Elements, Book 3

By Euclid

Edition: 0.0.0-dev | March 03, 2014

Authority: SCTA

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XX

BOOK III.

DEFINITIONS.

1

1 Equal circles are those the diameters of which are equal, or the radii of which are equal.

2

2 A straight line is said to touch a circle which, meeting the circle and being produced, does not cut the circle.

3

3 Circles are said to touch one another which, meeting one another, do not cut one another.

4

4 In a circle straight lines are said to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal.

5

5 And that straight line is said to be at a greater distance on which the greater perpendicular falls.

6

6 A segment of a circle is the figure contained by a straight line and a circumference of a circle.

7

7 An angle of a segment is that contained by a straight line and a circumference of a circle.

8

8 An angle in a segment is the angle which, when a point is taken on the circumference of the segment and straight lines are joined from it to the extremities of the straight line which is the base of the segment , is contained by the straight lines so joined.

9

9 And, when the straight lines containing the angle cut off a circumference, the angle is said to stand upon that circumference.

10

10 A sector of a circle is the figure which, when an angle is constructed at the centre of the circle, is contained by the straight lines containing the angle and the circumference cut off by them.

11

11 Similar segments of circles are those which admit equal angles, or in which the angles are equal to one another.

BOOK III. PROPOSITIONS.

PROPOSITION 1.

13 Let ABC be the given circle; thus it is required to find the centre of the circle ABC .

14 Let a straight line AB be drawn through it at random, and let it be bisected at the point D ; from D let DC be drawn at right angles to AB and let it be drawn through to E ; let CE be bisected at F ; I say that F is the centre of the circle ABC .

15 For suppose it is not, but, if possible, let G be the centre, and let GA , GD , GB be joined.

16 Then, since AD is equal to DB , and DG is common, the two sides AD , DG are equal to the two sides BD , DG respectively; and the base GA is equal to the base GB , for they are radii; therefore the angle ADG is equal to the angle GDB . [ I. 8 ]

17 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; [ I. Def. 10 ] therefore the angle GDB is right.

18 But the angle FDB is also right; therefore the angle FDB is equal to the angle GDB , the greater to the less: which is impossible. Therefore G is not the centre of the circle ABC .

19 Similarly we can prove that neither is any other point except F . Therefore the point F is the centre of the circle ABC .

PORISM.

20 From this it is manifest that, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line. Q. E. F.

PROPOSITION 2.

21 If on the circumference of a circle two points be taken at random , the straight line joining the points will fall within the circle .

22 Let ABC be a circle, and let two points A , B be taken at random on its circumference; I say that the straight line joined from A to B will fall within the circle.

23 For suppose it does not, but, if possible, let it fall outside, as AEB ; let the centre of the circle ABC be taken [ III. 1 ], and let it be D ; let DA , DB be joined, and let DFE be drawn through.

24 Then, since DA is equal to DB , the angle DAE is also equal to the angle DBE . [ I. 5 ] And, since one side AEB of the triangle DAE is produced, the angle DEB is greater than the angle DAE . [ I. 16 ]

25 But the angle DAE is equal to the angle DBE ; therefore the angle DEB is greater than the angle DBE . And the greater angle is subtended by the greater side; [ I. 19 ] therefore DB is greater than DE . But DB is equal to DF ; therefore DF is greater than DE , the less than the greater : which is impossible.

26 Therefore the straight line joined from A to B will not fall outside the circle.

27 Similarly we can prove that neither will it fall on the circumference itself; therefore it will fall within.

PROPOSITION 3.

29 If in a circle a straight line through the centre bisect a straight line not through the centre , it also cuts it at right angles; and if it cut it at right angles , it also bisects it .

30 Let ABC be a circle, and in it let a straight line CD through the centre bisect a straight line AB not through the centre at the point F ; I say that it also cuts it at right angles.

31 For let the centre of the circle ABC be taken, and let it be E ; let EA , EB be joined.

32 Then, since AF is equal to FB , and FE is common, two sides are equal to two sides; and the base EA is equal to the base EB ; therefore the angle AFE is equal to the angle BFE . [ I. 8 ]

33 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; [ I. Def. 10 ] therefore each of the angles AFE , BFE is right.

34 Therefore CD , which is through the centre, and bisects AB which is not through the centre, also cuts it at right angles.

35 Again, let CD cut AB at right angles; I say that it also bisects it. that is, that AF is equal to FB .

36 For, with the same construction, since EA is equal to EB , the angle EAF is also equal to the angle EBF . [ I. 5 ]

37 But the right angle AFE is equal to the right angle BFE , therefore EAF , EBF are two triangles having two angles equal to two angles and one side equal to one side, namely EF , which is common to them, and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [ I. 26 ] therefore AF is equal to FB .

PROPOSITION 4.

39 If in a circle two straight lines cut one another which are not through the centre , they do not bisect one another .

40 Let ABCD be a circle, and in it let the two straight lines AC , BD , which are not through the centre, cut one another at E ; I say that they do not bisect one another.

41 For, if possible, let them bisect one another, so that AE is equal to EC , and BE to ED ; let the centre of the circle ABCD be taken [ III. 1 ], and let it be F ; let FE be joined.

42 Then, since a straight line FE through the centre bisects a straight line AC not through the centre, it also cuts it at right angles; [ III. 3 ] therefore the angle FEA is right.

43 Again, since a straight line FE bisects a straight line BD , it also cuts it at right angles; [ III. 3 ] therefore the angle FEB is right.

44 But the angle FEA was also proved right; therefore the angle FEA is equal to the angle FEB , the less to the greater: which is impossible.

PROPOSITION 5.

47 If two circles cut one another , they will not have the same centre .

48 For let the circles ABC , CDG cut one another at the points B , C ; I say that they will not have the same centre.

49 For, if possible, let it be E ; let EC be joined, and let EFG be drawn through at random.

50 Then, since the point E is the centre of the circle ABC , EC is equal to EF . [ I. Def. 15 ]

51 Again, since the point E is the centre of the circle CDG , EC is equal to EG .

52 But EC was proved equal to EF also; therefore EF is also equal to EG , the less to the greater : which is impossible.

53 Therefore the point E is not the centre of the circles ABC , CDG .

PROPOSITION 6.

