Let and be two parallel lines and be a transversal. Show that the angle bisectors of a pair of two internal angles on the same side of the transversal are perpendicular to each other.

To Prove: ∠ 2 + ∠ 4 = 90°

The situation is shown in the diagram above

AB is parallel to CD and PQ is a traversal

From the diagram we can see that (∠1 + ∠ 2) is an interior angle and similarly ∠3 + ∠ 4 is another interior angle made on the same side

Now ∠1 = ∠ 2 (As they are angular bisector of ∠ AOQ

And similarly, ∠3 = ∠ 4 ( As they are bisector of ∠ BOQ)

From the straight line AB we know that angle made on straight line makes 180°

Therefore, ∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 = 180°

2 ( ∠ 2 + ∠ 4) = 180°

∠ 2 + ∠ 4 = 90°

Hence, Proved.