The circumradius $R$ of a triangle with side lengths $a, b, c$ satisﬁes $R =\frac{a\sqrt{bc}}{b+c}$. Find the angles of the triangle.
1. Consider triangle \(ABC\) with side lengths \(AB = c\), \(AC = b\), and \(BC = a\). Let \(M\) be the midpoint of \(BC\) and \(O\) be the circumcenter of triangle \(ABC\).

2. The circumradius \(R\) of a triangle with side lengths \(a, b, c\) is given by:
   \[
   R = \frac{a\sqrt{bc}}{b+c}
   \]

3. In triangle \(CMO\), since \(MO \perp BC\), we have:
   \[
   \frac{CM}{CO} = \cos{\angle MCO}
   \]
   Since \(M\) is the midpoint of \(BC\), \(CM = \frac{a}{2}\). Also, \(CO = R\). Therefore:
   \[
   \cos{\angle MCO} = \frac{\frac{a}{2}}{R}
   \]

4. Substituting \(R = \frac{a\sqrt{bc}}{b+c}\) into the equation, we get:
   \[
   \cos{\angle MCO} = \frac{\frac{a}{2}}{\frac{a\sqrt{bc}}{b+c}} = \frac{(b+c)}{2\sqrt{bc}}
   \]

5. By the AM-GM inequality, we know:
   \[
   \frac{b+c}{2} \geq \sqrt{bc}
   \]
   Therefore:
   \[
   \frac{b+c}{2\sqrt{bc}} \geq 1
   \]

6. Since \(\cos\) is always \(\leq 1\), we must have:
   \[
   \cos{\angle MCO} = 1 \implies \frac{b+c}{2\sqrt{bc}} = 1
   \]

7. Solving the equation \(\frac{b+c}{2\sqrt{bc}} = 1\), we get:
   \[
   b + c = 2\sqrt{bc}
   \]
   Squaring both sides:
   \[
   (b + c)^2 = 4bc
   \]
   Expanding and simplifying:
   \[
   b^2 + 2bc + c^2 = 4bc \implies b^2 - 2bc + c^2 = 0 \implies (b - c)^2 = 0 \implies b = c
   \]

8. Since \(b = c\), triangle \(ABC\) is isosceles with \(AB = AC\). Also, since \(O\) lies on \(BC\) and \(OB = OC\), \(M = O\) and \(BC\) is the diameter of the circumcircle of \(\triangle ABC\). Therefore, \(\triangle ABC\) is a right triangle with \(\angle BAC = 90^\circ\).

9. Since \(AC = AB\), \(\triangle ABC\) is an isosceles right triangle with angles \(45^\circ, 45^\circ, 90^\circ\).

The final answer is \( \boxed{ \triangle ABC } \) is a triangle with the angles \(90^\circ, 45^\circ, 45^\circ\).