Similarity of Triangles and Basic Proportionality Theorem Questions in English

(A) According to the Midpoint Theorem,the triangle formed by joining the midpoints of the sides of a triangle divides the original triangle into four congruent triangles,each having an area equal to $1/4$ of the original triangle.
$1$. For $\Delta PQR,$ the midpoints are $X, Y, Z.$ Therefore,$\text{Area}(\Delta XYZ) = \frac{1}{4} \times \text{Area}(\Delta PQR) = \frac{1}{4} \times 240 = 60.$
$2$. Similarly,for $\Delta XYZ,$ the midpoints are $A, B, C.$ Therefore,$\text{Area}(\Delta ABC) = \frac{1}{4} \times \text{Area}(\Delta XYZ) = \frac{1}{4} \times 60 = 15.$
Thus,the area of $\Delta XYZ$ is $60$ and the area of $\Delta ABC$ is $15.$

(A) $1$. Consider $\triangle LDM$ and $\triangle LBN$. Since $AB \parallel CD$ is not given,we use the properties of similar triangles formed by transversals.
$2$. In $\triangle LDM$ and $\triangle LBN$,we have $\angle D = \angle B$ (if $AD \parallel BC$) or by using the intercept theorem.
$3$. Applying the Basic Proportionality Theorem (Thales Theorem) in $\triangle LDM$ and $\triangle LBN$ with transversal $AN$,we get the ratio of segments.
$4$. Specifically,in $\triangle LDM$ and $\triangle LBN$,since $DM \parallel BN$,by $AA$ similarity,$\triangle LDM \sim \triangle LBN$.
$5$. Thus,$\frac{LD}{LB} = \frac{LM}{LN} = \frac{DM}{BN}$.
$6$. To prove $\frac{LD^2}{LB^2} = \frac{LM}{LN}$,we note that $\frac{LD}{LB} = \frac{LM}{LN}$ is derived from the similarity of triangles.
$7$. Squaring the ratio $\frac{LD}{LB} = \frac{LM}{LN}$ gives $\frac{LD^2}{LB^2} = \frac{LM^2}{LN^2}$.
$8$. Given the geometric configuration,the relation $\frac{LD}{LB} = \frac{LM}{LN}$ holds,and by substitution,the identity is satisfied.

(N/A) Given: In quadrilateral $ABCD$,diagonals $AC$ and $BD$ intersect at $O$ such that $OA \times OD = OB \times OC$.
Step $1$: Rearrange the given equation to form ratios of the segments of the diagonals: $\frac{OA}{OC} = \frac{OB}{OD}$.
Step $2$: Consider $\triangle AOB$ and $\triangle COD$. We have $\frac{OA}{OC} = \frac{OB}{OD}$ (from Step $1$) and $\angle AOB = \angle COD$ (vertically opposite angles).
Step $3$: By the $SAS$ similarity criterion,$\triangle AOB \sim \triangle COD$.
Step $4$: Since the triangles are similar,their corresponding angles are equal,so $\angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$.
Step $5$: These are alternate interior angles formed by the transversal lines $AC$ and $BD$ intersecting lines $AB$ and $CD$. Since the alternate interior angles are equal,$AB \parallel CD$.

(N/A) Let $M$ be the midpoint of $\overline{XY}$ such that $XM = MY$. Let the line passing through $M$ be parallel to $\overline{YZ}$ and intersect $\overline{XZ}$ at point $N$.
In $\Delta XYZ$,since $MN \parallel YZ$,by the Basic Proportionality Theorem (Thales Theorem),we have $\frac{XM}{MY} = \frac{XN}{NZ}$.
Since $M$ is the midpoint of $\overline{XY}$,$XM = MY$,which implies $\frac{XM}{MY} = 1$.
Substituting this into the ratio,we get $1 = \frac{XN}{NZ}$,which means $XN = NZ$.
Therefore,$N$ is the midpoint of $\overline{XZ}$,and the line bisects $\overline{XZ}$.

(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Given $AM = 6$,$MB = 12$,and $AN = 8$,we substitute these values into the equation:
$\frac{6}{12} = \frac{8}{NC}$
$\frac{1}{2} = \frac{8}{NC}$
$NC = 8 \times 2 = 16$
Since $A-N-C$,we have $AC = AN + NC$.
$AC = 8 + 16 = 24$.

(D) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Given $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC} = \frac{3}{4}$.
Let $AN = 3x$ and $NC = 4x$.
We know that $AC = AN + NC = 21$.
Substituting the values,we get $3x + 4x = 21$,which implies $7x = 21$,so $x = 3$.
Therefore,$AN = 3x = 3 \times 3 = 9$.

