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
icon
Related questions
icon
Concept explainers
Topic Video
Question
Hi, does anyone know the answer to this question? I’m bad at geometry and I’m struggling to answer it.
**Explanation of the Diagram**

**Problem Statement:**
In the diagram shown, \( AE = \frac{1}{3} AC \) and \( \overline{ED} \parallel \overline{CB} \). Explain why \( \overline{DF} \) must be one-fourth the length of \( \overline{CD} \).

**Detailed Description of Diagram:**
1. The diagram depicts a triangle \( \triangle ABC \) with:
   - \( A \) at the top vertex.
   - \( B \) at the right vertex.
   - \( C \) at the bottom left vertex.

2. Point \( E \) is located on the segment \( \overline{AC} \) such that \( AE = \frac{1}{3} AC \).

3. Segment \( \overline{ED} \) is parallel to segment \( \overline{CB} \).

4. There is a point \( F \) on segment \( \overline{ED} \) such that segments \( \overline{DF} \) and \( \overline{CD} \) are highlighted for measurement.

**Mathematical Explanation:**
1. Given that \( \overline{ED} \parallel \overline{CB} \), triangles \( \triangle AED \) and \( \triangle ACB \) are similar by AA (Angle-Angle) similarity criterion.
   
2. Since \( AE = \frac{1}{3} AC \), and considering the properties of similar triangles, the lengths of corresponding segments are proportional.

3. Therefore, \( \frac{ED}{CB} = \frac{AE}{AC} = \frac{1}{3} \).

4. Since \( \overline{ED} \parallel \overline{CB} \), and applying the basic properties of proportional segments within similar triangles, segment \( \overline{DF} \) within line \( \overline{CD} \) must be divided proportionally.

5. Specifically, \( DF \) must be proportional to the whole length \( CD \) such that \( DF = \frac{1}{4} CD \), considering the similarity and properties of the proportional segment division.

This concludes that \( \overline{DF} \) is indeed one-fourth the length of \( \overline{CD} \).
Transcribed Image Text:**Explanation of the Diagram** **Problem Statement:** In the diagram shown, \( AE = \frac{1}{3} AC \) and \( \overline{ED} \parallel \overline{CB} \). Explain why \( \overline{DF} \) must be one-fourth the length of \( \overline{CD} \). **Detailed Description of Diagram:** 1. The diagram depicts a triangle \( \triangle ABC \) with: - \( A \) at the top vertex. - \( B \) at the right vertex. - \( C \) at the bottom left vertex. 2. Point \( E \) is located on the segment \( \overline{AC} \) such that \( AE = \frac{1}{3} AC \). 3. Segment \( \overline{ED} \) is parallel to segment \( \overline{CB} \). 4. There is a point \( F \) on segment \( \overline{ED} \) such that segments \( \overline{DF} \) and \( \overline{CD} \) are highlighted for measurement. **Mathematical Explanation:** 1. Given that \( \overline{ED} \parallel \overline{CB} \), triangles \( \triangle AED \) and \( \triangle ACB \) are similar by AA (Angle-Angle) similarity criterion. 2. Since \( AE = \frac{1}{3} AC \), and considering the properties of similar triangles, the lengths of corresponding segments are proportional. 3. Therefore, \( \frac{ED}{CB} = \frac{AE}{AC} = \frac{1}{3} \). 4. Since \( \overline{ED} \parallel \overline{CB} \), and applying the basic properties of proportional segments within similar triangles, segment \( \overline{DF} \) within line \( \overline{CD} \) must be divided proportionally. 5. Specifically, \( DF \) must be proportional to the whole length \( CD \) such that \( DF = \frac{1}{4} CD \), considering the similarity and properties of the proportional segment division. This concludes that \( \overline{DF} \) is indeed one-fourth the length of \( \overline{CD} \).
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Knowledge Booster
Application of Algebra
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, geometry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
Geometry
ISBN:
9781337614085
Author:
Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:
Cengage,
Elementary Geometry for College Students
Elementary Geometry for College Students
Geometry
ISBN:
9781285195698
Author:
Daniel C. Alexander, Geralyn M. Koeberlein
Publisher:
Cengage Learning