Question

The side BC of ∆ABC is produced to point D, the bisector of ∠BAC intersects the side BC at E, prove that ∠ABC + ∠ACD = 2 ∠AEC.

Answer 1

Answer:

Hi , HERE'S YOUR ANSWER

≤AEC =≤ABC +≤BAE

Exterior angle=≤ABC+≤CAE(1)

≤BAE=≤CAE;(AE bisect ≤BAC )

In ∆AEC ,

≤ACD=AEC+≤CAE

Exterior angle theorem

=≤CAE = ≤ACD - ≤AEC ....(2)

From (1) and (2)

≤AEC =≤ABC + (≤ACD-≤AEC)

= 2≤AEC = ≤ABC+≤ACD

Hope it helps you ☺️