Q. 83.9( 8 Votes )

If APB and CQD ar A. a square

B. a rhombus

C. a rectangle

D. any other parallelogram

Answer :


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]


2MPQ = 2NQP [ 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]


2MQP + 2NQP = 180°


Dividing both sides by 2,


⇒ ∠MQP + NQP = 90°


⇒ ∠MQN = 90°


Hence, PMQN is a rectangle

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