In the adjacent figure, ABC is a right-angled triangle with right angl

1.	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» 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 triangles ABD, ABE and ABC, (using Pythagoras theorem) We have AD² = AB² + BD² ⇒ AD² = AB² + x² … (1) AE² = AB² + BE² ⇒ AB² + (2x)² ⇒ AE² = AB² + 4x² … (2) And AC² = AB² + BC² = AB² + (3x)² AC² = AB² + 9x² Now 8 AE² – 3 AC² – 5 AD² = 8 (AB² + 4x²) – 3 (AB² + 9x²) – 5 (AB² + x²) = 8AB² + 32x² – 3AB² – 27x² – 5AB² – 5x² = 0 ∴ 8 AE² – 3 AC² – 5 AD² = 0 8 AE² = 3 AC² + 5 AD². Hence it is proved.

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»

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 triangles ABD, ABE and ABC, (using Pythagoras theorem)

We have AD² = AB² + BD²

⇒ AD² = AB² + x² … (1)

AE² = AB² + BE²

⇒ AB² + (2x)²

⇒ AE² = AB² + 4x² … (2)

And AC² = AB² + BC² = AB² + (3x)²

AC² = AB² + 9x²

Now 8 AE² – 3 AC² – 5 AD² = 8 (AB² + 4x²) – 3 (AB² + 9x²) – 5 (AB² + x²)

= 8AB² + 32x² – 3AB² – 27x² – 5AB² – 5x²

= 0

∴ 8 AE² – 3 AC² – 5 AD² = 0

8 AE² = 3 AC² + 5 AD².

Hence it is proved.