Question

In the given Figure, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of $\angle$BPQ and $\angle$PQD respectively.
Prove that m $\angle$PTQ = 90 °.

Answer 1

AB || CD and PQ is a transversal line.
∠BPQ + ∠DQP = 180^∘ (Angles on the same side of a transversal line are supplementary angles)
$\Rightarrow \frac{1}{2}\angle \mathrm{BPQ}+\frac{1}{2}\angle \mathrm{DQP}=\frac{180°}{2}$
⇒ ∠QPT + ∠PQT = 90^∘
In △PQT
∠QPT + ∠PQT + ∠PTQ = 180^∘ (Angle sum property)
⇒ 90^∘ + ∠PTQ = 180^∘
⇒ ∠PTQ = 90^∘
Hence proved.