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Answer 1

Answer:

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.

Answer 2

Step-by-step explanation:

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)

Adding (i), (ii) , (iii), (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.

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