COROLLARIES. 1. From this it is manifest, if two right lines cut one another, they make the angles at the point of section equal to four right angles. 2. All the angles placed around one point are equal to four right angles. PROPOSITION XVI. THEOREM. One side of any triangle being produced, the exterior angle is greater than either of the interior and opposite angles. F a 15. 1. E b 4. 1. D Let ABC be a triangle, and let one of its sides BC be produced to D. The exterior angle ACD is greater than either of the interior and opposite angles; namely, the angles CBA and BAC. Bisect Ac in E, and BE being joined, produce it to F, and make EF equal to BE; join also Fc, and produce Ac to G. Therefore because A E is equal to Ec, and BE to EF, the two AE, EB, are equal to the two CE, EF, each to each; and the angle AEB is equal to the angle FEC, for they are vertically opposite. Therefore the base AB is equal to the base FC; and the triangle AEB to the triangle FEC; also the remaining angles to the remaining angles each to each, to which the equal sides are opposite. Therefore the angle BAE is equal to the angle ECF. the angle ACD is greater than ECF. Therefore the angle ACD is greater than the angle BAE. In like manner it may be shown, the right line BC being bisected, that the angle BCG, that is, ACD, is greater than the angle ABC. Therefore one side of a triangle being produced, &c. Q. E. D. PROPOSITION XVII. THEOREM. B E But Two angles of every triangle, howsoever taken, are less than two right angles. Let ABC be a triangle. Two angles of the triangle ABC, howsoever taken, are less than two right angles. Produce BC to D; and because the exterior angle ACD 16. 1. b 13. 1. ☐ 16. 1. b 5.1. a 5. 1. b 18. 1. B A -D of the triangle ACB is greater than Q. E. D. PROPOSITION XVIII. THEOREM. The greater side of every triangle subtends the greater angle. D Let ABC be a triangle having the side AC greater than the side AB. The angle ABC is greater than the angle вCA. For because AC is greater than AB, make AD equal to AB, and join BD. And because ADB is the exterior angle, it will be greater than the interior and opposite angle DCB.a But ADB is equal to ABD, because the side AB is equal to the side AD; therefore the angle ABD is greater than the angle ACB. Wherefore ABC will be much greater than ACB. Therefore the greater side of every triangle, &c. Q. E. D. PROPOSITION XIX. THEOREM. B The greater angle of every triangle subtends the greater side. Let ABC be a triangle having the angle ABC greater than the angle BCA. The side AC is greater than the side AB. For if AC is not greater, it is either equal to it or less; but it is not equal; for then the angle ABC would be equal to the angle ACB; but it is not. Therefore AC is not equal to AB. But neither is it less; for then the angle ABC would be less than the angle ACB, but it is not. There B fore AC is not less than AB. And it was shown that it is not equal. Whence AC is greater than AB. Therefore the greater angle, &c. Q. E. D. Deduction. If from a given point there be drawn any number of right lines to another right line given in position, the perpendicular is the shortest right line, and that which is nearer to the perpendicular is less than that which is more remote, and there can only be drawn two right lines which are equal to one another, one on each side of the perpendicular. PROPOSITION XX. THEOREM. Two sides of every triangle, howsoever taken, are greater than the third side. B D a 3. 1. b 5.1. For let ABC be a triangle, any two sides of the triangle ABC, howsoever taken, are greater than the third side; viz, the sides ba, AC, are greater than the side BC; the sides AB, BC, greater than AC; and the sides BC, CA, are greater than AB. For produce BA to the point D, and make ca equal to AD, and join DC. Therefore because DA is equal to AC, the angle ADC will be equal to the angle ACD. But the angle BCD is greater than the angle ACD; therefore the angle BCD is greater than the angle ADC. And because DCB is a triangle having the angle BCD greater than the angle BDC, and the greater side subtends the greater angle; the side DB will be greater 19. 1. than the side BC; but DB is equal to BA, AC; wherefore the sides BA, AC, will also be greater than BC. like manner we may show that the sides AB, BC, are greater than AC; and the sides BC, CA, greater than AB. Therefore two sides of every triangle, &c. Deductions. In 1. The difference of any two sides of a triangle is less than the third side. 2. Two sides of a triangle are together greater than twice the line from the vertex bisecting the base. с a 20. 1. b Ax. 4. 16. 1. PROPOSITION XXI. THEOREM. If from the ends of one side of a triangle two right lines be drawn within it, these will be less than the other two sides of the triangle, but will contain a greater angle. B E C For from the ends BC in one of the sides BC of the triangle ABC, let two right lines BD, DC, be drawn within it. The sides BD, DC, are less than the two sides BA, AC, of the triangle, but contain the angle BDC greater than the angle BAC. Produce BD to E; and because two sides of every triangle are greater than the third side, the two sides BA, AE, of the triangle ABE, are greater than BE. Add EC, which is common. Therefore BA, AC, are greater than BE, EC. Again, because CE, ED, two sides of the triangle CED, are greater than CD, add DB, which is common: wherefore CE, EB, are greater than CD, DB. But it has been shown that BA, AC, are greater than BE, EC; much more then are BA, AC, greater than BD, DC. Again, because the exterior angle of every triangle is greater than the interior and opposite angle, the exterior angle BDC of the triangle CDE will be greater than the interior and opposite angle CED. For the same reason, the exterior angle CEB of the triangle ABC is greater than BAC. But the angle BDC was shown to be greater than the angle CEB; much more then will the angle BDC be greater than the angle BAC. Wherefore if from the ends, &c. PROPOSITION XXII. PROBLEM. Q. E. D. To make a triangle of three right lines which shall be equal to three given right lines. But any two of these lines, howsoever taken, will be greater than the third; because any two sides of a triangle, howsoever taken, are greater than the third side. Let A, B, C, be three given right lines, two of which, howsoever taken, are greater than the third, viz. a, b, greater than c; A, C, greater than в; and B, C, greater than A. It is required to make a triangle whose three right distance GH describe another circle KLH, and join KF, с C circle DKL, FD will be equal to FK ; but FD is equal Def, 15. to A; therefore FK is also equal to A. Again, because the point G is the centre of the circle LKH, GH will be equal to GK, but GH is equal to c; therefore GK, will be equal to c. And FG is equal to B. Therefore the three right lines KF, FG, GK, are equal to the three A, B, Therefore the triangle KFG has been made whose three sides KF, FG, GH, are equal to the three right lines A, B, C. Q. E. F. Deduction. C. It is required to construct a rectilineal figure of any number of sides, having given the length of each side of all the triangles into which the figure is divided. PROPOSITION XXIII. PROBLEM.* To a given right line, and to a given point in it, to make a rectilineal angle equal to a given rectilineal angle. Let AB be the given right line, and a the given point in it; also DCE the given rectilineal angle. It is required therefore to the given right line AB, and to the point A in it, to make a rectilineal angle equal to the given rectilineal angle DCE. Take in each of them CD, CE, the * Apollonius has given a more simple and easy solution of this Problem; but as the demonstration of that method requires the assistance of Prop. 27, Book 3, Euclid could not introduce it into this place; since a well connected chain of consequences, and an uniform assumption of principles, previously demonstrated, were the main objects of Euclid's plan, and which, I think, constitute the beauty and superiority of his Elements. |