Book I. For GH, HC, are equal to EH, HC; and the bases GC, EC”, equal: Therefore the angles GHC, EHC, are equal b. a ver. 15. Each of them are right angles, and HC perpendicular to AB. Which was required. b PRO P. XIII. THE O R. HEN a right line stands upon a right line, making angles with it, thele angles are either two right angles, or, together, equal to tuo right angles. W". a def. 1o. b II. Cax. I. For, let a right line, AB, stand upon the right line CD, making angles CBA, ABD; these angles shall either be two right angles, or, together, equal to two right angles. For, if CBA, ABD, be equal, they are right anglesa; if not, from the point B draw BE, at right angles, to DC5: Therefore the two right angles CBE, EBD, are equal to the three angles ABC, ABE, EBD; but the two angles ABC, ABD, are equal to the same three angles: Therefore the two angles ABC, ABD, are equal to the two angles CBE, EBD; that is, equal to two right angles Wherefore, &c. Cor. If the two sides of an isosceles triangle ADE, be produced to B, C, the angles below the base will be equal to one another; for the angles ADE, EDB*, are equal to two right angles d, and AED, DEC, are equal to two right angles d; but the angles ADE, AED, above the base, are proved equal o ; Therefore the remaining angles BDE, DEC, are equal f: There: fore, in every isosceles triangle, the angles above the base are equal to one another; and, if the sides be produced, the angles below the base are likewise equal to one another, PRO P. XIV. THE O R. F to any right line, and point therein, two right lines be drawn qual to two right angles, these two right lines will make one contie nued right line. For, if to any right line AB, and point B therein, be drawn two right lines, CB, DB, from the opposite points, C, D, making the angles ABC, ABD, equal to two right angles; then DBC will be one right line. If not, let CBE be one right line; then the angles ABC, ABE, will be equal to two right anglesa; but ABC, ABD, are equal to two right angleş 6; Therefore the a. 13 two angles ABC, ABD, are equal to the two angles ABC, Book I, ABE. Take ABC from both ; then the angle ABE will be equal to the angle ABD, a part to the whole; which cannot be. Wherefore, &c. Cor. Hence two right lines CBD, CBE, cannot have common segment as CB; or BD, BE, cannot both be in a right line with CB. a PRO P. XV. THEO R. IE are equal. Let the right lines AB, CD, mutually cut each other in the point E, the angles AEC, DEB, will be equal; and likewise the angles CEB, AED, equal to one another.' For, because the right line CE falls upon the right line AB, the angles AEC, CEB, are equal to two right anglesa. For the same reason the a 13, angles AED, AEC, are equal to two right anglesa : Therefore the two angles AEC, CEB, are equal to the two angles AEC, AED 5. Take the common angle AEC from both, the remain-Ax. fo. | ing angles CEB, AED, are equal. Again, because AEC, AĚD, are equal to two right angles ?, and ÅED, DEB, equal to two right angles, take the common angle AED from both, the remaining angles AEC, DEB, will be equal Wherefore, &c. COR. 1. Hence, two right lines cutting each other, the angles at the section are equal to four right angles. 2. All the angles constitute about any point are equal to four right angles. Ć Ax. 3. PRO P. XVI. T H E O R. I Let ABC be a triangle, and one of its fides BC be produced to D), the outward angle ACD will be greater than the angle CBA, or BAC. For, bisect AC in Ea; join BE, which produce to F; make a 10. EF equal to EB, and join FC ; then the two fides AE, ÉB, are equal to the two sides FE, EC, and the angles AEB, FEC, equalo: Therefore the bases FC, AB, are equal; and the angies b 15. ECF, C4 Book. I. ECF, EAB, likewise equale : Therefore the angle ACD is greater than the angle BAC. In like manner, if the side BC is bifected in E, EF made equal to AE, and FC joind 15. ed, the angle BCG, or ACD d, is greater than ABC; but ACD is likewise proved greater than BAC. Wherefore, &c. T WO angles of any triangle, however taken, are, together, less than two right angles. a 16. b 13, Let ABC be the triangle, any two angles in it are less than two right angles. For, produce BC both ways to D, E; then, because the outward angle ACD is greater than ABC“, add ACB to both; then the angles ACD, ACB, are greater than ABC, ACB, or, BAC, ACB ; but ACD, ACB, are equal to two right angles b: Therefore ABC, ACB, or ACB, BAC, are lofs than two right angles. For the same