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Question

In a trapezium ABCD, if AB || CD, then (AC2 + BD2) = ?
(a) BC2 + AD2 + 2BC ⋅ AD
(b) AB2 + CD2 + 2AB ⋅ CD
(c) AB2 + CD2 + 2AD ⋅ BC
(d) BC2 + AD2 + 2AB ⋅ CD

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Solution

(c) BC2 + AD2 + 2AB.CD


Explanation:

Draw perpendicular from D and C on AB which meets AB at E and F, respectively.
∴​ DEFC is a parallelogram and EF = CD.

In ∆ABC, ∠B is acute.
∴ AC2= BC2 + AB2 - 2AB.AE
In ∆ABD, ∠A is acute.​
∴ ​BD2 = AD2 + AB2 - 2AB.AF
∴ ​AC2 + BD2 = (BC2 + AD2) + (AB2 + AB2 ) - 2AB(AE + BF)
= (BC2 + AD2) + 2AB(AB - AE - BF) [∵ AB = AE + EF + FB and AB - AE = BE]
= (BC2 + AD2) + 2AB(BE - BF)
= (BC2 + AD2) + 2AB.EF
∴ AC2 + BD2 = ​(BC2 + AD2) + 2AB.CD

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