推荐一本关于几何公理的在线书籍

hbghlyj

推荐一本网络书《几何概论》
web.mnstate.edu/peil/geometry/index.htm，里面有丰富的文本和图片，章节大纲，GeoGebra 探索，以及作者本人讲授的 mp4 讲座视频录像的链接

hbghlyj

Theorem 3.27. Every isometry of a Euclidean plane is the composition of at most three reflections. (For a dynamic diagram, GeoGebra or JavaSketchpad.)

Proof.  Let f be an isometry of a Euclidean plane, and let X, Y, and Z be any three noncollinear points. Further, let X' = f(X), Y' = f(Y), and Z' = f(Z). By Corollary 3.10, we only need to find a composition of three reflections that maps X, Y, and Z to X', Y', and Z', respectively. If X and X' are distinct, then by Theorem 3.23 there is a unique reflection R_l that maps X to X'. Let Y_l = R_l(Y) and Z_l = R_l(Z).  Similarly, if Y_l and Y' are distinct, there is a unique reflection R_m that maps Y_l to Y'. Since f and R_l are isometries, X'Y' = XY = X'Y_l. Hence, X' is on line m the perpendicular bisector of  and X' = R_m(X'). Thus, the composition  maps X to X', Y to Y', and Z to some point Z_lm. As before, if Z_lm and Z' are distinct, there is a unique reflection R_n that maps Z_lm to Z'. Since f, R_l, and R_m are isometries, X'Z' = XZ = X'Z_l = X' Z_lm and Y'Z' = YZ = Y_lZ_l = Y' Z_lm. Hence, X'  and Y' are on line n the perpendicular bisector of , X' = R_n(X') and Y' = R_n(Y'). Thus, the composition  maps X to X', Y to Y', and Z to Z'. If X = X', Y_l = Y', or Z_lm = Z', then omit R_l, R_m, or R_n, respectively, from the composition. By Theorem 3.20, the identity transformation is the composition of a reflection with itself. Therefore, every isometry of a Euclidean plane is the composition of no more than three reflections.//

hbghlyj

 In the introduction to projective geometry, we stated that in a later section we would consider a mapping between two pencils of points. We begin by showing that there exists a projectivity between two pencils of points. (The first video is the lecture and the second video is the construction in Geometer's Sketchpad.)
        Assume A, B, C are elements of a pencil with axis p and A', B', C' are elements of a pencil with axis p'.  Further, assume the points are distinct and the axes p and p' are distinct. We desire to find a projectivity so that ABC is projectively related to A'B'C', i.e. . Since a projectivity is a composition of perspectivities, we construct two perspectivities to map ABC to A'B'C'.

        To construct the first perspectivity, we define the center of a perspectivity that will map A to A' and C to itself with the image of B to be found. Let P be a point on AA' that is distinct from A and A'. (How do we know point P exists?) Let B₁ = BP · A'C. Thus .
     For the second perspectivity, we define the center of a perspectivity that maps A' to itself, B₁ to B', and C to C'. Let Q = B₁B' · CC'. Then . Since  and , we have . We have proven the following theorem.

 

Theorem 4.10. If A, B, C and A', B', C' are distinct elements in pencils of points with distinct axes p and p', respectively, then there exists a projectivity such that ABC is projectively related to A'B'C'.

 

        Further, the theorem and its constructive proof give a procedure to determine a corresponding point D' on axis p' by following the perspectivities when a fourth point D on axis p is given. That is, let D be an element of axis p. First, find D₁ on the pencil of points with A', B₁, and C by mapping D through center P; that is, let D₁ = DP · A'C. Next, map D₁ to D' by mapping D₁ through the center Q to p', i.e., D' = D₁ Q · p'. Then  and . Hence, .

 

Exercise 4.30.  Show that P in the constructive proof of Theorem 4.10 exists.

 

Exercise 4.31. (a)  State and prove the dual of Theorem 4.10. (Hint: principle of duality)   (b)  Assume a, b, c are elements of a pencil with center P and a', b', c' are elements of a pencil with center P'. Use dynamic geometry software to construct  and find a corresponding line d' of a line d.  (Hint: write the dual of the steps of the above construction for points.)

