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 (III. 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 FÈ be equal to EG, AB is equal to CD; for the same construction being made, it may, as before, be demonstrated, that AB is double AF, and CD double 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 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 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. A B E H DC From the centre draw EH, EK perpendiculars to BC, FG, and join EB, EC, EF; and K because AE is equal to EB, and ED to EC, AD is equal to EB, EC: but EB, EC are greater (I. 20) than BC; wherefore also AD is greater than BC. And because BC is nearer to the centre than FG, EH is less (IV. Def. 3) than EK; but, as was demonstrated in the preceding, BC is double BH, and FG double 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 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 EH less than EK. Wherefore, the diameter, &c. Q. E. D. PROP. XVI. THEOR. 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 without the circle. B F In 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 (I. 32); but the greater angle of any triangle is subtended by the greater side (I. 19), therefore DF is greater than DA; now DA is equal to DC, therefore DF is greater than DC, and the point F is therefore without the circle; and F is any point whatever in the 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; and because the angle DHA is a right angle, und the angle DAH is less than a right angle, the side DH of the triangle DAH is less than the side DA (I. 19); the point H, therefore, is within the circle, and therefore the straight line AG cuts the circle. B G E 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 (III. 2). 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 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 (III. 1) the centre E of the circle, and join AE; and from the centre E, at the distance EA, describe the circle AFG; from G F B the point D draw (I. 11) 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 sides AE, EB are equal to the two FE, ED, and they contain the angle at E common to the two triangles AEB, FED; therefore the base DF is equal to the base AB, and the triangle FED to the triangle AEB, and the other angles to the other angles (I. 4); 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 drawn from the centre; but a straight line drawn from the extremity of a diameter at right angles to it, touches the circle (III. 16, Cor.); 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 (III. 16, Cor.) PROP. XVIII. THEOR. If a straight line touch a circle, the straight line drawn from the centre to the point of contact is perpendicular to the line touching the circle. A Let the straight line DE touch the circle ABC in the point C take the centre F, and draw the straight line FC; FC is perpendicular to DE. For, if it be not, from the point F draw FBG perpendicular to DE; and because FGC is a right angle, GCF must be (I. 17) an acute angle; and to the greater angle the greater (I. 19) side is opposite; therefore FC is greater than FG: but FC is equal to FB; therefore FB is greater that FG, the less than D the greater, which is impossible; wherefore FG is not perpendicular to DE. In the same manner it may be shown that no other line than FC can be perpendicular to DE; FC is therefore perpendicular to DE. Therefore, if a straight line, &c. Q. E. D. PROP. XIX. THEOR. B C G E 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 is in that line. Let the straight line DE touch the circle ABC in C, and from Clet CA be drawn at right angles to DE; the centre of the circle is in CA. For, if not, let F be the centre, if possible, and join CF; because DE touches the circle ABC, and FC is drawn from the centre to the B D A point of contact, FC is perpendicular (III. 18) to DE; therefore FCE is a right angle; but ACE is also a right angle; therefore the angle FCE is equal to the angle ACE, the less to the greater, which is impossible; wherefore 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; that is, the centre is in CA. Therefore, if a straight line, &c. Q. E. D. The angle at the centre of a circle is double the angle at the circumference, upon the same base, that is, upon the same part of the circumference.* Let ABC be a circle, and BDC an angle at the centre, and BAC an angle at the circumference, which have the same circumference BC for their base; the angle BDC is double the angle BAC. First, Let D, the centre of the circle, be within the angle BAC, and join AD, and produce it to E. Because DA is equal to DB, the angle DAB is equal (I. 5) to the angle DBA ; therefore the angles DAB, DBA taken together B are double the angle DAB: but the angle BDE is equal (I. 32) to the angles DAB, DBA ; E A therefore also the angle BDE is double the angle DAB: for the same reason, the angle EDC is double the angle DAC; therefore the whole angle BDC is double the whole angle BAC. Again, Let D, the centre of the circle, be without the angle BAC, and join AD, and produce it to E. It may be demonstrated, as in the first case, that the angle EDC is double the angle DAC, and that EDB, a part E of the first, is double DAB, a part of the other; therefore the remaining angle BDC is double the remaining angle BAC. Therefore, the angle at the centre, &c. Q. E. D. B A The angles in the same segment of a circle are equal to one another.* Let ABCD be a circle, and BAD, BED angles in the same segment BAED; the angles BAD, BED are equal to one another. E D Take F, the centre of the circle ABCD; and, first, let the segment BAED be greater than a semicircle, and join BF, FD; and because the angle BFD is at the centre, and the B angle BAD at the circumference, both having the same part of the circumference, viz., BCD, for their base, therefore the angle BFD is double (III. 20) the * Sea Notes. C B E angle BAD. For the same reason, the angle BFD is double the angle BED; therefore the angle BAD is equal to the angle BED. But, if the segment BAED be not greater than a semicircle, let BAD, BED be angles in it; these also are equal to one another. Draw AF to the centre, and produce it to C, and join CE; therefore the segment BADC is greater than a semicircle; and the angles in it, BAC, BEC, are equal, by the first case. For the same reason, because CBED is greater than a semicircle, the angles CAD, CED are equal; therefore the whole angle BAD is equal to the whole angle BED. Wherefore, the angles in the same segment, &c. Q. E. D. F C The opposite angles of any quadrilateral figure described in a circle are together equal to two right angles. Let ABCD be a quadrilateral figure in the circle ABCD; any two of its opposite angles are together equal to two right angles. Join AC, BD; the angle CAB is equal (III. 21) to the angle CDB, because they are in the same segment BADC, and the angle ACB is equal to the angle ADB, because they are in the same segment ADCB; therefore the whole angle ADC is equal to the angle CAB, ACB. To each of these equals add the angle ABC; and A the angles ABC, ADC are equal to the angles ABC, CAB, BCA: but ABC, CAB, BCA B are equal to two right angles (I. 32); therefore also the angles ABC, ADC are equal to two right angles. In the same manner the angles BAD, DCB may be shown to be equal to two right angles. Therefore, the opposite angles, &c. Q. E. D. PROP. XXIII. THEOR. Upon the same straight line, and upon the same side of it, there cannot be two similar segments of circles not coinciding with one another. If it be possible, let the two similar segments of circles, viz., ACB, ADB be upon the same side of the same straight line AB, not coinciding with one another; then, because the circles ACB, ADB cut one another in the two points A, B, they cannot cut one another in any other point (III. 10); one of the segments must therefore fall within the other: let ACB fall within ADB, draw the straight line BCD, and join CA, DA; and because the segment ACB is similar to the segment ADB, and similar segments of circles contain (III. Def. 9) equal angles, the angle ACB is equal to the angle ADB, the exterior to the interior, which is impossible |