b UPON Let ABC be a triangle having the angle ABC equal to the angle Book I. ACB; the side AB is also equal to the side AC. For if AB be not equal to AC, one of them is greater than the o. A b. 4. 1. B С PROP. VII. THEOR. there cannot be two triangles that have their sides If it be poslible, let there be two triangles ACB, ADB upon the C D B a. 5.1. 2 d. 5. I. Book I. But if one of the Vertices, as D, be within the other triangle E than Therefore upon the fame base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity. Q.E.D. PROP. VIII. THEOR, bates equal; the angle which is contained by the two fides of the one shall be equal to the angle contained by the two lides equal to them, of the other. Let ABC, DEF be two triangles having the two sides AB, AC ęqual to the two sides DE, DF, each to each, viz. AB to DE, and AC to DF; and also the A D G For if the tri- СЕ F plied to DEF so that the point B be on E, and the straight line BC upon EF; the point ( shall also coincide with the point F, because BC is equal to EF. therefore BC coinciding with EF, BA and AC shall coincide with ED Book 1. and DF. for if the base BC coincides with the base EF, but the sides v 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 fides terminated in the other extremity. but this is impossible'. therefore if the base BC coincides with the base EF, the a. 7. 1. sides BA, AC cannot but coincide with the sides ED, DF; wherefore likewise the angle BAC coincides with the angle EDF, and is equal b to it. therefore if two triangles, &c. Q. E. D. b. 8. Ax, , PROP. IX. PROB. it into two equal angles. Take any point D in AB, and from AC cut off AE equal to 2. 3. 1. AD; join DE, and upon it defcribeban b. 1. 1. А equilateral triangle DEF, then join AF. the straight line AF bisects the angle BAC. Because AD is equal to AE, and AF D E is common to the two triangles DAF, EAF; the two sides DA, AF are eqnal to the two sides E A, AF, each to each; and B F the base DF is equal to the base EF; 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. с , PROP. X. PROB. it into two equal parts. Let AB be the given straight line; it is required to divide it into two equal parts. Describe upon it an equilateral triangle ABC, and bisect b the a. 1.1 a angle ACB by the straight line CD. AB is cut into two equal parts b. 9. 1. in the poiot D. T Araighe b. 1.1. PROP. XI. PROB. straight line, from a given point in the fame. See N. 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 Din AC, and make CE equal to CD, and upon F Because DC is equal to CE, С .qual to one another, each of them is called a right dangle; therefore d. 10. Def. 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 PE at right angles to AB; and because ABC is a straight line, the angle E B c. $. I. a. 10. Def. CBE is equal to the angle EBA; Book 1. in the same manner, because ABD E is a straight line, the angle DBE is equal to the angle EBA, wherefore the angle DBE is equal to the angle CBE, the less to the great D er; which is impossible. therefore two straight lines cannot have a A B common segment. PROP. XII. PROB. 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 C lar to AB from the point C. Take any point D upon the other side of AB, and from the E center C, at the distance CD, describe the circle EGF meet H b. 3. fot. ing AB in F, G; and bisect A • FG in H, and join CF, CH, CG, the 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 equal to the base 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 8.8. 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. ther upon one side of it, are either two right angles, or are together equal to two right angles. с B C. TO, I. 1. |