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# Q.5 ABCD is a quadrilateral. Is AB + BC + CD + DA < 2(AC + BD)?

$$\triangle$$AOB, we have AB < AO + BO …(i) [Any side of a triangle is less than the sum of other two sides]
In $$\triangle$$BOC, we have
BC < BO + CO …(ii)
In $$\triangle$$COD, we have
CD < CO + DO …(iii)
In $$\triangle$$AOD, we have
DA < DO + AO …(iv)
Adding eq. (i), (ii), (iii) and (iv), we have
AB + BC + CD + DA < 2AO + 2BO + 2CO + 2DO
Or, AB + BC + CD + DA < 2(AO + BO + CO + DO)
Or, AB + BC + CD + DA < 2 [(AO + CO) + (BO + DO)]
Or, AB + BC + CD + DA < 2(AC + BD)
Thus, AB + BC + CD + DA < 2(AC + BD)
Hence, proved.