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This book has grown out of that teaching experience. I assume only high-school geometry and some abstract algebra. The course begins in Chapter 1 with a critical examination of Euclid's Elements. Students are expected to read concurrently Books I-IV of Euclid's text, which must be obtained sepa rately. The remainder of the book is an exploration of questions that arise natu rally from this reading, together with their modern answers. To shore up the foundations we use Hilbert's axioms. The Cartesian plane over a field provides an analytic model of the theory, and conversely, we see that one can introduce coordinates into an abstract geometry. The theory of area is analyzed by cutting figures into triangles. The algebra of field extensions provides a method for deciding which geometrical constructions are possible. The investigation of the parallel postulate leads to the various non-Euclidean geometries. And in the last chapter we provide what is missing from Euclid's treatment of the five Platonic solids in Book XIII of the Elements. 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Euclidean Planes Chapter 3. Geometry over Fields 1 ་ 7 8 18 27 45 51 65 66 73 81 90 96 104 112 117 13. The Real Cartesian Plane 14. Abstract Fields and Incidence 15. Ordered Fields and Betweenness\u0026nbsp;...","page_url":"https://books.google.com.sg/books?id=EJCSL9S6la0C\u0026pg=PR9\u0026vq=intersection"},{"page_id":"PA19","page_number":"19","snippet_text":"... \u003cb\u003eintersections\u003c/b\u003e along the way , and ending with the desired figure . We will distinguish the construction , which is a series of applications of the ruler and compass to create a certain figure , from the proof that the figure con\u0026nbsp;..."},{"page_id":"PA21","page_number":"21","snippet_text":"... \u003cb\u003eintersections\u003c/b\u003e do not count as separate steps . Thus for example , the construction of the equilateral triangle ( 1.1 ) above takes four steps : The line segment AB is given 1. Draw circle with center A and radius AB . 2. Draw circle\u0026nbsp;..."},{"page_id":"PA29","page_number":"29","snippet_text":"... \u003cb\u003eIntersections\u003c/b\u003e of Circles and Lines As we read Euclid\u0026#39;s Elements let us note how well he succeeds in his goal of proving all his propositions by pure logical reasoning from first principles . We will find at times that he relies on\u0026nbsp;..."},{"page_id":"PA30","page_number":"30","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e point actually exist ? Today we will im- mediately think of continuity and the intermediate value ... \u003cb\u003eintersect\u003c/b\u003e the x - axis at some point in the interval . y y = f ( x ) 0 1 x However , we must bear in mind that the\u0026nbsp;..."},{"page_id":"PA31","page_number":"31","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e should exist in the statement of the proposition , namely that any two of the line segments should be greater than the third . However , he never alludes to this hypothesis in his proof , so that we do not see in what way\u0026nbsp;..."},{"page_id":"PA39","page_number":"39","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e of the lines m and n , and draw a line l \u0026#39; through P , making the alternate angle 3 equal to 1. This is possible by ( 1.23 ) , which belongs to neutral geometry . Then by ( I.27 ) , which also belongs to neutral geometry\u0026nbsp;..."},{"page_id":"PA49","page_number":"49","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e of MN with DO . 13. Circle MP , get Q. 14. Circle NP , get R. 15. Circle PR . 16. Circle RP , get S. 17. Circle QP , get T. 18. Line NS , get U. 19. Line MT , get V. Then D , M , N , U , V will be the vertices of the\u0026nbsp;..."},{"page_id":"PA61","page_number":"61","snippet_text":"... common point G. Hint : Let G be the \u003cb\u003eintersection\u003c/b\u003e of the first two circles , then show that CEGF is a cyclic quadrilateral ( due to A. Miquel , 1838 ) . A D G B E C F 5.15 ( Pappus\u0026#39;s theorem ) . Let A , B. 5. Some Newer Results 61."},{"page_id":"PA66","page_number":"66","snippet_text":"... \u003cb\u003eintersections\u003c/b\u003e . The points and lines are undefined objects . We simply postulate a set , whose ele- ments are called points , together with certain subsets , which we call lines . We do not say what the points are , nor which subsets\u0026nbsp;..."