BC is equal to EF. Therefore BC coinciding with EF; Book I. BA and AC shall coincide with ED and DF; for, if the base BC coincides with the base EF; but the sides BA, CA do not coincide with the sides ED, FD, but have a different situation as EG, FG, then, upon the same base EF, and upon the same side of it, there can be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise their sides terminated in the other extremity: But this is impossible; " 7. 1. therefore, if the base BC coincides with the base EF, the sides BA, AC cannot but coincide with the sides ED, DF; wherefore likewise the angle BAC coincides with the angle EDF, and is equalb to it. Therefore if two triangles, &c. & A x. Q. E. D. PROP. IX. PROB. To bisect a given rectilineal angle, that is, to divide it into two equal angles. 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 3. 1. to AD; join DE, and upon it describe b an equilateral triangle DEF; then join AF; the straight line AF bisects the angle BAC. A b 1.1. Because AD is equal to AE, and AF is common to the two triangles DAF, D 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; therefore the angle DAF is equal to the angle EAF; wherefore the given rectilineal angle BAC is bisected by the straight line AF. E F C 8.1. Which was to be done. PROP. X. PROB. 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. Describea upon it an equilateral triangle ABC, and bi- 1. 1. sect the angle ACB by the straight line CD. AB is cut 9. 1. into two equal parts in the point D. : See N. 3.1. PROP. XI. PROB. To draw a straight line at right angles to a given straight line, from a given point in the same. Let AB be a given straight line, and C a point given in it; it is required to draw a straight line from the point C at right angles to AB. Take any point D in AC, anda make CE equal to 1. 1. CD, and upon DE describeb the equilateral triangle DFE, CEB Because DC is equal to CE, and FC common to the A D two triangles DCF, ECF; the two sides DC, CF, are equal to the two EC, CF, each to each; and the base DF is equal to the base EF; there8. 1. fore the angle DCF is equal to the angle ECF; and they are adjacent angles. But, when the adjacent angles which one straight line makes with another straight line are d 10 Def. equal to one another, each of them is called a right angle; therefore each of the angles DCF, ECF, is a right angle. Wherefore, from the given point C, in the given straight line AB, FC has been drawn at right angles to AB. Which was to be done. 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 two straight lines ABC, ABD have the segment AB common to both of them. From the point B draw BE at rightangles to AB; and because ABC is the greater; which is impos segment. sible; therefore two straight lines cannot have a common PROP. XII PROB. To draw a straight line perpendicular to a given Let AB be the given straight line, which may be pro- out it. It is required to draw C a straight line perpendicular to AB from the point C. Take any point D upon E the other side of AB, and H distance CD, describe the A GB bs Post. circle EGF meeting AB in D 1 F, G; and bisects FG in H, and join CF, CH, CG; the 10.1... straight line CH, drawn from the given point C, is perpendicular to the given straight line AB. 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 each; and the base CF is equald 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 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; 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 an- 15 Def. 1. 8.1. Let the straight line AB make with CD, upon one side of it, the angles CBA, ABD: these are either two right angles, or are together equal to two right angles. For if the angle CBA be equal to ABD, each of them is a A 1 E A D B C B C D • Def. 10. rightä angle; but if not, from the point B draw BE at 11. 1. right anglesb to CD; therefore the angles CBE, EBD are two right anglesa; and because CBE is equal to the two angles CBA, ABE together, add the angle EBD to each of these equals; therefore the angles CBE, EBD are * 2 Ax. equal to the three angles CBA, ABE, EBD. Again, because the angle DBA is equal to the two angles DBE, EBA, add to these equals the angle ABC, 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 d 1 Ax. things that are equal to the same are equal d to one another; 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 when a straight line, &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 equal together to two right angles, BD is in the same straight line with CB. C 1 A E For, if BD be not in the same straight line with CB, let Book I. BE be in the same straight line with it; therefore, because the straight line AB makes angles with the straight line CBE, upon one side of it, the angles ABC, ABE are together equal a to two right angles; but the angles ABC, ABD 13.1. are likewise together equal to two right angles; therefore the angles CRA, ABE are equal to the angles CBA, ABD: Take away the common angle ABC, the remaining angle ABE is equalb to the remaining angle ABD, the less to the 63 Ax. 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. PROP. XV. THEOR. 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. to two right angles; and CEA, AED, have been demonstrated to be equal to two right angles; wherefore the angles CEA, AED, are equal to the angles AED, DEB. Take away the common angle AED, and the renaining angle CEA is equal to the remaining angle DEB. In the same 3 Ar. 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 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. C |