Question: As shown in the diagram, in the rhombus $ABCD$, $AB=8$, $\angle ADC=120^\circ$. What is the length of $AC$?

Solution:
\textbf{Solution:} Connect AC and BD, where AC intersects BD at point O. \\
Since quadrilateral ABCD is a rhombus, \\
it follows that AB $\parallel$ CD, AB = AD, AC$\perp$BD, and AO = OC = $\frac{1}{2}$AC, \\
Since $\angle$ADC = $120^\circ$, \\
it follows that $\angle$DAB = $180^\circ - 120^\circ = 60^\circ$, \\
Since AB = AD, \\
it follows that $\triangle$ABD is an equilateral triangle, \\
Since AC$\perp$BD, \\
it follows that $\angle$OAB = $\frac{1}{2}$$\angle$BAD = $30^\circ$, $\angle$AOB = $90^\circ$, \\
thus OB = $\frac{1}{2}$AB = 4, \\
thus AO = $\sqrt{AB^{2} - OB^{2}} = \sqrt{8^{2} - 4^{2}} = 4\sqrt{3}$, \\
thus AC = 2AO = $\boxed{8\sqrt{3}}$.


Question: As shown in the figure, the side length of square $ABCD$ is 10. Points $E$ and $F$ are inside the square where $AE=CF=8$, and $BE=DF=6$. What is the length of line segment $EF$?

Solution:
\textbf{Solution:} Extend DF to intersect AE at G, as shown in the figure,\\
$\because$ AB=10, AE=6, BE=8,\\
$\therefore$ AB$^{2}$=AE$^{2}$+BE$^{2}$,\\
$\therefore$ $\triangle$ABE is a right-angled triangle,\\
Similarly, $\triangle$DFC is a right-angled triangle,\\
$\because$ Quadrilateral ABCD is a square,\\
$\therefore$ $\angle$BAD=$\angle$ABC=$\angle$BCD=$\angle$ADC=$90^\circ$, AB= BC=CD=AD,\\
$\therefore$ $\angle$BAE+$\angle$DAG=$90^\circ$,\\
$\because$ In $\triangle$ABE and $\triangle$CDF,\\
$\begin{cases} AB=CD \\ AE=CF \\ BE=DF \end{cases}$,\\
$\therefore$ $\triangle$ABE$\cong$ $\triangle$CDF,\\
$\therefore$ $\angle$BAE=$\angle$DCF,\\
$\because$ $\angle$DCF+$\angle$CDF=$\angle$ADF+$\angle$CDF=$90^\circ$,\\
$\therefore$ $\angle$DCF=$\angle$ADG,\\
$\therefore$ $\angle$BAE=$\angle$ADG,\\
$\because$ $\angle$BAE+$\angle$DAG=$90^\circ$,\\
$\therefore$ $\angle$ADG+$\angle$DAG=$90^\circ$,\\
$\therefore$ $\angle$DGA=$90^\circ$, i.e., $\triangle$AGD is a right-angled triangle,\\
$\because$ In $\triangle$AGD and $\triangle$BAE,\\
$\begin{cases} \angle AGD=\angle BEA \\ \angle ADG=\angle BAE \\ AD=BA \end{cases}$,\\
$\therefore$ $\triangle$AGD$\cong$$\triangle$BAE,\\
$\therefore$ AG=BE=6, DG=AE=8,\\
$\therefore$ EG=8−6=2, GF=8−6=2,\\
$\because$ $\angle$DGA=$90^\circ$,\\
$\therefore$ $\angle$EGF=$180^\circ−90^\circ=90^\circ$,\\
$\therefore$ In $\triangle$EFG, $EF=\sqrt{2^2+2^2}=\boxed{2\sqrt{2}}$.


Question: As shown in the figure, point $P$ is a point on the angle bisector $AD$ of $\angle BAC$, where $AC=9$, $AB=5$, and $PB=3$. What could be the possible length of $PC$?

Solution:
\textbf{Solution:} Let's intercept AE=AB=5 on line AC, and connect PE. \\
Since AC=9, \\
thus CE=AC-AE=9-5=4, \\
Since point P is on the angle bisector AD of $\angle$BAC, \\
thus $\angle$CAD=$\angle$BAD, \\
In $\triangle APE$ and $\triangle APB$, $\left\{\begin{aligned} AE&=AB \\ \angle CAP&=\angle BAD \\ AP&=AP \end{aligned}\right.$, \\
thus $\triangle APE\cong \triangle APB$ (SAS), \\
thus PE=PB=3, \\
Since 4-3＜PC＜4+3, \\
We find that $1＜PC＜7$, \\
Upon examining the four options, the length of PC can be $\boxed{6}$, \\


Question: As shown in the diagram, it is known that the equilateral triangle $\triangle ABC$ has a side length of $12$, $D$ is a moving point on $AB$, a line is drawn through $D$ such that $DE \perp BC$ at point $E$, a line is drawn through $E$ such that $EF \perp AC$ at point $F$, and a line is drawn through $F$ such that $FG \perp AB$ at point $G$. When $G$ coincides with $D$, what is the length of $AD$?

Solution:
\textbf{Solution:} Let AD=x, \\
since $\triangle ABC$ is an equilateral triangle, \\
therefore $\angle A = \angle B = \angle C = 60^\circ$, \\
since DE$\perp$BC at point E, EF$\perp$AC at point F, FG$\perp$AB at point G, \\
therefore $\angle BDF = \angle DEB = \angle EFC = 90^\circ$, \\
therefore AF=2x, \\
therefore CF=12-2x, \\
therefore CE=2CF=24-4x, \\
therefore BE=12-CE=4x-12, \\
therefore BD=2BE=8x-24, \\
since AD+BD=AB, \\
therefore 8x-24+x=12, \\
therefore x=\boxed{4}, \\
therefore AD=4.


Question: As shown in the diagram, place the isosceles $\triangle ABC$ as in Figure 1, so that the base $BC$ coincides with the $x$-axis, at this moment the coordinates of point $A$ are $\left(2, \sqrt{5}\right)$. If this triangle is placed as in Figure 2, such that the side length $AB$ coincides with the $x$-axis, what are the coordinates of point $C$ then?

Solution:
\textbf{Solution:} As shown in Figure 1, draw $AD\perp OC$ through point A intersecting the x-axis at D,\\
since the coordinates of point A are $(2,\sqrt{5})$,\\
thus $BD=2$, $AD=\sqrt{5}$,\\
according to the Pythagorean theorem, $BA=\sqrt{BD^{2}+AD^{2}}=\sqrt{2^{2}+(\sqrt{5})^{2}}=3$,\\
since $\triangle ABC$ is an isosceles triangle,\\
thus $AB=AC=3$, $BC=2BD=4$,\\
As shown in Figure 2, draw $CE\perp AB$ through point C intersecting the x-axis at E,\\
let $BE=x$, then $AE=3-x$,\\
in $\triangle CBE$, $CE^{2}=BC^{2}-BE^{2}=16-x^{2}$,\\
in $\triangle CEA$, according to the Pythagorean theorem,\\
$CE^{2}+AE^{2}=CA^{2}$,\\
$16-x^{2}+(3-x)^{2}=3^{2}$,\\
$16+9-6x=0$,\\
$x=\frac{8}{3}$,\\
thus $CE=\sqrt{BC^{2}-BE^{2}}=\sqrt{4^{2}-\left(\frac{8}{3}\right)^{2}}=\frac{4\sqrt{5}}{3}$,\\
thus the coordinates of point C are $\left(\frac{8}{3},\frac{4\sqrt{5}}{3}\right)=\boxed{\left(\frac{8}{3},\frac{4\sqrt{5}}{3}\right)}$.


Question: As shown in the diagram, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=3$, $BC=4$. Points $E$ and $F$ are on the hypotenuse $AB$. The side $AC$ is folded along $CE$ making point $A$ fall on point $D$ on $AB$. Then, side $BC$ is folded along $CF$ making point $B$ fall on point $B'$ on the extension line of $CD$. What is the length of the segment $B'F$?

Solution:
\textbf{Solution:}

Since $\angle ACB=90^\circ, AC=3, BC=4$,

Hence, $AB=\sqrt{AC^2+BC^2}=5$,

By the property of folding, we have $B'F=BF, \angle AEC=\angle DEC=90^\circ, \angle ACE=\angle DCE, \angle B'CF=\angle BCF$,

Therefore, $S_{\triangle ABC}=\frac{1}{2}AB\cdot CE=\frac{1}{2}AC\cdot BC$, that is, $\frac{1}{2}\times 5CE=\frac{1}{2}\times 3\times 4$,

Solving this, we find $CE=\frac{12}{5}$,

Therefore, $AE=\sqrt{AC^2-CE^2}=\frac{9}{5}$,

Also, since $\angle ACE=\angle DCE, \angle B'CF=\angle BCF, \angle ACB=90^\circ$,

Therefore, $\angle DCE+\angle B'CF=\frac{1}{2}\angle ACB=45^\circ$, which means $\angle ECF=45^\circ$,

Therefore, right-angled $\triangle CEF$ is an isosceles right triangle, $EF=CE=\frac{12}{5}$,

Therefore, $BF=AB-AE-EF=\frac{4}{5}$,

Therefore, $B'F=BF=\boxed{\frac{4}{5}}$.


Question: As shown in the figure, a regular triangle, a square, and a regular pentagon are placed side by side with $AB$ as their common edge. What is the measure of $\angle CDE$?

Solution:
\textbf{Solution:}

Given that the sum of internal angles of a pentagon is $(5-2) \times 180^\circ = 540^\circ$,

it follows that $\angle BAC = \frac{1}{5} \times 540^\circ = 108^\circ$,

since the internal angle of a triangle ABE ($\angle BAE$) is $60^\circ$ and the internal angle of a square ($\angle BAD$) is $90^\circ$,

thus $\angle DAC = \angle BAC - \angle BAD = 108^\circ - 90^\circ = 18^\circ$, and $\angle DAE = \angle BAD - \angle BAE = 90^\circ - 60^\circ = 30^\circ$,

given AD = AC,

it implies $\angle ADC = \angle ACD = \frac{180^\circ - \angle DAC}{2} = \frac{180^\circ - 18^\circ}{2} = 81^\circ$,

also, given EA = AD,

it implies $\angle ADE = \angle AED = \frac{180^\circ - \angle DAE}{2} = \frac{180^\circ - 30^\circ}{2} = 75^\circ$,

thus $\angle CDE = \angle ADC + \angle ADE = 81^\circ + 75^\circ = \boxed{156^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $\angle B=30^\circ$, and $AD$ bisects $\angle BAC$. If $BC=9$, what is the distance from point D to line $AB$?

Solution:
\textbf{Solution:} As shown in the figure, draw $DE \perp AB$ at point E\\
$\therefore \angle DEB=90^\circ$\\
$\because \angle C=90^\circ$, $\angle B=30^\circ$\\
$\therefore \angle BAC=60^\circ$, $BD=2ED$, $CD\perp AC$\\
$\because AD$ bisects $\angle BAC$\\
$\therefore CD=DE$, $\angle DAE=\frac{1}{2}\angle BAC=30^\circ$\\
$\therefore \angle DAB=\angle B$\\
$\therefore BD=AD=2CD$\\
$\because BC=9=CD+BD=3ED$\\
$\therefore DE=\boxed{3}$


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC = \angle ABC = 42^\circ$. Through point C, draw $CD \perp AB$ at point D. Let point E be a point on $CD$. If $\triangle ACE$ is folded along $AE$ to obtain $\triangle AFE$ and points E, F, B are exactly on the same line, connect $CF$. What is the measure of $\angle CFA$?

Solution:
\textbf{Solution:} \\
Since $CD \perp AB$\\
$\therefore \angle ADC = 90^\circ$ \\
Since $\angle BAC = \angle ABC = 42^\circ$\\
$\therefore \angle ACD = 90^\circ - \angle BAC = 48^\circ$, $AC = BC$\\
Since $CD \perp AB$\\
$\therefore \angle BCD = \angle ACD = 48^\circ$ \\
In $\triangle ACE$ and $\triangle BCE$\\
$\left\{\begin{array}{l}
AC = BC\\
\angle ACD = \angle BCD\\
CE = CE
\end{array}\right.$ \\
$\therefore \triangle ACE \cong \triangle BCE$\\
$\therefore \angle CAE = \angle CBE$ \\
Since folding $\triangle ACE$ along $AE$ yields $\triangle AFE$,\\
$\therefore \angle AFE = \angle ACD = 48^\circ$, $\angle CAE = \angle FAE$, $AC=AF$\\
$\therefore \angle CAE = \angle FAE = \angle CBE$\\
Let $\angle CAE = \angle FAE = \angle CBE = x$\\
$\therefore \angle BAF = \angle BAC - \angle CAE - \angle FAE = 42^\circ - 2x$, $\angle ABF = \angle ABC - \angle CBE = 42^\circ - x$ \\
Since $\angle AFE = \angle BAF + \angle ABF = 42^\circ - 2x + (42^\circ - x) = 84^\circ - 3x$ \\
$\therefore 48^\circ = 84^\circ - 3x$\\
$\therefore x = 12^\circ$ \\
$\therefore \angle CAF = \angle CAE + \angle FAE = 24^\circ$ \\
Since $AC=AF$\\
$\therefore \angle CFA = \angle FCA = \frac{180^\circ - \angle CAF}{2} = \boxed{78^\circ}$


Question: As shown in the figure, point C is on the line segment BG. With BC and CG as sides, squares are constructed on both sides, with areas denoted by $S_{1}$ and $S_{2}$ respectively. The sum of the areas of the two squares is $S_{1}+S_{2}=40$. Given that BG=8, what is the area of the shaded part in the figure?

Solution:
\textbf{Solution:} Let $BC = a$ and $CG = b$. Then, $S_{1} = a^{2}$ and $S_{2} = b^{2}$, with $a + b = BG = 8$.\\
Hence, $a^{2} + b^{2} = 40$.\\
Since $(a + b)^{2} = a^{2} + b^{2} + 2ab = 64$,\\
it follows that $2ab = 64 - 40 = 24$,\\
and thus, $ab = 12$.\\
Therefore, the area of the shaded part is equal to $\frac{1}{2} ab = \frac{1}{2} \times 12 = \boxed{6}$.


Question: Given: In $\triangle ABC$, $AB=AC$, the perpendicular bisector of $AB$, $DE$, intersects $AB$ and $AC$ at $D$ and $E$ respectively. If $AD=3$ and $BC=5$, then what is the perimeter of $\triangle BEC$?

Solution:
\textbf{Solution:} Since $AB=AC$, and $DE$ is the perpendicular bisector of segment $AB$,\\
it follows that $AD=BD$, and $AE=BE$,\\
since $AD=3$,\\
then $AB=AC=2AD=6$,\\
since $BC=5$,\\
thus the perimeter of $\triangle BEC$ is $BC+BE+EC=BC+AE+EC=5+6=\boxed{11}$;


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle B=30^\circ$, $CD$ is the height on the hypotenuse $AB$, $AD=3$, then what is the length of $BD$?

Solution:
\textbf{Solution:} Since CD$\perp$AB and $\angle$ACB=$90^\circ$,\\
it follows that $\angle$ADC=$90^\circ=\angle$ACB,\\
since $\angle$B=$30^\circ$,\\
it follows that $\angle$A$=90^\circ-\angle$B$=60^\circ$,\\
thus $\angle$ACD$=90^\circ-\angle$A$=30^\circ$,\\
since AD$=3$,\\
it follows that AC$=2$AD$=6$,\\
thus AB$=2$AC$=12$,\\
hence BD$=AB-AD$=$12-3$=$\boxed{9}$


Question: As shown in the figure, $D$, $E$, and $F$ are the midpoints of $BC$, $AD$, and $BE$ respectively. If the area of $\triangle BFD$ is 6, what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Since D, E, and F are the midpoints of BC, AD, and BE, respectively,

$\therefore {S}_{\triangle ABC}=2{S}_{\triangle ABD}$, ${S}_{\triangle ABD}=2{S}_{\triangle BDE}$, and ${S}_{\triangle BDE}=2{S}_{\triangle BFD}$,

$\therefore {S}_{\triangle ABC}=8{S}_{\triangle BFD}=\boxed{48}$.


Question: As shown in the figure, in the isosceles right-angled $\triangle ABC$, where $AC=BC$ and $\angle ACB=90^\circ$, point D is a moving point on side $AC$ (not coinciding with A or C). A line is drawn from point A perpendicular to $BD$ at point E, and $AE$ is extended to intersect the extension of $BC$ at point F. $CE$ is connected. Then, what is the measure of $\angle BEC$?

