a 3. post b 1. poft.

C 15.def. d 1. ax.

e 23, def.

20. If one whole be double to another, and that which is taken away from the first be double to that which is taken away from the second, the remainder of the first shall be double to the remainder of the second.

The Citations are to be understood in this manner; When you meet with two numbers, the first shews the Propofition, the second the Book.; as by 4. 1. you are to understand the fourth Propofition of the first Book; and so of the reft, Moreover, ax. denotes Axiom, post. Poftulate, def. Definition, fch. Scholium, cor. Corollary.

PROPOSITION I. Plate I. Fig. 12.

Uequilateral triangle B

Pon a finite right-line given as AB, to describe an

From the centers A and B, at the distance of AB, or BA, (a) describe two circles cutting each other in the point C; from whence (b) draw two right-lines CA, CB. Then is AC (c) = AB (c) =LC (d)=AC, (c) Wherefore the triangle ACB is equilateral. Which was to be done.

a 3. post. b 1. post.

CI. I.

d 2. poft.

e 2. poft.

Scholium.

After the fame manner upon the line AB may be de. scribed an Isosceles triangle, if the distances of the equal circles be taken greater or less than the line AB,

PROP. II. Fig. 13.

From a point given A, to draw a right-line AG equal to a right-line given BC.

From the center C, at the distance of CB, (a) describe the circle CBE. (6) Join AC; upon which (c) raise the equilateral triangle ADC. (d) Produce DC to E. From the center D, at the distance of DE, describe the circle DEG; and let DA (e) be produced to the point G in the circumference thereof. Then AG СВ.

For DG (f)=DE, and DA (g) = DC. Wherefore AG (b) CE (k) = BC (1). Which was to be done. The putting of the point A within or without the line BC varies the cafes; but the construction, and the demonstration, are every where alike,

Schol.

The line AG might be taken with a pair of compasses; but the so doing answers to no poftulate, as Proclus well observes.

PROP. III. Plate I. Fig. 14.

Two right-lines, A and DC, being given, from the greater DC, to take away the right-line DE, equal to the leffer A.

From the point D (a) draw the right-line BD = A. The circle described from the center D at the distance of BD shall cut off DE (b)=BD (c) = A (d) = DE. Which was to be done.

PROP. IV. Fig. 15, and 16.

If two triangels BAC, EDF, have two fides of the one BA, AC, equal to two fides of the other ED, DF, each to its correspondent fide (that is BA=ED, and AC=DF) and have the angel A equal to the angle D contained under the equal right-lines; they shall have the base BC equal to the base EF; and the triangle BAC shall be equal to the triangle EDF; and the remaining angels B, C, shall be equal to the remaining angles, E, F, each to each, under which the equal des are fubtended.

a 2. 1.

b 15. df.
c constr.
d 1. ax.

If the point D be apply'd to the point A, and the right-line DE plac'd upon the right-line AB, the point E shall fall upon B, because DE (a)=AB, also the rightline DF shall fall upon AC, because the angle A (a)=D. Moreover the point F shall fall upon the point C, because AC (a)=DF. Therefore the right-lines EF, BC, shall agree, because they have the same terms, and consequently are equal. Wherefore the triangles, BAC, DEF, and the angles B, E, as also the angles C, F, do agree, and are equal. Which was to be demonstrated.

a hyp

PROP. V. Fig. 4.

The angles BAC, BCA, at the base of an Isofceles triangle ABC, are equal one to the other, And if the equal fides AB, BC, are produced, the angles CAD, ACE, under the base, shall be equal one to the other.

A 4

(a) Tak

1

(a) Take BE = BD; and (6) join CD, and AE. Because, in the triangles BCD, BAE, are AB (c) = BC, and BE, (d) = BD, and the angle B common to them both, (e) therefore is the angle BAE = BCD, and the angle AEB (e) = BDC, and the base AE (c)= CD; alfo EC (f) = DA. Therefore in the triangles AEC, ADC (g) will be the angle ECA DAC, Which was to be dem. Also the angle EAC = DCA, but the angle BAE (6) = BCD; therefore the angle BAC (k) BCA. Which was to be demonstrated,

Coroll.

Hence, every equilateral triangle is also equiangular.

PROP. VI. Plate I Fig. 4.

If two angles BCA, ACB of a triangle ABC, be equal the one to the other, the fides BC, AB, fubtended under the equal angles, shall also be equal one to the other.

If the fides be not equal, let one be bigger than the other, suppose BABC. (a) Make AF = BC, and (6) draw the line CF.

In the triangles FAC, BAC, because AF (c)=CB, and the fide, AC is common, and the angle BAC (d) = ACB, the triangles FAC, ACB (e) shall be equal the one to the other, a part to the whole, (f) Which is impossible.

