Let $ABC$ be a triangle with $\angle BAC=40^\circ $, $O$ be the center of its circumscribed circle and $G$ is its centroid. Point $D$ of line $BC$ is such that $CD=AC$ and $C$ is between $B$ and $D$. If $AD\parallel OG$, find $\angle ACB$.
1. **Given Information and Initial Setup:**
   - We have a triangle \(ABC\) with \(\angle BAC = 40^\circ\).
   - \(O\) is the center of the circumscribed circle of \(\triangle ABC\).
   - \(G\) is the centroid of \(\triangle ABC\).
   - Point \(D\) on line \(BC\) is such that \(CD = AC\) and \(C\) is between \(B\) and \(D\).
   - \(AD \parallel OG\).

2. **Understanding the Parallel Condition:**
   - Since \(AD \parallel OG\), we need to explore the implications of this parallelism.
   - The centroid \(G\) divides each median in the ratio \(2:1\).
   - The line \(OG\) is known as the Euler line of the triangle, which passes through several important points including the orthocenter, centroid, and circumcenter.

3. **Exploring the Geometry:**
   - Since \(CD = AC\), point \(D\) is such that \(D\) lies on the extension of \(BC\) beyond \(C\).
   - This implies that \(\triangle ACD\) is isosceles with \(AC = CD\).

4. **Considering the Isosceles Triangle:**
   - If \(AD\) is parallel to \(OG\), then \(AD\) must be parallel to the \(C\)-angle bisector of \(\triangle ABC\).
   - This implies that \(\triangle ABC\) is isosceles with \(AB = AC\).

5. **Calculating the Angles:**
   - Since \(\triangle ABC\) is isosceles with \(AB = AC\), we have \(\angle ABC = \angle ACB\).
   - Let \(\angle ACB = x\).
   - Then, \(\angle ABC = x\).
   - Using the angle sum property of a triangle, we have:
     \[
     \angle BAC + \angle ABC + \angle ACB = 180^\circ
     \]
     \[
     40^\circ + x + x = 180^\circ
     \]
     \[
     2x = 140^\circ
     \]
     \[
     x = 70^\circ
     \]

6. **Conclusion:**
   - Therefore, \(\angle ACB = 70^\circ\).

The final answer is \(\boxed{70^\circ}\).