## ELEMENTS

## by Euclid

Translated by Thomas L. HeathBook Six

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

[Reciprocally related figures. See note.]

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.

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

PROPOSITIONS

PROPOSITION 1.Triangles and parallelograms which are under the same height are to one another as their bases.

Let ABC, ACD be triangles and EC, CF parallelograms under the same height;

(5)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.

[Figure]For let BD be produced in both directions to the points H, L and let [any number of straight lines] BG, GH be

(10)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.

Then, since CB, BG, GH are equal to one another,

the triangles ABC, AGB, AHG are also equal to one

(15)another. [I. 38]

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

For the same reason,

(20)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] [p. 192] if the base HC is in excess of the base CL, the triangle AHC

(25)is also in excess of the triangle ACL, and, if less, less.

Thus, there being four magnitudes, two bases BC, CD and two triangles ABC, ACD, equimultiples have been taken of the base BC and the

(30)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

(35)is also in excess of the triangle ALC; if equal, equal; and, if less, less.

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

Next, since the parallelogram EC is double of the triangle

(40)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

(45)the parallelogram EC to the parallelogram FC.

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,

(50)therefore also, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram FC. [V. 11]

Therefore etc. Q. E. D. 1 2 3

PROPOSITION 2.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.

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.

For let BE, CD be joined.

Therefore the triangle BDE is equal to

[Figure]

the triangle CDE; for they are on the same base DE and in the same parallels DE, BC. [I. 38]And the triangle ADE is another area.

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.

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]

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

Therefore also, as BD is to DA, so is CE to EA. [V. 11] [p. 195]

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.

I say that DE is parallel to BC.

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

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

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

Therefore DE is parallel to BC.

Therefore etc. Q. E. D.

PROPOSITION 3.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 [p. 196] remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle.

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.

For let CE be drawn through C parallel to DA, and let BA

[Figure]

be carried through and meet it at E.Then, since the straight line AC falls upon the parallels AD, EC,

the angle ACE is equal to the angle CAD. [I. 29]But the angle CAD is by hypothesis equal to the angle BAD; therefore the angle BAD is also equal to the angle ACE.

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]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]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.

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

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.

For, with the same construction, since, as BD is to DC, so is BA to AC, [p. 197] 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]

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

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

Therefore etc. Q. E. D.

PROPOSITION 4.In equiangular triangles the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles. [p. 201]

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.

[Figure]For let BC be placed in a straight line with CE.

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]

Let them be produced and meet at F.

Now, since the angle DCE is equal to the angle ABC,

BF is parallel to CD. [I. 28]Again, since the angle ACB is equal to the angle DEC,

AC is parallel to FE. [I. 28]Therefore FACD is a parallelogram; therefore FA is equal to DC, and AC to FD. [I. 34]

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]

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

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] [p. 202]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]Therefore etc. Q. E. D.

PROPOSITION 5.If two triangles have their sides proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.

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.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] [p. 203]

therefore the remaining angle at A is equal to the remaining angle at G. [I. 32]Therefore the triangle ABC is equiangular with the triangle GEF.

[Figure]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.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]Therefore each of the straight lines DE, GE has the same ratio to EF;

therefore DE is equal to GE. [V. 9]For the same reason

DF is also equal to GF.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]Therefore the angle DFE is also equal to the angle GFE,

and the angle EDF to the angle EGF.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. [p. 204]

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.Therefore etc. Q. E. D.

PROPOSITION 6.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.

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.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]

[p. 205]Therefore the triangle ABC is equiangular with the triangle DGF.

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

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]

[Figure]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]Therefore the angle DFG is equal to the angle DFE,

and the angle DGF to the angle DEF.But the angle DFG is equal to the angle ACB; therefore the angle ACB is also equal to the angle DFE.

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.Therefore etc. Q. E. D.

[p. 206]

PROPOSITION 7.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.

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;

[Figure]

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

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]

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]

Therefore the triangle ABG is equiangular with the triangle DEF.

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

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] [p. 207]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]

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

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

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

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]

Therefore the triangle ABC is equiangular with the triangle DEF.

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.

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

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

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

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] [p. 208]

Therefore the triangle ABC is equiangular with the triangle DEF.

Therefore etc. Q. E. D.

PROPOSITION 8.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.

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. [p. 210]

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]

[Figure]

therefore the triangle ABC is equiangular with the triangle ABD.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]

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

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

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.

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

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.

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 [p. 211] 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]

Therefore etc.

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

Let AB be the given straight line; thus it is required to cut off from AB a prescribed part. [p. 212]

Let the third part be that prescribed.

(5)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]

[Figure]

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

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]

(15)But CD is double of DA;

therefore BF is also double of FA; therefore BA is triple of AF.Therefore from the given straight line AB the prescribed third part AF has been cut off. Q. E. F. 4

PROPOSITION 10.To cut a given uncut straight line similarly to a given cut straight line.

