几何原本吧 关注：416贴子：1,482
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Thethirteen books
ofEuclid's Elements
Στοιχεια
几何原本
几何原本
[古希腊] 欧几里得Ευκλειδης Euclid著
希思 Thomas Little Heath英译
利玛窦Matteo Ricci徐光启文言文汉译
兰纪正、朱恩宽 现代汉语汉译
David E. Joyce大卫伊乔伊斯
Euclid’s Elements
欧几里得原本
位于牛津大学自然历史博物馆的欧几里得石像
Euclid’s Elements
EUCLID原本
Euclid [‘ju:klid] 欧几里得
Euclid原本
大卫乔伊斯做了一件功德无量的事，我们大家都要感谢他。
David E. Joyce大卫伊乔伊斯
Professor of Mathematics and Computer ScienceClark University
克拉克大学数学和计算机科学教授
DavidJoyce, Ph.D.大卫乔伊斯，哲学博士
ProfessorDepartment of Mathematics and ComputerScience数学和计算机科学教授
Clark University克拉克大学
Worcester, MA 01610-1477伍斯特MA 01610-1477
Phone: (508) 793-7421电话: (508) 793-7421
Email: djoyce@clarku.edu电子邮箱: djoyce@clarku.edu
Professional Website专业网站
几何原本的网站如下：
http://aleph0.clarku.edu/~djoyce/java/elements/toc.html
Philosophy Presenting the Seven Liberal Arts to Boethius
Illuminated parchment at The J. Paul Getty Museum, Los Angeles
Date ca. 1460–1470, by Coetivy Master (Henry de Vulcop?), illuminator (French)
oMy Bio page from the Academic Catalog
oCourses
Spring 2018:
Math 114, Discrete Mathematics
Fall 2017:
Math 225, Modern Algebra I
Past courses
oMathematics web pages
Euclid's Elements
Compass geometry
a short course on trigonometry
a short course on complex numbers
Hilbert's address of 1900 and his 23 mathematical problems
my Numbers Page including notes on Richard Dedekind's Was sind und was sollen die Zahlen?.
William Story's lecture The Mathematician as a Type
An Algebraic Approach to Symmetry with Applications to Knot Theory, 1979
my Pix page on geometry, chaos, and fractals
a page on Mandelbrot and Julia sets, along with a generation form, and an Explorer on which you can point and click to get more refined image
Newton Basins
Millefiori! which generates a different kind of chaotic image
a page describing tiling the Hyperbolic Plane
a page on Wallpaper Groups
a Little Gallery of Roulettes
some kaleidoscopes
some Bowditch patterns, and other miscelaneous pictures
Dave's Puzzle page.
Humor
oTHE HUXLEY FILE: the works of T.H. Huxley.
oMoth: investigations into altruism
oPhylogeny and reconstructing phylogenetic trees
oOther stuff
Garden Irises
genealogy
________________________________________
This page has been on the web since June 1994.
https://mathcs.clarku.edu/~djoyce/java/elements/elements.html
这是克拉克大学的大卫乔伊斯（David.E.Joyce)于二十世纪九十年代创建的一个旨在从新点燃人们对《原本》的网站。网站是全英的，支持JAVA，英语不是太差的应该能看下去，实在不行就用翻译插件（百度翻译垃圾）。
在此衷心感谢大卫乔伊斯对《原本》的传播做出的贡献！
PS:该原本应该是大卫乔伊斯自己翻译的版本【最好译本】（十三卷本），并附有大量JAVA动画供大家理解（在浏览器上打开JAVA可改变图像），其中还有大量的指南，简直是超赞的网站！
顺带说一下，尽管网站是大卫费心费力做的，但网站完全免费！！！
A Quick Trip through the Elements
To get an idea of what's in the Elements, here are a few highlights in the order that they appear:
Book I on basic plane geometry
•Def. I.23, definition of parallel lines, one of many definitions in Book I
•Post. I.5, the parallel postulate
•Common notions, the axioms for magnitudes
•Prop. I.1, the first proposition which shows how to construct an equilateral triangle
•The congruence theorems for triangles: Prop. I.4, side-angle-side, Prop. I.8, side-side-side, and Prop. I.26, angle-side-angle
•Propositions on isosceles triangles: Prop. I.5, equal angles imply equal sides, and the converse, Prop. I.6, equal sides imply equal angles
•Prop. I.9 and Prop. I.10, constructions to bisect angles and line segments
•Prop. I.11 and Prop. I.12, constructions to draw perpendicular lines
•Prop. I.16, an exterior angle of a triangle is greater than either of the opposite interior angles (compare I.32 below)
•Prop. I.29, about angles made when a line crosses two parallel lines
•Prop. I.20, the triangle inequality (the sum of two sides is greater than the third)
•Prop. I.22, to construct a triangle with given sides
•Prop. I.32, an exterior angle of a triangle is the sum of the two opposite interior angles; the sum of the three interior angles equals two right angles.
