Similarity

Chapter 8

Similarity

Some Properties of Proportions

1. The Means-extremes Product Property:
   If $\dfrac{a}{b}=\dfrac{c}{d}$, then $ad=bc$
   where a,d are extremes and b,c are means.
   If $\dfrac{a}{b}=\dfrac{c}{d}$, then b is called the geometric mean of a and c.


2. Invertendo Property:
   If $\dfrac{a}{b}=\dfrac{c}{d}$ then $\dfrac{b}{a}=\dfrac{d}{c}$


3. Alternando Property:
   If $\dfrac{a}{b}=\dfrac{c}{d}$, then $\dfrac{a}{c}=\dfrac{b}{d}$ (or $\dfrac{d}{b}=\dfrac{c}{a}$)


4. Componendo Property:
   If $\dfrac{a}{b}=\dfrac{c}{d}$, then $\dfrac{a+b}{b}=\dfrac{c+d}{d}$


5. Dividendo Property:
   If $\dfrac{a}{b}=\dfrac{c}{d}$, then $\dfrac{a-b}{b}=\dfrac{c-d}{d}$


6. Componendo and Dividendo Property:
   If $\dfrac{a}{b}=\dfrac{c}{d}$, then $\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}$



8.1 Ideas of Similarities and Similar Triangles

Geometric figures that are of the same shape but not necessarily of the same size are said to be similar figures.
ပုံသဏ္ဌာန်ဆင်တူပြီးအရွယ်တူရန်မလိုသော geometry ပုံများကိုသဏ္ဌာန်တူပုံများဟုခေါ်သည်။

The conditions for two polygons to be similar are

1. corresponding angles must be equal, and
2. corresponding sides must be proportional.
ဗဟုဂံနှစ်ခုသဏ္ဌာန်တူရန်လိုအပ်ချက်များမှာ

1. လိုက်ဖက်ထောင့်များတူညီရမည်။
2. လိုက်ဖက်အနားများ၏အလျားများအချိုးတူရမည်။

Definition

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




8.2 The Basic Proportionality Theorem

Postulate 1

If three parallel lines intersect two transversals, then the lines divide the transversals proportionally.
မျဉ်းပြိုင်သုံးကြောင်းသည်ဖြတ်မျဉ်းနှစ်ကြောင်းကိုပိုင်းဖြတ်လျှင်ထိုမျဉ်းများသည်ဖြတ်မျဉ်းများကိုအချိုးတူစွာပိုင်းဖြတ်ကြ၏။

if l∥m∥n, then $\dfrac{a}{b}=\dfrac{c}{d}$.

Theorem 1
(The Basic Proportionality Theorem-BPT)

If a line intersecting the interior of a triangle is parallel to one side, then the line divides the other two sides proportionally. တြိဂံတစ်ခု၏အတွင်းပိုင်းကိုဖြတ်၍အနားတစ်ဖက်နှင့်အပြိုင်ဆွဲသောမျဉ်းသည်ကျန်အနားနှစ်ဖက်ကိုအချိုးတူစွာပိုင်းဖြတ်သည်။ In △ABC, if DE ∥BC, then $\dfrac{AD}{DB}=\dfrac{AE}{EC}$

Corollary 1.1

Using properties of proportions, it can be shown that the following three proportions are equalivalent that is they have the same value. အချိုးတူဂုဏ်သတ္တိများကိုအသုံးပြု၍အောက်ပါအချိုးသုံးခုသည်တန်ဖိုးအတူတူပင်ဖြစ်ကြောင်းပြသနိုင်သည်။ $\dfrac{AD}{DB}=\dfrac{AE}{EC}$ $\dfrac{AD}{AB}=\dfrac{AE}{AC}$ $\dfrac{AB}{DB}=\dfrac{AC}{EC}$ The following corollary is the converse of the BPT. Corollary 1.1 သည် BPT ၏အပြန်အလှန်ဖြစ်သည်။

Corollary 1.2 (CBPT)

If a line divides two sides of a triangle proportionally, then the line is parallel to the third side. တြိဂံတစ်ခု၏အနားနှစ်ဖက်ကိုအချိုးတူစွာပိုင်းဖြတ်သောမျဉ်းသည်ကျန်တတိယအနားနှင့်ပြိုင်၏။ In △PQR, $\dfrac{PX}{XQ}=\dfrac{PY}{YR}$ then XY ∥ PR.




8.3 Basic Theorems on Similar Triangles

Theorem 2
(The AAA Similarity Theorem)

If the angles of a triangle are equal to the angles of another triangles, then the two triangles are similar. တြိဂံတစ်ခု၏ထောင့်သုံးခုသည်အခြားတြိဂံ၏ထောင့်သုံးခုနှင့်ထပ်တူညီကြလျှင်ယင်းတြိဂံနှစ်ခုတို့သဏ္ဌာန်တူကြ၏။ In △ ABC and △ DEF, if ∠A = ∠D, ∠B = ∠E, ∠C = ∠F then △ ABC ∼ △ DEF.

Corollary 2.1
(The AA Corollary)

If two angles of a triangle are equal to two angles of another triangle, then the triangles are similar. တြိဂံတစ်ခု၏ထောင့်နှစ်ခုသည်အခြားတြိဂံ၏ထောင့်နှစ်ခုနှင့်ထပ်တူညီကြလျှင်ယင်းတြိဂံနှစ်ခုတို့သဏ္ဌာန်တူကြ၏။

Corollary 2.2

If a line parallel to one side of a triangle determines a second triangle, then the second triangle will be similar to the original triangle. တြိဂံတစ်ခု၏အနားတစ်ဖက်နှင့်အပြိုင်ဆွဲသောမျဉ်းကြောင့်ဖြစ်ပေါ်လာသောဒုတိယတြိဂံသည်မူလတြိဂံနှင့်သဏ္ဌာန်တူသည်။ If XY ∥ BC, then △ AXY ∼ △ ABC.

Theorem 3
(The SAS Similarity Theorem)

If an angle of a triangle is equal to an angle of another triangle, and the sides including these angles are proportional, then the triangles are similar. တြိဂံတစ်ခု၏ထောင့်တစ်ထောင့်သည်အခြားတြိဂံ၏ထောင့်တစ်ထောင့်နှင့်ထပ်တူညီပြီးယင်းထောင့်များကိုဆောင်သောအနားများ၏အလျားများအချိုးတူကြလျှင်ယင်းတြိဂံတို့သည်သဏ္ဌာန်တူကြ၏။ In △ ABC and △ DEF, if ∠A = ∠D and $\dfrac{AB}{DE}=\dfrac{AC}{DF}$, then △ ABC ∼ △ DEF.

Theorem 4
(The SSS Similarity Theorem)

If the corresponding sides of two triangles are proportional, then the triangles are similar. တြိဂံနှစ်ခု၏လိုက်ဖက်အနားများ၏အလျားများအချိုးတူကြလျှင်ယင်းတြိဂံတို့သဏ္ဌာန်တူကြ၏။ In △ ABC and △ DEF, if $\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{CA}{FD}$, then △ ABC ∼ △ DEF.




