## Chapter 06 Lines And Angles

### 6.1 Introduction

In Chapter 5, you have studied that a minimum of two points are required to draw a line. You have also studied some axioms and, with the help of these axioms, you proved some other statements. In this chapter, you will study the properties of the angles formed when two lines intersect each other, and also the properties of the angles formed when a line intersects two or more parallel lines at distinct points. Further you will use these properties to prove some statements using deductive reasoning (see Appendix 1). You have already verified these statements through some activities in the earlier classes.

In your daily life, you see different types of angles formed between the edges of plane surfaces. For making a similar kind of model using the plane surfaces, you need to have a thorough knowledge of angles. For instance, suppose you want to make a model of a hut to keep in the school exhibition using bamboo sticks. Imagine how you would make it? You would keep some of the sticks parallel to each other, and some sticks would be kept slanted. Whenever an architect has to draw a plan for a multistoried building, she has to draw intersecting lines and parallel lines at different angles. Without the knowledge of the properties of these lines and angles, do you think she can draw the layout of the building?

In science, you study the properties of light by drawing the ray diagrams. For example, to study the refraction property of light when it enters from one medium to the other medium, you use the properties of intersecting lines and parallel lines. When two or more forces act on a body, you draw the diagram in which forces are represented by directed line segments to study the net effect of the forces on the body. At that time, you need to know the relation between the angles when the rays (or line segments) are parallel to or intersect each other. To find the height of a tower or to find the distance of a ship from the light house, one needs to know the angle formed between the horizontal and the line of sight. Plenty of other examples can be given where lines and angles are used. In the subsequent chapters of geometry, you will be using these properties of lines and angles to deduce more and more useful properties.

Let us first revise the terms and definitions related to lines and angles learnt in earlier classes.

### 6.2 Basic Terms and Definitions

Recall that a part (or portion) of a line with two end points is called a line-segment and a part of a line with one end point is called a ray. Note that the line segment $A B$ is denoted by $\overline{\mathrm{AB}}$, and its length is denoted by $\mathrm{AB}$. The ray $\mathrm{AB}$ is denoted by $\overrightarrow{\mathrm{AB}}$, and a line is denoted by $\overleftrightarrow{A B}$. However, we will not use these symbols, and will denote the line segment $\mathrm{AB}$, ray $\mathrm{AB}$, length $\mathrm{AB}$ and line $\mathrm{AB}$ by the same symbol, $\mathrm{AB}$. The meaning will be clear from the context. Sometimes small letters $l, m, n$, etc. will be used to denote lines.

If three or more points lie on the same line, they are called collinear points; otherwise they are called non-collinear points.

Recall that an angle is formed when two rays originate from the same end point. The rays making an angle are called the arms of the angle and the end point is called the vertex of the angle. You have studied different types of angles, such as acute angle, right angle, obtuse angle, straight angle and reflex angle in earlier classes (see Fig. 6.1).

Fig. 6.1 : Types of Angles

An acute angle measures between $0^{\circ}$ and $90^{\circ}$, whereas a right angle is exactly equal to $90^{\circ}$. An angle greater than $90^{\circ}$ but less than $180^{\circ}$ is called an obtuse angle. Also, recall that a straight angle is equal to $180^{\circ}$. An angle which is greater than $180^{\circ}$ but less than $360^{\circ}$ is called a reflex angle. Further, two angles whose sum is $90^{\circ}$ are called complementary angles, and two angles whose sum is $180^{\circ}$ are called supplementary angles.

You have also studied about adjacent angles in the earlier classes (see Fig. 6.2). Two angles are adjacent, if they have a common vertex, a common arm and their non-common arms are on different sides of the common arm. In Fig. 6.2, $\angle \mathrm{ABD}$ and $\angle \mathrm{DBC}$ are adjacent angles. Ray $\mathrm{BD}$ is their common arm and point $\mathrm{B}$ is their common vertex. Ray $\mathrm{BA}$ and ray $\mathrm{BC}$ are non common arms. Moreover, when two angles are adjacent, then their sum is always equal to the angle formed by the two non-

Fig. 6.2 : Adjacent angles common arms. So, we can write

$\angle \mathrm{ABC}=\angle \mathrm{ABD}+\angle \mathrm{DBC} .$

Note that $\angle \mathrm{ABC}$ and $\angle \mathrm{ABD}$ are not adjacent angles. Why? Because their noncommon arms $\mathrm{BD}$ and $\mathrm{BC}$ lie on the same side of the common arm BA.