•On application of areas: Prop. I.42 to find a parallelogram equal in area to any given triangle, and Prop. I.45 to find a parallelogram equal in area to any given polygon
•Prop. I.47, the Pythagorean theorem and its converse Prop. I.48
Book II on geometric algebra
•Prop. II.4, a geometric version of the algebraic identity (x + y)2 = x2 + 2xy + y2
•Prop. II.5, a sample proposition showing how to factor the difference of two squares
•Prop. II.6, a geometric version to solve the quadratic equation (b – x)x = c
•Prop. II.11, construction to cut a line in the golden ratio
•Prop. II.12 and Prop. II.13, a pre-trigonometry version of the the law of cosines
•Prop. II.14, a final proposition on application of areas—to find a square equal in area to any given polygon
Book III on circles and angles
•Prop. III.1, how to find the center of a circle
•Prop. III.17, how to draw a line tangent to a circle
•Propositions on angles in circles: Prop. III.20, Prop. III.21, and Prop. III.22
•Prop. III.31, Thales' theorem that an angle inscribed in a semicircle is right, and similar statements giving acute and obtuse angles
•Prop. III.35, when two chords are drawn through a point inside a circle, then the product of the two segments of one chord equals the product of the two segments of the other chord
•Prop. III.36, if from a point outside a circle both a tangent and a secant are drawn, then the square of the tangent is the product of the whole secant and the external segment of the secant, and the converse in Prop. III.37
Book IV on constructions of regular polygons
•Inscribing and circumscribing circles and arbitrary triangles Prop. IV.2, Prop. IV.3, Prop. IV.4, and Prop. IV.5,
•Prop. IV.10, how to construct a particular triangle needed for regular pentagons
•Prop. IV.11, how to construct a regular pentagon
•Prop. IV.15, how to construct a regular hexagon
•Prop. IV.16, how to construct a regular 15-gon
Book V on Eudoxus’ abstract theory of ratio and proportion, abstract algebra
•Def. V.3, the definition and nature of ratio
•Def. V.5 and V.6, the definition of proportion (equality of ratios)
•Def. V.9, the definition of duplicate proportion (the square of a ratio)
•Prop. V.2, distributivity of multiplication over addition
•Prop. V.3, associativity of multiplication of whole numbers
•Prop. V.11, transitivity of equality of ratios
•Prop. V.16, alternate proportions
•Prop. V.22, ratios ex aequali
Book VI on similar figures and geometric proportions
•Def. VI.1, definition of similar figures
•Prop. VI.1, areas of triangles (also parallelograms) of the same height are proportional to their bases
•Prop. VI.2, a line parallel to the base of a triangle cuts the sides proportionally
•Propositions on similar triangles: Prop. VI.4, Prop. VI.5,
•Prop. VI.6, side-angle-side similarity theorem
•Prop. VI.9, to cut a line into a given number of equal segments
•Prop. VI.10, to cut a line into a specified ratio
•Constructions of fourth proportionals Prop. VI.12, and mean proportionals Prop. VI.13,
•Prop. VI.16, if four lines are proportional, w : x = y:z, then the rectangle contained by the extremes, w by z, has the same area as the rectangle contained by the means, x by y
•Prop. VI.19, on areas of similar triangles
•Prop. VI.25, on application of areas
To get an idea of what's in the Elements, here are a few highlights in the order that they appear:
Book I on basic plane geometry
•Def. I.23, definition of parallel lines, one of many definitions in Book I
•Post. I.5, the parallel postulate
•Common notions, the axioms for magnitudes
•Prop. I.1, the first proposition which shows how to construct an equilateral triangle
•The congruence theorems for triangles: Prop. I.4, side-angle-side, Prop. I.8, side-side-side, and Prop. I.26, angle-side-angle
•Propositions on isosceles triangles: Prop. I.5, equal angles imply equal sides, and the converse, Prop. I.6, equal sides imply equal angles
•Prop. I.9 and Prop. I.10, constructions to bisect angles and line segments
•Prop. I.11 and Prop. I.12, constructions to draw perpendicular lines
•Prop. I.16, an exterior angle of a triangle is greater than either of the opposite interior angles (compare I.32 below)
•Prop. I.29, about angles made when a line crosses two parallel lines
•Prop. I.20, the triangle inequality (the sum of two sides is greater than the third)
•Prop. I.22, to construct a triangle with given sides
•Prop. I.32, an exterior angle of a triangle is the sum of the two opposite interior angles; the sum of the three interior angles equals two right angles.
