Criteria for Similarity of Triangles Questions in English

(D) According to the theorem of the ratio of areas of two similar triangles,if $\Delta ABC \sim \Delta XZY$,then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
Specifically,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta XZY)} = \frac{AB^2}{XZ^2} = \frac{BC^2}{ZY^2} = \frac{AC^2}{XY^2}$.
Given the correspondence $ABC \leftrightarrow XZY$,the side $BC$ corresponds to the side $ZY$ (or $YZ$).
Therefore,the ratio of the areas is $BC^2 : YZ^2$.
Thus,the missing term is $YZ^2$.

(A) Given that in $\Delta ABC$ and $\Delta PQR$,$\angle A = \angle P$ and $\angle B = \angle R$.
By the Angle-Angle $(AA)$ similarity criterion,if two angles of one triangle are respectively equal to two angles of another triangle,then the two triangles are similar.
Since $\angle A$ corresponds to $\angle P$ and $\angle B$ corresponds to $\angle R$,the third angle $\angle C$ must correspond to $\angle Q$ (as the sum of angles in a triangle is $180^{\circ}$).
Therefore,the correspondence $ABC \leftrightarrow PRQ$ represents the similarity $\Delta ABC \sim \Delta PRQ$.

(C) Given that $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow RPQ$. This implies that $\angle A = \angle R$,$\angle B = \angle P$,and $\angle C = \angle Q$.
In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^{\circ}$.
Given $m \angle A + m \angle C = m \angle B$,we substitute this into the sum equation: $m \angle B + m \angle B = 180^{\circ}$,which gives $2(m \angle B) = 180^{\circ}$,so $m \angle B = 90^{\circ}$.
Since $\angle B = \angle P$ from the correspondence,it follows that $m \angle P = 90^{\circ}$.
Therefore,$\angle P$ is a right angle.

(D) Given that in $\Delta ABC$ and $\Delta PQR$,$\angle A = \angle R$ and $\angle B = \angle P$.
By the $AA$ similarity criterion,$\Delta ABC \sim \Delta R P Q$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AB}{RP} = \frac{BC}{PQ} = \frac{AC}{RQ}$.
We are given $AB = 6$,$BC = 8$,and $PR = 9$.
Substituting the values into the ratio $\frac{AB}{RP} = \frac{BC}{PQ}$:
$\frac{6}{9} = \frac{8}{PQ}$.
Simplifying the fraction $\frac{6}{9}$ gives $\frac{2}{3}$.
So,$\frac{2}{3} = \frac{8}{PQ}$.
$2 \times PQ = 8 \times 3$.
$2 \times PQ = 24$.
$PQ = \frac{24}{2} = 12$.
Therefore,$PQ = 12$.

(A) Given that $\angle A \cong \angle E$,so $m \angle A = m \angle E$.
Given the equation $m \angle A + m \angle B = m \angle D + m \angle E$.
Substituting $m \angle A = m \angle E$ into the equation,we get $m \angle E + m \angle B = m \angle D + m \angle E$.
Subtracting $m \angle E$ from both sides,we get $m \angle B = m \angle D$.
Since two angles of $\Delta ABC$ are congruent to two angles of $\Delta DEF$ (specifically $\angle A \cong \angle E$ and $\angle B \cong \angle D$),by the $AA$ similarity criterion,the triangles are similar.
The correspondence must match the congruent angles: $A$ corresponds to $E$,$B$ corresponds to $D$,and therefore $C$ must correspond to $F$.
Thus,the correspondence is $ABC \leftrightarrow EDF$.

(D) Given the ratio of the sides of the two triangles: $\frac{AB}{XZ} = \frac{BC}{XY} = \frac{AC}{YZ}$.
According to the $SSS$ (Side-Side-Side) similarity criterion,two triangles are similar if their corresponding sides are in the same ratio.
By observing the ratios:
$AB$ corresponds to $XZ$
$BC$ corresponds to $XY$
$AC$ corresponds to $YZ$
Therefore,the vertices $A, B, C$ of $\Delta ABC$ correspond to the vertices $X, Z, Y$ of $\Delta XYZ$ respectively.
Thus,the correspondence is $\Delta ABC \sim \Delta XZY$.

(A) Given that $\square ABCD$ is a quadrilateral where $\overline{AB} \parallel \overline{CD}$.
Since $\overline{AB} \parallel \overline{CD}$,the alternate interior angles are equal,i.e.,$\angle MAB = \angle MCD$ and $\angle MBA = \angle MDC$.
By the $AA$ similarity criterion,$\triangle MAB \sim \triangle MCD$.
For similar triangles,the ratio of corresponding sides is equal: $\frac{MA}{MC} = \frac{MB}{MD}$.
Substituting the given values: $\frac{6}{8} = \frac{9}{MD}$.
Cross-multiplying gives: $6 \times MD = 8 \times 9$.
$6 \times MD = 72$.
$MD = \frac{72}{6} = 12$.

