$a,b,c$ are positive reals and distinct with $a^2+b^2 -ab=c^2$. Prove $(a-c)(b-c)<0$ 
$a,b,c$ are positive reals and distinct with $a^2+b^2 -ab=c^2$.
Show that $(a-c)(b-c)<0$.

This is a question presented in the "Olimpiadas do Ceará 1987" a math contest held in Brazil. Sorry if this a duplicate.
Given the assumptions, it is easy to show that
$$0<(a-b)^2<a^2+b^2-ab=c^2.$$ But could not find a promising route to pursue.
Any hint or answer is welcomed.

Observe triangle with sides $a,b,c$ and angles $\alpha,\beta, \gamma$. 
By cosine theorem angle $\gamma = 60^{\circ}$, thus $\alpha +\beta =120^{\circ}$. 
So we can assume $\alpha \leq 60^{\circ}$ and $\beta \geq 60^{\circ}$. So $b\geq c\geq a$ and thus conclusion:
$$(a-c)(b-c)\leq 0$$
with equality iff $a=b=c$.