55 If two circles touch one another , they will not have the same centre .

56 For let the two circles ABC , CDE touch one another at the point C ; I say that they will not have the same centre.

57 For, if possible, let it be F ; let FC be joined, and let FEB be drawn through at random.

58 Then, since the point F is the centre of the circle ABC , FC is equal to FB .

59 Again, since the point F is the centre of the circle CDE , FC is equal to FE .

60 But FC was proved equal to FB ; therefore FE is also equal to FB , the less to the greater: which is impossible.

61 Therefore F is not the centre of the circles ABC , CDE .

PROPOSITION 7.

63 If on the diameter of a circle a point be taken which is not the centre of the circle , and from the point straight lines fall upon the circle , that will be greatest on which the centre is , the remainder of the same diameter will be least , and of the rest the nearer to the straight line through the centre is always greater than the more remote , and only two equal straight lines will fall from the point on the circle , one on each side of the least straight line .

64 Let ABCD be a circle, and let AD be a diameter of it; on AD let a point F be taken which is not the centre of the circle, let E be the centre of the circle, and from F let straight lines FB , FC , FG fall upon the circle ABCD ; I say that FA is greatest, FD is least, and of the rest FB is greater than FC , and FC than FG .

66 Then, since in any triangle two sides are greater than the remaining one, [ I. 20 ] EB , EF are greater than BF .

67 But AE is equal to BE ; therefore AF is greater than BF .

68 Again, since BE is equal to CE , and FE is common, the two sides BE , EF are equal to the two sides CE , EF .

69 But the angle BEF is also greater than the angle CEF ; therefore the base BF is greater than the base CF . [ I. 24 ]

71 Again, since GF , FE are greater than EG , and EG is equal to ED , GF , FE are greater than ED .

72 Let EF be subtracted from each; therefore the remainder GF is greater than the remainder FD .

73 Therefore FA is greatest, FD is least, and FB is greater than FC , and FC than FG .

74 I say also that from the point F only two equal straight lines will fall on the circle ABCD , one on each side of the least FD .

75 For on the straight line EF , and at the point E on it, let the angle FEH be constructed equal to the angle GEF [ I. 23 ], and let FH be joined.

76 Then, since GE is equal to EH , and EF is common, the two sides GE , EF are equal to the two sides HE , EF ; and the angle GEF is equal to the angle HEF ; therefore the base FG is equal to the base FH . [ I. 4 ]

77 I say again that another straight line equal to FG will not fall on the circle from the point F .

79 Then, since FK is equal to FG , and FH to FG , FK is also equal to FH , the nearer to the straight line through the centre being thus equal to the more remote: which is impossible.

80 Therefore another straight line equal to GF will not fall from the point F upon the circle; therefore only one straight line will so fall.

PROPOSITION 8.

82 If a point be taken outside a circle and from the point straight lines be drawn through to the circle , one of which is through the centre and the others are drawn at random , then , of the straight lines which fall on the concave circumference , that through the centre is greatest , while of the rest the nearer to that through the centre is always greater than the more remote , but , of the straight lines falling on the convex circumference , that between the point and the diameter is least , while of the rest the nearer to the least is always less than the more remote , and only two equal straight lines will fall on the circle from the point , one on each side of the least .

83 Let ABC be a circle, and let a point D be taken outside ABC ; let there be drawn through from it straight lines DA , DE , DF , DC , and let DA be through the centre; I say that, of the straight lines falling on the concave circumference AEFC , the straight line DA through the centre is greatest, while DE is greater than DF and DF than DC .; but, of the straight lines falling on the convex circumference HLKG , the straight line DG between the point and the diameter AG is least; and the nearer to the least DG is always less than the more remote, namely DK than DL , and DL than DH .

84 For let the centre of the circle ABC be taken [ III. 1 ], and let it be M ; let ME , MF , MC , MK , ML , MH be joined.

85 Then, since AM is equal to EM , let MD be added to each; therefore AD is equal to EM , MD .

86 But EM , MD are greater than ED ; [ I. 20 ] therefore AD is also greater than ED .

87 Again, since ME is equal to MF , and MD is common, therefore EM , MD are equal to FM , MD ; and the angle EMD is greater than the angle FMD ; therefore the base ED is greater than the base FD . [ I. 24 ]

88 Similarly we can prove that FD is greater than CD ; therefore DA is greatest, while DE is greater than DF , and DF than DC .

89 Next, since MK , KD are greater than MD , [ I. 20 ] and MG is equal to MK , therefore the remainder KD is greater than the remainder GD , so that GD is less than KD .

90 And, since on MD , one of the sides of the triangle MLD , two straight lines MK , KD were constructed meeting within the triangle, therefore MK , KD are less than ML , LD ; [ I. 21 ] and MK is equal to ML ; therefore the remainder DK is less than the remainder DL .

91 Similarly we can prove that DL is also less than DH ; therefore DG is least, while DK is less than DL , and DL than DH .

92 I say also that only two equal straight lines will fall from the point D on the circle, one on each side of the least DG .

93 On the straight line MD , and at the point M on it, let the angle DMB be constructed equal to the angle KMD , and let DB be joined.

94 Then, since MK is equal to MB , and MD is common, the two sides KM , MD are equal to the two sides BM , MD respectively; and the angle KMD is equal to the angle BMD ; therefore the base DK is equal to the base DB . [ I. 4 ]

95 I say that no other straight line equal to the straight line DK will fall on the circle from the point D .

96 For, if possible, let a straight line so fall, and let it be DN . Then, since DK is equal to DN ,

97 while DK is equal to DB , DB is also equal to DN , that is, the nearer to the least DG equal to the more remote: which was proved impossible.

98 Therefore no more than two equal straight lines will fall on the circle ABC from the point D , one on each side of DG the least.

PROPOSITION 9.

100 If a point be taken within a circle , and more than two equal straight lines fall from the point on the circle , the point taken is the centre of the circle .

101 Let ABC be a circle and D a point within it, and from D let more than two equal straight lines, namely DA , DB , DC , fall on the circle ABC ; I say that the point D is the centre of the circle ABC .

