RU-ALLRUSSIAN 2017 Problem 118

Given a convex quadrilateral $ABCD$ . We denote $I_A,I_B, I_C$ and $I_D$ centers of $\omega_A, \omega_B,\omega_C$ and $\omega_D$ ,inscribed In the triangles $DAB, ABC, BCD$ and $CDA$ , respectively.It turned out that $\angle BI_AA + \angle I_CI_AI_D = 180^\circ$ . Prove that $\angle BI_BA + \angle I_CI_BI_D = 180^{\circ}$ . (A. Kuznetsov)