},{"page_id":"PA74","page_number":"74","snippet_text":"... \u003cb\u003eintersect\u003c/b\u003e 1 . ( b ) Two points A , C not on I belong to the opposite sets ( one in S1 , the other in S2 ) if and only if the segment AC in- tersects 1 in a point . We will refer to the sets S1 , S2 as the two sides of 1 , and we will\u0026nbsp;..."},{"page_id":"PA75","page_number":"75","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e point would be the unique point in which the line AE meets 1 , namely D. In that case D would be between A and E. But we con- structed E so that D * A * E , so by ( B3 ) , D cannot lie between A and E. Thus AENIØ , so A E\u0026nbsp;..."},{"page_id":"PA76","page_number":"76","snippet_text":"... S2 , we define S1 and S2 to be the \u003cb\u003eintersections\u003c/b\u003e of S and S with 1. Then properties ( a ) and ( b ) follow immediately from the previous proposition . D A B C l E A l The only mildly nontrivial part is to show that S1. 76 2. Hilbert\u0026#39;s\u0026nbsp;..."},{"page_id":"PA80","page_number":"80","snippet_text":"... \u003cb\u003eintersections\u003c/b\u003e except at their endpoints . Then the union of these four segments is a simple closed quadrilateral . The segments AC and BD are the diag- onals of the quadrilateral . There are two cases to consider . Case 1 AC and BD meet\u0026nbsp;..."},{"page_id":"PA93","page_number":"93","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e point , we may assume that B , D , C lie on a line and BDC . On the other hand , re- placing B \u0026#39; , C \u0026#39; , D \u0026#39; by other points on the same rays , we may assume that AB≈ A\u0026#39;B \u0026#39; , and AC A\u0026#39;C \u0026#39; , and AD A\u0026#39;D \u0026#39; . We also have LBAD\u0026nbsp;..."},{"page_id":"PA97","page_number":"97","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e of lines and circles , so as to recover all of Euclid\u0026#39;s Elements , but it seems worthwhile to pause at this point and see how much of the geometry we can develop with this minimal set of axioms . The main reason for doing\u0026nbsp;..."},{"page_id":"PA102","page_number":"102","snippet_text":"... \u003cb\u003eintersect\u003c/b\u003e when they ought to , we cannot prove the existence of the triangle re- quired in this proposition . In fact , we will see later ( Exercise 16.11 ) that there are Hilbert planes in which a triangle with certain given sides\u0026nbsp;..."},{"page_id":"PA103","page_number":"103","snippet_text":"... Hilbert plane , prove that there exists a line 1 for which all the points are on the same side of 1 . A 4 . B C B \u0026#39; C \u0026#39; C D A B Αν • A , • A2 An 11 \u003cb\u003eIntersections\u003c/b\u003e of Lines and Circles In this section we. 10. Hilbert Planes 103."},{"page_id":"PA104","page_number":"104","snippet_text":"... \u003cb\u003eintersections\u003c/b\u003e of lines and circles in the Hilbert plane , and we will introduce the further axiom ( E ) , which will guarantee that lines and circles will \u003cb\u003eintersect\u003c/b\u003e when they \u0026quot; ought \u0026quot; to . With this axiom we can justify Euclid\u0026#39;s ruler\u0026nbsp;..."},{"page_id":"PA105","page_number":"105","snippet_text":"... point opposite A. So consider the line from O , perpendicular to 1 , meeting 1 at B. If BA , take a point C on the other side of B from A , so that AB = BC T B A ( axiom ( C1 ) ) . The AOBA AOBC. 11. \u003cb\u003eIntersections\u003c/b\u003e of Lines and Circles 105."},{"page_id":"PA107","page_number":"107","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e points . If D is a third point on l \u0026#39; and г \u0026#39; , then OD ≈ OA and O\u0026#39;D≈ O\u0026#39;A , so by ( I.7 ) , D must be equal to A or B. In the above ... \u003cb\u003eintersection\u003c/b\u003e property ) . Given. 11. \u003cb\u003eIntersections\u003c/b\u003e of Lines and Circles 107."},{"page_id":"PA108","page_number":"108","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e property LCI ) B A A In a Hilbert plane with the extra axiom ( E ) , if a line I contains a point A inside a circle I , then I will meet r ( necessarily in two points , because of ( 11.2 ) and ( 11.3 ) ) . Proof Suppose we\u0026nbsp;..."