Given that,

BC produced on both sides

We know that,

∠A + ∠ABC + ∠ACB = 180^o (i)

∠ABD = ∠A + ∠ACB (Exterior angle theorem) (ii)

∠ACE = ∠A + ∠ABC (Exterior angle theorem) (iii)

Adding (ii) and (iii), we get

∠ABD + ∠ACE = ∠A + (∠A + ∠ACB + ∠ACB)

∠ABD + ∠ACE = ∠A + 180^o

(∠ABD + ∠ACE) - ∠A = 180^o

Thus, between the sum of the exterior angles so formed and ∠A is 180^o.