lengths of angle bisectors

In any triangle, the $w_{a}$ , $w_{b}$ , $w_{c}$ of the angle bisectors opposing the sides $a$ , $b$ , $c$ , respectively, are

\displaystyle w_{a}=\frac{\sqrt{bc\,[(b\!+\!c)^{2}\!-\!a^{2}]\,}}{b\!+\!c},

(1)

\displaystyle w_{b}=\frac{\sqrt{ca\,[(c\!+\!a)^{2}\!-\!b^{2}]\,}}{c\!+\!a},

(2)

\displaystyle w_{c}=\frac{\sqrt{ab\,[(a\!+\!b)^{2}\!-\!c^{2}]\,}}{a\!+\!b}.

(3)

Proof. By the symmetry, it suffices to prove only (1).

According the angle bisector theorem, the bisector $w_{a}$ divides the side $a$ into the portions

\frac{b}{b\!+\!c}\cdot a\;=\;\frac{ab}{b\!+\!c},\qquad\frac{c}{b\!+\!c}\cdot a% \;=\;\frac{ca}{b\!+\!c}.

If the angle opposite to $a$ is $\alpha$ , we apply the law of cosines to the half-triangles by $w_{a}$ :

\displaystyle\begin{cases}2w_{a}b\cos\frac{\alpha}{2}\;=\;w_{a}^{2}\!+\!b^{2}% \!-\!\left(\frac{ab}{b+c}\right)^{2}\\ 2w_{a}c\cos\frac{\alpha}{2}\;=\;w_{a}^{2}\!+\!c^{2}\!-\!\left(\frac{ca}{b+c}% \right)^{2}\end{cases}

(4)

For eliminating the angle $\alpha$ , the equations (4) are divided sidewise, when one gets

\frac{b}{c}\;=\;\frac{w_{a}^{2}\!+\!b^{2}\!-\!\left(\frac{ab}{b+c}\right)^{2}}% {w_{a}^{2}\!+\!c^{2}\!-\!\left(\frac{ca}{b+c}\right)^{2}},

from which one can after some routine manipulations solve $w_{a}$ , and this can be simplified to the form (1).