Question

prove that in any triangle four times the sum of medians is greater than three times the perimeter​

Answer 1

Answer:

Let ABC be a triangle and D, E and F are mid points 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.

Step-by-step explanation:

Hence in ▶ ABD, 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)+(AC + BC)+(BC + AB)>2AD + CD +2BE

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

=AB+BC+AC>AD+BE+CF