Question -
Answer -
Yes, we have a quadrilateral ABCD.
In ∆AOB, we have AB < AO + BO …(i)
[Any side of a triangle is less than the sum of other two sides]
In ∆BOC, we have
BC < BO + CO …(ii)
[Any side of a quadrilateral is less than the sum of other two sides]
In ∆COD, we have
CD < CO + DO …(iii)
[Any side of a triangle is less than the sum of other two sides]
In ∆AOD, we have
DA < DO + AO …(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
∠2AO + ∠BO + ∠CO + ∠DO
∠2(AO + BO + CO + DO)
∠2 [(AO + CO) + (BO + DO)]
∠2(AC + BD)
Thus, AB + BC + CD + DA < 2(AC + BD)
Hence, proved.