Produce EF to intersect AC at K

Now,

∠DCE + ∠CEF = 35^o + 145^o

= 180^o

Therefore, EF ‖ CD (Since, sum of co-interior angles is 180^o) (i)

Now,

∠BAC = ∠ACD = 57^o

BA ‖ CD (Therefore, alternate angles are equal) (ii)

From (i) and (ii), we get

AB ‖ EF (Lines parallel to the same line are parallel to each other)

Hence, proved