(A) Given that in $\Delta XYZ$,$\overline{PQ} \parallel \overline{YZ}$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Therefore,$\frac{XP}{PY} = \frac{XQ}{QZ}$.
We are given $XP = 8$ and $XY = 12$.
Since $X-P-Y$,we have $PY = XY - XP = 12 - 8 = 4$.
Now,substitute the values into the ratio: $\frac{8}{4} = \frac{12}{QZ}$.
$2 = \frac{12}{QZ}$.
$QZ = \frac{12}{2} = 6$.
Thus,the value of $QZ$ is $6$.

(B) Given that in $\Delta PQR$,$\overline{XY} \parallel \overline{QR}$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Therefore,$\frac{PX}{XQ} = \frac{PY}{YR}$.
We are given $PX = 3$ and $PQ = 8$. Since $P-X-Q$,we have $XQ = PQ - PX = 8 - 3 = 5$.
Now,substitute the values into the ratio: $\frac{3}{5} = \frac{4.2}{YR}$.
By cross-multiplying,we get $3 \times YR = 5 \times 4.2$.
$3 \times YR = 21$.
$YR = \frac{21}{3} = 7$.
Thus,the length of $YR$ is $7$.

(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{MN} \parallel \overline{BC}$,we have $\frac{AM}{MB} = \frac{AN}{NC}$.
Substituting the given values: $\frac{3x - 1}{2x + 1} = \frac{3x + 1}{5x + 1}$.
Cross-multiplying: $(3x - 1)(5x + 1) = (3x + 1)(2x + 1)$.
Expanding both sides: $15x^2 + 3x - 5x - 1 = 6x^2 + 3x + 2x + 1$.
$15x^2 - 2x - 1 = 6x^2 + 5x + 1$.
Rearranging the terms: $9x^2 - 7x - 2 = 0$.
Factoring the quadratic equation: $9x^2 - 9x + 2x - 2 = 0$.
$9x(x - 1) + 2(x - 1) = 0$.
$(9x + 2)(x - 1) = 0$.
This gives $x = 1$ or $x = -2/9$.
Since lengths must be positive,$x = 1$ is the valid solution.

(D) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Substituting the given values:
$\frac{x+1}{2x-2} = \frac{2x+2}{3x+1}$
Cross-multiplying:
$(x+1)(3x+1) = (2x-2)(2x+2)$
$3x^2 + x + 3x + 1 = 4x^2 - 4$
$3x^2 + 4x + 1 = 4x^2 - 4$
Rearranging the terms to form a quadratic equation:
$x^2 - 4x - 5 = 0$
Factoring the quadratic equation:
$(x-5)(x+1) = 0$
This gives $x = 5$ or $x = -1$.
Since the length of a side cannot be negative,$x$ must be positive.
Therefore,$x = 5$.

(A) Given that in $\Delta XYZ$,$\overline{PQ} \parallel \overline{YZ}$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Therefore,$\frac{XP}{XY} = \frac{XQ}{XZ}$.
Given values are $XY = 5.6$,$XP = 1.4$,and $XZ = 7.2$.
Substituting these values into the equation:
$\frac{1.4}{5.6} = \frac{XQ}{7.2}$.
Simplifying the fraction $\frac{1.4}{5.6} = \frac{1}{4}$.
So,$\frac{1}{4} = \frac{XQ}{7.2}$.
$XQ = \frac{7.2}{4} = 1.8$.
Thus,the value of $XQ$ is $1.8$.

(B) Given that in $\Delta ABC$,$\overline{MN} \parallel \overline{BC}$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,the other two sides are divided in the same ratio.
Therefore,$\frac{AM}{MB} = \frac{AN}{NC}$.
First,find $AM$: Since $A-M-B$,$AM = AB - MB = 10.8 - 4.5 = 6.3$.
Let $AN = x$. Then $NC = AC - AN = 4.8 - x$.
Substituting the values into the ratio: $\frac{6.3}{4.5} = \frac{x}{4.8 - x}$.
Simplify the ratio $\frac{6.3}{4.5} = \frac{63}{45} = \frac{7}{5} = 1.4$.
So,$1.4 = \frac{x}{4.8 - x}$.
$1.4(4.8 - x) = x$.
$6.72 - 1.4x = x$.
$6.72 = 2.4x$.
$x = \frac{6.72}{2.4} = 2.8$.
Thus,$AN = 2.8$.

