In triangle $ABC$, $AB = 100$, $BC = 120$, and $CA = 140$. Points $D$ and $F$ lie on $\overline{BC}$ and $\overline{AB}$, respectively, such that $BD = 90$ and $AF = 60$. Point $E$ is an arbitrary point on $\overline{AC}$. Denote the intersection of $\overline{BE}$ and $\overline{CF}$ as $K$, the intersection of $\overline{AD}$ and $\overline{CF}$ as $L$, and the intersection of $\overline{AD}$ and $\overline{BE}$ as $M$. If $[KLM] = [AME] + [BKF] + [CLD]$, where $[X]$ denotes the area of region $X$, compute $CE$.

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