in the figure, the side BC of triangle ABC is produced to D. the bisector of angle A meets BC at L. prove that angle ABC+ angleACD= 2 angle ALC

Triangle ABL.

Let angle BAL = a and angle LBA = b.

Exterior angle ALC = angle BAL + angle LBA

= a + b.

Triangle ABC.

As AL bisects angle BAC, angle LAC = a.

Exterior angle ACD = angle BAC + angle ABC

= 2a + b.

Angle ABC + angle ACD

= b + (2a + b)

= 2(a + b)

= 2 angle ALC

**
**