Question

If $APB$ and $CQD$ are two parallel lines then the bisectors of $\angle APQ,\angle BPQ,\angle CQPand\angle PQD$ enclose a

(a) square

(b) rhombus

(c) rectangle

(d) kite

Answer 1

Consider the bisectors of angles $APQ$ and $CPQ$ meet at the point $M$ and the bisectors of angles $BPQ$ and $PQD$ meet at the point $N$.

Now join $PM,MQ,QN$ and $NP$

As $APB\parallel CQD$

$\angle APQ=\angle PQD$

As $NP$ and $PQ$ are angle bisectors

$2\angle MPQ=2\angle NQP$

$\Rightarrow \angle MPQ=\angle NQP$

$\Rightarrow PM\parallel QN$

In the same way,

$\angle BPQ=\angle CQP$

$\Rightarrow PN\parallel QM$

So $PNQM$ is a parallelogram.

We know that angles on a straight line is ${180}^{\circ }$

$\angle CQP+\angle CQP={180}^{\circ }$

$|Rightarrow2\angle MPQ+2\angle NQP={180}^{\circ }$

$\Rightarrow \angle MPQ+\angle NQP={90}^{\circ }$

$\Rightarrow MQN={90}^{\circ }$

Therefore, the bisectors of the angles $\angle APQ,\angle BPQ,\angle CQPand\angle PQD$ enclose a rectangle.

Hence, Option c is correct.