Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD
Given, ABCD is a quadrilateral
To Prove : AB + BC + CD + DA > AC + BD
Proof - Since, we know that sum of the two sides of a triangle is greater than the third side.
In ABC, AB + BC > AC
In BCD, BC + CD > BD
In CAD, AD + CD > AC
In BAD, BA + AD > BD
Adding all the above equations,
2(AB + BC + CA + AD) > 2(AC + BD)
2(AB + BC + CA + AD) > 2(AC + BD)
AB + BC + CA + AD > AC + BD
Hence, proved.