Question: As shown in the figure, the diameter $AB$ of $\odot O$ is perpendicular to the chord $CD$, with the foot of the perpendicular being $E$. If $\angle B = 60^\circ$ and $AC = 6$, what is the length of $CD$?

Solution:
\textbf{Solution:} Draw $AD$,\\
since the diameter $AB$ of the circle $O$ is perpendicular to the chord $CD$ at the foot $E$,\\
it follows that $AE=AE$, $CE=DE$, $\angle AED=\angle AEC=90^\circ$,\\
thus $\triangle AED\cong \triangle AEC$ (by SAS),\\
hence $AD=AC$,\\
since $\angle B=60^\circ$, $AC=6$,\\
it follows that $\angle BCE=30^\circ$,\\
since $\angle ACB=90^\circ$,\\
it follows that $\angle ACE=\angle ACB-\angle BCE=60^\circ$,\\
similarly, it can be found that: $\angle ADE=60^\circ$,\\
hence $\triangle ACD$ is an equilateral triangle,\\
therefore $CD=AC=\boxed{6}$.


Question: As shown in the figure, $\angle A$ is the inscribed angle of $\odot O$ and $\angle A = 40^\circ$, then what is the measure of $\angle BOC$?

Solution:
\textbf{Solution:} Given that $\angle A$ is the peripheral angle of $\odot O$, $\angle A=40^\circ$,\\
hence, $\angle BOC=2\angle A=\boxed{80^\circ}$.


Question: As shown in the figure, it is known that the central angle $\angle AOB$ is $100^\circ$. What is the measure of the inscribed angle $\angle ACB$?

Solution:
\textbf{Solution:} Let point E be a point on the major arc AB, and connect EA and EB. \\
$\because$ $\angle$AOB=100$^\circ$, \\
$\therefore$ $\angle$E= $ \frac{1}{2}\angle$AOB=50$^\circ$, \\
$\therefore$ $\angle$ACB=180$^\circ$−$\angle$E=\boxed{130^\circ}.


Question: As shown in the figure, the radius of the drainage pipe's cross-section is 5 decimeters, and the width of the water surface $AB=8$ decimeters, with $OC\perp AB$. What is the maximum depth $CD$ of the water?

Solution:
Solution: Since the diameter of circle O is 10 dm,\\
it follows that OA=5 (dm),\\
and since OD$\perp$AB, and AB=8 (dm),\\
it follows that AD=BD=4,\\
hence, OD$=\sqrt{OA^{2}-AD^{2}}=3$,\\
therefore, the maximum depth of the water CD=OC-OD=5-3=\boxed{2} (dm).


Question: As shown in the figure, AB is a chord of $\odot O$ (AB is not a diameter). Taking point A as the center and the length of AB as the radius, an arc is drawn to intersect $\odot O$ at point C. Connect AC, BC, OB, and OC. If $\angle ABC=65^\circ$, what is the degree measure of $\angle BOC$?

Solution:
\textbf{Solution:} From the given, we can derive: AB=AC, \\
$\because$ $\angle$ABC=65°, \\
$\therefore$ $\angle$ACB=65°, \\
$\therefore$ $\angle$A=50°, \\
$\therefore$ $\angle$BOC=\boxed{100^\circ}


Question: As shown in the diagram, in circle $O$, $\angle BAC=25^\circ$, $\angle CED=30^\circ$, then what is the degree measure of $\angle BOD$?

Solution:
\textbf{Solution:} As shown in the figure, connect OC,\\
Given that $\angle BAC=25^\circ$ and $\angle CED=30^\circ$,\\
by the Inscribed Angle Theorem, we have $\angle BOC=50^\circ$ and $\angle DOC=60^\circ$,\\
therefore, $\angle BOD=\angle BOC+\angle DOC=50^\circ+60^\circ=\boxed{110^\circ}$.


Question: As shown in the figure, it is known that the diameter $CD=8$ of the circle $\odot O$, $AB$ is a chord of $\odot O$, and $AB \perp CD$ at the foot point $M$, $OM=2$. What is the length of $AB$?

Solution:
\textbf{Solution:} Connect OB, \\
$\because$ the diameter $CD=8$, and $AB\perp CD$, $OM=2$\\
$\therefore$ BM=$\sqrt{OB^{2}-OM^{2}}$\\
=$\sqrt{4^{2}-2^{2}}$\\
=$2\sqrt{3}$,\\
According to the Perpendicular Diameter Theorem, we obtain\\
AB=2BM=$\boxed{4\sqrt{3}}$


Question: As shown in the figure, AB is the diameter of $\odot$O. If AC=4 and $\angle D=60^\circ$, what is the length of BC?

Solution:
\textbf{Solution:} Since AB is the diameter of the circle $O$,\\
it follows that $\angle ACB=90^\circ$,\\
since $\angle A=\angle D=60^\circ$,\\
thus, $\angle ABC=90^\circ-\angle A=30^\circ$,\\
since AC=4,\\
it follows that AB=2AC=8.\\
Therefore, $BC= \sqrt{AB^{2}-AC^{2}}=\sqrt{8^{2}-4^{2}}=\boxed{4\sqrt{3}}$.


Question: As shown in the figure, points A, B, C, and D are on the circle $\odot O$, with $\angle AOC=140^\circ$. Point B is the midpoint of the arc $\overset{\frown}{AC}$. What is the degree measure of $\angle D$?

Solution:
\textbf{Solution:} Draw line OB as shown in the figure,\\
since point B is the midpoint of $\widehat{AC}$, and $\angle$AOC=$140^\circ$,\\
therefore, $\angle$AOB=$\frac{1}{2}\angle$AOC=$70^\circ$,\\
by the Inscribed Angle Theorem, we have $\angle$D=$\frac{1}{2}\angle$AOB=$\boxed{35^\circ}$.


Question: As shown in the figure, the radius of $\odot O$ is $\sqrt{5}$. AB and CD are two parallel chords of $\odot O$, with $\angle CDE=30^\circ$ and $AD=2$. What is the length of chord BE?

Solution:
\textbf{Solution:} Connect OC, OE, BC, CE, and draw CH$\perp$BE intersecting BE at point H,\\
$\because$AB$\parallel$CD,\\
$\therefore$BC = AD = 2,\\
$\because$$\angle$CDE = $30^\circ$,\\
$\therefore$$\angle$COE= $60^\circ$, $\angle$CBE = $\angle$CDE = $30^\circ$,\\
$\therefore$$\triangle$OCE is an equilateral triangle,\\
$\therefore$CE= $\sqrt{5}$, in $Rt\triangle BCH$,\\
$CH=\frac{1}{2}BC=1$\\
$\therefore$BH= $\sqrt{2^2-1^2}=\sqrt{3}$, in $Rt\triangle$CEH\\
HE= $\sqrt{(\sqrt{5})^2-1^2}=2$\\
$\therefore$BE=2+ $\sqrt{3}$.\\
Therefore, the answer is: BE=\boxed{2+\sqrt{3}}.


Question: As shown in the figure, $PA$ and $PB$ are tangents to the circle $O$, with $A$ and $B$ being the points of tangency. Point $C$ is on the minor arc $\overset{\frown}{AB}$. If $\angle ACB = 110^\circ$, what is the measure of $\angle P$?

Solution:
\textbf{Solution:} \textit{Draw OA, OB,\\
since PA, PB are tangents to the circle $\odot O$,\\
it follows that $\angle OAP = \angle OBP = 90^\circ$,\\
since $\angle ACB = 110^\circ$,\\
the central angle corresponding to the major arc AB is $220^\circ$,\\
hence, $\angle AOB = 360^\circ - 220^\circ = 140^\circ$,\\
therefore, $\angle P = 180^\circ - \angle AOB = \boxed{40^\circ}$,}


Question: As shown in the figure, the diameter $AB$ of circle $\odot O$ is perpendicular to chord $CD$ at point $E$, and $BD$ is connected. If $CD=8$ and $OE=3$, what is the length of $BD$?

Solution:
\textbf{Solution:} Draw line $OD$, \\
since $AB \perp CD$, $AB$ passes through the center $O$, $CD=8$, \\
therefore $CE=DE=4$, $\angle OED=\angle DEB=90^\circ$, \\
since $OE=3$, \\
therefore $OD=\sqrt{OE^2+DE^2}=\sqrt{3^2+4^2}=5$, \\
therefore $OB=OD=5$, \\
therefore $BE=OB-OE=5-3=2$, \\
by the Pythagorean theorem, we get \\
therefore $BD=\sqrt{BE^2+DE^2}=\sqrt{2^2+4^2}=\boxed{2\sqrt{5}}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\odot O$, with $\angle BCD = 105^\circ$. What is the measure of $\angle BOD$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is inscribed in circle O,\\
it follows that $\angle A + \angle BCD = 180^\circ$,\\
therefore $\angle A = 180^\circ - \angle BCD = 180^\circ - 105^\circ = 75^\circ$,\\
hence $\angle BOD = 2\angle A = 2 \times 75^\circ = \boxed{150^\circ}$.


Question: As shown in the figure, points A, B, and C are on the circle with center O. If $\angle C=20^\circ$ and $\angle B=30^\circ$, what is the measure of $\angle A$?

Solution:
\textbf{Solution:} Let $\because$ $\angle C=20^\circ$,\\
$\therefore$ $\angle O=2\angle C=40^\circ$,\\
$\because$ $\angle O+\angle A=\angle B+\angle C$,\\
$\therefore$ $\angle A=30^\circ+20^\circ-40^\circ=\boxed{10^\circ}$


Question: As shown in the figure, $AB$ is the diameter of $\odot O$. If $\angle AOC=130^\circ$, what is the measure of $\angle D$ in degrees?

Solution:
\textbf{Solution:} Draw AD, \\
since AB is the diameter of circle O, and $\angle$AOC $= 130^\circ$, \\
therefore, $\angle$BDA $= 90^\circ$, and $\angle$CDA $= 65^\circ$, \\
hence, $\angle$BDC $=\boxed{25^\circ}$.


Question: As shown in the figure, AB is the diameter of $\odot O$. A tangent to $\odot O$ is drawn through the point C on $\odot O$, intersecting the extension of AB at point D. If $\angle D=40^\circ$, what is the magnitude of $\angle A$?

Solution:
\textbf{Solution:} Let:\\
Since $CD$ is the tangent to $\odot O$,\\
$\therefore$ $\angle OCD = 90^\circ$,\\
Since $\angle D = 40^\circ$,\\
$\therefore$ $\angle COD = 50^\circ$,\\
Since $AB$ is the diameter of $\odot O$,\\
$\therefore$ $\angle A$ and $\angle COD$ are respectively the inscribed angle and the central angle that subtend the same arc $\overset{\frown}{BC}$,\\
$\therefore$ $\angle A = \frac{1}{2} \angle COD = \boxed{25^\circ}$,


Question: As shown in the figure, after pouring some water into a cylindrical container with a diameter of 78 cm, the cross-section is as shown in the figure. If the width of the water surface $AB = 72cm$, what is the maximum depth of the water?

Solution:
\textbf{Solution:} Connect $OB$, and through point O draw $OC\perp AB$ at point D, intersecting $\odot O$ at point C, as shown in the diagram:\\
Then $BD=\frac{1}{2}AB=36\,\text{cm}$,\\
Since the diameter of $\odot O$ is $78\,\text{cm}$,\\
Therefore $OB=OC=39\,\text{cm}$,\\
In $\triangle OBD$ right-angled at D, $OD=\sqrt{OB^{2}-BD^{2}}=\sqrt{39^{2}-36^{2}}=15\,\text{cm}$,\\
Therefore, $CD=OC-OD=39-15=\boxed{24}\,\text{cm}$,\\
which means the maximum depth of water is $24\,\text{cm}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\odot O$. If the quadrilateral $ABCO$ is a rhombus, then what is the measure of $\angle D$ in degrees?

Solution:
\textbf{Solution:} Let $\angle ADC=\alpha$ and $\angle ABC=\beta$;\\
Since quadrilateral ABCO is a rhombus,\\
it follows that $\angle ABC=\angle AOC=\beta$;\\
Therefore $\angle ADC=\frac{1}{2}\beta$;\\
Since quadrilateral $ABCD$ is circumscribed within a circle,\\
it follows that $\alpha +\beta =180^\circ$,\\
therefore $\left\{\begin{array}{l}
\alpha +\beta =180^\circ\\
\alpha =\frac{1}{2}\beta
\end{array}\right.$,\\
From which we find: $\beta =120^\circ$, $\alpha =60^\circ$, thus $\angle ADC=\boxed{60^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of the semicircle $O$, and $C$ is a point on the semicircle $O$. Connect $AC$ and $CB$, where $D$ is a point on the arc $\widehat{BC}$, and connect $CD$ and $BD$. If $\angle ABC=40^\circ$, what is the degree measure of $\angle D$?

Solution:
\textbf{Solution:} Given that $AB$ is the diameter of semicircle O,\\
it follows that $\angle ACB=90^\circ$,\\
since $\angle ABC=40^\circ$,\\
it follows that $\angle CAB=50^\circ$,\\
since the quadrilateral ABCD is inscribed in semicircle O,\\
it follows that $\angle D=180^\circ-50^\circ=\boxed{130^\circ}$;


Question: As shown in the figure, $AB$ is a diameter of the circle $O$ with $AB=4$ and $CD=2\sqrt{2}$. The length of the minor arc $BC$ is twice that of the minor arc $BD$. What is the length of $AC$?

Solution:
\textbf{Solution:} As shown in the figure, connect $OC$, $OD$, and $BC$,

$\because AB=4$

$\therefore OC=OD=2$

$\because OC^2+OD^2=8$, $CD^2=8$

$\therefore OC^2+OD^2=CD^2$

$\therefore \triangle OCD$ is a right-angled triangle, and $\angle COD=90^\circ$

$\therefore \overset{\frown}{CB}=2\overset{\frown}{DB}$

$\therefore \overset{\frown}{BC}=\frac{2}{3}\overset{\frown}{CD}$

$\therefore \angle BOC=\frac{2}{3}\times \angle COD=60^\circ$

$\because OC=OB$

$\therefore \triangle OBC$ is an equilateral triangle

$\therefore BC=OC=2$

$\because AB$ is the diameter, $AB=4$

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

$\therefore AC=\sqrt{3}BC=2\sqrt{3}$

Thus, the answer is: $AC=\boxed{2\sqrt{3}}$


Question: As shown in the figure, in the $xy$-plane coordinate system, point $A(0, 3)$, point $B(2, 1)$, and point $C(2, -3)$. Through drawing operations, it can be known: what are the coordinates of the center of the circumscribed circle of $\triangle ABC$?

Solution: \textbf{Solution:} Since the circumcenter of $\triangle ABC$ is the intersection point of the perpendicular bisectors of the triangle's sides, as shown in the figure: the intersection point $O^{\prime}$ of EF and MN is the desired circumcenter of $\triangle ABC$, therefore, the coordinates of the circumcenter of $\triangle ABC$ are $\boxed{(-2,-1)}$.


Question: As shown in the figure, points A, B, and C are on the circle $\odot O$. If $\triangle OAB$ is an equilateral triangle, then what is the measure of $\angle ACB$?

Solution:
\textbf{Solution:} Since $\triangle OAB$ is an equilateral triangle,\\
it follows that $\angle AOB=60^\circ$,\\
therefore, $\angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 60^\circ = \boxed{30^\circ}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, with chord CDAB. The foot of the perpendicular from C to AB is point E. If the radius of $\odot O$ is 5, and CD equals 8, what is the length of AE?

