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In a right angled ∆BAC, angleBAC=90, segments AD, BE andCF are the medians.Prove that 2 (AD^2 + BE^2 +CF^2)=3BC2 |
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Answer» Draw right angle triangle ABC, draw medians. AD, BE, CF. Now Join DF.Since D and F are midpoints of sides AB and BC, DF will be parallel to AC and is equal to 1/2 AC. ADF, ABE, AFC are all right angle triangles. LHS = 2 (AD² + BE² + CF² ) = 2 [ (AF² + DF²) + (AB² + AE²) + (AF² + AC²) ] = 2 [ (AB²/4 + AC²/4) + (AB² + AC²) + (AC²/4 + AB²/4) ] = 2 [ BC² /2 + BC² ] = 3 ( BC² ] |
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