How can we prove that the following scalar product relation holds in any triangle?
$$\left [-\overrightarrow{AB}\tan B (\tan A +2\tan C)+\overrightarrow{AC}\tan C (\tan A+2\tan B)\right ]\cdot \left (\overrightarrow{AB}+\overrightarrow{AC}\right )=0 $$
I strived a lot but I only get to some long relations that overcomed me.
Let's define $\vec u=\overrightarrow{AB}$, $\vec v=\overrightarrow{AC}$, $\alpha=\angle A$, $\beta=\angle B$, $\gamma=\angle C$. From the sine rule one gets $u/\sin(\alpha+\beta)=v/\sin\beta$ and $u/\sin\gamma=v/\sin(\alpha+\gamma)$, whence: $$ \tan\beta={v\sin\alpha\over u-v\cos\alpha}, \quad \tan\gamma={u\sin\alpha\over v-u\cos\alpha}. $$ By substituting that into your expression one obtains: $$ \begin{align} &[-\vec u\tan\beta(\tan\alpha+2\tan\gamma)+\vec v\tan\gamma(\tan\alpha+2\tan\beta)](\vec u+\vec v)= \\ &\\ &(-u^2-uv\cos\alpha)\tan\beta(\tan\alpha+2\tan\gamma)+ (v^2+uv\cos\alpha)\tan\gamma(\tan\alpha+2\tan\beta)=\\ &\\ &(-u^2-uv\cos\alpha){v\sin\alpha\over u-v\cos\alpha}\left(\tan\alpha+2{u\sin\alpha\over v-u\cos\alpha}\right)+\\ &(v^2+uv\cos\alpha){u\sin\alpha\over v-u\cos\alpha} \left(\tan\alpha+2{v\sin\alpha\over u-v\cos\alpha}\right)=\\ &\\ &uv\sin\alpha\left[-{u+v\cos\alpha\over u-v\cos\alpha}\left(\tan\alpha+2{u\sin\alpha\over v-u\cos\alpha}\right)+ {v+u\cos\alpha\over v-u\cos\alpha}\left(\tan\alpha+2{v\sin\alpha\over u-v\cos\alpha}\right) \right]\\ &\\ &=uv\sin\alpha\tan\alpha\left( {v+u\cos\alpha\over v-u\cos\alpha}-{u+v\cos\alpha\over u-v\cos\alpha} \right) +2uv\sin^2\alpha{v^2-u^2\over(u-v\cos\alpha)(v-u\cos\alpha)} \\ &=0.\\ \end{align} $$