PROP. XXI. THEOR. Angles in the same segment of a circle are equal to one another. E Let ABCD be a circle, and BAD, BED angles in the same segment BAED: these angles are equal. Take (II. 1.) F the centre of the circle ABCD; and join BF, FD: draw also AF, and produce it to meet the circumference in C; and join EC. Then, because the angle BFC is at the centre, and the angle BAC at the circumference, and that they have the same arc BC for their base; therefore (III. 20.) the angle BFC is double of the angle BAC. For the same reason, the angle BFC is double of the angle BEC; therefore the angle BAC is equal (I. ax. 7.) to the angle BEC. In the same manner, it might be shown that the angles CAD, CED are equal. Therefore (1 ax. 2.) the whole angle BAD is equal to the whole angle BED. Wherefore angles in the same segment, &c. * PROP. XXII. THEOR. The opposite angles of any quadrilateral figure described in a circle, t 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. Then (I11.21.) the angles CAB, CDB are equal, because they are in the same segment BADC; and the angles ACB, ADB are equal, because they are in the same segment ADCB: therefore (I. ax. 2.) the whole angle ADC is equal to the angles CAB, ACB. To each of these equals add the angle ABC; therefore the angles ABC, CAB, BCA, are eqnal to the angles ABC, ADC: but (1. 32.) ABC, CAB, BCA are equal to two right angles; therefore also the angles ABC, ADC are equal to two right angles. In the same manner, the angles BAD, BCD D A BE * The demonstration here adopted is the same in substance as that yiven in Ingram's Euclid. It has the advantage of being applicable in all cases. In Simson's edition there are two cases, one in which the segment is greater than a semicircle, and the other in which it is not greater; the latter of which is wanting in the Greek. When the segment is greater than a semicircle, the proof is more simple than the one given above ; since (111. 20.) the angles BAD, BED are each half of BFD. + That is, having all its angular points on the circumference, according to the second definition of the fourth book. may be shown to be equal to two right angles. Therefore the opposite angles, &c. * Cor. If any side AB be produced to E, the exterior angle CBE is equal to the interior and opposite angle ADC. For CBA, CBE are together equal to CBA, CDA, each pair being equal to two right angles. Take away CBA, and there remains CBE equal to CDA. PROP. XXIII. THEOR. Upon the same straight line, and on 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, A DB, be upon the same side of the same straight line AB, not coinciding with one another: then (III. 10.) because the circle ACB cuts the circle ADB, in the two points A, B, they cannot cut one another in any other point. One of the segments must therefore fall within the other; let ACB fall within ADB, and draw the straight line BCD, cutting the arcs of the segments in C and D, and join CA, DA. Then, because the segment ACB is similar to the segment ADB, the angle ACB is equal (III. def. 8.) to ADB, the exterior to the interior and remote, which (I. 16.) is impossible. Therefore, upon the same straight line, &c. D B PROP. XXIV. THEOR. SIMILAR segments of circles upon equal bases are equal to one another, and have equal arcs. † Let A EB, CFD be similar segments of circles upon the equal straight lines or bases, AB, CD: the segments are equal ; and likewise the arcs AEB, CFD are equal. 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 on CD, the point B will coincide with D, because AB is equal to CD: therefore, the straight E А B * In proving that the angles ABC, ADC are together equal to two right angles, we may either prove that the two parts of ADC are respectively equal to two angles of the triangle ABC, or that the two parts of A BC are respectively equal to two angles of the triangle ADC, and ihe proof is completed in each instance by adding the remaining angle of the triangle. + The last clause, specifying that the segments have equal arcs, is improperly omitted in the Greek, and by Dr. Simson and others. line AB coinciding with CD, the segment AEB must coincide (111. 23.) with the segment CFD, and therefore is equal (1. ax. 8.) to it: and the arcs AEB, CFD are equal, because they coin. cide. Therefore similar segments, &c. PROP. XXV. PROB. A SEGMENT of a circle being given, to complete the circle of which it is a segment. Assume three points in the arc of the segment, and find (III. 1. schol.) the centre of the circle ; from that centre, at the distance between it and any point in the arc describe a circle ; it will be the one required. * PROP. XXVI. THEOR. D In equal circles, or in the same circle, t equal angles stand upon equal arcs, whether they are at the centres, or the circumferences. Let ABC, DEF be equal circles, having the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences : the arc BKC is equal to the arc ELF. Join BC, EF; and because the circles ABC, DEF are equal, their radii are equal: therefore