8.4 The Angle Bisector Theorem

If B is a point on the line containing segment AC, then $\dfrac{AB}{BC}$ is the ratio in which B divides AC.
B သည် မျဉ်းပြတ် AC ပေါ်ရှိအမှတ်တစ်ခုဖြစ်လျှင်အချိုး $\dfrac{AB}{BC}$ ဆိုသည်မှာ B က AC ကိုပိုင်းဖြတ်သောအချိုးပင်ဖြစ်သည်။
(i)

$\dfrac{AB}{BC}=\dfrac{2}{4}=\dfrac{1}{2}$
B divides AC internally in the ratio 1 ∶ 2.
B က AC ကို 1 ∶ 2 ဖြင့်အတွင်းပိုင်းပိုင်းဖြတ်သည်။

(ii)

$\dfrac{AB}{BC}=\dfrac{2}{4}=\dfrac{1}{2}$
B divides AC externally in the ratio 1 ∶ 2.
B က AC ကို 1 ∶ 2 ဖြင့်အပြင်ပိုင်းပိုင်းဖြတ်သည်။


Theorem 5
(The Angle Bisector Theorem- ABT)

The bisector of an interior (exterior) angle of a triangle divides the opposite side internally (externally) into a ratio equal to the ratio of the other two sides of the triangle. တြိဂံတစ်ခု၏အတွင်း(အပြင်)ထောင့်ကိုထက်ဝက်ပိုင်းသောမျဉ်းသည်မျက်နှာချင်းဆိုင်အနားကိုကျန်အနားနှစ်ဖက်တို့၏အချိုးနှင့်တူညီသောအချိုးဖြင့်အတွင်း(အပြင်)၌ပိုင်းဖြတ်၏။ In △ ABC, if AD bisects ∠BAC (B'AC), then $\dfrac{AB}{AC}=\dfrac{BD}{DC}$




8.5 The Pythagoras Theorem

A triangle with a right angle is called a right triangle and the sides which determine the right angle are called legs of the right triangle, and the side opposite the right angle is called the hypotenuse.
ထောင့်မှန်တစ်ခုပါသောတြိဂံကိုထောင့်မှန်တြိဂံဟုခေါ်သည်။ထောင့်မတ်ကျနေသောအနားများကိုထောင့်မှန်ဆောင်အနားများဟုခေါ်ပြီးထောင့်မှန်နှင့်မျက်နှာချင်းဆိုင်သောအနားကိုထောင့်မှန်ခံအနားဟုခေါ်သည်။



Theorem 6

The altitude to the hypotenuse of right triangle forms two triangles that are similar to each other and to the original triangle. ထောင့်မှန်တြိဂံတစ်ခုတွင်ထောင့်မှန်ခံအနားပေါ်သို့အမြင့်မျဉ်းဆွဲလျှင်ဖြစ်ပေါ်လာသောတြိဂံနှစ်ခုသည်အချင်းချင်းသဏ္ဌာန်တူပြီးမူလတြိဂံနှင့်လည်းသဏ္ဌာန်တူသည်။ In a triangle ABC with ∠ ADC=90°, if AD is an altitude then △ BAC ∼ △ADB ∼ △ADC.




Geometric Mean (GM)

If $\dfrac{b}{a}=\dfrac{c}{d}$ ie $b^{2}=ac$ then b is called the geometric mean between a and c.
Corollary 6.1

The altitude to the hypotenuse of a right triangle is the geometric mean of the segments into which it seperates the hypotenuse, and each leg of a right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to the leg. ထောင့်မှန်တြိဂံတစ်ခုတွင်ထောင့်မှန်ခံအနားပေါ်သို့ဆွဲသောအမြင့်မျဉ်း၏အလျားသည်ယင်းအမြင့်မျဉ်းကထောင့်မှန်ခံအနားကိုပိုင်းဖြတ်ထားသောမျဉ်းဖြတ်များ၏အလျားများအတွင်းရှိ geometric mean ဖြစ်၏။ ထောင့်မှန်ဆောင်အနားတစ်ခုစီ၏အလျားသည်ထောင့်မှန်ခံအနား၏အလျားနှင့်ထောင့်မှန်ခံအနားပေါ်ရှိယင်းထောင့်မှန်အနားနှင့်နီးစပ်သောမျဉ်းပြတ်အလျား၏ geometric mean ဖြစ်၏။ In a triangle ABC with ∠ ABC=90°, if BD is an altitude then $BD^{2}=AD . DC$ $AB^{2}=AD . AC$ $BC^{2}=CD . AC$

Theorem 7
(Pythagoras Theorem)

In any right triangle, the square of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. မည်သည့်ထောင့်မှန်တြိဂံတစ်ခုတွင်မဆိုထောင့်မှန်ခံအနား၏အလျားနှစ်ထပ်ကိန်းသည်ကျန်အနားနှစ်ဖက်၏အလျားများနှစ်ထပ်ကိန်းပေါင်းလဒ်နှင့်တူညီ၏။ $b^{2}=c^{2}+a^{2}$




Converse of Pythagoras Theorem

If a triangle has sides with lengths a, b, c and $a^{2}+b^{2}=c^{2}$, then the triangle is a right triangle.
တြိဂံတစ်ခု၏အနားများ၏အလျားများသည် a, b, c ဖြစ်ကြပြီး $a^{2}+b^{2}=c^{2}$ ဖြစ်လျှင်ယင်းတြိဂံသည်ထောင့်မှန်တြိဂံတစ်ခုဖြစ်၏။
∠ ACB=90° ဖြစ်၏။



8.6 Special Right Triangles

There are two special types of right triangles.
ထောင့်မှန်တြိဂံအထူးပုံစံနှစ်ခုရှိသည်။

(i)

The isosceles right triangle (45°- 45° right triangle):A triangle is formed by two sides and a diagonal of a square.
စတုရန်းတစ်ခုတွင်ထောင့်ဖြတ်မျဉ်းတစ်ကြောင်းဆွဲခြင်းဖြင့်နှစ်နားညီအထူးထောင့်မှန်တြိဂံတစ်ခုရရှိသည်။

(ii)

(30°- 60° right triangle):An altitude of an equilateral triangle determine the right triangle with acute angles of measures 30° and 60°.
30° နှင့် 60° ရှိသောထောင့်ကျဉ်းနှစ်ခုနှင့်သုံးနားညီတြိဂံ၏အမြင့်မျဉ်းဖြင့်ဖွဲ့စည်းထားသောအထူးထောင့်မှန်တြိဂံတစ်ခုဖြစ်သည်။


Theorem 8

In a 45°- 45° right triangle, the length of hypotenuse is equal to the length of each leg times $\sqrt{2}$. နှစ်နားညီထောင့်မှန်တြိဂံတစ်ခုတွင်ထောင့်မှန်ခံအနား၏အလျားသည်ထောင့်မှန်ဆောင်အနားတစ်ခုစီ၏အလျား၏ $\sqrt{2}$အဆရှိ၏။

Theorem 9

In 30°- 60° right triangle, the leg opposite the 30° angle is one-half the length of the hypotenuse, and the other leg is equal to the length of the hypotenuse time $\dfrac{\sqrt3}{2}$. 30°- 60° ထောင့်မှန်တြိဂံတစ်ခုတွင် 30° ထောင့်နှင့်မျက်နှာချင်းဆိုင်သောအနား၏အလျားသည်ထောင့်မှန်ခံအနား၏အလျားတစ်ဝက်ဖြစ်ပြီးကျန်အနားတစ်ဖက်၏အလျားသည်ထောင့်မှန်ခံအနားအလျား၏$\dfrac{\sqrt3}{2}$ အဆရှိ၏။




Problems

Example 1

In the following figure if AP ∥ BQ ∥ CR, find the value of x.




$A P \| B Q \| C R$, $\begin{aligned} \dfrac{x}{10}&=\dfrac{3}{15}\\ \mathrm{15x}&=30\\ \mathrm{x}&=2 \end{aligned}$





Example 2

Determine whether the following pairs of triangle are similar or not. If they are similar, state why.



(a) $\triangle A O B$ and $\triangle C O D$ are similar because $\angle A O B=\angle D O C$. $\begin{aligned} &\dfrac{A O}{D O}=\dfrac{32}{16}=\dfrac{2}{1}\\[2mm] &\dfrac{O B}{O C}=\dfrac{20}{10}=\dfrac{2}{1}\\ &\dfrac{A O}{D O}=\dfrac{O B}{O C}\\ &\triangle A B C \sim \triangle D O C\quad(SAS) \end{aligned}$ (b) $\triangle A B C$ and $\triangle D E F$ are similar because $\begin{aligned} &\dfrac{D E}{A B}=\dfrac{7}{14}=\dfrac{1}{2} \\ &\dfrac{E F}{B C}=\dfrac{8}{16}=\dfrac{1}{2} \\ &\dfrac{D F}{A C}=\dfrac{6}{12}=\dfrac{1}{2} \\ &\dfrac{D E}{A B}=\dfrac{E F}{B C}=\dfrac{D F}{A C} \\ &\triangle A B C \sim \triangle D E F \quad \text { (SSS) } \end{aligned}$ (c) $\triangle A B C$ and $\triangle D E F$ are not similar because $\begin{aligned} &\dfrac{A C}{E F}=\dfrac{16}{9} \\ &\dfrac{B C}{D F}=\dfrac{9}{5} \\ &\dfrac{A B}{E D}=\dfrac{5 \sqrt{7}}{2 \sqrt{14}}=\dfrac{5 \sqrt{2}}{4} \\ &\dfrac{A B}{E D} \neq \dfrac{B C}{D F}\\ &\dfrac{A C}{E F} \neq \dfrac{B C}{D F}\\ \end{aligned}$ (d) $\triangle A B C$ and $\triangle D E F$ are similar because $\begin{aligned} &\angle C=180^{\circ}-\left(45^{\circ}+78^{\circ}\right)=57^{\circ} \\ &\angle E=180^{\circ}-\left(45^{\circ}+57^{\circ}\right)=78^{\circ} \\ &\angle A=\angle D, \angle B=\angle E, \angle C=\angle F \\ &\triangle D E F \sim \triangle A B C \quad\text { (AAA) } \end{aligned}$





Example 3

Find the value of angle D.



