Given an acute triangle $ABC$ with $AB < AC$.Let $\Omega $ be the circumcircle of $ ABC$ and $M$ be centeriod of triangle $ABC$.$AH$ is altitude of $ABC$.$MH$ intersect with $\Omega $ at $A'$.prove that circumcircle of triangle $A'HB$ is tangent to $AB$.
A.I.Golovanov, A. Yakubov
Let $E,O$ be the orthocenter and the circumcenter . And let $AM$ intersect $BC$ at point $T$. and let$HM$ intersect the arc$AC$ at point $J$ . By a homegeneity with center $M$ and relation $-2$ the nine pointe circle goes to the circumcircle of $ABC$ so we can find that $(HM/MJ)=(MT/MA)=(1/2)$ so $AJ//BC$ and angles$ABC=JCB$ .we know that $JCB=BA'H$ so we find that $BA'H=ABH$ so the circumcircle of $BHA'$ is tangant to the line $AB$