Through an internal point $O$ of $\Delta ABC$ one draws 3 lines, parallel to each of the sides, intersecting in the points shown on the picture.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=289[/img]

Find the value of $\frac{|AF|}{|AB|}+\frac{|BE|}{|BC|}+\frac{|CN|}{|CA|}$.
1. **Identify the Parallelograms:**
   - Since \(OK \parallel AC\) and \(MN \parallel BC\), quadrilateral \(ONKC\) is a parallelogram. Therefore, \(CN = OK\).
   - Similarly, since \(AF \parallel OD\) and \(OF \parallel AD\), quadrilateral \(AFOD\) is a parallelogram. Therefore, \(AF = OD\).

2. **Establish Similar Triangles:**
   - Since \(\angle EOK = \angle BAC\) and \(\angle OKE = \angle ACB\), triangles \(\Delta ABC\) and \(\Delta OEK\) are similar by the AA (Angle-Angle) similarity criterion.
   - Similarly, since \(\angle ODN = \angle BAC\) and \(\angle DON = \angle ABC\), triangles \(\Delta ABC\) and \(\Delta ODN\) are also similar by the AA similarity criterion.

3. **Use Similarity Ratios:**
   - From the similarity of \(\Delta ABC\) and \(\Delta OEK\):
     \[
     \frac{CN}{CA} = \frac{OK}{CA} = \frac{EK}{BC}
     \]
   - From the similarity of \(\Delta ABC\) and \(\Delta ODN\):
     \[
     \frac{AF}{AB} = \frac{OD}{AB} = \frac{ON}{BC} = \frac{KC}{BC}
     \]

4. **Sum of Ratios:**
   - We need to find the value of \(\frac{AF}{AB} + \frac{BE}{BC} + \frac{CN}{CA}\).
   - From the similarity ratios, we have:
     \[
     \frac{AF}{AB} = \frac{KC}{BC}, \quad \frac{CN}{CA} = \frac{EK}{BC}
     \]
   - Since \(BE + EK + KC = BC\), we can write:
     \[
     \frac{BE}{BC} + \frac{EK}{BC} + \frac{KC}{BC} = \frac{BE + EK + KC}{BC} = \frac{BC}{BC} = 1
     \]

Therefore, the value of \(\frac{AF}{AB} + \frac{BE}{BC} + \frac{CN}{CA}\) is \(1\).

The final answer is \(\boxed{1}\)