Sorry, no pictures. Explanation here.
It is a fact of elementary geometry that the center of a triangle's incircle—known as its incenter—lies at the point of intersection of the triangle's angle bisectors. This leads to an elegant expression for the location of the incenter as a linear combination of the triangle's vertices, as stated in:
Proposition 1: Let $a$, $b$, $c$ be the lengths of the sides opposite to the vertices $A$, $B$, $C$ of the triangle $ABC$. Then the incenter, $O$, is expressed as a linear combination of the vertices as: \[ O = \frac{a}{p}A + \frac{b}{p}B + \frac{c}{p}C, \] where $p = a + b + c$ is the triangle's perimeter.
To make sense of this statement, you need to know something about the algebra of points. The following capsule summary is all that's needed:
The algebra of points: Consider the points $A$ and $B$ and a variable $t$ that takes values in the range 0 to 1. If $T$ is a point on the segment $AB$ such that $AT/AB = t$, we write $T = (1-t)A + tB$. Note that when $t=0$ we get $T=A$, and when $t=1$ we get $T=B$. As the value of $t$ ranges from 0 to 1, the point $T$ slides from $A$ to $B$.
The following elementary proposition is needed for the proof of Proposition 1. This one is a standard result and is likely to be found in Euclid's Elements but I haven't checked.
PS: After this web page was written, I found out that Proposition 2 appears in Euclid's Elements as Book VI, Proposition 3 with a simpler and more elegant proof! Nevertheless, I am retaining my original proof here as a not-so-elegant alternative.
Proposition 2: Let the bisector of the angle $C$ in the triangle $ABC$ meet the side $AB$ at point $P$. We have: \[ \frac{PA}{PB} = \frac{b}{a}, \] where $a$ and $b$ are as in Proposition 1.
Proof: From $P$ drop perpendicular $PM$ and $PN$ onto the sides $AC$ and $BC$; see the diagram below. The right triangles $CMP$ and $CNP$ are congruent because they share a common hypotenuse and their angles at $C$ are equal. Therefore, $PM = PN$.
Let $H$ be the foot of the altitude dropped from $C$. Let us note the area of the triangle $APC$ may be expressed in two different ways in terms of the altitudes $CH$ and $PM$: \[ \frac{1}{2} AP \cdot CH = \frac{1}{2} AC \cdot PM. \] Similarly, the area of the triangle $BPC$ may be expressed in two different way in terms of the altitudes $CH$ and $PN$: \[ \frac{1}{2} BP \cdot CH = \frac{1}{2} BC \cdot PN. \] Dividing these two equalities and recalling that $PM = PN$, we arrive at the desired assertion. QED
The diagram below shows the bisector $CC'$ of the angle $C$. It is known from elementary geometry that the incenter $O$ lies on the bisector. From Proposition 2 we have $C'A/C'B = b/a$. It follows that $AC'/AB = b/(a+b)$. In terms of the notation of the algebra of points introduced earlier in this page, this is expressed as: \[ C' = \Big(1 - \frac{b}{a+b} \Big) A + \frac{b}{a+b} B = \frac{a}{a+b} A + \frac{b}{a+b} B. \] Since $O = (1-t)C + tC'$ for some $t$, we arrive at: \[ O = (1-t)C + t\Big[ \frac{a}{a+b} A + \frac{b}{a+b} B \Big]. \] This expresses the triangle's incenter $O$ in terms of its vertices, sides lengths, and a yet unknown quantity $t$ that takes values in the range 0 to 1.
If we repeat the same calculation by replacing the vertex $C$ by vertex $A$, we arrive at: \[ O = (1-s)A + s\Big[ \frac{b}{b+c} B + \frac{c}{b+c} C \Big], \] where $s$ is yet another unknown that takes values in the range from 0 to 1.
Equating the two expressions for $O$ and collecting the terms, we arrive at: \[ \Big[\frac{ta}{a+b} + s - 1\Big] A + \Big[\frac{tb}{a+b} - \frac{sb}{b+c} \Big] B + \Big[1 - t - \frac{sc}{b+c} \Big] C = 0. \] In this equations, if the coefficient of $C$ is nonzero, then we may solve for $C$, obtaining $C = \alpha A + \beta B$ for some $\alpha$ and $\beta$. But this would mean that $C$ lies along the line $AB$ which would mean the the three points $A$, $B$ and $C$ are collinear, therefore the triangle $ABC$ is degenerate. We conclude that if the triangle is non-degenerate, then the coefficient of $C$ is zero. Similar arguments show that the coefficients of $A$ and $B$ are zero. Thus, the following system of three equations hold: \[ \frac{ta}{a+b} + s - 1 = 0, \quad \frac{tb}{a+b} - \frac{sb}{b+c} = 0, \quad 1 - t - \frac{sc}{b+c} = 0. \] Solving this system for the unknowns $t$ and $s$ we obtain: \[ t = \frac{a+b}{a+b+c}, \quad s = \frac{b+c}{a+b+c}. \] Substituting these in either of the expressions for $O$ we arrive at: \[ O = \frac{a}{a+b+C}A + \frac{b}{a+b+C}B + \frac{c}{a+b+C}C, \] which is equivalent to Proposition 1's assertion. QED
Propositions 1 extends to tetrahedra:
Proposition 3 Let $a$, $b$, $c$, $d$ be the areas of the faces opposite to the vertices $A$, $B$, $C$, $D$ of the tetrahedron $ABCD$. The the tetrahedron's incenter $O$ is given by: \[ O = \frac{a}{\mathcal{A}} A + \frac{b}{\mathcal{A}} B + \frac{c}{\mathcal{A}} C + \frac{d}{\mathcal{A}} D, \] where $\mathcal{A} = a + b + c + d$ is the tetrahedron's surface area.
This is proved with the aid of the following extension of Proposition 2:
Proposition 4 Let $a$, $b$, $c$, $d$ be the areas of the faces opposite to the vertices $A$, $B$, $C$, $D$ of the tetrahedron $ABCD$. Let the bisector plane of the (internal) dihedral angle of edge $CD$ intersect edge $AB$ at point $P$. The we have: \[ \frac{PA}{PB} = \frac{b}{a}. \]
I leave the proofs to you, the diligent reader. You will find useful information in a discussion in the sci.math newsgroup from January 17, 2006.
This applet was created by Rouben Rostamian using David Joyce's Geometry Applet on June 26, 2010.
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