A

CE

A

Because in the two triangles ABG and DEF, the angle at A is equal to the angle at D (Hyp.) and the angle ABG is equal to the angle DEF (Const.). Therefore the remaining angle AGB is equal to the remaining angle DFE (I. 32), and the triangle ABG is equiangular to DEF. Because these triangles are equiangular AB is to BG, as DE is to EF (VI. 4). But A B is to BC as DE is to EF (Hyp.). Therefore B AB is to BC, as AB is to BG (V. 11); and BG is equal to BC (V. 9). Wherefore the triangle GBC is isosceles, and the angle BGC is equal to the angle BCG. Therefore the angles BGC and BGA are both acute, or both not acute, according as the angles ACB and DFE, are both acute or both not acute. If they are both acute, as in the first figure, the angles BGC and BGA which BG makes with AC are together less than two right angles, which is impossible (I. 13). If they are both not acute, as in the second figure, the angles BGC and BCG of the triangle BGC, are together not less than two right angles, which is also impossible (I. 17). Therefore the angle ABC is not unequal to the angle DEF; that is, the angle ABC is equal to the angle DEF. Wherefore the remaining angle ACB is equal to the remaining angle DFE; and the triangle ABC is equiangular and similar to the triangle DEF. Q. E. D.

B

G

CE

F

In this proposition, we have considerably departed from Euclid, in order to make the demonstration more plain and easy to the learner. Besides, Euclid has three separate cases to demonstrate, which are here reduced to two, and the same proof is made, with a very slight exception, to apply to both. Corollary.-If two triangles have two sides in the one proportional to two sides in the other, and the angles opposite one pair of the homologous sides equal, the angles opposite the other pair of homologous sides are either equal or supplementary.

PROP. VIII. THEOREM.

In a right-angled triangle, if a perpendicular be drawn from the right angle to the opposite side, the triangles on each side of the perpendicular are similar to the whole triangle, and to one another.

Let A B C be a right-angled triangle, having the right angle BAC, and from the point A let AD be drawn perpendicular to the opposite side BC. The triangles ABD and ADC are similar to the whole triangle A B C, and to one another.

B

A

D C

Because the angle BAC is equal to the angle ADB, each of them being a right angle (I. Ax. 11), and the angle at B is common to the two triangles ABC and ABD. Therefore the remaining angle A CB is equal (I. 32) to the remaining angle BAD; and the triangle ABC is equiangular to the triangle ABD. But the sides about their equal angles are proportionals (VI. 4). Therefore the triangles are similar (VI. Def. 1). In like manner it may be shown that the triangle ADC is equiangular and similar to the triangle ABC. Because the triangles ABD and A CD, are both

Also, let the ratio of A to B, that is, the ratio of S to T, which is one of the first ratios, be equal to the ratio of e to g, which is compounded of the ratios of e to f, and f to g, which (Hyp.) are equal to the ratios of G to H, and K to L, two of the other ratios; and let the ratio of h to l be that which is compounded of the ratios of h to k, and k to l, which are equal to the remaining first ratios, viz., of C to D, and E to F.

Also, let the ratio of m top, be that which is compounded of the ratios of m to n, n to o, and o to p, which are equal, each to each, to the remaining other ratios, viz., of M to N, O to P, and Q to R.

The ratio of h to l is equal to the ratio of m to p; that is, h is to l, as m is to p.

h, k, l.

A. B; G, H; K, L; e, f, g.

C, D; E, F,

S, T, V, X.

M, N;. O, P, Q, R. Y, Z; a, b, c, d.
m, n, o, p.

Because e is to f, as G to H, that is, as Y to Z; and ƒ is to g, as K to L, that is, as Z to a. Therefore, ex æquali, e is to g, as Y to a (V. 22). But A is to B, that is, S is to T, as e is to g (Hyp.). Therefore S is to T, as Y is to a (V. 11). Wherefore, by inversion, T is to S, as a is to Y (V. B). But S is to X, as Y is to D (Hyp.). Therefore, ex equali, T is to X as a is to d. Because h is to k, as C is to D, that is, as Tis to V (Hyp.); and k is to las E is to F, that is, as V is to X. Therefore, ex æquali, h is to l, as T is to X. In like manner, it may be shown that m is to p, as a is to d. But it has been shown that T is to X, as a is to d. Therefore h is to l, as m is_to_p (V. 11). Wherefore, if there be any number of ratios, &c. Q. E. D.

