PROP. XXXIII, Plate III. Fig. 33.

If the proportion of the whole AB to the whole CD be greater than the proportion of the part taken away AE to the part taken away CF; then shall also the ratio of the remainder EB to the remainder FD be greater than that of the whole AB to the whole CD.

d 13.5

If there be any number of magnitudes, and others also equal to them in number; and the proportion of the first of the former to the first of the latter be greater than that of the second to the second, and that greater than the proportion of the third to the third, and fo forward: all the former magnitudes together shall have a greater ratio to all the latter together, than all the former, leaving out the first, ball have to the latter, leaving out the first; but less than that of the first of the former to the first of the latter; and lastly, greater than that of the last of the former to the last of the latter.

You may please to confult Interpreters for the demonstration hereof, we having for brevity fake omitted it, and because 'tis of no use in these Elements.

The End of the fifth Book.

The

2

Plate III. I.
Fig. 33.

Fig. 34.

Fig. 23.

Fig. 33.

a 20. def. 5.

b 15.5.

ELEMENTS.

L

Definitions.

Ike right-lined figures (ABC, DCE) are such whose several angles are equal one to the other, and also their fides about the equal angles, proportional.

The Angle BDCE, and AB: BC:: DCCE. Also the angle AD, and BA: AC:: CD: DE. Lastly the angle ACB=E, and BC: CA :: CE: ED.

II. Reciprocal figures are (BD, BF) when in each of the figures there are terms both antecedent and confequent (that is, AB: BG :: EB : BC.

III. A right-line AB is faid to be cut according tơ extreme and mean proportion, when as the whole AB is to the greater segment AC, so is the greater segment AC to the lefs CB (AB: AC:: AC: CB.)

IV. The altitude of any figure ABC, is a perpendicular line AG, drawn from the top A, to the base BC. V. A ratio is said to be compounded of ratio's, when the quantities of the ratio's, being multiplied into one another, produce a ratio, As the ratio of A to Cis A B

compounded of the ratio's of A to B and B to C. For-X

BC

Triangles ABC, ACD, and parallelograms BCAE, CDFA, which have the same height, are to each other, as their bases, BC, CD.

(a) Take

(a) Take as many as you please, BG, GH, equal to BC, and also DI=CD, and join AG, AH, ΑΙ.

(b) The triangles ACB, ABG, AGH, are equal, and (b) also the triangle ACD=ADI. Therefore the triangle ACH is the same multiple of the triangle ACB, as the base HC is of the base BC; and the triangle ACI the same multiple of the triangle ACD, as the base CI is of CD. But if HCCI, (c) then is likewise the triangle AHC,,ACI; and (d) therefore BC : CD: : the triangle ABC: ACD :: (e) Pgr. CE : CF. Which was to be demonftrated.

Schol. Plate III, Fig. 37.

Hence triangles, ABC, DEF, and Pgrs. AGBC, DEFH, whose bases BC, EF are equal, are to each other as their altitudes, AI, DK.

1

a 3. I.

b 38. 1.

c fch, 38.1. d 6. def. 5. e 41. 1.& 15.5.

a 3. 1. b 7.5. 7:: с 1. 6.

(a) Take IL=CB, and KM EF; and join LA, LG, MD, MH, then it is evident, that the triangle ABC: DEF: : (6) ALI: DKM :: (c) AI: DK:: (a) Pgr. d 41. 1.&

AGBC: DEFH. Which was to be demonftrated.

PROP. II. Fig. 36.

If to one fide BC of a triangle ABC, be drawn a parallel right-line DE, the fame shall cut the fides of the triangle proportionally (AB: BD :: AE: EC.) And if the fides of the triangle are cut proportionally (AD: BD :: AE : EC) then a right-line DE, joining the points of a section DE, shall be parallel to BC, the other fide of the triangle. Draw CD and BE.

1. Hyp. Because the triangle DEB (a) =DEC, (b) therefore shall be the triangle ADE: DBE:: ADE: ECD. But the triangle AED : DBE: : (c) AD: DB, and the triangle ADE: DEC :: AE: EC; (d) therefore AD : DB:: AE : EC.

2. Hyp. Because AD : DB:: AE: EC, (e) that is as the triangle ADE: DBE :: ADE: ECD; (f) therefore is the triangle DBE=ECD; and (g) therefore DE, BC are parallels. Which was to be demonstrated.

Schol.

If there are drawn several lines DE, FG parallel to one fide BC of a triangle, all the segments of the Ades shall be proportional.

