Incenter

Where a, b, c are the lengths of the sides opposite to vertices A, B, C respectively.

The incenter $I$ is equidistant from all three sides — it is the center of the inscribed circle. It is the intersection of the internal angle bisectors.

Derivation of the formula

Let $a = BC$, $b = CA$, $c = AB$ (side opposite to vertex $A$, $B$, $C$ respectively).

The angle bisector from $A$ divides $BC$ in the ratio $AB : AC = c : b$ (angle bisector theorem). So it meets $BC$ at:

$$D = \frac{b \cdot B + c \cdot C}{b + c} = \left(\frac{bx_2+cx_3}{b+c},\; \frac{by_2+cy_3}{b+c}\right)$$

The incenter divides $AD$ in the ratio $AB + AC : BC = (c+b) : a$ from $A$. Using the section formula:

$$I = \frac{a \cdot A + (b+c) \cdot D}{a + b + c}$$

Substituting $D$:

$$\boxed{I = \left(\frac{ax_1+bx_2+cx_3}{a+b+c},\; \frac{ay_1+by_2+cy_3}{a+b+c}\right)}$$

where $a = BC$, $b = CA$, $c = AB$.

Key facts

• The inradius $r = \Delta/s$ where $\Delta$ is the area and $s = (a+b+c)/2$ is the semi-perimeter.
• The incenter always lies inside the triangle.
• The excenters (centers of the three excircles) are given by similar weighted averages with one weight negated.