Elements, Book 6

By Euclid

Edition: 0.0.0-dev | March 03, 2014

Authority: SCTA

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

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BOOK VI.

DEFINITIONS.

1

1 Similar rectilineal figures are such as have their angles severally equal and the sides about the equal angles proportional.

3

3 A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less.

4

4 The height of any figure is the perpendicular drawn from the vertex to the base.

BOOK VI. PROPOSITIONS.

PROPOSITION 1.

6 Triangles and parallelograms which are under the same height are to one another as their bases .

7 Let ABC , ACD be triangles and EC , CF parallelograms under the same height; I say that, as the base BC is to the base CD , so is the triangle ABC to the triangle ACD , and the parallelogram EC to the parallelogram CF .

8 For let BD be produced in both directions to the points H , L and let [any number of straight lines] BG , GH be made equal to the base BC , and any number of straight lines DK , KL equal to the base CD ; let AG , AH , AK , AL be joined.

9 Then, since CB , BG , GH are equal to one another, the triangles ABC , AGB , AHG are also equal to one another. [ I. 38 ]

10 Therefore, whatever multiple the base HC is of the base BC , that multiple also is the triangle AHC of the triangle ABC .

11 For the same reason, whatever multiple the base LC is of the base CD , that multiple also is the triangle ALC of the triangle ACD ; and, if the base HC is equal to the base CL , the triangle AHC is also equal to the triangle ACL , [ I. 38 ] if the base HC is in excess of the base CL , the triangle AHC is also in excess of the triangle ACL , and, if less, less.

12 Thus, there being four magnitudes, two bases BC , CD and two triangles ABC , ACD , equimultiples have been taken of the base BC and the triangle ABC , namely the base HC and the triangle AHC , and of the base CD and the triangle ADC other, chance, equimultiples, namely the base LC and the triangle ALC ; and it has been proved that, if the base HC is in excess of the base CL , the triangle AHC is also in excess of the triangle ALC ; if equal, equal; and, if less, less.

13 Therefore, as the base BC is to the base CD , so is the triangle ABC to the triangle ACD . [ V. Def. 5 ]

14 Next, since the parallelogram EC is double of the triangle ABC , [ I. 41 ] and the parallelogram FC is double of the triangle ACD , while parts have the same ratio as the same multiples of them, [ V. 15 ] therefore, as the triangle ABC is to the triangle ACD , so is the parallelogram EC to the parallelogram FC .

15 Since, then, it was proved that, as the base BC is to CD , so is the triangle ABC to the triangle ACD , and, as the triangle ABC is to the triangle ACD , so is the parallelogram EC to the parallelogram CF , therefore also, as the base BC is to the base CD , so is the parallelogram EC to the parallelogram FC . [ V. 11 ]

PROPOSITION 2.

17 If a straight line be drawn parallel to one of the sides of a triangle , it will cut the sides of the triangle proportionally; and , if the sides of the triangle be cut proportionally , the line joining the points of section will be parallel to the remaining side of the triangle .

18 For let DE be drawn parallel to BC , one of the sides of the triangle ABC ; I say that, as BD is to DA , so is CE to EA .

20 Therefore the triangle BDE is equal to the triangle CDE ; for they are on the same base DE and in the same parallels DE , BC . [ I. 38 ]

22 But equals have the same ratio to the same; [ V. 7 ] therefore, as the triangle BDE is to the triangle ADE , so is the triangle CDE to the triangle ADE .

23 But, as the triangle BDE is to ADE , so is BD to DA ; for, being under the same height, the perpendicular drawn from E to AB , they are to one another as their bases. [ VI. 1 ]

24 For the same reason also, as the triangle CDE is to ADE , so is CE to EA .

25 Therefore also, as BD is to DA , so is CE to EA . [ V. 11 ]

26 Again, let the sides AB , AC of the triangle ABC be cut proportionally, so that, as BD is to DA , so is CE to EA ; and let DE be joined.

28 For, with the same construction, since, as BD is to DA , so is CE to EA , but, as BD is to DA , so is the triangle BDE to the triangle ADE , and, as CE is to EA , so is the triangle CDE to the triangle ADE , [ VI. 1 ] therefore also, as the triangle BDE is to the triangle ADE , so is the triangle CDE to the triangle ADE . [ V. 11 ]

29 Therefore each of the triangles BDE , CDE has the same ratio to ADE .

30 Therefore the triangle BDE is equal to the triangle CDE ; [ V. 9 ] and they are on the same base DE .

31 But equal triangles which are on the same base are also in the same parallels. [ I. 39 ]

PROPOSITION 3.

34 If an angle of a triangle be bisected and the straight line cutting the angle cut the base also , the segments of the base will have the same ratio as the remaining sides of the triangle; and , if the segments of the base have the same ratio as the remaining sides of the triangle , the straight line joined from the vertex to the point of section will bisect the angle of the triangle .

35 Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD ; I say that, as BD is to CD , so is BA to AC .

36 For let CE be drawn through C parallel to DA , and let BA be carried through and meet it at E .

37 Then, since the straight line AC falls upon the parallels AD , EC , the angle ACE is equal to the angle CAD . [ I. 29 ]

38 But the angle CAD is by hypothesis equal to the angle BAD ; therefore the angle BAD is also equal to the angle ACE .

39 Again, since the straight line BAE falls upon the parallels AD , EC , the exterior angle BAD is equal to the interior angle AEC . [ I. 29 ]

40 But the angle ACE was also proved equal to the angle BAD ; therefore the angle ACE is also equal to the angle AEC , so that the side AE is also equal to the side AC . [ I. 6 ]

41 And, since AD has been drawn parallel to EC , one of the sides of the triangle BCE , therefore, proportionally, as BD is to DC , so is BA to AE .

42 But AE is equal to AC ; [ VI. 2 ] therefore, as BD is to DC , so is BA to AC .

43 Again, let BA be to AC as BD to DC , and let AD be joined; I say that the angle BAC has been bisected by the straight line A.D .

44 For, with the same construction, since, as BD is to DC , so is BA to AC , and also, as BD is to DC , so is BA to AE : for AD has been drawn parallel to EC , one of the sides of the triangle BCE : [ VI. 2 ] therefore also, as BA is to AC , so is BA to AE . [ V. 11 ]

45 Therefore AC is equal to AE , [ V. 9 ] so that the angle AEC is also equal to the angle ACE . [ I. 5 ]

46 But the angle AEC is equal to the exterior angle BAD , [ I. 29 ] and the angle ACE is equal to the alternate angle CAD ; [ id .] therefore the angle BAD is also equal to the angle CAD .

47 Therefore the angle BAC has been bisected by the straight line AD .

PROPOSITION 4.

