Question

In the proof of BPT (\(DE \parallel BC\) in \(\triangle ABC\)), since \(\triangle BDE\) and \(\triangle CDE\) have equal areas, the ratio AD/DB is shown to be equal to ________.

• A. DB/AD
• B. AE/EC
• C. EC/AE
• D. AD/AE

Hint:

The statement of BPT is simply: "The ratio of the upper-left segment to the lower-left segment equals the ratio of the upper-right segment to the lower-right segment."
So, \(\frac{AD}{DB} = \frac{AE}{EC}\) is the core result of BPT.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question is a continuation of the theoretical understanding of the Basic Proportionality Theorem (BPT).
We need to identify the final proportional equality established at the end of the BPT proof.

Step 2: Key Formula or Approach:
According to BPT, a line parallel to one side of a triangle divides the other two sides in the same ratio.
Mathematically, for a triangle \(\triangle ABC\) where \(DE \parallel BC\):
\[ \frac{AD}{DB} = \frac{AE}{EC} \]

Step 3: Detailed Explanation:
In the proof of BPT, we establish two primary ratios of triangle areas:
First ratio:
\[ \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \]
Second ratio:
\[ \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \]
Since \(\triangle BDE\) and \(\triangle CDE\) are on the same base \(DE\) and between the same parallel lines \(DE\) and \(BC\), their areas are equal:
\[ \text{Area}(\triangle BDE) = \text{Area}(\triangle CDE) \]
Since their denominators are equal and their numerators are identical (\(\text{Area}(\triangle ADE)\)), the two area ratios are equal:
\[ \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} \]
Substituting the simplified fractions into this equality gives:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]

Step 4: Final Answer:
The ratio AD/DB is shown to be equal to AE/EC.