Question

D is any point on side BC of triangleABC.Prove that AB+BC+CA>2AD

Answer 1

AM is a median of a triangle ABC. Is AB+BC+CA>2AM?

In triangle ABC, AM is a median.

In triangle ABM, AB+BM>AM… (1) [the sum of two sides of a triangle are greater than the third side].

In triangle ACM, AC+CM>AM… (2) [the sum of two sides of a triangle are greater than the third side].

Add (1) and (2) to get

AB+BM+AC+CM>AM+AM, or

AB+(BM+CM)+AC>2AM, or

AB+BC+CA>2AM. Proved

Answer 2

Answer:

In ABD , By Inequality property of triangle

AB + BD > AD ----------(1)

And In ACD ,By Inequality property of triangle

DC + AC > AD ---------(2)

On Adding Eq (1) and Eq (2)

AB+BD+DC+AC> AD+AD

AB + BC + AC > 2 AD [Given BD+DC = BC]

 AB + BC + AC > 2 AD

hence Proved