Please remember: the status of this text is draft. You have been granted access through the generosity of the editor. Please use the comments to help make suggestions or corrections.

Edition: 0.0.0-dev | March 03, 2014

Authority: SCTA

License Availablity: free, Published under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 License

Sources:

X:

8 A plane angle is the inclination to one another of two lines in a plane which meet one another and do not lie in a straight line.

10 When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right , and the straight line standing on the other is called a perpendicular to that on which it stands.

15 A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure are equal to one another;

17 A diameter of the circle is any straight line drawn through the centre and terminated in both directions by the circumference of the circle, and such a straight line also bisects the circle.

18 A semicircle is the figure contained by the diameter and the circumference cut off by it. And the centre of the semicircle is the same as that of the circle.

19 Rectilineal figures are those which are contained by straight lines, trilateral figures being those contained by three, quadrilateral those contained by four, and multilateral those contained by more than four straight lines.

20 Of trilateral figures, an equilateral triangle is that which has its three sides equal, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle that which has its three sides unequal.

21 Further, of trilateral figures, a right-angled triangle is that which has a right angle, an obtuse-angled triangle that which has an obtuse angle, and an acuteangled triangle that which has its three angles acute.

22 Of quadrilateral figures, a square is that which is both equilateral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. And let quadrilaterals other than these be called trapezia.

35 On a given finite straight line to construct an equilateral triangle.

36 Let AB be the given finite straight line.

37 Thus it is required to construct an equilateral triangle on the straight line AB .

38 With centre A and distance AB let the circle BCD be described; [ Post. 3 ] again, with centre B and distance BA let the circle ACE be described; [ Post. 3 ] and from the point C , in which the circles cut one another, to the points A , B let the straight lines CA , CB be joined. [ Post. 1 ]

39 Now, since the point A is the centre of the circle CDB , AC is equal to AB . [ Def. 15 ]

40 Again, since the point B is the centre of the circle CAE , BC is equal to BA . [ Def. 15 ]

41 But CA was also proved equal to AB ; therefore each of the straight lines CA , CB is equal to AB .

42 And things which are equal to the same thing are also equal to one another; [ C.N. 1 ] therefore CA is also equal to CB .

43 Therefore the three straight lines CA , AB , BC are equal to one another.

44 Therefore the triangle ABC is equilateral; and it has been constructed on the given finite straight line AB .

45(Being) what it was required to do.

46 To place at a given point (as an extremity) a straight line equal to a given straight line.

47 Let A be the given point, and BC the given straight line.

48 Thus it is required to place at the point A (as an extremity) a straight line equal to the given straight line BC .

49 From the point A to the point B let the straight line AB be joined; [ Post. 1 ] and on it let the equilateral triangle DAB be constructed. [ I. 1 ]

50 Let the straight lines AE , BF be produced in a straight line with DA , DB ; [ Post. 2 ] with centre B and distance BC let the circle CGH be described; [ Post. 3 ] and again, with centre D and distance DG let the circle GKL be described. [ Post. 3 ]

51 Then, since the point B is the centre of the circle CGH , BC is equal to BG .

52 Again, since the point D is the centre of the circle GKL , DL is equal to DG .

53 And in these DA is equal to DB ; therefore the remainder AL is equal to the remainder BG. [ C.N. 3 ]

54 But BC was also proved equal to BG ; therefore each of the straight lines AL , BC is equal to BG .

55 And things which are equal to the same thing are also equal to one another; [ C.N. 1 ] therefore AL is also equal to BC .

56 Therefore at the given point A the straight line AL is placed equal to the given straight line BC .

57(Being) what it was required to do.

58 Given two unequal straight lines, to cut off from the greater a straight line equal to the less.

59 Let AB , C be the-two given unequal straight lines, and let AB be the greater of them.

60 Thus it is required to cut off from AB the greater a straight line equal to C the less.

61 At the point A let AD be placed equal to the straight line C ; [ I. 2 ] and with centre A and distance AD let the circle DEF be described. [ Post. 3 ] Now, since the point A is the centre of the circle DEF , AE is equal to AD . [ Def. 15 ] But C is also equal to AD . Therefore each of the straight lines AE , C is equal to AD ; so that AE is also equal to C . [ C.N. 1 ]

62 Therefore, given the two straight lines AB , C , from AB the greater AE has been cut off equal to C the less.

63(Being) what it was required to do.

64 If two triangles have the two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend.

65 Let ABC , DEF be two triangles having the two sides AB , AC equal to the two sides DE , DF respectively, namely AB to DE and AC to DF , and the angle BAC equal to the angle EDF .

66 I say that the base BC is also equal to the base EF , the triangle ABC will be equal to the triangle DEF , and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, that is, the angle ABC to the angle DEF , and the angle ACB to the angle DFE .

67 For, if the triangle ABC be applied to the triangle DEF , and if the point A be placed on the point D and the straight line AB on DE , then the point B will also coincide with E , because AB is equal to DE .

68 Again, AB coinciding with DE , the straight line AC will also coincide with DF , because the angle BAC is equal to the angle EDF ; hence the point C will also coincide with the point F , because AC is again equal to DF .

