Plane Geometry - Bruce E. Shapiro
Plane Geometry - Bruce E. Shapiro Plane Geometry - Bruce E. Shapiro
58 SECTION 13. INCIDENCE GEOMETRYFigure 13.5: Illustration of five point geometry: the points are representedby the symbols A, B, C, D.E and the lines are represented by any subset ofprecisely two points. Since lines {B, D} and {B, C} are both parallel to line{A, E} and both pass through point B, this model satisfies the hyperbolicparallel postulatehypothesis must be wrong, i.e., P = Q. Hence the point P is unique.Here are some other useful results that hold in incidence geometry.Theorem 13.10 Let l be a line. Then there exists at least one point Pthat does not lie on l.Theorem 13.11 Let P be a point. Then there exist at least two distinctlines that contain P .Proof. By Incidence Axiom 3 there exist at least three points R, S, T thatare non-collinear.Either P ∈ {R, S, T } or P ∉ {R, S, T }.If P ∈ {R, S, T } then without loss of generality we can relabel the points sothat P = R. Define l = ←→ P S and m = ←→ P T . Since P, S, T are non-collinear,then l and m are the desired lines.If P ∉ {R, S, T } then either P ∈ ←→ RS or P ∉ ←→ RS.If P ∈ ←→ RS then let l = ←→ RS and m = ←→ P T . By construction, T ∉ RS, elseR, S, T would be collinear. Since T ∈ m and T ∉ l, l ≠ m. Hence l and mare the two distinct lines that contain P .If P ∉ ←→ RS then let l = ←→ P R and m = ←→ P S. By construction R ∈ l, but« CC BY-NC-ND 3.0. Revised: 18 Nov 2012
SECTION 13. INCIDENCE GEOMETRY 59R ∉ m, else P, R, S would be collinear, and we have assumed otherwise.Hence l and m are distinct lines that contain P .Theorem 13.12 Let l be a line. Then there exist two distinct lines m andn that intersect l.Proof. Let l be a line.By incidence axiom 2 there are two points P, Q thatlie on l.By theorem 13.10 there exists a third point R that does lie on l.By incidence axiom 1, there exist lines m = ←→ P R and n = ←→ QR.Since P ∈ l and P ∈ m, l intersects m (definition of intersection).Since Q ∈ l and Q ∈ n, l intersects n (definition of intersection).Since R ∉ l then any line that contains R is different from l. The two linesm and n contain R, hence m ≠ l and n ≠ l.Suppose m = n. By definition of m, P ∈ m. By defintion of n, Q ∈ n.Hence if m = n, Q ∈ m, i.e., both P and Q line on m. Hence m = l bythe uniqueness part of incidence axiom 1. This contradicts the previousparagraph, m ≠ l. Hence the assumption m = n must be wrong.Hence m ≠ n, which means there are two distinc lines that intersect l.Theorem 13.13 Let P be a point. Then there exists at least one line thatdoes not contain P .Theorem 13.14 There exist three distinct lines such that no point lies onall three of them.Theorem 13.15 Let P be a point. Then there exist points Q and R suchthat P, Q, R are non-collinear.Proof. Let P and Q be points such that Q ≠ P . By incidence axiom 1,there is a unique line l that contains P and Q.By theorem 13.10 there exists at least one point R that does lie on l. Thepoints P, Q, R are non-collinear.Theorem 13.16 Let P ≠ Q be points. Then there exists a point Q suchthat P, Q, R are non-collinear.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.
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58 SECTION 13. INCIDENCE GEOMETRYFigure 13.5: Illustration of five point geometry: the points are representedby the symbols A, B, C, D.E and the lines are represented by any subset ofprecisely two points. Since lines {B, D} and {B, C} are both parallel to line{A, E} and both pass through point B, this model satisfies the hyperbolicparallel postulatehypothesis must be wrong, i.e., P = Q. Hence the point P is unique.Here are some other useful results that hold in incidence geometry.Theorem 13.10 Let l be a line. Then there exists at least one point Pthat does not lie on l.Theorem 13.11 Let P be a point. Then there exist at least two distinctlines that contain P .Proof. By Incidence Axiom 3 there exist at least three points R, S, T thatare non-collinear.Either P ∈ {R, S, T } or P ∉ {R, S, T }.If P ∈ {R, S, T } then without loss of generality we can relabel the points sothat P = R. Define l = ←→ P S and m = ←→ P T . Since P, S, T are non-collinear,then l and m are the desired lines.If P ∉ {R, S, T } then either P ∈ ←→ RS or P ∉ ←→ RS.If P ∈ ←→ RS then let l = ←→ RS and m = ←→ P T . By construction, T ∉ RS, elseR, S, T would be collinear. Since T ∈ m and T ∉ l, l ≠ m. Hence l and mare the two distinct lines that contain P .If P ∉ ←→ RS then let l = ←→ P R and m = ←→ P S. By construction R ∈ l, but« CC BY-NC-ND 3.0. Revised: 18 Nov 2012