VN-TST 2016 Problem 4

Given an acute triangle $ABC$ satisfying $\angle ACB<\angle ABC<\angle ACB+\dfrac{\angle BAC}{2}$ . Let $D$ be a point on $BC$ such that $\angle ADC=\angle ACB+\dfrac{\angle BAC}{2}$ . Tangent of circumcircle of $ABC$ at $A$ hits $BC$ at $E$ . Bisector of $\angle AEB$ intersects $AD$ and $(ADE)$ at $G$ and $F$ respectively, $DF$ hits $AE$ at $H.$ a) Prove that circle with diameter $AE,DF,GH$ go through one common point. b) On the exterior bisector of $\angle BAC$ and ray $AC$ given point $K$ and $M$ respectively satisfying $KB=KD=KM$ , On the exterior bisector of $\angle BAC$ and ray $AB$ given point $L$ and $N$ respectively satisfying $LC=LD=LN.$ Circle throughs $M,N$ and midpoint $I$ of $BC$ hits $BC$ at $P$ ( $P\neq I$ ). Prove that $BM,CN,AP$ concurrent.