Let $f(x)=x^2+y^2$ be the Euclidean square-norm and $A,B,C\in\mathbb{R}^2$ be vertices of a triangle $\Delta$ such that $f$ takes the maximum on $\Delta$ in $C$, the minimum in $A$ and takes the minimum on each edge in a vertex.
Does it hold that for $x$ on the edge $AB$, $f$ takes the minimum on the straight line segment $xC$ in $x$?
Update: I added an assumption later ($f$ takes the minimum on each edge in a vertex) and appologize for the confusion.
The squared distance from the origin $f(x,y)$ is a convex function and $\Delta$ is a triangle, hence a convex set, so we have that $f$ takes its maximum on $\partial\Delta$. We know that the maximum occurs in $C$ and the minimum on $A$, hence $C$ is the farthest point from the origin and $A$ is the closest one. However, $OA$ and $BC$ are free to be orthogonal: in such a case, the minimum of $f$ on $BC$ is attained in the projection of $A$ over $BC$, hence the answer to your question is negative.