In a trapezium ABCD, if AB || CD, then (AC2 + BD2) = ?

Draw perpendicular from D on AB meeting it on E and from C on AB meeting AB at F

DEFC will be a parallelogram and thus, EF = CD


Now, In ∆ABC


Since, B is acute


AC2 = BC2 + AB2 - 2AB × AE (i)


Similarly, In ∆ABD,


Since A is acute


BD2 = AD2 + AB2 - 2AB × AF (ii)


Adding (i) and (ii),


AC2 + BD2 = (BC2 + AD2) + (AB2 + AB2) - 2AB (AE + BF)


= (BC2 + AD2) + 2AB (AB - AE - BF) [Since, AB = AE + EF + FB and AB - AE = BE]


= (BC2 + AD2) + 2AB (BE - BF)


= (BC2 + AD2) + 2AB.EF


Now, we know that CD = EF


Thus, AC2 + BD2 = (BC2 + AD2) + 2AB.CD


Option D is correct

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