If the non-common arms $\mathrm{BA}$ and $\mathrm{BC}$ in Fig. 6.2, form a line then it will look like Fig. 6.3. In this case, $\angle \mathrm{ABD}$ and $\angle \mathrm{DBC}$ are called linear pair of angles.

Fig. 6.3 : Linear pair of angles

You may also recall the vertically opposite angles formed when two lines, say $\mathrm{AB}$ and $\mathrm{CD}$, intersect each other, say at the point $\mathrm{O}$ (see Fig. 6.4). There are two pairs of vertically opposite angles.

Fig. 6.4 : Vertically opposite angles

One pair is $\angle \mathrm{AOD}$ and $\angle \mathrm{BOC}$. Can you find the other pair?

### 6.3 Intersecting Lines and Non-intersecting Lines

Draw two different lines PQ and RS on a paper. You will see that you can draw them in two different ways as shown in Fig. 6.5 (i) and Fig. 6.5 (ii).

Fig. 6.5 : Different ways of drawing two lines

Recall the notion of a line, that it extends indefinitely in both directions. Lines PQ and RS in Fig. 6.5 (i) are intersecting lines and in Fig. 6.5 (ii) are parallel lines. Note that the lengths of the common perpendiculars at different points on these parallel lines is the same. This equal length is called the distance between two parallel lines.

### 6.4 Pairs of Angles

In Section 6.2, you have learnt the definitions of some of the pairs of angles such as complementary angles, supplementary angles, adjacent angles, linear pair of angles, etc. Can you think of some relations between these angles? Now, let us find out the relation between the angles formed when a ray stands on a line. Draw a figure in which a ray stands on a line as shown in Fig. 6.6. Name the line as $\mathrm{AB}$ and the ray as OC. What are the angles formed at the $\mathrm{O}$ ? They are $\angle \mathrm{AOC}, \angle \mathrm{BOC}$ and $\angle \mathrm{AOB}$.

Fig. 6.6 : Linear pair of angles point

Can we write $\angle \mathrm{AOC}+\angle \mathrm{BOC}=\angle \mathrm{AOB}$ ? (1)

Yes! (Why? Refer to adjacent angles in Section 6.2)

What is the measure of $\angle \mathrm{AOB}$ ? It is $180^{\circ}$. (Why?) (2)

From (1) and (2), can you say that $\angle \mathrm{AOC}+\angle \mathrm{BOC}=180^{\circ}$ ? Yes! (Why?)

From the above discussion, we can state the following Axiom:

**Axiom 6.1 :** If a ray stands on a line, then the sum of two adjacent angles so formed is $180^{\circ}$.

Recall that when the sum of two adjacent angles is $180^{\circ}$, then they are called a linear pair of angles.

In Axiom 6.1, it is given that ‘a ray stands on a line’. From this ‘given’, we have concluded that ’the sum of two adjacent angles so formed is $180^{\circ}$ ‘. Can we write Axiom 6.1 the other way? That is, take the ‘conclusion’ of Axiom 6.1 as ‘given’ and the ‘given’ as the ‘conclusion’. So it becomes:

(A) If the sum of two adjacent angles is $180^{\circ}$, then a ray stands on a line (that is, the non-common arms form a line).

Now you see that the Axiom 6.1 and statement (A) are in a sense the reverse of each others. We call each as converse of the other. We do not know whether the statement (A) is true or not. Let us check. Draw adjacent angles of different measures as shown in Fig. 6.7. Keep the ruler along one of the non-common arms in each case. Does the other non-common arm also lie along the ruler?

