In fig if D and E trisect BC prove that 8AE^2 = 3 AC^2 + 5 AD^2
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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.
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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