Law of Cosines/Proof 2

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

Then $c^2 = a^2 + b^2 - 2ab \cos C$.

Proof
Let $\triangle ABC$ be a triangle.

Using $AC$ as the radius, we construct a circle.

Now we extend:
 * $CB$ to $D$
 * $AB$ to $F$
 * $BA$ to $G$
 * $CA$ to $E$.

We join $D$ with $E$, and thus obtain this figure:


 * CosineRule.png

Using the Intersecting Chord Theorem we have:
 * $GB \cdot BF = CB \cdot BD$

$AF$ is a radius, so $AF = AC = b = GA$ and thus:
 * $GB = GA + AB = b + c$
 * $BF = AF - AB = b - c$

Thus:

Next:

As $CA$ is a radius, $CE$ is a diameter.

By Thales' Theorem, it follows that $\angle CDE$ is a right angle.

Then using the definition of cosine, we have