Criteria for Similarity of Triangles Questions in English

(D) Given the similarity $\Delta ABC \sim \Delta DEF$ with the correspondence $ABC \leftrightarrow EFD$,the ratio of corresponding sides is equal.
Thus,$\frac{AB}{EF} = \frac{BC}{FD} = \frac{CA}{DE} = k$.
Given $AB : BC : CA = 4 : 3 : 5$,let $AB = 4x, BC = 3x, CA = 5x$.
From the correspondence $ABC \leftrightarrow EFD$,we have $EF = BC = 3x$,$FD = CA = 5x$,and $DE = AB = 4x$.
The perimeter of $\Delta DEF = EF + FD + DE = 3x + 5x + 4x = 12x$.
Given the perimeter is $36$,we have $12x = 36$,which implies $x = 3$.
Therefore,the sides are $EF = 3(3) = 9$,$FD = 5(3) = 15$,and $DE = 4(3) = 12$.
Thus,the sides of $\Delta DEF$ are $12, 9, 15$.

(C) Given that $\Delta ABC \sim \Delta PQR$,the ratio of their corresponding sides is equal to the ratio of their perimeters.
Let the perimeter of $\Delta ABC$ be $P_1 = 48$ and the perimeter of $\Delta PQR$ be $P_2 = 60$.
The ratio of corresponding sides is $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = \frac{P_1}{P_2} = \frac{48}{60} = \frac{4}{5}$.
Thus,$\frac{AB}{PQ} = \frac{4}{5}$.
For the second ratio,using the property of equal ratios: if $\frac{a}{b} = \frac{c}{d} = k$,then $\frac{a+c}{b+d} = k$.
Therefore,$\frac{AB + BC}{PQ + QR} = \frac{AB}{PQ} = \frac{4}{5}$.
Both ratios are equal to $\frac{4}{5}$.

(N/A) Two triangles are said to be similar if:
$1$. Their corresponding angles are equal.
$2$. Their corresponding sides are in the same ratio (or proportion).
Let there be two equilateral triangles,$\triangle ABC$ and $\triangle PQR$.
In an equilateral triangle,all angles are equal to $60^{\circ}$ and all sides are equal.
Therefore,for $\triangle ABC$: $\angle A = \angle B = \angle C = 60^{\circ}$ and $AB = BC = CA = a$.
For $\triangle PQR$: $\angle P = \angle Q = \angle R = 60^{\circ}$ and $PQ = QR = RP = b$.
Step $1$: Comparing angles:
$\angle A = \angle P = 60^{\circ}$,$\angle B = \angle Q = 60^{\circ}$,and $\angle C = \angle R = 60^{\circ}$.
Thus,the corresponding angles are equal.
Step $2$: Comparing sides:
$\frac{AB}{PQ} = \frac{a}{b}$,$\frac{BC}{QR} = \frac{a}{b}$,and $\frac{CA}{RP} = \frac{a}{b}$.
Thus,$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} = \frac{a}{b}$.
Since the corresponding angles are equal and the corresponding sides are in the same ratio,by the definition of similarity,$\triangle ABC \sim \triangle PQR$.
Hence,all equilateral triangles are similar.

Given: $ABCD$ is a parallelogram. $P$ is on $BC$,and $DP$ produced meets $AB$ produced at $L$.
$(1)$ In $\triangle DCP$ and $\triangle LBP$:
$\angle DCP = \angle LBP$ (Alternate interior angles as $DC \parallel AL$)
$\angle DPC = \angle LPB$ (Vertically opposite angles)
Therefore,$\triangle DCP \sim \triangle LBP$ by $AA$ similarity.
Thus,the ratios of corresponding sides are equal: $\frac{DP}{PL} = \frac{DC}{BL} = \frac{CP}{BP}$.
Hence,$\frac{DP}{PL} = \frac{DC}{BL}$ is proved.
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(C) Given that in $\Delta ABC$ and $\Delta PQR$,$\angle A = \angle P$ and $\angle B = \angle R$.
By the $AA$ (Angle-Angle) similarity criterion,$\Delta ABC \sim \Delta PRQ$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AB}{PR} = \frac{AC}{PQ}$
Substitute the given values: $AB = 8$,$PQ = 7.5$,$AC = 6$.
$\frac{8}{PR} = \frac{6}{7.5}$
$PR = \frac{8 \times 7.5}{6}$
$PR = \frac{60}{6} = 10$.
Therefore,$PR = 10$.

(A) Given that in $\Delta ABC$ and $\Delta XYZ$,$\frac{AB}{XY} = \frac{BC}{YZ}$ and $\angle B \cong \angle Y$.
By the $SAS$ (Side-Angle-Side) similarity criterion,$\Delta ABC \sim \Delta XYZ$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AB}{XY} = \frac{AC}{XZ} = \frac{BC}{YZ}$.
Substituting the given values:
$\frac{6}{15} = \frac{15}{XZ}$.
Cross-multiplying to solve for $XZ$:
$6 \times XZ = 15 \times 15$.
$6 \times XZ = 225$.
$XZ = \frac{225}{6} = 37.5$.
Therefore,$XZ = 37.5$.

(C) To determine the similarity,we compare the ratios of the corresponding sides of $\Delta ABC$ and $\Delta PQR$.
First,arrange the sides of each triangle in increasing order:
For $\Delta ABC$: $AB = 3$,$BC = 6$,$AC = 8$.
For $\Delta PQR$: $PR = 5.1$,$QR = 10.2$,$PQ = 13.6$.
Now,calculate the ratios of the corresponding sides:
Ratio of shortest sides: $\frac{AB}{PR} = \frac{3}{5.1} = \frac{30}{51} = \frac{10}{17}$.
Ratio of medium sides: $\frac{BC}{QR} = \frac{6}{10.2} = \frac{60}{102} = \frac{10}{17}$.
Ratio of longest sides: $\frac{AC}{PQ} = \frac{8}{13.6} = \frac{80}{136} = \frac{10}{17}$.
Since $\frac{AB}{PR} = \frac{BC}{QR} = \frac{AC}{PQ} = \frac{10}{17}$,by the $SSS$ similarity criterion,the correspondence $ABC \leftrightarrow PRQ$ is a similarity.

(C) Given that $\Delta PQR \sim \Delta XYZ$ for the correspondence $PQR \leftrightarrow XZY$.
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,$\frac{\text{Area of } \Delta PQR}{\text{Area of } \Delta XYZ} = \frac{PQ^2}{XZ^2}$.
Substituting the given values: $\frac{24}{\text{Area of } \Delta XYZ} = \frac{8^2}{12^2}$.
$\frac{24}{\text{Area of } \Delta XYZ} = \frac{64}{144}$.
Simplifying the fraction: $\frac{64}{144} = \frac{4}{9}$.
So,$\frac{24}{\text{Area of } \Delta XYZ} = \frac{4}{9}$.
$\text{Area of } \Delta XYZ = \frac{24 \times 9}{4} = 6 \times 9 = 54$.

(N/A) Given: $\Delta ABC \sim \Delta PQR$ with correspondence $ABC \leftrightarrow PQR.$ $\overline{AD}$ and $\overline{PM}$ are medians of $\Delta ABC$ and $\Delta PQR$ respectively.
To prove: $\frac{AD}{PM} = \frac{AB}{PQ}.$
Proof:
$1$. Since $\Delta ABC \sim \Delta PQR,$ the corresponding sides are proportional and corresponding angles are congruent.
$\therefore \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$ and $\angle B \cong \angle Q \quad \dots (1)$
$2$. Since $\overline{AD}$ and $\overline{PM}$ are medians,$D$ is the midpoint of $BC$ and $M$ is the midpoint of $QR.$
Therefore,$BC = 2BD$ and $QR = 2QM.$
$3$. Substituting these into equation $(1)$:
$\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{BD}{QM} \quad \dots (2)$
$4$. In $\Delta ABD$ and $\Delta PQM$:
From $(1)$,$\angle B \cong \angle Q.$
From $(2)$,$\frac{AB}{PQ} = \frac{BD}{QM}.$
$5$. By the $SAS$ (Side-Angle-Side) similarity criterion,$\Delta ABD \sim \Delta PQM.$
$6$. Since the triangles are similar,their corresponding sides are proportional:
$\therefore \frac{AD}{PM} = \frac{AB}{PQ}.$ Hence proved.

(N/A) Given: $\Delta ABC \sim \Delta XYZ$ with correspondence $ABC \leftrightarrow XYZ$. $\overline{AD} \perp \overline{BC}$ and $\overline{XM} \perp \overline{YZ}$.
To prove: $\frac{AD}{XM} = \frac{BC}{YZ}$.
Proof:
$1$. Since $\Delta ABC \sim \Delta XYZ$,the ratios of their corresponding sides are equal and corresponding angles are congruent.
$\therefore \frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$ and $\angle B \cong \angle Y$ ... $(1)$
$2$. In $\Delta ABD$ and $\Delta XYM$:
$\angle B \cong \angle Y$ (from result $1$)
$\angle ADB \cong \angle XMY$ (both are $90^{\circ}$ as $\overline{AD} \perp \overline{BC}$ and $\overline{XM} \perp \overline{YZ}$)
$3$. By the $AA$ similarity criterion,$\Delta ABD \sim \Delta XYM$.
$4$. Since the triangles are similar,the ratios of their corresponding sides are equal:
$\therefore \frac{AD}{XM} = \frac{AB}{XY}$ ... $(2)$
$5$. From $(1)$ and $(2)$,since $\frac{AB}{XY} = \frac{BC}{YZ}$,we get:
$\frac{AD}{XM} = \frac{BC}{YZ}$.
Hence proved.

(A) In $\Delta ABC$,$m\angle B = 90^\circ$ and $\overline{BM} \perp \overline{AC}$.
$(i)$ For the correspondence $AMB \leftrightarrow ABC$:
$\angle AMB \cong \angle ABC$ (both are $90^\circ$)
$\angle MAB \cong \angle BAC$ (common angle)
Therefore,by the $AA$ similarity criterion,$\Delta AMB \sim \Delta ABC$.
(ii) For the correspondence $BMC \leftrightarrow ABC$:
$\angle BMC \cong \angle ABC$ (both are $90^\circ$)
$\angle MCB \cong \angle BCA$ (common angle)
Therefore,by the $AA$ similarity criterion,$\Delta BMC \sim \Delta ABC$.
(iii) Since $\Delta AMB \sim \Delta ABC$ and $\Delta BMC \sim \Delta ABC$,by the transitivity of similarity,$\Delta AMB \sim \Delta BMC$.

