123 Chapter 5 10. Given: C w D w is the perpendicular bisector of A w B w . E is a point on C w D w . Prove: EB 5 EA Proof: C w D w is the perpendicular bisector of A w B w .By definition of perpendicular bisector, D is the midpoint of A w B w .Thus, A w D w > B w D w 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 C w D w , / EDA and / EDB are right angles and are congruent. By the Reflexive Property, E w D w > E w D w .Thus, n EDA > n EDB by SAS. E w B w > E w A w because CPCTC, and by definition of congruence, EB 5 EA . 11. Given: n UVW is isosceles with vertex angle UVW . Y w V w is the bisector of / UVW . Prove: Y w V w is a median. Proof: 12. Given: G w L w is a median of n EGH . J w M w is a median of n IJK . n EGH > n IJK Prove: G w L w > J w M w Proof: 13. M w S w is an altitude of n MNQ ,so / MSQ is a right angle. m
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This note was uploaded on 12/19/2011 for the course MAC 1140 taught by Professor Dr.zhan during the Fall '10 term at UNF.