part is also $\frac{1}{4}$,

Because 2022 squares overlap in this manner, and the overlap area of every two squares is $\frac{1}{4}$, making a total of 2021 overlaps,

Therefore, the sum of the overlap areas formed by the 2022 squares overlapped in this way is $\boxed{\frac{2021}{4}}$.


Question: As shown in the diagram, in rectangle $OBAC$, the coordinates of point $A$ are $(5, 12)$. What is the length of $BC$?

Solution:
\textbf{Solution:} Draw line OA, and draw AM perpendicular to the x-axis at M, \\
since the coordinates of point A are (5, 12), \\
therefore, OM = 5, AM = 12, \\
by the Pythagorean theorem, we get: OA = $\sqrt{5^{2} + 12^{2}} = 13$, \\
since quadrilateral OBAC is a rectangle, \\
therefore, BC = OA = \boxed{13},


Question: As shown in the figure, points E and F are inside the square ABCD, with AE=CF=4, EF=6, and $\angle E=\angle F=90^\circ$. What is the area of square ABCD?

Solution:
\textbf{Solution:} Join AC and intersect EF at point O, as shown in the diagram: \\
$\because$ $\angle E=\angle F=90^\circ$ and $\angle AOE=\angle COF$, AE=CF, \\
$\therefore$ $\triangle AEO \cong \triangle CFO$, \\
$\therefore$ AO=OC, EO=OF, \\
$\because$ EF=6, \\
$\therefore$ EO=OF=3, \\
$\because$ AE=CF=4, \\
In $\triangle AOE$, according to the Pythagorean theorem, we get OA=5, \\
$\therefore$ AC=10, \\
In the square ABCD, AB=BC and $\angle ABC=90^\circ$, \\
According to the Pythagorean theorem, we have $AB^2+BC^2=AC^2$, \\
Solving this gives $AB^2=50$, \\
$\therefore$ The area of the square ABCD is $\boxed{50}$.


Question: As shown in the diagram, in $\square ABCD$, $\angle ABC=60^\circ$, points E and F are located on the extensions of AD and BA, respectively, with $CE \parallel BD$ and $EF \perp AB$, and $BC = 1$. What is the length of EF?

Solution:
\textbf{Solution:}

Since quadrilateral ABCD is a parallelogram, \\
$\therefore$AD = BC, and BC $\parallel$ AD, \\
Since CE $\parallel$ BD, \\
$\therefore$ quadrilateral BCED is a parallelogram, \\
$\therefore$ DE = BC = AD = 1, hence D is the midpoint of AE, \\
$\therefore$ AE = 2, \\
Since EF $\perp$ AB, \\
$\therefore$ $\angle$EFA = $90^\circ$, \\
Since AD $\parallel$ BC, \\
$\therefore$ $\angle$EAF = $\angle$ABC = $60^\circ$, $\angle$AEF = $30^\circ$, \\
$\therefore$ AF = $\frac{1}{2}$AE = 1, \\
$\therefore$ EF = $\sqrt{AE^{2} - AF^{2}} = \sqrt{2^{2} - 1^{2}} = \boxed{\sqrt{3}}$.


Question: As illustrated in the figure, point $P$ is on the diagonal $AC$ of rectangle $ABCD$. Draw $EF\parallel BC$ passing through point $P$ and intersecting $AB$ and $CD$ at points $E$ and $F$, respectively. Connect $PB$ and $PD$. If $AE=3$ and $PF=6$, what is the area of the shaded region in the figure?

Solution:
\textbf{Solution:} Let line PM be perpendicular to AD at point M and intersect BC at point N.\\
Since quadrilateral ABCD is a rectangle,\\
it follows that $\angle BAD = \angle ADC = \angle DCB = \angle ABC = 90^\circ$, AD$\parallel$BC, and AB$\parallel$CD,\\
therefore MN$\perp$BC.\\
Since EF$\parallel$BC,\\
it follows that EF$\perp$AB and EF$\perp$CD,\\
therefore $\angle AEP = \angle AMP = 90^\circ$,\\
hence quadrilateral AEPM is a rectangle.\\
Similarly, quadrilaterals DFPM, BEPN, and CFPN are rectangles,\\
thus $ {S}_{\triangle ADC}={S}_{\triangle ABC}$, ${S}_{\triangle AMP}={S}_{\triangle AEP}$, ${S}_{\triangle PBE}={S}_{\triangle PBN}$, ${S}_{\triangle PFD}={S}_{\triangle PDM}$, and ${S}_{\triangle PFC}={S}_{\triangle PCN}$,\\
therefore $ 2{S}_{\triangle PBE}={S}_{\triangle ABC}-{S}_{\triangle APE}-{S}_{\triangle PNC}$, $2{S}_{\triangle PDF}={S}_{\triangle ADC}-{S}_{\triangle APM}-{S}_{\triangle PFC}$,\\
thus $ 2{S}_{\triangle PBE}=2{S}_{\triangle PDF}$, i.e., ${S}_{\triangle PBE}={S}_{\triangle PDF}$,\\
Since $ AE=3$ and $PF=6$,\\
it follows that PM=AE=3,\\
thus $ {S}_{\triangle DFP}={S}_{\triangle PBE}=\frac{1}{2}\times 3\times 6=9$,\\
therefore, the area of the shaded part is $9\times 2=\boxed{18}$.


Question: As shown in the figure, the diagonals $AC$ and $BD$ of rectangle $ABCD$ intersect at point $O$. A line passing through point $O$ intersects $AD$ and $BC$ at points $E$ and $F$, respectively. Given $AB=2$ and $BC=3$, what is the area of the shaded region in the figure?

Solution:
\textbf{Solution:} Since the diagonals AC and BD of rectangle ABCD intersect at point O, \\
it follows that the area of the blank triangle within quadrilateral ABFE is equal to the area of the shaded triangle within quadrilateral EDCF.\\
Therefore, finding the area of the shaded part can be considered as finding the area of quadrilateral ABFE.\\
Hence, the area of the shaded part is: $\frac{2\times 3}{2}=\boxed{3}$.


Question: As shown in the figure, rectangle $ABCD$ is folded along $EF$, with point B exactly landing on point $B'$ on side $AD$. If $AE=2$, $DE=6$, and $\angle EFB=60^\circ$, what is the area of rectangle $ABCD$? The core question is: What is the area of rectangle $ABCD$?

Solution:
\textbf{Solution:} In rectangle $ABCD$, since $AD\parallel BC$, it follows that $\angle DEF=\angle EFB=60^\circ$,\\
since folding rectangle $ABCD$ along $EF$ precisely positions point $B$ at $B^{\prime}$ on side $AD$,\\
therefore $\angle EFB=\angle EFB=60^\circ$, $\angle B=\angle A^{\prime}B^{\prime}F=90^\circ$, $\angle A=\angle A^{\prime}=90^\circ$,\\
$AE=A^{\prime}E=2$, $AB=A^{\prime}B^{\prime}$,\\
in $\triangle EFB^{\prime}$, since $\angle DEF=\angle EFB=\angle EB^{\prime}F=60^\circ$, it follows that $\triangle EFB^{\prime}$ is an equilateral triangle,\\
in $\triangle A^{\prime}EB^{\prime}$, since $\angle A^{\prime}B^{\prime}E=90^\circ-60^\circ=30^\circ$, it follows that $B^{\prime}E=2A^{\prime}E$, and since $A^{\prime}E=2$,\\
thus $B^{\prime}E=4$, therefore $A^{\prime}B^{\prime}=2\sqrt{3}$, that is $AB=2\sqrt{3}$, since $AE=2$, $DE=6$,\\
thus $AD=AE+DE=2+6=8$, therefore the area of rectangle $ABCD$ = $AB\cdot AD=2\sqrt{3}\times 8=\boxed{16\sqrt{3}}$.


Question: As shown in the diagram, within rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $DF$ is perpendicular to and bisects $OC$, intersecting $AC$ at point $E$ and $BC$ at point $F$. $AF$ is connected. If $CD = 2\sqrt{3}$, then what is the length of $AF$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle, \\
it follows that AB=$CD=2\sqrt{3}$, and AO=CO=BO=DO, \\
since DF is perpendicular bisector of OC, \\
it follows that OD=DC, \\
thus OD=DC=OC, \\
therefore, $\triangle$ODC is an equilateral triangle, \\
thus OD=OC=$CD=2\sqrt{3}$, \\
therefore, AC=$4\sqrt{3}$, \\
thus, BC=$\sqrt{AC^{2}-AB^{2}}=6$, \\
since $\triangle$ODC is an equilateral triangle and DE$\perp$AC, \\
it follows that $\angle$CDE=$\angle$ODE=$30^\circ$, \\
thus, DC=$\sqrt{3}$CF=$2\sqrt{3}$, \\
therefore, CF=2, \\
thus, BF=4, \\
therefore, AF=$\sqrt{AB^{2}+BF^{2}}=\sqrt{12+16}=\boxed{2\sqrt{7}}$.


Question: As shown in the diagram, within rectangle $ABCD$, point $E$ is on $BC$, and $AE$ bisects $\angle BAC$, $AE=CE$, $BE=2$. What is the area of rectangle $ABCD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
$\therefore$ $\angle B=90^\circ$,\\
$\therefore$ $\angle BAC + \angle BCA = 90^\circ$,\\
Since AE bisects $\angle BAC$ and AE=CE,\\
$\therefore$ $\angle BAE = \angle EAC = \angle ECA$,\\
$\therefore$ $\angle BAE + \angle EAC + \angle ECA = 90^\circ$,\\
$\therefore$ $\angle BAE = \angle EAC = \angle ECA = 30^\circ$,\\
$\therefore$ AE=CE=2BE=4, AB=$\sqrt{4^2 - 2^2} = 2\sqrt{3}$,\\
$\therefore$ BC=BE+CE=6,\\
$\therefore$ The area of rectangle ABCD=AB×BC=$2\sqrt{3} \times 6 = \boxed{12\sqrt{3}}$;


Question: In $\triangle ABC$, $E$, $F$, $G$, and $H$ are the midpoints of $AC$, $CD$, $BD$, and $AB$, respectively. If $AD=3$ and $BC=8$, then what is the perimeter of the quadrilateral $EFGH$?

Solution:
\textbf{Solution:} Given that E, F, G, H are the midpoints of AC, CD, BD, AB respectively,\\
it follows that EF=$\frac{1}{2}$AD=1.5, GH=$\frac{1}{2}$AD=1.5, EH=$\frac{1}{2}$BC=4, FG=$\frac{1}{2}$BC=4,\\
therefore, the perimeter of the quadrilateral EFGH is EF+FG+GH+EH=1.5+4+1.5+4=\boxed{11}.


Question: As shown in the diagram, in square $ABCD$, $AB=9$, point $E$ is on side $CD$, and $DE=2CE$. Point $P$ is a moving point on diagonal $AC$. What is the minimum value of $PE+PD$?

Solution:
\textbf{Solution:} Draw BP and BE,\\
since quadrilateral ABCD is a square,\\
thus AB=AD, and $\angle$BAP=$\angle$DAP.\\
since AP=AP,\\
thus $\triangle$ABP$\cong$$\triangle$ADP,\\
thus BP=DP,\\
thus PE+PD=PE+PB$\geq$BE,\\
meaning the minimum value of PE+PD is the length of BE,\\
since AB=9, and DE=2CE,\\
thus CE=$\frac{1}{3}CD$=$\frac{1}{3}\times 9$=3,\\
thus BE=$\sqrt{BC^{2}+CE^{2}}$=$\sqrt{9^{2}+3^{2}}$=$3\sqrt{10}$,\\
thus the minimum value of PE+PD is \boxed{3\sqrt{10}}.


Question: As shown in the figure, the quadrilateral ABCD is a square. Perform the following steps:

Solution:
\textbf{Solution:} Based on the drawing process, it is known that $AD=AP=PD$,\\
$\therefore$ $\triangle ADP$ is an equilateral triangle,\\
$\therefore$ $\angle DAP=\angle ADP=\angle APD=60^\circ$,\\
$\because$ Quadrilateral $ABCD$ is a square,\\
$\therefore$ $AB=AD$, $\angle BAD=\angle ABC=90^\circ$,\\
$\therefore$ $AB=AP$, $\angle BAP=30^\circ$,\\
$\therefore$ $\angle ABP=\angle APB=\frac{180^\circ-30^\circ}{2}=75^\circ$,\\
$\therefore$ $\angle PBC=\angle ABC-\angle ABP=90^\circ-75^\circ=\boxed{15^\circ}$.


Question: As illustrated in the Cartesian coordinate system, given that the vertices O and B of the rhombus OABC are (0, 0) and (2, 2), respectively, what are the coordinates of the intersection point D of the diagonals of the rhombus after it has been rotated \(135^\circ\) counterclockwise about point O?

Solution:
\textbf{Solution:} Given the vertices of the rhombus OABC are O(0,0) and B(2,2), the coordinate of point D is found to be (1,1).\\
$\therefore OD=\sqrt{1^2+1^2}=\sqrt{2}$, the angle between OB and the positive y-axis is $45^\circ$,\\
and the angle between the positive y-axis and the negative x-axis is $90^\circ$,\\
$\therefore$ when the rhombus is rotated $135^\circ$ counterclockwise around point O, i.e., when line segment OD is rotated $135^\circ$ counterclockwise around point O, OD is then on the negative x-axis,\\
$\therefore$ the coordinates of the intersection of the diagonals of the rhombus D are ($-\sqrt{2}$, 0).\\
Thus, the answer is: $(−\sqrt{2}, 0) = \boxed{(-\sqrt{2}, 0)}$.


Question: As shown in the figure, the side length of square $ABCD$ is $6$. Point $P$ is a movable point on the side $BC$. With $P$ as the right angle vertex and $AP$ as the right angle side, an isosceles right $\triangle APE$ is constructed. $M$ is the midpoint of side $AE$. When point $P$ moves from point $B$ to point $C$, what is the length of the path that point $M$ moves?

