Please reason step by step, and put your final answer within \boxed{}.
Two identical equilateral triangles $ABC$ and $CDE$ with side length 1 are positioned such that they share exactly one common point $C$ and the angle $BCD$ is less than $60^{\circ}$. Point $K$ is the midpoint of $AC$, point $L$ is the midpoint of $CE$, and point $M$ is the midpoint of $BD$. The area of triangle $KLM$ is $\frac{\sqrt{3}}{5}$. Find $BD$.
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