102 For let AB , BC be joined and bisected at the points E , F , and let ED , FD be joined and drawn through to the points G , K , H , L .

103 Then, since AE is equal to EB , and ED is common, the two sides AE , ED are equal to the two sides BE , ED ; and the base DA is equal to the base DB ; therefore the angle AED is equal to the angle BED . [ I. 8 ]

104 Therefore each of the angles AED , BED is right; [ I. Def. 10 ] therefore GK cuts AB into two equal parts and at right angles.

105 And since, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line, [ III. 1, Por. ] the centre of the circle is on GK .

106 For the same reason the centre of the circle ABC is also on HL .

107 And the straight lines GK , HL have no other point common but the point D ; therefore the point D is the centre of the circle ABC .

PROPOSITION 10.

109 A circle does not cut a circle at more points than two .

110 For, if possible, let the circle ABC cut the circle DEF at more points than two, namely B , C , F , H ;

111 let BH , BG be joined and bisected at the points K , L , and from K , L let KC , LM be drawn at right angles to BH , BG and carried through to the points A , E .

112 Then, since in the circle ABC a straight line AC cuts a straight line BH into two equal parts and at right angles, the centre of the circle ABC is on AC . [ III. 1, Por. ]

113 Again, since in the same circle ABC a straight line NO cuts a straight line BG into two equal parts and at right angles, the centre of the circle ABC is on NO .

114 But it was also proved to be on AC , and the straight lines AC , NO meet at no point except at P ; therefore the point P is the centre of the circle ABC .

115 Similarly we can prove that P is also the centre of the circle DEF ; therefore the two circles ABC , DEF which cut one another have the same centre P : which is impossible. [ III. 5 ]

PROPOSITION 11.

117 If two circles touch one another internally , and their centres be taken , the straight line joining their centres , if it be also produced , will fall on the point of contact of the circles .

118 For let the two circles ABC , ADE touch one another internally at the point A , and let the centre F of the circle ABC , and the centre G of ADE , be taken; I say that the straight line joined from G to F and produced will fall on A .

119 For suppose it does not, but, if possible, let it fall as FGH , and let AF , AG be joined.

120 Then, since AG , GF are greater than FA , that is, than FH ,

121 let FG be subtracted from each; therefore the remainder AG is greater than the remainder GH .

122 But AG is equal to GD ; therefore GD is also greater than GH , the less than the greater: which is impossible.

123 Therefore the straight line joined from F to G will not fall outside; therefore it will fall at A on the point of contact.

PROPOSITION 12.

125 If two circles touch one another externally , the straight line joining their centres will pass through the point of contact .

126 For let the two circles ABC , ADE touch one another externally at the point A , and let the centre F of ABC , and the centre G of ADE , be taken; I say that the straight line joined from F to G will pass through the point of contact at A .

127 For suppose it does not, but, if possible, let it pass as FCDG , and let AF , AG be joined.

128 Then, since the point F is the centre of the circle ABC , FA is equal to FC .

129 Again, since the point G is the centre of the circle ADE , GA is equal to GD .

130 But FA was also proved equal to FC ; therefore FA , AG are equal to FC , GD , so that the whole FG is greater than FA , AG ; but it is also less [ I. 20 ]: which is impossible.

131 Therefore the straight line joined from F to G will not fail to pass through the point of contact at A ; therefore it will pass through it.

PROPOSITION 13.

133 A circle does not touch a circle at more points than one , whether it touch it internally or externally .

134 For, if possible, let the circle ABDC touch the circle EBFD , first internally, at more points than one, namely D , B .

135 Let the centre G of the circle ABDC , and the centre H of EBFD , be taken.

136 Therefore the straight line joined from G to H will fall on B , D . [ III. 11 ]

138 Then, since the point G is the centre of the circle ABCD , BG is equal to GD ; therefore BG is greater than HD ; therefore BH is much greater than HD .

139 Again, since the point H is the centre of the circle EBFD , BH is equal to HD ; but it was also proved much greater than it: which is impossible.

140 Therefore a circle does not touch a circle internally at more points than one.

141 I say further that neither does it so touch it externally.

142 For, if possible, let the circle ACK touch the circle ABDC at more points than one, namely A , C , and let AC be joined.

143 Then, since on the circumference of each of the circles ABDC , ACK two points A , C have been taken at random, the straight line joining the points will fall within each circle; [ III. 2 ] but it fell within the circle ABCD and outside ACK [ III. Def. 3 ]: which is absurd.

144 Therefore a circle does not touch a circle externally at more points than one.

145 And it was proved that neither does it so touch it internally.

PROPOSITION 14.

147 In a circle equal straight lines are equally distant from the centre , and those which are equally distant from the centre are equal to one another .

148 Let ABDC be a circle, and let AB , CD be equal straight lines in it; I say that AB , CD are equally distant from the centre.

149 For let the centre of the circle ABDC be taken [ III. 1 ], and let it be E ; from E let EF , EG be drawn perpendicular to AB , CD , and let AE , EC be joined.

150 Then, since a straight line EF through the centre cuts a straight line AB not through the centre at right angles, it also bisects it. [ III. 3 ] Therefore AF is equal to FB ; therefore AB is double of AF .

151 For the same reason CD is also double of CG ; and AB is equal to CD ; therefore AF is also equal to CG .

152 And, since AE is equal to EC , the square on AE is also equal to the square on EC . But the squares on AF , EF are equal to the square on AE , for the angle at F is right; and the squares on EG , GC are equal to the square on EC , for the angle at G is right; [ I. 47 ] therefore the squares on AF , FE are equal to the squares on CG , GE , of which the square on AF is equal to the square on CG , for AF is equal to CG ; therefore the square on FE which remains is equal to the square on EG , therefore EF is equal to EG .

153 But in a circle straight lines are said to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal; [ III. Def. 4 ] therefore AB , CD are equally distant from the centre.

