the side BC of triangle abc is produced such that D is on the ray BC the bisector of /_A meets BC IN L prove that angle abc+angle acd =2angle ALC
since AL is the angle bisector of ∠A.
let ∠BAC=2x
therefore ∠BAL=∠LAC=x
∠ACD=∠ABC+BAC [exterior angle is equal to the sum of opposite interior angles]
∠ACD=∠ABC+2x.........(1)
therefore
∠ABC+∠ACD=∠ABC+∠ABC+2x
=2∠ABC+2x
=2(∠ABC+x).......(2)
∠ALC=∠ABC+∠BAL=∠ABC+x....(3) [since ABL is a triangle and ALC is an exterior angle,exterior angle is equal to the sum of opposite interior angles]
by (2) and (3)
∠ABC+∠ACD=2∠ALC