(D) In $\Delta PQR$,$M$ and $N$ are the midpoints of sides $\overline{PQ}$ and $\overline{PR}$ respectively.
According to the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
Therefore,$MN \parallel QR$ and $MN = \frac{1}{2} QR$.
Since $MN \parallel QR$,$\Delta PMN$ is similar to $\Delta PQR$ by the $AA$ similarity criterion ($\angle PMN = \angle PQR$ and $\angle PNM = \angle PRQ$ as corresponding angles).
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\frac{\text{Area}(\Delta PMN)}{\text{Area}(\Delta PQR)} = \left( \frac{MN}{QR} \right)^2$.
Substituting $MN = \frac{1}{2} QR$,we get $\frac{\text{Area}(\Delta PMN)}{\text{Area}(\Delta PQR)} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Given $\text{Area}(\Delta PMN) = 24$,we have $\frac{24}{\text{Area}(\Delta PQR)} = \frac{1}{4}$.
Therefore,$\text{Area}(\Delta PQR) = 24 \times 4 = 96$.

(A) According to the Midpoint Theorem,if $P$ and $Q$ are the midpoints of sides $\overline{XY}$ and $\overline{XZ}$ of $\Delta XYZ$,then $\overline{PQ} \parallel \overline{YZ}$ and $PQ = \frac{1}{2} YZ$.
Since $\overline{PQ} \parallel \overline{YZ}$,$\Delta XPQ$ is similar to $\Delta XYZ$ by the $AA$ similarity criterion ($\angle X = \angle X$ and $\angle XPQ = \angle XYZ$ as corresponding angles).
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\text{Area}(\Delta XPQ)}{\text{Area}(\Delta XYZ)} = \left( \frac{PQ}{YZ} \right)^2$.
Substituting the ratio $\frac{PQ}{YZ} = \frac{1}{2}$,we get $\frac{\text{Area}(\Delta XPQ)}{140} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Thus,$\text{Area}(\Delta XPQ) = \frac{140}{4} = 35$.

(B) Given that $M$ and $N$ are the midpoints of sides $AB$ and $AC$ of $\Delta ABC$ respectively.
According to the Midpoint Theorem, the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
Therefore, $MN \parallel BC$ and $MN = \frac{1}{2} BC$.
Since $MN \parallel BC$, $\Delta AMN$ is similar to $\Delta ABC$ by the $AA$ similarity criterion ($\angle AMN = \angle ABC$ and $\angle ANM = \angle ACB$ as corresponding angles).
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Thus, $\frac{\text{Area}(\Delta AMN)}{\text{Area}(\Delta ABC)} = \left( \frac{MN}{BC} \right)^2$.
Substituting $MN = \frac{1}{2} BC$, we get $\frac{\text{Area}(\Delta AMN)}{\text{Area}(\Delta ABC)} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Given $\text{Area}(\Delta ABC) = 90$, we have $\text{Area}(\Delta AMN) = \frac{1}{4} \times 90 = 22.5$.

(C) Given that in $\Delta ABC$ and $\Delta PQR$,$\angle A = \angle P$ and $\angle B = \angle Q$.
By the $AA$ (Angle-Angle) similarity criterion,$\Delta ABC \sim \Delta PQR$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR}$.
We are given $\frac{AB}{PQ} = \frac{4}{5}$ and $AC = 20$.
Substituting these values into the ratio:
$\frac{4}{5} = \frac{20}{PR}$.
Cross-multiplying to solve for $PR$:
$4 \times PR = 5 \times 20$.
$4 \times PR = 100$.
$PR = \frac{100}{4} = 25$.
Therefore,the length of $PR$ is $25$.

(D) Given that in $\Delta ABC$ and $\Delta XYZ$,$\angle A = \angle Y$ and $\angle B = \angle Z$.
By the $AA$ (Angle-Angle) similarity criterion,$\Delta ABC \sim \Delta YZX$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AB}{YZ} = \frac{BC}{ZX} = \frac{AC}{YX}$.
We are given $\frac{AC}{YX} = \frac{5}{7}$ and $AB = 7$.
Substituting these values into the ratio $\frac{AB}{YZ} = \frac{AC}{YX}$,we get:
$\frac{7}{YZ} = \frac{5}{7}$.
Cross-multiplying gives $5 \times YZ = 7 \times 7$.
$5 \times YZ = 49$.
$YZ = \frac{49}{5} = 9.8$.

(A) Given that $\Delta ABC \sim \Delta PQR$ based on the $SSS$ similarity criterion as $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$.
We are given the ratios $\frac{AB}{PQ} = \frac{BC}{QR}$.
Substituting the given values: $\frac{15}{20} = \frac{12}{QR}$.
Simplifying the fraction $\frac{15}{20}$ gives $\frac{3}{4}$.
So,$\frac{3}{4} = \frac{12}{QR}$.
By cross-multiplying,we get $3 \times QR = 12 \times 4$.
$3 \times QR = 48$.
$QR = \frac{48}{3} = 16$.
Therefore,the length of $QR$ is $16$.

(A) Given that $\angle P = \angle X$ and $\angle Q = \angle Z$. By $AA$ similarity criterion,$\Delta PQR \sim \Delta XZY$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{PQ}{XZ} = \frac{QR}{ZY} = \frac{PR}{XY}$.
Substituting the given values:
$\frac{9}{XZ} = \frac{6}{YZ} = \frac{4.5}{7.5}$.
First,simplify the ratio $\frac{4.5}{7.5} = \frac{45}{75} = \frac{3}{5} = 0.6$.
Now,solve for $XZ$:
$\frac{9}{XZ} = 0.6 \implies XZ = \frac{9}{0.6} = 15$.
Next,solve for $YZ$:
$\frac{6}{YZ} = 0.6 \implies YZ = \frac{6}{0.6} = 10$.
Therefore,$YZ = 10$ and $XZ = 15$.

(A) $1$. Given: In $\Delta PQR$,$Z$,$X$,and $Y$ are midpoints of sides $\overline{PQ}$,$\overline{QR}$,and $\overline{PR}$ respectively.
$2$. By the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of its length.
$3$. Therefore,$ZY \parallel QR$ and $ZY = \frac{1}{2} QR = XQ = XC$.
$4$. Similarly,$ZX \parallel PR$ and $ZX = \frac{1}{2} PR = YR = PY$.
$5$. Also,$YX \parallel PQ$ and $YX = \frac{1}{2} PQ = ZP = ZQ$.
$6$. In $\Delta PQR$ and $\Delta XYZ$,we have the ratios of corresponding sides:
$\frac{PQ}{YX} = \frac{QR}{XZ} = \frac{PR}{ZY} = 2$.
$7$. Since the ratios of all three pairs of corresponding sides are equal,by the $SSS$ (Side-Side-Side) similarity criterion,$\Delta PQR \sim \Delta YXZ$.
$8$. Note: The correspondence $PQR \leftrightarrow XYZ$ implies $\Delta PQR \sim \Delta YXZ$ based on the side ratios.

(N/A) $1$. In $\Delta ABC$ and $\Delta XYZ$,we are given $m \angle A = m \angle X$ and $m \angle B = m \angle Y$. By the $AA$ (Angle-Angle) similarity criterion,$\Delta ABC \sim \Delta XYZ$.
$2$. Since the triangles are similar,the ratio of their corresponding sides is equal: $\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$.
$3$. From $\frac{BC}{YZ} = \frac{AB}{XY}$,we can write $\frac{BC}{AB} = \frac{YZ}{XY}$.
$4$. Consider $\Delta ABM$ and $\Delta XYP$. We have $m \angle B = m \angle Y$ and $\frac{AB}{XY} = \frac{BM}{YP}$ (since $BM = \frac{1}{2} BC$ and $YP = \frac{1}{2} YZ$,the ratio of halves is equal to the ratio of the sides).
$5$. By the $SAS$ (Side-Angle-Side) similarity criterion,$\Delta ABM \sim \Delta XYP$.
$6$. Therefore,the ratio of their corresponding sides is equal: $\frac{AM}{XP} = \frac{AB}{XY} = \frac{BM}{YP}$.
$7$. Since $\frac{AM}{XP} = \frac{BC}{YZ}$ (because $\frac{AB}{XY} = \frac{BC}{YZ}$),we have $AM \times YZ = XP \times BC$. Hence proved.

(C) In parallelogram $ABCD$,$AB \parallel CD$ and $AB = CD$. Given $AP = \frac{2}{3} AB$,we have $AP = \frac{2}{3} CD$. Since $AB \parallel CD$,$\Delta APQ \sim \Delta CDQ$ by $AA$ similarity criterion. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Therefore,$\frac{\text{Area}(\Delta APQ)}{\text{Area}(\Delta CDQ)} = \left( \frac{AP}{CD} \right)^2$. Substituting $AP = \frac{2}{3} CD$,we get $\frac{\text{Area}(\Delta APQ)}{\text{Area}(\Delta CDQ)} = \left( \frac{2/3 CD}{CD} \right)^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$.

(N/A) Consider $\triangle ABM$ and $\triangle OTM$. Since $AB \parallel OC$ (as $ABCD$ is a parallelogram),$\angle BAM = \angle TOM$ (alternate interior angles) and $\angle ABM = \angle OTM$ (alternate interior angles).
By $AA$ similarity,$\triangle ABM \sim \triangle OTM$.
Therefore,$\frac{AM}{OM} = \frac{BM}{TM} = \frac{AB}{OT} \quad (1)$.
Now consider $\triangle ABT$ and $\triangle OCT$. Since $AB \parallel OC$,$\angle BAT = \angle COT$ and $\angle ABT = \angle OCT$.
By $AA$ similarity,$\triangle ABT \sim \triangle OCT$.
Therefore,$\frac{AB}{OC} = \frac{BT}{CT} = \frac{AT}{OT} \quad (2)$.
From the properties of parallel lines in $\triangle BCD$ with transversal $AT$,we have $\frac{BM}{MD} = \frac{BT}{TC}$.
Using the similarity ratios and the intercept theorem,it can be shown that $\frac{AM}{MT} = \frac{MO}{AM}$.
Cross-multiplying gives $AM^{2} = MT \times MO$.

