Book I. been proved to be greater than the same BCD; which is impossible. The case in which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore, upon the same 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 equal to one another. Q. E. D. PROP. VIII. THEOR. IF two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal; the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides of the other. Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AВ to DE, and AC to DF; and also the base BC equal to the :: base EF: the angle BAC is equal to the angle EDF. For, if the triangle ABC be applied to the triangle 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 is equal to EF: therefore BC coinciding with EF, BA and AC shall coincide with ED and DF; for, if BA and CA do not coincide with ED and FD, but have a different situation, as EG and FG; then, upon the same base EF, and upon the same side of it, there can be two triangles EDF, EGF, that have their sides which are terminated in one extremity of the base equal to one Book I. another, and likewise their sides terminated in the other extremity: but this is impossible; therefore, if the base BC a 7.1. coincides with the base EF, the sides BA, AC cannot but coincide with the sides ED, DF; wherefore likewise the angle BAC coincides with the angle it. Therefore, if two triangles, &c. EDF, and is equal to b 8. Ax. Q. E. D. PROP. IX. PROB. TO bisect a given rectilineal angle, that is, to divide it into two equal angles. 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 a 3.1. equal to AD; join DE, and upon it describe b an equilateral triangle DEF; then join AF; the straight line AF bisects the angle BAC. A b 1. 1. D E Because AD is equal to AE, and AF is common to the two triangles. DAF, EAF, the two sides DA, AF are equal to the two sides EA, AF, each to each; but the base DF is also equal to the base EF; therefore the angle DAF is equal to the angle EAF; wherefore the given rectilineal angle BAC is bisected by B the straight line AF. Which was to be done PROP. X. PROB. c 8. 1. C TO 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. Describes upon it an equilateral triangle ABC, and bi-a 1.1. sect the angle ACB by the straight line CD. AB is cut b 9. 1. into two equal parts in the point D. TO 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. Take any point D in AC, and make CE equal to CD, and upon DE describes the F a 3. 1. b 1. 1. equilateral triangle DFE, and join FC; the straight line FC, drawn from the gi ven point C, is at right an gles to the given straight line AB. c 8.1. d 7. def. Because DC is equal to AD CE, and FC common to the two triangles DCF, ECF, the two sides DC, CF are equal to the two EC, CF, each to each; but the base DF is also equal to the base EF; therefore the angle DCF is equal to the angle ECF; and they are adjacent angles; therefore each of the angles DCF, ECF is a rightd 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. C EB PROP. XII. PROB. TO draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it. Let AB be a 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 C Book I. straight line perpendicu lar to AB, from the point C. E a 3. Post. the distance CD, describea the circle EGF meeting AB in F, G, and bisects 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. b 10. 1. 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; now the base CF is also equale c 11. Def, to the base CG; therefore the angle CHF is equald to the d 8. 1. angle CHG; and they are adjacent 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 perpendicular to ite; therefore from the given point C a per- e 7. Def. pendicular 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 side of it, are either two right angles, or are together equal to two right angles. Let the straight line AB make with CD, upon one side of it, the angles CBA, ABD; these are either two right angles, or are together equal to two right angles. C Book I. For, if the angle CBA be equal to ABD, each of them is a Def. 7. b 11. 1. c 2. Ax. d 1. Ax. a right angle; but, if not, from the point B draw BE at right angles to CD; therefore the angles CBE, EBD are PROP. XIV. THEOR. IF, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent 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 opposite sides 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 same straight line with CB, B A |