(B) In a right-angled triangle,the altitude drawn from the right-angle vertex to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.
By the property of geometric mean in a right triangle,the altitude $BD$ to the hypotenuse $AC$ satisfies the relation:
$BD^2 = AD \times CD$
Given $AD = 9$ and $CD = 4$,we have:
$BD^2 = 9 \times 4$
$BD^2 = 36$
Taking the square root on both sides:
$BD = \sqrt{36} = 6$
Thus,the length of $BD$ is $6$.

(N/A) Given: In $\Delta PQR$,$m \angle Q = 90^{\circ}$,$\overline{QD}$ is an altitude to the hypotenuse $PR$,and $PD = 9RD$.
To prove: $PQ = 3QR$.
Proof: In $\Delta PQR$,since $m \angle Q = 90^{\circ}$ and $\overline{QD} \perp \overline{PR}$,by the geometric mean theorem (or properties of similar triangles formed by an altitude to the hypotenuse):
$PQ^2 = PD \times PR$ and $QR^2 = RD \times PR$.
Dividing the two equations:
$\frac{PQ^2}{QR^2} = \frac{PD \times PR}{RD \times PR} = \frac{PD}{RD}$.
Substituting $PD = 9RD$ into the equation:
$\frac{PQ^2}{QR^2} = \frac{9RD}{RD} = 9$.
Taking the square root on both sides:
$\frac{PQ}{QR} = 3$.
Therefore,$PQ = 3QR$.

(A) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.
By the geometric mean theorem (or properties of similar triangles $\Delta AMB \sim \Delta BMC$),we have $BM^2 = AM \cdot MC$.
Given $AM = 12$ and $BM = 12$,we substitute these values into the equation:
$12^2 = 12 \cdot MC$
$144 = 12 \cdot MC$
$MC = \frac{144}{12} = 12$.
Since $AC = AM + MC$,we have:
$AC = 12 + 12 = 24$.

(D) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.
According to the geometric mean theorem (or properties of similar triangles),in a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other.
Specifically,$\Delta ABM \sim \Delta ACB$.
From the property of the altitude in a right triangle: $AB^2 = AM \cdot AC$.
Given $AB = 2\sqrt{10}$,we have $AB^2 = (2\sqrt{10})^2 = 4 \cdot 10 = 40$.
Given $AM = 5$,we substitute these values into the equation: $40 = 5 \cdot AC$.
Solving for $AC$: $AC = 40 / 5 = 8$.
Since $AC = AM + CM$,we have $8 = 5 + CM$.
Therefore,$CM = 8 - 5 = 3$.

(A) In a right-angled triangle,the altitude to the hypotenuse divides the triangle into two triangles similar to the original triangle and to each other.
By the geometric mean theorem,$BM^2 = AM \cdot CM$.
Substituting the given values: $BM^2 = (2x^2)(8x^2) = 16x^4$.
Taking the square root: $BM = 4x^2$.
Using the Pythagorean theorem in $\Delta ABM$: $AB^2 = AM^2 + BM^2 = (2x^2)^2 + (4x^2)^2 = 4x^4 + 16x^4 = 20x^4$.
Thus,$AB = \sqrt{20x^4} = 2\sqrt{5}x^2$.
Using the Pythagorean theorem in $\Delta BCM$: $BC^2 = CM^2 + BM^2 = (8x^2)^2 + (4x^2)^2 = 64x^4 + 16x^4 = 80x^4$.
Thus,$BC = \sqrt{80x^4} = 4\sqrt{5}x^2$.

(C) In a right-angled triangle $\Delta PQR$ where $\angle Q = 90^{\circ}$, $\overline{QM}$ is the altitude to the hypotenuse $PR$.
By the geometric mean theorem for right triangles, the altitude to the hypotenuse divides the triangle into two triangles similar to the original triangle and to each other.
Specifically, $\Delta QMR \sim \Delta PMQ$.
From the similarity $\Delta QMR \sim \Delta PMQ$, we have the ratio of corresponding sides: $\frac{QM}{RM} = \frac{PM}{QM}$.
Substituting the given values: $\frac{14}{7} = \frac{PM}{14}$.
$PM = \frac{14 \times 14}{7} = 2 \times 14 = 28$.
Now, in the right-angled triangle $\Delta PMQ$, by the Pythagorean theorem: $PQ^2 = PM^2 + QM^2$.
$PQ^2 = 28^2 + 14^2$.
$PQ^2 = 784 + 196 = 980$.
$PQ = \sqrt{980} = \sqrt{196 \times 5} = 14 \sqrt{5}$.