154 Next, let the straight lines AB , CD be equally distant from the centre; that is, let EF be equal to EG .

156 For, with the same construction, we can prove, similarly, that AB is double of AF , and CD of CG .

157 And, since AE is equal to CE , the square on AE is equal to the square on CE . But the squares on EF , FA are equal to the square on AE , and the squares on EG , GC equal to the square on CE . [ I. 47 ]

158 Therefore the squares on EF , FA are equal to the squares on EG , GC , of which the square on EF is equal to the square on EG , for EF is equal to EG ; therefore the square on AF which remains is equal to the square on CG ; therefore AF is equal to CG . And AB is double of AF , and CD double of CG ; therefore AB is equal to CD .

PROPOSITION 15.

160 Of straight lines in a circle the diameter is greatest , and of the rest the nearer to the centre is always greater than the more remote .

161 Let ABCD be a circle, let AD be its diameter and E the centre; and let BC be nearer to the diameter AD , and FG more remote; I say that AD is greatest and BC greater than FG .

162 For from the centre E let EH , EK be drawn perpendicular to BC , FG .

163 Then, since BC is nearer to the centre and FG more remote, EK is greater than EH . [ III. Def. 5 ]

164 Let EL be made equal to EH , through L let LM be drawn at right angles to EK and carried through to N , and let ME , EN , FE , EG be joined.

165 Then, since EH is equal to EL , BC is also equal to MN . [ III. 14 ]

166 Again, since AE is equal to EM , and ED to EN , AD is equal to ME , EN .

167 But ME , EN are greater than MN , [ I. 20 ] and MN is equal to BC ; therefore AD is greater than BC .

168 And, since the two sides ME , EN are equal to the two sides FE , EG , and the angle MEN greater than the angle FEG , therefore the base MN is greater than the base FG . [ I. 24 ]

170 Therefore the diameter AD is greatest and BC greater than FG .

PROPOSITION 16.

172 The straight line drawn at right angles to the diameter of a circle from its extremity will fall outside the circle , and into the space between the straight line and the circumference another straight line cannot be interposed; further the angle of the semicircle is greater , and the remaining angle less , than any acute rectilineal angle .

173 Let ABC be a circle about D as centre and AB as diameter; I say that the straight line drawn from A at right angles to AB from its extremity will fall outside the circle.

174 For suppose it does not, but, if possible, let it fall within as CA , and let DC be joined.

175 Since DA is equal to DC , the angle DAC is also equal to the angle ACD . [ I. 5 ]

176 But the angle DAC is right; therefore the angle ACD is also right: thus, in the triangle ACD , the two angles DAC , ACD are equal to two right angles: which is impossible. [ I. 17 ]

177 Therefore the straight line drawn from the point A at right angles to BA will not fall within the circle.

178 Similarly we can prove that neither will it fall on the circumference; therefore it will fall outside.

179 Let it fall as AE ; I say next that into the space between the straight line AE and the circumference CHA another straight line cannot be interposed.

180 For, if possible, let another straight line be so interposed, as FA , and let DG be drawn from the point D perpendicular to FA .

181 Then, since the angle AGD is right, and the angle DAG is less than a right angle, AD is greater than DG . [ I. 19 ]

182 But DA is equal to DH ; therefore DH is greater than DG , the less than the greater: which is impossible.

183 Therefore another straight line cannot be interposed into the space between the straight line and the circumference.

184 I say further that the angle of the semicircle contained by the straight line BA and the circumference CHA is greater than any acute rectilineal angle, and the remaining angle contained by the circumference CHA and the straight line AE is less than any acute rectilineal angle.

185 For, if there is any rectilineal angle greater than the angle contained by the straight line BA and the circumference CHA , and any rectilineal angle less than the angle contained by the circumference CHA and the straight line AE , then into the space between the circumference and the straight line AE a straight line will be interposed such as will make an angle contained by straight lines which is greater than the angle contained by the straight line BA and the circumference CHA , and another angle contained by straight lines which is less than the angle contained by the circumference CHA and the straight line AE .

187 therefore there will not be any acute angle contained by straight lines which is greater than the angle contained by the straight line BA and the circumference CHA , nor yet any acute angle contained by straight lines which is less than the angle contained by the circumference CHA and the straight line AE .—

PORISM.

188 From this it is manifest that the straight line drawn at right angles to the diameter of a circle from its extremity touches the circle. Q. E. D.

PROPOSITION 17.

189 From a given point to draw a straight line touching a given circle .

190 Let A be the given point, and BCD the given circle; thus it is required to draw from the point A a straight line touching the circle BCD .

191 For let the centre E of the circle be taken; [ III. 1 ] let AE be joined, and with centre E and distance EA let the circle AFG be described; from D let DF be drawn at right angles to EA , and let EF , AB be joined; I say that AB has been drawn from the point A touching the circle BCD .

192 For, since E is the centre of the circles BCD , AFG , EA is equal to EF , and ED to EB ; therefore the two sides AE , EB are equal to the two sides FE , ED : and they contain a common angle, the angle at E ; therefore the base DF is equal to the base AB , and the triangle DEF is equal to the triangle BEA , and the remaining angles to the remaining angles; [ I. 4 ] therefore the angle EDF is equal to the angle EBA .

193 But the angle EDF is right; therefore the angle EBA is also right.

194 Now EB is a radius; and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [ III. 16, Por. ] therefore AB touches the circle BCD .

195 Therefore from the given point A the straight line AB has been drawn touching the circle BCD .

PROPOSITION 18.

196 If a straight line touch a circle , and a straight line be joined from the centre to the point of contact , the straight line so joined will be perpendicular to the tangent .

197 For let a straight line DE touch the circle ABC at the point C , let the centre F of the circle ABC be taken, and let FC be joined from F to C ; I say that FC is perpendicular to DE .

198 For, if not, let FG be drawn from F perpendicular to DE .

199 Then, since the angle FGC is right, the angle FCG is acute; [ I. 17 ] and the greater angle is subtended by the greater side; [ I. 19 ] therefore FC is greater than FG .

200 But FC is equal to FB ; therefore FB is also greater than FG , the less than the greater: which is impossible.

202 Similarly we can prove that neither is any other straight line except FC ; therefore FC is perpendicular to DE .

PROPOSITION 19.

204 If a straight line touch a circle , and from the point of contact a straight line be drawn at right angles to the tangent , the centre of the circle will be on the straight line so drawn .

