Question

If ABP and CQD are 2 parallel lines then bisectors of angles APQ , BPQ , CQP & PQD forms a​

Answer 1

Step-by-step explanation:

ANGLE

I THINK IT'S CORRECT

Answer 2

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]

⇒ 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