1. **Problem Statement:** Show that if line segment PR is not parallel to BC in triangle ABC, then the ratio $\frac{|AP|}{|AB|} \neq \frac{|AR|}{|AC|}$. Then conclude that for any points P on AB and Q on AC, segment PQ is parallel to BC if and only if $\frac{|AP|}{|AB|} = \frac{|AQ|}{|AC|}$. 2. **Key Concept:** When a segment inside a triangle is parallel to one side, it divides the other two sides proportionally. This is the Triangle Proportionality Theorem. 3. **Step (i) Proof:** - Assume PR is not parallel to BC. - If $\frac{|AP|}{|AB|} = \frac{|AR|}{|AC|}$ were true, then by the converse of the Triangle Proportionality Theorem, PR would have to be parallel to BC. - Since PR is not parallel to BC, this equality cannot hold. - Therefore, $\frac{|AP|}{|AB|} \neq \frac{|AR|}{|AC|}$. 4. **Step (ii) Conclusion:** - If PQ is parallel to BC, then by the Triangle Proportionality Theorem, $\frac{|AP|}{|AB|} = \frac{|AQ|}{|AC|}$. - Conversely, if $\frac{|AP|}{|AB|} = \frac{|AQ|}{|AC|}$, then PQ must be parallel to BC. - Hence, PQ is parallel to BC if and only if $\frac{|AP|}{|AB|} = \frac{|AQ|}{|AC|}$. This completes the proof and conclusion.