Given an obtuse triangle $ABC$ with $\angle ABC$ obtuse, extend $\overline{AB}$ past $B$ to a point $D$ such that $\overline{CD}$ is perpendicular to $\overline{AB}$. Let $F$ be the point on line segment $\overline{AC}$ such that $\overline{BF}$ is perpendicular to $\overline{AB}$, and extend $\overline{BF}$ past $F$ to a point $E$ such that $\overline{BE}$ is perpendicular to $\overline{CE}$. Given that $\angle ECF = \angle BCD$, show that $\triangle ABC \sim \triangle BFC$.
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