Let $P$, $Q$ and $R$ be the midpoints of the sides $BC$, $AC$ and $AB$ of $\Delta ABC$. Let $S$ be the intersection of the perpendicular bisectors of $AC$ and $BC$.
Now I have to prove that given $\langle \vec{s}-\vec{p},\vec{a}-\vec{c}\rangle=0$ and $\langle \vec{s}-\vec{q},\vec{a}-\vec{c}\rangle=0$, we have $\langle\vec{s}-\vec{r},\vec{b}-\vec{c}\rangle=0$. Can somebody give a hint? The algebraic manipulation is pretty heavy, and I don't make any progress..
Vector approach. Let $P=(B+C)/2$, $Q=(A+C)/2$, $R=(A+B)/2$. We know that $$\langle S-P,B-C\rangle=0\implies \langle S,B-C\rangle=\langle (B+C)/2,B-C\rangle,$$ $$\langle S-Q,A-C\rangle=0\implies \langle S,A-C\rangle=\langle (A+C)/2,A-C\rangle.$$ By subtracting we get $$\langle S,B-A\rangle=\langle(B+C)/2,B-C\rangle -\langle (A+C)/2,A-C\rangle\\ =\frac{1}{2}(\langle B,B\rangle-\langle A,A\rangle).$$ Now $\langle S-R,B-A\rangle=0$ iff $$\langle S,B-A\rangle=\langle R,B-A\rangle=\langle(A+B)/2,B-A\rangle=\frac{1}{2}(\langle B,B\rangle-\langle A,A\rangle)$$ and we are done.
Geometric point of view. If $SP\perp CB$ then $|SC|=|SB|$. If $SQ\perp AC$ then $|SC|=|SA|$. It follows that $|SB|=|SA|$, which imply that $S$ is on the bisector of $AB$.