$a,b,c$ 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.

WLOG assume $a<b$. Then:
$$a^2+b^2-ab=c^2 \Rightarrow a(a-b)=(c-b)(c+b) \Rightarrow c<b$$
$$a^2+b^2-ab=c^2 \Rightarrow (a-c)(a+c)=b(a-b) \Rightarrow a<c.$$
