Proof that The Diagonals of a Rhombus are Perpendicular

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 \overline{AC} and \overline{BD}, then \overline{AC} is perpendicular to \overline{BD}.

parallelogram 2

What We Know

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

\overline{AB} \cong \overline{BC} \cong \overline{CD} \cong \overline{DA}

parallelogram

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 E is the intersection of the diagonals as shown in the figure

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

parallelogram 5

What We Want to Show

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

Proof

From the given, we can see that

\overline{AE} \cong \overline{CE}

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

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

So, by SSS congruence,

\triangle AEB \cong \triangle CEB.

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

\angle AEB and \angle CEB are corresponding angles

\angle AEB \cong \angle CEB.

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

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

In effect,

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.

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