Question

In a quadrilateral ABCD, show that (AB + BC + CD + DA) < 2 (BD + AC).

Answer 1

Consider △ AOB

We know that

AO + BO > AB ….. (1)

Consider △ BOC

We know that

BO + CO > BC ….. (2)

Consider △ COD

We know that

CO + DO > CD ….. (3)

Consider △ AOD

We know that

DO + AO > DA ….. (4)

By adding all the equations

AO + BO + BO + CO + CO + DO + DO + AO > AB + BC + CD + DA

So we get

2 (AO + CO) + 2 (BO + DO) > AB + BC + CD + DA

On further calculation

2AC + 2BD > AB + BC + CD + DA

By taking 2 as common

2 (AC + BD) > AB + BC + CD + DA

So we get

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

Therefore, it is proved that AB + BC + CD + DA < 2 (AC + BD)