69 But B also coincided with E ; hence the base BC will coincide with the base EF .

70 [For if, when B coincides with E and C with F , the base BC does not coincide with the base EF , two straight lines will enclose a space: which is impossible. Therefore the base BC will coincide with EF ] and will be equal to it. [ C.N. 4 ]

71 Thus the whole triangle ABC will coincide with the whole triangle DEF , and will be equal to it.

72 And the remaining angles will also coincide with the remaining angles and will be equal to them, the angle ABC to the angle DEF , and the angle ACB to the angle DFE .

73 Therefore etc.

74(Being) what it was required to prove.

75 In isosceles triangles the angles at the base are equal to one another, and, if the equal straight lines be produced further, the angles under the base will be equal to one another.

76 Let ABC be an isosceles triangle having the side AB equal to the side AC ; and let the straight lines BD , CE be produced further in a straight line with AB , AC . [ Post. 2 ]

77 I say that the angle ABC is equal to the angle ACB , and the angle CBD to the angle BCE .

78 Let a point F be taken at random on BD ; from AE the greater let AG be cut off equal to AF the less; [ I. 3 ] and let the straight lines FC , GB be joined. [ Post. 1 ]

79 Then, since AF is equal to AG and AB to AC , the two sides FA , AC are equal to the two sides GA , AB , respectively; and they contain a common angle, the angle FAG . Therefore the base FC is equal to the base GB , and the triangle AFC is equal to the triangle AGB , and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, that is, the angle ACF to the angle ABG , and the angle AFC to the angle AGB . [ I. 4 ]

80 And, since the whole AF is equal to the whole AG , and in these AB is equal to AC , the remainder BF is equal to the remainder CG .

81 But FC was also proved equal to GB ; therefore the two sides BF , FC are equal to the two sides CG , GB respectively; and the angle BFC is equal to the angle CGB , while the base BC is common to them; therefore the triangle BFC is also equal to the triangle CGB , and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; therefore the angle FBC is equal to the angle GCB , and the angle BCF to the angle CBG .

82 Accordingly, since the whole angle ABG was proved equal to the angle ACF , and in these the angle CBG is equal to the angle BCF , the remaining angle ABC is equal to the remaining angle ACB ; and they are at the base of the triangle ABC . But the angle FBC was also proved equal to the angle GCB ; and they are under the base.

83Therefore etc.

84Q. E. D.

85 If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.

86 Let ABC be a triangle having the angle ABC equal to the angle ACB ;

87 I say that the side AB is also equal to the side AC .

88 For, if AB is unequal to AC , one of them is greater.

89 Let AB be greater; and from AB the greater let DB be cut off equal to AC the less;

90 let DC be joined.

91 Then, since DB is equal to AC , and BC is common, the two sides DB , BC are equal to the two sides AC , CB respectively; and the angle DBC is equal to the angle ACB ; therefore the base DC is equal to the base AB , and the triangle DBC will be equal to the triangle ACB , the less to the greater: which is absurd. Therefore AB is not unequal to AC ; it is therefore equal to it.

92Therefore etc.

93Q. E. D.

94 Given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there cannot be constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.

95 For, if possible, given two straight lines AC , CB constructed on the straight line AB and meeting at the point C , let two other straight lines AD , DB be constructed on the same straight line AB , on the same side of it, meeting in another point D and equal to the former two respectively, namely each to that which has the same extremity with it, so that CA is equal to DA which has the same extremity A with it, and CB to DB which has the same extremity B with it; and let CD be joined.

96 Then, since AC is equal to AD , the angle ACD is also equal to the angle ADC ; [ I. 5 ] therefore the angle ADC is greater than the angle DCB ; therefore the angle CDB is much greater than the angle DCB .

97 Again, since CB is equal to DB , the angle CDB is also equal to the angle DCB . But it was also proved much greater than it: which is impossible.

98Therefore etc.

99Q. E. D.

100 If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they will also have the angles equal which are contained by the equal straight lines.

101 Let ABC , DEF be two triangles having the two sides AB , AC equal to the two sides DE , DF respectively, namely AB to DE , and AC to DF ; and let them have the base BC equal to the base EF ;

102 I say that the angle BAC is also equal to the angle EDF .

103 For, if the triangle ABC be applied to the triangle DEF , and if the point B be placed on the point E and the straight line BC on EF , the point C will also coincide with F , because BC is equal to EF .

104 Then, BC coinciding with EF , BA , AC will also coincide with ED , DF ; for, if the base BC coincides with the base EF , and the sides BA , AC do not coincide with ED , DF but fall beside them as EG , GF , then, given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there will have been constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it. But they cannot be so constructed. [ I. 7 ]

105 Therefore it is not possible that, if the base BC be applied to the base EF , the sides BA , AC should not coincide with ED , DF ; they will therefore coincide, so that the angle BAC will also coincide with the angle EDF , and will be equal to it.

106If therefore etc.

107Q. E. D.

108To bisect a given rectilineal angle.

109 Let the angle BAC be the given rectilineal angle.

110Thus it is required to bisect it.