Solution:
\textbf{Solution:} Connect OC, as shown in the diagram\\
Since AB is the diameter of $\odot$O, and CDAB is perpendicular at point E, with CD=8,\\
it follows that $CE=\frac{1}{2}CD=\frac{1}{2}\times 8=4$,\\
Since $AO=CO=5$,\\
it follows that $OE=\sqrt{CO^{2}-CE^{2}}=\sqrt{5^{2}-4^{2}}=3$,\\
therefore, $AE=5-3=\boxed{2}$;


Question: As shown in the figure, points A, B, and C are all on the circle $\odot O$. Connect OA, OB, AC, and BC. If OA $\perp$ OB, what is the degree measure of $\angle C$?

Solution:
\textbf{Solution:} Since $OA\perp OB$, \\
$\therefore$ $\angle AOB=90^\circ$, \\
$\therefore$ $\angle C=\frac{1}{2}\angle AOB=45^\circ$, \\
Thus, the answer is: $\boxed{45^\circ}$.


Question: As shown in the figure, in the cyclic pentagon $ABCDE$, if $\angle C + \angle CDE + \angle E + \angle EAB = 425^\circ$, what is the measure of $\angle CDA$?

Solution:
\textbf{Solution:} Since the sum of the interior angles of pentagon $ABCDE$ is $(5-2)\times 180^\circ=540^\circ$, \\
thus $\angle EAB+\angle B+\angle C+\angle CDE+\angle E=540^\circ$, \\
because $\angle EAB+\angle C+\angle CDE+\angle E=425^\circ$, \\
therefore $\angle B=540^\circ-425^\circ=115^\circ$, \\
since quadrilateral $ABCD$ is inscribed in circle $O$, \\
therefore $\angle B+\angle CDA=180^\circ$, \\
thus $\angle CDA=180^\circ-115^\circ=\boxed{65^\circ}$.


Question: As shown in the figure, two congruent circles $\odot O_1$ and $\odot O_2$ intersect at points A and B, and $\odot O_1$ passes through the center of $\odot O_2$. What is the measure of $\angle O_1AB$?

Solution:
\textbf{Solution:} Connect $O_1O_2$, $AO_2$, and $O_1B$, \\
because $O_1B= O_1A$ \\
thus, $\angle O_1AB=\angle O_1BA=\frac{1}{2}\angle AO_2O_1$ \\
because $\odot O_1$ and $\odot O_2$ are congruent circles, \\
thus, $AO_1=O_1O_2=AO_2$, \\
therefore, $\triangle AO_2O_1$ is an equilateral triangle, \\
thus, $\angle AO_2O_1=60^\circ$, \\
therefore, $\angle O_1AB= \frac{1}{2}\angle AO_2O_1 =\frac{1}{2}\times 60^\circ =\boxed{30^\circ}$.


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

Solution:
\textbf{Solution:} Since $\angle ABC$ and $\angle AOC$ are the inscribed angle and central angle subtended by the same arc AC, and $\angle AOC=90^\circ$,

$\therefore$ $\angle ABC=\frac{1}{2}\angle AOC=\boxed{45^\circ}$.


Question: As shown in the figure, the diameter $AB=6$ of circle $\odot O$, with chord $CD$ of $\odot O$ perpendicular to $AB$ at point $P$, and $BP:AP=1:5$, what is the length of $CD$?

Solution:
\textbf{Solution:} Given that the diameter $AB$ of the circle $\odot O$ equals 6,\\
it follows that $OB=\frac{1}{2}AB=3$,\\
since the ratio $BP:AP=1:5$,\\
we have $BP=\frac{1}{6}AB=\frac{1}{6} \times 6=1$,\\
thus $OP=OB-BP=3-1=2$,\\
by joining $OC$,\\
and since $CD\perp AB$,\\
it implies $CD=2PC$, and $\angle OPC=90^\circ$,\\
therefore, $PC=\sqrt{OC^{2}-OP^{2}}=\sqrt{3^{2}-2^{2}}=\sqrt{5}$,\\
resulting in $CD=2PC=2\sqrt{5}$.\\
Hence, the length of $CD$ is $\boxed{2\sqrt{5}}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, the chord $CD \perp AB$ at point $E$, and $AD$ is connected. If $AB=10$ and $CD=8$, what is the length of $AD$?

Solution:
\textbf{Solution:} As shown in the diagram, connect OD.

\[
\because AB \perp CD,
\]
\[
\therefore CE = ED = 4,
\]
\[
\because \angle OED = 90^\circ, \text{ and } OD = 5,
\]
\[
\therefore OE = \sqrt{CD^{2} - ED^{2}} = \sqrt{5^{2} - 4^{2}} = 3,
\]
\[
\therefore AE = OA + OE = 8,
\]
\[
\therefore AD = \sqrt{AE^{2} + DE^{2}} = \sqrt{8^{2} + 4^{2}} = \boxed{4\sqrt{5}}.
\]


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in $\odot O$, with $AB$ being the diameter of $\odot O$, and $BC=CD$. If $\angle DAB=40^\circ$, then what is the measure of $\angle ADC$?

Solution:
\textbf{Solution:} As shown in the figure, connect $AC$,\\
Since $BC=CD$,\\
Therefore, $\angle CAB=\frac{1}{2}\angle DAB=20^\circ$.\\
Therefore, $AB$ is the diameter of $\odot O$,\\
Therefore, $\angle ACB=90^\circ$.\\
Since $\angle DAB=40^\circ$,\\
Therefore, $\angle DCB=140^\circ$.\\
Therefore, $\angle DCA=140^\circ-90^\circ=50^\circ$.\\
Therefore, $\angle ADC=180^\circ-20^\circ-50^\circ=\boxed{110^\circ}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\odot O$. If $\angle ADC = 140^\circ$, what is the degree measure of $\angle AOC$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in circle $O$, and $\angle ADC=140^\circ$\\
Therefore, $\angle ABC=40^\circ$,\\
Therefore, $\angle AOC=2\angle ABC=80^\circ$,\\
Hence, $\angle AOC=\boxed{80^\circ}$


Question: As shown in the figure, it is known that $\odot O$ is the circumcircle of $\triangle ABD$, where AB is the diameter of $\odot O$, and CD is a chord of $\odot O$. If $\angle ABD = 58^\circ$, then what is the measure of $\angle BCD$?

Solution: \textbf{Solution:} Given that $\angle ABD=58^\circ$, we can then find that $\angle A=90^\circ-\angle ABD=32^\circ$. Furthermore, in the same or congruent circles, angles subtended by the same or congruent arcs are equal, $\therefore \angle BCD=\angle A=\boxed{32^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle A = 50^\circ$, what is the degree measure of $\angle OCB$?

Solution:
\textbf{Solution:} Since $\overset{\frown}{CB}=\overset{\frown}{CB}$ and $\angle A=50^\circ$,\\
it follows that $\angle BOC=2\angle A=100^\circ$\\
Because $OB=OC$,\\
it is concluded that $\angle OCB=\angle OBC=\frac{1}{2}(180^\circ-\angle BOC)=\boxed{40^\circ}$


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and C, D are two points on $\odot O$. If $\angle AOC=130^\circ$, then what is the value of $\angle D$?

Solution:
\textbf{Solution:} Since $\angle AOC=130^\circ$,\\
it follows that $\angle BOC=180^\circ-130^\circ=50^\circ$,\\
hence $\angle D=\frac{1}{2}\angle BOC=25^\circ$.\\
Therefore, the answer is $\boxed{25^\circ}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, and points C and D are on $\odot O$. If $\angle D=120^\circ$, what is the measure of $\angle CAB$?

Solution:
\textbf{Solution}: Since $\angle D + \angle B = 180^\circ$ and $\angle D = 120^\circ$,

$\therefore \angle B = 60^\circ$,

Since $AB$ is the diameter,

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

Hence, $\angle CAB = 90^\circ - \angle B = \boxed{30^\circ}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\odot O$. If $\angle B=108^\circ$, what is the measure of $\angle D$?

Solution:
\textbf{Solution:} Given that the quadrilateral $ABCD$ is inscribed in the circle $\odot O$, and $\angle B=108^\circ$,\\
$\therefore \angle D=180^\circ-\angle B=180^\circ-108^\circ=\boxed{72^\circ}$.


Question: As shown in the figure, $PA$ and $PB$ are tangent to the circle $\odot O$ at points $A$ and $B$ respectively. Point $C$ lies on circle $\odot O$, and lines $AC$ and $BC$ are connected. If $\angle ACB=60^\circ$, what is the degree measure of $\angle P$?

Solution:
\textbf{Solution:} Connect OA, OB,\\
since $\angle ACB=60^\circ$,\\
thus $\angle AOB=2\angle ACB =120^\circ$,\\
also since PA and PB are tangent to the circle O at points A and B respectively,\\
thus $\angle PAO=\angle PBO=90^\circ$,\\
therefore $\angle P=360^\circ-\angle PAO-\angle PBO-\angle AOB=\boxed{60^\circ}$.


Question: As shown in the figure, in rhombus $ABCD$, $AB=13$ and $AC=10$, what is the length of $BD$?

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rhombus,\\
hence BD = 2OB, AC = 2OA, and AC $\perp$ BD,\\
therefore OA = 5, and $\angle$AOB = 90$^\circ$\\
In right $\triangle$AOB, $BO = \sqrt{AB^{2}-OA^{2}} = \sqrt{13^{2}-5^{2}} = 12$\\
therefore BD = $2 \times 12 = \boxed{24}$.


Question: As shown in the figure, in the parallelogram $ABCD$, $BD$ is the diagonal, and point $O$ is the midpoint of $BD$. It is also given that $AD\parallel EO$ and $OF\parallel AB$. The perimeter of quadrilateral $BEOF$ is 10. What is the perimeter of the parallelogram $ABCD$?

Solution:
\textbf{Solution:} Since quadrilateral $ABCD$ is a parallelogram,\\
it follows that $AD \parallel BC$,\\
and since point $O$ is the midpoint of $BD$, and $AD \parallel EO$, $OF \parallel AB$,\\
it follows that $OE$, $OF$ are respectively the midsegments of triangle $ABD$, triangle $BCD$, with $BC \parallel EO$,\\
thus, quadrilateral $OEBF$ is a parallelogram, with $AD=2OE$, $CD=2OF$, $OE=BF$, $OF=BE$,\\
and since the perimeter of the quadrilateral $OEBF$ is $10$,\\
we have $OE+BE+BF+OF=10$,\\
therefore, $OE+OF=5$,\\
and since quadrilateral $ABCD$ is a parallelogram,\\
it follows that $AB=CD$, $AD=BC$,\\
therefore, the perimeter of the quadrilateral $ABCD$ is $AB+BC+AD+CD=2(AD+CD)=4(OE+OF)=\boxed{20}$.


Question: The great ancient Chinese mathematician Liu Hui divided the right-angled triangle (referred to as the Gougu shape by ancients) into one square and two pairs of congruent right-angled triangles. As shown in the figure, if $AF = 5$ and $CE = 12$, then what is the area of this triangle?

Solution:
Solution: Let the side length of the small square be $x$,\\
$\because$AF=5, CE=12,\\
$\therefore$AB=5+x, BC=12+x,\\
$\because$$\triangle$AFM$\cong$$\triangle$ADM, $\triangle$CDM$\cong$$\triangle$CEM,\\
$\therefore$AD=AF=5, CD=CE=12,\\
$\therefore$AC=AD+CD=5+12=17,\\
In $\triangle$ABC, $AC^{2}=AB^{2}+BC^{2}$,\\
$\therefore$17$^{2}=(5+x)^{2}+(12+x)^{2}$,\\
Solving, we get $x=3$ (negative values are discarded),\\
$\therefore$AB=8, BC=15,\\
$\therefore$The area of $\triangle$ABC is: $\frac{1}{2}\times 8\times 15=\boxed{60}$,


Question: As shown in the figure, point $P$ lies on the extension line of the diagonal $BD$ of square $ABCD$, point $M$ is on side $AD$. Lines $CP$, $BM$, and $MP$ are connected. Given that $AB=4$, $AM=1$, and $BM=PM$, what is the length of $CP$?

Solution:
\textbf{Solution:} As shown in the figure, draw ME$\perp$BP at E through point M and draw PF$\perp$BC intersecting the extended line of BC at F, \\
$\because$ Quadrilateral ABCD is a square, \\
$\therefore$ AD=AB=4, $\angle$MDE=45$^\circ$, $\angle$A=90$^\circ$ \\
$\therefore$ MD=AD−AM=3, $\angle$DME=$\angle$DBC=45$^\circ$, \\
$\therefore$ ME=DE, \\
$\because$ $M{D}^{2}=M{E}^{2}+D{E}^{2}$, \\
$\therefore$ ${3}^{2}=2M{E}^{2}$, \\
$\therefore$ ME=DE=$\frac{3\sqrt{2}}{2}$, \\
$\because$ $B{D}^{2}=A{B}^{2}+A{D}^{2}$, \\
$\therefore$ BD=$\sqrt{A{B}^{2}+A{D}^{2}}=4\sqrt{2}$, \\
$\therefore$ BE=BD−DE=$\frac{5\sqrt{2}}{2}$, \\
$\because$BM=PM, \\
$\therefore$ BP=2BE=$5\sqrt{2}$, \\
$\because$$\angle$PBC=45$^\circ$, $\angle$PFB=90$^\circ$, \\
$\therefore$$\angle$BPF=45$^\circ$, \\
$\therefore$BF=PF, $ B{P}^{2}=P{F}^{2}+B{F}^{2}$, \\
$\therefore$ ${\left(5\sqrt{2}\right)}^{2}=2P{F}^{2}$, \\
$\therefore$PF=BF=5, \\
$\therefore$CF=BF−BC=1, \\
$\therefore$ $PC=\sqrt{P{F}^{2}+C{F}^{2}}=\boxed{\sqrt{26}}$,


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

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\
$\therefore$OB=OD, AB=CD,\\
Since E is the midpoint of BC,\\
$\therefore$OE is the median of $\triangle BCD$,\\
$\therefore$CD=2EO,\\
Since EO=8,\\
$\therefore$CD=2EO=16,\\
$\therefore$AB=CD=\boxed{16}.


Question: As shown in the figure, a regular hexagon $ABCDEF$ is inscribed in circle $\odot O$. If the distance from the center to a side $OH=\sqrt{3}$, then what is the radius of circle $\odot O$?

Solution: \textbf{Solution:} Let's connect OD. Since ABCDEF is a regular hexagon and OH is the apothem, it follows that $\angle DOH = \frac{360^\circ}{12} = 30^\circ$ and $\angle OHD = 90^\circ$. Therefore, OD equals twice DH. In the right triangle $ODH$, we have $OH^{2}$ + $DH^{2}$ = $OD^{2}$. Hence, $3 + DH^{2} = 4DH^{2}$. Solving this gives: DH=1 (considering positive value). Therefore, DH=\boxed{2}.


Question: As shown in the figure, a regular hexagon (the shaded area in the figure) is drawn on a regular triangular dartboard. Assuming that the probability of the dart hitting any point on the board is equal (if the dart hits the boundary or misses the board, then rethrow once), what is the probability that a dart thrown at random hits the shaded area?

