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

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 Congruence, $\triangle ABE$ 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.

$\blacksquare$

Sources

• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): kite