angles in the other, each to each, and the side Ed, which is opposite to one of the equal angles in each, is common; therefore (I. 26.) AD is equal to DB. If a straight line, therefore, &c. Cor. 1. Hence, in an isosceles triangle, a straight line drawn from the vertex bisecting the base, is perpendicular to it; and a straight line drawn from the vertex perpendicular to the base, bisects it. Cor. 2. Let the straight line AB cut the concentric circles Then (III. 3.) AH is equal to HB, and DH Cor. 3. Any number of parallel chords in a circle are all bisected by a diameter perpendicular to them. F A B D H E PROP. IV. THEOR. Two chords of a circle, which are not both diameters, cannot bisect each other. Let ABCD be a circle, and let AC, BD, two chords in it, which are not both diameters, cut one another in the point E: they do not bisect one another. For, if it be possible, let AE be equal to EC, and BE to ED. If one of the lines pass through the centre, it is plain that it cannot be bisected by the other which does not pass through the centre. But if neither of them pass through the centre, take (III. 1.) F the centre, and join EF; and because FE, a straight line drawn through the centre, bisects the chord AC, which does not pass through the centre, it cuts it (III.3.) at right angles; wherefore FEA is a right angle. Again, because FE bisects BD, which does not pass through the centre, FEB is also (III. 3.) a right angle ; and FEA was shown to be a right angle: therefore FEA is equal to FEB, which (1. ax. 9.) is impossible. Therefore AC, BD do not bisect one another. Wherefore two chords, &c. A с PROP. V. THEOR. If two circles cut one another, they have not the same centre. Let the two circles ABC, DBE cut one another in the point B; they have not the same centre. E For, if it be possible, let F be their centre : join FB, and draw any straight line FCE meeting the circumferences in C and E: then (I. def. 30.) because F is the centre of the circle ABC, FB is equal to FC; and, because F is the centre of the circle DBE, FB is equal to FE. But FB was shown to be equal to FC: therefore FE is equal (I. ax. 1.) to FC, which (I. ax. 9.) is impossible. Therefore F is not the centre of the circles ABC, DBE; wherefore, if two circles, &c. A PROP. VI. THEOR. If one circle touch another internally, they have not the same centre. Let the circle ABC be touched internally by the circle DEC in the point C: they have not the same centre. For, if they can, let it be F; join FC, and draw any straight line FEB meeting the circumferences in E and B. Then, (I. def. 30.) because F is the centre of the circle ABC, FC is equal to FB; and, because F is the centre of the circle CDE, FC is equal to FE. But FC was shown to be equal to FB: therefore FE is equal to FB, which (I. ax. 9.) is impossible ; wherefore F is not the centre of the circles ABC, CDE. Therefore, if one circle, &c. B D PROP. VII. THEOR. If from any point within a circle, which is not the centre, straight lines be drawn to the circumference; (1.) the greatest is that which passes through the centre, and (2.) the continuation of that line to the circumfe. rence, in the opposite direction, is the least : (3.) of others, one nearer to the line passing through the centre is greater than one more remote : and (4.) from the same point there can be drawn only two equal straight lines, one upon each side of either the longest or shortest line, and making equal angles with that line. Let ABCD be a circle, E its centre, and AD a diameter, in which let any point F be taken which is not the centre : of all the straight lines FA, FB, FC, &c., that can be drawn from F to the circumference, FA is the greatest, and FD the least : and of the others, FB is greater than FC. E K H Н 1. Join BE, CE. Then (I. 20.) BE, EF are greater than BF; but AE is equal to EB; therefore AF, that is, AE, EF, is greater than BF. 2. Because CF, FE are greater (I. 20.) than EC, and EC is equal to ED; CF, FE are greater than ED. Take away the common part FE, and (I. ax. 5.) the remainder CF is greater than the remainder FD. 3. Again, because BE is equal to CE, and FE common to the triangles BEF, CEF; but the angle BEF is greater than CEF; therefore (I. 24.) the base BF is greater than the base CF. 4. Make (I. 23.) the angle FEK equal to FEC, and join Fú. Then, because CE is equal to HE, EF common to the two triangles CEF, HEF, and the angle CEF equal to the angle HEF; therefore (I. 4.) the base FC is equal to the base FH, and the angle EFC to the angle EFH. But, besides FH, no other straight line can be drawn from F to the circumference equal to FC. For, if there can, let it be FK; and because FK is equal to FC, and FC to FH, FK is equal to FH; that is, a line nearer to that which passes through the centre, is equal to one which is more remote ; which is impossible. Therefore, if from any point, &c. D PROP VIII. THEOR. IF from any point without a circle straight lines be drawn to the circumference; (1.) of those which fall upon the concave part of the circumference, the greatest is that which passes through the centre ; and (2.) of the rest, one nearer to