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?

Hint:Let's mark perpendicular bisector as (PB). 
If C is coincident on A or B then M and N will be coincident on A and B respectively and the (PB) of MN is exactly the (PB) of AB.Also when the triangle is isosceles MN is parallel with AB, so again (PB) of MN and AB are coincident. That is the point is on the (PB) of AB. To find this point extend the (PB) of MN to cross the the (PB) of AB at P. It can be shown that the ratio of $R=\frac{TP}{AB}$ is independent of position of C and is constant. The angle $(\alpha)$ between PT  and (PB) of MN is always equal to the angle $(\beta)$ between AB and MN(or their extensions),because their rays are perpendicular. If $\angle CAB$ or $\angle CBA$ is $90^o$ Then M or N locate on A or B respectively. Let's mark the intersection of (PB) of MN and AB as Q. The right triangles ABC and PQT are similar . Since AB is constant then TP must be constant due to sine law.That is the equality of angles $(\alpha)$ and $(\beta)$ requires that (PB) of MN always passes the point P.     
