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If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Given:
Diagonals of a parallelogram are equal.
To do:
We have to show that it is a rectangle.
Solution:
Consider triangles $ABD$ and $ACD$.
$AC = BD$ [Given]
$AB = DC$ [Opposite sides of a parallelogram are equal]
$AD = AD$ [Common side]
Therefore, by SSS congruence, we get,
$\triangle ABD \cong \triangle DCA$ [SSS congruence criterion]
This implies,
$\angle BAD = \angle CDA$ [Corresponding parts of congruent triangles are equal]
$\angle BAD + \angle CDA = 180^o$ [Adjacent angles of a parallelogram are supplementary]
So, $\angle BAD$ and $\angle CDA$ are right angles as they are congruent and supplementary.
Therefore, parallelogram $ABCD$ is a rectangle since a parallelogram with one right interior angle is a rectangle.
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