Question

Prove that, if the diagonals of a parallelogram are perpendicular to each other, the parallelogram is a rhombus.

Answer 1

Since the diagonals of a parallelogram bisect each other,

$BE$ and $DE$ are congruent and $AE$ is congruent to itself.

We are given that all four angles at point $E$ are ${90}^0$ and

are therefore congruent.

Triangles $ABE$ and $ADE$ are congruent by side-angle-side,

Therefore, $AB$ and $AD$ are congruent.

Since opposite sides of a parallelogram are congruent,

$AD$ and $BC$ are congruent, and $AB$ and $CD$ are congruent.

So all four sides are congruent.

Hence, $ABCD$ is a rhombus.