 

        In the above theorem, we have shown a projectivity exists between two pencils of points with three elements. But, is the projectivity unique? Since an arbitrary point was chosen in the construction, a different point would give different perspectivities. That is, would a different point  be determined, if the perspectivities were different? Click here to investigate the uniqueness of the perspectivity GeoGebra or JavaSketchpad constructed in the proof of the previous theorem. Also, to help answer these questions consider Axiom 6.

 

Axiom 6. If a projectivity on a pencil of points leaves three distinct points of the pencil invariant, it leaves every point of the pencil invariant.

 

      Axiom 6 implies that a projectivity on a pencil that leaves three elements of the pencil invariant is the identity mapping. What implications does this axiom have for distinct pencils of points? What implications does this axiom have for a projectivity on a pencil of points where no group of three points are mapped to themselves? Can this axiom extend the above theorem for constructing a projectivity between two pencils of points to more than three points? All of these questions are answered by the Fundamental Theorem of Projective Geometry, which has the surprising result that only three pairs of points are needed to determine a unique projectivity between two pencils of points.

 

Theorem 4.11. (Fundamental Theorem of Projective Geometry)  A projectivity between two pencils of points is uniquely determined by three pairs of corresponding points.

 

In other words, if A, B, C, D are in a pencil of points with axis p and A', B', C' are in a pencil of points with axis p', then there exists a unique point D'  on p' such that .

 

Proof.  Assume A, B, C, D are in a pencil of points with axis p and A', B', C' are in a pencil of points with axis p'. We have shown that there exists a point D' on p' such that . Suppose there is a projectivity and a point D" such that . Since  and , we have . Therefore, by Axiom 6, D' = D". //

hbghlyj

        Here, we will formalize the definitions and results from the investigation on the previous page.

Definition. Given two distinct pencils of points with axes p and p', the points A, B  on p and corresponding points A', B' on p' determine lines AB' and BA' called cross joins.

Your exploration of the problem should have led to conjecturing the following theorem known as the Theorem of Pappus after Pappus of Alexandria (290-350).

Theorem 4.15. (Theorem of Pappus) A projectivity between two distinct pencils of points determines a line, which contains the intersection of the cross joins of any two pairs of corresponding points.

Proof.  Let p and p'  be the axes of two distinct projectively related  pencils of points. Assume P = p · p' and  such that P is none of the six points A, B, C, A', B', C'. Note the line AA' and the cross joins from A and A' determine two elementary correspondences  and . Since ,  and , we have . Since line AA' maps to itself through the projectivity, the dual of Corollary 4.12 implies that there exists a line h such that this projectivity is a perspectivity with axis h, , and does not depend on the choice of B and C. Hence, the intersection of the cross joins containing A and A', such as A'B · AB' and A'C · AC',  are on h.
        We need to show the line h is unique and contains the intersections of all pairs of cross joins . Since the above derivation of h used the line AA', we need to show that h is independent of this choice. That is, we need to show that the choice of BB', CC', or some other corresponding pair of points would determine the same line h.
        Assume h_x is determined from XX' where X and X' are corresponding points from the projectivity that are distinct from P. Since P is common to both pencils p and p', let Q be the pre-image of P, i.e., Q' = P, and let R' be the image of P under the projectivity, i.e., P = R. Thus,  X'Q · XQ' = X'Q · XP = Q and X'R · XR' = X'P · XR' = R'. Hence, Q and R' are on line h_x. If Q and R' are distinct, then h_x = QR' for any choice of XX'. If Q and R' are not distinct, then Q = R' = P. Let S = X'Y · XY' which is on lines determined from XX' and YY'. Hence, h_x = PS = h_y; that is, the determined line is the same line for any choice of XX' and YY'. By the Fundamental Theorem, the image and pre-image of P are unique; therefore, by both cases, h is uniquely determined by Q and R'.
        The projectivity determines a unique line h that contains the intersection of the cross joins of any two pairs of corresponding points.//

Definition. The line containing the intersections of the cross joins between the two pencils of points is called the axis of homology. The point containing the intersection of the lines determined by the cross joins between two pencils of lines is called the center of homology.

Exercise 4.39. Construct the axis of homology for two projectively related pencils of points. Then use the construction to determine an image of an arbitrary point.  (May use dynamic geometry software.)

Exercise 4.40. Construct the center of homology for two projectively related pencils of lines. Then use the construction to determine an image of an arbitrary line.  (May use dynamic geometry software.)