Hence ÐDFC = ÐFDC = x^°.

Hence ÐFCD = (180-2x)^° (angle sum of triangle).

Now ÐEBC = ÐFDC = x^° (opposite angles of a parallelogram) and triangle EBC is isosceles (CE = CB). Hence ÐBEC = x^° and ÐECB = (180-2x)^°

Lines AD and BC are parallel and hence

ÐADC + ÐBCD = 180^°.

Therefore: x+2(180-2x) + 60 = 180

i.e.420-3x=180 i.e. 3x=240 i.e. x=80