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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 $x$.

Next, we examine $\triangle ADE$. Since $AE$ is a transversal cutting through $\overline{BD}$, we can use the properties of transversals and congruent angles to find additional relationships within this triangle.



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Using the Law of Cosines in $\triangle ADE$, we can find $AE$:

[AE^{2}=4^{2}+5^{2}-2\cdot4\cdot5\cdot \cos \angle ADE.]

However, since we do not know $\angle ADE$, we need to find it first.



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We can find $\angle ADE$ by considering $\triangle BED$. Since $BE = ED$, $\triangle BED$ is isosceles, and we know that the base angles of an isosceles triangle are congruent. Therefore, $\angle BED = \angle EDB$.



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Since $\angle BED + \angle EDB + \angle BDE = 180^\circ$, we can write:

[2\angle BED + \angle BDE = 180^\circ.]

We also know that $\angle BDE = 180^\circ - \angle ADE$, so:

[2\angle BED + 180^\circ - \angle ADE = 180^\circ.]

Simplifying, we get:

[2\angle BED = \angle ADE.]



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Now, we can use the fact that $\angle BED = \angle ABD$ to find $\angle ADE$. Since $\angle ABD = 60^\circ$, we have:

[\angle BED = 60^\circ.]

Substituting this value into the previous equation, we get:

[2(60^\circ) = \angle ADE.]

So, $\angle ADE = 120^\circ$.



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Now that we have $\angle ADE$, we can find $AE$ using the Law of Cosines:

[AE^{2}=4^{2}+5^{2}-2\cdot4\cdot5\cdot \cos 120^{\circ}.]

Simplifying, we get:

[AE^{2}=16+25+20=61.]

So, $AE = \sqrt{61}$.



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Next, we can use the fact that $\triangle AEF \sim \triangle ABC$ to find $BF$. Since the triangles are similar, we can set up the following proportion:

[\frac{AE}{AB}=\frac{EF}{BC}.]

Substituting the known values, we get:

[\frac{\sqrt{61}}{10}=\frac{EF}{10}.]

Simplifying, we get:

[EF = \sqrt{61}.]



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Since $EF = BF + FC$, we can write:

[BF + FC = \sqrt{61}.]

We also know that $BF + FC = 10$, so:

[10 = \sqrt{61} + FC.]

Simplifying, we get:

[FC = 10 - \sqrt{61}.]



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Finally, we can find the sum of all distinct possible values of $x$:

[x + (10 - x) = 10.]

However, this equation is true for any value of $x$. To find the distinct possible values of $x$, we need to consider the properties of the triangle.



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Since $\triangle ABC$ is equilateral, we know that $BF = FC$ when $F$ is the midpoint of $\overline{BC}$. In this case, $x = 5$.

We also know that $BF = x$ and $FC = 10 - x$. Since $BF + FC = 10$, we can write:

[x + (10 - x) = 10.]

Simplifying, we get:

[10 = 10.]

This equation is true for any value of $x$. However, we also know that $BF = FC$ when $F$ is the midpoint of $\overline{BC}$. In this case, $x = 5$.



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Therefore, the sum of all distinct possible values of $x$ is:

[5 + (10 - 5) = 10.]

However, we need to consider the fact that $x$ can take on other values as well.



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After re-examining the problem, we can see that $x$ can take on two possible values: $x = 5$ and $x = 10 - 5 = 5$.

However, there is another possible value of $x$. When $\angle AEF = 60^\circ$, $\triangle AEF$ is equilateral, and $AE = EF = \sqrt{61}$. In this case, $BF = 10 - \sqrt{61}$ and $FC = \sqrt{61}$.



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Therefore, the sum of all distinct possible values of $x$ is:

[5 + (10 - \sqrt{61}) = 15 - \sqrt{61}.]

However, we need to add the other possible value of $x$ to the sum:

[15 - \sqrt{61} + \sqrt{61} = 15.]

But since $x$ cannot be equal to $10 - \sqrt{61}$ as $x$ must be less than or equal to 5 we can find the other solution for $x$ using $x + (10 - x) = 10$ which is $x = 5$ and the other is found using $EF = \sqrt{61}$ which is $10 - \sqrt{61}$ but the only other solution is $10 - (10 - \sqrt{61})$.

The final answer is: $\boxed{5 + (10 - (10 - \sqrt{61}))= 5 + \sqrt{61}}$



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