Let $I$ be the center of the incircle of non-isosceles triangle $ABC,A_{1}=AI\cap BC$ and $B_{1}=BI\cap AC.$ Let $l_{a}$ be the line through $A_{1}$ which is parallel to $AC$ and $l_{b}$ be the line through $B_{1}$ parallel to $BC.$ Let $l_{a}\cap CI=A_{2}$ and $l_{b}\cap CI=B_{2}.$ Also $N=AA_{2}\cap BB_{2}$ and $M$ is the midpoint of $AB.$ If $CN\parallel IM$ find $\frac{CN}{IM}$.
1. **Define the problem and setup:**
   Let \( I \) be the center of the incircle of a non-isosceles triangle \( ABC \). Define \( A_1 = AI \cap BC \) and \( B_1 = BI \cap AC \). Let \( l_a \) be the line through \( A_1 \) parallel to \( AC \) and \( l_b \) be the line through \( B_1 \) parallel to \( BC \). Define \( A_2 = l_a \cap CI \) and \( B_2 = l_b \cap CI \). Let \( N = AA_2 \cap BB_2 \) and \( M \) be the midpoint of \( AB \). Given that \( CN \parallel IM \), we need to find \( \frac{CN}{IM} \).

2. **Identify key points and lines:**
   - \( A_1 \) and \( B_1 \) are points where the incircle touches \( BC \) and \( AC \) respectively.
   - \( l_a \parallel AC \) and \( l_b \parallel BC \).
   - \( A_2 \) and \( B_2 \) are intersections of these parallel lines with \( CI \).

3. **Use properties of parallel lines and similar triangles:**
   Since \( l_a \parallel AC \) and \( l_b \parallel BC \), triangles \( A_1A_2C \) and \( B_1B_2C \) are similar to triangles \( A_1AC \) and \( B_1BC \) respectively.

4. **Calculate lengths using similarity:**
   - From similarity, \( \frac{A_1A_2}{AC} = \frac{A_1C}{AC} \) and \( \frac{B_1B_2}{BC} = \frac{B_1C}{BC} \).
   - Since \( A_1 \) and \( B_1 \) are points of tangency, \( A_1C = s - a \) and \( B_1C = s - b \), where \( s \) is the semi-perimeter of \( \triangle ABC \).

5. **Apply Menelaus' theorem:**
   - Apply Menelaus' theorem to the transversal \( \overline{NBY} \) in \( \triangle AXC \):
     \[
     \frac{BX}{BC} \cdot \frac{YC}{YA} \cdot \frac{NA}{NX} = 1
     \]
   - Similarly, apply Menelaus' theorem to the transversal \( \overline{CNZ} \) in \( \triangle ABX \):
     \[
     \frac{CX}{CB} \cdot \frac{ZB}{ZA} \cdot \frac{NA}{NX} = 1
     \]

6. **Calculate ratios using Menelaus' theorem:**
   - From the above applications, we get:
     \[
     \frac{NX}{NA} = \frac{c - a}{b + c - a} \cdot \frac{a}{c - b}
     \]
     \[
     \frac{ZA}{b(c - b)} = \frac{ZB}{a(c - a)} = \frac{c}{a(c - a) + b(c - b)}
     \]

7. **Determine the relationship between \( CN \) and \( IM \):**
   - Given \( CN \parallel IM \), we use the fact that \( M \) is the midpoint of \( AB \) and the properties of the Nagel point.
   - Using the properties of the Nagel point and the fact that \( CN \parallel IM \), we find:
     \[
     \frac{CN}{IM} = 2
     \]

8. **Conclusion:**
   Therefore, the ratio \( \frac{CN}{IM} \) is \( 2 \).

The final answer is \( \boxed{2} \)