Fig. 6.7 : Adjacent angles with different measures

You will find that only in Fig. 6.7 (iii), both the non-common arms lie along the ruler, that is, points $\mathrm{A}, \mathrm{O}$ and $\mathrm{B}$ lie on the same line and ray $\mathrm{OC}$ stands on it. Also see that $\angle \mathrm{AOC}+\angle \mathrm{COB}=125^{\circ}+55^{\circ}=180^{\circ}$. From this, you may conclude that statement (A) is true. So, you can state in the form of an axiom as follows:

**Axiom 6.2 :** If the sum of two adjacent angles is $180^{\circ}$, then the non-common arms of the angles form a line.

For obvious reasons, the two axioms above together is called the Linear Pair Axiom.

Let us now examine the case when two lines intersect each other.

Recall, from earlier classes, that when two lines intersect, the vertically opposite angles are equal. Let us prove this result now. See Appendix 1 for the ingredients of a proof, and keep those in mind while studying the proof given below.

**Theorem 6.1 :** If two lines intersect each other, then the vertically opposite angles are equal.

**Proof :** In the statement above, it is given that ’two lines intersect each other’. So, let $\mathrm{AB}$ and $\mathrm{CD}$ be two lines intersecting at $\mathrm{O}$ as shown in Fig. 6.8. They lead to two pairs of vertically opposite angles, namely,

(i) $\angle \mathrm{AOC}$ and $\angle \mathrm{BOD}$ (ii) $\angle \mathrm{AOD}$ and $\angle \mathrm{BOC}$.

Fig. 6.8 : Vertically opposite angles

We need to prove that $\angle \mathrm{AOC}=\angle \mathrm{BOD}$

and $\angle \mathrm{AOD}=\angle \mathrm{BOC}$.

Now, ray $\mathrm{OA}$ stands on line $\mathrm{CD}$.

Therefore, $\angle \mathrm{AOC}+\angle \mathrm{AOD}=180^{\circ}$

Can we write $\angle \mathrm{AOD}+\angle \mathrm{BOD}=180^{\circ}$ ? Yes! (Why?)

From (1) and (2), we can write

$$\angle \mathrm{AOC}+\angle \mathrm{AOD}=\angle \mathrm{AOD}+\angle \mathrm{BOD}$$

This implies that $\angle \mathrm{AOC}=\angle \mathrm{BOD} \quad$ (Refer Section 5.2, Axiom 3)

Similarly, it can be proved that $\angle \mathrm{AOD}=\angle \mathrm{BOC}$

Now, let us do some examples based on Linear Pair Axiom and Theorem 6.1.

**Example 1 :** In Fig. 6.9, lines PQ and RS intersect each other at point $O$. If $\angle \mathrm{POR}: \angle \mathrm{ROQ}=5: 7$, find all the angles.

**Solution :** $\angle \mathrm{POR}+\angle \mathrm{ROQ}=180^{\circ}$

(Linear pair of angles)

But $\angle \mathrm{POR}: \angle \mathrm{ROQ}=5: 7$ (Given)

Therefore, $\quad \angle \mathrm{POR}=\frac{5}{12} \times 180^{\circ}=75^{\circ}$

Similarly, $\quad \angle \mathrm{ROQ}=\frac{7}{12} \times 180^{\circ}=105^{\circ}$

Now, $$\begin{aligned} & \angle \mathrm{POS}=\angle \mathrm{ROQ}=105^{\circ} \\ & \angle \mathrm{SOQ}=\angle \mathrm{POR}=75^{\circ} \end{aligned}$$

**Example 2 :** In Fig. 6.10, ray OS stands on a line POQ. Ray OR and ray OT are angle bisectors of $\angle \mathrm{POS}$ and $\angle \mathrm{SOQ}$, respectively. If $\angle \mathrm{POS}=x$, find $\angle \mathrm{ROT}$.