Solution:
\textbf{Solution:} As shown in the figure, draw $AN\perp BC$ passing through point $A$ and intersect $DI$ at point $M$. Let the side length of the regular hexagon be $a$,

since hexagon $DEFGHI$ is a regular hexagon,

therefore $DE=EF=FG=GH=HI=DI=a$,

the degree of each external angle is: $360^\circ\div6=60^\circ$,

therefore $\angle ADI=\angle AID=\angle BEF=\angle BFE=\angle CGH=\angle CHG=60^\circ$,

therefore $\triangle ADI$, $\triangle BFE$, and $\triangle CHG$ are all equilateral triangles, and their side lengths are all equal to $a$,

since $\triangle ABC$ is an equilateral triangle,

therefore $BC=CA=AB=AD+DE+BE=a+a+a=3a$,

$\angle BAC=\angle B=60^\circ$

In $\triangle ADI$, $\triangle BFE$, and $\triangle CHG$

$\begin{cases}
\angle ADI=\angle BFE=\angle CHG\\
DI=FE=HG\\
\angle AID=\angle BEF=\angle CGH
\end{cases}$

therefore $\triangle ADI\cong \triangle BFE\cong \triangle CHG\left(ASA\right)$,

therefore ${S}_{\triangle ADI}={S}_{\triangle BFE}={S}_{\triangle CHG}$,

since $\angle ADI=\angle B=60^\circ$,

therefore $DI\parallel BC$,

since $AN\perp BC$,

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

therefore $\angle AMD=\angle ANB=90^\circ$,

since $\triangle ADI$ and $\triangle ABC$ are both equilateral triangles and their side lengths are $a$ and $3a$ respectively,

therefore $DM=\frac{1}{2}a$, $BN=\frac{3}{2}a$,

therefore $AM=\sqrt{AD^{2}-DM^{2}}=\sqrt{a^{2}-\left(\frac{1}{2}a\right)^{2}}=\frac{\sqrt{3}}{2}a$,

$AN=\sqrt{AB^{2}-BN^{2}}=\sqrt{\left(3a\right)^{2}-\left(\frac{3}{2}a\right)^{2}}=\frac{3\sqrt{3}}{2}a$,

therefore ${S}_{\triangle ADI}=\frac{1}{2}DI\cdot AM=\frac{1}{2}a\times \frac{\sqrt{3}}{2}a=\frac{\sqrt{3}}{4}a^{2}$,

${S}_{\triangle ABC}=\frac{1}{2}BC\cdot AN=\frac{1}{2}\times 3a\times \frac{3\sqrt{3}}{2}a=\frac{9\sqrt{3}}{4}a^{2}$,

therefore ${S}_{\triangle BFE}={S}_{\triangle CHG}={S}_{\triangle ADI}=\frac{\sqrt{3}}{4}a^{2}$,

therefore ${S}_{\text{regular hexagon }DEFGHI}={S}_{\triangle ABC}-3{S}_{\triangle ADI}=\frac{9\sqrt{3}}{4}a^{2}-3\times \frac{\sqrt{3}}{4}a^{2}=\frac{3\sqrt{3}}{2}a^{2}$,

therefore the probability of the dart hitting the shaded area is $\frac{{S}_{\text{regular hexagon }DEFGHI}}{{S}_{\triangle ABC}}=\frac{\frac{3\sqrt{3}}{2}a^{2}}{\frac{9\sqrt{3}}{4}a^{2}}=\boxed{\frac{2}{3}}$.


Question: As shown in the figure, it is known that the side length of the regular hexagon inscribed in a circle is 6. What is the radius of the circle?

Solution:
\textbf{Solution:} Since $\angle AOB$ is the central angle of a regular hexagon,\\
it follows that $\angle AOB=\frac{1}{6}\times 360^\circ=60^\circ$,\\
and since $OA=OB$,\\
it follows that $\triangle AOB$ is an equilateral triangle,\\
hence $OA=AB=\boxed{6}$,\\
which means the radius of this regular hexagon is 6.


Question: As shown in the figure, circle $\odot O$ is inscribed in square $ABCD$ with $O$ as its center. Let $\angle MON=90^\circ$ with its two sides intersecting $BC$ and $CD$ at points $M$ and $N$ respectively, and intersecting $\odot O$ at points $E$ and $F$. If $CM+CN=6$, what is the length of arc $EF$?

Solution:
\textbf{Solution:} Let circle $O$ be tangent to side $CD$ of square $ABCD$ at point $G$ and to $BC$ at point $H$, as shown in the solution diagram. Connect $OG$ and $OH$. Then, quadrilateral $OHCG$ is a square. \\
Since $\angle GON+\angle NOH=90^\circ$ \\
and $\angle HOM+\angle NOH=90^\circ$, \\
thus $\angle GON=\angle HOM$, \\
Also, since $\angle OGN=\angle OHM=90^\circ$, \\
and $OG=OH$, \\
thus $\triangle OGN \cong \triangle OHM$ (ASA), \\
therefore $GN=HM$ \\
Hence, the radius of circle $O$ is $= \frac{1}{2}(CM+CN)=3$ \\
Therefore, $\overset{\frown}{EF}=\frac{90}{180}\times 3\pi =\boxed{1.5\pi}$.


Question: As shown in the figure, the pentagon $ABCDE$ is a regular pentagon. If $l_{1} \parallel l_{2}$, then what is the value of $\angle \alpha - \angle \beta$?

Solution:
\textbf{Solution} Let line $BF$ pass through point $B$ and be parallel to $l_{1}$, as shown in the figure:\\
Since $l_{1}\parallel l_{2}$,\\
it follows that $BF\parallel l_{1}\parallel l_{2}$,\\
thus, $\angle \beta =\angle CBF$ and $\angle ABF+\angle \alpha =180^\circ$,\\
therefore, $\angle \alpha =180^\circ-\angle ABF$,\\
also, since pentagon $ABCDE$ is a regular pentagon,\\
it follows that $\angle ABC=\frac{180^\circ\times (5-2)}{5}=108^\circ$,\\
also, since $\angle ABC=\angle ABF+\angle CBF$,\\
it follows that $108^\circ=\angle \beta +180^\circ-\angle \alpha$\\
therefore, $\angle \alpha -\angle \beta =\boxed{72^\circ}$.


Question: As shown in the figure, $BD$ is the diameter of the circle $\odot O$. A tangent to $\odot O$ at point A intersects the extension of line $BD$ at point C. Lines $AB$ and $AD$ are drawn. If $\angle ABC=30^\circ$, what is the measure of $\angle C$?

Solution:
\textbf{Solution:} Connect $OA$,

since $OA=OB$,

it follows that $\angle OAB=\angle OBA=30^\circ$,

therefore $\angle AOC=\angle OAB+\angle OBA=60^\circ$,

since $OA$ is the tangent to the circle $O$,

it follows that $\angle OAC=90^\circ$,

therefore $\angle C=180^\circ-\angle OAC-\angle AOC=180^\circ-90^\circ-60^\circ=\boxed{30^\circ}$.


Question: If a dart is thrown randomly onto the dartboard shown in the figure, where each point on the board has an equal chance of being hit, what is the probability that the dart will land in the white region?

Solution:
\textbf{Solution:} As shown in the diagram, let the area of each small triangle be $a$,\\
then the area of the shaded region is $6a$, the area of the hexagon is $18a$, and the area of the white region is $12a$,\\
$\therefore$ throwing a dart at the dartboard randomly, the probability of it landing in the white area is $\frac{12a}{18a}=\boxed{\frac{2}{3}}$.


Question: As shown in the diagram, within a regular pentagon, $BE$ is a diagonal, $AF \perp BE$ intersects $CD$ at point $F$, then what is the degree measure of $\angle BAF$?

Solution:
\textbf{Solution:} Given that the polygon ABCDE is a regular pentagon,\\
$\therefore \angle BAE=\frac{1}{5}\times \left(5-2\right)\times 180^\circ=108^\circ$,\\
$\because AF\perp BE$,\\
$\therefore \angle BAF=\frac{1}{2}\angle BAE=\frac{1}{2}\times 108^\circ=54^\circ$.\\
Hence, the answer is: $\boxed{54^\circ}$.


Question: As shown in the figure, points $A$, $B$, $C$, $D$ are all on the circle $\odot O$, $\angle BAC=15^\circ$, $\angle BOD=70^\circ$, and $DE$ is tangent to $\odot O$ at $D$. What is the measure of $\angle CDE$?

Solution:
\textbf{Solution:} Draw line $OC$, \\
since $\angle BAC=15^\circ$, \\
thus $\angle BOC=2\angle BAC=30^\circ$, \\
since $\angle BOD=70^\circ$, \\
thus $\angle COD=70^\circ-30^\circ=40^\circ$, \\
since $OC=OD$, \\
thus $\angle ODC=\angle OCD=\frac{1}{2}\left(180^\circ-40^\circ\right)=70^\circ$, \\
since $DE$ is tangent to circle $O$ at $D$, \\
thus $OD\perp DE$, \\
thus $\angle CDE=90^\circ-70^\circ=\boxed{20^\circ}$


Question:
As shown in Figure 1, select a fixed point O in the plane and draw a directed ray $Ox$. By choosing a unit length, the position of any point M on the plane can be determined by the degree $\theta$ of $\angle MOx$ and the length m of $OM$, forming an ordered pair $\left(\theta, m\right)$, which is referred to as the "polar coordinates" of point M. The coordinate system established in this manner is called the "polar coordinate system".

Application: Given a polar coordinate system as in Figure 2, if a regular hexagon has a side length of 2 and one side OA lies on the ray $Ox$, what would be the polar coordinates of the vertex C of the hexagon?

Solution:
\textbf{Solution:} As shown in the figure, draw CB$\perp$OA at point B,\\
$\therefore$ $\angle$OBC=$90^\circ$,\\
$\because$ the side length of the regular hexagon is 2,\\
$\therefore$ OA=CA=2, $\angle$OAC=$120^\circ$,\\
$\therefore$ $\angle$COA=$\angle$OCA=$\angle$ACB=$30^\circ$,\\
$\therefore$ AB=$\frac{1}{2}$AC=1, BC=$\sqrt{3}$,\\
$\therefore$ OC=2BC=2$\sqrt{3}$,\\
$\therefore$ the polar coordinates of vertex C of the regular hexagon should be $\boxed{(30^\circ, 2\sqrt{3})}$.


Question: As shown in the diagram, it is known that points $A$, $B$, $C$ are on the circle $\odot O$, with $\angle ABC = 30^\circ$. The minor arc $\overset{\frown}{BC}$ is folded along the straight line $CB$ to intersect chord $AB$ at point $D$. If $DB = 7$ and $AD = 4$, what is the length of $BC$?

Solution:
\textbf{Solution:} Let point D be on the circle $\odot O$ and its corresponding point be E. Connect CE, BE, CD, AC, and draw CF $\perp$ AD at point F, as shown in the figure,\\
$\because$ quadrilateral ABEC is inscribed in $\odot O$,\\
$\therefore$ $\angle A+\angle E=180^\circ$,\\
$\because$ the corresponding point of D on $\odot O$ is E,\\
$\therefore$ according to the property of folding: $\angle BEC=\angle BDC$,\\
$\because$ $\angle BDC+\angle CDA=180^\circ$,\\
$\therefore$ $\angle E+\angle CDA=180^\circ$,\\
$\because$ $\angle A+\angle E=180^\circ$,\\
$\therefore$ $\angle A=\angle ADC$,\\
$\therefore$ $\triangle ACD$ is an isosceles triangle,\\
$\because$ $CF\perp AD$, $AD=4$,\\
$\therefore$ $AF=FD=\frac{1}{2}AD=2$,\\
$\because$ $BD=7$,\\
$\therefore$ $BF=BD+DF=9$,\\
$\because$ $CF\perp AD$,\\
$\therefore$ $\triangle CFB$ is a right triangle,\\
$\because$ $\angle ABC=30^\circ$,\\
$\therefore$ in $Rt\triangle CFB$, $CF=\frac{1}{2}BC$,\\
$\because$ in $Rt\triangle CFB$, $CF^2+BF^2=BC^2$,\\
$\therefore$ $\left(\frac{1}{2}BC\right)^2+9^2=BC^2$,\\
$\therefore$ $BC=\boxed{6\sqrt{3}}$, (negative values are discarded).


Question: As shown in the figure, if the graph of the function $y=ax+b$ is as depicted, then what is the solution set of the inequality $ax+b<2$ with respect to $x$?

Solution:
\textbf{Solution:} From the graph, it is known that the coordinates of the intersection point of the function $y=ax+b$ with the $y$-axis are $(0,2)$,\\
hence, the solution set of the inequality $ax+b<2$ is $x>0$,\\
therefore, the answer is $\boxed{x>0}$.


Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ is as illustrated. What is the range of $y$ when $x \leq 0$?

Solution:
\textbf{Solution:} From the graph, we know that when $x=0$, $y=-2$. The graph of the function $y$ decreases as $x$ increases, \\
$\therefore$ when $x\leq 0$, $y\geq -2$; \\
Therefore, the answer is: $\boxed{y\geq -2}$.


Question: As shown in the figure, the line $ l_{1}: y = x + 1 $ intersects with the line $ l_{2}: y = mx + n $ at the point $ P(a, 2) $. What is the solution set of the inequality $ x + 1 \ge mx + n $ with respect to $x$?

Solution:
\textbf{Solution:} Since the line $l_{1}\colon y=x+1$ intersects with the line $l_{2}\colon y=mx+n$ at point P$(1,2)$,\\
$\therefore a+1=2$,\\
Solving this gives: $a=1$,\\
Observing the graph, we know that the solution set of the inequality $x+1\geq mx+n$ with respect to $x$ is $\boxed{x\geq 1}$.


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

Solution:
Solution: Since the line $y=kx+b$ intersects the axes at two points, $A(2,0)$ and $B(0,-3)$, we have:
\[
\begin{cases}
2k + b = 0 \\
b = -3
\end{cases}
\]
Thus,
\[
\begin{cases}
k=\frac{3}{2} \\
b=-3
\end{cases}
\]
Therefore, $\frac{3}{2}x+(-3)+3=\frac{3}{2}x>0$,\\
Implying $x>0$.\\
Hence, the answer is $\boxed{x>0}$.


Question: As shown in the figure, it is known that the line $l_1\colon y = 3x+1$ and the line $l_2\colon y = mx+n$ intersect at point $P(a, -8)$. What is the solution set of the inequality $3x+1 < mx+n$ with respect to $x$?

Solution:
\[
\textbf{Solution:} \text{Let} \because \text{line } l_1\colon y = 3x+1 \text{ and line } l_2\colon y = mx+n \text{ intersect at point } P(a, -8)\\
\therefore 3a+1=-8 \\
\therefore a=-3 \\
\therefore P(-3, -8)\\
\text{According to the problem, when } x<-3, 3x+1<mx+n\\
\therefore \text{the solution set of the inequality } 3x+1<mx+n \text{ with respect to } x \text{ is } \boxed{x<-3}
\]


Question: Based on the graph, it is known that the functions $y=3x+b$ and $y=ax-3$ intersect at point $P(-2,-5)$. Then, according to the graph, what is the solution set for the inequality $3x+b>ax-3$?

Solution:
\textbf{Solution:} Since the graphs of the functions $y=3x+b$ and $y=ax-3$ intersect at point $P(-2,-5)$,\\
it follows from the graphical representation that the solution set of the inequality $3x+b>ax-3$ is $x>-2$,\\
therefore, the answer is: $\boxed{x>-2}$.


Question: As shown in the figure, if the graph of the linear function $y_{1}=x+a$ intersects with the graph of the linear function $y_{2}=kx+b$ at point P(1,3), what is the solution set of the inequality $x+a\le kx+b$ with respect to $x$?

Solution:
\textbf{Solution:} According to the graph, when $x\leq 1$, $x+a\leq kx+b$,\\
that is, the solution set of the inequality $x+a\leq kx+b$ with respect to $x$ is $\boxed{x\leq 1}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the coordinates of vertex $A$ of the rhombus $OABC$ are $(0, 2)$. The vertices $B$ and $C$ are in the first quadrant, and the ordinate of point $C$ is $1$. What are the coordinates of point $B$?