},{"page_id":"PA110","page_number":"110","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e property is equivalent to the line - circle \u003cb\u003eintersection\u003c/b\u003e property . In an arbitrary Hilbert plane , the equivalence of these two statements follows from the classification theorem of Pejas ( cf. Section 43 ) , but I do not\u0026nbsp;..."},{"page_id":"PA111","page_number":"111","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e property ( 11.6 ) and hence the axiom ( E ) . ( Note that the other popular construction of the tan- gent line using ( III.31 ) requires the parallel axiom ! ) The other results of Book III , up to ( III.19 ) ( except ( III\u0026nbsp;..."},{"page_id":"PA112","page_number":"112","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e property stated as axiom ( E ) . Also , in the process of rewriting the foundations of geometry we have formulated a new notion , the Hilbert plane , which provides a minimum context in which to develop the beginnings of a\u0026nbsp;..."},{"page_id":"PA115","page_number":"115","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e ( E ) , and some , notably ( IV.10 ) , ( IV.11 ) , require ( P ) , ( E ) , and the theory of area . Thus we may regard the construction of the regular pentagon as the crowning result of the first four books of the Elements\u0026nbsp;..."},{"page_id":"PA119","page_number":"119","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e ( 0 , 0 ) the origin . P = ( a , b ) b x ( 0,0 ) a A line in this plane is the subset defined by a linear equation ax + by + c = 0 , with a , b not both zero . Among these are the vertical lines , which can be written as x\u0026nbsp;..."},{"page_id":"PA120","page_number":"120","snippet_text":"... \u003cb\u003eintersect\u003c/b\u003e this with the y - axis , we set x = 0 and obtain y = -ac / b . Now consider the line BC . It has slope -b / c , so the altitude through A will have slope c / b . Its equation becomes y = = ( x − a ) . B = ( 0,6 ) Setting x\u0026nbsp;..."},{"page_id":"PA122","page_number":"122","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e of two lines , we solve two linear equations , which can be done using only field operations . To \u003cb\u003eintersect\u003c/b\u003e a line with a circle , we solve the equations simultaneously , which requires solving a quadratic equation in x\u0026nbsp;..."},{"page_id":"PA130","page_number":"130","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e point E. Proof Since a composition of linear changes of variables is again one , we can proceed one step at a time . First , a change of the form x \u0026#39; = x - a , y \u0026#39; = y - b , will move the origin ( 0,0 ) to the point E = ( a\u0026nbsp;..."},{"page_id":"PA144","page_number":"144","snippet_text":"... \u003cb\u003eintersections\u003c/b\u003e of lines and circles . Recall from Section 11 the circle - circle \u003cb\u003eintersection\u003c/b\u003e property , which we called axiom ( E ) , and the line - circle \u003cb\u003eintersection\u003c/b\u003e property ( LCI ) , which was proved in ( 11.6 ) as a consequence of\u0026nbsp;..."},{"page_id":"PA145","page_number":"145","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e points of the circles have coordinates in F , so they exist in the plane II . We leave to the reader the troublesome verification that if I and I \u0026#39; satisfy the hypothesis of ( E ) , then the square roots we need will be\u0026nbsp;..."},{"page_id":"PA161","page_number":"161","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e of a line of II with Ilo , whenever that \u003cb\u003eintersection\u003c/b\u003e is nonempty . Take betweenness and congruence to have the same meaning as in II . Then Пlo is a Hilbert plane satisfying neither ( A ) nor ( P ) ( Exercise 18.3 ) . In\u0026nbsp;..."},{"page_id":"PA173","page_number":"173","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e of circles for ( 5.8 ) : See Exercise 19.2 . We also need ( 1.32 ) , which uses ( P ) but not ( E ) . Thus these two results hold in a Hilbert plane with ( P ) . We do not need ( E ) , nor did we ever use Archimedes \u0026#39; axiom\u0026nbsp;..."