111 Let a point D be taken at random on AB ; let AE be cut off from AC equal to AD ; [ I. 3 ] let DE be joined, and on DE let the equilateral triangle DEF be constructed; let AF be joined.

112 I say that the angle BAC has been bisected by the straight line AF .

113 For, since AD is equal to AE , and AF is common, the two sides DA , AF are equal to the two sides EA , AF respectively.

114 And the base DF is equal to the base EF ; therefore the angle DAF is equal to the angle EAF . [ I. 8 ]

115 Therefore the given rectilineal angle BAC has been bisected by the straight line AF .

116Q. E. F.

117To bisect a given finite straight line.

118 Let AB be the given finite straight line.

119 Thus it is required to bisect the finite straight line AB .

120 Let the equilateral triangle ABC be constructed on it, [ I. 1 ] and let the angle ACB be bisected by the straight line CD ; [ I. 9 ]

121 I say that the straight line AB has been bisected at the point D .

122 For, since AC is equal to CB , and CD is common, the two sides AC , CD are equal to the two sides BC , CD respectively; and the angle ACD is equal to the angle BCD ; therefore the base AD is equal to the base BD . [ I. 4 ]

123 Therefore the given finite straight line AB has been bisected at D .

124Q. E. F.

125 To draw a straight line at right angles to a given straight line from a given point on it.

126 Let AB be the given straight line, and C the given point on it.

127 Thus it is required to draw from the point C a straight line at right angles to the straight line AB .

128 Let a point D be taken at random on AC ; let CE be made equal to CD ; [ I. 3 ] on DE let the equilateral triangle FDE be constructed, [ I. 1 ] and let FC be joined;

129 I say that the straight line FC has been drawn at right angles to the given straight line AB from C the given point on it.

130 For, since DC is equal to CE , and CF is common, the two sides DC , CF are equal to the two sides EC , CF respectively; and the base DF is equal to the base FE ; therefore the angle DCF is equal to the angle ECF ; [ I. 8 ] and they are adjacent angles.

131 But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [ Def. 10 ] therefore each of the angles DCF , FCE is right.

132 Therefore the straight line CF has been drawn at right angles to the given straight line AB from the given point C on it.

133Q. E. F.

134 To a given infinite straight line, from a given point which is not on it, to draw a perpendicular straight line.

135 Let AB be the given infinite straight line, and C the given point which is not on it; thus it is required to draw to the given infinite straight line AB , from the given point C which is not on it, a perpendicular straight line.

136 For let a point D be taken at random on the other side of the straight line AB , and with centre C and distance CD let the circle EFG be described; [ Post. 3 ] let the straight line EG be bisected at H , [ I. 10 ] and let the straight lines CG , CH , CE be joined. [ Post. 1 ]

137 I say that CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it.

138 For, since GH is equal to HE , and HC is common, the two sides GH , HC are equal to the two sides EH , HC respectively; and the base CG is equal to the base CE ; therefore the angle CHG is equal to the angle EHC . [ I. 8 ] And they are adjacent angles.

139 But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands. [ Def. 10 ]

140 Therefore CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it.

141Q. E. F.

142 If a straight line set up on a straight line make angles, it will make either two right angles or angles equal to two right angles.

143 For let any straight line AB set up on the straight line CD make the angles CBA , ABD ;

144 I say that the angles CBA , ABD are either two right angles or equal to two right angles.

145 Now, if the angle CBA is equal to the angle ABD , they are two right angles. [ Def. 10 ]

146 But, if not, let BE be drawn from the point B at right angles to CD ; [ I. 11 ] therefore the angles CBE , EBD are two right angles.

147 Then, since the angle CBE is equal to the two angles CBA , ABE , let the angle EBD be added to each; therefore the angles CBE , EBD are equal to the three angles CBA , ABE , EBD . [ C. N . 2]

148 Again, since the angle DBA is equal to the two angles DBE , EBA , let the angle ABC be added to each; therefore the angles DBA . ABC are equal to the three angles DBE , EBA , ABC . [ C. N . 2]

149 But the angles CBE , EBD were also proved equal to the same three angles; and things which are equal to the same thing are also equal to one another; [ C. N . 1] therefore the angles CBE , EBD are also equal to the angles DBA , ABC . But the angles CBE , EBD are two right angles; therefore the angles DBA , ABC are also equal to two right angles.

150Therefore etc.

151Q. E. D.

152 If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with one another.

153 For with any straight line AB , and at the point B on it, let the two straight lines BC , BD not lying on the same side make the adjacent angles ABC , ABD equal to two right angles;

154 I say that BD is in a straight line with CB .

155 For, if BD is not in a straight line with BC , let BE be in a straight line with CB .

156 Then, since the straight line AB stands on the straight line CBE , the angles ABC , ABE are equal to two right angles. [ I. 13 ] But the angles ABC , ABD are also equal to two right angles; therefore the angles CBA , ABE are equal to the angles CBA , ABD . [ Post. 4 and C.N. 1 ]

157 Let the angle CBA be subtracted from each; therefore the remaining angle ABE is equal to the remaining angle ABD , [ C.N. 3 ] the less to the greater: which is impossible. Therefore BE is not in a straight line with CB .