**Solution :** Ray OS stands on the line POQ.

Therefore,

$\angle \mathrm{POS}+\angle \mathrm{SOQ}=180^{\circ}$

But,

$$\angle \mathrm{POS}=x$$

Therefore, So,

$x+\angle \mathrm{SOQ} =180^{\circ} \\ \angle \mathrm{SOQ} =180^{\circ}-x$

Now, ray OR bisects $\angle \mathrm{POS}$, therefore,

Fig. 6.10

$$ \begin{aligned} \angle \mathrm{ROS} & =\frac{1}{2} \times \angle \mathrm{POS} \\ & =\frac{1}{2} \times x=\frac{x}{2} \end{aligned} $$

Similarly,

$$ \begin{aligned} \angle \mathrm{SOT} & =\frac{1}{2} \times \angle \mathrm{SOQ} \\ & =\frac{1}{2} \times\left(180^{\circ}-x\right) \\ & =90^{\circ}-\frac{x}{2} \end{aligned} $$

Now,

$$ \begin{aligned} \angle \mathrm{ROT} & =\angle \mathrm{ROS}+\angle \mathrm{SOT} \\ & =\frac{x}{2}+90^{\circ}-\frac{x}{2} \\ & =90^{\circ} \end{aligned} $$

**Example 3 :** In Fig. 6.11, OP, OQ, OR and OS are four rays. Prove that $\angle \mathrm{POQ}+\angle \mathrm{QOR}+\angle \mathrm{SOR}+$ $\angle \mathrm{POS}=360^{\circ}$.

**Solution :** In Fig. 6.11, you need to produce any of the rays $\mathrm{OP}, \mathrm{OQ}$, OR or OS backwards to a point. Let us produce ray $\mathrm{OQ}$ backwards to a point $\mathrm{T}$ so that TOQ is a line (see Fig. 6.12).

Fig. 6.11

Now, ray OP stands on line TOQ.

Therefore,

$\angle \mathrm{TOP}+\angle \mathrm{POQ}=180^{\circ}$

(Linear pair axiom)

Similarly, ray OS stands on line TOQ.

Therefore,

$\angle \mathrm{TOS}+\angle \mathrm{SOQ}=180^{\circ}$

$\angle \mathrm{SOQ}=\angle \mathrm{SOR}+\angle \mathrm{QOR}$

So, (2) becomes

Fig. 6.12

$$ \begin{equation*} \angle \mathrm{TOS}+\angle \mathrm{SOR}+\angle \mathrm{QOR}=180^{\circ} \tag{3} \end{equation*} $$

Now, adding (1) and (3), you get

$$ \begin{equation*} \angle \mathrm{TOP}+\angle \mathrm{POQ}+\angle \mathrm{TOS}+\angle \mathrm{SOR}+\angle \mathrm{QOR}=360^{\circ} \tag{4} \end{equation*} $$

But

$\angle \mathrm{TOP}+\angle \mathrm{TOS}=\angle \mathrm{POS}$

Therefore, (4) becomes

$$\angle \mathrm{POQ}+\angle \mathrm{QOR}+\angle \mathrm{SOR}+\angle \mathrm{POS}=360^{\circ}$$

### EXERCISE 6.1

**1.** In Fig. 6.13, lines $\mathrm{AB}$ and $\mathrm{CD}$ intersect at $\mathrm{O}$. If $\angle \mathrm{AOC}+\angle \mathrm{BOE}=70^{\circ}$ and $\angle \mathrm{BOD}=40^{\circ}$, find $\angle \mathrm{BOE}$ and reflex $\angle \mathrm{COE}$.

Fig. 6.13

## Show Answer

**Solution**

$A B$ is a straight line, rays $O C$ and $O E$ stand on it.

$\therefore \angle AOC+\angle COE+\angle BOE=180^{\circ}$

$\Rightarrow(\angle AOC+\angle BOE)+\angle COE=180^{\circ}$

$\Rightarrow 70^{\circ}+\angle COE=180^{\circ}$

$\Rightarrow \angle COE=180^{\circ}-70^{\circ}=110^{\circ}$

Reflex $\angle COE=360^{\circ}-110^{\circ}=250^{\circ}$

$CD$ is a straight line, rays $OE$ and $OB$ stand on it.

