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Book III. will pass through the point A. If not, let F, G, be the cene

ters of the two circles, and the right line FG joining them j

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 b 20.1. likewise less b; which is impossible. Wherefore, &c.

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O

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

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The circles ABC, BFD, cannot touch one another in. wardly in more than one point; for, if possible, 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 a; therefore BH is equal to HD; and likewise greater; which is impossible : 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 possible ; join AC; then is AC within the circle ACK b, and likewile without the famec; which is im, poflible. Wherefore two circles, &c.

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PRO P. XIV. THEO R.

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Nr number of equal right lines, drawn in a circle, are

equally distant from the center; and, if they are equally distant from the center, they are equal to one another.

A

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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 a; the square of AE is equal to the squares of AF, FEb; and the square of EC equal to the squares of CG, GE; therefore the squares of AF, FE, are equal to the squares of CG, GE «; but the square of AF is equal to the square of CG d; therefore the square of FE is equal to the square of LG*; therefore AB, CD, are equally distant fron the center

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2dly, if EF be equal to EG, then AB will be equal to CD;Book III: for the squares of AF, FE, are equal to the squares of CG, GE; but the square of EF is equal to the square of EG; therefore the square of AF is equal to the square of CG ; bute Ax. 5. AB is double AF, and CD double CG; therefore AB is equal a 3. to CD f. Wherefore, &c.

f Ax. 6.

PRO P. XV. THEOR.

H E 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 side of the diameter only two right lines can be drawn equal to one another.

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Let ABC be a circle, whose diameter is AD; let MN, FG, be drawn any how in the circle ; tħen 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. r. AD is greater than MN; likewise ME, EN, are equal to FE, EG, and the angle MEN greater than FEG; therefore MN is greater than FGb: So likewise on each side of AD only b 24. I. 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 MNC. 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. Where. fore, &c.

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A

Line drawn from the extreme point of the diameter of a out the fame; and between that right line and the circumfer rence no right line can be drawn.

Let ABC be a circle, whose diameter is AB ; at the ex. tremity 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

Book III. be likewise a right angle, for DA is equal to DC a, that is, two

angles in a triangle equal to two right angles ; which cannot a 3. I. be b; neither can it fall upon the circlec; therefore it must fall b 17. I. 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 possible, let FA be drawn, then ĎAF 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 less than a right angle, DA is greater 19. I. than DG 4; but DA is equal to DH ; therefore DH is great

er than DG, and likewise less; which is impossible: There. fore, betwixt the circumference and right line AE no other Fight line can be drawn. Wherefore, &c.

Cor. I. Hence the angle between the right line and circumference is the least of all acute angles; and the angle betwixt the diameter and circumference is the greatest acute angle poflible.

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.

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PRO P. XVII. PROB.

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

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b. II. I.

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, describe the circle AFG; from the point D draw DF at right angles to DE b, 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 C4...

EBA to EDF; but EDF is a right angle; therefore ABE is likewise a right angle: Therefore AB is 'a tangent to the

circle in the point B d, and drawn from the point A. Which d 4. cor. 16. was required.

Book III.

PRO P. XVIII. and XIX. THEOR.

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Fany right line touches a circle, and from the center to the

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a 12. I.

at right angles to the tangent ; and if, from the point of contact, a right line be drawn, paling 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.

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. TB greater than FG, a part greater than the whole; which is c 19. 1. impossible. For the same 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 angled; but ACE is, d 18. a right angle e; therefore HCE is equal to ACE, a part toe Hyp: the whole, which is absurd. Wherefore, &c.

PROP. XX. THEOR.

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

For, join AE, and produce it to F; then, because EA is equal to EB, the angle EAB is equal to EBAa; but EAB, a 5. 1. EBA, are double EAB; and BEF is equal to EAB, EBAb,6 32. I. or double EAB. For the same reason, FEC is double EAC; therefore the whole angle BEC is double BAC. Again, lec 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 a. For the fame reason, 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.

PRO P. XXI. PRO B.

Segment of a circle being given, to describe the circle

whereof it segment.

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It is required to describe the circle, whereof ABC is a fegment, bisect AC in D, draw DB at right angles to AC, and join AB; then the angle BAD will be either equal, greater, or less than the angle ABD: First, let them be equal; then the fide AD is equal to DB a, and DC to ADb; therefore Dc 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 side 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 side AE is equal to ECc; 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 segment is greater than a semicircle, if without, less; if upon the base of the segment, then it is a femicircle. Wherefore, &c.

C. 4. In

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PRO P. XXII. THE OR.

NGLES that are in the fame segment of a circle are es

each other.

2 21.

b 20.

Let BAED be the segment of a circle, either greater or Sce fig. 21.

less than a semicircle ; and BAD, BED, angles in the fame, these angles are equal to one another. Complete the circle ABCDE a, and to the center F; draw BF, FD; then the

angle BFD is double the angle BADb, and likewise double C Ax. 7.1. BED; therefore BAD, BED, are equal to one another c,

If the segment is less than a semicircle, join AE, complete the circle a, and, to the center F, draw AF, FE ; then

the angle AFE is double the angle ABE or ADED; but the d 15. 1.

angle AGB is equal to EGDd; therefore the remaining Car

. 32. 1, angle BAD is equal 10 BEDWherefore, &c.

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