205 For let a straight line DE touch the circle ABC at the point C , and from C let CA be drawn at right angles to DE ; I say that the centre of the circle is on AC .

206 For suppose it is not, but, if possible, let F be the centre, and let CF be joined.

207 Since a straight line DE touches the circle ABC , and FC has been joined from the centre to the point of contact, FC is perpendicular to DE ; [ III. 18 ] therefore the angle FCE is right.

208 But the angle ACE is also right; therefore the angle FCE is equal to the angle ACE , the less to the greater: which is impossible.

210 Similarly we can prove that neither is any other point except a point on AC .

PROPOSITION 20.

212 In a circle the angle at the centre is double of the angle at the circumference , when the angles have the same circumference as base .

213 Let ABC be a circle, let the angle BEC be an angle at its centre, and the angle BAC an angle at the circumference, and let them have the same circumference BC as base; I say that the angle BEC is double of the angle BAC .

215 Then, since EA is equal to EB , the angle EAB is also equal to the angle EBA ; [ I. 5 ] therefore the angles EAB , EBA are double of the angle EAB .

216 But the angle BEF is equal to the angles EAB , EBA ; [ I. 32 ] therefore the angle BEF is also double of the angle EAB .

217 For the same reason the angle FEC is also double of the angle EAC .

218 Therefore the whole angle BEC is double of the whole angle BAC .

219 Again let another straight line be inflected, and let there be another angle BDC ; let DE be joined and produced to G .

220 Similarly then we can prove that the angle GEC is double of the angle EDC , of which the angle GEB is double of the angle EDB ; therefore the angle BEC which remains is double of the angle BDC .

PROPOSITION 21.

222 In a circle the angles in the same segment are equal to one another .

223 Let ABCD be a circle, and let the angles BAD , BED be angles in the same segment BAED ; I say that the angles BAD , BED are equal to one another.

224 For let the centre of the circle ABCD be taken, and let it be F ; let BF , FD be joined.

225 Now, since the angle BFD is at the centre, and the angle BAD at the circumference, and they have the same circumference BCD as base, therefore the angle BFD is double of the angle BAD . [ III. 20 ]

226 For the same reason the angle BFD is also double of the angle BED ; therefore the angle BAD is equal to the angle BED .

PROPOSITION 22.

228 The opposite angles of quadrilaterals in circles are equal to two right angles .

229 Let ABCD be a circle, and let ABCD be a quadrilateral in it; I say that the opposite angles are equal to two right angles.

231 Then, since in any triangle the three angles are equal to two right angles, [ I. 32 ] the three angles CAB , ABC , BCA of the triangle ABC are equal to two right angles.

232 But the angle CAB is equal to the angle BDC , for they are in the same segment BADC ; [ III. 21 ] and the angle ACB is equal to the angle ADB , for they are in the same segment ADCB ; therefore the whole angle ADC is equal to the angles BAC , ACB .

233 Let the angle ABC be added to each; therefore the angles ABC , BAC , ACB are equal to the angles ABC , ADC . But the angles ABC , BAC , ACB are equal to two right angles; therefore the angles ABC , ADC are also equal to two right angles.

234 Similarly we can prove that the angles BAD , DCB are also equal to two right angles.

PROPOSITION 23.

236 On the same straight line there cannot be constructed two similar and unequal segments of circles on the same side .

237 For, if possible, on the same straight line AB let two similar and unequal segments of circles ACB , ADB be constructed on the same side; let ACD be drawn through, and let CB , DB be joined.

238 Then, since the segment ACB is similar to the segment ADB , and similar segments of circles are those which admit equal angles, [ III. Def. 11 ] the angle ACB is equal to the angle ADB , the exterior to the interior: which is impossible. [ I. 16 ]

PROPOSITION 24.

240 Similar segments of circles on equal straight lines are equal to one another .

241 For let AEB , CFD be similar segments of circles on equal straight lines AB , CD ; I say that the segment AEB is equal to the segment CFD .

242 For, if the segment AEB be applied to CFD , and if the point A be placed on C and the straight line AB on CD , the point B will also coincide with the point D , because AB is equal to CD ; and, AB coinciding with CD , the segment AEB will also coincide with CFD .

243 For, if the straight line AB coincide with CD but the segment AEB do not coincide with CFD , it will either fall with it, or outside it; or it will fall awry, as CGD , and a circle cuts a circle at more points than two : which is impossible. [ III. 10 ]

244 Therefore, if the straight line AB be applied to CD , the segment AEB will not fail to coincide with CFD also; therefore it will coincide with it and will be equal to it.

PROPOSITION 25.

246 Given a segment of a circle , to describe the complete circle of which it is a segment .

247 Let ABC be the given segment of a circle; thus it is required to describe the complete circle belonging to the segment ABC , that is, of which it is a segment.

248 For let AC be bisected at D , let DB be drawn from the point D at right angles to AC , and let AB . be joined; the angle ABD is then greater than, equal to, or less than the angle BAD .

249 First let it be greater; and on the straight line BA , and at the point A on it, let the angle BAE be constructed equal to the angle ABD ; let DB be drawn through to E , and let EC be joined.

250 Then, since the angle ABE is equal to the angle BAE , the straight line EB is also equal to EA . [ I. 6 ]

251 And, since AD is equal to DC , and DE is common, the two sides AD , DE are equal to the two sides CD , DE respectively; and the angle ADE is equal to the angle CDE , for each is right; therefore the base AE is equal to the base CE .

252 But AE was proved equal to BE ; therefore BE is also equal to CE ; therefore the three straight lines AE , EB , EC are equal to one another.

253 Therefore the circle drawn with centre E and distance one of the straight lines AE , EB , EC will also pass through the remaining points and will have been completed. [ III. 9 ]

254 Therefore, given a segment of a circle, the complete circle has been described.

255 And it is manifest that the segment ABC is less than a semicircle, because the centre E happens to be outside it.

256 Similarly, even if the angle ABD be equal to the angle BAD , AD being equal to each of the two BD , DC , the three straight lines DA , DB , DC will be equal to one another, D will be the centre of the completed circle, and ABC will clearly be a semicircle.

