Question

Show that the bisector of angles of a parallelogram from a rectangle.

Answer 1

Let $ABCD$ is a parallelogram

In figure,

$AR,BR,CP$ and $DP$ are the bisectors of angles $\angle A,\angle B,\angle C$ and $\angle D$ respectively.

The bisectors $AR,BR,CP$ and $DP$ formed a quadrilateral $PQRS$.

To prove: $PQRS$ is a parallelogram.

Proof:

we know that,

In parallelogram, sum of the consecutive angles is supplementary.

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

$\Rightarrow \frac{\angle A+\angle B}{2}=\frac{{180}^{\circ }}{2}$

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

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

Now, In $\Delta ARB$,

$\frac{\angle A}{2}+\frac{\angle B}{2}+\angle ARB={180}^{\circ }$ [Since, angle sum property of a triangle]

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

$\Rightarrow \angle ARB={180}^{\circ }-{90}^{\circ }$

$\Rightarrow \angle ARB={90}^{\circ }$ ---(1)

Similarly,

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

$\Rightarrow \frac{\angle C+\angle D}{2}={90}^{\circ }$

$\Rightarrow \frac{\angle C}{2}+\frac{\angle D}{2}={90}^{\circ }$

In $\Delta CPD$,

$\Rightarrow \frac{\angle C}{2}+\frac{\angle D}{2}+\angle CPD={180}^{\circ }$

$\Rightarrow {90}^{\circ }+\angle CPD={180}^{\circ }$

$\Rightarrow \angle CPD={180}^{\circ }-{90}^{\circ }$

$\Rightarrow \angle CPD={90}^{\circ }$ ---(2)

Now,

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

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

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

In $\Delta ASD$

$\Rightarrow \frac{\angle A}{2}+\frac{\angle D}{2}+\angle ASD={180}^{\circ }$

$\Rightarrow {90}^{\circ }+\angle ASD={180}^{\circ }$

$\Rightarrow \angle ASD={180}^{\circ }-{90}^{\circ }$

$\Rightarrow \angle ASD={90}^{\circ }$

$\Rightarrow \angle RSP=\angle ASD={90}^{\circ }$ [Since, vertically opposite angles] ---(3)

In Quadrilateral $PQRS$,

$\angle P+\angle S+\angle R+\angle Q={360}^{\circ }$

$\Rightarrow {90}^{\circ }+{90}^{\circ }+{90}^{\circ }+\angle Q={360}^{\circ }$

$\Rightarrow {270}^{\circ }+\angle Q={360}^{\circ }$

$\Rightarrow \angle Q={360}^{\circ }-{270}^{\circ }$

$\Rightarrow \angle Q={90}^{\circ }$ ---(4)

From (1), (2), (3) and (4),

In Quadrilateral $PQRS$,

$\angle P=\angle Q=\angle R=\angle S={90}^{\circ }$

$\Rightarrow$ $PQRS$ is a rectangle.

Hence, proved.