In triangle \(ABC\), \(\measuredangle CBA=72^\circ\), \(E\) is the midpoint of side \(AC\), and \(D\) is a point on side \(BC\) such that \(2BD=DC\); \(AD\) and \(BE\) intersect at \(F\). Find the ratio of the area of triangle \(BDF\) to the area of quadrilateral \(FDCE\). The original answer is in \(\frac{k}{m}\) format, please give the value of k + m.

Let's think step by step and output the final answer within \boxed{}.
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