Solution:
\textbf{Solution:} As shown in the diagram below, connect AC and BD which intersect at point O, connect EC, and draw ET$\perp$BC intersecting the extension line of BC at T.\\
$\because$ $\triangle APE$ is an isosceles right triangle,\\
$\therefore \angle APE=90^\circ$, $AP=PE$\\
$\therefore \angle APB+\angle TPE=90^\circ$.\\
$\because$ Quadrilateral ABCD is a square, ET$\perp$BC,\\
$\therefore \angle ABP=90^\circ$, $\angle PTE=90^\circ$.\\
$\therefore \angle ABP=\angle PTE$, $\angle BAP+\angle APB=90^\circ$.\\
$\therefore \angle BAP=\angle TPE$.\\
$\therefore \triangle ABP\cong \triangle PTE$ (AAS).\\
$\therefore AB=PT$, $PB=ET$.\\
$\because$ Quadrilateral ABCD is a square,\\
$\therefore AB=BC$.\\
$\therefore BC=PT$.\\
$\therefore BC-PC=PT-BC$, that is, PB=CT.\\
$\therefore PB=CT=ET$.\\
$\therefore \angle TEC=\angle TCE=45^\circ$.\\
$\because$ In square ABCD, AC and BD intersect at point O,\\
$\therefore O is the midpoint of AC$, $\angle DBC=45^\circ$.\\
$\therefore \angle DBC=\angle TCE$.\\
$\therefore OD\parallel CE$.\\
$\because$ M is the midpoint of AE,\\
$\therefore$ OM is the median line of $\triangle ACE$.\\
$\therefore OM\parallel CE$.\\
$\therefore$ Point M lies on the line OD.\\
$\because$ Point P moves along side BC,\\
$\therefore$ The trajectory of point M is OD.\\
$\because$ The side length of square ABCD is 6, and AC, BD intersect at point O,\\
$\therefore AB=6$, $AD=6$, O is the midpoint of BD.\\
$\therefore BD=\sqrt{AB^2+AD^2}=6\sqrt{2}$.\\
$\therefore OD=\frac{1}{2}BD=3\sqrt{2}$.\\
Hence, the answer is: $\boxed{3\sqrt{2}}$.


Question: As shown in the diagram, for the rhombus $ABCD$ with diagonals $AC=6\,cm$ and $BD=8\,cm$, and $AH \perp BC$ at point $H$, what is the length of $AH$ in $cm$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rhombus, \\
$\therefore$ CO = $\frac{1}{2}$AC = 3cm, BO = $\frac{1}{2}$BD = 4cm, AO$\perp$BO, \\
$\therefore$ BC=$\sqrt{OC^{2}+BO^{2}}$=5cm, \\
$\therefore$ Area of rhombus ABCD = $\frac{BD\cdot AC}{2}$ = $\frac{1}{2}\times 6\times 8=24\text{cm}^{2}$, \\
Since the area of rhombus ABCD = BC$\times$AH, \\
$\therefore$ BC$\times$AH = 24, \\
$\therefore$ AH=$\boxed{\frac{24}{5}}$cm.


Question: As shown in the figure, the rectangle $ABCD$ is folded along the line $AE$, with the vertex $D$ precisely landing on point $F$ on side $BC$. Given that $CE=3\,\text{cm}$ and $AB=8\,\text{cm}$, what is the area of the shaded section?

Solution:
\textbf{Solution:} Given the properties of folding, $EF=DE=CD-CE=5$, $AD=AF=BC$,\\
by the Pythagorean theorem, $CF=4$, $AF^2=AB^2+BF^2$,\\
thus $AD^2=8^2+(AD-4)^2$,\\
Solving gives, $AD=10$,\\
$\therefore BF=6$,\\
The area of the shaded part in the diagram is $S_{\triangle ABF}+S_{\triangle CEF}=\boxed{30\text{cm}^2}$.


Question: As shown in the figure, place $\square ABCO$ in the Cartesian coordinate system $xOy$, where $O$ is the origin. If the coordinate of point $A$ is $(4,0)$ and the coordinate of point $C$ is $(1,3)$, then what are the coordinates of point $B$?

Solution:
\textbf{Solution:} Extend BC to intersect the y-axis at point D,\\
$\because$ the coordinates of point A are (4,0),\\
$\therefore$ OA = 4,\\
$\because$ quadrilateral ABCD is a parallelogram,\\
$\therefore$ BC $\parallel$ OA, BC = OA = 4,\\
$\because$ the coordinates of point C are (1,3),\\
$\therefore$ OD = 3, DC = 1,\\
$\therefore$ BD = DC + BC = 1 + 4 = 5,\\
$\therefore$ the coordinates of point B are \boxed{(5,3)},


Question: As shown in the figure, in the Cartesian coordinate system xOy, the lines $l_{1}$ and $l_{2}$ are respectively the graphs of the functions $y=k_{1}x+b_{1}$ and $y=k_{2}x+b_{2}$. Then, what is the estimated solution set for the inequality $k_{1}x+b_{1}>k_{2}x+b_{2}$ with respect to $x$?

Solution:
\textbf{Solution:} When $x<-2$, $k_1x+b_1>k_2x+b_2$,\\
thus, the solution set of the inequality $k_1x+b_1>k_2x+b_2$ is $x<-2$.\\
Therefore, the answer is: $\boxed{x<-2}$.


Question: As shown in the figure, the graph of the linear function $y=-2x$ intersects with the graph of $y=kx+b$ at point $A(-2, 4)$. What is the solution of the equation $kx+b+2x=0$ with respect to $x$?

Solution:
\textbf{Solution:} Since the graph of the linear function $y=-2x$ intersects with the graph of $y=kx+b$ at point A ($-2$, $4$), \\
therefore, the solution to the equation in $x$, $-2x=kx+b$, for which $x=-2$, i.e., the solution to $kx+b+2x=0$ is $x=\boxed{-2}$.


Question: As shown in the figure, the graphs of the functions $y=x$ and $y=ax+4$ intersect at point $A(2, 2)$. What is the solution set of the inequality $x<ax+4$?

Solution:
\textbf{Solution:} By observing the graph of the function, we conclude that when $x<2$,\\
the graph of $y=x$ is entirely below the line $y=ax+4$,\\
$\therefore$ the solution set for the inequality $x<ax+4$ is $\boxed{x<2}$.


Question:
As shown in the figure, the graphs of the linear functions $y=k_1x+b_1$ and $y=kx+b$ intersect the $x$-axis at points $A(-1, 0)$ and $B(2, 0)$, respectively. What is the solution set of the system of inequalities in $x$ given by
\[
\begin{cases}
k_1x+b_1>0 \\
kx+b>0
\end{cases}
\]
?

Solution:
\textbf{Solution:} Let the equations of the linear functions be $y=k_1x+b_1$ and $y=kx+b$, with their graphs intersecting the $x$-axis at points $A(-1,0)$ and $B(2,0)$, respectively.\\
Based on the graphs, we know that the solution set for $y=k_1x+b_1>0$ is: $x>-1$,\\
and the solution set for $y=kx+b>0$ is: $x<2$,\\
Therefore, the solution set for the system of inequalities $\begin{cases}k_1x+b_1>0\\ kx+b>0\end{cases}$ is $-1<x<2$,\\
Thus, the answer is: $-1<x<2$.


Question: As shown in the graph, there is a linear function $y_1=x+a$ and $y_2=kx+b$ where $k\neq 0$, what is the solution set of the inequality $kx+b \geq x+a$ with respect to $x$?

Solution:
\textbf{Solution:} Since $\because kx+b \geq x+a$,\\
$\therefore y_2 > y_1$,\\
$\therefore$ combining with the graph, it is known that when $x \leq -2$, $y_2 > y_1$.\\
Thus, the answer is: $\boxed{x \leq -2}$.


Question: As shown in the diagram, within the same Cartesian coordinate system, the graphs of the functions $y_1=|x|$ and $y_2=kx+b$ intersect at points $A(-1, 1)$ and $B(2, 2)$. Based on the graph, what is the solution set of the inequality $|x|-kx-b\ge 0$?

Solution:
\textbf{Solution:} Observing the graph, we find: when $x\le -1$ or $x\ge 2$, the graph of the function $y_1=|x|$ is above or intersects with the graph of $y_2=kx+b$, \\
$\because |x|-kx-b\ge 0$, \\
$\therefore |x| \ge kx+b$, \\
$\therefore$ the solution set of the inequality $|x|-kx-b\ge 0$ is $x\le -1$ or $x\ge 2$. \\
Thus, the answer is: $\boxed{x\le -1$ or $x\ge 2}$


Question: As shown in the figure, the line $l_{1}: y_{1}=ax+b$ and the line $l_{2}: y_{2}=mx$ are depicted in the same Cartesian coordinate system. What is the solution set of the inequality $mx<ax+b$ with respect to $x$?

Solution:
\textbf{Solution:} From the graph, it is known that the two lines intersect at the point $(-1,2)$, \\
when $x > -1$, $mx < ax+b$,\\
$\therefore$ the solution set is $\boxed{x > -1}$.


Question: As shown in the figure, the graphs of the functions $y=2x$ and $y=ax+4$ intersect at point $A(3,m)$. What is the solution set of the inequality $2x<ax+4$?

Solution:
\textbf{Solution:} Observing the graph, we have: when $x<3$, then $2x<ax+4$,\\
which means the solution set of the inequality $2x<ax+4$ is $\boxed{x<3}$.


Question: Given the diagram, in the Cartesian coordinate system, the graph of the function $y=-3x$ intersects with the graph of $y=kx+b$ $(k>0)$ at point $P(m, 3)$. What is the solution set of the inequality $kx+b<-3x$?

Solution:
\textbf{Solution:} Solve: when $y=3$, we have $-3x=3$,\\
Solving this, we get $x=-1$,\\
From the graph, the solution set of the inequality $kx+b<-3$ is: $\boxed{x<-1}$.


Question: As shown in the figure, the graph of the direct proportion function $y=ax$ and the graph of the linear function $y=kx+2$ intersect at the point $(2, 1)$, what is the solution set of the inequality $kx+2\ge ax$?

Solution:
\textbf{Solution:} The intersection point of the two lines is at $(2,1)$,\\
when $x=2$, the values of the two functions are equal,\\
when $x<2$, the line $y=kx+2$ is above the line $y=ax$, thus $kx+2>ax$,\\
therefore, the solution set for the inequality $kx+2\ge ax$ is: $\boxed{x\leq 2}$.


Question: As shown in the figure, the graph of the linear function $y = kx + b$ passes through the point $P\left(-\frac{1}{2}, 1\right)$. What is the solution set of the inequality $kx + b < 1$?

Solution:
\textbf{Solution:} The graph of the linear function $y=kx+b$ passes through the point $P\left( -\frac{1}{2}, 1\right)$.\\
By analyzing the graph, we can find that the solution set of $kx+b<1$ is: $x<-\frac{1}{2}$,\\
Therefore, the answer is: $\boxed{x<-\frac{1}{2}}$.


Question: Given the figure, the line $y=x+1$ intersects the line $y=ax+b$ at point $A(m,3)$. What is the solution set of the inequality $x+1<ax+b$ with respect to $x$?

Solution:
\textbf{Solution:}

Given that the line $y=x+1$ intersects the line $y=ax+b$ at point $A(2,3)$,\\
$\therefore m+1=3$,\\
Solving this gives: $m=2$,\\
$\therefore A(2,3)$,\\
$\therefore$ the solution set of the inequality $x+1<ax+b$ with respect to $x$ is $\boxed{x<2}$.


Question: The graph of the linear function $y=kx+b$ (where $k$ and $b$ are constants and $k\neq 0$) is shown in the figure, and it passes through the point $(-2, 0)$. What is the solution set of the inequality $kx+b>0$ with respect to $x$?

Solution:
\textbf{Solution}: Since the linear function $y=kx+b$ passes through the point $(-2, 0)$, \\
it follows that the solution set for $kx+b>0$ is $\boxed{x<-2}$.


Question: As shown in the figure, the line $y=kx+b$ intersects with the x-axis and y-axis at points $A(-3, 0)$ and $B(0, 2)$ respectively. What is the solution set of the inequality $kx+b>0$?

Solution:
\[ \text{\textbf{Solution:}} \quad \text{According to the graph, it is known that the function values of y to the right of point A are all greater than 0. Therefore, when } x > -3, \text{ the function values are greater than 0, that is, } kx + b > 0; \]
\[ \text{Thus, the answer is: } \boxed{x > -3}. \]


Question: As shown in the figure, the graph of the linear function $y = -x + b$ intersects with the graph of $y = kx - 1$ at point P, and intersects the x-axis at point B. It is known that the ordinate of point P is 3, and the abscissa of point B is 4. What is the solution set of the inequality $-x + b > kx - 1$?

Solution:
\textbf{Solution:}

Since the linear function $y=-x+b$ intersects the x-axis at point B and the x-coordinate of point B is 4,

we have B(4,0).

Substituting B(4,0) into $y=-x+b$, we get: $0=-4+b$.

Solving this, we find $b=4$.

Therefore, the equation of line BP is $y=-x+4$.

Given that the y-coordinate of point $P$ is 3,

we therefore have $3=-x+4$.

Thus, $x=1$.

Hence, P(1,3).

From the graph, it is known that when $x<1$, the graph of the line $y=kx-1$ is below the graph of the function $y=-x+b$,

therefore, the solution set of the inequality $-x+b>kx-1$ is $x<1$,

so the answer is: $\boxed{x<1}$.


Question: As shown in the diagram, in the Cartesian coordinate system xOy, if the line $y_1=-2x+a$ and the line $y_2=bx-4$ intersect at point $P(1, -3)$, what is the solution set of the inequality $-2x+a<bx-4$ with respect to x?

Solution:
\textbf{Solution:} When $x>1$, the graph of the function $y_{1}=-2x+a$ is entirely below the graph of $y_{2}=bx-4$. Hence, the solution set of the inequality $-2x+a<bx-4$ is $x>1$;\\
Therefore, the answer is: $\boxed{x>1}$.


Question: As shown in the figure, the graph of the direct proportion function $y = kx \left(k \neq 0\right)$ passes through point $A\left(2, -4\right)$. The line $y = x + b$ is translated parallel to the y-axis, intersects the x-axis and y-axis at points $B$ and $C$, respectively, and intersects line $OA$ at point $D$. When the symmetric point $A'$ of point $A$ with respect to line $BC$ happens to fall on the y-axis, what is the area of $\triangle OBD$?

Solution:
\textbf{Solution:} Substitute point A(2, -4) into $y=kx$, we get $-4=2k$, thus $k=-2$,\\
$\therefore$ the equation of line OA is $y=-2x$,\\
In $y=x+b$, let $y=0$, then $x=-b$,\\
$\therefore$ B$(-b, 0)$, which means $OB=|b|$,\\
In $y=x+b$, let $x=0$, then $y=b$,\\
$\therefore$ C$(0, b)$, which means $OC=|b|$,\\
$\therefore$ OB=OC,\\
Also, $\because$ $\angle$BOC$=90^\circ$,\\
$\therefore$ $\angle$OCB=$\angle$OBC$=45^\circ$,\\
$\because$ the symmetric point of point A with respect to line BC precisely falls on the y-axis,\\
$\therefore$ CD perpendicularly bisects AA$^{\prime }$,\\
$\therefore$ CA=CA$^{\prime }$,\\
$\therefore$ $\angle$ACD=$\angle$OCB$=45^\circ$,\\
$\therefore$ $\angle$ACO$=90^\circ$,\\
$\therefore$ OC$=|y_{A}|=4$,\\
$\therefore$ OB=OC$=4$, that is, C$(0, -4)$,\\
Substitute point C$(0, -4)$ into $y=x+b$, we get $-4=0+b$,\\
$\therefore$ b$=-4$,\\
$\therefore$ the equation of line BC is $y=x-4$,\\
From $\begin{cases}y=-2x\\ y=x-4\end{cases}$, we obtain $\begin{cases}x=\frac{4}{3}\\ y=-\frac{8}{3}\end{cases}$,\\
$\therefore$ point D$\left(\frac{4}{3}, -\frac{8}{3}\right)$,\\
$\therefore$ the area of $\triangle OBD=\frac{1}{2}OB\cdot \left|y_{D}\right|=\frac{1}{2}\times 4\times \frac{8}{3}=\boxed{\frac{16}{3}}$.


