CONSTRUCTION Take any point D, upon the other side of AB, and from the centre C, at the distance CD, describe the circle EGF, meeting AB in F, G; (post. 3) bisect FG in H, (1. 10) and join CF, CH, CG. Then the straight line CH, drawn from the point C, is perpendicular to the given straight line AB. DEMONSTRATION Because FH is equal to HG, (constr.) 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; (def. 15) therefore the angle CHF is equal to the angle CHG; (1. 8) and they are adjacent angles. But when a straight line standing on a 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.' (def. 10.) Therefore CH is perpendicular to AB, and it has been drawn from the point C. Q. E. F. PROP. XIII. — THEOREM. 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. (References - Prop. I. 11; ax. 1, 2; def. 10.) Let the straight line AB make with CD, upon one side of it, the angles CBA, ABD. Then the angles CBA, ABD are either two right angles, or are together equal to two right angles. First, let them be equal to one another. DEMONSTRATION Then, since the straight line AB, standing on the straight line CD, makes the adjacent angles ABC, ABD equal to one another, therefore each of them is a right angle ; (def. 10) wherefore the angles ABC, ABD, are two right angles. From the point B draw BE at right angles to CD. (I. 11.) DEMONSTRATION Because BE is at right angles to CD, (constr.) therefore the angles CBE, EBD are two right angles ; (def. 10) and because the angle CBE is equal to the two angles CBA, ABE, add to each of these equals the angle EBD ; then the angles CBE, EBD, are equal to the three angles CBA, ABE, EBD. (ax. 2.) Again, because the angle DBA is equal to the two angles DBE, EBA, add to each of these equals the angle ABC; then the angles DBA, ABC, are equal to the angles DBE, EBA, ABC. (ax. 2.) But the angles CBE, EBD, have heen proved to be equal to the same three angles; and things which are equal to the same thing are equal to one another, therefore the angles CBE, EBD, are equal to the angles DBA, ABC; (ax, 1) but the angles CBE, EBD are two right angles ; (constr.) therefore the angles DBA, ABC, are together equal to two right angles. (ax. 1.) Wherefore when a straight line, &c. Q. E. D. PROP. XIV.-THEOREM. 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, then these two straight lines shall be in one and the same straight line, (References — Prop. I. 13; ax. 1, 3.) At the point B in the straight line AB, let BC, BD, upon the opposite sides of AB, make the adjacent angles ABC, ABD, equal to two right angles. Then CB shall be in the same straight Une with BD. For if BD be not in the same right line with BC, let BE be in the same right line with it. DEMONSTRATION Because CBE is assumed to be a straight line, and AB meets it in A, therefore the adjacent angles ABC, ABE, must be together equal to two right angles ; (1. 13) but the angles ABC, ABD, are together equal to two right angles ; (hyp.) therefore the angles CBA, ABE, must be equal to the angles CBA, ABD; (ax. 1) take away the common angle ABC, therefore the remaining angle ABE must be equal to the remaining angle ABD; (ax. 3) 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 proved, that no other can be in the same straight line with it but BD, therefore BD is in the same straight line with BC, Wherefore, if at a point, &c. Q. E. D. PROP. XV.-THEOREM. If two straight lines cut each other, then the vertical or opposite angles shall be equal (References -- Prop. 1. 13; ax. 1, 3.) Let the two straight lines AB, CD, cut each other in the point E. Then the anglo AEC shall be equal to the angle DEB, and the angle CEB to the angle AED. B A DEMONSTRATION Because the straight line AE makes with CD the angles CEA, AED, these angles are together equal to two right angles. (I. 13.) Again, because the straight line DE makes with AB the angles AED, DEB, these angles are also together equal to two right angles ; and CEA, AED, have been proved to be together equal to two right angles ; wherefore the angles CEA, AED, are equal to the angles AED, DEB ; (ax. 1) take away the common angle AED, then the remaining angle CEA is equal to the remaining angle DEB. (ax. 3.) In the same manner it can be shown, that the angle CEB is equal to the angle AED: 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 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. -THEOREM. If one side of a triangle be produced, then the exterior angle is greater than either of the interior opposite angles. (References - Prop. I. 3, 4, 10, 15; ax. 9.) Let ABC be a triangle, and let the side BC be produced to D. Then the exterior angle ACD shall be greater than either of the angles ABC or CAB. CONSTRUCTION Bisect AC in E, (1. 10) and join BE, produce BE to F, and make EF equal to BE; (1. 3) and join FC. DEMONSTRATION Because AE is equal to EC, and BE to EF; therefore the base AB is equal to the base CF; (L 4) 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 the angle BAE. In the same manner, if the side BC be bisected, and AC be produced to G, it may be demonstrated that the angle BCG, ABC. R. E. D. PROP. XVII. -THEOREM. Any two angles of a triangle are together less than two right angles. (References - Prop. 1. 13, 16; ax. 4.) Let ABC be any triangle. |