SOLUTION: Prove that in any triangle ABC the length x of the median from C is given by: {{{x=(1/2) sqrt(2a^2+2b^2-c^2)}}} I know it has to do with the law of cosines: {{{cos(C) = (a^2+b^2-

Algebra ->  Geometry-proofs -> SOLUTION: Prove that in any triangle ABC the length x of the median from C is given by: {{{x=(1/2) sqrt(2a^2+2b^2-c^2)}}} I know it has to do with the law of cosines: {{{cos(C) = (a^2+b^2-      Log On


   

Question 285807: Prove that in any triangle ABC the length x of the median from C is given by:
x=%281%2F2%29+sqrt%282a%5E2%2B2b%5E2-c%5E2%29
I know it has to do with the law of cosines: cos%28C%29+=+%28a%5E2%2Bb%5E2-c%5E2%29%2F2ab but I'm not sure how to simplify it to this other equation.
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
x=(1/2)sqrt(2a^2+b^2-c^2)
2x=sqrt(2a^2+2b^2-c^2)
4x^2= (2a^2+2b^2-c^2)
x^2= (2a^2+2b^2-c^2)/4
x^2=(a^2+b^2-c^2)/2
so to be equal x^2 must be ab
can you take it from here?