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Geometry: Euclid and Beyond - Robin Hartshorne - Google Books
<!DOCTYPE html><html><head><title>Geometry: Euclid and Beyond - Robin Hartshorne - Google Books</title><link rel="stylesheet" href="/books/css/_ff77d0f0508c7e0bad470e7d6b7f28fa/kl_viewport_kennedy_full_bundle.css" type="text/css" /><link rel="stylesheet"href="https://fonts.googleapis.com/css2?family=Product+Sans:wght@400"><script src="/books/javascript/v2_ff77d0f0508c7e0bad470e7d6b7f28fa__en.js"></script><script>_OC_Hooks = ["_OC_Page", "_OC_SearchReload", "_OC_TocReload", "_OC_EmptyFunc", "_OC_SearchPage", "_OC_QuotePage" ];for (var _OC_i = 0; _OC_i < _OC_Hooks.length; _OC_i++) {eval("var " + _OC_Hooks[_OC_i] + ";");}function _OC_InitHooks () {for (var i = 0; i < _OC_Hooks.length; i++) {var func = arguments[i];eval( _OC_Hooks[i] + " = func;");}}</script><link rel="canonical" href="https://books.google.com/books/about/Geometry_Euclid_and_Beyond.html?id=EJCSL9S6la0C"/><meta property="og:url" content="https://books.google.com/books/about/Geometry_Euclid_and_Beyond.html?id=EJCSL9S6la0C"/><meta name="title" content="Geometry: Euclid and Beyond"/><meta name="description" content="In recent years, I have been teaching a junior-senior-level course on the classi cal geometries. 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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A Note About Accuracy and Exactness of Constructions When carrying out ruler and compass constructions , we attempt to make our drawings as accurate as possible . Using a\u0026nbsp;..."},{"page_id":"PA36","page_number":"36","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of BC . Let the perpendicular to BC at D meet the angle bisector at A at the point E. Drop perpendiculars EF and EG to the sides of the triangle , and draw BE , CE . The triangles AEF and AEG have the side common and two angles\u0026nbsp;..."},{"page_id":"PA44","page_number":"44","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of DE , and let AF meet BC in G. Prove that G is the \u003cb\u003emidpoint\u003c/b\u003e of BC . ( Hint : Draw some extra lines to make parallelograms , and use ( I.43 ) . ) 3.8 Let I be a circle with center O. Let AB and AC be tangents to l from a point\u0026nbsp;..."},{"page_id":"PA45","page_number":"45","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of two sides of a triangle is parallel to the third side . ( Hint : Draw BE and DC . Show that the triangles BDC and BEC have the same content and then apply ( 1.39 ) . ) B E D C 4 Construction of the Regular Pentagon One of\u0026nbsp;..."},{"page_id":"PA52","page_number":"52","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of BC , then DE ≈ BF . Proof We begin with a slightly different construction . Let D be the \u003cb\u003emidpoint\u003c/b\u003e of AB , and draw lines through D parallel to AC and BC . Let them meet the opposite sides in points E \u0026#39; , F \u0026#39; . Since DE \u0026#39; is\u0026nbsp;..."},{"page_id":"PA53","page_number":"53","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of the opposite side ) of a triangle meet in a single point ( called the centroid of the triangle ) . Proof Let ABC be the triangle , let D , E be the \u003cb\u003emidpoints\u003c/b\u003e of AB and AC , and draw DE . Let the two medians BE and CD meet at\u0026nbsp;..."},{"page_id":"PA54","page_number":"54","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of the opposite side . Proof Follows from the proof of ( 5.4 ) . Proposition 5.6 The three altitudes ( lines through a vertex , perpendicular to the opposite side ) of a triangle meet in a single point ( the orthocenter of the\u0026nbsp;..."