In a triangle ABC, The side BC OF TRIANGLE ABC IS PRODUCED TO D. The bisector of Angle A meets BC in L. Prove that angleABC+AngleADC=2Angle ALC

In Δ ABL.
Let ∠BAL = a and ∠LBA = b.
Exterior ∠ALC = ∠BAL + ∠LBA = 

∠ALC = a + b  …………(i)

In Δ ABC,
As AL bisects ∠BAC,

∠LAC = a
Exterior ∠ACD = ∠BAC + ∠ABC = 

∠ACD = 2a + b

As ∠ABC = b, by adding ∠ABC and ∠ACD, we get
∠ABC + ∠ACD = b + (2a + b) = 2(a + b)

∠ABC + ∠ACD  = 2 ∠ALC  [ from (i) ]