Solution:
\textbf{Solution:} As shown in the diagram, draw $CH \perp AF$ at $H$, and $CG \perp BE$ at $G$,\\
$\therefore \angle AHC=\angle BGC=90^\circ$,\\
$\because \angle ACB=90^\circ$, and $AF\perp BE$,\\
$\therefore \angle AEB=\angle BCD=\angle BEF=90^\circ$,\\
Also, since $\angle ADE=\angle BDC$,\\
$\therefore \angle CAH=\angle CBG$,\\
Also, since $AC=BC$,\\
$\therefore \triangle AHC\cong \triangle BCG$ (AAS),\\
$\therefore CH=CG$,\\
Since $CH\perp EF$, and $CG\perp BE$,\\
$\therefore CE$ bisects $\angle BEF$,\\
$\therefore \angle BEC= \frac{1}{2}\angle BEF=\boxed{45^\circ}$.


Question: As shown in the figure, $AD$ is the perpendicular bisector of $BC$ with the foot of the perpendicular as $D$, $\angle BAC=45^\circ$, $CE\perp AB$ at $E$, intersecting $AD$ at $F$, and $BD=2$. What is the length of $AF$?

Solution:
\textbf{Solution:} Since $AD$ is the perpendicular bisector of $BC$, and $BD=2$, \\
thus, $BC=2BD=4$, \\
Since $CE\perp AB$, \\
thus, $\angle AEC=90^\circ$, \\
Since $\angle BAC=45^\circ$, \\
thus, $\angle ACE=90^\circ-45^\circ=45^\circ$, \\
thus, $\angle EAC=\angle ACE$, \\
thus, $AE=CE$, \\
Since $AB=AC$, and point $D$ is the midpoint of $BC$, \\
thus, $AD\perp BC$, \\
thus, $\angle ADB=90^\circ$, \\
thus, $\angle B+\angle BAD=90^\circ$, \\
Since $\angle B+\angle BCE=90^\circ$, \\
thus, $\angle BAD=\angle BCE$, \\
In $\triangle AEF$ and $\triangle CEB$, \\
$\left\{\begin{array}{l} \angle AEF=\angle CEB \\ AE=CE \\ \angle EAF=\angle BCE \end{array}\right.$, \\
thus, $\triangle AEF\cong \triangle CEB$ (ASA), \\
thus, $AF=BC=\boxed{4}$,


Question: As shown in the figure, in $\triangle ABC$, $DE$ is the perpendicular bisector of $AC$. If $AE=3$ and the perimeter of $\triangle ABD$ is $13$, what is the perimeter of $\triangle ABC$?

Solution:
\textbf{Solution:} Given that $DE$ is the perpendicular bisector of $AC$ and $AE=3$,

$\therefore DA=DC$, $AC=2AE=6$,

Since the perimeter of $\triangle ABD$ is $13$,

$\therefore AB+BD+AD=AB+BD+DC=AB+BC=13$,

$\therefore$ the perimeter of $\triangle ABC$ $=AB+BC+AC=13+6=\boxed{19}$.


Question: As shown in the figure, $\triangle ABC \cong \triangle DEF$, where point A corresponds to point D, and point B corresponds to point E. If $\angle A=70^\circ$ and $\angle B=50^\circ$, what is the measure of $\angle 1$?

Solution:
\textbf{Solution:} In $\triangle ABC$, $\angle A=70^\circ$, $\angle B=50^\circ$,\\
thus $\angle C=180^\circ-\angle A-\angle B=180^\circ-70^\circ-50^\circ=60^\circ$,\\
since $\triangle ABC \cong \triangle DEF$,\\
it follows that $\angle 1=\angle C=\boxed{60^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $\angle B=90^\circ$, the perpendicular bisector $l$ of $AC$ intersects $BC$ at point $D$, and $AD$ is connected. If $AB=3$, $BC=9$, what is the length of $BD$?

Solution:
\textbf{Solution:} Let BD $= x$, thus CD $= 9 - x$, \\
$\because$ line $l$ is the perpendicular bisector of AC, \\
$\therefore$ AD = DC = $9-x$, \\
In $\triangle ABD$, $AB^{2} + BD^{2} = AD^{2}$, that is, $3^{2} + x^{2} = (9-x)^{2}$, \\
Solving gives: $x = \boxed{4}$, which means the length of BD is 4.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, squares are constructed externally on sides $AB$, $CA$, $BC$. The area of square $ABIH$ is $25$, and the area of square $BDEC$ is $169$. What is the area of square $ACFG$?

Solution:
\textbf{Solution:} In the right triangle $ABC$, $\angle BAC=90^\circ$, \\
$\therefore AB^2+AC^2=BC^2$, \\
$\because$ the area of the square $ABIH$ is $25$, and the area of the square $BDEC$ is $169$, \\
$\therefore AB^2=25, BC^2=169$, \\
$\therefore AC^2=BC^2-AB^2=169-25=144$, \\
$\therefore$ the area of the square $ACFG$ is $AC^2=\boxed{144}$.


Question: As shown in the figure, it is known that $ \triangle ABC \cong \triangle DEF $, with B, E, C, F lying on the same straight line. If $ BF = 8\text{cm} $ and $ BE = 2\text{cm} $, then what is the length of $ CE $ in $\text{cm}$?

Solution:
\textbf{Solution:} Since $\triangle ABC \cong \triangle DEF$,\\
it follows that $BC = EF$,\\
therefore $BC - CE = EF - CE$,\\
hence $BE = CF$,\\
given that $BE = 2\text{cm}$,\\
it implies $CF = BE = 2\text{cm}$,\\
since $BF = 8\text{cm}$,\\
thus $CE = BF - BE - CF = 8 - 2 - 2 = \boxed{4}\text{cm}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle B=\angle C$, $D$ is a point on side $BC$, and $E$ is a point on side $AC$, with $\angle ADE=\angle AED$. If $\angle BAD=28^\circ$, what is the measure of $\angle CDE$?

Solution:
\textbf{Solution:} Given that $\angle ADC$ is an exterior angle of $\triangle ABD$, \\
it follows that $\angle ADC = \angle B + \angle BAD = \angle B + 28^\circ$, \\
and since $\angle AED$ is an exterior angle of $\triangle CDE$, \\
we have $\angle AED = \angle C + \angle EDC$, \\
considering $\angle B = \angle C$ and $\angle ADE = \angle AED$, \\
thus $\angle C + \angle EDC = \angle ADC - \angle EDC = \angle B + 28^\circ - \angle EDC$, \\
leading to the solution $\angle EDC = \boxed{14^\circ}$.


Question: As shown in the diagram, within $\triangle ABC$, $\triangle BDE$ and $\triangle FGH$ are two congruent equilateral triangles, and $BC=5$. What is the perimeter of the pentagon $DECHF$?

Solution:
\textbf{Solution:} Given that $\triangle GFH$ is an equilateral triangle,\\
thus $FH=GH$, $\angle FHG=60^\circ$,\\
therefore, $\angle AHF+\angle GHC=120^\circ$,\\
given that $\triangle ABC$ is an equilateral triangle,\\
thus $AB=BC=AC=5$, $\angle ACB=\angle A=60^\circ$,\\
given that $\angle AHF=180^\circ-\angle FHG-\angle GHC =120^\circ-\angle GHC$,\\
$\angle HGC=180^\circ-\angle C-\angle GHC =120^\circ-\angle GHC$,\\
therefore, $\angle AHF=\angle HGC$,\\
in $\triangle AFH$ and $\triangle CHG$ we have\\
$\left\{\begin{array}{l}
\angle A=\angle C \\
\angle AHF=\angle HGC \\
FH=GH
\end{array}\right.$,\\
therefore, $\triangle AFH\cong \triangle CHG$ by AAS,\\
thus, AF=CH.\\
Given that $\triangle BDE$ and $\triangle FGH$ are two congruent equilateral triangles,\\
thus, BE=FH,\\
therefore, the perimeter of pentagon DECHF=DE+CE+CH+FH+DF=BD+CE+AF+BE+DF,\\
=(BD+DF+AF)+(CE+BE),\\
=AB+BC=\boxed{10}.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$ and $\angle A=36^\circ$. An arc is drawn with $B$ as the center and the length of $BC$ as the radius, intersecting $AC$ at point $D$. When connecting $BD$, what is the measure of $\angle ABD$?

Solution:
\textbf{Solution:} Given $\because$ AB=AC, $\angle A=36^\circ$,\\
$\therefore \angle C= \frac{180^\circ-\angle A}{2}=72^\circ$,\\
$\because$ BC=BD,\\
$\therefore \angle BDC=\angle C=72^\circ$,\\
$\therefore \angle ABD=\angle BDC-\angle A=\boxed{36^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, point D is the midpoint of $AB$, and $CD$ is connected. If $AC=4$ and $BC=3$, then what is the length of $CD$?

Solution: \textbf{Solution:} Since in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=4$, and $BC=3$, by the Pythagorean theorem we have: $AB= \sqrt{AC^{2}+BC^{2}}=\sqrt{4^{2}+3^{2}}=5$. Since point D is the midpoint of AB, it follows that $CD= \frac{1}{2}AB= \frac{1}{2}\times 5=\boxed{2.5}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $AD$ is the height from $A$ to side $BC$, point $E$ is on $AD$, and $AE=\frac{1}{3}AD$. If the area of $\triangle ABC$ is $s$, what is the area of $\triangle ABE$?

Solution:
Given that in $\triangle ABC$, $\angle ACB=90^\circ$ and AB=AC, with AD being the altitude on the side BC,\\
it follows that AD is also the median on side BC, that is BD=DC,\\
Furthermore, given that the area of $\triangle ABC$ equals S,\\
thus, the area of $\triangle ABD$ equals $\frac{S}{2}$,\\
Since AE=$\frac{1}{3}AD$,\\
consequently, the area of $\triangle ABE$ equals $\boxed{\frac{S}{6}}$.


Question: As shown in the figure, it is known that $\triangle ABC$ is an equilateral triangle with a side length of 10. $D$ is the midpoint of $AC$, $\angle EDF=120^\circ$, $DE$ intersects line segment $AB$ at $E$, and $DF$ intersects the extension of $BC$ at $F$. If $AE=4BE$, then what is the length of $CF$?

Solution:
\textbf{Solution:} As shown in the figure, construct $DM\parallel BC$ intersecting $AB$ at $M$, \\
$\because$ $AE=4BE$, \\
$\therefore$ $BE=\frac{1}{5}AB=2$, \\
$\because$ $D$ is the midpoint of $AC$, and $\triangle ABC$ is an equilateral triangle, \\
$\therefore$ $MD$ is the median of $\triangle ABC$, \\
$\therefore$ $MD=DC=BM=\frac{1}{2}BC=5$, $\angle DME=120^\circ=\angle DCF$, $\angle ADM=60^\circ$, \\
$\therefore$ $ME=BM-BE=3$, $\angle MDC=120^\circ$, \\
$\because$ $\angle EDF=120^\circ$, $\angle MDC-\angle EDC=\angle EDF-\angle EDC$, \\
$\therefore$ $\angle MDE=\angle CDF$, \\
In $\triangle DEM$ and $\triangle DFC$, \\
$\because$ $\left\{\begin{array}{l}\angle MDE=\angle CDF \\ \angle DME=\angle DCF \\ DM=DC \end{array}\right.$, \\
$\therefore$ $\triangle DEM\cong \triangle DFC\left(AAS\right)$, \\
$\therefore$ $CF=ME=\boxed{3}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=4$, and $\angle B=\angle C=15^\circ$. What is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Draw $CD \perp AB$ through point $C$ intersecting the extended line of $BA$ at $D$,\\
$\because$ $\angle B = \angle ACB = 15^\circ$,\\
$\therefore$ $\angle CAD = \angle B + \angle ACB = 15^\circ + 15^\circ = 30^\circ$,\\
$\because$ $AC = 4$ and $CD$ is the height from the base $AB$,\\
$\therefore$ $CD = \frac{1}{2} AC = \frac{1}{2} \times 4 = 2$,\\
$\therefore$ the area of $\triangle ABC = \frac{1}{2} \times 4 \times 2 = \boxed{4}$


Question: As shown in the figure, it is known that $\triangle ABC$ is an equilateral triangle with a side length of 4, and $\triangle DBC$ is an isosceles triangle with a vertex angle of $120^\circ$. The moving points $E$ and $F$ are on the sides $AB$ and $AC$, respectively, and $\angle EDF=60^\circ$. What is the perimeter of $\triangle AEF$?

Solution:
\textbf{Solution:} Extend $EB$ to $G$, such that $BG=FC$, and connect $DG$,\\
since $\triangle ABC$ is an equilateral triangle,\\
thus $\angle ABC = \angle ACB = 60^\circ$,\\
furthermore, since $BD=CD$,\\
thus $\angle DCB = \angle DBC = \frac{180^\circ - \angle BDC}{2} = 30^\circ$,\\
therefore $\angle DCF = \angle DBE = 90^\circ$,\\
in right-angled $\triangle DCF$ and $\triangle DBG$,\\
$\left\{\begin{array}{l}DC=DB \\ \angle DCF=\angle DBG \\ CF=BG \end{array}\right.$,\\
therefore $\triangle DCF \cong \triangle DBG$,\\
thus $DG=DF$, $\angle FDC = \angle GDB$,\\
therefore $\angle GDF = \angle BDC = 120^\circ$,\\
furthermore, since $\angle EDF = 60^\circ$,\\
thus $\angle EDG = 60^\circ$,\\
in $\triangle EDG$ and $\triangle EDF$,\\
$\left\{\begin{array}{l}DG=DF \\ \angle EDG=\angle EDF \\ DE=DE \end{array}\right.$,\\
thus $\triangle EDG \cong \triangle EDF$,\\
thus $EF=EG=EB+GB=EB+FC$,\\
therefore, the perimeter of $\triangle AEF$ is: $AE+AF+EF=AE+AF+BE+FC=AB+AC=\boxed{8}$,\\


Question: In the figure, within $\triangle ABE$, the perpendicular bisector MN of AE intersects BE at point C, and AC is connected. If $AB=AC$, $CE=5$, and $BC=6$, then what is the perimeter of $\triangle ABC$?

Solution:
\textbf{Solution:} Since $MN$ is the perpendicular bisector of $AE$ and $CE=5$,\\
it follows that $AC=CE=5$,\\
Since $AB=AC$,\\
and given that $AB=5$,\\
and $BC=6$,\\
thus, the perimeter of $\triangle ABC$ is $AB+AC+BC=5+5+6=\boxed{16}$.


Question: As shown in the figure, a set of triangles is stacked together. What is the value of $\angle \alpha$ in the figure?

Solution:
\textbf{Solution:} As shown in the figure, $\angle C=90^\circ$, $\angle DAE=45^\circ$, $\angle BAC=60^\circ$,\\
$\therefore \angle CAO=\angle BAC−\angle DAE=60^\circ−45^\circ=15^\circ$,\\
$\therefore \angle \alpha =\angle C+\angle CAO=90^\circ+15^\circ=\boxed{105^\circ}$.


Question: As shown in the figure, in the rectangle $OABC$, point $A$ is on the y-axis, and point $C$ is on the x-axis. We have $OA=BC=4$, and $AB=OC=8$. Point $D$ is on side $AB$, and point $E$ is on side $OC$. The rectangle is folded along the straight line $DE$ so that point $B$ coincides with point $O$. What are the coordinates of point $D$?

Solution:
\textbf{Solution:} Let AD = x. By the property of folding, we know that OD = BD = 8 - x,\\
In right triangle $\triangle OAD$,\\
$\because$ OA$^{2}$ + AD$^{2}$ = OD$^{2}$,\\
$\therefore$ $4^{2}$ + x$^{2}$ = (8 − x)$^{2}$,\\
$\therefore$ x = 3,\\
$\therefore$ D$(3,4)$,\\
Therefore, the answer is: D$\boxed{\left(3,4\right)}$.


Question: As shown in the figure, fold the triangular paper $ABC$ along $DE$. When point $A$ falls outside of the quadrilateral $BCED$, the measurements show that $\angle 1 = 70^\circ$ and $\angle 2 = 152^\circ$. What is the measure of $\angle A$?

