1. **Problem Statement:** Given a triangle ABC with points P on AB and Q on AC such that the line segment PQ is parallel to BC, and the ratio $\frac{|AP|}{|PB|} = \frac{|AQ|}{|QC|}$, show that this is equivalent to $\frac{|AP|}{|AB|} = \frac{|AQ|}{|AC|}$. 2. **Relevant Theorem and Formula:** When a line parallel to one side of a triangle intersects the other two sides, it divides those sides proportionally. This is known as the Basic Proportionality Theorem (or Thales' theorem). 3. **Step-by-step Solution:** - Given $\frac{|AP|}{|PB|} = \frac{|AQ|}{|QC|}$. - Note that $|AB| = |AP| + |PB|$ and $|AC| = |AQ| + |QC|$. - Express $|PB|$ and $|QC|$ in terms of $|AP|$, $|AQ|$, $|AB|$, and $|AC|$: $$|PB| = |AB| - |AP|$$ $$|QC| = |AC| - |AQ|$$ - Substitute into the given ratio: $$\frac{|AP|}{|AB| - |AP|} = \frac{|AQ|}{|AC| - |AQ|}$$ - Cross-multiply: $$|AP| (|AC| - |AQ|) = |AQ| (|AB| - |AP|)$$ - Expand both sides: $$|AP| \cdot |AC| - |AP| \cdot |AQ| = |AQ| \cdot |AB| - |AQ| \cdot |AP|$$ - Add $|AP| \cdot |AQ|$ to both sides to cancel terms: $$|AP| \cdot |AC| = |AQ| \cdot |AB|$$ - Rearrange to isolate ratios: $$\frac{|AP|}{|AB|} = \frac{|AQ|}{|AC|}$$ 4. **Conclusion:** We have shown that the given ratio $\frac{|AP|}{|PB|} = \frac{|AQ|}{|QC|}$ implies $\frac{|AP|}{|AB|} = \frac{|AQ|}{|AC|}$, as required. This confirms the equivalence using the properties of parallel lines and segment ratios in triangles.