If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
Step-by-Step Solution
Key Concept: Basic Proportionality Theorem (Thales theorem): In a triangle, a line drawn parallel to one side divides the other two sides proportionally.
1. Construction: Let \(\triangle ABC\) be given. Draw a line \(DE\) parallel to side \(BC\) such that \(D\) lies on \(AB\) and \(E\) lies on \(AC\).
2. Identify angles: Because \(DE \parallel BC\), the corresponding angles are equal:
- \(\angle ADE = \angle ABC\)
- \(\angle AED = \angle ACB\)
- \(\angle DAE = \angle BAC\) (common angle).
3. Similar triangles: From the equal angles, \(\triangle ADE\) and \(\triangle ABC\) are similar (AA similarity).
4. Write the proportion from similarity:
$$\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$$
5. Express the remaining parts: On side \(AB\), \(DB = AB - AD\); on side \(AC\), \(EC = AC - AE\).
6. Derive the required ratio:
From \(\frac{AD}{AB} = \frac{AE}{AC}\) we have \(AD \cdot AC = AE \cdot AB\).
Substituting \(AB = AD + DB\) and \(AC = AE + EC\):
$$AD(AE+EC) = AE(AD+DB)$$
Expanding and cancelling \(AD\cdot AE\) gives \(AD\cdot EC = AE\cdot DB\).
Hence
$$\frac{AD}{DB} = \frac{AE}{EC}$$
7. Conclusion: The line \(DE\) parallel to \(BC\) divides sides \(AB\) and \(AC\) in the same ratio.
Result: \(AD:DB = AE:EC\).
Correct Answer: AD:DB = AE:EC (the two sides are divided in the same ratio).