Assignment 31 - Copy - that are equidistant from the end...

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

View Full Document Right Arrow Icon
Assignment 31 You aren’t responsible for 6.6! 2a) BAC b) same c) same side d) same side e) opposite sides 1) a) suppose the triangle is isosceles b) Suppose there are two lines which contain the given point and perpendicular to the given line c) Suppose the point P does not lie on the perpendicular bisector of a seg. d) Suppose two coplanar lines which are perpendicular to the same line intersect at a given point. e) Suppose, in the given plane, there are two lines perpendicular to the given line at the given point of that line. f) Suppose 2 is a rational number g) Suppose zero has a reciprocal. 2) The perpendicular bisector of a segment is the line which is perpendicular to the segment at its midpoint. 3) The perpendicular bisector of a segment, in a plane, is the set of all points of the plane
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 ≠ ....
View Full Document

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.

Ask a homework question - tutors are online