1. **Problem Statement:** Show that for several parallels $P_1Q_1$, $P_2Q_2$, $P_3Q_3$, ... to side $BC$ of triangle $ABC$, the ratios satisfy $$\frac{|AP_1|}{|AQ_1|} = \frac{|AP_2|}{|AQ_2|} = \frac{|AP_3|}{|AQ_3|} = \cdots = \frac{|AB|}{|AC|}.$$ 2. **Key Idea:** If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally (Basic Proportionality Theorem or Thales' theorem). 3. **Applying the theorem:** Since $P_1Q_1$ is parallel to $BC$, by the Basic Proportionality Theorem, $$\frac{|AP_1|}{|P_1B|} = \frac{|AQ_1|}{|Q_1C|}.$$ 4. **Expressing segments:** Note that $|AB| = |AP_1| + |P_1B|$ and $|AC| = |AQ_1| + |Q_1C|$. 5. **Rewrite the ratio:** From step 3, $$\frac{|AP_1|}{|P_1B|} = \frac{|AQ_1|}{|Q_1C|} \implies \frac{|AP_1|}{|AQ_1|} = \frac{|P_1B|}{|Q_1C|}.$$ 6. **Using the whole sides:** Since $|AB| = |AP_1| + |P_1B|$ and $|AC| = |AQ_1| + |Q_1C|$, rearranging gives $$\frac{|AP_1|}{|AQ_1|} = \frac{|AB| - |AP_1|}{|AC| - |AQ_1|}.$$ 7. **Repeating for other parallels:** Similarly, for $P_2Q_2$ and $P_3Q_3$ parallel to $BC$, the same proportionality holds: $$\frac{|AP_2|}{|AQ_2|} = \frac{|AB|}{|AC|}, \quad \frac{|AP_3|}{|AQ_3|} = \frac{|AB|}{|AC|}.$$ 8. **Conclusion:** Thus, all these ratios are equal to the ratio of the sides $AB$ and $AC$: $$\frac{|AP_1|}{|AQ_1|} = \frac{|AP_2|}{|AQ_2|} = \frac{|AP_3|}{|AQ_3|} = \cdots = \frac{|AB|}{|AC|}.$$