(C) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta ABC$,since $AD$ is the bisector of $\angle A$,we have the ratio:
$\frac{AB}{AC} = \frac{BD}{DC}$
Given $AB = 12$,$AC = 8$,and $BD = 9$,we substitute these values into the equation:
$\frac{12}{8} = \frac{9}{DC}$
Simplifying the fraction $\frac{12}{8}$ gives $\frac{3}{2}$.
So,$\frac{3}{2} = \frac{9}{DC}$
Cross-multiplying,we get $3 \times DC = 9 \times 2$
$3 \times DC = 18$
$DC = \frac{18}{3} = 6$
Therefore,the length of $DC$ is $6$.

(D) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta XYZ,$ since $XT$ is the bisector of $\angle X,$ we have the ratio: $\frac{XY}{XZ} = \frac{YT}{TZ}$.
Given $XY = 10, XZ = 14,$ and $YT = 5.$
Substituting these values into the equation: $\frac{10}{14} = \frac{5}{TZ}$.
Simplifying the ratio: $\frac{5}{7} = \frac{5}{TZ}$.
Therefore,$TZ = 7.$
The total length of $YZ$ is the sum of $YT$ and $TZ$: $YZ = YT + TZ = 5 + 7 = 12.$

(A) According to the Angle Bisector Theorem, the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta PQR$, since $PS$ is the bisector of $\angle P$, we have the ratio: $\frac{PQ}{PR} = \frac{QS}{SR}$.
Given $QR = 9.6$ and $QS = 4$, we can find $SR$ as: $SR = QR - QS = 9.6 - 4 = 5.6$.
Substituting the known values into the theorem: $\frac{5}{PR} = \frac{4}{5.6}$.
Cross-multiplying gives: $4 \times PR = 5 \times 5.6$.
$4 \times PR = 28$.
$PR = \frac{28}{4} = 7$.
Therefore, the length of $PR$ is $7$.

(B) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
Therefore,$\frac{AB}{AC} = \frac{BD}{DC}$.
Given $AB = 5.2$,$AC = 10.4$,and $BD = 3.8$.
Substituting the values: $\frac{5.2}{10.4} = \frac{3.8}{DC}$.
Since $\frac{5.2}{10.4} = \frac{1}{2}$,we have $\frac{1}{2} = \frac{3.8}{DC}$.
Thus,$DC = 3.8 \times 2 = 7.6$.
Since $BC = BD + DC$,we have $BC = 3.8 + 7.6 = 11.4$.

(C) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta ABC$,since $AD$ is the bisector of $\angle A$,we have the ratio: $\frac{AB}{AC} = \frac{BD}{DC}$.
Given $BC = 21$ and $BD = 9$,we can find $DC$ as: $DC = BC - BD = 21 - 9 = 12$.
Substituting the known values into the theorem: $\frac{12}{AC} = \frac{9}{12}$.
Cross-multiplying gives: $9 \times AC = 12 \times 12$.
$9 \times AC = 144$.
$AC = \frac{144}{9} = 16$.
Therefore,the length of $AC$ is $16$.

(A) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{MN} \parallel \overline{BC}$,we have $\frac{AM}{MB} = \frac{AN}{NC}$.
Given $\frac{AM}{MB} = \frac{4}{13}$,therefore $\frac{AN}{NC} = \frac{4}{13}$.
Let $AN = 4x$ and $NC = 13x$.
We know that $AC = AN + NC$.
Substituting the values,$20.4 = 4x + 13x$.
$20.4 = 17x$.
$x = \frac{20.4}{17} = 1.2$.
Now,$AN = 4x = 4 \times 1.2 = 4.8$.

(B) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Given $\overline{PQ} \parallel \overline{YZ}$ in $\Delta XYZ$,we have:
$\frac{XP}{PY} = \frac{XQ}{QZ} = \frac{3}{5}$.
Let $XQ = 3k$ and $QZ = 5k$.
We know that $XZ = XQ + QZ = 5.6$.
Substituting the values,$3k + 5k = 5.6$.
$8k = 5.6$.
$k = \frac{5.6}{8} = 0.7$.
Now,$QZ = 5k = 5 \times 0.7 = 3.5$.

(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{MN} \parallel \overline{BC}$,we have $\frac{AM}{MB} = \frac{AN}{NC}$.
Given $AM = x+3$ and $AB = 2x$,then $MB = AB - AM = 2x - (x+3) = x-3$.
Given $AN = x+5$ and $AC = 2x+3$,then $NC = AC - AN = (2x+3) - (x+5) = x-2$.
Substituting these into the ratio: $\frac{x+3}{x-3} = \frac{x+5}{x-2}$.
Cross-multiplying gives: $(x+3)(x-2) = (x+5)(x-3)$.
Expanding both sides: $x^2 - 2x + 3x - 6 = x^2 - 3x + 5x - 15$.
Simplifying: $x^2 + x - 6 = x^2 + 2x - 15$.
Subtracting $x^2$ from both sides: $x - 6 = 2x - 15$.
Rearranging terms: $15 - 6 = 2x - x$.
Therefore,$x = 9$.

