hey guys please answer my question and I'll mark u brainliest
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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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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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Ɨ₮
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