Plane Geometry - Bruce E. Shapiro
Plane Geometry - Bruce E. Shapiro Plane Geometry - Bruce E. Shapiro
130 SECTION 28. CHARACTERIZATION OF BISECTORSgles. Therefore, d(P, ←→ AB) = P D = P E = d(P, ←→ AC).( ⇐)Let P be a point in the interior of ∠BAC, and assume that d(P, ←→ AB) =d(P, ←→ AC).Drop perpendiculars and call the feet D and E as before. Hence P D = P E.We also have ∠P EA = ∠P DA = 90, and both triangles △AP D and△AP E share a common side AP . Hence by the hypoteneuse-leg theoremthey are congruent.Hence ∠P AE = ∠P AD, which means P is on the angle bisector of ∠BAC.Figure 28.2: Theorem 28.2.Theorem 28.2 (Pointwise Characterization of Perpendicular Bisector)Let A and B be distinct points. A point P lies on the perpendicularbisector of AB if and only if P A = P B.Proof. (⇒) Let P be a point on the perpendicular bisector of AB (figure28.2).Let M be the intersection of the bisector and AB.Then AM = BM by definition of bisector and ∠AMP = ∠BMP = 90.Since P M ∼ = P M, by SAS △AMP ∼ = △BMP .Hence P A = P B.(⇐) Let P be a point that satisfies P A = P B. Then △P AB is isoscelesand by the isosceles triangle theorem ∠P AB = ∠P BA.Drop a perpendicular from P to AB and call the foot M. Then∠P AM = ∠P AB = ∠P BA = ∠P BM« CC BY-NC-ND 3.0. Revised: 18 Nov 2012
SECTION 28. CHARACTERIZATION OF BISECTORS 131Since △P AM and △P BM share a side, are right triangles, and have congruenthypotenuses, they are congruent by the Hypotenuse Leg Theorem.Hence AM = BM, meaning that ←−→ P M is a bisector of AB. Since by construction←−→ P M ⊥ AB it is a perpendicular bisector.Figure 28.3: Theorem 28.3.xyf(x)f(y)Theorem 28.3 (Continuity of Distance) Let A, B, and C be noncollinearand let d = AB. For each x ∈ [0, d], there exists a unique pointD x ∈ AB such that AD x = x. Define the function f : [0, d] ↦→ [0, ∞) byf(x) = CD x . Then the function f is continuous.Proof. Let A, B, C, f be as defined in the statement of theorem 28.3 (seefigure 28.3), and let ɛ > 0 be given. Choose δ = ɛ.Suppose y ∈ [0, d] such thatBy the triangle inequalityD x D y = |x − y| < δ = ɛCD x < CD y + D x D y < CD y + ɛandHenceCD y < CD x + D x D y < CD x + ɛCD x − ɛ < CD y < CD x + ɛf(x) − ɛ < f(y) < f(x) + ɛ−ɛ < f(y) − f(x) < ɛ|f(y) − f(x)| < ɛRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.
- Page 84 and 85: 80 SECTION 16. ANGLESFigure 16.1: T
- Page 86 and 87: 82 SECTION 16. ANGLESCorollary 16.6
- Page 88 and 89: 84 SECTION 16. ANGLESFigure 16.6: I
- Page 90 and 91: 86 SECTION 16. ANGLESD cannot be in
- Page 92 and 93: 88 SECTION 17. THE CROSSBAR THEOREM
- Page 94 and 95: 90 SECTION 17. THE CROSSBAR THEOREM
- Page 96 and 97: 92 SECTION 18. LINEAR PAIRSFigure 1
- Page 98 and 99: 94 SECTION 18. LINEAR PAIRSγ + ∠
- Page 100 and 101: 96 SECTION 19. ANGLE BISECTORSwith
- Page 102 and 103: 98 SECTION 19. ANGLE BISECTORSAngle
- Page 104 and 105: 100 SECTION 20. THE CONTINUITY AXIO
- Page 106 and 107: 102 SECTION 20. THE CONTINUITY AXIO
- Page 108 and 109: 104 SECTION 21. SIDE-ANGLE-SIDEBirk
