and CD, GH, are equimultiples of CK, GL; wherefore, by Cor. Prop. 4, Book v, AB is to CD, as M to GH, And, by the hypothesis, AB is to CD, as EF to GH; therefore M is equal to EF by Prop. 9, Book v. and consequently, EF is the same multiple of GL that AB is of CK. This is the method by which Simson shews that the Geometrical definition of proportion is a consequence of the Arithmetical definition, and conversely. It may however be shewn by employing the equation a с = and taking ma, mc any equimultiples of a and c the first and third, and nb, nd any equimultiples of b and d the second and fourth. Prop. XVIII, being the converse of Prop. XVII, has been demonstrated indirectly. Let AE, EB, CF, FD be proportionals, that is, as AE to EB, so let CF be to FD. Then these shall be proportionals also when taken jointly; that is, as AB to BE, so shall CD be to DF. For if the ratio of AB to BE be not the same as the ratio of CD to DF; the ratio of AB to BE is either greater than, or less than the ratio of CD to DF. First, let AB have to BE a greater ratio than CD has to DF; and let DQ be taken so that AB has to BE the same ratio as CD to DQ. are also proportionals when taken separately; (v. 17.) but, by the hypothesis, AE has to EB the same ratio as CF to FD; therefore the ratio of C'Q to QD is the same as the ratio of CF to FD. (v. 11.) And when four magnitudes are proportionals, if the first be greater than the second, the third is greater than the fourth; and if equal, equal; and if less, less; (v. 14.) but CQ is less than CF, therefore QD is less than FD; which is absurd. Wherefore the ratio of AB to BE is not greater than the ratio of CD to DF; that is, AB has the same ratio to BE as CD has to DF. Secondly. By a similar mode of reasoning, it may likewise be shewn, that AB has the same ratio to BE as CD has to DF, if AB be assumed to have to BE a less ratio than CD has to DF. For further information on the very important subject of Ratio and Proportion, reference may be made to Dr. Barrow's Mathematical Lectures; Professor De Morgan's Connexion of Number and Magnitude; and the fourth chapter of Professor Peacock's Algebra. BOOK VI. DEFINITIONS. I. SIMILAR rectilineal figures are those which have their several angles equal, each to each, and the sides about the equal angles proportionals. II. "Reciprocal figures, viz. triangles and parallelograms, are such as have their sides about two of their angles proportionals in such a manner, that a side of the first figure is to a side of the other, as the remaining side of this other is to the remaining side of the first.” III. A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment, as the greater segment is to the less. IV. The altitude of any figure is the straight line drawn from its vertex perpendicular to the base. PROPOSITION I. THEOREM. Triangles and parallelograms of the same altitude are one to another as their bases. Let the triangles ABC, ACD, and the parallelograms EC, CF, have the same altitude, viz. the perpendicular drawn from the point A to BD or BD produced. As the base BC is to the base CD, so shall the triangle ABC be to the triangle ACD, and the parallelogram EC to the parallelogram CF. E A F HGB C Produce BD both ways to the points H, L, and take any number of straight lines BG, GH, each equal to the base BC; (1. 3.) and DK, KL, any number of them, each equal to the base CD; and join AG, AH, AK, AL. Then, because CB, BG, GH, are all equal, the triangles AHG, AGB, ABC, are all equal: (1. 38.) therefore, whatever multiple the base HC is of the base BC, the same multiple is the triangle AHC of the triangle ABC: for the same reason, whatever multiple the base LC is of the base CD, the same multiple is the triangle ALC of the triangle ADC: and if the base HC be equal to the base CL, the triangle AHC is also equal to the triangle ALC: (1. 