D Then, because D B is equal to the right line A c, and the right line BC is a common fide, there will be two right lines D B, BC equal to two right lines A C, CB, each to each; and the angle DBC is equal to the angle ACB: wherefore the bafe DC will be equal to the bafe A B, and the triangle ABC will be equal to the triangle DCB; the greater to the lefs; which is abfurd. 1. C Therefore the right line A B is not unequal to the right line A C; they are confequently equal. Therefore if two angles of a triangle be equal, the fides oppofite to the equal angles will be equal. Which was to be demonftrated. Two right lines cannot be conftituted on the fame right line equal to two other right lines, each to each, at different points on the fame fide, and having the fame ends with the first right lines". For, if poffible, let two right lines A D, DB be conftituted upon the right line A B, equal each to each, to two right lines A C, CB, at different points C, D, fituated on the fame fide the line A B, the lines A D, DE, having the fame ends A, B with the two firft lines A C, CB; fo that CA be equal to D A, both having the fame end A and CB be equal to DB, both having the fame end B: for join the right line CD. Then because AC is equal to AD, the angle A CD is equal to the angle ADC [by prop. 5.] wherefore the angle ADC is greater than the angle DC B, and the angle CDB is much greater than the angle DC B. Again, because the right line CB is equal to the right line DB, the angle CDB will be equal to the angle DC B. But it has been proved to A be much greater, which is impoffible. C B Therefore two right lines cannot be conftituted upon the fame right line equal to two other right lines, each to each, at See Commandines Euclid. at different points on the fame fide, and having the same ends with the first right lines. Which was to be demonstrated. Although this 7th propofition may be thought to be of no great importance in itfelf, yet its ufe in demonftrating the 8th propofition makes it both valuable and neceffary. It is as much one of the pillars to fupport the great elementary building of our author, as any the most beautiful ones of them all. A B PROP. VIII. THEOR. D G E F If two triangles have two fides of the one equal to two fides of the other, each to each, and have alfo equal bafes, then shall the angle contained by the equal fides of the one triangle, be equal to the angle contained by the equal fides of the other". Let there be two triangles A B C, DEF, having the two fides A B, A C of the one, equal to the two fides DE, DF of the other, each to each; that is, the fide A B equal to DE, and the fide A C equal to DF; and let also the bafe B C of the one be equal to the bafe E F of the other. I fay, the angle B A C is equal to the angle EDF. For if the triangle ABC be applied to the triangle D EF, and the point в put upon the point E; and the right line. BC upon the right line EF; the point c will agree with the point F; because the right line BC is equal to the right line E F. But fince the right line B C agrees with the right line EF, the right lines B A, AC will alfo agree with the right lines E D, DF: for if the bafe B C agrees with the base EF, but the fides B A, A C do not agree with the fides E D, DF, but change their fituation, as EG, GF, then upon the fame right line can be constituted two right lines equal to two other right lines, each to each, at different points on the fame fide, and having the fame ends with the first right lines. But [by prop. 7.] they cannot be fo conftituted. Therefore the bafe B C agreeing with the base E F, the fides BA, AC cannot difagree with the fides E D, D F; they will therefore agree: and fo the angle B A C will agree with the angle EDF, and be equal to it. A If therefore two triangles have two fides of the one equal to two fides of the other, each to each, and have also equal bafes; then fhall the angle contained under the equal right lines of the one, be equal to the angle contained under the equal right lines of the other. Which was to be demonftrated. B F A Proclus gives the following direct theorem. Let the bafe в c be conceived to be put upon the base EF, so that the triangle A B C falls not upon the triangle EDF, but on the contrary way, as does the triangle A E F. Then the two fides DF, FA either both make one right line, viz. when the angles c, F are right angles, or not: if they both make one right line, as DA, the propofition is thus proved. Becaufe in the triangle A E D, the two fides A E, DB are put equal (for now the right line A E is the fame with AB, which by the fuppofition is equal to the right line D E) the angles A and D upon the base AD, will be [by 5. 