Question

in the adjoining figure the side BC of triangle ABC is produced to a point is such that the bisectors of angle abc and angle ACB meet at D prove that angle BAC is equal to half angle BAC​

Answer 1

Answer

Considering the question to be:

Show that ∠ABC+∠ACD=2∠AEC

Consider ∠B=α,∠BAE=∠EAC=x

Using exterior angle property, we have,

∠AEC=x+α

Similarly

∠ACD=2x+α

∴∠ABC+∠ACD

=α+(2x+α)

=2x+2α

=2(x+α)

=2∠AEC

∴∠ABC+∠ACD=2∠AEC