Editing Incircle and excircles (section)

===Incircle and its radius properties===

====Distances between vertex and nearest touchpoints====
The distances from a vertex to the two nearest touchpoints are equal; for example:<ref name=":0">''Mathematical Gazette'', July 2003, 323-324.</ref>
:<math display=block>d\left(A, T_B\right) = d\left(A, T_C\right) = \tfrac12(b + c - a) = s - a.</math>

====Other properties====
If the [[altitude (triangle)|altitudes]] from sides of lengths <math>a</math>, <math>b</math>, and <math>c</math> are <math>h_a</math>, <math>h_b</math>, and <math>h_c</math>, then the inradius <math>r</math> is one-third of the [[harmonic mean]] of these altitudes; that is,<ref>{{harvtxt|Kay|1969|p=203}}</ref>
:<math display=block> r = \frac{1}{\dfrac{1}{h_a} + \dfrac{1}{h_b} + \dfrac{1}{h_c}}.</math>

The product of the incircle radius <math>r</math> and the [[circumcircle]] radius <math>R</math> of a triangle with sides <math>a</math>, <math>b</math>, and <math>c</math> is{{sfn|Johnson|1929|p=189, #298(d)}}
:<math display=block>rR = \frac{abc}{2(a + b + c)}.</math>

Some relations among the sides, incircle radius, and circumcircle radius are:<ref name=Bell/>
:<math display=block>\begin{align}
     ab + bc + ca &=  s^2 +  (4R + r)r, \\
  a^2 + b^2 + c^2 &= 2s^2 - 2(4R + r)r.
\end{align}</math>

Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its incircle). There are either one, two, or three of these for any given triangle.<ref>Kodokostas, Dimitrios, "Triangle Equalizers", ''Mathematics Magazine'' 83, April 2010, pp. 141-146.</ref>

The incircle radius is no greater than one-ninth the sum of the altitudes.<ref>Posamentier, Alfred S., and Lehmann, Ingmar. ''[[The Secrets of Triangles]]'', Prometheus Books, 2012.</ref>{{rp|289}}

The squared distance from the incenter <math>I</math> to the [[circumcenter]] <math>O</math> is given by<ref name=Franzsen>{{cite journal
|last=Franzsen
|first=William N.
|journal=Forum Geometricorum
|mr=2877263
|pages=231–236
|title=The distance from the incenter to the Euler line
|volume=11
|year=2011
|url=http://forumgeom.fau.edu/FG2011volume11/FG201126.pdf
|access-date=2012-05-09
|url-status=dead
|archive-url=https://web.archive.org/web/20201205220605/http://forumgeom.fau.edu/FG2011volume11/FG201126.pdf
|archive-date=2020-12-05
}}.</ref>{{rp|232}}
:<math display=block>\overline{OI}^2 = R(R - 2r) = \frac{a\,b\,c\,}{a+b+c}\left [\frac{a\,b\,c\,}{(a+b-c)\,(a-b+c)\,(-a+b+c)}-1 \right ]</math>

and the distance from the incenter to the center <math>N</math> of the [[nine point circle]] is<ref name=Franzsen/>{{rp|232}}
:<math display=block>\overline{IN} = \tfrac12(R - 2r) < \tfrac12 R.</math>

The incenter lies in the [[medial triangle]] (whose vertices are the midpoints of the sides).<ref name=Franzsen/>{{rp|233, Lemma 1}}

====Relation to area of the triangle====
{{Redirect|Inradius|the three-dimensional equivalent|Inscribed sphere}}
The radius of the incircle is related to the [[area]] of the triangle.<ref>Coxeter, H.S.M. "Introduction to Geometry'' 2nd ed. Wiley, 1961.''</ref> The ratio of the area of the incircle to the area of the triangle is less than or equal to <math>\pi \big/ 3\sqrt3</math>,
with equality holding only for [[equilateral triangle]]s.<ref>Minda, D., and Phelps, S., "Triangles, ellipses, and cubic polynomials", ''[[American Mathematical Monthly]]'' 115, October 2008, 679-689: Theorem 4.1.</ref>

Suppose <math>\triangle ABC</math> has an incircle with radius <math>r</math> and center <math>I</math>. Let <math>a</math> be the length of <math>\overline{BC}</math>, <math>b</math> the length of <math>\overline{AC}</math>, and <math>c</math> the length of <math>\overline{AB}</math>.

Now, the incircle is tangent to <math>\overline{AB}</math> at some point <math>T_C</math>, and so <math>\angle AT_CI</math> is right. Thus, the radius <math>T_CI</math> is an [[altitude (triangle)|altitude]] of <math>\triangle IAB</math>.

Therefore, <math>\triangle IAB</math> has base length <math>c</math> and height <math>r</math>, and so has area <math>\tfrac12 cr</math>.

Similarly, <math>\triangle IAC</math> has area <math>\tfrac12 br</math> and <math>\triangle IBC</math> has area <math>\tfrac12 ar</math>.

Since these three triangles decompose <math>\triangle ABC</math>, we see that the area <math>\Delta \text{ of} \triangle ABC</math> is:
:<math display=block>\Delta = \tfrac12 (a + b + c)r = sr,</math>

{{spaces|4}} and {{spaces|4}}<math>r = \frac{\Delta}{s},</math>

where <math>\Delta</math> is the area of <math>\triangle ABC</math> and <math>s = \tfrac12(a + b + c)</math> is its [[semiperimeter]].

For an alternative formula, consider <math>\triangle IT_CA</math>. This is a right-angled triangle with one side equal to <math>r</math> and the other side equal to <math>r \cot \tfrac{A}{2}</math>. The same is true for <math>\triangle IB'A</math>. The large triangle is composed of six such triangles and the total area is:{{Citation needed|date=May 2020}}
:<math display=block>\Delta = r^2 \left(\cot\tfrac{A}{2} + \cot\tfrac{B}{2} + \cot\tfrac{C}{2}\right).</math>