Yes, AB + BC + CD + DA < 2="" (ac="" +="" bd)="" is="" true.="" 2="" (ac="" +="" bd)="" is="" />

In this problem, we are given a quadrilateral ABCD and we need to determine whether the sum of the lengths of its sides AB, BC, CD, and DA is greater than twice the length of its diagonal AC or BD.


To evaluate whether AB + BC + CD + DA > 2(AC + BD), we need to consider two cases separately.


In this case, we can apply the triangle inequality theorem to the triangles ABC and CDA.

• Triangle ABC: AB + BC > AC


• Triangle CDA: CD + DA > AC

Adding these two inequalities, we get AB + BC + CD + DA > 2AC.


Similarly, in this case, we apply the triangle inequality theorem to the triangles ABD and BCD.

• Triangle ABD: AB + DA > BD


• Triangle BCD: BC + CD > BD

Adding these two inequalities, we get AB + BC + CD + DA > 2BD.


Based on the two cases, we can conclude that in any quadrilateral ABCD, the sum of the lengths of its sides AB, BC, CD, and DA will always be greater than twice the length of its longer diagonal AC or BD.

Therefore, AB + BC + CD + DA > 2(AC + BD) holds true for any quadrilateral ABCD.