In the given figure, prove that CD + DA + AB + BC > 2 AC.

In ∆ABC, the sum of two sides of a trangle is greater than the third side,

∴AB + BC > AC …(i) (1)

In ∆ACD, the sum of two sides of a triangle is greater than the third side,

∴CD + DA > AC . …(ii) (1)

On adding Eqs. (i) and (ii), we get

CD + DA+ AB + BC > 2AC.