"The propositions G and K are usually, for the sake of brevity, expressed in the same terms with propositions F and H: and therefore it was proper to show the true meaning of them when they are so expressed; especially since they are very frequently made use of by geometers."

After having laboured to correct this book of Euclid, Dr. Simson says, "I most readily agree with what the learned Dr. Barrow says, that there is nothing in the whole body of the Elements of a more subtile invention, nothing more solidly established, and more accurately handled, than the doctrine of proportionals.' And there is some ground to hope, that geometers will think that this could not have been said with as good reason, since Theon's time (A.D. 380), till the present." The modesty of this remark is only surpassed by its truth. Most editors, since Dr. Simson's time, have only rendered the doctrine of proportion more obscure; as a remarkable example, see Professor De Morgan's "Connection of Number and Magnitude."

BOOK VI.

DEFINITIONS.

I.

SIMILAR rectilineal figures are those which have their several angle equal, each to each, and the sides about the equal angles proportionals.

да

In the case of triangles, this definition is redundant. For it is proved in Prop. IV. of this Book, that the sides about the equal angles of equiangular triangles are proportionals. In the case of quadrilaterals, or polygons, however, the definition is necessary. According to this definition, all equilateral triangles, squares, and regular polygons are similar rectilineal figures.

II.

Triangles and parallelograms are said to have their sides reciprocally proportional, when the sides about two of their angles are proportionals in such a manner, that a side of the first figure is to a side of the second, as the remaining side of the second is to the remaining side of the first.

Two magnitudes of any kind may be said to be reciprocally proportional to other two of the same kind, when one of the first pair is to one of the second pair, as the remaining one of the second pair is to the remaining one of the first.

III.

A straight 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.

A straight line is said to be cut in harmonical ratio, when the whole is to one of the extreme segments as the other extreme segment is to the middle segment. For brevity's sake, this mode of dividing a straight line is called harmonical section; and the mode of dividing a straight line explained in Euclid's definition is called medial section.

IV..

The altitude of any figure is the straight line drawn from its vertex perpendicular to the base.

Altitude is a term synonymous with perpendicular. By the vertex of a figure here, is meant that angular point of the figure which is most remote from any side of the figure assumed at the base, the degree of remoteness being measured by a perpendicular drawn to that side or that side produced, from the said vertex; in other words, the longest perpendicular drawn from any angular point to the base, or base produced, is the altitude.

PROP. I. THEOREM.

Triangles and parallelograms of the same altitude are to one another as their bases.

Let the triangles ABC and A CD, and the parallelograms EC and CF, have the same altitude, viz., the perpendicular drawn from the

Take of AB, BE, CD, and DF any equimultiples whatever GH, HK, LM, and MN; and of BE, DF, any other equimultiples whatever K O and NP.

Because KO and NP are equimultiples of BE and DF; and KH and NM are likewise equimultiples of BE and DF. If KO, the multiple of BE, be greater than KH, which is also a multiple of B E. Therefore, NP, the multiple of DF, is also greater than NM, the multiple of the same D F; if K O be equal to K H, NP is equal to Ñ M; and if less, less.

First, if KO be not greater than KH; NP is not greater than NM. Because, GH and HK are equimultiples of AB and BE, and AB is greater than BE. Therefore G H is greater than HK (Ax. 3); but KO is not greater than KH (Hyp.). Therefore G H is greater than K O. In like manner it may be shown, that LM is greater than NP. Therefore if K O be not greater than KH, GH, the multiple of A B, is greater than K O, the multiple of B E; and likewise L M, the multiple of CD, is greater than NP, the multiple of DF. Next, let K O be greater than K. H. Therefore, as has been shown, NP

