Question

In a quadrilateral ABCD, AC and BD are diagonals. Prove that AB + BC + CD + DA > 2AC.

Answer 1

Given,

ABCD is a quadrilateral. AC and BD are diagonals.

Consider ΔABC,

We know that

AB + BC > AC  ...(1) (Sum of any two sides of a triangle is greater than the third side

Consider ΔACD

We know that

AD + CD > AC  ...(2)  (Sum of any two sides of a triangle is greater than the third side)

By adding both the equations (1) and (2), we get

AB + BC + AD + CD > AC + AC

So, we get,

AB + BC + AD + CD > 2AC

Therefore, it is proved that AB + BC + AD + CD > 2AC.