Angle Bisectors

Given the sides of a triangle, what is the length of the angle bisector? Consider the following image of triangle ABC, where the sides opposite points A, B, C have lengths a, b, c. We want to find length d.

First, we can apply the Angle Bisector Theorem to see that \frac{a}{x} = \frac{b}{y}, so that x = \frac{ay}{b} and y = \frac{bx}{a}. Plugging these into the equation c = x + y yields c = x + \frac{bx}{a} = \frac{a + b}{a} x and c = \frac{ay}{b} + y = \frac{a + b}{b} y, so that x = \frac{ca}{a+b} and y = \frac{cb}{a+b}.

We can now apply Stewart’s Theorem to see that a^2 y + b^2 x = c(d^2 + xy). Plugging in our expressions for x and y, we see that
ab = \frac{ab (a + b)}{a + b} = \frac{a^2 b}{a + b} + \frac{b^2 a}{a + b} = \frac{a^2 y + b^2 x}{c} = d^2 + xy = d^2 + \frac{c^2 ab}{(a + b)^2}.

Therefore, the length of the angle bisector at vertex C is
\displaystyle \sqrt{ab - \frac{c^2 ab}{(a + b)^2}} = \sqrt{ab \left( 1 - \frac{c^2}{(a + b)^2}\right)}.
The lengths of the other two angle bisectors can be found analogously and are given by a simple permutation of coordinates.

Here’s another way to compute the length of the angle bisector that might be a bit simpler: Let \angle C be 2\alpha, so that the angle bisector separates two angles with measure \alpha. The area of triangle ABC is the sum of the areas of the two smaller triangles, which can be expressed as \frac12 ad \sin \alpha + \frac12 bd \sin \alpha = \frac12 ab \sin 2\alpha = ab \sin \alpha \cos \alpha. Simplifying and rearranging, we see that (a + b) d = 2 ab \cos \alpha, so the length of the angle bisector is therefore
\displaystyle d = \frac{2ab}{a + b} \cos \alpha.
This is simpler than what we had before, though it also involves an angle.

We can use this expression for the length of the angle bisector to show that the sum of lengths of the angle bisectors is less than the perimeter. The length of the angle bisector to angle C is d = \frac{2ab}{a + b} \cos \alpha < \frac{2ab}{a + b}, and the other angle bisectors are similar. Therefore, it is sufficient to show that
\displaystyle \text{sum of angle bisectors} < \frac{2ab}{a + b} + \frac{2ac}{a + c} + \frac{2bc}{b + c} \le a + b + c = \text{perimeter}.

This inequality actually follows as a simple application of the arithmetic mean – harmonic mean (AM-HM) inequality, which states that
\displaystyle \frac{2ab}{a+b} = \frac{2}{\frac1a + \frac1b} \le \frac{a  +b}{2}.
Applying this three times yields
\displaystyle \frac{2ab}{a + b} + \frac{2ac}{a + c} + \frac{2bc}{b + c} \le \frac{a + b}{2} + \frac{a + c}{2} + \frac{b+c}{2} = a + b + c,
which is what we wanted to show. Therefore, we've proven that the sum of lengths of angle bisectors in a triangle is less than the perimeter.

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One response to “Angle Bisectors

  1. Américo Tavares April 18, 2010 at 3:26 pm

    Your formula

    d=\sqrt{ab-\dfrac{c^{2}ab}{\left( a+b\right) ^{2}}}

    when applied to the following figure (based on a sample problem from the The New York Times’ article U.S. Falls Short in Measure of Future Math Teachers)

    \setlength{\unitlength}{1cm}\begin{picture}(6,5)\thicklines\put(1,1){\line(1,2){1}}\put(1,1){\line(3,2){3}}\put(1,1){\line(1,0){4}}\put(5,1){\line(1,2){1}}\put(5,1){\line(-1,2){1}}\put(2,3){\line(1,0){4}}\put(0.6,0.8){\textit{A}}\put(5.05,0.8){\textit{B}}\put(1.6,3.1){\textit{D}}\put(3.85,3.2){\textit{M}}\put(6.05,3.1){\textit{C}}\thinlines\put(3,1){\line(1,2){1}}\put(3,0.7){\textit{N}}\put(3,1){\line(3,2){3}}\put(3,1){\line(-1,2){1}}\end{picture}

    where

    MN\ //\ BC\ //\ AD

    B\widehat{C}M=M\widehat{B}C=D\widehat{A}N=60^\circ

    MB=MC

    AN=NB=BC=CM=MD=DA=1

    DN=MB=1,

    confirms that AM=\sqrt{3}:

    AM/2=\sqrt{AD\cdot AN-\dfrac{DN^{2}\cdot AD\cdot AN}{\left( AD+AN\right)^{2}}}

    AM=2\sqrt{1\cdot 1-\dfrac{1^{2}\cdot 1\cdot 1}{\left( 1+1\right) ^{2}}}=2\sqrt{\dfrac{3}{4}}=\sqrt{3}

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