158 Similarly we can prove that neither is any other straight line except BD . Therefore CB is in a straight line with BD .

159Therefore etc.

160Q. E. D.

161 If two straight lines cut one another, they make the vertical angles equal to one another.

162 For let the straight lines AB , CD cut one another at the point E ;

163 I say that the angle AEC is equal to the angle DEB , and the angle CEB to the angle AED .

164 For, since the straight line AE stands on the straight line CD , making the angles CEA , AED , the angles CEA , AED are equal to two right angles [ I. 13 ]

165 Again, since the straight line DE stands on the straight line AB , making the angles AED , DEB , the angles AED , DEB are equal to two right angles. [ I. 13 ]

166 But the angles CEA , AED were also proved equal to two right angles; therefore the angles CEA , AED are equal to the angles AED DEB . [ Post. 4 and C. N. 1 ] Let the angle AED be subtracted from each; therefore the remaining angle CEA is equal to the remaining angle BED . [ C. N. 3 ]

167 Similarly it can be proved that the angles CEB , DEA are also equal.

168 Therefore etc. Q. E. D.

170 In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles.

171 Let ABC be a triangle, and let one side of it BC be produced to D ;

172 I say that the exterior angle ACD is greater than either of the interior and opposite angles CBA , BAC .

173 Let AC be bisected at E [ I. 10 ], and let BE be joined and produced in a straight line to F ;

174 let EF be made equal to BE [ I. 3 ], let FC be joined [ Post. 1 ], and let AC be drawn through to G [ Post. 2 ].

175 Then, since AE is equal to EC , and BE to EF , the two sides AE , EB are equal to the two sides CE , EF respectively; and the angle AEB is equal to the angle FEC , for they are vertical angles. [ I. 15 ] Therefore the base AB is equal to the base FC , and the triangle ABE is equal to the triangle CFE , and the remaining angles are equal to the remaining angles respectively, namely those which the equal sides subtend; [ I. 4 ] therefore the angle BAE is equal to the angle ECF .

176 But the angle ECD is greater than the angle ECF ; [ C. N . 5] therefore the angle ACD is greater than the angle BAE .

177 Similarly also, if BC be bisected, the angle BCG , that is, the angle ACD [ I. 15 ], can be proved greater than the angle ABC as well.

178Therefore etc.

179Q. E. D.

180 In any triangle two angles taken together in any manner are less than two right angles.

181 Let ABC be a triangle; I say that two angles of the triangle ABC taken together in any manner are less than two right angles.

182 For let BC be produced to D . [ Post. 2 ]

183 Then, since the angle ACD is an exterior angle of the triangle ABC ,

184 it is greater than the interior and opposite angle ABC . [ I. 16 ] Let the angle ACB be added to each; therefore the angles ACD , ACB are greater than the angles ABC , BCA . But the angles ACD , ACB are equal to two right angles. [ I. 13 ]

185 Therefore the angles ABC , BCA are less than two right angles.

186 Similarly we can prove that the angles BAC , ACB are also less than two right angles, and so are the angles CAB , ABC as well.

187Therefore etc.

188Q. E. D.

189 In any triangle the greater side subtends the greater angle.

190 For let ABC be a triangle having the side AC greater than AB ;

191 I say that the angle ABC is also greater than the angle BCA .

192 For, since AC is greater than AB , let AD be made equal to AB [ I. 3 ], and let BD bejoined.

193 Then, since the angle ADB is an exterior angle of the triangle BCD ,

194 it is greater than the interior and opposite angle DCB . [ I. 16 ]

195 But the angle ADB is equal to the angle ABD , since the side AB is equal to AD ; therefore the angle ABD is also greater than the angle ACB ; therefore the angle ABC is much greater than the angle ACB .

196Therefore etc.

197Q. E. D.

198 In any triangle the greater angle is subtended by the greater side.

199 Let ABC be a triangle having the angle ABC greater than the angle BCA ;

200 I say that the side AC is also greater than the side AB .

201 For, if not, AC is either equal to AB or less.

202 Now AC is not equal to AB ; for then the angle ABC would also have been equal to the angle ACB ; [ I. 5 ] but it is not; therefore AC is not equal to AB .

203 Neither is AC less than AB , for then the angle ABC would also have been less than the angle ACB ; [ I. 18 ] but it is not; therefore AC is not less than AB .

204 And it was proved that it is not equal either. Therefore AC is greater than AB .

205Therefore etc.

206Q. E. D.

207 In any triangle two sides taken together in any manner are greater than the remaining one.

208 For let ABC be a triangle; I say that in the triangle ABC two sides taken together in any manner are greater than the remaining one, namely BA , AC greater than BC , AB , BC greater than AC , BC , CA greater than AB .

209 For let BA be drawn through to the point D , let DA be made equal to CA , and let DC be joined.

210 Then, since DA is equal to AC , the angle ADC is also equal to the angle ACD ; [ I. 5 ] therefore the angle BCD is greater than the angle ADC . [ C.N. 5 ]

211 And, since DCB is a triangle having the angle BCD greater than the angle BDC , and the greater angle is subtended by the greater side, [ I. 19 ] therefore DB is greater than BC .