•On application of areas: Prop. I.42 to find a parallelogram equal in area to any given triangle, and Prop. I.45 to find a parallelogram equal in area to any given polygon
•Prop. I.47, the Pythagorean theorem and its converse Prop. I.48
Book II on geometric algebra
•Prop. II.4, a geometric version of the algebraic identity (x + y)2 = x2 + 2xy + y2
•Prop. II.5, a sample proposition showing how to factor the difference of two squares
•Prop. II.6, a geometric version to solve the quadratic equation (b – x)x = c
•Prop. II.11, construction to cut a line in the golden ratio
•Prop. II.12 and Prop. II.13, a pre-trigonometry version of the the law of cosines
•Prop. II.14, a final proposition on application of areas—to find a square equal in area to any given polygon
Book III on circles and angles
•Prop. III.1, how to find the center of a circle
•Prop. III.17, how to draw a line tangent to a circle
•Propositions on angles in circles: Prop. III.20, Prop. III.21, and Prop. III.22
•Prop. III.31, Thales' theorem that an angle inscribed in a semicircle is right, and similar statements giving acute and obtuse angles
•Prop. III.35, when two chords are drawn through a point inside a circle, then the product of the two segments of one chord equals the product of the two segments of the other chord
•Prop. III.36, if from a point outside a circle both a tangent and a secant are drawn, then the square of the tangent is the product of the whole secant and the external segment of the secant, and the converse in Prop. III.37
Book IV on constructions of regular polygons
•Inscribing and circumscribing circles and arbitrary triangles Prop. IV.2, Prop. IV.3, Prop. IV.4, and Prop. IV.5,
•Prop. IV.10, how to construct a particular triangle needed for regular pentagons
•Prop. IV.11, how to construct a regular pentagon
•Prop. IV.15, how to construct a regular hexagon
•Prop. IV.16, how to construct a regular 15-gon
Book V on Eudoxus’ abstract theory of ratio and proportion, abstract algebra
•Def. V.3, the definition and nature of ratio
•Def. V.5 and V.6, the definition of proportion (equality of ratios)
•Def. V.9, the definition of duplicate proportion (the square of a ratio)
•Prop. V.2, distributivity of multiplication over addition
•Prop. V.3, associativity of multiplication of whole numbers
•Prop. V.11, transitivity of equality of ratios
•Prop. V.16, alternate proportions
•Prop. V.22, ratios ex aequali
Book VI on similar figures and geometric proportions
•Def. VI.1, definition of similar figures
•Prop. VI.1, areas of triangles (also parallelograms) of the same height are proportional to their bases
•Prop. VI.2, a line parallel to the base of a triangle cuts the sides proportionally
•Propositions on similar triangles: Prop. VI.4, Prop. VI.5,
•Prop. VI.6, side-angle-side similarity theorem
•Prop. VI.9, to cut a line into a given number of equal segments
•Prop. VI.10, to cut a line into a specified ratio
•Constructions of fourth proportionals Prop. VI.12, and mean proportionals Prop. VI.13,
•Prop. VI.16, if four lines are proportional, w : x = y:z, then the rectangle contained by the extremes, w by z, has the same area as the rectangle contained by the means, x by y
•Prop. VI.19, on areas of similar triangles
•Prop. VI.25, on application of areas
Proposition 26.
If two triangles have two angles equal to two angles respectively, and oneside equal to one side, namely, either the side adjoining the equal angles, orthat opposite one of the equal angles, then the remaining sides equal theremaining sides and the remaining angle equals the remaining angle.
