![]() ![]() The easy way to construct a perpendicular bisector #PQ# to segment #AB# is pictured below. Here the centers of these circles are the endpoints of a given segment #AB# and their radiuses must be the same. The only condition is for these circles is the existence two points of intersection, #P# and #Q#. For this the radius can be any, as long as it's greater than half of the length of #AB#. The simple method is to choose it to be equal to the length of #AB#. ![]() Wednesday 9/30/98 Topic: Similar Triangles and Pythagorean Theorem Criteria for similarity of triangles will be used to prove several.Monday, 9/28/98 Topic: Congruent Triangles SAS, SSS and ASA criteria for triangle congruence.So, we have proven that #M# is a modpoint of #AB# and #PM_|_AB#.Math 444 Syllabus and Topic Outline Math 444 Syllabus and Topic Outline #=>/_AMP=/_BMP# as angles of congruent triangles lying across congruent sides #AP# and #BP#=> AM=BM# as sides of congruent triangles lying across congruent angles #/_APM=/_BPM#=> Delta APM = Delta BPM# by side-angle-side theorem #=>/_APQ=/_BPQ# as angles of congruent triangles lying across congruent sides #AQ# and #BQ#Delta APQ = Delta BPQ# - by side-side-side theorem #AP=BP=AQ=BQ# - each is a radius, which we have chosen What's more interesting is to prove that this construction delivers the perpendicular bisector.Īssume that #M# is an intersection of #AB# and #PQ#. Math 487 Lab #1 Wednesday 9/28/98 Introduction to Geometer's Sketchpad Students will be introduced to the Sketchpad software by workingĬhapter 2 of GTC will also be used, so bringĮmail List Everyone should be on the email list.Results, including the Pythagorean Theorem. Getting email from the list this week you are on otherwise, ANSWERS TO GSP5 CONSTRUCTING PERPENDICULAR BISECTORS SOFTWARE
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