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![[parent]](http://images.planetmath.org:8080/images/uparrow.png)
lengths of angle bisectors
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(Corollary)
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In any triangle, the lengths $w_a$ , $w_b$ , $w_c$ of the angle bisectors opposing the sides $a$ , $b$ , $c$ , respectively, are
![$\displaystyle w_a = \frac{\sqrt{bc\,[(b+c)^2-a^2]\,}}{b+c},$](http://images.planetmath.org:8080/cache/objects/11108/js/img1.png "$\displaystyle w_a = \frac{\sqrt{bc\,[(b+c)^2-a^2]\,}}{b+c},$") |
(1) |
![$\displaystyle w_b = \frac{\sqrt{ca\,[(c+a)^2-b^2]\,}}{c+a},$](http://images.planetmath.org:8080/cache/objects/11108/js/img2.png "$\displaystyle w_b = \frac{\sqrt{ca\,[(c+a)^2-b^2]\,}}{c+a},$") |
(2) |
![$\displaystyle w_c = \frac{\sqrt{ab\,[(a+b)^2-c^2]\,}}{a+b}.$](http://images.planetmath.org:8080/cache/objects/11108/js/img3.png "$\displaystyle w_c = \frac{\sqrt{ab\,[(a+b)^2-c^2]\,}}{a+b}.$") |
(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 $$\frac{b}{b+c}\cdot a = \frac{ab}{b+c}, \quad \frac{c}{b+c}\cdot a = \frac{ca}{b+c}.$$ If the angle opposite to $a$ is $\alpha$ , we apply the law of cosines to the
half-triangles separated by $w_a$ :
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(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).
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"lengths of angle bisectors" is owned by pahio.
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Cross-references: equations, law of cosines, opposite, angle, divides, bisector, angle bisector theorem, symmetry, proof, sides, angle bisectors, triangle
This is version 5 of lengths of angle bisectors, born on 2008-09-29, modified 2008-09-30.
Object id is 11108, canonical name is LengthsOfAngleBisectors.
Accessed 509 times total.
Classification:
| AMS MSC: | 51M05 (Geometry :: Real and complex geometry :: Euclidean geometries and generalizations) |
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Pending Errata and Addenda
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