Perpendicular bisector that pass through a fixed point Points A and B fixed, and point C moves on circle such that ABC acute triangle. $AT = BT$ and $TM \perp AC, \, TN \perp BC$. How can I proove that all the middle perpendiculars (perpendicular bissector) to $MN$ passes through a fixed point?

Let the tangent at $A$ and the tangent at $B$ of the circumscribing circle $\mathcal{C}$ of $\triangle ABC$ meet at $U$ and let $V$ be the midpoint of $TU$.  We assert that $V$ is the required fixed point through which the perpendicular bisector of $MN$ passes as $C$ varies over $\mathcal{C}$.
First we claim that $CU \perp MN$.  This is equivalent to proving that $\angle NCT=\angle UCM$.  To this end, we must verify that $CP$ bisects $\angle TCU$ where $P$ is the midpoint of the arc $AB$ of $\mathcal{C}$ not containing $C$ (note also that $CP$ bisects $\angle ACB$).  This means we have to show that $\mathcal{C}$ is the circle of Apollonius of the segment $TU$.  In other words, if $Q$ is the point such that line segment $PQ$ is a diameter of $\mathcal{C}$ we must show that
$$\frac{|PU|}{|PT|}=\frac{|QU|}{|QT|}.$$
Now if $r$ is the radius of $\mathcal{C}$ and $h$ is the length $|TB|$, then it follows easily that $|OT|=\sqrt{r^2-h^2}$
and $|OU|=\frac{r^2}{\sqrt{r^2-h^2}}$  where $O$ is the center of $\mathcal{C}$.  So 
$$|PU|=\frac{r^2}{\sqrt{r^2-h^2}}-r,$$
$$|PT|=r-\sqrt{r^2-h^2},$$
$$|QU|=\frac{r^2}{\sqrt{r^2-h^2}}+r,$$
$$|QT|=r+\sqrt{r^2-h^2}.$$
That is
$$\frac{|PU|}{|PT|}=\frac{r}{\sqrt{r^2-h^2}}=\frac{|QU|}{|QT|},$$
finishing the claim.  (An alternative proof is to notice that $$\angle PBT=90^\circ-\angle OTB = 90^\circ-\left(90^\circ-\frac12\angle TOB\right)=\frac12\angle TOB$$
and $$\angle UBP=\angle UBO-\angle OBP=90^\circ-\left(90^\circ-\frac12\angle TOB\right)=\frac12\angle TOB,$$
so $PB$ bisects $\angle UBT$ internally.  Because $Q$ is on the line $UT$ and $\angle PBQ=90^\circ$, $QB$ bisects $\angle UBT$ externally.)
Now the dilation $d$ about $T$ with factor $+1/2$ sends the straight line $CU$ to a line $\ell$ perpendicular to the segment $MN$.  Since $\ell$ passes through the midpoint of $CT$, which is the center of the circumscribing circle of the cyclic quadrilateral $CMTN$, $\ell$ must be a perpendicular bisector of $MN$.  Because $d(U)=V$, $\ell$ passes through $V$ as asserted.
