PROPOSITION XXII. THEOREM. The opposite angles of any quadrilateral figure, inscribed in a circle, are together equal to two right angles. B Let ABCD be a quadrilateral fig. inscribed in a O. Then must each pair of its opposite s be together equal to two rt. LS. Draw the diagonals AC, BD. Then ADB = ▲ ACB, in the same segment, III. 21. and 4 BDC= 1 BAC, in the same segment; III. 21. .. sum of 4s ADB, BDC= sum of 48 ACB, BAC; Then 48 ADC, ABC together= sum of 8 ACB, BAC, ABC; and.. 8 ADC, ABC together=two rights. I. 32. Similarly, it may be shewn, that 8 BAD, BCD together=two right 4 s. Q. E. D. NOTE. Another method of proving this proposition is given on page 209. Ex. 1. If one side of a quadrilateral figure inscribed in a circle be produced, the exterior angle is equal to the opposite angle of the quadrilateral. Ex. 2. If the sides AB, DC of a quadrilateral inscribed in a triangle be produced to meet in E, then the triangles EBC, EAD will be equiangular. Ex. 3. Shew that a circle cannot be described about a rhombus. Ex. 4. The lines, bisecting any angle of a quadrilateral figure inscribed in a circle and the opposite exterior angle, meet in the circumference of the circle. Ex. 5. AB, a chord of a circle, is the base of an isosceles triangle, whose vertex C is without the circle, and whose equal sides meet the circle in D, E: shew that CD is equal to CE. Ex. 6. If in any quadrilateral the opposite angles be together equal to two right angles, a circle may be described about that quadrilateral. Propositions XXIII and XXIV., not being required in the method adopted for proving the subsequent propositions in this book, are removed to the Appendix. Proposition xxv. has been already proved. NOTE III. On the Method of Superposition, as In Props. XXVI. XXVII. XXVIII. XXIX. we prove certain relations existing between chords, arcs and angles in equal circles. As we shall employ the Method of Superposition, we must state the principles which render this method applicable, as a test of equality, in the case of figures with circular boundaries. 8. E. II. 11 DEF. XIII. Equal circles are those, of which the radii are equal. R Α' B C For suppose ABC, A'B'C' to be circles, of which the radii are equal. Then if A'B'C' be applied to ○ ABC, so that O', the centre of A'B'C', coincides with O, the centre of ABC, it is evident that any particular point A' in the Oce of the former must coincide with some point A in the Oce of the latter, because of the equality of the radii O'A' and OA. Hence Oce A'B'C' must coincide with Oce ABC, that is, A'B'C' = © ABC. Further, when we have applied the circle A'B'C' to the circle ABC, so that the centres coincide, we may imagine ABC to remain fixed while A'B'C' revolves round the common centre. Hence we may suppose any particular point B' in the circumference of A'B'C' to be made to coincide with any particular point B in the circumference of ABC. Again, any radius O'A' of the circle A'B'C' may be made to coincide with any radius OA of the circle ABC. Also, if A'B' and AB be equal arcs, they may be made to coincide. Again, every diameter of a circle divides the circle into equal segments. For let AOB be a diameter of the circle ACBD, of which O is the centre. Suppose the segment ACB to be applied to the segment ADB, so as to keep AB a common boundary: then the arc ACB must coincide with the arc ADB, because every point in each is equally distant from 0. B PROPOSITION XXVI. THEOREM. In equal circles, the arcs, which subtend equal angles, whether they be at the centres or at the circumferences, must be equal. K H Let ABC, DEF be equal circles, and let 48 BGC, EHF at their centres, and 48 BAC, EDF at their Oces, be equal. Then must arc BKC=arc ELF. For, if ABC be applied to O DEF, so that G coincides with H, and GB falls on HE, then, : GB=HE, .. B will coincide with E. And :: ▲ BGC= ▲ EHF, :. GC will fall on HF; Then B coincides with E and C with F, .. arc BKC will coincide with and be equal to arc ELF, Q. E. D. NOTE. This and the three following Propositions are, and will hereafter be assumed to be, true for the same circle as well as for equal circles. PROPOSITION XXVII. THEOREM. In equal circles, the angles, which are subtended by equal arcs, whether they are at the centres or at the circumferences must be equal. A A Let ABC, DEF be equal circles, and let 28 BGC, EHF at their centres, and ▲ 8 BAC, EDF at their Oces, be subtended by equal arcs BKC, ELF. Then must BGC = ▲ EHF, and ▲ BAC= 2 EDF. For, if ABC be applied to DEF, so that G coincides with H, and GB falls on HE, then ·· GB=HE, .. B will coincide with E; and arc BKC=arc ELF, :. C will coincide with F. Then Hence, GC will coincide with HF. BG coincides with EH, and GC with HF, .. BGC will coincide with and be equal to Again, :: ¿BAC=half of ▲ BGC, and LEDF-half of EHF, LBAC= ▲ EDF. EHF. III. 20. III. 20. 1. Ax. 7. Q. E. D. Ex. 1. If, in a circle, AB, CD be two arcs of given magnitude, and AC, BD be joined to meet in E, shew that the angle AEB is invariable. Ex. 2. The straight lines joining the extremities of the chords of two equal arcs of the same circle, towards the same parts, are parallel to each other. Ex. 3. If two equal chords, in a given circle, cut one another, the segments of the one shall be equal to the segments of the other, each to each. |