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]

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

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] [p. 214]

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

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]

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

PROPOSITION 11.To two given straight lines to find a third proportional.

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.

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]

[Figure]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]

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

Therefore to two given straight lines AB, AC a third proportional to them, CE, has been found. Q. E. F. 5

[p. 215]

PROPOSITION 12.To three given straight lines to find a fourth proportional.

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

[Figure]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]

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]

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

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

PROPOSITION 13.To two given straight lines to find a mean proportional.

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

Let them be placed in a straight line, and let the semicircle ADC be described on AC;

[Figure]

let BD be drawn from the point B at right angles to the straight line AC, and let AD, DC be joined.Since the angle ADC is an angle in a semicircle, it is right. [III. 31]

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.]

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

PROPOSITION 14.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. [p. 217]

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]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.

[Figure]For let the parallelogram FE be completed.

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]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]

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

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

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]

Therefore etc. Q. E. D.

PROPOSITION 15.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.

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.

[p. 220]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]

Let BD be joined.

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]

[Figure]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.]

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

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

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.

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]

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

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

Therefore etc. Q. E. D.

PROPOSITION 16.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.

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.

[Figure]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.

Let the parallelograms BG, DH be completed.

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.

Therefore in the parallelograms BG, DH the sides about the equal angles are reciprocally proportional. [p. 222]

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.

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.

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.

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.And they are equiangular

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

Therefore, as AB is to CD, so is CH to AG.

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

Therefore etc. Q. E. D.

PROPOSITION 17If 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.

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.

[Figure]Let D be made equal to B.

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.

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

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

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.

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. [p. 229]

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.

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

Therefore, as A is to B, so is D to C.

But B is equal to D;

therefore, as A is to B, so is B to C.Therefore etc. Q. E. D.

PROPOSITION 18.On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure.

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.

[Figure]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]

Therefore the remaining angle CFD is equal to the angle AGB; [I. 32]

therefore the triangle FCD is equiangular with the triangle GAB.Therefore, proportionally, as FD is to GB, so is FC to GA, and CD to AB. [p. 230]

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]

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]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.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.

For the same reason

the angle CDE is also equal to the angle ABH.And the angle at C is also equal to the angle at A,

and the angle at E to the angle at H.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]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.Similar triangles are to one another in the duplicate ratio of the corresponding sides.

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

(5)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.

[Figure]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]

(10)and let AG be joined.

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] [p. 233]

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]

(15)Therefore in the triangles ABG, DEF the sides about the equal angles are reciprocally proportional.

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]

(20)therefore the triangle ABG is equal to the triangle DEF.

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

(25)has to EF.

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.

(30)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.

Therefore etc.

PORISM.From this it is manifest that, if three straight

(35)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.

6

PROPOSITION 20.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

(5)side.

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

(10)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.

Let BE, EC, GL, LH be joined.

[Figure]Now, since the polygon ABCDE is similar to the polygon

(15)FGHKL, the angle BAE is equal to the angle GFL;

and, as BA is to AE, so is GF to FL. [VI. Def. 1]Since then ABE, FGL are two triangles having one angle equal to one angle and the sides about the equal angles

(20)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. [p. 236]

(25)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.

And, since, because of the similarity of the triangles ABE,

(30)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]

(35)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

(40)LGH. [VI. 4 and Def. 1]

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

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

(45)multitude.

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

(50)has to the polygon FGHKL a ratio duplicate of that which the corresponding side has to the corresponding side, that is AB to FG.

For let AC, FH be joined.

Then since, because of the similarity of the polygons,

(55)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.

(60)And, since the angle BAM is equal to the angle GFN, and the angle ABM is also equal to the angle FGN, [p. 237] therefore the remaining angle AMB is also equal to the remaining angle FNG; [I. 32] therefore the triangle ABM is equiangular with the triangle

(65)FGN.

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

Therefore, proportionally, as AM is to MB, so is FN to NG,

(70)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.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

(75)bases. [VI. 1]

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.

(80)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.

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

(85)GLH.

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,

(90)so is the triangle BEC to the triangle GLH.

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.

And since, as the triangle ABE is to the triangle FGL,

(95)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. [p. 238]

(100)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]

Therefore the polygon ABCDE also has to the polygon

(105)FGHKL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG.

Therefore etc.

PORISM.Similarly also it can be proved in the case of quadrilaterals that they are in the duplicate ratio of the

(110)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.

7 8

PROPOSITION 21.Figures which are similar to the same rectilineal figure are also similar to one another.

For let each of the rectilineal figures A, B be similar to C; I say that A is also similar to B. [p. 240]

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

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

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.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.

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.

