Diagonals of Kite are Perpendicular

Theorem
Let $ABCD$ be a kite such that $AC$ and $BD$ are its diagonals.

Then $AC$ and $BD$ are perpendicular.

Proof

 * Diagonals-of-Kite.png

Let $AC$ and $BD$ meet at $E$.

Consider the triangles $\triangle ABD$ and $\triangle CBD$.

We have that:
 * $AB = CB$
 * $AD = CD$
 * $BD$ is common.

Hence by Triangle Side-Side-Side Equality, $\triangle ABD$ and $\triangle CBD$ are congruent.

Consider the triangles $\triangle ABE$ and $\triangle CBE$.

We have from the congruence of $\triangle ABD$ and $\triangle CBD$ that:
 * $\angle ABE = \angle CBE$
 * $AB = CB$

and $BE$ is common.

Hence by Triangle Side-Angle-Side Equality, $\triangle ABE$ and $\triangle CBE$ are congruent.

We have that $AC$ is a straight line.

We have from the congruence of $\triangle ABE$ and $\triangle CBE$ that:
 * $\angle BEC = \angle BEA$

From Two Angles on Straight Line make Two Right Angles, $\angle BEC + \angle BEA$ make two right angles.

Thus:


 * $2 \angle BEC = 2 \angle BEA = 2$ right angles

and so:


 * $\angle BEC = \angle BEA$ are both right angles.

That is, $AC$ and $BD$ are perpendicular.