257 But, if the angle ABD be less than the angle BAD , and if we construct, on the straight line BA and at the point A on it, an angle equal to the angle ABD , the centre will fall on DB within the segment ABC , and the segment ABC will clearly be greater than a semicircle.

258 Therefore, given a segment of a circle, the complete circle has been described. Q. E. F.

PROPOSITION 26.

259 In equal circles equal angles stand on equal circumferences , whether they stand at the centres or at the circumferences .

260 Let ABC , DEF be equal circles, and in them let there be equal angles, namely at the centres the angles BGC , EHF , and at the circumferences the angles BAC , EDF ; I say that the circumference BKC is equal to the circumference ELF .

262 Now, since the circles ABC , DEF are equal, the radii are equal.

263 Thus the two straight lines BG , GC are equal to the two straight lines EH , HF ; and the angle at G is equal to the angle at H ; therefore the base BC is equal to the base EF . [ I. 4 ]

264 And, since the angle at A is equal to the angle at D , the segment BAC is similar to the segment EDF ; [ III. Def. 11 ] and they are upon equal straight lines.

265 But similar segments of circles on equal straight lines are equal to one another; [ III. 24 ] therefore the segment BAC is equal to EDF . But the whole circle ABC is also equal to the whole circle DEF ; therefore the circumference BKC which remains is equal to the circumference ELF .

PROPOSITION 27.

267 In equal circles angles standing on equal circumferences are equal to one another , whether they stand at the centres or at the circumferences .

268 For in equal circles ABC , DEF , on equal circumferences BC , EF , let the angles BGC , EHF stand at the centres G , H , and the angles BAC , EDF at the circumferences; I say that the angle BGC is equal to the angle EHF , and the angle BAC is equal to the angle EDF .

269 For, if the angle BGC is unequal to the angle EHF , one of them is greater. Let the angle BGC be greater : and on the straight line BG , and at the point G on it, let the angle BGK be constructed equal to the angle EHF . [ I. 23 ]

270 Now equal angles stand on equal circumferences, when they are at the centres; [ III. 26 ] therefore the circumference BK is equal to the circumference EF .

271 But EF is equal to BC ; therefore BK is also equal to BC , the less to the greater : which is impossible.

272 Therefore the angle BGC is not unequal to the angle EHF ; therefore it is equal to it.

273 And the angle at A is half of the angle BGC , and the angle at D half of the angle EHF ; [ III. 20 ] therefore the angle at A is also equal to the angle at D .

PROPOSITION 28.

275 In equal circles equal straight lines cut off equal circumferences , the greater equal to the greater and the less to the less .

276 Let ABC , DEF be equal circles, and in the circles let AB , DE be equal straight lines cutting off ACB , DFE as greater circumferences and AGB , DHE as lesser; I say that the greater circumference ACB is equal to the greater circumference DFE , and the less circumference AGB to DHE .

277 For let the centres K , L of the circles be taken, and let AK , KB , DL , LE be joined.

278 Now, since the circles are equal, the radii are also equal; therefore the two sides AK , KB are equal to the two sides DL , LE ; and the base AB is equal to the base DE ; therefore the angle AKB is equal to the angle DLE . [ I. 8 ]

279 But equal angles stand on equal circumferences, when they are at the centres; [ III. 26 ] therefore the circumference AGB is equal to DHE .

280 And the whole circle ABC is also equal to the whole circle DEF ; therefore the circumference ACB which remains is also equal to the circumference DFE which remains.

PROPOSITION 29.

282 In equal circles equal circumferences are subtended by equal straight lines .

283 Let ABC , DEF be equal circles, and in them let equal circumferences BGC , EHF be cut off; and let the straight lines BC , EF be joined; I say that BC is equal to EF .

284 For let the centres of the circles be taken, and let them be K , L ; let BK , KC , EL , LF be joined.

285 Now, since the circumference BGC is equal to the circumference EHF , the angle BKC is also equal to the angle ELF . [ III. 27 ]

286 And, since the circles ABC , DEF are equal, the radii are also equal; therefore the two sides BK , KC are equal to the two sides EL , LF ; and they contain equal angles; therefore the base BC is equal to the base EF . [ I. 4 ]

PROPOSITION 30.

289 Let ADB be the given circumference; thus it is required to bisect the circumference ADB .

290 Let AB be joined and bisected at C ; from the point C let CD be drawn at right angles to the straight line AB , and let AD , DB be joined.

291 Then, since AC is equal to CB , and CD is common, the two sides AC , CD are equal to the two sides BC , CD ; and the angle ACD is equal to the angle BCD , for each is right; therefore the base AD is equal to the base DB . [ I. 4 ]

292 But equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less; [ III. 28 ] and each of the circumferences AD , DB is less than a semicircle; therefore the circumference AD is equal to the circumference DB .

293 Therefore the given circumference has been bisected at the point D . Q. E. F.

PROPOSITION 31.

294 In a circle the angle in the semicircle is right , that in a greater segment less than a right angle , and that in a less segment greater than a right angle; and further the angle of the greater segment is greater than a right angle , and the angle of the less segment less than a right angle .

295 Let ABCD be a circle, let BC be its diameter, and E its centre, and let BA , AC , AD , DC be joined; I say that the angle BAC in the semicircle BAC is right, the angle ABC in the segment ABC greater than the semicircle is less than a right angle, and the angle ADC in the segment ADC less than the semicircle is greater than a right angle.

296 Let AE be joined, and let BA be carried through to F .

297 Then, since BE is equal to EA , the angle ABE is also equal to the angle BAE . [ I. 5 ]

298 Again, since CE is equal to EA , the angle ACE is also equal to the angle CAE . [ I. 5 ]

299 Therefore the whole angle BAC is equal to the two angles ABC , ACB .

300 But the angle FAC exterior to the triangle ABC is also equal to the two angles ABC , ACB ; [ I. 32 ] therefore the angle BAC is also equal to the angle FAC ; therefore each is right; [ I. Def. 10 ] therefore the angle BAC in the semicircle BAC is right.

301 Next, since in the triangle ABC the two angles ABC , BAC are less than two right angles, [ I. 17 ] and the angle BAC is a right angle, the angle ABC is less than a right angle; and it is the angle in the segment ABC greater than the semicircle.