For let there be taken a third proportional O to AB, CD, and a third proportional P to EF, GH. [VI. 11] [p. 241]

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]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]

[Figure]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.

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]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.

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]Therefore MF has the same ratio to each of the figures NH, SR;

therefore NH is equal to SR. [V. 9]But it is also similar and similarly situated to it;

therefore GH is equal to QR.

[p. 242]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.

Therefore etc. Q. E. D.

PROPOSITION 23.Equiangular parallelograms have to one another the ratio compounded of the ratios of their sides.

Let AC, CF be equiangular parallelograms having the angle BCD equal to the angle ECG;

(5)I say that the parallelogram AC has to the parallelogram CF the ratio compounded of the ratios of the sides.

[Figure]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.

(10)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] [p. 248]Then the ratios of K to L and of L to M are the same

(15)as the ratios of the sides, namely of BC to CG and of DC to CE.

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

(20)of the sides.

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]

(25)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]

(30)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

(35)CF.

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.

(40)Therefore etc. Q. E. D. 1, 6, 19, 36 10

PROPOSITION 24.In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another.

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.

[Figure]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]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]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]Therefore in the parallelograms ABCD, EG, the sides about the common angle BAD are proportional.

And, since GF is parallel to DC, [p. 252]

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.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.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.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]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]

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.

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.

Therefore etc. Q. E. D.

PROPOSITION 25.To construct one and the same figure similar to a given rectilineal figure and equal to another given rectilineal figure.

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.

[Figure]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] [p. 254]

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

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]

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.

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

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]

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

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

And KGH is also similar to ABC.

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. 11

PROPOSITION 26.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

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;

[Figure]

I say that ABCD is about the same diameter with AF.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]

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

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]Therefore GA has the same ratio to each of the straight lines AK, AE. [p. 256]

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

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

Therefore etc. Q. E. D.

[p. 257]

PROPOSITION 27.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.

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;

[Figure]

I say that, of all the parallelograms applied to AB and deficient by parallelogrammic figures similar and similarly situated to DB, AD is greatest.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.

For, since the parallelogram DB is similar to the parallelogram FB,

they are about the same diameter. [VI. 26]Let their diameter DB be drawn, and let the figure be described.

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

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

Therefore GC is also equal to EK.

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.Therefore etc.

PROPOSITION 28.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.

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; [p. 261] 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.

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.

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.

[Figure]But, if not, let HE be greater than C.

Now HE is equal to GB;

therefore GB is also greater than C.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]

But D is similar to GB;

therefore KM is also similar to GB. [VI. 21]Let, then, KL correspond to GE, and LM to GF.

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.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.Therefore GQ is also similar to GB; [VI. 21] therefore GQ is about the same diameter with GB. [VI. 26] [p. 262]

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

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.

And, since PR is equal to OS,

let QB be added to each;

therefore the whole PB is equal to the whole OB.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.Let OS be added to each;

therefore the whole TS is equal to the whole, the gnomon VWU.But the gnomon VWU was proved equal to C;

therefore TS is also equal to C.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.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.

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.

[Figure]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]

Let KH correspond to FL and KG to FE.

Now, since GH is greater than FB, therefore KH is also greater than FL, and KG than FE. [p. 266]

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

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

Let EL be subtracted from each;

therefore the remainder, the gnomon XWV, is equal to C.Now, since AE is equal to EB,

AN is also equal to NB [I. 36], that is, to LP [I. 43].Let EO be added to each;

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

therefore AO is also equal to C.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.To cut a given finite straight line in extreme and mean ratio.

Let AB be the given finite straight line;

thus it is required to cut AB in extreme and mean ratio.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]

Now BC is a square;

therefore AD is also a square.

[Figure]And, since BC is equal to CD, let CE be subtracted from each;

therefore the remainder BF is equal to the remainder AD.

[p. 268]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.But FE is equal to AB, and ED to AE.

Therefore, as BA is to AE, so is AE to EB.

And AB is greater than AE;

therefore AE is also greater than EB.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.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.

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.

Let AD be drawn perpendicular.

Then since, in the right-angled triangle ABC, AD has [p. 269] 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]

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

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.]

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

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.

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.

Therefore etc. Q. E. D.

PROPOSITION 32.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.

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. [p. 271]

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]

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.

[Figure]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.But the angle ACD was also proved equal to the angle BAC;

therefore the whole angle ACE is equal to the two angles ABC, BAC.Let the angle ACB be added to each; therefore the angles ACE, ACB are equal to the angles BAC, ACB, CBA.

But the angles BAC, ABC, ACB are equal to two right angles; [I. 32]

therefore the angles ACE, ACB are also equal to two right angles.Therefore with a straight line AC, and at the point C on it, the two straight lines BC, CE not lying on the same side make the adjacent angles ACE, ACB equal to two right angles;

therefore BC is in a straight line with CE. [I. 14]Therefore etc. Q. E. D.

