from A C cut off A E equal to A D (I. 3). Join D E. On the side of DE remote from A describe the equilateral Then the straight line A F shall bisect the angle B A C. and AF is common to the two triangles D AF, EAF; and the base D F is equal to the base E F (constr.) therefore the angle D A F is equal to the angle EAF (I. 8). Therefore the angle B A C is bisected by A F. Proposition X. Problem. Q. E. F. To bisect a given finite straight line, that is, to divide it into two equal parts. Let A B be the given straight line, It is required to bisect it. On A B describe the equilateral triangle A B C (I. 1). by the straight line CD meeting A B in the point D (I. 9). Then A B shall be bisected in the point D. Because A C is equal to C B (constr.), and CD is common to the two triangles AC D, B C D, the sides A C, C D are equal to the sides B C, C D, each to each, and the angle A C D is equal to the angle BCD (constr.). Therefore the base A D is equal to the base D B (I. 4). Therefore the straight line A B is bisected in the point D. Q. E. F. Proposition XI. Problem. To draw a straight line at right angles to a given straight line, from a given point in the same. Let A B be the given straight line, and C a given point in it. It is required to draw a straight line from the point C at right angles to A B. In A C take any point, D. Make C E equal to CD (I. 3). On D E describe the equilateral triangle D E F (I. 1). Then C F shall be at right angles to A B. and F C is common to the two triangles D C F, E CF, therefore D C, C F are equal to E C, C F, each to each, and the base D F is equal to the base E F (constr.), therefore the angle D C F is equal to the angle E C F (I. 8), and these are adjacent angles. But when a straight line standing on another straight line makes the adjacent angles equal to one another, each of these angles is called a right angle (Def. 10). Therefore each of the angles DC F, E C F is a right angle, and, from the given point C, in the given straight line A B, a straight line, F C, has been drawn at right angles to A B. Q. E. F. COR. By help of this Problem it may be proved that two straight lines cannot have a common segment. If it be possible, let the two straight lines A B C, A B D have the segment A B common to both of them. E B From the point B, draw B E at right angles to A B (I. 11). Then, because A B C is a straight line, therefore the angle A B E is equal to the angle E B C (Def. 10). And, because A B D is a straight line, therefore the angle A B E is equal to the angle E B D. Therefore the angles E B D, E B C are each equal to the angle ABE therefore the angle E B D is equal to the angle E B C, which is impossible. Therefore two straight lines cannot have a common segment. Proposition XII Problem. To draw a straight line perpendicular to a given straight line of unlimited length, from a given point without it. Let A B be the given straight line, which may be produced both ways to any length, and let C be a point without it. It is required to draw a straight line perpendicular to A B from the point C. H E D Take any point, D, on the other side of A B. From the centre C, at the distance C D, describe the circle E G F, meeting A B in F and G (Post. 3). Bisect F G in H (I. 10). Join C H. Then, the straight line CH, drawn from the given point C, shall be perpendicular to the given straight line A B. Join C F and C G. Then, because F H is equal to H G (constr.), and H C is common to the two triangles F H C, G H C ; FH, H C are equal to G H, H C, each to each. Also the base C F is equal to the base C G (Def. 15), Therefore the angle F H C is equal to the angle G H C (I. 8), and these are adjacent angles. But when a straight line standing on another straight line makes the adjacent angles equal to one another, each of these angles is called a right angle, and the straight line which stands upon the other is called a perpendicular to it (Def. 10). Therefore, from the given point C a perpendicular CH has been drawn to the given straight line A B. Proposition XIII. Theorem. Q. E. F. 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 A B make with C D, upon one side of it, the angles CB A, A B D. Then these two angles are either two right angles or are together equal to two right angles. For, if the angle C B A be equal to the angle A B D, But if the angle C B A be not equal to the angle A B D, And because the angle CBE is equal to the angles CBA, ABE, to each of these equals add the angle E B D ; therefore the angles C B E, E B D are equal to the three angles CBA, ABE, EBD (Ax. 2). Again, because the angle D BA is equal to the angles D B E, EBA, to each of these equals add the angle A B C ; therefore the angles D B A, A B C are equal to the three angles D B E, E B A, A B C (Ax. 2). But the angles CBE, EBD have been proved equal to the same three angles, and things which are equal to the same thing are equal to one another (Ax. 1). Therefore the angles CBE, EBD, are equal to the angles D BA, ABC. But the angles CBE, E B D are two right angles; therefore the angles D B A, A B C are together equal to two right angles. Therefore, the angles which one straight line, etc. Q. E. D. Proposition 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, 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 B C, B D upon the opposite sides of A B make the adjacent angles A B C, A B D together equal to two right angles. Then B D shall be in the same straight line with B C. |