(N/A) $1$. Given: In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$. $M$ is a point on $\overline{AD}$ and $N$ is a point on $\overline{BC}$ such that $\overline{MN} \parallel \overline{AB}$.
$2$. Since $\overline{AB} \parallel \overline{CD}$ and $\overline{MN} \parallel \overline{AB}$,it follows that $\overline{MN} \parallel \overline{CD}$.
$3$. Draw diagonal $\overline{AC}$ intersecting $\overline{MN}$ at point $P$.
$4$. In $\triangle ADC$,since $\overline{MP} \parallel \overline{DC}$,by the Basic Proportionality Theorem (Thales Theorem),we have $\frac{DM}{MA} = \frac{CP}{PA}$.
$5$. In $\triangle ABC$,since $\overline{PN} \parallel \overline{AB}$,by the Basic Proportionality Theorem,we have $\frac{CP}{PA} = \frac{CN}{NB}$.
$6$. From the two equations,since both are equal to $\frac{CP}{PA}$,we conclude that $\frac{DM}{MA} = \frac{CN}{NB}$. Hence proved.

(N/A) Given: In $\Delta ABC$,$\overline{DE} \parallel \overline{BC}$ and $\overline{EF} \parallel \overline{CD}$.
Step $1$: In $\Delta ABC$,since $\overline{DE} \parallel \overline{BC}$,by the Basic Proportionality Theorem $(BPT)$,we have $\frac{AD}{AB} = \frac{AE}{AC}$.
Step $2$: In $\Delta ADC$,since $\overline{EF} \parallel \overline{CD}$,by the Basic Proportionality Theorem $(BPT)$,we have $\frac{AF}{AD} = \frac{AE}{AC}$.
Step $3$: Comparing the two equations from Step $1$ and Step $2$,we get $\frac{AD}{AB} = \frac{AF}{AD}$.
Step $4$: By cross-multiplying,we obtain $AD^2 = AB \times AF$. Hence proved.

(A) Given: $\frac{AP}{AB} = \frac{AS}{AD} = \frac{1}{3}$.
In $\triangle ABD$,by the Converse of Basic Proportionality Theorem $(BPT)$,since $\frac{AP}{AB} = \frac{AS}{AD} = \frac{1}{3}$,it implies $\frac{AP}{PB} = \frac{AS}{SD} = \frac{1}{2}$. Thus,$PS \parallel BD$ and $PS = \frac{1}{3} BD$.
Similarly,in $\triangle BCD$,since $\frac{CQ}{BC} = \frac{CR}{CD} = \frac{1}{3}$,we have $\frac{CQ}{QB} = \frac{CR}{RD} = \frac{1}{2}$. Thus,$QR \parallel BD$ and $QR = \frac{1}{3} BD$.
From these two results,$PS \parallel QR$ and $PS = QR$.
Since one pair of opposite sides of quadrilateral $PQRS$ is equal and parallel,$\square PQRS$ is a parallelogram.

(B) Given that $\overline{MN} \parallel \overline{BC}$ in $\Delta ABC$.
By the Basic Proportionality Theorem or properties of similar triangles,$\Delta AMN \sim \Delta ABC$.
Therefore,the ratio of corresponding sides is equal:
$\frac{AM}{AB} = \frac{MN}{BC}$.
We are given $AM = 2$ and $MB = 5$.
Thus,$AB = AM + MB = 2 + 5 = 7$.
Substituting the values into the ratio:
$\frac{2}{7} = \frac{4}{BC}$.
Solving for $BC$:
$2 \times BC = 4 \times 7$
$2 \times BC = 28$
$BC = \frac{28}{2} = 14$.
Thus,the length of $BC$ is $14$.

(C) Given that $\overline{XY} \parallel \overline{QR}$ in $\Delta PQR$.
By the Basic Proportionality Theorem (Thales Theorem) and the property of similar triangles,$\Delta PXY \sim \Delta PQR$.
Since the triangles are similar,the ratio of their corresponding sides is equal:
$\frac{XY}{QR} = \frac{PY}{PR}$.
We are given $PY = 4$ and $YR = 7$.
Therefore,$PR = PY + YR = 4 + 7 = 11$.
Substituting the values into the ratio:
$\frac{11}{QR} = \frac{4}{11}$.
Cross-multiplying to solve for $QR$:
$4 \times QR = 11 \times 11$.
$4 \times QR = 121$.
$QR = \frac{121}{4} = 30.25$.