In $\triangle A B C$ and $\triangle D E F$, $\begin{aligned} &\dfrac{A C}{D E}=\dfrac{4}{8}=\dfrac{1}{2} \\ &\dfrac{A B}{D E}=\dfrac{3.5}{7}=\dfrac{1}{2} \\ &\dfrac{B C}{E F}=\dfrac{5.5}{11}=\dfrac{1}{2} \\ &\dfrac{A C}{D F}=\dfrac{A B}{D E}=\dfrac{B C}{E F} \\ &\triangle A B C \sim \triangle D E F\quad(S S S) \\ &\angle D=80^{\circ} \end{aligned}$





Exercise (8.1)

1. State why the two polygons are, or are not similar.
In the given figure, find the values of x and y.



(a) $\begin{aligned} \dfrac{A B}{P Q}&=\dfrac{6}{3}=\dfrac{2}{1}\\ \dfrac{B C}{Q R}&=\dfrac{2}{2}=\dfrac{1}{1} \\ \dfrac{A B}{P Q} &\neq \dfrac{B C}{Q R}\\ \end{aligned}$ $\text{ABCD and PQRS are not similar.}$ (b) $\angle A=\angle P, \angle B=\angle Q, \angle C=\angle R, \angle D=\angle S$ $\dfrac{A B}{P Q}=\dfrac{B C}{Q R}=\dfrac{C D}{R S}=\dfrac{D A}{S P}=\dfrac{3}{7}$ $\text{ABCD and PQRS are similar.}$ (c) $\begin{aligned} \angle A &=\angle P, \angle B=\angle Q, \angle C=\angle R=60^{\circ} \\ \dfrac{A B}{P Q} &=\dfrac{B C}{Q R}=\dfrac{A C}{P R}=\dfrac{3}{5}\\ \end{aligned}$ $\triangle A B C$ and $\triangle P Q R$ are similar. (d) ABCDE is a pentagon. PQRSTU is a hexagon. They are not similar.




2. Complete the proportions.



(a) If $\triangle A B C \sim \triangle D E F$ then $\dfrac{A B}{?}=\dfrac{B C}{?}=\dfrac{?}{D F}$. (b) If $\triangle G H I \sim \triangle K L M$ then $\dfrac{?}{H I}=\dfrac{?}{G H}=\dfrac{?}{G I}$. solution (a) If $\triangle A B C \sim \triangle D E F$, then $\dfrac{A B}{D E}=\dfrac{B C}{E F}=\dfrac{A C}{D F}$. (b) If $\triangle G H I \sim \triangle K L M$, then $\dfrac{L M}{H I}=\dfrac{K L}{G H}=\dfrac{K M}{G I}$.




3. State whether the proportions are correct for the indicated similar triangles.



(a) $\triangle A B C \sim \triangle X Y Z$ $\dfrac{A B}{X Y}=\dfrac{B C}{Y Z} \quad$ (correct) (b) $\triangle D E F \sim \triangle H I J$ $\dfrac{D E}{H I}=\dfrac{E F}{I J}\quad$(correct) (c) $\triangle R S T \sim \triangle L M K$ $\dfrac{R T}{L M}=\dfrac{S T}{M K}\quad$(incorrect) (d) $\triangle X Y Z \sim \Delta U V W$ $\dfrac{X Y}{U V}=\dfrac{X Z}{V W}\quad$(incorrect)



$\begin{aligned} 4.\text{ Given } :& △ P Q R ∼ △ U V W \\ &\text{ and lengths of sides are as marked }.\\ \text{ Find } :& \text{ The values of x and y }. \end{aligned}$




$\triangle P Q R \sim \triangle U V W$ $\begin{aligned} \dfrac{P R}{U W}&=\dfrac{P Q}{U V}=\dfrac{Q R}{V W} \\ \dfrac{12}{9}&=\dfrac{7}{x}=\dfrac{10}{y} \\ \dfrac{12}{9}&=\dfrac{7}{x} \\ 12 {x}&=63\\ \therefore\ x&=\dfrac{21}{4}=5.25,\\ \dfrac{12}{9}&=\dfrac{10}{y} \\ 12 y&=90 \\ y&=\dfrac{15}{2}=7.5\\ \end{aligned}$



5.The measures of two angles of △ XYZ are 82° and 16°. Find the measures of the angles of a triangle similar to △ XYZ.



The measures of $\triangle X Y Z$ are $82^{\circ}, 16^{\circ}$ and $180^{\circ}-\left(82^{\circ}+16^{\circ}\right)=180^{\circ}-98^{\circ}=82^{\circ}$ The measures of the angles of a triangle similar to $\triangle X Y Z$ are $82^{\circ}, 16^{\circ}$ and $82^{\circ}$.





Exercise (8.2)

1. Use the Basic Proportionality Theorem (BPT) and its corollary to complete the proportions for the adjoining figures.



In $\triangle A O B, T S \| O A$. (a) $\dfrac{O T}{T B}=?$ (b) $\dfrac{S A}{B A}=$ ? solution (a) $\dfrac{O T}{T B}=\dfrac{A S}{S B}$ (b) $\dfrac{S A}{B A}=\dfrac{T O}{B O}$ In $\triangle R V L, E U \|L V$. (c) $\dfrac{E L}{R E}=$ ? (d) $\dfrac{R U}{R V}=?$ solution (c) $\dfrac{E L}{R E}=\dfrac{U V}{R U}$ (d) $\dfrac{R U}{R V}=\dfrac{R E}{R L}$




2. To find the value of x in the figure, a student wrote the proportions $\dfrac{x}{10}=\dfrac{4}{8}$.
   (a) Is this correct?
   (b) Another student wrote the proportion $\dfrac{x}{x-4}=\dfrac{4}{8}$.
   Is this correct?
   (c) Write a simpler proportion that will give the correct answer.



(a) $\dfrac{x}{10}=\dfrac{4}{8}$ is incorrect. (b) $\dfrac{x}{x-4}=\dfrac{4}{8}$ is incorrect. (c) $\begin{aligned} \dfrac{x}{10}&=\dfrac{4}{4+8}\\ \dfrac{x}{10}&=\dfrac{1}{3} \end{aligned}$




3. Find the value of x in each of the figure below.(They are not drawn to scales.)



(a) $\dfrac{x}{4}=\dfrac{3}{5}$ $x=\dfrac{12}{5}=2.4$ (b) $\begin{aligned} \dfrac{x}{24}&=\dfrac{15}{15+21}\\ x &=\dfrac{15}{36} \times 24 \\ &=10\\ \end{aligned}$ (c) $\dfrac{x}{4}=\dfrac{16}{x}$ $\begin{aligned} &x^{2}=64 \\ &x=8\\ \end{aligned}$ (d) $\dfrac{x}{3}=\dfrac{6+4}{4}$ $x=\dfrac{30}{4}=7.5$




4. Given △ PQR with ST∥PQ and lengths of segments are marked. Which of the following proportions are correct?



(a) $\dfrac{b}{a}=\dfrac{d}{c}$  (correct)(Theorem1 ) (b) $\dfrac{a+b}{a}=\dfrac{c+d}{d}$  (incorrect) (c) $\dfrac{c}{d+c}=\dfrac{a}{b+a}$ (correct)(Corollary 1.1 ) (d) $\dfrac{a}{c}=\dfrac{b}{d}$  (correct)(alternando) (e) $\dfrac{a}{b}=\dfrac{c}{d} $ (correct)(Theorem1 ) (f) $\dfrac{a-b}{b}=\dfrac{c-d}{d}$ (incorrect)