I

15.5.

a 37. 1.
b 7.5.
с 1. 6.
d 11. 5.
e 1. 6.

f 9. 5

g 39. I.

For

à 2.6.

a 5. 1.
b 32. 1.

c byp.

d 27. 1. e 2. 6.

f 2. 6.

g 29. 1.

h 5. 1

k 1. ax.

For DF: FA (a) :: EG: GA; and compounding and inverting, FA: DA::GA: EA; (a) and DA : DỄ :: EA: EC; therefore by equality DF: DB:: EG: EC. Which was to be demonstrated.

Coroll.

If DF: DB:: EG: EC; (a) then BC, DE, FG, shall be parallels.

PROP. III. Plate III. Fig. 38.

If an angle BAC of a triangle BCA be bisected, and the right-line AD, that bisects the angle, cut the base also; then pall the segments of the base have the fame ratio that the other fides of the triangle have, (BD: DC::AB: AC.) And if the segments of the base have the same ratio, that the other fides of the triangle have (BD: DC::AB: AC) then a right-line AD drawn from the top A to the fection D, shall bisect that angle BAC of the triangle.

Produce BA, and make AE=AC, and join CE.

1. Hyp. Because AE = AC, therefore is the angle ACE (a) = E (6) = half BAC (c) = DAC ; (d) therefore DA, CE are parallels. (e) Wherefore BA: AE (AC) : : BD : DC,

2. Hyp. Because BA: AC (AE) :: BD : DC, (f) therefore are DA, CE parallels; and (g) therefore is the angle BAD=E; and the angle DAC (g)=ACE (b) = E, (k) therefore the angle BAD DAC. Wherefore the angle BAC is bisected. Which was to be demonstrated.

PROP. IV. Fig. 39.

Of equiangular triangles ABC, DCE, the fides are proportional which are about the equal angles, B, DCE, (AB : BC:: DC: CE, &c.) And the fides AB, DC, &c. which are fubtended under the equal angles ACB, E, &c. are homologous, or of like ratio.

Set the fide BC in a direct line to the side CE, and

a 32. 1. & produce BA and ED till they (a) meet in F.

Because the angle B (b) =ECD, (c) therefore BF, CD are parallel: Alto because the angle BCA (6) = CED, (e) therefore are CA, EF parallel. Therefore the figure CAFD is a Pgr. (d) therefore AF=CD, and AC = (d) FD. Whence it is evident, that AB: AF (CD): : (e)

BC

BC: CE. (f) By permutation therefore AB: BC::CD: CE, alfo BC: CE :: FD (AC): DE. (f) And thence by permutation BC:AC::CE: DE. (8) Wherefore also by equality AB: AC:: CD: DE. Therefore, &c.

Corolla

Hence AB: DC:: BC:CE :: AC: DE.
Schol.

Hence, if in a triangle FBE there be drawn AC, a parallel to one fide FE, the triangle ABC shall be fimilar or like to the whole FBE.

PROP, V. Plate III. Fig. 40, 41.

If two triangles ABC, DEF, have their fides propor= tional (AB: BC:: DE: EF, and AC: BC:: DF: EE, and also AB : AC:: DE: DF.) those triangles are equiangular, and those angles equal, under which are subtended the homologous fides.

At the fide EF (a) make the angle FEG=B, and the angle EFG=C; (6) whence the angle G=A. Therefore GE: EF (c) :: AB: BC:: (d) DE: EF. (e) And therefore GE=DE. Likewife GF: FE (c):: AC: C3 :: (d) DF: FE; (e) therefore GF=DF. Therefore the triangles DEF, GEF, are mutually equilateral. (f) Therefore the angle D=G=A, and the angle FED (f) =FEG-B, and (g) consequently the angle DFE C. Therefore, &c.

PROP. VI. Fig. 40, 41.

If two triangles ABC, DEF have one angle B equal to one angle DEF, and the fides about the equal angles B, DEF proportional (AB: BC :: DE: EF) then those triangles ABC, DEF, are equiangular, and have those angles equal, under which are fubtended the homologous fides.

At the fide EF make the angle FEG = B, and the angle EFG=C; (a) then will the angle G=A. Therefore GE: EF:: (6) AB: BC:: (c) DE: EF, (d) and therefore DE=GE. But the angle DEF (e) = B (f) =GEF; therefore the angle D (g) =G=A, (b) and confequently the angle EFDC. Which was to be demonstrated.

f 16. 5. g 22. 5.

a 23. 1. b 32. 1. C 4. 6.

d hyp.

e 11. 5.&

9.5.

f 8. 1.

g 32. I.