Question

In a triangle ABC, medians AD,BE,CF intersect each other at point G .prove that 3(AB+BC+CA) is greater than 2(AD+BE+CF) ​

Answer 1

Answer:

Step-by-step explanation:

In a triangle ABC

AB>AD

CA>CF

BC>BE

then

AB+BC+CA>AD+BE+CF

multiply both sides on 3 then

3(AB+BC+CA)>3(AD+BE+CF)

hence proved