# One Diagonal of Kite Bisects the Other

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## 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 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.

$\blacksquare$

## Sources

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