Take any point D upon the other side of AB, and * 3 Post. from the centre C, at the distance CD, describe the *10. 1. circle EGF meeting AB in F, G; and 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. * 8. 1. 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 *15 Def. 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. PROP. 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 be either two right angles, or shall be together equal to two right angles. For if the angle CBA be equal to ABD, each of * 10 Def. them is a right* A A angle: but if not, from the point B draw BE at right * 11. 1. angles to CD; therefore the angles *10 Def. CBE, EBD are two right angles; and because the angle CBE is equal to the two angles CBA, ABE, add the angle EBD to each of these equals; therefore the angles CBE, EBD are equal to the three angles CBA, * 2 Ax. ABE, EBD. Again, because the angle DBA is equal to the two angles DBE, EBA, add the angle ABC to each of these equals; therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC; 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; * 1 Ax. 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, &c. Q. E. D. PROP. XIV. 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 sides of AB, make the adjacent angles ABC, ABD together equal to two right angles; then BD shall be in the same straight line with CB. B E D 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; * 13. 1. but the angles ABC, ABD are likewise together equal to two right angles; therefore the angles CBA, ABE are equal to the angles CBA, ABD: take away the common angle ABC, and the remaining angle ABE is *3 Ax. equal to the remaining angle ABD, the less to the greater, 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. If two straight lines cut one another, the vertical, or opposite, angles shall be equal. Let the two straight lines AB, CD cut one another in the point E; the angle AEC shall be equal to the angle DEB, and CEB to AED. Because the straight line AE makes with CD the angles CEA, AED, these с * 13. 1. angles are together equal to two right angles. Again, be- A cause the straight line DE E B 13. 1. AED, DEB, these also are together equal to two right angles; and the angles CEA, AED have been demonstrated to be together equal to two right angles; wherefore the angles CEA, AED are equal to the angles AED, DEB. Take away the common angle * 3 Ax. AED, and the remaining angle CEA is equal to the remaining angle DEB. In the same manner it can be demonstrated that the angles CEB, AED are equal. Therefore, if two straight lines, &c. 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. PROP. 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 in BE produced 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 an B A * 10. 1. F * 15. 1. gles; therefore the base AB is equal to the base CF, * 4. 1. and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite; wherefore the angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle ECF; therefore the angle ACD is greater than BAE. In like manner, if the side BC be bisected, and AC be produced to G, it may be demonstrated that the angle BCG, that is, * * 15. 1. the angle ACD, is greater than the angle ABC. Therefore, if one side, &c. Q. E. D. PROP. 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. Produce BC to D; and because ACD is the exterior angle of the triangle ABC, ACD is * 16. 1. greater than the interior and opposite angle ABC; to each of * these add the angle ACB; there B fore the angles ACD, ACB are greater than the angles 13. 1. ABC, ACB; but ACD, ACB are together equal to two right angles; therefore the angles ABC, BCA are less than two right angles. In like manner, may be demonstrated, that BAC, ACB, as also CAB, ABC, are less than two right angles. Therefore any two angles, &c. Q. E. D. 3. 1. * 5. 1. 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 Because AC is greater than AB, B C make* AD equal to AB, and join BD; and because ADB is the exterior angle of the triangle BDC, it is 16. 1. greater than the interior and opposite angle DCB; but ADB is equal to ABD, because the side AB is equal to the side AD; therefore the angle ABD is likewise greater than the angle ACB; much more then is the angle ABC greater than ACB. Therefore the greater side, &c. Q. E. D. PROP. XIX. THEOR. The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it. Let ABC be a triangle of which the angle ABC is |