The three bisector lines of the inner angles of a triangle (the bisectrices)
intersect in a unique point, the incentre. Every point of the bisectrix
of the angle with vertex at *P* is equidistant from the sides *PQ*
and *PR*. Also every point of the bisectrix of *Q* is equidistant
from the sides *QR* and *QP*. Hence its intersection *I*
is simultaneously equidistant from the three sides, that is, *I* is
unique and it is the centre of the circle inscribed into the triangle.
In order to calculate the equation of a bisectrix, we take the sum of both
unitary vectors of adjacent sides:

The incentre *I* is the intersection
of the bisectrix passing through *P*, whose directing vector is *u*,
and that passing through *Q*, with directing vector *v*:

*I* = *P* + *k u*
= *Q* + *m v
k*, *m* real

Arranging terms we find *PQ*
as a linear combination of *u* and *v*:

*k u* - *m v* = *Q*
- *P* = *PQ*

The coefficient *k* is:

Since all outer products are equal because they are twice the triangle area, this expression is simplified:

The centre of the inscribed circle is:

By taking common denominator and simplifying, we arrive at:

For example, let us calculate the centre of the circle inscribed in the triangle with vertices:

*P* = ( 0, 0 )
*Q* = ( 0, 3 )
*R* = ( 4, 0 )

| *PQ | *= 3
| *QR | *= 5
| *RP | *= 4