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Part of 2019 Ukraine Team Selection Test

Problems(1)

incenter lies on P_1T_1, <ABC=60^o, PI //BC, TI //AB, AP_1 = B , CT_1 = BT

Source: 2019 Ukraine TST 1.1

11/17/2020
In a triangle ABCABC, ∠ABC=60o\angle ABC= 60^o, point II is the incenter. Let the points PP and TT on the sides ABAB and BCBC respectively such that PI∥BCPI \parallel BC and TI∥ABTI \parallel AB , and points P1P_1 and T1T_1 on the sides ABAB and BCBC respectively such that AP1=BPAP_1 = BP and CT1=BTCT_1 = BT. Prove that point II lies on segment P1T1P_1T_1.
(Anton Trygub)
geometryincenterparallelequal segmentscollinear