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Unformatted text preview: that are equidistant from the end points of the segment. 4) no 5)a) F b) T c) F d) F 6) Theorem 6-4 7)1) AD=AB and CD=CB (given) b) AC is the perpendicular bisector of DB (corollary 6-2.1) 3) Let M be the intersection of AC and DB (A and C lie on opposite sides of a line DB . 4) MD=MB (M lies on AC and DB ) 5) AC bisects DB ( def. of bisect for segments) 9) The two vertices are each equidistant from the end points of the base, so the line they determine is perpendicular to the base by corollary 6-2.1 10) if AB AC ≠ then the median of AM were perpendicular to BC we would have MB=CM, AM=AM and ∠ AMB AMC ∠ 2245 . Therefore AMC AMB ∆ 2245 ∆ by SAS, so AB=AC contradicting the hypothesis that AB AC ≠ ....
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This note was uploaded on 05/16/2011 for the course ALGEBRA 098 taught by Professor Johnson during the Spring '09 term at Grand Valley State University.
- Spring '09