Show that in a quadrilateral ABCD
AB+BC+CD+DA>AC+BD.
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