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In the given figure, line segment \(DF\) intersect the side \(AC\) of a triangle \(ABC\) at the point \(E\) such that \(E\) is the mid-point of \(CA\) and \(\angle AEF = \angle AFE\). Prove that \(\frac{BD}{CD} = \frac{BF}{CE}\). [Hint: Take point \(G\) on \(AB\) such that \(CG||DF\).]
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