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

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