Question

$ABCD$ is a rectangle in which diagonal $AC$ bisects $\angle A$ as well as $\angle C$. Show that:
(i) $ABCD$ is a square (ii) diagonal $BD$ bisects $\angle B$ as well as $\angle D$

Answer 1

$\square ABCD$ is a rectangle ...given

$\therefore$ All the angles are equal to ${90}^{\circ }$ ...by definition

$\therefore \angle A=\angle C={90}^o$

So, $\frac{\angle A}{2}=\frac{\angle C}{2}={45}^o$ ...($AC$ bisects $\angle A$ as well as $\angle C$)

So, $\angle x=\angle y=\angle z=\angle w={45}^o$ ....(i)

In $△ADC$,
$\angle x=\angle y$
$AD=CD$ ....from (i) and converse of isosceles triangle theorem

So, $\square ABCD$ is a square.

(ii) Diagonals bisect the angles in a square.

$\Rightarrow BD$ bisects $\angle B$ as well as $\angle D$. ....from(i)