5. If PQ ∥ LM ∥ AB, and the lengths are as shown, how long are the segments MQ and BM?



P Q ∥ L M $\begin{aligned} \dfrac{M Q}{Q C}& =\dfrac{L P}{P C} \\ \dfrac{M Q}{6} &=\dfrac{5}{5} \\ M Q& =6\\ \end{aligned}$ L M ∥ A B $\begin{aligned} \dfrac{B M}{M C}& =\dfrac{A L}{L C} \\ \dfrac{B M}{6+6}& =\dfrac{3}{10} \\ B M &=\dfrac{18}{5}=3.6\\ \end{aligned}$




6. If the segments in the figure have the lengths indicated, is UV ∥ RT? Justify your answer.



$\begin{aligned} \text { US }&=9-7=2 \\ \text { VT }&=18-4=14 \\ \dfrac{\text { US }}{R U}&=\dfrac{2}{7} \\ \dfrac{\text { VS }}{\text { VT }}&=\dfrac{4}{14}=\dfrac{2}{7} \\ \dfrac{\text { US }}{\text { RU }}&=\dfrac{\text { VS }}{\text { VT }} \\ \end{aligned}$ $\therefore \text { UV } \| R T$




7. Given the figure as marked with ST ∥ PQ, find the lengths of the segments PS, SR, RT and TQ.



$S T ∥ P Q$ $\begin{aligned} \frac{R S}{S P}&=\frac{R T}{T Q}\\ \frac{x+3}{2 x}&=\frac{3 x-6}{2 x+7}\\ (x+3)(2 x+7)&=2 x(3 x-6) \\ 2 x^{2}+{13 x+21}&=6 x^{2}-12 x \\ 4 x^{2}-25 x-21&=0 \\ (4 x+3)(x-7)&=0 \\ 4 x+3&=0 \text { (or) } x-7=0 \\ x&=-\frac{3}{4} \text { (or) } x=7 \\ &(impossible) \end{aligned}$ $\begin{aligned} x &=7 \\ P S &=2 x=14 \\ S R &=x+3=10 \\ R T &=3 x-6=15 \\ T Q &=2 x+7=21 \end{aligned}$



$\begin{aligned} 8.\text{ Given } :&\dfrac{AD}{DB}=\dfrac{2}{1}\\ \text{ Prove } :&\dfrac{AX}{XB}=\dfrac{3}{1}\\ \end{aligned}$




$\begin{aligned} \text { Given } :&\dfrac{A D}{D B}=\dfrac{2}{1} \\ \text { Prove } :& \dfrac{A X}{X B}=\dfrac{3}{1} \\ \text { Proof } :& \text{ D E } ∥ \text{ B C } \\ &\dfrac{A E}{E C}=\dfrac{A D}{D B}=\dfrac{2}{1}\\ &\text{ B E } ∥ \text{ X C }\\ &\dfrac{A B}{B X}=\dfrac{A E}{E C}=\dfrac{2}{1} \\ &\dfrac{A B+B X}{B X}=\dfrac{2+1}{1}\quad(\text { componendo } ) \\ &\dfrac{A X}{X B}=\dfrac{3}{1} \end{aligned}$



$\begin{aligned} 9.\text{ Given } :& A E = E B, \dfrac{BD}{DC}=\dfrac{2}{3}\\ \text{ Find } :& \text{ The ratio } \dfrac{CH}{CE}\\ \end{aligned}$





$\begin{aligned} \text { Given } :& A E=E B, \dfrac{B D}{D C}=\dfrac{2}{3} \\ \text { Find } :& \dfrac{C H}{C E}=\text { ? } \\ \textbf { solution } :& \text { Draw } E F \| A D \text { cut } B D \text { at } F . \\ &\dfrac{B F}{F D}=\dfrac{B E}{E A}=1 \\ & B F = F D \\ &\text { Let } B D=2 x, D C=3 x \text { where } x \text { is constant. } \\ & B F=F D=x \\ &\text { In } \triangle C E F, \\ &\text HD \| E F . \\ & \dfrac{C H}{C E}=\dfrac{C D}{C F}=\dfrac{3 x}{4 x}=\dfrac{3}{4}\\ \end{aligned}$





Exercise (8.3)

1. Use the given information to tell whether each pair of triangles is similar or not. Give a reason for each answer.



(a) $\begin{aligned} \angle A &=180^{\circ}-\left(20^{\circ}+80^{\circ}\right)=80^{\circ} \\ \angle D &=180^{\circ}-\left(70^{\circ}+80^{\circ}\right)=30^{\circ} \end{aligned}$ Only one angle is equal and the other two angles are not equal. $\triangle A B C$ is not similar to $\triangle D E F$. (b) $ \begin{aligned} \angle A B C &=90^{\circ}-38^{\circ}=52^{\circ} \\ \angle A B C &=\angle D \\ \angle B C A &=\angle D C E=90^{\circ} \\ \triangle A B C & \sim \triangle E D C \quad(A A \operatorname{Corollary}) \end{aligned}$ (c) $\begin{aligned} \angle D A C &=\angle C A B=55^{\circ} \\ \dfrac{A B}{A C} &=\dfrac{36}{24}=\dfrac{3}{2} \\ \dfrac{A C}{A D} &=\dfrac{24}{16}=\dfrac{3}{2} \\ \dfrac{A B}{A C} &=\dfrac{A C}{A D} \\ \triangle A B C & \sim \triangle A C D \quad\text { (SAS) } \end{aligned}$ (d) $ \begin{aligned} \dfrac{A B}{D E} &=\dfrac{21}{7}=\dfrac{3}{1} \\ \dfrac{B C}{E F} &=\dfrac{24}{8}=\dfrac{3}{1} \\ \dfrac{C A}{F D} &=\dfrac{18}{6}=\dfrac{3}{1} \\ \dfrac{A B}{D E} &=\dfrac{B C}{E F}=\dfrac{C A}{F D} \\ \triangle A B C & \sim \triangle D E F\quad(S S S) \end{aligned}$ (e) $ \begin{aligned} \angle A &=\angle A\quad( \text{ common angle } ) \\ \angle B &=\angle A D E=50^{\circ} \\ \end{aligned}$ $\triangle A B C \sim \triangle A D E $  (A A Corollary) (f) $ \begin{aligned} \quad \frac{A B}{A D} &=\frac{48}{16}=\frac{3}{1} \\ \dfrac{A C}{A E} &=\dfrac{30}{10}=\dfrac{3}{1} \\ \dfrac{A B}{A D} &=\dfrac{A C}{A E} \\ \angle B A C &=\angle D A E \quad( \text{ vertically opposite angle } ) \\ \end{aligned}$ $\triangle A B C \sim \triangle A D E \quad\text { (SAS) }$ (g) $\begin{aligned} \angle A &=180^{\circ}-\left(34^{\circ}+8 1^{\circ}\right)=64^{\circ} .\\ \angle E &=180^{\circ}-\left(36^{\circ}+8 1^{\circ}\right)=62^{\circ} \end{aligned}$ Only one angle is equal and the other two angles are not equal. $ \triangle A B C$ is not similar to $\triangle D C E$. (h) $\begin{aligned} \angle B A C&=\angle D A E \quad (\text{ vertically ooposite angle })\\ \dfrac{A B}{A D}&=\dfrac{3}{5}\\ \dfrac{A C}{A E}&=\dfrac{3}{5}\\ \dfrac{A B}{A D}&=\dfrac{A C}{A E}\\ \end{aligned}$ $\triangle A B C \sim \triangle A D E$  (SAS)



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



(a) $DE ∥BC$ $\begin{aligned} \dfrac{x}{5}&=\dfrac{2}{6} \\ x&=\dfrac{10}{6}=1 \dfrac{2}{3} \end{aligned}$ $\triangle A D E \sim \triangle A B C$ ( Corollary $2.2$ ) $\begin{aligned} \dfrac{y}{4}&=\dfrac{6}{6+2} \\ y&=\dfrac{24}{8}=3 \end{aligned}$ (b) $DE ∥ B C$ $\triangle A D E \sim \triangle A B C$ ( Corollary $2.2$ ) $\begin{aligned} \dfrac{w}{w+5} &=\dfrac{2}{6} \\ 6 w &=2 w+10 \\ 4 w &=10 \\ w &=2 \dfrac{1}{2} \\ 1+z &=3 \\ z &=2 \end{aligned}$ (c) $\begin{aligned} \angle A &=\angle A\quad( \text{ common angle } ) \\ \angle C &=\angle A E D \\ \triangle A B C & \sim \triangle A D E \quad(A A \text { Corollary }) \\ \dfrac{A C}{A D} &=\dfrac{A C}{A E} \\ \dfrac{4+v}{6} &=\dfrac{6+2}{4} \\ 16+4 v &=48 \\ 4 v &=32 \\ v &=8 \end{aligned}$



