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Complete the proof. m B. Given: AB = BC m is the I bisector of AB. n is the I bisector of CD. Prove: AP = DQ Statements Reasons 1. Given 1. AB = BC, m is the L bisector of AB, and n is the I bisector of CD. 2. а. 2. AB = CD 3. Definition of 3. АР - -AB bisector 4. b. 4. DQ = CD 2 5. AP = DQ 5. Substitution 6. C. 6. d.

Complete the proof. m B. Given: AB = BC m is the I bisector of AB. n is the I bisector of CD. Prove: AP = DQ Statements Reasons 1. Given 1. AB = BC, m is the L bisector of AB, and n is the I bisector of CD. 2. а. 2. AB = CD 3. Definition of 3. АР - -AB bisector 4. b. 4. DQ = CD 2 5. AP = DQ 5. Substitution 6. C. 6. d.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.3: Lines
Problem 8E
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Question
# Geometric Proof ## Complete the Proof ### Diagram Explanation: The diagram consists of two sets of intersecting lines. In the first set, line \( m \) is the perpendicular bisector of segment \( \overline{AB} \) at point \( P \). In the second set, line \( n \) is the perpendicular bisector of segment \( \overline{CD} \) at point \( Q \). ### Given: - \( \overline{AB} \cong \overline{BC} \) - \( m \) is the perpendicular bisector of \( \overline{AB} \). - \( n \) is the perpendicular bisector of \( \overline{CD} \). ### Prove: - \( \overline{AP} \cong \overline{DQ} \) ### Proof Steps: | **Statements** | **Reasons** | |---------------------------------------------------------------|---------------------------| | 1. \( \overline{AB} \cong \overline{BC}, m \) is the perpendicular bisector of \( \overline{AB}, \) and \( n \) is the perpendicular bisector of \( \overline{CD} \). | 1. Given | | 2. \( AB = CD \) | 2. a. | | 3. \( AP = \frac{1}{2} AB \) | 3. Definition of bisector | | 4. \( DQ = \frac{1}{2} CD \) | 4. b. | | 5. \( AP = DQ \) | 5. Substitution | | 6. c. | 6. d. |
Transcribed Image Text:# Geometric Proof ## Complete the Proof ### Diagram Explanation: The diagram consists of two sets of intersecting lines. In the first set, line \( m \) is the perpendicular bisector of segment \( \overline{AB} \) at point \( P \). In the second set, line \( n \) is the perpendicular bisector of segment \( \overline{CD} \) at point \( Q \). ### Given: - \( \overline{AB} \cong \overline{BC} \) - \( m \) is the perpendicular bisector of \( \overline{AB} \). - \( n \) is the perpendicular bisector of \( \overline{CD} \). ### Prove: - \( \overline{AP} \cong \overline{DQ} \) ### Proof Steps: | **Statements** | **Reasons** | |---------------------------------------------------------------|---------------------------| | 1. \( \overline{AB} \cong \overline{BC}, m \) is the perpendicular bisector of \( \overline{AB}, \) and \( n \) is the perpendicular bisector of \( \overline{CD} \). | 1. Given | | 2. \( AB = CD \) | 2. a. | | 3. \( AP = \frac{1}{2} AB \) | 3. Definition of bisector | | 4. \( DQ = \frac{1}{2} CD \) | 4. b. | | 5. \( AP = DQ \) | 5. Substitution | | 6. c. | 6. d. |
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