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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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In the figure , the parallelogram ABCD is \u003cb\u003eequal\u003c/b\u003e to the parallelogram BCEF . Clearly , the ... \u003cb\u003econtent\u003c/b\u003e , to avoid confusion with other notions of \u003cb\u003eequality\u003c/b\u003e or congruence . We do 1 not want to use the word area\u0026nbsp;..."},{"page_id":"PA41","page_number":"41","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e . ( d ) Ditto for subtraction , noting that equality of content of the difference does not depend on where the equal pieces were removed . ( e ) Halves of figures of \u003cb\u003eequal content\u003c/b\u003e have \u003cb\u003eequal content\u003c/b\u003e ( used in the proof of\u0026nbsp;..."},{"page_id":"PA42","page_number":"42","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e to the square on the hypotenuse . The proof goes like this . The triangle ABF is one half of the square ABFG . This triangle ABF has \u003cb\u003eequal content\u003c/b\u003e with the triangle BFC by ( 1.37 ) . The triangle BFC is con- gruent to\u0026nbsp;..."},{"page_id":"PA43","page_number":"43","snippet_text":"... \u003cb\u003econtent equal\u003c/b\u003e to the triangle ABC , and with one side \u003cb\u003eequal\u003c/b\u003e to DE . A B 3.4 Given a rectangle , construct a square with the same \u003cb\u003econtent\u003c/b\u003e . 3.5 Given a line 1 and given two points A , B not on 1 , construct a circle passing through A\u0026nbsp;..."},{"page_id":"PA44","page_number":"44","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e to the square on OB ) . = A F E G B B A 0 A A A F 3.9 Let ABC be a right triangle , and let AD be the altitude from the right angle A to the hypotenuse BC . Prove that AD2 BD × DC ( in the sense of content ) . == 3.10\u0026nbsp;..."},{"page_id":"PA45","page_number":"45","snippet_text":"... \u003cb\u003econtent\u003c/b\u003e and then apply ( 1.39 ) . ) B E D C 4 Construction of the Regular Pentagon One of the most beautiful results ... \u003cb\u003eequal\u003c/b\u003e to twice the vertex angle . Construction ( ( II.11 ) , ( IV.10 ) ) Let A , B be two points chosen at\u0026nbsp;..."},{"page_id":"PA46","page_number":"46","snippet_text":"... \u003cb\u003eequal\u003c/b\u003e to twice the angle at A. J C A K B H E G Proof From a modern point of view , it would seem that some theory of ... \u003cb\u003econtent\u003c/b\u003e - cf . Section 3 . Book II contains a number of results of geometrical algebra , as described above\u0026nbsp;..."},{"page_id":"PA113","page_number":"113","snippet_text":"... equal . If not , con- struct a line l \u0026#39; through A making an angle a with n ( axiom ( C4 ) ) . By ( 1.27 ) , l \u0026#39; will be parallel to m . But then I and l \u0026#39; are two lines through A parallel ... \u003cb\u003eequal content\u003c/b\u003e . (. 12. Euclidean Planes 113."},{"page_id":"PA114","page_number":"114","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e have \u003cb\u003eequal content\u003c/b\u003e . ( 3 ) Differences of figures with \u003cb\u003eequal content\u003c/b\u003e have \u003cb\u003eequal content\u003c/b\u003e . ( 4 ) Halves of figures with \u003cb\u003eequal content\u003c/b\u003e have \u003cb\u003eequal content\u003c/b\u003e . ( 5 ) The whole is greater than the part . ( 6 ) If two squares\u0026nbsp;..."},{"page_id":"PA195","page_number":"195","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e by saying that two figures have \u003cb\u003eequal content\u003c/b\u003e if we can transform one figure into the other by adding and subtracting congruent tri- angles ( Section 22 ) . We can prove all the properties of area that Euclid uses , except\u0026nbsp;..."},{"page_id":"PA196","page_number":"196","snippet_text":"... \u003cb\u003eequal\u003c/b\u003e area . \u0026quot; The area we are familiar with from high school attaches a number to each figure . So the area of a ... \u003cb\u003econtent\u003c/b\u003e . \u0026quot; To begin with , let us be precise about our terminology . We presuppose the axioms of a Hilbert plane\u0026nbsp;..."