One Diagonal of Kite Bisects the Other

Theorem
Let $ABCD$ be a kite such that:
 * $AC$ and $BD$ are its diagonals
 * $AB = BC$
 * $AD = DC$

Then $BD$ is the perpendicular bisector of $AC$.

Proof

 * Diagonals-of-Kite.png

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

From Diagonals of Kite are Perpendicular, $AC$ and $BD$ are perpendicular.

That is:
 * $\angle AEB = \angle CEB$

both being right angles.

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

We have that:
 * $\angle AEB = \angle CEB$ are both right angles.
 * $AB = BC$
 * $BE$ is common

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

Thus:


 * $AE = CE$

and it has already been established that $BD$ is perpendicular to $AC$.

Hence the result by definition of perpendicular bisector.