Mix Examples - Triangles Questions in English

(N/A) Given that $\triangle ABC \sim \triangle DEF$.
Therefore,the ratios of their corresponding sides are equal:
$\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD}$
Substituting the given expressions:
$\frac{2x - 1}{18} = \frac{2x + 2}{3x + 9} = \frac{3x}{6x}$
Taking the first and third ratios:
$\frac{2x - 1}{18} = \frac{3x}{6x}$
$\frac{2x - 1}{18} = \frac{1}{2}$
Cross-multiplying:
$2(2x - 1) = 18$
$4x - 2 = 18$
$4x = 20$
$x = 5$
Now,calculating the side lengths:
For $\triangle ABC$:
$AB = 2(5) - 1 = 9 \, cm$
$BC = 2(5) + 2 = 12 \, cm$
$CA = 3(5) = 15 \, cm$
For $\triangle DEF$:
$DE = 18 \, cm$
$EF = 3(5) + 9 = 24 \, cm$
$FD = 6(5) = 30 \, cm$

(D) Given,$\angle A = \angle C$,$AB = 6 \, cm$,$BP = 15 \, cm$,$AP = 12 \, cm$,and $CP = 4 \, cm$.
In $\triangle APB$ and $\triangle CPD$:
$\angle A = \angle C$ [Given]
$\angle APB = \angle CPD$ [Vertically opposite angles]
Therefore,$\triangle APB \sim \triangle CPD$ [By $AA$ similarity criterion]
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AP}{CP} = \frac{BP}{PD} = \frac{AB}{CD}$
Substituting the given values:
$\frac{12}{4} = \frac{15}{PD} = \frac{6}{CD}$
Taking the first two terms:
$\frac{12}{4} = \frac{15}{PD} \implies 3 = \frac{15}{PD} \implies PD = \frac{15}{3} = 5 \, cm$
Taking the first and last terms:
$\frac{12}{4} = \frac{6}{CD} \implies 3 = \frac{6}{CD} \implies CD = \frac{6}{3} = 2 \, cm$
Hence,the length of $PD = 5 \, cm$ and the length of $CD = 2 \, cm$.

(N/A) Given,$\triangle ABC \sim \triangle EDF$. Since the triangles are similar,their corresponding sides are in the same ratio.
i.e.,$\frac{AB}{ED} = \frac{AC}{EF} = \frac{BC}{DF}$ ...... $(i)$
Given values are $AB = 5 \, cm$,$AC = 7 \, cm$,$DF = 15 \, cm$,and $DE = 12 \, cm$.
Substituting these values in Eq. $(i)$,we get:
$\frac{5}{12} = \frac{7}{EF} = \frac{BC}{15}$
Taking the first and second terms:
$\frac{5}{12} = \frac{7}{EF}$
$\Rightarrow EF = \frac{7 \times 12}{5} = \frac{84}{5} = 16.8 \, cm$
Taking the first and third terms:
$\frac{5}{12} = \frac{BC}{15}$
$\Rightarrow BC = \frac{5 \times 15}{12} = \frac{75}{12} = 6.25 \, cm$
Hence,the lengths of the remaining sides are $EF = 16.8 \, cm$ and $BC = 6.25 \, cm$.

(N/A) Let a $\triangle ABC$ in which a line $DE$ parallel to $BC$ intersects $AB$ at $D$ and $AC$ at $E$. To prove: $DE$ divides the two sides in the same ratio,i.e.,$\frac{AD}{DB} = \frac{AE}{EC}$.
Construction: Join $BE, CD$ and draw $EF \perp AB$ and $DG \perp AC$.
Proof: Here,$\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EF}{\frac{1}{2} \times DB \times EF} = \frac{AD}{DB}$ ......$(i)$ [Since,area of triangle $= \frac{1}{2} \times \text{base} \times \text{height}$]
Similarly,$\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DG}{\frac{1}{2} \times EC \times DG} = \frac{AE}{EC}$ ......$(ii)$
Now,since $\triangle BDE$ and $\triangle DEC$ lie between the same parallel lines $DE$ and $BC$ and on the same base $DE$,therefore,$\text{ar}(\triangle BDE) = \text{ar}(\triangle DEC)$ ......$(iii)$
From Eqs. $(i), (ii)$ and $(iii)$,we get $\frac{AD}{DB} = \frac{AE}{EC}$.
Hence proved.

(N/A) Given: $PQRS$ is a parallelogram,so $PQ \parallel SR$ and $PS \parallel QR$. Also,$AB \parallel PS$.
To prove: $OC \parallel SR$.
Proof: In $\triangle OPS$ and $\triangle OAB$,since $PS \parallel AB$:
$\angle POS = \angle AOB$ (Common angle)
$\angle OSP = \angle OBA$ (Corresponding angles)
Therefore,$\triangle OPS \sim \triangle OAB$ (by $AAA$ similarity criterion).
Thus,$\frac{PS}{AB} = \frac{OS}{OB}$ ..........$(i)$
In $\triangle CQR$ and $\triangle CAB$,since $QR \parallel PS \parallel AB$:
$\angle QCR = \angle ACB$ (Common angle)
$\angle CQR = \angle CAB$ (Corresponding angles)
Therefore,$\triangle CQR \sim \triangle CAB$ (by $AA$ similarity criterion).
Thus,$\frac{QR}{AB} = \frac{CR}{CB}$.
Since $PQRS$ is a parallelogram,$PS = QR$. Substituting this,we get:
$\frac{PS}{AB} = \frac{CR}{CB}$ ..........$(ii)$
From equations $(i)$ and $(ii)$:
$\frac{OS}{OB} = \frac{CR}{CB} \Rightarrow \frac{OB}{OS} = \frac{CB}{CR}$.
Subtracting $1$ from both sides:
$\frac{OB}{OS} - 1 = \frac{CB}{CR} - 1$
$\frac{OB - OS}{OS} = \frac{CB - CR}{CR}$
$\frac{BS}{OS} = \frac{BR}{CR}$.
By the converse of the Basic Proportionality Theorem $(BPT)$,$SR \parallel OC$ (or $OC \parallel SR$). Hence proved.

