# 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).

Title lengths of angle bisectors LengthsOfAngleBisectors 2013-03-22 18:26:50 2013-03-22 18:26:50 pahio (2872) pahio (2872) 9 pahio (2872) Corollary msc 51M05 Incenter AngleBisectorAsLocus LengthsOfMedians