Class 9 Parallel Lines and a Transversal and Lines Parallel to the Same Line for CBSE-NCERT

### Topic covered

color{red} ♦Parallel Lines and a Transversal
color{red} ♦Lines Parallel to the Same Line

### Parallel Lines and a Transversal

As we know that a line which intersects two or more lines at distinct points is called a "transversal" (see Fig. 6.18).

Line l intersects lines m and n at points P and Q respectively. Therefore, line l is a transversal for lines m and n. Observe that four angles are formed at each of the points P and Q.

Let us name these angles as angle 1, angle 2, . . ., angle 8 as shown in Fig. 6.18.

angle 1, angle 2, angle 7 and angle 8 are called exterior angles, while angle 3, angle 4, angle 5 and angle 6 are called interior angles.

Recall that in the earlier classes, you have named some pairs of angles formed when a transversal intersects two lines. These are as follows:

(a) Corresponding angles :

(i) angle 1 and angle 5 (ii) angle 2 and angle 6
(iii) angle 4 and angle 8 (iv) angle 3 and angle 7

(b) Alternate interior angles :

(i) angle 4 and angle 6 (ii) angle 3  and angle 5

(c) Alternate exterior angles:
(i) angle 1 and angle 7 (ii) angle 2  and angle 8

(d) Interior angles on the same side of the transversal:

(i) angle 4 and angle 5 (ii) angle 3 and angle 6

Interior angles on the same side of the transversal are also referred to as consecutive interior angles or allied angles or co-interior angles. Further, many a times, we simply use the words alternate angles for alternate interior angles.

Now, let us find out the relation between the angles in these pairs when line m is parallel to line n. You know that the ruled lines of your notebook are parallel to each other.

So, with ruler and pencil, draw two parallel lines along any two of these lines and a transversal to intersect them as shown in Fig. 6.19.

Now, measure any pair of corresponding angles and find out the relation between them. You may find that : angle 1 = angle 5, angle 2 = angle 6, angle 4 = angle 8 and angle 3 = angle 7. From this, you may conclude the following axiom.

color{blue} text(Axiom 6.3) :

If a transversal intersects two parallel lines, then each pair of corresponding angles is equal.

Axiom 6.3 is also referred to as the corresponding angles axiom. Now, let us discuss the converse of this axiom which is as follows:

If a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines are parallel.

Does this statement hold true? It can be verified as follows: Draw a line AD and mark points B and C on it. At B and C, construct angle ABQ and angle BCS  equal to each other as shown in Fig. 6.20 (i).

Produce QB and SC on the other side of AD to form two lines PQ and RS [see Fig. 6.20 (ii)]. You may observe that the two lines do not intersect each other.

You may also draw common perpendiculars to the two lines PQ and RS at different points and measure their lengths.

You will find it the same everywhere. So, you may conclude that the lines are parallel. Therefore, the converse of corresponding angles axiom is also true. So, we have the following axiom:

color{blue} text(Axiom 6.4) :

If a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines are parallel to each other.

Can we use corresponding angles axiom to find out the relation between the alternate interior angles when a transversal intersects two parallel lines? In Fig. 6.21, transveral PS intersects parallel lines AB and CD at points Q and R respectively.

Is angle BQR = angle QRC and angle AQR = angle QRD ?

You know that angle PQA = angle QRC (1)

(Corresponding angles axiom)

Is angle PQA = angle BQR ? Yes! (Why ?) (2)

So, from (1) and (2), you may conclude that

angle BQR = angle QRC.

Similarly, angle AQR = angle QRD.

This result can be stated as a theorem given below:

color {blue} text(Theorem 6.2 :)

If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.

Now, using the converse of the corresponding angles axiom, can we show the two lines parallel if a pair of alternate interior angles is equal? In Fig. 6.22, the transversal PS intersects lines AB and CD at points Q and R respectively such that angle BQR = angle QRC.

Is AB || CD?

angle BQR = angle PQA (1)

But, angle BQR = angle QRC (Given) (2)

So, from (1) and (2), you may conclude that

angle PQA = angle QRC

But they are corresponding angles.

So, AB || CD (Converse of corresponding angles axiom)

This result can be stated as a theorem given below:

 color{blue} text(Theorem 6.3) :

If a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.

In a similar way, you can obtain the following two theorems related to interior angles on the same side of the transversal.

 color{blue} text(Theorem 6.4) :

If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

 color{blue} text(Theorem 6.5) :

If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.

You may recall that you have verified all the above axioms and theorems in earlier classes through activities. You may repeat those activities here also.

### 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. 6.23 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, angle 2 = angle 3

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

Therefore, you can say that

Line m | | Line n

(Converse of corresponding angles axiom)

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

color {blue} text( 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.
Q 3210467319

In Fig. 6.24, if PQ || RS , angle MXQ = 135° and angle MYR = 40°, find angle XMY.

Class 9 Chapter 6 Example 4
Solution:

Here, we need to draw a line AB parallel to line PQ, through point M as shown in Fig. 6.25. Now, AB || PQ  and PQ || RS.

Therefore, AB || RS (Why ?)

Now, angle QXM + angle XMB = 180°

(AB || PQ, Interior angles on the same side of the transversal X M)

But angle QXM = 135°

So, 135° + angle XMB = 180°

Therefore, angle XMB = 45° (1)

Now, angle BMY = angle MYR (AB || RS, Alternate angles)

Therefore, angle BMY = 40° (2)

Adding (1) and (2), you get

angle XMB + angle BMY = 45° + 40°

That is, angle XMY = 85°
Q 3210567410

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.
Class 9 Chapter 6 Example 5
Solution:

In Fig. 6.26, a transversal AD intersects two lines PQ and RS at points B

and C respectively. Ray BE is the bisector of angle ABQ and ray CG is the bisector of

angle BCS ; and BE || CG.

We are to prove that PQ || RS.

It is given that ray BE is the bisector of angle ABQ.

Therefore, angle ABE = 1/2 angle ABQ (1)

Similarly, ray CG is the bisector of angle BCS.

Therefore, angle BCG = 1/2 angle BCS (2)

But BE || CG  and AD is the transversal.

Therefore, angle ABE = angle BCG

(Corresponding angles axiom) (3)

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

 1/2 angle ABQ = 1/2 angle BCS

That is, angle ABQ = angle 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)
Q 3220567411

In Fig. 6.27, AB || CD and CD || EF. Also EA bot AB. If angle BEF = 55°, find the values of x, y and z.
Class 9 Chapter 6 Example 6
Solution:

y + 55° = 180°

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

Therefore, y = 180º – 55º = 125º

Again x = y

(AB || CD, Corresponding angles axiom)

Therefore x = 125º

Now, since AB || CD and CD || EF, therefore, AB || EF.

So, angle EAB + angle FEA = 180° (Interior angles on the same

side of the transversal EA)

Therefore, 90° + z + 55° = 180°

Which gives  z = 35°