In ABCDEF is regular hexagon then prove that AB vector +AC vector+AE vector +AF= 3AD vector?
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in ABCDEF is regular hexagon then prove that AB vector +AC vector+AE vector +AF= 3AD vector Vector means sign of vector here
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Answer:
Correction : Asker missed an (AD↑) in the LHS. ......................................… Draw a diagram. Note that in Parallelogram ABDE, sum of the adjacent sides AB↑ and AE↑ is equal to the diagonal AD↑. ∴ (AB↑) + (AE↑) = (AD↑) ... (1) ......................................… Similarly, in Parallelogram ACDF, ... (AC↑) + (AF↑) = (AD↑) ... (2). ......................................… ∴ (AB↑) + (AC↑) + (AD↑) + (AE↑) + (AF↑) = [ (AB↑) + (AE↑) ] + (AD↑) + [ (AC↑) + (AF↑) ] = (AD↑) + (AD↑) + (AD↑) = 3(AD↑) ............................. Q.E.D. ______________________________________… Happy To Help ! ______________________________________…
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Other answers
Interior angle of hexagon - 120 degrees, so angle ABF- 30 degrees. Let midpoint of BF be O. BO = AB (cos 30) BF = 2 x BO = 2AB (cos 30) Let midpoint of CE be P AO = AB (sin 30) = DP OP = BC = AB AD = 2 x AO + OP = 2AB Therefore, 3AD = 6AB AC = AE = BF = 2AB cos 30 AC = 1.7 AB By round up, AC = 2AB AC+AB+AE+AF = 2AB+AB+2AB+AB = 6AB = 3AD Therefore, AC+AB+AE+AF = 3AD (approximately) I hope this helps. Good luck!
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