ABCD is quadrilateral is AB + BC + CD +AD <2(AC+BD)
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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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