triangle ABC is right angled triangle at B.side BC is trisected at points D and E.Prove that 8AE2=3AC2+5AD2
Given: ABC is a triangle right angles at B and D and E are points at trisection of BC.
Let BD = DE = EC = x
Then BE = 2x and BC = 3x
In ΔABD
AD2 = AB2 + BD2 = AB2 + x2
In ΔABE
AE2 = AB2 + BE2 = AB2 + (2x)2
= AB2 + 4x2
In ΔABC
AC2 = AB2 + BC2 = AB + (3x)2 = AB2 + 9x2
Now,
3AC2 + 5AD2 = 3(AB2 + 9x2) + 5(AB2 + x2)
= 8AB2 + 32x2
= 8(AB2 + 4x2)
= 8AE2