212 But DA is equal to AC ; therefore BA , AC are greater than BC .

213 Similarly we can prove that AB , BC are also greater than CA , and BC , CA than AB .

214Therefore etc.

215Q. E. D.

216 If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle.

217 On BC , one of the sides of the triangle ABC , from its extremities B , C , let the two straight lines BD , DC be constructed meeting within the triangle;

218 I say that BD , DC are less than the remaining two sides of the triangle BA , AC , but contain an angle BDC greater than the angle BAC .

219 For let BD be drawn through to E .

220 Then, since in any triangle two sides are greater than the remaining one, [ I. 20 ] therefore, in the triangle ABE , the two sides AB , AE are greater than BE .

221 Let EC be added to each; therefore BA , AC are greater than BE , EC .

222 Again, since, in the triangle CED , the two sides CE , ED are greater than CD , let DB be added to each; therefore CE , EB are greater than CD , DB .

223 But BA , AC were proved greater than BE , EC ; therefore BA , AC are much greater than BD , DC .

224 Again, since in any triangle the exterior angle is greater than the interior and opposite angle, [ I. 16 ] therefore, in the triangle CDE , the exterior angle BDC is greater than the angle CED .

225 For the same reason, moreover, in the triangle ABE also, the exterior angle CEB is greater than the angle BAC . But the angle BDC was proved greater than the angle CEB ; therefore the angle BDC is much greater than the angle BAC .

226Therefore etc.

227Q. E. D.

228 Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one. [ I. 20 ]

229 Let the three given straight lines be A , B , C , and of these let two taken together in any manner be greater than the remaining one, namely A , B greater than C , A , C greater than B , and B , C greater than A ; thus it is required to construct a triangle out of straight lines equal to A , B , C .

230 Let there be set out a straight line DE , terminated at D but of infinite length in the direction of E , and let DF be made equal to A , FG equal to B , and GH equal to C . [ I. 3 ]

231 With centre F and distance FD let the circle DKL be described; again, with centre G and distance GH let the circle KLH be described; and let KF , KG be joined;

232 I say that the triangle KFG has been constructed out of three straight lines equal to A , B , C .

233 For, since the point F is the centre of the circle DKL , FD is equal to FK .

234 But FD is equal to A ; therefore KF is also equal to A .

235 Again, since the point G is the centre of the circle LKH , GH is equal to GK .

236 But GH is equal to C ; therefore KG is also equal to C . And FG is also equal to B ; therefore the three straight lines KF , FG , GK are equal to the three straight lines A , B , C .

237 Therefore out of the three straight lines KF , FG , GK , which are equal to the three given straight lines A , B , C , the triangle KFG has been constructed.

238Q. E. F.

239 On a given straight line and at a point on it to construct a rectilineal angle equal to a given rectilineal angle.

240 Let AB be the given straight line, A the point on it, and the angle DCE the given rectilineal angle;

241 thus it is required to construct on the given straight line AB , and at the point A on it, a rectilineal angle equal to the given rectilineal angle DCE .

242 On the straight lines CD , CE respectively let the points D , E be taken at random; let DE be joined, and out of three straight lines which are equal to the three straight lines CD , DE , CE let the triangle AFG be constructed in such a way that CD is equal to AF , CE to AG , and further DE to FG .

243 Then, since the two sides DC , CE are equal to the two sides FA , AG respectively, and the base DE is equal to the base FG , the angle DCE is equal to the angle FAG . [ I. 8 ]

244 Therefore on the given straight line AB , and at the point A on it, the rectilineal angle FAG has been constructed equal to the given rectilineal angle DCE .

245Q. E. F.

246 If two triangles have the two sides equal to two sides respectively, but have the one of the angles contained by the equal straight lines greater than the other, they will also have the base greater than the base.

247 Let ABC , DEF be two triangles having the two sides AB , AC equal to the two sides DE , DF respectively, namely AB to DE , and AC to DF , and let the angle at A be greater than the angle at D ;

248 I say that the base BC is also greater than the base EF .

249 For, since the angle BAC is greater than the angle EDF , let there be constructed, on the straight line DE , and at the point D on it, the angle EDG equal to the angle BAC ; [ I. 23 ] let DG be made equal to either of the two straight lines AC , DF , and let EG , FG be joined.

250 Then, since AB is equal to DE , and AC to DG , the two sides BA , AC are equal to the two sides ED , DG , respectively; and the angle BAC is equal to the angle EDG ; therefore the base BC is equal to the base EG . [ I. 4 ]

251 Again, since DF is equal to DG , the angle DGF is also equal to the angle DFG ; [ I. 5 ] therefore the angle DFG is greater than the angle EGF .

252 Therefore the angle EFG is much greater than the angle EGF .

253 And, since EFG is a triangle having the angle EFG greater than the angle EGF , and the greater angle is subtended by the greater side, [ I. 19 ] the side EG is also greater than EF .

