point A, and let F be the centre of the circle ABC, and G the centre of the circle ADE ; the straight line wbich joins the centres F, G, being produced, passes through the point A. Н. For, if not, let it fall otherwise, if D possible, as FGDH, and join AF, AG : And because AG, GF are greater (20. 1.) than FA, that is, than FH, for FA is equal to FH, being radii of the same circle ; take away the common part FG, and the E remainder AG is greater than the remainder GH. But AĞ is equal to GD, there. fore GD is greater than GH; and it is B also less, which is impossible. Therefore the straight line which joins the points F and G cannot fall otherwise than on the point A ; that i, it must pass through A. Therefore, if two circles, &c. Q. E. D. c PROP. XII. THEOR. If two circles touch each other externally, the straight line which joins their centres will pass through the point of contact. Let the two circles ABC, ADE, touch each other externally ia 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. For, if not, let it pass otherwise, if possible, as FCDG, and join PA, AG : and because F is the centre of the circle ABC, AF is equal to FC: Also because G is the centre of the B circle, ADE, AG is E equal to GD.' Therefore FA,AG are equal where A fore the whole FG is greater than FA, AG; CAD 75 G but it is also less (20. 1.), which is impossible : Therefore the straight line which joins the points F, G cannot pass otherwise than through the point of contact A ; that is, it passes through A. 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 to FC, DG ; 2 more points than one, and first on the inside, in the points B, D ; join BD, and draw (10. 11. 1.) GH, bisecting BD at right angles : Therefore because the points B, D are in the circumference of each of the H В. D BV D : o circles, the straight line BD falls within each (2. 3.) of them: and therefore their 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 the points B, D are without the straight line GH, which is absurd : Therefore one circle cannot touch another in the ioside 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, because the two points A,C are K in the circumference of the circle ACK, the straight line AC which joins them shall fall within the circle ACK: And the circle ACK is without the circle ABC; and therefore the straight line AC is also without ABC ; 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 a circle cannot touch apotber on the outside in more than one B point: and it bas been shewn, that a circle cannot touch another on the inside in more than one point. Therefore, one circle, &c. Q E. D. QE a a PROP. XIV. THEOR. 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. G Take E the centre of the circle ABDC, and from it draw EF, EG, perpendiculars to AB, CD; join AE and EC. 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 ; Where- Α. E fore AF is equal to FB, and AB double of AF. For the same reason, CD is double of CG : But AB is equal to CD; E therefore AF is equal to CG : And be. cause AE is equal to EC, the square of AE is equal to the square of EC: Now B D the squares of AF, FÈ 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 : 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 EF 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 (3. 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 game construction being made it may, as before, be demonstrated, 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 : But AB is double of AF, and CD double of CG; wherefore AB is equal to CD. Therefore equal straight lines, &c. Q. E. D. PROP. XV. THEOR. 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 than the less. Let ABCD be a circle, of which the di А. ameter is AD, and the centre E; and let B 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 to BC, FG, and join EB, EC, EF; ED and because AE is equal to EB, and ED to EC, AD is equal to EB, EC: But EB, EC are greater (20. 1.) than BC; wherefore, also AD is greater than BC. D GD a Н And, because BC is nearer to the centre than FG, EH is less (4. 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 ; there. fore the square of BH is greaer 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 KF ; but 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 EŇ less than EK. Wherefore, the diameter, &c. Q. E. D. 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 of the diameter, so as not to cut the circle. Let ABC be a circle, the centre of which is D, and the diameter AB : and let AE be drawn from A perpendicular to AB, AE shall fall witbout the circle. lo AE take any point F, join DF, and let DF meet the circle in C. Because DAF is a right angle, it is greater than the angle AFD (32. 1.); E but the greater angle of any triangle is subtended by the greater side (19. 1.), therefore DF is greater than DA; T Dow DA is equal to DC, therefore DF is greater than DC, and the point F is therefore without the circle. B And F is any point whatever in the D A line AE, therefore AE falls without the circle. Again, between the straight line AE and the circumference, no straight line can be drawn from the point A, which does not cut the circle. Let AG be drawn, in the angle DAE, from D draw DH at right angles to AG; E and because the angle DHA is a right angle, and the angle DAH less than a right angle, the side DH of the triangle DAH is less than the side DA (19. 1.). The point H, therefore, is within the H) circle, and therefore the straight line AG cuts the circle. B D 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 the circle in the same point. PROP. XVII. PROB. To draw a straight line from a given point, either without or in the cir cumference, 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.) DF at right angles to EA, join EBF, and draw AB. AB touches the circle BCD. Because E is the centre of the circles BCD, AFG, EA is equal to EF, and ED to EB; therefore the two sidės AE, EB are equal to the two FE, ED, and they contain the angle at E common to the two triangles AEB, E FED; Therefore the base DF is equal B to the base AB, and the triangle FED to the triangle AEB, and the other angles to the other angles (4. 1.): Therefore the angle EBA is equal to the angle EDF; but EDF is a right angle, wherefore EBA is a right angle ; and EB is a drawn from the centre: but a straight line drawn from the extremity of a diameter, at right angles to it, touches the circle (Cor. 16. 3.): Therefore AB touches the circle ; and is drawn from the given point A. Which was to be done. But if the given point be in the circumference of the circle, as the point D, draw DE to the centre E, and DF at right angles to DE; DF touches the circle (Cor. 16. 3.) |