(A) In a right-angled triangle $\Delta XYZ$ where $\angle Y = 90^{\circ}$,$\overline{YM}$ is the altitude to the hypotenuse $\overline{XZ}$.
According to the geometric mean theorem for right triangles,the altitude to the hypotenuse divides the triangle into two triangles similar to the original triangle and to each other.
Specifically,$YM^2 = XM \cdot MZ$.
Given $YM = 12$ and $XM = 8$,we substitute these values into the equation:
$12^2 = 8 \cdot MZ$
$144 = 8 \cdot MZ$
$MZ = 144 / 8 = 18$.
The length of the hypotenuse $XZ$ is the sum of the segments $XM$ and $MZ$:
$XZ = XM + MZ = 8 + 18 = 26$.
Therefore,the correct option is $A$.

(A) In a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other.
By the geometric mean theorem for the altitude: $BM^2 = AM \cdot CM$.
Substituting the given values: $BM^2 = 4 \cdot 5 = 20$.
Therefore,$BM = \sqrt{20} = 2\sqrt{5}$.
In $\Delta ABM$,by the Pythagorean theorem: $AB^2 = AM^2 + BM^2$.
$AB^2 = 4^2 + (2\sqrt{5})^2 = 16 + 20 = 36$.
Therefore,$AB = \sqrt{36} = 6$.

(C) In a right-angled triangle,the altitude drawn from the right-angle vertex to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.
According to the geometric mean theorem,the altitude $\overline{BM}$ to the hypotenuse $\overline{AC}$ satisfies the relation: $BM^2 = AM \times CM$.
Given $AM = 12$ and $CM = 3$.
Substituting the values: $BM^2 = 12 \times 3 = 36$.
Taking the square root of both sides: $BM = \sqrt{36} = 6$.

(A) In a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other. By the geometric mean theorem,$BM^2 = AM \cdot CM$.
Given $AM = 4$ and $CM = 12$,$BM^2 = 4 \cdot 12 = 48$,so $BM = \sqrt{48} = 4\sqrt{3}$.
In $\Delta ABM$,by the Pythagorean theorem,$AB^2 = AM^2 + BM^2 = 4^2 + (4\sqrt{3})^2 = 16 + 48 = 64$. Thus,$AB = \sqrt{64} = 8$.
In $\Delta CBM$,by the Pythagorean theorem,$BC^2 = CM^2 + BM^2 = 12^2 + (4\sqrt{3})^2 = 144 + 48 = 192$. Thus,$BC = \sqrt{192} = 8\sqrt{3}$.

(D) In a right-angled triangle $\Delta PQR$ with $\angle Q = 90^{\circ}$,$\overline{QM}$ is the altitude to the hypotenuse $\overline{PR}$.
By the geometric mean theorem (or properties of similar triangles),we have $QM^2 = PM \cdot MR$.
Let $PM = x$. Since $PR = 10$,then $MR = 10 - x$.
Given $QM = 4$,we substitute these values into the equation:
$4^2 = x(10 - x)$
$16 = 10x - x^2$
$x^2 - 10x + 16 = 0$
Factoring the quadratic equation:
$(x - 2)(x - 8) = 0$
Thus,$x = 2$ or $x = 8$.
Therefore,$PM$ can be $2$ or $8$.

(C) In a right-angled triangle,if an altitude is drawn from the right-angle vertex to the hypotenuse,the triangle on either side of the altitude is similar to the original triangle and to each other.
Specifically,$\Delta RMQ \sim \Delta RQP$.
From the property of similar triangles,the ratio of corresponding sides is equal:
$\frac{RM}{RQ} = \frac{RQ}{RP}$.
Given $QR = 9$ and $PR = 13.5$.
Substituting the values into the equation:
$\frac{RM}{9} = \frac{9}{13.5}$.
$RM = \frac{9 \times 9}{13.5}$.
$RM = \frac{81}{13.5}$.
$RM = 6$.

(B) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
Here,$\overline{BE}$ is the median to the hypotenuse $\overline{AC}$ because $\angle B = 90^{\circ}$.
Therefore,$BE = \frac{1}{2} AC$.
Given $BE = 8.5$,we have $8.5 = \frac{1}{2} AC$,which implies $AC = 17$.
In $\Delta ABC$,by the Pythagorean theorem,$AB^2 + BC^2 = AC^2$.
Substituting the known values: $15^2 + BC^2 = 17^2$.
$225 + BC^2 = 289$.
$BC^2 = 289 - 225 = 64$.
$BC = \sqrt{64} = 8$.

(C) Let the two equal sides of the isosceles right-angled triangle be $x$.
According to the Pythagorean theorem, $x^2 + x^2 = (\text{hypotenuse})^2$.
Given the hypotenuse is $24$, we have $2x^2 = 24^2$.
$2x^2 = 576$.
$x^2 = 288$.
The area of a right-angled triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Here, base $= x$ and height $= x$, so Area $= \frac{1}{2} \times x \times x = \frac{1}{2}x^2$.
Substituting the value of $x^2$, we get Area $= \frac{1}{2} \times 288 = 144$.

