Prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals.

Given: Let ABCD is a quadrilateral with AC and BD as its diagonals

To Prove: Sum of all the sides of a quadrilateral is greater than the sum of its diagonals


Proof: Consider a quadrilateral ABCD where AC and BD are the diagonals


AB+BC > AC (i) (Sum of two sides is greater than the third side)


AD+DC > AC (ii)


AB+AD > BD (iii)


DC+BC > BD (iv)


Adding (i), (ii), (iii), and (iv)


AB+BC+AD+DC+AB+AD+DC+BC > AC+AC+BD+BD


2(AB+BC+CD+DA) > 2(AC+BD)


AB+BC+CD+DA > AC+BC


Hence, proved that the Sum of all the sides of a quadrilateral is greater than the sum of its diagonals


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