the outward angle is greater than either of the inward and opposite angles. Which was to be demonftrated, PROP. XVII. THEOR. Any two angles of every triangle taken together, are lefs than two right angles. Let there be a triangle ABC: I fay, any two angles of the triangle A B C taken together, are lefs than two right angles. For [by poft, 2.] produce в C to D. Then becaufe the outward angle A C D of the triangle A CB, is [by prop. 16.] greater than the inward and oppofite angle' A B C; add the common angle A CB; then the angles A B D ACD, ACB are greater than the angles A B C, B C A. But the angles A CD, ACB [by prop. 13.] are equal to two right angles wherefore the angles A B C, B C A are less than two right angles. After the fame way we demonftrate that the angles B A C, A C B are less than two right angles; as alfo the angles C A B, A B C. Therefore any two angles of every triangle taken together, are less than two right angles, Which was to be demonftrated. PROP. XVIII. THEOR. The greater angle of every triangle is opposite to the greater fide. Let there be a triangle A B C, having the fide A c greater than the fide AB: I fay, the angle ABC fhall also be greater than the angle B C A. For because the fide A c is A greater than the fide A B, put [by prop. 3.] the right line A D equal to the fide A B, and draw the right line BD: Then because ADB is an outward angle of the triangle BDC, itis greater [by prop. 16.] C4 D B than than the inward oppofite angle D C B. But [by prop. 5.] the angle A D B is equal to the angle A B D, because the fide AB is equal to the fide A D; and therefore the angle ABD is greater than the angle A CB; confequently the angle ABC is much greater than the angle AC B. Therefore the greater angle of every triangle is oppofite to the greater fide. Which was to be demonstrated. The greater fide of every triangle is oppofite to the greater angle. Let there be a triangle ABC, having the angle A C greater than the angle BCA: I fay, the fide AC is greater than the fide A B. A B For if it be not greater, the fide AC is either equal to the fide A B, or less than it. But the fide A C is not equal to the fide A B; for then the angle A B C would be equal [by prop. 5.] to the angle AC B. But it is not equal to it: therefore the Neither is But it fide AC will not be equal to the fide A B. the fide A c less than the fide A B; for then the angle A B C would be [by prop. 18.] lefs than the angle A C B. is not lefs; nor therefore will the fide A C be less than the fide A B. But it has been demonstrated not to be equal to it; wherefore the fide A C is greater than the fide A B, Wherefore the greater fide of every triangle is oppofite to the greater angle. Which was to be demonftrated. PROP. XX. THEOR. Any two fides of every triangle taken together, are greater than the remaining fide f. For let there be a triangle ABC: I fay, any two fides of this triangle taken together, are greater than the remaining fide; that is, BA, AC together, greater than the fide BC; AB, BC together, greater than the fide A C; and BC, CA together, greater than the fide A. For For produce the fide B A to the point D, and [by prop. 3.] put D A equal to the right line C A, and join D C. D Then because D A is equal to the right line AC, the ingle ADC [by prop. 5.] will be equal to the angle ACL 5 but [by ax. 9.] the angle B C D is greater than the angle ACD; therefore the angle ACD is greater than the angle A DC. Alfo, because the triangle D C B has the angle B C D greater than the angle BDC, and [by prop. 19.] the greater fide is oppofite to the B A C greater angle, the fide D B will be greater than the fide B C. But the right line D B is equal to the fides, A B, A C ; therefore the fides B A, AC together, are greater than the fide BC, After the fame manner we demonstrate, that the fides A B, BC together, are greater than the fide C A, and the fides B C, CA together, greater than the fide A B. Therefore any two fides of every triangle, taken together, are greater than the third fide. Which was to be demonftrated. Some think this propofition too felf evident to require a demonftration. But Euclid thought the feweft axioms, or indemonftrable propofitions was beft; and therefore has every where given demonflration when he could. If from the ends of any fide of a triangle, two right lines be conftituted within the triangle, thefe lines fball indeed be less than the two remaining fides of the triangle, but will contain a greater angle. For from the ends B, C, of one of the fides B C of the triangle ABC, let two right lines B D, CD be conftituted within the triangle: I fay, the right lines B D, DC are less than the two remaining fides B A, A C of the triangle, but yet do contain an angle B D C greater than the angle B A C. For continue out the right line B D to the point E. Then because [by prop. 20.] the two fides of every triangle are greater than the third fide remaining, the two des A B, AE are greater than the fide BE. Add the right line line E C, which is common: therefore the fides B A, A C are greater than the right lines BE, EC. Again, because the two fides c E, E D of the triangle CED are greater than the fide c D, let D B, which is common, be added; then the right lines CE, EB are greater But the fides B A, A C, than the right lines CD, D B. have been proved to be greater than the right lines в E, E C: wherefore B A, A C are much greater than BD DC. Again, because the angle, which is without any triangle, is [by prop. 16.] greater than either of the internal and oppofite angles, the angle B D C, which is without the triangle CD E, is greater than the angle c E D. By the fame reafon the angle CEB, which is without the triangle A B E, is greater than the angle B A C. But the angle B D C has been proved to be greater than the angle CEB: therefore the angle B D C is much greater than the angle B AC. If therefore from the ends of any fide of a triangle, two right lines be constituted within the triangle, these lines fhall indeed be less than the two remaining fides of the triangle, but will contain a greater angle. Which was to be demonftrated. PROP. XXII. PROBL. To conftitute a triangle, whofe fides fhall be equal to three given right lines: but any two of these taken together, must be greater than that which remains. Let the given right lines be A, B, C, whereof any two taken together are greater than that remaining; viz. A and B greater than C; alfo A and c greater than B; and B and c greater than A; it is required to constitute a triangle, whofe three fides fhall be equal to A, B, C. Let a right line D E be put finite at D, but infinite towards E, and put [by prop. 3.] D F equal to the right line A, the right line F G equal to the right line B, and let GH be equal to the right line c, and about the centre F, with the distance » D, let a circle D K L be described [by post. 3.], And lines A, B, C. A B C F is the centre of the circle D K L, the right line F D is [by def. 15.] equal to the right line F K; but FD is equal to the right line A: therefore alfo the right line F K is equal to A. Again, because the point G is the centre of the circle L K H, therefore G H is equal to the right line GK; but GH is equal to the right line c: therefore alfo G K is equal to the right line c, and F G is alfo equal to the right line B wherefore the three right lines KF, FG, G K, are equal to the three right lines A, B, C. Therefore the triangle K F G is conftituted, whose three fides K F, F G, G K are equal to the three given right lines A, B, C. Which was to be done. PROP. XXIII. PROBL To make a right-lined angle at a given right line, and at a given point in it, equal to a given right-lined angle 8. Let A B be the given right line, and A the point given therein; and let the right-lined angle given be DCE; it is required to make a right-lined angle at the given point A, with the line A B, equal to the given right-lined angle DCE. In each of the right lines CD, CE take any two points D, E, and draw the right line DE, and of three right linesequal to the three right lines CD, DE, CE, A conftitute [by prop. 22.] E a triangle AF G; fo that the right line CD be equal G B to AF, the right line CE to A G, and the right line D E to the right line FG. |