therefore Wherefore 2. The angle ACD is equal to the angle BCD; A D 3. The base AD is equal to the base BD (I. 4). 4. The straight line AB is divided into two equal parts in the point D. Q.E.F. PROPOSITION 11.-Problem. To draw a straight line at right angles to a given straight line, from a given point in the same. Let AB be the given straight line, and C a given point in it. It is required to draw a straight line from the point Cat right angles to AB. and A. D C Construction. In AC take any point D, and make CE equal to CD (I. 3); upon DE describe the equilateral triangle DEF (I. 1), and join CF. B Then CF, drawn from the point C, shall be at right angles to AB. Proof. Because DC is equal to EC, and FC is common to the two triangles DCF, ECF, 1. The two sides DC, CF are equal to the two sides EC, CF, each to each: therefore 2. The base DF is equal to the base EF (constr.); 3. The angle DCF is equal to the angle ECF (I. 8), and these two angles are adjacent angles. But when the two adjacent angles which one straight line makes with another straight line are equal to one another, each of them is called a right angle (Def. 10); therefore A 4. Each of the angles DCF, ECF is a right angle. Wherefore, from the given point C, in the given straight line AB, 5. FC has been drawn at right angles to AB. Q.E.F. Cor. By help of this problem, it may be demonstrated that two straight lines cannot have a common segment. If it be possible, let the segment AB be common to the two straight lines ABC, ABD. B Construction. From the point B, draw BE at right angles to AB (I. 11). Demonstration. Then, because ABC is a straight line, therefore 1. The angle ABE is equal to the angle EBC (Def. 10). Similarly, because ABD is a straight line, therefore 2. The angle ABE is equal to the angle EBD; but the angle ABE is equal to the angle EBC; wherefore 3. The angle EBD is equal to the angle EBC (Ax. 1), the less equal to the greater angle, which is impossible. Therefore, 4. Two straight lines cannot have a common segment. PROPOSITION 12.-Problem. To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it. Let AB be the given straight line, which may be produced any length both ways, and let C be a point without it. It is required to draw a straight line perpendicular to AB from the point C. Construction. Upon the other side of AB take any point D, and from the centre C, at the distance CD, describe the circle EGF, nieeting AB, produced if necessary, in F and G (Post. 3); bisect FG in H (I. 10), and join CF, CH, CG. E Then the straight line CH drawn from the given point C, shall be perpendicular to the given straight line AB. Proof. Because FH is equal to HG (constr.), and HC is commón to the two triangles FHC, GHC; and 1. The two sides FH, HC are equal to the two GH, HC, each to each, 2. The base CF is equal to the base CG (Def. 15); 3. The angle FHC is equal to the angle GHC (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 them is 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, therefore 4. A perpendicular CH has been drawn to the given straight line AB. Q.E.F. PROPOSITION 13.-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. Let the straight line AB make with CD upon one side of it, the angles CBA, ABD. Then the angles CBA, ABD shall be either two right angles, or shall together equal to two right angles. First, let them be equal to one another. A D C A B Demonstration. Then, since the straight line AB, standing on the straight line CD, makes the adjacent angles ABC, ABD equal to one another, each of them is a right angle (Def. 10); therefore 1. The angles ABC, ABD are two right angles. Secondly, let them not be equal to one another. B Construction. From the point B draw BE at right angles to CD (I. 11). Demonstration. Because BE is at right angles to CD (constr.), therefore 1. The angles CBE, EBD are two right angles (Def. 10). And because the angle CBE is equal to the angles CBA, ABE, add the angle EBD to each of these equals, therefore 2. 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. Therefore 3. The angles DBA, ABC are equal to the three angles DBE, EBA, ABC. But the angles CBE, EDD, have been proved equal to the same three angles, and things which are equal to the same thing are equal to one another. Therefore EBD 4. The angles CBE, EDD are equal to the angles DBA, ABC, but the angles CBE, EBD are two right angles, therefore 5. The angles DBA, ABC are together equal to two right angles (Ax. 1). Wherefore, the angles which one straight line, &c. Q.E.D. PROPOSITION 14.-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. 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 BC. B Construction. For, if BD be not in the same straight line with BC; if possible, let BE be in the same straight line with it. Demonstration. Then because AB meets the straight line CBE, therefore 1. The adjacent angles CBA, ABE are equal to two right angles (I. 13); but the angles CBA, ABD, are equal to two right angles (hyp.), therefore take 2. The angles CBA, ABE are equal to the angles CBA, ABD (Ax. 1); away from these equals the common angle CBA, therefore 3. The remaining angle ABE is equal to the remaining angle ABD (Ax. 3), the less angle equal to the greater, which is impossible; therefore 5. BD is in the same straight line with BC. |