Let $E,F,G$ be tangency points of the incircle with $BC,CA,AB$ respectively. It suffices to show $BD=AG+FC$. WLOG let inradius be $1$, then $AD=\frac{1}{\sin(\frac{A}{2})\cos(\frac{A}{2})}$, $AG=\frac{1}{\tan(\frac{A}{2})}$, and $BD=\frac{1}{\tan(\frac{B}{2})}$. From the equation $2AD=AG+BD$, we get $\frac{2-\cos^2(\frac{A}{2})}{\sin(\frac{A}{2})\cos(\frac{A}{2})}=\frac{1}{\tan(\frac{B}{2})}$. Then use algebra to get $\frac{1}{\tan(\frac{B}{2})}=\frac{1}{\tan(\frac{A}{2})}+\tan(\frac{A+B}{2})$.