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Prove that three times the sum of the squares of the sides of a triangle is equal to four timesthe sum of the squares of the medians of the triangle.17. |
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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+ AC²= 2BD²+ 2AD² = 2 × (½BC)²+ 2AD² = ½ BC²+ 2AD² ∴ 2AB²+ 2AC²= 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+ 4AC²= AB²+ BC²+ AC²+ 4AD²+ 4BE²+ 4CF²3(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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