Solution:
\[
\text{Solution:} \\
\text{Based on the fold, we know } \angle A'=\angle A, \\
\because \angle 1=70^\circ, \\
\therefore \angle A'DA=180^\circ-\angle 1=110^\circ, \\
\therefore \text{according to the external angle of a triangle } \angle A'=\angle 2-\angle A'DA=152^\circ-110^\circ=42^\circ, \\
\therefore \angle A=\boxed{42^\circ}.
\]


Question: As shown in the figure, points D and E are located on line segments BC and AC, respectively, and lines AD and BE are connected. If $\angle A=35^\circ$, $\angle B=30^\circ$, and $\angle C=45^\circ$, what is the measure of $\angle AFB$?

Solution:
\textbf{Solution:} Given $\because$ $\angle A=35^\circ$, $\angle B=30^\circ$, $\angle C=45^\circ$\\
$\therefore$ $\angle ADC=180^\circ-\angle A-\angle C=180^\circ-35^\circ-45^\circ=100^\circ$,\\
$\angle BEC=180^\circ-\angle B-\angle C=180^\circ-30^\circ-45^\circ=105^\circ$,\\
In quadrilateral CDFE, $\angle DFE=360^\circ-\angle FEC-\angle FDC-\angle C=360^\circ-105^\circ-100^\circ-45^\circ=110^\circ$,\\
$\therefore$ $\angle AFB=\angle DFE=\boxed{110^\circ}$.


Question: As shown in the figure, in an equilateral $\triangle ABC$, $D$ and $E$ are points on $BC$ and $AC$ respectively, and $BD = CE$, $AD$ intersects $B$ at point $P$. What is the measure of $\angle APE$ in degrees?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is an equilateral triangle,\\
we have $AB=BC$ and $\angle ABC=\angle C=60^\circ$,\\
Given $BD=CE$,\\
it follows that $\triangle ABD\cong \triangle BCE$ (SAS),\\
hence $\angle BAD=\angle CBE$,\\
Since $\angle ABP+\angle CBE=\angle ABC=60^\circ$,\\
we get $\angle ABP+\angle BAD=60^\circ$,\\
thus $\angle APE = \angle ABP+\angle BAD=60^\circ$;\\
Therefore, the answer is: $\boxed{60^\circ}$.


Question: As shown in the diagram, in $\triangle ABC$, $AB=AC$, $\angle ABC=45^\circ$, $AD\perp BC$ at point D. If $BC=8$, what is the length of $AD$?

Solution:
\textbf{Solution:} Given that $AB=AC$ and $AD\perp BC$,\\
it follows that $BD=DC=4$,\\
since $\angle ABC=45^\circ$ and $AD\perp BC$,\\
it results in $\angle ABD=\angle BAD=45^\circ$,\\
thus $AD=BD=\boxed{4}$.


Question: As shown in the figure, $D$ is a point on side $BC$ of $\triangle ABC$, $\angle B = \angle 1$, $\angle 3 = 80^\circ$, $\angle BAC = 70^\circ$. What is the measure of $\angle 2$?

Solution:
\textbf{Solution:} Given $\angle 3$ is an exterior angle of $\triangle ABD$,\\
thus $\angle 3 = \angle B + \angle 1$,\\
since $\angle 3 = 80^\circ$ and $\angle B = \angle 1$,\\
it follows that $\angle 1 = \frac{1}{2} \angle 3 = \frac{1}{2} \times 80^\circ = 40^\circ$,\\
given $\angle BAC = 70^\circ$,\\
thus $\angle 2 = \angle BAC - \angle 1 = 70^\circ - 40^\circ = \boxed{30^\circ}$.


Question: As shown in the figure, $\angle ABD = \angle CBE = 90^\circ$, $AB = BD$, $\angle CAB = \angle E$. If $BE = 10$, $AD = 4\sqrt{2}$ , what is the value of $\frac{AC}{DE}$?

Solution:
\textbf{Solution:} \\
Since $\angle ABD=90^\circ$ and $AB=BD$, $AD=4\sqrt{2}$, \\
therefore $2AB^{2}=2BD^{2}=AD^{2}$, \\
hence $2AB^{2}=32$, solving this yields: $AB=BD=4$ (taking the positive value); \\
Since $\angle ABD=\angle CBE=90^\circ$, \\
it follows that $\angle ABD-\angle CBD=\angle CBE-\angle CBD$, \\
therefore $\angle ABC=\angle DBE$, \\
and since $\angle CAB=\angle E$, \\
it can be concluded that $\triangle ABC \sim \triangle EBD$, \\
therefore $\frac{AC}{DE}=\frac{AB}{BE}=\frac{4}{10}=\boxed{\frac{2}{5}}$.


Question: As shown in the figure of a newly launched shoe rack from a Taobao store, it can be abstracted into a diagram where $l_1\parallel l_2\parallel l_3$. Lines $AC$ and $DF$ are intersected by $l_1$, $l_2$, $l_3$. If $AB=30\,\text{cm}$, $BC=50\,\text{cm}$, and $EF=40\,\text{cm}$, what is the length of $DE$?

Solution:
Solution: Since $l_1 \parallel l_2 \parallel l_3$,\\
therefore $\frac{AB}{BC}=\frac{DE}{EF}$\\
Therefore $\frac{30}{50}=\frac{DE}{40}$\\
Solving, we find that $DE=\boxed{24}$.


Question: As shown in the figure, in $\triangle ABC$, points D, E, and F are located on sides AB, AC, and BC respectively, with $DE \parallel BC$ and $EF \parallel AB$, and $AD = 2BD$. What is the value of $\frac{CF}{CB}$?

Solution:
\textbf{Solution:} Since $AD=2BD$, \\
it follows that $BD: AB=1: 3$, \\
Since $DE\parallel BC$, \\
it follows that $CE: AC=BD: AB=1: 3$, \\
Since $EF\parallel AB$, \\
it follows that $CF: CB=CE: AC=1: 3$. \\
Therefore, $\frac{CF}{CB} = \boxed{\frac{1}{3}}$


Question: As shown, $\triangle ABO \sim \triangle CDO$, if $BO = 10$, $DO = 5$, and $AC = 9$, then what is the length of $OC$?

Solution:
\textbf{Solution:} Since $\triangle ABO\sim \triangle CDO$,\\
it follows that $\frac{OB}{OD}=\frac{OA}{OC}$,\\
therefore $\frac{10}{5}=\frac{9-OC}{OC}$,\\
solving this yields $OC=\boxed{3}$.


Question: As shown in the diagram, in $ \triangle ABC$, $ \angle ACB=90^\circ$, $ \angle ABC=60^\circ$, and $BC=2\,cm$. Point D is the midpoint of BC. If a moving point E starts from point A and moves in the direction of A→B→A at a speed of $1\,cm/s$, let the movement time of point E be $t\,s$ ($0 \leq t < 6$). Connect DE. When $ \triangle BDE$ is similar to $ \triangle ABC$, what is the value of t?

Solution:
\textbf{Solution:} Given that in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle ABC=60^\circ$, and $BC=2cm$,\\
$\therefore$AB=2BC=4(cm),\\
Since $BC=2cm$, and $D$ is the midpoint of $BC$, and a moving point $E$ starts from $A$ at a speed of 1cm/s,\\
$\therefore$BD=$\frac{1}{2}$BC=1(cm), BE=AB−AE=4−t(cm),\\
When $\triangle BDE$ is similar to $\triangle ABC$, $\triangle BDE$ is a right triangle,\\
If $\angle BED=90^\circ$,\\
As A moves towards B, since $\angle ABC=60^\circ$,\\
$\therefore$$\angle BDE=30^\circ$,\\
$\therefore$BE=$\frac{1}{2}$BD=$\frac{1}{2}$(cm),\\
$\therefore$t=3.5,\\
As B moves towards A, t=4+0.5=4.5.\\
If $\angle BDE=90^\circ$,\\
As A moves towards B, since $\angle ABC=60^\circ$,\\
$\therefore$$\angle BED=30^\circ$,\\
$\therefore$BE=2BD=2(cm),\\
$\therefore$t=4−2=$2$,\\
As B moves towards A, t=4+2=6 (discard).\\
In summary, the value of $t$ is either \boxed{2, 3.5, or 4.5}.


Question: As shown in the diagram, in $\triangle ABC$, $D$ and $E$ are points on $AB$ and $AC$ respectively, and $DE \parallel BC$. If $AE:EC = 1:4$, then what is the value of $\frac{S_{\triangle ADE}}{S_{\triangle BEC}}$?

Solution:
\textbf{Solution:} Since $\frac{AE}{EC}=\frac{1}{4}$,\\
thus $\frac{{S}_{\triangle ABE}}{{S}_{\triangle EBC}}=\frac{1}{4}$,\\
thus ${S}_{\triangle ABE}=\frac{1}{4}{S}_{\triangle EBC}$,\\
Since DE$\parallel$BC,\\
thus $\frac{AD}{DB}=\frac{AE}{EC}=\frac{1}{4}$,\\
thus $\frac{{S}_{\triangle ADE}}{{S}_{\triangle BDE}}=\frac{1}{4}$,\\
thus ${S}_{\triangle BDE}=4{S}_{\triangle ADE}$,\\
Also, since ${S}_{\triangle BDE}={S}_{\triangle ABE}-{S}_{\triangle ADE}$,\\
thus $4{S}_{\triangle ADE}=\frac{1}{4}{S}_{\triangle EBC}-{S}_{\triangle ADE}$,\\
thus $\frac{{S}_{\triangle ADE}}{{S}_{\triangle EBC}}=\boxed{\frac{1}{20}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, $AB=6$, $AC=8$. Point $P$ is any point on $BC$, connect $PA$, with $PA$ and $PC$ as adjacent sides, construct $\square PAQC$, connect $PQ$, then what is the minimum value of $PQ$?

Solution:
\textbf{Solution:} Let $PQ$ intersect $AC$ at point $O$, and draw $O{P}^{\prime} \perp BC$ at ${P}^{\prime}$.\\
In $\triangle ABC$, $BC=\sqrt{AC^{2}+AB^{2}}=\sqrt{6^{2}+8^{2}}=10$,\\
$\because \angle OC{P}^{\prime} = \angle ACB$, $\angle O{P}^{\prime}C = \angle CAB$,\\
$\therefore \triangle CO{P}^{\prime} \sim \triangle CBA$,\\
$\therefore \frac{CO}{CB}=\frac{O{P}^{\prime}}{AB}$,\\
$\therefore \frac{4}{10}=\frac{O{P}^{\prime}}{6}$,\\
$\therefore O{P}^{\prime}=\frac{12}{5}$,\\
When $P$ coincides with ${P}^{\prime}$, the value of $PQ$ is minimized, and the minimum value of $PQ=2O{P}^{\prime}=\boxed{\frac{24}{5}}$.


Question: As shown in the figure, in rectangle $ABCD$, point $E$ is the midpoint of side $AB$. Lines $AC$ and $DE$ intersect at point $F$. If the area of $\triangle CDF$ is 4, what is the area of $\triangle AED$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\
it follows that AB $\parallel$ CD and AB = CD,\\
which means that $\triangle AEF \sim \triangle CDF$,\\
given that point E is the midpoint of AB,\\
hence CD = AB = 2AE,\\
thus $\frac{EF}{DF} = \frac{AE}{CD} = \frac{1}{2}$,\\
given that the area of $\triangle CDF$ is 4,\\
it follows that the area of $\triangle AEF$ is 1,\\
hence the area of $\triangle ADF$ is 2,\\
therefore, the area of $\triangle AED$ is $\boxed{3}$,


Question: As shown in the figure, in parallelogram $ABCD$, $E$ is the midpoint of side $AD$. Line segments $AC$ and $BE$ intersect at point $F$. If the area of $\triangle AEF$ is 2, then what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Given $\because$ parallelogram ABCD\\
$\therefore$ $AD\parallel BC$ and $AD=BC$\\
$\because$ E is the midpoint of side AD\\
$\therefore$ BC=2AE\\
$\because$ $AE\parallel BC$\\
$\therefore$ $\angle EAC=\angle BCA$\\
Also, $\because$ $\angle EFA=\angle BFC$\\
$\therefore$ $\triangle AEF\sim \triangle CBF$\\
As illustrated, draw $FH\perp AD$ at point H and $FG\perp BC$ at point G,\\
then $\frac{EF}{BF}=\frac{AF}{CF}=\frac{AE}{BC}=\frac{HF}{FG}=\frac{1}{2}$,\\
$\therefore$ $\frac{S_{\triangle AEF}}{S_{\triangle ABC}}=\frac{\frac{1}{2}\cdot AE\cdot FH}{\frac{1}{2}\cdot BC\cdot HG}=\frac{\frac{1}{2}BC\cdot FH}{BC\cdot 3FH}=\frac{1}{6}$,\\
$\because$ the area of $\triangle AEF$ is 2\\
$\therefore$ $S_{\triangle ABC}=6S_{\triangle AEF}=6\times 2=\boxed{12}$


Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is on side $AD$ and $AE=2ED$. $EC$ intersects the diagonal $BD$ at point $F$ and the area of $\triangle DEF$ is $2$. What is the area of $\triangle BCF$?

Solution:
\textbf{Solution:} Since $AE=2ED$, and $AD=AE+DE=3DE$,\\
thus $\frac{ED}{AD}=\frac{1}{3}$,\\
Since quadrilateral ABCD is a parallelogram,\\
thus $AD\parallel BC$, and $BC=AD$,\\
thus $\triangle EDF\sim \triangle CBF$,\\
thus $\frac{ED}{BC}=\frac{EF}{CF}=\frac{ED}{AD}=\frac{1}{3}$,\\
thus $\frac{{S}_{\triangle EDF}}{{S}_{\triangle BCF}}=\frac{ED^{2}}{BC^{2}}=\left(\frac{1}{3}\right)^{2}=\frac{1}{9}$,\\
Since ${S}_{\triangle EDF}=2$,\\
thus ${S}_{\triangle BCF}=9\times 2=\boxed{18}$.


Question: As shown in the figure, $AB \parallel CD$, and $AC$ and $BD$ intersect at point $O$. If $AB = 6$, $CD = 3$, and $DO = 4$, what is the length of $BO$?

Solution:
\textbf{Solution:} Since $AB\parallel CD$, \\
it follows that $\triangle ABO\sim \triangle CDO$, \\
which implies $\frac{OB}{OD}=\frac{AB}{CD}$, \\
thus $\frac{OB}{4}=\frac{6}{3}$, \\
solving yields: $OB=\boxed{8}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, point C is a point on the semicircle above AB, point D is the midpoint of the semicircle below AB. The segments AC, BC, and AD are connected, and through point D, DE is drawn parallel to AB and intersects the extension of CB at point E. If AD equals $5 \sqrt{2}$, what is the maximum value of the product of AC and CE?

Solution:
\textbf{Solution:} As shown in the figure, connect CD and BD, \\
$\because$ DE$\parallel$AB, \\
$\therefore$ $\angle$ABC=$\angle$E, \\
$\because$ $\angle$ABC=$\angle$CDA, \\
$\therefore$ $\angle$E =$\angle$CDA, \\
$\because$ point D is the midpoint of the semicircle below AB, \\
$\therefore$ $\angle$ACD=$\angle$DCE, \\
$\therefore$ $\triangle ACD\sim \triangle DCE$, \\
$\therefore$ $\frac{CD}{CE}=\frac{AC}{CD}$, \\
$\therefore$ $CD^{2}=AC\cdot CE$, \\
$\because$ point D is the midpoint of the semicircle below AB, \\
$\therefore$ $\angle$ADB=$90^\circ$, BD=AD=$5\sqrt{2}$, \\
$\therefore$ $AB=\sqrt{AD^{2}+BD^{2}}=10$, \\
$\because$ $CD\le AB=10$, \\
$\therefore$ $AC\cdot CE\le 10^{2}=\boxed{100}$.


Question: As shown in the figure, $AB$ and $CD$ are two chords of the circle $\odot O$, and they intersect at point $P$. Connect $AD$ and $BD$. Given that $AD=BD=4$ and $PC=6$, what is the length of $CD$?

Solution:
\textbf{Solution:} Connect AC, \\
according to the inscribed angle theorem, we have $\angle C = \angle B$, \\
since AD = BD, \\
it follows that $\angle B = \angle DAB$, \\
thus, $\angle DAP = \angle C$, \\
therefore $\triangle DAP \sim \triangle DCA$, \\
hence, AD : CD = DP : AD, \\
we obtain $AD^{2} = DP \cdot CD = CD \cdot (CD - PC)$, \\
substituting AD = 4, PC = 6 into the equation, we get, \\
$CD^{2} - 6CD - 16 = 0$, \\
solving this, we find CD = \boxed{8} or CD = -2 (discard the latter).


