A triangle, showing the "incircle" and the partitioning of the sides. The angle bisectors meet at the incenter, which is the center of the incircle.
By the above reasoning, all six parts are as shown.
In trigonometry, the law of cotangents is a relationship among the lengths of the sides of a triangle and the cotangents of the halves of the three angles.
Just as three quantities whose equality is expressed by the law of sines are equal to the diameter of the circumscribed circle of the triangle (or to its reciprocal, depending on how the law is expressed), so also the law of cotangents relates the radius of the inscribed circle of a triangle (the inradius) to its sides and angles.
Using the usual notations for a triangle (see the figure at the upper right), where a, b, c are the lengths of the three sides, A, B, C are the vertices opposite those three respective sides, α, β, γ are the corresponding angles at those vertices, s is the semi-perimeter, that is, s = a + b + c/2, and r is the radius of the inscribed circle, the law of cotangents states that
In the upper figure, the points of tangency of the incircle with the sides of the triangle break the perimeter into 6 segments, in 3 pairs. In each pair the segments are of equal length. For example, the 2 segments adjacent to vertex A are equal. If we pick one segment from each pair, their sum will be the semiperimeter s. An example of this is the segments shown in color in the figure. The two segments making up the red line add up to a, so the blue segment must be of length s − a. Obviously, the other five segments must also have lengths s − a, s − b, or s − c, as shown in the lower figure.
By inspection of the figure, using the definition of the cotangent function, we have
and similarly for the other two angles, proving the first assertion.
A number of other results can be derived from the law of cotangents.
Heron's formula. Note that the area of triangle ABC is also divided into 6 smaller triangles, also in 3 pairs, with the triangles in each pair having the same area. For example, the two triangles near vertex A, being right triangles of width s − a and height r, each have an area of 1/2r(s − a). So those two triangles together have an area of r(s − a), and the area S of the whole triangle is therefore
This gives the result S = √s(s − a)(s − b)(s − c) as required.