# Similarity of Triangles

Two triangles whose corresponding angles are equal and whose corresponding sides are proportional are said to be similar.

လိုက်ဖက်ထောင့်များ တူညီပြီး လိုက်ဖက်အနားများ အချိုးညီသော တြိဂံနှစ်ခုကို သဏ္ဌာန်တူ တြိဂံများဟုခေါ်သည်။

Exercise (8.3)

1. Use the given information to tell whether each pair of triangles is similar. Give a reason for each answer.

(a)

$\begin{array}{l} \text{ In}\ \triangle ABC,\ \angle B=80^{\circ} .\\\\ \text{ In}\ \triangle XYZ,\ \angle X=30^{\circ} .\\\\ \therefore \ \ \triangle ABC\ \text{and}\ \triangle XYZ\ \text{are not similar.} \end{array}$

(b)

$\begin{array}{l} \text{ In}\ \triangle ABE,\ \angle AEB=52^{\circ} .\\\\ \text{ In}\ \triangle DBC,\ \angle CDB=36^{\circ} .\\\\ \therefore \ \ \angle A = \angle D, \angle AEB=\angle C\\\\ \therefore \ \ \triangle ABE \sim \triangle DBC \end{array}$

(c)

$\begin{array}{l} \quad\quad\displaystyle\frac{AB}{BC}=\displaystyle\frac{16}{24}=\displaystyle\frac{2}{3}\\\\ \quad\quad\displaystyle\frac{BC}{BD}=\displaystyle\frac{24}{36}=\displaystyle\frac{2}{3}\\\\ \therefore\quad\displaystyle\frac{AB}{BC}=\displaystyle\frac{BC}{AD}\\\\ \quad\quad \angle ABC = \angle CBD = 55^{\circ}\\\\ \therefore\quad \triangle ABC \sim \triangle CBD\quad \text{(SAS)} \end{array}$

(d)

$\begin{array}{l} \quad\quad\displaystyle\frac{AC}{DF}=\displaystyle\frac{18}{6}=3\\\\ \quad\quad\displaystyle\frac{AB}{DE}=\displaystyle\frac{21}{7}=3\\\\ \quad\quad\displaystyle\frac{BC}{EF}=\displaystyle\frac{24}{8}=3\\\\ \therefore\quad\displaystyle\frac{AC}{DF}=\displaystyle\frac{AB}{DE}=\displaystyle\frac{BC}{EF}\\\\ \therefore\quad \triangle ABC \sim \triangle DEF\quad \text{(SSS)} \end{array}$

(e)

$\begin{array}{l} \quad\quad \text{ In}\ \triangle PQR\ \text{and}\ \triangle PST\\\\ \quad\quad \angle P = \angle P\quad (\text{common}\ \angle)\\\\ \quad\quad \angle PQR = \angle PST = 50^{\circ} \quad (\text{given})\\\\ \therefore\quad \triangle PQR \sim \triangle PST\quad \text{(AA)} \end{array}$

(f)

$\begin{array}{l} \quad\quad \text{ In}\ \triangle ABC\ \text{and}\ \triangle ADE\\\\ \quad\quad\displaystyle\frac{AB}{AD}=\displaystyle\frac{48}{16}=3\\\\ \quad\quad\displaystyle\frac{AC}{AE}=\displaystyle\frac{30}{10}=3\\\\ \therefore\quad\displaystyle\frac{AB}{AD}=\displaystyle\frac{AC}{AE}\\\\ \quad\quad \angle BAC = \angle DAE\quad (\text{vertically opposite}\ \angle\text{s})\\\\ \therefore\quad \triangle ABC \sim \triangle ADE\quad \text{(SAS)} \end{array}$

(g)

$\begin{array}{l} \quad\quad \text{ In}\ \triangle PQR,\\\\ \quad\quad\angle Q=180^{\circ}-\left(34^{\circ}+82^{\circ} \right)= 64^{\circ}\\\\ \quad\quad \text{ In}\ \triangle PST, \\\\ \quad\quad\angle TPS=82^{\circ}\\\\ \therefore\quad \angle PST=180^{\circ}-\left(36^{\circ}+82^{\circ} \right)= 62^{\circ}\\\\ \therefore\quad \triangle PQR\ \text{and}\ \triangle PST\ \text{are not similar}. \end{array}$

