Book III. will pass through the point A. If not, let F, G, be the cens ters of the two circles, and the right line FG joining them cut the circles in C, D; join FA, AG. Because F is the a def. 15. 1. center of the circle ABC, FA is equal to FC a; and, because G is the center of the circle ADE, GA is equal to GD; add DC; then the whole FG is greater than FA, AG; and likewife lefs b; which is impoffible. Wherefore, &c.

b 20.1.

a II.

b 2.
c byp.

a 3:
b 47. I.

c Ax. I.

d Ax. 7:

PRO P. XIII. THE OR.

NE circle cannot touch another, either outwardly or inwardly, in more than one point.

The circles ABC, BFD, cannot touch one another inwardly in more than one point; for, if poffible, let them touch in B, D; let G be the center of the one circle, and H of the other; then BG is equal to GD, and greater than HD; therefore BH is much greater than HD; but H is the center of the circle BDF; therefore BH is equal to HD; and likewife greater; which is impoffible: Therefore the circles ABC, BFD, cannot touch one another inwardly in the points B, D. Let the circle AKC touch the circle ABC outwardly in the points A, C, if poffible; join AC; then is AC within the circle ACK, and likewife without the famec; which is im, poffible. Wherefore two circles, &c.

A

PRO P. XIV. T H E OR.

NY number of equal right lines, drawn in a circle, are equally distant from the center; and, if they are equally diftant from the center, they are equal to one another.

Let AB, CD, be two equal right lines, drawn in the circle ABD; find E, the center of the circle, and from it let fall the perpendiculars EF, EG, they will be equal to one another; for, join EA, EC; then are the right lines AB, CD, bifected by the right lines EF, EG; the fquare of AE is equal to the fquares of AF, FEb; and the fquare of EC equal to the fquares of CG, GE; therefore the fquares of AF, FE, are equal to the fquares of CG, GE; but the fquare of AF is equal to the fquare of CG d; therefore the square of FE is equal to the fquare of EG; therefore AB, CD, are equally diftant from the center,

2dly, if EF be equal to EG, then AB will be equal to CD ; Book III. for the fquares of AF, FE, are equal to the fquares of CG, GE; but the fquare of EF is equal to the fquare of EG; therefore the fquare of AF is equal to the fquare of CG; bute Ax. 5. AB is double AF, and CD double CG; therefore AB is equal, to CD f. Wherefore, &c.

T

PROP. XV. THEO R.

HE diameter of a circle is the greatest right line in it, and the line nearest to the diameter is greater than that more remote; and on each fide of the diameter only two right lines can be drawn equal to one another.

a 3.

f Ax. 6.

Let ABC be a circle, whofe diameter is AD; let MN, FG, be drawn any how in the circle; then AD is the greatest line; AD greater than MN, and MN greater than FG. Find the center E; draw EM, EN, EF, EG; then AE, ED, are equal to ME, EN; but ME, EN, are greater than MN a; therefore a 20. f. AD is greater than MN; likewife ME, EN, are equal to FE, EG, and the angle MEN greater than FEG; therefore MN

is

greater than FGb: So likewife on each fide of AD only b 24. 16 two right lines can be drawn equal to one another, viz. upon which the equal perpendiculars fall. For, let fall a perpendicular EL upon MN, and draw EH equal to it, and BC at right angles to EH; then BC is equal to MN c. If any other c 14. right line can be equal to MN, or BC, let it be FG, a line nearer to the diameter equal to that more remote. Wherefore, &c.

A

PROP. XVI. THEOR.

Line drawn from the extreme point of the diameter of a circle, at right angles to that diameter, shall fall without the fame and between that right line and the circumfe rence no right line can be drawn.

Let ABC be a circle, whofe diameter is AB; at the extremity of which, if a right line is drawn at right angles, it fhall fall without the circle.

If not, let it fall within the circle, as AC; find the center, and join CD. Then, because DAC is a right angle, DCA will

be

a 5. 1. b 17. I.

Book III. be likewife a right angle, for DA is equal to DCa, that is, two angles in a triangle equal to two right angles; which cannot be b; neither can it fall upon the circle c; therefore it must fall without the circle, which let be AE; and betwixt the right * Def. 4. I. line AE, and circumference CHA, no right line can be drawn. If poffible, let FA be drawn; then DAF is less than a right angle. From the point D, to the right line FA, a line can be drawn at right angles to FA, falling without the circle; which let be DG; then, because DGA is a right angle, and DAG lefs than a right angle, DA is greater than DG d; but DA is equal to DH; therefore DH is greater than DG, and likewife lefs; which is impoffible: There, fore, betwixt the circumference and right line AE no other right line can be drawn. Wherefore, &c.

d 19. I.

C 2.

a I.

b. II. I.

€ 4. X.

d 4. cor. 16.

COR. I. Hence the angle between the right line and circumference is the leaft of all acute angles; and the angle betwixt the diameter and circumference is the greatest acute angle poffible.

