**Question 8: In the given Figure, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of**

**∠**

**BPQ and**

**∠**

**PQD respectively. Prove that m**

**∠**

**PTQ = 90 °.**

Answer:

AB || CD and PQ is a transversal line.

∠BPQ + ∠DQP = 180∘ (Angles on the same side of a transversal line are supplementary angles)

⇒ ∠QPT + ∠PQT = 90∘

In △PQT

∠QPT + ∠PQT + ∠PTQ = 180∘ (Angle sum property)

⇒ 90∘ + ∠PTQ = 180∘

⇒ ∠PTQ = 90∘

Hence proved.

∠BPQ + ∠DQP = 180∘ (Angles on the same side of a transversal line are supplementary angles)

⇒ ∠QPT + ∠PQT = 90∘

In △PQT

∠QPT + ∠PQT + ∠PTQ = 180∘ (Angle sum property)

⇒ 90∘ + ∠PTQ = 180∘

⇒ ∠PTQ = 90∘

Hence proved.