Solution:
\textbf{Solution:} Extend BC to intersect the x-axis at D,\\
$\because$ the coordinates of point A are (0, 2),\\
$\therefore$ OA = 2,\\
$\because$ quadrilateral OABC is a rhombus,\\
$\therefore$ AO = OC = BC = 2,\\
$\because$ BC $\parallel$ y-axis,\\
$\therefore$ BD $\perp$ x-axis,\\
In $\triangle$OCD,\\
$\because$ the ordinate of point C is 1,\\
$\therefore$ CD = 1,\\
$\therefore$ OD = $\sqrt{OC^{2}-CD^{2}} = \sqrt{2^{2}-1^{2}} = \sqrt{3}$,\\
$\because$ BD = BC + CD = 2 + 1 = 3,\\
$\therefore$ the coordinates of point B are $\boxed{(\sqrt{3}, 3)}$.


Question: As shown in the figure, the line $y=\frac{1}{2}x-2$ intersects the x-axis at point A. An isosceles right triangle $OAB$ is constructed above the x-axis with $OA$ as its hypotenuse. When the line is translated to the left along the x-axis until point B lies on the translated line, what is the distance by which the line has been translated?

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

Draw a perpendicular line from B to the x-axis, with the foot at D, let the intersection of the translated line with the x-axis be C,

For the line $y=\frac{1}{2}x-2$, let y=0, solving for x gives x=4, $\therefore$ the coordinates of point A are $(4,0)$

$\therefore$ OA=4

$\because$ $\triangle$OAB is an isosceles right triangle, BD$\perp$x-axis

$\therefore$ it is easy to find that OD=2, BD=2

$\therefore$ B$(2,2)$;

Let the equation of the translated line be: $y=\frac{1}{2}x+b$, substituting B$(2,2)$ into it gives $2=1+b$, solving for b gives $b=1$,

Therefore, the equation of the translated line is $y=\frac{1}{2}x+1$, setting its y=0 gives

$0=\frac{1}{2}x+1$

Solving it gives x=$-2$

$\therefore$ C$(0,-2)$,

$\therefore$ OC=2

$\therefore$ The distance of translation is OA+OC=4+2=\boxed{6}.


Question: As shown in the figure, it is known that the graphs of the functions $y=ax+b$ and $y=kx$ intersect at point $P$. According to the graph, what is the solution set of the linear inequality in one variable $ax+b>kx$ with respect to $x$?

Solution:
\textbf{Solution:} Since the graph of the function indicates that when $x>-3$, the linear function $y=ax+b$ is above the graph of the linear function $y=kx$, \\
therefore, the solution to the inequality in $x$, $ax+b>kx$, is $\boxed{x>-3}$.


Question: As shown in the figure, the straight line $y=kx+b$ ($k\neq 0$) passes through points $(2,0)$ and $(0,-3)$. What is the range of values of $x$ when $y>0$?

Solution:
\textbf{Solution:} Let the equation of the line be $y=kx+b$. This line intersects the x-axis at the point $(2,0)$, and it passes through the first, third, and fourth quadrants. From the graph, it is evident that\\
when $x>2$, the value of $y$ is greater than $0$,\\
therefore, the range of $x$ is $\boxed{x>2}$.


Question: As shown in the figure, fold the equilateral triangle $\triangle ABC$ such that point $C$ falls on point $D$ on side $AB$, where $EF$ is the crease. If $\angle ADE = 90^\circ$ and $AD = 1$, what is the length of $AC$?

Solution:
\textbf{Solution:} Given an equilateral $\triangle ABC$ is folded such that point C falls on the point D on side AB,\\
therefore $DE = CE$, and $\angle A = 60^\circ$,\\
since $\angle ADE = 90^\circ$,\\
it follows that $\angle AED = 30^\circ$,\\
also, given $AD = 1$,\\
thus $AE = 2AD = 2$,\\
therefore $DE = \sqrt{AE^2 - AD^2} = \sqrt{3}$,\\
hence $AC = AE + CE = AE + DE = 2 + \sqrt{3}$,\\
thus, the answer is: $AC = \boxed{2 + \sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=5$, $AC<BC$, and the perimeter of $\triangle ABC$ is $14$. Construct the perpendicular bisector $l$ of $AB$, which intersects $BC$ at point $E$. Connect $AE$. What is the perimeter of $\triangle ACE$?

Solution:
\textbf{Solution:} Given that $AB=5$ and the perimeter of $\triangle ABC$ is 14, \\
thus $AC+BC=14-5=9$, \\
since $DE$ is the perpendicular bisector of the segment $AB$, \\
it follows that $EA=EB$, \\
hence the perimeter of $\triangle ACE$ is $EA+EC+AC=EB+EC+AC=BC+AC=9$, \\
therefore, the answer is: $\boxed{9}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $\angle ABC = \angle ADC = 90^\circ$, $E$ is the midpoint of diagonal $AC$, connecting $BE$, $ED$, $BD$. If $\angle BAD = 58^\circ$, what is the degree measure of $\angle EBD$?

Solution:
\textbf{Solution:} Since $\angle ABC = \angle ADC = 90^\circ$,\\
points A, B, C, D lie on the same circle with E as the center and AC as the diameter.\\
Since $\angle BAD = 58^\circ$,\\
$\angle DEB = 116^\circ$.\\
Since DE = BE = $\frac{1}{2}$AC,\\
$\angle EBD = \angle EDB = \boxed{32^\circ}$.


Question: As shown in the figure, $BE\perp AC$ at point D, and $AD = CD$, $BD = ED$. If $\angle E = 35^\circ$, what is the value of $\angle ABC$?

Solution:
\textbf{Solution:} Given that $BE\perp AC$ and $AD=DC$,\\
it follows that $BA=BC$,\\
which implies $\angle ABD=\angle CBD=\frac{1}{2}\angle ABC$,\\
Considering $\triangle ADB$ and $\triangle CDE$, we have\\
\[
\begin{cases}
AD=DC\\
\angle ADB=\angle EDC\\
BD=DE
\end{cases},
\]
hence, $\triangle ADB\cong \triangle CDE$,\\
which means $\angle E=\angle ABD=35^\circ$,\\
thus, $\angle ABC=2\angle ABD=\boxed{70^\circ}$.


Question: As shown in the figure, it is known that in $\triangle ABC$, $AD=BD$, $F$ is the intersection of the height $AD$ and $BE$, $CD=2$, $AF=3$. What is the length of segment $BC$?

Solution:
\textbf{Solution:}\\
Since $AD$ and $BE$ are the altitudes of the triangle,\\
it follows that $\angle C+\angle DBF=90^\circ$, $\angle C+\angle CAD=90^\circ$, $AD\perp BC$,\\
which means that $\angle DBF=\angle DAC$, $\angle ADB=\angle ADC=90^\circ$,\\
Considering triangles $\triangle ACD$ and $\triangle BFD$,\\
we have $\left\{\begin{array}{l}
\angle DBF=\angle DAC\\
AD=BD\\
\angle BDF=\angle ADC=90^\circ
\end{array}\right.$,\\
therefore, $\triangle ACD\cong \triangle BFD$ (by ASA),\\
which implies that $CD=FD$,\\
Given that $CD=2$, $AF=3$,\\
it follows that $BD=AD=FD+AF=2+3=5$,\\
thus $BC=5+2=\boxed{7}$.


Question: As shown in the figure, within $\triangle ABC$, G is any point on side BC, and D, E, and F are the midpoints of AG, BD, and CE respectively. If the area of $\triangle ABC$ is $48$, then what is the value of the area of $\triangle DEF$?

Solution:
Solution: Draw the line CD as shown in the figure: \\
$\because$ Point D is the midpoint of AG,\\
$\therefore S_{\triangle ABD}=\frac{1}{2}S_{\triangle ABG}$, $S_{\triangle ACD}=\frac{1}{2}S_{\triangle AGC}$,\\
$\therefore S_{\triangle ABD}+S_{\triangle ACD}=\frac{1}{2}S_{\triangle ABC}=24$,\\
$\therefore S_{\triangle BCD}=\frac{1}{2}S_{\triangle ABC}=24$,\\
$\because$ Point E is the midpoint of BD,\\
$\therefore S_{\triangle CDE}=\frac{1}{2}S_{\triangle BCD}=12$,\\
$\because$ Point F is the midpoint of CE,\\
$\therefore S_{\triangle DEF}=\frac{1}{2}S_{\triangle CDE}=\boxed{6}$.


Question: As shown in the figure, AD is the bisector of $\angle BAC$ in $\triangle ABC$, DE$\perp$AB meeting AB at point E, and DF$\perp$AC meeting AC at point F. If DE$=2$ and AC$=4$, what is the area of $\triangle ADC$?

Solution:
Solution: Since $AD$ is the bisector of $\angle BAC$, and $DE \perp AB$, $DF \perp AC$,

$\therefore DF = DE$,

Since $DE = 2$,

$\therefore DF = 2$,

Therefore, the area of $\triangle ADC$ = $\frac{1}{2} \times AC \times DF = \frac{1}{2} \times 4 \times 2 = \boxed{4}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=60^\circ$, $\angle BCE=40^\circ$, $AD$ bisects $\angle BAC$, $CE \perp AB$ at point E, then what is the degree measure of $\angle ADC$?

Solution:
\textbf{Solution:} Since $CE\perp AB$,
\[\therefore \angle BEC=90^\circ,\]
Since \(\angle BCE=40^\circ\),
\[\therefore \angle B=50^\circ,\]
Since \(\angle BAC=60^\circ\) and $AD$ bisects \(\angle BAC\),
\[\therefore \angle BAD=\frac{1}{2}\angle BAC=30^\circ,\]
\[\therefore \angle ADC=\angle B+\angle BAD\]
\[=50^\circ+30^\circ\]
\[=\boxed{80^\circ}.\]


Question: As shown in the figure, $CD$ is the altitude from the right angle of $\triangle ABC$ to the hypotenuse $AB$. By folding $\triangle BCD$ along $CD$ such that point $B$ coincides precisely with the midpoint $E$ of $AB$, what is the measure of $\angle A$?

Solution:
\textbf{Solution:} Given that $\triangle BCD$ is folded along CD, \\
it follows that $BC = CE$, \\
since E is the midpoint of AB and $\angle ACB=90^\circ$, \\
it follows that $CE = BE = AE$, \\
thus, $\triangle BEC$ is an equilateral triangle, \\
hence, $\angle B = 60^\circ$, \\
therefore, $\angle A = 90^\circ - 60^\circ = \boxed{30^\circ}$


Question: As shown in the figure, if a set of triangular boards are stacked as illustrated, what is the degree measure of $\angle 1$?

Solution:
\textbf{Solution:} As shown in the figure,\\
$\because$ $\angle ABD=90^\circ$, $\angle ABC=45^\circ$, $\angle D=60^\circ$,\\
$\therefore$ $\angle CDB=\angle ABD-\angle ABC=90^\circ-45^\circ=45^\circ$,\\
$\because$ $\angle 1=\angle D+\angle CBD$,\\
$\therefore$ $\angle 1=60^\circ+45^\circ=\boxed{105^\circ}$.


Question: As shown in the figure, $\triangle ABC$ has a vertex angle of $120^\circ$ and $AB = AC$, with $EC = 4$. By folding $\triangle ABC$ so that point $B$ coincides with point $A$, with the crease being $DE$, what is the length of $DE$?

Solution:
\textbf{Solution:} Given that by folding triangle $ABC$ such that point $B$ coincides with point $A$, the fold line is $DE$,\\
it follows that $BE=AE$, and $\angle B = \angle BAE$,\\
given $AB=AC$ and $\angle BAC=120^\circ$,\\
it follows that $\angle B = \angle C = \angle BAE = \frac{1}{2}(180^\circ-120^\circ)=30^\circ$,\\
therefore, $\angle EAC = 120^\circ - 30^\circ = 90^\circ$,\\
in right triangle $AEC$, $AE = \frac{1}{2}EC = 2$;\\
in right triangle $AED$, $DE = \frac{1}{2}AE = \boxed{1}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, D is a point on AC, if $DA=DB=5$, the area of $\triangle ABD$ is $10$, what is the length of CD?

Solution:
\textbf{Solution:} Given that the area of $\triangle ABD$ is 10,\\
it follows that $\frac{1}{2}AD\cdot BC=10$, thus $\frac{1}{2}\times 5BC=10$. Solving this gives $BC=4$,\\
in the right-angled $\triangle BCD$, we have $DC=\sqrt{BD^{2}-CB^{2}}=\sqrt{5^{2}-4^{2}}=\boxed{3}$.


Question: As shown in the figure, $ \triangle ABC\cong \triangle AED$, with point E on line segment BC, and $ \angle 1=56^\circ$, what is the measure of $ \angle AED$?

Solution:
\textbf{Solution:} Since $\triangle ABC\cong \triangle AED$,\\
then $\angle BAC=\angle EAD$, and $AB=AE$,\\
thus $\angle BAE=\angle 1=56^\circ$,\\
therefore $\angle B=\angle AEB=\frac{1}{2}(180^\circ-56^\circ)=62^\circ$,\\
hence $\angle AED=\angle B=\boxed{62^\circ}$.


Question: As shown in the figure below, $\angle A=70^\circ$, BP and CP bisect $\angle ABC$ and $\angle ACB$ respectively. What is the measure of $\angle P$?

Solution:
\textbf{Solution:} In $\triangle ABC$, $\angle A = 70^\circ$,\\
$\therefore \angle ABC + \angle ACB = 180^\circ - \angle A = 180^\circ - 70^\circ = 110^\circ$,\\
$\because BP$ and $CP$ bisect $\angle ABC$ and $\angle ACB$ respectively,\\
$\therefore \angle PBC = \frac{1}{2} \angle ABC$, $\angle PCB = \frac{1}{2} \angle ACB$,\\
In $\triangle PBC$,\\
$\angle P + \angle PBC + \angle PCB = 180^\circ$,\\
$\therefore \angle P = 180^\circ - (\angle PBC + \angle PCB) = 180^\circ - \frac{1}{2} (\angle ABC + \angle ACB) = 180^\circ - \frac{1}{2} \times 110^\circ = \boxed{125^\circ}$.


Question: As shown in the figure, in the regular pentagon $ABCDE$, connect $AD$. What is the measure of $\angle 1$?

Solution:
\textbf{Solution:} From the polygon interior angle sum theorem, we can deduce that the sum of the interior angles of a pentagon is $(5-2)\times 180^\circ=540^\circ$,\\
$\therefore$ $\angle AED=540^\circ\div 5=108^\circ$,\\
$\because$ in a regular pentagon $AE=DE$,\\
$\therefore$ $\angle 1=\angle EDA=\left(180^\circ-108^\circ\right)\div 2=\boxed{36^\circ}$


Question: As shown in the figure, there are two congruent triangles. The letters in the figure represent the lengths of the sides of the triangles. What is the degree measure of $\angle 1$?

Solution:
\textbf{Solution:} As shown in the figure,\\
by the triangle interior angle sum theorem, $ \angle 2=180^\circ-60^\circ-54^\circ=66^\circ$,\\
since the two triangles are congruent,\\
hence $ \angle 1=\angle 2=\boxed{66^\circ}$.


Question: As shown in the figure, in $ \triangle ABC$, $\angle C=90^\circ$, D is a point on $AC$. If $DA=DB=15$ and the area of $\triangle ABD$ is $90$, what is the length of AC?

Solution:
\textbf{Solution:} Since the area of $\triangle ABD$ is $90$, and $\angle C=90^\circ$\\
Therefore, $\frac{1}{2}AD\cdot BC=90$\\
Therefore, $BC=\frac{90\times 2}{AD}=12$\\
In $Rt\triangle ABC$, $CD=\sqrt{BD^{2}-BC^{2}}=\sqrt{15^{2}-12^{2}}=9$\\
Therefore, $AC=AD+CD=24$\\
Hence, the answer is: $AC=\boxed{24}$.


