Question -
Answer -
Yes, we have a quadrilateral ABCD.
┬а
In тИЖAOB, we have AB < AO + BO тАж(i)
[Any side of a triangle is less than the sum of other two sides]
In тИЖBOC, we have
BC < BO + CO тАж(ii)
[Any side of a quadrilateral is less than the sum of other two sides]
In тИЖCOD, we have
CD < CO + DO тАж(iii)
[Any side of a triangle is less than the sum of other two sides]
In тИЖAOD, we have
DA < DO + AO тАж(iv)
[Any side of a triangle is less than the sum of other two sides]
Adding eq. (i), (ii), (iii) and (iv), we have
AB + BC + CD + DA
тИа2AO + тИаBO + тИаCO + тИаDO
тИа2(AO + BO + CO + DO)
тИа2 [(AO + CO) + (BO + DO)]
тИа2(AC + BD)
Thus, AB + BC + CD + DA < 2(AC + BD)
Hence, proved.