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In an equilateral triangle ABC, the side BC is trisected at D, then 9AD2 is .......(1) 7AB2 (2) 8BC2 (3) 4AC2(4) 3/2 AB2 |
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Answer» Correct option is (1) 7AB2 Let A(0, 0), B(2a, 0), C(a, √3 a), Then D \((\frac{4a}{3},\frac{2\sqrt{3}}{3})\) ⇒ 9 AD2 = 28a2 = 7 AB2 |
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