This problem is about proving the relationship between the segments in a triangle when a line intersects the sides and is parallel to the third side. Here's a breakdown of the proof:

Problem:

Given a triangle $\triangle ABC$, a line intersects sides $AB$ and $AC$ at points $D$ and $E$ respectively, and this line is parallel to side $BC$. We need to prove that: $\frac{AD}{AB} = \frac{AE}{AC}$ This is related to the Basic Proportionality Theorem (also called Thales' Theorem).

Solution:

1. Given: The line $DE \parallel BC$, and it intersects $AB$ at $D$ and $AC$ at $E$.

2. By the Basic Proportionality Theorem (Theorem 6.1 as mentioned): Since $DE \parallel BC$, it follows that: $\frac{AD}{DB} = \frac{AE}{EC}$

3. Rearrange this equation to express it differently: $\frac{DB}{AD} = \frac{EC}{AE}$

4. Adding 1 on both sides: $\frac{DB}{AD} + 1 = \frac{EC}{AE} + 1$ This simplifies to: $\frac{AB}{AD} = \frac{AC}{AE}$ (since $AB = AD + DB$ and $AC = AE + EC$).

5. Taking the reciprocal on both sides gives the desired result: $\frac{AD}{AB} = \frac{AE}{AC}$ This completes the proof.

This is a direct application of the Basic Proportionality Theorem, where the line parallel to one side of a triangle divides the other two sides in the same ratio.

Would you like a more detailed explanation of any step?

Here are 5 related questions to deepen your understanding:

1. How is the Basic Proportionality Theorem applied in different triangle configurations?
2. Can this theorem be extended to other polygons or non-Euclidean geometry?
3. What happens if the line is not parallel to side $BC$?
4. How can this theorem help in solving real-life geometric problems?
5. How would the proof change if instead of proving $\frac{AD}{AB} = \frac{AE}{AC}$, we had to find the length of one segment given others?

Tip:

The Basic Proportionality Theorem is fundamental for many geometric constructions and proofs, especially in coordinate geometry and similarity theorems.