let it touch II. Next, let the circle ACK touch the circle ABC externally in the point A. Then ÅCK cannot touch ABC in any other point. CONSTRUCTION.—If it be possible, let ACK tauch ABC in If possible another point C. Join AC. in C also; PROOF.—Because the points A, C are K in the circumference of the circle ACK, the straight line AC must fall within the circle ACK (III. 2). But the circle ACK is without the circle ABC (Hyp.); Therefore the straight line AC is without the circle ABC. But because the two points A, C are in B ABC, the circumference of the circle ABC, the which is straight line AC falls within the circle ABC (III. 2); which absurd. is absurd. Therefore one circle cannot touch another externally in more points than one. And it has been shown that one circle cannot touch another internally in more points than one. Therefore, one circle, &c. Q.E.D. O then AC falls without the circle Proposition 14.—Theorem. Equal straight lines in a circle are equally distant from the centre; and, conversely, those which are equally distant from the centre tire equal to one another. Let the straight lines AB, CD, in the circle ABDC, be equal to one another. Then they shall be equally distant from the centre. CONSTRUCTION:—Take É," the centre of the circle ABDC (III. 1). From E draw EF, EG, perpendiculars to AB, CD (I. 12). Prook. Because the straight line EF, passing through he centre, cuts the straight line AB, which does not pass through the centre, at right angles, it also bisects it (III. 3). Therefore AF is equal to FB, and AB is double of AF. А F and D B в AF = CG, But AB is equal to CD (Hyp.); therefore AF is equal to CG (Ax. 7). And because AE is equal to CE, the square on AE is equal to the square on CE. But the squares on AF, FE are equal to the square on AE, because the angle AFE is a right angle (I. 47). For the like reason the squares on CG, GE are equal to the square on CE; Therefore the squares on AF, FE are equal to the squares on CG, GE (Ax. 1). But the square on AF is equal to the square on CG, because AF is equal to CG; Therefore the remaining square on FE is equal to the remaining square on GE (Ax. 3); ..EF=EG. And therefore the straight line EF is equal to the straight line 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 (III. Def. 4); Therefore AB, CD are equally distant from the centre. Conversely, let the straight lines AB, CD be equally distant from the centre, that is, let EF be equal to EG; Then AB shall be equal to CD. EF = EG, PROOF.—The same construction being made, it may be AF2+EF2 demonstrated, as before, that AB is double AF, and CD EGCG2 + double of CG, and that the squares on EF, FA are equal to the squares on EG, GC. But the square on EF is equal to the square on EG, because EF is equal to EG (Hyp:); Therefore the remaining square on FA is equal to the remaining square on GC (Ax. 3), And therefore the straight line AF is equal to the straight CG, &c. line CG. But AB was shown to be double of AF, and CD double of CG; Therefore AB is equal to CD (Ax. 6); Here and .. AF = B K H BE + EC or Proposition 15.-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 than the less. Let ABCD be a circle, of which AD is a diameter, and E the centre; and let BC be nearer to the centre than FG. Then AD shall be greater than any straight line CB, which is not a diameter; and BC shall be greater than FG. CONSTRUCTION.—From the centre E draw EH EK perpendiculars to BC, FG (I. 12), and join EB, EC, EF. Proof.—Because AE is equal to BE, and ED to EC, Therefore AD is equal to BE, EC. AD > BC, But BE, EC are greater than BC (I. 20); Therefore also AD is greater than BC. And because BC is nearer to the centre than FG and (Hyp.), EH is less than EK (III. Def. 5). But, as was demonstrated in the preceding proposition, BC is double of BH, and FG double of FK, and the squares EH2+ HD? on EH, HB are equal to the squares on EK, KF. KF2, But the square on EH is less than the square on EK, because EH is less than EK; Therefore the square on HB is greater than the square on And therefore BC is greater than FG. Then BC shall be nearer to the centre than FG, that is, the same construction being made, EH shall be less than EK. PROOF.—Because BC is greater than FG, BH is greater than FK. But the squares on BH, HE are equal to the squares on FK, KE; And the square on BH is greater than the square on FK, because BH is greater than FK; Therefore the square on HE is less than the square on KE, and the straight line EH less than EK; EH <EK. ., since EK? + HB> FK. And therefore BC is nearer to the centre than FG (III. def. 5). Therefore, the diameter, &c. Q.E.D. Take any Proposition 16.-Theorem. The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle; and a straight line, making an acute angle with the diameter at its extremity, cuts the circle. Let ABC be a circle, of which D is the centre, and AB a diameter, and AE a line drawn from A perpendicular to AB. The straight line AE shall fall without the circle. CONSTRUCTION.—In AE take any point F; join DF, and point F in let DF meet the circle in C. AE, PROOF.-Because DAF is a right angle, it is greater than the angle AFD (I. 17); Therefore DF is greater than DA (I. 19). But DA is equal to DC; therefore DF is greater than DC. Therefore the point F is without the circle. In the same manner it may be shown that any other point in AE, except the point A, is without the circle. Therefore AE falls without the circle. Again, let AG make with the diameter the angle DAG less than a right angle. The line AG shall fall within the circle, and hence cut it. CONSTRUCTION.–From D draw DH at at right right angles to AG, and meeting the cirangles to HG, then cumference in K (I. 12). DH<DA, PROOF.—Because DHA is a right angle, and DAH less than a right angle; Therefore the side DH is less than the side DA (I. 19). But DK is equal to DA; therefore DH is less than DK. Therefore the point H is within the circle, then Draw DH H) and .. DK. Bi D A Therefore the straight line AG cuts the circle. COROLLARY.–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 (III. Def. 2); and that it touches it only in one point, because if it did meet the circle in two points it would fall within it (III. 2). Also it is evident that there can be but one straight line which touches the circle in the same point. Proposition 17.—Problem. To draw a straight line from a given point, either without or in the circumference, which shall touch a given circle. First, let the given point A be without the given circle BCD. It is required to draw from A a straight line which shall touch the given circle. CONSTRUCTION.—Find the centre E of the circle (III. 1), and join AE. From the centre E, at the distance EA, describe the circle AFG. From the point D draw DF at right angles to EA (I. 11), and join EBF and AB. Then AB shall touch the circle BCD. PROOF.—Because E is the centre of the circles AFG, BCD, EA is equal to EA, ED reEF, and ED to EB; spectively Therefore the two sides AE, EB are equal to the two sides and E = EF, EB FE, ED, each to each; And the angle at E is common to the two triangles AEB, F B common; TED; Therefore the base AB is equal to the base FD, and the triangle AEB to the triangle FED, and the other angles to the other angles, each to each, to which the equal sides are opposite (I. 4); Therefore the angle ABE is equal to the angle FDE. a right But the angle FDE is a right angle (Const.); angle. .: LABL = 6 FDE |