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

Given:

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

To do :

We have to show that

(i) $ABCD$ is a square

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

Solution :

(i) A square is a rectangle when all sides are equal.

In the above figure, $AC$ bisects $\angle A$ as well as $\angle C$.

Therefore, 

$\angle DAC = \angle BAC$............(i)

$\angle DCA = \angle BCA$............(ii)

In a rectangle, the opposite sides are parallel.

So,  $AD \parallel BC$

$AC$ is transversal.

Therefore, 

$\angle DAC = \angle BCA$...........(iii)                      [Alternate interior angles]

From (i) and (iii),

$\angle BCA = \angle BAC$..........(iv)

In $\Delta ABC$,

$\angle BCA = \angle BAC$

So, $AB = BC$...........(v)

We already know that,

In rectangle $ABCD$, $AB = CD, BC = DA$.............(vi)

From (v) and (vi), 

$AB = BC = CD = DA$.

Therefore, $ABCD$ is a square.

(ii) $ABCD$ is a square, 

Diagonals of a square bisect its angles.

Therefore, Diagonal $BD$ bisects $\angle B$ as well as $\angle D$. 

Hence proved. 

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Updated on: 10-Oct-2022

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