prove that perimeter of triangle is greater than sum of the 3 medians
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a,b,c are the sides of triangle
M1=1/2*√(2a^2+2b^2-c^2)
M2=1/2*√(2b^2+2c^2-a^2)
M3=1/2*√(2c^2+2a^2-b^2)
M1+M2+M3=1/2*√3(a^2+b^2+c^2)
if it's equalateral triangle
with side a
M1+M2+M3=3a/2
perimeter=a+b+c=3a
so,
perimeter is greater than sum of median
M1=1/2*√(2a^2+2b^2-c^2)
M2=1/2*√(2b^2+2c^2-a^2)
M3=1/2*√(2c^2+2a^2-b^2)
M1+M2+M3=1/2*√3(a^2+b^2+c^2)
if it's equalateral triangle
with side a
M1+M2+M3=3a/2
perimeter=a+b+c=3a
so,
perimeter is greater than sum of median
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