In the acute triangle $ ABC $, the point $ I $ is the center of the inscribed the circle, the point $ O $ is the center of the circumscribed circle and the point $ I_a $ is the center the excircle tangent to the side $ BC $ and the extensions of the sides $ AB $ and $ AC $. Point $ A'$ is symmetric to vertex $ A $ with respect to the line $ BC $. Prove that $ \angle IOI_a = \angle IA'I_a $.