the greatest is greater than one more remote. (3.) But of those which fall upon the convex part, the least is that which, when produced, passes through the centre; and (4.) of the rest, one nearer to the least is less than one more remote. And (5.) only two equal straight lines can be drawn from the point to either part of the circumference, one upon each side of the line passing through the centre, and making equal angles with it. Let ABF be a circle, M its centre, and D any point without it, from which let the straight lines DA, DE, DF be drawn to the circunference. Of those which fall upon the concave part of the circumference AEF, the greatest is DMA, which passes through the centre; and a line DE nearer to it is greater than D DF, one more remote. But of those which fall upon the convex circumference LKG, the least is DG, the external part of DMA; and a line DK nearer to it is less than DL, one more remote. 1. Join ME, MF, MK, ML; and because MA is equal to ME, add MD to each ; therefore AD is equal to EM, MD: but (I. 20.) EM, MD are greater than ED; therefore also AD is greater than ED. 2. Because ME is equal to MF, and MD common to the triangles EMD, FMD, but the angle EMD is greater than FMD; therefore (I. 24.) the base ED is greater than the base FD. 3. Because (I. 20.) MK, KD are greater than MD, and MK is equal to MG, the remainder KD is greater (1. ax. 5.) than the remainder GD; that is, GÒ is less than KD. 4. Because MK is equal to ML, and MD common to the triangles KMD, LMD, but the angle DMK less than DML; therefore the base DK is less (I. 24.) than the base DL. 5. Make (I. 23.) the angle DMB equal to DMK, and join DB. Then, because MK is equal to MB, MD common to the triangles KMD, BMD, and the angle KMD equal to BMD; therefore (I. 4.) the base DK is equal to the base DB, and the angle MDK to the angle MDB. 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, a line nearer to the least equal to one more remote, which is impossible. If from any point, therefore, &c. F M A PROP. IX. THEOR. A point in a circle, from which more than two equal straight lines fall to the circumference, is the centre. For, from any point which is not the centre, only two equal straight lines (III. 7.) can be drawn to the circumference, and therefore a point from which more than two equal straight lines can be drawn to the circumference, cannot be any other than the Therefore a point, &c. * centre. * Dr. Simson's proof of this proposition is essentially wrong; as, according to it, a point within the circle from which two equal straight lines could be drawn to the circumference would appear to be the centre; and therefore any point whatever within the circle would, from it, appear to be the centre, since (111. 7.) from any point within the circle two equal straight lines can be drawn to the circumference. PROP. X. THEOR. B One circle cannot cut another in more than two points. If it be possible, let the circle FAB cut the circle DEF in more than two points, viz., in B, G, F: take (III. 1.) the centre H, of the circle ABC, and join HB, HG, HF. Then, because within the circle DEF there is taken the point H, from which there fall to the circumference DEF more than two equal straight lines HB, HG, HF, H is (III. 9.) the centre of the circle DEF; but H is also the centre of the circle ABC; therefore the same point is the centre of two circles that cut one another, which (III. 5.) is impossible. Therefore one circle, &c. PROP. XI. THEOR. If one circle touch another internally in any point, the straight line which joins their centres, being produced, passes through that point. Let the circle ADE touch the circle ABC internally in the point A, and let F be the centre of ABC, and G the centre of ADE: the straight line which joins F and G, being produced, passes through A. For, if not, let it fall otherwise, if possible, as FGDH, and join AF, AG. Then, because (I. 20.) AG, GF are greater than FA, that is, than FH, for FA is equal to FH, both being radii of ABC; take away the common part FG; therefore (I. ax. 5.) the remainder AG is greater than the remainder GH; but AG is equal (1. def. 30.) to GD; therefore GD is greater than GH, which (I. ax. 9.) is impossible. The straight line, therefore, which joins the points F, G cannot fall otherwise than upon A, that is, it must pass through it. Therefore, if one circle, &c. GF B * Dr. Simson's enunciation is as follows: “If two circles touch each other internally, the straight line which joins their centres, being produced, shall pass through the point of contact.' This is liable to two objections. The circles do not “touch each other internally," the interior one being touched externally. The expression, " the point of contact," seems to assume that there is but one point of contact, which is proved in a following proposition, the 13th. Mr. Walker proposes to say a point of contact ; but this would seem to imply that there may be more points of contact than one. In the enunciation here given, these objections are obviated. A similar change is made in one part of the enunciation of the 12th proposition, and also of the 6th. |