Question

Question 8
If APB and COD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form
(A) a square
(B) a rhombus
(C) a rectangle
(D) any other parallelogram

Answer 1

Given, APB and CQD are two parallel lines.

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

Join PM. MQ, QN and NP.

Since, APB ∥ CQD

Then, $\angle APQ=\angle PQD$ [alternate interior angles]
$\Rightarrow \angle MPQ=2\angle NQP$
[Since, PM and NQ are the angle bisectors of $\angle APQand\angle DQP$ respectively]
$\Rightarrow \angle MPQ=\angle NQP$ [Dividing both sides by 2]
{since, alternate interior angles are equal.]
$\therefore PM\parallel QN$
Similarly , $\angle BPQ=\angle CQP$ [alternate interior angles]
PN ∥ QM
So, quadrilateral PMQN is a parallelogram.
$\because \angle CQD={180}^{\circ }$ [Since , CQD is a line]
$\Rightarrow \angle CQP+\angle DQP={180}^{\circ }\Rightarrow 2\angle MQP+2\angle NQP={180}^{\circ }$
[Since , MQ and NQ are the bisectors of the angles CQP and DQP]
$\Rightarrow 2(\angle MCQ+\angle NQP)={180}^{\circ }\Rightarrow \angle MQN={90}^{\circ }$
Hence, PMQN is a rectangle.