(C) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
First,calculate the length of the hypotenuse $PR$ using the Pythagorean theorem: $PR^2 = PQ^2 + QR^2$.
$PR^2 = 20^2 + 21^2 = 400 + 441 = 841$.
$PR = \sqrt{841} = 29$.
Since $\overline{QM}$ is the median to the hypotenuse $PR$,its length is $QM = \frac{1}{2} \times PR$.
$QM = \frac{29}{2} = 14.5$.

(A) In $\Delta GBS$,$\angle B = 90^{\circ}$ and $BM \perp GS$.
By the property of similar triangles in a right-angled triangle,when an altitude is drawn to the hypotenuse,the triangles formed on both sides of the altitude are similar to the original triangle and to each other.
Specifically,$\Delta GMB \sim \Delta GBS$ and $\Delta B M S \sim \Delta GBS$.
From $\Delta GMB \sim \Delta GBS$,we have $\frac{GB}{GS} = \frac{GM}{GB}$,which implies $GB^{2} = GM \cdot GS$ (Equation $1$).
From $\Delta BMS \sim \Delta GBS$,we have $\frac{BS}{GS} = \frac{SM}{BS}$,which implies $BS^{2} = SM \cdot GS$ (Equation $2$).
Dividing Equation $1$ by Equation $2$:
$\frac{GB^{2}}{BS^{2}} = \frac{GM \cdot GS}{SM \cdot GS} = \frac{GM}{SM}$.
Hence,the result is proved.

(N/A) $1$. In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $\overline{QD} \perp \overline{PR}$.
$2$. By the property of similarity in right-angled triangles,$\Delta PDQ \sim \Delta PQR$ and $\Delta RDQ \sim \Delta RQP$.
$3$. Specifically,$\Delta PDQ \sim \Delta QDR$ is not directly applicable,but we use the property that in a right triangle,the altitude to the hypotenuse divides the triangle into two triangles similar to the original and to each other.
$4$. Thus,$\Delta PDQ \sim \Delta QDR$.
$5$. From the similarity $\Delta PDQ \sim \Delta QDR$,we have the ratio of corresponding sides: $\frac{PD}{QD} = \frac{QD}{RD} = \frac{PQ}{QR}$.
$6$. We are given $PQ = 3QR$,which implies $\frac{PQ}{QR} = 3$.
$7$. From $\frac{PD}{QD} = \frac{PQ}{QR}$,we get $PD = 3QD$.
$8$. From $\frac{QD}{RD} = \frac{PQ}{QR}$,we get $QD = 3RD$.
$9$. Substituting $QD = 3RD$ into $PD = 3QD$,we get $PD = 3(3RD) = 9RD$.
$10$. Hence,$PD = 9RD$ is proved.

(N/A) In $\Delta XYZ$,$\angle Y = 90^{\circ}$ and $\overline{YM} \perp \overline{XZ}$.
According to the Geometric Mean Theorem (or properties of similar triangles),in a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other.
Specifically,$\Delta XMY \sim \Delta YMZ$.
From the similarity $\Delta XMY \sim \Delta YMZ$,we have the ratio of corresponding sides:
$\frac{XY}{YZ} = \frac{XM}{YM} = \frac{YM}{ZM}$.
From $\frac{XY}{YZ} = \frac{YM}{ZM}$,we get $YM^2 = XY \cdot ZM$ (not directly useful here).
Alternatively,using the property of the altitude in a right triangle:
$XY^2 = XM \cdot XZ$ and $YZ^2 = ZM \cdot XZ$.
Dividing these two equations:
$\frac{XY^2}{YZ^2} = \frac{XM \cdot XZ}{ZM \cdot XZ} = \frac{XM}{ZM}$.
Given $XM = 16ZM$,we substitute this into the ratio:
$\frac{XY^2}{YZ^2} = \frac{16ZM}{ZM} = 16$.
Taking the square root of both sides:
$\frac{XY}{YZ} = \sqrt{16} = 4$.
Therefore,$XY = 4YZ$.

(N/A) In $\Delta ABC$,$\angle B = 90^\circ$ and $\overline{BM} \perp \overline{AC}$.
By the property of geometric mean in a right-angled triangle,we have $\Delta AMB \sim \Delta BMC$.
Therefore,the ratio of their corresponding sides is equal: $\frac{AB}{BC} = \frac{AM}{BM} = \frac{BM}{CM}$.
From $\frac{AB}{BC} = \frac{BM}{CM}$,we get $BM = \frac{AB \cdot CM}{BC}$.
Also,from $\Delta AMB \sim \Delta ABC$,we have $\frac{AB}{AC} = \frac{AM}{AB}$,which implies $AB^2 = AM \cdot AC$.
Similarly,from $\Delta BMC \sim \Delta ABC$,we have $\frac{BC}{AC} = \frac{CM}{BC}$,which implies $BC^2 = CM \cdot AC$.
Dividing the two equations: $\frac{AB^2}{BC^2} = \frac{AM \cdot AC}{CM \cdot AC} = \frac{AM}{CM}$.
Given $AM = 4CM$,we substitute this into the ratio: $\frac{AB^2}{BC^2} = \frac{4CM}{CM} = 4$.
Taking the square root of both sides: $\frac{AB}{BC} = \sqrt{4} = 2$.
Thus,$AB = 2BC$.

