# $\mathrm{ABCD}$ 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 $\angle \mathrm{D}$.

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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.

Updated on 10-Oct-2022 13:41:01