49 In equiangular triangles the sides about the equal angles are proportional , and those are corresponding sides which subtend the equal angles .

50 Let ABC , DCE be equiangular triangles having the angle ABC equal to the angle DCE , the angle BAC to the angle CDE , and further the angle ACB to the angle CED ; I say that in the triangles ABC , DCE the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.

52 Then, since the angles ABC , ACB are less than two right angles, [ I. 17 ] and the angle ACB is equal to the angle DEC , therefore the angles ABC , DEC are less than two right angles; therefore BA , ED , when produced, will meet. [ I. Post. 5 ]

54 Now, since the angle DCE is equal to the angle ABC , BF is parallel to CD . [ I. 28 ]

55 Again, since the angle ACB is equal to the angle DEC , AC is parallel to FE . [ I. 28 ]

56 Therefore FACD is a parallelogram; therefore FA is equal to DC , and AC to FD . [ I. 34 ]

57 And, since AC has been drawn parallel to FE , one side of the triangle FBE , therefore, as BA is to AF , so is BC to CE . [ VI. 2 ]

58 But AF is equal to CD ; therefore, as BA is to CD , so is BC to CE , and alternately, as AB is to BC , so is DC to CE . [ V. 16 ]

59 Again, since CD is parallel to BF , therefore, as BC is to CE , so is FD to DE . [ VI. 2 ]

60 But FD is equal to AC ; therefore, as BC is to CE , so is AC to DE , and alternately, as BC is to CA , so is CE to ED . [ V. 16 ]

61 Since then it was proved that, as AB is to BC , so is DC to CE , and, as BC is to CA , so is CE to ED ; therefore, ex aequali , as BA is to AC , so is CD to DE . [ V. 22 ]

PROPOSITION 5.

63 If two triangles have their sides proportional , the triangles will be equiangular and will have those angles equal which the corresponding sides subtend .

64 Let ABC , DEF be two triangles having their sides proportional, so that, as AB is to BC , so is DE to EF , as BC is to CA , so is EF to FD , and further, as BA is to AC , so is ED to DF ; I say that the triangle ABC is equiangular with the triangle DEF , and they will have those angles equal which the corresponding sides subtend, namely the angle ABC to the angle DEF , the angle BCA to the angle EFD , and further the angle BAC to the angle EDF .

65 For on the straight line EF , and at the points E , F on it, let there be constructed the angle FEG equal to the angle ABC , and the angle EFG equal to the angle ACB ; [ I. 23 ] therefore the remaining angle at A is equal to the remaining angle at G . [ I. 32 ]

66 Therefore the triangle ABC is equiangular with the triangle GEF .

67 Therefore in the triangles ABC , GEF the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles; [ VI. 4 ] therefore, as AB is to BC , so is GE to EF .

68 But, as AB is to BC , so by hypothesis is DE to EF ; therefore, as DE is to EF , so is GE to EF . [ V. 11 ]

69 Therefore each of the straight lines DE , GE has the same ratio to EF ; therefore DE is equal to GE . [ V. 9 ]

71 Since then DE is equal to EG , and EF is common, the two sides DE , EF are equal to the two sides GE , EF ; and the base DF is equal to the base FG ; therefore the angle DEF is equal to the angle GEF , [ I. 8 ] and the triangle DEF is equal to the triangle GEF , and the remaining angles are equal to the remaining angles, namely those which the equal sides subtend. [ I. 4 ]

72 Therefore the angle DFE is also equal to the angle GFE , and the angle EDF to the angle EGF .

73 And, since the angle FED is equal to the angle GEF , while the angle GEF is equal to the angle ABC , therefore the angle ABC is also equal to the angle DEF .

74 For the same reason the angle ACB is also equal to the angle DFE , and further, the angle at A to the angle at D ; therefore the triangle ABC is equiangular with the triangle DEF .

PROPOSITION 6.

76 If two triangles have one angle equal to one angle and the sides about the equal angles proportional , the triangles will be equiangular and will have those angles equal which the corresponding sides subtend .

77 Let ABC , DEF be two triangles having one angle BAC equal to one angle EDF and the sides about the equal angles proportional, so that, as BA is to AC , so is ED to DF ; I say that the triangle ABC is equiangular with the triangle DEF , and will have the angle ABC equal to the angle DEF , and the angle ACB to the angle DFE .

78 For on the straight line DF , and at the points D , F on it, let there be constructed the angle FDG equal to either of the angles BAC , EDF , and the angle DFG equal to the angle ACB ; [ I. 23 ] therefore the remaining angle at B is equal to the remaining angle at G . [ I. 32 ]

79 Therefore the triangle ABC is equiangular with the triangle DGF .

80 Therefore, proportionally, as BA is to AC , so is GD to DF . [ VI. 4 ]

81 But, by hypothesis, as BA is to AC , so also is ED to DF ; therefore also, as ED is to DF , so is GD to DF . [ V. 11 ]

82 Therefore ED is equal to DG ; [ V. 9 ] and DF is common; therefore the two sides ED , DF are equal to the two sides GD , DF ; and the angle EDF is equal to the angle GDF ; therefore the base EF is equal to the base GF , and the triangle DEF is equal to the triangle DGF , and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [ I. 4 ]

83 Therefore the angle DFG is equal to the angle DFE , and the angle DGF to the angle DEF .

84 But the angle DFG is equal to the angle ACB ; therefore the angle ACB is also equal to the angle DFE .

85 And, by hypothesis, the angle BAC is also equal to the angle EDF ; therefore the remaining angle at B is also equal to the remaining angle at E ; [ I. 32 ] therefore the triangle ABC is equiangular with the triangle DEF .

PROPOSITION 7.

87 If two triangles have one angle equal to one angle , the sides about other angles proportional , and the remaining angles either both less or both not less than a right angle , the triangles will be equiangular and will have those angles equal , the sides about which are proportional .

88 Let ABC , DEF be two triangles having one angle equal to one angle, the angle BAC to the angle EDF , the sides about other angles ABC , DEF proportional, so that, as AB is to BC , so is DE to EF , and, first, each of the remaining angles at C , F less than a right angle; I say that the triangle ABC is equiangular with the triangle DEF , the angle ABC will be equal to the angle DEF , and the remaining angle, namely the angle at C , equal to the remaining angle, the angle at F .

89 For, if the angle ABC is unequal to the angle DEF , one of them is greater.

90 Let the angle ABC be greater; and on the straight line AB , and at the point B on it, let the angle ABG be constructed equal to the angle DEF . [ I. 23 ]

91 Then, since the angle A is equal to D , and the angle ABG to the angle DEF , therefore the remaining angle AGB is equal to the remaining angle DFE . [ I. 32 ]

92 Therefore the triangle ABG is equiangular with the triangle DEF .

93 Therefore, as AB is to BG , so is DE to EF [ VI. 4 ]

94 But, as DE is to EF , so by hypothesis is AB to BC ; therefore AB has the same ratio to each of the straight lines BC , BG ; [ V. 11 ] therefore BC is equal to BG , [ V. 9 ] so that the angle at C is also equal to the angle BGC . [ I. 5 ]

95 But, by hypothesis, the angle at C is less than a right angle; therefore the angle BGC is also less than a right angle; so that the angle AGB adjacent to it is greater than a right angle. [ I. 13 ]

96 And it was proved equal to the angle at F ; therefore the angle at F is also greater than a right angle.

97 But it is by hypothesis less than a right angle : which is absurd.

98 Therefore the angle ABC is not unequal to the angle DEF ; therefore it is equal to it.

99 But the angle at A is also equal to the angle at D ; therefore the remaining angle at C is equal to the remaining angle at F . [ I. 32 ]

100 Therefore the triangle ABC is equiangular with the triangle DEF .

101 But, again, let each of the angles at C , F be supposed not less than a right angle; I say again that, in this case too, the triangle ABC is equiangular with the triangle DEF .

102 For, with the same construction, we can prove similarly that BC is equal to BG ; so that the angle at C is also equal to the angle BGC . [ I. 5 ]

103 But the angle at C is not less than a right angle; therefore neither is the angle BGC less than a right angle.

104 Thus in the triangle BGC the two angles are not less than two right angles: which is impossible. [ I. 17 ]

105 Therefore, once more, the angle ABC is not unequal to the angle DEF ; therefore it is equal to it.

106 But the angle at A is also equal to the angle at D ; therefore the remaining angle at C is equal to the remaining angle at F . [ I. 32 ]

107 Therefore the triangle ABC is equiangular with the triangle DEF .

PROPOSITION 8.

109 If in a right-angled triangle a perpendicular be drawn from the right angle to the base , the triangles adjoining the perpendicular are similar both to the whole and to one another .

110 Let ABC be a right-angled triangle having the angle BAC right, and let AD be drawn from A perpendicular to BC ; I say that each of the triangles ABD , ADC is similar to the whole ABC and, further, they are similar to one another.

111 For, since the angle BAC is equal to the angle ADB , for each is right, and the angle at B is common to the two triangles ABC and ABD , therefore the remaining angle ACB is equal to the remaining angle BAD ; [ I. 32 ] therefore the triangle ABC is equiangular with the triangle ABD .

112 Therefore, as BC which subtends the right angle in the triangle ABC is to BA which subtends the right angle in the triangle ABD , so is AB itself which subtends the angle at C in the triangle ABC to BD which subtends the equal angle BAD in the triangle ABD , and so also is AC to AD which subtends the angle at B common to the two triangles. [ VI. 4 ]

113 Therefore the triangle ABC is both equiangular to the triangle ABD and has the sides about the equal angles proportional.

114 Therefore the triangle ABC is similar to the triangle ABD . [ VI. Def. 1 ]

115 Similarly we can prove that the triangle ABC is also similar to the triangle ADC ; therefore each of the triangles ABD , ADC is similar to the whole ABC .

116 I say next that the triangles ABD , ADC are also similar to one another.

117 For, since the right angle BDA is equal to the right angle ADC , and moreover the angle BAD was also proved equal to the angle at C , therefore the remaining angle at B is also equal to the remaining angle DAC ; [ I. 32 ] therefore the triangle ABD is equiangular with the triangle ADC .

118 Therefore, as BD which subtends the angle BAD in the triangle ABD is to DA which subtends the angle at C in the triangle ADC equal to the angle BAD , so is AD itself which subtends the angle at B in the triangle ABD to DC which subtends the angle DAC in the triangle ADC equal to the angle at B , and so also is BA to AC , these sides subtending the right angles; [ VI. 4 ] therefore the triangle ABD is similar to the triangle ADC . [ VI. Def. 1 ]

PORISM.

120 From this it is clear that, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the straight line so drawn is a mean proportional between the segments of the base. Q. E. D.

PROPOSITION 9.

121 From a given straight line to cut off a prescribed part .

122 Let AB be the given straight line; thus it is required to cut off from AB a prescribed part.

124 Let a straight line AC be drawn through from A containing with AB any angle; let a point D be taken at random on AC , and let DE , EC be made equal to AD . [ I. 3 ]

125 Let BC be joined, and through D let DF be drawn parallel to it. [ I. 31 ]

126 Then, since FD has been drawn parallel to BC , one of the sides of the triangle ABC , therefore, proportionally, as CD is to DA , so is BF to FA . [ VI. 2 ]

127 But CD is double of DA ; therefore BF is also double of FA ; therefore BA is triple of AF .

128 Therefore from the given straight line AB the prescribed third part AF has been cut off. Q. E. F.

PROPOSITION 10.

129 To cut a given uncut straight line similarly to a given cut straight line .

130 Let AB be the given uncut straight line, and AC the straight line cut at the points D , E ; and let them be so placed as to contain any angle; let CB be joined, and through D , E let DF , EG be drawn parallel to BC , and through D let DHK be drawn parallel to AB . [ I. 31 ]

131 Therefore each of the figures FH , HB is a parallelogram; therefore DH is equal to FG and HK to GB . [ I. 34 ]

132 Now, since the straight line HE has been drawn parallel to KC , one of the sides of the triangle DKC , therefore, proportionally, as CE is to ED , so is KH to HD . [ VI. 2 ]

133 But KH is equal to BG , and HD to GF ; therefore, as CE is to ED , so is BG to GF .

134 Again, since FD has been drawn parallel to GE , one of the sides of the triangle AGE , therefore, proportionally, as ED is to DA , so is GF to FA . [ VI. 2 ]

135 But it was also proved that, as CE is to ED , so is BG to GF ; therefore, as CE is to ED , so is BG to GF , and, as ED is to DA , so is GF to FA .

136 Therefore the given uncut straight line AB has been cut similarly to the given cut straight line AC . Q. E. F.

PROPOSITION 11.

137 To two given straight lines to find a third proportional .

138 Let BA , AC be the two given straight lines, and let them be placed so as to contain any angle; thus it is required to find a third proportional to BA , AC .

139 For let them be produced to the points D , E , and let BD be made equal to AC ; [ I. 3 ] let BC be joined, and through D let DE be drawn parallel to it. [ I. 31 ]

140 Since, then, BC has been drawn parallel to DE , one of the sides of the triangle ADE , proportionally, as AB is to BD , so is AC to CE . [ VI. 2 ]

141 But BD is equal to AC ; therefore, as AB is to AC , so is AC to CE .

142 Therefore to two given straight lines AB , AC a third proportional to them, CE , has been found. Q. E. F.

PROPOSITION 12.

