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, $AB = AD = BC = DC$.

, 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 $\angle BAD$.

From Bisector of Apex of Isosceles Triangle also Bisects Base, $AC$ bisects $BD$.

From Bisector of Apex of Isosceles Triangle is Perpendicular to Base, $AC$ bisects $BD$ at right angles.

Hence the result.