Let ABC be a triangle having the angle ABC equal to the angle Book I. ACB; the fide AB is alfo equal to the fide AC. For if AB be not equal to AC, one of them is greater than the o A D ther, let AB be the greater, and from it cut off DB equal to AC, the a. 2.1. lefs, and join DC. therefore because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two fides DB, BC are equal to the two AC, CB, each to each; and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC is equal to the triangle ACB, the less to the greater; which is abfurd. Therefore AB is not unequal to AC, that is, it is equal to it. Wherefore if two angles, &c. Q. E. D. B COR. Hence every equiangular triangle is alfo equilateral. UPON PROP. VII. THEOR. AC b. 4. 1. PON the fame base, and on the fame fide of it, see N. there cannot be two triangles that have their fides which are terminated in one extremity of the bafe equal to one another, and likewife thofe which are terminated in the other extremity. If it be poffible, let there be two triangles ACB, ADB upon the faine base AB, and upon the fame fide of it, which have their fides CA, DA, terminated in the extremity A of the bafe, equal to one another, and likewise their fides CB, DB that are terminated in B. Join CD; then, in the cafe in which the Vertex of each of the triangles is without the other triangle, becaufe AC is equal to AD, the angle ACD is equal to the angle ADC. but the angle ACD is greater than the angle C D BCD, therefore the angle ADC is great- A B er also than BCD; much more then is the angle BDC greater than the angle BCD. again, because CB is equal to DB, the angle BDC is equal to the angle BCD; but it has been demonftrated to be greater than it; which is impoffible. Book I. But if one of the Vertices, as D, be within the other triangle ACB; produce AC, AD to E, F. therefore because AC is equal to AD in the triangle E F a. 5. I. to ACD, the angles ECD, FDC upon the o- a the angle BCD; but BDC has been proved to be greater than the fame BCD, which is impoffible. The cafe in which the Vertex of one triangle is upon a fide of the other, needs no demonstration. Therefore upon the fame base, and on the fame fide of it, there cannot be two triangles that have their fides which are terminated in one extremity of the bafe equal to one another, and likewife thofe which are terminated in the other extremity. Q.E. D. PROP. VIII. THEOR, F two triangles have two fides of the one equal to two Afides of the other, each to each, and have likewise their bates equal; the angle which is contained by the two fides of the one shall be equal to the angle contained by the two fides equal to them, of the other. A DG Let ABC, DEF be two triangles having the two fides AB, AC equal to the two fides DE, DF, each to each, viz. AB to DE, and AC to DF; and alfo the bafe BC equal to the bafe EF. The angle BAC is equal to the angle EDF. For if the tri angle ABC be ap- B plied to DEF fo that the point B be on E, and the straight line BC upon EF; the point C fhall alfo coincide with the point F, because BC is equal to EF. therefore BC coinciding with EF, BA and AC fhall coincide with ED Book I. and DF. for if the bafe BC coincides with the base 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 base 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 base equal to one another, and likewise their fides terminated in the other extremity. but this is impoffible. therefore if the bafe BC coincides with the base EF, the a. 7. 1. 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. O PROP. IX. PROB. To bifect a given rectilineal angle, that is, to divide it into two equal angles. a b. 8. Ax. b. I. I. 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 a. 3. 1. AD; join DE, and upon it defcribeban equilateral triangle DEF, then join AF. the straight 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 equal to the bafe EF; c D E B F therefore the angle DAF is equal to the angle EAF. wherefore the c. 8. 1. given rectilineal angle BAC is bifected by the straight line AF. Which was to be done. PROP. X. PROB. To bifect a given finite ftraight line, that is, to divide Τ O it into two equal parts. Let AB be the given straight line; it is required to divide it into two equal parts. a # Describe upon it an equilateral triangle ABC, and bifect b the a. 1. 1: angle ACB by the straight line CD. AB is cut into two equal parts b. 9. 1. in the point D. Book I. C. 4. 8. Becaufe 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 DB, and the ftraight line AB is divided into two equal parts in the point D. Which was to be done. A C See N. 2. 3. 7. b. I. I. T PROP. XI. PROB. O draw a straight line at right angles to a given ftraight 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 defcribe the equilateral tri angle DFE, and join FC. the F Becaufe DC is equal to CE, and FC common to the two tri- A D C angles DCF, ECF; the two fides EB c. 8. I. d. 10. Def. J. c DC, CF are equal to the 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 straight line makes with another straight line are 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 ftraight 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 demonftrated 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 PE at right angles to AB; and because ABC is a straight line, the angle common fegment. To PROP. XII. PROB. O 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 a point without it. It is required to draw a straight line perpendicu lar 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, C D FG in H, and join CF, CH, pendicular to the given straight line AB. Because 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, e Book I. a. 10. Def. 2. each to each; and the bafe CF is equal to the bafe CG; therefore d. 15. Det. 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. 8. I. 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 another upon one fide of it, are either two right angles, or are together equal to two right angles. |