The problem is asking to find the value of $\angle ACB$ in the given figure where:

• $\triangle DEF$ is shown.
• Points $A$, $B$, and $C$ are located on line segments $DE$, $DF$, and $EF$, respectively.
• It is given that $EC = AC$ and $CF = BC$.
• $\angle D = 40^\circ$, and the task is to calculate $\angle ACB$.

Solution Outline:

1. Isosceles Triangle Property: Since $EC = AC$ and $CF = BC$, triangles $\triangle AEC$ and $\triangle BCF$ are isosceles. This property can help us in deducing other angles in the triangle.

2. Use of Known Angles: The given angle $\angle D = 40^\circ$ will also help us compute other angles using triangle sum properties or exterior angle properties in the triangle.

Let me calculate and derive the value of $\angle ACB$.The value of $\angle ACB$ is $140^\circ$.

Let me know if you would like more details or have any questions.

Here are five related questions to expand this topic:

1. What is the triangle sum property and how does it apply here?
2. Can the exterior angle theorem help in solving this problem?
3. What other properties of isosceles triangles can we apply in geometric problems like this?
4. How do we calculate unknown angles when additional information, like angle bisectors, is given?
5. How would the solution change if $\angle D$ were different?

Tip: When dealing with isosceles triangles, remember that two sides being equal implies two equal angles, which can significantly simplify your angle calculations.