143 To three given straight lines to find a fourth proportional .

144 Let A , B , C be the three given straight lines; thus it is required to find a fourth proportional to A , B , C .

145 Let two straight lines DE , DF be set out containing any angle EDF ; let DG be made equal to A , GE equal to B , and further DH equal to C ; let GH be joined, and let EF be drawn through E parallel to it. [ I. 31 ]

146 Since, then, GH has been drawn parallel to EF , one of the sides of the triangle DEF , therefore, as DG is to GE , so is DH to HF . [ VI. 2 ]

147 But DG is equal to A , GE to B , and DH to C ; therefore, as A is to B , so is C to HF .

148 Therefore to the three given straight lines A , B , C a fourth proportional HF has been found. Q. E. F.

PROPOSITION 13.

149 To two given straight lines to find a mean proportional .

150 Let AB , BC be the two given straight lines; thus it is required to find a mean proportional to AB , BC .

151 Let them be placed in a straight line, and let the semicircle ADC be described on AC ; let BD be drawn from the point B at right angles to the straight line AC , and let AD , DC be joined.

152 Since the angle ADC is an angle in a semicircle, it is right. [ III. 31 ]

153 And, since, in the right-angled triangle ADC , DB has been drawn from the right angle perpendicular to the base, therefore DB is a mean proportional between the segments of the base, AB , BC . [ VI. 8, Por. ]

154 Therefore to the two given straight lines AB , BC a mean proportional DB has been found. Q. E. F.

PROPOSITION 14.

155 In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal .

156 Let AB , BC be equal and equiangular parallelograms having the angles at B equal, and let DB , BE be placed in a straight line; therefore FB , BG are also in a straight line. [ I. 14 ]

157 I say that, in AB , BC , the sides about the equal angles are reciprocally proportional, that is to say, that, as DB is to BE , so is GB to BF .

159 Since, then, the parallelogram AB is equal to the parallelogram BC , and FE is another area, therefore, as AB is to FE , so is BC to FE . [ V. 7 ]

160 But, as AB is to FE , so is DB to BE , [ VI. 1 ] and, as BC is to FE , so is GB to BF . [ id .] therefore also, as DB is to BE , so is GB to BF . [ V. 11 ]

161 Therefore in the parallelograms AB , BC the sides about the equal angles are reciprocally proportional.

162 Next, let GB be to BF as DB to BE ; I say that the parallelogram AB is equal to the parallelogram BC .

163 For since, as DB is to BE , so is GB to BF , while, as DB is to BE , so is the parallelogram AB to the parallelogram FE , [ VI. 1 ] and, as GB is to BF , so is the parallelogram BC to the parallelogram FE , [ VI. 1 ] therefore also, as AB is to FE , so is BC to FE ; [ V. 11 ] therefore the parallelogram AB is equal to the parallelogram BC . [ V. 9 ]

PROPOSITION 15.

165 In equal triangles which have one angle equal to one angle the sides about the equal angles are reciprocally proportional; and those triangles which have one angle equal to one angle , and in which the sides about the equal angles are reciprocally proportional , are equal .

166 Let ABC , ADE be equal triangles having one angle equal to one angle, namely the angle BAC to the angle DAE ; I say that in the triangles ABC , ADE the sides about the equal angles are reciprocally proportional, that is to say, that, as CA is to AD , so is EA to AB .

167 For let them be placed so that CA is in a straight line with AD ; therefore EA is also in a straight line with AB . [ I. 14 ]

169 Since then the triangle ABC is equal to the triangle ADE , and BAD is another area, therefore, as the triangle CAB is to the triangle BAD , so is the triangle EAD to the triangle BAD . [ V. 7 ]

170 But, as CAB is to BAD , so is CA to AD , [ VI. 1 ] and, as EAD is to BAD , so is EA to AB . [ id .]

171 Therefore also, as CA is to AD , so is EA to AB . [ V. 11 ]

172 Therefore in the triangles ABC , ADE the sides about the equal angles are reciprocally proportional.

173 Next, let the sides of the triangles ABC , ADE be reciprocally proportional, that is to say, let EA be to AB as CA to AD ; I say that the triangle ABC is equal to the triangle ADE .

174 For, if BD be again joined, since, as CA is to AD , so is EA to AB , while, as CA is to AD , so is the triangle ABC to the triangle BAD , and, as EA is to AB , so is the triangle EAD to the triangle BAD , [ VI. 1 ] therefore, as the triangle ABC is to the triangle BAD , so is the triangle EAD to the triangle BAD . [ V. 11 ]

175 Therefore each of the triangles ABC , EAD has the same ratio to BAD .

176 Therefore the triangle ABC is equal to the triangle EAD . [ V. 9 ]

PROPOSITION 16.

178 If four straight lines be proportional , the rectangle contained by the extremes is equal to the rectangle contained by the means; and , if the rectangle contained by the extremes be equal to the rectangle contained by the means , the four straight lines will be proportional .

179 Let the four straight lines AB , CD , E , F be proportional, so that, as AB is to CD , so is E to F ; I say that the rectangle contained by AB , F is equal to the rectangle contained by CD , E .

180 Let AG , CH be drawn from the points A , C at right angles to the straight lines AB , CD , and let AG be made equal to F , and CH equal to E .

182 Then since, as AB is to CD , so is E to F , while E is equal to CH , and F to AG , therefore, as AB is to CD , so is CH to AG .

183 Therefore in the parallelograms BG , DH the sides about the equal angles are reciprocally proportional.

184 But those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal; [ VI. 14 ] therefore the parallelogram BG is equal to the parallelogram DH .

185 And BG is the rectangle AB , F , for AG is equal to F ; and DH is the rectangle CD , E , for E is equal to CH ; therefore the rectangle contained by AB , F is equal to the rectangle contained by CD , E .

186 Next, let the rectangle contained by AB , F be equal to the rectangle contained by CD , E ; I say that the four straight lines will be proportional, so that, as AB is to CD , so is E to F .

187 For, with the same construction, since the rectangle AB , F is equal to the rectangle CD , E , and the rectangle AB , F is BG , for AG is equal to F , and the rectangle CD , E is DH , for CH is equal to E , therefore BG is equal to DH .

189 But in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional. [ VI. 14 ]

191 But CH is equal to E , and AG to F ; therefore, as AB is to CD , so is E to F .

PROPOSITION 17

193 If three straight lines be proportional , the rectangle contained by the extremes is equal to the square on the mean ; and , if the rectangle contained by the extremes be equal to the square on the mean , the three straight lines will be proportional .

