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

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.

$\blacksquare$

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