302 Next, since ABCD is a quadrilateral in a circle, and the opposite angles of quadrilaterals in circles are equal to two right angles, [ III. 22 ] while the angle ABC is less than a right angle, therefore the angle ADC which remains is greater than a right angle; and it is the angle in the segment ADC less than the semicircle.

303 I say further that the angle of the greater segment, namely that contained by the circumference ABC and the straight line AC , is greater than a right angle; and the angle of the less segment, namely that contained by the circumference ADC and the straight line AC , is less than a right angle.

304 This is at once manifest. For, since the angle contained by the straight lines BA , AC is right, the angle contained by the circumference ABC and the straight line AC is greater than a right angle.

305 Again, since the angle contained by the straight lines AC , AF is right, the angle contained by the straight line CA and the circumference ADC is less than a right angle.

PROPOSITION 32.

307 If a straight line touch a circle , and from the point of contact there be drawn across , in the circle , a straight line cutting the circle , the angles which it makes with the tangent will be equal to the angles in the alternate segments of the circle .

308 For let a straight line EF touch the circle ABCD at the point B , and from the point B let there be drawn across, in the circle ABCD , a straight line BD cutting it; I say that the angles which BD makes with the tangent EF will be equal to the angles in the alternate segments of the circle, that is, that the angle FBD is equal to the angle constructed in the segment BAD , and the angle EBD is equal to the angle constructed in the segment DCB .

309 For let BA be drawn from B at right angles to EF , let a point C be taken at random on the circumference BD , and let AD , DC , CB be joined.

310 Then, since a straight line EF touches the circle ABCD at B , and BA has been drawn from the point of contact at right angles to the tangent, the centre of the circle ABCD is on BA . [ III. 19 ]

311 Therefore BA is a diameter of the circle ABCD ; therefore the angle ADB , being an angle in a semicircle, is right. [ III. 31 ]

312 Therefore the remaining angles BAD , ABD are equal to one right angle. [ I. 32 ]

313 But the angle ABF is also right; therefore the angle ABF is equal to the angles BAD , ABD .

314 Let the angle ABD be subtracted from each; therefore the angle DBF which remains is equal to the angle BAD in the alternate segment of the circle.

315 Next, since ABCD is a quadrilateral in a circle, its opposite angles are equal to two right angles. [ III. 22 ]

316 But the angles DBF , DBE are also equal to two right angles; therefore the angles DBF , DBE are equal to the angles BAD , BCD , of which the angle BAD was proved equal to the angle DBF ; therefore the angle DBE which remains is equal to the angle DCB in the alternate segment DCB of the circle.

PROPOSITION 33.

318 On a given straight line to describe a segment of a circle admitting an angle equal to a given rectilineal angle .

319 Let AB be the given straight line, and the angle at C the given rectilineal angle; thus it is required to describe on the given straight line AB a segment of a circle admitting an angle equal to the angle at C .

321 First let it be acute, and, as in the first figure, on the straight line AB , and at the point A , let the angle BAD be constructed equal to the angle at C ; therefore the angle BAD is also acute.

322 Let AE be drawn at right angles to DA , let AB be bisected at F , let FG be drawn from the point F at right angles to AB , and let GB be joined.

323 Then, since AF is equal to FB , and FG is common, the two sides AF , FG are equal to the two sides BF , FG ; and the angle AFG is equal to the angle BFG ; therefore the base AG is equal to the base BG . [ I. 4 ]

324 Therefore the circle described with centre G and distance GA will pass through B also.

325 Let it be drawn, and let it be ABE ; let EB be joined.

326 Now, since AD is drawn from A , the extremity of the diameter AE , at right angles to AE , therefore AD touches the circle ABE . [ III. 16, Por. ]

327 Since then a straight line AD touches the circle ABE , and from the point of contact at A a straight line AB is drawn across in the circle ABE , the angle DAB is equal to the angle AEB in the alternate segment of the circle. [ III. 32 ]

328 But the angle DAB is equal to the angle at C ; therefore the angle at C is also equal to the angle AEB .

329 Therefore on the given straight line AB the segment AEB of a circle has been described admitting the angle AEB equal to the given angle, the angle at C .

330 Next let the angle at C be right; and let it be again required to describe on AB a segment of a circle admitting an angle equal to the right angle at C .

331 Let the angle BAD be constructed equal to the right angle at C , as is the case in the second figure; let AB be bisected at F , and with centre F and distance either FA or FB let the circle AEB be described.

332 Therefore the straight line AD touches the circle ABE , because the angle at A is right. [ III. 16, Por. ]

333 And the angle BAD is equal to the angle in the segment AEB , for the latter too is itself a right angle, being an angle in a semicircle. [ III. 31 ]

334 But the angle BAD is also equal to the angle at C .

335 Therefore the angle AEB is also equal to the angle at C .

336 Therefore again the segment AEB of a circle has been described on AB admitting an angle equal to the angle at C .

337 Next, let the angle at C be obtuse; and on the straight line AB , and at the point A , let the angle BAD be constructed equal to it, as is the case in the third figure; let AE be drawn at right angles to AD , let AB be again bisected at F , let FG be drawn at right angles to AB , and let GB be joined.

338 Then, since AF is again equal to FB , and FG is common, the two sides AF , FG are equal to the two sides BF , FG ; and the angle AFG is equal to the angle BFG ; therefore the base AG is equal to the base BG . [ I. 4 ]

339 Therefore the circle described with centre G and distance GA will pass through B also; let it so pass, as AEB .

340 Now, since AD is drawn at right angles to the diameter AE from its extremity, AD touches the circle AEB . [ III. 16, Por. ]

341 And AB has been drawn across from the point of contact at A ; therefore the angle BAD is equal to the angle constructed in the alternate segment AHB of the circle. [ III. 32 ]

343 Therefore the angle in the segment AHB is also equal to the angle at C .

344 Therefore on the given straight line AB the segment AHB of a circle has been described admitting an angle equal to the angle at C . Q. E. F.

PROPOSITION 34.

345 From a given circle to cut off a segment admitting an angle equal to a given rectilineal angle .

346 Let ABC be the given circle, and the angle at D the given rectilineal angle; thus it is required to cut off from the circle ABC a segment admitting an angle equal to the given rectilineal angle, the angle at D .

