Euclid's Elements of geometry, transl. To which are added, algebraic demonstrations to the second and fifth books; also deductions in the first six, eleventh and twelfth books, with notes, by G. Phillips. Part 1, containing book i-vi

EUCLID'S ELEMENTS.

BOOK VI.

DEFINITIONS.

1. Similar rectilineal figures are those which have their angles equal each to each, and the sides about the equal angles proportional.

2. Reciprocal figures are such as have their sides about two of their angles proportional in such a manner that a side of the first figure is to a side of the other, as the remaining side of this other is to the remaining side of the first.

3. A right line is said to be cut in extreme and mean ratio when the whole is to the greater segment as the greater segment is to the less.

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

PROPOSITION I.

THEOREM.

Triangles, and parallelograms, which have the same altitude, are to one another as their bases.

Let there be the triangles ABC, ACD, also the parallelograms EC, CF, which have the same altitude, viz. the perpendicular drawn from the point A to BD. Then 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.

Produce BD both ways to the points H, L, and take BG, GH, any number of times equal to the base BC; also DK, KL, any number of times equal to the base CD; and join AG, AH, AK, AL. Therefore because CB, BG, GH, are equal to one another, the triangles AHG, AGB, ABC, will be equal to one another. Therefore the base нC PART I.

HG B C D K. L

a 38. 1.

is the same multiple of the base BC, as the triangle AHC is of the triangle ABC. For the same reason the base LC is the same multiple of the base CD as the triangle ALC is of the triangle ACD; and if the base HC be equal to the base CL, the triangle AHC is equal to the triangle ALC: if the base HC be greater than the base CL, the triangle AHC is also greater than the triangle ALC; and if less, less. Therefore there are four magnitudes; viz. the bases BC, CD, and the two triangles ABC, ACD, such that if equimultiples of the base BC and the triangle ABC be taken, viz. the base нc and the triangle AHC; also of the base CD and the triangle ACD any other equimultiples, viz. the base CD and the triangle ALC. And it has been shown that if the base HC be greater than the base CL, the triangle AHC will be greater than the triangle ALC, if equal, equal, and if less, less; therefore as the base BC is to the base CD so b5 Def. 5. is the triangle ABC to the triangle ACD.b

41. 1.

d 15. 5.

e 11. 5.

And because the parallelogram EC is double of the triangle ABC also the parallelogram Fc is double of the triangle ACD; and magnitudes have the same ratio which their equimultiples have ;d therefore as the triangle ABC is to the triangle ACD So is the parallelogram Ec to the parallelogram Fc. Hence because it has been shown, that as the base BC is to the base CD so is the triangle ABC to the triangle ACD; but as the triangle ABC is to the triangle ACD so is the parallelogram BC to the parallelogram FC; and, therefore, as the base BC is to the base CD so is the parallelogram EC to the parallelogram Fc. Therefore, triangles, &c. Q. E. d.

Deductions.

1. Triangles and parallelograms having equal bases, are to each other as their altitudes.

2. If four right lines be proportional, their squares shall also be proportional.

PROPOSITION II.

THEOREM.

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

For draw DE parallel to one of the sides of the triangle ABC, viz. to BC; then as BD is to DA so is CE

to EA.

Join BE, CD.

a a

b 7.5.

e 1.6.

Then the triangle BDE is equal to the triangle CDE‚a a 37. 1. for they are upon the same base DE and between the same parallels DE, BC. But ADE is any other triangle; also equal magnitudes have the same ratio to the same magnitude; therefore as the triangle BDE is to the triangle ADE SO is the triangle CDE to the triangle E ADE. But as the triangle BDE is to the triangle ADE so is BD to DA. For they having the same altitude, viz. the perpendicular drawn from the point E to AB are to one another as their bases. For the same reason as the triangle CDE is to the triangle ADE SO is CE to EA; and hence as BD is to DA SO is CE to EA. But also if the sides 11. 5. AB, AC, of the triangle ABC be cut proportionally in the points D, E, as BD is to DA SO is CE to EA, and join DE; then is DE parallel to BC.

