a 32.1. 21. 3. given point, shall cut off a greater circumference than the two, which are more remote. 2. Given the segments of the base of a triangle made by a perpendicular drawn from the vertex and the vertical angle to construct the triangle. PROPOSITION XXII. The opposite angles of quadrilateral figures which are inscribed in a circle are equal to two right angles. Let ABDC be a circle, and ABDC a quadrilateral figure in it. Then any two opposite angles of it are equal to two right angles; join AD, BC. Therefore because the three angles of every triangle are equal to two right angles, the angles CAB, ABC, CBA, are equal to two right angles. But the n angle ABC is equal to the angle ADB, for they are in the same segment, ABDC. And the angle ACB will be equal to the angle ADB, because they are in the same segment, ABCD: therefore the whole angle BDC is equal to the angles: ABC, ACB. Take away the angle BAC, which is common the angles BAC, ABC, ACB, are equal to the, angles BAC, BDC. But the angles BAC, ABC, ACB, are equal to two right angles: wherefore also the angles BAC, BDC, are equal to two right angles. In like manner we can demonstrate that the angles ABD, ACD, are also equal to two right angles. Therefore the opposite angles, &c. Q. E. D. Deduction. If two opposite angles of any trapezium be equal to two right angles, the other two angles are equal to two right angles, and a circle may be described about it. PROPOSITION XXIII. THEOREM. Upon the same straight line, and on the same side of it, two similar segments of circles cannot be described which do not coincide with each other. For, if it be possible, on the same right line, AB, let two similar segments of circles, ACB, ADB, be de A Ba 10. 3. scribed, which do not coincide with each other. Let the circumferences ACB, ADB, meet one another at the points A, B, and they have no other points common except A, B. But between A, B, one will be interior, and the other exterior. In the interior take any point c, and join AC, which, produced, will meet the exterior in D, and join CB, BD. Therefore because ACB is a segment similar to the segment ADB, and segments of circles are similar which contain equal angles, the "Def. 10.3. angle ACB will be equal to the angle ADB, the exterior to the interior, which is impossible. Therefore upon the same right line, &c. Q. E. D. Deduction. Of unequal segments described upon the same base and towards the same parts, the circumference of that which contains the greater angle will be the interior. PROPOSITION XXIV. THEOREM. Similar segments of circles, upon equal right lines, are equal to one another. E For let the similar segments of circles AEB, CFD, be upon the equal right lines AB, CD. The segment AEB is equal to the segment CFD. For the segment AEB coinciding with the segment CFD, and the point A with the point c; but the right line AB, with the right line CD, A the point в shall also coincide with the point D, because AB is equal to CD; but AB coinciding with CD, the segment AEB shall coincide with the F B segment CFD. For if the right line AB coinciding with the right line CD, the segment AEB does not coincide with the segment CFD, it will fall either within or without it, which is impossible. Therefore the right line AB a 23.3. coinciding with the right line CD, the segment AEB cannot but coincide with the segment CFD, and as it coincides with it, it is consequently equal. Therefore similar segments, &c. Q. E. D. a 10. 1. b 10. 1. e 23. 1. d 6.1. e 4. 1. 19. 3. PROPOSITION XXV. PROBLEM. The segments of a circle being given to describe the circle of which it is a segment. Let ABC be a given segment of a circle. It is required to describe the circle of which ABC is a segment. Bisect Aca in D, and from the point D draw AC at right angles to DB, and join AB. Therefore the angle ABD is either greater than the angle BAD, or less, or equal to it. Let it be greater, and at the right line AB draw EC, and at the given point a in it, make the angle BAE equal to the angle ABD; but BD, AE, being produced, will meet one another in E, and join EC. Therefore because the angle ABE is equal to the angle BAE, the right line BEd will also be equal to AE, and DE is common: the two AD, DE, are equal to the two CD, DE, each to each, and the angle ADE to the angle CDE, for each of them is a right angle. Wherefore also the base BE is equal to the base EC. But AE has been shown to be equal to EB: wherefore also BE is equal to EC, and consequently the three right lines AE, EB, EC, are equal to one another. Therefore from the centre E, with the distance of any of them, AE, EB, EC, the circle so described will pass through the remaining points, and it will be the circle to be described. Wherefore the segment of a circle being given, the circle of which it is given is described. But it is also evident that the segment ABC is less than a semicircle, because its centre falls without it. In like manner the angle ABD is also equal to the angle BAD, the right line AD will be equal to each of the right lines BD, DC. Therefore the three right lines AD, DB, DC, will be equal to one another, because D will be the centre of the circle described, and the segment ABC a semicircle. But if the angle ABD be less than the angle BAD, describe at the right line BA, and at the given point a in it, the angle ABD equal to the angle BAE, within the segment ABC, the centre E will be in DB, and the segment ABC will be greater than a semicircle. Therefore the segment of a circle being given, the circle is described of which it is a segment. Q. E. F. PROPOSITION XXVI.* THEOREM. In equal circles equal angles stand upon equal circumferences, whether they stand at the centre or the circumference. Let ABC, DEF, be equal circles, and BGC, EHF, equal angles in them at the centre; also BAC, EDF, at the circumference. The circumference BAC is equal to the circumference EDF, for B K G E H a Def. 1. EH, HF, and the angle at join BC, EF. Because the circles ABC, DEF, are equal, the righta lines drawn from their centres shall also be equal: therefore the two BG, GC, are equal to the two G equal to the angle at H. Bcb is equal to the base EF. at A is equal to the angle at be similar to the segment EDF, and they are upon Def. 11. equal right lines BC, EF. But similar segmentsd d Cor. 24.3. standing upon equal bases have equal circumferences. Therefore the segment BAC is equal to the segment EDF. But the whole ABC is equal to the whole EDF: © Ex hyp. therefore the remaining segment BKC is equal to the remaining segment ELF. Therefore in equal circles, &c. Q. E. D. Deductions. 1. If two equal circles cut each other, and from either point of intersection a line be drawn meeting the circumferences, the part of it intercepted between the circumferences will be bisected by the circle whose diameter is the common chord of the equal circles. 2. The arcs of circles intercepted between two parallel chords are equal to one another. 3. In equal circles the greater angle stands upon the greater circumference. * This and the three succeeding propositions will hold good, if in the same circle. с a 23. 1. 26.3. € 20. 3. 4. In equal circles, two equal right lines terminated in a point in the circumference of the one being equal to two other right lines terminated in a point in the circumference of the other, then shall the intercepted arcs be equal to one another. PROPOSITION XXVII. THEOREM. In equal circles angles which stand upon equal circumferences are equal to one another, whether they stand at the centre or the circumference. For in the equal circles ABC, DEF, let the angles BGC, EHF, at the centre, also BAC, EDF, at the circumference, stand upon equal circumferences BC, EF. The angle BGC is equal to the angle EHF, and the angle BAC to the angle EDF. If the angles BGC, EHF, be not equal, one of them will be greater than the other. Let BGC be the greater, and make at the right line BG, and at the point G in it, the angle BGK circumference BK is equal to the circumference EF. But the circumference EF is equal to the circumference BC wherefore also BK is equal to BC, the less to the greater, which is impossible. Therefore the angle BGC is not unequal to the angle EHF; wherefore it is equal to it. But the angle which is at A is half of the angle BGC; also the angle at D is half of the angle EHF: wherefore the angle which is at A is equal to the angle which is at D. Therefore in equal circles, &c. Q. E. D. Deduction. In equal circles, the greater of two circumferences subtends the greater angle, whether those angles be at the centre or the circumference. |