Angle bisector in a triangle

Question

For the angle bisector $I_a$ in a triangle $ABC$ it holds $$I_a^2 = \frac{bc}{(b+c)^2}[(b+c)^2 - a^2]$$ If $I$ is the incenter, I wonder if there exist similar formula for the part $AI^2$.

Answer 1

Put weights $a$, $b$, $c$ on the three vertices $A$, $B$, $C$. Their center of gravity is $I$. The center of gravity of $B$ and $C$ is the intersection of $AI$ with $BC$.

Therefore, $AI : I_a = \dfrac{b+c}{a+b+c}$.