Question

In triangle ABC, right angled at B, points D and E trisect the side BC. Prove that 8 AE^2=3AC^2+5AD^2.

Answer 1

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Answer 2

Explanation:

ABC is a triangle right angled at B, and D and E are points of trisection of BC.

Let BD = DE = EC = x

Then BE = 2x and BC = 3x

In Δ ABD,

AD² = AB² + BD²

AD² = AB² + x²

In Δ ABE,

AE² = AB² + BE²

AE² = AB² + (2x)²

AE² = AB² + 4x²

In Δ ABC,

AC² = AB² + BC²  

AC² = AB + (3x)²

AC² = AB² + 9x²

Now,

3AC² + 5AD² = 3(AB² + 9x²) + 5(AB² + x²) 

8AB² + 32x²

8(AB² + 4x²)

= 8AE²

⇒ 8AE² = 3AC² + 5AD²

Hence proved