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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