in a triangle BCA there are three medians b,c,a. Prove that

Question

3(A^2+B^2+C^2)=4(a^2+b^2+c^2)
(sides) = (medians)

Answer 1

The standard notation for the length of medians:

a=(1/2)*sqrt(2C^2+2B^2-A^2)
b=(1/2)*sqrt(2C^2+2A^2-B^2)
c=(1/2)*sqrt(2A^2+2B^2-C^2)

square both sides:

a^2=(1/4)*(2C^2+2B^2-A^2)
b^2=(1/4)*(2C^2+2A^2-B^2)
c^2=(1/4)*(2A^2+2B^2-C^2)

multiply 4 on both sides:

4a^2=(2C^2+2B^2-A^2)
4b^2=(2C^2+2A^2-B^2)
4c^2=(2A^2+2B^2-C^2)

add them all vertically:

you get:

4(a^2+b^2+c^2)=3(A^2+B^2+C^2)