T

Coroll.

Hence, every equiangular triangle is also equilateral,

PROP. VII. Fig. 4.

Upon the fame right-line AC two right-lines being drawn AB, BC, trvo other right-lines equal to the former, AF, CF, each to each (viz. AF=AB, and CF=BC) cannot be drawn from the same points A, C, on the fame fide B, to several points, as B and F, but only to B.

1. Cafe. If the point F be set in the line AB, it is plain that AF is (a) not equal to AB.

ag ax.

Fig. 17.

b 5. 1.

c fuppof.

2. Cafe. If the point D be placed within the triangle ACB, then draw the line CD, and produce BDF, and ECF. Now you would have AD-AC, then the angle ALC (b) ACD; as also, because BD (c) BC, the angle FDC=

(6) (6) ECD, therefore is the angle FDC

(d) ACD, that d 9 ax.

is, the angle FDC - ADC. (d) Which is impossible.

3. Cafe. If D (Fig. 18.) falls without the triangle ACB, let CD be joined.

Again, the angle ACD (e) =ADC, and the angle BCD (c) =BDC. (f) Therefore the angle ACDBDC, viz. the angle ADC-BDC. Which is impossible. Therefore, &c.

PROP. VIII. Plate I. Fig. 15, and 16.

If two triangles ABC, DEF have two fides AB, AC, equal to tawo fides DE, DF, each to each, and the base BC equal to the baje EF, then the angles contained under the equal right-lines shall be equal, viz. A to D.

e 5. 1. f 9. ax.

a hyp.
b ax. 8.

c hyp.

Because, BC (a) =EF, if the base BC be laid on the base EF, (b) they will agree: therefore whereas AB (c)= DE, and AC=DF, the point A will fall on D (for it cannot fall on any other point, by the precedent propofition) and so the fides of the angles A and D are coincident; (d) wherefore those angles are equal. Which was to be de- d 8. ax.

1. Hence, triangles mutually equilateral are also mutually (x) equiangular.

2. Triangles mutually equilateral (x) are equal one to the other.

PROP. IX. Fig. 19.

To bisect, or divide into two equal parts, a right-lined angle given BAC.

(a) Take AD to AE, and draw the line DE; upon which (6) make an equilateral triangle DFE, draw the right-line AF; it shall bisect the Angle.

For AD (c)=AE, and the fide AF is common, and the base DF (c)=FE. (d) therefore the angle DAF EAF. Which was to be done,

Coroll.

Hence it appears how an angle may be cut into 4, 8, 16, 32, &c. equal parts, to wit, by bisecting each part again.

The method of cutting angles into any number of equal parts required, by a Rule and Compass, is as yet unknown to Geometricians,

PROP,

x 4. 1.

a 3.1. b1.1.

c conftr.

d 8. 1.

a 1. 1.

b 9. 1.

c conftr.

d 4.1.

a 3. 1. b 1. 1.

c conftr. d 8. 1.

e 10. def.

a 3 poft. b 10. 1.

cconftr. d 8. 1.

e 10. def.

PROP. X. Plate I. Fig. 20.

To bisect a right-line given AB.

Upon the line given AB (a) erect an equilateral triangle ABC; and (6) bisect the angle C with the rightline CD. That line shall also bisect the line given AB.

For AC (c) BC, and the fide CD is common, and the angle ACD (c) =BCD. therefore AD (d) =BD. Which was to be done.

The practice of this and the preceding propofition is easily shewn by the Construction of the ist propofition of this Book.

PROP. XI. Fig. 20.

From a point D in a right-line given AB to erect a rightline DC at right-angles.

(a) Take on either fide of the point given DA DB, upon the right-line AB (6) erect an equilateral triangle, draw the line CD, and it will be the perpendicular required.

For the triangles DAC, DCB are mutually (c) equilateral, (d) therefore the angle ADC=CDB; (e) therefore DC is perpendicular. Which was to be done.

The practice of this and the following is easily performed by the help of a square.

PROP. XII. Fig. 20.

Upon an infinite right-line given AB, from a point given that is not in it, to let fall a perpendicular right-line CD.

From the center C (a) describe a circle cutting the right-line given in the points A and B. Then (6) bisect AB in D, and draw the right-line in CD, which will be the perpendicular required.

Let the lines CA, CB be drawn. The triangles ADC CDB are mutually (c) equilateral; (d) therefore the angles ADC, BDC are equal, and by (e) consequence right. (e) Wherefore DC is a perpendicular. Which was to be done.

PROP. XIII. Fig. 1.

When a right-line DG standing upon a right-line AB maketh angles DGA, DGB; it maketh either two rightangles, or two angles equal to two right-angles.