Question

$ABCD$ is a rectangle in which diagonal AC bisects $\angle A$ as well as$\angle C$.

Show that,
(i) $ABCD$ is square
(ii) diagonal $BD$ bisects $\angle B$ as well as $\angle D$.

Answer 1

$\left(i\right)$ABCD is a rectangle , in which diagonal AC bisect $\angle A$ as well as $\angle C$. Therefore,

$\angle DAC=\angle CAB\to \left(1\right)$

$\angle DCA=\angle BCA\to \left(2\right)$

A square is a rectangle when all sides are equal. Now,

$AD\parallel BC$ & $AC$ is transversal, therefore

$\angle DAC=\angle BCA$ [Alternate angles]

From (1), $\angle CAB=\angle BCA\to \left(3\right)$

In $△ABC$,

$\angle CAB=\angle BCA$ , therefore

$BC=AB$ $\to \left(4\right)$[sides opposite to equal angles]

But $BC=AD$ & $AB=DC$$\to \left(5\right)$ [Opposite sides of rectangle]

Therefore from (4)& (5),

$AB=BC=CD=AD$

Hence, ABCD is a square.

$\left(ii\right)$ ABCD is a square and we know that diagonals of a square bisect its

angles.

Hence, BD bisects $\angle B$ as well as $\angle D$.