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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 64 7)1) AD=AB and CD=CB (given) b) AC is the perpendicular bisector of DB (corollary 62.1) 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 62.1 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
 Johnson
 Algebra

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