Angle BAC = $70^o$ , Angle ABE = Angle EBC. Find angle BOC
Let angle ABE = $\alpha$ and angle BOC = $\beta$
Angle ACB = $110^o - 2\alpha$
Angle AEB = $110^o-\alpha $
Angle FOE = $\beta$
Angle AFO = $180^0-\beta +\alpha$
Angle BFC = $\beta - \alpha$
Then what to do?
The angle can be anything between $70°$ and $180°$ degrees, exclusive.
With so little information, there is a huge amount of flexibility in the shape. By placing $B$ a long way away from $AC$, we get $\angle BEC$ to just above $70°$, and $\angle EBC$ to almost $0°$, allowing us to choose $F$ to give an angle anywhere between $70°$ and $180°$.
This image shows $B$ a reasonably large distance away from $AC$, and $F$ in varying locations. $\angle EBC$ here is about $3°$, so $\angle BOC$ can be anywhere from $73°$ to $177°$ just by moving $F$.