Question: Given the functions $y_{1}=2x$ and $y_{2}=ax+4$ ($a\neq 0$) intersect at point $A(m,3)$, what is the solution set for the inequality $2x<ax+4$ in terms of $x$?

Solution:
Solution: The graph of the function $y_1=2x$ intersects with the graph of $y_2=ax+4$ ($a\neq 0$) at point A$\left(\frac{3}{2}, 3\right)$, \\
$\therefore 3=2\times\frac{3}{2}$, \\
$\therefore m=\frac{3}{2}$, \\
$\therefore$ point A$\left(\frac{3}{2}, 3\right)$, \\
$\because 2x<ax+4$, that is $y_1<y_2$, \\
$\therefore \boxed{x<\frac{3}{2}}$.


Question: As shown in the figure, it is the graph of the linear function $y=mx+n$. What is the solution set of the inequality $mx+n>0$ with respect to $x$?

Solution:
\textbf{Solution:} From the graph, we know: \\
The solution set of the inequality $mx+n>0$ with respect to $x$ is: $x>-1$. \\
Thus, the answer is: $\boxed{x>-1}$.


Question: As shown in the figure, the line $y=kx+b$ ($k\neq 0$) passes through point $A(1, 2)$. What is the solution set of the inequality $kx+b<2x$ with respect to $x$?

Solution:
\textbf{Solution:} Given that $\because x=1$ leads to $y=2$,\\
$\therefore$ the graph of the function $y=2x$ also passes through point A $(1, 2)$,\\
$\therefore$ the graphs of the function $y=kx+b$ (where $k<0$) and $y=2x$ both pass through point A $(1, 2)$.\\
Upon observing the graphs, we find that for $x>1$, $kx+b<2x$.\\
Hence, the answer is: $\boxed{x>1}$.


Question: As shown in the figure, the line $l_{1}\colon y=ax-b$ and the line $l_{2}\colon y=kx$ are in the same Cartesian coordinate system. What is the solution set for the inequality $ax-b>kx$ in terms of $x$?

Solution:
\textbf{Solution:} When $x < -1$, we have $ax - b > kx$. \\
Therefore, the solution set of the inequality $ax - b > kx$ with respect to $x$ is $x < -1$, \\
thus the answer is: $\boxed{x < -1}$.


Question: As shown in the figure, in right-angled triangle $\triangle ABC$, $\angle ACB=90^\circ$, $AC=6$, and $AB=10$. Point D moves from point B in the direction of ray BC at a speed of $5$ units per second, and point F moves from point A towards point B along edge AB at the same speed. When F reaches point B, point D stops moving. Let the movement time of point D be $t$ seconds. A square $DEFG$ is constructed with $DF$ as its diagonal. During the movement, if one side of square $DEFG$ happens to coincide with one side of the right-angled triangle $ABC$, then what is the value of $t$?

Solution:
\textbf{Solution:}

Given that in the right triangle $ABC$, $AC=6$ and $AB=10$, \\
$\therefore BC=\sqrt{AB^2-AC^2}=8$, \\
because point $D$ moves from point $B$ in the direction of ray $BC$ at a speed of 5 units per second, and point $F$ moves from point $A$ towards point $B$ along edge $AB$ at the same speed, \\
$\therefore AF=DB=5t$, \\
As shown in the figure, taking $B$ as the origin and the line along $BC$ as the x-axis, a Cartesian coordinate system is established, \\
$\therefore A(8,6)$ \\
Let the equation of line $BA$ be $y=kx$, then $6=8k$ \\
Solving gives $k=\frac{3}{4}$, \\
$\therefore$ the equation of line $BA$ is $y=\frac{3}{4}x$, \\
Let $F(m,\frac{3}{4}m)$, $0\le m\le 8$, \\
(1) As shown in the figure, when $DE$ is on side $BC$, draw $FM\perp AC$ at $M$. \\
$\therefore FM=8-m$, $AM=6-\frac{3}{4}m$, \\
$\therefore AF^2=AM^2+FM^2=\left(8-m\right)^2+\left(6-\frac{3}{4}m\right)^2=\left(\frac{5m-40}{4}\right)^2$, \\
because $AF=5t$, \\
$\therefore 5t=\left|\frac{5m-40}{4}\right|$, \\
Solving gives $t=\left|\frac{m-8}{4}\right|$, \\
because $0\le m\le 8$, \\
$\therefore t=\frac{8-m}{4}$ \\
$\therefore$ $FM=EC=4t$, $AM=3t$, $CM=EF=DE=6-3t$, \\
because $BD+DE+EC=8$, \\
$\therefore$ $5t+6-3t+4t=8$, \\
Solving gives $t=\boxed{\frac{1}{3}}$, \\
(2) As shown in the figure, when $FG$ is on side $AB$, \\
In right triangle $\triangle BGD$, with $DB=5t$, similar to (1) we can conclude $DG=FG=3t$, then $BG=4t$, \\
because $BG+FG+AF=10$, \\
$\therefore$ $4t+3t+5t=10$, \\
Solving gives $t=\frac{5}{6}$, \\
(3) When $DG$ is on side $BC$, \\
then $FG=DG=6-3t$, $BG=8-4t$, \\
because $BD=BG+DG=5t$, \\
$\therefore$ $8-4t+6-3t=5t$, \\
Solving gives $t=\frac{7}{6}$; \\
(4) When $EF$ is on side $AB$, \\
Similar to (1) we can conclude $BE=4t$, $DE=EF=3t$, \\
because $BE-EF=BF$, \\
$\therefore$ $4t-3t=10-5t$, \\
Solving gives $t=\frac{5}{3}$; \\
In summary, $t=\frac{1}{3}$ or $\frac{5}{6}$ or $\frac{7}{6}$ or $\frac{5}{3}$.


Question: As shown in the figure, in the Cartesian coordinate system, with O as the center and the length of OP as the radius, an arc is drawn intersecting the negative x-axis at point A. Thus, the coordinates of point A are ($- \sqrt{17}$, 0). Given that the y-coordinate of point P is $-1$, what are the coordinates of point P?

Solution:
\textbf{Solution:} Construct line $PB\perp OA$ at point B, \\
$\because$ the coordinates of point A are ($-\sqrt{17}$, 0), \\
$\therefore$ OP=OA=$\sqrt{17}$, \\
$\because$ the ordinate of point P is $-1$, \\
$\therefore$ PB=1, \\
$\therefore$ OB=$\sqrt{OP^{2}-PB^{2}}=4$, \\
$\therefore$ the coordinates of point P are $\boxed{(-4, -1)}$.


Question: As shown in the figure, it is known that the graph of the function $y=2x+b$ intersects with the graph of the function $y=kx-3$ at point $P$. What is the solution set of the inequality $kx-3>2x+b$?

Solution:
\textbf{Solution}: Observing the graph, we find: when $x<4$, the graph of the function $y=2x+b$ is below the graph of the function $y=kx-3$, \\
$\therefore$ the solution set of the inequality $kx-3>2x+b$ is $\boxed{x<4}$.


Question: As shown in the figure, given the functions $y=ax+b$ and $y=kx$ intersect at point $P$, according to the graph, what is the solution to the system of linear equations $\begin{cases}y=ax+b,\\ y=kx?\end{cases}$

Solution:
\textbf{Solution:} Since the intersection point P of the graphs of the functions $y=ax+b$ and $y=kx$ is at ($-2$, $-1$), \\
therefore, the solution to the system of linear equations $\begin{cases}y=ax+b\\ y=kx\end{cases}$ is $\boxed{\begin{cases}x=-2\\ y=-1\end{cases}}$.


Question: The coordinates of the intersection point of the lines $y_{1}=k_{1}x+a$ and $y_{2}=k_{2}x+b$ are $(1, 2)$. What is the range of values of $x$ for which $y_{1} \ge y_{2}$?

Solution:
\textbf{Solution:} Given the graph of the function, the inequality $y_{1}\ge y_{2}$, which translates to $k_{1}x+a\ge k_{2}x+b$, corresponds to the set of x-values for which the graph of $y_{1}=k_{1}x+a$ is above the graph of $y_{2}=k_{2}x+b$. Therefore, the solution set to the inequality $y_{1}\ge y_{2}$ is $x\geq 1$, \\
hence the answer is: $\boxed{x\geq 1}$.


Question: Given the graph of the linear function $y=kx-b$ as shown in the figure, what is the solution set of the inequality $kx-1>b$ with respect to $x$?

Solution:
\textbf{Solution:} Given $\because kx-1>b$,\\
$\therefore kx-b>1$,\\
$\therefore$ the graph of the linear function lies above the line $y=1$. Observing the graph, we find that $x>4$,\\
$\therefore$ the solution set for the inequality $kx-1>b$ in terms of $x$ is $\boxed{x>4}$.


Question: As shown in the figure, in the Cartesian coordinate system, side BC of the parallelogram ABCD is on the x-axis, with point A at $(0, 3)$ and point B at $(-1, 0)$. If the line $y=-2x+4$ perfectly bisects the area of the parallelogram ABCD, what are the coordinates of point D?

Solution:
\textbf{Solution:} Let's consider the diagram, where we assume that the line $y=-2x+4$ intersects with AD and BC at points H and G, respectively.\\
Since ABCD is a parallelogram with A at (0, 3),\\
it follows that H is at $\left(\frac{1}{2}, 3\right)$, and G is at (2, 0), with AD $\parallel$ BC and AD=BC,\\
Let's assume D is at (d, 3),\\
thus DH=AD−AH=d−$\frac{1}{2}$, and BG=2−(−1)=3,\\
Since the line $y=-2x+4$ precisely bisects the area of the parallelogram ABCD into two congruent trapezoids,\\
it follows that DH=BG,\\
thus d−$\frac{1}{2}$=3,\\
which implies d=$\frac{7}{2}$,\\
Therefore, the coordinates of D are $\boxed{\left(\frac{7}{2}, 3\right)}$.


Question: Based on the diagram, the triangle \(ABC\) is placed inside a grid with each small square having a side length of 1. The points \(A\), \(B\), \(C\) all lie on the grid points. How do we determine the value of \(\tan A\)?

Solution:
\textbf{Solution:}

Consider $BD \perp AC$ at point $D$, \\
$\because BC = 2$, $AC = \sqrt{3^{2}+3^{2}} = 3\sqrt{2}$, the distance from point $A$ to $BC$ is 3, $AB=\sqrt{3^{2}+1^{2}}=\sqrt{10}$,\\
$\therefore \frac{AC \cdot BD}{2}=\frac{BC\cdot 3}{2}$, that is $\frac{3\sqrt{2}\times BD}{2}=\frac{2\times 3}{2}$,\\
Solving for it, we get: $BD=\sqrt{2}$,\\
$\therefore AD= \sqrt{AB^{2}-BD^{2}}=\sqrt{(\sqrt{10})^{2}-(\sqrt{2})^{2}}=2\sqrt{2}$,\\
$\therefore \tan A=\frac{BD}{AD}=\frac{\sqrt{2}}{2\sqrt{2}}=\boxed{\frac{1}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC<BC$, $CD$ bisects $\angle ACB$ and intersects $AB$ at point $D$. A circle $\odot O$ is drawn with $DB$ as its diameter, intersecting $CD$ and $BC$ at points $E$ and $F$, respectively. Lines $BE$ and $EF$ are drawn. What is the measure of $\angle EBF$? If $DE=DC$ and $BC=8$, what is the length of $EF$?

Solution:
\textbf{Solution:} Connect DF, and draw EG$\perp$BC at point G,\\
$\because$BD is a diameter,\\
$\therefore \angle CEB=90^\circ$,\\
$\because \angle ACB=90^\circ$, CD bisects $\angle ACD$,\\
$\therefore \angle DCF= \frac{1}{2}\angle ACB=45^\circ$,\\
$\therefore \angle EBF=90^\circ-\angle DCF=90^\circ-45^\circ=\boxed{45^\circ}$;\\
$\because$BD is a diameter,\\
$\therefore \angle DFG=90^\circ$,\\
$\therefore DF\perp BC$,\\
$\therefore DF\parallel FG$,\\
$\because DE=DC$,\\
$\therefore CF=FG$,\\
$\because \angle FCG=\angle EBC=45^\circ$,\\
$\therefore EC=BE$,\\
In $\triangle CEB$, $\angle EBC=45^\circ$, BC=8,\\
$\therefore BE=CB\sin\angle EBC=8\sin45^\circ= 8\times \frac{\sqrt{2}}{2}=4\sqrt{2}$;\\
In $\triangle EBG$, EG=CG=BE$\sin\angle EBC= 4\sqrt{2}\sin45^\circ= 4\sqrt{2}\times \frac{\sqrt{2}}{2}=4$,\\
$\therefore FG=CG-4$,\\
$\therefore FG=2$\\
In $\triangle EFG$, $EF=\sqrt{FG^{2}+EG^{2}}=\sqrt{2^{2}+4^{2}}=\boxed{2\sqrt{5}}$.


Question: As shown in the figure, point $P(12, a)$ lies on the graph of the inverse proportion function $y=\frac{60}{x}$. Line $PH$ is perpendicular to the $x$-axis at point $H$. What is the value of $\tan \angle POH$?

Solution:
\textbf{Solution:} Since $P(12,a)$ lies on the graph of the inverse proportion function $y=\frac{60}{x}$,\\
thus $a=\frac{60}{12}=5$,\\
therefore, $P(12,5)$,\\
therefore, $OH=12$, $PH=5$,\\
thus, $\tan \angle POH=\frac{PH}{OH}=\boxed{\frac{5}{12}}$.


Question: As shown in the figure, the slope ratio of the water-facing slope $AB$ of the river embankment is $1\colon \sqrt{3}$. Given that the height of the embankment $BC=10m$, what is the length of the slope $AB$ in meters?