Proposition 26
Iftwo triangles have two angles equal to two angles respectively, and one sideequal to one side, namely, either the side adjoining the equal angles, or thatopposite one of the equal angles, then the remaining sides equal the remainingsides and the remaining angle equals the remaining angle.
Let ABC and DEF betwo triangles having the two angles ABC and BCA equalto the two angles DEF and EFD respectively,namely the angle ABC to the angle DEF, and theangle BCA to the angle EFD, and let them alsohave one side equal to one side, first that adjoining the equal angles,namely BC equal to EF.
I say that the remainingsides equal the remaining sides respectively, namely AB equals DE and AC equals DF, andthe remaining angle equals the remaining angle, namely the angle BAC equalsthe angle EDF.
If AB doesnot equal DE, then one of them is greater.
I.3, I.Post.1
Let AB begreater. Make BG equal to DE, and join GC.
I.4
C.N.1
Since BG equals DE, and BC equals EF, thetwo sides GB and BC equal the two sides DE and EF respectively,and the angle GBC equals the angle DEF, thereforethe base GC equals the base DF, the triangle GBC equalsthe triangle DEF, and the remaining angles equalthe remaining angles, namely those opposite the equal sides. Therefore theangle GCB equals the angle DFE. But theangle DFE equals the angle ACB by hypothesis.Therefore the angle BCG equals the angle BCA, theless equals the greater, which is impossible.
Therefore AB isnot unequal to DE, and therefore equals it.
I.4
But BC alsoequals EF. Therefore the two sides AB and BC equalthe two sides DE and EF respectively, and theangle ABC equals the angle DEF. Therefore the base AC equalsthe base DF, and the remaining angle BAC equalsthe remaining angle EDF.
Next, let sides oppositeequal angles be equal, as AB equals DE.
I say again that theremaining sides equal the remaining sides, namely AC equals DF and BC equals EF, andfurther the remaining angle BAC equals the remainingangle EDF.
If BC isunequal to EF, then one of them is greater.
I.3, I.Post.1
Let BC begreater, if possible. Make BH equal to EF, andjoin AH.
I.4
Since BH equals EF, and AB equals DE, thetwo sides AB and BH equal the two sides DE and EF respectively,and they contain equal angles, therefore the base AH equalsthe base DF, the triangle ABH equals thetriangle DEF, and the remaining angles equal the remainingangles, namely those opposite the equal sides. Therefore the angle BHA equalsthe angle EFD.
C.N.1
I.16
But the angle EFD equalsthe angle BCA, therefore, in the triangle AHC, theexterior angle BHA equals the interior and opposite angle BCA, whichis impossible.
Therefore BC isnot unequal to EF, and therefore equals it.
I.4
But AB alsoequals DE. Therefore the two sides AB and BC equalthe two sides DE and EF respectively, andthey contain equal angles. Therefore the base AC equalsthe base DF, thetriangle ABC equals the triangle DEF, and theremaining angle BAC equals the remaining angle EDF.
Therefore if twotriangles have two angles equal to two angles respectively, and one side equalto one side, namely, either the side adjoining the equal angles, or thatopposite one of the equal angles, then the remaining sides equal the remainingsides and the remaining angle equals the remaining angle.
Q.E.D.
Proposition 27.
If a straight line falling on two straight lines makes the alternateangles equal to one another, then the straight lines are parallel to oneanother.
Proposition 27
Ifa straight line falling on two straight lines makes the alternate angles equalto one another, then the straight lines are parallel to one another.
Let the straight line EF fallingon the two straight lines AB and CD make thealternate angles AEF and EFD equal to oneanother.
I say that AB isparallel to CD.
If not, AB and CD whenproduced meet either in the direction of B and D ortowards A and C.
Let them be produced andmeet, in the direction of B and D, at G.
I.16
Then, in the triangle GEF, theexterior angle AEF equals the interior and oppositeangle EFG, which is impossible.
Therefore AB and CD whenproduced do not meet in the direction of B and D.
Similarly it can be provedthat neither do they meet towards A and C.
I.Def.23
But straight lines which donot meet in either direction are parallel. Therefore AB isparallel to CD.
Therefore if astraight line falling on two straight lines makes the alternate angles equal toone another, then the straight lines are parallel to one another.
Q.E.D.
Proposition 28.
If a straight line falling on two straight lines makes the exterior angle外角 equal to the interior and opposite angle on the same side同侧内对角【同位角】, or the sum of the interior angles on thesame side equal to two right angles, then the straight lines are parallel toone another.