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





(a) $E F∥ B C$ $\triangle A B C \sim \triangle A E F$ ( Corollary $2.2$ ) $\begin{aligned} \dfrac{A C}{A F} &=\dfrac{BC }{E F} \\ \dfrac{12+a}{12} &=\dfrac{9}{6} \\ 12+a &=18 \\ a &=6 \end{aligned}$ $F G ∥ A D$ $ \triangle C A D \sim \triangle C F G$ ( Corollary $2.2$ ) $\begin{aligned} \dfrac{A D}{F G} &=\dfrac{C A}{C F} \\ \dfrac{b}{2} &=\dfrac{a+12}{a} \\ \dfrac{b}{2} &=\dfrac{18}{6} \\ b &=6 \end{aligned}$ (b) $D E∥ B C$ $\begin{aligned} \triangle B C F &\sim \triangle E D F\quad(\text { Corollary }2.2) \\ \dfrac{B C}{E D}&=\dfrac{C F}{D F}=\dfrac{B F}{E F} \\ \dfrac{c}{4}&=\dfrac{5}{2}=\dfrac{d}{3} \\ \dfrac{c}{4}&=\dfrac{5}{2} \\ c&=10\\ \end{aligned}$ $\begin{aligned} \dfrac{d}{3} &=\dfrac{5}{2} \\ d &=\dfrac{15}{2}=7.5 \\ \triangle A B C & \sim \triangle A D E \quad(\text { Corollary }2.2)\\ \dfrac{A B}{A D} &=\dfrac{B C}{D E} \\ \dfrac{5+e}{5} &=\dfrac{c}{4} \\ 4(5+e) &=5 c \\ 20+4 e &=50 \\ 4 e &=30 \\ e &=7.5 \end{aligned}$ (c) $D E ∥B C$ $ \triangle A D E \sim \triangle A C B$( Corollary $2.2$ ) $\begin{aligned} \dfrac{A E}{A B}&=\dfrac{B E}{C B}\\ \dfrac{f}{f+5}&=\dfrac{4}{6}\\ 6 f&=4 f+20\\ 2 f&=20\\ f&=10\\ \end{aligned}$ $D E ∥ F A$ $\begin{aligned} \dfrac{B D}{B F}&=\dfrac{B E}{B A}\\ \dfrac{h}{h+11}&=\dfrac{5}{5+f}\\ 15 h&=5 h+55\\ 10 \mathrm{~h}&=55\\ h&=5.5\\ \end{aligned}$ $\begin{aligned} \dfrac{5}{f+5} &=\dfrac{4}{9} \\ \dfrac{5}{15} &=\dfrac{4}{9} \\ 5 g &=60 \\ g &=12 \\ \dfrac{A D}{A C} &=\dfrac{D E}{C B} \\ \dfrac{k}{k+7} &=\dfrac{4}{6} \\ 6 k &=4 k+28 \\ 2 k &=28 \\ k &=14 \end{aligned}$ (d) $D E ∥ B C$ $\begin{aligned} \triangle A D E & \sim \triangle A B C\quad(\text { Corollary }2.2) \\ \dfrac{A D}{A B} &=\dfrac{D E}{B C} \\ \dfrac{8}{8+m} &=\dfrac{10}{15} \\ 80+10 m &=120 \\ 10 m &=40 \\ m &=4 \\ D G &\| B F \\ \triangle A D G &\sim \triangle A B F\quad(\text { Corollary }2.2) \\ \dfrac{A G}{A F} &=\dfrac{A D}{A B} \\ \dfrac{p}{6} &=\dfrac{8}{8+m} \\ p &=\dfrac{8}{12} \times 6=4 \end{aligned}$ (e) Since BCDE is parallelogram, $\begin{aligned} E D&=B C=10\\ E G &\| D C \\ \triangle F E G& \sim \triangle F D C\quad(\text { Corollary }2.2)\\ \dfrac{E G}{D C}&=\dfrac{F E}{F D}\\ \dfrac{q}{14}&=\dfrac{6}{6+10}\\ \dfrac{q}{7}&=\dfrac{3}{4}\\ q&=\dfrac{21}{4}=2.25\\ \end{aligned}$ $A B C F$ is a parallelogram, $\begin{aligned} F A&=C B=10\\ F G & \| A B \\ \dfrac{F G}{A B} & =\dfrac{E F}{E A} \\ \dfrac{r}{12} & =\dfrac{6}{6+10} \\ r & =\dfrac{6}{16}\times 12\\ r & =4.5 \end{aligned}$



4. In the figure, XY ∥ PR and VT ∥ QR. If $\dfrac{PT}{TR}=\dfrac{3}{2}$, $\dfrac{QY}{YR}=\dfrac{2}{1}$ and $PQ=15 cm$, calcaulate
   (a) the lengths of PV, PX and XV.
   (b) the numerical vaalues of $\dfrac{YW}{WX}$ and $\dfrac{VW}{QY}$.
   






(a) $\begin{aligned} V T &\| Q R\\ \dfrac{P V}{V Q}&=\dfrac{P T}{T R}\\ \dfrac{P V}{V Q}&=\dfrac{3}{2}\\ \dfrac{P V}{P V+V Q}&=\dfrac{3}{3+2}\\ \dfrac{P V}{P Q}&=\dfrac{3}{5}\\ P V&=\dfrac{3}{5} \times 15=9\mathrm{cm }\\ X Y &\| P R\\ \dfrac{Q X}{X P}&=\dfrac{Q Y}{Y R}\\ \dfrac{Q X}{X P}&=\dfrac{2}{1}\\ \dfrac{Q X+X P}{X P} &=\dfrac{2+1}{1} \\ \dfrac{P Q}{P X} &=\dfrac{3}{1} \\ \dfrac{15}{P X} &=\dfrac{3}{1} \\ P X &=5 \mathrm{~cm} \\ X V &=P V-P X \\ &=9-5\\ &=4 \mathrm{~cm} \end{aligned}$ (b) $\begin{aligned} V W &\| Q Y\\ \dfrac{Y W}{W X} &=\dfrac{Q V}{V X} \\ &=\dfrac{15-9}{4}=\dfrac{6}{4}=\dfrac{3}{2} \\ \triangle X V W & \sim \triangle X Q Y \quad(\text { Corollary }2.2)\\ \dfrac{V W}{Q Y} &=\dfrac{X V}{X Q} \\ &=\dfrac{4}{15-5}=\dfrac{4}{10}=\frac{2}{5} \end{aligned}$


$\begin{aligned} 5.\text{ Given } :&\text{ Parallelogram B I R D }:\\ &\text{ I G bisects } \angle \text{ B I R }.\\ \text{ Prove } :&\dfrac{B E}{E I}=\dfrac{R G}{G I} \end{aligned}$






$\begin{aligned} \text { Proof: } & \beta=\beta_{1}\quad(B I R D \text { is a parallelogram) }\\ & \alpha=\alpha_{1}\quad(I G \text { bisects } \angle B I R) \\ & \triangle I B E \sim \triangle I R G\quad(A A \text { Corollary }) \\ & \dfrac{B E}{R G}=\dfrac{E I}{G I} \\ & \dfrac{B E}{E I}=\dfrac{R G}{G I} \end{aligned}$


$\begin{aligned} 6.\text{ Given } :& R Q ⊥ P Q, P Q ⊥ P T, S T ⊥ P R.\\ \text{ Prove } :& S T. R Q = P S. P Q \end{aligned}$






