Question

prove that three times the sum of the squares of the sides of triangle is equal to the four times the sum of the squares of the medians of the triangle

Answer 1

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.

Answer 2

Answer:

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.

Step-by-step explanation: