math
posted by POD on .
in a triangle BCA there are three medians b,c,a. Prove that
3(A^2+B^2+C^2)=4(a^2+b^2+c^2)
(sides) = (medians)

The standard notation for the length of medians:
a=(1/2)*sqrt(2C^2+2B^2A^2)
b=(1/2)*sqrt(2C^2+2A^2B^2)
c=(1/2)*sqrt(2A^2+2B^2C^2)
square both sides:
a^2=(1/4)*(2C^2+2B^2A^2)
b^2=(1/4)*(2C^2+2A^2B^2)
c^2=(1/4)*(2A^2+2B^2C^2)
multiply 4 on both sides:
4a^2=(2C^2+2B^2A^2)
4b^2=(2C^2+2A^2B^2)
4c^2=(2A^2+2B^2C^2)
add them all vertically:
you get:
4(a^2+b^2+c^2)=3(A^2+B^2+C^2)