Question: As shown in the figure, in $\triangle ABC$, $AB = AC$, $EF$ is the perpendicular bisector of $AC$, intersecting $AC$ at point $E$ and $AB$ at point $F$. $D$ is the midpoint of side $BC$, and $M$ is a moving point on segment $EF$. If $BC = 4$ and the area of $\triangle ABC$ is 16, what is the minimum perimeter of $\triangle CDM$?

Solution:
\textbf{Solution:}

Since the perimeter of $\triangle CDM$ is equal to $CM+MD+CD$, \\
then, when $CM+MD$ is at its minimum, the perimeter of $\triangle CDM$ is also at its minimum. \\
Since EF is the perpendicular bisector of AC, \\
then points A and C are symmetric with respect to EF, \\
hence $CM+MD=AM+MD\ge AD$. \\
By connecting AD, and intersecting EF at point M, at this point, the perimeter of $\triangle CDM$ is minimized. \\
Since AB=AC and D is the midpoint of side BC, \\
then $AD\perp BC$, \\
thus, the area of $\triangle ABC$ is $\frac{1}{2}BC\cdot AD=\frac{1}{2}\times 4AD=16$, \\
hence, $AD=8$, \\
therefore, the perimeter of $\triangle CDM$ is $AD+CD=8+2=\boxed{10}$.


Question: As shown in the figure, it is known that $\triangle ABC$ is an equilateral triangle, and points $B$, $C$, $D$, $E$ are on the same line, with $CG=CD$, $DF=DE$. What is the measure of $\angle EFD$?

Solution:
\textbf{Solution:} Since $\triangle ABC$ is an equilateral triangle,\\
it follows that $\angle ACB=60^\circ$.\\
Since $\angle ACB=\angle CGD+\angle CDG$,\\
it follows that $\angle CGD+\angle CDG=60^\circ$.\\
Since $CG=CD$,\\
it follows that $\angle CGD=\angle CDG=30^\circ$.\\
Since $\angle CDG=\angle DFE+\angle E$,\\
it follows that $\angle DFE+\angle E=30^\circ$.\\
Since $DF=DE$,\\
it follows that $\angle DFE=\angle E=15^\circ$.\\
Therefore, $\angle EFD=\boxed{15^\circ}$.


Question: As shown in the figure below, what is the sum of $ \angle A+\angle B+\angle C+\angle D+\angle E$?

Solution:
Extend $BE$ to meet $AC$ at point $G$, as shown in the figure,\\
$\because \angle BGC=\angle A+\angle B$ and $\angle GFC=\angle D+\angle E$,\\
$\therefore \angle A+\angle B+\angle C+\angle D+\angle E=\angle BGC+\angle GFC+\angle C=180^\circ$.\\
Hence, the answer is: $\boxed{180^\circ}$.


Question: The "Zhao Shuang Chord Diagram" is acclaimed as the totem of Chinese mathematics. It consists of four congruent right-angled triangles and a smaller square within a larger square, as shown in the diagram. By connecting $AE$, if the area of the large square $ABCD$ is $25$ and the area of $\triangle ABE$ is $8$, what is the area of the smaller square $EFGH$?

Solution:
\textbf{Solution:} Let $AG=BH=b$ and $AH=BE=a$,

since the area of the large square ABCD is 25,

thus $AB=5$,

therefore $a^{2}+b^{2}=25$,

since the area of $\triangle ABE$ is 8,

thus $\frac{1}{2}a^{2}=8$,

therefore $a=4$ (negative value discarded),

thus $b=\sqrt{25-16}=3$,

therefore, the area of the small square EFGH $=\left(a-b\right)^{2}=\left(4-3\right)^{2}=\boxed{1}$.


Question: As shown in the figure, $DE$ is the perpendicular bisector of side $AC$ of $\triangle ABC$, intersecting $BC$ at point $D$ and intersecting $AC$ at point $E$. If $BC = 8\text{cm}$ and $AD = 5\text{cm}$, what is the length of $BD$?

Solution:
\textbf{Solution:} Since $DE$ is the perpendicular bisector of side $AC$ in $\triangle ABC$, and $AD=5\text{cm}$, \\
then $DC=AD=5\text{cm}$, \\
thus $BD=BC-DC=8-5=\boxed{3}\text{cm}$.


Question: As shown in the figure, in the isosceles $\triangle ABC$, where $AB=AC$ and $\angle A=36^\circ$. $BD$ bisects $\angle ABC$. What is the measure of $\angle BDC$?

Solution:
\textbf{Solution:} Given that $AB = AC$ and $\angle A = 36^\circ$,\\
it follows that $\angle ABC = \frac{180^\circ - 36^\circ}{2} = 72^\circ$,\\
since BD bisects $\angle ABC$,\\
we have $\angle ABD = \frac{1}{2} \angle ABC = 36^\circ$,\\
therefore, $\angle BDC = \angle A + \angle ABD = \boxed{72^\circ}$.


Question: As shown in the figure, the right angle of the set square $DEF$ is placed inside $\triangle ABC$, where the two right-angle sides of the set square pass through points B and C, respectively. If $\angle A=55^\circ$, what is the sum of $\angle ABD+\angle ACD$?

Solution:
\textbf{Solution:} In $\triangle ABC$, $\angle A=55^\circ$, \\
$\therefore \angle ABC+\angle ACB=180^\circ-\angle A=125^\circ$, \\
In $\triangle DBC$, $\because \angle BDC=90^\circ$, \\
$\therefore \angle DBC+\angle DCB=180^\circ-\angle BDC=90^\circ$, \\
$\angle ABD+\angle ACD=\angle ABC+\angle ACB-(\angle DBC+\angle DCB)=125^\circ-90^\circ=\boxed{35^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $\angle A=120^\circ$, and $BC=15$ cm. The perpendicular bisector of $AB$ intersects $AB$ at point D and intersects $BC$ at point E; the perpendicular bisector of $AC$ intersects $AC$ at point G, and intersects $BC$ at point F. What is the length of $EF$?

Solution:
\textbf{Solution:} Let us connect $AE$ and $AF$,\\
since the perpendicular bisector of $AB$ intersects $AB$ at point D and $BC$ at point E; the perpendicular bisector of $AC$ intersects $AC$ at point G,\\
it follows that $BE=AE$, $CF=AF$,\\
therefore $\angle EAB=\angle B$, $\angle CAF=\angle C$,\\
since $\angle BAC=120^\circ$ and $AB=AC$,\\
it follows that $\angle B=\angle C=30^\circ$,\\
therefore $\angle BAE+\angle CAF=60^\circ$, $\angle AEF=\angle AFE=60^\circ$,\\
thus $\triangle AEF$ is an equilateral triangle,\\
hence $AE=AF=EF$,\\
consequently, $BE=EF=FC$,\\
since $BC=15=BE+EF+FC=3EF$,\\
it follows that $EF=\boxed{5}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, $AC = 4$, $BC = 8$, $ED$ bisects $AB$ perpendicularly at point $D$ and intersects $BC$ at $E$. What is the perimeter of $\triangle ACE$?

Solution:
\textbf{Solution:} Given that $ED$ is the perpendicular bisector of side $AB$,\\
it follows that $EA=EB$,\\
thus, the perimeter of $\triangle ACE$, which is $AC+EC+EA=AC+EC+EB=AC+BC$,\\
given $AC=4$ and $BC=8$,\\
therefore, the perimeter of $\triangle ACE$ is $AC+BC=4+8=\boxed{12}$.


Question: As shown in the figure, fold $\triangle ABC$ along the line $BD$, so that point $A$ falls exactly on point $E$ on side $BC$. If $CE=DE$ and $\angle C=32^\circ$, what is the degree measure of $\angle DBE$?

Solution:
\textbf{Solution}: Given folding $\triangle ABC$ along the line $BD$ such that point A precisely falls on point E on side $BC$,\\
$\therefore \angle DBE=\angle ABD=\frac{1}{2}\angle ABC$, $\angle A=\angle BED$,\\
$\because CE=DE$, $\angle C=32^\circ$\\
$\therefore \angle C=\angle EDC=32^\circ$,\\
$\therefore \angle BED=\angle C+\angle EDC=64^\circ$,\\
$\therefore \angle A=64^\circ$,\\
$\therefore \angle ABC=180^\circ-\angle A-\angle C=180^\circ-64^\circ-32^\circ=84^\circ$,\\
$\therefore \angle DBE=\frac{1}{2}\angle ABC=\boxed{42^\circ}$.


Question: As shown in the figure, it is known that $AB \parallel CD$, $AE$ and $CE$ bisect $\angle BAC$ and $\angle ACD$ respectively, $AE$ and $CE$ intersect at point $E$, draw line $ED \perp CD$ with the foot of the perpendicular being $D$, intersecting $AB$ at point $B$. If $AC=8, BD=6$, what is the area of $\triangle ACE$?

Solution:
\textbf{Solution:} Draw $EF\perp AC$ at point F from point E,\\
since $AB\parallel CD$ and $ED\perp CD$,\\
therefore $EB\perp AB$,\\
since $AE$ and $CE$ bisect $\angle BAC$ and $\angle ACD$ respectively,\\
therefore $BE=EF=ED$,\\
thus point E is the midpoint of $BD$,\\
since $AC=8$ and $BD=6$,\\
therefore $EF=\frac{1}{2}BD=3$,\\
thus, the area of $\triangle ACE$ is $\frac{1}{2}\times AC\times EF=\frac{1}{2}\times 8\times 3=\boxed{12}$


Question: As shown in the figure, in an equilateral $\triangle ABC$ with side length 6, $AD$ is the median on side $BC$, and $M$ is a moving point on $AD$, $E$ is a point on side $AC$. If $AE=2$, what is the minimum value of $EM+CM$?

Solution:
\textbf{Solution:} Draw line BE intersecting AD at point M, and draw CM. Construct EH$\perp$BC at point H,\\
$\therefore$ $\angle EHC=\angle BHE=90^\circ$,\\
$\because$ $\triangle ABC$ is an equilateral triangle, and AD is the median to side BC,\\
$\therefore$ AD$\perp$BC, BD=CD,\\
$\therefore$ Points B and C are symmetric about AD,\\
$\therefore$ BM=CM,\\
$\therefore$ EM+CM=BM+ME=BE,\\
The shortest distance between two points is along the straight line between them, the minimum value of EM+CM is the length of BE,\\
$\because$ $\triangle ABC$ is an equilateral triangle, AE=2, BC=AC=6,\\
$\therefore$ CE=4, $\angle ECH=60^\circ$, $\angle CEH=90^\circ-60^\circ=30^\circ$,\\
$\therefore$ $CH=\frac{1}{2}CE=2$,\\
$\therefore$ BH=6-2=4,\\
$\therefore$ $EH=\sqrt{CE^2-CH^2}=\sqrt{4^2-2^2}=2\sqrt{3}$;\\
In $\triangle BEH$,\\
$BE=\sqrt{BH^2+EH^2}=\sqrt{4^2+(2\sqrt{3})^2}=2\sqrt{7}$.\\
Therefore, the minimum value of EM+CM is $\boxed{2\sqrt{7}}$.


Question: As illustrated, AD is the median on the base BC of the isosceles $\triangle ABC$, BE bisects $\angle ABC$ and intersects AD at point E. Given $AC=12$ and $DE=3$, what is the area of $\triangle ABE$?

Solution:
\textbf{Solution:} Draw line EF $\perp$ AD at point F, \\
since AD is the median of the base BC of the isosceles $\triangle ABC$, \\
therefore EF $\perp$ AB, \\
since BE bisects $\angle ABC$, \\
therefore EF=ED=3, \\
therefore $S_{\triangle ABE}=\frac{1}{2}AB\cdot EF=\frac{1}{2}\times 12\times 3=\boxed{18}$.


Question: As shown in the figure, fold the rectangular paper \(ABCD\) such that side \(DC\) falls onto the diagonal \(AC\), with the fold line being \(CE\), and point \(D\) falls on the diagonal at point \(D'\). If \(AB=3\) and \(AD=4\), what is the length of \(ED\)?

Solution: \textbf{Solution:} Given that $AB=3$, $AD=4$, thus $DC=3$, $BC=4$. Therefore, $AC=\sqrt{AB^{2}+BC^{2}}=5$. By the principle of folding, we have $\triangle DEC\cong \triangle D'EC$, hence $D'C=DC=3$, $DE=D'E$. Let $ED=x$, then $D'E=x$, $AD'=AC-CD'=2$, $AE=4-x$. In the triangle $\triangle AED'$: $(AD')^{2}+(ED')^{2}=AE^{2}$, therefore $2^{2}+x^{2}=(4-x)^{2}$. Solving this, we find $x=\boxed{\frac{3}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, it is known that points D, E, and F are the midpoints of sides BC, AD, and CE, respectively, and the area of $\triangle ABC$ is $8\,\text{cm}^2$. What is the area of the shaded region?

Solution:
\textbf{Solution:} Since $AD$ is the median of $\triangle ABC$,\\
then $S_{\triangle ABD}=S_{\triangle ACD}=\frac{1}{2}S_{\triangle ABC}=4\,\text{cm}^2$,\\
Since point $E$ is the midpoint of $AD$,\\
then $S_{\triangle BDE}=\frac{1}{2}S_{\triangle ABD}=2\,\text{cm}^2$, $S_{\triangle CDE}=\frac{1}{2}S_{\triangle CAD}=2\,\text{cm}^2$,\\
thus $S_{\triangle BCE}=S_{\triangle BDE}+S_{\triangle CDE}=2+2=4\,\text{cm}^2$,\\
Since point $F$ is the midpoint of $CE$,\\
then $S_{\triangle BFE}=\frac{1}{2}S_{\triangle BCE}=2\,\text{cm}^2$.\\
Therefore, the area of the shaded part is $\boxed{2\,\text{cm}^2}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ is the angle bisector, and $AE$ is the height. Given that $\angle B=50^\circ$ and $\angle C=60^\circ$, then what is the degree measure of $\angle EAD$?

Solution:
\[ \textbf{Solution:} \]
\[ \because \angle BAC = 180^\circ - \angle B - \angle C, \]
\[ \therefore \angle BAC = 180^\circ - 50^\circ - 60^\circ = 70^\circ; \]
\[ \because \text{AD is the bisector of the angle}, \]
\[ \therefore \angle DAC = \frac{1}{2} \angle BAC = 35^\circ; \]
\[ \because \text{AE is the altitude}, \]
\[ \therefore \angle AEC = 90^\circ, \]
\[ \therefore \angle EAC = 90^\circ - \angle C = 90^\circ - 60^\circ = 30^\circ, \]
\[ \therefore \angle EAD = \angle DAC - \angle EAC = 35^\circ - 30^\circ = \boxed{5^\circ}. \]


Question: As shown in the figure, in $\triangle ABC$, $\angle BAC=90^\circ$, $AB=8$, $AC=6$, and point D is the midpoint of $BC$. What is the length of $AD$?

Solution:
\textbf{Solution}: Given that in $\triangle ABC$, $\angle BAC=90^\circ$, AB=8, AC=6, \\
therefore, $BC=\sqrt{AB^2+AC^2}=\sqrt{8^2+6^2}=10$, \\
since point D is the midpoint of BC, \\
thus, AD is the median of $\triangle ABC$, \\
therefore, AD=$\frac{1}{2}BC=\frac{1}{2}\times 10=\boxed{5}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=90^\circ$, $CD$ bisects $\angle ACB$, through point B draw $BD\perp CD$ with the foot of the perpendicular at point D, connect AD. If $AB=3$ and $BC=4$, what is the area of $\triangle ABD$?

