$F$ is the Fermat pt. of $\triangle ABC$, so using SAS congruence $\triangle DAC \cong \triangle BAE$, and since the incenter lies on the angle bisectors, $180 = \angle DAE + \angle DEA + \angle ADE = 240 - \angle A + \angle AEB + \angle ADC = 240 - \angle A + \angle ACF + \angle ABF $, and using the property of the Fermat pt. that $\angle AFB = \angle BFC = \angle CFA = 120$, $180 = 240 - \angle A + 60 - \angle BAF + 60 - \angle FAC = 360 - 2 \angle A$, therefore $ \angle BAC = 90$.