To solve this problem, let's break down the given information step by step:

1. Triangle Structure:

   • We have triangle $DEF$.
   • Points $A$, $B$, and $C$ are taken on segments $DE$, $DF$, and $EF$ respectively.

2. Conditions Given:

   • $EC = AC$
   • $CF = BC$
   • $\angle D = 40^\circ$

3. Objective:

   • Find the measure of $\angle ACB$.

4. Analysis:

   • Since $EC = AC$ and $CF = BC$, triangles $AEC$ and $BCF$ are isosceles.
   • By the isosceles triangle property, this implies that:
     • $\angle EAC = \angle ECA$
     • $\angle FCB = \angle FBC$

5. Angle Calculation:

   • Let’s consider triangle $DEF$ as a whole.
   • The sum of the angles in any triangle is $180^\circ$, so: $\angle D + \angle E + \angle F = 180^\circ$
   • Given $\angle D = 40^\circ$, we have: $\angle E + \angle F = 140^\circ$

6. Symmetry and Isosceles Properties:

   • With the isosceles properties of triangles $AEC$ and $BCF$, and symmetry around the lines through $A$ and $B$, we can deduce that the configuration likely forms a scenario where the interior angle $\angle ACB$ adds up to maintain the total of 180 degrees in triangle $ACB$ based on equal splitting.

7. Answer: After considering symmetry and structure, the answer is likely one of the given options. Testing all, the probable answer is: [ \angle ACB = 100^\circ