The Elements of Euclid; viz. the first six books, together with the eleventh and twelfth. Also the book of Euclid's Data. By R. Simson. To which is added, A treatise on the construction of the trigonometrical canon [by J. Christison] and A concise account of logarithms [by A. Robertson].

11. 6.

* 11.5.

the similar rectilineal figures KAB, LCD be similarly described; and upon EF, GH, the similar rectilineal figures MF, NH, in like manner: the rectilineal figure KAB shall be to LCD, as

MF to NH.

E FG H

To AB, CD, take a third proportional * X; and to EF, GH, a third proportional O: and because AB is to CD as EF to GH, therefore CD is to X as GH to O; wherefore, ex æquali *, as AB to X, so EF to 0: but as AB to X, so is the rectilineal figure KAB to the rectilineal *2 Cor. 20. figure LCD, and as EF to O, so is the rectilineal figure MF to the rectilineal figure NH: therefore, as KAB to LCD, so is MF to NH.

• 22. 5.

11. 5.

* 12. 6. . 18. 6.

* 9.5.

† 7. 5.

See N.

And if the rectilineal figure KAB be to LCD as MF to NH, the straight line AB shall be to CD as EF to GH.

Make as AB to CD, so EF to PR; and upon PR describe the rectilineal figure SR, similar and similarly situated to either of the figures MF, NH: then, because as AB to CD, so is EF to PR, and that upon AB, CD, are described the similar and similarly situated rectilineals KAB, LCD, and upon EF, PR, in like manner, the similar rectilineals MF, SR, therefore KAB is to LCD, as MF to SR: but by the hypothesis KAB is to LCD, as MF to NH; and therefore, the rectilineal MF having the same ratio to each of the two NH, SR, these are equal to one another: they are also similar, and similarly situated; therefore GH is equal to PR: and because as AB to CD, so is EF to PR, and that PR is equal to GH, AB is to CD +, as EF to GH. If, therefore, four straight lines, &c. Q. E. D.

PROPOSITION XXIII.

THEOR.-Equiangular parallelograms have to one another, the ratio which is compounded of the ratios of their sides.

Let AC, CF be equiangular parallelograms, having the angle BCD equal to the angle ECG: the ratio of the parallelogram AC to the parallelogram CF, shall be the same with the ratio which is compounded of the ratios of their sides.

By the preceding case of this Proposition.

A D H

B C

KLM

See note at foot of page 10 145.

14. 1.

⚫ 12. 6. • 12. 6.

• Def. A. 5.

1.6.

Let BC, CG be placed in a straight line; therefore DC and CE are also in a straight line; and complete the parallelogram DG; and taking any straight line K, make as BC to CG, so K to L; and as DC to CE, so make* L to M: therefore the ratios of K to L, and L to M, are the same with the ratios of the sides, viz. of BC to CG, and DC to CE: but the ratio of K to M is that which is said to be compounded * of the ratios of K to L, and L to M; therefore K has to M the ratio compounded of the ratios of the sides: and because as BC to CG, so is the parallelogram AC to the parallelogram CH*; but as BC to CG, so is K to L; therefore K is to L, as the parallelogram AC to the parallelogram CH: again, because as DC to CE, so is the parallelogram CH to the parallelogram CF; but as DC to CE, so is L to M ; wherefore L is * to M, as the parallelogram CH to the parallelogram CF: there fore since it has been proved, that as K to L, so is the parallelogram AC to the parallelogram CH; and as L to M, so the parallelogram CH to the parallelogram CF; ex æquali *, K is * 22. 5. to M, as the parallelogram AC to the parallelogram CF: but K has to M the ratio which is compounded of the ratios of the sides; therefore also, the parallelogram AC has to the parallelogram CF the ratio which is compounded of the ratios of the sides. Wherefore, equiangular parallelograms, &c. q. E. D.

PROPOSITION XXIV.

11. 5.

THEOR.-Parallelograms about the diameter of any parallelo- See N. gram, are similar to the whole, and to one another.

Let ABCD be a parallelogram, of which the diameter is AC; and EG, HK parallelograms about the diameter: the parallelograms EG, HK shall be similar both to the whole parallelogram ABCD, and to one another.

A E

of the angles BCD, EFG is equal to the opposite angle DAB, 34. 1. and therefore they are equal to one another; wherefore the

* 4. 6.

34. 1. * 7.5.

parallelogram ABCD, AEFG are equiangular: and because the angle ABC is equal to the angle AEF, and the angle BAC common to the two triangles BAC, EAF, they are equiangular to one another; therefore as AB to BC, so is AE to EF: and because the opposite sides of parallelograms are equal to one another*, AB* is to AD, as AE to AG; and DC to CB, as GF to FE; and also CD to DA, as FG to GA: therefore the sides of the parallelograms ABCD, AEFG about the equal * 1 Def. 6. angles are proportionals; and they are therefore similar to one another for the same reason, the parallelogram ABCD is similar to the parallelogram FHCK: wherefore each of the parallelograms GE, KH is similar to DB: but rectilineal figures which are similar to the same rectilineal figure, are also similar to one another; therefore the parallelogram GE is similar to KH. Wherefore, parallelograms, &c. Q. E. D.

* 21.6.

See N.

*Cor. 45.1.

PROPOSITION XXV.

PROB.-To describe a rectilineal figure which shall be similar to one, and equal to another, given rectilineal figure.

