If in Fig. 6.11,

It is given to us –

∠EAB and ∠ABH are a pair of alternate interior angles.

AP and BQ are the bisectors of the angles ∠EAB and ∠ABH respectively.

⇒ ∠EAP = ∠PAB, and ∠ABQ = ∠QBH - - - - (i)

Also, AP || BQ

We have to show that l || m.

Now, since t is a transversal intersecting two parallel lines AP and BQ at points A and B,

∠PAB = ∠ABQ (alternate interior angles) - - - - (ii)

Now,

∠EAB = ∠EAP + ∠PAB, and ∠ABH = ∠ABQ + ∠QBH

⇒ ∠EAB = 2 × ∠PAB, and ∠ABH = 2 × ∠ABQ [From equation (i)]

⇒ ∠EAB = 2 × ∠PAB, and ∠ABH = 2 × ∠PAB

⇒ ∠EAB = ∠ABH - - - - (iii)

From the figure and the equation (iii), we can say that –

t is a transversal intersecting lines l and m such that the alternate angles, ∠EAB and ∠ABH are equal.

Thus, it is true that l || m.