Vertex \(E\) of equilateral triangle \(ABE\) is in the interior of square \(ABCD\), and \(F\) is the point of intersection of diagonal \(BD\) and line segment \(AE\). If length \(AB\) is \(\sqrt{1+\sqrt{3}}\), then find the area of \(\triangle ABF\). The original answer is in the form \(\frac{\sqrt{k}}{m}\), please give the value of k + m.

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