Irish Mathematical Olympiad

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