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A nice(Hard) problem from Russia!

Source: All Russian 2017,Grade 11,day 2,P8

May 1, 2017
geometryincenterAngle Chasingincircle

Problem Statement

Given a convex quadrilateral ABCDABCD. We denote IA,IB,ICI_A,I_B, I_C and IDI_D centers of ωA,ωB,ωC\omega_A, \omega_B,\omega_C and ωD\omega_D,inscribed In the triangles DAB,ABC,BCDDAB, ABC, BCD and CDACDA, respectively.It turned out that BIAA+ICIAID=180\angle BI_AA + \angle I_CI_AI_D = 180^\circ. Prove that BIBA+ICIBID=180\angle BI_BA + \angle I_CI_BI_D = 180^{\circ}. (A. Kuznetsov)