Question

If a line intersects sides \( AB \) and \( AC \) of a triangle \( ABC \) at \( D \) and \( E \) respectively and is parallel to \( BC \), prove that \( \frac{AD}{AB} = \frac{AE}{AC} \).

Hint:

In a triangle, if a line is parallel to one side, it divides the other two sides in the same ratio. This is known as the Basic Proportionality Theorem.

Answer 1

We are given a triangle \( ABC \) and a line passing through points \( D \) and \( E \) on sides \( AB \) and \( AC \), respectively, such that the line is parallel to \( BC \). By the Basic Proportionality Theorem (also known as Thales' Theorem), if a line is parallel to one side of a triangle and intersects the other two sides, then it divides those two sides in the same ratio. Therefore, we have: \[ \frac{AD}{AB} = \frac{AE}{AC}. \] Thus, the required result is proved.
Conclusion: \[ \frac{AD}{AB} = \frac{AE}{AC}. \]