Solution:
\textbf{Solution:} As shown in the figure, extend $BD$ to intersect $AC$ at point $E$. In the right triangle $\triangle ABC$, $AB=3$, $BC=4$,\\
$\therefore AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{3^{2}+4^{2}}=5$,\\
$\because CD$ bisects $\angle ACB$, and $BD\perp CD$,\\
$\therefore BC=CE=4$, $AE=AC-CE=1$, $BD=ED$,\\
$\therefore \frac{S_{\triangle ABE}}{S_{\triangle ABC}}=\frac{AE}{AC}=\frac{1}{5}$,\\
$\therefore S_{\triangle ABE}=\frac{1}{5}S_{\triangle ABC}=\frac{1}{5}\times\frac{1}{2}\times3\times4=\frac{6}{5}$,\\
$\therefore S_{\triangle ABD}=\frac{1}{2}S_{\triangle ABE}=\frac{1}{2}\times\frac{6}{5}=\boxed{\frac{3}{5}}$.


Question: As shown in the figure, point D is on the extension line of BC, DE $\perp$ AB at point E, and intersects AC at point F. If $\angle A = 35^\circ$ and $\angle D = 15^\circ$, what is the degree measure of $\angle ACB$?

Solution:
\textbf{Solution:} Since $DE \perp AB$ and $\angle A=35^\circ$,\\
it follows that $\angle CFD = \angle AFE = 90^\circ - \angle A = 90^\circ - 35^\circ = 55^\circ$,\\
Also, since $\angle D = 15^\circ$,\\
it follows that $\angle ACB = \angle D + \angle CFD = 15^\circ + 55^\circ = \boxed{70^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, and point D is the midpoint of AB. Connect CD. If $CD+AB=7.5$, what is the length of $CD$?

Solution:
\textbf{Solution:} Since $\angle ACB=90^\circ$ and point D is the midpoint of AB, \\
it follows that CD=$\frac{1}{2}$AB, \\
also, since CD+AB=7.5, \\
it follows that 3CD=7.5, \\
therefore, CD=\boxed{2.5}.


Question: As shown in the figure, in $\triangle ABC$, $D$ and $E$ are the midpoints of line segments $BC$ and $BA$ respectively. Let the area of $\triangle ABC$ be $S_{1}$, and the area of $\triangle EBD$ be $S_{2}$. What is the value of $\frac{S_{2}}{S_{1}}$?

Solution:
\textbf{Solution:} Since $D$ and $E$ are the midpoints of line segments $BC$ and $BA$ respectively, \\
thus $DE$ is the midline of $\triangle ABC$, \\
therefore $DE=\frac{1}{2}AC$ and $DE\parallel AC$, \\
therefore $\triangle EBD\sim \triangle ABC$, \\
therefore $\frac{S_{2}}{S_{1}}=\frac{S_{\triangle EBD}}{S_{\triangle ABC}}=\left(\frac{DE}{AC}\right)^{2}=\left(\frac{1}{2}\right)^{2}=\boxed{\frac{1}{4}}$.


Question: As shown in the figure, $AB \parallel CD \parallel EF$, and $AF$ intersects with $BE$ at point $G$. If $AG=2$, $GD=1$, $DF=5$, $BC=4$, then what is the length of $BE$?

Solution:
\textbf{Solution:} Given that $AB\parallel CD\parallel EF$,\\
it follows that $\frac{AD}{DF}=\frac{BC}{CE}$,\\
since $AD=AG+GD$, where $AG=2$, $GD=1$, $DF=5$, and $BC=4$,\\
we have $\frac{2+1}{5}=\frac{4}{CE}$,\\
which leads to $CE=\frac{20}{3}$,\\
thus, $BE=BC+CE=4+\frac{20}{3}=\boxed{\frac{32}{3}}$;


Question: As shown in the diagram, within $ \triangle ABC$, $DE \parallel BC$, $AD=9$, $DB=3$, and $DE=6$. What is the length of $BC$?

Solution:
\textbf{Solution:} Given that $DE \parallel BC$, \\
then $\triangle ADE \sim \triangle ABC$, \\
hence $\frac{AD}{AB} = \frac{DE}{BC}$, \\
since $AD = 9$, and $DB = 3$, \\
then AB = AD + BD = 12, \\
also, because $DE = 6$, \\
therefore, $\frac{9}{12} = \frac{6}{BC}$, \\
thus, BC = \boxed{8}.


Question: As shown in the diagram, D is a point on side AB, with $AD:DB = 3:4$ and $DE \parallel AC$, intersecting BC at point E. What is the value of $\frac{S_{\triangle BDE}}{S_{\triangle AEC}}$?

Solution:
\textbf{Solution:} Since $DE\parallel AC$,\\
it follows that $\triangle BDE\sim \triangle BAC$, and $AD\colon DB=3\colon 4$,\\
thus $BD\colon AB=DE\colon AC=4\colon 7$, $S_{\triangle BDE}\colon S_{\triangle BAC}=16\colon 49$,\\
therefore $S_{\triangle BDE}\colon S_{\text{quadrilateral }DECA}=16\colon 33$,\\
since $DE\colon AC=4\colon 7$, and the heights of $\triangle ADE$ and $\triangle ACE$ are equal,\\
it follows that $S_{\triangle ADE}\colon S_{\triangle ACE}=4\colon 7=12\colon 21$,\\
therefore $S_{\triangle BDE}\colon S_{\triangle AEC}=\boxed{16\colon 21}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC = 90^\circ$, $\angle A = 60^\circ$. Let D be a point on side $AC$, with $\angle DBC = 30^\circ$, and $AC = 12\,cm$. What is the perimeter of $\triangle ABD$?

Solution:
\textbf{Solution:} As shown in the diagram, since quadrilateral ABCD is a parallelogram, \\
therefore, AD $\parallel$ BC and AD = BC, \\
therefore, $\triangle$ADQ $\sim$ $\triangle$RCD, \\
therefore, AD $\colon$ RC = DQ $\colon$ CQ, that is AD $\colon$ RC $=3\colon1$, \\
therefore, AD = 3RC, \\
Similarly, $\triangle$ADP $\sim$ $\triangle$RBP, \\
therefore, AD $\colon$ BR = AP $\colon$ PR, that is AD $\colon$ (AD + RC) = AP $\colon$ PR, \\
therefore, 3RC $\colon$ 4RC = AP $\colon$ PR, that is AP $\colon$ PR $=\boxed{3\colon4}$.


Question: As shown in the diagram, in $\triangle ABC$, points $D$ and $E$ are on $AB$ and $AC$ respectively. If $\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{3}$, and the area of $\triangle ADE$ is 9, what is the area of quadrilateral $BCED$?

Solution:
\textbf{Solution}: Since $\because \frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{3}$ and $\angle A=\angle A$,\\
$\therefore \triangle ADE \sim \triangle ABC$ with a similarity ratio of $k=\frac{1}{3}$,\\
$\therefore S_{\triangle ADE} : S_{\triangle ABC} = 1 : 9$,\\
$\because S_{\triangle ADE} = 9$,\\
$\therefore S_{\triangle ABC} = 81$,\\
$\therefore$ the area of the quadrilateral $BCED$ is $S_{\triangle ABC} - S_{\triangle ADE} = 81 - 9 = \boxed{72}$.


Question: As shown in the figure, the centroid of $\triangle ABC$ is $G$, and point $D$ is on the extension line of $CB$, with $BD = \frac{1}{2}BC$. The line passing through points $D$ and $G$ intersects $AC$ at point $E$. What is the value of $\frac{AE}{AC}$?

Solution:
\textbf{Solution:} As shown in the diagram, extend CG and intersect AB at F, extend AG and intersect BC at H, connect FH which intersects DE at N,\\
$\because$ G is the centroid of $\triangle ABC$,\\
$\therefore$ $FH$ is the midline of $\triangle ABC$,\\
$\therefore$ $FH\parallel AC$,\\
$\because$ $BD=\frac{1}{2}BC$,\\
$\therefore$ $BD=BH=CH$,\\
$\because$ $NH\parallel EC$,\\
$\therefore$ $\frac{DH}{DC}=\frac{NH}{EC}=\frac{2}{3}$,\\
$\therefore$ $EC=\frac{3}{2}NH$,\\
$\because$ $NH\parallel AE$,\\
$\therefore$ $\frac{NH}{EA}=\frac{HG}{AG}=\frac{1}{2}$, thus $AE=2NH$,\\
$\therefore$ $\frac{AE}{CE}=\frac{2NH}{\frac{3}{2}NH}=\frac{4}{3}$,\\
$\therefore$ $\frac{AE}{AC}=\boxed{\frac{4}{7}}$


Question: As shown in the figure, it is known that the side length of rhombus $ABCD$ is 4, $E$ is the midpoint of $BC$, $AF$ bisects $\angle EAD$ and intersects $CD$ at point $F$, $FG \parallel AD$ and intersects $AE$ at point $G$. If $AB=AE$, what is the length of $FG$?

Solution:
\textbf{Solution:} As shown in the figure, we draw $AH$ perpendicular to $BC$ at H, extend $AE$ and $DC$ to meet at point M,\\
since $AB=AE$,\\
therefore $BH=EH$, $\angle ABE=\angle AEB$,\\
since the side length of the rhombus $ABCD$ is 4,\\
therefore $AB=AE=AD=BC=4$, $AB\parallel DC$,\\
since $E$ is the midpoint of $BC$,\\
therefore $BE=CE=2$,\\
therefore $BH=EH=\frac{1}{2}BE=1$,\\
since $AB\parallel DC$,\\
therefore $\angle B=\angle MCE$,\\
since $\angle AEB=\angle MEC$,\\
therefore $\angle MEC=\angle MCE$,\\
therefore $AE=AB=EM=CM=4$,\\
since $FG\parallel AD$,\\
therefore $\angle DAF=\angle AFG$,\\
since $AF$ bisects $\angle EAD$,\\
therefore $\angle GAF=\angle DAF$,\\
therefore $\angle GAF=\angle AFG$,\\
therefore $AG=GF$,\\
let $GF=x$,\\
then $AG=x$, $GE=4-x$, $MG=GE+EM=8-x$,\\
since $GF\parallel AD\parallel BC$,\\
therefore $\triangle MGF\sim \triangle MEC$,\\
therefore $\frac{EC}{FG}=\frac{EM}{MG}$,\\
therefore $\frac{2}{x}=\frac{4}{8-x}$,\\
solving for $x$ yields $x=\boxed{\frac{8}{3}}$.


Question: As shown in the figure, $l_{1}\parallel l_{2}\parallel l_{3}$, lines $a$ and $b$ intersect $l_{1}$, $l_{2}$, and $l_{3}$ at points $A$, $B$, $C$ and $D$, $E$, $F$ respectively. Given that $\frac{AB}{BC} = \frac{2}{3}$ and $DE = 4$, what is the length of $EF$?

Solution:
\textbf{Solution.} Given: $\because$ ${l}_{1}\parallel {l}_{2}\parallel {l}_{3}$\\
$\therefore$ $\frac{AB}{BC}=\frac{DE}{EF}$, that is $\frac{2}{3}=\frac{4}{EF}$, solving gives $EF=\boxed{6}$


Question: As shown in the figure, in rectangle $ABCD$, we have $AB=6$ and $AD=4$. Points $E$ and $F$ are the midpoints of $BC$ and $CD$, respectively. Lines $BF$ and $DE$ intersect at point $G$. Through point $E$, draw $EH \parallel CD$, intersecting $BF$ at point $H$. What is the length of segment $GH$?

Solution:
\textbf{Solution:} Given that quadrilateral $ABCD$ is a rectangle, $AB=6$, $AD=4$,\\
thus $DC=AB=6$, $BC=AD=4$, $\angle C=90^\circ$,\\
since points E and F are the midpoints of $BC$ and $CD$, respectively,\\
thus $DF=CF=\frac{1}{2}DC=3$, $CE=BE=\frac{1}{2}BC=2$,\\
since $EH\parallel CD$,\\
thus $FH=BH$,\\
since $BE=CE$,\\
thus $EH=\frac{1}{2}CF=\frac{3}{2}$,\\
By the Pythagorean theorem, $BF=\sqrt{BC^2+CF^2}=\sqrt{4^2+3^2}=5$,\\
thus $BH=FH=\frac{1}{2}BF=\frac{5}{2}$,\\
since $EH\parallel CD$,\\
thus $\triangle EHG\sim \triangle DFG$,\\
thus $\frac{EH}{DF}=\frac{GH}{FG}$,\\
thus $\frac{\frac{3}{2}}{3}=\frac{GH}{\frac{5}{2}-GH}$,\\
solving for: $GH=\boxed{\frac{5}{6}}$.


Question: As shown in the figure, two lines are intersected by parallel lines $l_{1}$, $l_{2}$, $l_{3}$, with points A, B, C, D, E, F being the points of intersection. Given that $AB=5$, $BC=6$, $EF=4$, what is the length of $DE$?

Solution:
\textbf{Solution:} Given that two lines are intersected by the parallel lines $l_{1}$, $l_{2}$, $l_{3}$,\\
it follows that $\frac{AB}{DE}=\frac{BC}{CF}$,\\
which means $\frac{5}{DE}=\frac{6}{4}$,\\
solving for $DE$ gives $DE=\boxed{\frac{10}{3}}$;


Question: As shown in the figure, the musical staff is composed of five equidistant and equal-length parallel horizontal lines. There are three points, A, B, and C, all located on the lines. If the length of segment AB equals \( \frac{3}{2} \), what is the length of segment BC?

Solution:
\textbf{Solution:} Draw a perpendicular line through point A parallel to the horizontal line, intersecting the horizontal line through point B at D and the horizontal line through point C at E,\\
Given that:\\
AD＝2DE, $ \frac{AB}{BC}=\frac{AD}{DE}$,\\
Since AB＝$\frac{3}{2}$,\\
Therefore, $\frac{\frac{3}{2}}{BC}$ ＝2,\\
Thus, BC＝$\boxed{\frac{3}{4}}$.


Question: As shown in the figure, it is known that AD is the median of side BC in $\triangle ABC$. A line is drawn through the centroid G, $GE\parallel AC$, intersecting BC at point E. If $DE=2$, what is the length of BC?

Solution:
\textbf{Solution:} Since $G$ is the centroid,\\
hence $DG : DA = 1 : 3$,\\
since $GE \parallel AC$,\\
hence $DE : DC = DG : DA = 1 : 3$,\\
since $DE = 2$,\\
hence $CD = 6$,\\
therefore $BC = 2CD = \boxed{12}$


Question: As shown in the figure, $G$ is the centroid of $\triangle ABC$, extending $BG$ to intersect $AC$ at point $D$, and extending $CG$ to intersect $AB$ at point $E$. $P$ and $Q$ are respectively the centroids of $\triangle BCE$ and $\triangle BCD$, the length of $BC$ is $12$, then what is the length of $PQ$?

Solution:
\textbf{Solution:} Draw line EP and extend EP to intersect BC at point F, connect DE,\\
since point G is the centroid of $\triangle ABC$,\\
therefore, BE and CE are the medians of $\triangle ABC$,\\
therefore, DE is the midsegment of $\triangle ABC$,\\
therefore, DE=$\frac{1}{2}BC=\frac{1}{2}\times 12=6$,\\
since P and Q are the centroids of $\triangle BCE$ and $\triangle BCD$ respectively,\\
therefore, the extensions of EP and DQ intersect at point F, and point F is on BC,\\
therefore, $\frac{PF}{EF}=\frac{FQ}{DF}=\frac{1}{3}$,\\
since $\angle PFQ=\angle DFE$,\\
therefore, $\triangle PFQ\sim \triangle DFE$,\\
therefore, $\frac{PF}{EF}=\frac{PQ}{DE}=\frac{1}{3}=\frac{PQ}{6}$,\\
Solving gives: $PQ=\boxed{2}$


Question: As shown in the figure, the straight lines $l_1 \parallel l_2 \parallel l_3$. If $AB = 3, BC = 5, EF = 4$, then what is the length of DE?

