Question

Prove that the perimeter of a quadrilateral is greater than twice of any side.

Answer 1

For any quadrilateral ABCD,ABCD\,, we can easily prove that
AB+BC+CD+DA>AC+BD……AB+BC+CD+DA \gt AC+BD……\tag{1}

Now, three cases arise. Either, AC=BDorAC>BDorACBD⟹AC+BD>2BDAC \gt BD \implies AC+BD \gt 2BD and then the result follows from (1).

If BD>AC⟹AC+BD>2AC BDandAC+BD $\geq$2AC and then the result follows from (1).

Answer 2

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