In a triangle $ABC$, $\measuredangle BAC = 3.\measuredangle ACB$ and the point $D$ lies on the side $BC$ so that $\measuredangle ADC = 2.\measuredangle ACB$. Prove $BC = AD+AB$.
My guess is to use laws of sines and cosines.
First I drew the angle bisector of $\angle ADC$ and showed that the triangles $\triangle ADC$ and $\triangle AED$ are similar. hence $DC = AD(1 - 2\cos{2\theta})$ where $\theta = \measuredangle ACD$.
I don't know how to continue.
Im going to asumme you meant $\angle BAD$ instead of $\angle ABC$.
Denote angle $\angle ACB$ as $\theta$ then $\angle BAC=3\theta$ and $\angle BAD=2\theta$ so $\angle DAC=3\theta - 2\theta=\theta$ and $\angle BDA=180-(180-2\theta)=2\theta$ so AD = DC and AB = BD $ $ BC = BD + DC =AD + AB.