Also, as the circumference BC to EF, so shall the sector BGC be to the sector EHF. B x C E FM Join BC, CK, and in the circumferences, BC, CK, take any points X, O, and join BX, XC, CO, OK. Then, because in the triangles GBC, GCK, the two sides BG, GC are equal to the two CG, GK each to each, and that they contain equal angles; and the triangle Ġ BC to the triangle GCK: and because the circumference BC is equal to the circumference CK, the remaining part of the whole circumference of the circle ABC, is equal to the remaining part of the whole circumference of the same circle: (ax. 3.) therefore the angle BXC is equal to the angle COK; (III. 27.) and the segment BXC is therefore similar to the segment COK; (111. def. 11.) and they are upon equal straight lines, BC, CK: but similar segments of circles upon equal straight lines, are equal to one another: (III. 24.) therefore the segment BXC is equal to the segment COK: and the triangle BĞC was proved to be equal to the triangle CGK; therefore the whole, the sector BGC, is equal to the whole, the sector CGK: for the same reason, the sector KGL is equal to each of the sectors BGC, CGK: in the same manner, the sectors EHF, FHM, MHN may be proved equal to one another: therefore, what multiple soever the circumference BL is of the circumference B C, the same multiple is the sector BGL of the sector BGC; and for the same reason, whatever multiple the circumference EN is of EF, the same multiple is the sector EHN of the sector EHF: and if the circumference BL be equal to EN, the sector BGL is equal to the sector EHN; and if the circumference BL be greater than EN, the sector BGL is greater than the sector EÉN; and if less, less ; since, then, there are four magnitudes, the two circumferences BC, EF, and the two sectors BGC, EHF, and that of the circumference BC, and sector BGC, the circumference BL and sector BGL are any equimultiples whatever; and of the circumference EF, and sector EHF, the circumference EN, and sector EHN are any equimultiples whatever ; and since it has been proved, that if the circumference BL be greater than EN, the sector BGL is greater than the sector EHN; and if equal, equal; and if less, less : therefore, as the circumference B C is to the circumference EF, so is the sector BGC to the sector EHF. (v. def. 5.) Wherefore, in equal circles, &c. Q. E. D. PROPOSITION B. THEOREM. If an angle of a triangle be bisected by a straight line which likewise cuts the base ; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle. Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD. The rectangle BA, AC shall be equal to the rectangle BD, DC, together with the square on AD. Describe the circle ACB about the triangle, (IV. 5.) and produce AD to the circumference in E, and join EC. Then because the angle BAD is equal to the angle CAE, (hyp.) and the angle ABD to the angle AEC, (III. 21.) for they are in the same segment; the triangles ABD, AEC are equiangular to one another: (1. 32.) therefore as B A to AD, so is EA to AC; (VI. 4.) and consequently the rectangle BA, AC is equal to the rectangle EA, AD, (VI. 16.) that is, to the rectangle ED, DA, together with the square on AD; (II. 3.) but the rectangle ED, DA is equal to the rectangle BD, DC; (III. 35.) therefore the rectangle BA, À C is equal to the rectangle BD, DC, together with the square on AD. Wherefore, if an angle, &c. Q. E. D. PROPOSITION C. THEOREM. If from any angle of a triangle, a straight line be drauon perpendicular to the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle, Let ABC be a triangle, and AD the perpendicular from the angle A to the base BC. The rectangle BA, AC shall be equal to the rectangle contained by AD and the diameter of the circle described about the triangle. E Describe the circle ACB about the triangle, (IV. 5.) and draw its diameter AE, and join EC. Because the right angle BDA is equal to the angle ECA in a semicircle, (III. 31.) and the angle ABD equal to the angle AEC in the same segment; (III. 21.) the triangles ABD, AEC are equiangular: therefore as BA to AD, so is EA to AC; (VI. 4.) and consequently the rectangle BA, AC is equal to the rectangle EA, AD. (vi. 16.) If therefore from any angle, &c. Q.E. D. PROPOSITION D. THEOREM. The rectangle contained by the diagonals of a quadrilateral figure inscribed in a circle, is equal to both the rectangles contained by its opposite sides. Let ABCD be any quadrilateral figure inscribed in a circle, and join AC, BD. The rectangle contained by AC, BD shall be equal to the two rectangles contained by AB, CD, and by AD, BC. Make the angle ABE equal to the angle DBC: (1. 23.) then the angle ABD is equal to the angle EBC: and the angle BDĂ is equal to the angle BCE, because they are in the same segment: (111. 21.) therefore the triangle ABD is equiangular to the triangle BCE: wherefore, as BC is to CE, so is BD to DA; (VI. 4.) and consequently the rectangle BC, AD is equal to the rectangle BD, CE: (vi. 16.) again, because the angle ABE is equal to the angle DBC, and the angle BAE to the angle BDC, (III. 21.) therefore as BA to AE, so is B D to DC; wherefore the rectangle BA, DC is equal to the rectangle BD, AE: but the rectangle BC, AD has been shewn to be equal to the rectangle BD, CE; therefore the whole rectangle AC, BD is equal to the rectangle AB, DC, together with the rectangle AD, BC. (11. 