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therefore DG is greater than DC, DC than DB, and DB Book III than DA; but DČ, DB, DA, are equal; and likewife not e- un. qual; which is impoffible; therefore no point but D is the cen- a 7. ter of the circle.

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Wherefore, &c.

PROP. X. THE O R.

NE circle cannot cut another in more than two points.

For, if poffible, let the circle ABC cut the circle DEF in the points B, G, F; let K be the center of the circle ABC; join BK, KG, KF. Now, becaufe K is a point within the circle DEF, from which there is drawn to the circumference the right lines BK, KG, KF, equal to one another; therefore K is the center of both circles; which is impoffible. Wherefore,

&c.

b Hyp.

a 9. b 5.

I

PRO P. XI. THE OR.

F two circles touch one another inwardly, a line joining their centers will fall on the point of contact.

Let the two circles ABC, ADE, touch each other inwardly in the point A ; let F and G be the centers of the circles ABC, ADE; then the line joining the centers F, G, will pafs through the point A. If not, let the right line joining the centers F, G, cut the circles in the points D, H; join GA; then, becaufe F is the center of the circle ABC, FA is equal to FHa. For the a def. 15. 1. fame reason, GD is equal to GA; but GA, GF, are greater than AF; therefore DF is greater than AF; therefore greater b 20. 1. than HF; and likewife lefs; which is impoffible; therefore the Ax. 9. 1. line joining the centers will not pass through any other point than A. Wherefore, &c.

I

PRO P. XII. THE OR.

F two circles touch one another outwardly, a line joining their centers will pass through the point of contact.

Let the two circles ABC, ADE, touch one another outwardly in the point A; a right line, joining their centers, will pafs F through

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Book III, through the point A. If not, let F, G, be the centers of the two circles, and the right line FG joining them; cut the circles in C, D; join FA, AG. Becaufe F is the center of a def. 15. 1. the circle ABC, FA is equal to FC 2; 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 "; which is impoffible. Wherefore, &c.

b 20. I.

a II,

a 31

b 47. I.

I.

5.

O

PROP. XIII. THE O R.

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 BDFa; therefore BH is equal to HD; and likewife greater; which is impoflible: 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 within the one circle, and without the other; which is impoffible. Wherefore two circles, &c.

A

PRO P. XIV. THE 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 F, EG, they will be equal to one another; for, join EA, EC, then are the right lis AB, CD, biccted by the right lines EF, EG; the fquare of AE is equal to the fquares of AF, FE, and the fquare of EC equal to the fquares cf 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 G the fo e the fquare of 1 is equal to the fquare of EG; herefore AL, VD, are equally diftent from the center. y be equ. to EG, then a will be equal to CD; Quares of AF, F, are equal to the fquares of CG;

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1

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GE; but the fquare of EF is equal to the fquare of EG; there- Book III. fore the fquare of AF is equal to the fquare of CC; but AB is double AFa, and CD double CG; therefore AB is equal e Ax. s. to CD f. Wherefore, &c.

a 3.

f Ax. 6.

TH

PRO P. XV. THE OR.

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 file of the diameter only two right lines can be drawn equal to one another.

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; therefore ADa 20. 1. is greater than MN; likewife ME, EN, are equal to FE, EG, and the angle MEN greater than FEG; therefore MN is greater than FG : So likewife on each fide of AD only two right lines b 24. 1; 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. If any other right line can be equal toc 14. 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 circumference 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 shall 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 s. 1.

b 17. I.

Book III. be likewife a right angle, for DA is equal to DC, that is, two angles in a triangle equal to two right angles; which cannot beb; neither can it fall upon the circle; therefore it must fall withc Def. 4. 1. out the circle, which let be AE; and betwixt the right 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, becaufe DGA is a right angle, and DAG lefs than a right angle, DA is greater than DG; but DA is equal to DH; therefore DH is greater than DG, and likewife lefs; which is impoffible: Therefore, betwixt the circumference. and right line AE no other right line can be drawn. Wherefore, &c.

d 19. 1.

2.

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, likewife, 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.

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•T

PRO P. XVII PR O 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; join AE, cutting the circle BCD in D. About the center E, with the diftance EA, describe the circle AFG 2: the point D; draw DF at right angles to DE, cutting the circle AFG in F; join EF; cutting DRC in B, and join AB; then is AB the tangent required.

For, becaufe 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; 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 B, and drawn from the point A. Which was required.

PROB.

Book III.

PRO P. XVIII. and XIX. THE OR.

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 DEa.

a 12. I.

Now, because FGC is a right angle, FCG will be lefs than a right angle; therefore FC is greater than FG ; that is, FBb 17. 1. greater than FG, a part greater than the whole; which is impof- c 19. 1. fible. For the fame reafon, no right line but FC can be perpen

dicular 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 angled; but ACE is a right angle;d r§. therefore HCE is equal to AČE, a part to the whole; which is Hyp. abfurd. Wherefore, &c.

T

PRO P. XX. THE OR.

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

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 base of both.

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For, join AE, and produce it to F; then, because EA is equal to EB, the angle EAB is equal to EBA2; but EAB, EBA, are a 5. 1. double EДB; and BEF is equal to EAB, EBA, or double EAB.ь 32. 5. For the fame reafon, 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

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