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.
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 --- (**)
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?