Here, I have a very interesting problem that I found.
In triangle $ABC,$ we have $\angle B = 2\angle A.$ Prove that $b^2 = a(a+c).$
Here is how far I have gotten:
Here, $\angle B= 2 \angle A$ and therefore, $\angle C = 180- 3 \angle A$. Now, I use the Law of Cosines because if I distribute out the RHS of the equation we need to prove, we get $b^2 = a^2 + ac$, which is close to what I would get by the Law of Cosines.
If I use Law of Cosines, I get the following possibilities: $$ a^2 = b^2+c^2 -2bc\cdot \cos A\\ \boxed{b^2 = a^2+c^2 -2ac\cdot \cos B} \\ c^2 = a^2+b^2 -2ab\cdot \cos C.$$ Now what would I do? Thanks in advance for any help.