(h)

$\begin{array}{l} \quad\quad \text{ In}\ \triangle PQR\ \text{and}\ \triangle PST\\\\ \quad\quad\displaystyle\frac{PQ}{PS}=\displaystyle\frac{5}{3}\\\\ \quad\quad\displaystyle\frac{PR}{PT}=\displaystyle\frac{5}{3}\\\\ \therefore\quad\displaystyle\frac{PQ}{PS}=\displaystyle\frac{PR}{PT}\\\\ \quad\quad \angle QPR = \angle SPT\quad (\text{vertically opposite}\ \angle\text{s})\\\\ \therefore\quad \triangle PQR \sim \triangle PST\quad \text{(SAS)} \end{array}$

2. In each of the following triangles, the lengths of certain segments are marked. Find the value of $x$, $y$, $z$, $w$ and $v$.

(a)

$\begin{array}{l} \quad\quad\text{Since}\ DE\parallel BC,\\\\ \quad\quad\displaystyle\frac{CE}{EA}=\displaystyle\frac{BD}{DA}\\\\ \therefore\quad\displaystyle\frac{x}{5}=\displaystyle\frac{2}{6}\\\\ \therefore\quad x=\displaystyle\frac{5}{3}\\\\ \quad\quad\text{Since}\ DE\parallel BC,\triangle ADE \sim \triangle ABC\\\\ \therefore\quad\displaystyle\frac{DE}{BC}=\displaystyle\frac{AD}{AB}\\\\ \therefore\quad\displaystyle\frac{y}{4}=\displaystyle\frac{6}{8}\\\\ \therefore\quad y=3 \end{array}$

(b)

$\begin{array}{l} \quad\quad\text{Since}\ ST\parallel QR,\\\\ (1)\quad \displaystyle\frac{PS}{SQ}=\displaystyle\frac{PT}{TR}\\\\ (2)\quad \triangle PST \sim \triangle PQR\\\\ \therefore\quad\displaystyle\frac{PT}{PR}=\displaystyle\frac{ST}{QR}\\\\ \quad\quad\displaystyle\frac{w}{w+5}=\displaystyle\frac{2}{6}\\\\ \quad\quad 3w=w+5\\\\ \quad\quad w=\displaystyle\frac{5}{2}\\\\ \quad\quad\text{From}\ (1),\\\\ \quad\quad \displaystyle\frac{1}{z}=\displaystyle\frac{w}{5}\\\\ \therefore\quad z=\displaystyle\frac{5}{w}\\\\ \quad\quad z=\displaystyle\frac{5}{5/2}\\\\ \therefore\quad z=2 \end{array}$

(c)

$\begin{array}{l} \quad\quad\text{In}\ \triangle ABC\ \text{and}\ \triangle AED,\\\\ \quad\quad\angle A= \angle A\quad (\text{common}\ \angle)\\\\ \quad\quad\angle ACB = \angle ADE \quad (\text{given})\\\\ \therefore\quad \triangle ABC\sim \triangle AED\\\\ \therefore\quad\displaystyle\frac{AB}{AE}=\displaystyle\frac{AC}{AD}\\\\ \quad\quad\displaystyle\frac{v+4}{6}=\displaystyle\frac{8}{4}\\\\ \therefore\quad v+4 = 12\\\\ \therefore\quad v=8 \end{array}$

3. Find the marked lengths in each of the figures.

(a)

$\begin{array}{l} \quad\quad\text{Since}\ PQ\parallel AD, \triangle ABD\sim \triangle PBQ\\\\ \therefore\quad\displaystyle\frac{BD}{BQ}=\displaystyle\frac{AD}{PQ}\\\\ \quad\quad\displaystyle\frac{12+a}{12}=\displaystyle\frac{9}{6}\\\\ \therefore\quad 12+a = 18\\\\ \therefore\quad a = 6\\\\ \quad\quad\text{Similarly}\ QR\parallel BC, \triangle BCD\sim \triangle QRD\\\\ \therefore\quad\displaystyle\frac{BC}{QR}=\displaystyle\frac{BD}{QD}\\\\ \quad\quad\displaystyle\frac{b}{2}=\displaystyle\frac{12+a}{a}\\\\ \therefore\quad b = 6\\\\ \end{array}$

