Given $\triangle ABC$ with $\angle C = 60^\circ$, show $a^2+b^2-ab=c^2$ without trigonometry I have the following problem: 

In $\triangle ABC$, the angle at vertex $C$ is $60^{\circ}$. Prove that $a^2+b^2-ab=c^2$.

Of course, it is easy if you use cosine rule. I believe there exists a beautiful proof as well without using trigonometry. Anyone?

Here's a proof using Ptolemy's Theorem:

$$\begin{align}
|\overline{AB}| |\overline{A^\prime B^\prime}| &= |\overline{AA^\prime}||\overline{BB^\prime}|+|\overline{AB^\prime}||\overline{A^\prime B}| \tag{1} \\[4pt]
c^2 &= a b + ( a - b )^2 \tag{2} \\[4pt]
&= a^2 + b^2 - a b \tag{3}
\end{align}$$

Incidentally, the Wikipedia entry mentions that one can use Ptolemy to prove the Law of Cosines generally. However, the description there isn't as clear as it might be. All one needs to do is observe that, in general, 
$$|\overline{AA^\prime}| = 2b\sin\frac{C}{2} \qquad |\overline{BB^\prime}| = 2a\sin\frac{C}{2} \tag{4}$$
so that the counterpart of $(2)$ is
$$c^2 = 4ab\sin^2\frac{C}{2} + ( a - b )^2 \tag{5}$$
and we have
$$c^2 = a^2 + b^2 - 2 a b \left( 1 - 2\sin^2\frac{C}{2} \right) = a^2 + b^2 - 2 a b \cos C \tag{6}$$
