Question

in triangle abc right angled at B , points D and E trisect BC prove that 8AE²=3AC²+5AD²

Answer 1

Given right DABC, right angled at B.
D and E are points of trisection of the side BC
Let 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)2
AC2= AB2 + 9k2
Consider, 3AC2 + 5AD2 = 3(AB2 + 9k2) + 5(AB2 + 4k2)
= 8AB2 + 32k2
= 8(AB2 + 4k2)
∴ 3AC2 + 5AD2 = 8AE2

Answer 2

Hope it will be helpful to you