},{"page_id":"PA197","page_number":"197","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e if there are other figures Q , Q \u0026#39; such that : ( 1 ) P and Q are nonoverlapping . ( 2 ) P \u0026#39; and Q \u0026#39; are nonoverlapping . ( 3 ) Q and Q \u0026#39; are equidecomposable . ( 4 ) PUQ and P\u0026#39;UQ \u0026#39; are equidecomposable . 3 4 Example 22.1.1\u0026nbsp;..."},{"page_id":"PA198","page_number":"198","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e . Indeed , if we let P = ABCD and P \u0026#39; = CDEF , take Q = Q \u0026#39; = triangle BGE . Then PUQ and P\u0026#39;UQ \u0026#39; are the unions of the congruent triangles ACE and BDF and the equal triangles CDG and CDG . Example 22.1.3 If two figures P\u0026nbsp;..."},{"page_id":"PA199","page_number":"199","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e equiva- lent ? We will see that the answer is yes in any Hilbert plane with ( P ) and ( A ) , by using a measure of area function with values in the field of segment arith- metic ( 24.7.3 ) . I do not know any purely\u0026nbsp;..."},{"page_id":"PA200","page_number":"200","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e has the following properties : ( a ) \u003cb\u003eEqual content\u003c/b\u003e is an equivalence relation . ( b ) Equidecomposable figures have \u003cb\u003eequal content\u003c/b\u003e . ( c ) Nonoverlapping unions of figures of \u003cb\u003eequal content\u003c/b\u003e have \u003cb\u003eequal content\u003c/b\u003e . ( d ) If Q P\u0026nbsp;..."},{"page_id":"PA201","page_number":"201","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e is obviously reflexive and sym- metric . The nontrivial part is to show that it is transitive . So suppose figures P and P \u0026#39; have \u003cb\u003eequal content\u003c/b\u003e , and P \u0026#39; and P \u0026quot; have \u003cb\u003eequal content\u003c/b\u003e . Then there are equidecomposable figures\u0026nbsp;..."},{"page_id":"PA202","page_number":"202","snippet_text":"... content depends on ( Z ) ( Exercise 22.7 ) . We will also see in the exercises that ( Z ) implies the other two ... \u003cb\u003eequal content\u003c/b\u003e has properties 1 , 2 , 3 at the beginning of this section . In a Euclidean plane with ( Z ) , 4 , 5\u0026nbsp;..."},{"page_id":"PA203","page_number":"203","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e . So by tran- sitivity it follows that the triangles ABC and DBC have \u003cb\u003eequal content\u003c/b\u003e . Continuing with our examination of Euclid\u0026#39;s results , ( 1.38 ) follows by tran- sitivity . Its converse ( 1.39 ) uses \u0026quot; the whole is\u0026nbsp;..."},{"page_id":"PA204","page_number":"204","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e with a rectangle P \u0026#39; of given fixed side AB . 22.8 In a Euclidean plane with ( Z ) show that \u0026quot; halves of equals are equal , \u0026quot; in the follow- ing sense : If P and Q are figures with \u003cb\u003eequal content\u003c/b\u003e , and if P = P1 UP2 is a\u0026nbsp;..."},{"page_id":"PA206","page_number":"206","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e , then a ( P ) = α ( P \u0026#39; ) . ( d ) If a figure Q is contained in a figure P , and P - Q has nonempty interior , then a ( Q ) \u0026lt; x ( P ) . In particular , P and Q cannot have \u003cb\u003eequal content\u003c/b\u003e , so ( Z ) holds ( cf. Sec- tion 22 )\u0026nbsp;..."},{"page_id":"PA210","page_number":"210","snippet_text":"... equality \u0026quot; of figures to mean \u003cb\u003eequal content\u003c/b\u003e in the sense of Section 22 . Proof We have already seen in Section 22 that all these results follow from the definition of \u003cb\u003eequal content\u003c/b\u003e plus the statement ( Z ) . In this section we saw that\u0026nbsp;..."},{"page_id":"PA211","page_number":"211","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e if and only if x ( P ) a ( Q ) . = Proof If P and Q have \u003cb\u003eequal content\u003c/b\u003e , then a ( P ) = a ( Q ) by ( 23.1 ) . Conversely , suppose α ( P ) = a ( Q ) . By ( I.44 ) we can find rectangles P \u0026#39; and Q \u0026#39; with content equal to P\u0026nbsp;..."