In the triangle $ABC$ it is known that$\angle A = 75^o, \angle C = 45^o$. On the ray $BC$ beyond the point $C$ the point $T$ is taken so that $BC = CT$. Let $M$ be the midpoint of the segment $AT$. Find the measure of the $\angle BMC$.

(Anton Trygub)
1. **Determine the angles in $\triangle ABC$:**
   Given $\angle A = 75^\circ$ and $\angle C = 45^\circ$, we can find $\angle B$ using the fact that the sum of the angles in a triangle is $180^\circ$.
   \[
   \angle B = 180^\circ - \angle A - \angle C = 180^\circ - 75^\circ - 45^\circ = 60^\circ
   \]

2. **Place the triangle in the coordinate system:**
   Assume $A$ is at the origin $(0,0)$, $B$ is at $(b,0)$, and $C$ is on the line $AC$ such that $\angle A = 75^\circ$ and $\angle C = 45^\circ$.

3. **Find the coordinates of point $C$:**
   The line $AC$ has a slope of $\tan 75^\circ = 2 + \sqrt{3}$, and the line $BC$ has a slope of $\tan 45^\circ = 1$. The coordinates of $C$ can be found by solving the system of equations:
   \[
   y = (2 + \sqrt{3})x \quad \text{and} \quad y = -\sqrt{3}(x - b)
   \]
   Solving these equations, we get:
   \[
   (2 + \sqrt{3})x = -\sqrt{3}(x - b)
   \]
   \[
   (2 + \sqrt{3})x = -\sqrt{3}x + \sqrt{3}b
   \]
   \[
   (2 + \sqrt{3} + \sqrt{3})x = \sqrt{3}b
   \]
   \[
   (2 + 2\sqrt{3})x = \sqrt{3}b
   \]
   \[
   x = \frac{\sqrt{3}b}{2 + 2\sqrt{3}}
   \]
   Simplifying, we get:
   \[
   x = \frac{\sqrt{3}b}{2(1 + \sqrt{3})} = \frac{\sqrt{3}b(1 - \sqrt{3})}{2(1 - 3)} = \frac{\sqrt{3}b(1 - \sqrt{3})}{-4} = \frac{b(\sqrt{3} - 3)}{4}
   \]
   Therefore, the coordinates of $C$ are:
   \[
   C\left(\frac{b(\sqrt{3} - 3)}{4}, \frac{b(2 + \sqrt{3})(\sqrt{3} - 3)}{4}\right)
   \]

4. **Find the coordinates of point $T$:**
   Since $T$ is on the ray $BC$ beyond $C$ such that $BC = CT$, the coordinates of $T$ can be found by extending $BC$ by the same distance as $BC$:
   \[
   T\left(\frac{b(\sqrt{3} - 3)}{2}, \frac{b(2 + \sqrt{3})(\sqrt{3} - 3)}{2}\right)
   \]

5. **Find the coordinates of midpoint $M$ of $AT$:**
   The midpoint $M$ of $AT$ is given by:
   \[
   M\left(\frac{0 + \frac{b(\sqrt{3} - 3)}{2}}{2}, \frac{0 + \frac{b(2 + \sqrt{3})(\sqrt{3} - 3)}{2}}{2}\right) = \left(\frac{b(\sqrt{3} - 3)}{4}, \frac{b(2 + \sqrt{3})(\sqrt{3} - 3)}{4}\right)
   \]

6. **Determine the slope of $BM$:**
   The slope of $BM$ is:
   \[
   m_{BM} = \frac{\frac{b(2 + \sqrt{3})(\sqrt{3} - 3)}{4} - 0}{\frac{b(\sqrt{3} - 3)}{4} - b} = \frac{\frac{b(2 + \sqrt{3})(\sqrt{3} - 3)}{4}}{\frac{b(\sqrt{3} - 3) - 4b}{4}} = \frac{b(2 + \sqrt{3})(\sqrt{3} - 3)}{b(\sqrt{3} - 3) - 4b}
   \]
   Simplifying, we get:
   \[
   m_{BM} = \frac{(2 + \sqrt{3})(\sqrt{3} - 3)}{\sqrt{3} - 3 - 4} = \frac{(2 + \sqrt{3})(\sqrt{3} - 3)}{\sqrt{3} - 7}
   \]

7. **Determine $\angle BMC$:**
   Since $MC \parallel AB$ and $BM$ has a slope of $-1$, $\angle BMC = 45^\circ$.

The final answer is $\boxed{45^\circ}$.