therefore BC coinciding with EF, BA and ACshall coincide with ED Book I. and DF. for if the base BC coincides with the base EF, but the sides BA, CA do not coincide with the fides ED, FD, but have a different fituation, as EG, FG; then upon the fame base EFand upon the same fide of it there can be two triangles that have their fides 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. therefore if the base BC coincides with the base EF, a. 7. 1. the fides BA, AC cannot but coincide with the fides ED, DF; wherefore likewise the angle BAC coincides with the angle EDF, and is equal to it. therefore if two triangles, &c. Q. E. D. b. 8. Az. PROP. IX. PRO В. a given rectilineal angle, that is, to divide To bifect it into two equal angles. Let BAC be the given rectilineal angle, it is required to bisect it. • AD; join DE, and upon it describe b 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 base DF is equal to the base B EF; therefore the angle DAF is equal A DE F b. I. I. • to the angle EAF. wherefore the given rectilineal angle BAC C. 8. 1. is bisected by the straight line AF. T 10 bisect Which was to be done. PROP. X. PRO B. a given finite straight line, that is, to di vide it into two equal parts. Let AB be the given straight line; it is required to divide it into two equal parts. Describe a upon it an equilateral triangle ABC, and bisect ba. I. I. b. g. 1. the angle ACB by the straight line CD. AB is cut into two equal parts in the point D. Book I. C. 4. I. See N. a. 3. 1. b. J. 1. 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 base AD is equal to the base c DB, and the straight line AB is divided C into two equal parts in the point D. A. DB Which was to be done. T PROP. XI. PRO B. O draw a straight line at right angles to a given straight line, from a given point in the fame. 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, and make CE equal to CD, and upon DE describe the equilateral triangle DFE, and join FC. the Because DC is equal to CE, 4 c. 8. 1. I. angles DCF, ECF; the two fides AD DC, CF are equal to the two EC, CF, each to each; and the base DF is equal to the base EF; therefore 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 equal to one d. 10. Def. another each of them is called a right d 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. CEB 1 1 COR. By help of this Problem it may be demonstrated that two straight lines cannot have a common fegment. If it be poffible, let the two straight 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, the T O draw B PROP. XII. PRO В. L D a straight line perpendicular to a given straight line of an unlimited length, from given point without it. a Let AB be the given straight 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 center C, at the distance CD, defcribe the circle EGF meet C M b. 3. Poft ing AB in F, G; and bisect FG in H, and join CF, CH, CG. AF the straight line CH drawn from H G D the given point C, is perpendicular to the given straight line AB. Because FHis equal toHG, and HCcommon 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 base CG; therefore d.15.Def.r. the angle CHF is equal to the angle CHG; and they are adjacent e. 8. 1. 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 perpen- f. 10. Def.r. dicular to it. therefore from the given point Ca perpendicular CHhas. been drawn to the given straight line AB. Which was to be done. T PROP. XIII. THEOR. HE angles which one straight line makes with another upon one fide of it, are either two right angles, or are together equal to two right angles. Baok I. Let the straight line AB make with CD, upon one fide of it, the angles CBA, ABD; these are either two right angles, or are together equal to two right angles. a. Def. 10. For if the angle CBA be equal to ABD, each of them is a right A E A b. II. I. d. 1. Ax. angle. but if not, from the point B draw BE at right angles to CD. therefore the angles CBE, EBD are two right angles 2. 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 c. 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 demonftrated to be equal to the same three angles; and things that are equal to the fame 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. 1 Fat a point in a straight line, two other straight cent 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 oppofite fides of AB, make the adjacent angles ABC, ABD equal together to two right angles. BD is in the same straight line with CB. For if BD be not in the fame C straight line with CB, let BE be B کر A E D in the same straight line with it. therefore because the straight line Book I. AB makes angles with the straight line CBE, upon one fide of it, the angles ABC, ABE are together equal a to two right angles; a. 13. 1. 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 angle ABD, the less to the b. 3. As 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. I E. D. F two straight lines cut one another, the vertical, opposite, angles shall be equal. of 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. Because the straight line AE makes with CD the angles CEA, AED, these angles are together equal a to two right angles. again, because the straight line DE makes with AB the angles AED, DEB; C these also are together equal to two right angles. and CEA, AED A B a. 13. 1. have been demonstrated to be e qual to two right angles; where-. fore the angles ĆEA, AED.are equal to the angles AED, DEB. take away the common angle AED, and the remaining angle CEA is equal to the remaining b. 3. As. angle DEB. In the fame 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. |