Question

The bisectors of angles $B$ and $C$ of a parallelogram $ABCD$ meet at point $O.$ If the triangle $OBC$ formed is an isosceles right triangle, then $ABCD$ is a rectangle.
If the above statement is true then mention the answer as $1,$ else mention $0$ if false.

Answer 1

Given:

$ABCD$ is a parallelogram. $OB$ and $OC$ bisect $\angle B$ and $\angle C$, $OBC$ is a right isosceles triangle.

To prove:

$ABCD$ is a rectangle.

Proof:

If we can prove that adjacent angles of parallelogram $ABCD$ are of ${90}^{\circ }$ then that will implies that $ABCD$ is a rectangle.

In $△OBC$,

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

And,

$OB=OC$

By using the theorem that, angles opposite to equal sides are equal we can assume that,

$\angle OBC=\angle OCB=x$

Now, by applying angle sum property,

$\angle OBC+\angle OCB+\angle BOC={180}^{\circ }$

$\Rightarrow x+x+{90}^{\circ }={180}^{\circ }$

$\Rightarrow 2x={90}^{\circ }$

$\Rightarrow x={45}^{\circ }$

Hence,

$\angle OBC=\angle OCB={45}^{\circ }$

But it is given that, $OB$ and $OC$ bisect $\angle B$ and $\angle C$ that implies $\angle B=\angle C={90}^{\circ }$

Since, adjacent angles are right angles. $ABCD$ is a rectangle.