For the same construction being made, because it is 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, but as CE is to 1. 6. EA So is the triangle CDE to the triangle ADE; and therefore as the triangle BDE is to the triangle ADE SO is the triangle CDE to the triangle ADE. Hence each of the triangles BDE, CDE, has the same ratio to the triangle ADE. Therefore the triangle BDE is equal to the triangle CDE, and they are upon the same base DE. f 9.5. But equal triangles constituted upon the same base are also between the same parallels. Therefore DE is 39. 1. parallel to Bc. If, therefore, triangles, &c. Q. E. D.*

From this and the 18th proposition of the fifth book, it may be demonstrated that the sides of the triangles AED, ACB, containing the angle CAB are proportional, viz. as AR is to AD so is AC to AB.

a 31. 1.

b 29. 1.

c 6. 1.

d 2.6.

• 7.5.

PROPOSITION III.

THEOREM.

If the angle of a triangle be bisected, and the right line cutting the angle cuts the base also; the segments of the base shall have the same ratio which the remaining sides of the triangle have; and if the segments of the base have the same ratio which the remaining sides of the triangle have, the right line drawn from the vertex to the point of section bisects the vertical angle of the triangle.

Let ABC be a triangle and bisect the angle BAC by the right line AD; then as BD is to DC so is BA to ac. For through c draw CE parallela to DA and BA produced will meet with it in E.

And because the right line AC falls upon the parallel right lines AD, EC, therefore the angle ACE is equal to the angle CAD. But CAD is supposed equal to BAD, therefore BAD is also equal to ACE. Again, because the right line BAE falls upon the parallels AD, EC, the exterior angle BAD is equal to the interior angle AEC. But it has been shown that ACE is equal to BAD, and hence the angle B ACE is equal to AEC; wherefore, also, the side AE is equal to the side Ac. And because AD is drawn parallel to one of the sides EC of the triangle BCE, therefore, proportionally, as BD is to DC so is BA to AE. And AE is equal to AC; therefore as BD is to DC so is BA to Ac.e

But let BD be to DC as BA is to AC, and join AD; then is the angle BAC bisected by the right line AD.

For the same construction being made, because as BD is to DC so is BA to AC, but also as BD is to DC so is BA to AE; for AD is drawn parallel to one of the sides Ec of the triangle BCE; therefore, also, as BA is to AC so is BA to AE; hence AC is equal to AE; wherefore the angle AEC is equal to the angle ACE. But the angle AEC is equal to the exterior angle BAD, also the angle ACE to the alternate angle CAD, and therefore BAD is equal to CAD. Hence the angle BAC is bisected by the right line AD. If, therefore, the angle of a triangle, &c. Q. E. D.

PROPOSITION IV.

THEOREM.

The sides of the equiangular triangles about the equal angles are proportional, and the homologous sides subtend the equal angles.

Let ABC, DCE, be equiangular triangles, having the angle BAC equal to CDE, also ACB to DEC, and consequently ABC to DCE; then the sides of the triangles * 32. 1. ABC, DCE, about the equal angles, are proportional, and the homologous sides subtend the equal angles. For put BC in a direct line with CE, and because the angles ABC, ACB, are less than two right angles,b but ACB is equal to DEC, therefore ABC, DEC, are less than two right angles; hence BA, ED, produced, will meet. Let them be produced, and let them meet in F.

b 17. 1.

€ 12 Ax. 1.

34. 1.

2. 6.

And because the angle DCE is equal to ABC, therefore BF is parallel to CD. Again, because the angle 28. 1. ACB is equal to DEC, AC is parallel to FE; hence FACD is a parallelogram; therefore FA is equal to DC, also AC to FD. And because AC is drawn parallel to one of the sides, FE, of the triangle FBE, therefore, as BA is to AF so is BC to CE. But AF is equal to CD; hence as BA is to CD so is BC to CE, and alternately, as AB is to BC so is DC to CE. Again, because CD is parallel to BF, therefore as BC is to CE so is FD to DE. But FD is equal to AC; hence as BC is to CE so is AC to ED, and alternately, as BC is to CA so is CE to ED. And because it has been shown as AB is to BC so is DC to CE; also as BC is to CA so is CE to ED; therefore, by equality, as BA is to AC so is CD to DE. Therefore the sides of equiangular triangles about, &c. q. E. D.*

Deductions.

1. In isosceles triangles, which are equiangular, the perpendiculars drawn from the vertices to the bases are proportional to the sides of the triangle.

Hence if in a triangle FBE, there be drawn AC, parallel to the side FE, the triangle ABC shall be similar to the whole FBE.

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Euclid's Elements of geometry, transl. To which are added, algebraic ...