1.) Therefore the rectangle, &c. Q. E. D. This is a Lemma of Cl. Ptolemæus, in page 9 of his Mɛyádn Eúvtatis, NOTES TO BOOK VI. In this Book, the theory of proportion exhibited in the Fifth Book, is applied to the comparison of the sides and areas of plane rectilineal figures, both of those which are similar, and of those which are not similar. Def. 1. In defining similar triangles, one condition is sufficient, namely, that similar triangles are those which have their three angles respectively equal; as in Prop. 4, Book vi, it is proved that the sides about the equal angles of equiangular triangles are proportionals. But in defining similar figures of more than three sides, both of the conditions stated in Def. 1, are requisite, as it is obvious, for instance, in the case of a square and a rectangle, which have their angles respectively equal, but have not their sides about their equal angles proportionals. The following definition has been proposed : “Similar rectilineal figures of more than three sides, are those which may be divided into the same number of similar triangles.” This definition would, if adopted, require the omission of a part of Prop. 20, Book vi. Def. III. To this definition may be added the following: A straight line is said to be divided harmonically, when it is divided into three parts, such that the whole line is to one of the extreme segments, as the other extreme segment is to the middle part. "Three lines are in harmonical proportion, when the first is to the third, as the difference between the first and second, is to the difference between the second and third ; and the second is called a harmonic mean between the first and third. The expression 'harmonical proportion' is derived from the following fact in the Science of Acoustics, that three musical strings of the same material, thickness and tension, when divided in the manner stated in the definition, or numerically as 6, 4, and 3, produce a certain musical note, its fifth, and its octave. Def. iv. The term altitude, as applied to the same triangles and parallelograms, will be different according to the sides which may be assumed as the base, unless they are equilateral. Prop. I. In the same manner may be proved, that triangles and parallelograms upon equal bases, are to one another as their altitudes. Prop. A. When the triangle ABC is isosceles, the line which bisects the exterior angle at the vertex is parallel to the base. In all other cases, if the line which bisects the angle BAC cut the base BC in the point G, then the straight line BD is harmonically divided in the points G, C. For BG is to GC as BA is to AC; (v1. 3.) therefore BD is to DC as BG is to GC, but BG = BD – DG, and GC = GD - DC. Wherefore BD is to DC as BD – DG is to GD - DC. Hence BD, DG, DC, are in harmonical proportion. Prop. IV is the first case of similar triangles, and corresponds to the third case of equal triangles, Prop. 26, Book 1. Sometimes the sides opposite to the equal angles in two equiangular triangles, are called the corresponding sides, and these are said to be proportional, which is simply taking the proportion in Euclid alternately. The term homologous (óóroyos), has reference to the places the sides of the triangles have in the ratios, and in one sense, homologous sides may be considered as corresponding sides. The homologous sides of any two similar rectilineal figures will be found to be those which are adjacent to two equal angles in each figure. Prop. v, the converse of Prop. iv, is the second case of similar triangles, and corresponds to Prop. 8, Book 1, the second case of equal triangles. Prop. vi is the third case of similar triangles, and corresponds to Prop. 4, Book 1, the first case of equal triangles. The property of similar triangles, and that contained in Prop. 47, Book I, are the most important theorems in Geometry. Prop. VII is the fourth case of similar triangles, and corresponds to the fourth case of equal triangles demonstrated in the note to Prop. 26, Book 1. Prop. ix. The learner here must not forget the different meanings of the word part, as employed in the Elements. The word here has the same meaning as in Euc. v. def. 1. It may be remarked, that this proposition is a more simple case of the next, namely, Prop. x. Prop. xi. This proposition is that particular case of Prop. xii, in which the second and third terms of the proportion are equal. These two problems exhibit the same results by a Geometrical construction, as are obtained by numerical multiplication and division. Prop. xil. The difference in the two propositions Euc. II. 14, and Euc. vi. 13, is this : in the Second Book, the problem is, to make a rectangular figure or square equal in area to an irregular rectilinear figure, in which the idea of ratio is not introduced. In the Prop, in the Sixth Book, the problem relates to ratios only, and it requires to divide a line into two parts, so that the ratio of the whole line to the greater segment may be the same as the ratio of the greater segment to the less. The result in this proposition obtained by a Geometrical construction, is analogous to that which is obtained by the multiplication of two numbers, and the extraction of the square root of the product. It may be observed, that half the sum of AB and BC is called the Arithmetic mean between these lines ; also that BD is called the Geometric mean between the same lines. To find two mean proportionals between two given lines is impossible by the straight line and circle. Pappus has given several solutions of this problem in Book 111, of his Mathematical Collections; and Eutocius has given, in his Commentary on the Sphere and Cylinder of Archimedes, ten different methods of solving this problem. Prop. xiv depends on the same principle as Prop. xv, and both may easily be demonstrated from one diagram. Join DF, FE, EG in the fig. to Prop. xiv, and the figure to Prop. xv is formed. We may add, that there does not appear any reason why the properties of the triangle and parallelogram should be here separated, and not in the first proposition of the Sixth Book. Prop. xv holds good when one angle of one triangle is equal to the defect from what the corresponding angle in the other wants of two right angles. This theorem will perhaps be more distinctly comprehended by the learner, if he will bear in mind, that four magnitudes are reciprocally |