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.
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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.
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.
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