Question

In a quadrilateral ABCD: prove that:
(1) AB + BC + CD > DA
(ii) AB + BC + CD + DA > 2AC
(ii) AB + BC + CD + DA > 2BD​

Answer 1

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Answer 2

Answer:

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.