In the previous post, we have proved the **converse of the Pythagorean Theorem**. In this post, we will prove that the diagonals of a rhombus are perpendicular to each other. That is, if we have parallelogram *ABCD* with diagonal $latex \overline{AC}$ and $latex \overline{BD}$, then $latex \overline{AC}$ is perpendicular to $latex \overline{BD}$.

**What We Know**

A rhombus is a parallelogram with four congruent sides. So, all sides of rhombus *ABCD* are congruent. That is

$latex \overline{AB} \cong \overline{BC} \cong \overline{CD} \cong \overline{DA}$. Continue reading