(D) Let $AC$ be the ladder of length $5\,m$ and $BC = 4\,m$ be the height of the wall where the ladder is placed. If the foot of the ladder is moved $1.6\,m$ towards the wall,i.e.,$AD = 1.6\,m$,then the ladder slides upward to a new position $ED$.
In right-angled $\triangle ABC$,by Pythagoras theorem:
$AC^2 = AB^2 + BC^2$
$5^2 = AB^2 + 4^2$
$25 = AB^2 + 16$
$AB^2 = 9 \Rightarrow AB = 3\,m$.
Now,the new distance of the foot from the wall is $DB = AB - AD = 3 - 1.6 = 1.4\,m$.
In right-angled $\triangle EBD$,by Pythagoras theorem:
$ED^2 = EB^2 + DB^2$
$5^2 = EB^2 + (1.4)^2$
$25 = EB^2 + 1.96$
$EB^2 = 25 - 1.96 = 23.04$
$EB = \sqrt{23.04} = 4.8\,m$.
The distance by which the top of the ladder slides upwards is $EC = EB - BC = 4.8 - 4 = 0.8\,m$.

(A) Given,$AC \perp CB$,$AC = 2x \, km$,$CB = 2(x+7) \, km$ and $AB = 26 \, km$.
From the figure,we have a right-angled $\triangle ACB$ with the right angle at $C$.
In $\triangle ACB$,by the Pythagoras theorem:
$AB^2 = AC^2 + BC^2$
$(26)^2 = (2x)^2 + \{2(x+7)\}^2$
$676 = 4x^2 + 4(x^2 + 49 + 14x)$
$676 = 4x^2 + 4x^2 + 196 + 56x$
$676 = 8x^2 + 56x + 196$
$8x^2 + 56x - 480 = 0$
Dividing by $8$,we get:
$x^2 + 7x - 60 = 0$
$x^2 + 12x - 5x - 60 = 0$
$x(x + 12) - 5(x + 12) = 0$
$(x + 12)(x - 5) = 0$
$x = -12$ or $x = 5$.
Since distance cannot be negative,$x = 5$ (as $x \neq -12$).
Now,$AC = 2x = 2(5) = 10 \, km$ and $BC = 2(x+7) = 2(5+7) = 24 \, km$.
The distance covered to reach city $B$ from city $A$ via city $C$ is $AC + BC = 10 + 24 = 34 \, km$.
The distance covered to reach city $B$ from city $A$ after the construction of the highway is $AB = 26 \, km$.
Therefore,the distance saved is $34 - 26 = 8 \, km$.

(B) Let $BC = 18 \, m$ be the flag pole and its shadow be $AB = 9.6 \, m$. The distance of the top of the pole $(C)$ from the far end of the shadow $(A)$ is $AC$.
In right-angled $\triangle ABC$,by Pythagoras theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = (9.6)^2 + (18)^2$
$AC^2 = 92.16 + 324$
$AC^2 = 416.16$
$AC = \sqrt{416.16} = 20.4 \, m$
Hence,the required distance is $20.4 \, m$.

(C) Let $A$ be the position of the street bulb fixed on a pole $AB = 6 \, m$ and $CD = 1.5 \, m$ be the height of a woman and her shadow be $ED = 3 \, m$. Let the distance between the pole and the woman be $x \, m$.
Here,the woman and the pole are both standing vertically.
So,$CD \parallel AB$.
In $\triangle CDE$ and $\triangle ABE$,$\angle E = \angle E$ [common angle].
$\angle ABE = \angle CDE = 90^{\circ}$ [each equal to $90^{\circ}$].
Therefore,$\triangle CDE \sim \triangle ABE$ [by $AAA$ similarity criterion].
Then,$\frac{ED}{EB} = \frac{CD}{AB}$.
$\frac{3}{3 + x} = \frac{1.5}{6}$.
$3 \times 6 = 1.5(3 + x)$.
$18 = 4.5 + 1.5x$.
$1.5x = 18 - 4.5$.
$1.5x = 13.5$.
$x = \frac{13.5}{1.5} = 9 \, m$.
Hence,she is at a distance of $9 \, m$ from the base of the pole.

(N/A) Given,$\triangle ABC$ in which $\angle B = 90^{\circ}$ and $BD \perp AC$.
Also,$AD = 4 \, cm$ and $CD = 5 \, cm$.
In $\triangle ADB$ and $\triangle CDB$,$\angle ADB = \angle CDB = 90^{\circ}$.
Also,$\angle BAD = \angle DBC$ (since both are equal to $90^{\circ} - \angle C$).
Therefore,$\triangle ADB \sim \triangle CDB$ by $AA$ similarity criterion.
Thus,$\frac{BD}{AD} = \frac{CD}{BD}$.
$\Rightarrow BD^2 = AD \times CD = 4 \times 5 = 20$.
$\Rightarrow BD = \sqrt{20} = 2\sqrt{5} \, cm$.
Now,in right-angled $\triangle ADB$,by Pythagoras theorem:
$AB^2 = AD^2 + BD^2 = 4^2 + (2\sqrt{5})^2 = 16 + 20 = 36$.
$\Rightarrow AB = \sqrt{36} = 6 \, cm$.
Hence,$BD = 2\sqrt{5} \, cm$ and $AB = 6 \, cm$.

