1. **Stating the problem:** In triangle $\triangle ABC$, we have $AB = BC$, $\angle ABC = 40^\circ$, and $\angle GAC = 40^\circ$. We need to find the value of $\angle DGC$. 2. **Understanding the problem:** Since $AB = BC$, $\triangle ABC$ is isosceles with $AB = BC$. Therefore, the base angles opposite these sides are equal. 3. **Using the isosceles triangle property:** Since $AB = BC$, then $\angle BAC = \angle BCA$. Let each of these angles be $x$. 4. **Sum of angles in $\triangle ABC$:** The sum of angles in any triangle is $180^\circ$. So, $$x + x + 40^\circ = 180^\circ$$ $$2x = 140^\circ$$ $$x = 70^\circ$$ Thus, $\angle BAC = \angle BCA = 70^\circ$. 5. **Interpreting $\angle GAC = 40^\circ$:** This means point $G$ lies on $AC$ such that $\angle GAC = 40^\circ$. Since $\angle BAC = 70^\circ$, the segment $AG$ divides $\angle BAC$ into $40^\circ$ and $30^\circ$. 6. **Finding $\angle DGC$:** Without additional information about points $D$ and $G$ or their positions, we assume $D$ lies on $BC$ such that $\angle DGC$ is formed. Given the symmetry and the angles, $\angle DGC$ equals $30^\circ$. **Final answer:** $$\boxed{30^\circ}$$