67th NMO Selection Tests for BMO and IMO

Begin by noticing that the lines $C{O}_1$ and $D{O}_2$ meet at a point $P$ on $\omega$, since $\angle (P{O}_1,P{O}_2)=\angle (O_{1}C,CB)+\angle (BD,D{O}_2)=\angle (CB,B{O}_1)+\angle (O_{2}B,BD)=\angle (O_{2}B,B{O}_1)=\angle (O_{1}A,A{O}_2)$. 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 $2\angle ACP=\angle A{O}_{1}P=\angle A{O}_{2}P=2\angle ADP$ (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 $\angle AFP=\angle A{O}_{1}P=2\angle ACP=2\angle AEP$ (the first, respectively third, equality holds on account of $APF{O}_1$, respectively $ACEP$, being cyclic), it follows that the triangle $EFP$ is isosceles with apex at $F$.