$\begin{aligned} \text { Proof } :& R Q \perp P Q \text { and } P Q \perp P T \text {. } \\ &\therefore \text { TP } \| {R Q }\\ &\therefore\alpha=\beta \\ &\text { In } \triangle T S P \text { and } \triangle P Q R, \\ &\alpha=\beta \quad\text { ( proved) } \\ &\angle T S P=\angle P Q R \quad\text { ( given) } \\ &\Delta T S P \sim \triangle P Q R \quad\text { (AA Corollary) } \\ &\dfrac{S T}{P Q}=\dfrac{P S}{R Q} \\ &S T . R Q=P S . P Q \end{aligned}$


$\begin{aligned} 7.\text{ Given } :& \text{ Parallelogram } A B C D ; P Q ∥ M B\\ \text{ Prove } :& △ A B M ∼ △C Q P. \end{aligned}$






$\begin{aligned} \text { Proof } :&\alpha=\beta\quad\text { (A D }\| \text {B C)}\\ &\gamma=\beta \quad\text { (PQ }\| \text {BM)}\\ &\alpha=\gamma \\ &\text { In } \triangle A B M and \triangle C Q P, \\ &\alpha=\gamma \quad\text { (proved) } \\ &\angle A=\angle C \quad \text { (ABCD is a parallelogram) } \\ &\triangle A B M \sim \triangle C Q P\quad(A A \text{ Corollary })\\ \end{aligned}$


8. △ ABC and △ CAD are drawn on opposite sides of AC such that AB : BC : CA= CA : AD : DC.
Prove that DC ∥ AB.





$\begin{aligned} \text { Given: }& A B: B C: C A=C A: A D: D C \\ \text { Prove : }& D C \| A B \\ \text { Proof : }& A B: B C: C A=C A: A D: D C \\ &\dfrac{A B}{C A}=\dfrac{B C}{A D}=\dfrac{C A}{D C} \\ & \triangle A B C \sim \triangle C A D \quad\text { (SSS) } \\ & \angle B A C=\angle A C D \\ & \therefore D C \| A B \end{aligned}$




Exercise (8.4)

1. Which of the following proportions follow from the fact that AE bisects ∠ WAV in △ WAV?






AE bisects $\angle W A V .$ (a) $\dfrac{W E}{E V}=\dfrac{W A}{A V}$  (True) (b) $\dfrac{W E}{E V}=\dfrac{V A}{A W}$  (False) (c) $\dfrac{W E}{W A}=\dfrac{E V}{A V}$  (True) (d) $\dfrac{A V}{A W}=\dfrac{V E}{E W}$  (True)



2. AX bisects ∠ CAB. Complete the following statements:
   (a) A C : A B =-----
   (b) A B : A C =-----
   (c) X C : X B =-----






AX bisects $\angle C A B .$ $\dfrac{A C}{A B}=\dfrac{C X}{X B}$ (a) $A C: A B=C X: X B$ (b) $A B: A C=B X: X C$ (c) $X C: X B=A C: A B$



3. PT bisects ∠ RPS. Complete the following statements:
   (a) P Q : P R =-----
   (b) T R : P R =-----
   (c) Q R : T R =-----






PT bisects $\angle R P S .$. $\dfrac{P Q}{P R}=\dfrac{Q T}{T R}$ (a) $P Q: P R=Q T: T R$ (b) $T R: P R=Q T: P Q$  ( Alternando Property ) (c) $Q R: T R=(P Q-P R): P R$  ( Dividendo Property )



4. What can you say about the rays AD, BE and CF?
   





$\begin{aligned} \dfrac{A B}{A C}&=\dfrac{2 \dfrac{2}{3}+3 \dfrac{1}{3}}{3+5}=\dfrac{6}{8}=\dfrac{3}{4}\\ \dfrac{B D}{D C}&=\dfrac{4 \frac{2}{7}}{5 \dfrac{5}{7}}=\dfrac{30}{40}=\dfrac{3}{4}\\ \dfrac{A B}{A C}&=\dfrac{B D}{D C}\\ \end{aligned}$ $ A D \text { bisects }\angle B A C$. $\begin{aligned} \dfrac{B A}{B C}&=\dfrac{3 \dfrac{1}{3}+2 \dfrac{2}{3}}{4 \dfrac{2}{7}+5 \dfrac{5}{7}}=\dfrac{6}{10}=\dfrac{3}{5} \\ \dfrac{A E}{E C}&=\dfrac{3}{5} \\ \dfrac{B A}{B C}&=\dfrac{A E}{E C} \\ \end{aligned}$ $B E \text { bisects } \angle A B C .$ $\begin{aligned} \dfrac{C A}{C B}&=\dfrac{5+3}{5 \dfrac{5}{7}+4 \dfrac{2}{7}}=\dfrac{8}{10}=\dfrac{4}{5} \\ \dfrac{A F}{F B}&=\dfrac{2 \dfrac{2}{3}}{3 \dfrac{1}{3}}=\dfrac{8}{10}=\dfrac{4}{5} \\ \dfrac{C A}{C B}&=\dfrac{A F}{F B} \\ \end{aligned}$ $C F \text { bisects } \angle A C B .$



5. If AD and AE are bisectors of the interior and exterior angles at A of ABC, then which of the following are true?

(a) ∠ D A E = 90°
(b) B D : D C =B C : C E
(c) B D : D C =B E : C E
(d) A D : A E =D C : C E






(a) $\angle D A E=90^{\circ} \quad$ (True) (b)$B D: D C=B C: C E\quad$ (False) (c) $B D: D C=B E: C E\quad$ (True) (d) $A D: A E=D C: C E\quad$ (False)



6. Find the value of x in each of the following figures.





(a) $\begin{aligned} \dfrac{x}{3} &=\dfrac{10}{6}\\ & x=5 \end{aligned}$ (b) $\begin{aligned} \dfrac{x}{9}&=\dfrac{13}{8}\\ x&=\frac{117}{8}=14 \frac{5}{8} \end{aligned}$ (c) $\begin{aligned} \dfrac{x}{4-x}&=\dfrac{8}{6}\\ 6 x&=32-8 x\\ 14 x &=32 \\ x &=\dfrac{16}{7}=2 \dfrac{2}{7} \end{aligned}$ (d) $\begin{aligned} \dfrac{x}{6-x} &=\dfrac{4}{3}\\ 3 x &=24-4 x\\ 7 x &=24\\ x &=3 \dfrac{3}{7}\\ \end{aligned}$ (e) $\begin{aligned} \dfrac{x-2}{2}&=\dfrac{7}{3}\\ 3 x-6 &=14 \\ 3 x &=20 \\ x &=6 \dfrac{2}{3}\\ \end{aligned}$ (f) $\begin{aligned} \dfrac{2.3+x}{x} &=\dfrac{3.8}{2.3}\\ 2.3 \times 2.3+2.3 x &=3.8 x \\ 1.5 x &=5.29\\ x &=\dfrac{5.29}{1.5}\\ &=\dfrac{52.9}{15} \\ &=3.53\\ \end{aligned}$



7. Find the unknown marked lengths in the figure.






DC bisects $\angle A C B$. $\begin{aligned} \dfrac{x}{24} &=\dfrac{20}{30} \\ x &=16\\ \end{aligned}$ AE bisects $\angle F A C$. $\begin{aligned} \dfrac{30+y}{y} &=\dfrac{x+24}{20} \\ \dfrac{30+y}{y} &=\dfrac{40}{20} \\ 30+y &=2 y \\ y &=30\\ \end{aligned}$



8. AB= 12cm, BC= 9cm, CA= 7cm.
   BD bisects B and AG= AD, CH= CD.
   Calculate BG, BH. Does GH ∥ AC?






$\begin{aligned} \text { Given } :& A B=12 \mathrm{~cm}, B C=9 \mathrm{~cm}, C A=7 \mathrm{~cm}.\\ & BD \text { bisects } \angle B \text { and } A G=A D, C H=C D\\ \text { To find } :& B G, B H=?\\ & G H \| A C ?\\ \text{ solution } :& B D \text{ bisects } ∠ C B A.\\ &\dfrac{A B}{B C}=\dfrac{12}{9}=\dfrac{4}{3} \\ &\text { Let } A D=4 x, D C=3 x \\ & A D+D C=C A \\ & 4 x+3 x=7 \\ & 7 x=7 \\ & x=1\\ & A G =A D=4 \mathrm{~cm}, H C=D C=3 \mathrm{~cm} \\ & B G =A B-A G=12-4=8 \mathrm{~cm} \\ & B H =B C-H C=9-3=6 \mathrm{~cm} \\ &\dfrac{B G}{G A} =\dfrac{8}{4}=\frac{2}{1} \\ &\dfrac{B H}{H C} =\dfrac{6}{3}=\frac{2}{1} \\ &\dfrac{B G}{G A} =\dfrac{B H}{H C} \\ &\therefore G H \| A C . \end{aligned}$



