How do I prove that a triangle with sides a, b, c, has an angle bisector (bisecting angle A) is of length:

$$\frac{2 \sqrt{bcs(s-a)}}{b+c}$$

I have tried using the sine and cosine rule but have largely failed. A few times I have found a way but they are way too messy to work with.

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  • $\begingroup$ Hint: Stewart's theorem: en.wikipedia.org/wiki/Stewart%27s_theorem $\endgroup$ – Michael Hoppe Oct 25 '13 at 12:25
  • $\begingroup$ Michael, I would like a bit more hint. Angle bisector theorem I know, just to tell you. The main problem I am having is that I cannot eliminate either of m or n. $\endgroup$ – Sawarnik Oct 25 '13 at 13:39
  • $\begingroup$ Angle bisector theorem: en.wikipedia.org/wiki/Angle_bisector_theorem $\endgroup$ – Michael Hoppe Oct 25 '13 at 13:50
  • $\begingroup$ That is what I said, I am aware of that theorem. I just want to ask you how do I eliminate one of m or n. As far my case goes, that theorem helps me to eliminate one of them. Looking at the formula I realize that it may be some way connected to the area, law of sines, or sin(A/2), isn't it? Sorry for the late reply, I was answering a simpler question on SE. $\endgroup$ – Sawarnik Oct 25 '13 at 14:15
  • $\begingroup$ Perhaps, if you would post a detailed answer, I would be happy to look into it. $\endgroup$ – Sawarnik Oct 25 '13 at 14:18

Refering tp http://en.wikipedia.org/wiki/Stewart%27s_theorem we have $n=\frac{b}{b+c}a$ and $m=\frac{c}{b+c}$, hence $$b^2\frac{ac}{b+c}+c^2\frac{ab}{b+c}=a\left(d^2+\frac{ab}{b+c}\frac{ac}{b+c}\right),$$ where $d$ denotes the length of the bisector. From here we can conclude $$d^2=bc\Bigl( 1-\frac{a^2}{(b+c)^2}\Bigr)$$ using $\frac{b^2c}{b+c}+\frac{bc^2}{b+c}=bc$. Nice formula. Can you get from here to your version?

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  • $\begingroup$ I want to know how did you get n = b/(b+c) and m=c/(b+c), the rest is crystal clear. $\endgroup$ – Sawarnik Oct 25 '13 at 14:59
  • $\begingroup$ Thank you very much. The angle bisector theorem states that the bisector divides the opposite side in the ratio of these sides: if, e.g., $b=2c$ then $n=2m$, hence $n=2/3a$ and $m=1/3a$. Or, if $b=3$ and $c=7$, then $n=\frac{3}{10}a$ and $m=\frac{7}{10}a$. Is that (crystal) clear enough? BTW: I wouldn't let go $s$ into the formula; it kind of disturbs the purity. $\endgroup$ – Michael Hoppe Oct 25 '13 at 18:33

A method where no trigonometry is used.

Consider triangle $ABC$. Let $AD$, the angle bisector, intersect the circumcircle at $L$. Join $LC$. Consider triangle $ABD$ and triangle $ALC$.

Triangle $ABD$ is similar to triangle $ALC$ (by A.A similarity theorem). Therefore, $$\frac{AD}{AC} = \frac{AB}{AL}$$

i.e, $$AD\cdot AL = AC\cdot AB$$ $$= AD(AD+DL) = AC\cdot AB$$ $$=AD\cdot AD + AD\cdot DL = AC \cdot AB \text{ ... (1)}$$

By power of point of point result: $$AD\cdot DL = BD\cdot DC$$ $$BD = BC\cdot \frac{AB}{AB+AC}$$ $$DC = BC\cdot\frac{AC}{AB+AC}$$

In $(1)$ , $$AD\cdot AD = AC\cdot AB-BC^2\cdot AB\cdot \frac{AC}{(AB+AC)^2}$$ $$AD^2 = AC\cdot AB\Bigl(1-\frac{BC^2}{(AB+AC)^2}\Bigr)$$ If $AB = c , BC = a , AC = b$: $$AD^2 = bc\Bigl( 1-\frac{a^2}{(b+c)^2}\Bigr)$$ Hence proved.

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  • $\begingroup$ I made some edits to make it more readable :) Can you check if its alright, and its a nice method! $\endgroup$ – Sawarnik Apr 20 '14 at 8:12

Assuming the usage of Trigonometry is allowed,

Let $AD$ be the bisector of $\angle BAC$

$\triangle ABC=\frac12bc\sin A $

$\triangle ABD=\frac12\cdot c\cdot AD\sin\frac A2$ and $\triangle ADC=\frac12\cdot b\cdot AD\sin\frac A2$

$\triangle ABC=\triangle ABD+\triangle ADC$

$\sin A=2\sin\frac A2\cos\frac A2$

As $\displaystyle 0<A<\pi,0<\frac A2<\frac\pi2\implies \cos\frac A2>0\implies \cos \frac A2=+\sqrt{\frac{1+\cos A}2}$ as $\cos A=2\cos^2\frac A2-1$

Use $\displaystyle\cos A=\frac{b^2+c^2-a^2}{2bc}$ and $2s=a+b+c$

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  • $\begingroup$ I get your answer, very simple! Looking into Michael's answer. $\endgroup$ – Sawarnik Oct 25 '13 at 14:55
  • $\begingroup$ @Sawarnik, is trigonometry allowed? $\endgroup$ – lab bhattacharjee Oct 25 '13 at 15:27
  • $\begingroup$ Yes, of course. $\endgroup$ – Sawarnik Oct 25 '13 at 15:49
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    $\begingroup$ @Sawarnik, cool :) $\endgroup$ – lab bhattacharjee Oct 25 '13 at 15:54
  • $\begingroup$ At the 4th line, there shouldn't be $\triangle{BDA}$ instead of $\triangle{ABC}$? $\endgroup$ – mnulb Oct 10 '17 at 13:03

The length of the angle bisector of a triangle is (2bc*cosa/2)/b+c Just prove this using cosine rule and you will be done because $\cos a= \sqrt{s(s-a)bc}$

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