Answer :

Here, ABCD is a quadrilateral and AC and BD are its diagonals.

Now, As we that, sum of two sides of a triangle is greater than the third side.


In Δ ACB,


AB + BC > AC (i)


In Δ BDC,


CD + BC > BD (ii)


In Δ BAD,


AB + AD>BD (iii)


In Δ ACD,


AD + DC > AC (iv)


Now, adding (i), (ii), (iii) and (iv):


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


2AB + 2BC + 2CD + 2AD > 2AC + 2BD


Thus, AB + BC + CD + AD > AC + BD


Hence, proved


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