Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
Related questions
Question
![**Problem Statement:**
32) In triangle ABC, D is a point on AB and E is a point on AC such that \( \overline{DE} \parallel \overline{BC} \). If \(AD = 2\), \(DB = x - 1\), \(AE = x\), and \(EC = x + 2\), find \(AE\).
**Solution Explanation:**
Since \( \overline{DE} \parallel \overline{BC} \), by the Basic Proportionality Theorem (or Thales's theorem), we have:
\[\frac{AD}{DB} = \frac{AE}{EC}\]
Given:
- \( AD = 2 \)
- \( DB = x - 1 \)
- \( AE = x \)
- \( EC = x + 2 \)
Substitute the given values into the proportion:
\[\frac{2}{x - 1} = \frac{x}{x + 2}\]
Cross-multiplying to solve for \(x\):
\[2(x + 2) = x(x - 1)\]
\[2x + 4 = x^2 - x\]
Rearrange into a standard quadratic equation form:
\[x^2 - 3x - 4 = 0\]
Factorizing the quadratic equation:
\[(x - 4)(x + 1) = 0\]
So, \(x\) can be either \(4\) or \(-1\).
Since lengths cannot be negative, \(x = 4\).
Therefore, the length of \(AE\) is:
\[AE = x = 4\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3e7bbe56-6e0f-4e60-be65-bd82ac2b3970%2Fdea59efb-2f8d-4507-811d-435c27634211%2Fobgvvg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
32) In triangle ABC, D is a point on AB and E is a point on AC such that \( \overline{DE} \parallel \overline{BC} \). If \(AD = 2\), \(DB = x - 1\), \(AE = x\), and \(EC = x + 2\), find \(AE\).
**Solution Explanation:**
Since \( \overline{DE} \parallel \overline{BC} \), by the Basic Proportionality Theorem (or Thales's theorem), we have:
\[\frac{AD}{DB} = \frac{AE}{EC}\]
Given:
- \( AD = 2 \)
- \( DB = x - 1 \)
- \( AE = x \)
- \( EC = x + 2 \)
Substitute the given values into the proportion:
\[\frac{2}{x - 1} = \frac{x}{x + 2}\]
Cross-multiplying to solve for \(x\):
\[2(x + 2) = x(x - 1)\]
\[2x + 4 = x^2 - x\]
Rearrange into a standard quadratic equation form:
\[x^2 - 3x - 4 = 0\]
Factorizing the quadratic equation:
\[(x - 4)(x + 1) = 0\]
So, \(x\) can be either \(4\) or \(-1\).
Since lengths cannot be negative, \(x = 4\).
Therefore, the length of \(AE\) is:
\[AE = x = 4\]
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