Question: As shown in the figure, quadrilateral $ABCD$ and quadrilateral $AEFG$ are similar figures with point A as the center of similarity, and $AC: AF = 2:3$, then what is the ratio of the area of quadrilateral $ABCD$ to the area of quadrilateral $AEFG$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ and quadrilateral $AEFG$ are similar figures with point $A$ as the center of similitude and the ratio of $AC \colon AF = 2 \colon 3$, \\
therefore, the similarity ratio between quadrilateral $ABCD$ and quadrilateral $AEFG$ is $2 \colon 3$, \\
therefore, the ratio of the areas of quadrilateral $ABCD$ to quadrilateral $AEFG$ is $\boxed{4 \colon 9}$.


Question: As shown in the figure, in $ \triangle ABC$, $ DE\parallel BC$, $ AD=9$, $ DB=3$, $ CE=2$, what is the length of $ AC$?

Solution:
\textbf{Solution:} Since DE $\parallel$ BC, with AD = 9, DB = 3, CE = 2,\\
it follows that $\frac{AD}{DB} = \frac{AE}{EC}$, i.e., $\frac{9}{3} = \frac{AE}{2}$,\\
Solving this, we find AE = 6,\\
Therefore, AC = AE + EC = \boxed{8}


Question: As shown in the figure, D and E are points on sides AB and AC of $\triangle ABC$ respectively, with DE $\parallel$ BC, and the area ratio $S_{\triangle ADE} : S_{\triangle ABC} = 1 : 9$. What is the value of the ratio AD : BD?

Solution:
\textbf{Solution:} Since $DE\parallel BC$,\\
it follows that $\triangle ADE\sim \triangle ABC$,\\
which implies $\frac{S_{\triangle ADE}}{S_{\triangle ABC}}=\frac{AD^2}{AB^2}=\frac{1}{9}$\\
Therefore, $\frac{AD}{AB}=\frac{1}{3}$\\
Consequently, $\frac{AD}{BD}=\boxed{\frac{1}{2}}$


Question: As shown in the figure, two right-angled triangles $\triangle ABC$ and $\triangle BCD$ are placed as shown in the figure, with $\angle BCA=45^\circ$, $\angle D=30^\circ$, and the two hypotenuses intersecting at point O. What is the ratio of the areas of $\triangle AOB$ to $\triangle COD$?

Solution:
\textbf{Solution:} Let $BC=x$,\\
In right triangle $\triangle ABC$,\\
$\because$ $\angle BCA=45^\circ$,\\
$\therefore \tan \angle BCA = \frac{AB}{BC}$,\\
$\therefore AB=BC=x$,\\
In right triangle $\triangle BCD$, $\angle D=30^\circ$,\\
$\therefore \tan D = \frac{BC}{DC}=\frac{\sqrt{3}}{3}$,\\
$\therefore DC= \sqrt{3}BC=\sqrt{3}x$,\\
Furthermore, $\because AB\parallel CD$,\\
$\therefore \triangle AOB \sim \triangle COD$,\\
$\therefore \frac{S_{\triangle AOB}}{S_{\triangle COD}}=\left(\frac{AB}{CD}\right)^2=\left(\frac{x}{\sqrt{3}x}\right)^2=\boxed{\frac{1}{3}}$,\\


Question: In the figure, in $\triangle OAB$, $\angle OAB=90^\circ$, point A is on the x-axis, the graph of the inverse proportion function $y=\frac{k}{x}$ ($x>0$) passes through the midpoint D of hypotenuse OB and intersects AB at point C. If the area of $\triangle OBC$ is 3, what is the value of $k$?

Solution:
\textbf{Solution:} Draw $DE \perp OA$ at point $E$, then $S_{\triangle ODE}=S_{\triangle OAC}=\frac{1}{2}|k|$,

$\because$ D is the midpoint of OB,

$\therefore$ OD=BD=$\frac{1}{2}$OB,

$\because$ DE$\perp$OA and $\angle$OAB$=90^\circ$,

$\therefore$ DE$\parallel$AB,

$\therefore$ $\triangle ODE \sim \triangle OBA$,

$\therefore$ $\frac{S_{\triangle ODE}}{S_{\triangle OBA}}=\left(\frac{OD}{OB}\right)^2=\frac{1}{4}$,

$\therefore$ $S_{\triangle OAB}=4S_{\triangle ODE}=2|k|$,

$\therefore$ $S_{\triangle OBC}=3=S_{\triangle OAB}-S_{\triangle OAC}=\frac{3}{2}|k|$,

Again, $\because k>0$,

$\therefore k=\boxed{2}$,


Question: In the figure, within $\triangle ABC$, point D is a point on side $AB$. If $\angle ACD = \angle B$, $AD = 1$, $AC = 2$, and the area of $\triangle ADC$ is $1$, what is the area of $\triangle BCD$?

Solution:
\textbf{Solution:} Given that $\angle ACD = \angle B$, \\
and $\angle A = \angle A$, \\
thus $\triangle ACD \sim \triangle ABC$, \\
hence $\frac{S_{\triangle ACD}}{S_{\triangle ABC}}=\left(\frac{AD}{AC}\right)^2=\frac{1}{4}$, \\
since $S_{\triangle ACD} = 1$, \\
it follows that $S_{\triangle ABC} = 4$, \\
therefore, the area of $\triangle BCD = S_{\triangle ABC} - S_{\triangle ACD} = \boxed{3}$.


Question: As shown in the figure, AB and CD are two chords of $\odot O$ that intersect at point P. AD and BD are connected, and it is known that AD = BD = 4 and PC = 6. What is the length of CD?

Solution:
\textbf{Solution:} As shown in the figure, connect AC,\\
according to the inscribed angle theorem, we know that $\angle C = \angle B$,\\
since $AD=BD$,\\
therefore $\angle B = \angle DAB$,\\
thus $\angle DAP = \angle C$,\\
hence $\triangle DAP \sim \triangle ACA$,\\
therefore $AD \colon CD = DP \colon AD$,\\
it follows that $AD^2 = DP \cdot CD = CD(CD - PC)$,\\
substituting $AD = 4$, $PC = 6$ into the equation yields,\\
$CD = \boxed{8}$.


Question: As shown in the figure, in $\triangle ABC$, BD and CE are two medians. Then, what is the ratio of the area of $\triangle ADE$ to the area of $\triangle DEF$?

Solution:
\textbf{Solution}:
Given that in $\triangle ABC$, the two medians $BD$ and $CE$ intersect at point $F$,\\
then $DE$ is the midline, and the area of $\triangle CDE$ equals the area of $\triangle ADE$,\\
thus $DE \parallel BC$ and $DE = \frac{1}{2}BC$,\\
which implies $\triangle DEF \sim \triangle BCF$,\\
therefore $\frac{EF}{CF}=\frac{DE}{BC}=\frac{1}{2}$,\\
since $CF = 2EF$,\\
thus the area of $\triangle DEF$ is $\frac{1}{2}$ the area of $\triangle DCF$,\\
hence the area of $\triangle DEF$ is $\frac{1}{3}$ the area of $\triangle CDE$,\\
and therefore, the area of $\triangle DEF$ is $\frac{1}{3}$ the area of $\triangle ADE$,\\
thus the ratio of the areas $\triangle ADE$ to $\triangle DEF$ is $\boxed{3:1}$.


Question: As shown in the figure, $AD \parallel EF \parallel BC$, point $G$ is the midpoint of $EF$, $\frac{EF}{BC} = \frac{3}{5}$. If $EF = 6$, what is the length of $AD$?

Solution:
\textbf{Solution:} Given: $EF\parallel BC$\\
$\therefore$ $\triangle AEF\sim \triangle ABC$\\
$\therefore$ $\frac{AE}{AB}=\frac{EF}{BC}=\frac{3}{5}$\\
$\therefore$ $\frac{BE}{BA}=\frac{2}{5}$\\
Given: $EF\parallel AD$\\
$\therefore$ $\triangle BEG\sim \triangle BAD$\\
$\therefore$ $\frac{EG}{AD}=\frac{BE}{BA}=\frac{2}{5}$\\
Given: Point G is the midpoint of EF, and EF=6\\
$\therefore$ EG=3\\
$\therefore$ $AD=\frac{5}{2}EG=\boxed{\frac{15}{2}}$


Question: As shown in the figure, in the Cartesian coordinate system, if point A has the coordinates $(3,1)$, what is the value of $\sin\alpha$?

Solution:
\textbf{Solution:} As shown in the diagram: draw AB $\perp$ x-axis at point B,\\
$\because$ the coordinates of point A are (3, 1),\\
$\therefore$ OB = 3, and AB = 1,\\
In right triangle ABO, according to the Pythagorean theorem, AO = $\sqrt{OB^{2}+AB^{2}}=\sqrt{3^{2}+1^{2}}=\sqrt{10}$,\\
$\therefore \sin\alpha = \frac{AB}{OA}=\frac{1}{\sqrt{10}}=\frac{\sqrt{10}}{10}$,\\
Therefore, $\sin\alpha = \boxed{\frac{\sqrt{10}}{10}}$


Question: In the Cartesian coordinate system, the ray OA in the first quadrant forms an angle $\alpha$ with the positive x-axis. If $\sin\alpha =\frac{3}{5}$, what could be the coordinates of point P located on ray OA?

Solution:
\textbf{Solution:} Draw PB $\perp$ OB at point B through point P.\\
A, $\because$ point P (3, 5), $\therefore$ OB = 3, PB = 5,\\
$\therefore$ OP = $\sqrt{34}$ (by the Pythagorean theorem),\\
$\therefore \sin\alpha = \frac{PB}{OP} = \frac{5}{\sqrt{34}} = \frac{5\sqrt{34}}{34} \ne \frac{3}{5}$, thus this option does not meet the criteria;\\
B, $\because$ point P (5, 3), $\therefore$ OB = 5, PB = 3,\\
$\therefore$ OP = $\sqrt{34}$ (by the Pythagorean theorem),\\
$\therefore \sin\alpha = \frac{PB}{OP} = \frac{3}{\sqrt{34}} = \frac{3\sqrt{34}}{34} \ne \frac{3}{5}$, thus this option does not meet the criteria;\\
C, $\because$ point P (4, 3), $\therefore$ OB = 4, PB = 3,\\
$\therefore$ OP = $5$ (by the Pythagorean theorem),\\
$\therefore \sin\alpha = \frac{PB}{OP} = \frac{3}{5}$, thus this option meets the criteria;\\
D, $\because$ point P (3, 4), $\therefore$ OB = 3, PB = 4,\\
$\therefore$ OP = $5$ (by the Pythagorean theorem),\\
$\therefore \sin\alpha = \frac{PB}{OP} = \frac{4}{5} \ne \frac{3}{5}$, thus this option does not meet the criteria;\\
Therefore, the answer is: \boxed{C}.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$. Point D is on the extension line of AB, and CD is connected. If $AB = 2BD$ and $\tan\angle BCD = \frac{2}{3}$, what is the value of $\frac{AC}{BC}$?

Solution:
\textbf{Solution:} Draw $DM\perp BC$ from point D, intersecting the extension of BC at point M,\\
$\because$ $\angle ACB=\angle DMB=90^\circ$, $\angle ABC=\angle DBM$,\\
$\therefore$ $\triangle ABC\sim \triangle DBM$,\\
$\therefore$ $\frac{BD}{AB}=\frac{BM}{BC}=\frac{DM}{AC}$,\\
$\because$ AB＝2BD,\\
$\therefore$ $\frac{BD}{AB}=\frac{BM}{BC}=\frac{DM}{AC}=\frac{1}{2}$,\\
In $\triangle CDM$,\\
since $\tan\angle MCD=\frac{2}{3}=\frac{DM}{CM}$, let DM＝2k, then CM＝3k,\\
also $\because$ $\frac{BM}{BC}=\frac{1}{2}=\frac{DM}{AC}$,\\
$\therefore$ BC＝2k, AC＝4k,\\
$\therefore$ $\frac{AC}{BC}=\frac{4k}{2k}=\boxed{2}$.


Question: As shown in the figure, two tangents $PA$ and $PB$ are drawn from a point $P$ outside the circle $O$, and the points of tangency are $A$ and $B$, respectively. If $\angle APB=60^\circ$ and $PA=8$, then what is the radius of circle $O$?

Solution:
\textbf{Solution:} Draw OA, OP.\\
Since PA, PB are tangents to the circle $O$ and $\angle APB=60^\circ$,\\
it follows that $\angle OPA= \frac{1}{2}\angle APB=30^\circ$ and OA$\perp$OP,\\
thus OA=AP$\cdot \tan \angle OPA=8\times \frac{\sqrt{3}}{3}=\frac{8\sqrt{3}}{3}$,\\
Therefore, the radius of circle $O$ is $\boxed{\frac{8\sqrt{3}}{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, and $AD$ is drawn perpendicular to $BC$ at point $D$. If $\frac{BD}{CD} = \frac{4}{3}$, what is the value of $\tan C$?

Solution:
\textbf{Solution:}

Given $\frac{BD}{CD} = \frac{4}{3}$,\\
let $BD=4x$, and $CD=3x$.\\
Since $AD \perp BC$,\\
we have $\angle ADB = \angle ADC = \angle BAC = 90^\circ$,\\
which implies $\angle BAD + \angle CAD = \angle C + \angle CAD = 90^\circ$,\\
hence $\angle C = \angle BAD$,\\
leading to $\triangle BAD \sim \triangle ACD$,\\
thus $\frac{AD}{BD} = \frac{CD}{AD}$, i.e., $AD^2 = BD \cdot CD = 3x \cdot 4x = 12x^2$,\\
therefore, $AD = 2\sqrt{3}x$,\\
from which it follows $\tan C = \frac{AD}{CD} = \frac{2\sqrt{3}x}{3x} = \boxed{\frac{2\sqrt{3}}{3}}$.


Question: As shown in the figure, the vertices of $\triangle ABC$ are all on the vertices of small squares with equal side lengths. What is the value of $\cos\angle BAC$?

Solution:
\[ \textbf{Solution:} \text{As shown in the figure, construct } CD\perp AB \text{ at point } D, \]
\[ \text{Given that, } S_{\triangle ABC}=\frac{1}{2}\times 2\times 2=2, \quad AB=\sqrt{2^{2}+4^{2}}=2\sqrt{5}, \quad AC=2\sqrt{2}, \]
\[ \therefore S_{\triangle ABC}=\frac{1}{2}AB\cdot CD=2, \quad CD=\frac{2S_{\triangle ABC}}{AB}=\frac{4}{2\sqrt{5}}=\frac{2\sqrt{5}}{5}, \]
\[ \text{In } \triangle ADC, \quad AD=\sqrt{AC^{2}-CD^{2}}=\frac{6\sqrt{5}}{5}, \]
\[ \therefore \cos\angle BAC=\frac{AD}{AC}=\frac{\frac{6\sqrt{5}}{5}}{2\sqrt{2}}=\boxed{\frac{3\sqrt{10}}{10}}. \]


Question: As shown in the figure, $\triangle AOB$ in the Cartesian coordinate system is rotated $90^\circ$ clockwise around point $O$ to get $\triangle A^\prime OB^\prime$. Given that $\angle AOB=60^\circ$, $\angle B=90^\circ$, and $AB= \sqrt{3}$, what are the coordinates of point $B^\prime$?

Solution:
\textbf{Solution:} As shown in the figure, draw line $B^{\prime}C \perp x$-axis at point C, \\
since $\triangle AOB$ rotates $90^\circ$ clockwise around point O to become $\triangle A^{\prime}OB^{\prime}$,\\
therefore, $OB^{\prime} = OB$, and $\angle BOB^{\prime} = 90^\circ$,\\
since $\angle AOB = 60^\circ$, and $AB= \sqrt{3}$,\\
therefore, $OB = 1$,\\
therefore, $OB^{\prime} = 1$,\\
$\angle B^{\prime}OC = 180^\circ - \angle AOB - \angle BOB^{\prime} = 180^\circ - 60^\circ - 90^\circ = 30^\circ$,\\
therefore, $OC = OB^{\prime}\cos30^\circ = 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$,\\
$B^{\prime}C = OB^{\prime}\sin30^\circ = 1 \times \frac{1}{2} = \frac{1}{2}$,\\
therefore, the coordinates of $B^{\prime}$ are $\boxed{\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)}$.


