the same ratio.
OR
In Figure-6, in an equilateral triangle ABC, AD 1 BC, BE 1 AC and
CF I AB. Prove that 4 (AD2 + BE2 + CF2) = 9 AB2.
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SE
B
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Figure-6
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4 (AD² + BE² + CF²) = 9 AB²
Step-by-step explanation:
ABC is an equilateral triangle
=> AB = BC = AC = x
=> AD ⊥ BC is also medain => BD = CD = BC/2 = x/2
AD² = AB² - BD²
=> AD² = x² -(x/2)²
=> AD² = x² - x²/4
=> AD² = 3x²/4
Similarly BE² = CF² = 3x²/4
4 (AD² + BE² + CF²) = 9 AB²
LHS = 4 (AD² + BE² + CF²)
= 4(3x²/4 + 3x²/4 + 3x²/4)
= 3x² + 3x² + 3x²
= 9x²
= 9 AB²
= RHS
QED
Proved
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