{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Pre-Calculus Homework Solutions 121

# Pre-Calculus Homework Solutions 121 - 10 Given CD is the...

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

123 Chapter 5 10. Given: CD is the perpendicular bisector of AB . E is a point on CD . Prove: EB EA Proof: CD is the perpendicular bisector of AB . By definition of perpendicular bisector, D is the midpoint of AB . Thus, AD BD by the Midpoint Theorem. CDA and CDB are right angles by definition of perpendicular. Since all right angles are congruent, CDA CDB . Since E is a point on CD , EDA and EDB are right angles and are congruent. By the Reflexive Property, ED ED . Thus, EDA EDB by SAS. EB EA because CPCTC, and by definition of congruence, EB EA . 11. Given: UVW is isosceles with vertex angle UVW . YV is the bisector of UVW . Prove: YV is a median. Proof: 12. Given: GL is a median of EGH . JM is a median of IJK . EGH IJK Prove: GL JM Proof: 13. MS is an altitude of MNQ , so MSQ is a right angle. m 1 m 2 ¬ 90 3 x 11 7 x 9 ¬ 90 10 x 20 ¬ 90 10 x ¬ 70 x ¬ 7 m 2 ¬ 7(7) 9 ¬ 58 14. MS is a median of MNQ , so QS SN . 3 a 14 ¬ 2 a 1 a 14 ¬ 1 a ¬ 15 m MSQ ¬ 7 a 1 ¬ 7(15) 1 ¬ 106 MS is not an altitude of MNQ because m MSQ 106. I K M J E H L G U Y W V A D C E B Statements Reasons 1. UVW
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online