Question: As shown in the diagram, in $\triangle ABC$, $\angle C=90^\circ$, $AC=4$, and $BC=8$, then what is the value of $\tan B$?

Solution:
\textbf{Solution:} Since $\angle C=90^\circ$, $AC=4$, and $BC=8$, \\
therefore $\tan B=\frac{AC}{BC}=\boxed{\frac{1}{2}}$


Question: As shown in the figure, the vertices of $\triangle ABC$ are lattice points of a square grid. What is the value of $\sin B$?

Solution:
\textbf{Solution:} By the Pythagorean theorem, we have: $AB=\sqrt{3^2+3^2}=3\sqrt{2}$\\
$\therefore \sin B=\frac{3}{3\sqrt{2}}=\frac{\sqrt{2}}{2}$\\
Therefore, $\sin B=\boxed{\frac{\sqrt{2}}{2}}$


Question: As shown in the figure, the angle of elevation to the top of the flagpole A at point C is measured to be $30^\circ$. Moving forward 10 meters to point D, the angle of elevation to point A is measured to be $60^\circ$. What is the height of the flagpole? Meters.

Solution:
\textbf{Solution:} Given from the problem, $\angle C=30^\circ$ and $\angle ADB=60^\circ$,\\
$\therefore \angle DAC=\angle ADB-\angle C=30^\circ$,\\
$\therefore \angle DAC=\angle C$,\\
$\therefore AD=DC=10$ meters,\\
In $\triangle ADB$, $\sin \angle ADB= \frac{AB}{AD}$,\\
Then $AB=AD \cdot \sin \angle ADB=10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ (meters),\\
Therefore, the height of the flagpole is $AB=\boxed{5\sqrt{3}}$ (meters).


Question: As shown in the diagram, in the Cartesian coordinate system, the rhombus OABC has side OA on the x-axis, with point $A(10, 0)$, and $\sin\angle COA=\frac{4}{5}$. If the inverse proportion function $y=\frac{k}{x}$ ($k>0, x>0$) passes through point C, what is the value of $k$?

Solution:
\[ \text{{\bfseries Solution:}} \text{ As shown in the figure, draw } CE \perp OA \text{ at point E,} \]
\[ \because \text{ the side OA of the rhombus OABC lies on the x-axis, with point } A(10,0), \]
\[ \therefore OC = OA = 10, \]
\[ \because \sin\angle COA = \frac{4}{5} = \frac{CE}{OC}, \]
\[ \therefore CE = 8, \]
\[ \therefore OE = \sqrt{CO^2 - CE^2} = 6, \]
\[ \therefore \text{ the coordinates of point C are }(6, 8), \]
\[ \because \text{ if the inverse ratio function } y = \frac{k}{x} (k>0, x>0) \text{ passes through point C,} \]
\[ \therefore k = 6 \times 8 = \boxed{48} \]


Question: As shown in the figure, to facilitate pedestrians crossing a certain footbridge, the municipal government constructed ramps at both ends of the 10-meter-high footbridge. The design of the slope satisfies $\sin A = \frac{1}{3}$. What is the length of the ramp AC?

Solution:
\textbf{Solution:} In $\triangle ABC$, where $\angle ABC = 90^\circ$ and $BC = 10$ meters, with $\sin A = \frac{1}{3}$,\\
then $\frac{BC}{AC} = \frac{1}{3}$, that is, $\frac{10}{AC} = \frac{1}{3}$,\\
Solving this, we get: $AC = \boxed{30}$ (meters).


Question: As shown in the diagram, the sunlight strikes the ground at an angle of $80^\circ$. Given that window $AB$ is 2 meters high, to install a horizontal sunshade $DC$ at point $D$, which is 0.2 meters above the window, such that the sunlight cannot directly enter the room, what is the minimum length of sunshade $DC$?

Solution:
\textbf{Solution}: Given that $DA=0.2$ meters, $AB=2$ meters,\\
thus $DB=DA+AB=2.2$ meters,\\
since the angle between the ray and the ground is $80^\circ$, thus $\angle BCD=80^\circ$.\\
Also, since $\tan \angle BCD= \frac{DB}{DC}$,\\
thus $DC= \frac{DB}{\tan \angle BCD}=\frac{2.2}{\tan 80^\circ}=\boxed{\frac{2.2}{\tan 80^\circ}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=3$, $BC=4$, then what is the value of $\cos A$?

Solution:
\textbf{Solution:} In the right triangle $\triangle ABC$, with $\angle C=90^\circ$, $AC=3$, and $BC=4$,\\
by the Pythagorean theorem, we obtain\\
$AB= \sqrt{AC^{2}+BC^{2}}= 5$.\\
$\cos A= \frac{AC}{AB} = \boxed{\frac{3}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, $CD \perp AB$ at D. If $AC=2\sqrt{3}$ and $AB=3\sqrt{2}$, then what is the value of $\tan\angle BCD$?

Solution:
\textbf{Solution:} Given that $\angle ACB=90^\circ$, $AC=2\sqrt{3}$, $AB=3\sqrt{2}$, \\
hence $BC=\sqrt{(3\sqrt{2})^{2}-(2\sqrt{3})^{2}}=\sqrt{6}$, and $\angle B+\angle A=90^\circ$, \\
since $CD\perp AB$ at $D$, \\
thus $\angle B+\angle BCD =90^\circ$, \\
therefore $\angle BCD=\angle A$, \\
thus $\tan\angle BCD=\tan\angle A=\frac{BC}{AC}=\frac{\sqrt{6}}{2\sqrt{3}}=\boxed{\frac{\sqrt{2}}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=13$ and $BC=10$, what is the value of $\sin B$?

Solution:
\textbf{Solution:} As shown in the figure, draw AD $\perp$ BC at point D,\\
$\therefore$ $\angle$ADB$=90^\circ$,\\
$\because$ AB=AC=13, BC=10,\\
$\therefore$ BD=5,\\
$\therefore$ AD=$\sqrt{13^2-5^2}=12$,\\
$\therefore$ $\sin B=\frac{AD}{AB}=\boxed{\frac{12}{13}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC=2\sqrt{5}$, $\angle BAC=\alpha^\circ$, $\tan\angle ABC=\frac{1}{2}$, $G$ is the midpoint of $BC$, and $D$ is a moving point in the plane with $DG=\frac{\sqrt{5}}{5}$. If line segment $BD$ rotates counterclockwise by $\alpha^\circ$ around point $D$ to obtain $DB'$, what is the maximum area of quadrilateral $BACB'$?

Solution:
\textbf{Solution:} As shown in the figure, connect AD, AG, and draw GH perpendicular to AB at point H through point G,\\
since $AB=AC=2\sqrt{5}$ and $BG=GC$,\\
thus, $AG\perp BC$,\\
since $\tan\angle ABC=\frac{AG}{BG}=\frac{1}{2}$,\\
thus, $AG=2$ and $BG=4$,\\
since $\sin\angle ABG=\sin\angle GBH$,\\
thus, $\frac{GH}{BC}=\frac{AG}{AB}$,\\
thus, $\frac{GH}{4}=\frac{2}{2\sqrt{5}}$,\\
thus, $GH=\frac{4\sqrt{5}}{5}$,\\
since $AB=AC$, $DB=DB'$, $\angle BAC=\angle BDB'$,\\
thus, $\angle ABC=\angle DBB'$, $\frac{AB}{BC}=\frac{BD}{BB'}$,\\
thus, $\angle ABD=\angle CBB'$,\\
thus, $\triangle ABD\sim \triangle CBB'$,\\
thus, $\frac{S_{\triangle ABD}}{S_{\triangle CBB'}}=\left(\frac{AB}{BC}\right)^2=\left(\frac{2\sqrt{5}}{4}\right)^2=\frac{5}{16}$,\\
since $DG=\frac{\sqrt{5}}{5}$,\\
thus, the motion trajectory of point G is a circle with G as the center and $\frac{\sqrt{5}}{5}$ as the radius. When point D is on the extension line of HG, the area of $\triangle ABD$ is maximized, with the maximum value $=\frac{1}{2}\times 2\sqrt{5}\times \left(\frac{4\sqrt{5}}{5}+\frac{\sqrt{5}}{5}\right)=5$,\\
thus, the maximum area of $\triangle BCB'$ is $16$,\\
thus, the maximum area of quadrilateral $BACB'$ is $\frac{1}{2}\times 8\times 2+16=\boxed{24}$.


Question: As shown in the figure, the slope of the waterside slope AB of the river embankment cross section is 1:2, and the slope length AB = $6\sqrt{5}$. What is the height of the embankment?

Solution:
Solution: Since the slope of the upstream slope AB is 1:2, it means that $\tan A = \frac{BC}{AC} = \frac{1}{2}$,\\
therefore $AC = 2BC$. Also, in the right triangle $\triangle ABC$, $BC^{2}+AC^{2}=AB^{2}$, $AB=6\sqrt{5}$,\\
thus $BC^{2}+(2BC)^{2}=(6\sqrt{5})^{2}$. Solving this yields: $BC=\boxed{6}$ (taking the positive value).


Question: As shown in the figure, in $\triangle ABC$, $\cos B = \frac{\sqrt{2}}{2}$, $\sin C = \frac{3}{5}$, and $AC=5$, what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Let $AH\perp BC$,\\
then $AH=AC\sin C=5\times\frac{3}{5}=3$,\\
$\therefore$ $CH=\sqrt{AC^{2}-AH^{2}}=4$,\\
$\because$ $\cos B=\frac{\sqrt{2}}{2}$,\\
$\therefore$ $\angle B=45^\circ$,\\
$\therefore$ $BH=AH=3$,\\
$\therefore$ $BC=BH+CH=3+4=7$,\\
$\therefore$ $S_{\triangle ABC}=\frac{1}{2}\times BC\times AH=\frac{1}{2}\times7\times3=\boxed{\frac{21}{2}}$.


Question: As shown in the figure, circle $ \odot A$ passes through the points O, D, C, and B. Line segments CO and CD are drawn. Given that $\angle DCO = 30^\circ$ and $B(4, 0)$, what are the coordinates of point D?

Solution:
\textbf{Solution:} As shown in the figure, connect DB,\\
since $\angle DCO=30^\circ$ and $\angle DOB=90^\circ$,\\
therefore, $\angle OBD=30^\circ$, DB is the diameter,\\
since B(4,0),\\
therefore, $OB=4$,\\
in $\triangle DOB$, $\tan\angle OBD= \frac{OD}{OB}$,\\
thus $\frac{OD}{4}=\frac{\sqrt{3}}{3}$,\\
therefore, $OD=\frac{4\sqrt{3}}{3}$,\\
therefore, the coordinates of point D are $\boxed{(0, \frac{4\sqrt{3}}{3})}$.


Question: As shown in the figure, in square $ABCD$ with a side length of $4$, point $E$ is a point on side $CD$, and point $F$ is symmetric to point $D$ with respect to line $AE$. Connect $AF$ and $BF$. If $\tan\angle ABF=2$, what is the length of $DE$?

Solution:
\textbf{Solution:} Construct FN$\perp$AB at point N, and extend NF to intersect CD at point M,\\
since AB$\parallel$CD,\\
therefore MN$\perp$CD,\\
therefore $\angle$FME＝$90^\circ$,\\
since $\tan\angle$ABF＝2,\\
therefore $\frac{FN}{BN} = 2$,\\
Let BN＝x, then FN＝2x,\\
therefore AN＝4−x,\\
since point F is the symmetric point of point D with respect to line AE,\\
therefore DE＝EF, DA＝AF＝4,\\
since AE＝AE,\\
therefore $\triangle$ADE$\cong\triangle$AFE (SSS),\\
therefore $\angle$D＝$\angle$AFE＝$90^\circ$,\\
since AN$^{2}$＋NF$^{2}$＝AF$^{2}$,\\
therefore $(4−x)^{2}＋(2x)^{2}=4^{2}$,\\
therefore $x_{1}=0$ (discard), $x_{2}=\frac{8}{5}$,\\
therefore AN＝4−x＝4−$\frac{8}{5}$＝$\frac{12}{5}$, MF＝4−2x＝4−$\frac{16}{5}$＝$\frac{4}{5}$,\\
since $\angle$EFM+$\angle$AFN＝$\angle$AFN+$\angle$FAN＝$90^\circ$,\\
therefore $\angle$EFM＝$\angle$FAN,\\
therefore $\cos\angle$EFM＝$\cos\angle$FAN,\\
therefore $\frac{FM}{EF} = \frac{AN}{AF}$, that is $\frac{\frac{4}{5}}{EF}=\frac{\frac{12}{5}}{4}$,\\
therefore EF＝$\boxed{\frac{4}{3}}$,\\
therefore DE＝EF＝$\frac{4}{3}$.


Question: In the coordinate system shown, let $O(0,0)$, $A(3,3\sqrt{3})$, and $B(6,0)$. If $\triangle OAB$ is folded along line $CD$ such that point $A$ precisely falls on point $E$ on segment $OB$ and given that $OE=\frac{6}{5}$, what is the value of $AC: AD$?

Solution:
\textbf{Solution:} Draw AF$\perp$OB at F, as shown in the figure: \\
Since A(3, 3$\sqrt{3}$), B(6, 0),\\
therefore AF $=$ 3$\sqrt{3}$, OF $=$ 3, OB $=$ 6,\\
therefore BF $=$ 3,\\
therefore OF $=$ BF,\\
therefore AO $=$ AB,\\
since $\tan\angle AOB = \frac{AF}{OF} = \sqrt{3}$,\\
therefore $\angle AOB = 60^\circ$,\\
therefore $\triangle AOB$ is an equilateral triangle,\\
therefore $\angle AOB = \angle ABO = 60^\circ$,\\
since folding $\triangle OAB$ along the line CD such that point A precisely falls on the point E on segment OB,\\
therefore $\angle CED = \angle OAB = 60^\circ$,\\
therefore $\angle OCE = \angle DEB$,\\
therefore $\triangle CEO \sim \triangle EDB$,\\
therefore $\frac{OE}{BD} = \frac{CE}{ED} = \frac{CO}{BE}$,\\
since OE $= \frac{6}{5}$,\\
therefore BE $=$ OB - OE $= 6 - \frac{6}{5} = \frac{24}{5}$,\\
Let CE $= a$, then CA $= a$, CO $= 6-a$, ED $= b$, then AD $= b$, DB $= 6-b$,\\
then $\frac{\frac{6}{5}}{6-b} = \frac{a}{b}$, $\frac{6-a}{\frac{24}{5}} = \frac{a}{b}$,\\
therefore $6b = 30a - 5ab$ (1), $24a = 30b - 5ab$ (2),\\
(2) - (1) gives: $24a - 6b = 30b - 30a$,\\
therefore $\frac{a}{b} = \frac{2}{3}$,\\
that is AC$\colon$AD $=$ $\boxed{\frac{2}{3}}$.


Question: As shown in the figure, $\odot O$ is a circle centered at the origin $O$ with a radius of $4\sqrt{2}$. The coordinates of point $P$ are $(2, 2)$. If the chord $AB$ passes through point $P$, what is the minimum area of the shaded region in the figure?

Solution:
\textbf{Solution:} Given that when OP$\perp$A'B', the area of the shaded region is minimized,\\
$\because$P(2,2),\\
$\therefore$OP = 2 $\sqrt{2}$,\\
$\because$OA' = OB' = 4 $\sqrt{2}$,\\
$\therefore \cos \angle$A'OP = $\cos \angle$B'OP = $\frac{1}{2}$,\\
$\therefore$$\angle$A'OP = $\angle$B'OP = 60$^\circ$,\\
$\therefore$$\angle$A'OB' = 120$^\circ$, A'$\text{P}$ = 4 $\sqrt{2}$ $\times$ $\frac{\sqrt{3}}{2}$ = 2 $\sqrt{6}$,\\
$\therefore$A'$\text{B}'$ = 4 $\sqrt{6}$,\\
$\therefore$S$_{\text{shaded}}$ = S$_{\text{sector}}$$_{OA'B'}$ - S$_{\triangle}$$_{A'OB'}$ = $\frac{120\pi \times {(4\sqrt{2})}^2}{360}$ - $\frac{1}{2} \times 4\sqrt{6} \times 2\sqrt{2}$ = $\boxed{\frac{32\pi}{3} - 8\sqrt{3}}$.


Question: As shown in the figure, points A, B, and C are all located at the vertices of the small square, and the side length of each small square is 1. What is the value of $\cos\angle BAC$?

