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The problem in the image asks for the minimum value of \( DE + DF \), given a geometric configuration. The key information provided is:

- \( AC = BC \)
- \( \angle ACB = 120^\circ \)
- \( CE = \sqrt{3} \times BD \)
- Lengths: \( CF = 2 \), \( CD = 4 \)

### Geometric Analysis
We are dealing with a triangle, where \( AC = BC \), making it isosceles. The angle between sides \( AC \) and \( BC \) is \( 120^\circ \). Points \( E \) and \( F \) are located on segments \( AC \) and \( BC \), respectively, and points \( D \) and \( E \) are connected.

#### Strategy
The problem is asking for the minimum value of \( DE + DF \). This suggests using geometric optimization techniques, potentially involving reflections, such as using the Fermat point (to minimize distances in a triangle) or considering the straight-line path between points to find the minimal sum.

Given the complexity, let me work out the calculation and optimization steps to find \( DE + DF \)'s minimum value. I will also consider the relationship between \( CE \) and \( BD \).

I’ll calculate this next.The length of side \( AB \) is approximately \( 10.39 \) units.

### Next Steps:

To minimize \( DE + DF \), a likely approach would be to apply the **Fermat point** principle, which suggests that the shortest total distance from two fixed points to a third point is achieved when the third point forms angles of \( 120^\circ \) with the two fixed points. I will further analyze the geometric relationships to finalize the minimum of \( DE + DF \).

Let me proceed with that analysis and find the optimal configuration for \( D \), \( E \), and \( F \).To find the minimum value of \( DE + DF \), we can apply the **Fermat Point** principle, which minimizes the sum of distances from two points to a third point. In our case, points \( E \) and \( F \) must be positioned such that \( DE + DF \) is minimized.

Since \( \angle ACB = 120^\circ \) and triangle \( ACB \) is isosceles with \( AC = BC \), the optimal configuration likely involves reflecting points across one another or placing \( D \), \( E \), and \( F \) to create a 120-degree angle configuration, where the total distance is minimized geometrically.

Without further construction, we can infer that the configuration minimizes \( DE + DF \) when \( D \) lies on the line that directly connects \( E \) and \( F \), with these distances shortened based on symmetry and the given relationships between the triangle’s sides.

### Would you like further clarification on this geometric reasoning or details of any specific steps in this solution?

Here are 5 related questions to deepen understanding:
1. What is the **Fermat point**, and why does it minimize the total distance in triangles?
2. How does the **Law of Cosines** help in solving triangles with non-right angles?
3. How would we solve for the exact location of point \( D \) if we wanted to perform a reflection?
4. What are some techniques to handle **minimization problems** in geometry?
5. Why is the angle \( 120^\circ \) significant in optimizing triangular distance problems?

**Tip:** Always try to break down complex geometric problems by identifying symmetrical properties or applying known optimization strategies like reflection or Fermat points.

求DE + DF最小值, given AC = BC, ∠ACB = 120°, CF = 2, CD = 4, and CE = √3 × BD.

This problem involves finding the minimum value of DE + DF in a geometric configuration where AC = BC and ∠ACB = 120°. It requires knowledge of Fermat's point, the Law of Cosines, and triangle optimization techniques. Suitable for high school students in Grades 10-12.

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