## Geometry Proof Exercise **Given:** \( \angle 1 = \angle 6 \) and \( CF = DE \). Prove that \( \triangle FOC \equiv \triangle EOD \) and hence that \( CO = DO \) and \( FO = EO \). **Diagram Explanation:** The diagram displays two intersecting triangles, \( \triangle FOC \) and \( \triangle EOD \), which share a common vertex \( O \). The lines are labeled with congruent and supplementary angles. **Proof:** ### Statements and Reasons | Statements | Reasons | |-----------------------------|-------------------------------------------------------------------------| | 1. \( \angle 1 = \angle 6 \), \( CF = DE \) | Given | | 2. \( \angle 1 + \angle FCD = \) straight angle \( \angle 6 + \angle EDC = \) straight angle | The sum of all the angles about a point on one side of a straight line is a straight angle. | | 3. \( \angle 1 + \angle FCD = \angle 6 + \angle EDC \) | ? | | 4. \( \angle FCD = \angle EDC \) | ? | | 5. \( CD = DC \) | ? | | 6. \( \triangle FCD \equiv \triangle EDC \) | ? | | 7. \( \angle 3 = \angle 4 \) | ? | **Note:** The question marks in the reasons column represent the steps in the proof that need justification based on geometric theorems or properties such as congruent parts of congruent triangles (CPCTC), properties of supplementary angles, etc. Each proof step usually requires a known theorem or postulate to justify the equivalence or equality. ### Explanation of Diagram: - Triangles \( FOC \) and \( EOD \) are depicted with a unique intersection at point \( O \), signifying a connection that assists in proving congruence. - Important angles and line segments that are used in the proof are labeled to facilitate understanding of congruence relations and equivalent subdivisions of the larger geometric setup. This exercise demonstrates a structured approach to proving congruency between two triangles and understanding subsequent implications for line segments derived from congruent triangles.

## Geometry Proof Exercise Given: \( \angle 1 = \angle 6 \) and \( CF = DE \). Prove that \( \triangle FOC \equiv \triangle EOD \) and hence that \( CO = DO \) and \( FO = EO \). Diagram Explanation: The diagram displays two intersecting triangles, \( \triangle FOC \) and \( \triangle EOD \), which share a common vertex \( O \). The lines are labeled with congruent and supplementary angles. Proof: ### Statements and Reasons | Statements | Reasons | |-----------------------------|-------------------------------------------------------------------------| | 1. \( \angle 1 = \angle 6 \), \( CF = DE \) | Given | | 2. \( \angle 1 + \angle FCD = \) straight angle \( \angle 6 + \angle EDC = \) straight angle | The sum of all the angles about a point on one side of a straight line is a straight angle. | | 3. \( \angle 1 + \angle FCD = \angle 6 + \angle EDC \) | ? | | 4. \( \angle FCD = \angle EDC \) | ? | | 5. \( CD = DC \) | ? | | 6. \( \triangle FCD \equiv \triangle EDC \) | ? | | 7. \( \angle 3 = \angle 4 \) | ? | Note: The question marks in the reasons column represent the steps in the proof that need justification based on geometric theorems or properties such as congruent parts of congruent triangles (CPCTC), properties of supplementary angles, etc. Each proof step usually requires a known theorem or postulate to justify the equivalence or equality. ### Explanation of Diagram: - Triangles \( FOC \) and \( EOD \) are depicted with a unique intersection at point \( O \), signifying a connection that assists in proving congruence. - Important angles and line segments that are used in the proof are labeled to facilitate understanding of congruence relations and equivalent subdivisions of the larger geometric setup. This exercise demonstrates a structured approach to proving congruency between two triangles and understanding subsequent implications for line segments derived from congruent triangles.

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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