},{"page_id":"PA57","page_number":"57","snippet_text":"... \u003cb\u003emidpoints\u003c/b\u003e of the three sides , the feet of the three altitudes , and the \u003cb\u003emidpoints\u003c/b\u003e of the segments joining the three vertices to the orthocenter all lie on a circle . Proof Let ABC be the given triangle . Let D , E , F be the\u0026nbsp;..."},{"page_id":"PA58","page_number":"58","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of DQ . Therefore , P and F also lie on г. Finally , MDPF is a cyclic quadrilateral for the same reasons as above , so M is also on г. Proposition 5.10 ( The orthic triangle ) Let ABC be any acute triangle , and let K , L , M\u0026nbsp;..."},{"page_id":"PA59","page_number":"59","snippet_text":"... \u003cb\u003emidpoints\u003c/b\u003e of the four sides is a parallelogram . 5.6 In any triangle , show that the center X of the nine - point circle lies on the Euler line ( Proposition 5.7 ) , and is the \u003cb\u003emidpoint\u003c/b\u003e of the segment OH joining the circumcenter O to\u0026nbsp;..."},{"page_id":"PA88","page_number":"88","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of AB in the sense that A * E * B and AE ≈ EB , and if F is a \u003cb\u003emidpoint\u003c/b\u003e of CD , then AE = CF. ( Note that we have not yet said anything about the existence of a \u003cb\u003emidpoint\u003c/b\u003e : That will come later ( Section 10 ) . ) Conclude\u0026nbsp;..."},{"page_id":"PA100","page_number":"100","snippet_text":"... rest of Euclid\u0026#39;s proof then works to show that a \u003cb\u003emidpoint\u003c/b\u003e of the segment exists . For ( 1.11 ) we can also use ( 10.2 ) to construct a line perpendicular to a line at a point . By the way , this also proves. 100 2. Hilbert\u0026#39;s Axioms."},{"page_id":"PA101","page_number":"101","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of AC ( 1.10 ) , and extend BE to F so that BEEF ( axiom ( C1 ) ) . Draw the line CF. Now the vertical angles at E are equal ( 1.15 ) , so by SAS ( C6 ) , the tri- angles AABE and ACFE are congruent . Hence LA LECF . ܚܙ\u0026nbsp;..."},{"page_id":"PA102","page_number":"102","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of a given line segment . We used Hilbert\u0026#39;s axioms to prove corresponding existence results in a Hilbert plane . However , we can reinterpret these existence results as constructions if we imagine tools corresponding to certain\u0026nbsp;..."},{"page_id":"PA103","page_number":"103","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of a given segment ( par = 4 ) . 10.3 Construct with Hilbert\u0026#39;s tools a line perpendicular to a given line 1 ... \u003cb\u003emidpoints\u003c/b\u003e of AB and CD ; then use ( 1.27 ) . 10.11 Given a finite set of points A1 , ... , An in a Hilbert\u0026nbsp;..."},{"page_id":"PA134","page_number":"134","snippet_text":"... first three , D , E , F are the \u003cb\u003emidpoints\u003c/b\u003e of the sides . In the last , they are one - third of the way along each side . ( Par = 20 to 25 steps each . ) 14.10 D A € F 14.11 14.12 14.13 B E D A יד F D. 134 3. Geometry over Fields."},{"page_id":"PA182","page_number":"182","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of A1A2 . Drop a perpendicular from A1 to O1B , and a perpendicular from A2 to O2B , and let these two lines meet at P. Then the perpendicular from P to O1O2 is the required radical axis of the two circles . A B D 0 . P A\u0026nbsp;..."},{"page_id":"PA183","page_number":"183","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of AC . Then find E on AC such that DB . Find F on AB such that AF . Prove that DE AE ( AF ) 2 = ( AB ) ( BF ) in the field of segment arithmetic . Hint : Use Proposition 20.6 . We say that AB has been divided in extreme and\u0026nbsp;..."},{"page_id":"PA184","page_number":"184","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of OB . Let CD be the angle bisector of LACO . Let DE be perpendicular to OA . Then prove that AE is a side of the regular pentagon inscribed in the circle . Hint : Use Exercise 20.2 . How many steps would it take to construct\u0026nbsp;..."