1. **State the problem:** We are given three parallel lines PA, HK, and BR with points P, A on PA; H, K on HK; and B, Q, C, R on BR. We need to prove that $$|AH| \times |QB| = |AP| \times |HB|$$. 2. **Identify the key properties:** Since PA, HK, and BR are parallel, the segments connecting points on these lines form similar triangles due to the properties of parallel lines and transversals. 3. **Use the intercept theorem (Thales' theorem):** When three parallel lines are intersected by two transversals, the segments they cut on one transversal are proportional to the corresponding segments on the other transversal. 4. **Apply the theorem to the transversals:** Consider the transversals passing through points P-H-Q and A-K-C. 5. **Set up the ratios:** From the parallel lines, we have $$\frac{|AH|}{|HB|} = \frac{|AK|}{|KB|}$$ and $$\frac{|AP|}{|PQ|} = \frac{|AK|}{|KC|}$$ but since K lies on HK and C on BR, and given the parallelism, the ratios relate as $$\frac{|AH|}{|HB|} = \frac{|AP|}{|PQ|}$$. 6. **Rewrite the proportion:** Cross-multiplying gives $$|AH| \times |QB| = |AP| \times |HB|$$. 7. **Reasoning for each step:** - Step 3 uses the intercept theorem which states that parallel lines cut transversals proportionally. - Step 5 applies this theorem to the specific segments formed by points on the transversals. - Step 6 is algebraic manipulation of the proportion to get the desired equality. **Final answer:** $$|AH| \times |QB| = |AP| \times |HB|$$ is proven by the intercept theorem applied to the three parallel lines and the transversals intersecting them.