254 But EG is equal to BC . Therefore BC is also greater than EF .

255Therefore etc.

256Q. E. D.

257 If two triangles have the two sides equal to two sides respectively, but have the base greater than the base, they will also have the one of the angles contained by the equal straight lines greater than the other.

258 Let ABC , DEF be two triangles having the two sides AB , AC equal to the two sides DE , DF respectively, namely AB to DE , and AC to DF ; and let the base BC be greater than the base EF ;

259 I say that the angle BAC is also greater than the angle EDF .

260For, if not, it is either equal to it or less.

261 Now the angle BAC is not equal to the angle EDF ; for then the base BC would also have been equal to the base EF , [ I. 4 ] but it is not; therefore the angle BAC is not equal to the angle EDF .

262 Neither again is the angle BAC less than the angle EDF ; for then the base BC would also have been less than the base EF , [ I. 24 ] but it is not; therefore the angle BAC is not less than the angle EDF .

263 But it was proved that it is not equal either; therefore the angle BAC is greater than the angle EDF .

264Therefore etc.

265Q. E. D.

266 If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle to the remaining angle.

267 Let ABC , DEF be two triangles having the two angles ABC , BCA equal to the two angles DEF , EFD respectively, namely the angle ABC to the angle DEF , and the angle BCA to the angle EFD ; and let them also have one side equal to one side, first that adjoining the equal angles, namely BC to EF ;

268 I say that they will also have the remaining sides equal to the remaining sides respectively, namely AB to DE and AC to DF , and the remaining angle to the remaining angle, namely the angle BAC to the angle EDF .

269 For, if AB is unequal to DE , one of them is greater.

270 Let AB be greater, and let BG be made equal to DE ; and let GC be joined.

271 Then, since BG is equal to DE , and BC to EF , the two sides GB , BC are equal to the two sides DE , EF respectively; and the angle GBC is equal to the angle DEF ; therefore the base GC is equal to the base DF , and the triangle GBC is equal to the triangle DEF , and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [ I. 4 ] therefore the angle GCB is equal to the angle DFE . But the angle DFE is by hypothesis equal to the angle BCA ; therefore the angle BCG is equal to the angle BCA , the less to the greater: which is impossible. Therefore AB is not unequal to DE , and is therefore equal to it.

272 But BC is also equal to EF ; therefore the two sides AB , BC are equal to the two sides DE , EF respectively, and the angle ABC is equal to the angle DEF ; therefore the base AC is equal to the base DF , and the remaining angle BAC is equal to the remaining angle EDF . [ I. 4 ]

273 Again, let sides subtending equal angles be equal, as AB to DE ;

274 I say again that the remaining sides will be equal to the remaining sides, namely AC to DF and BC to EF , and further the remaining angle BAC is equal to the remaining angle EDF .

275 For, if BC is unequal to EF , one of them is greater.

276 Let BC be greater, if possible, and let BH be made equal to EF ; let AH be joined.

277 Then, since BH is equal to EF , and AB to DE , the two sides AB , BH are equal to the two sides DE , EF respectively, and they contain equal angles; therefore the base AH is equal to the base DF , and the triangle ABH is equal to the triangle DEF , and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [ I. 4 ] therefore the angle BHA is equal to the angle EFD .

278 But the angle EFD is equal to the angle BCA ; therefore, in the triangle AHC , the exterior angle BHA is equal to the interior and opposite angle BCA : which is impossible. [ I. 16 ]

279 Therefore BC is not unequal to EF , and is therefore equal to it.

280 But AB is also equal to DE ; therefore the two sides AB , BC are equal to the two sides DE , EF respectively, and they contain equal angles; therefore the base AC is equal to the base DF , the triangle ABC equal to the triangle DEF , and the remaining angle BAC equal to the remaining angle EDF . [ I. 4 ]

281Therefore etc.

282Q. E. D.

283 If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another.

284 For let the straight line EF falling on the two straight lines AB , CD make the alternate angles AEF , EFD equal to one another;

285 I say that AB is parallel to CD .

286 For, if not, AB , CD when produced will meet either in the direction of B , D or towards A , C .

287 Let them be produced and meet, in the direction of B , D , at G .

288 Then, in the triangle GEF , the exterior angle AEF is equal to the interior and opposite angle EFG : which is impossible. [ I. 16 ]

289 Therefore AB , CD when produced will not meet in the direction of B , D .

290 Similarly it can be proved that neither will they meet towards A , C .

291 But straight lines which do not meet in either direction are parallel; [ Def. 23 ] therefore AB is parallel to CD .

292Therefore etc.

293Q. E. D.

294 If a straight line falling on two straight lines make the exterior angle equal to the interior and opposite angle on the same side, or the interior angles on the same side equal to two right angles, the straight lines will be parallel to one another.

