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

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

We also know that the diagonals of a parallelogram bisect each other.  Since a rhombus is a parallelogram, it has also this property. Therefore, if point $latex E$ is the intersection of the diagonals as shown in the figure

$latex \overline{AE} \cong \overline{EC}$ and $latex \overline{DE} \cong \overline{EB}$.

What We Want to Show

Again, we want to show that $latex AC$ is perpendicular to $latex BD$. Now, if we can show that $latex AEB = 90^{\circ}$, then, we will have proven the statement above.

Proof

From the given, we can see that

$latex \overline{AE} \cong \overline{CE}$

$latex \overline{BE} \cong \overline{BE}$ by reflexive property of congruence. A segment is congruent to itself.

$latex \overline{AB} \cong \overline{CB}$.

So, by SSS congruence,

$latex \triangle AEB \cong \triangle CEB$.

Now, we know that corresponding angles of congruent triangles are congruent. Since

$latex \angle AEB$ and $latex \angle CEB$ are corresponding angles

$latex \angle AEB \cong \angle CEB$.

But since they are supplementary angles (Can you see why?)

$latex \angle AEB + \angle CEB = 180 ^\circ$.

In effect,

$latex m \angle AEB = m \angle CEB = 90^{\circ}$.

Therefore, the diagonals of ABCD are perpendicular to each other, which is what we want to prove.