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

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