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.

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 Congruence, $\triangle ABD$ 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 Congruence, $\triangle ABE$ 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:

$\angle BEC = \angle BEA$

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.

$\blacksquare$