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In the given figure a triangle ABC is right angled at B. Side BC is trisected at pointD and E. Prove that 8AE2 3AC2 + 4AD28 |
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Answer» Solution ABC is a triangle right angled at B, and D and E are points of trisection of BC.Let BD = DE = EC= xThen BE = 2x and BC = 3xInΔ ABD,AD² = AB²+BD² AD² = AB²+x²InΔ ABE,AE² = AB² + BE²AE² = AB² + (2x)²AE² = AB² + 4x²InΔ ABC,AC² = AB² + BC² AC² =AB + (3x)² AC² =AB²+ 9x² Now,3AC²+5AD² = 3(AB² + 9x²) + 5(AB² + x²)8AB² + 32x²8(AB² + 4x²)= 8AE²⇒ 8AE² = 3AC² + 5AD²Hence proved. Please let me know if you have understood and like this solution |
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