{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Pre-Calculus Homework Solutions 123

# Pre-Calculus Homework Solutions 123 - CEB 6 SAS 7 A w E...

This preview shows page 1. Sign up to view the full content.

125 Chapter 5 29. m ( ( y x 2 2 y x 1 1 ) ) 7 0 3 3 or 4 3 30. No, RX is not an altitude of RST . The slope of ST is 1. The product of the slopes of ST and RX is 4 3 , not 1. Thus, the segments are not perpendicular. 31. Given: CA CB , AD BD Prove: C and D are on the perpendicular bisector of AB . Proof: 32. Given: BAC , P is in the interior of BAC , PD PE Prove: AP is the angle bisector of BAC Proof: A B D P E C C D E B A y x ( 1, 6) (3, 3) R S (1, 8) T y x ( 1, 6) (3, 3) R S (1, 8) T Statements Reasons 1. BAC , P is in the interior of BAC , PD PE 1. Given 2. PD PE 2. Def. of 3. PD AB , PE AC 3. Distance from a point to a line is measured along segment from the point to the line. 4. ADP and AEP are rt. 4. Def. of 5. ADP and AEP are rt. s 5. Def. of rt. 6. AP AP 6. Reflexive Property 7. ADP AEP 7. HL 8. DAP EAP 8. CPCTC 9. AP is the angle bisector of BAC 9. Def. of bisector Statements Reasons 1. CA CB , AD BD 1. Given 2. CD CD 2. Reflexive Property 3. ACD
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CEB 6. SAS 7. A w E w > B w E w 7. CPCTC 8. E is the midpoint of A w B w . 8. Def. of midpoint 9. / CEA > / CEB 9. CPCTC 10. / CEA and / CEB form a linear pair. 10. Def. of linear pair 11. / CEA and / CEB are supplementary. 11. Supplement Theorem 12. m / CEA 1 m / CEB 5 180 12. Def. of suppl. ? 13. m / CEA 1 m / CEA 5 180 13. Substitution 14. 2( m / CEA ) 5 180 14. Substitution 15. m / CEA 5 90 15. Division Property 16. / CEA and / CEB are rt. ? . 16. Def. of rt. / 17. C w D w ' A w B w 17. Def. of ' 18. C w D w is the perpendicular bisector of A w B w . 18. Def. of ' bisector 19. C and D are on the perpendicular bisector of A w B w . 19. Def. of points on a line...
View Full Document

{[ snackBarMessage ]}