Question

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

Options

• a square

• a rhombus

• a rectangle

• any other parallelogram

Answer 1

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

Explanation:

Given, APB and CQD are two parallel lines.

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

Join PM, MQ, QN and NP.

Since, APB || CQD

Then, ∠APQ = ∠PQD   ...[Alternate interior angles]

⇒ ∠MPQ = 2∠NQP   ...[Since, PM and NQ are the angle bisectors of ∠APQ and ∠DQP respectively]

⇒ ∠MPQ = ∠NQP  ...[Dividing both sides by 2] [Since, alternative interior angles are equal]

∴ PM || QN    

Similarly, ∠BPQ = ∠CQP   ...[Alternate interior angles]

∴ PN || QM

So, quadrilateral PMQN is a parallelogram.

∵ ∠CQD = 180°   ...[Since, CQD is a line]

⇒ ∠CQP + ∠DQP = 180°

⇒ 2∠MQP + 2∠NQP = 180°   ...[Since, MQ and NQ are the bisectors of the angles CQP and DQP]

⇒ 2(∠MQP + ∠NQP) = 180°

⇒ ∠MQN = 90°

Hence, PMQN is a rectangle.