Solution:
\textbf{Solution:} Since $l_{1}\parallel l_{2}\parallel l_{3}$,\\
it follows that $\frac{AB}{BC}=\frac{DE}{EF}$,\\
since $AB=3$, $BC=5$, $EF=4$,\\
it follows that $\frac{3}{5}=\frac{DE}{4}$, hence $DE=\boxed{\frac{12}{5}}$


Question: As shown in the diagram, point \(P\) is the centroid of \(\triangle ABC\). Through point \(P\), draw \(DE\parallel AC\) intersecting \(BC, AB\) at \(D, E\) respectively, and \(EF\parallel BC\) intersecting \(AC\) at point \(F\). If \(BC=11\), what is the length of \(EF\)?

Solution:
Solution: Connect BP extending it to meet AC at point G,\\
since DE$\parallel$AC, and EF$\parallel$BC,\\
thus quadrilateral EFCD is a parallelogram,\\
therefore EF=CD;\\
since point P is the centroid,\\
therefore $\frac{BP}{PG}=2$,\\
since ED$\parallel$AC,\\
thus $\frac{BP}{PG}=\frac{BD}{CD}=2$,\\
therefore $\frac{BD}{EF}=2$\\
since BD+CD=BC=11 which implies 2EF+EF=11\\
Solving gives: $EF=\boxed{\frac{11}{3}}$


Question: As shown in the figure, in $\triangle ABC$, points D and E are on segments AB and AC, respectively, with $DE\parallel BC$, and $AD: AB = 1: 3$. If the area of $\triangle ADE$ is 1, what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Given that $DE \parallel BC$,

$\therefore$ $\triangle ADE \sim \triangle ABC$,

$\therefore$ $\frac{S_{\triangle ADE}}{S_{\triangle ABC}}=\left(\frac{AD}{AB}\right)^2=\left(\frac{1}{3}\right)^2$,

$\therefore$ $\frac{1}{S_{\triangle ABC}}=\frac{1}{9}$,

$\therefore$ $S_{\triangle ABC}=\boxed{9}$.


Question: As shown in the diagram, locations A and B are separated by a pond. Xiao Ming used the following method to measure the distance between A and B: He selected a point C outside line $AB$, then on lines $AC$ and $BC$ he found points M and N, respectively, such that $AM=2CM$ and $BN=2CN$. He measured the length of $MN$ to be $12m$. Based on this, what is the distance between A and B?

Solution:
\textbf{Solution:} \\
Since $AM=2CM$ and $BN=2CN$, \\
it follows that $\frac{AM}{CM}=2$ and $\frac{BN}{CN}=2$, thus $\frac{AC}{CM}=3$ \\
Because $\angle ACB=\angle MCN$, \\
it implies that $\triangle ACB\sim \triangle MCN$, \\
hence $\frac{AB}{MN}=3$, \\
Given that $MN=12m$, \\
it follows that $AB=\boxed{36m}$.


Question: As shown in the figure, in square $ABCD$, point $G$ is located on $BC$ and $\frac{GC}{BG}=\frac{1}{2}$. Line $DG$ intersects the diagonal $AC$ at point $F$. A perpendicular line from $D$ to $DG$ intersects the extension of line $CA$ at point $E$. If $AE=5$, what is the length of $DF$?

Solution:
\textbf{Solution:} Draw line $EH \perp AD$ through point $E$, extending it to intersect the extended line of $DA$ at $H$,\\
$\therefore \angle H=90^\circ$,\\
In square $ABCD$, $AB=BC=CD=AD$, $\angle BAD=\angle B=\angle BCD=\angle ADC=90^\circ$,\\
$\therefore \angle 2+\angle 3=90^\circ$, $\angle H=\angle BCD$,\\
Because $DE\perp DG$,\\
$\therefore \angle EDG=90^\circ$,\\
$\therefore \angle 2+\angle 1=90^\circ$,\\
$\therefore \angle 1=\angle 3$,\\
$\therefore \triangle DEH\sim \triangle DGC$,\\
$\therefore \frac{EH}{GC}=\frac{DH}{DC}$,\\
Since $\frac{GC}{BG}=\frac{1}{2}$,\\
$\therefore$ let $GC=x$, then $BG=2x$, $DC=BC=3x$,\\
$\therefore \frac{EH}{GC}=\frac{DH}{3x}$,\\
$\therefore DH=3EH$,\\
Because $AC$ is the diagonal of square $ABCD$,\\
$\therefore \angle DAC=45^\circ$,\\
Since $\angle EAH=\angle DAC=45^\circ$,\\
$\therefore \angle HEA=45^\circ$,\\
$\therefore EH=HA$,\\
$\therefore EH^2+HA^2=25$,\\
$\therefore EH=HA=\frac{5\sqrt{2}}{2}$,\\
$\therefore DH=3EH=\frac{15\sqrt{2}}{2}$,\\
$\therefore AD=DH−AH=5\sqrt{2}$,\\
$\therefore GC=\frac{1}{3}BC=\frac{1}{3}AD=\frac{5\sqrt{2}}{3}$,\\
$\therefore DG=\sqrt{CD^2+CG^2}=\frac{10\sqrt{5}}{3}$,\\
Since in square $ABCD$, $AD\parallel BC$,\\
$\therefore \frac{CG}{AD}=\frac{GF}{DF}=\frac{1}{3}$,\\
$\therefore DF=3GF$,\\
$\therefore DF=\frac{3}{4}DG=\frac{3}{4}\times \frac{10\sqrt{5}}{3}=\boxed{\frac{5\sqrt{5}}{2}}$;


Question: As shown in the figure, point C is the midpoint of segment AB, point D is the midpoint of segment AC, and DB equals 9. What is the length of AB?

Solution:
\textbf{Solution:} Since point C is the midpoint of AB,\\
it follows that AC=BC,\\
Since point D is the midpoint of the segment AC,\\
it follows that AC=BC=2AD=2DC\\
Since DB=9,\\
it follows that DC+BC=9,\\
it follows that DC+2DC=9,\\
Solving this gives: DC=AD=3,\\
Therefore, AB=AD+BD=3+9=\boxed{12}.


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

Solution:
\textbf{Solution:} Given that $AB=24$ and point C is the midpoint of AB,\\
it follows that $BC= \frac{1}{2} AB= \frac{1}{2} \times 24=12$,\\
since $AD\colon CB=1\colon 3$,\\
thus $AD= \frac{1}{3} \times 12=4$,\\
therefore $DB=AB−AD=24−4=\boxed{20}$.


Question: As shown in the diagram, it is known that line AD $\parallel$ BC, and BE bisects $\angle ABC$ and intersects line DA at point E. If $\angle DAB=54^\circ$, what is the value of $\angle E$?

Solution:
\textbf{Solution}: Since AD$\parallel$BC,\\
it follows that $\angle$ABC=$\angle$DAB=54$^\circ$, $\angle$EBC=$\angle$E,\\
since BE bisects $\angle$ABC,\\
it follows that $\angle$EBC= $\frac{1}{2}$ $\angle$ABC=27$^\circ$,\\
hence, $\angle$E=\boxed{27^\circ}.


Question: As shown in the figure, $AB \parallel CD$, $DE$ bisects $\angle BDC$, $\angle AED = 110^\circ$. What is the degree measure of $\angle B$?

Solution:
\textbf{Solution:} Given that $AB\parallel CD$,\\
it follows that $\angle AED + \angle CDE = 180^\circ$, $\angle B + \angle BDC = 180^\circ$,\\
since $\angle AED = 110^\circ$,\\
it follows that $\angle CDE = 70^\circ$,\\
since DE bisects $\angle BDC$,\\
it follows that $\angle BDC = 2\angle CDE = 140^\circ$,\\
therefore $\angle B = \boxed{40^\circ}$


Question: As shown in the figure, $AB=18\,cm$, C is the midpoint of $AB$. Point D is on line segment $AC$, and $AD:CB=1:3$, what is the length of $DC$?

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


Question: As shown in the figure, point C is the midpoint of segment AB, and CD is $\frac{1}{3}$ of AC. If AD equals 2 cm, what is the length of AB?

Solution:
\textbf{Solution:} Let $CD=x$,\\
since $CD=\frac{1}{3}AC$,\\
therefore $AC=3x$\\
since $AD=2cm$,\\
therefore $AD=AC-CD=3x-x=2x=2$\\
therefore $x=1$\\
therefore $AC=3cm$\\
since point C is the midpoint of the line segment AB,\\
therefore $AB=2AC= \boxed{6cm}$


Question: As shown in the figure, point $O$ is a point inside $\triangle ABC$, with $\angle A=60^\circ$. $BO$ and $CO$ are the angle bisectors of $\angle ABC$ and $\angle ACB$, respectively. What is the measure of $\angle BOC$?

Solution:
\textbf{Solution:} Since $\angle A=60^\circ$, \\
therefore $\angle ABC+\angle ACB=180^\circ-60^\circ=120^\circ$, \\
since $BO$ and $CO$ are the angle bisectors of $\angle ABC$ and $\angle ACB$ respectively, \\
therefore $\angle OBC+\angle OCB=\frac{1}{2}\left(\angle ABC+\angle ACB\right)=60^\circ$, \\
hence $\angle BOC=180^\circ-\left(\angle OBC+\angle OCB\right)=180^\circ-60^\circ=\boxed{120^\circ}$.


Question: As shown in the figure, it is known that $AD \parallel BC$, $\angle B=35^\circ$, and $DB$ bisects $\angle ADE$, what is the measure of $\angle DEC$?

Solution:
\textbf{Solution:} Since $AD\parallel BC$ and $\angle B=35^\circ$, \\
it follows that $\angle ADB=\angle B=35^\circ$, \\
Since DB bisects $\angle ADE$, \\
it follows that $\angle ADE=2\angle ADB=70^\circ$, \\
Since $AD\parallel BC$, \\
it follows that $\angle DEC =\angle ADE =\boxed{70^\circ}$.


Question: As shown in the figure, OC is the bisector of $\angle AOB$, OD bisects $\angle AOC$, and $\angle COD=20^\circ$. What is the measure of $\angle AOB$?

Solution:
\textbf{Solution:} Given that $OD$ bisects $\angle AOC$ and $\angle COD = 20^\circ$, we can deduce that $\angle AOC = 2\angle COD = 40^\circ$, \\
and since $OC$ is the bisector of $\angle AOB$, it follows that $\angle AOB = 2\angle AOC = 80^\circ$. \\
Therefore, $\angle AOB = \boxed{80^\circ}$.


Question: Given: As shown in the figure, the line $BO \perp AO$ at point $O$, $OB$ bisects $\angle COD$, $\angle AOC=68^\circ$, what is the degree measure of $\angle DOA$?

Solution:
\textbf{Solution:} Since $AO \perp BO$,\\
it follows that $\angle AOB = 90^\circ$,\\
Since $\angle AOC = 68^\circ$,\\
it follows that $\angle BOC = 90^\circ - 68^\circ = 22^\circ$,\\
Since $OB$ bisects $\angle COD$,\\
it follows that $\angle COD = 2\angle BOC = 44^\circ$,\\
thus $\angle DOA = \angle DOC + \angle AOC = 44^\circ + 68^\circ = \boxed{112^\circ}$, therefore C meets the requirements of the problem.


Question: As shown in the figure, $OB$ is the bisector of $\angle AOC$, $\angle COD = \frac{1}{3}\angle BOD$, $\angle COD = 17^\circ$. What is the measure of $\angle AOD$?

Solution:
\textbf{Solution:} Given $\because$ $\angle COD = \frac{1}{3} \angle BOD$, and $\angle COD = 17^\circ$,\\
$\therefore$ $\angle BOC = 2\angle COD = 2 \times 17^\circ = 34^\circ$,\\
since OB is the bisector of $\angle AOC$,\\
$\therefore$ $\angle AOC = 2\angle BOC = 2 \times 34^\circ = 68^\circ$,\\
$\therefore$ $\angle AOD = \angle AOC + \angle COD = 68^\circ + 17^\circ = \boxed{85^\circ}$


Question: As shown in the figure, $\angle AOD = 84^\circ$, $\angle AOB = 18^\circ$, and OB bisects $\angle AOC$. What is the measure of $\angle COD$?

Solution:
\textbf{Solution:} Given that $OB$ bisects $\angle AOC$, and $\angle AOB=18^\circ$,\\
it follows that $\angle AOC=2\angle AOB=36^\circ$,\\
Also, given that $\angle AOD=84^\circ$,\\
it results in $\angle COD=\angle AOD-\angle AOC=84^\circ-36^\circ=\boxed{48^\circ}$.


Question: As shown in the figure, point D is the midpoint of line segment AC, $BC = \frac{1}{2}AB$, and $BD = 1$cm. What is the length of AB?

Solution:
\textbf{Solution:} Let $BC=x$ and $BC=\frac{1}{2}AB$,\\
thus $AB=2BC=2x$, $AC=AB+BC=3x$,\\
since point D is the midpoint of line segment AC,\\
thus $AD=CD=\frac{3}{2}x$ and $DC=DB+BC$, $DB=1$,\\
hence $\frac{3}{2}x=1+x$,\\
Solving for $x$ gives: $x=2$,\\
thus $AB=2x=\boxed{4}$.


Question: As shown in the figure, it is known that the length of line segment AB is 12. Point C is on line segment AB with BC being $\frac{1}{3}$ of AB. Point D is the midpoint of AB. What is the length of CD?

Solution:
\textbf{Solution:} Since BC = $\frac{1}{3}$ AB = 4, \\
Since point D is the midpoint of AB, \\
Therefore, BD = 6, \\
Therefore, CD = BD - BC = \boxed{2}.


Question: As shown in the diagram, this is the net of a solid. What is the solid?

Solution:
\textbf{Solution:} From the unfolded figure, it can be seen that the solid is composed of three rectangular faces and two triangular faces,\\
therefore, the solid is a triangular prism.\\
Hence, the answer is: \boxed{B}.


Question: As shown in the figure, point $O$ is on line $AB$, rays $OC$ and $OD$ are on the same side of line $AB$ (where $0^\circ<\angle AOC<90^\circ$, $0^\circ<\angle BOD<90^\circ$), ray $OE$ bisects $\angle AOC$, and ray $OF$ bisects $\angle BOD$. If $\angle EOD$ and $\angle COF$ are complementary, what is the angle of $\angle COD$?

Solution:
\textbf{Solution:} Since $\angle EOD$ and $\angle COF$ are complementary,\\
it follows that $\angle EOD+\angle COF=180^\circ$,\\
thus $\angle EOF+\angle COD=180^\circ$,\\
and since $\angle EOF+\angle AOE+\angle BOF=180^\circ$,\\
it results in $\angle COD=\angle AOE+\angle BOF$,\\
given that ray OE bisects $\angle AOC$ and ray OF bisects $\angle BOD$,\\
we have $\angle AOE=\angle COE$, $\angle BOF=\angle DOF$,\\
thus $\angle COE+\angle DOF=\angle COD$,\\
which implies $\angle COD=180^\circ\div 3=\boxed{60^\circ}$.


Question: As shown in the figure, $OA$ is the radius of the circle $\odot O$, and $BC$ is a chord of $\odot O$, with $OA=6$. If $OA$ is perpendicular bisector of $BC$, what is the length of $BC$?

