1. **Problem statement:** Given triangle ABC with points P, Q, R, and S such that QS is parallel to BA, QR is parallel to CA, and PQ = 10 cm. We need to find the product $PB \times PC$. 2. **Understanding the problem:** Since QS || BA and QR || CA, triangles PQS and PBA are similar, and triangles PQR and PCA are similar by the AA similarity criterion. 3. **Using similarity:** From the similarity of triangles PQS and PBA, corresponding sides are proportional: $$\frac{PQ}{PB} = \frac{QS}{BA}$$ From the similarity of triangles PQR and PCA: $$\frac{PQ}{PC} = \frac{QR}{CA}$$ 4. **Key insight:** Since QS || BA and QR || CA, points Q and S divide sides BC and AB proportionally. Also, since PQ = 10 cm, and P lies on the extension of line segment from B, the length PQ relates to PB and PC. 5. **Using the intercept theorem (basic proportionality theorem):** Because QS || BA and QR || CA, the segments satisfy: $$\frac{PB}{PQ} = \frac{PQ}{PC}$$ Rearranged: $$PB \times PC = PQ^2$$ 6. **Calculate the product:** Given $PQ = 10$ cm, $$PB \times PC = 10^2 = 100$$ **Final answer:** $$\boxed{100}$$