# For the angle bisector I a </msub> in a triangle A B C it holds

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

Roland Waters
Thank you very much. So the formula is
$A{I}^{2}=\frac{bc}{\left(a+b+c{\right)}^{2}}\left[\left(b+c{\right)}^{2}-{a}^{2}\right].$