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 $CE$.

[i]Proposed by Lewis Chen
[/i]
1. **Understanding the problem and given conditions:**
   - We have a triangle \(ABC\) with sides \(AB = 100\), \(BC = 120\), 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 \(\overline{AC}\).
   - We need to find \(CE\) given that \([KLM] = [AME] + [BKF] + [CLD]\), where \([X]\) denotes the area of region \(X\).

2. **Rewriting the given condition:**
   - The condition \([KLM] = [AME] + [BKF] + [CLD]\) can be interpreted as the sum of the areas of the smaller triangles formed by the intersections of the cevians being equal to the area of the triangle \(ABC\).

3. **Using the area ratios:**
   - Let \(AE = x\). The area of \(\triangle AEB\) is proportional to the length \(AE\) along \(AC\).
   - The area of \(\triangle AEB\) is \(\frac{x}{140} \times [ABC]\).
   - The area of \(\triangle BFC\) is proportional to the segment \(BF\) along \(AB\). Since \(AF = 60\), \(BF = 40\), and the area of \(\triangle BFC\) is \(\frac{40}{100} \times [ABC] = \frac{2}{5} \times [ABC]\).
   - The area of \(\triangle CDA\) is proportional to the segment \(CD\) along \(BC\). Since \(BD = 90\), \(CD = 30\), and the area of \(\triangle CDA\) is \(\frac{30}{120} \times [ABC] = \frac{1}{4} \times [ABC]\).

4. **Setting up the equation:**
   - According to the given condition, we have:
     \[
     [ABC] = [AEB] + [BFC] + [CDA]
     \]
   - Substituting the areas in terms of \([ABC]\):
     \[
     [ABC] = \frac{x}{140} \times [ABC] + \frac{2}{5} \times [ABC] + \frac{1}{4} \times [ABC]
     \]

5. **Solving for \(x\):**
   - Dividing both sides by \([ABC]\):
     \[
     1 = \frac{x}{140} + \frac{2}{5} + \frac{1}{4}
     \]
   - Converting the fractions to a common denominator:
     \[
     1 = \frac{x}{140} + \frac{8}{20} + \frac{5}{20}
     \]
     \[
     1 = \frac{x}{140} + \frac{13}{20}
     \]
   - Subtracting \(\frac{13}{20}\) from both sides:
     \[
     1 - \frac{13}{20} = \frac{x}{140}
     \]
     \[
     \frac{7}{20} = \frac{x}{140}
     \]
   - Solving for \(x\):
     \[
     x = \frac{7}{20} \times 140 = 49
     \]

6. **Finding \(CE\):**
   - Since \(AE = 49\), we have:
     \[
     CE = AC - AE = 140 - 49 = 91
     \]

The final answer is \(\boxed{91}\)