find find

Asked by Aadya Ambastha | 26th Jan, 2014, 04:31: PM

Expert Answer:

It is given that ABCD is a trapezium with AB parallel to CD.

Draw DE and CF perpendicular to AB.

In triangle ABC, angle BAC is acute, so we have:

BC2 = AC2 + AB2 - 2AF. AB              ... (1)

In triangle BDA angle DBA is acute, so we have:

AD2 = BD2 + AB2 - 2BE. AB             ... (1)

Adding (1) and (2), we have:

BC2 + AD2 = AC2 + BD2 + 2AB2 - 2AF. AB - 2BE. AB

BC2 + AD2 = AC2 + BD2 + 2AB[AB - AF - BE]

BC2 + AD2 = AC2 + BD2 + 2AB[AB - (AE + EF) - (BF + EF)]

BC2 + AD2 = AC2 + BD2 + 2AB[AB - (AE + EF + BF + EF)]

BC2 + AD2 = AC2 + BD2 + 2AB[AB - (AB + CD)]

BC2 + AD2 = AC2 + BD2 - 2AB. CD

AC2 + BD2 = BC2 + AD2 + 2AB. CD

Answered by  | 26th Jan, 2014, 11:24: PM

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