An **excircle** of a triangle is a circle that has as tangents one side of the triangle and the other two sides extended. There are three such circles, one corresponding to each side of the triangle.

The centre of each such circle, an **excentre** of the triangle, is at the intersection of the bisector of the angle opposite the side tangent to the circle and the external bisectors of the other two angles of the triangle. The proof is similar to that for the location of the incentre.

Let the sides of a triangle be a, b and c and the angles opposite be A, B and C respectively. Denote the area of the triangle by Δ. Write s = ^{1}⁄_{2}(a+b+c).

If the radius of the excircle touching side a is r_{a}, then , with similar expressions for the other two excircles. If r is the inradius, we have

If the vertices of a triangle are ABC and the excircle touching side BC does so at point D, then AB+BD = AC+CD. (This is easily proved using the theorem that the two tangents from a point to a circle are equal in length.) Both these expressions are s, the semiperimeter.

The square of the distance of the excentre corresponding to side a from the circumcentre is R(R+2r_{a}), with similar expressions for the other two excentres. Also,

The area of the triangle is

Three other expressions for r_{a} are

with similar expressions for r_{b} and r_{c}.

It can readily be proved from the second of these expressions that if r_{a} = r + r_{b} + r_{c} then A is a right angle.

The distance of I_{a} from the three vertices A, B, C are

respectively, with similar expressions for the other excentres.