Proposition 28
Ifa straight line falling on two straight lines makes the exterior angle equal tothe interior and opposite angle on the same side, or the sum of the interiorangles on the same side equal to two right angles, then the straight lines areparallel to one another.
Let the straight line EF fallingon the two straight lines AB and CD make theexterior angle EGB equal to the interior and oppositeangle GHD, or the sum of the interior angles on the same side,namely BGH and GHD, equal to two rightangles.
I say that AB isparallel to CD.
I.15
C.N.1. I.27
Since the angle EGB equalsthe angle GHD, and the angle EGB equals theangle AGH, therefore the angle AGH equals the angle GHD. Andthey are alternate错角, therefore AB isparallel to CD.
I.13
C.N.1
I.Post.4
Next, since the sum of theangles BGH and GHD equals two right angles,and the sum of the angles AGH and BGH alsoequals two right angles, therefore the sum of the angles AGH and BGH equalsthe sum of the angles BGH and GHD.
C.N.3
I.27
Subtract the angle BGH fromeach. Therefore the remaining angle AGH equals the remainingangle GHD. And they are alternate, therefore AB isparallel to CD.
Therefore if astraight line falling on two straight lines makes the exterior angle equal tothe interior and opposite angle on the same side, or the sum of the interiorangles on the same side equal to two right angles, then the straight lines areparallel to one another.
Q.E.D.
Proposition 29.
A straight line falling on parallel straight lines makes the alternateangles equal to one another, the exterior angle equal to the interior andopposite angle, and the sum of the interior angles on the same side equal totwo right angles.
Proposition 29
Astraight line falling on parallel straight lines makes the alternate anglesequal to one another, the exterior angle equal to the interior and oppositeangle, and the sum of the interior angles on the same side equal to two rightangles.
Let the straight line EF fallon the parallel straight lines AB and CD.
I say that it makes thealternate angles AGH and GHD equal,the exterior angle EGB equalto the interior and opposite angle GHD, andthe sum of the interior angles on the same side, namely BGH and GHD, equalto two right angles.
If the angle AGH doesnot equal the angle GHD, then one of them is greater. Let theangle AGH be greater.
Add the angle BGH toeach. Therefore the sum of the angles AGH and BGH isgreater than the sum of the angles BGH and GHD.
I.13
But sum of the angles AGH and BGH equalstwo right angles. Therefore the sum of the angles BGH and GHD isless than two right angles.
Post.5
But straight lines producedindefinitely from angles less than two right angles meet. Therefore AB and CD, ifproduced indefinitely, will meet. But they do not meet, because they are byhypothesis parallel.
Therefore the angle AGH isnot unequal to the angle GHD, and therefore equals it.
I.15, C.N.1
Again, the angle AGH equalsthe angle EGB. Therefore the angle EGB alsoequals the angle GHD.
C.N.2
Add the angle BGH toeach. Therefore the sum of the angles EGB and BGH equalsthe sum of the angles BGH and GHD.
I.13
C.N.1
But the sum of theangles EGB and BGH equals two right angles.Therefore the sum of the angles BGH and GHD alsoequals two right angles.
Therefore astraight line falling on parallel straight lines makes the alternate anglesequal to one another, the exterior angle equal to the interior and oppositeangle, and the sum of the interior angles on the same side equal to two rightangles.
Q.E.D.
Proposition 30.
Straight lines parallelto the same straight line are also parallel to one another.
Proposition 30
Straightlines parallel to the same straight line are also parallel to one another.
Let each of the straightlines AB and CD be parallel to EF.
I say that AB isalso parallel to CD.
I.29
Let the straight line GK fallupon them. Since the straight line GK falls on the parallelstraight lines AB and EF, therefore theangle AGK equals the angle GHF.
I.29
Again, since the straightline GK falls on the parallel straight lines EF and CD, thereforethe angle GHF equals the angle GKD.
C.N.1
But the angle AGK wasalso proved equal to the angle GHF. Therefore the angle AGK alsoequals the angle GKD, and they are alternate.
Therefore AB isparallel to CD.
Therefore straightlines parallel to the same straight line are also parallel to one another.
Q.E.D.
If two triangles have two angles equal to two angles respectively, and oneside equal to one side, namely, either the side adjoining the equal angles, orthat opposite one of the equal angles, then the remaining sides equal theremaining sides and the remaining angle equals the remaining angle.
