KöMaL 2024 B5421

The incenter and the inradius of the acute triangle $ABC$ is $I$ and $r$, respectively. The excenter and exradius relative to vertex $A$ is $I_a$ and $r_a$, respectively. Let $R$ denote the circumradius. Prove that if $II_a=r_a+R-r$, then $\angle BAC = 60^{\circ}$.