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 

Answer 1

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