},{"page_id":"PA212","page_number":"212","snippet_text":"... \u003cb\u003eequal\u003c/b\u003e to the area of the whole triangle . Hint : Review the proof of Proposition 5.4 and use similar triangles ... \u003cb\u003econtent\u003c/b\u003e to the sum of the two rectangles made by opposite sides of the quadrilateral . 23.7 In a Euclidean plane\u0026nbsp;..."},{"page_id":"PA213","page_number":"213","snippet_text":"... \u003cb\u003econtent\u003c/b\u003e , are also true in the stronger sense of dis- section . We will prove the theorem , due to Bolyai and Gerwien , that in an Archimedean plane , any two figures of \u003cb\u003eequal\u003c/b\u003e area ( Section 23 ) are equivalent by dissection . The\u0026nbsp;..."},{"page_id":"PA215","page_number":"215","snippet_text":"... \u003cb\u003eequal\u003c/b\u003e area are equivalent by dissection . The proof parallels ( 23.7 ) , where the analogous result was proved for \u003cb\u003econtent\u003c/b\u003e . So suppose figures P and Q have \u003cb\u003eequal\u003c/b\u003e area : ( P ) = x ( Q ) . By ( 24.6 ) , they can each be dissected into\u0026nbsp;..."},{"page_id":"PA216","page_number":"216","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e if and only if they are equidecomposable ( cf. 22.1.4 ) . Proposition 24.8 Assume ( E ) . Let ABCD be a rectangle with sides a = AB and b = AC satisfying a≤ b ≤ 4a . Then there is a segment c = AE such that the rectangle\u0026nbsp;..."},{"page_id":"PA217","page_number":"217","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e to the square on the hypotenuse . Using Archimedes \u0026#39; axiom ( A ) one can improve Euclid\u0026#39;s results about parallelograms ( 1.35 ) and triangles ( 1.37 ) to hold also for dissection . Then Euclid\u0026#39;s proof will show that ( 1.47 )\u0026nbsp;..."},{"page_id":"PA226","page_number":"226","snippet_text":"... \u003cb\u003econtent\u003c/b\u003e for plane figures , achieved by adding and subtracting congruent pieces of a dissection of the figure . To ... \u003cb\u003eequal\u003c/b\u003e volume . Our attention will be focused on how Euclid proves this \u003cb\u003eequality\u003c/b\u003e . First we fix our terminology\u0026nbsp;..."},{"page_id":"PA227","page_number":"227","snippet_text":"... \u003cb\u003eequal\u003c/b\u003e to twice ( in the sense of \u003cb\u003econtent\u003c/b\u003e ) the triangular base of the other , and if they have the same height , then they have \u003cb\u003eequal\u003c/b\u003e volume . The method is to double each to get parallelepipeds , which have \u003cb\u003eequal\u003c/b\u003e vol- ume by the\u0026nbsp;..."},{"page_id":"PA228","page_number":"228","snippet_text":"... \u003cb\u003eequal\u003c/b\u003e bases ( in the sense of \u003cb\u003econtent\u003c/b\u003e ) and \u003cb\u003eequal\u003c/b\u003e height have \u003cb\u003eequal\u003c/b\u003e volume . Proof For the proof , Euclid uses the method of exhaustion , attributed to Eudoxus . The idea is to write both figures as infinite unions of subfigures of\u0026nbsp;..."},{"page_id":"PA229","page_number":"229","snippet_text":"... \u003cb\u003eequal\u003c/b\u003e in \u003cb\u003econtent\u003c/b\u003e , it follows that parallelepipeds on the bases of T1 and T of the same height will have \u003cb\u003eequal\u003c/b\u003e volume ( XI.39 ) , and hence their halves T1 and T1 will have \u003cb\u003eequal\u003c/b\u003e volume . So T1 + T2 and T + T1⁄2 have \u003cb\u003eequal\u003c/b\u003e volumes\u0026nbsp;..."},{"page_id":"PA230","page_number":"230","snippet_text":"... \u003cb\u003econtent\u003c/b\u003e ) and the same height , ( XII.5 ) applies to show that P and P \u0026#39; have the same volume . Gauss , in a letter to Gerling in 1844 , says that it is too bad that the \u003cb\u003eequality\u003c/b\u003e of volume of two symmetrical but not congruent figures\u0026nbsp;..."},{"page_id":"PA232","page_number":"232","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e - cf . Section 22 ) have the same invariant 8. We will compute 8 of a tetrahedron , which will be nonzero , and 8 of a cube , which will be zero . This will show that a tetrahedron is not equivalent to any cube by\u0026nbsp;..."