1. **Problem statement:** Given points A, B, C, A', B', C' on concurrent lines 𝓛, 𝓜, 𝓝 meeting at O, with AB \parallel A'B' and BC \parallel B'C', prove using Thales theorem that $$\frac{|OA|}{|OC|} = \frac{|OA'|}{|OC'|}$$ 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 AB \parallel A'B':** Lines 𝓛 and 𝓜 are transversals intersected by the parallel lines AB and A'B'. Thus, $$\frac{|OA|}{|OA'|} = \frac{|OB|}{|OB'|}.$$ 4. **Apply Thales theorem to BC \parallel B'C':** Lines 𝓜 and 𝓝 are transversals intersected by the parallel lines BC and B'C'. Thus, $$\frac{|OB|}{|OB'|} = \frac{|OC|}{|OC'|}.$$ 5. **Combine the two equalities:** From steps 3 and 4, $$\frac{|OA|}{|OA'|} = \frac{|OB|}{|OB'|} = \frac{|OC|}{|OC'|}.$$ 6. **Rearrange to get the desired ratio:** $$\frac{|OA|}{|OC|} = \frac{|OA'|}{|OC'|}.$$ 7. **Deduce AC \parallel A'C':** Since the ratios of segments on 𝓛 and 𝓝 are equal and points B and B' lie on 𝓜, by converse of Thales theorem, the segment AC is parallel to A'C'. This completes the proof using Thales theorem.