- Page 110 and 111: 106 SECTION 21. SIDE-ANGLE-SIDEFigu
- Page 112 and 113: 108 SECTION 22. NEUTRAL GEOMETRYEll
- Page 114 and 115: 110 SECTION 22. NEUTRAL GEOMETRY«
- Page 116 and 117: 112 SECTION 23. ANGLE-SIDE-ANGLEFig
- Page 118 and 119: 114 SECTION 23. ANGLE-SIDE-ANGLE«
- Page 120 and 121: 116 SECTION 24. EXTERIOR ANGLESTheo
- Page 122 and 123: 118 SECTION 24. EXTERIOR ANGLES« C
- Page 124 and 125: 120 SECTION 25. ANGLE-ANGLE-SIDEBy
- Page 126 and 127: 122 SECTION 26. SIDE-SIDE-SIDECase
- Page 128 and 129: 124 SECTION 26. SIDE-SIDE-SIDEBy si
- Page 130 and 131: 126 SECTION 27. SCALENE AND TRIANGL
- Page 132 and 133: 128 SECTION 27. SCALENE AND TRIANGL
- Page 136 and 137: 132 SECTION 28. CHARACTERIZATION OF
- Page 138 and 139: 134 SECTION 29. TRANSVERSALSFigure
- Page 140 and 141: 136 SECTION 29. TRANSVERSALSCorolla
- Page 142 and 143: 138 SECTION 30. TRIANGLES IN NEUTRA
- Page 144 and 145: 140 SECTION 30. TRIANGLES IN NEUTRA
- Page 146 and 147: 142 SECTION 30. TRIANGLES IN NEUTRA
- Page 148 and 149: 144 SECTION 30. TRIANGLES IN NEUTRA
- Page 150 and 151: 146 SECTION 31. QUADRILATERALS IN N
- Page 152 and 153: 148 SECTION 31. QUADRILATERALS IN N
- Page 154 and 155: 150 SECTION 31. QUADRILATERALS IN N
- Page 156 and 157: 152 SECTION 31. QUADRILATERALS IN N
- Page 158 and 159: 154 SECTION 31. QUADRILATERALS IN N
- Page 160 and 161: 156 SECTION 32. THE EUCLIDEAN PARAL
- Page 162 and 163: 158 SECTION 32. THE EUCLIDEAN PARAL
- Page 164 and 165: 160 SECTION 32. THE EUCLIDEAN PARAL
- Page 166 and 167: 162 SECTION 32. THE EUCLIDEAN PARAL
- Page 168 and 169: 164 SECTION 32. THE EUCLIDEAN PARAL
- Page 170 and 171: 166 SECTION 32. THE EUCLIDEAN PARAL
- Page 172 and 173: 168 SECTION 33. RECTANGLESLemma 33.
- Page 174 and 175: 170 SECTION 33. RECTANGLESFigure 33
- Page 176 and 177: 172 SECTION 33. RECTANGLES« CC BY-
- Page 178 and 179: 174 SECTION 34. THE PARALLEL PROJEC
- Page 180 and 181: 176 SECTION 34. THE PARALLEL PROJEC
- Page 182 and 183: 178 SECTION 35. SIMILAR TRIANGLESBy
SECTION 28. CHARACTERIZATION OF BISECTORS 131Since △P AM and △P BM share a side, are right triangles, and have congruenthypotenuses, they are congruent by the Hypotenuse Leg Theorem.Hence AM = BM, meaning that ←−→ P M is a bisector of AB. Since by construction←−→ P M ⊥ AB it is a perpendicular bisector.Figure 28.3: Theorem 28.3.xyf(x)f(y)Theorem 28.3 (Continuity of Distance) Let A, B, and C be noncollinearand let d = AB. For each x ∈ [0, d], there exists a unique pointD x ∈ AB such that AD x = x. Define the function f : [0, d] ↦→ [0, ∞) byf(x) = CD x . Then the function f is continuous.Proof. Let A, B, C, f be as defined in the statement of theorem 28.3 (seefigure 28.3), and let ɛ > 0 be given. Choose δ = ɛ.Suppose y ∈ [0, d] such thatBy the triangle inequalityD x D y = |x − y| < δ = ɛCD x < CD y + D x D y < CD y + ɛandHenceCD y < CD x + D x D y < CD x + ɛCD x − ɛ < CD y < CD x + ɛf(x) − ɛ < f(y) < f(x) + ɛ−ɛ < f(y) − f(x) < ɛ|f(y) − f(x)| < ɛRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.
Change language