1. **Problem Statement:** Given points A, B, C, A', B', C' on concurrent lines L, M, N meeting at point O, with AB \parallel A'B' and BC \parallel B'C', prove using Thales' theorem that $$\frac{|OA|}{|OA'|} = \frac{|OC|}{|OC'|}$$ and deduce that AC \parallel A'C'. 2. **Recall Thales' Theorem:** If a set of parallel lines intersect two transversals, then the segments they cut on the transversals are proportional. 3. **Apply Thales' Theorem to lines L and M:** Since AB \parallel A'B', and points A, A' lie on line L, points B, B' lie on line M, the segments satisfy: $$\frac{|OA|}{|OA'|} = \frac{|OB|}{|OB'|}$$ 4. **Apply Thales' Theorem to lines M and N:** Since BC \parallel B'C', and points B, B' lie on line M, points C, C' lie on line N, the segments satisfy: $$\frac{|OB|}{|OB'|} = \frac{|OC|}{|OC'|}$$ 5. **Combine the two equalities:** From steps 3 and 4: $$\frac{|OA|}{|OA'|} = \frac{|OB|}{|OB'|} = \frac{|OC|}{|OC'|}$$ Therefore, $$\frac{|OA|}{|OA'|} = \frac{|OC|}{|OC'|}$$ 6. **Deduce AC \parallel A'C':** Triangles ABC and A'B'C' are in perspective from O, and the ratios above imply that the segments on lines L and N are proportional. By the converse of Thales' theorem, since the ratios of segments on L and N are equal, the segment AC is parallel to A'C'. **Final answer:** $$\frac{|OA|}{|OA'|} = \frac{|OC|}{|OC'|} \quad \text{and} \quad AC \parallel A'C'$$