},{"page_id":"PA180","page_number":"180","snippet_text":"... theorem ) Let ABC be any triangle , and let a line I cut the sides of the triangle ( extended if nec- essary ) in points D , E , F. Then AD BF CE = 1 . BD CF AE D 20.4 If two circles \u003cb\u003eintersect\u003c/b\u003e in two points ,. 180 4. Segment Arithmetic."},{"page_id":"PA181","page_number":"181","snippet_text":"Robin Hartshorne. D 20.4 If two circles \u003cb\u003eintersect\u003c/b\u003e in two points , the. Proof Draw a line through A parallel to BC , and let it meet 1 at G. Then the tri- angle ADG is similar to BDF , and the triangle AEG is similar to CEF . From this we\u0026nbsp;..."},{"page_id":"PA182","page_number":"182","snippet_text":"... \u003cb\u003eintersect\u003c/b\u003e , this latter property defines a straight line that is taken to be the radical axis in that case ... \u003cb\u003eintersection\u003c/b\u003e points , show that the fol- lowing construction ( which can be done with Hilbert\u0026#39;s tools ) gives the\u0026nbsp;..."},{"page_id":"PA184","page_number":"184","snippet_text":"... ( \u003cb\u003eintersection\u003c/b\u003e of the altitudes ) of these four triangles are collinear . 20.15 ( Trigonometry ) . In a Hilbert plane with ( P ) , suppose that you are given a right triangle ABC with sides a , b , c and angle a at A. Define B C a a b a\u0026nbsp;..."},{"page_id":"PA186","page_number":"186","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e property ( LCI ) is equivalent to the circle - circle \u003cb\u003eintersection\u003c/b\u003e prop- erty ( E ) , because we have shown that both of these are equivalent to the Eucli- dean condition on the field F , ( 16.2 ) . We do not know any\u0026nbsp;..."},{"page_id":"PA187","page_number":"187","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e point O the origin . On each axis choose a point 1 , and ly such that the segments Ol , and Oly both represent 1 in the field F. These then define the positive rays on the x - axis and the y - axis . Now for any point P in\u0026nbsp;..."},{"page_id":"PA193","page_number":"193","snippet_text":"... by its center A and a point B , and given a line 1 , construct with ruler alone an \u003cb\u003eintersection\u003c/b\u003e point of A and 1 ( par = 54 ) . X B r A l 21.9 Now prove the theorem of Poncelet - Steiner ,. 21. Introduction of Coordinates 193."},{"page_id":"PA197","page_number":"197","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e of any two figures is a figure . The union of any two figures is a figure . The complement of one figure inside another figure ( plus the line seg- ments that form its sides ) is a figure . In particular , any finite union\u0026nbsp;..."},{"page_id":"PA199","page_number":"199","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e has a nonempty interior , it will be a figure ( Exercise 22.1 ) that can be written as a union of triangles Tins = √ Uk . k = 1 ijk . Now let : Ti → T be a rigid motion ( Exercise 17.10 ) taking the triangle T to the\u0026nbsp;..."},{"page_id":"PA200","page_number":"200","snippet_text":"... refine decompositions appropriately . For each i , consider the \u003cb\u003eintersections\u003c/b\u003e TP1 and TP2 . We can write each as unions of triangles ( 22.1 ) TOP1 = U ; Sij , Tin P2 = U ; Sij2 . Use rigid motions : T → T to transport these 200 5. Area."},{"page_id":"PA203","page_number":"203","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e property ( 11.6 ) is used in ( II.11 ) to divide a line such that the square on the larger piece has equal content to the rectangle on the whole and the smaller piece . It is also used in ( II.14 ) to construct a square\u0026nbsp;..."},{"page_id":"PA210","page_number":"210","snippet_text":"... \u003cb\u003eintersections\u003c/b\u003e of the T and T can be further subdivided into triangles Uijk , as in the proof of ( 22.2 ) . Now applying Lemma 23.4 to each Ti = Ujk Uijk and to each T = Ui , k Uijk , we find that Σa ( Ti ) = Σa ( Uijk ) = Σa ( T\u0026nbsp;..."},{"page_id":"PA216","page_number":"216","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e point . ) Now let ABCD be the given rectan- gle , let AE be congruent to the segment c , and let AEFG be the square on side AE . Since the angle in a semicircle is a right angle ( III.31 ) , the angles a , ẞ in the circle\u0026nbsp;..."