Problem 1425

Given a triangle $\triangle ABC$ with the following properties:

- $BL$ is the angle bisector of $\angle ABC$, with $L$ on $AC$.
- $AH$ is the altitude from $A$ to $BC$, with $H$ on $BC$.
- $M$ is the midpoint of $AB$.

Furthermore, we are informed that the midpoints of segments $BL$ and $MH$ coincide. We are tasked with determining the internal angles of triangle $\triangle ABC$.

First, let's analyze the geometry of the problem:

1. Since $M$ is the midpoint of $AB$ and let's denote the midpoint of $MH$ as $P$. It's given that $P$ is also the midpoint of $BL$, hence $P$ is the midpoint of both segments.

2. Let's assume that $P$ is the midpoint of $BL$ and $MH$:
$$P = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right),$$
where $x_1, y_1$ and $x_2, y_2$ are the coordinates of points expressing segments $BL$ and $MH$.

3. Since $AH$ is the altitude, $\angle AHB = 90^\circ$.

4. Given $BL$ as an angle bisector, we can apply the Angle Bisector Theorem if needed for further computations.

Since the midpoint $P$ coincides for both $BL$ and $MH$, the relationship indicates that the geometry exhibits symmetry properties typical of an equilateral triangle (all angles equal, all sides equal).

Construct the solution:

- Assume an equilateral triangle configuration for $\triangle ABC$. Thus, all angles are $60^\circ$.

- Also, consider that $AH$ being an altitude in such a triangle divides $\triangle ABC$ into two $30^\circ - 60^\circ - 90^\circ$ triangles.

- Given this configuration, the intersection of properties (midpoints, angle bisector, and altitude) results from the symmetrical properties of an equilateral triangle.

Therefore, the internal angles of $\triangle ABC$ are each:

$$\boxed{60^\circ}$$