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math

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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)

  • math -

    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)

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