In Figure 3.11, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of ∠BPQ and ∠PQD respectively.

Prove that ∠PTQ = 90°.

Given: AB || CD, line PQ is the tranversal

Ray PT and Ray QT are bisectors of ∠BPQ and ∠PQD

To prove: ∠PTQ = 90°

Proof: Since, Ray PT and Ray QT are bisectors of ∠BPQ and ∠PQD

∠TPQ = ∠BPQ/2 ……..(1)

∠PQT = ∠PQD/2 ………(2)

Since, two parallel lines are intersected by a transversal, the interior angles on either side of the transversal are supplementary.

So, ∠BPQ + ∠PQD = 180°

Dividing both sides by 2, we get

⇒ (∠BPQ + ∠PQD)/2 = 180°/2

⇒ ∠BPQ/2 + ∠PQD/2 = 90°

In ΔPQT,

∠TPQ + ∠PQT + ∠PTQ = 180°

Substituting ∠TPQ and ∠PQT from (1) and (2) respectively

⇒ ∠BPQ/2 + ∠PQD/2 + ∠PQT = 180°

⇒ 90° + ∠PQT = 180°

⇒ ∠PQT = 180° - 90°

⇒ ∠PQT = 90°

Hence, proved.