By the angle bisector theorem,we have $\frac{BI}{IE}=\frac{BC}{CE}$ and $\frac{CI}{IF}=\frac{BC}{BF}$,so we have $BC.IE=BI.CE$ and $BC.IF=CI.BF$,as $IE=IF$ we evidently have $\frac{BI}{BF}=\frac{CI}{CE}$.Let $\angle BFC=x,\angle CEB=y,\angle BIF=\angle CIE=a$.By sine law on $\bigtriangleup BIF$ and $\bigtriangleup CIE$,$\frac{\sin x}{\sin a}=\frac{\sin y}{\sin a}$,so we have $x=y$ or $x=180^{\circ}-y$,since $AB \ne AC$,we must have $x+y=180^{\circ}$,which implies $A,F,I,E$ are concyclic.It is well known that $\angle FIE=90^{\circ}+\frac{1}{2}\angle A$,using the cyclicity,we have $\angle BAC=60^{\circ}$$\blacksquare$

CEE FIGURE

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Well, it may be little bit different too! Best regards, sunken rock

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Ireland 2009