Question

Prove that 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.

Hint:

While proving theorems, always provide a neat diagram and clearly label "Given", "To Prove", and "Construction".

Answer 1

Step 1: Understanding the Concept:
This is the Basic Proportionality Theorem (Thales Theorem). We prove it using the areas of triangles with common heights.
Step 2: Detailed Explanation:
Given: In \( \triangle ABC \), a line \( DE \parallel BC \) intersects \( AB \) at \( D \) and \( AC \) at \( E \).
To Prove: \( \frac{AD}{DB} = \frac{AE}{EC} \).
Construction: Join \( BE \) and \( CD \). Draw \( DM \perp AC \) and \( EN \perp AB \).
Proof:
Area(\( \triangle ADE \)) = \( \frac{1}{2} \times \text{base } AD \times \text{height } EN \).
Area(\( \triangle BDE \)) = \( \frac{1}{2} \times \text{base } DB \times \text{height } EN \).
Therefore, \( \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{AD}{DB} \) (1)
Similarly:
Area(\( \triangle ADE \)) = \( \frac{1}{2} \times \text{base } AE \times \text{height } DM \).
Area(\( \triangle CED \)) = \( \frac{1}{2} \times \text{base } EC \times \text{height } DM \).
Therefore, \( \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CED)} = \frac{AE}{EC} \) (2)
Now, \( \triangle BDE \) and \( \triangle CED \) are on the same base \( DE \) and between the same parallels \( DE \parallel BC \).
So, Area(\( \triangle BDE \)) = Area(\( \triangle CED \)) (3)
From (1), (2), and (3), we get:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]
Step 3: Final Answer:
The ratio of the segments is equal. Hence Proved.