NCERT Class 9 Mathematics Textbook for Blind Students made Screen Readable by Professor T K Bansal.

If two lines are parallel to the same line, will they be parallel to each other? Let us check it. See Figure 6.23 in which line m ∥ line l and line n ∥ line l.

Figure 6.23

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, ∠1 = ∠2 and ∠1 = ∠3 (Corresponding angles axiom)

So, ∠2 = ∠3 (Why?)

But ∠2 and ∠3 are corresponding angles, and they are equal.

Therefore, we can say that

Line m ∥ Line n (Converse of corresponding angles axiom)

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

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 as well.

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

Example 4

In Figure 6.24, if PQ ∥ RS, ∠MXQ = 135° and ∠MYR = 40°, find ∠XMY.

Figure 6.24

Solution :

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

Figure 6.25

Therefore, AB ∥ RS (Why?)

Now, ∠QXM + ∠XMB = 180°
(AB ∥ PQ, Interior angles on the same side of the transversal XM)
But ∠QXM = 135°
So, 135° + ∠XMB = 180°
Therefore, ∠XMB = 45° (1)

Now, ∠BMY = ∠MYR (AB ∥ RS, Alternate angles)
Therefore, ∠BMY = 40° (2)
Adding (1) and (2), we get
∠XMB + ∠BMY = 45° + 40°

That is, ∠XMY = 85°

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 Figure 6.26, a transversal AD intersects two lines PQ and RS at points B and C respectively. Ray BE is the bisector of ∠ABQ and ray CG is the bisector of ∠BCS; and BE ∥ CG.

Figure 6.26

We are to prove that PQ ∥ RS.

It is given that ray bE is the bisector of ∠ABQ.
Therefore, ∠ABE = 1/2∠ABQ (1)

Similarly, ray CG is the bisector of ∠BCS.
Therefore, ∠BCG =1/2 ∠BCS (2)

But B E ∥ CG and AD is the transversal. Therefore, ∠ABE = ∠BCG (Corresponding angles axiom) (3)
Substituting (1) and (2) in (3), we get 1/2 ∠ABQ =1/2 ∠BCS
That is, ∠ABQ = ∠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 Figure 6.27, AB ∥ CD and CD ∥ eF. Also, EA perpendicular AB. If ∠BEF = 55°, find the values of x, y and z.

Figure 6.27

Solution :

y + 55° = 180° (Interior angles on the same side of the transversal ED)

Therefore, y = 180° − 55° = 125°

Again x = y (AB ∥ CD, Corresponding angles axiom)
Therefore x = 125°

Now, since AB ∥ CD and CD ∥ eF, therefore, AB ∥ eF.

So, ∠EAB + ∠FEA = 180° (Interior angles on the same side of the transversal EA)

Therefore, 90° + z + 55° = 180°
Which gives z = 35°

EXERCISE 6.2

Q1. In Figure 6.28, find the values of x and y and then show that AB ∥ CD (AB parallel to CD).

Figure 6.28

A1. 130°, 130°

Q2. In Figure 6.29, if AB ∥ CD, CD ∥ eF and y ∶ z = 3 ∶ 7, find x.

Figure 6.29

A2. 126°

Q3. In Figure 6.30, if AB ∥ CD, eF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Figure 6.30

A3. 126°, 36°, 54°

Q4. In Figure 6.31, if PQ ∥ ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

Figure 6.31

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

A4. 60°

Q5. In Figure 6.32, if AB ∥ CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Figure 6.32

A5. 50°, 77°

Q6. In Figure 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB ∥ CD.

Figure 6.33

A6. ∠of incidence = ∠of reflection. At point B, draw BE perpendicular to PQ and at point C, draw CF perpendicular to RS.