Proof of the law of cosines with the dot product

Proof

We expand the square of the length of a difference of two vectors.

1. Let $u = A - C$ and $v = B - C$. With the Euclidean distance and the length, $\norm{u} = b$ and $\norm{v} = a$.
2. We have $u - v = A - B$, so $c^2 = \norm{u - v}^2 = (u - v) \cdot (u - v)$, with the dot product.
3. The dot product is bilinear and symmetric. So $(u - v) \cdot (u - v) = u \cdot u - 2 \, u \cdot v + v \cdot v = b^2 + a^2 - 2 \, u \cdot v$.
4. The vertices of the triangle are not collinear, so they are distinct. Thus $u \neq 0$ and $v \neq 0$, and the angle $\gamma$ at $C$ is defined.
5. By the definition of the angle at $C$, $\cos \gamma = \dfrac{u \cdot v}{\norm{u} \, \norm{v}} = \dfrac{u \cdot v}{ab}$. So $u \cdot v = ab \cos \gamma$.
6. Steps 2, 3 and 5 give $c^2 = a^2 + b^2 - 2ab \cos \gamma$.