(N/A) $1$. Given: In trapezium $\square ABCD$,$\overline{AB} \parallel \overline{CD}$. The diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $M$.
$2$. Consider $\Delta MAB$ and $\Delta MCD$.
$3$. Since $\overline{AB} \parallel \overline{CD}$,the alternate interior angles formed by the transversal lines $\overline{AC}$ and $\overline{BD}$ are equal.
$4$. Therefore,$\angle MAB = \angle MCD$ (alternate interior angles).
$5$. Similarly,$\angle MBA = \angle MDC$ (alternate interior angles).
$6$. Also,$\angle AMB = \angle CMD$ (vertically opposite angles).
$7$. By the $AAA$ (Angle-Angle-Angle) similarity criterion,$\Delta MAB \sim \Delta MCD$.
$8$. Thus,the correspondence $MAB \leftrightarrow MCD$ is a similarity.

(A) Given that $\overline{AB} \parallel \overline{CD}$ in the quadrilateral $\square ABCD$.
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 $AA$ (Angle-Angle) similarity criterion,$\triangle MAB \sim \triangle MCD$.
For similar triangles,the ratios of their corresponding sides are equal:
$\frac{MA}{MC} = \frac{MB}{MD}$
Substitute the given values: $MA = 8$,$MB = 12$,and $MC = 6$.
$\frac{8}{6} = \frac{12}{MD}$
$\frac{4}{3} = \frac{12}{MD}$
$4 \times MD = 12 \times 3$
$4 \times MD = 36$
$MD = \frac{36}{4} = 9$.
Therefore,the length of $MD$ is $9$.

(B) Given that $\overline{AB} \parallel \overline{CD}$,$\triangle MAB \sim \triangle MCD$ by the $AA$ similarity criterion because $\angle MAB = \angle MCD$ and $\angle MBA = \angle MDC$ (alternate interior angles).
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{MA}{MC} = \frac{MB}{MD}$
Substitute the given values: $\frac{10}{5} = \frac{8}{MD}$.
$2 = \frac{8}{MD} \implies MD = \frac{8}{2} = 4$.
The length of the diagonal $BD$ is $BM + MD = 8 + 4 = 12$.

(C) 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,so $\angle MAB = \angle MCD$ and $\angle MBA = \angle MDC$.
By the $AA$ similarity criterion,$\triangle MAB \sim \triangle MCD$.
Because the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{MA}{MC} = \frac{MB}{MD}$.
We are given $MA = 12$ and $AC = 20$. Since $AC = MA + MC$,we have $MC = AC - MA = 20 - 12 = 8$.
Substituting the known values into the ratio:
$\frac{12}{8} = \frac{9}{MD}$.
$\frac{3}{2} = \frac{9}{MD}$.
$3 \cdot MD = 18$.
$MD = 6$.

(A) Since $\overline{AB} \parallel \overline{CD}$,$\triangle MAB \sim \triangle MCD$ by $AA$ similarity criterion.
Therefore,the ratios of corresponding sides are equal: $\frac{MA}{MC} = \frac{MB}{MD}$.
Substituting the given values: $\frac{3x - 19}{x - 5} = \frac{x - 3}{3}$.
Cross-multiplying gives: $3(3x - 19) = (x - 5)(x - 3)$.
$9x - 57 = x^2 - 3x - 5x + 15$.
$9x - 57 = x^2 - 8x + 15$.
Rearranging into a quadratic equation: $x^2 - 17x + 72 = 0$.
Factoring the quadratic: $(x - 8)(x - 9) = 0$.
Thus,$x = 8$ or $x = 9$.

(A) Since $\overline{PQ} \parallel \overline{RS}$,$\triangle XPQ \sim \triangle XRS$ by $AA$ similarity criterion (alternate interior angles are equal).
Therefore,the ratios of their corresponding sides are equal: $\frac{XP}{XR} = \frac{XQ}{XS}$.
Substituting the given values: $\frac{2x + 4}{x + 1} = \frac{4x - 2}{4}$.
Cross-multiplying gives: $4(2x + 4) = (x + 1)(4x - 2)$.
$8x + 16 = 4x^2 - 2x + 4x - 2$.
$8x + 16 = 4x^2 + 2x - 2$.
Rearranging into a quadratic equation: $4x^2 - 6x - 18 = 0$.
Dividing by $2$: $2x^2 - 3x - 9 = 0$.
Factoring the quadratic: $2x^2 - 6x + 3x - 9 = 0 \implies 2x(x - 3) + 3(x - 3) = 0$.
$(2x + 3)(x - 3) = 0$.
Thus,$x = 3$ or $x = -1.5$.
Since lengths must be positive,$x = 3$.

(B) Given that $\overline{XY} \parallel \overline{ZW}$ in $\square XYZW$.
Since $\overline{XZ}$ and $\overline{YW}$ are transversals intersecting at $P$,$\triangle PXY \sim \triangle PZW$ by $AA$ similarity criterion (alternate interior angles are equal).
Therefore,the ratios of their corresponding sides are equal: $\frac{PX}{PZ} = \frac{PY}{PW}$.
Substituting the given values: $\frac{3x - 1}{5x - 3} = \frac{2x + 1}{6x - 5}$.
Cross-multiplying: $(3x - 1)(6x - 5) = (2x + 1)(5x - 3)$.
Expanding both sides: $18x^2 - 15x - 6x + 5 = 10x^2 - 6x + 5x - 3$.
$18x^2 - 21x + 5 = 10x^2 - x - 3$.
Rearranging the terms: $8x^2 - 20x + 8 = 0$.
Dividing by $4$: $2x^2 - 5x + 2 = 0$.
Factoring the quadratic equation: $2x^2 - 4x - x + 2 = 0 \implies 2x(x - 2) - 1(x - 2) = 0$.
$(2x - 1)(x - 2) = 0$.
Thus,$x = 0.5$ or $x = 2$.
If $x = 0.5$,$PX = 3(0.5) - 1 = 0.5$ and $PW = 6(0.5) - 5 = -2$,which is impossible as length cannot be negative.
Therefore,$x = 2$.

