PROP. VII. Plate III. Fig. 43, 44.

If two triangles ABC, DEF have one angle A equal to one angle D, and the fides about the other angles ABC, E, proportional (AB: BC:: DE: EF) and if they have the remaining angles C, F, either both less or both greater than a right-angle; then shall the triangles ABC, DEF, be equiangular, and have those angles equal about which the proportional fides are.

AB: BG

For, if it can be, let the angle ABC - E, and make the angle ABG=E. Therefore, whereas the angle A (a) =D, (b) thence is the angle AGB=F. Therefore AB: BG (c) :: DE: EF: : (d) AB: BC, (e) therefore BG-BC; (f) therefore the angle BGC= BCG. (g) Therefore BGC, or C, is less than a rightangle, and (b) consequently AGB or F is greater than a right: Therefore the angles C and F are not of the fame species or kind, which is against the Hypothesis.

PROP. VIII. Fig. 42.

If a line AD be drawn from the right-angle A, of a right-angled triangle ABC, perpendicular to the base BC; then the triangles ADB, ADC, on each side the perpendicular, are fimilar both to the whole ABC, and to one another.

For because BAC, ADB are (a) right-angles, (b) and so equal, and B common; the triangles BAC, ADB, DB (c) are like. By the fame way of arguing BAC, ADC, are like; (d) whence also ADB, ADC will be like. Which was to be demonstrated.

Coroll.

Hence, 1. BD: DA (e) :: DA: DC.

2. BC:AC:: AC: DC, and CB: BA:: BA: BD.

PROP. IX. Fig. 45.

From a right-line given AB to cut off any part required, as one third (AG.)

From the point A draw an infinite line AC any wife, in which (a) take any three equal parts AD, DE, EF; join FB, to which, from D, (6) draw the parallel DG, and the thing is done.

For

For GB:AG:: (c) FD: AD; whence, by (d) compofition, AB: AG::AF: AD; therefore since AD= one third of AF, AG shall be one third of AB. Which was to be done.

PROP. X. Plate III. Fig. 46.

To divide a given undivided right-line, AB (in F, G) as another given right-line is divided (in D, E.)

Let a right-line BC join the extremities of the line divided, and of the line not divided; and parallel to this, from the points E, D, (a) draw EG, DF, meeting with the right-line which is to be cut in G and F; then the thing is done.

For let DH be (a) drawn parallel to AB. Then AD: DE:: (6) AF: FG,and DE: EC (6) :: DI : IH :: (c) FG: GB. Which was to be done.

Schol. Fig. 47.

Hence we learn to cut a right-line given AB, into as many equal parts as we please (Suppose 5 ;) which will be more casily performed thus.

Draw an infinite line AD, and another BH parallel to it, and infinite also. Of these take equal parts, AR, RS, SV, VN; and BZ, ZX, XT, TL; in each line less parts by one, then are required in AB; then let the right-lines LR, TS, XV, ZN, be drawn; these lines fo drawn shall cut the right-line given AB into five equal parts.

For RL, ST, VX, NZ, are (a) parallels; therefore whereas AR, RS, SV, VN are (b) equal; (c) thence AM, MO, OP, PQ, are equal also. Likewise, because that BZ = ZX, therefore is BQ=PQ, and therefore A B is cut into five equal parts. Which was to be done.

PROP. XI. Fig. 48.

Two right-lines being given AB, AD, to find out a third in proportion to them (DE.)

Join BD, and from AB, being produced, take BC= AD. Through C draw CE parallel to BD; with which let AD produced meet in E, then is DE the proportional required.

For AB: BC (AD) (0) :: AD: DE. Which was to

be done.

c 2. 6. d 18.5

2 31. 1.

b 2. 6,

C 34. 1. 7.5

a 33. 1 b constr. с 2. 6.

a 2.6.

Or thus: make the angle BDA (Fig. 42.) right, and bi.cor.8.6. also the angle BAC right, then (6) BD: DA:: DA :

a 2. 6.

3 31.3.

b cor. 8. 6.

DC.

PROP. XII. Plate III. Fig. 49.

Three right-lines being given, DE, EF, DG, to find out a fourth proportional, GH.

Join EG, and through F draw FH parallel to EG; with which let DG, produced to H, meet. Then it is evident that DE: EF (a) :: DG:GH. Which was to be done.

PROP. XIII. Fig. 50.

Two right-lines being given AE, EB, to find a mean proportional, EF.

Upon the whole line AB, as a diameter, describe a femicircle AFB, and from E erect a perpendicular FE meeting with the periphery in F, then AE: EF::EF: EB. For let AF and FB be drawn; (a) then from the right-angle of the right-angled triangle AFB is drawn a right-line FE, perpendicular to the base. (6) There fore AE: FE:: FE: EB. Which was to be done.

Coroll.

Hence, a right-line drawn in a circle from any point of a diameter, perpendicular to that diameter, and produced to the circumference, is a mean proportional betwixt the two segments of that diameter.

PROP. XIV. Fig. 34.