Solution:
\textbf{Solution:} Given that the slope ratio of the upstream slope $AB$ is $1\colon \sqrt{3}$,\\
it follows that $\frac{BC}{AC}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$\\
Therefore, $\tan A=\frac{BC}{AC}=\frac{\sqrt{3}}{3}$\\
Thus, $\angle A=30^\circ$\\
Hence, $AB=2BC=\boxed{20}\text{m}$


Question: As shown in the figure, the slope of a hillside is $i=1\colon \sqrt{3}$. If Xiaoming climbs from point A to point B and the straight-line distance AB is 10 meters, then what is the vertical ascent height CB in meters?

Solution:
\textbf{Solution:} Given that the slope of the hill, $i = 1\colon\sqrt{3}$,\\
it follows that $\tan\angle BAC = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$,\\
therefore, $\angle BAC = 30^\circ$,\\
and given $AB = 10$,\\
it follows that $BC = \frac{1}{2} AB = \frac{1}{2} \times 10 = \boxed{5}$ (meters),


Question: As shown in the figure, a ship is located at point $A$, which is at the northeast direction $45^\circ$ from the lighthouse $C$ and at a distance of $60$ nautical miles from it. After sailing southward for a while, it arrives at point $B$, which is at the southeast direction $30^\circ$ from the lighthouse $C$. What is the distance between the ship and the lighthouse $C$ at this point? (Keep the square root in the result.)

Solution:
\textbf{Solution:} Draw CD $\perp$ AB. In the right triangle $ACD$, $\angle A=45^\circ$ and AC=60 nautical miles,\\
$\therefore \triangle ACD$ is an isosceles right triangle,\\
$\therefore$ AD=CD=$\frac{\sqrt{2}}{2}$AC=30$\sqrt{2}$ nautical miles,\\
In right triangle $BCD$, $\angle B=30^\circ$,\\
$\therefore$ BC=2CD=2×30$\sqrt{2}$=60$\sqrt{2}$ nautical miles\\
Thus, at this time, the distance from the ship to the lighthouse is \boxed{60\sqrt{2}} nautical miles.


Question: As shown in the figure, in rectangle $ABCD$, $AD=6$, $AB=12$, and $N$ is a moving point on side $AB$. Triangle $AND$ is folded along $DN$ to obtain triangle $DMN$. Connect $BM$ and $CM$ respectively. If $\triangle BMC$ is an isosceles triangle, what is the length of $AN$?

Solution:
\textbf{Solution:} \\
Since $MD = AD = 6$, \\
it follows that the locus of M is a circle with D as the center and 6 as the radius, \\
(1) When $MC = MB$, then M lies on the perpendicular bisector of BC, \\
thus $DP = \frac{1}{2}DM$, \\
hence $\angle DMP = 30^\circ$, \\
thus $\angle PDM = 60^\circ$, \\
hence $\angle ADN = \angle MDN = 30^\circ$, \\
thus $AN = \frac{\sqrt{3}}{3}AD = 2\sqrt{3}$; \\
(2) When $CM = CB = 6$, we have $DM = AD = 6$, hence quadrilateral ANMD is a square, \\
thus $AN = AD = 6$; \\
(3) Since the minimum value of BM is $6\sqrt{5} - 6 = 6(\sqrt{5} - 1) > 6$, \\
MB cannot be equal to BC; \\
Therefore, the answer is: $AN = \boxed{2\sqrt{3}}$ or $6$.


Question: As depicted in the figure, in square $ABCD$, $E$ is a point on $AB$. Line $DE$ is drawn, and $\triangle DAE$ is rotated $90^\circ$ counterclockwise around point $A$ to $\triangle BAF$ (with $E$ corresponding to $F$ and $D$ to $B$). Lines $FE$ and $FC$ are drawn. If $AB=4$ and $FE$ bisects $\angle BFC$, what is the length of $AE$?

Solution:
\textbf{Solution:} Let DE and CF intersect at O, as shown in the diagram: \\
$\because$ Rotating $\triangle DAE$ $90^\circ$ counterclockwise around point A to $\triangle BAF$, \\
$\therefore AF=AE$, $\angle ADE=\angle ABF$ \\
$\therefore \angle AFE=\angle AEF=45^\circ$, \\
$\because FE$ bisects $\angle BFC$, \\
$\therefore \angle EFB=\angle EFC$, \\
$\therefore \angle OFD=45^\circ-\angle EFC$, $\angle ADE=\angle ABF=45^\circ-\angle EFB$, \\
$\therefore \angle OFD=\angle ADE$, that is $\angle CFD=\angle ADE$, \\
$\therefore \tan\angle CFD=\tan\angle ADE$, that is $\frac{CD}{DF}=\frac{AE}{AD}$, \\
Let $AE=AF=x$, \\
$\therefore \frac{4}{4+x}=\frac{x}{4}$, \\
$\therefore x=\boxed{2\sqrt{5}-2}$ or $-2\sqrt{5}-2$ (discard the latter), \\
$\therefore AE=2\sqrt{5}-2$.


Question: As shown in the figure, a semicircle $O$ with diameter $AB=6$ is rotated counterclockwise about endpoint $A$. When the intersection point $H$ on the circular arc and the diameter satisfies $BH:AH=1:2$, what is the value of $\tan\angle B'AB$?

Solution:
\textbf{Solution:} Connect $HB'$,

$\because$ When semi-circle $O$ with diameter $AB=6$ is rotated counterclockwise around endpoint $A$,

$\therefore$ $AB'=AB=6$, $\angle AHB'=90^\circ$,

$\because$ $AH+BH=6$, $BH\colon AH=1\colon 2$,

$\therefore$ $BH=2$, $AH=4$

$\therefore$ $B'H=\sqrt{AB'^2-AH^2}=\sqrt{6^2-4^2}=2\sqrt{5}$;

In $\triangle AHB'$,

$\tan\angle B'AB=\frac{B'H}{AH}=\frac{2\sqrt{5}}{4}=\boxed{\frac{\sqrt{5}}{2}}$.


Question: As shown in the figure, within a square grid, the quadrilateral ABCD is a rhombus. What is the value of $\tan\frac{\angle BAD}{2}$?

Solution:
\textbf{Solution}: From the properties of a rhombus, we know that $\frac{\angle BAD}{2}=\angle BAC$\\
From the figure, we have $\tan\angle BAC=\frac{3}{4}$\\
$\therefore \tan\frac{\angle BAD}{2}=\tan\angle BAC=\boxed{\frac{3}{4}}$


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=2$, and $AB=3$, what is the value of $\sin B$?

Solution:
\textbf{Solution:} Given that in the right-angled triangle $ABC$, $\angle C=90^\circ$, $AC=2$, $AB=3$,\\
therefore, $\sin B=\frac{AC}{AB}=\boxed{\frac{2}{3}}$


Question: As shown in the figure, $ \triangle ABC$ is an equilateral triangle with $O$ as the center of its circumcircle, and point $D$ is on the arc $ABC$. Connect $AD$ and $BD$. $AE$ is a chord on the right side of $AD$, and $\angle DAE=30^\circ$. Connect $CE$. What is the angle between the lines containing $BD$ and $CE$? When $BD=\sqrt{3}$, and $CE=4$, what is the radius of circle $O$?

Solution:
\textbf{Solution:} Extend BD and CE to intersect at point F, connect OB, OC, CD, and BE,\\
since $\triangle ABC$ is an equilateral triangle,\\
therefore $\angle BAC=60^\circ$,\\
since $\angle DBC=\angle DAC$, $\angle ECB=\angle BAE$,\\
therefore $\angle BFC=180^\circ-\angle FBC-\angle FCB$\\
$=180^\circ-\angle DAC-\angle BAE$\\
$=180^\circ-(\angle DAC+\angle BAD+\angle DAE)$\\
$=180^\circ-(\angle BAC+\angle DAE)$\\
$=180^\circ-(60^\circ+30^\circ)$\\
$=90^\circ$,\\
therefore the angle between the lines containing BD and CE is $\boxed{90^\circ}$;\\
since $\angle DAE=30^\circ$,\\
therefore $\angle DBE=\angle ECD=30^\circ$,\\
in triangle $\triangle BEF$, let DF$=m$, EF$=n$,\\
therefore BF$=BD+DF=\sqrt{3}+m$,\\
since $\angle FBE=30^\circ$,\\
therefore $\tan 30^\circ=\frac{EF}{BF}$,\\
therefore $\frac{\sqrt{3}}{3}=\frac{n}{\sqrt{3}+m}$,\\
therefore $n=1+\frac{\sqrt{3}}{3}m$,\\
in triangle $\triangle CDF$, CE$=4$,\\
therefore CF$=CE+EF=4+n$,\\
since $\angle FCD=30^\circ$,\\
therefore $\tan 30^\circ=\frac{DF}{CF}$,\\
therefore $\frac{\sqrt{3}}{3}=\frac{m}{4+n}$,\\
therefore $m=\frac{\sqrt{3}}{3}n+\frac{4\sqrt{3}}{3}$,\\
therefore $m=\frac{\sqrt{3}}{3}\left(1+\frac{\sqrt{3}}{3}m\right)+\frac{4\sqrt{3}}{3}$,\\
solving for $m$ yields: $m=\frac{5\sqrt{3}}{2}$,\\
therefore $n=\frac{7}{2}$,\\
therefore $BF=\sqrt{3}+m=\sqrt{3}+\frac{5\sqrt{3}}{2}=\frac{7\sqrt{3}}{2}$,\\
$CF=4+n=4+\frac{7}{2}=\frac{15}{2}$,\\
therefore $BC=\sqrt{BF^2+CF^2}=\sqrt{\left(\frac{7\sqrt{3}}{2}\right)^2+\left(\frac{15}{2}\right)^2}=\sqrt{93}$,\\
draw line OH perpendicular to BC at point H through point O,\\
since OB$=$OC,\\
therefore $BH=\frac{1}{2}BC=\frac{\sqrt{93}}{2}$,\\
since $\angle BOC=2\angle BAC=120^\circ$,\\
therefore $\angle BOH=60^\circ$,\\
therefore $\sin 60^\circ=\frac{BH}{OB}$,\\
therefore $OB=\frac{BH}{\frac{\sqrt{3}}{2}}=\boxed{\sqrt{31}}$,\\
thus, the radius of circle $O$ is $\boxed{\sqrt{31}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AB \parallel CD$, $AD=4$, and $AB=BC=BD=6$, what is the sine of $\angle ACD$?

Solution:
\textbf{Solution:} Given that $AB=BC=BD=6$,\\
thus points A, D, C are on the circle with center B and radius equal to the length of AB, as shown in the diagram, extend AB to meet the circle B at point E,\\
since $AB\parallel CD$,\\
therefore $\angle CDB=\angle ABD$, $\angle DCB=\angle CBE$,\\
since BD=BC,\\
therefore $\angle CDB=\angle BCD$,\\
thus $\angle ABD=\angle CBE$,\\
therefore $AD=CE=4$,\\
since AE is the diameter of circle B,\\
therefore $\angle ACE=90^\circ$,\\
in the right triangle $\triangle ACE$, AE=$2AB=12$,\\
therefore $\sin\angle CAE=\frac{CE}{AE}=\frac{4}{12}=\frac{1}{3}$,\\
since $\angle ACD=\angle CAE$,\\
therefore $\sin\angle ACD=\boxed{\frac{1}{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $AC=6$, $BC=8$, and $D$, $E$ are on $CA$, $CB$ respectively, with point $F$ inside $\triangle ABC$. If quadrilateral $CDFE$ is a square with a side length of $2$, then what is $\sin\angle FBA$?

Solution:
\textbf{Solution:} Construct $FG \perp AB$ at G, and connect AF. \\
Since quadrilateral CDFE is a square with side length 2, \\
thus $CD = CE = DF = EF = 2$, $\angle C = \angle ADF = 90^\circ$, \\
since $AC = 6$, $BC = 8$, \\
thus $AD = 4$, $BE = 6$, \\
therefore $AB = \sqrt{AC^{2} + BC^{2}} = 10$, $AF = \sqrt{AD^{2} + DF^{2}} = 2\sqrt{5}$, $BF = \sqrt{BE^{2} + EF^{2}} = 2\sqrt{10}$, \\
let $BG = x$, \\
since $FG^{2} = AF^{2} - AG^{2} = BF^{2} - BG^{2}$, \\
thus $(2\sqrt{5})^{2} - (10 - x)^{2} = (2\sqrt{10})^{2} - x^{2}$, solving gives: $x = 6$, \\
therefore $FG = \sqrt{BF^{2} - BG^{2}} = 2$, \\
therefore $\sin\angle FBA = \frac{FG}{BF} = \boxed{\frac{\sqrt{10}}{10}}$.


Question: In the square grid, the position of $\triangle ABC$ is shown in the figure, what is the value of $\sin\angle BAC$?

Solution:
\textbf{Solution:} Connect the lattice points DC, BD.\\
In $\triangle ABD$,\\
$\because CD=3$, $AD=4$,\\
$\therefore AC= \sqrt{CD^2+AD^2}=5$.\\
$\therefore \sin\angle BAC= \frac{CD}{AC}=\boxed{\frac{3}{5}}$.


Question: As shown in the figure, points C and D are the trisection points of line segment AB. Construct an equilateral triangle $\triangle OCD$ upwards with CD as its base. With O as the center and the length of OA as the radius, draw an arc that intersects the extensions of OC and OD at points E and F, respectively. If $AB = 6$, what is the area of the shaded region?

Solution:
\textbf{Solution:} Connect OA, OB, and construct $OH\perp AB$ at point H as shown in the diagram,\\
$\because AB=6$, C and D are the trisection points of the line segment AB, and $\triangle COD$ is an equilateral triangle,\\
$\therefore \angle OHA=90^\circ$, $\angle COH=30^\circ$, $AC=CD=DB=2$, $AH=BH=3$, $\angle OCH=60^\circ$,\\
$\therefore OH=2\times \sin 60^\circ=\sqrt{3}$,\\
$\therefore OA=\sqrt{3^2+(\sqrt{3})^2}=2\sqrt{3}$, $\tan \angle OAH=\frac{OH}{AH}=\frac{\sqrt{3}}{3}$, $\therefore \angle OAH=30^\circ$, $\therefore \angle AOH=60^\circ$,\\
$\therefore \angle AOC=30^\circ$,\\
$\therefore$ the area of the shaded part is: $\frac{2\times \sqrt{3}}{2}+\left[\frac{30\times \pi \times (2\sqrt{3})^{2}}{360}-\frac{2\times \sqrt{3}}{2}\right]\times 2=\boxed{2\pi -\sqrt{3}}$


Question: As shown in the figure, square $ABCD$ has a side length of $2$. Points $E$ and $F$ are two moving points on side $AD$, satisfying $AE=DF$. Lines $BE$ and $CF$ are drawn, and $BE$ intersects the diagonal $AC$ at point $G$. Line $DG$ is drawn and intersects $CF$ at point $H$. When line $BH$ is drawn, what is the minimum value of $BH$?