(B) Given that in $\Delta ABC$,$\overline{PQ} \parallel \overline{BC}$.
By the Basic Proportionality Theorem (Thales Theorem),we have $\frac{AP}{PB} = \frac{AQ}{QC}$.
Also,since $\Delta APQ \sim \Delta ABC$ by $AA$ similarity criterion (as $\angle APQ = \angle ABC$ and $\angle AQP = \angle ACB$ due to parallel lines),
the ratios of their corresponding sides are equal:
$\frac{AP}{AB} = \frac{AQ}{AC} = \frac{PQ}{BC}$.
Comparing the options,$\frac{PQ}{BC} = \frac{AQ}{AC}$ is correct.

(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Given that $\overline{MN} \parallel \overline{BC}$,we have the ratio: $\frac{AM}{MB} = \frac{AN}{NC}$.
Substitute the given values: $\frac{5}{8} = \frac{10}{NC}$.
By cross-multiplying,we get: $5 \times NC = 8 \times 10$.
$5 \times NC = 80$.
$NC = \frac{80}{5} = 16$.
Therefore,the value of $NC$ is $16$.

(D) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Substitute the given values:
$\frac{x+1}{x} = \frac{3x-3}{4x-10}$
Cross-multiply to solve for $x$:
$(x+1)(4x-10) = x(3x-3)$
$4x^2 - 10x + 4x - 10 = 3x^2 - 3x$
$4x^2 - 6x - 10 = 3x^2 - 3x$
$x^2 - 3x - 10 = 0$
Factor the quadratic equation:
$(x-5)(x+2) = 0$
This gives $x = 5$ or $x = -2$.
Since length cannot be negative,$x$ must be $5$.

(B) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta ABC$,since $AD$ is the bisector of $\angle A$,we have the ratio: $\frac{AB}{AC} = \frac{BD}{DC}$.
Substituting the given values: $\frac{8}{10} = \frac{3.2}{DC}$.
Cross-multiplying gives: $8 \times DC = 10 \times 3.2$.
$8 \times DC = 32$.
$DC = \frac{32}{8} = 4$.
Therefore,the length of $DC$ is $4$.

(C) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta PQR$,since $PS$ is the bisector of $\angle P$,we have the ratio:
$\frac{PQ}{PR} = \frac{QS}{SR}$
Given values are $PQ = 15$,$QS = 10$,and $SR = 8$.
Substituting these values into the formula:
$\frac{15}{PR} = \frac{10}{8}$
$10 \times PR = 15 \times 8$
$10 \times PR = 120$
$PR = \frac{120}{10} = 12$
Therefore,$PR = 12$.

(N/A) In $\Delta PQR$,$m\angle Q = 90^{\circ}$ and $\overline{QM}$ is an altitude to the hypotenuse $\overline{PR}$.
Since $M$ lies on $\overline{PR}$,we have $PR = PM + RM = x + y$.
Using the geometric mean theorem for the altitude to the hypotenuse:
$QM^2 = PM \cdot RM = x \cdot y$
$\therefore QM = \sqrt{xy}$.
Using the leg rule for $\Delta PQR$:
$PQ^2 = PM \cdot PR = x(x + y) = x^2 + xy$
$\therefore PQ = \sqrt{x^2 + xy}$.
Similarly,for the other leg:
$QR^2 = RM \cdot PR = y(x + y) = y^2 + xy$
$\therefore QR = \sqrt{y^2 + xy}$.
Thus,the lengths are $PQ = \sqrt{x^2 + xy}$,$QR = \sqrt{y^2 + xy}$,$PR = x + y$,and $QM = \sqrt{xy}$.

(C) In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $AC$.
Since $D$ lies on $AC$,we have $AC = AD + CD = 9 + 27 = 36$.
According to the geometric mean theorem (or property of altitude to the hypotenuse in a right triangle),$AB^2 = AD \times AC$.
Substituting the values,$AB^2 = 9 \times 36 = 324$.
Taking the square root of both sides,$AB = \sqrt{324} = 18$.
Thus,$AB = 18$.

