1. **Problem statement:** Given triangle ABC with points P on AB and Q on AC such that PQ is parallel to BC, prove that $$\frac{|AP|}{|AB|} = \frac{|AQ|}{|AC|} = \frac{|PQ|}{|BC|}.$$ 2. **Recall Thales' theorem:** If a line parallel to one side of a triangle intersects the other two sides, it divides those sides proportionally. Formally, if PQ is parallel to BC, then $$\frac{|AP|}{|PB|} = \frac{|AQ|}{|QC|}.$$ 3. **First part:** From Thales' theorem, $$\frac{|AP|}{|PB|} = \frac{|AQ|}{|QC|}.$$ Adding 1 to both sides of each ratio, we get: $$\frac{|AP| + |PB|}{|PB|} = \frac{|AQ| + |QC|}{|QC|} \Rightarrow \frac{|AB|}{|PB|} = \frac{|AC|}{|QC|}.$$ 4. **Invert both sides:** $$\frac{|PB|}{|AB|} = \frac{|QC|}{|AC|}.$$ 5. **Rewrite the original ratio:** $$\frac{|AP|}{|AB|} = 1 - \frac{|PB|}{|AB|} = 1 - \frac{|QC|}{|AC|} = \frac{|AQ|}{|AC|}.$$ Thus, $$\frac{|AP|}{|AB|} = \frac{|AQ|}{|AC|}.$$ 6. **To prove $$\frac{|AQ|}{|AC|} = \frac{|PQ|}{|BC|}$$, consider point R on BC such that QR is parallel to AB (as per the hint). 7. **Apply Thales' theorem to triangle BQC with line QR parallel to AB:** $$\frac{|BQ|}{|QC|} = \frac{|BR|}{|RC|}.$$ 8. **Since QR is parallel to AB, triangle PQR is similar to triangle ABC. Therefore, corresponding sides are proportional:** $$\frac{|PQ|}{|BC|} = \frac{|AQ|}{|AC|}.$$ 9. **Combining all results:** $$\frac{|AP|}{|AB|} = \frac{|AQ|}{|AC|} = \frac{|PQ|}{|BC|}.$$ This completes the proof.