Question

In a trapezium ABCD in which AB parallel CD then Anwer is​

Answer 1

Answer:

Incomplete question

Answer 2

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

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