(N/A) Given,$\Delta PQR$ in which $\angle Q = 90^{\circ}$,$QS \perp PR$,$PQ = 6 \, cm$,and $PS = 4 \, cm$.
In right-angled $\Delta PSQ$,by Pythagoras theorem:
$PQ^2 = PS^2 + QS^2$
$6^2 = 4^2 + QS^2$
$36 = 16 + QS^2$
$QS^2 = 20$
$QS = \sqrt{20} = 2\sqrt{5} \, cm$.
Since $\Delta PSQ \sim \Delta QSR$,we have:
$\frac{PS}{QS} = \frac{QS}{RS}$
$QS^2 = PS \times RS$
$20 = 4 \times RS$
$RS = 5 \, cm$.
In right-angled $\Delta QSR$,by Pythagoras theorem:
$QR^2 = QS^2 + RS^2$
$QR^2 = 20 + 5^2$
$QR^2 = 20 + 25 = 45$
$QR = \sqrt{45} = 3\sqrt{5} \, cm$.
Thus,$QS = 2\sqrt{5} \, cm$,$RS = 5 \, cm$,and $QR = 3\sqrt{5} \, cm$.

(N/A) Given: In $\triangle PQR$,$PD \perp QR$,$PQ = a$,$PR = b$,$QD = c$,and $DR = d$.
To prove: $(a+b)(a-b) = (c+d)(c-d)$.
Proof: In right-angled $\triangle PDQ$,by Pythagoras theorem:
$PQ^2 = PD^2 + QD^2$
$a^2 = PD^2 + c^2$
$PD^2 = a^2 - c^2$ ...... $(i)$
In right-angled $\triangle PDR$,by Pythagoras theorem:
$PR^2 = PD^2 + DR^2$
$b^2 = PD^2 + d^2$
$PD^2 = b^2 - d^2$ ...... $(ii)$
From equations $(i)$ and $(ii)$:
$a^2 - c^2 = b^2 - d^2$
$a^2 - b^2 = c^2 - d^2$
$(a - b)(a + b) = (c - d)(c + d)$
Hence,$(a + b)(a - b) = (c + d)(c - d)$ is proved.

(N/A) Given: $A$ quadrilateral $ABCD$ in which $\angle A + \angle D = 90^{\circ}$.
To prove: $AC^{2} + BD^{2} = AD^{2} + BC^{2}$.
Construction: Produce $AB$ and $DC$ to meet at point $E$.
Proof: In $\triangle AED$,$\angle A + \angle D = 90^{\circ}$ (given).
Since the sum of angles in a triangle is $180^{\circ}$,we have $\angle E = 180^{\circ} - (\angle A + \angle D) = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
In right-angled $\triangle AED$,by Pythagoras theorem: $AD^{2} = AE^{2} + DE^{2}$ ... $(i)$
In right-angled $\triangle BEC$,by Pythagoras theorem: $BC^{2} = BE^{2} + CE^{2}$ ... $(ii)$
Adding $(i)$ and $(ii)$,we get: $AD^{2} + BC^{2} = AE^{2} + DE^{2} + BE^{2} + CE^{2}$ ... $(iii)$
In right-angled $\triangle AEC$,by Pythagoras theorem: $AC^{2} = AE^{2} + CE^{2}$ ... $(iv)$
In right-angled $\triangle BED$,by Pythagoras theorem: $BD^{2} = BE^{2} + DE^{2}$ ... $(v)$
Adding $(iv)$ and $(v)$,we get: $AC^{2} + BD^{2} = AE^{2} + CE^{2} + BE^{2} + DE^{2}$ ... $(vi)$
Comparing $(iii)$ and $(vi)$,we get: $AC^{2} + BD^{2} = AD^{2} + BC^{2}$.
Hence proved.

(N/A) Given $l \parallel m$ and line segments $AB, CD$ and $EF$ are concurrent at point $P$.
To prove: $\frac{AE}{BF} = \frac{AC}{BD} = \frac{CE}{FD}$.
Proof:
In $\triangle APC$ and $\triangle BPD$,
$\angle APC = \angle BPD$ [Vertically opposite angles]
$\angle PAC = \angle PBD$ [Alternate interior angles,as $l \parallel m$]
Therefore,$\triangle APC \sim \triangle BPD$ [By $AA$ similarity criterion]
Then,$\frac{AP}{PB} = \frac{AC}{BD} = \frac{PC}{PD}$ ......$(i)$
In $\triangle APE$ and $\triangle BPF$,
$\angle APE = \angle BPF$ [Vertically opposite angles]
$\angle PAE = \angle PBF$ [Alternate interior angles]
Therefore,$\triangle APE \sim \triangle BPF$ [By $AA$ similarity criterion]
Then,$\frac{AP}{PB} = \frac{AE}{BF} = \frac{PE}{PF}$ ......$(ii)$
In $\triangle PEC$ and $\triangle PFD$,
$\angle EPC = \angle FPD$ [Vertically opposite angles]
$\angle PCE = \angle PDF$ [Alternate interior angles]
Therefore,$\triangle PEC \sim \triangle PFD$ [By $AA$ similarity criterion]
Then,$\frac{PE}{PF} = \frac{PC}{PD} = \frac{CE}{FD}$ ......$(iii)$
From equations $(i), (ii)$ and $(iii)$,
$\frac{AP}{PB} = \frac{AC}{BD} = \frac{AE}{BF} = \frac{PE}{PF} = \frac{CE}{FD}$
Thus,$\frac{AE}{BF} = \frac{AC}{BD} = \frac{CE}{FD}$. Hence proved.

