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

Given:

$ABCD$ is a rectangle.

Diagonal $AC$ bisects $\angle A$ as well as $\angle C$.

To prove:

(i). $ABCD$ is a square.

(ii). Diagonal $BD$ bisects $\angle B$ as well as $\angle D$.

Proof:

Since $AC$ bisects $\angle A$ as well as $\angle C$

$\Rightarrow ,\angle 1=\angle 2$ and $\angle 3=\angle 4$

Since, in rectangle all the angles are ${90}^{\circ }$.

in the rectangle $ABCD$

$\therefore \angle 1=\angle 2=\angle 3=\angle 4$

In $\Delta ABC\angle 2=\angle 4$

$\Rightarrow AB=BC$ [Angles are opposite to equal sides of triangle]---(1)

In rectangle,

$AB=CD$ and $AD=BC$.

i.e., $AB=CD=BC=AD$

Thus, the rectangle $ABCD$ is square.---(i)

(ii)

In square , diagonal bisects the vertical angles.

So $BD$ bisects $\angle B$ as well as $\angle D$.