PROP. XXVII. THEOR. In equal circles, the angles which stand upon equal arcs are equal to one another, whether they be at the centres or at the circumferences. Let ABC, DEF be equal circles, and let the angles BAC, EDF, whether at the centres or at the circumferences stand upon equal arcs BGC, EHF; the angle BAC shall be equal to the angle EDF. For if not let one of them, as BAC, be the greater, and make the angle BAK equal to EDF. Then, because in the equal circles ABC, DEF the angle *Const. #26.3. +Hyp. BAK is equal to EDF, the arcs BGK, EHF upon which they stand are equal; but BGC† also is equal to EHF; therefore BGK is equal to BGC, a part to the whole, which is impossible; therefore the angles BAC, EDF cannot be the one greater than the other: that is, they are equal. PROP. XXVIII. THEOR. In equal circles, equal chords cut off equal arcs, the greater equal to the greater, and the less to the less. Let ABC, DEF be equal circles, and BC, EF equal chords in them, which cut off the two greater arcs BAC, EDF, and the two less BGC, EHF: the greater BAC shall be equal to the greater EDF, and the less BGC to the less EHF. #1. 3. Take* K, L, the centres of the circles, and join BK, KC, EL, LF: and because the circles are +12 Def. 1. equal, the straight lines from their centres aret equal: therefore BK, KC are equal to EL, LF, each to each: and D +Hyp. the base BC is equalt to the base EF; there #8. 1. fore the angle BKC is B #26. 3. +Hyp. +3 Ax. stand upon equal arcs, when they are at the centres; therefore the arc BGC is equal to the arc EHF: but the whole circle ABC is equal to the whole EDF; therefore the remaining part of the circumference, viz. BAC, is equal to the remaining part EDF. Therefore, in equal circles, &c. Q. E. D. PROP. XXIX. THEOR. In equal circles, equal arcs are subtended by equal chords. Let ABC, DEF be equal circles, and let the arcs BGC, EHF also be equal; and join BC, EF: the chord BC shall be equal to the chord EF. If the arcs are semicircumferences, then BC, EF are diameters; and as the circles are equal, these diameters must be equal. But if the arcs be not semicircumferences, then take* K, L, #1.3. the centres of the circles, and join BK, KC, EL, LF: and because the arc BGC is equal to the arc EHF, the angle BKC is equal to the angle ELF: and because the circles ABC, DEF, are equal, *27 3 To bisect a given are, that is, to divide it into two equal *10. 1. parts. Let ADB be the given arc; it is required to bisect it. Join AB, and bisect* it in C; from the point +11. 1. C draw CD at right anglest to AB: the arc ADB shall be bisected in the point D. Join AD, DB: and because AC is equal to CB, and CD common to the triangles ACD, BCD, the two sides AC, CD are equal to the two BC, CD, each to each; and the angle ACD is equal to the angle BCD, because each of them is a right angle: therefore the base AD is equal* to *4. 1. *28. 3. equal to the greater, and the less to the less; and AD, DB are each of them less than a semicircle, *Cor. 1. 3. because DC* passes through the centre: therefore the arc AD is equal to the arc DB. Therefore Which was to be done. the given arc is bisected in D. PROP. XXXI. THEOR. In a circle, the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle. Let ABCD be a circle, of which the diameter is BC, and centre E: and draw CA, dividing the circle into the segments ABC, ADC, and join BA, AD, DC: the angle in the semicircle BAC shall be a right angle; and the angle in the segment ABC, which is greater than a semicircle, shall be less than a right angle; and the angle in the segment ADC, which is less than a semicircle, shall be greater than a right angle. #5. 1. F Join AE, and produce BA to F: and because BE is +12. Def. 1. equal to EA, the angle EAB is equal to EBA; also, because AE is equal to EC, the angle EAC is equal to ECA; wherefore the whole angle BAC is equal to the two angles ABC, ACB: but FAC, the exterior angle of the triangle ABC, is equal to the two angles ABC, ACB; therefore the angle BAC is equal to the 12 Ax. #32. 1. +1 Ax. #8 Def. 1. angle FAC: and therefore each of them is a right* angle: wherefore the angle BAC in a semicircle is a right angle. #17.1. And because the two angles ABC, BAC of the triangle ABC are together less than two right angles, and that BAC has been proved to be a right angle; therefore ABC must be less than a right angle: and therefore the angle in a segment ABGC, greater than a semicircle, is less than a right angle. And because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are equal to two right angles: therefore the angles ABC, ADC, $22. 3. are equal to two right angles: and ABC has been proved to be less than a right angle; wherefore the other ADC is greater than a right angle. This prop. may be easily demonstrated by taking BADG for the greater arc, and BAD for the less, and joining B, G; C, G; for then the opposite angles at G and A of the inscribed quadrilateral are equal to two right angles; and these angles, being in equal segments or standing upon equal arcs, are themselves equal; each is therefore a right angle; and the angle whose sides terminate in B, G, the extremities of the greater arc is less, and that whose sides terminate in B, D, the extremities of the less arc is greater than this right angle. PROP. XXXII. THEOR. If a straight line touches a circle, and from the point of contact a straight line be drawn, cutting the circle, the angles which this line makes with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle. Let the straight line EF touch the circle ABCD in B, and from the point B let the straight line BD be drawn, cutting the circle: the angles which BD makes with the touching line EF shall be equal to the angles in the alternate segments of the circle; that is, the angle DBF shall be equal to the angle which is in the segment DAB, and the angle DBE to that in the segment DCB. *11. 1. From the point B draw* BA at right angles to EF, and take any point C in the arc DCB, and join AD, DC, CB: and because the straight line EF touches the circle ABCD in the point B, and BA is drawn at right angles to the touching line from *19. 3. the point of contact B, the centre of the *#31.3. |