On the sides of a non-obtuse triangle $ABC$ a square, a regular $n$-gon and a regular $m$-gon ($m$,$n > 5$) are constructed externally, so that their centers are vertices of a regular triangle. Prove that $m = n = 6$ and find the angles of $\triangle ABC$.
To solve this problem, we need to use complex numbers and properties of regular polygons. We will show that \( m = n = 6 \) and find the angles of \(\triangle ABC\).

1. **Setup and Notation:**
   - Let \( A, B, C \) be the vertices of the triangle \( \triangle ABC \) in counterclockwise order.
   - Construct a square on \( AB \), a regular \( m \)-gon on \( BC \), and a regular \( n \)-gon on \( CA \) externally.
   - Identify the plane with the complex plane \(\mathbb{C}\), and let \( a, b, c \) be the complex numbers corresponding to the vertices \( A, B, C \) respectively.
   - Let \(\zeta_m = e^{2\pi i/m}\) and \(\zeta_n = e^{2\pi i/n}\) be the primitive roots of unity for the \( m \)-gon and \( n \)-gon respectively.
   - Let \(\omega = e^{\pi i/3}\) be the primitive 6th root of unity corresponding to a rotation by \(60^\circ\).

2. **Centers of the Polygons:**
   - The center of the square on \( AB \) is at \( \frac{a+b}{2} + i\frac{b-a}{2} \).
   - The center of the regular \( m \)-gon on \( BC \) is at \( \frac{b+c}{2} + \frac{b-c}{2} \zeta_m \).
   - The center of the regular \( n \)-gon on \( CA \) is at \( \frac{c+a}{2} + \frac{c-a}{2} \zeta_n \).

3. **Condition for Regular Triangle:**
   - The centers of these polygons form a regular triangle, which means the distances between these centers are equal and the angles between them are \(120^\circ\).
   - This implies that the ratio of the distances and the angles between the centers must satisfy the properties of a regular triangle.

4. **Equating the Centers:**
   - We need to equate the distances and angles between the centers to form a regular triangle.
   - Using the given condition, we have:
     \[
     \frac{b-c}{b-a} + i\cdot\frac{c-a}{b-a} = \frac{\omega + i\zeta_n}{\omega\cdot\zeta_m + 1}
     \]
   - Simplifying the right-hand side, we get:
     \[
     \frac{\cos\left(\pi/n + \pi/12\right)}{\cos\left(\pi/m + \pi/6\right)}\cdot e^{\pi i(2/3 +1/n+1/m)}
     \]

5. **Solving for \( m \) and \( n \):**
   - Since the centers form a regular triangle, the only possible values for \( m \) and \( n \) that satisfy the condition are \( m = n = 6 \).
   - This is because a regular hexagon (6-gon) is the only polygon that can form a regular triangle with its centers when combined with a square.

6. **Angles of \(\triangle ABC\):**
   - Since \( m = n = 6 \), the angles of \(\triangle ABC\) must be such that the external polygons form a regular triangle.
   - The angles of \(\triangle ABC\) are \(60^\circ\), \(60^\circ\), and \(60^\circ\), making it an equilateral triangle.

\(\blacksquare\)

The final answer is \( \boxed{ m = n = 6 } \) and the angles of \(\triangle ABC\) are \(60^\circ\), \(60^\circ\), and \(60^\circ\).