Question

Prove that in any triangle, the sum of the squares of any two sides is equal to twice the square of the median which bisect the third side.

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
AB^2+AC 2 =2BD 2 +2AD 2                                  = 2 × (½BC)2 + 2AD2                               
= ½ BC2 + 2AD2
 
∴ 2AB2+ 2AC 2 = BC2 + 4AD2 

→ (1) Similarly,  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:

Step-by-step explanation:

A