The side BC of a triangle ABC is produced, such that D is on ray BC. The bisector of ∠A meets BC at E. Show that ∠ABC + ∠ACD = 2 ∠AEC.

∠1 = ∠2
Since ∠ACD is the exterior angle of triangle ABC 
.^.. ∠ACD = ∠A + ∠B ….(i)
Now in ΔABE, we have
    ∠AEC = ∠1 + ∠6
⇒ 2∠AEC = 2∠1 + 2∠6
⇒ 2∠AEC = ∠A + 2∠B
⇒ 2∠AEC = (∠ACD - ∠B) + 2∠B
⇒ 2∠AEC = ∠ACD + ∠B
           = ∠ACD +∠ABC