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