triangle proof: intersection of $w_\alpha$ and $m_a$ is outside $\Delta ABC$ I try to proove, that the intersection $M$ of $w_\alpha$ (angle bisector) and $m_a$ (perpendicular bisectors of $a$) of any triangle $\Delta ABC$ is always outside $\Delta ABC$, except $\Delta ABC$ is equilateral (the same for the other possibilities)
My thoughts: If $M$ is outside $\Delta ABC$, then $|AM|>|AD|$ for $D$ being the  perpendicular foot of $h_a$.
But I had no idea how to start. A geometric or an analytic proof is both fine.

Hint: The points $A$, $B$, $C$, and $M$ are in the same circle. (See figure 1).

Figure 1
