The side BC of a △ABC produced, such that D is on ray BC. The bisector of ∠A meets BC in L. Prove that ∠ABC + ∠ACD = 2∠ALC. [3 MARKS]
Process : 2 Marks
Proof : 1 Mark
In ΔABC,
∠ACD = ∠B + ∠A
∠ACD = ∠B + 2∠1 ..........(i)
In ΔABL,
∠ALC = ∠B + ∠BAL
∠ALC = ∠B + ∠1
2∠ALC = 2∠B +2∠1 ......(ii)
Subtracting (i) from (ii), we get
2∠ALC - ∠ACD = ∠B
∠ACD + ∠B = 2∠ALC
∠ACD + ∠ABC = 2∠ALC