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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?

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  • $\begingroup$ Please provide a figure for a geometric problem. $\endgroup$ – Jean Marie Dec 4 '17 at 8:29
  • $\begingroup$ "altitude of triangle BEC from B=altitude of triangle BFC from C". Why? $\endgroup$ – quasi Dec 4 '17 at 8:30
  • $\begingroup$ Note that OBC is isosceles. altitude of triangle BEC from B=altitude of triangle OBC from B=altitude of triangle OBC from C=altitude of triangle BFC from B $\endgroup$ – Stupid Girl Dec 4 '17 at 12:47
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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.

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