Let's look at an application of the Sine rule. Suppose we bisect one angle of a triangle and see where the bisector meets the opposite edge, cutting our triangle into two. Let the halves of the bisected angle each be α; let the edges out of it have lengths u and v and the parts the bisector cuts the opposite edge into have lengths h on u's side and k on v's side. The two angles that the bisector makes with the edge opposite 2.α add up to a half turn, so have the same sine: call this s. The sine rule then tells us, from the two parts of the triangle:

- h/sin(α) = u/s and
- k/sin(α) = v/s

from which we obtain h/u = k/v and h/k = u/v. Thus the angle bisector cuts the opposite edge into pieces whose lengths are proportional to the lengths of the sides that connect their ends to the bisected angle.

When working with vector representations of the positions of the corners, this lets us obtain the point where the angle bisector meets the opposite edge as a weighted average of the ends of that opposite edge, each end's weight being the length of the opposite side of the triangle.

This, in turn, leads to a simple formula for the co-ordinates of the incentre of the triangle, in terms of the co-ordinates of its vertices and lengths of its edges. Every point on the bisector is a weighted average of the meeting point just obtained and the vertex whose angle we bisected; this can be re-expressed as a weighted average of the triangle's three vertices, in which the weights of the other two vertices are proportional to the lengths of the sides opposite them, with the weight of the bisected vertex varying along the line. This being true for all three angle bisector lines, with the incentre at their intersection, the incentre's co-ordinates are simply given by a weighted average of the three vertices, with the weight of each proportional to the length of its opposite side.

Next, for a given line segment BC and point D on it, consider the set of points A for which BAC's bisector cuts BC at D. We can chose co-ordinates in our plane such that D is the origin, B is [-b, 0] and C is [c, 0], with b and c being the lengths of BD and DC. Then the locus of A satisfies length(AB)/b = length(AC)/c; writing A as [x, y] this implies

- c.c.((x +b).(x +b) +y.y) = b.b.((x −c).(x −c) +y.y), implying
- 0 = (c.c −b.b)(x.x +y.y) +2.b.b.c.x +2.b.c.c.x, or
- 0 = (c +b).(c −b)(x.x +y.y) +2.b.(b +c).c.x

in which we can divide out the factor of b +c (since b and c are lengths, hence positive, hence so is their sum). If c = b, this just reduces to x = 0, the perpendicular bisector of BC; otherwise, introduce a = b.c/(b −c) = 1/(1/c −1/b), to obtain (dividing the previous by c.c −b.b)

- 0 = x.x +y.y −2.a.x, whence (completing the square)
- a.a = x.x +y.y −2.a.x +a.a = (x −a).(x −a) +y.y

and our locus is a circle, centred at [a, 0] with radius ±a. As 1/a +1/b = 1/c, with both b and c positive, 1/a is smaller than the larger of 1/b and 1/c, whence a is bigger than (albeit possibly with the opposite sign to) the smaller of b and c, i.e. either 0 < a > c < b or c > b < −a > 0. As a is negative for c > b and positive for b > c, the centre [a, 0] lies on the same side of D as the closer of B and C. Thus the circle's centre lies further from D than this end of BC does.

From 1/a +1/b = 1/c we can infer b +a = a.b/c and, via 1/b = 1/c −1/a, a.c/b = a −c multiplying these, we get (a +b).(a −c) = a.a, in which a +b and a −c are the (possibly both negated) distances of B and C from the circle's centre. Thus B and C are mutual images under central inversion in this circle.

So our locus is a circle, centred on a point on the extension of BC through whichever of its ends is closer to D. Its centre's coordinates can be expressed as a weighted average of those of B and C, giving weight c.c to B and −b.b to C. (Negating the signs of both weights, i.e. toggling which one is negated, won't change the average. The fact that their signs differ is what puts the weighted average outside the segment BC.)

Written by Eddy.