9. In ABC, DE ∥ BC, AD=2.7cm. DB=1.8cm and BC=3cm.
   Prove that BE bisects ∠ ABC.






$\begin{aligned} \text { Given } &: In \triangle A B C, D E \| B C,\\ &A D=2.7 \mathrm{~cm}, D B=1.8 \mathrm{~cm}, B C=3 \mathrm{~cm}\\ \text { Prove } &: B E \text { bisects } \angle A B C.\\ \text { Proof } &: D E \| B C\\ &\dfrac{A E}{E C} =\dfrac{A D}{D B}=\dfrac{2.7}{1.8}=\frac{3}{2} \\ &\dfrac{A B}{B C} =\dfrac{2.7+1.8}{3}=\dfrac{4.5}{3}=\frac{3}{2} \\ &\dfrac{A E}{E C} =\dfrac{A B}{B C} \\ &\therefore B E \text { bisects } ∠ A B C . \end{aligned}$



10. In a parallelogram ABCD, AB= 3.6 cm, BC= 2.7 cm. AX= 3.2 cm, XC= 2.4 cm. Prove that △ BCY is isosceles.






$\begin{aligned} \text { Given } :& \text { In a parallelogram } A B C D ,\\ &A B=3.6 \mathrm{~cm}, B C=2.7 \mathrm{~cm}, A X=3.2 \mathrm{~cm}, X C=2.4 \mathrm{~cm}\\ \text { Prove } :& \triangle B C Y \text { is isosceles. } \\ \text { Proof } :& \dfrac{A B}{B C}=\dfrac{3.6}{2.7}=\dfrac{4}{3} \\ &\dfrac{A X}{X C}=\dfrac{3.2}{2.4}=\dfrac{4}{3} \\ &\dfrac{A B}{B C}=\dfrac{A X}{X C} \\ &\therefore B X \text { bisects } \angle A B C .\\ &\therefore \alpha=\beta \\ &B u t \alpha=\gamma \quad( A B \| D C) \\ &\gamma=\beta \\ &B C=C Y \\ &\therefore \triangle B C Y \text { is isosceles. }\\ \end{aligned}$



11. Calculate BD and DC in terms of a, b, c.






$A D$ bisects $\angle C A B$. $\begin{aligned} \dfrac{B D}{D C}&=\dfrac{A B}{A C} \\ \dfrac{B D}{a-B D} &=\dfrac{c}{b} \\ b . B D &=a c-c .B D \\ (b+c) B D &=a c \\ B D &=\dfrac{a c}{b+c} \\ D C &=a-B D \\ &=a-\dfrac{ac}{b+c} \\ &=\dfrac{a(b+c)-a c}{b+c} \\ &=\dfrac{a b+a c-a c}{b+c} \\ &=\dfrac{a b}{b+c} \end{aligned}$


$\begin{aligned} 12.\text{ Given } :& AH \text{ bisects } ∠ BAC \text{ in } △ABC.\\ & EH ∥ AC.\\ \text{ Prove } :& \dfrac{BE}{EA}=\dfrac{BA}{AC} \end{aligned}$






$\begin{aligned} \text { Given } &: A H \text { bisects } \angle B A C \text{ in } \triangle A B C.\\ &EH \|AC.\\ \text { Prove } &: \dfrac{B E}{E A} =\dfrac{B A}{A C}\\ \text { Proof } &: AH \text { bisects } \angle B A C.\\ & \dfrac{B A}{A C}=\dfrac{B H}{H C}----(1) \\ & E H \| A C \\ & \dfrac{B E}{E A}=\dfrac{B H}{H C}----(2) \\ &\text{From }\ eq^{n} (1) and (2),\\ & \dfrac{B E}{E A}=\dfrac{B A}{A C} \end{aligned}$


$\begin{aligned} 13.\text{ Given } :&\text{ In }\triangle A B C, B M= M C;\\ & M X \text{ bisects } \angle A M B\\ & M Y \text{ bisects } \angle A M C.\\ \text{ Prove } :& X Y \| B C. \end{aligned}$





$\begin{aligned} \text{ Given } :&\text{ In }\triangle A B C, B M= M C;\\ &M X \text{ bisects } \angle A M B\\ &M Y \text{ bisects } \angle A M C.\\ \text{ Prove } :&X Y \| B C\\ \text{ Proof } :& M X \text { bisects } \angle A M B\\ & \dfrac{A X}{X B}=\dfrac{A M}{B M}----(1) \\ & M Y \text{ bisects } \angle A M C . \\ & \dfrac{A Y}{Y C}=\dfrac{A M}{C M}=\dfrac{A M}{B M}----(2) \\ &\text{From}\ eq^{n} (1) and (2),\\ & \dfrac{A X}{X B}=\dfrac{A Y}{Y C} \\ &\therefore X Y \| B C . \end{aligned}$


$\begin{aligned} 14.\text{ Given } :&\text{ In }\triangle A B C, \angle A= 2 \angle C;\\ &A D \text{ bisects } \angle B A C\\ &D E \text{ bisects } \angle AD B.\\ \text{ Prove } :&\dfrac{BE}{EA}=\dfrac{BA}{AC}. \end{aligned}$





$\begin{aligned} \text{ Given } :&\text{ In }\triangle A B C, \angle A= 2 \angle C\\ &A D \text{ bisects } \angle B A C\\ &D E \text{ bisects } \angle AD B\\ \text{ Prove } :&\dfrac{BE}{EA}=\dfrac{BA}{AC}\\ \text{ Proof } :&\angle D A C= \angle D A B ( A D \text { biects } \angle B A C) \\ &\angle A=2 \angle C \text{(given)}\\ &2 \angle D A C=2 \angle C \\ &\angle D A C=\angle C \\ & D C=D A \\ & A D \text { bisects } \angle B A C . \\ & \dfrac{B A}{A C}=\dfrac{B D}{D C} \\ &\dfrac{B A}{A C}=\dfrac{B D}{D A}----(1)\\ &D E \text{ bisects } \angle A D B .\\ & \dfrac{B E}{E A}=\dfrac{B D}{D A}----(2) \\ &\text{From}\ eq^{n} (1) and (2),\\ & \dfrac{B E}{E A}=\dfrac{B A}{A C} \end{aligned}$




Exercise (8.5)

1. In the figure ∠ PQR= 90°, QS ⊥ PR. Complete each of the following true statements.
   (a) △ PQR ∼ △? ∼ △?
   (b) QS is the geometric mean between ? and ?
   (c) QR is the geometric mean between ? and ?
   (d) $\dfrac{?}{PQ}= \dfrac{PQ}{?}$






(a) $\triangle P Q R \sim \triangle P S Q \sim \triangle Q S R$ (b) $ Q S$ is the geometric mean between $ P S $and $S R$. (c) $Q R$ is the geometric mean between $P R$ and $S R$. (d) $\dfrac{P R}{P Q}=\dfrac{P Q}{P S}$



2. In the figure CD ⊥ AB and C=90°. If DE ⊥ BC, DF ⊥ CA, write out all the triangles that are similar to △ABC.





$\triangle A C D, \triangle C B D, \triangle A D F, \triangle D C F, \triangle C D E, \triangle D B E$ are similar to $\triangle A B C$.



