VN-TST 2015 Problem 5

Let $ABC$ be a triangle with an interior point $P$ such that $\angle APB = \angle APC = \alpha$ and $\alpha > 180^o-\angle BAC$ . The circumcircle of triangle $APB$ cuts $AC$ at $E$ , the circumcircle of triangle $APC$ cuts $AB$ at $F$ . Let $Q$ be the point in the triangle $AEF$ such that $\angle AQE = \angle AQF =\alpha$ . Let $D$ be the symmetric point of $Q$ wrt $EF$ . Angle bisector of $\angle EDF$ cuts $AP$ at $T$ . a) Prove that $\angle DET = \angle ABC, \angle DFT = \angle ACB$ . b) Straight line $PA$ cuts straight lines $DE, DF$ at $M, N$ respectively. Denote $I, J$ the incenters of the triangles $PEM, PFN$ , and $K$ the circumcenter of the triangle $DIJ$ . Straight line $DT$ cut $(K)$ at $H$ . Prove that $HK$ passes through the incenter of the triangle $DMN$ .

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