Begin by noticing that the lines $CO_1$ and $DO_2$ meet at a point $P$ on $\omega$, since $\angle(PO_1, PO_2) = \angle(O_1C, CB) + \angle(BD, DO_2) = \angle(CB, BO_1) + \angle(O_2B, BD) = \angle(O_2B, BO_1) = \angle(O_1A, AO_2)$. In what follows, we consider the case where $O_1$ and $O_2$ lie on the segments $CP$ and $DP$, respectively; the other cases are similar.

Since the angles $PCE$ and $PDE$ are both right, and $2\angle ACP = \angle AO_1P = \angle AO_2P = 2\angle ADP$ (the equality in the middle holds on account of $P$ lying on $\omega$), the points $A$, $C$, $D$, $E$, $P$ all lie on the circle on diameter $EP$, so $FP$ is a diameter of $\omega$, and it is therefore sufficient to show that $EF = FP$. Finally, since $\angle AFP = \angle AO_1P = 2\angle ACP = 2\angle AEP$ (the first, respectively third, equality holds on account of $APFO_1$, respectively $ACEP$, being cyclic), it follows that the triangle $EFP$ is isosceles with apex at $F$.