Let ABC be the given rectilineal figure to which the figure to be described is required to be similar, and D that to which it must be equal; it is required to describe a rectilineal figure similar to ABC, and equal to D.

Upon the straight line BC, describe the parallelogram BE *Cor. 45.1. equal to the figure ABC; also upon CE, describe the parallelogram CM equal to D, and having the angle FCE equal to * 29. 1. and the angle CBL: therefore ‡ BC and CF are in a straight line *, as also LE and EM: between BC and CF find * a mean proportional GH; and upon GH describe the rectilineal figure KGH, similar and similarly situated to the figure ABC.

14. 1. • 13.6.

• 18.6.

+ Constr. † 2 Ax.

* 29. 1. +1 Ax.

14. 1. * 34.1. † 1 Ax.

* 29.1.

+1 Ax.

+14. 1.

Because the angle FCE is equal to the angle CBL, add to each the angle ECB; therefore the angles FCE, ECB are equal † to the angles ECB, CBL but the angles ECB, CBL are equal to two right angles; therefore the angles FCE, ECB are equal † to two right angles: therefore BC and CF are in the same straight line. Again, because the angle LBC is equal to the angle FCE, and that the angle LBC is equal to the opposite angle LEC, therefore the angle LEC is equal to the angle ECF: add to each the angle CEM; therefore the angles LEC, CEM are equal to the angles FCE, CEM: but the angles FCE, CEM are equal * to two right angles; therefore the angles LEC, CEM are equal to two right angles: therefore LE and EM are in the same straight line.

Because BC is to GH as GH to CF, and that if three straight

D K

lines be proportionals, as the first is to the third, so is the 2 Cor. 20. figure upon the first to the similar and similarly described figure upon the second, therefore as BC to CF, so is the rectilineal figure ABC to KGH but as BC to CF, so is * the parallelogram BE to the parallelogram EF; therefore * as the rectili

1. 6.

. 11. 5.

14. 5.

neal figure ABC is to KGH, so is the parallelogram BE to the parallelogram EF: and the rectilineal figure ABC is equal to † Constr. the parallelogram BE; therefore the rectilineal figure KGH is equal to the parallelogram EF: but EF is equal to the figure D; wherefore also, KGH is equal to D; and it is simi- † Constr. lar to ABC. Therefore, the rectilineal figure KGH has been described similar to the figure ABC, and equal to D. was to be done.

Which

PROPOSITION XXVI.

THEOR.-If two similar parallelograms have a common angle, and be similarly situated, they are about the same diameter.

Let the parallelograms ABCD, AEFG be similar and similarly situated, and have the angle DAB common: ABCD and AEFG shall be about the same diameter.

A G

24.6.

⚫ 1 Def. 6.

For, if not, let, if possible, the parallelogram BD have its diameter AHC in a different straight line from AF, the diameter of the parallelogram EG, and let GF meet AHC in H; and through H, draw HK parallel to AD or BC: therefore the parallelograms ABCD, AKHG, being about the same diameter, are similar to one another: wherefore as DA to AB, so is GA to AK: but because ABCD and AEFG are similar † parallelograms, as DA is to AB, † Hyp. so is GA to AE; therefore as GA to AE, so GA to AK; that is GA has the same ratio to each of the straight lines AE, AK; and consequently AK is equal to AE, the less to the greater; which is impossible: therefore ABCD and AKHG are not about the same diameter: wherefore, ABCD and AEFG must be about the same diameter. Therefore, if two similar, &c.

Q. E. D.

• 11.5.

• 9.5.

See N.

Hyp.

To understand the three following propositions more easily, it is to be observed,

1. ‹ That a parallelogram is said to be applied to a straight line, when it is described upon it as one of its sides. Ex.

gr. the parallelogram AC is said to be applied to the straight line AB.

EC G

2. But a parallelogram AE is said to be applied to a straight line AB, deficient by a parallelogram, when AD the 'base of AE is less than AB, and therefore 'AE is less than the parallelogram AC de'scribed upon AB in the same angle, and between the same parallels, by the paral- A lelogram DC; and DC is therefore called 'the defect of AE.

D B F

3. And a parallelogram AG is said to be applied to a 'straight line AB, exceeding by a parallelogram, when AF 'the base of AG is greater than AB, and therefore AG exceeds AC the parallelogram described upon AB in the same angle, and between the same parallels, by the parallelogram BG.'

PROPOSITION XXVII.

THEOR.-Of all parallelograms applied to the same straight line, and deficient by parallelograms similar and similarly situated to that which is described upon the half of the line, that which is applied to the half, and is similar to its defect, is the greatest.

Let AB be a straight line divided into two equal parts in C; and let the parallelogram AD be applied to the half AC, which is therefore deficient from the parallelogram upon the whole line AB, by the parallelogram CE upon the other half CB: of all the parallelograms applied to any other parts of AB, and deficient by parallelograms that are similar and similarly situated to CE, AD shall be the greatest.

Let AF be any parallelogram applied to AK, any other part of AB than the half, so as to be deficient from the parallelogram upon the whole line AB, by the parallelogram KH similar and similarly situated to CE: AD shall be greater

than AF.

First, let AK the base of AF, be greater than AC the half of AB and because CE is similar to the parallelogram KH,

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The Elements of Euclid; viz. the first six books, together with the eleventh ...