Let BAC be the given rectilineal angle: it is required to bisect it. Take any point D in AB, and from AC cut off AE equal to AD; [I. 3. join DE, and on DE, on the side remote from A, describe the equilateral triangle DEF. [I. 1. Join AF. The straight line AF shall bisect the angle BAO. Because AD is equal to AE, Construction. and AF is common to the two triangles DAF, EAF, the two sides DA, AF are equal to the two sides EA, AF, each to each; and the base DF is equal to the base EF; [Definition 24. therefore the angle DAF is equal to the angle EAF. [I. 8. Wherefore the given rectilineal angle BAC is bisected by the straight line AF. Q.E.F. PROPOSITION 10. PROBLEM. To bisect a given finite straight line, that is to divide it into two equal parts. Let AB be the given straight line : it is required to divide it into two equal parts. Describe on it an equilateral triangle ABC, [I. 1. and bisect the angle ACB by the straight line CD, meeting AB at D. [I. 9. AB shall be cut into two equal parts at the point D. Because AC is equal to CB, [Definition 24. and CD is common to the two triangles ACD, BCD, the two sides AC, CD are equal to the two sides BC, CD, each to each; and the angle ACD is equal to the angle BCD; [Constr. therefore the base AD is equal to the base DB. [I. 4. Wherefore the given straight line AB is divided into two equal parts at 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 the given point in it: it is required to draw from the point C a straight line at right angles A b ó to AB. Take any point D in AC, and make CE equal to CD. [I. 3. On DE describe the equilateral triangle DFE, [I. 1. and join CF. The straight line CF drawn from the given point C shall be at right angles to the given straight line AB. Because DC is equal to CE, [Construction. and CF is common to the two triangles DCF, ECF; the two sides DC, CF are equal to the two sides ÉC, CF, each to each ; and the base DF is equal to the base EF; [Definition 24. therefore the angle DCF is equal to the angle ECF; (1. 8. 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 the angles is called a right angle; [Definition 10. therefore each of the angles DCF, ECF is a right angle. Wherefore from the given point ( in the given straight line AB, CF has been drawn at right angles to AB. Q.E.F. Corollary. By the help of this problem it may be shewn that two straight lines cannot have a common segment. E If it be possible, let the two straight lines ABC, ABD have the segment AB common to both of them. From the point B draw в BE at right angles to AB. Then, because ABC is a straight line, [Hypothesis. the angle CBE is equal to the angle EBA. [Definition 10. Also, because ABD is a straight line, (Hypothesis. the angle DBE is equal to the angle EBA. Therefore the angle DBE is equal to the angle CBE, [Ax. 1. the less to the greater; which is impossible. [Axiom 9. Wherefore 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 to any length both ways, and let C be the given point without it: it is required to draw from the point C a straight line perpendicular to AB. Take any point Don the other side of AB, and from the centre C, at the distance CD, describe the circle EGF, meeting AB at F and G. [Postulate 3. Bisect FG at H, [I. 10. 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. [Construction. and HC is common to the two triangles FHC, GHC; the two sides FH, HC are equal to the two sides GH, HC, each to each ; and the base CF is equal to the base CG; [Definition 15. therefore the angle CHF is equal to the angle CHG ; [I. 8. 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 the angles is called a right angle, and the straight line which stands on the other is called a perpendicular to it. [Def. 10. Wherefore a perpendicular CH has been drawn to the given straight line AB from the given point C with: out it. Q.E.F. GB PROPOSITION 13. THEOREM. The angles which one straight line makes with another straight line on one side of it, either are two right angles, or are together equal to two right angles. Let the straight line AB make with the straight line CD, on one side of it, the angles CBA, ABD: these either are two right angles, or are together equal to two right angles. For if the angle CBA is equal to the angle ABD, each of them is a right angle. [Definition 10. But if not, from the point B draw BE at right angles to [I. 11. therefore the angles CBE, EBD are two right angles.[Def.10. Now the angle CBE is equal to the two angles CBA, ABE; to each of these equals add the angle EBD; therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. [Axiom 2. Again, the angle DBA is equal to the two angles DBE, EBA; to each of these equals add the angle ABC; therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC. [Axiom 2. But the angles CBE, EBD have been shewn to be equal to the same three angles. Therefore the angles CBE, EBD are equal to the angles DBA, ABC. . [Axiom 1. But CBE, EBD are two right angles; therefore DBA, ABC are together equal to two right angles. Wherefore, the angles &c. Q.E.D. PROPOSITION 14. THEOREM. If, at a point in a straight line, two other straight lines, on 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, on the opposite sides of AB, make the adjacent angles ABC, ABD together equal to two right angles: BD shall be in the same straight line with CB. For if BD be not in the same straight line with CB, let BE be in the same straight line with it. Then because the straight line AB makes with the straight line CBE, on one side of it, the angles ABC, ABE, these angles are together equal to two right angles. [I. 13. But the angles ABC, ABD are also together equal to two right angles. [Hypothesis. Therefore the angles ABC, ABE are equal to the angles ABC, ABD. From each of these equals take away the common angle ABC, and the remaining angle ABE is equal to the remaining angle ABD, [Axiom 3. the less to the greater; which is impossible. Therefore BE is not in the same straight line with CB. And in the same manner it may be shewn that no other can be in the same straight line with it but BD; therefore BD is in the same straight line with CB. Wherefore, if at a point &c. Q.E.D. PROPOSITION 15. THEOREM. If two straight lines cut one another, the vertical, or opposite, angles shall be equal. |