},{"page_id":"PA185","page_number":"185","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of BC . Hint : Use Ceva\u0026#39;s theorem ( Exercise 20.7 ) . B D F E C 20.19 ( a ) Given a line segment BC and its \u003cb\u003emidpoint\u003c/b\u003e F , construct with ruler alone a line through a given point D parallel to the line BC ( par = 6 ) . ( b )\u0026nbsp;..."},{"page_id":"PA192","page_number":"192","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of a given segment ( par = 15 ) . The diagram is given as a hint of one possible con- struction . A P В 21.5 Given I and O , construct with ruler alone a line parallel to a given line 1 and passing through a given point P ( par\u0026nbsp;..."},{"page_id":"PA202","page_number":"202","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of AB , and draw a line n through E , parallel to 1. Let this K E H G F L n line meet DC at F. Then , from ( 5.1 ) ( cf. Exercise 5.3 ) it follows that F is the \u003cb\u003emidpoint\u003c/b\u003e of DC . Draw a line through B , parallel to AC , to meet\u0026nbsp;..."},{"page_id":"PA212","page_number":"212","snippet_text":"... \u003cb\u003emidpoints\u003c/b\u003e of AB and AC . B F A L E с E 23.5 To get a feeling for ordered abelian groups , try this one . Let Q ( √2 ) { a + b√2 a , b e Q } , and similarly Q ( √3 ) . C ( a ) Show that Q ( √2 ) and Q ( √3 ) are isomorphic as\u0026nbsp;..."},{"page_id":"PA213","page_number":"213","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of AC . Draw lines through D parallel to BC and through C parallel to AB , meeting at F. Then AADE ACDF by ( ASA ) . So AABC can be dissected into the parallel- B ogram BCEF . Lemma 24.2 Let ABC be a triangle , and suppose that\u0026nbsp;..."},{"page_id":"PA215","page_number":"215","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of BD , so that G is above the \u003cb\u003emidpoint\u003c/b\u003e of AC , and so F is below L. This follows from the hypothesis AB \u0026lt; AE \u0026lt; 2AB , because the line from C to the \u003cb\u003emidpoint\u003c/b\u003e of BD would meet the line AB at a point M with AM = 2AB . Proposition\u0026nbsp;..."},{"page_id":"PA216","page_number":"216","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e , draw a circle with center O passing through the segment\u0026#39;s endpoints , and let c be the segment cut off on a perpendicular dividing the seg- ment into a + b , from the segment to where it meets the circle . ( Here we use ( E )\u0026nbsp;..."},{"page_id":"PA218","page_number":"218","snippet_text":"... \u003cb\u003emidpoints\u003c/b\u003e of the sides of the square BCRS , and draw lines through them parallel to AB and AC , to form the figure shown . We claim that 1 \u0026#39; , 2 \u0026#39; , 3 \u0026#39; , 4 \u0026#39; , 5 \u0026#39; are respectively congruent to 1,2,3,4,5 , which will complete the proof\u0026nbsp;..."},{"page_id":"PA228","page_number":"228","snippet_text":"... \u003cb\u003emidpoints\u003c/b\u003e of the sides . Then the pyramid P is decomposed into four pieces : two smaller pyramids P1 = AEFG and P2 = FBHK , which are con- gruent to each other and have edges equal to the edges of P ; and two trian- gular prisms T1\u0026nbsp;..."},{"page_id":"PA257","page_number":"257","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of BF , and let the circle HB meet OA in J. Take K to be the \u003cb\u003emidpoint\u003c/b\u003e of OG . Let the circle with center J and radius OK meet OB in L. Then KL is the side of the inscribed 34 - gon . See ( Exer- cise 29.4 ) . Using the ideas\u0026nbsp;..."