reason, ABE, ABC, are greater than BAC, ABC. Wherefore, &c. Cor. Hence, if a right line fall upon two right lines, making the inward angles on the same fue less than two right angles, these lines will meet one another on that fide where the angles are less than right ones. PRO P. XVIII. THEO R. HE greater side of every triangle subtends the greater. angle. a 3. Let ABC be a triangle, and the side AC greater than AB; then the angle ABC will be greater than the angle ACB. For, from the greater AC cut off AD, equal to AB?, join DB ; then, because ADB is greater than ACB , ABD is likewise greater than ACB, and ABC much greater. Wherefore, &c. PRO P. XIX. TH E O R. HE greater angle of every triangle is subtended by the In the triangle ABC let the angle ABC be greater than the Book I. angle BCA; the side AC will be greater than AB. If not, let m AC be either equal or less than AB. If equal, then the angle ABC is equal to ACB ; but it is not b: Therefore AC is not a s. equal to AB. If AC is less than AB, the angle ABC is less o Hyp. than ACB °; but it is not : Therefore AC is not less than AB.C 18. It is therefore greater, since it has been proved neither equal nor less. Wherefore, &c. T W o sides of any triangle, however taken, are greater In any triangle, ABC, two sides of it, however taken, are greater than the third, viz. AB, AC, greater than BC; AC, BC, greater than AB; or BC, AB, greater than AC. For, produce any fide, as BA, to D; make AD equal to ACa; and a 3. join DC: Then, because AD is equal to AC, the angles ADC, ACD, are equalb; but the angle BCD is greater than ACD, 5. that is, than ADC : Therefore the fide BD is greater than BC^; but BD is equal to BA, AC: Therefore the sides BA, AC, are greater than BC. Wherefore, &c. C 19 PRO P. XXI. THEOR. I. of a triangle, to a point within the same, these two right lines will be less than the sides of the triangle, but contain a greater angle. From the extreme points of the right line BC, let the two right lines BD, CD, be drawn to the point D, within the same; these lines shall be less than the sides BA, AC; butthe angle BDC will be greater than BAC. For, produce BD to E; then the two fides BA, AE, are greater than the third side BE * ; add EC to both; then BA, AC, are a 20. greater than BE, ECb. For the fame reason, BE, EC, are b, Ax. 4; greater than BD, DC; but BA, AC, are greater than BE, EC; therefore much greater than BD, DC. But the angle BDC is greater than BAC; for the angle BDC is greater than BEC; and BEC is greater than BAČ “: Therefore BDC is c 16 much greater than BAC. Wherefore, &c. COB, Book I. Cor. Hence BD, DC, are not equal to BA, AC, each to each. Wherefore, if in any case it is thought necessary to prove that part of Prop. VII. when the one point falls within the triengle, it is evident from this. PRO P. XXII. PRO B. O make a triangle, whose sides are equal to three given right lines, if any two of them, however taken, are greater than the third. Let A, B, C, be the three given right lines, any two of which are greater than the third. Take any right line bounded at D, but not bounded at E, from which cut off DF equal to A, FG equal to B, and make GH equal to C; then, with the center F, and distance DF, describe the circle DKL; with the center G, and distance GH, describe the circle KLH; from the point K, where the circles cut each other, draw the right lines FK, KG; then FD is equal to FK"; but FD is equal to A; therefore FK is equal to A. For the same reason GK is equal to C, and FG is equal to B: Therefore the three fides FK, FG, GK, of the triangle FKG, are equal to the three given right lines, A, B, C. Wherefore there is constitute, &c, & Def, 15. P R O P. XXIII. PRO B, T a given point, in any right line, to make an angle en qualto a given right lined angle. Let A be the given point in the right line AB; it is required to make an angle equal to the right lined angle DCE. Assume any points D, E, in the right lines CD, CE, and join DE. At the point A, in the line AB, make a triangle AFG, whose fides are equal to the three right lines CD, CE, DE; then, because the two sides GA, AF, are equal to the two sides CE, CD, each to each, and the bases GF, ED, equal, the angles GAF, ECD, are equal b. Wherefore there is constitute, &c. PRO P. XXIV. T H E O R. IS F two triangles have two sides of the one equal to two sides of the other, each to each, and the angle contained by the two fides of the one greater than the angle contained by the correspond |