the angle-bisector length theorem

When the bisector of an angle of a triangle is drawn down to the opposite side, it does not usually land at the midpoint — instead it lands at a point that splits the opposite side in proportion to the two sides forming the angle. The bisector divides the far side into pieces whose ratio matches the ratio of the two adjacent sides. A longer neighbouring side claims a longer piece.

In triangle ABC, let the bisector of angle A meet side BC at point D. Then BD/DC = AB/AC. So if the two sides meeting at A are in the ratio 3 : 5, the bisector cuts the opposite side BC into pieces in that same 3 : 5 ratio. To find an unknown piece, set up the proportion and solve: if AB = 6, AC = 9, and BC = 10, then BD/DC = 6/9 = 2/3, and since BD + DC = 10, you get BD = 4 and DC = 6. (The proof draws a line through C parallel to the bisector and applies the side-splitter theorem.)

This is the similarity-flavoured partner of the midpoint and median ideas: a median always hits the midpoint, but an angle bisector hits a proportional point that only coincides with the midpoint when the triangle is isosceles at A. There is also an external version for the bisector of the exterior angle, which divides the opposite side externally in the same ratio. The theorem is a favourite in olympiad geometry and in any problem that mixes an angle bisector with side lengths.