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Prompt: In triangle $ABC$, $\angle A = 60^\circ$, $\angle B = 60^\circ$, and $AB = 10$. Point $D$ is on $\overline{AC}$ such that $AD = 4$ and $CD = 6$. Point $E$ is on $\overline{BD}$ such that $BE = 5$ and $ED = 5$. Let $F$ be the intersection of $\overline{AE}$ and $\overline{BC}$. If $BF = x$ and $FC = 10 - x$, find the sum of all distinct possible values of $x$.

We first analyze the given triangle $ABC$ and observe that $\angle A = \angle B = 60^\circ$, which means that triangle $ABC$ is isosceles with $AB = AC$.



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Since $AB = 10$, we can conclude that $AC = 10$ as well, because $ABC$ is an isosceles triangle.



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Next, we examine triangle $ADC$. Given that $AD = 4$ and $CD = 6$, we can apply the Law of Cosines to find $\angle ADC$: $$\cos \angle ADC = \frac{4^2 + 6^2 - 10^2}{2 \cdot 4 \cdot 6} = -\frac{1}{2}.$$ Since $\cos 120^\circ = -\frac{1}{2}$, we can determine that $\angle ADC = 120^\circ$.



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Furthermore, we notice that $AE$ is the angle bisector of $\angle DAC$ because $\angle DAE = \angle EAC = 30^\circ$.



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Using the Angle Bisector Theorem, we can express the relationship between the lengths $BF$ and $FC$ as follows: $$\frac{BF}{FC} = \frac{BD}{DC}.$$ Since $BD = BE + ED = 5 + 5 = 10$, we can simplify the expression: $$\frac{BF}{FC} = \frac{10}{6} = \frac{5}{3}.$$



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We are given that $BF = x$ and $FC = 10 - x$. Substituting these values into the previous expression, we obtain: $$\frac{x}{10-x} = \frac{5}{3}.$$ Solving this equation, we have: $$3x = 50 - 5x \Rightarrow 8x = 50 \Rightarrow x = \frac{25}{4}.$$



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However, we also need to consider the case where $F$ coincides with $B$. In this case, $BF = 0$ and $FC = 10$.



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Additionally, we should consider the case where $F$ coincides with $C$. In this case, $BF = 10$ and $FC = 0$.



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Now, we need to find the sum of all distinct possible values of $x$, which are $0$, $\frac{25}{4}$, and $10$.



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Therefore, the sum of all distinct possible values of $x$ is $0 + \frac{25}{4} + 10 = \frac{65}{4}$.

The final answer is: $\boxed{\frac{65}{4}}$



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