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Let $ABC$ be a triangle and $I$ its incenter. It is given that $BC=a$, $CA=b$, $AB=c$, $\angle A=\alpha$, $\angle B=\beta$ and $\angle C=\gamma$. Find the distance $AI$ in terms of these given values.

I've tried everything. I managed to find $AD=\frac{2bc\cos (\alpha/2)}{b+c}$, where $D$ is the point of intersection of line $AI$ with side $BC$. However this doesn't seem to help unless I can also find $ID$, because then $AI$ is given by $AD-ID$.

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  • $\begingroup$ Its simple, consider the triangle $AIE$, where $E$ is a point on $AB$ such that $IE\perp AB$. You have a triangle with all angles known and a side as well, apply some trig and you are done. Should I write that as an elaborated answer? $\endgroup$ – Sawarnik Feb 2 '15 at 19:23
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let us look at the triangle $AIB.$ the $\angle IAB = \alpha/2, \angle IBA = \beta/2, AB = c.$ applying the rule of sines you get $$\dfrac{AI}{\sin (\beta/2)} = \dfrac{c}{\sin(\alpha+\beta)/2)} = \dfrac{c}{\cos (\gamma/2)}$$

that is $$AI = \dfrac{c\sin(\beta/2)}{\cos(\gamma/2)}. $$

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  • $\begingroup$ Although the question didn't ask for it, it's worth noting that, using the Law of Sines on $\triangle ABC$ with $d$ the circumdiameter, you can write $c = d\;\sin\gamma = 2d\;\sin(\gamma/2)\;\cos(\gamma/2)$, so that $$AI = 2 d\;\sin(\beta/2)\;\sin(\gamma/2)$$ This representation has better "balance". $\endgroup$ – Blue Feb 2 '15 at 20:00
  • $\begingroup$ @Blue, i know that i could replace $\sin c$ by the diameter. i did not want to make it more complicated than what the op asked for. $\endgroup$ – abel Feb 2 '15 at 20:02
  • $\begingroup$ I know that you know. :) Still, I thought my comment might be helpful to someone out there. $\endgroup$ – Blue Feb 2 '15 at 20:06

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