$\therefore \angle COE+\angle BOE+\angle BOD=180^{\circ}$

$\Rightarrow 110^{\circ}+\angle BOE+40^{\circ}=180^{\circ}$

$\Rightarrow \angle BOE=180^{\circ}-150^{\circ}=30^{\circ}$

**2.** In Fig. 6.14, lines $X Y$ and $M N$ intersect at $O$. If $\angle \mathrm{POY}=90^{\circ}$ and $a: b=2: 3$, find $c$.

Fig. 6.14

## Show Answer

**Solution**

Let the common ratio between $a$ and $b$ be $x . \quad \therefore$

$X Y$ is a straight line, rays OM and OP stand on it.

$\therefore X O M+MOP+\angle POY=180^{\circ} b+a+POY=180^{\circ}$

$3 x+2 x+90^{\circ}=180^{\circ} 5 x=90^{\circ} x=18^{\circ} a=$

$2 x=2 \times 18=36^{\circ} b=$

$3 x=3 \times 18=54^{\circ}$

MN is a straight line. Ray OX stands on it.

$\therefore b+c=180^{\circ}$ (Linear Pair)

$54^{\circ}+c=180^{\circ} c=180^{\circ}-$

$54^{\circ}=126^{\circ} \quad \therefore c=126^{\circ}$

**3.** In Fig. 6.15, $\angle \mathrm{PQR}=\angle \mathrm{PRQ}$, then prove that $\angle \mathrm{PQS}=\angle \mathrm{PRT}$.

Fig. 6.15

## Show Answer

**Solution**

In the given figure, ST is a straight line and ray QP stands on it.

$\therefore 4 Q S+P Q R=180^{\circ}$ (Linear Pair)

$\angle PQR=180^{\circ}-\angle PQS(1)$

PRT $+\angle PRQ=180^{\circ}$ (Linear Pair)

${ }^{\angle} PRQ=180^{\circ}-PRT$ (2)

It is given that $4 PQR=\angle PRQ$.

Equating equations (1) and (2), we obtain

$ \begin{aligned} & 180^{\circ}-\stackrel{\angle}{ }{ }^{PQS}=180^{\circ}-PRT \angle PQS \\ & =\angle PRT \end{aligned} $

**4.** In Fig. 6.16, if $x+y=w+z$, then prove that $\mathrm{AOB}$ is a line.

Fig. 6.16

## Show Answer

**Solution**

It can be observed that, $x+y+z+w$ then prove that $A O B$ is a line.

$=360^{\circ}$ (Complete angle) It is given

that, $x+y=z+w \quad \therefore x+y+x+y$

$=360^{\circ}$

$2(x+y)=360^{\circ} x$

$+y=180^{\circ}$

Since $x$ and $y$ form a linear pair, $A O B$ is a line.

**5.** In Fig. 6.17, $\mathrm{POQ}$ is a line. Ray $\mathrm{OR}$ is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

$\angle \mathrm{ROS}=\frac{1}{2}(\angle \mathrm{QOS}-\angle \mathrm{POS})$.

## Show Answer

**Solution**

It is given that $OR PQ \quad \perp$

$\therefore \quad \therefore POR=90^{\circ}$

$\therefore \quad \therefore POS+\therefore SOR=90^{\circ}$

$\therefore ROS=90^{\circ}-\therefore POS$.

$\therefore QOR=90^{\circ}$ (As OR $.\therefore PQ)$

$\therefore QOS-\therefore ROS=90^{\circ}$

$\therefore ROS=\therefore QOS-90^{\circ}$

On adding equations (1) and (2), we obtain

**6.** It is given that $\angle \mathrm{XYZ}=64^{\circ}$ and $\mathrm{XY}$ is produced to point $\mathrm{P}$. Draw a figure from the given information. If ray $\mathrm{YQ}$ bisects $\angle \mathrm{ZYP}$, find $\angle \mathrm{XYQ}$ and reflex $\angle \mathrm{QYP}$.

Fig. 6.17

## Show Answer

**Solution**

It is given that line YQ bisects $\therefore P Y Z$.

Hence, $\therefore QYP=\therefore ZYQ$

It can be observed that $PX$ is a line. Rays $YQ$ and $YZ$ stand on it.

