Question

in a given figure ABC is a right angle triangle at B. D, E are points on BC trisect it. prove that 8AEsquare = 3ACsquare + 5AD square

Answer 1

Given right DABC, right angled at B. D and E are points of trisection of the side BCLet BD = DE = EC = k Hence we get BE = 2k and BC = 3k In ΔABD, by Pythagoras theorem, we get AD2 = AB2 + BD2 AD2 = AB2 + k2 Similarly, in ΔABE we get AE2 = AB2 + BE2 Hence AE2 = AB2 + (2k)2 = AB2+ 4k2 and AC2 = AB2 + BC2  = AB + (3k)2AC2= AB2 + 9k2 Consider, 3AC2 + 5AD2 = 3(AB2 + 9k2) + 5(AB2 + 4k2) = 8AB2 + 32k2= 8(AB2 + 4k2) ∴ 3AC2 + 5AD2 = 8AE2