Deductive geometry O is a point in triangle ABC, and equidistant from B and C.
BO extended meets AC at F.
CO extended meets AB at E.
If AE=AF, prove that BE=CF.
I've seen a proof for the case that ∠BEC and ∠BFC are acute.
Assume BE>CF.
Consider triangle ABC. Longer side faces larger angle.
BE>CF <=> ∠ACB>∠ABC --- (*)
altitude of triangle BEC from B=altitude of triangle BFC from C
BE sin∠BEC=BF sin∠BFC
When ∠BEC and ∠BFC are acute,
BE>CF <=> ∠BFC>∠BEC --- (**)
(*)+(**),
∠ACB+∠BFC>∠ABC+∠BEC
As the angle sum of triangle BFC and BEC must be 180 degrees, there is contradiction.
Similarly, there is contradiction when BE< CF.
Therefore, we must have BE=CF.
This proof does not work when ∠BEC and ∠BFC are obtuse. Then, how can the obtuse scenario be proved?
 A: For simplicity of notation, scale triangle $ABC$ so that $AE=AF=1$.

Let $x = BE,\;y = CF$.

Suppose $x \ne y$.

Our goal is to derive a contradiction.

Without loss of generality, assume $x > y$.

For a given triangle $T$, let $k(T)$ denote the area of $T$.

Letting $K = k(ABC)$, we get
$$
\frac{k(BEC)}{k(CFB)} 
=
\frac{{\large{\frac{x}{x+1}}}K}{{\large{\frac{y}{y+1}}}K}
=
\frac{xy + x}{xy + y} > 1
$$
As you explained in your comments, since triangle $BOC$ is isosceles, the altitude from $B$ in triangle $BEC$ is equal to the altitude from $C$ in triangle $CFB$, hence

$$
\frac{k(BEC)}{k(CFB)} 
=
\frac{CE}{BF}
$$
Then since  ${\large{\frac{k(BEC)}{k(CFB)}}} >1$, we get $CE > BF$, hence $OE > OF$.

Choosing $G$ on segment $OE$ so that $OG=OF$, we get $\triangle CBG \cong \triangle  BCF\;\;($by $\text{SAS})$.
\begin{align*}
\text{Then}\;\;&\triangle CBG \cong \triangle  BCF\\[4pt]
\implies\;&\angle CBG = \angle BCF\\[4pt]
\implies\;&\angle CBE > \angle BCF\qquad\text{[since $\angle CBE > \angle CBG$]}\\[4pt]
\implies\;&\angle CBA > \angle BCA\\[4pt]
\implies\;&AC > AB\\[4pt]
\implies\;&y+1 > x+1\\[4pt]
\implies\;&y > x\\[4pt]
\end{align*}
contradiction.
