1. **Stating the problem:** We have triangle ABC with points P and Q on sides AB and AC respectively, and segment PQ parallel to BC. Point R lies on BC. We want to analyze the relationships in this configuration. 2. **Key theorem:** When a line segment (PQ) is drawn parallel to one side (BC) of a triangle (ABC), it divides the other two sides proportionally. This is known as the Basic Proportionality Theorem (Thales' theorem). 3. **Formula:** $$\frac{AP}{PB} = \frac{AQ}{QC}$$ 4. **Explanation:** Since PQ is parallel to BC, the ratios of the segments on AB and AC are equal. 5. **Intermediate work:** If we denote lengths as $AP = x$, $PB = y$, $AQ = m$, and $QC = n$, then: $$\frac{x}{y} = \frac{m}{n}$$ 6. **Using point R on BC:** If R lies on BC, and PQ is parallel to BC, then triangle APQ is similar to triangle ABC by AA similarity. 7. **Conclusion:** The segments are proportional, and the smaller triangle APQ is similar to ABC, preserving angle measures and proportional side lengths. This is the fundamental property to solve problems involving such a configuration.