(N/A) Given: $AB = 6 \, cm$,$BC = 9 \, cm$,$CD = 12 \, cm$,and $PS = 36 \, cm$.
Also,$PA$,$QB$,$RC$,and $SD$ are all perpendiculars to line $l$,which implies $PA \parallel QB \parallel RC \parallel SD$.
By the intercept theorem (or Thales' theorem application for parallel lines),the ratio of intercepts on one transversal is equal to the ratio of intercepts on another transversal:
$PQ : QR : RS = AB : BC : CD$
$PQ : QR : RS = 6 : 9 : 12$
Let $PQ = 6x$,$QR = 9x$,and $RS = 12x$.
Since the total length $PS = 36 \, cm$:
$PQ + QR + RS = 36$
$6x + 9x + 12x = 36$
$27x = 36$
$x = \frac{36}{27} = \frac{4}{3}$
Now,calculating the lengths:
$PQ = 6x = 6 \times \frac{4}{3} = 8 \, cm$
$QR = 9x = 9 \times \frac{4}{3} = 12 \, cm$
$RS = 12x = 12 \times \frac{4}{3} = 16 \, cm$

(N/A) Given: $ABCD$ is a trapezium with $AB \parallel DC$. Diagonals $AC$ and $BD$ intersect at $O$. $PQ \parallel AB \parallel DC$.
To prove: $PO = QO$.
Proof: In $\triangle ADC$,$PO \parallel DC$ (since $PQ \parallel DC$).
By the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{AP}{PD} = \frac{AO}{OC}$ ........$(i)$
In $\triangle ABC$,$OQ \parallel AB$.
By $BPT$,we have:
$\frac{BQ}{QC} = \frac{AO}{OC}$ ........$(ii)$
From $(i)$ and $(ii)$,we get:
$\frac{AP}{PD} = \frac{BQ}{QC}$
Adding $1$ to both sides:
$\frac{AP}{PD} + 1 = \frac{BQ}{QC} + 1$
$\frac{AP + PD}{PD} = \frac{BQ + QC}{QC}$
$\frac{AD}{PD} = \frac{BC}{QC}$
$\Rightarrow \frac{PD}{AD} = \frac{QC}{BC}$ ........$(iii)$
Now,in $\triangle ABD$,$PO \parallel AB$. Therefore,$\triangle POD \sim \triangle ABD$ (by $AA$ similarity).
So,$\frac{PO}{AB} = \frac{PD}{AD}$ ........$(iv)$
In $\triangle ABC$,$OQ \parallel AB$. Therefore,$\triangle OQC \sim \triangle ABC$ (by $AA$ similarity).
So,$\frac{OQ}{AB} = \frac{QC}{BC}$ ........$(v)$
From $(iii)$,$(iv)$,and $(v)$,we get:
$\frac{PO}{AB} = \frac{OQ}{AB}$
$\Rightarrow PO = OQ$. Hence proved.

(N/A) Given: In $\triangle ABC$,$E$ is the mid-point of $CA$ and $\angle AEF = \angle AFE$.
To prove: $\frac{BD}{CD} = \frac{BF}{CE}$.
Construction: Draw a line $CG$ parallel to $EF$ such that $G$ lies on $AB$.
Proof: Since $E$ is the mid-point of $CA$,we have $CE = AE$ $(i)$.
In $\triangle ACG$,since $CG \parallel EF$ and $E$ is the mid-point of $CA$,by the converse of the Mid-point Theorem,$F$ is the mid-point of $AG$. Thus,$GF = AF$ $(ii)$.
Also,in $\triangle ACG$,since $CG \parallel EF$,by the Mid-point Theorem,$EF = \frac{1}{2} CG$. However,this is not needed here. Instead,consider $\triangle BDF$ and $\triangle BCG$. Since $CG \parallel EF$,by the Basic Proportionality Theorem in $\triangle BDF$,we have $\frac{BD}{CD} = \frac{BF}{GF}$.
Since $\angle AEF = \angle AFE$,in $\triangle AEF$,we have $AE = AF$. From $(i)$ and $(ii)$,$CE = AE = AF = GF$. Therefore,$CE = GF$.
Substituting $GF = CE$ in the ratio $\frac{BD}{CD} = \frac{BF}{GF}$,we get $\frac{BD}{CD} = \frac{BF}{CE}$.
Hence proved.

(N/A) Let $ABC$ be a right triangle,right-angled at $B$ with $AB = y$ and $BC = x$.
Three semicircles are drawn on the sides $AB$,$BC$,and $AC$ respectively,with diameters $AB$,$BC$,and $AC$.
Let the areas of the semicircles with diameters $AB$,$BC$,and $AC$ be $A_1$,$A_2$,and $A_3$ respectively.
To prove: $A_3 = A_1 + A_2$.
Proof: In $\triangle ABC$,by the Pythagoras theorem,
$AC^2 = AB^2 + BC^2$
$AC^2 = y^2 + x^2$
$AC = \sqrt{y^2 + x^2}$
We know that the area of a semicircle with diameter $d$ is $\frac{\pi}{2} \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{8}$.
Therefore,the area of the semicircle drawn on $AC$ is:
$A_3 = \frac{\pi (AC)^2}{8} = \frac{\pi (y^2 + x^2)}{8}$ .....$(i)$
Now,the area of the semicircle drawn on $AB$ is:
$A_1 = \frac{\pi (AB)^2}{8} = \frac{\pi y^2}{8}$ .....$(ii)$
And the area of the semicircle drawn on $BC$ is:
$A_2 = \frac{\pi (BC)^2}{8} = \frac{\pi x^2}{8}$ .....$(iii)$
Adding equations $(ii)$ and $(iii)$:
$A_1 + A_2 = \frac{\pi y^2}{8} + \frac{\pi x^2}{8} = \frac{\pi (y^2 + x^2)}{8}$
From equation $(i)$,we see that $A_1 + A_2 = A_3$.
Hence,the area of the semicircle on the hypotenuse is equal to the sum of the areas of the semicircles on the other two sides.