Solution:
\textbf{Solution:} As shown in the figure, \\
since square ABCD, it follows that $AB=CD$, $\angle BAE=\angle CDF=90^\circ$ thus $\triangle ABE \cong \triangle DCF$ (SAS), hence $\angle ABE=\angle DCF$. Since the median line where AC is located is the axis of symmetry of the square, it follows that $\angle ABE=\angle ADG=\angle DCF$. Since $\angle ADG + \angle GDC = 90^\circ$, it implies $\angle DCF + \angle GDC = 90^\circ$, hence $\angle DHC = 90^\circ$ which means $DH \perp CF$. Therefore, point H moves on the circle with CD as its diameter. When point H moves to the midpoint of AC, BH is the shortest. Connect BD and intercept AC at point O, since $BH + DH \geq BD$, when points B, H, and D are on the same line, since the perpendicular segment is the shortest, hence when BH $\perp$ AC, that is, when point H coincides with point O, the value of BH is minimum. In right $\triangle BOC$, $BO=BC\sin\angle BCO=2\times \sin45^\circ=2\times \frac{\sqrt{2}}{2}=\sqrt{2}$. Therefore, the minimum value of BH is $\boxed{\sqrt{2}}$.


Question: The cross-section of a certain flood control embankment is the trapezoid ABCD as shown in the figure, with the dam height $AH=3$ meters and the slope of the water-facing side AB being $i=1\colon3$. What is the length of the slope AB in meters?

Solution:
\textbf{Solution:} Given that the slope of embankment AB has a slope $i=1\colon 3$,\\
$AH=3$ meters,\\
$\therefore BH=9$ meters.\\
By the Pythagorean theorem, it follows that $AB=\sqrt{BH^{2}+AH^{2}}=3\sqrt{10}$ meters.\\
Therefore, the answer is: $AB=\boxed{3\sqrt{10}}$ meters.


Question: As shown in the figure, in a grid where the side length of each small square is 1, the vertices A, B, and C of $\triangle ABC$ all fall on the grid points. What is the value of $\cos \angle ABC$?

Solution:
\textbf{Solution:} As shown in the figure, draw $AD\perp BC$ with the foot of the perpendicular at D,\\
$AD=3$,\\
$BD=4$,\\
$\therefore AB=\sqrt{AD^{2}+BD^{2}}=5$,\\
$\therefore \cos\angle ABC=\frac{BD}{AB}=\boxed{\frac{4}{5}}$.


Question: As shown in the diagram, in rectangle $ABCD$, $AB=4$, $BC=3$. The triangle $\triangle BCD$ is translated along ray $BD$ by a distance $a$ ($a>0$) to obtain $\triangle B'C'D'$. When line segments $AB'$ and $AD'$ are connected, what is the value of $a$ when $\triangle AB'D'$ forms a right-angled triangle?

Solution:
\textbf{Solution:} Consider two cases:\\
(1) As shown in Figure 1, $\angle D'AB' = 90^\circ$, extend C'B' to meet AB at G, and draw D'H$\perp$AB, intersecting the extension of BA at H,\\
$\therefore \angle H = \angle AGB' = \angle BGB' = 90^\circ$,\\
$\because$ quadrilateral ABCD is a rectangle,\\
$\therefore \angle BAD = \angle C = 90^\circ$, AD = BC = 3,\\
$\because \tan \angle ABD = \frac{AD}{AB} = \frac{B'G}{BG}$, thus $\frac{B'G}{BG} = \frac{3}{4}$,\\
Let B'G = 3x, BG = 4x,\\
$\therefore BB' = a = 5x$,\\
By translation: DD' = BB' = 5x,\\
$\therefore D'H = 3 + 3x$, AH = BG = 4x,\\
$\therefore AG = AB = BG = 4 - 4x$,\\
$\because \angle D'AB' = \angle HAD' + \angle BAB' = 90^\circ$,\\
$\angle AD'H + \angle HAD' = 90^\circ$,\\
$\therefore \angle AD'H = \angle GAB'$,\\
$\because \angle H = \angle AGB' = 90^\circ$,\\
$\therefore \triangle D'HA \sim \triangle AGB'$,\\
$\therefore \frac{D'H}{AG} = \frac{AH}{B'G}$, i.e.,\\
$\frac{3 + 3x}{4 - 4x} = \frac{4x}{3x}$,\\
$\therefore x = \frac{7}{25}$,\\
$\therefore a = 5 \times \frac{7}{25} = \boxed{\frac{7}{5}}$;\\
(2) As shown in Figure 2, $\angle AB'D' = 90^\circ$, extend C'B' to meet AB at M, then C'M$\perp$AB,\\
$\therefore \angle AMB' = 90^\circ$,\\
By translation: B'C' = BC = 3,\\
Similarly, let B'M = 3m, BM = 4m, then BB' = a = 5m,\\
$\therefore AM = 4 - 4m$,\\
$\because \angle AB'M + \angle D'B'C' = 90^\circ$, $\angle MAB' + \angle AB'M = 90^\circ$,\\
$\therefore \angle D'B'C' = \angle MAB'$,\\
$\because \angle C' = \angle AMB' = 90^\circ$,\\
$\therefore \triangle D'C'B' \sim \triangle B'MA$,\\
$\therefore \frac{C'D'}{MB'} = \frac{B'C'}{AM}$, i.e.,\\
$\frac{4}{3m} = \frac{3}{4 - 4m}$,\\
$\therefore m = \frac{16}{25}$,\\
$\therefore a = 5m = 5 \times \frac{16}{25} = \boxed{\frac{16}{5}}$;\\
In summary, the value of a is $\frac{7}{5}$ or $\frac{16}{5}$.


Question: In the Cartesian coordinate system $xOy$, a line $y = \frac{1}{2}x + b$ (where $b$ is a constant and $b < 2$) is drawn through point $P(0, 2)$ perpendicular to it, with the foot of the perpendicular at point $Q$. What is the value of $\sin\angle OPQ$?

Solution:
\textbf{Solution}: Given $\because \angle PBQ = \angle ABO$,\\
$PQ \perp AQ$, $BO \perp AO$\\
$\therefore \angle PBQ + \angle OPQ = \angle ABO + \angle OAB$\\
$= 90^\circ$\\
$\therefore \angle OPQ = \angle OAB$\\
From $y=\frac{1}{2}x+b$, setting $x=0$, $y=b$,\\
Setting $y=0$,\\
$x=-2b$\\
$\therefore AO=2b$, $BO=b$\\
In $\triangle AOB$ right,\\
$AB=\sqrt{AO^2 + BO^2} = \sqrt{5}b$\\
$\therefore \sin\angle OPQ =\sin\angle BAO =\frac{BO}{AB} =\frac{b}{\sqrt{5}b} = \boxed{\frac{\sqrt{5}}{5}}$


Question: As shown in the figure, building AD is 30m high. There is a distant tower BC. Someone at the bottom of the building A measures the elevation angle to the top of the tower to be $60^\circ$. After climbing to the top of the building D, the elevation angle to the top of the tower is measured to be $30^\circ$. What is the height of the tower BC in meters?

Solution:
\textbf{Solution:} According to the problem, we have: $BC=\frac{AC}{\tan 30^\circ}=\sqrt{3}AC$,\\
since $BE=DE\tan 30^\circ=AC\tan 30^\circ=\frac{\sqrt{3}}{3}AC$,\\
therefore, the height of the building $AD=BC-BE=\left(\sqrt{3}-\frac{\sqrt{3}}{3}\right)AC=30$.\\
Thus, we find: $AC=15\sqrt{3}$.\\
Hence, $BC=\sqrt{3}AC=\boxed{45}$.


Question: \textbf{Definition:} A convex quadrilateral in which one pair of opposite sides is equal and the other pair is unequal is called an "equilateral quadrilateral". As shown in the figure, in $Rt\triangle PBC$, $\angle PCB=90^\circ$, point A is on side BP, and point D is on side CP. If $BC=11$, $\tan\angle PBC=\frac{12}{5}$, $AB=13$, and quadrilateral ABCD is an "equilateral quadrilateral", then what is the length of CD?

Solution:
\textbf{Solution:} As shown in the diagram, the position of point D is illustrated as follows: \\
(1) If $CD=AB$, at this time point D is at the position $D_{1}$, $CD_{1}=AB=13$; \\
(2) If $AD=BC=11$, at this time point D is at the positions $D_{2}$, $D_{3}$, $AD_{2}=AD_{3}=BC=11$, \\
Through point A, draw $AE\perp BC$, $AF\perp PC$, with E and F as the feet of the perpendiculars, respectively, \\
Let $BE=x$, \\
$\because$ $\tan \angle PBC = \frac{12}{5}$, \\
$\therefore$ $AE=\frac{12}{5}x$, \\
In $\triangle ABE$, $AE^{2}+BE^{2}=AB^{2}$, \\
i.e., $x^{2}+\left(\frac{12}{5}x\right)^{2}=13^{2}$, \\
Solving yields: $x_{1}=5$, $x_{2}=-5$ (discard), \\
$\therefore$ $BE=5$, $AE=12$, \\
$\therefore$ $CE=BC-BE=6$, \\
Since quadrilateral $AECF$ is a rectangle, we have $AF=CE=6$, $CF=AE=12$, \\
In $\triangle AFD_{2}$, $FD_{2}=\sqrt{AD_{2}^{2}-AF^{2}}=\sqrt{85}$, \\
$\therefore$ $CD_{2}=CF-FD_{2}=12-\sqrt{85}$, \\
$CD_{3}=CF+FD_{2}=12+\sqrt{85}$, \\
In summary, the length of CD can be $13$, $12-\sqrt{85}$, or $12+\sqrt{85}$. \\
Hence, the answer is: $\boxed{13}$, $12-\sqrt{85}$, or $12+\sqrt{85}$.


Question: As shown in the rectangle $ABCD$, where $AB=4$ and $AD=8$, points $E$ and $F$ are located on sides $AD$ and $BC$ respectively, with $AE=3$. Perform the following steps: First, fold along line $EF$ such that the corresponding point of $A$ falls exactly on the diagonal $AC$, and the corresponding point of $B$ is $B'$. What is the value of $\tan\angle AEF$? Second, take points $M$ on $EF$ and $N$ on $A'B'$, and continue folding along line $MN$ so that point $F$ coincides with point $E$. What is the length of segment $MN$?

Solution:
\textbf{Solution:} Let point $O$ be the intersection of $AC$ and $EF$,\\
since quadrilateral $ABCD$ is a rectangle,\\
therefore $CD=AB=4$,\\
since along the line $EF$ when folded, the corresponding point of $A$ exactly falls on the diagonal $AC$, and the corresponding point of $B$ is $B$,\\
therefore $\angle AOE=\angle AOE=90^\circ$,\\
therefore $\angle AOE=\angle ADC=90^\circ$,\\
since $\angle EAO=\angle CAD$,\\
therefore $\triangle AOE\sim \triangle ADC$,\\
therefore $\angle AEO=\angle ACD$,\\
therefore $\tan\angle AEF=\tan\angle ACD=\frac{AD}{CD}=\frac{8}{4}=\boxed{2}$,\\
Connect NE, NF, and draw $FT\perp AD$ at point $T$ through point $F$,\\
since $\triangle AOE\sim \triangle ADC$,\\
therefore $\angle FET=\angle ACD$,\\
since $\angle FTE=\angle ADC=90^\circ$,\\
therefore $\triangle FTE\sim \triangle ADC$,\\
therefore $\frac{FT}{AD}=\frac{ET}{CD}$, that is, $\frac{4}{8}=\frac{ET}{4}$,\\
therefore $ET=2$,\\
therefore $BF=AT=AE-ET=3-2=1$,\\
Based on the folding property: $BF=BF=1$, $MN\perp EF$, $NF=NE$,\\
Let $BN=m$,\\
then $NF^2=1^2+m^2=NE^2=3^2+(4-m)^2$,\\
Solving we get: $m=3$,\\
therefore $NF=\sqrt{10}$,\\
since $EF=\sqrt{2^2+4^2}=2\sqrt{5}$,\\
therefore $MF=\sqrt{5}$,\\
therefore $MN=\sqrt{5}$.


Question: As shown in the figure, place $ \angle BAC$ on a $5\times 5$ square grid. If the vertices A, B, and C are all on the grid points, then what is the value of the tangent of $ \angle BAC$?

Solution:
\textbf{Solution:} As shown in the figure, connect BC,\\
then $AB=BC=\sqrt{1^2+3^2}=\sqrt{10}$, $AC=\sqrt{2^2+4^2}=2\sqrt{5}$,\\
$\therefore AB^2+BC^2=10+10=20=AC^2$,\\
$\therefore \triangle ABC$ is an isosceles right triangle, and $\angle ABC=90^\circ$,\\
$\therefore \angle BAC=45^\circ$,\\
then $\tan\angle BAC=\boxed{1}$.


Question: As shown in the diagram, the side $OB$ of the rhombus $OBAC$ lies on the x-axis. The point $A$ is located at $(8,4)$, and $\tan\angle COB=\frac{4}{3}$. If the graph of the inverse proportion function $y=\frac{k}{x}\left(k\neq 0\right)$ passes through point $C$, what is the equation of the inverse proportion function?

Solution:
\textbf{Solution:} Draw a perpendicular from point C to $OB$, intersecting at $M$, as shown in the figure.\\
Since quadrilateral $OBAC$ is a rhombus, with $A(8,4)$,\\
it follows that $CM=4$,\\
Also, since $\tan\angle COB=\frac{CM}{OM}=\frac{4}{3}$,\\
it follows that $OM=3$,\\
Therefore, the coordinates of point C are $C(3,4)$,\\
Since point $C(3,4)$ lies on the graph of the inverse proportion function $y=\frac{k}{x}$,\\
it follows that $k=3\times 4=12$,\\
Therefore, the equation of the inverse proportion function is $\boxed{y=\frac{12}{x}}$.


Question: In the square grid paper shown in the figure, each small quadrilateral is an identical square. Points $A$, $B$, $C$, and $D$ are all located at the grid points, and $AB$ intersects with $CD$ at point $O$. What is the value of $\tan\angle AOC$?

Solution:
\textbf{Solution:} Let $BE$ be connected. We have $OE= \sqrt{1^2+1^2}=\sqrt{2}$ and $BE= \sqrt{3^2+3^2}=3\sqrt{2}$, and $BO= \sqrt{4^2+2^2}=2\sqrt{5}$.\\
Therefore, $BO^2=20=EO^2+EB^2=2+18=20$.\\
Hence, $\triangle OBE$ is a right-angled triangle.\\
Therefore, $\tan\angle AOC=\tan\angle BOE= \frac{BE}{EO}=\frac{3\sqrt{2}}{\sqrt{2}}=\boxed{3}$.


