1. **Problem Statement:** Given two triangles ABC and A'B'C' on concurrent lines L, M, N meeting at point O, with AB \parallel A'B' and BC \parallel B'C', use Thales theorem to show $$\frac{|OA|}{|OC|} = \frac{|OA'|}{|OC'|}$$ and deduce that AC \parallel A'C'. 2. **Recall Thales Theorem:** If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally. 3. **Apply Thales Theorem to triangles OAB and OA'B':** Since AB \parallel A'B', by Thales theorem on lines L, M, N, $$\frac{|OA|}{|OA'|} = \frac{|OB|}{|OB'|}$$ 4. **Apply Thales Theorem to triangles OBC and OB'C':** Since BC \parallel B'C', similarly, $$\frac{|OB|}{|OB'|} = \frac{|OC|}{|OC'|}$$ 5. **Combine the two equalities:** From steps 3 and 4, $$\frac{|OA|}{|OA'|} = \frac{|OB|}{|OB'|} = \frac{|OC|}{|OC'|}$$ 6. **Invert the ratios to get the desired form:** $$\frac{|OA|}{|OC|} = \frac{|OA'|}{|OC'|}$$ 7. **Deduce AC \parallel A'C':** Since the ratios of segments on lines N and L are equal and points A, C and A', C' lie on these lines, by converse of Thales theorem, the segment AC is parallel to A'C'. **Final answer:** $$\frac{|OA|}{|OC|} = \frac{|OA'|}{|OC'|} \quad \text{and} \quad AC \parallel A'C'$$