(N/A) Let $\triangle ABC$ be a right-angled triangle with $\angle A = 90^\circ$,$AC = y$,and $AB = x$.
Three equilateral triangles $\triangle AEC$,$\triangle AFB$,and $\triangle CBD$ are drawn on sides $AC$,$AB$,and $BC$ respectively.
Let the areas of these equilateral triangles be $A_1$,$A_2$,and $A_3$ respectively.
We need to prove that $A_3 = A_1 + A_2$.
In $\triangle ABC$,by Pythagoras theorem: $BC^2 = AC^2 + AB^2 = y^2 + x^2$.
The area of an equilateral triangle with side $s$ is given by $\frac{\sqrt{3}}{4} s^2$.
$A_1 = \text{Area}(\triangle AEC) = \frac{\sqrt{3}}{4} AC^2 = \frac{\sqrt{3}}{4} y^2$ ... $(i)$
$A_2 = \text{Area}(\triangle AFB) = \frac{\sqrt{3}}{4} AB^2 = \frac{\sqrt{3}}{4} x^2$ ... $(ii)$
$A_3 = \text{Area}(\triangle CBD) = \frac{\sqrt{3}}{4} BC^2 = \frac{\sqrt{3}}{4} (y^2 + x^2)$ ... $(iii)$
From $(i)$,$(ii)$,and $(iii)$:
$A_3 = \frac{\sqrt{3}}{4} y^2 + \frac{\sqrt{3}}{4} x^2 = A_1 + A_2$.
Hence,the area of the equilateral triangle on the hypotenuse is equal to the sum of the areas of the equilateral triangles on the other two sides.

(N/A) In $\Delta ABC$,the ratio of the angles is $m \angle A : m \angle B : m \angle C = 15 : 8 : 7$.
Let the angles be $15t$,$8t$,and $7t$ respectively.
Since the sum of angles in a triangle is $180^\circ$,we have $15t + 8t + 7t = 180^\circ$.
$30t = 180^\circ$,which gives $t = 6^\circ$.
Therefore,$m \angle A = 15 \times 6^\circ = 90^\circ$,$m \angle B = 8 \times 6^\circ = 48^\circ$,and $m \angle C = 7 \times 6^\circ = 42^\circ$.
Given the similarity correspondence $ABC \leftrightarrow QPR$,the corresponding angles are equal.
Thus,$m \angle Q = m \angle A = 90^\circ$,$m \angle P = m \angle B = 48^\circ$,and $m \angle R = m \angle C = 42^\circ$.
Hence,the angles of $\Delta PQR$ are $m \angle P = 48^\circ$,$m \angle Q = 90^\circ$,and $m \angle R = 42^\circ$.

(C) Given that the correspondence $ABC \leftrightarrow XYZ$ between $\Delta ABC$ and $\Delta XYZ$ is a similarity.
Therefore,the ratios of their corresponding sides are equal:
$\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$
From the given equation $4AB = 5XY$,we can write:
$\frac{AB}{XY} = \frac{5}{4}$
Since the triangles are similar,we have:
$\frac{AB}{XY} = \frac{BC}{YZ}$
Substituting the known values:
$\frac{5}{4} = \frac{12}{YZ}$
Solving for $YZ$:
$YZ = \frac{12 \times 4}{5} = \frac{48}{5} = 9.6$
Thus,$YZ = 9.6$.

(D) Since $\Delta ABC \sim \Delta PQR$,the corresponding angles are equal: $m \angle A = m \angle P$,$m \angle B = m \angle Q$,and $m \angle C = m \angle R = 100^{\circ}$.
In $\Delta ABC$,the sum of angles is $180^{\circ}$,so $m \angle A + m \angle B + m \angle C = 180^{\circ}$.
Substituting $m \angle C = 100^{\circ}$,we get $m \angle A + m \angle B = 80^{\circ}$.
Given $2 m \angle P = 3 m \angle Q$,and since $m \angle P = m \angle A$ and $m \angle Q = m \angle B$,we have $2 m \angle A = 3 m \angle B$,which implies $m \angle A = 1.5 m \angle B$.
Substitute this into $m \angle A + m \angle B = 80^{\circ}$:
$1.5 m \angle B + m \angle B = 80^{\circ}$
$2.5 m \angle B = 80^{\circ}$
$m \angle B = 80 / 2.5 = 32^{\circ}$.

(A) Given that the correspondence $PQR \leftrightarrow ZYX$ is a similarity,we have $\Delta PQR \sim \Delta ZYX$.
From the property of similar triangles,the ratios of corresponding sides are equal:
$\frac{PQ}{ZY} = \frac{QR}{YX} = \frac{PR}{ZX}$.
We are given $\frac{PQ}{ZY} = \frac{5}{3}$ and $PR = 10$.
Substituting these values into the ratio $\frac{PQ}{ZY} = \frac{PR}{ZX}$,we get:
$\frac{5}{3} = \frac{10}{ZX}$.
Cross-multiplying,we get $5 \times ZX = 3 \times 10$.
$5 \times ZX = 30$.
$ZX = \frac{30}{5} = 6$.
Therefore,the length of $ZX$ is $6$.

(A) Given that $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow QRP$.
This implies that the corresponding angles are equal:
$m \angle A = m \angle Q$
$m \angle B = m \angle R$
$m \angle C = m \angle P$
Given $m \angle A = 80^{\circ}, m \angle B = 40^{\circ}, m \angle C = 60^{\circ}$.
Substituting these values:
$m \angle Q = 80^{\circ}$
$m \angle R = 40^{\circ}$
$m \angle P = 60^{\circ}$
Thus,the angles of $\Delta PQR$ are $m \angle P = 60^{\circ}, m \angle Q = 80^{\circ}, m \angle R = 40^{\circ}$.

(C) Given that $\Delta XYZ \sim \Delta DEF$ with the correspondence $XYZ \leftrightarrow EDF$.
This implies the ratio of corresponding sides is equal: $\frac{XY}{ED} = \frac{YZ}{DF} = \frac{ZX}{FE} = k$.
Given $XY = 3, YZ = 4, ZX = 6$ and $DF = 12$.
From the ratio $\frac{YZ}{DF} = \frac{4}{12} = \frac{1}{3}$,the scale factor $k = \frac{1}{3}$.
Thus,$\frac{XY}{ED} = \frac{1}{3} \implies \frac{3}{ED} = \frac{1}{3} \implies ED = 9$.
And $\frac{ZX}{FE} = \frac{1}{3} \implies \frac{6}{FE} = \frac{1}{3} \implies FE = 18$.
The perimeter of $\Delta DEF = ED + DF + FE = 9 + 12 + 18 = 39$.