(D) In $\Delta ABC$,we are given $m \angle B = 90^{\circ}$ and $m \angle C = 30^{\circ}$.
Since the sum of angles in a triangle is $180^{\circ}$,$m \angle A = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ}$.
In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle,the side opposite to $30^{\circ}$ is half the hypotenuse.
Here,$AB$ is opposite to $30^{\circ}$ (at $C$),so $AB = \frac{1}{2} AC$.
Given $AB = 5$,we have $5 = \frac{1}{2} AC$,which implies $AC = 10$.
In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
Therefore,$BD = \frac{1}{2} AC = \frac{1}{2} \times 10 = 5$.

(A) In a right-angled triangle $\Delta ABC$ with $\angle B = 90^\circ$,if $\overline{BM}$ is the altitude to the hypotenuse $AC$,then by the Geometric Mean Theorem (or properties of similar triangles $\Delta BMC \sim \Delta BCA$),we have the relation: $BC^2 = CM \cdot AC$.
Given $BC = 2x + 2$,$CM = x + 1$,and $AC = 5x$.
Substituting these values into the equation: $(2x + 2)^2 = (x + 1)(5x)$.
Expanding the left side: $(2(x + 1))^2 = 4(x + 1)^2 = 4(x^2 + 2x + 1)$.
So,$4(x^2 + 2x + 1) = 5x^2 + 5x$.
$4x^2 + 8x + 4 = 5x^2 + 5x$.
Rearranging the terms to form a quadratic equation: $x^2 - 3x - 4 = 0$.
Factoring the quadratic: $(x - 4)(x + 1) = 0$.
This gives $x = 4$ or $x = -1$.
Since $x$ represents a length component and $BC, CM, AC$ must be positive,we discard $x = -1$.
Therefore,$x = 4$.

(A) In a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other.
By the geometric mean theorem for the altitude: $BM^2 = AM \cdot CM$.
Substituting the given values: $BM^2 = 4 \cdot 5 = 20$.
Therefore,$BM = \sqrt{20} = 2\sqrt{5}$.
Using the Pythagorean theorem in $\Delta ABM$: $AB^2 = AM^2 + BM^2 = 4^2 + (2\sqrt{5})^2 = 16 + 20 = 36$.
Therefore,$AB = \sqrt{36} = 6$.
Using the Pythagorean theorem in $\Delta CBM$: $BC^2 = CM^2 + BM^2 = 5^2 + (2\sqrt{5})^2 = 25 + 20 = 45$.
Therefore,$BC = \sqrt{45} = 3\sqrt{5}$.

(A) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $BE \perp AC$.
By the property of similarity in right-angled triangles,if an altitude is drawn to the hypotenuse,the triangles on both sides of the altitude are similar to the whole triangle and to each other.
$1$. $\Delta ABE \sim \Delta ABC$ (by $AA$ similarity).
Therefore,$\frac{AB}{AC} = \frac{AE}{AB}$,which implies $AB^2 = AE \cdot AC$ --- $(1)$.
$2$. $\Delta CBE \sim \Delta ABC$ (by $AA$ similarity).
Therefore,$\frac{BC}{AC} = \frac{CE}{BC}$,which implies $BC^2 = CE \cdot AC$ --- $(2)$.
Dividing equation $(1)$ by equation $(2)$:
$\frac{AB^2}{BC^2} = \frac{AE \cdot AC}{CE \cdot AC}$.
Canceling $AC$ from the numerator and denominator,we get:
$\frac{AB^2}{BC^2} = \frac{AE}{CE}$.
Hence proved.

(N/A) In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $QD \perp PR$.
By the property of geometric mean in a right-angled triangle,we have $\Delta PDQ \sim \Delta QDR$.
From the similarity of triangles,the ratio of corresponding sides is equal: $\frac{PQ}{QR} = \frac{PD}{QD} = \frac{QD}{DR}$.
From $\frac{PQ}{QR} = \frac{QD}{DR}$,we get $PQ^2 = QR^2 \cdot \frac{PD}{DR}$.
Given $PD = 25 DR$,so $\frac{PD}{DR} = 25$.
Substituting this value,$PQ^2 = QR^2 \cdot 25$.
Taking the square root of both sides,we get $PQ = 5 QR$.

