In equilateral triangle ΔDEF, X and Y are points on the side DE and DF respectively, such that DX = FY, EY and FX intersect at Z. The measure (in degrees) of ∠FZE is ?

120°

ΔEFY ≅ ΔDFX [by SAS congruency]

So, the three angles of these two triangles are the same.

Let ∠YEF = θ ⇒ ∠DFX is also θ

Now ∠EYF = 180° - (60° + θ) = 120° - θ

Now, in ΔFYZ

Exterior angle ∠FZY = (120° + θ) + θ = 120°