In equilateral ΔABC, D and E are points on the side AB and AC, respectively, such that AD = CE. BE and CD intersect at F. The measure (in degrees) of ∠CFB is :

Solution (By Examveda Team)

ΔADC and ΔEBC
AD = EC
∠A = ∠C = 60°
AC = BC
ΔADC ∽ ΔEBC   [By SAS]
Let ∠ECF = ∠EBC = x
∠FCB = 60° - x
In ΔBFC
∠B + ∠C + ∠F = 180°
x + 60° - x + ∠BFC = 180°
∠BFC = 180° - 60° = 120°
∠CFB = 120°