3. Find the length of each marked segment.



(a) $\begin{aligned} B D^{2} &=A D \cdot D C\quad(\text{ Corollary } 6.1) \\ x^{2} &=5 \times 10=50 \\ x &=5 \sqrt{2} \end{aligned}$ (b) $ \begin{aligned} B C^{2}&=C D \cdot C A\quad(\operatorname{Corollary} 6.1)\\ y^{2}&=3 \times 9=27\\ y&=3 \sqrt{3} \end{aligned}$ (c) $ \begin{aligned} E F^{2}&=E H \cdot E G\quad(\operatorname{Corollary} 6.1)\\ 8^{2}&=x \times 12\\ x&=\dfrac{64}{12}=5 \dfrac{1}{3}\\ \end{aligned}$ (d) $ \begin{aligned} E G^{2}&=D G \cdot G F\quad(\text{ Corollary } 6.1)\\ 6^{2}&=(13-x) x\\ x^{2}-13 x+36&=0\\ (x-9)(x-4)&=0\\ x-9&=0 (or) x-4=0\\ x&=9 (or) x=4 \end{aligned}$ (e) $\triangle A B D$ is a $45^{\circ}-45^{\circ}$ right triangle. $\begin{aligned} A D &=B D \\ x &=6 \\ A D &=\sqrt{2} B D \\ u &=6 \sqrt{2}\\ \end{aligned}$ $\triangle B D C$ is a $30^{\circ}-60^{\circ}$ right triangle. $\begin{aligned} D C& =2 B D \\ \quad 2 &=2(6)=12 \\ B C &=\sqrt{3} B D \\ y &=6 \sqrt{3} \end{aligned}$ (f) $\triangle M Q F$ is a $30^{\circ}-60^{\circ}$ right triangle. $\begin{aligned} F Q &=\dfrac{1}{2} M F=\dfrac{1}{2}(12)=6 \\ M Q &=\dfrac{\sqrt{3}}{2} M F=\dfrac{\sqrt{3}}{2}(12)=6 \sqrt{3} \\ \angle F R Q &=45^{\circ}\\ \end{aligned}$ $\triangle F R Q$ is a $45^{\circ}-45^{\circ}$ right triangle. $\begin{aligned} y &=F Q =6 \\ x+y &=M Q \\ x+6 &=6 \sqrt{3} \\ x &=6 \sqrt{3}-6 \\ z &=\sqrt{2} \mathrm{FQ} \\ z &=6 \sqrt{2} \end{aligned}$



4. In the figure, if AD= 10 cm, AB= 8 cm, BC= 4 cm, find the length of CD.





Draw $C E$ ⊥ $A D$. $\begin{aligned} A D&=10-4=6 \mathrm{~cm}\\ E C&=A B=8 \mathrm{~cm}\\ \end{aligned}$ By the Pythagoras Theorem, $\begin{aligned} C D^{2}&=D E^{2}+E C^{2} \\ &=6^{2}+8^{2}=100\\ C D&=10 \mathrm{~cm}\\ \end{aligned}$



5. In the figure, find the distance of D from A.





Draw A D and A B E ⊥ D E. $\begin{aligned} B E&=C D=1 \\ A E&=A B+B E=3 \\ E D&=B C=4\\ \end{aligned}$ By the Pythagoras Theorem, $\begin{aligned} A D^{2}&=A E^{2}+E D^{2}=25 \\ A D^{2}&=3^{2}+4^{2}=25 \\ A D &= 5 \\ \end{aligned}$



6. A parallelogram with sides 8 cm and 15 cm has a diagonal of 17 cm. Is it a rectangle?



$\begin{aligned} A D^{2}+A B^{2}&=8^{2}+15^{2}=289 \\ B D^{2}&=17^{2}=289 \\ B D^{2}&=A D^{2}+A B^{2} \\ \angle A&=90^{\circ} \\ \end{aligned}$ Since $ A B C D $ is a parallelogram and $ \angle A=90^{\circ}$, $A B C D $ is a rectangle.



7. In the figure, △ ABC, △ ACD, △ ADE are right triangles and AB= BC= CD= DE. Show that AE= 2 AB.





$\begin{aligned} \text{ Given } :& \triangle A B C, \triangle A C D, \triangle A D E \text{ are right triangles }.\\ & A B=B C=C D=D E\\ \text{ Prove } :& A E=2 A B\\ \text{ Proof } :& \text{ Let } A B=B C=C D=D E=x\\ &\text{ In rt } \triangle A B C,\\ &A C^{2} =A B^{2}+B C^{2}=x^{2}+x^{2}=2 x^{2}\\ &\text{ In rt } \triangle A C D,\\ &A D^{2} =A C^{2}+C D^{2}=2 x^{2}+x^{2}=3 x^{2}\\ &\text{ In rt } \triangle A D E,\\ &A E^{2} =A D^{2}+D E^{2}=3 x^{2}+x^{2}=4 x^{2} \\ &A E =2 x \\ &A E =2 A B \end{aligned}$


$\begin{aligned} 8.\text{ Given } :& A D ⊥ B C\\ \text{ Prove } :& A B²- A C²= B D²- D C² \end{aligned}$





$\begin{aligned} \text { Given } :& A D \perp B C \\ \text { Prove } :& A B^{2}-A C^{2}=B D^{2}-D C^{2} \\ \text { Proof } :& \text{ In rt } \triangle A B D \text {, } \\ &A B^{2}=A D^{2}+B D^{2}---(1) \\ &\text { In rt } \triangle A D C \text {, } \\ &A C^{2}=A D^{2}+D C^{2}---(2) \\ &e q^{n}(1)-(2), \\ &A B^{2}-A C^{2}=B D^{2}-D C^{2} \end{aligned}$


$\begin{aligned} 9.\text{ Given } :& ∠ BAC= 90°, D \text{ is any point on } A B.\\ \text{ Prove } :& B C²+ A D²= A B²+ C D² \end{aligned}$





$\begin{aligned} \text { Given } :& \angle B A C=90^{\circ} \\ &D \text { is any point on } A B \text {. } \\ \text { Prove } :& B C^{2}+A D^{2}=A B^{2}+C D^{2} \\ \text { Proof } :&\text { In rt } \triangle A B C , \\ &B C^{2}= A B^{2}+A C^{2}---(1) \\ &\text { In rt } \triangle A C D,&\\ &A D^{2}=C D^{2}-A C^{2}---(2) \\ & e q^{n}(1)+(2) \\ &B C^{2} +A D^{2}=A B^{2}+C D^{2} \end{aligned}$


$\begin{aligned} 10.\text{ Given } :& ∠Q P R= 90°, P S ⊥ QR.\\ \text{ Prove } :& \dfrac{1}{PS^{2}}=\dfrac{1}{PQ^{2}}+\dfrac{1}{PR^{2}} \end{aligned}$





$\begin{aligned} \text{ Given } :& \angle Q P R=90^{\circ}, P S \perp Q R\\ \text{ Prove } :& \dfrac{1}{P S^{2}}=\dfrac{1}{P Q^{2}}+\dfrac{1}{P R^{2}}\\ \end{aligned}$ $\begin{aligned} \text{ Proof } :\text{ Since } \angle Q P R&=90^{\circ} \text{ and } P S \perp Q R \\ P S^{2} &=Q S \cdot S R\quad( \operatorname{ Corollary } 6.1) \\ P Q^{2} &=Q S \cdot Q R\quad( \operatorname{ Corollary } 6.1) \\ P R^{2} &=S R \cdot Q R\quad( \operatorname{ Corollary } 6.1) \\ \dfrac{1}{P Q^{2}}+\dfrac{1}{P R^{2}} &=\dfrac{1}{Q S \cdot Q R}+\dfrac{1}{S R \cdot Q R} \\ &=\dfrac{S R+Q S}{Q S \cdot Q R \cdot S R} \\ &=\dfrac{Q R}{Q S \cdot Q R \cdot S R} \\ &=\dfrac{1}{P S^{2}} \end{aligned}$


$\begin{aligned} 11.\text{ Given } :& A D ⊥ B C, A D = 9 \mathrm{~cm},\\ & B D = 13.5 \mathrm{~cm}, D C = 6 \mathrm{~cm}.\\ \text{ Prove } :& ∠ B A C = 90° \end{aligned}$





$\begin{aligned} \text { Given } :& A D \perp B C, \mathrm{AD}=9 \mathrm{~cm}, \mathrm{BD}=13.5 \mathrm{~cm}, \mathrm{DC}=6 \mathrm{~cm}.\\ \text { Prove } :& \angle B A C=90^{\circ} \\ \text { Proof } :& B C^{2}=(13.5+6)^{2}=380.25\\ &A B^{2}=(13.5)^{2}+9^{2}=263.25 \\ &A C^{2}=6^{2}+9^{2}=117 \\ &A B^{2}+A C^{2}=263.25+117=380.25 \\ & B C^{2}=A B^{2}+A C^{2} \\ &\therefore \angle B A C=90^{\circ} \end{aligned}$