SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Denote $M$ as the midpoint of $BC$. Since $\angle AEH=\angle AFH=90^{\circ }$ then $A,H,E,F$ belong to the same circle and the center of this circle is midpoint $N$ of $AH$. It is easy to check that $NE=NH$ and $ME=MB$, so $$\angle MEN=\angle MEB+\angle NEB=\angle MBE+\angle NHE=\angle EAH+\angle AHE=90^{\circ }.$$ Then $ME$ is the tangent line of $(AEF)$ at $E$.

By same way, we have $FM$ is the tangent line of $(AEF)$ at $F$. These imply that the tangent lines of $(AEF)$ at $E,F$ meet at the fixed point $M$.



2) Consider the diameter $A{A}^{\prime }$ of $(O)$ then it is easy to see that $BHC{A}^{\prime }$ is a parallelogram and $H,M,A^{\prime }$ are collinear.

Suppose that $MH\cap (O)=D\ne A^{\prime }$ then $\angle ADH=\angle AD{A}^{\prime }=90^{\circ }$ which means $D\in (AEF)$.

By consider three radical axis of three circles $(O)$, $(BFEC)$, $(AEHF)$, we have $AD,EF,BC$ are concurrent at $T$, which is the radical center. In triangle $ATM$, we have $AH\perp TM$ and $MH\perp AT$, then $H$ is the orthocenter. From this, we can conclude that $TH\perp AM$. $\square$