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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 $AI^2$.

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

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    $\begingroup$ Thank you very much. So the formula is $$AI^2 = \frac{bc}{(a+b+c)^2}[(b+c)^2 - a^2].$$ $\endgroup$ – martin Jan 2 '14 at 11:25

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