Question: As shown in the figure, the grid is composed of small squares with a side length of 1, arranged in a $4\times 4$ pattern. What is the value of $\tan\angle BAC$?

Solution:
Solution: Connect BC,\\
By Pythagoras' theorem, it is known that:\\
$AC = \sqrt{1^2 + 2^2} = \sqrt{5}$,\\
$BC = \sqrt{2^2 + 4^2} = 2\sqrt{5}$,\\
$AB = \sqrt{3^2 + 4^2} = 5$,\\
Since $5^2 = (\sqrt{5})^2 + (2\sqrt{5})^2$,\\
Hence $AB^2 = AC^2 + BC^2$,\\
Therefore, $\triangle ABC$ is a right-angled triangle,\\
Therefore, $\tan\angle BAC = \frac{BC}{AC} = \frac{2\sqrt{5}}{\sqrt{5}} = \boxed{2}$.


Question: As shown in the figure, circle $ \odot A$ with a radius of 3 passes through the origin $O$ and point $C(0, 2)$. Point $B$ is a point on the major arc of $\odot A$ on the left side of the y-axis. What is the value of $\tan\angle OBC$?

Solution:
\textbf{Solution:} As shown in the diagram, connect CD,\\
$\because$ $\angle$DOC=$90^\circ$,\\
$\therefore$ CD is the diameter of circle A,\\
$\because$ the circle $A$ with radius $3$ passes through the origin $O$ and the point $C(0,2)$,\\
$\therefore$ CD=6, OC=2,\\
$\therefore$ DO=$\sqrt{CD^{2}-OC^{2}}=\sqrt{6^{2}-2^{2}}=4\sqrt{2}$,\\
$\therefore$ $\tan\angle$ODC=$\frac{OC}{OD}=\frac{2}{4\sqrt{2}}=\frac{\sqrt{2}}{4}$,\\
$\because$ $\angle$ODC=$\angle$OBC,\\
$\therefore$ $\tan\angle$OBC=$\boxed{\frac{\sqrt{2}}{4}}$.


Question: As shown in the figure, in $ \square ABCD$, $ \angle D=100^\circ$. AC is the diagonal. If $ \triangle ACD$ is rotated clockwise by a certain angle around point A to obtain $ \triangle AFE$, such that the corresponding point E of point D falls on side AB, and the corresponding point F of point C falls on the extension of side CB, then what is the degree measure of $ \angle EFB$?

Solution:
\textbf{Solution:} In parallelogram $ABCD$, $\angle D=100^\circ$,\\
$\therefore \angle DAB=180^\circ - \angle D=80^\circ$,\\
$\because \triangle ACD$ is rotated clockwise around point $A$ by a certain angle to obtain $\triangle AEF$,\\
$\therefore AF=AC$, $\angle FAE=\angle CAD$, $\angle AFE=\angle ACD$,\\
$\therefore \angle FAC=\angle FAE+\angle BAC=\angle CAD+\angle BAC=\angle BAD=80^\circ$\\
$\therefore \angle AFC=\angle ACF=\frac{1}{2}\left(180^\circ-\angle FAC\right)=50^\circ$\\
$\because AD \parallel BC$,\\
$\therefore \angle DAC=\angle ACF=50^\circ$,\\
$\therefore \angle ACD=180^\circ-\angle D-\angle CAD=180^\circ-100^\circ-50^\circ=30^\circ$,\\
$\therefore \angle AFE=\angle ACD=30^\circ$,\\
$\therefore \angle EFB=\angle AFC-\angle AFE=50^\circ-30^\circ=\boxed{20^\circ}$


Question: As shown in the figure, the coordinates of the two endpoints of line segment AB are A(6, 6) and B(8, 2), respectively. Using the origin O as the center of similarity, and reducing the line segment AB to half of its original size within the first quadrant, what are the coordinates of endpoint C?

Solution:
\textbf{Solution:} Let segments AB have end points A(6,6) and B(8,2). By taking the origin O as a center of similarity and scaling down segment AB to half its size within the first quadrant, we obtain segment CD.\\
Therefore, both the x and y coordinates of endpoint C become half of those of point A,\\
Thus, the coordinates of endpoint C are: $\boxed{(3,3)}$.


Question: As shown in the figure, it is known that the side length of the square $ABCD$ is $a$, and point $E$ is a moving point on side $AB$. Connect $ED$, and rotate $ED$ clockwise by $90^\circ$ around point $E$ to $EF$. Connect $DF$ and $CF$. Then, when the sum of $DF+CF$ takes the minimum value, what is the perimeter of $\triangle DCF$? (Express in an algebraic equation involving $a$)

Solution:
\textbf{Solution:} Draw BF, and through F, draw FG $\perp$ to the extension of AB at point G, \\
since rotating ED clockwise by $90^\circ$ around point E gives EF, \\
therefore EF$\perp$DE, EF=DE, \\
therefore $\angle$DEA + $\angle$FEG = $\angle$DEA + $\angle$ADE = $90^\circ$, \\
therefore $\angle$ADE = $\angle$FEG, \\
therefore $\triangle$AED $\cong$ $\triangle$GFE (AAS), \\
therefore FG=AE, AD=EG, \\
therefore AD=EG=AB, \\
therefore BG=AE=FG, \\
therefore $\angle$CBF=$\angle$GBF+$45^\circ$, i.e., point F moves along the angle bisector of $\angle$CBG. \\
Construct the symmetric point C$^{\prime }$ of C with respect to BF, then point C$^{\prime }$ lies on the extension of AB, hence when D, F, and C are collinear, DF+CF=DC is minimal. \\
since AD=a, AC$^{\prime }$=2a, \\
therefore DC$^{\prime }$= $\sqrt{5}$a, \\
therefore the minimum value of DF+CF is $\sqrt{5}$a, \\
therefore the perimeter of $\triangle$DCF is DF+CF+CD=$\sqrt{5}$a+a=($\sqrt{5}$+1)a. \\
Thus, the answer is: $\boxed{(\sqrt{5}+1)a}$.


Question: As shown in Figure (1), in the Cartesian coordinate system, the rectangle $ABCD$ is in the first quadrant, and $BC\parallel x$-axis. The line $y=2x+1$ translates in the positive direction of the $x$-axis. During the translation process, the length of the segment cut by the line from rectangle $ABCD$ is $a$, and the distance by which the line translates on the $x$-axis is $b$. The graph of the function relationship between $a$ and $b$ is shown in Figure (2). What is the area of rectangle $ABCD$?

Solution:
\textbf{Solution:} As shown in the figure, draw lines parallel to $y=2x+1$ through points $B$ and $D$, intersecting $AD$ and $BC$ at points $E$ and $F$, respectively.\\
From the diagram and the given information, we have $AE=4-3=1$, $CF=8-7=1$, $BE=DF=\sqrt{5}$, and $BF=DE=7-4=3$.\\
Thus, $AB=\sqrt{BE^2-AE^2}=\sqrt{5-1}=2$ and $BC=BF+CF=3+1=4$.\\
Therefore, the area of rectangle $ABCD$ is $AB\cdot BC=2\times 4=\boxed{8}$.


Question: As shown in the figure, in rectangle $ABCD$, $AB=8$, $AD=4$. Point $E$ is a moving point inside rectangle $ABCD$, and $\angle BEC = 90^\circ$. Point $P$ is a moving point on side $AB$. With $PD$ and $PE$ connected, what is the minimum value of $PD+PE$?

Solution:
Solution: Let the point O be the midpoint of BC. According to the problem statement, it is known that point E moves on a semicircle O with BC as the diameter.\\
Construct the symmetric figure of semicircle O and line segment BC with respect to AB (semicircle O'), where the symmetric point of O is O' and the symmetric point of E is E'.\\
Connect O'E' and PE'. Then, PE=PE'.\\
It is easy to see that when points D, P, E', and O' are collinear, the sum of PD+PE is minimized and equals the length of DE'.\\
As shown in the figure,\\
in $\triangle DCO'$, we have $CD=AB=8$, $CO'=6$,\\
$\therefore DO'=10$\\
Furthermore $\because O'E'=2$\\
$\therefore DE'=DO'-O'E'=8$, that is, the minimum value of $PD+PE$ is $\boxed{8}$


Question: As shown in the figure, in the square $ABCD$ with side length 2, point $E$ is the midpoint of $AD$. Connect $BE$ and fold $\triangle ABE$ along $BE$ to obtain $\triangle FBE$. $BF$ intersects $AC$ at point $G$. What is the length of $CG$?

Solution:
\textbf{Solution:} Extend $BF$ to meet $CD$ at $H$, and connect $EH$,\\
since $\triangle FBE$ is obtained by folding $\triangle ABE$ along $BE$,\\
therefore $EA=EF$, $\angle EFB=\angle EAB=90^\circ$,\\
since $E$ is the midpoint of $AD$, and the side length of square $ABCD$ is $2$,\\
therefore $EA=ED=1$,\\
therefore $ED=EF=1$,\\
since quadrilateral $ABCD$ is a square,\\
therefore $\angle D=\angle EFB=\angle EFH=90^\circ$,\\
in $Rt\triangle EDH$ and $Rt\triangle EFH$,\\
$\left\{\begin{array}{l}ED=EF\\ EH=EH\end{array}\right.$,\\
therefore $Rt\triangle EDH \cong Rt\triangle EFH\ (HL)$,\\
therefore $\angle DEH=\angle FEH$,\\
also since $\angle AEB=\angle FEB$,\\
therefore $\angle DEH+\angle AEB=90^\circ$,\\
since $\angle ABE+\angle AEB=90^\circ$,\\
therefore $\angle ABE=\angle DEH$,\\
therefore $\triangle DHE \sim \triangle AEB$,\\
therefore $\frac{DH}{DE}=\frac{AE}{AB}=\frac{1}{2}$,\\
therefore $DH=\frac{1}{2}$,\\
therefore $CH=CD−DH=2−\frac{1}{2}=\frac{3}{2}$,\\
since $CH\parallel AB$,\\
therefore $\triangle HGC \sim \triangle BGA$,\\
therefore $\frac{CG}{AG}=\frac{CH}{AB}=\frac{3}{4}$,\\
therefore $CG=\frac{3}{4}AG=\frac{3}{4}(AC−CG)$,\\
since $AB=2$, $CB=2$, $\angle CBA=90^\circ$,\\
therefore $AC=2\sqrt{2}$,\\
therefore $CG=\frac{3}{4}(2\sqrt{2}−CG)$,\\
therefore $CG=\boxed{\frac{6}{7}\sqrt{2}}$.


Question: As shown in the rectangle $ABCD$, it is known that $AD=18$ and $AB=24$. Point $E$ is a moving point on side $DC$, and $\triangle AD'E$ and $\triangle ADE$ are symmetrical with respect to line $AE$. When $\triangle CD'E$ is a right-angled triangle, what is the length of $DE$?

Solution:
\textbf{Solution:}

(1) When $\angle CED'=90^\circ$, as shown in Figure (1),\\
$\because \angle CED'=90^\circ$,\\
according to the property of axial symmetry, $\angle AED=\angle AED'=\frac{1}{2}\times90^\circ=45^\circ$,\\
$\because \angle D=90^\circ$,\\
$\therefore \triangle ADE$ is an isosceles right triangle,\\
$\therefore DE=AD=18$;\\
(2) When $\angle ED'A=90^\circ$, as shown in Figure (2),\\
according to the property of axial symmetry, $\angle AD'E=\angle D=90^\circ$, $AD'=AD$, $DE=D'E$,\\
$\triangle CD'E$ is a right triangle,\\
that is $\angle CD'E=90^\circ$,\\
$\therefore \angle AD'E+\angle CD'E=180^\circ$,\\
$\therefore A$, $D'$, $C$ are collinear,\\
according to the Pythagorean theorem, $AC=\sqrt{AD^{2}+CD^{2}}=30$,\\
$\therefore CD'=30-18=12$,\\
let $DE=D'E=x$, then $EC=CD-DE=24-x$,\\
in $\triangle D'EC$, $D'E^{2}+D'C^{2}=EC^{2}$,\\
that is $x^{2}+144=(24-x)^{2}$,\\
solving for $x$ gives $x=9$,\\
that is $DE=9$;\\
In summary: The length of $DE$ is either 9 or 18;\\
Hence, the answer is: \boxed{9 \text{ or } 18}.


Question: As illustrated, $ \triangle ABC$ and $ \triangle A_1B_1C_1$ are similar with the center of similarity being point $O$. Given that $OA: OA_1 = 1:2$ and the area of $ \triangle ABC$ is 3, what is the area of $ \triangle A_1B_1C_1$?

Solution:
\textbf{Solution:} Since $\triangle ABC \sim \triangle A_1B_1C_1$,\\
it follows that $\triangle ABC \sim \triangle A_1B_1C_1$, $AC \parallel A_1C_1$,\\
thus $\triangle AOC \sim \triangle A_1OC_1$,\\
therefore $\frac{AC}{A_1C_1} = \frac{OA}{OA_1} = \frac{1}{2}$,\\
thus, the area ratio of $\triangle ABC$ to $\triangle A_1B_1C_1$ is $1\colon 4$,\\
since the area of $\triangle ABC$ is $3$,\\
the area of $\triangle A_1B_1C_1$ is $\boxed{12}$.


Question: As shown in the figure, rotate the rectangle $ABCD$ counterclockwise around point $A$ by an angle of $\alpha$ $(0^\circ<\alpha<90^\circ)$, and connect $EC$ and $ED$. At what value of $\alpha$ will $EC=ED$?

Solution:
\textbf{Solution:} Draw $BE$, and through point $E$, draw $EH \perp CD$ at point $H$, intersecting $AB$ at point $M$, as shown in the diagram below,\\
To ensure $EC=ED$, then $EH\perp CD$, and $CH=CD$,\\
$\because AB\parallel CD$,\\
$\therefore EF\perp AB$,\\
$\therefore \angle BMH=\angle AMH=\angle MBC=\angle BCH=\angle ADH=\angle MAD=90^\circ$,\\
$\therefore$ Quadrilateral $ADHM$ and quadrilateral $BCHM$ are both rectangles,\\
$\therefore BM=CH=DH=AM$,\\
$\therefore EM$ bisects $AB$ perpendicularly,\\
$\therefore AE=BE$,\\
By the property of rotation, $AB=AE$,\\
$\therefore AB=AE=BE$,\\
$\therefore \triangle ABE$ is an equilateral triangle,\\
$\therefore \angle BAE=60^\circ$,\\
Therefore, when $\alpha$ is $\boxed{60^\circ}$, then $EC=ED$.


