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In a triangle ABC right angled …

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In a triangle ABC right angled at B . Side BC is trisected at a point D and E Prove that: 8AE^2 = 3AC^2 +5AD^2
  • 1 answers

Aditi .D 6 years ago

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.
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