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°.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» ∠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°

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»

∠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°