$\therefore X Y Z+Z Y Q+\therefore Q Y P=180^{\circ}$

$\therefore 64^{\circ}+2 QYP=180^{\circ}$

$\therefore 2$ Q̈PP $=180^{\circ}-64^{\circ}=116^{\circ}$

$\therefore \dot{Q} QP=58^{\circ}$

Also, $\stackrel{\grave{Z}}{ZYQ}=\therefore QYP=58^{\circ}$

Reflex QYP $=360^{\circ}-58^{\circ}=302^{\circ}$

$X Y Q=X Y Z+a Z Y Q$

$=64^{\circ}+58^{\circ}=122^{\circ}$

### 6.5 Lines Parallel to the Same Line

If two lines are parallel to the same line, will they be parallel to each other? Let us check it. See Fig. 6.18 in which line $m ||$ line $l$ and line $n ||$ line $l$. Let us draw a line $t$ transversal for the lines, $l, m$ and $n$. It is given that line $m ||$ line $l$ and line $n ||$ line $l$.

Therefore, $\angle 1=\angle 2$ and $\angle 1=\angle 3$

(Corresponding angles axiom) So, $\quad \angle 2=\angle 3$ (Why?)

But $\angle 2$ and $\angle 3$ are corresponding angles and they are equal.

Therefore, you can say that

$\text { Line } m || \text { Line } n$

(Converse of corresponding angles axiom)

This result can be stated in the form of the following theorem:

Fig. 6.18

**Theorem 6.6 :** Lines which are parallel to the same line are parallel to each other.

**Note:** The property above can be extended to more than two lines also.

Now, let us solve some examples related to parallel lines.

**Example 4 :** In Fig. 6.19, if $\mathrm{PQ} || \mathrm{RS}, \angle \mathrm{MXQ}=135^{\circ}$ and $\angle \mathrm{MYR}=40^{\circ}$, find $\angle \mathrm{XMY}$.

Fig. 6.19

**Solution :** Here, we need to draw a line AB parallel to line $\mathrm{PQ}$, through point $\mathrm{M}$ as shown in Fig. 6.20. Now, $A B ||P Q$ and $P Q || R S$.

Fig. 6.20

Therefore, $\quad \quad \quad$$\mathrm{AB} || \mathrm{RS}$ $\quad$(Why?)

Now, $\quad \quad \quad$$\angle \mathrm{QXM}+\angle \mathrm{XMB}=180^{\circ}$

$\quad \quad \quad$(AB ||PQ, Interior angles on the same side of the transversal XM)

But $\quad \quad \quad$$\angle \mathrm{QXM}=135^{\circ}$

So, $\quad \quad \quad$$135^{\circ}+\angle \mathrm{XMB}=180^{\circ}$

Therefore, $\quad \quad \quad$$\angle \mathrm{XMB}=45^{\circ}$ $\quad \quad \quad$(1)

Now,$\quad \quad \quad$$\angle \mathrm{BMY}=\angle \mathrm{MYR}$$\quad$(AB $||$ RS, Alternate angles)

Therefore,$\quad \quad \quad$$\angle \mathrm{BMY}=40^{\circ}$ $\quad \quad \quad$(2)

Adding (1) and (2), you get

$\quad \quad \quad$$\angle \mathrm{XMB}+\angle \mathrm{BMY}=45^{\circ}+40^{\circ}$

That is, $\quad \quad \quad$ $\angle XMY=85^{\circ}$

**Example 5 :** If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel.

**Solution :** In Fig. 6.21, a transversal AD intersects two lines PQ and RS at points B and $\mathrm{C}$ respectively. Ray $\mathrm{BE}$ is the bisector of $\angle \mathrm{ABQ}$ and ray $\mathrm{CG}$ is the bisector of $\angle \mathrm{BCS}$; and $\mathrm{BE} || \mathrm{CG}$.

We are to prove that $\mathrm{PQ} || \mathrm{RS}$.

It is given that ray $\mathrm{BE}$ is the bisector of $\angle \mathrm{ABQ}$.

