Q8 of 56 Page 73

If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form


Let the bisectors of the angles APQ and CQP meet at M and bisectors of the angles BPQ and PQD meet at N.


Join PM, MQ, QN and NP.


∠APQ = ∠PQD [∵APB ∥ CQD]


⇒ 2∠MPQ = 2∠NQP [∵ NP and PQ are angle bisectors]


Dividing both sides by 2,


⇒ ∠MPQ = ∠NQP


⇒ PM ∥ QN


Similarly,


⇒ ∠BPQ = ∠CQP


⇒ PN ∥ QM


∴ PNQM is a parallelogram


Now,


∠CQP + ∠CQP = 180° [Angles on a straight line]


⇒ 2∠MQP + 2∠NQP = 180°


Dividing both sides by 2,


⇒ ∠MQP + ∠NQP = 90°


⇒ ∠MQN = 90°


Hence, PMQN is a rectangle

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