Irska

Let $M$ and $N$ be the midpoints of $BD$ and $DC$ respectively. Let $AD$ meet $EF$ at $L^{\prime }$. Then $F{L}^{\prime }\parallel BD$ and $|F{L}^{\prime }|=\frac{1}{2}|BD|$. Similarly $|L^{\prime }E|=\frac{1}{2}|DC|$, hence $|F{L}^{\prime }|=|L^{\prime }E|$, i.e. $L^{\prime }=L$ is the midpoint of $FE$.



Let $P$ be the midpoint of $OD$. Then $PL\parallel OA$ and $|PL|=\frac{1}{2}|OA|$. Similarly $|PM|=\frac{1}{2}|OB|$ and $|PN|=\frac{1}{2}|OC|$. Since $|OA|=|OB|=|OC|$ then $|PL|=|PM|=|PN|$. So the circle with centre $P$ and radius $|PM|$ passes through $L$. As $A$ moves on the circumference of the circle, the points $O$, $P$, $D$ and $M$ remain fixed. Thus $L$ will always lie on the circumference of the circle centre $P$ and radius equal to half the radius of the circumcircle.