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 $latex \overlineBD$.

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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.


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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