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# CONCURRENCE

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ADDITIONAL REMARKS ON CONCURRENCE In an equilateral triangle the centroid circumcenter orthocenter and incenter all coincide The reason for this is simple If we are given an equilateral triangle ABC and the midpoints of BC AC and AB are given by D E and F respectively then i the ray AD is also the angle bisector for 6 BAC and the line AD is also the altitude from A to BC and the perpendicular bisector of BC all because d A B d A C ii similarly for the rays BE and CF as well as the lines BE and CF However in general the centroid circumcenter orthocenter and incenter of a triangle are all distinct and in fact one can prove that if two of these points coincide then the triangle is equilateral The proof splits into six cases corresponding to the following hypotheses 1 2 3 4 5 6 The The The The The The centroid and orthocenter coincide centroid and incenter coincide centroid and circumcenter coincide incenter and orthocenter coincide circumcenter and incenter coincide circumcenter and orthocenter coincide Preliminaries Before proving the theorem stated above we shall establish some auxiliary results that we shall need PROPOSITION If L is a line x is a positive real number and X is a point not on L then there are at most two points on L whose distance from X is equal to x Proof Suppose that B C D are three points on L such that d X B d X C d X D x Relabeling the points if necessary we may assume that B C D holds Let E be the midpoint of BC and let F be the midpoint of CD Since X is equidistant from B C D it follows that XE is the perpendicular bisector of BC and XF is the perpendicular bisector of CD However we know that there is only one perpendicular from X to L so this is a contradiction The source of this contradiction is our assumption that there are three points on L which are equidistant from X and therefore we conclude that there are at most two such points The second result analyzes pairs of triangles which satisfy SSA as noted in Section II 4 there is no general

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