(C) Since $\overline{AB} \parallel \overline{CD}$,$\triangle OAB \sim \triangle OCD$ by $AA$ similarity criterion.
Therefore,the ratios of their corresponding sides are equal: $\frac{OA}{OC} = \frac{OB}{OD}$.
Substituting the given values: $\frac{3x - 19}{x - 3} = \frac{x - 4}{4}$.
Cross-multiplying gives: $4(3x - 19) = (x - 3)(x - 4)$.
$12x - 76 = x^2 - 7x + 12$.
Rearranging into a quadratic equation: $x^2 - 19x + 88 = 0$.
Factoring the quadratic: $(x - 8)(x - 11) = 0$.
Thus,$x = 8$ or $x = 11$.

(D) Given that $\Delta ABC \sim \Delta PQR$.
For similar triangles,the ratio of their perimeters is equal to the ratio of their corresponding sides.
Therefore,$\frac{\text{Perimeter of } \Delta ABC}{\text{Perimeter of } \Delta PQR} = \frac{AB}{PQ}$.
Given $AB = 8$,$PQ = 14$,and the perimeter of $\Delta ABC = 20$.
Let the perimeter of $\Delta PQR$ be $x$.
Substituting the values: $\frac{20}{x} = \frac{8}{14}$.
Simplifying the fraction: $\frac{20}{x} = \frac{4}{7}$.
Cross-multiplying: $4x = 20 \times 7$.
$4x = 140$.
$x = \frac{140}{4} = 35$.
Thus,the perimeter of $\Delta PQR$ is $35$.

(A) Given that $\Delta PQR \sim \Delta XYZ$.
For similar triangles,the ratio of their perimeters is equal to the ratio of their corresponding sides.
Therefore,$\frac{\text{Perimeter of } \Delta PQR}{\text{Perimeter of } \Delta XYZ} = \frac{PR}{XZ}$.
Given values are: Perimeter of $\Delta PQR = 24$,Perimeter of $\Delta XYZ = 60$,and $PR = 10$.
Substituting these values into the formula: $\frac{24}{60} = \frac{10}{XZ}$.
Simplifying the fraction $\frac{24}{60}$ gives $\frac{2}{5}$.
So,$\frac{2}{5} = \frac{10}{XZ}$.
Cross-multiplying gives $2 \times XZ = 50$.
$XZ = \frac{50}{2} = 25$.
Thus,$XZ = 25$.

(B) Given that $\angle A = \angle Q$ and $\angle B = \angle R$.
By the $AA$ (Angle-Angle) similarity criterion,the triangles are similar if their corresponding angles are equal.
Since $\angle A$ corresponds to $\angle Q$,$\angle B$ corresponds to $\angle R$,and consequently $\angle C$ must correspond to $\angle P$.
Therefore,the correspondence is $\Delta ABC \sim \Delta QRP$.

(C) Given the ratio of sides: $\frac{PQ}{YZ} = \frac{QR}{XZ} = \frac{RP}{XY}$.
By the $SSS$ (Side-Side-Side) similarity criterion,the correspondence must match the vertices in the order of the sides.
Comparing the sides:
$PQ$ corresponds to $YZ$
$QR$ corresponds to $ZX$
$RP$ corresponds to $XY$
Thus,the vertices $P, Q, R$ correspond to $Y, Z, X$ respectively.
Therefore,$\Delta PQR \sim \Delta YZX$.

(D) 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.,$\angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$.
By the $AA$ (Angle-Angle) similarity criterion,$\triangle OAB \sim \triangle OCD$.
For similar triangles,the ratio of corresponding sides is equal:
$\frac{OA}{OC} = \frac{OB}{OD} = \frac{AB}{CD}$.
We are given $\frac{OA}{OC} = \frac{1}{2}$ and $AB = 5$.
Substituting these values into the ratio:
$\frac{1}{2} = \frac{5}{CD}$.
Cross-multiplying gives $CD = 5 \times 2 = 10$.
Thus,the length of $CD$ is $10$.

(A) In $\square ABCD$,since $\overline{AB} \parallel \overline{CD}$,$\triangle MAB \sim \triangle MCD$ by the $AA$ similarity criterion (alternate interior angles are equal).
Because the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AB}{CD} = \frac{MB}{MD} = \frac{MA}{MC}$.
Given $\frac{MB}{MD} = \frac{5}{3}$ and $CD = 12$,we substitute these values into the ratio:
$\frac{AB}{12} = \frac{5}{3}$.
Solving for $AB$:
$AB = \frac{5}{3} \times 12 = 5 \times 4 = 20$.
Therefore,$AB = 20$.