Solution:
\textbf{Solution:} Let's connect $OB$ and suppose $OA$ intersects $BC$ at point $D$.\\
Since $OA$ is perpendicular bisector to $BC$,\\
it follows that $BC=2BD$ and $OD=\frac{1}{2}OA=3$,\\
Also, since $OB=OA=6$,\\
then $BD=\sqrt{OB^{2}-OD^{2}}=\sqrt{6^{2}-3^{2}}=3\sqrt{3}$,\\
thus, $BC=2BD=6\sqrt{3}$.\\
Therefore, the length of $BC$ is $\boxed{6\sqrt{3}}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\odot O$, where $\angle A=100^\circ$. What is the measure of $\angle C$?

Solution:
\textbf{Solution}: Since quadrilateral ABCD is inscribed in a circle, \\
$\therefore$ $\angle A + \angle C = 180^\circ$, \\
Since $\angle A = 100^\circ$, \\
$\therefore$ $\angle C = 180^\circ - \angle A = 180^\circ - 100^\circ = \boxed{80^\circ}$


Question: A cylindrical oil tank is partially filled with oil, with its cross-section shown in the figure. It is known that the radius of the cross-section $\odot O$ is $5\,\text{cm}$, and the width of the oil surface $AB$ is $6\,\text{cm}$. If more oil is added into the tank, and the width of the oil surface becomes $8\,\text{cm}$, by how many centimeters does the oil surface $AB$ rise?

Solution:
\textbf{Solution:} Solve in two cases:\\
(1) As shown in Figure 1, for the oil surface CD with a width of 8cm, draw $ON\perp AB$ intersecting CD and AB at points M and N respectively.\\
From the problem statement, we know\\
$OM\perp CD$,\\
$CM=MD=\frac{1}{2}CD=4\,\text{cm}$,\\
$AN=BN=\frac{1}{2}AB=3\,\text{cm}$\\
In $\triangle BON$, by the Pythagorean theorem, we have\\
$ON=\sqrt{OB^{2}-BN^{2}}=4\,\text{cm}$\\
In $\triangle DOM$, by the Pythagorean theorem, we have\\
$OM=\sqrt{OD^{2}-DM^{2}}=3\,\text{cm}$\\
$\therefore MN=ON-OM=1\,\text{cm}$\\
(2) As shown in Figure 2, for the oil surface EF with a width of 8cm, draw $PN\perp EF$ intersecting AB and EF at points N and P respectively, and connect OB.\\
From the problem statement, we know\\
$PN\perp AB$,\\
$EP=PF=\frac{1}{2}EF=4\,\text{cm}$,\\
$AN=BN=\frac{1}{2}AB=3\,\text{cm}$\\
In $\triangle BON$, by the Pythagorean theorem, we have\\
$ON=\sqrt{OB^{2}-BN^{2}}=4\,\text{cm}$\\
In $\triangle EPO$, by the Pythagorean theorem, we have\\
$OP=\sqrt{OE^{2}-EP^{2}}=3\,\text{cm}$\\
$\therefore NP=ON+OP=7\,\text{cm}$\\
$\therefore$ The oil surface AB rises to CD by 1cm, and the oil surface AB rises to EF by \boxed{7}cm;


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $O$. If $\angle AOB = 40^\circ$ and $BC \parallel OA$, then what is the measure of $\angle ADC$?

Solution:
\textbf{Solution:} Given $\because \angle AOB=40^\circ$, and OA=OB,\\
$\therefore \angle ABO=70^\circ$,\\
$\because BC\parallel OA$,\\
$\therefore \angle OBC=\angle AOB=40^\circ$,\\
$\therefore \angle ABC=110^\circ$,\\
$\therefore \angle ADC=180^\circ-110^\circ=\boxed{70^\circ}$.


Question: As shown in the figure, in circle $O$, $OA \perp BC$, $\angle AOC = 50^\circ$, then what is the degree measure of $\angle ADB$?

Solution:
\textbf{Solution:} Connect OB\\
Since $OA\perp BC$, and $\angle AOC=50^\circ$,\\
then $\angle AOB=50^\circ$,\\
thus $\angle ADB=\frac{1}{2}\angle AOB=\boxed{25^\circ}$


Question: As shown in the figure, in the Cartesian coordinate system, the line $l\colon y=-x+\sqrt{2}$ intersects the $x$-axis at point $B$ and the $y$-axis at point $A$. With $AB$ as the right angle side, an isosceles right triangle $\triangle ABC$ is constructed to the right, and with $AO$ as the hypotenuse, an isosceles right triangle $\triangle ADO$ is constructed to the left. Connect $DC$ to intersect line $l$ at point $E$. What are the coordinates of point $E$?

Solution:
\textbf{Solution:} For the line $l\colon y=-x+\sqrt{2}$,

when $x=0$, $y=\sqrt{2}$; when $y=0$, $x=\sqrt{2}$.

$\therefore A\left(0,\sqrt{2}\right), B\left(\sqrt{2},0\right)$.

$\therefore OA=OB=\sqrt{2}$.

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

$\therefore AB=\sqrt{2}OA=2$, $\angle ABO=\angle BAO=45^\circ$.

Since $\triangle ABC$ is an isosceles right triangle, $\angle ABC=90^\circ$,

$\therefore BC=AB=2$, $\angle BAC=\angle BCA=45^\circ$.

$\therefore AC=\sqrt{2}AB=2\sqrt{2}$.

Since $\angle CAB=\angle ABO=45^\circ$,

$\therefore AC\parallel x$ axis.

$\therefore C\left(2\sqrt{2},\sqrt{2}\right)$.

Since $\triangle ADO$ is an isosceles right triangle, $\angle ADO=90^\circ$,

$\therefore AD=OD$.

Draw $DF\perp x$ axis at point F, thus $DF=OF=\frac{1}{2}AO=\frac{\sqrt{2}}{2}$.

$\therefore D\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$.

Let the function of line CD be $y=kx+b$,

according to the question, we get

$\left\{\begin{array}{l}
-\frac{\sqrt{2}}{2}k+b=\frac{\sqrt{2}}{2} \\
2\sqrt{2}k+b=\sqrt{2}
\end{array}\right.$.

Solving it, we find $\left\{\begin{array}{l}
k=\frac{1}{5} \\
b=\frac{3\sqrt{2}}{5}
\end{array}\right.$.

$\therefore$ The function of line CD is $y=\frac{1}{5}x+\frac{3\sqrt{2}}{5}$.

Combining the functions of line CD and AB, we get the system of equations

$\left\{\begin{array}{l}
y=\frac{1}{5}x+\frac{3\sqrt{2}}{5} \\
y=-x+\sqrt{2}
\end{array}\right.$,

solving it, we find $\left\{\begin{array}{l}
x=\frac{\sqrt{2}}{3} \\
y=\frac{2\sqrt{2}}{3}
\end{array}\right.$.

$\therefore E\left(\frac{\sqrt{2}}{3},\frac{2\sqrt{2}}{3}\right)$.

Thus, the coordinates of point $E$ are $\boxed{\left(\frac{\sqrt{2}}{3},\frac{2\sqrt{2}}{3}\right)}$.


Question: As shown in the figure, the position of the line $y=kx+b$ in the Cartesian coordinate system is illustrated. What is the solution set for the inequality $kx+b\le 2$?

Solution:
\textbf{Solution:} According to the graph, the line $y=kx+b$ passes through the points (0,1) and (2,0),\\
Substituting these two points into $y=kx+b$ yields $\left\{\begin{array}{l}b=1\\ 2k+b=0\end{array}\right.$, \\
Solving this system gives $\left\{\begin{array}{l}b=1\\ k=-\frac{1}{2}\end{array}\right.$, \\
$\therefore$ the equation of the line is: $y=-\frac{1}{2}x+1$, \\
Substituting y=2 we get $2=-\frac{1}{2}x+1$, \\
This gives x=$-2$, \\
$\therefore$ the solution set of the inequality $kx+b\le 2$ is $\boxed{x\ge -2}$.


Question: As shown in the figure, the intersection point of the lines $y=-x+a$ and $y=x+b$ has an abscissa of $-2$, and the abscissas of the points where the two lines intersect the $x$-axis are $-1$ and $-3$, respectively. What is the solution set of the inequality $-x+a > x+b > 0$ with respect to $x$?

Solution:
\textbf{Solution:} Let us assume that the intersection point of the line $y = -x + a$ and the line $y = x + b$ has an abscissa of $-2$. The abscissas where the two lines intersect the x-axis are $-1$ and $-3$, respectively.\\
Therefore, the solution set of the inequality in $x$, $-x + a > x + b > 0$, corresponds to the range of $x$ values for which the line $y = -x + a$ is above the line $y = x + b$, i.e., $-3 < x < -2$.\\
Hence, the solution set is $x \in \boxed{(-3, -2)}$.


Question: As shown in the graph, the line $y=-2x-4$ intersects the x-axis and y-axis at points A and B, respectively. Points C and D are the midpoints of line segment $AB$ and $OB$, respectively. Point M is a moving point on $OA$. What are the coordinates of point M when the value of $MC+MD$ is minimized?

Solution:
Solution:Given the problem, we know $A(-2,0)$, $B(0,-4)$,\\
then $C(-1,-2)$, $D(0,-2)$,\\
Construct the symmetric point of D with respect to the x-axis, $E(0,2)$, as shown below:\\
From this, we know: $MD=ME$,\\
$MC+MD=MC+ME$,\\
$\therefore$When points $C$, $M$, and $E$ are collinear, $MC+MD$ is minimized,\\
Let the line through $CE$ be $y=kx+b$, substituting $C(-1,-2)$, $E(0,2)$ into the equation, we get\\
$\left\{\begin{array}{l}
b=2 \\
-k+b=-2
\end{array}\right.$, solving this yields $\left\{\begin{array}{l}
b=2 \\
k=4
\end{array}\right.$,\\
$y=4x+2$,\\
Setting $y=0$, we solve to get $x=-\frac{1}{2}$, hence $M\left(-\frac{1}{2},0\right)$,\\
Therefore, the answer is $M\left(-\frac{1}{2},0\right)=\boxed{\left(-\frac{1}{2},0\right)}$


Question: As shown in the figure, the graph of the direct proportion function $y=kx$ ($k \neq 0$) intersects with the graph of the linear function $y=ax+b$ ($a \neq 0$) at point A(1, 1). What is the solution set of the inequality $kx \ge ax+b$?

Solution:
\textbf{Solution:} When $x\ge 1$, the graph of $y=kx$ is above the graph of $y=ax+b$,\\
$\therefore kx\ge ax+b$,\\
which means the solution set for $kx\ge ax+b$ is: $x\ge \boxed{1}$.


Question: As shown in the figure, the intersection of the line $y=-x+m$ and $y=nx+4n\ (n\neq 0)$ has an abscissa of $-2$. What are the integer solutions of the inequality $-x+m>nx+4n>0$ with respect to $x$?

Solution:
\textbf{Solution:} From the given condition, for $y=nx+4n\ (n\neq 0)$, then $x=-4$. Thus, the solution set for $nx+4n>0$ is $x>-4$.\\
Since the $x$-coordinate of the intersection point between $y=-x+m$ and $y=nx+4n\ (n\neq 0)$ is $-2$, by observing the graph, the solution set for $-x+m>nx+4n$ is $x<-2$.\\
Therefore, the solution set for $-x+m>nx+4n>0$ is $-4<x<-2$.\\
Since $x$ is an integer, thus $x=\boxed{-3}$.


Question: As shown in the figure, the intersection of the lines $y = -x + m$ and $y = nx + 4n$ ($n \neq 0$) has an x-coordinate of $-3$. What is the solution set for the inequality $-x + m > nx + 4n$ with respect to $x$?

Solution:
\textbf{Solution}: When $x<-3$, we have $-x+m>nx+4n$, \\
$\therefore$ the solution set for the inequality $-x+m>nx+4n$ with respect to $x$ is $\boxed{x<-3}$.


Question: In the Cartesian coordinate system, the position of the rhombus OABC is shown as in the figure, with $B(16, 8)$. What are the coordinates of point C?

Solution:
\textbf{Solution:} As shown in the figure, draw $BD\perp x$ axis at point $D$ through point $B$,

$\because B(16,8)$,

$\therefore D(16,0)$,

i.e., $OD=16$, $BD=8$,

$\because$ for diamond $OABC$, let $OA=x$, then $AB=OA=x$, $AD=16-x$,

In $\triangle ABD$ right, $AB^{2}=AD^{2}+BD^{2}$,

$x^{2}=(16-x)^{2}+8^{2}$,

Solving for $x$ yields $x=10$,

$\therefore BC=10$,

$\therefore C(6,8)$.

Thus, the answer is: $C(6,8)=\boxed{(6,8)}$.


Question: As shown in the figure, the intersection point of the lines $y=-x+a$ and $y=x+b$ has an abscissa of $-2$, and the abscissas of the points where the two lines intersect with the $x$ axis are $-1$ and $-3$, respectively. What is the solution set of the inequality $-x+a>x+b>0$ with respect to $x$?

Solution:
\textbf{Solution:} Given $\because$ the intersection point of the line $y = -x + a$ and $y = x + b$ has an x-coordinate of $-2$, and the x-intercepts of the two lines are $-1$ and $-3$ respectively, \\
$\therefore$ the solution set for the inequality in $x$, $-x + a > x + b > 0$, is the range of $x$ values for which the line $y = -x + a$ is above the line $y = x + b$, that is, $-3 < x < -2$.\\
Thus, the answer is: $\boxed{-3 < x < -2}$.


Question: As shown in the figure, the graph of the linear function $y=-\frac{1}{2}x+3$ intersects the x-axis at point A and the y-axis at point B, and intersects the graph of the direct proportionality function $y=x$ at point C. What is the ratio of the areas of $\triangle BOC$ to $\triangle AOC$?

Solution:
\textbf{Solution:} Given that the graph of the linear function $y=-\frac{1}{2}x+3$ intersects the x-axis at point A and the y-axis at point B, \\
thus $A\left(6,0\right)$, and $B\left(0,3\right)$, \\
Given that the linear function $y=-\frac{1}{2}x+3$ intersects the graph of the direct proportionality function $y=x$ at point C, \\
thus solving $\left\{\begin{array}{l}y=-\frac{1}{2}x+3 \\ y=x\end{array}\right.$ gives $C\left(2,2\right)$, \\
thus, the area of $\triangle BOC$ is $\frac{1}{2}\times 3\times 2=3$, and the area of $\triangle AOC$ is $\frac{1}{2}\times 6\times 2=6$, \\
thus, the ratio of the areas of $\triangle BOC$ and $\triangle AOC$ is $\boxed{\frac{1}{2}}$.


Question: As shown in Figure (1), within rectangle $ABCD$, a moving point $P$ starts from point $B$, travels along $BC$, $CD$, and $DA$ to stop at point $A$. Let the distance traveled by point $P$ be $x$, and the area of $\triangle ABP$ be $y$. If the graph of $y$ as a function of $x$ is as shown in Figure (2), what is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} According to the analysis of the function graph, when $0<x<3$, the value of $y$ continuously increases, and $P$ moves along $BC$; when $3\le x\le 7$, the value of $y$ does not change, so $P$ moves along $CD$; when $7<x<10$, the value of $y$ continuously decreases, and $P$ moves along $DA$;\\
Hence $BC=3$, and $CD=4$\\
Because quadrilateral $ABCD$ is a rectangle\\
It follows that $AB=CD=4$\\
The area of $\triangle ABC=\frac{1}{2}AB\cdot BC=\frac{1}{2}\times 3\times 4=\boxed{6}$