(D) Given that $\Delta ABC \sim \Delta XYZ$ under the correspondence $ABC \leftrightarrow ZYX$.
This implies that the ratios of corresponding sides are equal: $\frac{AB}{ZY} = \frac{BC}{YX} = \frac{CA}{XZ} = k$.
From the given values,$AB = 3, BC = 5, CA = 6$ and $XY = 6$.
Using the correspondence $ABC \leftrightarrow ZYX$,we have the ratios: $\frac{AB}{ZY} = \frac{BC}{YX} = \frac{CA}{XZ}$.
Substituting the known values: $\frac{3}{ZY} = \frac{5}{6} = \frac{6}{XZ}$.
From $\frac{5}{6} = \frac{3}{ZY}$,we get $ZY = \frac{3 \times 6}{5} = \frac{18}{5} = 3.6$.
From $\frac{5}{6} = \frac{6}{XZ}$,we get $XZ = \frac{6 \times 6}{5} = \frac{36}{5} = 7.2$.
The perimeter of $\Delta XYZ = XY + YZ + ZX = 6 + 3.6 + 7.2 = 16.8$.

(A) In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^{\circ}$.
Given $m \angle A + m \angle B = 130^{\circ}$,substituting this into the sum gives $130^{\circ} + m \angle C = 180^{\circ}$,so $m \angle C = 50^{\circ}$.
Given $m \angle B + m \angle C = 125^{\circ}$,substituting $m \angle C = 50^{\circ}$ gives $m \angle B + 50^{\circ} = 125^{\circ}$,so $m \angle B = 75^{\circ}$.
Since $m \angle A + m \angle B = 130^{\circ}$,we have $m \angle A + 75^{\circ} = 130^{\circ}$,so $m \angle A = 55^{\circ}$.
Given the correspondence $ABC \leftrightarrow QPR$,the corresponding angles are $\angle A \leftrightarrow \angle Q$,$\angle B \leftrightarrow \angle P$,and $\angle C \leftrightarrow \angle R$.
Therefore,$m \angle Q = m \angle A = 55^{\circ}$.

(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).

(C) Given that $\Delta ABC \sim \Delta RPQ$.
By the property of similar triangles,the corresponding angles are equal.
The correspondence is $A \leftrightarrow R$,$B \leftrightarrow P$,and $C \leftrightarrow Q$.
Therefore,$m\angle A = m\angle R = 35^{\circ}$ and $m\angle B = m\angle P = 50^{\circ}$.
In $\Delta ABC$,the sum of angles is $180^{\circ}$.
$m\angle C = 180^{\circ} - (m\angle A + m\angle B) = 180^{\circ} - (35^{\circ} + 50^{\circ}) = 180^{\circ} - 85^{\circ} = 95^{\circ}$.
Since $C \leftrightarrow Q$,we have $m\angle C = m\angle Q$.
Thus,$m\angle Q = 95^{\circ}$.

(D) Given that $\Delta ABC \sim \Delta XYZ$ under the correspondence $ABC \leftrightarrow XZY$.
This implies that the ratio of the corresponding sides is equal to the ratio of their perimeters.
Therefore,$\frac{AB}{XZ} = \frac{BC}{ZY} = \frac{AC}{XY} = \frac{\text{Perimeter of } \Delta ABC}{\text{Perimeter of } \Delta XYZ}$.
Given: $\text{Perimeter of } \Delta ABC = 45$,$\text{Perimeter of } \Delta XYZ = 30$,and $AB = 21$.
Substituting the values into the ratio: $\frac{21}{XZ} = \frac{45}{30}$.
Simplifying the fraction $\frac{45}{30}$ gives $\frac{3}{2}$.
So,$\frac{21}{XZ} = \frac{3}{2}$.
Cross-multiplying gives $3 \times XZ = 21 \times 2$.
$3 \times XZ = 42$.
$XZ = \frac{42}{3} = 14$.
Thus,the length of $XZ$ is $14$.

(A) Since $\Delta PQR \sim \Delta XYZ$,the ratios of their corresponding sides are equal:
$\frac{PQ}{XY} = \frac{QR}{YZ} = \frac{PR}{XZ}$.
From the given equation $\frac{PQ}{3} = \frac{XY}{5}$,we can rearrange it to get $\frac{PQ}{XY} = \frac{3}{5}$.
Substituting this ratio into the similarity condition $\frac{PQ}{XY} = \frac{QR}{YZ}$,we get:
$\frac{3}{5} = \frac{9}{YZ}$.
Cross-multiplying,we find $3 \times YZ = 5 \times 9$,which simplifies to $3 \times YZ = 45$.
Therefore,$YZ = \frac{45}{3} = 15$.

(A) Given that $\Delta ABC \sim \Delta PQR$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$.
Substituting the given values:
$\frac{3}{PQ} = \frac{4}{6} = \frac{5}{PR}$.
First,solve for $PQ$:
$\frac{3}{PQ} = \frac{4}{6} \implies \frac{3}{PQ} = \frac{2}{3} \implies 2 \cdot PQ = 9 \implies PQ = 4.5$.
Next,solve for $PR$:
$\frac{4}{6} = \frac{5}{PR} \implies \frac{2}{3} = \frac{5}{PR} \implies 2 \cdot PR = 15 \implies PR = 7.5$.
Thus,$PQ = 4.5$ and $PR = 7.5$.