},{"page_id":"PA262","page_number":"262","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of DE , and let G be the \u003cb\u003emidpoint\u003c/b\u003e of AE . Then FG is perpendicular to AE , so by ( SAS ) the triangles EFG and AFG are congruent . Now the new angle EOB = a is equal to LAEO by parallel lines and to LEAF by congruent triangles\u0026nbsp;..."},{"page_id":"PA268","page_number":"268","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of FA ( dotted in the diagram ) , then we recognize triangles HGF and FCR satisfying the conditions of ( 30.5 ) . Let us fix OA = 1. Then OF = FR = 1. If S is the \u003cb\u003emidpoint\u003c/b\u003e of FR , then CS = √3 , because it is the altitude of\u0026nbsp;..."},{"page_id":"PA303","page_number":"303","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of AB . From C drop perpendiculars to 1 and m , and extend each an equal distance on the far side to obtain D and E. Show that C , D , E are not collinear , and then use Bolyai\u0026#39;s axiom to prove that I and m must meet . ( b )\u0026nbsp;..."},{"page_id":"PA306","page_number":"306","snippet_text":"... \u003cb\u003emidpoints\u003c/b\u003e of AB and CD , the midline , is perpendicular to both . Proof Given ABCD as above , let E be the \u003cb\u003emidpoint\u003c/b\u003e of AB and let I be the per- pendicular to AB at E. Since I is the per- pendicular bisector of AB , the points A , C lie\u0026nbsp;..."},{"page_id":"PA310","page_number":"310","snippet_text":"... \u003cb\u003emidpoints\u003c/b\u003e of AB and AC , and draw the line DE , which we call the midline of the triangle . Drop perpendiculars BF , AG , CH to DE . Now , AD = DB , and the vertical angles at D are equal , so by ( AAS ) the triangles ADG and BDF are\u0026nbsp;..."},{"page_id":"PA319","page_number":"319","snippet_text":"... \u003cb\u003emidpoints\u003c/b\u003e of two sides of a tri- angle is orthogonal to the perpendicular bisector of the third side . 35 ... \u003cb\u003emidpoint\u003c/b\u003e of AB will bisect the angle ẞ and will make a right angle at C. Since it is just as good to prove the lemma\u0026nbsp;..."},{"page_id":"PA332","page_number":"332","snippet_text":"... \u003cb\u003emidpoints\u003c/b\u003e of the sides , and let AG be the altitude that bisects DE and BC . In a semihyperbolic plane with ( A ) : ( a ) Show that AG \u0026lt; 2AF . ( b ) Show that BC \u0026gt; 2DE . 13 A E F G с ( c ) In the Euclidean case , the area of ABC would\u0026nbsp;..."},{"page_id":"PA350","page_number":"350","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e E. Draw the circle with center E passing through A and B ( 1 step ) . To find the tangent from P to this circle , bisect PE , using F , G ( 3 steps ) and get H. Draw the circle with center H through P , E ( 1 step ) and let it\u0026nbsp;..."},{"page_id":"PA371","page_number":"371","snippet_text":"... \u003cb\u003emidpoints\u003c/b\u003e of the sides of the triangle are equal to the Euclidean \u003cb\u003emidpoints\u003c/b\u003e , and then use the Euclidean theorem about medians in the ambi- ent plane . A B C 39.23 In the Cartesian plane over the field F , let I be a circle of radius r\u0026nbsp;..."},{"page_id":"PA377","page_number":"377","snippet_text":"... \u003cb\u003emidpoints\u003c/b\u003e of AC and BD will be perpendicular to both 1 and m , by ( 34.1 ) . If ABCD , we may assume CD \u0026gt; AB , and we proceed as follows . Take E on CD such that AB = ED . Let n be a ray through E making the same angle with ED as I\u0026nbsp;..."},{"page_id":"PA378","page_number":"378","snippet_text":"... \u003cb\u003emidpoints\u003c/b\u003e of FH and GK will be perpen- dicular to both 1 and m . с A n } H P } E D B G m K It remains to show the uniqueness of the line perpendicular to both I and m . Suppose to the contrary that AB and CD were two common\u0026nbsp;..."