remainder KD is greater (Ax. 5.) than the remainder GD, that is, GD is less than KD: And because MK, DK are drawn to the point K within the triangle MLD, from M, D, the extremities of its side MD, therefore MK, KD are less (I. 21.) than ML, LD, whereof MK is equal to ML; therefore the remainder DK is less than the remainder DL: In like manner it may be shown, that DL is less than DH: Therefore DG is the least, and DK less than DL, and DL than DH: Also there can be drawn only two equal straight lines from the point D to the circumference, one upon each side of the least: At the point м, in the straight line MD, make (1. 23.) the angle DMB equal to the angle DMK, and join DB: And because MK is equal to MB, and MD common to the triangles KMD, BMD, the two sides KM, MD are equal to the two BM, MD; and the angle KMD is equal to the angle BMD; therefore the base DK is equal (1. 4.) to the base DB: But, besides DB, there can be no straight line drawn from D to the circumference equal to DK: For, if there can, let it be DN; and because DK is equal to DN, and also to DB; therefore DB is equal to DN, that is, the nearer to the least equal to the more remote, which is impossible. If, therefore, any point, &c. Q. E. D. PROP. IX. PROB. If a point be taken within a circle, from which there fall more than two equal straight lines to the circumference, that point is the centre of the circle. Let the point D be taken within the circle ABC, from which to the circumference there fall more than two equal straight lines, viz. DA, DB, DC, the point D is the centre of the circle. For, if not, let E be the centre, join DE and pro duce it to the circumference in B G DB, and DB than DA: But they are likewise equal (Hyp.), which is impossible: Therefore E is not the centre of the circle ABC: In like manner it may be demonstrated that no other point but D is the centre; D therefore is the centre. Wherefore, if a point be taken, &c. Q. E. D. One circumference of a circle cannot cut another in more than two points. B A D H If it be possible, let the circumference FAB cut the circumference DEF in more than two points, viz. in B, G, F; take the centre K of the circle ABC, and join KB, KG, KF: And because within the circle DEF there is taken the point E K, from which to the circumference DEF fall more than two equal straight lines K B, KG, KF, the point K is (III. 9.) the K centre of the circle DEF: But K is also the centre of the circle ABC: therefore the same point is the centre of two circles that cut one another, which is impossible (III. 5.). Therefore one circumference of a circle cannot cut another in more than two points. Q. E. D. If two circles touch each other internally, the straight line which joins their centres being produced shall pass through the point of contact. A Let the two circles ABC, ADE, touch each other internally in the point a, and let F be the centre of the circle ABC, and G the centre of the circle ADE: The straight line which joins the centres F, G, being produced, passes through the point A. For, if not, let it fall otherwise, if possible, as FGDH, and join AF, AG: And because a G, GF are greater (1. 20.) than FA, that is, than FH, for FA is equal H D E to FH, both being from the same centre; take away the common part FG; therefore the remainder AG is greater than the remainder GH: But AG is equal to GD; therefore GD is greater than GH, the less than the greater, which is impossible. Therefore the straight line which joins the points F, G cannot fall otherwise than upon the point A, that is, it must pass through it. Therefore, if two circles, &c. Q. E. D. PROP. XII. THEOR. If two circles touch each other externally, the straight line which joins their centres shall pass through the point of contact. Let the two circles ABC, ADE touch each other externally in the point A and let F be the centre of the circle ABC, and G the centre of ADE: The straight line which joins the points F, G shall pass through the point of contact a. B E For, if not, let it pass otherwise, if possible, as FCDG, and join FA, AG: And because F is the centre of the circle ABC, AF is equal to FC: Also because G is the centre of the circle ADE, AG is equal to GD: Therefore FA, AG are equal to FC, DG; wherefore the whole F A FG is greater than FA, AG; But it is also less (1. 20.); which is impossible: Therefore the straight line which joins the points F, G shall not pass otherwise than through the point of contact A; that is it must pass through it. Therefore, if two circles, &c. Q. E. D. PROP. XIII. THEOR. One circle cannot touch another in more points than one, whether it touches it on the inside or outside. For, if it be possible, let the circle EBF touch the circle ABC in more points than one, and first on the inside, in the points B, D; join BD, and draw (1. 10, 11.) GH bisecting BD at right angles: Therefore, because the points B, D are in the circumference of each of the circles, the straight line BD falls within each (III. 2.) of them: And their centres are (II. 1. Cor.) in the straight line GH which bisects BD at right angles; Therefore GH passes through the point of contact (III. 11.); but it does not pass through it, because the points B, D are without the straight line GH, which is absurd: Therefore one circle cannot touch another on the inside in more points than one. K Nor can two circles touch one another on the outside in more than one point: For, if it be possible, let the circle ACK touch the circle ABC in the points A, C, and join AC: Therefore, because the two points A, C are in the circumference of the circle ACK, the straight line AC which joins them shall fall within (III. 2.) the circle ACK: And the circle ACK is without the circle ABC; and therefore the straight line AC is without this last circle; but, because the points A, C are in B the circumference of the circle A ABC, the straight line AC must be within (111. 2. ̧ the same circle, which is absurd: Therefore one circle cannot touch another on the outside in more than one point: And it has been shown, that they cannot touch on the inside in more points than one: Therefore, one circle, &c. Q. E. D. Equal straight lines in a circle are equally distant from the centre; and those which are equally distant from the centre, are equal to one another. Let the straight lines AB, CD, in the circle ABDC, be equal to one another; they are equally distant from the centre. Take E the centre of the circle ABDC, and from it draw EF, EG perpendiculars to AB, CD: Then, because the straight line EF passing through the |