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 and , then is perpendicular to .

**What We Know**

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

.

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 is the intersection of the diagonals as shown in the figure

and .

**What We Want to Show**

Again, we want to show that is perpendicular to . Now, if we can show that , then, we will have proven the statement above.

**Proof**

From the given, we can see that

by reflexive property of congruence. A segment is congruent to itself.

.

So, by SSS congruence,

.

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

and are corresponding angles

.

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

.

In effect,

.

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