the two sides BG, GC are equal to the two EH, HF; and (hyp.) the angles G and H are equal : therefore (I. 4.) the base BC is equal to the base EF. Then, because the angles A and D are equal, the segment BAC is similar (III. def. 8.) to the segment EDF; and they are upon equal straight lines BC, EF; but (III. 24.) similar segments of circles upon equal straight lines have equal arcs; therefore the arc BAC is equal to the arc EDF. But the whole circumference ABC is equal to the whole DEF, because the circles are equal; therefore the remaining arc BKC is equal (I. ax. 3.) to the remaining arc ELF. Wherefore, in equal circles, &c. Cor. Hence, in a circle the arcs intercepted between parallel H B с E K к L • This short and easy solution is substituted instead of the long one given by Euclid. + In this proposition and the three following, the words, “ or in the same circle," are introduced in this edition ; as what is proved in these propositions must evidently hold respecting the same circle, as well as in equal circles : and they are often thus applied by Euclid himself, chords are equal. For, if a straight line be drawn transversely, joining two extremities of the chords, it will (I. 29.) make equal angles with the chords; and therefore the arcs on which these stand are equal. PROP. XXVII. THEOR. In equal circles, or in the same circle, the angles which stand upon equal arcs are equal to one another, whether they are at the centres, or the circumferences. Let the angles BGC, EHF at the centres, and BAC, EDF at the circumferences, of the equal circles ABC, DEF, stand upon the equal arcs BC, EF: the angle BGC is equal to EHF, and BAC to EDF. If the angle BGC be equal to EHF, it is manifest (III. 20.) that BAC is also equal to EDF. But, if BGC, EHF be not equal, one of them is the greater; let BGC, if possible, be the greater, and (I. 23.) make BGK equal to EHF. Then (III. 26.) equal angles stand upon equal arcs, when they are the centres ; therefore the arc BK is equal to the arc EF: but EF is equal (hyp.) to BC; therefore also BK is equal to BC, which (I. ax. 9 ) is impossible. Therefore the angle BGC is not unequal to EHF; that is, it is equal to it: and the angle A is half (III. 20.) of BGC, and the angle D half of EHF: therefore the angle A is equal (I. ax. 7.) to the angle D. Wherefore, in equal circles, &c. G B E с K F PROP. XXVIII. THEOR. In equal circles, or in the same circle, equal chords divide the circumferences into parts which are equal, each to each. + Let ABC, DEF be equal circles, and BC, EF egual chords in them : the arc BAC is equal to EDF, and BGC to EHF. If BC and EF be diameters, they bisect the circumferences ; and those being equal, the parts of the one are equal to the parts of the other. It is plain from the next proposition, that the converse of this corollary is true. + Dr. Simson and most other editors omit, after the Greek, the case in which the chords are diameters. Some editors exclude this case in their enunciations, but give it in their proofs. The corresponding case in the next proposition is also omitted in most editions. A D к L But if BC and EF be not diameters, take (III. 1.) K, L, the centres of the circles, and join BK, KC, EL, LF. Then, because the circles are equal, BK, KC are equal to EL, LF; and the base BC is equal (byp.) to the base EF; therefore (1. 8.) the angle BKC is equal to ELF. Byt (III. 26.) equal angles stand upon equal arcs, when they are at the centres ; therefore the less arc BGC, in the one circle, is equal to the less arc EHF, in the other. But the whole circumference ABGC is equal to the whole DEHF; the remaining part therefore of the circumference, viz., the greater arc BAC, in the one circle, is equal to the remaining part, viz., the greater arc EDF in the other. Therefore in equal circles, &c. B с E E H PROP. XXIX. THEOR. A D In equal circles, or in the same circle, equal arcs have equal chords. Let ABC, DEF be equal circles, and let the arcs BGC, EHF, and consequently BAC, EDF be equal; and join BC, EF: the chords BC, EF are equal. If BC and EF pass through the centres, they are equal, being diameters of equal circles. But if they do not pass through the centres, take (III. 1.) K, L, the centres of the circles, and join BK, KC, EL, LF. Then, because the arc BGC is equal (hyp.) to the arc EHF, the angle BKC is equal (III. 27.) to ELF. Also, because the circles ABC, DEF are equal, their radii are equal : therefore BK, KC are equal to EL, LF, and they contain equal angles: wherefore (I. 4.) the base BC is equal to the base EF. Therefore in equal circles, &c. K L с E PROP. XXX. PROB. To bisect a given arc of a circle. Join AB, and (i. 10. and 11.) bisect it in C, by the perpendicular CD; the arc ABD is bisected in the point D. Join AD, DB. Then, because AC is equal to CB, CD com mon to the triangles ACD, BCD, and the angle ACD equal to BCD, each of them being a right angle; therefore (I. 4.) AD is equal to BD. But (III. 28.) equal straight lines cut off equal arcs, D A |