D is any point on the side BC of

ABC. Let’s prove that AB+BC+CA>2AD

Let us first draw a figure for above question.

Given:

D is any point on the segment BC

We know one basic fact about triangles that the sum of any two sides of a triangle is always greater than the third side.

So let us first consider Δ ADB

AB + BD > AD …………………….. (i)

Similarly let us consider the Δ ADC

AC + DC >AD …………………….. (ii)

Let us now add the equations (i) and (ii)

AB + AC + BD + DC > AD + AD

From the figure we observe that total length BC = BD + DC

AB + AC + BC > 2 × AD

Hence Proved.