38.) therefore, since there are four magnitudes, viz. the two bases BC, CD, and the two triangles ABC, ACD; and of the base BC, and the triangle ABC the first and third, any equimultiples whatever have been taken, viz. the base HC and the triangle AHC; and of the base CD and the triangle ACD, the second and fourth, have been taken any equimultiples whatever, viz. the base CL and the triangle ALC; and since it has been shewn, that, if the base HC be greater than the base CL, the triangle AHC is greater than the triangle ALC; and if equal, equal; and if less, less: therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD. (v. def. 5.) And because the parallelogram CE is double of the triangle ABC, (1.41.) and the parallelogram CF double of the triangle ACD, and that magnitudes have the same ratio which their equimultiples have; (v. 15.) as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF; and because it has been shewn, that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD; and as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF; therefore, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram CF. (v. 11.) Wherefore, triangles, &c. Q.E.D. COR. From this it is plain, that triangles and parallelograms that have equal altitudes, are one to another as their bases. Let the figures be placed so as to have their bases in the same straight line; and having drawn perpendiculars from the vertices of the triangles to the bases, the straight line which joins the vertices is parallel to that in which their bases are, (1. 33.) because the perpendiculars are both equal and parallel to one another. (1. 28.) Then, if the same construction be made as in the proposition, the demonstration will be the same. PROPOSITION II. THEOREM. If a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or these produced, proportionally: and conversely, if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section shall be parallel to the remaining side of the triangle. Let DE be drawn parallel to BC, one of the sides of the triangle ABC. Then BD shall be to DA, as CE to EA. Then the triangle BDE is equal to the triangle CDE, (1.37.) because they are on the same base DE, and between the same parallels DE, BC: but ADE is another triangle; and equal magnitudes have the same ratio to the same magnitude; (v. 7.) therefore, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE: but as the triangle BDE to the triangle ADE, so is BD to DA, (vI. 1.) because, having the same altitude, viz. the perpendicular drawn from the point E to AB, they are to one another as their bases; and for the same reason, as the triangle CDE to the triangle ADE, so is CE to EA: therefore, as BD to DA, so is CE to EA._(v. 11.) Next, let the sides AB, AC of the triangle ABC, or these sides produced, be cut proportionally in the points D, E, that is, so that BD may be to DA as CE to EA, and join DE. Then DE shall be parallel to BC. The same construction being made, because as BD to DA, so is CE to EA; and as BD to DA, so is the triangle BDE to the triangle ADE; (v1.1.) and as CE to EA, so is the triangle CDE to the triangle ADE; therefore the triangle BDE is to the triangle ADE, as the triangle CDE to the triangle ADE; (v. 11) that is, the triangles BDE, CDE have the same ratio to the triangle ADE: therefore the triangle BDE is equal to the triangle CDE: (v.9.) and they are on the same base DE: but equal triangles on the same base and on the same side of it, are between the same parallels; (1. 39.) therefore DE is parallel to BC. Wherefore, if a straight line, &c. Q. E.D. PROPOSITION III. THEOREM. If the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; the segments of the base shall have the same ratio which the other sides of the triangle have to one another and conversely, if the segments of the base have the same ratio which the other sides of the triangle have to one another; the straight line drawn from the vertex to the point of section, divides the vertical angle into two equal angles. Let ABC be a triangle, and let the angle BAC be divided into two equal angles by the straight line AD. Then BD shall be to DC, as BA to AC. E B D Through the point C draw CE parallel to DA, and let BA produced meet CE in E. (I. 31.) Because the straight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate angle CAD: (1. 29.) but CAD, by the hypothesis, is equal to the angle BÀD; wherefore BAD is equal to the angle ACE. (ax. 1.) Again, because the straight line BAE meets the parallels AD, EC, the outward angle BAD is equal to the inward and opposite angle AEC: (1.29.) but the angle ACE has been proved equal to the angle BAD; therefore BD is to DC, as BA to AE: (VI. 2.) therefore, as BD to DC, so is BA to AC. (v.7.) Next, let BD be to DC, as BA to AC, and join AD. Then the angle BAC shall be divided into two equal angles by the straight line AD. |