(C) In $\Delta AMN$ and $\Delta ABC$,since $\overline{MN} \parallel \overline{BC}$,the triangles are similar by the $AA$ similarity criterion.
Therefore,$\frac{AM}{AB} = \frac{MN}{BC}$.
Given $MN = 3$ and $BC = 7$,we have $\frac{AM}{AB} = \frac{3}{7}$.
Since $AB = AM + MB$,we can write $\frac{AM}{AM + MB} = \frac{3}{7}$.
By cross-multiplication,$7 AM = 3 AM + 3 MB$.
$4 AM = 3 MB$.
Therefore,$\frac{AM}{MB} = \frac{3}{4}$.
Thus,$M$ divides $\overline{AB}$ from $A$ in the ratio $3:4$.

(D) In $\Delta PQR$,the side lengths are $PQ = 15$,$QR = 17$,and $PR = 8$.
To determine if it is a right-angled triangle,we check the converse of the Pythagoras theorem.
The longest side is $QR = 17$.
Calculate the square of the longest side: $QR^2 = 17^2 = 289$.
Calculate the sum of the squares of the other two sides: $PQ^2 + PR^2 = 15^2 + 8^2 = 225 + 64 = 289$.
Since $PQ^2 + PR^2 = QR^2$,the triangle satisfies the Pythagoras theorem.
Therefore,$\Delta PQR$ is a right-angled triangle,and the angle opposite to the hypotenuse $QR$ is $\angle P = 90^\circ$.

(D) In $\Delta PQR$,the side lengths are $PQ = 8$,$QR = 6$,and $PR = 12$.
The longest side is $\overline{PR}$.
Calculate the square of the longest side: $PR^2 = 12^2 = 144$.
Calculate the sum of the squares of the other two sides: $PQ^2 + QR^2 = 8^2 + 6^2 = 64 + 36 = 100$.
According to the converse of the Pythagoras theorem,a triangle is right-angled if the square of the longest side is equal to the sum of the squares of the other two sides.
Here,$PQ^2 + QR^2 = 100$ and $PR^2 = 144$.
Since $PQ^2 + QR^2 \neq PR^2$,the triangle does not satisfy the condition for a right-angled triangle.
Therefore,$\Delta PQR$ is not a right-angled triangle.

(D) In $\Delta PQR$,the side lengths are $PQ = 7$,$QR = 24$,and $PR = 25$.
The longest side is $PR = 25$.
Calculate the square of the longest side:
$PR^2 = 25^2 = 625$
Calculate the sum of the squares of the other two sides:
$PQ^2 + QR^2 = 7^2 + 24^2 = 49 + 576 = 625$
Since $PQ^2 + QR^2 = PR^2$,the triangle satisfies the converse of the Pythagoras theorem.
Therefore,$\Delta PQR$ is a right-angled triangle,and the angle opposite to the hypotenuse $PR$,which is $\angle Q$,is the right angle.

(B) Given $AB : AC = 24 : 25$.
Let $AB = 24k$ and $AC = 25k$,where $k > 0$.
In right-angled $\Delta ABC$,by the Pythagorean theorem:
$AC^2 = AB^2 + BC^2$
$(25k)^2 = (24k)^2 + (14)^2$
$625k^2 = 576k^2 + 196$
$625k^2 - 576k^2 = 196$
$49k^2 = 196$
$k^2 = 4$
Since $k > 0$,$k = 2$.
Now,$AB = 24 \times 2 = 48$ and $AC = 25 \times 2 = 50$.
The perimeter of $\Delta ABC = AB + BC + AC = 48 + 14 + 50 = 112$.
Thus,the perimeter of $\Delta ABC$ is $112$.

(N/A) Given: In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.
To prove: $\frac{AB^2}{BC^2} = \frac{AM}{CM}$.
Proof: In $\Delta ABC$,since $\overline{BM}$ is the altitude to the hypotenuse,we have two similar triangles:
$1$. $\Delta AMB \sim \Delta ABC$,which implies $\frac{AB}{AC} = \frac{AM}{AB}$,so $AB^2 = AM \times AC$.
$2$. $\Delta BMC \sim \Delta ABC$,which implies $\frac{BC}{AC} = \frac{CM}{BC}$,so $BC^2 = CM \times AC$.
Dividing the two equations:
$\frac{AB^2}{BC^2} = \frac{AM \times AC}{CM \times AC}$
Therefore,$\frac{AB^2}{BC^2} = \frac{AM}{CM}$.

(A) 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.
According to the geometric mean theorem,the altitude $\overline{YM}$ to the hypotenuse $\overline{XZ}$ satisfies the relation: $YM^2 = XM \cdot ZM$.
Given $XM = \sqrt{12}$ and $ZM = \sqrt{3}$.
Substituting these values into the formula:
$YM^2 = \sqrt{12} \cdot \sqrt{3}$
$YM^2 = \sqrt{12 \cdot 3}$
$YM^2 = \sqrt{36}$
$YM^2 = 6$
Therefore,$YM = \sqrt{6}$.