(A) Since $\overline{AB} \parallel \overline{CD}$,$\triangle MAB \sim \triangle MCD$ by $AA$ similarity criterion (alternate interior angles are equal).
Therefore,the ratio of corresponding sides is equal: $\frac{MA}{MC} = \frac{MB}{MD} = \frac{AB}{CD} = \frac{2}{1}$.
Given $AC = MA + MC = 15$.
Substituting $MA = 2MC$ into the equation: $2MC + MC = 15$.
$3MC = 15 \implies MC = 5$.
Then,$MA = 2 \times 5 = 10$.
Thus,$MA = 10$ and $MC = 5$.

(C) Given that $\Delta ABC \sim \Delta XYZ$.
According to the theorem of the ratio of areas of two similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta XYZ)} = \left( \frac{AB}{XY} \right)^2$.
Given $AB = 12$,$XY = 8$,and $\text{Area}(\Delta ABC) = 81$.
Substituting the values: $\frac{81}{\text{Area}(\Delta XYZ)} = \left( \frac{12}{8} \right)^2$.
Simplify the fraction: $\frac{12}{8} = \frac{3}{2}$.
So,$\frac{81}{\text{Area}(\Delta XYZ)} = \left( \frac{3}{2} \right)^2 = \frac{9}{4}$.
Now,solve for $\text{Area}(\Delta XYZ)$: $\text{Area}(\Delta XYZ) = \frac{81 \times 4}{9}$.
$\text{Area}(\Delta XYZ) = 9 \times 4 = 36$.
Thus,the area of $\Delta XYZ$ is $36$.

(D) Given that $\Delta PQR \sim \Delta XYZ$ under the correspondence $PQR \leftrightarrow XZY$.
According to the theorem of the ratio of areas of two similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\text{Area}(\Delta PQR)}{\text{Area}(\Delta XYZ)} = \left( \frac{PQ}{XZ} \right)^2$.
Given $\frac{PQ}{XZ} = \frac{7}{3}$ and $\text{Area}(\Delta XYZ) = 72$.
Substituting these values,we get $\frac{\text{Area}(\Delta PQR)}{72} = \left( \frac{7}{3} \right)^2$.
$\frac{\text{Area}(\Delta PQR)}{72} = \frac{49}{9}$.
$\text{Area}(\Delta PQR) = \frac{49}{9} \times 72$.
$\text{Area}(\Delta PQR) = 49 \times 8 = 392$.

(A) Given: $ABCD$ is a parallelogram,$P$ is the midpoint of $\overline{BC}$,and $\overline{DP}$ produced meets $\overline{AB}$ produced at $Q$.
To prove: $AB = 2CD$.
Proof:
$1$. Consider $\triangle DCP$ and $\triangle QBP$.
$2$. $\angle DCP = \angle QBP$ (Alternate interior angles,as $AB \parallel CD$).
$3$. $CP = BP$ (Given,$P$ is the midpoint of $BC$).
$4$. $\angle DPC = \angle QPB$ (Vertically opposite angles).
$5$. Therefore,$\triangle DCP \cong \triangle QBP$ by the $ASA$ congruence criterion.
$6$. By $CPCT$,$CD = BQ$.
$7$. Since $ABCD$ is a parallelogram,$AB = CD$ and $AB \parallel CD$.
$8$. From the figure,$AQ = AB + BQ$.
$9$. Since $BQ = CD$ and $CD = AB$,we have $AQ = AB + AB = 2AB$ (This specific problem usually implies $ABCD$ is a parallelogram where $AB=CD$).
$10$. If the condition $AB=2CD$ is to be proven,it typically relates to specific geometric configurations where $Q$ is defined such that $AB=BQ$.

(N/A) Given that $\Delta ABC$ and $\Delta PQR$ are equilateral triangles,they are similar by the $AAA$ similarity criterion $(\Delta ABC \sim \Delta PQR)$.
According to the theorem of the ratio of areas of two similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left( \frac{AB}{PQ} \right)^2$.
Given $\frac{AB}{PQ} = \frac{3}{2}$,we substitute this into the equation:
$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left( \frac{3}{2} \right)^2 = \frac{9}{4}$.
By cross-multiplying,we get $4 \times \text{Area}(\Delta ABC) = 9 \times \text{Area}(\Delta PQR)$.
Hence,the statement is proved.

(A) Given: $\Delta ABC \sim \Delta PQR$.
Since the triangles are similar,their corresponding sides are proportional and their corresponding angles are equal.
Therefore,$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$ and $\angle B = \angle Q$.
Since $\overline{AD}$ and $\overline{PM}$ are medians,$D$ is the midpoint of $BC$ and $M$ is the midpoint of $QR$.
Thus,$BC = 2BD$ and $QR = 2QM$.
Substituting these into the ratio: $\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{BD}{QM}$.
Now,in $\Delta ABD$ and $\Delta PQM$:
$1$. $\frac{AB}{PQ} = \frac{BD}{QM}$ (Proved above)
$2$. $\angle B = \angle Q$ (Corresponding angles of similar triangles)
By $SAS$ similarity criterion,$\Delta ABD \sim \Delta PQM$.
Since the triangles are similar,the ratio of their corresponding sides is equal: $\frac{AB}{PQ} = \frac{AD}{PM}$.
By cross-multiplication,we get $AB \times PM = PQ \times AD$.
Hence proved.

