In the figure given below .Is $ A B+B C+C D+D A
answer.
"
Solution:
Let the diagonals AC and BD intersect in point O.
Since the sum of lengths of any two sides in a triangle should be greater than the length of the 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.
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