Question

In quadilateral ABCD
show that i AB + BC + CD + DA <2(BD + AC)
(i) AB + BC + CD + DA
> (BD + AC)

please answer this​

Answer 1

Answer:

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Step-by-step explanation:

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

Step-by-step explanation:

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