* 3 Post. * 10. 1. + Constr. 15 Def. *8. 1. † 10 Def. † 10 Def. * 11. 1. * 10 Def. 2 Ax. Let AB be the given straight line, which may be produced to any length both ways, and let C be a point without it; it is required to draw a straight line perpendi cular to AB, from the point C. Take any point D upon the other side of AB, and from the centre C, at the distance A GB in F, G; bisect* FG in H, and join CH: the straight line CH, drawn from the given point C, shall be perpendicular to the given straight line AB. Join CF, CG and because FH is equal† to HG, and HC common to the two triangles FHC, GHC, the two sides FH, HC are equal to the two GH, HC, each to each; and the base CF is equal to the base CG: therefore the angle CHF is equal to the angle CHG; and they are adjacent angles: but when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of them is a right angle, and the straight line. which stands upon the other, is called a perpendicular to it: therefore, from the given point C, a perpendicular CH has been drawn to the given straight line AB. Which was to be done. PROPOSITION XIII. THEOR.-The angles which one straight line makes with another upon one side of it, are either two right angles, or are together equal to two right angles. Let the straight line AB make with CD, upon one side of it, the angles CBA, ABD: these shall either be two right angles, or shall together be equal to two right angles. For if the angle CBA be equal to ABD, each of them is a right angle: but if not, from the point B draw BE at right angles to CD; therefore the angles CBE, EBD are two right angles and because the angle CBE is equal to the each of these equals; therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. Again, because the angle DBA is equal to the two angles DBE, EBA, add to each of these equals the angle ABC; therefore the angles * 1 Ax. DBA, ABC are equal to the three angles DBE, EBA, ABC: † 2 Ax. but the angles CBE, EBD have been demonstrated to be equal to the same three angles; and things that are equal to the same thing, are equal to one another; therefore the angles CBE, EBD are equal to the angles DBA, ABC: but CBE, EBD are two right angles; therefore, DBA, ABC are together equal to two right angles. Wherefore, the angles which +1 Ax. one straight line, &c. Q. E. D. THEOR.-If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line. At the point B, in the straight line AB, let the two straight lines BC, BD upon the opposite side of AB, make the adjacent angles ABC, ABD equal together to two right angles: BD shall be in the same C straight line with CB. A †1 * 13. 1. † 1 Ax. For if BD be not in the same straight line with CB, let BE be in the same straight line with it: therefore, because the straight line AB makes with the straight line CBE, upon one side of it, the angles ABC, ABE, these angles are together equal to two right angles; but the angles ABC, ABD are likewise together equal † to two right angles; therefore, † Hyp. the angles CBA, ABE are equal + to the angles CBA, ABD: take away the common angle ABC, and the remaining angle ABE is equal to the remaining angle ABD, the less to the greater, 3 Ax. which is impossible; therefore BE is not in the same straight line with BC. And in like manner, it may be demonstrated, that no other can be in the same straight line with it but BD, which, therefore, is in the same straight line with CB. Wherefore, if at a point, &c. Q. E. D. * PROPOSITION XV. THEOR If two straight lines cut one another, the vertical, or opposite, angles shali be equal, Let the point Et wo straight lines AB, CD cut one another in the CEB to AED. * 13. 1. * 13. 1. † 1 Ax. * 3 Ax. Because the straight line AE makes with CD the angles CEA, AED, these angles are together equal to two right angles. Again, because the straight line DE makes with AB the angles AED, DEB, these also are * together equal to two right angles; and Q. E. D. COR. 1. From this it is manifest, that if two straight lines cut one another, the angles which they make at the point where they cut, are together equal to four right angles. COR. 2. And consequently, that all the angles made by any number of lines meeting in one point, are together equal to four right angles. * 10. 1. † 3. 1. + Constr. * 15. 1. * 4. 1. † 9 Ax. PROPOSITION XVI. THEOR-If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. Let ABC be a triangle, and let its side BC be produced to D: the exterior angle ACD shall be greater than either of the interior opposite angles CBA, BAC. Bisect * AC in E; join BE and produce it to F, and make EF equal to BE, and join FC. Because AE is equal † to EC, and BE † to EF; AE, EB are equal to CE, EF, each to each; and the angle AEB is equal * to the angle CEF, because they are opposite vertical angles; therefore the base AB is equal to the base CF, and the triangle F E B D G AEB to the triangle CEF, and the remaining angles to the re- be demonstrated, that the angle BCG, that is, the angle * ACD, is greater than the angle ABC. Therefore, if one side, &c. Q. E. D. * 15. 1. PROPOSITION XVII. THEOR.-Any two angles of a triangle are together less than two right angles. Let ABC be any triangle: any two of its angles together shall be less than two right angles. * 16. 1. B † 4 Ax. Produce BC to D: and because ACD is the exterior angle of the triangle ABC, ACD is greater than the interior and opposite angle ABC; to each of these add the angle ACB; therefore the angles ACD, ACB are greater + than the angles ABC, ACB: but ACD, ACB are together equal to two right angles; therefore the angles 13. 1. ABC, BCA are less than two right angles. In like manner, it may be demonstrated, that BAC, ACB, as also CAB, ABC, are less than two right angles. Therefore, any two angles, &c. Q. E. D. * PROPOSITION XVIII. THEOR.-The greater side of every triangle is opposite to the greater angle. Let ABC be a triangle, of which the side AC is greater than the side AB: the angle ABC shall be greater than the angle BCA. Because AC is greater than AB, make* B AD equal to AB, and join BD: and because D * 3. 1. 16. 1. * ADB is the exterior angle of the triangle BDC, it is greater than the interior and opposite angle DCB; but ADB is equal 5. 1. to ABD, because the side AB is equal † to the side AD; there- † Constr. fore the angle ABD is likewise greater than the angle ACB; therefore much more is the angle ABC greater than ACB. Therefore, the greater side, &c. Q. E. D. PROPOSITION XIX. THEOR.-The greater angle of every triangle is subtended by C * 5. 1. + Hyp. * 18. 1. See N. * 3. 1. * 5. 1. † 9 Ax. * 19. 1. See N. *Constr. † 2 Ax. than the angle BCA: the side AC shall be greater than the side AB. For if it be not greater, AC must either be equal to AB, or less than it it is not equal, because then the angle ABC would be equal to the angle CAB; but it is not; therefore AC is not equal to AB: but it is not; therefore the side AC is not B it is not equal to AB: therefore AC is greater than AB. Q. E. D. PROPOSITION XX. THEOR.-Any two sides of a triangle are together greater than the third side. Let ABC be a triangle: any two sides of it together shall be greater than the third side; viz. the sides BA, AC greater than the side BC; and AB, BC greater than AC; and BC, CA greater than AB. Produce BA to the point D, and make * Because DA is equal to AC, the angle * B D the angle BCD is greater than the angle ADC: and because the angle BCD of the triangle DCB, is greater than its angle BDC, and that the greater angle is subtended by the greater side, therefore the side DB is greater than the side BC: but DB is equal to BA and AC; therefore the sides BA, AC are greater than BC. In the same manner it may be demonstrated, that the sides AB, BC are greater than CA; and BC, CA greater than AB. Therefore, any two sides, &c. Q. E. D. PROPOSITION XXI. THEOR.-If from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle. Let ABC be a triangle, and from the points B, C, the ends Because AD is equal to AC, add BA to each, therefore the whole BD is equal to the two BA, AC. |