Question

ABCD is a quadrilateral.prove that AB+BC+CD+DA<2(AC+BD).

Answer 1

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side

Therefore,

In Δ AOB, AB < OA + OB ……….(i)

In Δ BOC, BC < OB + OC ……….(ii)

In Δ COD, CD < OC + OD ……….(iii)

In Δ AOD, DA < OD + OA ……….(iv)

⇒ AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD

⇒ AB + BC + CD + DA < 2[(AO + OC) + (DO + OB)]

⇒ AB + BC + CD + DA < 2(AC + BD)

Hence, it is proved.

Hope that this answer will help you ✌️