Let $T_1$ and $T_2$ be the points of tangency of the excircles of a triangle $ABC$ with its sides $BC$ and $AC$ respectively. It is known that the reflection of the incenter of $ABC$ across the midpoint of $AB$ lies on the circumcircle of triangle $CT_1T_2$. Find $\angle BCA$.
1. **Define the coordinates and parameters:**
   - Let \( A = (1, 0, 0) \), \( B = (0, 1, 0) \), \( C = (0, 0, 1) \).
   - Let \( a = BC \), \( b = CA \), \( c = AB \), and let \( s \) be the semiperimeter of \(\triangle ABC\).
   - The incenter \( I \) of \(\triangle ABC\) has coordinates \( I = (a : b : c) \).
   - The midpoint \( M \) of \( AB \) has coordinates \( M = \left(\frac{1}{2}, \frac{1}{2}, 0\right) \).

2. **Find the reflection of \( I \) over \( M \):**
   - The reflection \( I' \) of \( I \) over \( M \) can be found using the midpoint formula. Since \( M \) is the midpoint, we have:
     \[
     M = \left(\frac{a + b+c}{2}, \frac{a + b+c}{2}, 0\right)
     \]
     The reflection \( I' \) of \( I \) over \( M \) is:
     \[
     I' = 2M - I = \left(2 \cdot \frac{a+b+c}{2} - a, 2 \cdot \frac{a+b+c}{2} - b, 0 - c\right) = (b+c, a+c, -c)
     \]

3. **Determine the coordinates of \( T_1 \) and \( T_2 \):**
   - The excircle opposite \( A \) touches \( BC \) at \( T_1 \), so \( T_1 = (0 : s-b : s-c) \).
   - The excircle opposite \( B \) touches \( AC \) at \( T_2 \), so \( T_2 = (s-a : 0 : s-c) \).

4. **Find the equation of the circumcircle of \( \triangle CT_1T_2 \):**
   - The general equation of a circle in barycentric coordinates is:
     \[
     a^2yz + b^2zx + c^2xy - (x+y+z)(ux + vy + wz) = 0
     \]
   - Substituting the coordinates of \( C \), \( T_1 \), and \( T_2 \):
     \[
     C = (0, 0, 1), \quad T_1 = (0, s-b, s-c), \quad T_2 = (s-a, 0, s-c)
     \]
   - The equation becomes:
     \[
     a^2yz + b^2zx + c^2xy - (x+y+z)(b(s-c)x + a(s-c)y) = 0
     \]

5. **Verify that \( I' \) lies on the circumcircle:**
   - Substitute \( I' = (b+c, a+c, -c) \) into the circle equation:
     \[
     a^2(a+c)(-c) + b^2(-c)(b+c) + c^2(b+c)(a+c) - (a+b+c)(b(s-c)(b+c) + a(s-c)(a+c)) = 0
     \]
   - Simplify the equation:
     \[
     a^2(a+c)(-c) + b^2(-c)(b+c) + c^2(b+c)(a+c) = (a+b+c)(b(s-c)(b+c) + a(s-c)(a+c))
     \]
     \[
     a^2(-ac - c^2) + b^2(-bc - c^2) + c^2(ab + ac + bc + c^2) = (a+b+c)(b(s-c)(b+c) + a(s-c)(a+c))
     \]
     \[
     -a^2c - a^2c^2 - b^2c - b^2c^2 + c^2ab + c^2ac + c^2bc + c^4 = (a+b+c)(b(s-c)(b+c) + a(s-c)(a+c))
     \]
     \[
     \frac{1}{2} (a+b+c)(a+b+2c)(c^2 - a^2 - b^2) = 0
     \]
     \[
     c^2 - a^2 - b^2 = 0
     \]

6. **Conclude the angle:**
   - Since \( c^2 = a^2 + b^2 \), \(\triangle ABC\) is a right triangle with \(\angle C = 90^\circ\).

The final answer is \(\boxed{90^\circ}\).