(N/A) $1$. In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $\overline{QD} \perp \overline{PR}$.
$2$. By the property of similarity in right-angled triangles,$\Delta PDQ \sim \Delta QDR \sim \Delta PQR$.
$3$. From $\Delta PDQ \sim \Delta QDR$,we have the ratio of corresponding sides: $\frac{PD}{QD} = \frac{QD}{RD} = \frac{PQ}{QR}$.
$4$. From the ratio $\frac{PD}{QD} = \frac{PQ}{QR}$,we get $PD = QD \cdot \frac{PQ}{QR}$.
$5$. From the ratio $\frac{QD}{RD} = \frac{PQ}{QR}$,we get $QD = RD \cdot \frac{PQ}{QR}$.
$6$. Substituting $QD$ in the expression for $PD$: $PD = (RD \cdot \frac{PQ}{QR}) \cdot \frac{PQ}{QR} = RD \cdot (\frac{PQ}{QR})^2$.
$7$. Given $PQ = 4QR$,so $\frac{PQ}{QR} = 4$.
$8$. Therefore,$PD = RD \cdot (4)^2 = 16RD$.

(N/A) Given: In $\Delta PQR$,$\overline{PM} \perp \overline{QR}$ and $PM^2 = QM \times RM$.
Step $1$: From the given equation,we have $\frac{PM}{QM} = \frac{RM}{PM}$.
Step $2$: In $\Delta PMQ$ and $\Delta RMP$,$\angle PMQ = \angle RMP = 90^\circ$ (since $\overline{PM}$ is an altitude).
Step $3$: By the $SAS$ similarity criterion,$\Delta PMQ \sim \Delta RMP$.
Step $4$: Since the triangles are similar,their corresponding angles are equal: $\angle QPM = \angle PRM$ and $\angle PQM = \angle RPM$.
Step $5$: In $\Delta PQR$,the sum of angles is $180^\circ$. Thus,$\angle P + \angle Q + \angle R = 180^\circ$.
Step $6$: Substituting the angles,we get $\angle QPM + \angle RPM + \angle Q + \angle R = 180^\circ$.
Step $7$: Since $\angle QPM = \angle R$ and $\angle RPM = \angle Q$,we have $\angle R + \angle Q + \angle Q + \angle R = 180^\circ$,which simplifies to $2(\angle Q + \angle R) = 180^\circ$,so $\angle Q + \angle R = 90^\circ$.
Step $8$: Therefore,$\angle P = 180^\circ - 90^\circ = 90^\circ$. Hence,$\Delta PQR$ is a right-angled triangle.

(N/A) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BD} \perp \overline{AC}$.
By the property of geometric mean in a right-angled triangle,we have $BD^2 = AD \cdot CD$.
Given $AC = 5 CD$. Since $AC = AD + CD$,we have $AD + CD = 5 CD$,which implies $AD = 4 CD$.
Substituting $AD = 4 CD$ into the geometric mean equation:
$BD^2 = (4 CD) \cdot CD$
$BD^2 = 4 CD^2$
Taking the square root of both sides:
$BD = \sqrt{4 CD^2} = 2 CD$.
Thus,it is proved that $BD = 2 CD$.

(C) In a right-angled triangle,the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.
Specifically,$\Delta AMB \sim \Delta BMC$.
From the similarity $\Delta AMB \sim \Delta ABC$,we have the property: $BM^2 = AM \cdot MC$.
Given $AC = 25$ and $AM = 16$,we find $MC = AC - AM = 25 - 16 = 9$.
Substituting these values into the formula: $BM^2 = 16 \cdot 9$.
$BM^2 = 144$.
Taking the square root,$BM = \sqrt{144} = 12$.

(B) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.
Consider $\Delta ABM$ and $\Delta ABC$.
In $\Delta ABM$,$\angle AMB = 90^{\circ}$. Thus,$\angle BAM + \angle ABM = 90^{\circ}$.
In $\Delta ABC$,$\angle B = 90^{\circ}$. Thus,$\angle BAM + \angle C = 90^{\circ}$.
Comparing these two equations,we get $\angle ABM = \angle C$ (or $\angle MCB$).
Also,in $\Delta BMC$,$\angle BMC = 90^{\circ}$,so $\angle MBC + \angle C = 90^{\circ}$.
Since $\angle BAM + \angle C = 90^{\circ}$ and $\angle MBC + \angle C = 90^{\circ}$,it follows that $\angle BAM = \angle MBC$.

