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Assume Length of AB = x so AB = CD = x and BC = 3x
so distance between AD = \(5x = \frac{1}{3}-\frac{1}{2} = \frac{1}{6}\) or \(x= \frac{1}{30}\) now AC = \(4x+\frac{1}{3} = \frac{4}{30} + \frac{1}{3} = \frac{14}{30} or \frac{7}{15}\)
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66% (02:35) correct 