(b)

$\begin{array}{l} \quad\quad\text{Since}\ DE\parallel BC\\\\ (1)\quad \triangle ADE\sim \triangle ABC\\\\ (2)\quad \triangle BFC\sim \triangle EFD\\\\ \therefore\quad \displaystyle\frac{BF}{EF}=\frac{FC}{FD}=\frac{BC}{DE}\\\\ \therefore\quad \displaystyle\frac{d}{3}=\frac{5}{2}=\frac{c}{4}\\\\ \therefore\quad \displaystyle\frac{d}{3}=\frac{5}{2}\Rightarrow d=\frac{15}{2}=7.5\\\\ \quad\quad \displaystyle\frac{c}{4}=\frac{5}{2}\Rightarrow c=10\\\\ \quad\quad\text{Since}\ \triangle ADE\sim \triangle ABC, \\\\ \quad\quad\displaystyle\frac{AD}{AE}=\frac{DE}{BC}\\\\ \quad\quad\displaystyle\frac{5}{5+e}=\frac{4}{10}\\\\ \therefore\quad 5+e=\displaystyle\frac{25}{2}\\\\ \therefore\quad e=\displaystyle\frac{15}{2}=7.5\\\\ \end{array}$

(c)

ပုံပါပေးထားချက်များအရ လက်တွေ့တွင် အမှန်တကယ်ဆွဲရန် မဖြစ်နိုင်သော degenerated figure ဖြစ်နေသာကြောင့် အဖြေရှာတွက်ပြနိုင်သော်လည်း ထိုအဖြေများ လက်တွေ့တွင် မရှိပါ။ ထို့ကြောင့် ပုစ္ဆာဖြေရှင်းခြင်းသည် အဓိပ္ပာယ်မရှိတော့ပါ။

(d)

$\begin{array}{l} \quad\quad\text{Since}\ DE\parallel BC\\\\ \quad\quad \triangle ADE\sim \triangle ABC\\\\ \therefore\quad \displaystyle\frac{AD}{AB}=\frac{DE}{BC}\\\\ \therefore\quad \displaystyle\frac{8}{8+m}=\frac{10}{15}\\\\ \therefore\quad 8+m=12\\\\ \therefore\quad m=4\\\\ \quad\quad\text{Since}\ DP\parallel BQ\\\\ \quad\quad \triangle ADP\sim \triangle ABQ\\\\ \therefore\quad \displaystyle\frac{AP}{AQ}=\frac{AD}{AB}\\\\ \therefore\quad \displaystyle\frac{p}{6}=\frac{8}{8+m}\\\\ \therefore\quad \displaystyle\frac{p}{6}=\frac{8}{12}\\\\ \therefore\quad p=4\\\\ \end{array}$

(e)

$\begin{array}{l} \quad\quad\text{Since}\ DC\parallel AB, AD\parallel BE, AF\parallel BC\\\\ (1)\quad ABED\ \text{is a parallelogram.}\\\\ \therefore\quad DE=10\\\\ (2)\quad ABCF\ \text{is a parallelogram.}\\\\ \therefore\quad CF=10\\\\ (3)\quad \triangle FEO\sim \triangle FDA\\\\ \therefore\quad \displaystyle\frac{EO}{DA}=\frac{FE}{FD}\\\\ \quad\quad \displaystyle\frac{q}{14}=\frac{6}{6+10}\\\\ \quad\quad q=\displaystyle\frac{21}{4}=5.25\\\\ (4)\quad \triangle EOF\sim \triangle EBC \\\\ \therefore\quad \displaystyle\frac{OF}{BC}=\frac{EF}{EC}\\\\ \quad\quad \displaystyle\frac{r}{12}=\frac{6}{6+10}\\\\ \quad\quad r=\displaystyle\frac{9}{2}=4.5 \end{array}$

4. In the figure, $X Y \parallel P R$ and $V T \parallel Q R$. If $\displaystyle \frac{P T}{T R}=\displaystyle \frac{3}{2}, \displaystyle \frac{Q Y}{Y R}=\displaystyle \frac{2}{1}$ and $P Q=15 \mathrm{~cm}$, calculate

(a) the lengths of $P V, P X$ and $X V$.

