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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 $\angle D$.
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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