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

$w_a={\displaystyle \frac{\sqrt{bc[{(b+c)}^2-a^2]}}{b+c}},$(1)

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

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

$$\frac{b}{b+c}\cdot a=\frac{ab}{b+c},\frac{c}{b+c}\cdot a=\frac{ca}{b+c}.$$

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

$\{\begin{array}{cc}2w_{a}b\cos \frac{\alpha }{2}=w_a^2+b^2-{\left(\frac{ab}{b+c}\right)}^2\hfill & \\ 2w_{a}c\cos \frac{\alpha }{2}=w_a^2+c^2-{\left(\frac{ca}{b+c}\right)}^2\hfill & \end{array}$(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).