Given: The side BC of ∆ABC is produced to D. The bisector of ∠A meets BC in L.
To Prove: ∠ABC + ∠ACD = 2 ∠ALC.
Proof: ∠ABC + ∠ACD
= ∠ABC + (∠ABC + ∠BAC)
| Exterior angle theorem
= 2 ∠ABC + ∠BAC
= 2 ∠ABC + 2 ∠BAL
| ∵ AL is the bisector of ∠A
= 2 (∠ABC + ∠BAL)
= 2 ∠ALC | Exterior angle theorem