P and Q are points on sides AB and AC respectively of `DeltaABC` . If AP = 3 cm , PB = 6 cm , AQ = 5 cm and QC = 10 , show that BC =3PQ .

Answer

To show that \( BC = 3PQ \) given the points \( P \) and \( Q \) on the sides \( AB \) and \( AC \) of triangle \( ABC \), we can use the concept of similar triangles and the properties of proportional segments.

### Step-by-Step Solution:

1. **Identify the segments**:
- Given \( AP = 3 \, \text{cm} \), \( PB = 6 \, \text{cm} \), \( AQ = 5 \, \text{cm} \), and \( QC = 10 \, \text{cm} \).
- Therefore, \( AB = AP + PB = 3 + 6 = 9 \, \text{cm} \) and \( AC = AQ + QC = 5 + 10 = 15 \, \text{cm} \).

2. **Set up the ratios**:
- The ratio of the segments on side \( AB \) is \( \frac{AP}{PB} = \frac{3}{6} = \frac{1}{2} \).
- The ratio of the segments on side \( AC \) is \( \frac{AQ}{QC} = \frac{5}{10} = \frac{1}{2} \).

3. **Use the property of similar triangles**:
- Since \( \frac{AP}{PB} = \frac{AQ}{QC} \), by the Basic Proportionality Theorem (or Thales' theorem), the line segment \( PQ \) is parallel to \( BC \).

4. **Express \( BC \) in terms of \( PQ \)**:
- By the properties of similar triangles, we have:
\[ \frac{PQ}{BC} = \frac{AP}{AB} = \frac{3}{9} = \frac{1}{3} \]
- Rearranging this gives:
\[ BC = 3 \cdot PQ \]

5. **Conclusion**:
- We have shown that \( BC = 3PQ \).