If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Step-by-Step Solution
Key Concept: The statement is the converse of the Basic Proportionality Theorem (also called Thales’ theorem). If a line cuts two sides of a triangle proportionally, the line must be parallel to the third side. The proof uses similarity of triangles.
1. Let the triangle be \(\triangle ABC\) and let a line intersect \(AB\) at \(D\) and \(AC\) at \(E\) such that
$$\frac{AD}{DB}=\frac{AE}{EC}=k \quad (k>0).$$
2. Express the ratios with respect to the whole sides:
\[\frac{AD}{AB}=\frac{AD}{AD+DB}=\frac{k}{k+1},\qquad \frac{AE}{AC}=\frac{AE}{AE+EC}=\frac{k}{k+1}.\]
Hence
$$\frac{AD}{AB}=\frac{AE}{AC} \tag{1}$$
3. Consider triangles \(\triangle ADE\) and \(\triangle ABC\).
- They share the angle \(\angle A\).
- From (1) we have the proportion of the sides surrounding \(\angle A\):
$$\frac{AD}{AB}=\frac{AE}{AC}.$$
By the SAS (Side‑Angle‑Side) similarity criterion, the two triangles are similar:
$$\triangle ADE \sim \triangle ABC.$$
4. Corresponding angles are equal. From the similarity we get
$$\angle ADE = \angle ABC \quad\text{and}\quad \angle AED = \angle ACB.$$
5. Parallelism. Since a pair of interior angles are equal, the lines forming those angles are parallel. Thus
$$DE \parallel BC.$$
6. Conclusion. The line that divides the two sides \(AB\) and \(AC\) in the same ratio is indeed parallel to the third side \(BC\).
Correct Answer: The line is parallel to the third side; i.e., \(DE \parallel BC\).