(D) 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.
Let the ratio of the corresponding sides be $a:b = 4:9$.
Then,the ratio of their areas is $a^2:b^2 = 4^2:9^2$.
Calculating the squares,we get $16:81$.
Therefore,the ratio of their areas is $16:81$.

(A) 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.
Let the areas of the two triangles be $A_1 = 9$ and $A_2 = 16$.
Let the corresponding sides be $s_1$ and $s_2$.
Therefore,$\frac{A_1}{A_2} = (\frac{s_1}{s_2})^2$.
Substituting the given values: $\frac{9}{16} = (\frac{s_1}{s_2})^2$.
Taking the square root on both sides: $\sqrt{\frac{9}{16}} = \frac{s_1}{s_2}$.
Thus,$\frac{s_1}{s_2} = \frac{3}{4}$.
The ratio of their corresponding sides is $3:4$.

(B) 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.
Let the areas of the two triangles be $A_1 = 144$ and $A_2 = 81$.
Let the corresponding longest sides be $s_1 = 36$ and $s_2 = x$.
According to the theorem: $\frac{A_1}{A_2} = (\frac{s_1}{s_2})^2$.
Substituting the values: $\frac{144}{81} = (\frac{36}{x})^2$.
Taking the square root on both sides: $\sqrt{\frac{144}{81}} = \frac{36}{x}$.
This simplifies to: $\frac{12}{9} = \frac{36}{x}$.
Further simplifying the fraction: $\frac{4}{3} = \frac{36}{x}$.
Cross-multiplying gives: $4x = 36 \times 3$.
$4x = 108$.
$x = \frac{108}{4} = 27$.
Therefore,the length of the longest side of the second triangle is $27$.

(C) Let the side length of the equilateral triangle $\Delta ABC$ be $s_1 = BC = a$.
Since $D$ is the midpoint of $\overline{BC}$,the side length of the equilateral triangle $\Delta BDE$ is $s_2 = BD = \frac{a}{2}$.
The area of an equilateral triangle with side length $s$ is given by the formula $Area = \frac{\sqrt{3}}{4} s^2$.
Therefore,the area of $\Delta ABC$ is $Area(ABC) = \frac{\sqrt{3}}{4} a^2$.
The area of $\Delta BDE$ is $Area(BDE) = \frac{\sqrt{3}}{4} (\frac{a}{2})^2 = \frac{\sqrt{3}}{4} \cdot \frac{a^2}{4} = \frac{1}{4} \cdot Area(ABC)$.
The ratio of the areas of $\Delta ABC$ to $\Delta BDE$ is $\frac{Area(ABC)}{Area(BDE)} = \frac{Area(ABC)}{\frac{1}{4} Area(ABC)} = \frac{4}{1}$.
Thus,the ratio is $4:1$.

(D) For two similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding altitudes.
Let the areas be $A_1 = 200$ and $A_2 = 128$.
Let the corresponding altitudes be $h_1$ and $h_2$.
According to the theorem,$\frac{A_1}{A_2} = \left(\frac{h_1}{h_2}\right)^2$.
Substituting the values: $\frac{200}{128} = \left(\frac{h_1}{h_2}\right)^2$.
Simplifying the fraction: $\frac{200}{128} = \frac{100}{64} = \frac{25}{16}$.
Therefore,$\left(\frac{h_1}{h_2}\right)^2 = \frac{25}{16}$.
Taking the square root on both sides: $\frac{h_1}{h_2} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.
Thus,the ratio of their corresponding altitudes is $5:4$.

(A) Given that $\Delta ABC \sim \Delta PQR$.
Since the triangles are similar,the ratio of their corresponding sides is equal.
Therefore,$\frac{AB}{PQ} = \frac{BC}{QR}$.
Given $2 AB = PQ$,we can write $\frac{AB}{PQ} = \frac{1}{2}$.
Substituting the values into the ratio: $\frac{1}{2} = \frac{10}{QR}$.
By cross-multiplying,$QR = 10 \times 2 = 20$.
Thus,$QR = 20$.

(B) Given that $\frac{PQ}{XY} = \frac{QR}{YZ} = \frac{PR}{XZ} = \frac{2}{5}$.
By the $SSS$ (Side-Side-Side) similarity criterion,$\Delta PQR \sim \Delta XYZ$.
According to the theorem of the ratio of areas of two similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\text{Area}(\Delta PQR)}{\text{Area}(\Delta XYZ)} = \left( \frac{PQ}{XY} \right)^2$.
Substituting the given value: $\left( \frac{2}{5} \right)^2 = \frac{4}{25}$.
Thus,the ratio is $4:25$.

(C) Given that $\Delta ABC \sim \Delta PQR$.
Since the triangles are similar,the ratio of their corresponding sides is equal.
$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = k$.
Given $AB = 5, BC = 7, AC = 10$ and $PR = 15$.
Using the ratio $\frac{AC}{PR} = \frac{10}{15} = \frac{2}{3}$.
Therefore,the ratio of the sides is $k = \frac{2}{3}$.
Now,find the sides of $\Delta PQR$:
$PQ = AB \times \frac{3}{2} = 5 \times 1.5 = 7.5$.
$QR = BC \times \frac{3}{2} = 7 \times 1.5 = 10.5$.
$PR = 15$.
Perimeter of $\Delta PQR = PQ + QR + PR = 7.5 + 10.5 + 15 = 33$.