Provide a proof for the following situation. Given: DB || EC and ZD & ZE are right angles. Prove : Δ ADBΔ BEC E 12 B C 5.

Elementary Geometry For College Students, 7e
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Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
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Provide a proof for the following situation. Please help! 

**Problem Statement:**

Provide a proof for the following situation.

**Given:** \( \overline{DB} \parallel \overline{EC} \) and \( \angle D \) & \( \angle E \) are right angles.

**Prove:** \( \triangle ADB \sim \triangle BEC \)

**Diagram Explanation:**

The diagram shows two right triangles: \(\triangle ADB\) and \(\triangle BEC\). 

- Triangle \(\triangle ADB\) is a smaller triangle with:
  - \( \overline{AD} = 4 \)
  - \( \overline{DB} = 5 \)
  - \( \angle ADB\) is a right angle at \( \angle D \).

- Triangle \(\triangle BEC\) is a larger triangle with:
  - \( \overline{BE} = 12 \)
  - \( \overline{EC} \) is parallel to \( \overline{DB} \)
  - \( \angle BEC \) is a right angle at \( \angle E \).

**Proof of Similarity:**

Triangles are considered similar if their corresponding angles are equal and their corresponding sides are proportional.

1. **Right Angles:**
   - Given \( \angle D \) and \( \angle E \) are right angles, \( \triangle ADB \) and \( \triangle BEC \) both have right angles at \(D\) and at \(E\) respectively.

2. **Parallel Lines:**
   - Given \( \overline{DB} \parallel \overline{EC} \), we know that \( \angle ADB \) and \( \angle BEC \) create corresponding angles because they are cut by the transversal line \( \overline{AB} \) or \( \overline{BC} \).

3. **Angle-Angle (AA) Similarity:**
   - Since \( \overline{DB} \parallel \overline{EC} \), we can infer that \( \angle BAD \) in \( \triangle ADB \) is congruent to \( \angle CBE \) in \( \triangle BEC \). 
   - Therefore, \( \angle ADB = \angle BEC = 90^\circ \).

Since both corresponding angles in \(\triangle ADB\) and \(\triangle B
Transcribed Image Text:**Problem Statement:** Provide a proof for the following situation. **Given:** \( \overline{DB} \parallel \overline{EC} \) and \( \angle D \) & \( \angle E \) are right angles. **Prove:** \( \triangle ADB \sim \triangle BEC \) **Diagram Explanation:** The diagram shows two right triangles: \(\triangle ADB\) and \(\triangle BEC\). - Triangle \(\triangle ADB\) is a smaller triangle with: - \( \overline{AD} = 4 \) - \( \overline{DB} = 5 \) - \( \angle ADB\) is a right angle at \( \angle D \). - Triangle \(\triangle BEC\) is a larger triangle with: - \( \overline{BE} = 12 \) - \( \overline{EC} \) is parallel to \( \overline{DB} \) - \( \angle BEC \) is a right angle at \( \angle E \). **Proof of Similarity:** Triangles are considered similar if their corresponding angles are equal and their corresponding sides are proportional. 1. **Right Angles:** - Given \( \angle D \) and \( \angle E \) are right angles, \( \triangle ADB \) and \( \triangle BEC \) both have right angles at \(D\) and at \(E\) respectively. 2. **Parallel Lines:** - Given \( \overline{DB} \parallel \overline{EC} \), we know that \( \angle ADB \) and \( \angle BEC \) create corresponding angles because they are cut by the transversal line \( \overline{AB} \) or \( \overline{BC} \). 3. **Angle-Angle (AA) Similarity:** - Since \( \overline{DB} \parallel \overline{EC} \), we can infer that \( \angle BAD \) in \( \triangle ADB \) is congruent to \( \angle CBE \) in \( \triangle BEC \). - Therefore, \( \angle ADB = \angle BEC = 90^\circ \). Since both corresponding angles in \(\triangle ADB\) and \(\triangle B
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