Prove that three times the sum of the squares, Mathematics

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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+ BE2 + 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.

1619_triangle1.png

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)


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