Let ABC be a triangle having the angle ABC equal to the Book 1. angle ACB ; the fide AB is also equal to the side AC. For, if AB be not equal to AC, one of them is greater than the other : Let AB be the greater, and from it cut off DB e• a 3. I. qual to AC, the less, and join DC; therefore, because in the triangles DBC, ACB, A DB is equal to AC, and BC common to D both, the two sides 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 6 ACB, the less to the greater ; which is absurd. Therefore B AB is not unequal to AC, that is, it is equal to it. Wherefore, if two angles, &c. Q E. D. Cor. Hence every equiangular triangle is also equilateral. C 64. s. U PON the same base, and on the same side of it, See N. there cannot be two triangles that have their fides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity. If it be possible, let there be two triangles ACB, ADB, upon the fame base AB, and upon the same side of it, which have their fides CA, DA, terminated in the extremity A of the base, equal to one another, and like. wise their fides CB, DB that are ter minated in B. Join CD; then, in the case in which the vertex of each of the triangles is without the other triangle, because AC is equal to AD, the angle a 5. I. ACD is equal to the angle ADC: But the angle ACD is greater than the angle BCD; therefore the angle A B ADC is greater 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 demonstrated to be greater than it ; which is impoflible. But, B 2 Bok 1. F; F a s. I. But, if one of the vertices, as D, be within the other tri, angle ACB; produce AC, AD to E, therefore, because AC is equal to AD in the triangle ACD, the angles ECD, FDC upon the other side of the base CD are equala to one another, but the angle ECD is greater than the angle BCD ; wherefore the angle FDC is likewise greater than BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal A B to DB, the angle BDC is equal to the angle BCD; but BDC has been proved to be greater than the same BCD; which is impossible. The case in which the ver. tex of one triangle is upon a side of the other, needs 110 de. monstration. Therefore upon the same base, and on the same ide of it, there cannot be two triangles that have their fides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity. 0. E. D. (F two triangles have two sides of the one equal to two their bases equal ; the angle which is contained by the two sides of the one shall be equal to the angle con tained by the two fides equal to them, of the other. Let ABC, DEF be two triangles having the two sides AB, D G For, if the tri. F plied to DEF, so that the point B be on E, and the straight line BC upon EF; the point C shall also coincide with the point F, Because BC CE BC is equal to EF; therefore BC coinciding with EF, BA and Book 1: AC shall coincide with ED and DF; for, if the base BC caincides with the base EF, but the sides BA, CA do not coin. cide with the Gides ED, FD, but have a different Gtuation, as EG, FG; then, upon the fame base EF, and 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 Gides terminated in the other extremity : But this is impofliblea; therefore, if the base BC coin- * 7. 1. cides with the base EF, the sides 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. There. b. 8. Ax' tore if two triangles, &c. Q. E. D: T: O bisect a given rectilineal angle, that is, to divide it into two equal angles. Let BAC be the given rect lineal angle, it is required to bifeet it. Take any point D in AB, and from AC cut a off AE e-a 3. I. qual to AD; join DE, and upon it defcribe an equilateral triangle DEF; A bi. I. 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 D E DAF, EAF; the two sides DA, AF, are equal to the two sides EA, AF, each to each ; and the base DF is e- B qual to the base EF ; therefore the F angle DAF is equal to the angle c 8. 1. EAF; wherefore the given rectilineal angle BAC is bisected by the straight line AF. Which was to be done. O 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. Describe a upon it an equilateral triangle ABC, and bifect a f. the angle ACB by the straight line CD. AB is cut into two b 9. s. equal parts in the point D. B 3 Because See N. b. I. 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. a 3. I. Take any point D in AC, and a make CE equal to CD, and upon DE describe b the equila- F Because DC is equal to CE, C С E B fides DC, CF, are equal to the two EC, CF, each to each ; and the base DF is equal to the c 8. 1. base EF; therefore the angle DCF is equal c 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 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. 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 bare the segment AB common to both of them. From the point B draw BE at right angles to AB; and because ABC is a Itraight line, 1 a 1. line, the angle CBE is equal a Book 1. E to the angle EBA; in th fame manner, because ABD is a 10. Def. 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 D greater ; wbich is impossible ; therefore two straight lines can. A B not have a common segment. PRO P. XII. PRO B. 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. Ic is required to draw a straight С line perpendicular to AB from the point C. Take any point Dupon the E other Gde of AB, and from the centre C, at the distance H CD, describe the circle EGF b 3. Polt. meeting AB in F, G; and bi. A F D sect c 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 (wo GH, HC, each to each ; and the base CF is equal d to the d 15. Def. base CG; therefore the angle CHP is equal e to the angle CHG; and they are adjacent angles; but when a straight line standing on a Itraight 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. PRO P. XIII. THEO R. B I. "HE angles which one straight line makes with an. other upon the one side of it, are either two right angles, or are together equal to two right angles. B4 Let T" |