In the given figure, side BC of Δ ABC is produced to form ray BD and CE || BA. Then ∠ ACD is equal to :

From above given figure , we can see that
CE || BA and AC is the transversal.
∴ ∠4 = ∠1 (alternate interior ∠s) .......... ( 1 )
Again, CE || BA and BD is the transversal.
∴ ∠5 = ∠2 (corresponding ∠s) ................( 2 )
Adding equations ( 1 ) and ( 2 )
∴ ∠4 + ∠5 = ∠1 + ∠2
∴∠ACD = ∠A + ∠B.