Question

prove that Semiperimeter of a ∆ is less than the sum of it's medians

Answer 1

Step-by-step explanation:

Let ABC be the triangle and D. E and F are midpoints of BC, CA and AB respectively.

Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side,

Hence in ΔABD, AD is a median

⇒ AB + AC > 2(AD)

Similarly, we get

BC + AC > 2CF

BC + AB > 2BE

On adding the above inequations, we get

(AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE

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

∴ AB + BC + AC > AD + BE + CF

Hence, we can say that the perimeter of a triangle is greater than the sum of the medians.

Answer