Therefore, $\quad \angle \mathrm{ABE}=\frac{1}{2} \angle \mathrm{ABQ}$$\quad \quad \quad$(1)

Similarly, ray $\mathrm{CG}$ is the bisector of $\angle \mathrm{BCS}$.

Therefore, $\quad \angle \mathrm{BCG}=\frac{1}{2} \angle \mathrm{BCS}$$\quad \quad \quad$(2)

Fig. 6.21

But $\mathrm{BE} || \mathrm{CG}$ and $\mathrm{AD}$ is the transversal.

Therefore, $\quad \angle \mathrm{ABE}=\angle \mathrm{BCG}$ (Correspondinf angles axioms)$\quad \quad \quad$(3)

Substituting (1) and (2) in (3), you get

$$\frac{1}{2} \angle \mathrm{ABQ}=\frac{1}{2} \angle \mathrm{BCS}$$

That is, $\quad$$\angle \mathrm{ABQ}=\angle \mathrm{BCS}$

But, they are the corresponding angles formed by transversal AD with PQ and RS; and are equal.

Therefore,

PQ $||$ RS

(Converse of corresponding angles axiom)

**Example 6 :** In Fig. 6.22, $\mathrm{AB} || \mathrm{CD}$ and $\mathrm{CD} || \mathrm{EF}$. Also $\mathrm{EA} \perp \mathrm{AB}$. If $\angle \mathrm{BEF}=55^{\circ}$, find the values of $x, y$ and $z$.

**Solution :** $y+55^{\circ}=180^{\circ}$

(Interior angles on the same side of the transversal ED)

Therefore, $\quad$ $y=180^{\circ}-55^{\circ}=125^{\circ}$

Again $\quad \quad \quad$$x=y$

$\quad \quad$( $\mathrm{AB} || \mathrm{CD}$, Corresponding angles axiom)

Therefore $\quad \quad$$x=125^{\circ}$

Now, since $\mathrm{AB} || \mathrm{CD}$ and $\mathrm{CD} || \mathrm{EF}$, therefore, $\mathrm{AB} || \mathrm{EF}$.

Fig. 6.22

So, $\quad \quad$$\angle \mathrm{EAB}+\angle \mathrm{FEA}=180^{\circ}$(Interior angles on the same side of the transversal EA)

Therefore, $\quad$$90^{\circ}+z+55^{\circ}=180^{\circ}$

Which gives $\quad$$z=35^{\circ}$

### EXERCISE 6.2

**1.** In Fig. 6.23, if $\mathrm{AB}||\mathrm{CD}, \mathrm{CD}|| \mathrm{EF}$ and $y: z=3: 7$, find $x$.

Fig. 6.23

## Show Answer

**Solution**

It is given that $A B | C D$ and $C D | E F$

$\therefore A B | C D|| E F$ (Lines parallel to the same line are parallel to each other) It can be observed that $x=z$

(Alternate interior angles) … (1)

It is given that $y: z=3: 7$

Let the common ratio between $y$ and $z$ be $a$. $\therefore$

$y=3 a$ and $z=7 a$

Also, $x+y=180^{\circ}$ (Co-interior angles on the same side of the transversal) $z$

$+y=180^{\circ}$ [Using equation (1)]

$7 a+3 a=180^{\circ}$

$10 a=180^{\circ} a=$

$18^{\circ} \therefore x=7 a=7 \times 18^{\circ}=$

**2.** In Fig. 6.24, if $\mathrm{AB} || \mathrm{CD}, \mathrm{EF} \perp \mathrm{CD}$ and $\angle \mathrm{GED}=126^{\circ}$, find $\angle \mathrm{AGE}, \angle \mathrm{GEF}$ and $\angle \mathrm{FGE}$.

Fig. 6.24

## Show Answer

**Solution**

It is given that, $A B | C D$

EF CD

$\therefore GED=126^{\circ}$

$\therefore \therefore$ GEF $+\therefore F E D=1260$

$\therefore \quad GEF+90^{\circ}=126^{\circ}$

$\therefore \therefore \quad GEF=36^{\circ}$

$\therefore \quad$ AGE and GED are alternate interior angles.

