Question

Prove that, if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

• A. Proof Completed
• B. Statement is False
• C. Proof is Incomplete
• D. Cannot be Determined

Hint:

Be clear in your construction description. Draw the perpendicular heights \(DM\) and \(EN\) carefully on your diagram.
This theorem is highly fundamental and serves as the proof foundation for many other properties of triangles in geometry.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks us to state and prove the Basic Proportionality Theorem (also known as Thales' Theorem).

Step 2: Key Formula or Approach:
We will draw a triangle \(ABC\) with a line \(DE\) parallel to side \(BC\) such that \(D\) lies on \(AB\) and \(E\) lies on \(AC\).
We need to prove:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]
We will use the properties of areas of triangles sharing the same height or base.

Step 3: Detailed Explanation:
(----------paste image here--------------)
1.

Given: In \(\Delta ABC\), \(DE \parallel BC\) where \(D\) is on \(AB\) and \(E\) is on \(AC\).
2.

To Prove:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]
3.

Construction:
- Join \(BE\) and \(CD\).
- Draw \(DM \perp AC\) and \(EN \perp AB\).
4.

Proof:
- Consider the area of \(\Delta ADE\) taking \(AD\) as the base. The height is \(EN\):
\[ \text{Area}(\Delta ADE) = \frac{1}{2} \times AD \times EN \quad \text{--- (Equation 1)} \]
- Consider the area of \(\Delta BDE\) taking \(BD\) as the base. The height is the same altitude \(EN\):
\[ \text{Area}(\Delta BDE) = \frac{1}{2} \times BD \times EN \quad \text{--- (Equation 2)} \]
- Divide Equation 1 by Equation 2:
\[ \frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta BDE)} = \frac{AD}{BD} \quad \text{--- (Equation 3)} \]
- Now, consider the area of \(\Delta ADE\) taking \(AE\) as the base. The height is \(DM\):
\[ \text{Area}(\Delta ADE) = \frac{1}{2} \times AE \times DM \quad \text{--- (Equation 4)} \]
- Consider the area of \(\Delta CDE\) taking \(EC\) as the base. The height is the same altitude \(DM\):
\[ \text{Area}(\Delta CDE) = \frac{1}{2} \times EC \times DM \quad \text{--- (Equation 5)} \]
- Divide Equation 4 by Equation 5:
\[ \frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta CDE)} = \frac{AE}{EC} \quad \text{--- (Equation 6)} \]
5.

Relate the denominators:
Note that \(\Delta BDE\) and \(\Delta CDE\) lie on the same base \(DE\) and between the same parallel lines \(DE\) and \(BC\).
According to geometry theorems, triangles on the same base and between the same parallel lines are equal in area:
\[ \text{Area}(\Delta BDE) = \text{Area}(\Delta CDE) \]
6. Since the denominators of the LHS of Equation 3 and Equation 6 are equal, and their numerators are identical, we have:
\[ \frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta BDE)} = \frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta CDE)} \]
Therefore, their RHS values must also be equal:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]
The theorem is successfully proven.

Step 4: Final Answer:
The Basic Proportionality Theorem is proven by comparing the areas of the triangles. Thus, the proof is completed.