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lengths of angle bisectors

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In any triangle, the \PMlinkescapetext{lengths} $w_a$, $w_b$, $w_c$ of the angle bisectors opposing the sides $a$, $b$, $c$, respectively, are
w_a = \frac{\sqrt{bc\,[(b\!+\!c)^2\!-\!a^2]\,}}{b\!+\!c},
w_b = \frac{\sqrt{ca\,[(c\!+\!a)^2\!-\!b^2]\,}}{c\!+\!a},
w_c = \frac{\sqrt{ab\,[(a\!+\!b)^2\!-\!c^2]\,}}{a\!+\!b}.

{\em 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 \PMlinkescapetext{separated} by $w_a$:
2w_ab\cos\frac{\alpha}{2} \;=\; w_a^2\!+\!b^2\!-\!\left(\frac{ab}{b+c}\right)^2\\
2w_ac\cos\frac{\alpha}{2} \;=\; w_a^2\!+\!c^2\!-\!\left(\frac{ca}{b+c}\right)^2
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).