If in Fig. 6.11, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m.
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.