Question

In a quadrilateral ABCD in which diagonal AC and BD intersect at 0, show that
AB + BC + CD + DA < 2(AC+ BD).​

Answer 1

Answer:

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Step-by-step explanation:

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)

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Answer 2

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