295 For let the straight line EF falling on the two straight lines AB , CD make the exterior angle EGB equal to the interior and opposite angle GHD , or the interior angles on the same side, namely BGH , GHD , equal to two right angles;

296 I say that AB is parallel to CD .

297 For, since the angle EGB is equal to the angle GHD , while the angle EGB is equal to the angle AGH , [ I. 15 ] the angle AGH is also equal to the angle GHD ; and they are alternate; therefore AB is parallel to CD . [ I. 27 ]

298 Again, since the angles BGH , GHD are equal to two right angles, and the angles AGH , BGH are also equal to two right angles, [ I. 13 ] the angles AGH , BGH are equal to the angles BGH , GHD .

299 Let the angle BGH be subtracted from each; therefore the remaining angle AGH is equal to the remaining angle GHD ; and they are alternate; therefore AB is parallel to CD . [ I. 27 ]

300Therefore etc.

301Q. E. D.

302 A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the interior angles on the same side equal to two right angles.

303 For let the straight line EF fall on the parallel straight lines AB , CD ;

304 I say that it makes the alternate angles AGH , GHD equal, the exterior angle EGB equal to the interior and opposite angle GHD , and the interior angles on the same side, namely BGH , GHD , equal to two right angles.

305 For, if the angle AGH is unequal to the angle GHD , one of them is greater.

306 Let the angle AGH be greater.

307 Let the angle BGH be added to each; therefore the angles AGH , BGH are greater than the angles BGH , GHD .

308 But the angles AGH , BGH are equal to two right angles; [ I. 13 ] therefore the angles BGH , GHD are less than two right angles.

309 But straight lines produced indefinitely from angles less than two right angles meet; [ Post. 5 ] therefore AB , CD , if produced indefinitely, will meet; but they do not meet, because they are by hypothesis parallel.

310 Therefore the angle AGH is not unequal to the angle GHD , and is therefore equal to it.

311 Again, the angle AGH is equal to the angle EGB ; [ I. 15 ] therefore the angle EGB is also equal to the angle GHD . [ C.N. 1 ]

312 Let the angle BGH be added to each; therefore the angles EGB , BGH are equal to the angles BGH , GHD . [ C.N. 2 ]

313 But the angles EGB , BGH are equal to two right angles; [ I. 13 ] therefore the angles BGH , GHD are also equal to two right angles.

314Therefore etc.

315Q. E. D.

316 Straight lines parallel to the same straight line are also parallel to one another.

317 Let each of the straight lines AB , CD be parallel to EF ; I say that AB is also parallel to CD .

318 For let the straight line GK fall upon them;

319 Then, since the straight line GK has fallen on the parallel straight lines AB , EF , the angle AGK is equal to the angle GHF . [ I. 29 ]

320 Again, since the straight line GK has fallen on the parallel straight lines EF , CD , the angle GHF is equal to the angle GKD . [ I. 29 ]

321 But the angle AGK was also proved equal to the angle GHF ; therefore the angle AGK is also equal to the angle GKD ; [ C.N. 1 ] and they are alternate.

322 Therefore AB is parallel to CD .

323Q. E. D.

324 Through a given point to draw a straight line parallel to a given straight line.

325 Let A be the given point, and BC the given straight line; thus it is required to draw through the point A a straight line parallel to the straight line BC .

326 Let a point D be taken at random on BC , and let AD be joined; on the straight line DA , and at the point A on it, let the angle DAE be constructed equal to the angle ADC [ I. 23 ]; and let the straight line AF be produced in a straight line with EA .

327 Then, since the straight line AD falling on the two straight lines BC , EF has made the alternate angles EAD , ADC equal to one another, therefore EAF is parallel to BC . [ I. 27 ]

328 Therefore through the given point A the straight line EAF has been drawn parallel to the given straight line BC .

329Q. E. F.

330 In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of the triangle are equal to two right angles.

331 Let ABC be a triangle, and let one side of it BC be produced to D ;

332 I say that the exterior angle ACD is equal to the two interior and opposite angles CAB , ABC , and the three interior angles of the triangle ABC , BCA , CAB are equal to two right angles.

333 For let CE be drawn through the point C parallel to the straight line AB . [ I. 31 ]

334 Then, since AB is parallel to CE , and AC has fallen upon them, the alternate angles BAC , ACE are equal to one another. [ I. 29 ]

335 Again, since AB is parallel to CE , and the straight line BD has fallen upon them, the exterior angle ECD is equal to the interior and opposite angle ABC . [ I. 29 ]

336 But the angle ACE was also proved equal to the angle BAC ; therefore the whole angle ACD is equal to the two interior and opposite angles BAC , ABC .

337 Let the angle ACB be added to each; therefore the angles ACD , ACB are equal to the three angles ABC , BCA , CAB .

338 But the angles ACD , ACB are equal to two right angles; [ I. 13 ] therefore the angles ABC , BCA , CAB are also equal to two right angles.

339Therefore etc.

340Q. E. D.

341 The straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are themselves also equal and parallel.

342 Let AB , CD be equal and parallel, and let the straight lines AC , BD join them (at the extremities which are) in the same directions (respectively); I say that AC , BD are also equal and parallel.

