Question

d,e trisects the side bcof a triangle abc prove that 8ae2=3ac2+5ad2

Answer 1

Answer:

Step-by-step explanation:

since D and E are the points of trisection of BC therefore,

BD = DE = CE

Let BD = DE = CE = x, then, BE=2x and BC=3x

In right angle triangle ABD, ABE and ABC, we have

=>  AD^2 = AB^2 + BD^2

      AD^2 = AB^2 + x^2

=>   AE^2 = AB^2 + BE^2

      AE^2 = AB^2 + 4x^2

=>  AC^2 = AB^2 + BC^2

     AC^2 = AB^2 + 9x^2

Now,

      8 AE^2 - 3AC^2 - 5 AD^2

      =8(AB^2 + 4x^2) - 3(AB^2 + 9x^2) - 5( AB^2 + x^2)

      = 8 AE^2 - 3AC^2 - 5 AD^2 = 0

      =8 AE^2 = 3AC^2 + 5 AD^2

HENCE PROVED