Question

Prove that the perimeter of a triangle is greater than the sum of its three medians.

Answer 1

Given: ΔABC is on which AD, BE and CF are its medians.

To Prove: We know that the sum of any two sides of a triangle is greater than twice the median bisecting the third side.

Therefore, AD is the median bisecting BC

⇒ AB + AC > 2AD    ...(i)

BE is the median bisecting AC   ...(ii)

And CF is the median bisecting AB

⇒ BC + AC > 2EF    ...(iii)

Adding (i), (ii) and (iii), we get

(AB + AC) + (AB + BC) + (BC + AC) > 2. AD + 2 . BE + 2 . BE + 2 . CF

⇒ 2 (AB + BC + AC) > 2 (AD + BE + CF)

⇒ AB + BC + AC > AD + BE + CF.