16th Eötvös 1910 Problem 3 ABC is a triangle with angle C = 120 o . Find the length of the angle bisector of angle C in terms of BC and CA.
Let the bisector be CX with length k. Take AC, BC to have lengths b, a as usual. Then area ACX = ½bk sin 60 o , area BCX = ½ak sin 60 o , and area ABC = ½ab sin 60 o . But area ABC = area ACX + area BCX, so k = ab/(a+b). 17th Eötvös 1910 © John Scholes jscholes@kalva.demon.co.uk 29 Oct 2003 Last corrected/updated 29 Oct 03