Editing Incircle and excircles (section)

====Derivation of the formula stated above====
Use the [[Law of sines]] in the triangle <math>\triangle IAB</math>.

We get <math>\frac{\overline{AI}}{\sin \frac{B}{2}} = \frac{c}{\sin \angle AIB}</math>.
We have that <math>\angle AIB = \pi - \frac{A}{2} - \frac{B}{2} = \frac{\pi}{2} + \frac{C}{2}</math>.

It follows that <math>\overline{AI} = c \ \frac{\sin \frac{B}{2}}{\cos \frac{C}{2}}</math>.

The equality with the second expression is obtained the same way.

The distances from the incenter to the vertices combined with the lengths of the triangle sides obey the equation<ref>
{{citation
|last1=Allaire |first1=Patricia R.
|last2=Zhou |first2=Junmin
|last3=Yao |first3=Haishen
|date=March 2012
|journal=[[Mathematical Gazette]]
|pages=161–165
|title=Proving a nineteenth century ellipse identity
|volume=96
|doi=10.1017/S0025557200004277
|s2cid=124176398
}}.</ref>
:<math display=block>\frac{\overline{IA} \cdot \overline{IA}}{\overline{CA} \cdot \overline{AB}} + \frac{\overline{IB} \cdot \overline{IB}}{\overline{AB} \cdot \overline{BC}} + \frac{\overline{IC} \cdot \overline{IC}}{\overline{BC} \cdot \overline{CA}} = 1.</math>

Additionally,<ref>{{citation |last=Altshiller-Court |first=Nathan |author-link=Nathan Altshiller Court |title=College Geometry |publisher=Dover Publications |year=1980}}. #84, p.&nbsp;121.</ref>
:<math display=block>\overline{IA} \cdot \overline{IB} \cdot \overline{IC} = 4Rr^2,</math>

where <math>R</math> and <math>r</math> are the triangle's [[circumradius]] and [[inradius]] respectively.