Question

In the adjoining 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°.

Given: line AB || line CD and line PQ is the transversal. 

ray PT and ray QT are the bisectors of ∠BPQ and ∠PQD respectively. 

To prove: m∠PTQ = 90°

Answer 1

∠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] 

∴ \(\frac{1}2\)(∠BPQ) + \(\frac{1}2\)(∠PQD) =\(\frac{1}2\) x 180° [Multiplying both sides by \(\frac{1}2\)] 

∠TPQ + ∠TQP = 90°

In ∆PTQ, 

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

∴ 90° + ∠PTQ = 180° [From (iii)] 

∴ ∠PTQ = 180° – 90°

 = 90° 

∴ m∠PTQ = 90°