$a,b,c$ 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.

If $a<b$, then
$$
c^2 = a^2+b^2-ab = a^2+b(b-a) > a^2 \\
c^2 = a^2+b^2-ab = b^2-a(b-a) < b^2
$$
which means $a<c<b$ and $(a-c)(b-c)<0$.
If $a>b$, then
$$
c^2 = a^2+b^2-ab = a^2-b(a-b) < a^2 \\
c^2 = a^2+b^2-ab = b^2+a(a-b) > b^2
$$
which means $b<c<a$ and $(a-c)(b-c)<0$.