},{"page_id":"PA247","page_number":"247","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e points of the line with I have coor- dinates in K. l P ( c ) Now give a ruler and compass construction for the \u003cb\u003eintersection\u003c/b\u003e points of a line m with the parabola having focus F and directrix 1. ( Par = 9 to get one point\u0026nbsp;..."},{"page_id":"PA260","page_number":"260","snippet_text":"... \u003cb\u003eintersections\u003c/b\u003e of conics in the plane . New tools were invented , such as the marked ruler that could slide to cut off a given distance between two curves . There was even a classification of problems according to the methods needed for\u0026nbsp;..."},{"page_id":"PA264","page_number":"264","snippet_text":"... \u003cb\u003eintersect\u003c/b\u003e it with lines . Indeed , by rigid motions and similarities , any application of the marked ruler can be reduced to finding an \u003cb\u003eintersection\u003c/b\u003e point of the conchoid with a certain line . As an application of the marked ruler , we\u0026nbsp;..."},{"page_id":"PA267","page_number":"267","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e of the line HI with the circle . First we will show that OQ is parallel to AI . By ( VI.2 ) = ( 20.2 ) , it will be sufficient to show that HQ HO = HI HA Using ( III.36 ) we have HQ HI HB HA . Therefore , . HQ HI НВ . НА\u0026nbsp;..."},{"page_id":"PA276","page_number":"276","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e of the splitting field with the real numbers . Now a Є E , and by the fundamental theorem of Galois theory ( 32.4 ) , the chain of subgroups G ; will give a chain of field extensions . F = En En - 1 S SE1 = E where each\u0026nbsp;..."},{"page_id":"PA278","page_number":"278","snippet_text":"... \u003cb\u003eintersect\u003c/b\u003e it with lines and circles , are precisely equivalent of those one can achieve with compass and marked ruler . 31.8 The problem of Alhazen . Given a cir- cle I , and given two points A , B , find a point Cer such that the lines\u0026nbsp;..."},{"page_id":"PA303","page_number":"303","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e would make the same angle with both of them , so they would be equal . By the same reasoning it is clear that there can be only one parallel to a given line through a given point . If two lines make the same angle with a\u0026nbsp;..."},{"page_id":"PA336","page_number":"336","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e points ) , then y is transformed into itself by circular inversion in T. Conversely , if a circle y contains a single pair A , A \u0026#39; of inverse points , then y is perpendicular to I and is sent into itself . Proof First\u0026nbsp;..."},{"page_id":"PA338","page_number":"338","snippet_text":"... \u003cb\u003eintersect\u003c/b\u003e they have the same angle at both \u003cb\u003eintersections\u003c/b\u003e . For this proof we need to observe that a line and a circle , or two circles , are tangent if and only if they have just one point in common . Hence the property of tangency is\u0026nbsp;..."},{"page_id":"PA344","page_number":"344","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e is A \u0026#39; . ( This works equally well if A is inside г. ) P R A P 3 4 . .0 S Q ท R A \u0026#39; 37.13 ( a ) Given four points A , B , P , Q , if you permute A and B , or if you permute P and Q , the cross - ratio is replaced by its\u0026nbsp;..."},{"page_id":"PA348","page_number":"348","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e point of two circles , and let r be a circle with center O. Perform a circular inversion in the circle г. Then the two circles through O are transformed into straight lines , while the third circle becomes another circle\u0026nbsp;..."},{"page_id":"PA352","page_number":"352","snippet_text":"... \u003cb\u003eintersect\u003c/b\u003e m at B , C ( 1 step ) . Bisect BC ( 3 steps ) using A , D , get E. Circle with center E , ra- dius AB to get F , G ( 1 step ) . Draw FG ( 1 step ) and get H. This is the midline . Make HI = radius of y ( 1 step ) 352 7. Non\u0026nbsp;..."},{"page_id":"PA362","page_number":"362","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e property ( E ) holds in the Poincaré model over a Euclidean ordered field F. Proof Since P - lines and ... \u003cb\u003eintersections\u003c/b\u003e falling outside г. For the next proposition it will be convenient to introduce the notion of a\u0026nbsp;..."