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32) In triangle ABC, D is a point on AB and E is a point on AC such that DE || BC. If AD = 2, DB = x - 1, AE = x, and EC = x + 2, find AE.

32) In triangle ABC, D is a point on AB and E is a point on AC such that DE || BC. If AD = 2, DB = x - 1, AE = x, and EC = x + 2, find AE.

Elementary Geometry For College Students, 7e
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
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**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: \[AE = x = 4\]
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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