Law of Cosines/Proof 3/Acute Triangle

Theorem
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that:
 * $a$ is opposite $A$
 * $b$ is opposite $B$
 * $c$ is opposite $C$.

Let $\triangle ABC$ be an acute triangle.

Then:
 * $c^2 = a^2 + b^2 - 2a b \cos C$

Proof
Let $\triangle ABC$ be an acute triangle.
 * CosineRule-Proof3-acute.png

Let $BD$ be dropped perpendicular to $AC$.

Then $\triangle CDB$ and $\triangle ADB$ are right triangles.

So: