BC is equal to EF; therefore BC coinciding with EF, BA and Book I. AC fhall epincide with ED and DF; for, if the bafe BC coincides with the bafe EF, but the fides BA, CA do not coincide with the fides ED, FD, but have a different fituation, as EG, FG; then, upon the fame bafe EF, and upon the fame fide of it, there can be two triangles that have their fides which are terminated in one extremity of the bafe equal to one another, and likewise their fides terminated in the other extremity: But this is impoffible ; therefore, if the base BC coin 7. 1. cides with the bafe EF, the fides BA, AC cannot but coincide with the fides ED, DF; wherefore likewife the angle BAC coincides with the angle EDF, and is equal to it. There- b. 8. Ax fore if two triangles, &c. Q. E. D. T PROP. IX. PRO B. O bifect a given rectilineal angle, that is, to divide Let BAC be the given rectilineal angle, it is required to bi fect it. A b I. I. Take any point D in AB, and from AC cut off AE e- a 3. 1. qual to AD; join DE, and upon it defcribe an equilateral triangle DEF; then join AF; the ftraight line AF bifects the angle BAC. Because AD is equal to AE, and AF is common to the two triangles DAF, EAF; the two fides DA, AF, are equal to the two fides EA, AF, each to each; and the bafe DF is e- B qual to the bafe EF; therefore the D E F. C c 8. I. EAF; wherefore the given rectilineal angle BAC is bifected by the flraight line AF. T Which was to be done. PROP. X. PRO B. 'O bisect a given finite ftraight line, that is, to divide Let AB be the given ftraight line; it is required to divide it into two equal parts. Defcribe a upon it an equilateral triangle ABC, and bifect a 1. 1. the angle ACB by the ftraight line CD. AB is cut into two b 9. 1. equal varts in the poin, D. B 3 Becaufe Book I. € 4. I. See N. a 3. I. b f. I. Because AC is equal to CB, and CD common to the two triangles ACD, BCD; the two fides AC, CD are equal to BC, CD, each to each; and the angle ACD is equal to the angle BCD; therefore the bafe AD is equal to the bafe c DB, and the ftraight line AB is divided into two equal parts in the point D. Which was to be done. A PROP. XI. PROB. D B O draw a ftraight line at right angles to a given straight line, from a given point in the fame. T& a Let AB be a given ftraight 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. F Take any point D in AC, and a make CE equal to CD, and upon DE describe the equilateral triangle DFE, and join FC; the ftraight line FC drawn. from the given point C is at right angles to the given ftraight line AB. Because DC is equal to CE, and FC common to the two triangles DCF, ECF; the two A fides DC, CF, are equal to the c 8. I. two EC, CF, each to each; and the bafe DF is equal to the bafe EF; therefore the angle DCF is equal to the angle ECF; and they are adjacent angles. But, when the adjacent angles which one ftraight line makes with another ftraight line are 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 I. to be done. COR. By help of this problem, it may be demonstrated, that two ftraight lines cannot have a common fegment. If it be poffible, let the two ftraight lines ABC, ABD have the fegment AB common to both of them. From the point B draw BE at right angles to AB; and because ABC is a straight line, line, the angle CBE is equal a T PROP. XII. E D B PROB. O draw a ftraight line perpendicular to a given straight line of an unlimited length, from a given point without it. Let AB be the given ftraight line, which may be produced to 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. Take any point D upon the other fide of AB, and from the centre C, at the distance CD, defcribe the circle EGF meeting AB in F, G; and bi. A fect FG in H, and join CF, C E H D CH, CG; the ftraight line CH, drawn from the given point C, is perpendicular to the given ftraight line AB. e Book I. 10. Def. I. b 3. Poft. C IO. I. Becaufe FH is equal to HG, and HC common to the two triangles FHC, GHC, the two fides FH, HC are equal to the two GH, HC, each to each; and the bafe CF is equal d to the d 15. Def. bafe CG; therefore the angle CHF is equal to the angle CHG; and they are adjacent angles; but when a ftraight line standing on a ftraight line makes the adjacent angles equal to one another, each of them is a right angle, and the ftraight line which ftands upon the other is called a perpendicular to it; therefore trom the given point C a perpendicular CH has been drawn to the given ftraight line AB. Which was to be done. TH PROP. XIII. THEOR. HE angles which one straight line makes with another upon the one fide of it, are either two right angles, or are together equal to two right angles. Book I. Let the ftraight line AB make with CD, upon one side of it, the angles CBA, ABD; thefe 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 b II. I. c 2. Ax. a def. fo. right angle; but, if not, from the point B draw BE at right angles to CD; therefore the angles CBE, EBD are two right angles a; and becaufe CBE is equal to the two angles CBA, ABE together, add the angle EBD to each of thefe equals; therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. Again, becaufe 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 demonftrated to be equal to the fame three angles; and things 1. Ax. that are equal to the fame are equal 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. I , at a point in a ftraight line, two other ftraight lines,. upon the opposite fides of it, make the adjacent angles together equal to two right angles, these two straight lines fhall be in one and the fame ftraight line. At the point B in the straight line AB, let the two ftraight lines BC, BD upon the oppofite fides of AB, make the adjacent angles ABC, ABD equal together to two right angles. BD is in the fame ftraight line with CB. For, if BD be not in the fame ftraight line with CB, let BE be A E B D in in the fame ftraight line with it; therefore, because the ftraight Book I. line AB makes angles with the ftraight line CBE, upon one n fide of it, the angles ABC, ABE are together equal a to two a 13. 1. right angles; but the angles ABC, ABD are likewife together equal to two right angles; therefore the angles CBA, ABE are equal to the angles CBA, ABD: Take away the common angle ABC, the remaining angle ABE is equal to the remaining b 3. Ax. angle ABD, the lefs to the greater, which is impoffible; there fore BE is not in the fame ftraight line with BC. And, in like manner, it may be demonftrated, that no other can be in the fame ftraight line with it but BD, which therefore is in the fame ftraight line with CB. Wherefore, if at a point, &c. Q. E. D. IF two straight lines cut one another, the vertical, or oppofite, angles fhall be equal. Let the two ftraight lines AB, CD cut one another in the point E; the angle AEC fhall be equal to the angle DEB, and CEB to ALD. E B D Because the ftraight line AE makes with CD the angles CEA, AED, thefe angles are together equal to two right angles. Again, because the ftraight line DE makes with AB the angles A AED, DEP, thefe alfo are together equal a to two right angles; and CEA, AED have been demonftrated to be equal to two right angles; wherefore the angles CEA, to the angles AED, DEB. Take away the AED, and the remaining angle CEA is equal ing angle DEB. In the fame manner it can be demonftrated that the angles CEB, AED are equal. Therefore, if two Araight lines, &c. Q. E. D. a 13. I. AED are equal COR. 1. From this it is manifeft, that, if two ftraight lines cut one another, the angles they make at the point where they cut, are together equal to four right angles. COR. 2. And confequently that all the angles made by any number of lines meeting in one point, are together equal to four right angles. PROP. |