Question

please prove it...........

Answer 1

Given : ABCD is rectangle



To prove : AB + BC + CD + AD < 2 ( AC + BD)


Proof:

In ∆AOD

AD < OD + AO__________________eq 1

{sum of two side of ∆ is always greater than third side}


In ∆AOB

AB < OA + OB__________________eq 2

{sum of two side of ∆ is always greater than third side}

In ∆BOC

BC < OC + BO___________________eq 3

{sum of two side of ∆ is always greater than third side}


In ∆ COD

CD < OD + OC___________________eq 4

{sum of two side of ∆ is always greater than third side}



On adding eq 1, 2, 3, 4

AB + BC + CD + AD < OD + AO + OB + OA + OC + OB + OC + OD

AB + BC + CD + AD < 2 ( OD + OB + OC + OA)

AB + BC + CD + AD < 2 ( AC + BD)

{ OA + OC = AC, OD + OB = AD, from figure}



Hence proved.
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