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IA_2=IB_2=R where A_1, B_1 touchpoints of incircle (I), prove AA_2=BB_2=OI

Source: SRMC 2013

September 2, 2018
geometryincirclecircumcircle

Problem Statement

Circle with center II, inscribed in a triangle ABCABC , touches the sides BCBC and ACAC at points A1A_1 and B1B_1 respectively. On rays A1IA_1I and B1IB_1I, respectively, let be the points A2A_2 and B2B_2 such that IA2=IB2=RIA_2=IB_2=R, where RRis the radius of the circumscribed circle of the triangle ABCABC. Prove that: a) AA2=BB2=OIAA_2 = BB_2 = OI where OO is the center of the circumscribed circle of the triangle ABCABC, b) lines AA2AA_2 and BB2BB_2 intersect on the circumcircle of the triangle ABCABC.