4 drawn thro' the centre into two equal parts: I fay, it will cut the fame at right angles. For [by 1. 3.] find the centre of the circle A BC, which let be E, and join E A, E B. C Then because AF is equal to FB, and FE is common, two fides AF, EF will be equal to two fides BF, FE, and the base EA is equal to the base EB; therefore the angle AFE [by 8. 1.] will be equal to the angle BFE. But fince a right line standing upon a right line makes the adjoining angles equal to one another, each of the equal angles is a right angle: therefore each of the angles A FE, BFE is a right angle, and fo the right line C D drawn thro' A the centre cutting the right line AB, not drawn thro' the centre, into two equal parts, cuts it at right angles. B F D Now if the right line C D cuts the right line A B at right angles: I fay, it will divide the fame into two equal parts AF, BF. For the fame conftruction remaining, because the right line E A drawn from the centre is equal to E B, the angle EAF will be [by 5. 1.] equal to the angle E B F. But the right angle AFE is equal to the right angle BFE: therefore the two triangles E AF, E B F have two angles of the one equal to two angles of the other, and the fide EF of the one equal to the fide E F of the other, viz. the common fide oppofite to one of the equal angles: therefore the remaining fides of the one triangle will be [by 26. 1.] equal to the remaining fides of the other: Wherefore AF is equal to B F. If therefore a right line drawn through the centre of a circle divides a right line not drawn through the centre into two equal parts, it will cut the fame at right angles; and if it cuts the fame at right angles, it will cut it into two equal parts. Which was to be demonftrated. PROP. IV. THEOR. In a circle if two right lines not drawn through the centre, do interfect each other; they will not cut one another into two equal parts. Let there be a circle A B C D, and in it let two right lines A C, B D not drawn through the centre interfect one another in the point E: I fay, these will not cut one another into two equal parts. For, if poffible, let them bifect each other fo that A E be equal to EC, and B E equal to E D. Find [by 1. 3.] the centre F of the circle A B C D, and join F E. Then because the right line F E drawn thro' the centre cuts the right line AC not drawn through the centre into two equal parts; it will [by 3. 3.] cut the fame at right angles: wherefore FEA will be a right angle. Also because the right line F E cuts the right line BD not drawn through the centre into A B E two equal parts, it will alfo cut it at right angles therefore FEB is a D right angle. But it has been also proved that FEA is a right angle. Therefore the angle FEA will be equal to the angle FE B, the leffer equal to the greater, which is impoffible. Therefore A C, B D do not cut one another into equal parts. Wherefore if in a circle two right lines not drawn thro' the centre do interfect one another, they will not cut one another into two equal parts. Which was to be demonstrated. PROP. V. THEOR. If two circles interfect one another, they will not both bave the fame centre. G For let two circles A B C, C D G interfect one another in the points B, c: I fay, they have not both the fame centre. For, if poffible, let E be the centre of them both. Join EC, and draw any right line EF G. Then becaufe E is the centre of the circle A B C, the right line EC will be [by 15. def. 1.] equal to EF; and because E is the centre of the circle CDG, the right line E C will be equal to EG. But E c has been proved to be equal to EF, wherefore E F will D E F B be be equal to E G, the leffer equal to the greater, which is impoffible. Therefore the point E is not the centre of both the circles ABC, CDG. Which was to be demonftrated. Therefore if two circles cut one another, they will not both have the fame centre. PROP. VI. THEOR. If two circles touch one another inwardly, they will not both have the fame centre 1. For let two circles A B C, CDE touch one another inwardly in the point c: I fay, they have not both the fame centre. For if they have, let the fame be F. Join F c, and draw F B any how. C Then becaufe F is the centre of the circle A B C, the line F C is equal to F B. Also be'cause F is the centre of the circle CDE, the line F C will be equal to F E: but FC has been proved to be equal to FB: therefore F E is equal to F B, the leffer to the greater, which is impoffible. Wherefore F is not the centre of both the circles A B C, CDE. Therefore if two circles touch one D F EB another inwardly, they will not both have