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If G is the CENTROID of triangle ABC then prove that

AB∧2 +BC∧2+AC∧2=3(GA∧2+GB∧2+GC∧2) BEST OF LUCK FOR THE EQUATION

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AB∧2 +BC∧2+AC∧2=3(GA∧2+GB∧2+GC∧2) BEST OF LUCK FOR THE EQUATION

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Using cosine rule in a triangle ABC and triangle ABD (D = mid point of BC, E = mid point of AC, and F = mid point of AB)

cos B = (AB² + BC² - AC² ) / 2 AB * AC

cos B = (AB² + BD² - AD² ) / 2 AB * AD

equating both sides, replace BD = BC/2 we get

2 AB² + 2 AC² = 4 AD² + BC² write now AD = GA + GD = (3/2) GA

= 4 (9/4 GA² ) + BC² = 9 GA²+ BC²

similarly 2 AB² + 2 BC² =9 GB2 + AC2

2 BC² + 2 AC² = 9 GC2 + AB2

Add the three equations

you get answer

cos B = (AB² + BC² - AC² ) / 2 AB * AC

cos B = (AB² + BD² - AD² ) / 2 AB * AD

equating both sides, replace BD = BC/2 we get

2 AB² + 2 AC² = 4 AD² + BC² write now AD = GA + GD = (3/2) GA

= 4 (9/4 GA² ) + BC² = 9 GA²+ BC²

similarly 2 AB² + 2 BC² =9 GB2 + AC2

2 BC² + 2 AC² = 9 GC2 + AB2

Add the three equations

you get answer