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Question
In the adjacent figure, ABC is a right angled triangle with right angle at B and points D, E trisect BC. Prove that 8AE2 = 3AC2 + 5AD2
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Solution
Since Points D, E trisect BC.
BD = DE = CE
Let BD = DE = CE = x
BE = 2x and BC = 3x
In the right ∆ABD,
AD2 = AB2 + BD2
AD2 = AB2 + x2 ...(1)
In the right ∆ABE,
AE2 = AB2 + 2BE2
AE2 = AB2 + 4x2 ...(2) ...(BE = 2x)
In the right ∆ABC
AC2 = AB2 + BC2
AC2 = AB2 + 9x2 ...(3) ...(BC = 3x)
R.H.S. = 3AC2 + 5AD2
= 3[AB2 + 9x2] + 5[AB2 + x2] ...[From (1) and (3)]
= 3AB2 + 27x2 + 5AB2 + 5x2
= 8AB2 + 32x2
= 8(AB2 + 4x2)
= 8AE2 ...[From (2)]
= R.H.S.
∴ 8AE2 = 3AC2 + 5AD2
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