Given: D is mid point of BC. AD produced to E, AD = DE
In △ABD and △EDC
∠ADB=∠EDC (Vertically Opposite angles)
BD=CD (D is mid point of BC)
AD=DE (Given)
Thus, △ABD≅△ECD (SAS rule)
Thus, ∠BAD=∠DEC (By cpct)
Since, these are alternate angles w.r.t AB and EC, thus, AB∥EC