How do I find the length of an angle bisector if three sides are given?

Answer 1

Triangle ABC, A on top, B on the left, C on the right

Length of AB = c, length of AC=b, and length of BC=a, as usual

Three unknowns: angle A = 2* theta, as bisected by ray X. M is the point where ray X intersects BC.

We seek AM=x.

Law of cosines says: a^2 = b^2 + c^2 - 2bc* cos(A)

Solving for the angle gives: A = inv_cos( (a^2 - b^2 - c^2)/(-2bc))

Next, theta = (1/2)A, and we got the angle that is biesected.

Two remaining unknowns, y and x

Let y = MB; then a-y = MC;

Law of cosines, twice more, gives:

y^2 = c^2 + x^2 - 2cx * cos(theta)

(a-y)^2 = b^2 + x^2 - 2bx * cos(theta)

This system can be solved for y and x