Problem 1942

a) We rewrite the first part of the problem as following. Let the acute triangle $ABC$ inscribed in the circle $(O)$. $AD$ is the altitude and $H$ is orthocenter of the triangle $ABC$. $M$ is the midpoint of $BC$. The circle with the diameter $AH$ cuts $(O)$ again at $G$. $GD$ cuts $(O)$ again at $K$. The straight line passing through $K$ and perpendicular with $BC$ cuts $AM$ at $L$. Then, four points $B$, $C$, $L$, $H$ belong to the same circle.

Since $G$ lies on the circle with the diameter $AH$, $GH$ cuts $(O)$ again at $E$ and then $AE$ is the diameter of $(O)$. From that, the quadrilateral $HBEC$ is the parallelogram, and $HE$ goes through $M$. Let $N$ be the intersection of $AM$ and $(O)$. Since the quadrilaterals $AGDM$ and $AGKN$ are inscribed, we have
$$\angle \text{GDM} = 180^{\circ} - \angle \text{AGD} = \angle \text{GKN}.$$
This implies that $KN \parallel BC$, or the quadrilateral $BCNK$ is an isosceles trapezoid. We have $M$ is the midpoint of $BC$ so $MK = MN$. It is easy to see that the triangle $LKN$ is a right angle at $K$. We decide that $M$ is the midpoint of $LN$ or $L$ is the reflection of $K$ through $BC$. Note that the reflection of $H$ through $BC$ belongs to $(O)$ then $H$, $L$, $B$, $C$ all belong to reflection circle of $(O)$ through $BC$. The first part of the problem follows.

b) For solution of the second part, we rewrite the statement as following. Let the triangle $ABC$ be inscribed in the circle $(O)$, and $H$ be the orthocenter of $ABC$. The circle with diameter $AH$ cuts $(O)$ again at $G$. A circle touches $AG$ at $A$ cuts $CA$ and $AB$ again at $E$ and $F$, respectively. Then, $AO$ bisects the segment $EF$.

Let $D$ be the intersection of $GH$ and $(O)$ then $AD$ is the diameter of $(O)$. The quadrilateral $HBDC$ is the parallelogram so the line $HD$ goes through the midpoint $M$ of $BC$. Let $P$ be the intersection of $AD$ and the circumcircle of the triangle $AEF$. The line $AP$ cuts $EF$ at $N$. It is easy to see that
$$\angle \text{FPN} = \angle \text{FAE} = \angle \text{GAB} = \angle \text{BDM}$$
and
$$\angle \text{PFE} = \angle \text{PAE} = \angle \text{DBM}.$$
Therefore, the triangles $DBM$ and $PFM$ are similar. Analogously, the triangles $DCM$ and $PEN$ are similar. Because $M$ is the midpoint of $BC$, we have $N$ is the midpoint of $EF$. The second part of the problem follows.