(b) the numerical values of $\displaystyle \frac{Y W}{W X}$ and $\displaystyle \frac{V W}{Q Y}$.

$\begin{array}{ll} \text{(a)} & \quad\quad\text{Since}\ VT\parallel QR, \\\\ & \quad\quad\displaystyle\frac{PV}{VQ}=\frac{PT}{TR}\\\\ & \quad\quad\displaystyle\frac{PV}{15-PV}=\frac{3}{2}\\\\ & \quad\quad2 PV=45 - 3 PV\\\\ & \therefore\quad5 PV=45\\\\ & \quad\quad\therefore \quad PV=9\ \text{cm}\\\\ & \quad\quad\text{Since}\ XY\parallel PR, \\\\ & \quad\quad\displaystyle\frac{XQ}{PX}=\frac{QY}{YR}\\\\ & \quad\quad\displaystyle\frac{15-PX}{PX}=\frac{2}{1}\\\\ & \quad\quad 2PX=15-PX\\\\ & \quad\quad 3 PX=15\\\\ & \therefore\quad PX=5\ \text{cm}\\\\ & \therefore\quad XV=PV - PX= 9-5=4\ \text{cm}\\\\ & \therefore\quad QV=15-PV= 15-9=6\ \text{cm}\\\\ \text{(b)} & \text{Since}\ VT\parallel QR, \\\\ & \quad\quad\displaystyle\frac{YW}{WX}=\frac{QV}{VX}\\\\ & \therefore\quad\displaystyle\frac{YW}{WX}=\frac{6}{4}=\frac{3}{2}\\\\ & \text{Since}\ VT\parallel QR, \\\\ & \quad\quad \triangle XVW\sim \triangle XQY\\\\ & \therefore\quad\displaystyle\frac{VW}{QY}=\frac{XV}{XQ}\\\\ & \therefore\quad\displaystyle\frac{VW}{QY}=\frac{4}{10}=\frac{2}{5}\\\\ \end{array}$

5. Given: Parallelogram $B I R D$,

$I G$ bisects $\angle B I R$

Prove: $\displaystyle\frac{B E}{E I}=\frac{R G}{G I}$

$\begin{array}{l} \quad\quad \text{Since}\ BIRD\ \text{is a parallelogram}, \angle B = \angle R\\\\ \quad\quad \text{Since}\ IG\ \text{bisects}\ \angle BIR, \angle BIE = \angle RIG\\\\ \therefore\quad \triangle BIE\sim \triangle RIG\\\\ \therefore\quad\displaystyle\frac{BE}{RG}=\frac{EI}{GI}\\\\ \therefore\quad\displaystyle\frac{BE}{EI}=\frac{RG}{GI}\\\\ \end{array}$

6. Given : $R Q \perp P Q$

$P Q \perp P T$

$S T \perp P R$

Prove: $S T \cdot R Q=P S \cdot P Q$

$\begin{array}{l} \quad\quad \text{Since}\ P Q \perp P T\ \text{and}\ P Q \perp RQ,\\\\ \quad\quad PT\parallel QR\\\\ \therefore\quad \angle TPS = \angle PRQ\quad (\text{alternate}\ \angle\text{s})\\\\ \quad\quad \angle TSP = \angle PQR\quad (\text{alternate}\ \angle\text{s})\\\\ \therefore\quad \triangle PST\sim \triangle RQP\\\\ \therefore\quad\displaystyle\frac{ST}{PQ}=\frac{PS}{QR}\\\\ \therefore\quad ST\cdot RQ = PS\cdot PQ \end{array}$

7. Given : Parallelogram $A B C D$ ; $P Q \parallel M B$

Prove : $\triangle A B M \sim \triangle C Q P$ .