Proposition 26
Iftwo triangles have two angles equal to two angles respectively, and one sideequal to one side, namely, either the side adjoining the equal angles, or thatopposite one of the equal angles, then the remaining sides equal the remainingsides and the remaining angle equals the remaining angle.
Let ABC and DEF betwo triangles having the two angles ABC and BCA equalto the two angles DEF and EFD respectively,namely the angle ABC to the angle DEF, and theangle BCA to the angle EFD, and let them alsohave one side equal to one side, first that adjoining the equal angles,namely BC equal to EF.
I say that the remainingsides equal the remaining sides respectively, namely AB equals DE and AC equals DF, andthe remaining angle equals the remaining angle, namely the angle BAC equalsthe angle EDF.
If AB doesnot equal DE, then one of them is greater.
I.3, I.Post.1
Let AB begreater. Make BG equal to DE, and join GC.
I.4
C.N.1
Since BG equals DE, and BC equals EF, thetwo sides GB and BC equal the two sides DE and EF respectively,and the angle GBC equals the angle DEF, thereforethe base GC equals the base DF, the triangle GBC equalsthe triangle DEF, and the remaining angles equalthe remaining angles, namely those opposite the equal sides. Therefore theangle GCB equals the angle DFE. But theangle DFE equals the angle ACB by hypothesis.Therefore the angle BCG equals the angle BCA, theless equals the greater, which is impossible.
Therefore AB isnot unequal to DE, and therefore equals it.
I.4
But BC alsoequals EF. Therefore the two sides AB and BC equalthe two sides DE and EF respectively, and theangle ABC equals the angle DEF. Therefore the base AC equalsthe base DF, and the remaining angle BAC equalsthe remaining angle EDF.
Next, let sides oppositeequal angles be equal, as AB equals DE.
I say again that theremaining sides equal the remaining sides, namely AC equals DF and BC equals EF, andfurther the remaining angle BAC equals the remainingangle EDF.
If BC isunequal to EF, then one of them is greater.
I.3, I.Post.1
Let BC begreater, if possible. Make BH equal to EF, andjoin AH.
I.4
Since BH equals EF, and AB equals DE, thetwo sides AB and BH equal the two sides DE and EF respectively,and they contain equal angles, therefore the base AH equalsthe base DF, the triangle ABH equals thetriangle DEF, and the remaining angles equal the remainingangles, namely those opposite the equal sides. Therefore the angle BHA equalsthe angle EFD.
C.N.1
I.16
But the angle EFD equalsthe angle BCA, therefore, in the triangle AHC, theexterior angle BHA equals the interior and opposite angle BCA, whichis impossible.
Therefore BC isnot unequal to EF, and therefore equals it.
I.4
But AB alsoequals DE. Therefore the two sides AB and BC equalthe two sides DE and EF respectively, andthey contain equal angles. Therefore the base AC equalsthe base DF, thetriangle ABC equals the triangle DEF, and theremaining angle BAC equals the remaining angle EDF.
Therefore if twotriangles have two angles equal to two angles respectively, and one side equalto one side, namely, either the side adjoining the equal angles, or thatopposite one of the equal angles, then the remaining sides equal the remainingsides and the remaining angle equals the remaining angle.
Q.E.D.
Proposition 27.
If a straight line falling on two straight lines makes the alternateangles equal to one another, then the straight lines are parallel to oneanother.
Proposition 27
Ifa straight line falling on two straight lines makes the alternate angles equalto one another, then the straight lines are parallel to one another.
Let the straight line EF fallingon the two straight lines AB and CD make thealternate angles AEF and EFD equal to oneanother.
I say that AB isparallel to CD.
If not, AB and CD whenproduced meet either in the direction of B and D ortowards A and C.
Let them be produced andmeet, in the direction of B and D, at G.
I.16
Then, in the triangle GEF, theexterior angle AEF equals the interior and oppositeangle EFG, which is impossible.
Therefore AB and CD whenproduced do not meet in the direction of B and D.
Similarly it can be provedthat neither do they meet towards A and C.
I.Def.23
But straight lines which donot meet in either direction are parallel. Therefore AB isparallel to CD.
Therefore if astraight line falling on two straight lines makes the alternate angles equal toone another, then the straight lines are parallel to one another.
Q.E.D.
Proposition 28.