is greater than NM. Because the whole GH is the same multiple of the whole AB, that HK is of BE. Therefore the remainder GK is the same multiple of the remainder AE that GH is of AB (V. 5), which is the same that L M is of CD. In like manner, because LM is the same multiple of CD, that MN is of DF. Therefore the remainder LN is the same multiple of the remainder CF, that the whole L M is of the whole CD (V. 5). But it was shown that L M is the same multiple of CD, that G Kis of A E. Therefore GK is the same multiple of AE, that LN is of CF; that is, GK and LN are equimultiples of AE and CF. But KO and NP are equimultiples of BE and DF; and if from K O and NP there be taken HK and MN, which are likewise equimultiples of BE and DF. Therefore, the remainders HO and MP are either equal to BE and DF, or equimultiples of them (V. 6).

First, let H O and M P be equal to BE and DF. Because AE is to EB, as CF to FD (Hyp.), and GK and L N are equimultiples of AE and CF. Therefore GK is to EB, as LN to FD (V. 4, Cor.). But HO is equal to EB, and M P to FD. Therefore G K is to HO, as LN to MP. Wherefore if GK be greater than HO, LN is greater than MP; if equal, equal; and if less, less (V. A).

Next, let HO and MP be equimultiples of E B and FD. Because

AE is to EB, as CF to FD (Hyp.), and of AE and CF are taken equimultiples GK and LN; and of EB and FD, the equimultiples

HO and MP. If GK be greater than HO, LN is greater than MP; if equal, equal; and if less, less (V. Def. 5); which was likewise shown in the preceding case. But if GH be greater than K O, taking KH from both, GK is greater than HO (Ï. Ax. 5). Therefore also LN is greater than MP. By adding NM to these unequals, LM is greater than NP (I. Ax. 4). Therefore, if G H be greater than K O, LM is greater than NP. In like manner it may be shown, that if GH be equal to KO, LM is equal to NP; and if less, less. But in the case in which K O is not greater than K H, it has been shown that GH is always greater than K O, and likewise LM greater than NP. And GH and LM are any equimultiples whatever of AB and CD (Const.), also KO and NP are any whatever of BE and DF. Therefore, as AB is to BE, so is CD to DF (V. Def. 5). If four magnitudes, &c. Q. E. D.

The term composition used in the enunciation of this proposition, signifies simply the addition of the two magnitudes, or the finding of one magnitude equal to both.

PROP. XIX. THEOREM.

If there be two magnitudes such that the first is to the second, as a part of the first is to a part of the second; the remainder is to the remainder as the first is to the second.

There- A

E

C F D

B

Let AB be to CD, as AE a part of AB is to CF a part of CD. The remainder EB is to the remainder FD, as AB is to CD. Because AB is to CD, as AE to CF. fore alternately, BA is to A E, as DC to CF (V. 16). Because EB is the excess of AB above A E, and DF the excess of CD above CF. Therefore, as BE is to EA, so is DF to FC (V. 17). But alternately, as BE is to D F, so is EA to FC (V. 16), and as A E to CF, so is AB to to CD (Hyp.). Therefore BE is to DF, as AB is to CD (V. 11). Wherefore, if there be two magnitudes such, &c. Q. E. D.

COROLLARY.—If there be two magnitudes such that the first is to the second as a part of the first is to a part of the second; the remainder is to the remainder, as the part of the first is to the part of the second. The demonstration is contained in the preceding.

PROP. E. THEOREM.

If four magnitudes be proportionals, they are also proportionals by conversion ; that is, the first is to its excess above the second, as the third to its excess above the fourth.

E B

Let A B be to BE, as CD to DF. And let AE be the excess of A B above BE, and CF the excess of CD above DF. AB is to AE, as DC to CF. Because AB is to BE, as CD to DF (Hyp.). Therefore, by division, A E is to EB, as CF to A FD (V. 17). But, by inversion, BE is to EA, as DF to FC (V. B). Wherefore, by composition, BA is to c to CF (V. 18). If therefore four, &c.

A E, as DC is

Q. E. D.

F D

This proposition was added by Dr. Simson, as a substitute for a corollary to the next proposition given in the original Greek, which he declares to be vitiated.