It is given to us –

BA || ED

BC || EF

We have to show that ∠ABC = ∠DEF

Let us extend DE to intersect BC at P. Then, the figure becomes –

Since, BA || ED

⇒ BA || DP

We have two parallel lines BA and DP, and BP is a transversal intersecting BA and DP at points B and P respectively.

⇒ ∠ABC = ∠DPC (corresponding angles) - - - - (i)

Similarly, we have two parallel lines BC and EF, and DP is a transversal intersecting BC and EF at points E and P respectively.

⇒ ∠DPC = ∠DEF (corresponding angles) - - - - (ii)

From (i) and (ii), we have

∠ABC = ∠DPC, and ∠DPC = ∠DEF

⇒ ∠ABC = ∠DEF

Hence, proved.