$a,b,c$ 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.

we get $$c=\sqrt{a^2+b^2-ab}$$ since $c>0$ and $$a^2+b^2>ab$$ from here we get
the product
$$(a-\sqrt{a^2+b^2-ab})(b-\sqrt{a^2+b^2-ab})<0$$
we have two cases.
1) $$a>\sqrt{a^2+b^2-ab}$$ and $$b<\sqrt{a^2+b^2-ab}$$ after squaring we get $$a>b$$
2) $$a<\sqrt{a^2+b^2-ab}$$ and $$b>\sqrt{a^2+b^2-ab}$$
after squaring we get $$a<b$$ and our Statement is true.
