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