194 Let the three straight lines A , B , C be proportional, so that, as A is to B , so is B to C ; I say that the rectangle contained by A , C is equal to the square on B .

196 Then, since, as A is to B , so is B to C , and B is equal to D , therefore, as A is to B , so is D to C .

197 But, if four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means. [ VI. 16 ]

198 Therefore the rectangle A , C is equal to the rectangle B , D .

199 But the rectangle B , D is the square on B , for B is equal to D ; therefore the rectangle contained by A , C is equal to the square on B .

200 Next, let the rectangle A , C be equal to the square on B ; I say that, as A is to B , so is B to C .

201 For, with the same construction, since the rectangle A , C is equal to the square on B , while the square on B is the rectangle B , D , for B is equal to D , therefore the rectangle A , C is equal to the rectangle B , D .

202 But, if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportional. [ VI. 16 ]

204 But B is equal to D ; therefore, as A is to B , so is B to C .

PROPOSITION 18.

206 On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure .

207 Let AB be the given straight line and CE the given rectilineal figure; thus it is required to describe on the straight line AB a rectilineal figure similar and similarly situated to the rectilineal figure CE .

208 Let DF be joined, and on the straight line AB , and at the points A , B on it, let the angle GAB be constructed equal to the angle at C , and the angle ABG equal to the angle CDF . [ I. 23 ]

209 Therefore the remaining angle CFD is equal to the angle AGB ; [ I. 32 ] therefore the triangle FCD is equiangular with the triangle GAB .

210 Therefore, proportionally, as FD is to GB , so is FC to GA , and CD to AB .

211 Again, on the straight line BG , and at the points B , G on it, let the angle BGH be constructed equal to the angle DFE , and the angle GBH equal to the angle FDE . [ I. 23 ]

212 Therefore the remaining angle at E is equal to the remaining angle at H ; [ I. 32 ] therefore the triangle FDE is equiangular with the triangle GBH ; therefore, proportionally, as FD is to GB , so is FE to GH , and ED to HB . [ VI. 4 ]

213 But it was also proved that, as FD is to GB , so is FC to GA , and CD to AB ; therefore also, as FC is to AG , so is CD to AB , and FE to GH , and further ED to HB .

214 And, since the angle CFD is equal to the angle AGB , and the angle DFE to the angle BGH , therefore the whole angle CFE is equal to the whole angle AGH .

215 For the same reason the angle CDE is also equal to the angle ABH .

216 And the angle at C is also equal to the angle at A , and the angle at E to the angle at H .

217 Therefore AH is equiangular with CE ; and they have the sides about their equal angles proportional; therefore the rectilineal figure AH is similar to the rectilineal figure CE . [ VI. Def. 1 ]

218 Therefore on the given straight line AB the rectilineal figure AH has been described similar and similarly situated to the given rectilineal figure CE . Q. E. F.

PROPOSITION 19.

219 Similar triangles are to one another in the duplicate ratio of the corresponding sides .

220 Let ABC , DEF be similar triangles having the angle at B equal to the angle at E , and such that, as AB is to BC , so is DE to EF , so that BC corresponds to EF ; [ V. Def. 11 ] I say that the triangle ABC has to the triangle DEF a ratio duplicate of that which BC has to EF .

221 For let a third proportional BG be taken to BC , EF , so that, as BC is to EF , so is EF to BG ; [ VI. 11 ] and let AG be joined.

222 Since then, as AB is to BC , so is DE to EF , therefore, alternately, as AB is to DE , so is BC to EF . [ V. 16 ]

223 But, as BC is to EF , so is EF to BG ; therefore also, as AB is to DE , so is EF to BG . [ V. 11 ]

224 Therefore in the triangles ABG , DEF the sides about the equal angles are reciprocally proportional.

225 But those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal; [ VI. 15 ] therefore the triangle ABG is equal to the triangle DEF .

226 Now since, as BC is to EF , so is EF to BG , and, if three straight lines be proportional, the first has to the third a ratio duplicate of that which it has to the second, [ V. Def. 9 ] therefore BC has to BG a ratio duplicate of that which CB has to EF .

227 But, as CB is to BG , so is the triangle ABC to the triangle ABG ; [ VI. 1 ] therefore the triangle ABC also has to the triangle ABG a ratio duplicate of that which BC has to EF .

228 But the triangle ABG is equal to the triangle DEF ; therefore the triangle ABC also has to the triangle DEF a ratio duplicate of that which BC has to EF .

PORISM.

230 From this it is manifest that, if three straight lines be proportional, then, as the first is to the third, so is the figure described on the first to that which is similar and similarly described on the second. Q. E. D.

PROPOSITION 20.

231 Similar polygons are divided into similar triangles , and into triangles equal in multitude and in the same ratio as the wholes , and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding side .

232 Let ABCDE , FGHKL be similar polygons, and let AB correspond to FG ; I say that the polygons ABCDE , FGHKL are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which AB has to FG .

234 Now, since the polygon ABCDE is similar to the polygon FGHKL , the angle BAE is equal to the angle GFL ; and, as BA is to AE , so is GF to FL . [ VI. Def. 1 ]

235 Since then ABE , FGL are two triangles having one angle equal to one angle and the sides about the equal angles proportional, therefore the triangle ABE is equiangular with the triangle FGL ; [ VI. 6 ] so that it is also similar; [ VI. 4 and Def. 1 ] therefore the angle ABE is equal to the angle FGL .

236 But the whole angle ABC is also equal to the whole angle FGH because of the similarity of the polygons; therefore the remaining angle EBC is equal to the angle LGH .

237 And, since, because of the similarity of the triangles ABE , FGL , as EB is to BA , so is LG to GF , and moreover also, because of the similarity of the polygons, as AB is to BC , so is FG to GH , therefore, ex aequali , as EB is to BC , so is LG to GH ; [ V. 22 ] that is, the sides about the equal angles EBC , LGH are proportional; therefore the triangle EBC is equiangular with the triangle LGH , [ VI. 6 ] so that the triangle EBC is also similar to the triangle LGH . [ VI. 4 and Def. 1 ]

238 For the same reason the triangle ECD is also similar to the triangle LHK .

239 Therefore the similar polygons ABCDE , FGHKL have been divided into similar triangles, and into triangles equal in multitude.

240 I say that they are also in the same ratio as the wholes, that is, in such manner that the triangles are proportional, and ABE , EBC , ECD are antecedents, while FGL , LGH , LHK are their consequents, and that the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which the corresponding side has to the corresponding side, that is AB to FG .

