Question

The bisectors of the angle of a parallelogram enclose a

(a) parallelogram

(b) rhombus

(c) rectangle

(d) square

Answer 1

The correct choice is (c). rectangle.

We have ABCD, a parallelogram given below:

Therefore, we have $AD\parallel BC$

Now, $AD\parallel BC$ and transversal AB intersects them at A and B respectively. Therefore,

The Sum of consecutive interior angles is supplementary. That is;

$\angle A+\angle B={180}^{\circ }$

$\frac{1}{2}\angle A+\frac{1}{2}\angle B={90}^{\circ }$

We have AR and BR as bisectors of $\angle A$ and $\angle B$ respectively.

$\angle RAB+\angle RBA={90}^{\circ }$ …… (i)

Now, in $△ABR$, by angle sum property of a triangle, we get:

$\angle RAB+\angle RBA+\angle ARB={180}^{\circ }$

From equation (i), we get:

${90}^{\circ }+\angle ARB={180}^{\circ }$

$\angle ARB={90}^{\circ }$

Similarly, we can prove that $\angle DPC={90}^{\circ }$.

Now, $AB\parallel DC$ and transversal AD intersects them at A and D respectively. Therefore,

The Sum of consecutive interior angles is supplementary. That is;

$\angle A+\angle D={180}^{\circ }$

$\frac{1}{2}\angle A+\frac{1}{2}\angle D={90}^{\circ }$

We have AR and DP as bisectors of $\angle A$ and $\angle D$ respectively.

$\angle DAR+\angle ADP={90}^{\circ }$ …… (ii)

Now, in $△ADR$, by angle sum property of a triangle, we get:

$\angle DAR+\angle ADP+\angle AQD={180}^{\circ }$

From equation (i), we get:

${90}^{\circ }+\angle AQD={180}^{\circ }$

$\angle AQD={90}^{\circ }$

We know that $\angle AQD$ and $\angle PQR$ are vertically opposite angles, thus

$\angle PQR={90}^{\circ }$

Similarly, we can prove that $\angle PSR={90}^{\circ }$.

Therefore, PQRS is a rectangle.

Hence, the correct choice is (c).