Theincircle of a triangle is the unique circle that has the three sides of the triangle as tangents. It is the largest circle lying entirely within a triangle.

Its centre, theincentre of the triangle, is at the intersection of the bisectors of the three angles of the triangle. This can be explained as follows:

• The bisector of${\displaystyle \mathrm{\angle }ABC}$ is the set of points equidistant from the line${\displaystyle \overline{AB}}$ and${\displaystyle \overline{BC}}$.
• The bisector of${\displaystyle \mathrm{\angle }ACB}$ is the set of points equidistant from the line${\displaystyle \overline{BC}}$ and${\displaystyle \overline{AC}}$.
• The point where those two lines intersect is both equidistant from${\displaystyle \overline{AB}}$ and${\displaystyle \overline{BC}}$ and equidistant form${\displaystyle \overline{BC}}$ and${\displaystyle \overline{AC}}$. So the intersection point's distance from${\displaystyle \overline{AB}}$ is the same as its distance to${\displaystyle \overline{BC}}$ is the same as its distance to${\displaystyle \overline{AC}}$. So it is equidistant from${\displaystyle \overline{AB}}$ and${\displaystyle \overline{AC}}$ and hence lies on the bisector of${\displaystyle \mathrm{\angle }BAC}$.
• Where the three bisectors cross is equidistant from${\displaystyle \overline{AB}}$,${\displaystyle \overline{BC}}$ and${\displaystyle \overline{AC}}$. Hence a circle with that point as centre can be tangent to all three lines.

Its radius, theinradius (usually denoted by r) is given by r = K/s, where K is the area of the triangle and s is the semiperimeter (a+b+c)/2 (a, b and c being the sides). To prove this, note that the lines joining the angles to the incentre divide the triangle into three smaller triangles, with bases a, b and c respectively and each with height r. The total area of these three triangles, hence the area K of the original triangle, is ar/2 + br/2 + cr/2. Rearranging, the result follows.

Applying Heron's theorem,${\displaystyle r=\sqrt{(s-a)(s-b)(s-c)/s}}$

If thecircumradius of the triangle is R, K${\displaystyle R=\frac{abc}{4}}$. Combining this with the formula for r,${\displaystyle Rr=\frac{abc}{4s}}$.

The distance of the incentre from A is 4Rsin(^B⁄₂)sin(^C⁄₂), and similarly for the other vertices.

The square of the distance between the circumcentre and incentre is R(R-2r). It follows that R > 2r unless the two centres coincide (which only happens for an equilateral triangle).

Let I be the incentre. Consider the triangle BIC. Let D be the point where the incircle touches BC; the angles IDB, IDC are right angles.

Angle IBD =^B⁄₂ and angle ICD =^C⁄₂.

BD =r cot(^B⁄₂); CD =r cot(^C⁄₂); BD+CD = BC =a.

${\displaystyle r\left(\cot \left(\frac{B}{2}\right)+\cot \left(\frac{C}{2}\right)\right)=a}$

${\displaystyle r=\frac{a}{\cot (\frac{B}{2})+\cot (\frac{C}{2})}=a\frac{\sin (\frac{B}{2})\sin (\frac{C}{2})}{\cos (\frac{A}{2})}}$

By symmetry, there are two other formulae involvingb andc respectively.

Substitutinga = 2Rsin(A), it follows that

${\displaystyle r=4R\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\sin \left(\frac{C}{2}\right)}$.

^r⁄_R thus equals¹⁄₂ for an equilateral triangle, and it can be shown that it is less than this for any other triangle.

Consider a straight line and a point X not on that line. Choose points A, B, C, D, E, F ... such that the triangles XAB, XBC, XCD, XDE, XEF, ... have equal inradii. Then the triangles XAC, XBD, XCE, XDF, ... will have inradii equal to each other (though larger than the inradius of XAB). Similarly, the triangles XAD, XBE, XCF, will have inradii equal to each other and so on.

There are three points where the angle bisectors intersect the opposite sides. These three points define a circle that will, in general, cut each side twice, defining three chords of the circle. (In an isosceles triangle, the base is a tangent to the circle; in an equilateral triangle, all three sides are tangents.) The length of the longest chord equals the sum of the lengths of the other two chords.

• Book:Trigonometry