Solution:
Solution: Connect BC. Since the side length of each small square is 1, we have AB=$\sqrt{5}$, BC=$\sqrt{5}$, AC=$\sqrt{10}$. Since $(\sqrt{5})^{2}+(\sqrt{5})^{2}=(\sqrt{10})^{2}$, triangle ABC is a right-angled triangle. Therefore, $\cos\angle BAC=\frac{AB}{AC}=\frac{\sqrt{5}}{\sqrt{10}}=\frac{\sqrt{2}}{2}$.

Hence, $\cos\angle BAC = \boxed{\frac{\sqrt{2}}{2}}$.


Question: As shown in the figure, in $ \triangle ABC$, $ BD \perp AB$, with BD and AC intersecting at point D, $ AD = \frac{4}{7}AC$, $ AB = 4$, and $ \angle ABC = 150^\circ$. What is the area of $ \triangle DBC$?

Solution:
\textbf{Solution:} As shown in the diagram, draw a perpendicular from point C to BD, intersecting the extension of BD at point E, \\
then $ \angle E=90^\circ$, \\
since $BD\perp AB$ and $CE\perp BD$, \\
it follows that $AB\parallel CE$, $\angle ABD=90^\circ$, \\
hence $\triangle ABD\sim \triangle CED$, \\
therefore $\frac{AD}{CD}=\frac{AB}{CE}=\frac{BD}{DE}$, \\
since $AD=\frac{4}{7}AC$, \\
it follows that $\frac{AD}{CD}=\frac{4}{3}$, \\
furthermore, since $AB=4$, \\
it follows that $\frac{4}{CE}=\frac{4}{3}=\frac{BD}{DE}$, \\
therefore $CE=3$, $BD=\frac{4}{7}BE$, \\
since $\angle ABC=150^\circ$, $\angle ABD=90^\circ$, \\
it follows that $\angle CBE=60^\circ$, \\
therefore, in $Rt\triangle BCE$, $\tan\angle CBE=\frac{CE}{BE}=\sqrt{3}$, \\
hence $BE=\frac{\sqrt{3}}{3}CE=\sqrt{3}$, \\
thus $BD=\frac{4}{7}BE=\frac{4\sqrt{3}}{7}$, \\
therefore, the area of $\triangle BCD=\frac{1}{2}\cdot BD\cdot CE=\frac{1}{2}\times 3\times \frac{4\sqrt{3}}{7}=\boxed{\frac{6\sqrt{3}}{7}}$.


Question: As shown in the figure, it is known that $O$ is the circumcircle of $\triangle ABC$, and $AD$ is the diameter of $O$ with $CD$ connected. If $AD=3$ and $AC=2$, what is the value of $\cos B$?

Solution:
\textbf{Solution:} Given that $AD$ is the diameter of circle $O$,\\
it follows that $\angle ACD=90^\circ$,\\
and therefore $DC=\sqrt{AD^{2}-CD^{2}}=\sqrt{3^{2}-2^{2}}=\sqrt{5}$,\\
Given that arc $AC$ equals arc $AC$,\\
it implies that $\angle B = \angle D$,\\
thus $\cos B = \cos D = \frac{DC}{AD} = \frac{\sqrt{5}}{3}$.\\
Hence, $\cos B = \boxed{\frac{\sqrt{5}}{3}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $E$ and $F$ are the midpoints of $AB$ and $AD$ respectively. If $EF=2$, $BC=5$, and $CD=3$, then what is the value of $\sin C$?

Solution:
\textbf{Solution:} Draw line $BD$, \\
since points E and F are the midpoints of $AB$ and $AD$ respectively, \\
therefore, $EF=\frac{1}{2}BD$, \\
since $EF=2$, \\
therefore, $BD=2EF=4$, \\
also, since $BC=5$ and $CD=3$, \\
therefore, $BD^{2}+CD^{2}=BC^{2}$, \\
thus, $\triangle BCD$ is a right-angled triangle with $\angle BDC=90^\circ$, \\
therefore, $\sin C=\frac{BD}{BC}=\boxed{\frac{4}{5}}$.


Question: As shown in the figure, the vertices of $\triangle ABC$ are all on the grid points of the graph paper. What is the value of $\sin A$?

Solution:
\textbf{Solution:} Given the diagram, we have $AC=\sqrt{1^2+3^2}=\sqrt{10}$, \\
$\therefore \sin A=\frac{1}{AC}=\frac{1}{\sqrt{10}}=\boxed{\frac{\sqrt{10}}{10}}$


Question: As shown in the diagram, the slope ratio of the cross-section of the river embankment $BC: AC$ is $1: \sqrt{3}$, with $AC = 6\,m$. What is the length of the slope surface $AB$?

Solution:
\textbf{Solution:} Given the slope ratio $BC:AC$ is $1:\sqrt{3}$, and $AC=6$m, \\
it follows that $\frac{BC}{6} = \frac{1}{\sqrt{3}}$, \\
solving this gives $BC=2\sqrt{3}$ (m), \\
by the Pythagorean theorem, $AB=\sqrt{AC^{2}+BC^{2}} =\boxed{4\sqrt{3}}$ (m).


Question: As shown in the figure, in the given grid, the side length of the small square is 1, and the points $A$, $B$, and $O$ are all on the lattice points. Then, what is the value of $\cos \angle A$?

Solution:
\textbf{Solution:} As shown in the figure, according to the problem we have: $OC=2$, $AC=4$, \\
Based on the Pythagorean theorem, we get $AO=\sqrt{AC^2+OC^2}=\sqrt{4^2+2^2}=2\sqrt{5}$, \\
$\therefore$ $\cos \angle A=\frac{AC}{AO}=\frac{4}{2\sqrt{5}}=\boxed{\frac{2\sqrt{5}}{5}}$


Question: As shown in the figure, $\triangle ABC$ is an inscribed triangle of $\odot O$ with $AB=BC$, $\angle BAC=30^\circ$, and $AD$ is the diameter with $AD=8$. What is the length of $AC$?

Solution:
\textbf{Solution:} As shown in the figure, connect OB, \\
since $\triangle ABC$ is an inscribed triangle of $\odot O$, \\
therefore OB perpendicularly bisects AC, \\
therefore $AM=CM=\frac{1}{2}AC$, $OM\perp AM$, \\
furthermore, since $AB=BC$, $\angle BAC=30^\circ$, \\
therefore, $\angle BCA=30^\circ$, \\
therefore, $\angle BOA=60^\circ$, \\
furthermore, since AD=8, \\
therefore, AO=4, \\
therefore, $\sin 60^\circ =\frac{AM}{AO}=\frac{AM}{4}=\frac{\sqrt{3}}{2}$, \\
solving gives: $AM=2\sqrt{3}$, \\
therefore, $AC=2AM=4\sqrt{3}$. \\
Thus, the answer is: $AC=\boxed{4\sqrt{3}}$.


Question: As shown in the figure, the side length of the square $ABCD$ is $1$. Extend $BA$ to $E$ so that $AE=1$. Connect $EC$ and $ED$. Then, what is the value of $\sin\angle CED$?

Solution:
\textbf{Solution:} Connect AC, draw EF $\perp$ CA, meeting the extension of CA at point F.\\
$\because$ Quadrilateral ABCD is a square, $\therefore$ $\angle$BAC = 45$^\circ$, $\angle$DAE = $\angle$DAB = 90$^\circ$.\\
$\because$ AD = AE = 1, $\therefore$ $\angle$AED = $\angle$ADE = 45$^\circ$, that is $\angle$DEA = $\angle$CAB = 45$^\circ$, $\therefore$ AC $\parallel$ ED, $\therefore$ $\angle$CED = $\angle$ECA.\\
$\because$ AE = 1, $\therefore$ by the Pythagorean theorem: EF = AF = $\frac{\sqrt{2}}{2}$.\\
$\because$ in $\triangle$EBC, by the Pythagorean theorem: CE$^{2}$ = $1^{2} + 2^{2} = 5$, $\therefore$ CE = $\sqrt{5}$, $\therefore$ $\sin\angle$CED = $\sin\angle$ECF = $\frac{EF}{CE} = \frac{\frac{\sqrt{2}}{2}}{\sqrt{5}} = \boxed{\frac{\sqrt{10}}{10}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C = 90^\circ$, point D is on $AC$, and $\angle DBC = \angle A$. If $AC=4$ and $\cos A = \frac{4}{5}$, what is the length of $BD$?

Solution:
\textbf{Solution:} Since $\angle C=90^\circ$,\\
hence $\cos A=\frac{AC}{AB}$,\\
since $AC=4$, $\cos A=\frac{4}{5}$,\\
thus $AB=5$,\\
by the Pythagorean theorem, we get $BC= \sqrt{AB^{2}-AC^{2}}=3$,\\
since $\angle DBC=\angle A$,\\
thus $\cos \angle DBC=\cos A= \frac{4}{5}$,\\
therefore $\cos \angle DBC= \frac{BC}{BD}=\frac{4}{5}$, which means $\frac{3}{BD}=\frac{4}{5}$\\
hence $BD=\boxed{\frac{15}{4}}$


Question: As shown in the figure, in $\triangle ABC$, $\angle A = 90^\circ$, $AD \perp BC$ at point D. If $BD = 3$ and $CD = 2$, then what is the value of $\tan B$?

Solution:
\textbf{Solution:} Given $\because$ $\angle BAC=90^\circ$, $AD\perp BC$,\\
$\therefore$ $\angle BAC=\angle ADC=\angle ADB=90^\circ$,\\
$\therefore$ $\angle B+\angle BAD=\angle BAD+\angle CAD=90^\circ$,\\
$\therefore$ $\angle B=\angle CAD$,\\
$\therefore$ $\triangle ADB\sim \triangle CAD$,\\
$\therefore$ $\frac{AD}{CD}=\frac{BD}{AD}$,\\
$\therefore$ $AD^2=BD\cdot CD$,\\
$\because$ $BD=3$, $CD=2$,\\
$\therefore$ $AD=\sqrt{6}$,\\
$\therefore$ $\tan B=\frac{AD}{BD}=\boxed{\frac{\sqrt{6}}{3}}$;


Question: As shown in the diagram, in rhombus ABCD, DE $\perp$ AB, $\cos A = \frac{3}{5}$, and BE = 2, then what is the value of $\tan \angle DBE$?

Solution:
\textbf{Solution:} Let the side length of rhombus $ABCD$ be $t$.\\
$\because BE = 2$,\\
$\therefore AE = t - 2$.\\
$\therefore \cos A = \frac{3}{5} = \frac{AE}{AD} = \frac{AB - BE}{AD}$,\\
$\therefore \frac{3}{5} = \frac{t - 2}{t}$,\\
$\therefore t = 5$.\\
$\therefore AE = 5 - 2 = 3$.\\
$\therefore DE = \sqrt{AD^2 - AE^2} = \sqrt{5^2 - 3^2} = 4$.\\
$\therefore \tan \angle DBE = \frac{DE}{BE} = \frac{4}{2} = \boxed{2}$.


Question: As shown in the figure, in rectangle $ABCD$, $DE\perp AC$ at $E$, let $\angle ADE=\alpha$ and $\cos\alpha =\frac{3}{5}$, with $AB=5$. What is the length of $AD$?

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle, it follows that $\angle B=\angle BAC=90^\circ$, BC=AD. Therefore, $\angle BAC+\angle DAE=90^\circ$.\\
Since $DE\perp AC$, it follows that $\angle ADE+\angle DAE=90^\circ$, and therefore, $\angle BAC= \angle ADE=\alpha$.\\
In right triangle $\triangle ABC$, since $\cos\alpha =\frac{3}{5}$ and $AB=5$, it follows that $AC=\frac{AB}{\cos\alpha}=\frac{25}{3}$.\\
Therefore, AD=BC= $\sqrt{AC^{2}-AB^{2}}=\sqrt{\left(\frac{25}{3}\right)^{2}-5^{2}}=\boxed{\frac{20}{3}}$.


Question: As shown in the Cartesian coordinate system, a circular arc passes through the lattice points A, B, and C. What are the coordinates of the center of the circle?

Solution:
\textbf{Solution:} Based on the corollary of the Perpendicular Diameter Theorem: the perpendicular bisector of a chord always passes through the center of the circle,\\
one can construct the perpendicular bisectors of chords AB and BC, and their intersection point will be the center of the circle.\\
As shown in the figure, thus the center of the circle is $\boxed{(2,0)}$.


Question: As shown in the figure, points A, B, and C are all on circle $O$. When $\angle OAC = 40^\circ$, what is the measure of $\angle B$?

Solution:
\textbf{Solution:} Let's select any point D on the major arc AB, and connect AD and CD.\\
Since OA=OC, $\angle$OAC=40$^\circ$,\\
it follows that $\angle$OAC=$\angle$OCA=40$^\circ$,\\
which implies $\angle$AOC=180$^\circ$-40$^\circ$-40$^\circ$=100$^\circ$,\\
thus $\angle$D=$\frac{1}{2}$$\angle$AOC=50$^\circ$,\\
Since quadrilateral ABCD is circumscribed around $\odot$O,\\
it follows that $\angle$D+$\angle$B=180$^\circ$,\\
therefore $\angle$B=180$^\circ$-50$^\circ$=\boxed{130^\circ},


Question: As shown in the figure, the diameter $CD$ of $O$ is perpendicular to the chord $AB$ at point $E$, and $CE=2$, $OB=4$. What is the length of $AB$?

Solution: \textbf{Solution:} Because the diameter CD of the circle $\odot$O is perpendicular to the chord AB at point E, therefore $AB=2BE$. Because $CE=2$, and $OB=4$, therefore $OE=4-2=2$, hence $BE= \sqrt{OB^{2}-OE^{2}}=\sqrt{4^{2}-2^{2}}=2\sqrt{3}$, therefore $AB=\boxed{4\sqrt{3}}$.


Question: As shown in the figure, points $A$, $B$, $C$ are on the circle $\odot O$, $\angle ACB=105^\circ$, then what is the value of $\angle \alpha$?

Solution:
\textbf{Solution:} Let's take any point $D$ on the major arc $AB$ and connect $AD$ and $BD$,\\
since quadrilateral $ACBD$ is inscribed in circle $O$, and $\angle ACB=105^\circ$,\\
it follows that $\angle ADB=180^\circ-\angle C=180^\circ-105^\circ=75^\circ$,\\
hence $\angle AOB=2\angle ADB=2\times 75^\circ=150^\circ$.\\
Therefore, the answer is: $\boxed{150^\circ}$.


Question: Given: As shown in the figure, $OA$ and $OB$ are two radii of $\odot O$, and $OA \perp OB$. Point $C$ is on $\odot O$. What is the measure of $\angle ACB$?

Solution:
\textbf{Solution:} Since $OA \perp OB$,\\
it follows that $\angle AOB = 90^\circ$,\\
which means $\angle ACB = \frac{1}{2} \angle AOB = \boxed{45^\circ}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\odot O$, with $DE$ being the diameter of $\odot O$. Connect $BD$. If $\angle BCD = 2\angle BAD$, then what is the measure of $\angle BDE$?

Solution:
\textbf{Solution:} Connect BE, \\
$\because$ Quadrilateral ABCD is inscribed in circle $O$, \\
$\therefore \angle BCD + \angle BAD = 180^\circ$, \\
$\because \angle BCD = 2\angle BAD$, \\
$\therefore \angle BAD = 60^\circ$, \\
By the inscribed angle theorem, we have: $\angle BED = \angle BAD = 60^\circ$, \\
$\because DE$ is the diameter of circle $O$, \\
$\therefore \angle EBD = 90^\circ$, \\
$\therefore \angle BDE = 90^\circ - 60^\circ = \boxed{30^\circ}$.


Question: As shown in the figure, point $I$ is the incenter of $\triangle ABC$. Line segment $AI$ is extended to intersect the circumcircle of $\triangle ABC$ at point $D$. Point $E$ is the midpoint of chord $AC$. Lines $CD$, $EI$, and $IC$ are connected. When $AI=2CD$, $IC=6$, and $ID=5$, what is the length of $IE$?