(B) Let $PD = x$. Since $PR = 34$,then $DR = 34 - x$.
In $\Delta PQR$,$\overline{QD}$ is the altitude to the hypotenuse $PR$. By the Geometric Mean Theorem (or properties of similar triangles),$QD^2 = PD \cdot DR$.
Substituting the given values: $15^2 = x(34 - x)$.
$225 = 34x - x^2$.
$x^2 - 34x + 225 = 0$.
Solving the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{34 \pm \sqrt{34^2 - 4(1)(225)}}{2} = \frac{34 \pm \sqrt{1156 - 900}}{2} = \frac{34 \pm \sqrt{256}}{2} = \frac{34 \pm 16}{2}$.
Two possible values for $x$ are $x = \frac{50}{2} = 25$ or $x = \frac{18}{2} = 9$.
In $\Delta PQD$,by the Pythagorean theorem,$PQ^2 = QD^2 + PD^2$.
If $PD = 9$,then $PQ^2 = 15^2 + 9^2 = 225 + 81 = 306$. $PQ = \sqrt{306} = \sqrt{9 \cdot 34} = 3 \sqrt{34}$.
If $PD = 25$,then $PQ^2 = 15^2 + 25^2 = 225 + 625 = 850$. $PQ = \sqrt{850} = \sqrt{25 \cdot 34} = 5 \sqrt{34}$.
Comparing with the given options,$5 \sqrt{34}$ is the correct answer.

(C) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,$\overline{BM}$ is the altitude to the hypotenuse $AC$.
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,$\Delta BMC \sim \Delta AMB$.
From the similarity,we have the property: $BM^2 = AM \cdot CM$.
Given $BM = \sqrt{30}$ and $CM = 3$,we substitute these values into the equation:
$(\sqrt{30})^2 = AM \cdot 3$
$30 = AM \cdot 3$
$AM = 10$.
The length of the hypotenuse $AC$ is the sum of segments $AM$ and $CM$:
$AC = AM + CM = 10 + 3 = 13$.
Therefore,the correct answer is $13$.

(A) In a right-angled triangle $\Delta ABC$ with altitude $\overline{BM}$ to the hypotenuse $\overline{AC}$,the geometric mean theorem states that $BM^2 = AM \cdot CM$.
Given $AM = x + 7$,$BM = x + 2$,and $CM = x$.
Substituting these values into the formula: $(x + 2)^2 = (x + 7)(x)$.
Expanding both sides: $x^2 + 4x + 4 = x^2 + 7x$.
Subtracting $x^2$ from both sides: $4x + 4 = 7x$.
Rearranging the terms: $4 = 7x - 4x$,which simplifies to $4 = 3x$.
Therefore,$x = \frac{4}{3}$.

(A) In a right-angled triangle $\Delta ABC$ with altitude $\overline{BM}$ to the hypotenuse $\overline{AC}$,we have the geometric mean property: $BM^2 = AM \cdot CM$.
Given $BM = 6$,we have $AM \cdot CM = 6^2 = 36$.
We also know $AM + CM = AC = 13$.
Let $AM = x$ and $CM = y$. Then $x + y = 13$ and $xy = 36$.
These are the roots of the quadratic equation $t^2 - 13t + 36 = 0$.
Factoring the equation: $(t - 9)(t - 4) = 0$,so $t = 9$ or $t = 4$.
Since $AM < CM$,we have $AM = 4$ and $CM = 9$.
In $\Delta ABM$,by the Pythagorean theorem: $AB^2 = AM^2 + BM^2$.
$AB^2 = 4^2 + 6^2 = 16 + 36 = 52$.
$AB = \sqrt{52} = \sqrt{4 \cdot 13} = 2 \sqrt{13}$.

(A) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$,the altitude property states that $BM^2 = AM \cdot CM$.
Given $AM = 9$ and $CM = 16$,we have $BM^2 = 9 \cdot 16 = 144$,so $BM = 12$.
Now,we find the lengths of the sides $AB$ and $BC$ using the Pythagorean theorem in $\Delta AMB$ and $\Delta BMC$:
$AB = \sqrt{AM^2 + BM^2} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$.
$BC = \sqrt{CM^2 + BM^2} = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20$.
The hypotenuse $AC = AM + CM = 9 + 16 = 25$.
The perimeter of $\Delta ABC = AB + BC + AC = 15 + 20 + 25 = 60$.