343 Let BC be joined.

344 Then, since AB is parallel to CD , and BC has fallen upon them, the alternate angles ABC , BCD are equal to one another. [ I. 29 ]

345 And, since AB is equal to CD , and BC is common, the two sides AB , BC are equal to the two sides DC , CB ; and the angle ABC is equal to the angle BCD ; therefore the base AC is equal to the base BD , and the griangle ABC is equal to the triangle DCB , and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; [ I. 4 ] therefore the angle ACB is equal to the angle CBD .

346 And, since the straight line BC falling on the two straight lines AC , BD has made the alternate angles equal to one another, AC is parallel to BD . [ I. 27 ]

347And it was also proved equal to it.

348Therefore etc.

349Q. E. D.

350 In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.

351 Let ACDB be a parallelogrammic area, and BC its diameter; I say that the opposite sides and angles of the parallelogram ACDB are equal to one another, and the diameter BC bisects it.

352 For, since AB is parallel to CD , and the straight line BC has fallen upon them, the alternate angles ABC , BCD are equal to one another. [ I. 29 ]

353 Again, since AC is parallel to BD , and BC has fallen upon them, the alternate angles ACB , CBD are equal to one another. [ I. 29 ]

354 Therefore ABC , DCB are two triangles having the two angles ABC , BCA equal to the two angles DCB , CBD respectively, and one side equal to one side, namely that adjoining the equal angles and common to both of them, BC ; therefore they will also have the remaining sides equal to the remaining sides respectively, and the remaining angle to the remaining angle; [ I. 26 ] therefore the side AB is equal to CD , and AC to BD , and further the angle BAC is equal to the angle CDB .

355 And, since the angle ABC is equal to the angle BCD , and the angle CBD to the angle ACB , the whole angle ABD is equal to the whole angle ACD . [ C.N. 2 ] And the angle BAC was also proved equal to the angle CDB .

356 Therefore in parallelogrammic areas the opposite sides and angles are equal to one another.

357 I say, next, that the diameter also bisects the areas.

358 For, since AB is equal to CD , and BC is common, the two sides AB , BC are equal to the two sides DC , CB respectively; and the angle ABC is equal to the angle BCD ; therefore the base AC is also equal to DB , and the triangle ABC is equal to the triangle DCB . [ I. 4 ]

359 Therefore the diameter BC bisects the parallelogram ACDB .

360Q. E. D.

361 Parallelograms which are on the same base and in the same parallels are equal to one another.

362 Let ABCD , EBCF be parallelograms on the same base BC and in the same parallels AF , BC ; I say that ABCD is equal to the parallelogram EBCF .

363 For, since ABCD is a parallelogram, AD is equal to BC . [ I. 34 ]

364 For the same reason also EF is equal to BC , so that AD is also equal to EF ; [ C.N. 1 ] and DE is common; therefore the whole AE is equal to the whole DF . [ C.N. 2 ]

365 But AB is also equal to DC ; [ I. 34 ] therefore the two sides EA , AB are equal to the two sides FD , DC respectively, and the angle FDC is equal to the angle EAB , the exterior to the interior; [ I. 29 ] therefore the base EB is equal to the base FC , and the triangle EAB will be equal to the triangle FDC . [ I. 4 ]

366 Let DGE be subtracted from each; therefore the trapezium ABGD which remains is equal to the trapezium EGCF which remains. [ C.N. 3 ]

367 Let the triangle GBC be added to each; therefore the whole parallelogram ABCD is equal to the whole parallelogram EBCF . [ C.N. 2 ]

368Therefore etc.

369Q. E. D.

370 Parallelograms which are on equal bases and in the same parallels are equal to one another.

371 Let ABCD , EFGH be parallelograms which are on equal bases BC , FG and in the same parallels AH , BG ; I say that the parallelogram ABCD is equal to EFGH .

372 For let BE , CH be joined.

373 Then, since BC is equal to FG while FG is equal to EH , BC is also equal to EH . [ C.N. 1 ]

374But they are also parallel.

375 And EB , HC join them; but straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are equal and parallel. [ I. 33 ]

376 Therefore EBCH is a parallelogram. [ I. 34 ]

377 And it is equal to ABCD ; for it has the same base BC with it, and is in the same parallels BC , AH with it. [ I. 35 ]

378 For the same reason also EFGH is equal to the same EBCH ; [ I. 35 ] so that the parallelogram ABCD is also equal to EFGH . [ C.N. 1 ]

379Therefore etc. Q. E. D.

380 Triangles which are on the same base and in the same parallels are equal to one another.

381 Let ABC , DBC be triangles on the same base BC and in the same parallels AD , BC ; I say that the triangle ABC is equal to the triangle DBC .

382 Let AD be produced in both directions to E , F ; through B let BE be drawn parallel to CA , [ I. 31 ] and through C let CF be drawn parallel to BD . [ I. 31 ]

383 Then each of the figures EBCA , DBCF is a parallelogram; and they are equal,

384 for they are on the same base BC and in the same parallels BC , EF . [ I. 35 ]

385 Moreover the triangle ABC is half of the parallelogram EBCA ; for the diameter AB bisects it. [ I. 34 ]

386 And the triangle DBC is half of the parallelogram DBCF ; for the diameter DC bisects it. [ I. 34 ]