Question

In a quadrilateral ABCD, show that (AB + BC + CD + DA) > (AC + BD).

Answer 1

Consider △ ABC

We know that

AB + BC > AC ….. (1)

Consider △ ACD

We know that

DA + CD > AC ….. (2)

Consider △ ADB

We know that

DA + AB > BD ….. (3)

Consider △ BDC

We know that

BC + CD > BD ….. (4)

By adding all the equations

AB + BC + DA + CD + DA + AB + BC + CD > AC + AC + BD + BD

So we get

2 (AB + BD + CD + DA) > 2 (AC + BD)

Dividing by 2 both the sides

AB + BD + CD + DA > AC + BD

Therefore, it is proved that AB + BD + CD + DA > AC + BD.