prove equilateral triangle Recently I have encountered such a proving problem. 
As shown below, given a $\triangle ABC$, $AD$ intersects $BC$ at $D$ so that $AD$ is perpendicular to $BC$, $BE$ intersects $AC$ at $E$ so that $BE$ is the angle bisector of $\angle ABC$, $CF$ intersects $AB$ at $F$ so that $CF$ is the median of $AB$, $AD$, $BE$, $CF$ intersect at a common point.

We need to prove that triangle ABC is an equilateral triangle. I have tried several approaches, but it seems useless in the end.
Any thoughts? Or is it actually not an equilateral triangle?
 A: I naively thought such a triangle would have to be equilateral, but when I tried to demonstrate it by construction I ended up producing a counterexample pretty easily.
Start by fixing points $A$ and $B$. We also fix a third point $C^\prime$ and require that $C$ lie on the ray $\vec{BC^\prime}$. Drop a perpendicular from $A$ down to $BC^\prime$ and call the point of intersection $D$. Next construct the angle bisector of $\angle ABC^\prime$; denote the intersection of the angle bisector with the line $AD$ by $G$. Finally, construct the midpoint $F$ of $AB$, and extend a line through $F$ and $G$. The intersection of $FG$ and $BC^\prime$ determines the third vertex of the triangle, $C$.
The result is that for any angle $\angle ABC^\prime$, there is a point $C$ such that the triangle $\triangle ABC$ satisfies all of our hypotheses. And of course, if $\angle ABC^\prime$ is not necessarily $60^\circ$, then $\triangle ABC$ is not necessarily equilateral.

EDIT: The last paragraph is not entirely correct. The first sentence should read: for any $\color{red}{\text{non-obtuse}}$ angle $\angle ABC^\prime$, there is a point $C$ such that the triangle $\triangle ABC$ satisfies all of our hypotheses.