},{"page_id":"PA383","page_number":"383","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e at right angles . Now it follows from ( 34.1 ) that this line continued will meet AB at its \u003cb\u003emidpoint\u003c/b\u003e F , at right angles . с E Remark 40.10.1 A B To make a more. 40. Hyperbolic Geometry 383."},{"page_id":"PA392","page_number":"392","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of OA . Then the line perpendicular to ( 0 , ∞ ) at B will have an end ẞ with the property ẞ2 = a . Thus a has a square root in F. ß2 To relate this newly constructed field of ends with the geometry of the plane , our first\u0026nbsp;..."},{"page_id":"PA408","page_number":"408","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of one side . Its angle a is π / n , its hypotenuse is r , and its side we call bn . Its area is An / 2n , where An is the area of the inscribed polygon . Hence sin л / n cosл / n sin bn sin An 2n ( 1 - sin ) . π / n bn\u0026nbsp;..."},{"page_id":"PA440","page_number":"440","snippet_text":"... Lastly , to make a dodecahedron , take the icosahedron previously con- structed . For each vertex of the icosahedron , join the five \u003cb\u003emidpoints\u003c/b\u003e of the tri- angles meeting at that vertex . This makes a regular. 440 8. Polyhedra."},{"page_id":"PA441","page_number":"441","snippet_text":"... \u003cb\u003emidpoints\u003c/b\u003e of the faces of the icosa- hedron . These are all equidistant from the center of the sphere containing the icosahedron , so the vertices of the dodecahedron lie on a new smaller sphere , which is inscribed in the icosahedron\u0026nbsp;..."},{"page_id":"PA469","page_number":"469","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of the side 23 , and reflecting the figure in that line . This induces the permutation ( 23 ) on the vertices . Reflections in two other axes give the permutations ( 12 ) and ( 13 ) . A symmetry of the triangle is completely\u0026nbsp;..."},{"page_id":"PA471","page_number":"471","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of the side 13 and the \u003cb\u003emidpoint\u003c/b\u003e of the side 24 , a rotation of л around this axis will induce this symmetry . So in this case , the composition of rotations around two different axes is equal to a rotation around a third axis\u0026nbsp;..."},{"page_id":"PA473","page_number":"473","snippet_text":"... \u003cb\u003emidpoints\u003c/b\u003e of opposite edges has order 2 . So there are 1 x 30 = 15 elements of order 2. Summing up , we have identity 1 elements of order 5 24 elements of order 3 20 elements of order 2 15 60 : Next , let us look at subgroups of G 47\u0026nbsp;..."},{"page_id":"PA474","page_number":"474","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of a face is of order 3 , and the stabilizer of the \u003cb\u003emidpoint\u003c/b\u003e of an edge is of order 2 . If we consider an axis through two opposite vertices , the stabilizer of this line includes the stabilizer of one vertex , but allows also\u0026nbsp;..."},{"page_id":"PA511","page_number":"511","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of segment , 23 of 11 - gon , 277 of 13 - gon , 269 of 17 - gon , 250-259 of 19 - gon , 277 of circular inverse , 335 , 344 of configurations , 134 , 135 of heptagon , 264-269 of inverse , 343 of limiting parallel , 396-398 of\u0026nbsp;...","page_url":"https://books.google.com.sg/books?id=EJCSL9S6la0C\u0026pg=PA511\u0026vq=midpoint"},{"page_id":"PA518","page_number":"518","snippet_text":"... \u003cb\u003emidpoint\u003c/b\u003e of segment , 88 by Hilbert\u0026#39;s tools , 103 existence , 100 Millay , Edna St. Vincent , 50 minimal polynomial , 259 , 283 , 292 Miquel point , 61 Miquel , A. , 61 mirror image , 227 , 230 , 231 , 451 mirror symmetry . See symmetry\u0026nbsp;...","page_url":"https://books.google.com.sg/books?id=EJCSL9S6la0C\u0026pg=PA518\u0026vq=midpoint"}],"search_query_escaped":"midpoint"},{});</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>