},{"page_id":"PA372","page_number":"372","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e property ( E ) holds for circles 7,8 or- thogonal to г. Hint : Write the equations of y , d , and show that the square root needed to find their \u003cb\u003eintersection\u003c/b\u003e exists because of condition ( * d ) . ( b ) Conclude that axiom\u0026nbsp;..."},{"page_id":"PA374","page_number":"374","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e axiom ( E ) nor Archimedes \u0026#39; axiom ( A ) . Instead , we use Hilbert\u0026#39;s axioms of incidence , betweenness , and congruence plus the fol- lowing hyperbolic axiom ( L ) : L. For each line I and each point A not on 1 , there are\u0026nbsp;..."},{"page_id":"PA382","page_number":"382","snippet_text":"... \u003cb\u003eintersections\u003c/b\u003e ( in the diagram V , W ) will have the angles in corresponding posi- tions , so that by ( 1.28 ) the lines BC and AC will be parallel . This contradicts their meeting at the point C. Thus 1 cannot meet any side of the\u0026nbsp;..."},{"page_id":"PA394","page_number":"394","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e of the lines ( a , -a ) and ( b , ∞o ) . However , we have found that 0 calculations seem to work out better if we continue to think of a line as given by coordinates ( u1 , u2 ) , and a point as given by an equation\u0026nbsp;..."},{"page_id":"PA398","page_number":"398","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e of ( V1 , V2 ) with ( 0 , 0 ) . So we compute V1 V2 = a -1b a - 1b + 1 , ( a - 1b - 1-1 ( ba ) ( 1 - ab ) = a - 1b - 1 + 1 ( b + a ) ( 1 + ab ) Observing that u1u2 = v12 and u1u2 \u0026lt; 0 , since b \u0026gt; a \u0026gt; 1 in our situation , and\u0026nbsp;..."},{"page_id":"PA399","page_number":"399","snippet_text":"... one of the other alti- tudes , find the \u003cb\u003eintersection\u003c/b\u003e of this line with ( 0 , ∞ ) , and then show that the third altitude passes through this same point . 0 6 a ī U2 Let AC be ( a1 , a2 ) and let. 41. Hilbert\u0026#39;s Arithmetic of Ends 399."},{"page_id":"PA420","page_number":"420","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e of the line Aẞ with I. Let ( B ) = b and ( y ) = c . Then ẞ and y have Cartesian coordinates B = b2-1 2b 1 + b2\u0026#39;1 + b2 c2-1 2c y = 1+ c2\u0026#39;1 + c2 To express the fact that A , B , y are collinear , we set the slopes of Aẞ and\u0026nbsp;..."},{"page_id":"PA422","page_number":"422","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e property ( E ) holds in a hyperbolic plane . Proof Indeed , the field of ends is Euclidean ( 41.4 ) , the hyperbolic plane is iso- morphic to the Poincaré model over that field , and ( E ) holds in the Poincaré model ( 39.9 )\u0026nbsp;..."},{"page_id":"PA424","page_number":"424","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e point is C , so С є По . E A B C ( b ) Given M a nonzero convex subgroup of G , fix a point O e II , let IIo be the set of points A of II such that the segment class [ OA ] is in M , and take for lines of IIo the nonempty\u0026nbsp;..."},{"page_id":"PA430","page_number":"430","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e of the altitudes of a triangle in neutral geometry . Theorem 43.15 In any Hilbert plane , if two of the altitudes of a triangle meet , then all three meet at the same point . Proof In the triangle ABC , let the altitudes BE\u0026nbsp;..."},{"page_id":"PA443","page_number":"443","snippet_text":"... \u003cb\u003eintersect\u003c/b\u003e the three faces with the unit sphere around V. We obtain a spherical triangle , which we call the vertex figure of the polyhedron at the vertex V. The sides of this spherical triangle are portions of great circles , subtending\u0026nbsp;..."},{"page_id":"PA451","page_number":"451","snippet_text":"... \u003cb\u003eintersect\u003c/b\u003e the faces of the polyhedron with a small sphere around the vertex . This produces a spherical polygon whose inte- rior angles are precisely the dihedral angles of the original polyhedron . We call it the vertex figure at the\u0026nbsp;..."