Dand E are the midpoints of the sides of A ABC. Find the following:

Elementary Geometry For College Students, 7e
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ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
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ChapterP: Preliminary Concepts
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### Midpoints in Triangles: Problem Set

**Diagram Explanation:**
The diagram shows a triangle \( \triangle ABC \) with points \( D \) and \( E \) marked on sides \( AB \) and \( AC \) respectively. Points \( D \) and \( E \) are midpoints of these sides, dividing \( AB \) and \( AC \) into two equal segments. The line segment \( DE \) runs parallel to \( BC \).

**Problems:**

1. **Given:**
   - \( BE = 2x \)
   - \( EC = x + 3 \)
   
   Find \( BC \).

2. **Given:**
   - \( DE = 3x + 5 \)
   - \( AC = 4x - 10 \)
   
   Find \( AC \).

3. **Given:**
   - \( AD = -6x + 14 \)
   - \( AB = 18x + 10 \)
   
   Find \( DB \).

4. **Given:**
   - \( BC = 7x \)
   - \( EC = 4x - 9 \)
   
   Find \( BE \). 

---

**Tips for Solving:**

- Use the property that \( DE \) is parallel to \( BC \) and \( D \) and \( E \) are midpoints to set up proportional relationships.
- Remember that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long.

---

**Example Solution (a):**

1. \( BE = 2x \)
2. \( EC = x + 3 \)
   
   Since \( E \) is the midpoint of \( AC \), \( AE \) is equal to \( EC \). Therefore, \( EC \) represents half of \( AC \).

   Substitute \( EC \) into the equation for \( EC \):
   \[
   x + 3 = \frac{1}{2}AC
   \]

   To find \( BC \), consider the proportional relationship due to \( D \) and \( E \) being midpoints, giving \( BC \):
   \[
   BC = 2BE
   \]

   Hence,
   \[
   BC = 2(2x) = 4x
   \]

Include these examples and tips as references for solving similar problems
Transcribed Image Text:### Midpoints in Triangles: Problem Set **Diagram Explanation:** The diagram shows a triangle \( \triangle ABC \) with points \( D \) and \( E \) marked on sides \( AB \) and \( AC \) respectively. Points \( D \) and \( E \) are midpoints of these sides, dividing \( AB \) and \( AC \) into two equal segments. The line segment \( DE \) runs parallel to \( BC \). **Problems:** 1. **Given:** - \( BE = 2x \) - \( EC = x + 3 \) Find \( BC \). 2. **Given:** - \( DE = 3x + 5 \) - \( AC = 4x - 10 \) Find \( AC \). 3. **Given:** - \( AD = -6x + 14 \) - \( AB = 18x + 10 \) Find \( DB \). 4. **Given:** - \( BC = 7x \) - \( EC = 4x - 9 \) Find \( BE \). --- **Tips for Solving:** - Use the property that \( DE \) is parallel to \( BC \) and \( D \) and \( E \) are midpoints to set up proportional relationships. - Remember that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long. --- **Example Solution (a):** 1. \( BE = 2x \) 2. \( EC = x + 3 \) Since \( E \) is the midpoint of \( AC \), \( AE \) is equal to \( EC \). Therefore, \( EC \) represents half of \( AC \). Substitute \( EC \) into the equation for \( EC \): \[ x + 3 = \frac{1}{2}AC \] To find \( BC \), consider the proportional relationship due to \( D \) and \( E \) being midpoints, giving \( BC \): \[ BC = 2BE \] Hence, \[ BC = 2(2x) = 4x \] Include these examples and tips as references for solving similar problems
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