Equal Parallelograms BD, BF, having one angle ABC, squal to one EBG, have the fides which are about the equal angles reciprocal (AB: BG :: EB: BC;) and those parallelograms BD, BF, which have one angle ABC equal to one EBG, and the fides which are about the equal angles reciprocal, are equal.

For let the fides AB, BG, about the equal angles make & fch. 15. 1. one right-line; (a) wherefore EB, BC, shall do the fame. Let FG, DC, be produced till they meet.

b 1. 6.

C 7.5.

%.

1. Hyp. AB: BG (6) : : BD: BH: : (c) BF: BH:: (d) BE: BC, (8) therefore, &c.

1

2. Hyp. BD: BH :: (f) AB: BG:: (g) BE: BC :: (6) BF: BH. (k) Therefore the Pgr. BD=BF. Which was to be demonstrated.

PROP. XV. Plate III. Fig. 51.

Equal triangles having one angle ABC, equal to one DBE, their fides which are about the equal angles are reciprocal (AB: BE :: DB: BC.) And those triangles that have one angle ABC equal to one DBE, and have also the fides that are about the equal angles reciprocal (AB: BE :: DB: BC) are equal.

Let the fides CB, BD, which are about the equal angles be set in a strait-line; (a) therefore ABE is a rightline. Let CE be drawn.

1. Hyp. AB: BE :: (6) the triangle ABC: CВЕ (c) : : triangle DBE: CBE:: (4) DB: BC; (e) therefore, &c.

2. Hyp. The triangle ABC: CBE : : (f) AB : BE:: (z) DB: BC (b): : the triangle DBE: CBE. (A) Therefore the triangle ABC=DBE. Which was to be demonftrated.

PROP. XVI. Fig. 52.

If four right-lines are proportional (AB: FG :: EF: CB) the rectangle AC, comprehended under the extremes AB, CB, is equal to the rectangle EG, comprehended under the means FG, EF. And if the rectangle AC, comprehended under the extremes AB, CB, be equal to the rectangle EG, comprehended under the means FG, EF, then are the four rightlines proportional (AB: FG :: EF. CB.)

1. Hyp. The angles B and F are right, and (a) consequently equal, and by hypothesis AB: FG :: EF: CB, (b) therefore the rectangle AC=EG.

2. Hyp. The rectangle AC (c) =EG, and the angle B=F; (d) therefore AB: FG:: EF: CB. Which was to be demonstrated.

Coroll.

f 1.6. g byp. h 1. 6.

k 11. and

9.5.

a fch.15.1. b 1. 6.

C7.5. d1. d 1. 6. e 11.5. f 1.6. g hyp. h 1.6. k 11. and 9.5.

a 12.ax.

b 14.6.
c byp.
d 14.6.

Hence it is easy to apply a rectangle given EG to a right-line given AB, (viz.) (e) by making AB: EF: : fc 12.6.]

FG: BC.

PROP. XVII. Plate III. Fig. 52.

If three right-lines are proportional (AB: EF:: EF:CB) the rectangle AC, made under the extremes AB, CR, is equal to the Square EG, made of the middle EF. And if the rectangle A C, comprehended under the extremes AB, CB, be equal to the Square EG, made of the middle EF, then the three lines are proportional, (AB: EF:: EF: CB.)

Take FG=EF.

1. Hyp. AB: EF: : (a) EF (FG) : CB, therefore the rectangle AC (6) = EG (c) = EFq.

2. Hyp. The rectangle AC (d) = to the square EG EFq; (e) therefore AB: EF:: FG (EF): BC. Which was to be demonstrated.

Coroll.

Let AXB=Cq, then A: C::C: B.

PROP. XVIII, Fig. 53.

Upon a right-line given AB, to describe a right-lined figure AGHB, fimilar and alike fituate to a right-lined figure given CEFD.

Resolve the right-lined figure given into triangles; (a) Make the angle ABH =D, [a) and the angle BAH= DCF, (a) and the angle AHG=CFE, (a) and the angle HAG=FCE, then AGHB, shall be the right-lined figure sought,

For the angle B (6) = D, and the angle BAH (6)= DCF, (c) wherefore the angle AHB=CFD; (6) alfo the angle HAG=FCE, and the angle AHG (6) = CFE; (c) wherefore the angle G-E, and the whole angle GAB (d) = ECD, and the whole angle GAB (d) =EFD. The Polygons, therefore are mutually equiangular. Moreover because the triangles are equiangular, therefore AB: BH (e) :: CD: DF; and AG : GH (e) :: CE: EF, Likewife AG: AH: : (e) CE: CF, and AH:AB::CF: CD. (f) From whence by equality AG:AB:: CE: CD. After the fame manner GH': HB:: EF: FD. (g) Therefore the Polygons ABHG, CDFE are fimilar and alike situate. Which was to be done,

PROP. XIX. Fig. 54.

Like triangles ABC, DEF, are in dublicate ratio of their homologous fides, BC, EF.

(a) Let