I need to prove that $BO \perp CD$.
Two circles $ω_1$ and $ω_2$ intersect at points $A$ and $B$. A line passing through point $B$ intersects ω1 at point $C$ and $ω_2$ at point $D$. Line $AC$ intersects circle $ω_2$ a second time at point $F$, and line $AD$ intersects circle a second time $ω_1$ at point $E$. Let $O$ be the center of the circumcircle around $\triangle AEF$. Prove that $OB\perp CD$.
I proved that $\angle CBE=\angle CAE= \angle FAD= \angle FBD$. So now I need to prove only that $\angle EBO= \angle FBO$. But I don't know how do this.
You were almost done!
You have already proved: $\angle CBE=\angle CAE=\angle DBF$.
Now prolong the line $BF$ till the second intersection with $\omega$ at $G$. We have $\angle FGE=\angle FAE \implies \angle FGE=\angle FBD \implies (EG)\parallel (CD)\\ \implies \angle GEB=\angle CBE \implies \angle GEB=\angle EGB$.
That means $\triangle EBG$ is isosceles and both $B$ and $O$ lie on the perpendicular bisector to $EG$.