Question

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 1

Given: line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of ∠BPQ and ∠PQD respectively. 

To prove: m∠PTQ = 90°

Proof:

∴ ∠TPB = ∠TPQ = `1/2∠"BPQ"`      ...(i)[Ray PT bisects ∠BPQ]

∴ ∠TQD = ∠TQP = `1/2∠"PQD"`     ...(ii)[Ray QT bisects ∠PQD]

line AB || line CD and line PQ is their transversal.    ...(Given)

∴ ∠BPQ + ∠PQD = 180^∘     ...[Interior angles]

`1/2(∠"BPQ") + 1/2(∠"PQD") = 1/2 × 180°   ...["Multiplying both sides by" 1/2]`

∠TPQ + ∠TQP = 90°   ...(iii)

In ΔPTQ,

∠TPQ + ∠TQP + ∠PTQ = 180°     ...[Sum of the measures of the angles of a triangle is 180°]

∴ 90° + ∠PTQ = 180°                     ...[From (iii)]

∴ ∠PTQ = 180° − 90°

∴ ∠PTQ = 90°

∴ m∠PTQ = 90°

Hence proved.