242 Then since, because of the similarity of the polygons, the angle ABC is equal to the angle FGH , and, as AB is to BC , so is FG to GH , the triangle ABC is equiangular with the triangle FGH ; [ VI. 6 ] therefore the angle BAC is equal to the angle GFH , and the angle BCA to the angle GHF .

243 And, since the angle BAM is equal to the angle GFN , and the angle ABM is also equal to the angle FGN , therefore the remaining angle AMB is also equal to the remaining angle FNG ; [ I. 32 ] therefore the triangle ABM is equiangular with the triangle FGN .

244 Similarly we can prove that the triangle BMC is also equiangular with the triangle GNH .

245 Therefore, proportionally, as AM is to MB , so is FN to NG , and, as BM is to MC , so is GN to NH ; so that, in addition, ex aequali , as AM is to MC , so is FN to NH .

246 But, as AM is to MC , so is the triangle ABM to MBC , and AME to EMC ; for they are to one another as their bases. [ VI. 1 ]

247 Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [ V. 12 ] therefore, as the triangle AMB is to BMC , so is ABE to CBE .

248 But, as AMB is to BMC , so is AM to MC ; therefore also, as AM is to MC , so is the triangle ABE to the triangle EBC .

249 For the same reason also, as FN is to NH , so is the triangle FGL to the triangle GLH .

250 And, as AM is to MC , so is FN to NH ; therefore also, as the triangle ABE is to the triangle BEC , so is the triangle FGL to the triangle GLH ; and, alternately, as the triangle ABE is to the triangle FGL , so is the triangle BEC to the triangle GLH .

251 Similarly we can prove, if BD , GK be joined, that, as the triangle BEC is to the triangle LGH , so also is the triangle ECD to the triangle LHK .

252 And since, as the triangle ABE is to the triangle FGL , so is EBC to LGH , and further ECD to LHK , therefore also, as one of the antecedents is to one of the consequents so are all the antecedents to all the consequents; [ V. 12 therefore, as the triangle ABE is to the triangle FGL , so is the polygon ABCDE to the polygon FGHKL .

253 But the triangle ABE has to the triangle FGL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG ; for similar triangles are in the duplicate ratio of the corresponding sides. [ VI. 19 ]

254 Therefore the polygon ABCDE also has to the polygon FGHKL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG .

PORISM.

256 Similarly also it can be proved in the case of quadrilaterals that they are in the duplicate ratio of the corresponding sides. And it was also proved in the case of triangles; therefore also, generally, similar rectilineal figures are to one another in the duplicate ratio of the corresponding sides. Q. E. D.

PROPOSITION 21.

257 Figures which are similar to the same rectilineal figure are also similar to one another .

258 For let each of the rectilineal figures A , B be similar to C ; I say that A is also similar to B .

259 For, since A is similar to C , it is equiangular with it and has the sides about the equal angles proportional. [ VI. Def. 1 ]

260 Again, since B is similar to C , it is equiangular with it and has the sides about the equal angles proportional.

261 Therefore each of the figures A , B is equiangular with C and with C has the sides about the equal angles proportional; therefore A is similar to B . Q. E. D.

PROPOSITION 22.

262 If four straight lines be proportional , the rectilineal figures similar and similarly described upon them will also be proportional ; and , if the rectilineal figures similar and similarly described upon them be proportional , the straight lines will themselves also be proportional .

263 Let the four straight lines AB , CD , EF , GH be proportional, so that, as AB is to CD , so is EF to GH , and let there be described on AB , CD the similar and similarly situated rectilineal figures KAB , LCD , and on EF , GH the similar and similarly situated rectilineal figures MF , NH ; I say that, as KAB is to LCD , so is MF to NH .

264 For let there be taken a third proportional O to AB , CD , and a third proportional P to EF , GH . [ VI. 11 ]

265 Then since, as AB is to CD , so is EF to GH , and, as CD is to O , so is GH to P , therefore, ex aequali , as AB is to O , so is EF to P . [ V. 22 ]

266 But, as AB is to O , so is KAB to LCD , [ VI. 19, Por. ] and, as EF is to P , so is MF to NH ; therefore also, as KAB is to LCD , so is MF to NH . [ V. 11 ]

267 Next, let MF be to NH as KAB is to LCD ; I say also that, as AB is to CD , so is EF to GH .

268 For, if EF is not to GH as AB to CD , let EF be to QR as AB to CD , [ VI. 12 ] and on QR let the rectilineal figure SR be described similar and similarly situated to either of the two MF , NH . [ VI. 18 ]

269 Since then, as AB is to CD , so is EF to QR , and there have been described on AB , CD the similar and similarly situated figures KAB , LCD , and on EF , QR the similar and similarly situated figures MF , SR , therefore, as KAB is to LCD , so is MF to SR .

270 But also, by hypothesis, as KAB is to LCD , so is MF to NH ; therefore also, as MF is to SR , so is MF to NH . [ V. 11 ]

271 Therefore MF has the same ratio to each of the figures NH , SR ; therefore NH is equal to SR . [ V. 9 ]

272 But it is also similar and similarly situated to it; therefore GH is equal to QR .

273 And, since, as AB is to CD , so is EF to QR , while QR is equal to GH , therefore, as AB is to CD , so is EF to GH .

PROPOSITION 23.

275 Equiangular parallelograms have to one another the ratio compounded of the ratios of their sides .

276 Let AC , CF be equiangular parallelograms having the angle BCD equal to the angle ECG ; I say that the parallelogram AC has to the parallelogram CF the ratio compounded of the ratios of the sides.

277 For let them be placed so that BC is in a straight line with CG ; therefore DC is also in a straight line with CE .

278 Let the parallelogram DG be completed; let a straight line K be set out, and let it be contrived that, as BC is to CG , so is K to L , and, as DC is to CE , so is L to M . [ VI. 12 ]

279 Then the ratios of K to L and of L to M are the same as the ratios of the sides, namely of BC to CG and of DC to CE .

280 But the ratio of K to M is compounded of the ratio of K to L and of that of L to M ; so that K has also to M the ratio compounded of the ratios of the sides.

281 Now since, as BC is to CG , so is the parallelogram AC to the parallelogram CH , [ VI. 1 ] while, as BC is to CG , so is K to L , therefore also, as K is to L , so is AC to CH . [ V. 11 ]

282 Again, since, as DC is to CE , so is the parallelogram CH to CF , [ VI. 1 ] while, as DC is to CE , so is L to M , therefore also, as L is to M , so is the parallelogram CH to the parallelogram CF . [ V. 11 ]

283 Since then it was proved that, as K is to L , so is the parallelogram AC to the parallelogram CH , and, as L is to M , so is the parallelogram CH to the parallelogram CF , therefore, ex aequali , as K is to M , so is AC to the parallelogram CF .