(C) Since $\Delta ABC \sim \Delta DEF$,the ratios of their corresponding sides are equal:
$\frac{AB}{DE} = \frac{AC}{DF} = \frac{BC}{EF} = k$
Given $AB = 8$,$DE = 12$,$AC = 10$,and $EF = 18$.
The ratio of similarity $k = \frac{AB}{DE} = \frac{8}{12} = \frac{2}{3}$.
Now,find the remaining sides of $\Delta DEF$:
$\frac{AC}{DF} = \frac{2}{3} \implies \frac{10}{DF} = \frac{2}{3} \implies DF = \frac{10 \times 3}{2} = 15$.
$\frac{BC}{EF} = \frac{2}{3} \implies \frac{BC}{18} = \frac{2}{3} \implies BC = \frac{18 \times 2}{3} = 12$.
Now,find the perimeter of $\Delta DEF = DE + EF + DF = 12 + 18 + 15 = 45$.

(D) Given the correspondence $ABC \leftrightarrow XZY$,the triangles are similar,i.e.,$\Delta ABC \sim \Delta XZY$.
From the property of similar triangles,the ratio of corresponding sides is equal:
$\frac{AB}{XZ} = \frac{BC}{ZY} = \frac{AC}{XY}$.
Given $AB = 6$,$XY = 12$,$YZ = 6$,and $ZX = 9$.
Using the correspondence $ABC \leftrightarrow XZY$,the sides are proportional as follows:
$\frac{AB}{XZ} = \frac{BC}{ZY} = \frac{AC}{XY}$
$\frac{6}{9} = \frac{BC}{6} = \frac{AC}{12}$
First,find $BC$: $\frac{6}{9} = \frac{BC}{6} \implies BC = \frac{6 \times 6}{9} = \frac{36}{9} = 4$.
Next,find $AC$: $\frac{6}{9} = \frac{AC}{12} \implies AC = \frac{6 \times 12}{9} = \frac{72}{9} = 8$.
The perimeter of $\Delta ABC = AB + BC + AC = 6 + 4 + 8 = 18$.

(A) Given that $\Delta ABC \sim \Delta PQR.$
According to the properties of similar triangles,the ratio of their perimeters is equal to the ratio of their corresponding sides.
Therefore,$\frac{\text{Perimeter of } \Delta ABC}{\text{Perimeter of } \Delta PQR} = \frac{AB}{PQ}.$
Substituting the given values: $\frac{16}{24} = \frac{6}{PQ}.$
Simplifying the ratio $\frac{16}{24}$ gives $\frac{2}{3}.$
So,$\frac{2}{3} = \frac{6}{PQ}.$
Cross-multiplying gives $2 \times PQ = 18.$
Thus,$PQ = \frac{18}{2} = 9.$
The correct option is $A$.

(B) In $\Delta PQR$,the sum of angles is $180^{\circ}$.
$m \angle P + m \angle Q + m \angle R = 180^{\circ}$.
$80^{\circ} + 40^{\circ} + m \angle R = 180^{\circ}$.
$120^{\circ} + m \angle R = 180^{\circ} \implies m \angle R = 60^{\circ}$.
Given the correspondence $PQR \leftrightarrow YZX$,the corresponding angles are equal:
$m \angle P = m \angle Y = 80^{\circ}$,
$m \angle Q = m \angle Z = 40^{\circ}$,
$m \angle R = m \angle X = 60^{\circ}$.
Therefore,$m \angle Z = m \angle Q = 40^{\circ}$.

(C) Given that $\Delta XYZ \sim \Delta DEF$,the corresponding angles are equal: $m \angle X = m \angle D$,$m \angle Y = m \angle E$,and $m \angle Z = m \angle F$.
In $\Delta XYZ$,the sum of angles is $180^{\circ}$,so $m \angle X + m \angle Y + m \angle Z = 180^{\circ}$.
Given $2 m \angle X = 3 m \angle Y$,we have $m \angle X = 1.5 m \angle Y$.
Substituting the values into the sum equation: $1.5 m \angle Y + m \angle Y + 30^{\circ} = 180^{\circ}$.
$2.5 m \angle Y = 150^{\circ}$.
$m \angle Y = 150^{\circ} / 2.5 = 60^{\circ}$.
Since $\Delta XYZ \sim \Delta DEF$,$m \angle E = m \angle Y = 60^{\circ}$.

(D) Given that $\Delta PQR \sim \Delta XYZ$ under the correspondence $PQR \leftrightarrow ZYX$,the ratios of their corresponding sides are equal.
This implies that $\frac{PQ}{ZY} = \frac{QR}{YX} = \frac{PR}{ZX}$.
We are given $PQ = 6$,$QR = 8$,and $XY = 12$.
From the correspondence $PQR \leftrightarrow ZYX$,the side $QR$ corresponds to $YX$.
Thus,$\frac{PQ}{ZY} = \frac{QR}{YX}$.
Substituting the given values: $\frac{6}{YZ} = \frac{8}{12}$.
Simplifying the fraction $\frac{8}{12}$ gives $\frac{2}{3}$.
So,$\frac{6}{YZ} = \frac{2}{3}$.
Cross-multiplying gives $2 \times YZ = 6 \times 3$,which means $2 \times YZ = 18$.
Therefore,$YZ = \frac{18}{2} = 9$.

(N/A) Given that $\Delta PQR \sim \Delta DEF$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{PQ}{DE} = \frac{QR}{EF} = \frac{PR}{DF}$.
From the given equation $3 PQ = 2 DE$,we get $\frac{PQ}{DE} = \frac{2}{3}$.
Now,using the ratio $\frac{PQ}{DE} = \frac{QR}{EF}$:
$\frac{2}{3} = \frac{QR}{6} \implies QR = \frac{2 \times 6}{3} = 4$.
Next,using the ratio $\frac{PQ}{DE} = \frac{PR}{DF}$:
$\frac{2}{3} = \frac{8}{DF} \implies DF = \frac{8 \times 3}{2} = 12$.
Thus,$QR = 4$ and $DF = 12$.

