We here insert Euclid's proofs of Props. 23, 24 of Book III. first observing that he gives the following definition of similar segments : DEF. Similar segments of circles are those in which the angles are equal, or which contain equal angles. 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 each other. D If it be possible, on the same base AB, and on the same side of it, let there be two similar segments of Os, ABC, ABD, which do not coincide. Because ADB cuts © ACB in pts. A and B, they cannot cut one another in any other pt., and .. one of the segments must fall within the other. Let ADB fall within ACB. Draw the st. line BDC and join CA, DA. Then segment ADB is similar to segment ACB, .. LADB= L ACB. Or the extr. 4 of a ▲ the intr. and opposite, which is impossible; .. the segments cannot but coincide. Q. E. D. PROPOSITION XXIV. THEOREM. Similar segments of circles, upon equal straight lines, are equal to one another. Let ABC, DEF be similar segments of Os on equal st. lines AB, DE. Then must segment ABC=segment DEF. For if segment ABC be applied to segment DEF, so that A may be on D and AB on DE, then B will coincide with E, and AB with DE; .. segment ABC must also coincide with segment DEF ; .. segment ABC=segment DEF. III. 23. Ax. 8. Q. E. D. We gave one Proposition, C, page 150, as an example of the way in which the conceptions of Flat and Reflex Angles may be employed to extend and simplify Euclid's proofs. We here give the proofs, based on the same conceptions, of the important propositions XXII. and XXXI. 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 C. Then must each pair of its opposites be together equal to .. sum of 48 at 0=twice sum of 4 8 BAD, BCD. But sum of 4 s at 0=4 right 48; I. 15, Cor. 2. ..twice sum of 4 s BAD, BCD=4 rights; .. sum of 48 BAD, BCD=two right 4 s. Similarly, it may be shewn that sum of 48 ABC, ADC=two right ▲ s. Q. E. D. PROPOSITION XXXI. THEOREM. In a circle, the angle in a semicircle is a right angle; and 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. B Let ABC be a O, of which O is the centre and BC a diameter. Draw AC, dividing the C into the segments ABC, ADC. Join BA, AD, DC. Then must the ▲ in the semicircle BAC be a rt. ≤, and ▲ in segment ABC, greater than a semicircle, less than a rt. 4, and in segment ADC, less than a semicircle, greater than a rt. 4. First, the flat angle BOC=twice / BAC, .. BAC is a rt. 4. Next, BAC is a rt. 4, .. ABC is less than a rt. 4. Lastly, sum of 4 s ABC, ADC=two rt. 4 S, and ABC is less than a rt. 4, .. ADC is greater than a rt. 2. III. C. p. 150. I. 17. III. 22. Q. E. D. BOOK IV. INTRODUCTORY REMARKS. EUCLID gives in this Book of the Elements a series of Problems relating to cases in which circles may be described in or about triangles, squares, and regular polygons, and of the last-mentioned he treats of three only: the Pentagon, or figure of 5 sides, The Student will find it useful to remember the following Theorems, which are established and applied in the proofs of the Propositions in this Book. I. The bisectors of the angles of a triangle, square, or regular polygon meet in a point, which is the centre of the inscribed circle. II. The perpendiculars drawn from the middle points of the sides of a triangle, square, or regular polygon meet in a point, which is the centre of the circumscribed circle. III. In the case of a square, or regular polygon the inscribed and circumscribed circles have a common centre. IV. If the circumference of a circle be divided into any number of equal parts, the chords joining each pair of consecutive points form a regular figure inscribed in the circle, and the tangents drawn through the points form a regular figure described about the circle. |