In the adjoining figure, line AB || line CD and line PQ is the transve

1.	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°.
Answer» 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° Proof: ∠TPB = (1/2) ∠TPQ = ∠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) x 180° [Multiplying both sides by (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°

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°.

Answer»

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°

Proof: ∠TPB = (1/2) ∠TPQ = ∠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) x 180°

[Multiplying both sides by (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°