but a straight line drawn from the extremity of a diameter, at right angles to it, touches the circle: (11. 16. Cor.) therefore AB touches the circle; and it is drawn from the given point A. But if the given point be in the circumference of the circle, draw DE to the centre E, and DF at right angles to DE: PROPOSITION XVIII. THEOREM. If a straight line touches 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; For, if it be not, from the point F draw FBG perpendicular to DE. and to the greater angle the greater side is opposite: (1. 19.) but FC is equal to FB; (1. def. 15.) the less than the greater, which is impossible: Q.E.D. PROPOSITION XIX. THEOREM. If a straight line touches 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, 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 point of contact, therefore FC is perpendicular to DE; (III. 18.) therefore FCE is a right angle: but ACE is also a right angle; (hyp.) therefore the angle FCE is equal to the angle ACE, (ax. 1.) In the same manner it may be shewn, that no other point which is not in CA, is the centre; PROPOSITION XX, 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 part of the circumference. Let ABC be a circle, and BEC an angle at the centre, and BAC an angle at the circumference, which have the same circumference BC for their base. Then the angle BEC shall be double of the angle BAC. A First, let the B F Join AE, and produce it to F. centre of the circle be within the angle BAC. therefore the angle EAB is equal to the angle EBA; (1.5.) therefore the angles EAB, EBA are double of the angle EAB: but the angle BEF is equal to the angles EAB, EBA; (1. 32.) therefore also the angle BEF is double of the angle EAB: for the same reason, the angle FEC is double of the angle EAC: therefore the whole angle BEC is double of the whole angle BAC. Again, let the centre of the circle be without the angle BAC. E A It may be demonstrated, as in the first case, that the angle FEC is double of the angle FAC, and that FEB, a part of the first, is double of FAB, a part of the other; therefore the remaining angle BEC is double of the remaining angle BAC. Therefore the angle at the centre, &c. Q. E. D. PROPOSITION XXI. THEOREM. 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. Then the angles BAD, BED shall be equal to one another. First, let the segment BAED be greater than a semicircle. E B C Take F, the centre of the circle ABCD, (111. 1.) and join BF, FD. And because the angle BFD is at the centre, and the angle BAD at the circumference, and that they have the same part of the circumference, viz. BCD for their base; therefore the angle BFD is double of the angle BAD: (111. 20.) for the same reason the angle BFD is double of the angle BED: therefore the angle BAD is equal to the angle BED. (ax. 7.) Next, let the segment BAED be not greater than a semicircle. 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. (ax. 2.) Wherefore the angles in the same segment, &c. Q.E. D. PROPOSITION XXII. THEOREM. The opposite angles of any quadrilateral figure inscribed in a circle, are together equal to two right angles. Let ABCD be a quadrilateral figure in the circle ABCD. Then any two of its opposite angles shall together be equal to two right angles. B Join AC, BD. And because the three angles of every triangle are equal to two right angles, (1. 32.) the three angles of the triangle CAB, viz. the angles CAB, ABC, but the angle CAB is equal to the angle CDB, (III. 21.) and the angle ACB is equal to the angle ADB, to each of these equals add the angle ABC; therefore the three angles ABC, CAB, BCA are equal to the two angles ABC, ADC: (ax. 2.) but ABC, CAB, BCA, are equal to two right angles; therefore also the angles ABC, ADC are equal to two right angles. (ax. 1.) In the same manner, the angles BAD, DCB, may be shewn to be equal to two right angles. Therefore, the opposite angles, &c. Q.E.D. PROPOSITION XXIII. THEOREM. 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, upon the same straight line AB, and upon the same side of it, let there be two similar segments of circles, ACB, ADB, not coinciding with one another. D Then, because the circle ACB cuts the circle ADB in the two points A, B, they cannot cut one another in any other point: (III. 10.) therefore one of the segments must 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, (hyp.) and that similar segments of circles contain equal angles; (III. def. 11.) therefore the angle ACB is equal to the angle ADB, the exterior to the interior, which is impossible. (1. 16.) Therefore, there cannot be two similar segments of circles same side of the same line, which do not coincide. Q. E.D. upon the PROPOSITION XXIV. THEOREM. Similar segments of circles upon equal straight lines are equal to one another. Let AEB, CFD be similar segments of circles upon the equal straight lines AB, CD. Then the segment AEB shall be equal to the segment CFD. For if the segment AEB be applied to the segment CFD, so that the point A may be on C, and the straight line AB upon CD, then the point B shall coincide with the point D, because AB is equal to CD: therefore, the straight line AB coinciding with CD, the segment AEB must coincide with the segment CFD, (111. 23.) and therefore is equal to it. (ax. 8.) Wherefore similar segments, &c. Q. E.D. PROPOSITION XXV. PROBLEM. A segment of a circle being given, to describe the circle of which it is the segment. Let ABC be the given segment of a circle. It is required to describe the circle of which it is the segment. Bisect AC in D, (1. 10.) and from the point D draw DB at right angles to AC, (1. 11.) and join AB. First, let the angles ABD, BAD be equal to one another: B C then the straight line BD is equal to DA, (1.6.) and therefore, to DC; and because the three straight lines DA, DB, DC are all equal, therefore D is the centre of the circle. (III. 9.) From the centre D, at the distance of any of the three, DA, DB, DC, describe a circle ; this shall pass through the other points; and the circle of which ABC is a segment is described: and because the centre D is in AC, the segment ABC is a semicircle. But if the angles ABD, BAD are not equal to one another. At the point A, in the straight line AB, make the angle BAE equal to the angle ABD, (1.23.) and produce BD, if necessary, to meet AE in E, and join EC. And because the angle ABE is equal to the angle BAE, therefore the straight line BE is equal to EA: (1.6.) and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE, are equal to the two CD, DE, each to each; and the angle ADE is equal to the angle CDE, |