SAUDI ARABIAN MATHEMATICAL COMPETITIONS

1) Let $M$ be the midpoint of $BC$ then $G\in AM$ and $\frac{AG}{AM}=\frac{2}{3}$. Take the point $K^{\prime }$ on $BC$ such that $M$ is the midpoint of $H{K}^{\prime }$, then $AM$ is the median of triangle $AH{K}^{\prime }$ and $G$ is its centroid. Then $K^{\prime }G$ is the median of triangle $AH{K}^{\prime }$ or $K^{\prime }G$ passes through the midpoint of $AH$. This implies that $K\equiv K^{\prime }$ and we have $BH=CK$.



2) The line passes through $A$ and parallel to $BC$ intersects $(O)$ at $E$ different from $A$. Then by the symmetry through the perpendicular bisector of $BC$, it is easy to check that $AHKE$ is a rectangle. Since $\frac{GA}{GM}=\frac{AE}{HM}=2$, we have $H,G,E,L$ are collinear. Then $\angle BLH=\angle BLE=\angle BCE=\angle ABC$ which implies that $$\angle ABT=\angle ABC+\angle CBT=\angle BLH+\angle CBT=90^{\circ }.$$ Thus if we denote $D=BT\cap (O)$ then $AD$ is the diameter of $(O)$, then $O\in AD$. Therefore, $BT$ and $AO$ intersect at a point that belongs to $(O)$. $\square$