Solution:
\textbf{Solution:} Extend $ID$ to $M$, such that $DM=ID$, and connect $CM$.\\
Since $I$ is the incenter of $\triangle ABC$,\\
it follows that $\angle IAC=\angle IAB$, $\angle ICA=\angle ICB$,\\
Since $\angle DIC=\angle IAC+\angle ICA$, $\angle DCI=\angle BCD+\angle ICB$,\\
it follows that $\angle DIC=\angle DCI$,\\
thus $DI=DC=DM$,\\
therefore $\angle ICM=90^\circ$,\\
hence $CM=\sqrt{IM^{2}-IC^{2}}=8$,\\
Since $AI=2CD=10$,\\
it follows that $AI=IM$,\\
Since $AE=EC$,\\
it follows that $IE$ is the midline of $\triangle ACM$,\\
thus $IE=\frac{1}{2}CM=\boxed{4}$


Question: As shown in the figure, points $A$, $B$, and $C$ are all on the circle $\odot O$. If $\angle ACB = 36^\circ$, what is the measure of $\angle OAB$?

Solution:
\textbf{Solution:} Given that $\because \angle ACB=\frac{1}{2}\angle AOB$ and $\angle ACB=36^\circ$, \\
$\therefore \angle AOB=2\times \angle ACB=72^\circ$. \\
$\because OA=OB$, \\
$\therefore \triangle OAB$ is an isosceles triangle, \\
$\because \angle AOB+\angle OAB+\angle OBA=180^\circ$, \\
$\therefore \angle OAB=\frac{1}{2}\left(180^\circ−\angle AOB\right)=54^\circ$, \\
Thus, the answer is: $\boxed{54^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, $CD$ is a chord, and $\angle CAB=50^\circ$, then what is the degree measure of $\angle D$?

Solution:
\textbf{Solution:} Since AB is a diameter,\\
it follows that $\angle ACN=90^\circ$,\\
therefore, $\angle B=90^\circ-\angle CAB=90^\circ-50^\circ=40^\circ$;\\
Since arc AC equals arc AC,\\
it follows that $\angle B=\angle D=\boxed{40^\circ}$.


Question: As shown in the figure, $\triangle ABC$ is inscribed in $\odot O$. If $\angle A=50^\circ$ and $\angle B=70^\circ$, what is the length of $AB$?

Solution:
\textbf{Solution:}

As shown in the figure, connect OA and OB, and draw $OD\perp AB$,

$\because$ $\angle A=50^\circ$, $\angle B=70^\circ$,

$\therefore$ $\angle C=180^\circ-50^\circ-70^\circ=60^\circ$,

$\therefore$ $\angle AOB=2\angle C=120^\circ$,

$\because$ $OA=OB$,

$\therefore$ $\triangle AOB$ is an isosceles triangle,

$\therefore$ $\angle AOD=\frac{1}{2}\angle AOB=60^\circ$, $AD=BD=\frac{1}{2}AB$,

$\therefore$ $\angle DAO=30^\circ$,

$\therefore$ $OD=\frac{1}{2}$, $AD=\sqrt{OA^2-OD^2}=\sqrt{1^2-\left(\frac{1}{2}\right)^2}=\frac{\sqrt{3}}{2}$,

$\therefore$ $AB=2AD=\boxed{\sqrt{3}}$.


Question: As shown in the figure, $AB$ and $CD$ are chords of the circle $\odot O$, and $AB \parallel CD$. If $\angle AOC = 80^\circ$, what is the measure of $\angle BAD$?

Solution:
\textbf{Solution:} Since $\angle AOC = 80^\circ$,\\
it follows that $\angle ADC = \frac{1}{2} \angle AOC = 40^\circ$,\\
Since $AB \parallel CD$,\\
it follows that $\angle BAD = \angle ADC = \boxed{40^\circ}$.


Question: As shown in the figure, in the Cartesian coordinate system \(xOy\), given points \(A(1,0)\), \(B(3,0)\), and \(C\) as a moving point in the plane satisfying \(\angle ACB=90^\circ\), and \(D\) as a moving point on the line \(y=x\), what is the minimum length of the segment \(CD\)?

Solution:
\textbf{Solution:} Let E be the midpoint of AB. Draw a perpendicular line from point E to the line y=x, and let D be the foot of the perpendicular. \\
$\because$ Point A is (1,0), and point B is (3,0), \\
$\therefore$ OA=1, OB=3, \\
$\therefore$ OE=2, \\
$\therefore$ ED=2$\times\frac{\sqrt{2}}{2}=\sqrt{2}$, \\
$\because$ $\angle$ACB$=90^\circ$, \\
$\therefore$ point C lies on the circle with diameter AB, \\
$\therefore$ the minimum length of segment CD is $\boxed{\sqrt{2}-1}$.


Question: As shown in the figure, the circumcircle of $\triangle ABC$ has its center at $O$, and $\angle BOC=110^\circ$. What is the measure of $\angle A$?

Solution:
\textbf{Solution:} Since $\odot O$ is the circumcircle of $\triangle ABC$ and $\angle BOC=110^\circ$,
\[ \therefore \angle A=\frac{1}{2}\angle BOC=\frac{1}{2}\times 110^\circ=\boxed{55^\circ}. \]


Question: As shown in the diagram, $\triangle ABC$ is inscribed in circle $\odot O$, $CD$ is the diameter of $\odot O$, $\angle BCD=56^\circ$, what is the degree measure of $\angle A$?

Solution:
\textbf{Solution:} As shown in the figure, connect $BD$, \\
since $CD$ is the diameter of $\odot O$, \\
thus $\angle DBC=90^\circ$, \\
since $\angle BCD=56^\circ$, \\
therefore $\angle BDC=90^\circ-56^\circ=34^\circ$, \\
since $\overset{\frown}{BC}=\overset{\frown}{BC}$, \\
thus $\angle A=\boxed{34^\circ}$.


Question: In the figure, points A, B, and C are all on the circle $\odot O$, and point C is on the major arc corresponding to chord AB. If $\angle AOB=72^\circ$, what is the degree measure of $\angle ACB$?

Solution:
\textbf{Solution:} \\
Since the central angle $\angle AOB$ and the inscribed angle $\angle ACB$ both subtend the arc $\overset{\frown}{AB}$, \\
therefore $\angle ACB = \frac{1}{2}\angle AOB = \frac{1}{2} \times 72^\circ = \boxed{36^\circ}$


Question: As shown in the figure, the radius of the circumcircle of $\triangle ABC$ is $8$, and $\angle ACB=60^\circ$, then what is the length of $AB$?

Solution:
\textbf{Solution:} Connect OA and OB, and draw OH $\perp$ AB at H, \\
$\because$ $\angle$ACB=$60^\circ$, \\
$\therefore$ $\angle$AOB=$2\angle$ACB=$120^\circ$, \\
$\because$ OB=OA=8, \\
$\therefore$ $\angle$AOH=$\angle$BOH=$60^\circ$, \\
$\therefore$ $\angle$OAB=$30^\circ$, \\
$\therefore$ OH=$\frac{1}{2}$OA=4, \\
$\therefore$ AH=$\sqrt{OA^{2}-OH^{2}}=\sqrt{8^{2}-4^{2}}=4\sqrt{3}$, \\
$\therefore$ AB=2AH=$\boxed{8\sqrt{3}}$.


Question: As shown in the figure, $AB$ is the diameter of the semicircle $O$, and $C,D$ are two points on the semicircle, satisfying $\angle ADC=120^\circ$ and $BC=1$. What is the length of $BC$?

Solution:
\[
\textbf{Solution:} \text{As shown in the figure, connect OC}\\
\because \angle ADC=120^\circ, \\
\therefore \angle ABC=60^\circ, \\
\because \text{OB}=\text{OC}, \\
\therefore \angle OCB=\angle OBC=\angle B=60^\circ, \\
\text{OB}=\text{OC}=\text{BC}=1, \\
\therefore \text{the length of } \overset{\frown}{BC} \text{ is } \frac{60^\circ \pi \times 1}{180}=\boxed{\frac{1}{3}\pi}
\]


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\odot O$, with $AE$ being the diameter of $\odot O$. If the radius of $\odot O$ is $6$, and $\angle ADC - \angle ABC = 40^\circ$, what is the length of $\overset{\frown}{CE}$?

Solution:
\textbf{Solution:} As shown in the figure, connect $OC$.

Since quadrilateral $ABCD$ is inscribed in $\odot O$,

it follows that $\angle ADC+\angle ABC=180^\circ$,

and since $\angle ADC-\angle ABC=40^\circ$,

we have $\angle ADC=110^\circ$, $\angle ABC=70^\circ$.

Therefore, $\angle AOC=2\angle ABC=140^\circ$,

which leads to $\angle COE=180^\circ-\angle AOC=40^\circ$,

Since the radius of $\odot O$ is 6,

the length of the arc $\overset{\frown}{CE}$ is $\frac{40\pi \times 6}{180}=\boxed{\frac{4\pi }{3}}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, $AP$ is a tangent to $\odot O$, and $PB$ intersects $\odot O$ at point $C$. If point $D$ is on $\odot O$ and $\angle ADC=40^\circ$, what is the degree measure of $\angle P$?

Solution:
\textbf{Solution:} Given $\angle ADC=40^\circ$,\\
it follows that $\angle ABC=40^\circ$,\\
since $AB$ is the tangent of $\odot O$ and point $A$ is the point of tangency,\\
it follows that $\angle OAB=90^\circ$,\\
thus, $\angle P=90^\circ-\angle ABC=90^\circ-40^\circ=\boxed{50^\circ}$.


Question: As shown in the figure, it is known that CD is the diameter of the circle $\odot O$, and the chord DE passing through point D is parallel to the radius OA. If the degree of arc CE is $92^\circ$, what is the degree of $\angle C$?

Solution:
\textbf{Solution:} As shown in the figure, connect OE, \\
$\because$ the degree of arc CE is $92^\circ$, \\
$\therefore$ $\angle COE = 92^\circ$, \\
$\therefore$ $\angle CDE = \frac{1}{2}\angle COE = 46^\circ$, \\
$\because$ OA $\parallel$ DE, \\
$\therefore$ $\angle AOD = \angle CDE = 46^\circ$, \\
$\therefore$ $\angle C = \frac{1}{2}\angle AOD = \boxed{23^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of a semicircle, and $C$, $D$ are two points on the semicircle. Given that $ \angle ADC=105^\circ$, what is the measure of $ \angle ABC$?

Solution:
\textbf{Solution:} Since $A,B,C,D$ are points on the circle $\odot O$ and $\angle ADC=105^\circ$,

it follows that $\angle ABC=180^\circ-105^\circ=\boxed{75^\circ}$


Question: As shown in the figure, $ABCD$ is a quadrilateral inscribed in circle $O$, and $\angle ABC = 125^\circ$. What is the measure of $\angle AOC$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is inscribed in circle $O$,\\
it follows that $\angle D+\angle ABC=180^\circ$ \\
Since $\angle ABC=125^\circ$,\\
we have $\angle D=180^\circ−\angle A=180^\circ−125^\circ=55^\circ$,\\
By the Inscribed Angle Theorem, $\angle AOC=2\angle D= \boxed{110^\circ}$


Question: As shown in the figure, points A, B, and C are on the circle $\odot O$. If $\angle CAB=70^\circ$, then what is the measure of $\angle BOC$?

Solution:
\textbf{Solution:} Since $\angle CAB=70^\circ$,\\
it follows that $\angle BOC=2\angle CAB=\boxed{140^\circ}$


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=12$, and $BC=5$, what is the distance between the centroid $P$ and the circumcenter $Q$ of $\triangle ABC$?

Solution:
\textbf{Solution:} Given the centroid $P$ and circumcenter $Q$ of the right-angled triangle $ABC$, it's known that points $C$, $P$, and $Q$ are collinear.\\
In right-angled triangle $ABC$, $\angle C=90^\circ$, $AC=12$, $BC=5$,\\
$\therefore AB= \sqrt{AC^{2}+BC^{2}}=\sqrt{5^{2}+12^{2}}=13$\\
Since the circumcenter of right-angled triangle $ABC$ is $Q$,\\
$\therefore Q$ is the midpoint of hypotenuse $AB$,\\
$\therefore CQ= \frac{1}{2}AB= \frac{13}{2}$,\\
Since the centroid of right-angled triangle $ABC$ is $P$,\\
$\therefore PQ= \frac{1}{3}CQ= \boxed{\frac{13}{6}}$.


Question: As shown in the figure, within the circle $\odot O$, $CD$ is the diameter of $\odot O$, and $AB \perp CD$ at point $E$. If $AB=8$ and $CE=2$, what is the radius of $\odot O$?

Solution:
\textbf{Solution:} Let the radius of $\odot O$ be $r$, \\
since CD is the diameter of $\odot O$ and AB $\perp$ CD, with AB=8, \\
therefore AE=$\frac{1}{2}$AB=4, \\
in $\triangle OAE$, by the Pythagorean theorem we have: $AE^{2}+OE^{2}=OA^{2}$, \\
that is $4^{2}+(r-2)^{2}=r^{2}$, \\
solving this gives: $r=\boxed{5}$, \\
thus, the radius of $\odot O$ is 5.


Question: As shown in the figure, AB is the diameter of the circle $O$, and point C is on the circle. If $\angle ABC = 70^\circ$, what is the degree measure of $\angle BAC$?

Solution:
\textbf{Solution:} Given that $AB$ is the diameter of circle $O$,\\
it follows that $\angle C=90^\circ$,\\
since $\angle ABC=70^\circ$,\\
it follows that $\angle BAC=90^\circ-70^\circ=\boxed{20^\circ}$.


Question: As shown in the figure, points $A,B,C$ are on circle $\odot O$. If $\angle BAC=40^\circ$, what is the measure of $\angle BOC$?

Solution:
\textbf{Solution:} Given that $ \because \angle BOC=2\angle BAC$ and $\angle BAC=40^\circ$,\\
$\therefore \angle BOC=\boxed{80^\circ}$.


Question: As shown in the figure, point $P$ is outside circle $\odot O$, line $PA$ intersects circle $\odot O$ at point $C$, line $PB$ intersects circle $\odot O$ at point $D$, if $\overset{\frown}{AB}=80^\circ$, $\angle P=28^\circ$, then what is the degree measure of $\angle CAD$?

Solution:
\textbf{Solution:} Connect OA and OB, as shown in the figure,\\
$\because$ The degree of $\overset{\frown}{AB}$ is $80^\circ$,\\
$\therefore$ $\angle AOB=80^\circ$,\\
$\therefore$ $\angle ADB= \frac{1}{2}\angle AOB=40^\circ$,\\
$\because$ $\angle ADB=\angle PAD+\angle P$,\\
$\therefore$ $\angle PAD=40^\circ-28^\circ=12^\circ$,\\
that is, the degree of $\angle CAD$ is $\boxed{12^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, $AC$ is the tangent of $\odot O$ with $A$ being the point of tangency, and $BC$ intersects $\odot O$ at point $D$. Connect $OD$. If $\angle C=50^\circ$, what is the degree measure of $\angle AOD$?

Solution:
\textbf{Solution:} Since $AC$ is the tangent to the circle $\odot O$,\\
$\therefore \angle CAB=90^\circ$,\\
And since $\angle C=50^\circ$,\\
$\therefore \angle ABC=90^\circ-50^\circ=40^\circ$,\\
And since $OD=OB$,\\
$\therefore \angle BDO=\angle ABC=40^\circ$,\\
And since $\angle AOD=\angle OBD+\angle OBD$,\\
$\therefore \angle AOD=40^\circ+40^\circ=\boxed{80^\circ}$.


Question: As shown in the figure, within circle $\odot O$, $OC \perp AB$. If $\angle BOC=40^\circ$, what is the value of $\angle OAB$?

Solution:
\textbf{Solution:} In circle $\odot O$, $OA=OB$, \\
$\therefore \triangle AOB$ is an isosceles triangle,\\
$\because OC\perp AB$,\\
$\therefore \angle AOC=\angle BOC=40^\circ$,\\
$\therefore \angle AOB=80^\circ$,\\
$\therefore \angle OAB=(180^\circ-\angle AOB)\div 2=50^\circ$.\\
Thus, $\angle OAB=\boxed{50^\circ}$.


Question: As shown in the figure, AE is the diameter of the circumscribed circle $\odot O$ of the quadrilateral ABCD, $AD=CD$, $\angle B=50^\circ$, then what is the degree measure of $\angle DAE$?

Solution:
\textbf{Solution:} Draw OC and OD, \\
$\because$ $\angle B=50^\circ$, \\
$\therefore$ $\angle AOC=2\angle B=100^\circ$, \\
$\because$ AD=CD, \\
$\therefore$ $\overset{\frown}{AD}=\overset{\frown}{CD}$, \\
$\therefore$ $\angle AOD=\angle COD= \frac{1}{2}\angle AOC=50^\circ$, \\
$\because$ OA=OD, \\
$\therefore$ $\angle OAD=\angle ODA$, \\
$\therefore$ $\angle DAE=(180^\circ-50^\circ)\div2=\boxed{65^\circ}$.


