Let AB, CD and EF intersect at O
∠AOD = ∠COB (Vertically opposite angle)
b = e (i)
∠EOC = ∠DOF (Vertically opposite angle)
f = c (ii)
Adding (i) and (ii), we get
b + f = c + e (iii)
Now,
∠ADE + ∠EOC + ∠COB = 180°
a + f + e = 180°
a + c + e = 180°[From (ii)]