(B) Given that $\Delta ABC \sim \Delta PQR$.
According to the properties of similar triangles,the ratio of their perimeters is equal to the ratio of their corresponding sides.
Therefore,$\frac{\text{Perimeter of } \Delta ABC}{\text{Perimeter of } \Delta PQR} = \frac{AB}{PQ}$.
We are given that $\frac{AB}{PQ} = \frac{3}{4}$ and the perimeter of $\Delta PQR = 24$.
Let the perimeter of $\Delta ABC$ be $x$.
Substituting the values,we get $\frac{x}{24} = \frac{3}{4}$.
$x = \frac{3}{4} \times 24$.
$x = 3 \times 6 = 18$.
Thus,the perimeter of $\Delta ABC$ is $18$.

(C) Given that $\Delta PQR \sim \Delta XYZ$,the ratios of their corresponding sides are equal.
Therefore,$\frac{PQ}{XY} = \frac{QR}{YZ} = \frac{PR}{XZ}$.
From the given equation $3 QR = 4 YZ$,we can write the ratio $\frac{QR}{YZ} = \frac{4}{3}$.
Since the triangles are similar,the ratio of corresponding sides is constant,so $\frac{PR}{XZ} = \frac{QR}{YZ} = \frac{4}{3}$.
Substituting the given value $PR = 8$ into the equation,we get $\frac{8}{XZ} = \frac{4}{3}$.
Solving for $XZ$: $4 XZ = 8 \times 3$,which means $4 XZ = 24$.
Thus,$XZ = \frac{24}{4} = 6$.

(D) Since $\Delta ABC \sim \Delta XYZ$,the ratio of their corresponding sides is equal.
Therefore,$\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$.
Given that $\frac{AB}{XY} = \frac{4}{5}$ and $YZ = 20$.
Substituting the values into the ratio $\frac{AB}{XY} = \frac{BC}{YZ}$,we get:
$\frac{4}{5} = \frac{BC}{20}$.
Multiplying both sides by $20$,we get:
$BC = \frac{4}{5} \times 20$.
$BC = 4 \times 4 = 16$.
Thus,the length of $BC$ is $16$.

(A) Since $\Delta ABC \sim \Delta PQR$,the ratio of their corresponding sides is equal.
Therefore,$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = k$ (where $k$ is the scale factor).
From the properties of ratios,if $\frac{a}{b} = \frac{c}{d} = k$,then $\frac{a+c}{b+d} = k$.
Thus,$\frac{AB + BC}{PQ + QR} = \frac{AC}{PR}$.
Substituting the given values: $\frac{12}{15} = \frac{8}{PR}$.
Simplifying the fraction: $\frac{4}{5} = \frac{8}{PR}$.
Cross-multiplying gives: $4 \times PR = 5 \times 8$.
$4 \times PR = 40$.
$PR = \frac{40}{4} = 10$.
Hence,the length of $PR$ is $10$.

(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.

(N/A) In $\Delta ABC$,$A-P-B$,$A-Q-C$ and $\overleftrightarrow{PQ} \parallel \overleftrightarrow{BC}$.
By the Basic Proportionality Theorem (Thales Theorem):
$\frac{AP}{PB} = \frac{AQ}{QC}$
Given $AP = 3$,$PB = 5$,$AQ = 6$.
$\frac{3}{5} = \frac{6}{QC}$
$QC = \frac{6 \times 5}{3} = 10$.
Now,to find $AC$:
$AC = AQ + QC = 6 + 10 = 16$.
Alternatively,using the corollary $\frac{AP}{AB} = \frac{AQ}{AC}$:
$\frac{3}{3+5} = \frac{6}{AC}$
$\frac{3}{8} = \frac{6}{AC}$
$AC = \frac{6 \times 8}{3} = 16$.

(B) In $\Delta XYZ$,$XM$ is the bisector of $\angle X$,which intersects $\overline{YZ}$ at $M$.
According to the Angle Bisector Theorem,the ratio of the sides forming the angle is equal to the ratio of the segments of the opposite side.
$\therefore \frac{XY}{XZ} = \frac{YM}{MZ}$
Substituting the given values:
$\therefore \frac{6}{8} = \frac{4.2}{MZ}$
$\therefore MZ = \frac{4.2 \times 8}{6}$
$\therefore MZ = 0.7 \times 8 = 5.6$
Now,$YZ = YM + MZ$ (since $Y-M-Z$ are collinear).
$YZ = 4.2 + 5.6 = 9.8$.

(C) In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. According to the Angle Bisector Theorem:
$\frac{AB}{AC} = \frac{BD}{DC}$
Given $\frac{AB}{5} = \frac{AC}{3}$,we have $\frac{AB}{AC} = \frac{5}{3}$.
Substituting the values:
$\frac{5}{3} = \frac{4.5}{DC}$
$DC = \frac{4.5 \times 3}{5} = 0.9 \times 3 = 2.7$
Since $D$ lies on $BC$,$BC = BD + DC$.
$BC = 4.5 + 2.7 = 7.2$.

(N/A) In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$.
By the Angle Bisector Theorem,we have:
$\frac{AB}{AC} = \frac{BD}{DC}$
To find $BD$:
$\frac{AB}{AC} = \frac{BD}{DC}$
Using the property of Componendo,$\frac{AB + AC}{AC} = \frac{BD + DC}{DC} = \frac{BC}{DC}$
This does not directly give $BD$. Let's use $\frac{AB}{AC} = \frac{BD}{DC} \implies \frac{AB}{AB + AC} = \frac{BD}{BD + DC} = \frac{BD}{BC}$
$\therefore BD = \frac{BC \times AB}{AB + AC}$
To find $DC$:
$\frac{AB}{AC} = \frac{BD}{DC} \implies \frac{AC}{AB} = \frac{DC}{BD}$
Using Componendo,$\frac{AC + AB}{AB} = \frac{DC + BD}{BD} = \frac{BC}{BD}$
$\implies \frac{AC}{AC + AB} = \frac{DC}{BC}$
$\therefore DC = \frac{BC \times AC}{AB + AC}$