},{"page_id":"PA326","page_number":"326","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e if one can add to them equidecomposable figures so that the whole becomes equidecomposable . We showed in Section 22 that both of these notions are equivalence relations . In Section 23 we defined the notion of a measure\u0026nbsp;..."},{"page_id":"PA329","page_number":"329","snippet_text":"... content equal to the Saccheri quadrilateral FHBC , so they have \u003cb\u003eequal content\u003c/b\u003e to each other ( 22.3 ) . F H B In the Archimedean case , they are equidecomposable . Remark 36.4.1 Note how the Euclidean hypothesis \u0026quot; lying in the same\u0026nbsp;..."},{"page_id":"PA330","page_number":"330","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e . If furthermore we assume ( A ) , they are equidecomposable . Proof We use the area function of ( 36.2 ) , so that equal area means equal defect . First let us consider the case where one side of the first triangle is\u0026nbsp;..."},{"page_id":"PA332","page_number":"332","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e , they are equidecomposable - cf . ( 22.1.4 ) . Indeed , they will have the same area ( 23.1 ) , using the area function of ( 36.2 ) , and then the theorem applies . Exercises 36.1 Gauss , in a letter to W. ( = Farkas )\u0026nbsp;..."},{"page_id":"PA336","page_number":"336","snippet_text":"... Let A be another point of y and let OA meet y again at A \u0026#39; . Applying ( III.36 ) to y we obtain OP2 = OA OA \u0026#39; . ( Actually , Euclid meant that the square on OP has \u003cb\u003econtent equal\u003c/b\u003e to the rectangle formed. 336 7. Non - Euclidean Geometry."},{"page_id":"PA489","page_number":"489","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e \u0026quot; is sometimes called \u0026quot; equicomplementable \u0026quot; ( German : ergänzungs- gleich ) . What we call \u0026quot; equidecomposable \u0026quot; or \u0026quot; equivalent by dissection \u0026quot; is some- times called \u0026quot; equivalent by finite decomposition \u0026quot; ( German Notes 489."},{"page_id":"PA513","page_number":"513","snippet_text":"... \u003cb\u003econtent\u003c/b\u003e , 113 , 195 , 326 as equivalence relation , 200 definition , 197 does not imply equidecomposable , 198 equivalent to equidecomposable , 199 if same area , 211 implies \u003cb\u003eequal\u003c/b\u003e ... \u003cb\u003eequality\u003c/b\u003e of angles , 32 \u003cb\u003eequality\u003c/b\u003e , definition of\u0026nbsp;...","page_url":"https://books.google.com.sg/books?id=EJCSL9S6la0C\u0026pg=PA513\u0026vq=equal+content"},{"page_id":"PA514","page_number":"514","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e , 197 , 326 ordering of , 204 filling plane with triangles , 367 , 368 finite element , 159 finite field extension , 280-294 finite field , 157 finite geometry , 67 , 72 , 73 , 129 finite subgroups of SO ( 3 ) , 475\u0026nbsp;...","page_url":"https://books.google.com.sg/books?id=EJCSL9S6la0C\u0026pg=PA514\u0026vq=equal+content"},{"page_id":"PA524","page_number":"524","snippet_text":"... equal area , 211 of \u003cb\u003eequal content\u003c/b\u003e , 204 squaring of circle , 23 , 163 , 195 , 221–225 , 244 , 490 in hyperbolic plane , 409 , 414 squaring of parabola , 247 SSS ( side - side - side ) , 35 stabilizer subgroup , 473 , 474 standard\u0026nbsp;...","page_url":"https://books.google.com.sg/books?id=EJCSL9S6la0C\u0026pg=PA524\u0026vq=equal+content"},{"page_id":"PA526","page_number":"526","snippet_text":"... \u003cb\u003eequal content\u003c/b\u003e , 202 triangular dipyramid , 442 triangular dodecahedron , 457 triangular prism , 226 triangular pyramid , 226 volume of , 228 , 229 right isosceles , 231 , 239 tricapped triangular prism , 456 , Plate XVIII trichotomy\u0026nbsp;...","page_url":"https://books.google.com.sg/books?id=EJCSL9S6la0C\u0026pg=PA526\u0026vq=equal+content"}],"search_query_escaped":"equal content"},{});</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>