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. But BE, EC are greater than BC (I. 20); Therefore also AD is greater than BC. F B K H BE+ EC or AD > 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 on EH, HB are equal to the squares on EK, KF. 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 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; EHEK. ... since EH2+ HE? HB> FK. Take any And therefore BC is nearer to the centre than FG (III. def. 5). Therefore, the diameter, &c. Q.E.D. 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 АВ. The straight line AE shall fall without the circle. point Fin let DF meet the circle in C. ΛΕ, then Draw DH at right angles to HG, then DH < DA, and ... < DK. B D r 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 in AE, except the point A, is without the circle. Therefore AE falls without the circle. other point 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 right angles to AG, and meeting the circumference in K (I. 12). E A 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, Therefore the straight line AG cuts the circle. Q.E.D. 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 EF, and ED to EB; E EA, ED respectively = EF, EB Therefore the two sides AE, EB are equal to the two sides and E FE, ED, each to each; And the angle at E is common to the two triangles AEB, FED; 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. common; .. ABL = 4FDE a right angle. .. AB touches Therefore the angle ABE is a right angle (Ax. 1). But the straight line drawn at right angles to a diameter of a circle, from the extremity of it, touches the circle (III, 16, cor.); Therefore AB touches the circle, and it is drawn from the the circle, given point A. If not, suppose FG perpendi cular. Next, let the given point be in the circumference of the circle, at the point D. Draw DE to the centre E, and DF at right angles to DE; Therefore, from the given points A and D, straight lines, Proposition 18.-Theorem. If a straight line touch a circle, the straight line drawn from the centre to the point of contact shall be perpendicular to the line touching the circle. Let the straight line DE touch the circle ABC in the point C; take the centre F (III. 1), and draw the straight line FC. D Then must FC shall be perpendicular to DE. A C Ꮐ E PROOF.-Because FGC is a right angle (Hyp.), FCG is an acute angle (I. 17), and to the greater angle the greater side is opposite (I. 19); Therefore FC is greater than FG. But FC is equal to FB; therefore FB FB>FG. is greater than FG; the part greater than the whole, which is impossible. Therefore FG is not perpendicular to DE. In the same manner it may be shown that no other straight line from F is perpendicular to DE, but FC; therefore FC is perpendicular to DE. Therefore, if a straight line, &c. Q.E.D. Proposition 19.-Theorem. If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the centre of the circle shall be in that line. Let the straight line DE touch the circle ABC in C, and from C let CA be drawn at right angles to DE. The centre of the circle shall be in CA. CONSTRUCTION.-For, if not, if possible, let F be the centre, If not, and join CF. PROOF.-Because DE touches the circle ABC, and FC is drawn from the assumed centre to the point of contact, Therefore FC is perpendicular to DE (III. 18); Therefore FCE is a right angle. B But the angle ACE is also a right angle (Const.); take F the centre, out of the line. Therefore the angle FCE is equal to the angle ACE; the Then less to the greater, which is impossible. Therefore F is not the centre of the circle ABC. In the same manner it may be shown that no other point which is not in CA is the centre; therefore the centre is in CA. Therefore, if a straight line, &c. Q.E.D. Proposition 20.-Theorem. The angle at the centre of a circle is double of the angle at the circumference, upon the same base, that is, upon the same arc. Let ABC be a circle, and BEC an angle at the centre, and BAC an angle at the circumference, which have the same arc BC for their base. The angle BEC shall be double of the angle BAČ. CASE I.-First, let the centre E of the circle be within the angle BAC. CONSTRUCTION.Join AE, and produce it to the circum ference in F. LACE= ZFCE, being right angles. |