ABC is an equilateral triangle. The side BC is trisected at D such that BC=3 BD. What is the ratio of AD^{2} to AB^{2}?

Explanation

Let the side of the triangle be a. Draw an altitude AM perpendicular to BC. BM = \frac{a}{2} and AM = \frac{\sqrt{3}}{2}a. Since BD = \frac{a}{3}, DM = BM - BD = \frac{a}{2} - \frac{a}{3} = \frac{a}{6}. In right \Delta ADM, AD^2 = AM^2 + DM^2 = (\frac{\sqrt{3}}{2}a)^2 + (\frac{a}{6})^2 = \frac{3a^2}{4} + \frac{a^2}{36} = \frac{28a^2}{36} = \frac{7a^2}{9}. Thus, AD^2 : AB^2 = 7 : 9.

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