347 Let EF be drawn touching ABC at the point B , and on the straight line FB , and at the point B on it, let the angle FBC be constructed equal to the angle at D . [ I. 23 ]

348 Then, since a straight line EF touches the circle ABC , and BC has been drawn across from the point of contact at B , the angle FBC is equal to the angle constructed in the alternate segment BAC . [ III. 32 ]

349 But the angle FBC is equal to the angle at D ; therefore the angle in the segment BAC is equal to the angle at D .

350 Therefore from the given circle ABC the segment BAC . has been cut off admitting an angle equal to the given rectilineal angle, the angle at D . Q. E. F.

PROPOSITION 35.

351 If in a circle two straight lines cut one another , the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other .

352 For in the circle ABCD let the two straight lines AC , BD cut one another at the point E ; I say that the rectangle contained by AE , EC is equal to the rectangle contained by DE , EB .

353 If now AC , BD are through the centre, so that E is the centre of the circle ABCD , it is manifest that, AE , EC , DE , EB being equal, the rectangle contained by AE , EC is also equal to the rectangle contained by DE , EB .

354 Next let AC , DB not be through the centre; let the centre of ABCD be taken, and let it be F ; from F let FG , FH be drawn perpendicular to the straight lines AC , DB , and let FB , FC , FE be joined.

355 Then, since a straight line GF through the centre cuts a straight line AC not through the centre at right angles, it also bisects it; [ III. 3 ] therefore AG is equal to GC .

356 Since, then, the straight line AC has been cut into equal parts at G and into unequal parts at E , the rectangle contained by AE , EC together with the square on EG is equal to the square on GC ; [ II. 5 ]

357 Let the square on GF be added; therefore the rectangle AE , EC together with the squares on GE , GF is equal to the squares on CG , GF .

358 But the square on FE is equal to the squares on EG , GF , and the square on FC is equal to the squares on CG , GF ; [ I. 47 ] therefore the rectangle AE , EC together with the square on FE is equal to the square on FC .

359 And FC is equal to FB ; therefore the rectangle AE , EC together with the square on EF is equal to the square on FB .

360 For the same reason, also, the rectangle DE , EB together with the square on FE is equal to the square on FB .

361 But the rectangle AE , EC together with the square on FE was also proved equal to the square on FB ; therefore the rectangle AE , EC together with the square on FE is equal to the rectangle DE , EB together with the square on FE .

362 Let the square on FE be subtracted from each; therefore the rectangle contained by AE , EC which remains is equal to the rectangle contained by DE , EB .

PROPOSITION 36.

364 If a point be taken outside a circle and from it there fall on the circle two straight lines , and if one of them cut the circle and the other touch it , the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference will be equal to the square on the tangent .

365 For let a point D be taken outside the circle ABC , and from D let the two straight lines DCA , DB fall on the circle ABC ; let DCA cut the circle ABC and let BD touch it; I say that the rectangle contained by AD , DC is equal to the square on DB .

366 Then DCA is either through the centre or not through the centre.

367 First let it be through the centre, and let F be the centre of the circle ABC ; let FB be joined; therefore the angle FBD is right. [ III. 18 ]

368 And, since AC has been bisected at F , and CD is added to it, the rectangle AD , DC together with the square on FC is equal to the square on FD . [ II. 6 ]

369 But FC is equal to FB ; therefore the rectangle AD , DC together with the square on FB is equal to the square on FD .

370 And the squares on FB , BD are equal to the square on FD ; [ I. 47 ] therefore the rectangle AD , DC together with the square on FB is equal to the squares on FB , BD .

371 Let the square on FB be subtracted from each; therefore the rectangle AD , DC which remains is equal to the square on the tangent DB .

372 Again, let DCA not be through the centre of the circle ABC ; let the centre E be taken, and from E let EF be drawn perpendicular to AC ; let EB , EC , ED be joined.

374 And, since a straight line EF through the centre cuts a straight line AC not through the centre at right angles, it also bisects it; [ III. 3 ] therefore AF is equal to FC .

375 Now, since the straight line AC has been bisected at the point F , and CD is added to it, the rectangle contained by AD , DC together with the square on FC is equal to the square on FD . [ II. 6 ]

376 Let the square on FE be added to each; therefore the rectangle AD , DC together with the squares on CF , FE is equal to the squares on FD , FE .

377 But the square on EC is equal to the squares on CF , FE , for the angle EFC is right; [ I. 47 ] and the square on ED is equal to the squares on DF , FE ; therefore the rectangle AD , DC together with the square on EC is equal to the square on ED .

378 And EC is equal to EB ; therefore the rectangle AD , DC together with the square on EB is equal to the square on ED .

379 But the squares on EB , BD are equal to the square on ED , for the angle EBD is right; [ I. 47 ] therefore the rectangle AD , DC together with the square on EB is equal to the squares on EB , BD .

380 Let the square on EB be subtracted from each; therefore the rectangle AD , DC which remains is equal to the square on DB .

PROPOSITION 37.

382 If a point be taken outside a circle and from the point there fall on the circle two straight lines , if one of them cut the circle , and the other fall on it , and if further the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference be equal to the square on the straight line which falls on the circle , the straight line which falls on it will touch the circle .

383 For let a point D be taken outside the circle ABC ; from D let the two straight lines DCA , DB fall on the circle ACB ; let DCA cut the circle and DB fall on it; and let the rectangle AD , DC be equal to the square on DB .

385 For let DE be drawn touching ABC ; let the centre of the circle ABC be taken, and let it be F ; let FE , FB , FD be joined.

387 Now, since DE touches the circle ABC , and DCA cuts it, the rectangle AD , DC is equal to the square on DE . [ III. 36 ]

388 But the rectangle AD , DC was also equal to the square on DB ; therefore the square on DE is equal to the square on DB ; therefore DE is equal to DB .

389 And FE is equal to FB ; therefore the two sides DE , EF are equal to the two sides DB , BF ; and FD is the common base of the triangles; therefore the angle DEF is equal to the angle DBF . [ I. 8 ]

390 But the angle DEF is right; therefore the angle DBF is also right.