Školjka

%V0 Denote by $M$ midpoint of side $BC$ in an isosceles triangle $\triangle ABC$ with $AC = AB$. Take a point $X$ on a smaller arc $\widehat{MA}$ of circumcircle of triangle $\triangle ABM$. Denote by $T$ point inside of angle $BMA$ such that $\angle TMX = 90$ and $TX = BX$. Prove that $\angle MTB - \angle CTM$ does not depend on choice of $X$. Author: unknown author, Canada