In a triangle ABC, angle A is twice the angle B. show that a² =b(b+c),where a,b,and c are the sides opposite to angle A,B and,C respectively.

Dear Student

Given, angle A = 2*angle B,
Apply sine rule i.e..

a/(Sin A)= b/(Sin B)= c/(Sin c)

Take the first 2 entities as we were given information about A and B. Also, substitute A = 2B
Now, we have
a/(Sin 2B) = b/(Sin B)
Sin 2B = 2 *(Sin B)*(Cos B)

Substituting this in the above equation. Thus,
Sin B gets canceled because B is not equal to zero(because it is a triangle). By extension. Sin B is not equal to zero.

We get,
a/(2 Cos B) = b [now apply cosine rule]
Cos B = (a^2 + c^2 - b^2)/2*a*c
Substituting we get the result,
a^2 = b(b+c)=b^2+b*c
BC^2 = AC^2 + AC*AB

Regards