II. Hence, likewise, a right line, drawn at right angles, at the extreme point of the diameter of a circle, touches the circle only in one point; for, if it meet it in two points, it would fall within the circle c.

T

PROP. XVII. PRO B.

O draw a right line that will touch a given circle from a given point without the fame.

Let BCD be the circle, and A the point without it; it is required to draw a right line from the point A, that will touch the circle BCD. Find E the center of the circle a; join AE, cutting the circle BCD in D. About the center E, with the distance EA, defcribe the circle AFG; from the point D draw DF at right angles to DEb, cutting the circle AFG in F; join EF; cutting DBC in B, and join AB; then is AB the tangent required.

For, because E is the center of both circles, the right lines AE, EB, are equal to FE, ED, and the angle E common; therefore the triangle ABE is equal to FDE c; and the angle EBA to EDF; but EDF is a right angle; therefore ABE is likewife a right angle: Therefore AB is a tangent to the circle in the point Bd, and drawn from the point A. Which was required.

I

PRO P. XVIII. and XIX. THEOR.

F any right line touches a circle, and from the center to the point of contact a right line be drawn, that line will be at right angles to the tangent; and if, from the point of contact, a right line be drawn, paffing through the circle, at right angles to the tangent, the center of the circle will be in that line.

Let ABC be a circle, and DE a right line touching it in the point C; and if, from the center F, there be drawn a right line FC, that line will be perpendicular to the tangent. If not, let FG be drawn from the center F, at right angles to DE a.

Book III.

a 12. I.

Now, because FGC is a right angle, FCG will be less than a right angle b; therefore FC is greater than FG c; that is, b 17. 1. FB greater than FG, a part greater than the whole; which is c 19. . impoffible. For the fame reason, no right line but FC can be perpendicular to DE.

2dly, If, from the point of contact C, of the tangent DE, AC be drawn through the circle ABC, at right angles to DE, the center of the circle will be in AC. If not, let it be in H; join HC; then HCE is a right angle d; but ACE is d 18. a right angle e; therefore HCE is equal to ACE, a part toe Hyp the whole; which is abfurd. Wherefore, &c.

PROP. XX. THEOR.

HE angle at the center of the circle is double the angle at
the circumference, when the fame arc is the base of both.

TH

Let ABC be a circle, and E its center, the angle BEC, at the center, is double the angle BAC, at the circumference; the arc BC being the bafe of both.

For, join AE, and produce it to F; then, because EA is equal to EB, the angle EAB is equal to EBA; but EAB, a 5. 1. EBA, are double EAB; and BEF is equal to EAB, EBA, b 32. F. or double EAB. For the fame reason, FEC is double EAC; therefore the whole angle BEC is double BAC. Again, let there be another angle EDC; join EC, and produce DE to B; then the outward angle BEC is equal to EDC, ECD, or double EDC. For the fame reafon, the angle BEF is double the angle BDF; but the whole angle BEC is double BDC, and a part BEF is double a part BDF; therefore, the remainder FEC is double the remainder FDC. Wherefore, &c. PROP.

Book III.

a 6. I.
b Conft.

€ 9.

A

PROP. XXI. PRO B.

Segment of a circle being given, to defcribe the circle whereof it is the fegment.

See fig. 25. It is required to defcribe the circle, whereof ABC is a fegment, bifect AC in D, draw DB at right angles to AC, and join AB; then the angle BAD will be either equal, greater, or lefs than the angle ABD: First, let them be equal; then the fide AD is equal to DB2, and DC to AD; therefore De is the center of the circle. 2dly, If the angle BAD is greater or less than ABD, make the angle BAE equal to ABD, and join AE, EC; then the fide AE is equal to BE a, and because AD is equal to DC, and DE common, the angles ADE, CDE are right ones; therefore the fide AE is equal to EC; therefore the three lines EA, EB, EC are equal; therefore E is the center of the circle c. If the center E is within the fegment, then the fegment is greater than a femicircle, if without, lefs; if upon the base of the fegment, then it is a femicircle. Wherefore, &c.

c. 4. In

See fig. 21.

a 21.

b 20.

A

PROP. XXII. THEOR.

NGLES that are in the fame fegment of a circle are e qual to each other.

Let BAED be the fegment of a circle, either greater or lefs than a femicircle; and BAD, BED, angles in the fame, these angles are equal to one another. Complete the circle ABCDE, and to the center F; draw BF, FD; then the angle BFD is double the angle BAD, and likewise double CAX. 7. I. BED; therefore BAD, BED, are equal to one another c If the fegment is lefs than a femicircle, join AE, complete the circle, and, to the center F, draw AF, FE; then the angle AFE is double the angle ABE or ADE; but the angle AGB is equal to EGD; therefore the remaining e Cor. 32. 1, angle BAD is equal to BED. Wherefore, &c.

d is. I.