the fame centre. Which was to be demonftrated. f Euclid only propofed this theorem of circles touching one another inwardly; for when they touch outwardly, it is most manifeft, they will not both have the same centre. PROP. VII. THEOR. If any point except the centre be taken in the diameter of a circle, and any right lines drawn from it do fall upon the circumference of the circle, the greatest of them will be that wherein is the centre, and the leaft, the remainder of the fame line; and of all the other lines, that which is nearest the line I 2 paling paffing through the centre will always be greater than that more remote; and only two equal lines can fall from the aforefaid point, being one on one fide of the leaft line, and one on the other. Let A B C D be a circle, A D a diameter of it, and let any point F be taken in that diameter, not being the centre of the circle; let E be the centre of the circle, and let any right lines F B, F C F G drawn from the point F fall upon the circumference of the circle A B C D: I fay, FA is the greatest of these right lines, and FD is the leaft; and of the reft, FB is greater than F C, and c than FG. For join BE, CE, GE. Then because two fides of every triangle are [by 20. 1.1 greater than the remaining fide, the fides E B, EF will be greater than BF. But A E is equal to BE: therefore в E, B A E F are equal to A F: wherefore AF is greater than B F. Again, because B E is equal to c E, and FE is common, the two fides B E, EF áre equal to the two fides CE, E F. But the angle BEF is greater than the angle CEF: therefore [by 24. 1.] the bafe B F is greater than the base c F. By the fame reafon, CF is greater than E F D K H F G. Again, because GF, FE together are greater than E G, but E G is equal to E D, the fides GF, FE will be greater than ED. Take away FE, which is common, then the remainder GF is greater than the remainder F Dd. Therefore FA is the greatest of the right lines, and FD the leaft: but FB is greater than FC, and FC than F G. I fay moreover, there can but two equal right lines fall from the point F upon the circumference of the circle A B C D, one on one fide FD the leaft of the lines, and the other on the other fide of this line. For [by 23. 1.] make the angle FEH at the given point E in the right line E D, equal to the angle GEF, and draw F H; then because & E is equal to EH, and E F is common, the two fides GE, EF are equal to the two fides HE, E F, and the angle GEF [by conftr.] is equal to the angle HEF: therefore the bafe FG [by 4. 1.] will be equal to the base F H. I fay, no other right line can fall from F upon the circumference of the circle equal equal to F G; for if poffible, let F K fall upon it, equal to FG; then because F K is equal to F G, and FH is equal to FG: FK will be equal to FH too, viz. a line nearer to the line drawn thro' the centre, equal to a line which is more remote. Which is impoffible. Or thus: Join E K, and becaufe GE is equal to EK, and F E is common, and the base GF is equal to the bale F K; the angle GEF [by 8. 1.] will be equal to the angle KEF. But the angle GEF is equal to the angle HEF: therefore the angle HEF will be equal to the angle K E F, the leffer equal to the greater. Which cannot be: wherefore no other line can fall from the point F upon the circumference of the circle equal to GF. There is therefore but one only equal to it. If therefore any point except the centre, be taken in the diameter of a circle, &c. Which was to be demonstrated. PROP. VIII. THEOR. If any point be taken without a circle, and from that point be drawn any right lines to the circle, one of which paffes thro' the centre, and the others any bow at pleasure that paffing thro' the centre will be the greatest of all thofe that fall upon the concave part of the circumference of the circle, and of the rest that which is nearest it will always be greater than that which is more remote: but amongst these lines which fall upon the convex part of the circumference, the least is that which lies between the affumed point and the diameter: of the others that which is nearer to the leaft will always be less than that which is more remote; and only two equal right lines can fall upon the circumference of the circle, the one on one fide the leaft line, and the other on the other fide of it. Let ACB be a circle, and any point D be taken without it, from which any right lines D A, DE, DF, are drawn to the circumference of the circle, whereof D A is that paffing thro' the centre: I fay, the greatest of these lines that fall upon the concave part of the circumference |