Given $\triangle ABC$ with $\angle C = 60^\circ$, show $a^2+b^2-ab=c^2$ without trigonometry

Without using trigonometry, we can use Stewart's theorem.

We build an equilateral triangle as shown in the attached figure so we have

$$b^2(a-b)+c^2b=ab(a-b)+b^2a$$ which is equivalent to the given $$a^2+b^2-ab=c^2$$

Drop perpendicular from $C$ onto $AB$ and call foot of perpendicular as $D$. Then using pythagoras theorem, $$AD^2 + DC^2 = b^2 \tag{1}$$ $$DC^2 + DB^2 = a^2\tag{2}$$ and from side $c$ $$AD + DB = c \tag{3}$$

Finally we need a relation that comes from the angle $A$ being $60^\circ$. Consider the triangle $ADC.$ Make a midpoint $E$ on $AC$ and join $ED$. Thus from what Arnaldo showed, we can show $$AD = \frac{b}{2}\tag{4}$$

So you have to just eliminate between these equations.