$C$ is the midpoint of $BT$, so $CM$ and $AB$ are parallel. Let $H$ be the orthogonal projection $A$ to $BC$, and $\angle ABC = 180^\circ - 45^\circ - 75^\circ = 60^\circ$, so if $AB = 2a$, then $BH = a$, $AH = a\sqrt3 = HC$, $AC = a\sqrt6$. Let $D$ be the intersection of $AC$ and $BM$. From $AB \parallel CM$ we also have $\frac{AD}{DC}=\frac{AB}{CM}=2$, so $$AD \cdot AC = \frac{2}{3}AC^2 = \frac{2}{3} \cdot 6a^2 = 4a^2 = (2a)^2 = AB^2,$$so $AB$ is tangent to $(BDC)$, giving us $\angle ABD = 45^\circ = \angle BMC$

Let us draw altitude from vertex $A$ to $BC$ and call intersection point $H$.Then $\angle BAH = 30^\circ$ and $\angle HAC = 45^\circ$ , So if $BH = x$ then $AB = 2x , AH = x\sqrt3 , HC = x\sqrt3 , AC = x\sqrt6$.Since $BC=CT , AM=MT\implies$ $MC$ is midsegment of $\triangle ABT(MC =x)$.Thus $BM$ and $AC$ are medians.Let $K$ be the intersection point of $BM$ and $AC\implies$ $K$ is centroid. Then $\frac{AK}{KC} = 2\implies$ $AK = \frac{2x\sqrt2}{\sqrt3}$ and $KC = \frac{x\sqrt2}{\sqrt3}$.In $\triangle KMC$ from Law of Cosines$\implies$ $KM^2 = KC^2 + CM^2 - 2\cdot KC \cdot CM \cdot \cos75$, also we have $\cos75 = \frac{\sqrt 6 - \sqrt2}{4}$. Thus(after simplyf the expression) $KM^2 =\frac{x^2(2 + \sqrt3)}{3}\implies$ $KM = \frac{x\sqrt{2 + \sqrt3}}{\sqrt3}$ In $\triangle KMC $ from Law of Sines $\implies$ $\frac{\sin75}{KM} = \frac{\sin \angle KMC}{KC}$ also we have $\sin75 = \frac{\sqrt 6 + \sqrt2}{4}$. Thus $\sin \angle KMC = \frac{1 + \sqrt3}{2\sqrt{2 + \sqrt3}}$ İf we multiply both numerator and denominator by $\sqrt{2 - \sqrt3}$ , we get that $\sin \angle KMC = \frac{\sqrt2}{2}$ , Hence $\angle KMC = 45^\circ = \angle BMC$ ... so we are done