1. **Problem statement:** Given triangle ABC, points P on AB and Q on AC such that PQ is parallel to BC. By Thales' theorem, we have $ \frac{|AP|}{|PB|} = \frac{|AQ|}{|QC|} $. We need to prove: $$\frac{|AP|}{|AB|} = \frac{|AQ|}{|AC|} = \frac{|PQ|}{|BC|}$$ 2. **Recall Thales' theorem:** If a line parallel to one side of a triangle intersects the other two sides, it divides those sides proportionally. 3. **Step 1: Use the given parallelism** Since $PQ \parallel BC$, by Thales' theorem: $$\frac{|AP|}{|PB|} = \frac{|AQ|}{|QC|}$$ 4. **Step 2: Introduce point R on BC such that $QR \parallel AB$** By the hint, consider point R on BC with $QR \parallel AB$. 5. **Step 3: Apply Thales' theorem to triangle AQC with line QR parallel to AB** In triangle AQC, since $QR \parallel AB$, Thales' theorem gives: $$\frac{|AQ|}{|AC|} = \frac{|PQ|}{|BC|}$$ 6. **Step 4: Combine the ratios** From step 3, we have: $$\frac{|AQ|}{|AC|} = \frac{|PQ|}{|BC|}$$ From step 1, since $PQ \parallel BC$, the segments are proportional: $$\frac{|AP|}{|AB|} = \frac{|AQ|}{|AC|}$$ 7. **Step 5: Final equality** Combining these results: $$\frac{|AP|}{|AB|} = \frac{|AQ|}{|AC|} = \frac{|PQ|}{|BC|}$$ This completes the proof. **Summary:** By applying Thales' theorem twice, first on triangle ABC with line PQ parallel to BC, and then on triangle AQC with line QR parallel to AB, we establish the proportionality of the segments as required.