ABC is an equilateral triangle. D is on the side AB and E is on the side AC such that DE touches the incircle. Show that AD/DB + AE/EC = 1.
Solution
Put BD = x, CE = y, BC = a. Then since the two tangents from B to the incircle are of equal length, and similarly the two tangents from D and E, we have ED + BC = BD + CE, or ED = x + y - a. By the cosine law, ED2 = AE2 + AD2 - AE.AD. Substituting and simplifying, we get a = 3xy/(x + y). Hence AD/DB = (2y - x)/(x + y) and AE/EC = (2x - y)/(x + y) with sum 1.
© John Scholes
jscholes@kalva.demon.co.uk
31 July 2002