In the ahead post, we have proved the converse the the Pythagorean Theorem. In this post, we will prove the the diagonals that a rhombus are perpendicular to each other. The is, if we have actually parallelogram ABCD v diagonal \$latex \overlineAC\$ and also \$latex \overlineBD\$, climate \$latex \overlineAC\$ is perpendicular to \$latex \overlineBD\$.

You are watching: Prove that the diagonals of a rhombus are perpendicular What us Know

A rhombus is a parallelogram with 4 congruent sides. So, all sides that rhombus ABCD room congruent. The is

\$latex \overlineAB \cong \overlineBC \cong \overlineCD \cong \overlineDA\$. We also know the the diagonals that a parallel bisect each other. Since a rhombus is a parallelogram, the has additionally this property. Therefore, if point \$latex E\$ is the intersection that the diagonals as presented in the figure

\$latex \overlineAE \cong \overlineEC\$ and \$latex \overlineDE \cong \overlineEB\$. What We want to Show

Again, we want to display that \$latex AC\$ is perpendicular to \$latex BD\$. Now, if we can present that \$latex AEB = 90^\circ\$, then, we will have actually proven the explain above.

Proof

From the given, we deserve to see that

\$latex \overlineAE \cong \overlineCE\$

\$latex \overlineBE \cong \overlineBE\$ by reflexive home of congruence. A segment is congruent come itself.

\$latex \overlineAB \cong \overlineCB\$.

So, by SSS congruence,

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

Now, we know that matching angles that congruent triangles are congruent. Since

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

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

But since they room supplementary angles (Can you check out why?)

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

In effect,

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

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

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