Q.5 ABCD is a quadrilateral.

Is AB + BC + CD + DA < 2(AC + BD)?

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.