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 °.

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)
⇒12∠BPQ+12∠DQP=180°2$\frac{\mathrm{}}{\mathrm{}}\frac{\mathrm{}}{\mathrm{}}\frac{\mathrm{}}{\mathrm{}}$
⇒ ∠QPT + ∠PQT = 90∘
In △PQT
∠QPT + ∠PQT + ∠PTQ = 180∘   (Angle sum property)
⇒ 90∘ + ∠PTQ = 180∘         
⇒ ∠PTQ = 90∘ 
Hence proved.