PROPOSITION 33.In equal circles angles have the same ratio as the circumferences on which they stand, whether they stand at the centres or at the circumferences.

Let ABC, DEF be equal circles, and let the angles BGC, EHF be angles at their centres G, H, and the angles BAC, EDF angles at the circumferences; I say that, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

[Figure]For let any number of consecutive circumferences CK, KL be made equal to the circumference BC, and any number of consecutive circumferences FM, MN equal to the circumference EF; and let GK, GL, HM, HN be joined.

Then, since the circumferences BC, CK, KL are equal to one another, the angles BGC, CGK, KGL are also equal to one another; [III. 27] [p. 274] therefore, whatever multiple the circumference BL is of BC, that multiple also is the angle BGL of the angle BGC.

For the same reason also, whatever multiple the circumference NE is of EF, that multiple also is the angle NHE of the angle EHF.

If then the circumference BL is equal to the circumference EN, the angle BGL is also equal to the angle EHN; [III. 27] if the circumference BL is greater than the circumference EN, the angle BGL is also greater than the angle EHN; and, if less, less.

There being then four magnitudes, two circumferences BC, EF, and two angles BGC, EHF, there have been taken, of the circumference BC and the angle BGC equimultiples, namely the circumference BL and the angle BGL, and of the circumference EF and the angle EHF equimultiples, namely the circumference EN and the angle EHN.

And it has been proved that, if the circumference BL is in excess of the circumference EN, the angle BGL is also in excess of the angle EHN; if equal, equal; and if less, less.

Therefore, as the circumference BC is to EF, so is the angle BGC to the angle EHF. [V. Def. 5]

But, as the angle BGC is to the angle EHF, so is the angle BAC to the angle EDF; for they are doubles respectively.

Therefore also, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

Therefore etc. Q. E. D.

NOTES1 Under the same height. The Greek text has “under the same height AC,” with a figure in which the side AC common to the two triangles is perpendicular to the base and is therefore itself the “height.” But, even if the two triangles are placed contiguously so as to have a common side AC, it is quite gratuitous to require it to be perpendicular to the base. Theon, on this occasion making an improvement, altered to “which are (honta) under the same height, (namely) the perpendicular drawn from A to BD.” I have ventured to alter so far as to omit “AC” and to draw the figure in the usual way.

2 ABC, AGB, AHG. Euclid, indifferent to exact order, writes “AHG, AGB, ABC.”

3 Since then it was proved that, as the base BC is to CD, so is the triangle ABC to the triangle ACD. Here again words have to be supplied in translating the extremely terse Greek epei oun edeichthw, 'wV men 'w basiV BG proV twn GD, outwV to ABG trigwnon proV to AGD trigwnon, literally “since was proved, as the 'base BC to CD, so the triangle ABC to the triangle ACD.” Cf. note on V. 16, p. 165.

4 any angle. The expression here and in the two following

PROPOSITIONs is tuchousa gônia, corresponding exactly to tuchon sêmeion which I have translated as “a point (taken) at random”; but “an angle (taken) at random” would not be so appropriate where it is a question, not of taking any angle at all, but of drawing a straight line casually so as to make any angle with another straight line.

5 to find. The Greek word, here and in the next two

PROPOSITIONs, is proseurein, literally “to find in addition.”

6 and such that, as AB is to BC, so is DE to EF, literally “(triangles) having the angle at B equal to the angle at E, and (having), as AB to BC, so DE to EF.”

7 in the same ratio as the wholes. The same word homologos is used which I have generally translated by “corresponding.” But here it is followed by a dative, homologa tois holois “homologous with the wholes,” instead of being used absolutely. The meaning can therefore here be nothing else but “in the same ratio with” or “proportional to the wholes”; and Euclid seems to recognise that he is making a special use of the word, because he explains it lower down (I. 46): “the triangles are homologous to 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.”

8 hepomena autôn, “their consequents,” is a little awkward, but may be supposed to indicate which triangles correspond to which as consequent to antecedent.

1, 6, 19, 36 the ratio compounded of the ratios of the sides, logon ton sunkeimenon ek tôn pleurôn which, meaning literally “the ratio compounded of the sides,” is negligently written here and commonly for logon ton sunkeimenon ek tôn pleurôn (sc. logôn).

10 let it be contrived that, as BC is to CG, so is K to L. The Greek phrase is of the usual terse kind, untranslatable literally : kai gegonetw 'wsmen 'w BG pro twn GÊ, 'outwV 'w K proV to L, the words meaning “and let (there) be made, as BC to CG, so K to L,” where L is the straight line which has to be' constructed.

11 to which the figure to be constructed must be similar, literally “to which it is required to construct (one) similar,” 'wi dei 'omoion sustwsasqai.

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