Let $M$, $N$, $O$ be the midpoints of $EA$, $EB$, $EC$, respectively and denote $L'$ the midpoint of $MO$. As the common chord of two circles is perpendicular to and bisected by the line connecting the centres, we get (regarding the circumcircles of $\triangle{ABE}$, $\triangle{BCE}$ and $\triangle{CEA}$, the latter having $F$ as its midpoint) in particular $M\in FO_1$, $O\in FO_2$ and $O_1O_2\perp BD$. The latter implies that $O_1O_2\| AC$, together with $MO\| AC$ we know that $MO\| O_1O_2$. Therefore, a dilation with centre $F$ and scaling factor $\frac{FM}{FO_1}$ maps $L$ to $L'$ and it suffices to show that $F$, $E$ and $L'$ are collinear. But this is obvious as $FMEO$ is a rectangular and $L'$ the midpoint of its diagonal $MO$.

Let $FO_1 \cap AE=G$ and $FO_2 \cap EC=K$, $\angle O_2O_1G=180^{\circ}-\angle AEB=\angle ACE=\angle AFO_1$ $\implies$ $O_1O_2$ $||$ $AC$ $||$ $GK$ and since, $EF$ bisects $GK$ $\implies$ $EF$ bisects $O_1O_2$