},{"page_id":"PA453","page_number":"453","snippet_text":"... \u003cb\u003eintersect\u003c/b\u003e the polyhedron with a small sphere , and look at the resulting vertex figure . This is a convex spherical polygon , and its vertices inherit markings + or - from the edges , which by construction correspond to the increase or\u0026nbsp;..."},{"page_id":"PA458","page_number":"458","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e of any plane with the sphere . We measure angles between lines and circles by the angle between their tangent lines in 3 - space . We measure length of a line segment by the angle ( in radians ) that it subtends at the\u0026nbsp;..."},{"page_id":"PA483","page_number":"483","snippet_text":"... \u003cb\u003eintersect\u003c/b\u003e , they do not have the same center . 6. If two circles are tangent , they do not have the same center . 10. Two circles can \u003cb\u003eintersect\u003c/b\u003e in at most two points . 11 , 12. If two circles are tangent , their centers lie in a line\u0026nbsp;..."},{"page_id":"PA503","page_number":"503","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e ( Section 11 ) Archimedes \u0026#39; axiom ( Section 12 ) А EADZ L ( C6 ) . Dedekind\u0026#39;s axiom ( Section 12 ) de Zolt\u0026#39;s axiom ( Section 22 ) existence of limiting parallel rays ( Section 40 ) A Hilbert plane ( Section 10 ) is a\u0026nbsp;...","page_url":"https://books.google.com.sg/books?id=EJCSL9S6la0C\u0026pg=PA503\u0026vq=intersection"},{"page_id":"PA509","page_number":"509","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e , 144 configurations in , 132-135 congruence in , 140–148 Desargues\u0026#39;s theorem , 133 ERM in , 151 Hilbert\u0026#39;s axioms in , 148 incidence axioms in , 129 inversion in , 334 is Euclidean , 153 is Hilbert plane , 153 non\u0026nbsp;...","page_url":"https://books.google.com.sg/books?id=EJCSL9S6la0C\u0026pg=PA509\u0026vq=intersection"},{"page_id":"PA510","page_number":"510","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e , 108 , 112 equivalent to ( LCI ) , 110 , 145 , 423 , 431 in Cartesian plane , 144 in hyperbolic plane , 422 in Poincaré model , 362 circle area in hyperbolic plane , 407 , 413 , 414 area of , 221 , 333 circumscribed , 25\u0026nbsp;...","page_url":"https://books.google.com.sg/books?id=EJCSL9S6la0C\u0026pg=PA510\u0026vq=intersection"},{"page_id":"PA512","page_number":"512","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e edge of polyhedron , 438 eigenvector , 472 Einschiebung , 490 Eisenstein\u0026#39;s criterion , 255 , 292 elementary polyhedron , 464 elliptic geometry , 311 enclosing line , 367 , 378 , 384 end of graph , 449 end , 378 , 390\u0026nbsp;...","page_url":"https://books.google.com.sg/books?id=EJCSL9S6la0C\u0026pg=PA512\u0026vq=intersection"},{"page_id":"PA516","page_number":"516","snippet_text":"... 205 of quadrilateral , 80 , 81 of triangle , 77 , 80 , 196 intermediate value theorem , 30 , 247 , 457 interpretation of undefined notions , 355 \u003cb\u003eintersection\u003c/b\u003e of circles , 29 of conics , 260 of figures , 197 , 203 516 Index.","page_url":"https://books.google.com.sg/books?id=EJCSL9S6la0C\u0026pg=PA516\u0026vq=intersection"},{"page_id":"PA517","page_number":"517","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e Legendre\u0026#39;s axiom , 322 , 324 , 431 Legendre , Adrien Marie , 227 , 305 , 319 , 322 , 323 , 439 Leistner , Joseph , 224 , 225 length , 2 , 28 , 42 , 165 absolute in Poincaré model , 366 in field of segment arithmetic , 175\u0026nbsp;...","page_url":"https://books.google.com.sg/books?id=EJCSL9S6la0C\u0026pg=PA517\u0026vq=intersection"},{"page_id":"PA519","page_number":"519","snippet_text":"... \u003cb\u003eintersection\u003c/b\u003e with circle , 247 squaring of , 247 to trisect angle , 277 parallel axiom , 97 false in Poincaré model , 357 in Cartesian plane , 130 parallel lines , 38 by Hilbert\u0026#39;s tools , 103 common orthogonal to , 366 , 377\u0026nbsp;...","page_url":"https://books.google.com.sg/books?id=EJCSL9S6la0C\u0026pg=PA519\u0026vq=intersection"}],"search_query_escaped":"intersection"},{});</script></div></div></div><script>(function() {var href = window.location.href;if (href.indexOf('?') !== -1) {var parameters = href.split('?')[1].split('&');for (var i = 0; i < parameters.length; i++) {var param = parameters[i].split('=');if (param[0] == 'focus') {var elem = document.getElementById(param[1]);if (elem) {elem.focus();}}}}})();</script>