Question: As shown in the figure, an equilateral triangle $ABC$ with a side length of 4 is translated along ray $BC$ to obtain triangle $DEF$. Points $M$ and $N$ are the midpoints of $AC$ and $DF$, respectively. Point $P$ is the midpoint of segment $MN$. When connecting $PA$ and $PC$, if triangle $APC$ forms a right-angled triangle, what is the length of $BE$?

Solution:
\textbf{Solution:} As shown in Figure 1, when $\angle APC=90^\circ$, \\
$\because$AM=MC, AC=4,\\
$\therefore$PM is the median to the hypotenuse of $\triangle APC$,\\
$\therefore$AM=CM=PM=2,\\
$\therefore$PN=2,\\
$\therefore$MN=4,\\
Thus, translating $\triangle ABC$ 4 units to the right is sufficient,\\
$\therefore$BE=\boxed{4};\\
As shown in Figure 2, when $\angle ACP=90^\circ$,\\
$\because$$\triangle ABC$ is an equilateral triangle, AM=MC,\\
$\therefore$$\angle BMC=90^\circ$,\\
$\therefore$$\angle BMC=\angle ACP$,\\
$\therefore$BM$\parallel$CP,\\
$\because$$\triangle ABC$ is an equilateral triangle, $\triangle DEF$ is an equilateral triangle, M and N are the midpoints of AC and DF respectively,\\
$\therefore$$\angle ACB=\angle DFE= 60^\circ$, CM=NF,\\
$\therefore$MC$\parallel$NF,\\
$\therefore$Quadrilateral MCFN is a parallelogram,\\
$\therefore$MP$\parallel$BC,\\
$\therefore$Quadrilateral BCPM is a parallelogram,\\
$\therefore$PM=4,\\
$\therefore$PN=4,\\
$\therefore$MN=8,\\
Thus, translating $\triangle ABC$ 8 units to the right is sufficient,\\
$\therefore$BE=\boxed{8};


Question: As shown in the figure, points $P$, $Q$, and $R$ are located on the graph of the inverse proportion function $y=\frac{k}{x}$ (where the constant $k>0$ and $x>0$), respectively. Lines parallel to the $x$-axis and the $y$-axis through these three points are drawn, forming shaded regions in the figure with areas denoted as ${S}_{1}$, ${S}_{2}$, and ${S}_{3}$ from left to right. If $OE=ED=DC$ and ${S}_{1}{S}_{3}=27$, what is the value of ${S}_{2}$?

Solution:
\textbf{Solution:} Let us assume $CD=DE=OE=a$,\\
Hence, $P\left(\frac{k}{3a}, 3a\right)$, $Q\left(\frac{k}{2a}, 2a\right)$, $R\left(\frac{k}{a}, a\right)$,\\
Thus, $CP=\frac{k}{3a}$, $DQ=\frac{k}{2a}$, $ER=\frac{k}{a}$,\\
Therefore, $OG=AG$, $OF=2FG$, $OF=\frac{2}{3}GA$,\\
Consequently, $S_{1}=2S_{2}$, $S_{3}=3S_{2}$,\\
Given $S_{1}+S_{3}=27$,\\
It implies $2S_{2}+3S_{2}=27$,\\
Hence, $S_{2}=\boxed{\frac{27}{5}}$.


Question: As shown in the figure, the graph of the inverse proportion function $y=\frac{1}{x}$ intersects with the graph of the direct proportion function $y=x$ at points A and C. $AB\perp x$-axis and $CD\perp x$-axis, with the perpendicular feet at points B and D, respectively. What is the area of the quadrilateral ABCD?

Solution:
\textbf{Solution:} Solving the system of equations for the two functions gives \(\begin{cases}y=\frac{1}{x}, \\ y=x, \end{cases}\) which results in \(\begin{cases}x=1, \\ y=1, \end{cases}\) or \(\begin{cases}x=-1, \\ y=-1. \end{cases}\)

\(\therefore A(1, 1), C(-1, -1).\)

\(\because AB\perp x\) axis, \(CD\perp x\) axis,

\(\therefore B(1, 0), D(-1, 0),\)

\(\therefore BD=2, AB=CD=1.\)

\(\because AB\perp BD, CD\perp BD\)

\(\therefore AB\parallel CD\)

\(\therefore\) Quadrilateral ABCD is a parallelogram,

\(\therefore\) The area of quadrilateral ABCD \(=AB\cdot BD=1\times 2=\boxed{2}\).


Question: As shown in the figure, the graph of the inverse proportion function $y=\frac{k}{x}$ $(x>0)$ passes through points A and B. Draw $AC\perp y$ axis at point C through A, and draw $BD\perp y$ axis at point D through B. Also, draw $BE\perp x$ axis at point E through B, and connect AD. It is known that $AC=1$, $BE=1$, and the area of rectangle $BDOE$ is $4$. Then, what is the area of $\triangle ACD$?

Solution:
\textbf{Solution:} Construct AH $\perp$ x-axis at point H, intersecting BD at point F, \\
$\therefore$ Quadrilateral ACOH, quadrilateral ACDF, quadrilateral ODBE are all rectangles, \\
$\because$ The area of rectangle BDOE is 4, and the inverse proportion function $y=\frac{k}{x}$ (x＞0) passes through point B, with k＞0, \\
$\therefore$k＝4 \\
$\therefore$The area of rectangle ACOH is 4, \\
$\because$AC＝1 \\
$\therefore$OC＝4÷1＝4, \\
$\therefore$CD＝OC−OD＝OC−BE＝4−1＝3, \\
$\therefore$The area of rectangle ACDF＝1×3＝3, \\
$\therefore$${S}_{\triangle ACD}=\frac{1}{2}{S}_{\text{rectangle }CDFA}=\boxed{\frac{3}{2}}$.


Question: As shown in the figure, in the Cartesian coordinate system, given a rhombus OABC, where the coordinates of point $A$ are $(10, 0)$, and the hyperbola $y=\frac{k}{x} (x>0)$ passes through point $C$, and $OB \cdot AC = 160$, what is the value of $k$?

Solution:
\textbf{Solution:} Construct $CD\perp OA$ through point $C$.\\
Since quadrilateral OABC is a rhombus,\\
and $OB\cdot AC=160$,\\
therefore, the area of the rhombus OABC is 80,\\
which means $OA\cdot CD=80$.\\
Since $OA=OC=10$\\
therefore, $CD=8$.\\
In $\triangle OCD$, $OC=10$, $CD=8$, \\
therefore, $OD=\sqrt{10^{2}-8^{2}}=6$, that is $C(6, 8)$,\\
thus the value of $k$ is $\boxed{48}$.


Question: As shown in the figure, in the rhombus $ABCD$, $\angle C = 108^\circ$, the perpendicular bisector of $AD$ intersects the diagonal $BD$ at point $P$, and the foot of the perpendicular is $E$. Connecting $AP$, what is the measure of $\angle APB$?

Solution:
\textbf{Solution:} Given: Since $ABCD$ is a rhombus,\\
therefore $\angle C = \angle DAB = 108^\circ$, and $AD = AB$\\
therefore $\angle ADP = (180^\circ - 108^\circ) \div 2 = 36^\circ$,\\
since the perpendicular bisector of $AD$ intersects diagonal $BD$ at point $P$,\\
therefore $AP = DP$,\\
therefore $\angle ADP = \angle DAP = 36^\circ$,\\
therefore $\angle APB = \angle ADP + \angle DAP = 36^\circ + 36^\circ = \boxed{72^\circ}$.


Question: As shown in the figure, in square $ABCD$, $E$ is any point on diagonal $BD$. A line is drawn through point $E$ so that $EF\perp BC$ at point $F$, and $EG\perp CD$ at point $G$. If the perimeter of the square $ABCD$ is $a$, what is the perimeter of quadrilateral $EFCG$?

Solution:
\textbf{Solution:} Given square $ABCD$, \\
$\therefore \angle BDC=45^\circ$, $\angle C=90^\circ$, \\
$\because$ $EF \perp BC$, $EG \perp CD$, \\
$\therefore \angle C = \angle CGE = \angle CFE = \angle DGE = 90^\circ$, \\
$\therefore \angle EDG = \angle DEG = 45^\circ$, quadrilateral $CGEF$ is a rectangle, \\
$\therefore EG = DG$, \\
$\therefore CG+EG=CG+DG=DC$ \\
$\because$ The perimeter of square $ABCD$ is $a$, \\
$\therefore$ The perimeter of quadrilateral $EFCG$ is $2DC = \boxed{\frac{1}{2}a}$.


Question: As shown in the figure, for square $ABCD$ with side length $3$, point $E$ is on side $AB$, and $BE=1$. If point $P$ moves along diagonal $BD$, what is the minimum value of $PA+PE$?

Solution: \textbf{Solution:} Given the figure, since ABCD is a square, thus points A and C are symmetrical about BD. Draw line CE intersecting BD at point P, and connect AP. Therefore, $AP=CP$, which implies $AP+PE=CP+PE=CE$. Hence, the minimum value of $AP+PE$ is the length of CE; in $\triangle BCE$, we have $CE=\sqrt{BE^2+BC^2}=\sqrt{3^2+1^2}=\sqrt{10}$. Consequently, the answer is $\boxed{\sqrt{10}}$.


Question: As shown in the figure, in the isosceles right-angled triangle $OAB$ with a hypotenuse length of 1, an inscribed square ${A}_{1}{B}_{1}{C}_{1}{D}_{1}$ is constructed (with the vertices of the square lying on the sides of $\triangle OAB$). In the isosceles right-angled triangle $O{A}_{1}{B}_{1}$, another inscribed square ${A}_{2}{B}_{2}{C}_{2}{D}_{2}$ is constructed; in the isosceles right-angled triangle $O{A}_{2}{B}_{2}$, an inscribed square ${A}_{3}{B}_{3}{C}_{3}{D}_{3}$ is constructed; and so on. Continuing this process, what is the side length of the fifth square ${A}_{5}{B}_{5}{C}_{5}{D}_{5}$?

Solution:
\textbf{Solution:} Given an isosceles right triangle OAB with a hypotenuse of 1, constructing a square $A_{1}B_{1}C_{1}D_{1}$,

\(\therefore \angle A = \angle B = \angle AA_{1}D_{1} = \angle BB_{1}C_{1} = 45^\circ\), AO=BO, $A_{1}D_{1} = D_{1}C_{1} = B_{1}C_{1}$,

\(\therefore AD_{1} = A_{1}D_{1} = D_{1}C_{1} = A_{1}B_{1} = C_{1}B\),

\(\therefore\) The edge length of the square $A_{1}B_{1}C_{1}D_{1}$ is $A_{1}B_{1} = \frac{1}{3}AB = \frac{1}{3}\);

Similarly, the edge length of the square $A_{2}B_{2}C_{2}D_{2}$ is $A_{2}B_{2} = \left(\frac{1}{3}\right)^{2}$;

The edge length of the square $A_{3}B_{3}C_{3}D_{3}$ is $A_{3}B_{3} = \left(\frac{1}{3}\right)^{3}$;

\(\therefore\) The edge length of the 5th square $A_{5}B_{5}C_{5}D_{5}$ is \(\boxed{\left(\frac{1}{3}\right)^{5}}\).


Question: As shown in the figure, rectangle $ABCD$ is folded along crease $EF$. The corresponding edge of $BC$, denoted as $B'C'$, intersects $CD$ at point $M$. If $\angle BMD = 50^\circ$, then what is the measure of $\angle BEF$ in degrees?

Solution:
\textbf{Solution:} Since we fold the rectangle ABCD along the crease EF,\\
it follows that $\angle C^{\prime} = \angle A = 90^\circ$, $\angle DMB^{\prime} = \angle C^{\prime}MF = 50^\circ$, $\angle EFC = \angle C^{\prime}FE$, and $DC \parallel AB$\\
therefore, $\angle C^{\prime}FM = 90^\circ - 50^\circ = 40^\circ$, $\angle MFE = \angle FEB$, and $\angle CFE + \angle FEB = 180^\circ$,\\
thus, $\angle C^{\prime}FE = 40^\circ + \angle MFE$, $\angle EFC = 180^\circ - \angle MFE$,\\
it leads to $40^\circ + \angle MFE = 180^\circ - \angle MFE$,\\
Solving gives: $\angle MFE = 70^\circ$,\\
hence, $\angle BEF = \boxed{70^\circ}$.


Question: As shown in the figure, ABCD is a square with a side length of 4. Point H is on the extension of CD, and CEFH is also a square. What is the area of $\triangle DBF$?

Solution:
\textbf{Solution:} Let the side length of the square $CEFH$ be $a$,

according to the problem, we have: $S_{\triangle BDF}=S_{\text{square }ABCD}+S_{\text{square }CEFH}-S_{\triangle ABD}-S_{\triangle DHF}-S_{\triangle BEF}$,

$=4^2+a^2 -\frac{1}{2}\times4\times4-\frac{1}{2}a(a-4)-\frac{1}{2}a(a+4) =\boxed{8}$.


Question: As shown in the figure, place two strips of paper with parallel opposite sides and equal widths crosswise at will. By rotating one of them, the overlapping part forms a quadrilateral. What can this quadrilateral be?

Solution:
\textbf{Solution:} As shown in the figure, draw heights AE and AF from point D to sides AB and BC, respectively,\\
since quadrilateral ABCD is formed by overlapping two strips of equal width,\\
therefore AB$\parallel$CD, and AD$\parallel$BC,\\
therefore, quadrilateral ABCD is a parallelogram,\\
since DE$\perp$AB, and DF$\perp$BC,\\
therefore DE=DF,\\
since the area of parallelogram ABCD, $S_{\text{parallelogram } ABCD}$=AB$\cdot$DE=BC$\cdot$DF,\\
therefore AB=BC,\\
therefore, quadrilateral ABCD is a rhombus.\\
Hence, the answer is: Quadrilateral ABCD is a \boxed{rhombus}.


Question: As shown in the diagram, in quadrilateral $ABCD$, $AB= BC$, and $\angle ABC=\angle CDA=90^\circ$, $BE\perp AD$ at point $E$, and the area of quadrilateral $ABCD$ is 8. What is the length of $BE$?

Solution:
\textbf{Solution:} As shown in the diagram, draw line BF $\perp$ CD, intersecting the extension of DC at point F, \\
$\because$ BE $\perp$ AD, \\
$\therefore$ $\angle$AEB = $\angle$BED = $\angle$F = 90$^\circ$, \\
$\because$ $\angle$CDA = 90$^\circ$, \\
$\therefore$ Quadrilateral DEBF is a rectangle, \\
$\therefore$ $\angle$EBF = $\angle$ABC = 90$^\circ$, \\
$\therefore$ $\angle$ABE = $\angle$CBF, \\
$\because$ AB = BC, \\
$\therefore$ $\triangle$BCF $\cong$ $\triangle$BAE (AAS), \\
$\therefore$ BE = BF, \\
$\therefore$ Quadrilateral DEBF is a square, \\
$\therefore$ Area of Quadrilateral ABCD = Area of Square BEDF = 8, \\
$\therefore$ BE = \boxed{2\sqrt{2}}.


