# Diagonals of Rhombus Bisect Each Other at Right Angles

## Theorem

Let $ABCD$ be a rhombus.

The diagonals $AC$ and $BD$ of $ABCD$ bisect each other at right angles.

## Proof

By the definition of a rhombus, $AB = AD = BC = DC$.

Without loss of generality, consider the diagonal $BD$.

Thus:

$\triangle ABD$ is an isosceles triangle whose apex is $A$ and whose base is $BD$.

By Diagonals of Rhombus Bisect Angles, $AC$ bisects $\angle BAD$.

Hence the result.

$\blacksquare$