Question

Prove that 3 times the sum of the squares of the sides of a triangle is equal to the 4 times the sum of the squares of the medians of a triangle

Answer 1

Sol:

Apollonius theorem states that the sum of the squares of two sides of a triangle is equal to twice the square of the median on the third side plus half the square of the third side.

Hence AB2 + AC 2 = 2BD 2 + 2AD 2
                                  = 2 × (½BC)2 + 2AD2
                                  = ½ BC2 + 2AD2

∴ 2AB2 + 2AC 2 = BC2 + 4AD2  → (1)

Similarly, we get

2AB2 + 2BC2 = AC2 + 4BE2   → (2)

2BC2 + 2AC2 = AB2 + 4CF2   → (3)

Adding (1) (2) and (3), we get

4AB2 + 4BC2 + 4AC 2 = AB2 + BC2 + AC2 + 4AD2 + 4BE2 + 4CF2  
   
3(AB2 + BC2 + AC2) = 4(AD2 + BE2 + CF2)      

Hence, three times the sum of squares of the sides of a triangle is equal to four times the sum of squares of the medians of the triangle.
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