1.] equal to one another. Which was to be demonftrated. Now if neither D F, B F A, nor D E, E A, both make one right line, from D to A draw a right line DA; which will either fall within the triangles or without them. First, let it fall within them, which will happen when the angles at E, F are acute. Then becaufe in the triangle A E D the two fides A E, ED are put equal, the two angles E A D, EDA at the bafe AD will [by 5. 1.] be equal B to one another. By the fame reason, fince the two fides a F, DF are equal [by fuppofition] the two angles FA D, FDA upon the bafe D A will be equal. If therefore thefe equals be added to thofe, the whole angles EAF, EDF will [by axiom 2.] be equal. Which was to be demonstrated. Again, let the right line D A fall without the triangles, which will happen when the angles at F are obtufe; then because in the the triangle AED the two fides AE, DE are fuppofed to be equal. The angles E AD, EDA [by 5. 1.] will be equal. By the fame reason, became the two fides A F, FD of the triangle AFD are equal, the angles FAD, FDA upon the bafe AD will be equal. These therefore being taken away from the former, and there will remain [by axiom 3.] the angles E A F, E D F equal. Which was to be demonftrated. .. PROP. IX. PRO BL. To bifect or cut a given right-lined angle into two equal parts. Let the given right-lined angle be BAC; it is required to bifect or cut the fame into two equal parts. Let there be taken any point, as D in the right line в A, and from the right line AC [by: prop. 3.] cut off the right line A E equal to the right line AD; and draw the right. line DE; and upon the right line DE [by: prop. 1.] describe the equilateral triangle DEF, and draw the right line A F: Ifay, the angle B A C is cut into two equal parts by the right line a F. D B F For because the right line A D is equal to the right line A E, and A F is common, therefore the two fides DA, AF will be equal to the two fides E A, A F, each to each; and the bafe DF is equal to the bafe E F. Therefore the angle BAF is equal to the angle EAF [by prop. 8.]. Therefore the given right-lined angle B A C is cut into two equal parts by the right line AF. Which was to be done. a The operation would be fomewhat more elegant, if an ifofceles triangle DEF were defcribed or conftructed upon the right line DE, having one of its equal fidés DF or E F equal to AD Or A E. Hence it appears, how any given angle may be divided into 4, 8, 16, 32, &c. equal parts, viz. by bifecting each part again. But from Euclid's poftulatums cannot be obtained the divifion generally of an angle into any other number of equal parts, as 3, 5, 6, 7, 5c. PRO P. PROP. X. PROBL. To cut a given finite right line into two equal parts v. Let A B be the given finite right line, which is to be cut into two equal parts. C Upon A B let there be made the equilateral triangle ABC [by prop. 1.] and cut the angle A CB into two equal parts [by prop. 9.] by the right line c D. I fay, the right line AB is cut into two equal parts at D. D B For because the fide A C is equal to the fide C B, and the fide CD is common; the two fides A C, C D are equal A to the two fides B C, C D, each to each; and the angle A C D is equal to the angle B CD: therefore the bafe AD [by prop. 4.] is equal to the bafe BD. Therefore the given finite right line A B is cut into two equal parts in the point D. Which was to be done. The folution of this problem is really no more than defcribing two equilateral triangles, the one on one fide the given right line A B, and the other on the other fide that right line; and joining the angles of each of thefe triangles, which are oppofite to the given right line A B, by a right line, which will cut the given right line A в into two equal parts. PROP. XI. PROB L. From a point given in a given right line, to draw a right line at right angles to it. Let there be a given right line A B, and let the given point in the fame be c; it is required to draw a right line from the given point c at right angles to the right line A B. In the right line AB let there be taken any point D; and [by prop. 3.] put the right line C E equal to the right line CD; and [by prop. 1.] upon the right line D E defcribe the equilateral triangle F D E, and draw the right line F C : I fay, from the given point c in the given right line AB is drawn the right line C F at right angles to A B. For because the right line C D is equal to the right line. CE, and the right line C F is common; there are two right lines D C, GF, equal to the two right lines E C, C F, each C to |