Hence $\angle \mathrm{DFC}=\angle \mathrm{FDC}=x^{\circ }$.

Hence $\angle \mathrm{FCD}=(180-2x{)}^{\circ }$ (angle sum of triangle).

Now $\angle \mathrm{EBC}=\angle \mathrm{FDC}=x^{\circ }$ (opposite angles of a parallelogram) and triangle EBC is isosceles (CE = CB). Hence $\angle \mathrm{BEC}=x^{\circ }$ and $\angle \mathrm{ECB}=(180-2x{)}^{\circ }$

Lines $\mathrm{AD}$ and $\mathrm{BC}$ are parallel and hence

$\angle \mathrm{ADC}+\angle \mathrm{BCD}={180}^{\circ }$.

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

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