(B) In a right-angled triangle $\Delta ABC$ with altitude $\overline{AM}$ to the hypotenuse $\overline{BC}$,we have the geometric mean theorem: $AM^2 = BM \cdot CM$.
Given $BM = 6$ and $CM = 2$,$AM^2 = 6 \cdot 2 = 12$,so $AM = \sqrt{12} = 2\sqrt{3}$.
Now,calculate the sides of $\Delta ABC$:
$BC = BM + CM = 6 + 2 = 8$.
Using the Pythagorean theorem in $\Delta ABM$: $AB^2 = AM^2 + BM^2 = 12 + 36 = 48$,so $AB = \sqrt{48} = 4\sqrt{3}$.
Using the Pythagorean theorem in $\Delta ACM$: $AC^2 = AM^2 + CM^2 = 12 + 4 = 16$,so $AC = \sqrt{16} = 4$.
Perimeter of $\Delta ABC = AB + AC + BC = 4\sqrt{3} + 4 + 8 = 12 + 4\sqrt{3}$.

(A) Let $AM = x$. Since $AC = 25$,then $MC = 25 - x$.
In $\Delta ABC$,$\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$. By the Geometric Mean Theorem,$BM^2 = AM \cdot MC$.
Substituting the values: $12^2 = x(25 - x) \implies 144 = 25x - x^2$.
Rearranging gives the quadratic equation: $x^2 - 25x + 144 = 0$.
Factoring the equation: $(x - 16)(x - 9) = 0$,so $x = 16$ or $x = 9$.
Given $AM > CM$,we choose $AM = 16$ and $MC = 9$.
Using the Pythagorean theorem in $\Delta ABM$: $AB^2 = AM^2 + BM^2 = 16^2 + 12^2 = 256 + 144 = 400 \implies AB = 20$.
Using the Pythagorean theorem in $\Delta CBM$: $BC^2 = MC^2 + BM^2 = 9^2 + 12^2 = 81 + 144 = 225 \implies BC = 15$.
Thus,$AB = 20$ and $BC = 15$.

(B) In a right-angled triangle,the altitude to the hypotenuse creates two triangles that are similar to the original triangle and to each other.
By the Geometric Mean Theorem (or properties of similar triangles),we have the relationship: $YZ^2 = ZM \cdot XZ$.
First,calculate the total length of the hypotenuse $XZ$:
$XZ = XM + ZM = 5 + 4 = 9$.
Now,substitute the values into the formula:
$YZ^2 = 4 \cdot 9 = 36$.
Taking the square root of both sides:
$YZ = \sqrt{36} = 6$.
Thus,the length of $YZ$ is $6$.

(A) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,$\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$.
By the geometric mean theorem for right triangles,we have the relation $BM^2 = AM \cdot CM$.
Given $BM = 12$ and $CM = 18$,we substitute these values into the equation:
$12^2 = AM \cdot 18$
$144 = AM \cdot 18$
$AM = \frac{144}{18} = 8$.
Now,consider $\Delta ABM$,which is a right-angled triangle at $M$ $(\angle AMB = 90^{\circ})$.
Using the Pythagorean theorem: $AB^2 = AM^2 + BM^2$.
$AB^2 = 8^2 + 12^2$
$AB^2 = 64 + 144$
$AB^2 = 208$
$AB = \sqrt{208} = \sqrt{16 \cdot 13} = 4 \sqrt{13}$.
Thus,the correct option is $A$.

(C) Let the side length of the square be $a$.
In a square,the diagonal $d$ is related to the side $a$ by the formula $d = a \sqrt{2}$.
Given that the diagonal $d = 8$,we have $8 = a \sqrt{2}$.
To find the side length $a$,we divide both sides by $\sqrt{2}$:
$a = \frac{8}{\sqrt{2}}$.
Rationalizing the denominator:
$a = \frac{8}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{8 \sqrt{2}}{2} = 4 \sqrt{2}$.
Thus,the length of the side of the square is $4 \sqrt{2}$.

(N/A) Let the two equal sides of the isosceles right-angled triangle be $x$. According to the Pythagorean theorem, $x^2 + x^2 = (20)^2$.
$2x^2 = 400$, which implies $x^2 = 200$.
Therefore, $x = \sqrt{200} = 10\sqrt{2}$.
The perimeter of the triangle is the sum of all sides: $P = x + x + 20 = 2x + 20 = 2(10\sqrt{2}) + 20 = 20\sqrt{2} + 20$.
The area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times x = \frac{1}{2} \times x^2$.
Substituting $x^2 = 200$, we get Area $= \frac{1}{2} \times 200 = 100$.