Therefore, In AOB, AB < OA + OB ……….(i)
In ∆BOC, BC < OB + OC ……….(ii)
In ∆COD, CD < OC + OD ……….(iii)
In ∆AOD, DA < OD + OA ……….(iv)
Adding equations (i), (ii), (iii) and (iv), we get
AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD + OA
⇒ 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.

Class 7 Maths Chapter 6 Exercise 6.4

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