(D) In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $\overline{QD} \perp \overline{PR}$.
By the property of geometric mean in a right-angled triangle,we have $QD^2 = PD \times DR$.
Also,in $\Delta PQD$ and $\Delta PQR$,by similarity criteria,we have $PQ^2 = PD \times PR$.
Given $PD = 9 DR$,we can write $PR = PD + DR = 9 DR + DR = 10 DR$.
Substituting these into the expression for $PQ^2$:
$PQ^2 = (9 DR) \times (10 DR) = 90 DR^2$.
Similarly,in $\Delta QDR$ and $\Delta PQR$,we have $QR^2 = DR \times PR$.
$QR^2 = DR \times (10 DR) = 10 DR^2$.
Now,find the ratio $\frac{PQ^2}{QR^2} = \frac{90 DR^2}{10 DR^2} = 9$.
Taking the square root on both sides,we get $\frac{PQ}{QR} = \sqrt{9} = 3$.
Therefore,$PQ = 3 \times QR$.

(B) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$,the triangle is divided into two triangles $\Delta AMB$ and $\Delta BMC$ which are similar to the original triangle $\Delta ABC$ and to each other.
By the property of similarity,$\Delta AMB \sim \Delta ABC$.
This implies that the ratio of corresponding sides is equal: $\frac{AM}{AB} = \frac{AB}{AC}$.
Cross-multiplying gives $AB^2 = AM \cdot AC$.
Therefore,$AB$ is the geometric mean of $AM$ and $AC$.

(C) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,if $\overline{BM}$ is the altitude to the hypotenuse $AC$,then by the properties of similar triangles $(\Delta BMC \sim \Delta BCA)$:
$\frac{BC}{AC} = \frac{MC}{BC}$
Cross-multiplying gives:
$BC^2 = MC \cdot AC$
This implies that $BC$ is the geometric mean of $CM$ and $AC$.

(D) In a right-angled triangle,the altitude drawn to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.
Specifically,$\Delta AMB \sim \Delta BMC$.
From the similarity of these triangles,we have the ratio of corresponding sides:
$\frac{AM}{BM} = \frac{BM}{CM}$.
This leads to the geometric mean theorem: $BM^2 = AM \times CM$.
Given $AM = 8$ and $CM = 2$,we substitute these values into the equation:
$BM^2 = 8 \times 2 = 16$.
Taking the square root of both sides,we get $BM = \sqrt{16} = 4$.
Therefore,the length of $BM$ is $4$.

(C) In a right-angled triangle,the altitude drawn to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.
Specifically,$\Delta AMB \sim \Delta BMC$.
From the property of similarity,the ratio of corresponding sides is equal:
$\frac{AM}{BM} = \frac{BM}{CM}$.
This implies $BM^2 = AM \times CM$.
Given $AM = 4$ and $CM = 6$,we have:
$BM^2 = 4 \times 6 = 24$.
Taking the square root of both sides:
$BM = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$.

(A) In $\Delta PQR$,$\angle P = 90^{\circ}$ and $\overline{PX} \perp \overline{QR}$.
According to the geometric mean theorem (or property of altitude to the hypotenuse),in a right-angled triangle,the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.
Specifically,$\Delta QXP \sim \Delta QPR$ and $\Delta PXR \sim \Delta QPR$.
From the similarity $\Delta QXP \sim \Delta QPR$,we have $\frac{QX}{PQ} = \frac{PQ}{QR}$.
Thus,$PQ^2 = QX \cdot QR$.
Given $QX = 4$ and $RX = 5$,we have $QR = QX + RX = 4 + 5 = 9$.
Substituting these values,$PQ^2 = 4 \cdot 9 = 36$.
Therefore,$PQ = \sqrt{36} = 6$.

(A) In $\Delta XYZ$,the ratio of angles is $m \angle X : m \angle Y : m \angle Z = 2 : 3 : 5$.
Let the angles be $2k, 3k, 5k$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $2k + 3k + 5k = 180^{\circ}$,which gives $10k = 180^{\circ}$,so $k = 18^{\circ}$.
Thus,$m \angle X = 2(18^{\circ}) = 36^{\circ}$,$m \angle Y = 3(18^{\circ}) = 54^{\circ}$,and $m \angle Z = 5(18^{\circ}) = 90^{\circ}$.
The correspondence $XYZ \leftrightarrow EFD$ is a similarity,which implies $\Delta XYZ \sim \Delta EFD$.
Therefore,the corresponding angles are equal: $m \angle X = m \angle E$,$m \angle Y = m \angle F$,and $m \angle Z = m \angle D$.
Since $m \angle Z = 90^{\circ}$,it follows that $m \angle D = 90^{\circ}$.
Thus,$\angle D$ is a right angle.

