and join DB : And bec?use 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 (4. 1.) 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. Proposition IX. Theorene. 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 produce it to the circumference in F, G; then FG is a diameter of the circle ABC : And because in FG, the diameter of the circle ABC, there is taken the point D, DE which is not the centre, DG shall be the greatest F G line from it to the circumference, and DC greater c (7. 3.) than DB, and DB than DA: but they are B likewise equal, 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. Proposition X. Theorem. One circumference of a circle cannot cut another in more than two points. If it be possible, let the circumference FAB cut the circumference DEF in more than two points, B U D viz. in B, G, F; take the centre K of the circle K ABC, and join KB, KG, KF: and because within the circle DEF there is taken the point K, from which to the circumference DEF fal more than G two equal straight lines KB, KG, KF, the point K is (9. 3.) the centre of the circle DEF: But K is also the oentre of the circle ABC; therefore the same point is the centre of two circles that cut one another, which is impossible. (5. 3.) Therefore one circumference of a circle cannot cut another in more thai. two points. Q. E. D. Proposition XI. Theorem. If two circles touch each other internally, the straight line which joins their centres being produced, shall pass through the point of contact. 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. H E B 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. Proposition XII, Theorem. 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. 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 : where E fore the whole FG is greater than FA, AG: But it is also less ; (20. 1.) which is impossible : Therefore the straight line which joins G 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. Proposition XIII. Theorem. One circle cannot touch another in more points than one, whether it touches it on the ivside 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, H B D; join BD, and draw (10.11. A E A 1.) GH bisecting BD at right E BI angles : Therefore, because the H points B, D are in the circumference of each of the circles, the straight line BD falls within cach (2.3.) of them : And their B A הן B H B centres are (Cor. 1. 3.) in the straight line GH which bisects BD at right angles : Therefore GH passes through the point of contact; (11.3.) but it does not pass through it, because ihe 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. 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, cause the two points A, C, are in the circumference of the circle ACK, the straight line AC which joins them shall tall within (2. 3.) 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 the circumference of the circle ABC, the straight line AC must be within (2. 3.) the same circle, which is absurd : therefore one circle cannot touch another on the outside in more than one point : And it has been shewn, that they cannot touch on the inside in more points than one. Therefore, one circle, &c. Q. E. D. Proposition XIV. Theorem. 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 centre, cuts the straight line AB, which does not pass through the centre at right angles, it also bisects (3. 3.) it: Wherefore AF is equal to FB, and AB double of AF. For the same reason CD is double of CG, and AB is equal to CD: Therefore AF is equal to CG. And because AE is equal to EC, the square of AE is equal to the square of EC : But the squares of AF, FE are equal (47. 1.) to the square of AE, because the angle AFE is a right angle; and for the like reason, the squares of EG, GC, are equal to the square of EC: B Therefore the squares,of AF, FE are equal to the squares of CG, GE, of which the square of AF is equal to the square of CG, because AF is equal to CG ; therefore the remaining square of FE is equal to the remaining square of EG, and the straight line FE is therefore equal to EG : But straight lines in a circle are said to be equally distant from the centre, when the perpendiculars drawn to them from the centre are equal: (4. Def. 3.) Therefore AB, CD are equally distant from the centre. Next, if the straight lines AB, CD be equally distant from the centre, that is, if FE be equal to EG, AB is equal to CD: For the same construction being made, it may, as before, be demonstrated, AB E that AB is double of AF, and CD double of CG, and that the squares of EF, FA are equal to the squares of EG, GC ; of which the square of FE is equal to the square of EG, because FE is equal to EG; therefore the remaining square of AF is equal to the remaining square of CG; and the straight line AF is therefore equal to CG: And AB is double of AF, and CD double of CG: wherefore AB is equal to CD. Therefore equal straight lines, &c. Q. E. D. Proposition XV. Theorem. The diameter is the greatest straight line in a circle : and, of all others, that which is nearer to the centre is always greater than one more remote ; and the greater is nearer to the centre ihan the less. Let ABCD be a circle of which the diameter is AD, and the centre E; and let BC be nearer to the centre than FG; AD is greater than any straight line BC which is not a diameter, and BC greater than FG. From the centre draw EH, EK perpendiculars E H And, because BC is nearer to the centre than FG, EH is less (5. Def. 3.) than EK : But, as was demonstrated in the preceding, BC is double of BH, and FG double of FK, and the squares of EH, HB are equal to the squares of EK, KF, of which the square of EH is less than the square of EK, because EH is less than EK; therefore the square of BH is greater than the square of FK, and the straight line BH greater than FK; and therefore BC is greater than FG. Next, let BC be greater than FG; BC is nearer to the centre than FG, that is, the same construction being made, EH is less than EK : Because BC is greater than FG, BH likewise is greater than FK: And the squares of BH, HE are equal to the squares of FK, KE, of which the square of BH is greater than the square of FK, because BH is greater than FK ; therefore the square of EH is less than the square of EK, and the straight line Ek less than EK. Wherefore the diameter, &c. Q. E. D. Proposition. XVI. Theorem. The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle, and no straight line can be drawn between that straight line and the circumference from the extremity, so as not to cut the circle; or, which is the same thing, no straight line can make so great an acute angle with the diameter at its extremity, or so small an angle with the straight line which is at ļight angles to it, as not to cut the circie. Let ABC be a circle, the centre of which is D, and the diameter AB: The straight line drawn at right angles to A B from its extremity A, shall fall without the circle. P For, if it does not, let it fall, if possible, within the circle, as AC, and draw DC to the point C where it meets the circumference : And because DA is equal to DC, the angle DAC is equal (5.1.) B В to the angle ACD; but DAC is a right angle, therefore ACD is a right angle, and the angles DAC, ACD are therefore equal to two right angles ; which is impossible. (17. 1.) Therefore the straight line drawn from A at right angles to BA does not fall within the circle: In the same manner, it may be demon'strated, that it does not fall upon the circumference; therefore it must fall without the circle, as AE. And between the straight line AE and the circumference no straight line can be drawn from the point A, which does not cut the circle: For, if possible, let FA be between them, and from the point D draw (12. 1.) DG perpendicular to FA, and let it meet the circumference in H: And because AGD is a right angle, and DAG less (19. 1.) than a right angle: DA is greater (19. 1.) than DG: But DA is equal to DH ; therefore DH is greater than DG, the less to the greater, which is impossible : Therefore no F E straight line can be drawn from the point A between AE and the circuinference, which does not cut the HX circle, or, which amounts to the same thing, howe ever great an acute angle a straight line makes with the diameter at the point A, or, however small an angle it makes with AE, the circumference passes between that straight line and the perpendicular AE. "And this is all that is to be understood, when, in the Greek text and translations from it, the angle of the semicircle is said to be greater than any acute rectilineal angle, and the remaining angle less than any rectilineal angle.' Cor. From this it is manifest, that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it, touches the circle ; and that it touches it only in one point, because, if it did meet the circle in two, it would fall within it. (2. 3.) • Also it is evident that there can be but one straight line which touches tha circle in the same point. Proposition. XVII. Problem. To draw a straight line from a given point, either without or in the circumference, which shall touch a given circle. First, let A be a given point without the given circle BCD; it is required to draw a straight line from A which shall touch the circle. Find (1. 3.) the centre E of the circle, and join AE; and from the centre E, at the distance EA, describe the circle AFG; from the point D draw (11. 1.) DP at right angles to EA, and join EBF, AB. AB touches the circle BCD |