In ABC, B =C. The points D, E, F are taken on AB, BC and CA
such that DEF is an equilateral triangle. Suppose that the measures
of the angles DEB, ADF and CFE are ,  and  respectively.
Show that

Let angle B = angle C = xLet Alpha be a, beta be b and gamma be cUsing linear pair,Angle {BDE + EDF + ADF} = 180°So angle BDE = 120°- bSimilarly angle FEC = 120°- aNow in ∆ BDE, using angle sum property, we getx + a + 120°-b = 180=> a- b = 60° - x -----------(1)Similarly from ∆ EFC, we getc-a = 60° - x -------------(2)Equating (1) & (2), we geta =(b+c)/2 ----------(3)Using sine rule,Sin x/DE =sin a/BDSince DE = EFSin x/EF = sin a /BD = sin (120°-a)/CFWe know that a =(b+c)/2Hence proved that {sin (b+c)/2}/{sin (120°-a)} = BD/CF

Approved