(C) Given that $\Delta ABC \sim \Delta XYZ$ under the correspondence $ABC \leftrightarrow ZXY$.
This implies the ratio of corresponding sides is equal:
$\frac{AB}{ZX} = \frac{BC}{XY} = \frac{CA}{YZ}$.
From the properties of ratios,we have $\frac{AB}{ZX} = \frac{BC + CA}{XY + YZ}$.
Substituting the given values: $AB = 12, BC = 8, CA = 10, ZX = 10$.
$\frac{12}{10} = \frac{8 + 10}{XY + YZ}$.
$\frac{6}{5} = \frac{18}{XY + YZ}$.
$XY + YZ = \frac{18 \times 5}{6} = 3 \times 5 = 15$.
Thus,$XY + YZ = 15$.

(B) For two similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Let the areas be $A_1 = 25$ and $A_2 = 16$.
Let the corresponding sides be $s_1$ and $s_2$.
Then,$\frac{A_1}{A_2} = (\frac{s_1}{s_2})^2$.
$\frac{25}{16} = (\frac{s_1}{s_2})^2$.
Taking the square root on both sides,$\frac{s_1}{s_2} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.
The ratio of the perimeters of two similar triangles is equal to the ratio of their corresponding sides.
Therefore,the ratio of their perimeters is $5: 4$.

(C) Given that $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow PQR$.
According to the theorem of areas of similar triangles,the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left(\frac{AB}{PQ}\right)^2$.
Substituting the given values: $\frac{36}{64} = \left(\frac{12}{PQ}\right)^2$.
Taking the square root on both sides: $\sqrt{\frac{36}{64}} = \frac{12}{PQ}$.
$\frac{6}{8} = \frac{12}{PQ}$.
Simplifying the fraction: $\frac{3}{4} = \frac{12}{PQ}$.
Cross-multiplying: $3 \times PQ = 12 \times 4$.
$3 \times PQ = 48$.
$PQ = \frac{48}{3} = 16$.

(D) Given that $\Delta ABC \sim \Delta XYZ$ for the correspondence $ABC \leftrightarrow XYZ$.
According to the theorem of areas of similar triangles,the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta XYZ)} = \left(\frac{BC}{YZ}\right)^2$.
Substituting the given values: $\frac{72}{\text{Area}(\Delta XYZ)} = \left(\frac{6}{10}\right)^2$.
$\frac{72}{\text{Area}(\Delta XYZ)} = \frac{36}{100}$.
$\text{Area}(\Delta XYZ) = \frac{72 \times 100}{36}$.
$\text{Area}(\Delta XYZ) = 2 \times 100 = 200$.

(C) In $\Delta ABC$,$D$,$E$,and $F$ are the midpoints of $\overline{AB}$,$\overline{BC}$,and $\overline{CA}$ respectively.
By the Midpoint Theorem,$EF \parallel AB$ and $EF = \frac{1}{2} AB$,$DF \parallel BC$ and $DF = \frac{1}{2} BC$,and $DE \parallel AC$ and $DE = \frac{1}{2} AC$.
Therefore,$\frac{DE}{AC} = \frac{EF}{AB} = \frac{DF}{BC} = \frac{1}{2}$.
By the $SSS$ similarity criterion,$\Delta DEF \sim \Delta CAB$ (since $D$ corresponds to $C$,$E$ corresponds to $A$,and $F$ corresponds to $B$).
Thus,the correspondence $DEF \leftrightarrow CAB$ is a similarity.

(A) In $\Delta ADB$ and $\Delta BDC$:
$\angle ADB = \angle BDC = 90^{\circ}$ (since $\overline{BD} \perp \overline{AC}$).
In $\Delta ABC$,$\angle A + \angle C = 90^{\circ}$.
In $\Delta ADB$,$\angle A + \angle ABD = 90^{\circ}$.
Therefore,$\angle ABD = \angle C$.
By the $AA$ similarity criterion,the correspondence $A \leftrightarrow B$,$D \leftrightarrow D$,and $B \leftrightarrow C$ holds.
Thus,$\Delta ADB \sim \Delta BDC$ is incorrect; the correct correspondence is $\Delta ADB \sim \Delta BDC$ is not the standard notation,but based on the vertices,$\Delta ADB \sim \Delta BDC$ implies $A \leftrightarrow B, D \leftrightarrow D, B \leftrightarrow C$.
Looking at the options,the correspondence $ADB \leftrightarrow BDC$ is the correct similarity.

(A) Given that the correspondences $ABC \leftrightarrow BAC$ and $ABC \leftrightarrow ACB$ are similarities.
From $ABC \leftrightarrow BAC$,we have $\angle A = \angle B$,$\angle B = \angle A$,and $\angle C = \angle C$.
From $ABC \leftrightarrow ACB$,we have $\angle A = \angle A$,$\angle B = \angle C$,and $\angle C = \angle B$.
Combining these,we get $\angle A = \angle B$ and $\angle B = \angle C$.
Therefore,$\angle A = \angle B = \angle C$.
Since all three angles are equal,the triangle must be equiangular.
An equiangular triangle is always an equilateral triangle.
Thus,$\Delta ABC$ is an equilateral triangle.