geometry

Since ABCD is a square, each of its corner angles are right angles, each = 90^o.

Since DE and DF trisect angle(ADC), angle(ADE) = 30^o, angle(EDF) = 30^o, angle(FDC) = 30^o.

 

Triangle(DCF) is a 30^o - 60^o - 90^o right triangle.

 

Since DC = 1, CF = 1/sqrt(3) and DF = 2/sqrt(3).

Similarly, DE = 2/sqrt(3).

 

Using this formula to find the area of triangle(EDF):  area = ½·a·b·sin(C)

--->   area  =  ½ · 2/sqrt(3) · 2/sqrt(3) ·sin(30^o)

--->   area  =  ½ · 2/sqrt(3) · 2/sqrt(3) · 0.50

--->   area  =  1/3