284 But K has to M the ratio compounded of the ratios of the sides; therefore AC also has to CF the ratio compounded of the ratios of the sides.

PROPOSITION 24.

286 In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another .

287 Let ABCD be a parallelogram, and AC its diameter, and let EG , HK be parallelograms about AC ; I say that each of the parallelograms EG , HK is similar both to the whole ABCD and to the other.

288 For, since EF has been drawn parallel to BC , one of the sides of the triangle ABC , proportionally, as BE is to EA , so is CF to FA . [ VI. 2 ]

289 Again, since FG has been drawn parallel to CD , one of the sides of the triangle ACD , proportionally, as CF is to FA , so is DG to GA . [ VI. 2 ]

290 But it was proved that, as CF is to FA , so also is BE to EA ; therefore also, as BE is to EA , so is DG to GA , and therefore, componendo , as BA is to AE , so is DA to AG , [ V. 18 ] and, alternately, as BA is to AD , so is EA to AG . [ V. 16 ]

291 Therefore in the parallelograms ABCD , EG , the sides about the common angle BAD are proportional.

292 And, since GF is parallel to DC , the angle AFG is equal to the angle DCA ; and the angle DAC is common to the two triangles ADC , AGF ; therefore the triangle ADC is equiangular with the triangle AGF .

293 For the same reason the triangle ACB is also equiangular with the triangle AFE , and the whole parallelogram ABCD is equiangular with the parallelogram EG .

294 Therefore, proportionally, as AD is to DC , so is AG to GF , as DC is to CA , so is GF to FA , as AC is to CB , so is AF to FE , and further, as CB is to BA , so is FE to EA .

295 And, since it was proved that, as DC is to CA , so is GF to FA , and, as AC is to CB , so is AF to FE , therefore, ex aequali , as DC is to CB , so is GF to FE . [ V. 22 ]

296 Therefore in the parallelograms ABCD , EG the sides about the equal angles are proportional; therefore the parallelogram ABCD is similar to the parallelogram EG . [ VI. Def. 1 ]

297 For the same reason the parallelogram ABCD is also similar to the parallelogram KH ; therefore each of the parallelograms EG , HK is similar to ABCD .

298 But figures similar to the same rectilineal figure are also similar to one another; [ VI. 21 ] therefore the parallelogram EG is also similar to the parallelogram HK .

PROPOSITION 25.

300 To construct one and the same figure similar to a given rectilineal figure and equal to another given rectilineal figure .

301 Let ABC be the given rectilineal figure to which the figure to be constructed must be similar, and D that to which it must be equal; thus it is required to construct one and the same figure similar to ABC and equal to D .

302 Let there be applied to BC the parallelogram BE equal to the triangle ABC [ I. 44 ], and to CE the parallelogram CM equal to D in the angle FCE which is equal to the angle CBL . [ I. 45 ]

303 Therefore BC is in a straight line with CF , and LE with EM .

304 Now let GH be taken a mean proportional to BC , CF [ VI. 13 ], and on GH let KGH be described similar and similarly situated to ABC . [ VI. 18 ]

305 Then, since, as BC is to GH , so is GH to CF , and, if three straight lines be proportional, as the first is to the third, so is the figure on the first to the similar and similarly situated figure described on the second, [ VI. 19, Por. ] therefore, as BC is to CF , so is the triangle ABC to the triangle KGH .

306 But, as BC is to CF , so also is the parallelogram BE to the parallelogram EF . [ VI. 1 ]

307 Therefore also, as the triangle ABC is to the triangle KGH , so is the parallelogram BE to the parallelogram EF ; therefore, alternately, as the triangle ABC is to the parallelogram BE , so is the triangle KGH to the parallelogram EF . [ V. 16 ]

308 But the triangle ABC is equal to the parallelogram BE ; therefore the triangle KGH is also equal to the parallelogram EF .

309 But the parallelogram EF is equal to D ; therefore KGH is also equal to D .

311 Therefore one and the same figure KGH has been constructed similar to the given rectilineal figure ABC and equal to the other given figure D . Q. E. D.

PROPOSITION 26.

312 If from a parallelogram there be taken away a parallelogram similar and similarly situated to the whole and having a common angle with it , it is about the same diameter with the whole

313 For from the parallelogram ABCD let there be taken away the parallelogram AF similar and similarly situated to ABCD , and having the angle DAB common with it; I say that ABCD is about the same diameter with AF .

314 For suppose it is not, but, if possible, let AHC be the diameter of ABCD >, let GF be produced and carried through to H , and let HK be drawn through H parallel to either of the straight lines AD , BC . [ I. 31 ]

315 Since, then, ABCD is about the same diameter with KG , therefore, as DA is to AB , so is GA to AK . [ VI. 24 ]

316 But also, because of the similarity of ABCD , EG , as DA is to AB , so is GA to AE ; therefore also, as GA is to AK , so is GA to AE . [ V. 11 ]

317 Therefore GA has the same ratio to each of the straight lines AK , AE .

318 Therefore AE is equal to AK [ V. 9 ], the less to the greater : which is impossible.

319 Therefore ABCD cannot but be about the same diameter with AF ; therefore the parallelogram ABCD is about the same diameter with the parallelogram AF .

PROPOSITION 27.

321 Of all the parallelograms applied to the same straight line and deficient by parallelogrammic figures similar and similarly situated to that described on the half of the straight line , that parallelogram is greatest which is applied to the half of the straight line and is similar to the defect .

322 Let AB be a straight line and let it be bisected at C ; let there be applied to the straight line AB the parallelogram AD deficient by the parallelogrammic figure DB described on the half of AB , that is, CB ; I say that, of all the parallelograms applied to AB and deficient by parallelogrammic figures similar and similarly situated to DB , AD is greatest.

323 For let there be applied to the straight line AB the parallelogram AF deficient by the parallelogrammic figure FB similar and similarly situated to DB ; I say that AD is greater than AF .

324 For, since the parallelogram DB is similar to the parallelogram FB , they are about the same diameter. [ VI. 26 ]

325 Let their diameter DB be drawn, and let the figure be described.

326 Then, since CF is equal to FE , [ I. 43 ] and FB is common, therefore the whole CH is equal to the whole KE .

327 But CH is equal to CG , since AC is also equal to CB . [ I. 36 ]

329 Let CF be added to each; therefore the whole AF is equal to the gnomon LMN ; so that the parallelogram DB , that is, AD , is greater than the parallelogram AF .

PROPOSITION 28.

331 To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to a given one : thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect .

332 Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, not being greater than the parallelogram described on the half of AB and similar to the defect, and D the parallelogram to which the defect is required to be similar; thus it is required to apply to the given straight line AB a parallelogram equal to the given rectilineal figure C and deficient by a parallelogrammic figure which is similar to D .

333 Let AB be bisected at the point E , and on EB let EBFG be described similar and similarly situated to D ; [ VI. 18 ] let the parallelogram AG be completed.

334 If then AG is equal to C , that which was enjoined will have been done; for there has been applied to the given straight line AB the parallelogram AG equal to the given rectilineal figure C and deficient by a parallelogrammic figure GB which is similar to D .

336 Now HE is equal to GB ; therefore GB is also greater than C .

337 Let KLMN be constructed at once equal to the excess by which GB is greater than C and similar and similarly situated to D . [ VI. 25 ]

338 But D is similar to GB ; therefore KM is also similar to GB . [ VI. 21 ]

340 Now, since GB is equal to C , KM , therefore GB is greater than KM ; therefore also GE is greater than KL , and GF than LM .

341 Let GO be made equal to KL , and GP equal to LM ; and let the parallelogram OGPQ be completed; therefore it is equal and similar to KM .

342 Therefore GQ is also similar to GB ; [ VI. 21 ] therefore GQ is about the same diameter with GB . [ VI. 26 ]

343 Let GQB be their diameter, and let the figure be described.

344 Then, since BG is equal to C , KM , and in them GQ is equal to KM , therefore the remainder, the gnomon UWV , is equal to the remainder C .

345 And, since PR is equal to OS , let QB be added to each; therefore the whole PB is equal to the whole OB .

346 But OB is equal to TE , since the side AE is also equal to the side EB ; [ I. 36 ] therefore TE is also equal to PB .

347 Let OS be added to each; therefore the whole TS is equal to the whole, the gnomon VWU .

348 But the gnomon VWU was proved equal to C ; therefore TS is also equal to C .

349 Therefore to the given straight line AB there has been applied the parallelogram ST equal to the given rectilineal figure C and deficient by a parallelogrammic figure QB which is similar to D . Q. E. F.

PROPOSITION 29.

350 To a given straight line to apply a parallelogram equal to a given rectilineal figure and exceeding by a parallelogrammic figure similar to a given one .

351 Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, and D that to which the excess is required to be similar; thus it is required to apply to the straight line AB a parallelogram equal to the rectilineal figure C and exceeding by a parallelogrammic figure similar to D .

352 Let AB be bisected at E ; let there be described on EB the parallelogram BF similar and similarly situated to D ; and let GH be constructed at once equal to the sum of BF , C and similar and similarly situated to D . [ VI. 25 ]

354 Now, since GH is greater than FB , therefore KH is also greater than FL , and KG than FE .

355 Let FL , FE be produced, let FLM be equal to KH , and FEN to KG , and let MN be completed; therefore MN is both equal and similar to GH .

356 But GH is similar to EL ; therefore MN is also similar to EL ; [ VI. 21 ] therefore EL is about the same diameter with MN . [ VI. 26 ]

357 Let their diameter FO be drawn, and let the figure be described.

358 Since GH is equal to EL , C , while GH is equal to MN , therefore MN is also equal to EL , C .

359 Let EL be subtracted from each; therefore the remainder, the gnomon XWV , is equal to C .

360 Now, since AE is equal to EB , AN is also equal to NB [ I. 36 ], that is, to LP [ I. 43 ].

361 Let EO be added to each; therefore the whole AO is equal to the gnomon VWX .

362 But the gnomon VWX is equal to C ; therefore AO is also equal to C .

363 Therefore to the given straight line AB there has been applied the parallelogram AO equal to the given rectilineal figure C and exceeding by a parallelogrammic figure QP which is similar to D , since PQ is also similar to EL [ VI. 24 ]. Q. E. F.

PROPOSITION 30.

364 To cut a given finite straight line in extreme and mean ratio .

365 Let AB be the given finite straight line; thus it is required to cut AB in extreme and mean ratio.

366 On AB let the square BC be described; and let there be applied to AC the parallelogram CD equal to BC and exceeding by the figure AD similar to BC . [ VI. 29 ]

368 And, since BC is equal to CD , let CE be subtracted from each; therefore the remainder BF is equal to the remainder AD .

369 But it is also equiangular with it; therefore in BF , AD the sides about the equal angles are reciprocally proportional; [ VI. 14 ] therefore, as FE is to ED , so is AE to EB .

372 And AB is greater than AE ; therefore AE is also greater than EB .

373 Therefore the straight line AB has been cut in extreme and mean ratio at E , and the greater segment of it is AE . Q. E. F.

PROPOSITION 31.

374 In right - angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle .

375 Let ABC be a right-angled triangle having the angle BAC right; I say that the figure on BC is equal to the similar and similarly described figures on BA , AC .

377 Then since, in the right-angled triangle ABC , AD has been drawn from the right angle at A perpendicular to the base BC , the triangles ABD , ADC adjoining the perpendicular are similar both to the whole ABC and to one another. [ VI. 8 ]

378 And, since ABC is similar to ABD , therefore, as CB is to BA , so is AB to BD . [ VI. Def. 1 ]

379 And, since three straight lines are proportional, as the first is to the third, so is the figure on the first to the similar and similarly described figure on the second. [ VI. 19, Por. ]

380 Therefore, as CB is to BD , so is the figure on CB to the similar and similarly described figure on BA .

381 For the same reason also, as BC is to CD , so is the figure on BC to that on CA ; so that, in addition, as BC is to BD , DC , so is the figure on BC to the similar and similarly described figures on BA , AC .

382 But BC is equal to BD , DC ; therefore the figure on BC is also equal to the similar and similarly described figures on BA , AC .

PROPOSITION 32.

384 If two triangles having two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining sides of the triangles will be in a straight line .

385 Let ABC , DCE be two triangles having the two sides BA , AC proportional to the two sides DC , DE , so that, as AB is to AC , so is DC to DE , and AB parallel to DC , and AC to DE ; I say that BC is in a straight line with CE .

386 For, since AB is parallel to DC , and the straight line AC has fallen upon them, the alternate angles BAC , ACD are equal to one another. [ I. 29 ]

387 For the same reason the angle CDE is also equal to the angle ACD ; so that the angle BAC is equal to the angle CDE .

388 And, since ABC , DCE are two triangles having one angle, the angle at A , equal to one angle, the angle at D , and the sides about the equal angles proportional, so that, as BA is to AC , so is CD to DE , therefore the triangle ABC is equiangular with the triangle DCE ; [ VI. 6 ] therefore the angle ABC is equal to the angle DCE .