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.
Proof
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:
- $\angle BEC = \angle BEA$
From Two Angles on Straight Line make Two Right Angles, $\angle BEC + \angle BEA$ make two right angles.
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.
$\blacksquare$
Sources
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): kite