If a straight line falling on two straight lines makes the exterior angle外角 equal to the interior and opposite angle on the same side同侧内对角【同位角】, or the sum of the interior angles on thesame side equal to two right angles, then the straight lines are parallel toone another.
Proposition 28
Ifa straight line falling on two straight lines makes the exterior angle equal tothe interior and opposite angle on the same side, or the sum of the interiorangles on the same side equal to two right angles, then the straight lines areparallel to one another.
Let the straight line EF fallingon the two straight lines AB and CD make theexterior angle EGB equal to the interior and oppositeangle GHD, or the sum of the interior angles on the same side,namely BGH and GHD, equal to two rightangles.
I say that AB isparallel to CD.
I.15
C.N.1. I.27
Since the angle EGB equalsthe angle GHD, and the angle EGB equals theangle AGH, therefore the angle AGH equals the angle GHD. Andthey are alternate错角, therefore AB isparallel to CD.
I.13
C.N.1
I.Post.4
Next, since the sum of theangles BGH and GHD equals two right angles,and the sum of the angles AGH and BGH alsoequals two right angles, therefore the sum of the angles AGH and BGH equalsthe sum of the angles BGH and GHD.
C.N.3
I.27
Subtract the angle BGH fromeach. Therefore the remaining angle AGH equals the remainingangle GHD. And they are alternate, therefore AB isparallel to CD.
Therefore if astraight line falling on two straight lines makes the exterior angle equal tothe interior and opposite angle on the same side, or the sum of the interiorangles on the same side equal to two right angles, then the straight lines areparallel to one another.
Q.E.D.
Proposition 29.
A straight line falling on parallel straight lines makes the alternateangles equal to one another, the exterior angle equal to the interior andopposite angle, and the sum of the interior angles on the same side equal totwo right angles.
Proposition 29
Astraight line falling on parallel straight lines makes the alternate anglesequal to one another, the exterior angle equal to the interior and oppositeangle, and the sum of the interior angles on the same side equal to two rightangles.
Let the straight line EF fallon the parallel straight lines AB and CD.
I say that it makes thealternate angles AGH and GHD equal,the exterior angle EGB equalto the interior and opposite angle GHD, andthe sum of the interior angles on the same side, namely BGH and GHD, equalto two right angles.
If the angle AGH doesnot equal the angle GHD, then one of them is greater. Let theangle AGH be greater.
Add the angle BGH toeach. Therefore the sum of the angles AGH and BGH isgreater than the sum of the angles BGH and GHD.
I.13
But sum of the angles AGH and BGH equalstwo right angles. Therefore the sum of the angles BGH and GHD isless than two right angles.
Post.5
But straight lines producedindefinitely from angles less than two right angles meet. Therefore AB and CD, ifproduced indefinitely, will meet. But they do not meet, because they are byhypothesis parallel.
Therefore the angle AGH isnot unequal to the angle GHD, and therefore equals it.
I.15, C.N.1
Again, the angle AGH equalsthe angle EGB. Therefore the angle EGB alsoequals the angle GHD.
C.N.2
Add the angle BGH toeach. Therefore the sum of the angles EGB and BGH equalsthe sum of the angles BGH and GHD.
I.13
C.N.1
But the sum of theangles EGB and BGH equals two right angles.Therefore the sum of the angles BGH and GHD alsoequals two right angles.
Therefore astraight line falling on parallel straight lines makes the alternate anglesequal to one another, the exterior angle equal to the interior and oppositeangle, and the sum of the interior angles on the same side equal to two rightangles.
Q.E.D.
Proposition 30.
Straight lines parallelto the same straight line are also parallel to one another.
Proposition 30
Straightlines parallel to the same straight line are also parallel to one another.
Let each of the straightlines AB and CD be parallel to EF.
I say that AB isalso parallel to CD.
I.29
Let the straight line GK fallupon them. Since the straight line GK falls on the parallelstraight lines AB and EF, therefore theangle AGK equals the angle GHF.
I.29
Again, since the straightline GK falls on the parallel straight lines EF and CD, thereforethe angle GHF equals the angle GKD.
C.N.1
But the angle AGK wasalso proved equal to the angle GHF. Therefore the angle AGK alsoequals the angle GKD, and they are alternate.
Therefore AB isparallel to CD.
Therefore straightlines parallel to the same straight line are also parallel to one another.
Q.E.D.
