Prove that three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangle.

Ans:    To prove 3(AB² + BC² + CA²) = 4 (AD²+ B^E2 + CF²)

In any triangle sum of squares of any two sides is equal to twice the square of half of third side, together with twice the square of median bisecting it.

If AD is the median

AB² + AC² = 2 { AD²  + BC²/4}

2(AB² + AC²) = 4AD² + BC²

Similarly by taking BE & CF as medians we get

⇒ 2 (AB² + BC²) = 4BE² + AC²

& 2 (AC² + BC²) = 4CF² + AB2

Adding we get

⇒ 3(AB² + BC² + AC²) = 4 (AD²+ BE² + CF²)