Incenter Coordinate Formula

A vector proof that the incenter is the side-weighted average of the vertices, I = (aA + bB + cC)/(a + b + c), from two applications of the angle-bisector theorem and the section formula.

Keywords: incenter, angle-bisector theorem, section formula, barycentric coordinates, weighted average

Prerequisites: angle-bisect-prop-proof, incenter-proof · Difficulty: advanced

The incenter has a clean vector address: the average of the three vertices, each weighted by the length of the opposite side,

I = \frac{a A + b B + c C}{a + b + c},

with the usual convention $a = B C$ , $b = C A$ , $c = A B$ .

Strategy: read the vertices as position vectors and apply the section formula twice. The bisector from $A$ splits $B C$ at $D$ with $B D : D C = c : b$ , then the bisector from $B$ splits $A D$ at $I$ with $A I : I D = (b + c) : a$ .

Step 1: Triangle and Incircle

The incenter $I$ is where the three angle bisectors meet, and the center of the incircle. We will pin down its position as a weighted average of the vertices.

Step 2: Vectors and the Section Formula

Read each vertex as a position vector from any fixed origin. The one tool we need is the section formula: the point dividing segment $P Q$ in the ratio $m : n$ , measured from $P$ , is

\frac{n P + m Q}{m + n},

so the nearer endpoint carries the larger weight. Side lengths follow the convention $a = B C$ , $b = C A$ , $c = A B$ .

Step 3: $I = \frac{a A + b B + c C}{a + b + c}$

Claim. The incenter is the average of the vertices weighted by the opposite side lengths,

I = \frac{a A + b B + c C}{a + b + c} .

Two strokes of the angle-bisector theorem prove it: one to split $B C$ , one to split the bisector from $A$ .

Step 4: Split $B C$ with the Bisector from $A$

The bisector from $A$ meets $B C$ at $D$ . By the angle-bisector theorem $\frac{B D}{D C} = \frac{A B}{A C} = \frac{c}{b}$ , so the section formula gives $D = \frac{b B + c C}{b + c}$ .

Step 5: Split $A D$ with the Bisector from $B$

Inside triangle $A B D$ , the bisector from $B$ is $B I$ and it meets $A D$ at the incenter $I$ . The angle-bisector theorem again gives

\frac{A I}{I D} = \frac{B A}{B D} = \frac{b + c}{a},

since $B D = \frac{a c}{b + c}$ (from $B D : D C = c : b$ with $B D + D C = a$ ). By the section formula,

I = \frac{a A + ( b + c ) D}{a + b + c} = \frac{a A + b B + c C}{a + b + c} .

Step 6: The Incenter Coordinates

So the incenter is the side-weighted average of the vertices,

I = \frac{a A + b B + c C}{a + b + c} .

In the special case of the circumcenter $O$ at the origin, expanding $∣ I ∣^{2}$ collapses to $∣ O I ∣^{2} = R^{2} - 2 R r$ , Euler's formula.

The incenter is the side-weighted average of the vertices,

I = \frac{a A + b B + c C}{a + b + c} .

This is the general coordinate (barycentric) formula for $I$ . In the special case of the circumcenter $O$ at the origin, expanding it collapses to Euler's formula $∣ O I ∣^{2} = R^{2} - 2 R r$ , derived in the introduction notes and proved synthetically in the companion scene. ∎

Notes

A special origin: Euler's distance formula

The address holds from any origin, but one choice pays off. Take the special case of the circumcenter $O$ as origin, so all three vertices have length $R$ . Each chord length then fixes a dot product: from $c^{2} = ∣ A - B ∣^{2}$ we get $A \cdot B = R^{2} - \frac{c ^{2}}{2}$ , and likewise for $B \cdot C$ and $C \cdot A$ . Expanding the square,

∣ a A + b B + c C ∣^{2} = R^{2} (a + b + c)^{2} - ab c (a + b + c),

so dividing by $(a + b + c)^{2}$ leaves $∣ O I ∣^{2} = R^{2} - \frac{ab c}{a + b + c}$ . Finally $ab c = 4 R [△]$ and $[△] = r s$ shrink the last term to $2 R r$ , giving Euler's formula $∣ O I ∣^{2} = R^{2} - 2 R r$ .