Question: As shown in the figure, AD is the diameter of the circle $O$, with $AD=8$ and $\angle DAC = \angle ABC$. What is the length of AC?

Solution:
\textbf{Solution:} Let's connect CD\\
$\because \angle DAC=\angle ABC$\\
$\therefore AC=DC$\\
Also, $\because AD$ is the diameter of the circle $O$\\
$\therefore \angle ACD=90^\circ$\\
$\therefore AC^{2}+DC^{2}=AD^{2}$\\
$\therefore 2AC^{2}=AD^{2}$\\
$\therefore AC=\frac{\sqrt{2}}{2}AD=\frac{\sqrt{2}}{2}\times 8=4\sqrt{2}$\\
Therefore, the answer is: $AC=\boxed{4\sqrt{2}}$


Question: As shown in the figure, the radius $OD$ of circle $O$ is perpendicular to chord $AB$ at point $C$. Extend $AO$ to meet circle $O$ at point $E$, and connect $EC$. If $AB=8$ and $OC=3$, what is the length of $EC$?

Solution:
\textbf{Solution:} Connect BE, \\
$\because$AE is the diameter of $\odot$O, \\
$\therefore$$\angle$ABE$=90^\circ$, \\
$\because$OD$\perp$AB and OD passes through O, \\
$\therefore$AC$=$BC$=\frac{1}{2}$AB$=\frac{1}{2}\times 8=4$, \\
$\because$AO$=$OE, \\
$\therefore$BE$=2$OC, \\
$\because$OC$=3$, \\
$\therefore$BE$=6$, \\
In the right triangle $\triangle$CBE, EC$=\sqrt{BE^{2}+CB^{2}}=\sqrt{4^{2}+6^{2}}=\boxed{2\sqrt{13}}$.


Question: The protractor with center $O$ and the right-angled triangle $ABC$ are placed as shown in the figure, with the 0-degree line of the protractor coinciding with the hypotenuse $AB$. Point $D$ is a point on the hypotenuse $AB$. A ray $CD$ is drawn intersecting the arc $AB$ at point $E$. If the reading corresponding to point $E$ is $50^\circ$, what is the measure of $\angle BDE$?

Solution:
\textbf{Solution:} As shown in the figure, connect $OE$,\\
the reading corresponding to point $E$ is $50^\circ$,\\
$\therefore \angle AOE=50^\circ$,\\
$\because AB$ is the diameter, $\angle ACB=90^\circ$,\\
$\therefore$ point $C$ lies on $\odot O$,\\
$\therefore \angle ACE=\frac{1}{2}\angle AOE=\frac{1}{2}\times 50^\circ=25^\circ$,\\
$\therefore \angle BCE=90^\circ-25^\circ=65^\circ$,\\
$\because \angle BDE$ is the exterior angle of $\triangle BDC$,\\
$\therefore \angle BDE=\angle BCE+\angle DBC=65^\circ+45^\circ=\boxed{110^\circ}$.


Question: As shown in the figure, \(AB\) is the diameter of the circle \(\odot O\), and \(CD\) is a chord. If \(\angle BCD=34^\circ\), then what is the measure of \(\angle ABD\)?

Solution:
\textbf{Solution:} Given that AB is the diameter of circle O, \\
thus, $\angle ADB=90^\circ$, \\
therefore, $\angle DAB + \angle ABD = 90^\circ$, \\
since $\angle DAB = \angle BCD = 34^\circ$, \\
thus, $\angle ABD = 90^\circ - 34^\circ = \boxed{56^\circ}$.


Question: As shown in the figure, inside the circle $ \odot O$, $ \overset{\frown}{AB}=\overset{\frown}{AC}$, and $ \angle C=75^\circ$, what is the degree measure of $ \angle A$?

Solution:
\textbf{Solution:} In $\odot O$, $\overset{\frown}{AB}=\overset{\frown}{AC}$, \\
$\therefore \angle B=\angle C=75^\circ$, \\
$\because \angle A+\angle B+\angle C=180^\circ$, \\
$\therefore \angle A=180^\circ-\angle B-\angle C=\boxed{30^\circ}$


Question: Steel balls are commonly used in engineering to measure the width of small circular holes in parts. Suppose the diameter of the steel ball is 10mm, and the distance from the top of the steel ball to the surface of the part is measured to be 8mm, as shown in the figure. What is the length of the width $AB$ of this small circular hole?

Solution:
\textbf{Solution:} As shown in the figure, point O is the center of the circle, and through point O, draw OC$\perp$AB,\\
According to the perpendicular diameter theorem, it follows that: AC=BC,\\
$\because$The diameter is 10mm,\\
$\therefore$OA=5mm, OC=8−5=3mm,\\
In $\triangle AOC$, $\angle OCA=90^\circ$,\\
$\therefore$ $AC=\sqrt{OA^{2}-OC^{2}}=\sqrt{5^{2}-3^{2}}=4$,\\
$\therefore$AB=2AC=8mm,\\
Thus, the answer is: $AB=\boxed{8}$mm.


Question: As shown in the figure, the circumcircle of $\triangle ABC$, with point D on the arc $\overset{\frown}{BC}$, connect AD and CD. If $\angle CAD=25^\circ$, then what is the measure of $\angle ACD$?

Solution:
\textbf{Solution:} Given $\triangle ABC$ is an equilateral triangle,\\
$\therefore$ $\angle ACB = \angle ABC = \angle BAC = 60^\circ$,\\
Given $\angle CAD = 25^\circ$,\\
$\therefore$ $\angle BAD = \angle BAC - \angle CAD = 35^\circ$,\\
Since arc $\overset{\frown}{BD} = \overset{\frown}{BD}$,\\
$\therefore$ $\angle BCD = \angle BAD = 35^\circ$,\\
$\therefore$ $\angle ACD = \angle ACB + \angle BCD = \boxed{95^\circ}$


Question: As shown in the figure, points A, B, and C are on the circle $\odot O$. If $\angle AOB=130^\circ$, what is the measure of $\angle ACB$?

Solution:
\textbf{Solution:} As shown in the figure, take a point \(D\) on the major arc \(AB\), and connect \(AD\), \(DB\).\\
Since \(\angle ADB = \frac{1}{2} \angle AOB\), \(\angle AOB = 130^\circ\),\\
Therefore, \(\angle ADB = 65^\circ\),\\
Since \(\angle ACB + \angle ADB = 180^\circ\),\\
Therefore, \(\angle ACB = \boxed{115^\circ}\).


Question: As shown in the figure, AB is the diameter of $\odot O$, point C is on the extension of AB, and CD is tangent to $\odot O$ at point D. If $\angle CDA=118^\circ$, what is the degree measure of $\angle C$?

Solution:
Solution:As shown in the diagram, connect OD, \\
$\because$CD is tangent to $\odot$O at point D,\\
$\therefore \angle$ODC$=90^\circ$,\\
$\because \angle$CDA$=118^\circ$,\\
$\therefore \angle$ODA$=\angle$CDA$-\angle$ODC$=118^\circ-90^\circ=28^\circ$,\\
$\because$OD$=$OA,\\
$\therefore \angle$OAD$=\angle$ODA$=28^\circ$,\\
$\therefore \angle$DOC$=2\angle$ODA$=56^\circ$,\\
$\therefore \angle$C$=90^\circ-\angle$DOC$=\boxed{34^\circ}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, CD is a chord of $\odot O$, $\angle CAB=55^\circ$, then what is the degree measure of $\angle D$?

Solution:
\textbf{Solution:} Given that $AB$ is the diameter of $\odot O$,\\
it follows that $\angle ACB=90^\circ$,\\
given $\angle CAB=55^\circ$,\\
thus $\angle B=90^\circ-\angle CAB=35^\circ$,\\
therefore, $\angle D=\angle B=\boxed{35^\circ}$.


Question: As shown in the figure, OA is the radius of the circle $\odot O$, with chord BC $\perp$ OA at point D. AC is connected. If BC = $4\sqrt{2}$ and AC = 3, then what is the length of the radius of $\odot O$?

Solution:
\textbf{Solution:} As illustrated in the diagram, connect OC.\\
Since BC$\perp$OA,\\
it follows that $\angle ADC=\angle ODC=90^\circ$, $CD=BD=\frac{1}{2}BC=2\sqrt{2}$,\\
therefore $AD=\sqrt{AC^{2}-CD^{2}}=1$.\\
Let $OA=OC=r$, then $OD=OA-AD=r-1$,\\
since $OD^{2}+CD^{2}=OC^{2}$,\\
it can be derived that $(r-1)^{2}+\left(2\sqrt{2}\right)^{2}=r^{2}$,\\
solving this gives $r=\boxed{\frac{9}{2}}$.


Question: As shown in the figure, it is known that $AD$, $BE$, $CF$ are the three altitudes of $\triangle ABC$ (with $D$, $E$, $F$ being the feet of the perpendiculars), $\angle ABC=45^\circ$, $\angle C=60^\circ$, then what is the value of $\frac{DE}{DF}$?

Solution:
\textbf{Solution:} Given $\because \angle ABC=45^\circ$,\\
$\therefore \angle BAD=45^\circ$, $\angle BCF=45^\circ$,\\
$\therefore \triangle ABD$ and $\triangle BCF$ are both isosceles right triangles,\\
$\because \frac{BF}{BC}=\frac{BD}{AB}=\frac{\sqrt{2}}{2}$, $\angle ABC=\angle ABC$,\\
$\therefore \triangle BFD\sim \triangle BCA$,\\
$\therefore \frac{DF}{AC}=\frac{\sqrt{2}}{2}$,\\
Similarly, it can be derived that $\triangle CDE\sim \triangle CBA$,\\
$\therefore \frac{DE}{AB}=\frac{CD}{AC}=\frac{1}{2}$,\\
Hence, $DF= \frac{\sqrt{2}}{2} AC$, $DE= \frac{1}{2} AB$,\\
$\therefore \frac{DE}{DF}=\frac{\frac{1}{2}AB}{\frac{\sqrt{2}}{2}AC}=\frac{\sqrt{2}AB}{2AC}$,\\
Let $AD=a$, then $AB= \sqrt{2} a$, $AC= \frac{2a}{\sqrt{3}}$,\\
$\therefore \frac{DE}{DF}=\frac{\sqrt{2}AB}{2AC}=\frac{\sqrt{2}\times \sqrt{2}a}{2\times \frac{2a}{\sqrt{3}}}=\boxed{\frac{\sqrt{3}}{2}}.$


Question: As shown in the diagram, in the Cartesian coordinate system, the squares $ABCD$ and $BEFG$ are similar figures with the origin $O$ as the center of similarity, and the ratio of similarity is $\frac{1}{3}$. The points $A$, $B$, and $E$ are located on the $x$-axis. If the side length of square $BEFG$ is 6, what are the coordinates of point $C$?

Solution:
\textbf{Solution:} Given that square ABCD and square BEFG are similar figures with center at the origin O, and the ratio of similarity is $\frac{1}{3}$,\\
$\therefore$ $\frac{BC}{EF}=\frac{1}{3}$,\\
Given that EF = 6,\\
$\therefore$ BC=2,\\
Given that $\triangle OBC\sim \triangle OEF$,\\
$\therefore$ $\frac{OB}{OE}=\frac{BC}{EF}$,\\
$\therefore$ $\frac{OB}{OB+6}=\frac{1}{3}$,\\
$\therefore$ OB=3,\\
$\therefore$ the coordinates of point C are $\boxed{(3, 2)}$.


Question: As shown in the figure, in $\triangle ABC$, points D and E are on the sides AB and AC respectively, and $\frac{AE}{AB} = \frac{AD}{AC} = \frac{1}{2}$, then what is the ratio of the perimeter of $\triangle ADE$ to the perimeter of $\triangle ABC$?

Solution:
\textbf{Solution:} Since $\frac{AE}{AB} = \frac{AD}{AC}$, and $\angle DAE = \angle CAB$,\\
therefore $\triangle ADE \sim \triangle ACB$,\\
thus, the ratio of the perimeter of $\triangle ADE$ to that of $\triangle ABC$ is $\boxed{\frac{1}{2}}$.


Question: As shown in the figure, $\triangle ABC$ is similar to $\triangle DEF$ with the center of similarity at point O. If $OC:OF=1:3$, then what is the ratio of the perimeters of $\triangle ABC$ to $\triangle DEF$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is similar to $\triangle DEF$ with OC:OF=1:3,\\
it follows that $\triangle ABC\sim \triangle DEF$, and AB:DE=1:3,\\
thus, the ratio of the perimeters of $\triangle ABC$ to $\triangle DEF$ is: $1:3$,\\
Therefore, the ratio of the perimeters is $\boxed{1:3}$.


Question: As shown in the figure, points $D$ and $E$ are on $AB$ and $AC$ respectively, with $DE \parallel BC$. Given that $AD = 6$, $AE = 4$, and $CE = 2$, what is the length of $BD$?

Solution:
\textbf{Solution:} Since DE$\parallel$BC, \\
it follows that $\frac{AD}{BD}=\frac{AE}{CE}$, \\
given AD=6, AE=4, CE=2, \\
thus $\frac{6}{BD}=\frac{4}{2}$, \\
solving gives $BD=\boxed{3}$.


Question: As shown in the figure, an isosceles triangle with a vertex angle of $36^\circ$ and the ratio of its base to its legs equal to $k$ is known as a golden triangle. Given that the leg $AB=1$ in the $\triangle ABC$ which is the first golden triangle, $\triangle BCD$ is the second golden triangle, and $\triangle CDE$ is the third golden triangle, and so on, what is the perimeter of the third golden triangle?

Solution:
\textbf{Solution:} Since $\triangle ABC$ and $\triangle BCD$ are both isosceles triangles with a vertex angle of $36^\circ$,
\[
\therefore \triangle ABC \sim \triangle BCD,
\]
\[
\therefore \text{the perimeter of } \triangle BCD : \text{the perimeter of } \triangle ABC = BC: AB,
\]
Given that $AB=AC=1$, and $\frac{BC}{AB}=k$,
\[
\therefore BC=k,
\]
\[
\therefore \text{the perimeter of } \triangle ABC = AB+AC+BC=2+k,
\]
\[
\therefore \text{the perimeter of } \triangle BCD : \text{the perimeter of } \triangle ABC = k,
\]
\[
\therefore \text{the perimeter of } \triangle BCD = k(2+k),
\]
By the same reasoning, \text{the perimeter of } $\triangle CDE$ : \text{the perimeter of } $\triangle BCD = k$,
\[
\therefore \text{the perimeter of } \triangle CDE = k^2(2+k) = \boxed{k^2(2+k)}.
\]


Question: As shown in the figure, D and E are the midpoints of AB and AC, respectively. What is the value of $\frac{S_{\triangle ADE}}{S_{\triangle ABC}}$?

Solution:
\textbf{Solution:} Since D and E are the midpoints of AB and AC respectively,\\
it follows that DE is the midline of the triangle,\\
therefore, the ratio DE: BC = 1: 2,\\
therefore, the area ratio $S_{\triangle ADE} : S_{\triangle ABC}$ = 1: 4.\\
Hence, the answer is: $S_{\triangle ADE} : S_{\triangle ABC} = \boxed{\frac{1}{4}}$.


Question: As shown in the figure, $\triangle ABC$ and $\triangle ADE$ are both equilateral triangles, with point D on side BC, and DE intersecting AC at point F. If AB=6, AD=5, and CD=4, what is the length of EF?

Solution:
\textbf{Solution:} Since $\triangle ABC$ and $\triangle ADE$ are both equilateral triangles,\\
it follows that $\angle B=\angle C=\angle ADE=60^\circ$, and $DE=AD=5$,\\
Since $\angle ADC=\angle ADE+\angle CDE=\angle B+\angle BAD$,\\
it follows that $\angle CDE=\angle BAD$,\\
therefore, $\triangle ABD\sim \triangle DCF$,\\
thus, $\frac{AB}{DC}=\frac{AD}{DF}$,\\
which means $\frac{6}{4}=\frac{5}{DF}$,\\
therefore, $DF=\frac{10}{3}$,\\
therefore, $EF=DE-DF=\boxed{\frac{5}{3}}$.