Nice. I like this solution (# means angle) :
Triangle BCE: it is given that #ABC = 80 and #BCE = 60; then #BEC = 40
Triangle CDF: this triangle being equilateral, then #CFD = 60;
since #BFC = 80, this leaves #DFE = 180-80-60 = 40
Triangle DEF: #EDF = #DEF, so #CED + 40 = 70, hence #CED = 30