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Vietnam TST 2016 Problem 4

Source: Vietnam TST 2016

March 26, 2016
geometrycircumcircle

Problem Statement

Given an acute triangle ABCABC satisfying ∠ACB<∠ABC<∠ACB+∠BAC2\angle ACB<\angle ABC<\angle ACB+\dfrac{\angle BAC}{2}. Let DD be a point on BCBC such that ∠ADC=∠ACB+∠BAC2\angle ADC=\angle ACB+\dfrac{\angle BAC}{2}. Tangent of circumcircle of ABCABC at AA hits BCBC at EE. Bisector of ∠AEB\angle AEB intersects ADAD and (ADE)(ADE) at GG and FF respectively, DFDF hits AEAE at H.H. a) Prove that circle with diameter AE,DF,GHAE,DF,GH go through one common point. b) On the exterior bisector of ∠BAC\angle BAC and ray ACAC given point KK and MM respectively satisfying KB=KD=KMKB=KD=KM, On the exterior bisector of ∠BAC\angle BAC and ray ABAB given point LL and NN respectively satisfying LC=LD=LN.LC=LD=LN. Circle throughs M,NM,N and midpoint II of BCBC hits BCBC at PP (P≠IP\neq I). Prove that BM,CN,APBM,CN,AP concurrent.
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