Angle in triangle

Question

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?

With no information about line AF or how O is located, there is not enough information to solve this problem. — user317176, Apr 18 '23 at 09:35
That's the whole question so no information about exact location of point F and E and also no information about other angles besides what has been written in the question — Magenta, Apr 18 '23 at 11:31
What was whited out? In particular, there seems like there used to be a point on $\overline{BC}$; are you sure that is not part of this question? — Brian Tung, Apr 18 '23 at 23:08
Well I am not sure since the question given already has some parts whited out — Magenta, Apr 19 '23 at 03:14

score 1 · Accepted Answer · answered Apr 18 '23 at 23:03

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 $F$.

Angle in triangle

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