Question: As shown in the figure, $O$ is the intersection point of the diagonals of rectangle $ABCD$, $DE \parallel AC$, $CE \parallel BD$, and $DE$ intersects $CE$ at point $E$. Given that $AB=4$ and $AD=6$, what is the perimeter of quadrilateral $OCED$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
it follows that $\angle BAD=90^\circ$, OB=OD=OC=OA,\\
thus BD= $\sqrt{AB^{2}+AD^{2}}= 2\sqrt{13}$,\\
since OD= $\frac{1}{2}$BD= $\sqrt{13}$,\\
and DE$\parallel$ AC, CE$\parallel$ BD,\\
it follows that quadrilateral OCED is a parallelogram,\\
since OC=OD,\\
thus quadrilateral OCED is a rhombus,\\
hence the perimeter of quadrilateral OCED $=4$OD$= 4\sqrt{13}=\boxed{4\sqrt{13}}$.


Question: As shown in the figure, the diagonals $AC$ and $BD$ of the rhombus $ABCD$ intersect at point $O$. Given that $AC=8$ and $BD=6$, what is the length of the side $AB$ of the rhombus $ABCD$? What is its area? If a perpendicular line $OH$ is drawn from point $O$ to side $AB$, with the foot of the perpendicular being point $H$, what is the distance $OH$ from point $O$ to side $AB$?

Solution: \textbf{Solution:} Since quadrilateral $ABCD$ is a rhombus, $\therefore$ $AC\perp BD$, $OB=OD=3$, $OA=OC=4$. Therefore, $AB= \sqrt{OA^{2}+OB^{2}}=\boxed{5}$, $S_{\text{rhombus}ABCD}= \frac{1}{2}AC\cdot BD=\boxed{24}$, $S_{\triangle AOB}= \frac{1}{2}OA\cdot OB= \frac{1}{2}AB\cdot OH$. Therefore, $OH= \frac{OA\cdot OB}{AB}=\boxed{2.4}$.


Question: As shown in the diagram, it is known that in rhombus ABCD, one interior angle $\angle BAD=80^\circ$, the diagonals AC and BD intersect at point O, and point E is on AB, with BE=BO. What is the measure of $\angle EOA$ in degrees?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\
it follows that AC$\perp$BD, $\angle$BAO= $\frac{1}{2}\angle$BAD=$40^\circ$,\\
hence $\angle$ABO=$90^\circ$−$\angle$BAO=$50^\circ$,\\
since BE=BO,\\
it follows that $\angle$BOE= $\frac{180^\circ−\angle ABO}{2}$=$65^\circ$,\\
therefore $\angle$EOA=$\angle$AOB−$\angle$BOE=$90^\circ−65^\circ=\boxed{25^\circ}$.


Question: As shown in the figure, in rectangle $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. If $\angle DOC = 110^\circ$, what is the measure of $\angle DAC$ in degrees?

Solution:
\textbf{Solution}: Given rectangle ABCD,\\
hence DO=AO,\\
thus $\angle$ADO=$\angle$DAO,\\
since $\angle$DOC=110$^\circ$,\\
thus 2$\angle$DAO=110$^\circ$, i.e., $\angle$DAO=55$^\circ$,\\
therefore, $\angle$DAC=\boxed{55^\circ}.


Question: As shown in the figure, in rectangle $ABCD$, $AE \perp BD$ at point $E$, and $\angle DAE = 2\angle BAE$, what is the measure of $\angle EAC$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\
it follows that OA=OC=OB=OD, and $\angle$BAD=$90^\circ$,\\
therefore, $\angle$ABO=$\angle$BAO,\\
given that $\angle$DAE=2$\angle$BAE, and $\angle$DAE+$\angle$BAE=$90^\circ$,\\
it follows that $\angle$BAE=$30^\circ$,\\
given that AE$\perp$BD, that is $\angle$AEB=$90^\circ$,\\
it follows that $\angle$ABO=$60^\circ$,\\
therefore, $\angle$BAO=$60^\circ$,\\
therefore, $\angle$EAC=$\angle$BAO−$\angle$BAE=$60^\circ−30^\circ=\boxed{30^\circ}$.


Question: As shown in the figure, in rectangle $ABCD$, $E$ is the midpoint of $AD$, and $AE = 1\,\text{cm}$. The perpendicular bisector of $BE$, $MN$, exactly passes through point $C$. What is the length of $AB$ in centimeters?

Solution:
\textbf{Solution:} As shown in the figure, connect EC, and BE intersects MN at point F, \\
since rectangle ABCD, E is the midpoint of AD, and AE=1cm, \\
therefore AB=CD, AD=BC=2AE=2cm, ED=1cm, \\
since FC is perpendicular bisector of BE, \\
therefore EC=BC=2cm, $\angle$BFC=$\angle$EFC=$90^\circ$, \\
therefore in Rt$\triangle$EFC, CD= $\sqrt{EF^{2}-ED^{2}}=\sqrt{2^{2}-1^{2}}=\sqrt{3}$, \\
therefore AB=CD= $\boxed{\sqrt{3}}$cm.


Question: As shown in the diagram, in the Cartesian coordinate system, the points are \(A_{1}(1,0)\), \(A_{2}(3,0)\), \(A_{3}(6,0)\), \(A_{4}(10,0)\), and so forth. The first square \(A_{1}C_{1}A_{2}B_{1}\) is constructed with \(A_{1}A_{2}\) as its diagonal, the second square \(A_{2}C_{2}A_{3}B_{2}\) with \(A_{2}A_{3}\) as its diagonal, the third square \(A_{3}C_{3}A_{4}B_{3}\) with \(A_{3}A_{4}\) as its diagonal, and so on, with vertices \(B_{1}\), \(B_{2}\), \(B_{3}\), etc., all in the first quadrant. Following this pattern, what are the coordinates of the point \(B_n\)?

Solution:
\textbf{Solution}: Draw perpendiculars from points $B_1$, $B_2$, $B_3$ to the x-axis at points $D$, $E$, $F$ respectively, such that $B_1D\perp x$-axis, $B_2E\perp x$-axis, $B_3F\perp x$-axis, Since $A_1(1,0)$,\\
Therefore $A_1A_2=3-1=2$, $A_1D=1$, $OD=2$,\\
$B_1D=A_1D=1$,\\
We obtain $B_1(2,1)$, Since $A_2(3,0)$,\\
Therefore $A_3A_2=6-3=3$,\\
$EB_2=\frac{3}{2}$,\\
$B_2E=EA_2=\frac{3}{2}$,\\
$OE=6-\frac{3}{2}=\frac{9}{2}$,\\
We obtain $B_2\left(\frac{9}{2},\frac{3}{2}\right)$,\\
Similarly, we find: $B_3(8,2)$, $B_4\left(\frac{25}{2},\frac{5}{2}\right)$, $\dots$,\\
Since the x-coordinates of $B_1$, $B_2$, $B_3$, $\dots$ are respectively: $\frac{4}{2}$, $\frac{9}{2}$, $\frac{16}{2}$, $\frac{25}{2}\dots$,\\
Therefore, the x-coordinate of point $B_n$ is: $\frac{(n+1)^2}{2}$,\\
Since the y-coordinates of $B_1$, $B_2$, $B_3$, $\dots$ are respectively: 1, $\frac{3}{2}$, 2, $\frac{5}{2}$, $\dots$,\\
Therefore, the y-coordinate of point $B_n$ is: $\frac{n+1}{2}$,\\
Therefore, the coordinates of point $B_n$ are: $\left[\frac{(n+1)^2}{2},\frac{n+1}{2}\right]$.\\
Hence, the answer is: $\boxed{\left[\frac{(n+1)^2}{2},\frac{n+1}{2}\right]}$.


Question: As shown in the figure, the quadrilateral $OABC$ is a rectangular sheet placed in the Cartesian coordinate system, where $O$ is the origin, point $A$ is on the positive semi-axis of the $x$-axis, and point $C$ is on the positive semi-axis of the $y$-axis, with $OA=5$; $OC=4$. A point $D$ is taken on side $OC$, and the sheet is folded along $AD$ so that point $O$ falls on point $E$ on side $BC$. What are the coordinates of point $D$?

Solution:
\textbf{Solution:} Since quadrilateral OABC is a rectangle, with OA=5 and OC=4.

Hence, $OA=BC=5$, $OC=AB=4$,

$\angle AOC=\angle B=\angle OCB=90^\circ$,

By the property of folding, we get: OA=AE=5, OD=DE,

In $\triangle ABE$, $BE=\sqrt{AE^2-AB^2}=3$,

Therefore, $CE=BC-BE=2$,

Let D$(0,x)$, then OD=DE=$x$, CD=$4-x$,

Therefore, in $\triangle DCE$, by the Pythagorean theorem, we get: $(4-x)^2+2^2=x^2$,

Solving this gives: $x=\frac{5}{2}$,

Hence, $D\boxed{(0,2.5)}$.


Question: As shown in the figure, a square is placed on a desk. What is the side length of this square?

Solution:
\textbf{Solution:} As shown in the diagram, \\
$\because$ $\angle B = \angle D = \angle ACE = 90^\circ$, \\
$\angle BAC + \angle ACB = 90^\circ$, $\angle ECD + \angle ACB = 90^\circ$, \\
$\therefore$ $\angle BAC = \angle ECD$, \\
$\because$ AC = CE, \\
$\therefore$ $\triangle ABC \cong \triangle CBE$, \\
$\therefore$ BC = DE = 2, \\
$\therefore$ AC = $\sqrt{AB^2 + BC^2} = \sqrt{1^2 + 2^2} = \sqrt{5}$, \\
$\therefore$ the length of the side of the square is $\boxed{\sqrt{5}}$.


Question: As shown in the figure, $C$ is a moving point on line segment $BD$, with perpendiculars $AB \perp BD$ and $ED \perp BD$ drawn through points $B$ and $D$, respectively. The lines $AC$ and $EC$ are also drawn. Given that $AB=5$, $DE=1$, and $BD=8$, and letting $CD=x$, what is the minimum value of $AC+CE$?

Solution:
\textbf{Solution:} As shown in the diagram: \\
When points A, C, and E are collinear, AC + CE is minimized;\\
When points A, C, and E are on the same line,\\
extend AB and construct line EF $\perp$ AB at point F,\\
$\because$AB = 5, DE = 1,\\
$\therefore$AF = 6,\\
$\because$$\angle$ABD = $90^\circ$,\\
$\therefore$$\angle$FBD = $90^\circ$,\\
$\because$$\angle$BDE = $\angle$BFE = $90^\circ$,\\
$\therefore$Quadrilateral BFED is a rectangle,\\
$\therefore$BD = EF = 8,\\
$\therefore$AE = $ \sqrt{AF^{2} + EF^{2}}$ = $ \sqrt{6^{2} + 8^{2}}$ = \boxed{10}.


Question: As shown in the figure, five squares with a side length of 1cm are arranged as illustrated. The points $A_1$, $A_2$, ..., $A_n$ represent the centers of the squares, respectively. What is the total area of the overlapping regions formed by the overlap of the five squares?

Solution:
As shown in the diagram, through the center O of the square ABCD, draw OM$\perp$CD at M, and draw ON$\perp$BC at N,\\
then $\angle EOM=\angle FON$, and OM=ON,\\
in $\triangle OEM$ and $\triangle OFN$,\\
$\left\{\begin{array}{l}
\angle OME=\angle ONF \\
OM=ON \\
\angle EOM=\angle FON
\end{array}\right.$,\\
$\therefore \triangle OEM\cong \triangle OFN$ (ASA),\\
thus, the area of the quadrilateral OECF is equal to the area of the square OMCN,\\
if the side length of the square ABCD is 1, then the area of OMCN is $\frac{1}{4}\text{cm}^2$,\\
$\therefore$ the area of the shaded part is equal to $\frac{1}{4}$ of the square's area,\\
$\therefore$ the total area of the shaded parts (overlaps) in 5 such squares is $\frac{1}{4} \times 4 = \boxed{1\text{cm}^2}$.


Question: As shown in the figure, DE is the median of $\triangle ABC$, the angle bisector of $\angle ABC$ intersects DE at point F, $AB=8$, $BC=12$, then what is the length of $EF$?

Solution:
\textbf{Solution:} Given that $DE$ is the median line of $\triangle ABC$,\\
it follows that $DE \parallel BC$ and $DE = \frac{1}{2}BC = 6$, $BD = AD = \frac{1}{2}AB = 4$,\\
thus $\angle DFB = \angle FBC$,\\
since $BF$ bisects $\angle ABC$,\\
it follows that $\angle DBF = \angle FBC$,\\
thus $\angle DFB = \angle DBF$,\\
therefore $DF = BD = 4$,\\
hence $EF = DE - DF = 6 - 4 = \boxed{2}$


Question: As shown in the figure, in $\triangle ABC$, points D, E, and F are the midpoints of the sides, respectively. If the area of $\triangle ABC$ is $16\,cm^2$, then what is the area of $\triangle DEF$ in $cm^2$?

Solution:
\textbf{Solution:}

Since points D and F are the midpoints of AB and AC, respectively, \\
$\therefore$DF$\parallel $BC and DF$ = \frac{1}{2}$BC, \\
$\therefore$DF$\parallel $BE, \\
Since E is the midpoint of BC, \\
$\therefore$BE$ = \frac{1}{2}$BC, \\
$\therefore$DF$ = $BE, \\
$\therefore$Quadrilateral BEFD is a parallelogram, \\
$\therefore$BD$ = $EF, \\
In $\triangle $BDE and $\triangle $FED, \\
$ \left\{\begin{array}{l}DF=BE \\ BD=EF \\ DE=DE \end{array}\right.$ \\
$\therefore$$\triangle $BDE$\cong $$\triangle $FED \text{ (SSS)}, \\
Similarly, it can be proven that $\triangle $DAF$\cong $$\triangle $FED and $\triangle $EFC$\cong $$\triangle $FED, \\
i.e., $\triangle $BDE$\cong $$\triangle $DAF$\cong $$\triangle $EFC$\cong $$\triangle $FED, \\
$\therefore$Area of $\,\triangle $DEF$ = \frac{1}{4}$Area of $\,\triangle $ABC$ = \frac{1}{4} \times 16 = \boxed{4} \text{ (cm}^2\text{)}$.