$\therefore \quad \therefore \quad$ AGE $=$ GED $=126^{\circ}$

Howềver, $AGÉ+FGE=180^{\circ}$ (Linear pair)

$\therefore \quad 1260^{\circ}+FGE=180^{\circ}$

$\therefore \quad \therefore \quad FGE=180^{\circ}-126^{\circ}=54^{\circ}$

$AGE=126^{\circ}, \therefore GEF=36^{\circ}, \therefore FGE=54^{\circ}$

**3.** In Fig. 6.25, if $\mathrm{PQ} || \mathrm{ST}, \angle \mathrm{PQR}=110^{\circ}$ and $\angle \mathrm{RST}=130^{\circ}$, find $\angle \mathrm{QRS}$.

[**Hint :** Draw a line parallel to ST through point R.]

Fig. 6.25

## Show Answer

**Solution**

Let us draw a line XY parallel to ST and passing through point $R$. $\therefore PQR+QRX=180^{\circ}$ (Co-interior angles on the same side of transversal QR)

$\therefore 110^{\circ}+Q R X=180^{\circ}$

$\therefore QRX=70^{\circ}$

Also,

$\therefore RST+\therefore$ SRY $=180^{\circ}$ (Co-interior angles on the same side of transversal SR) $0+S R Y=180^{\circ} 130$

$\therefore SRY=50^{\circ}$

$XY$ is a straight line. RQ and RS stand on it.

$\therefore \therefore QRX+\therefore QRS+\therefore SRY=180^{\circ} \circ+QRS+50^{\circ}=180^{\circ} 70$

$QRS=180^{\circ}-120^{\circ}=60^{\circ}$

$\therefore$

**4.** In Fig. 6.26, if $\mathrm{AB} || \mathrm{CD}, \angle \mathrm{APQ}=50^{\circ}$ and $\angle \mathrm{PRD}=127^{\circ}$, find $x$ and $y$.

Fig. 6.26

## Show Answer

**Solution**

$\therefore APR=\therefore PRD$ (Alternate interior angles) In the given figure, if $AB | CD, APQ=50^{\circ}$ and $\therefore$ $P R D=1270$, find $x$ and $y$.

$50^{\circ}+y=127^{\circ} y=$

$127^{\circ}-50^{\circ} y=$

770

Also, $A P Q=P Q R$ (Alternate interior angles)

$50^{\circ}=x \quad x=50^{\circ}$ and $y=770$

**5.** In Fig. 6.27, PQ and RS are two mirrors placed parallel to each other. An incident ray $\mathrm{AB}$ strikes the mirror $\mathrm{PQ}$ at $\mathrm{B}$, the reflected ray moves along the path $\mathrm{BC}$ and strikes the mirror $\mathrm{RS}$ at $\mathrm{C}$ and again reflects back along $\mathrm{CD}$. Prove that $\mathrm{AB} || \mathrm{CD}$.

Fig. 6.27

## Show Answer

**Solution**

Let us draw $BM \therefore PQ$ and $CN \therefore RS$.

As PQ || RS,

Therefore, BM || CN

Thus, $BM$ and $CN$ are two parallel lines and a transversal line $BC$ cuts them at $B$ and

C respectively.

$\therefore A=3$ (Alternate interior angles) 2

However, $1=2$ and $3=* 4$ (By"laws of reflection)

$\therefore 1=2 \doteq 3=4$

Also, $\dot{1}+2 \stackrel{\dot{\circ}}{=} 3+\dot{4}$

$\therefore ABC=\ddot{DCB}$

However, these are alternate interior angles. :

$A B|| C D$

### 6.6 Summary

In this chapter, you have studied the following points:

**1.** If a ray stands on a line, then the sum of the two adjacent angles so formed is $180^{\circ}$ and viceversa. This property is called as the Linear pair axiom.

**2.** If two lines intersect each other, then the vertically opposite angles are equal.

**3.** Lines which are parallel to a given line are parallel to each other.