CinarArslan wrote: In a triangle $\Delta ABC$, $|AB|=|AC|$. Let $M$ be on the minor arc $AC$ of the circumcircle of $\Delta ABC$ different than $A$ and $C$. Let $BM$ and $AC$ meet at $E$ and the bisector of $\angle BMC$ and $BC$ meet at $F$ such that $\angle AFB=\angle CFE$. Prove that the triangle $\Delta ABC$ is equilateral. It's easy to see that $\triangle ABF \sim+ \triangle ECF$. So we have $$\frac{AB}{EC}=\frac{BF}{CF}=\frac{BM}{CM}.$$ But $\angle ABM=\angle ECM$. Therefore, $\triangle ABM \sim+ \triangle ECM$. Hence, $\angle ACB=\angle AMB=\angle EMC=\angle BAC$. So $\triangle ABC$ must be equilateral.