U PROP. VII. THEOR. Book I. 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 base equal to one another, and likewife those which are terminated in the other extremity, equal to one another. If it be possible, let there be two triangles ACB, ADB, upon the fame base AB, and upon the fame fide of it, which have their fides CA, DA, terminated in A equal to one another, and likewise their fides CB, DB, terminated in B, equal to one another. C D Join CD; then, in the cafe in which the vertex of each of the triangles is without the other triangle, because AC is equal to AD, the angle ACD is equal a to the angle ADC: But the angle ACD is greater than the angle BCD; therefore the A B angle 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 a to the angle BCD; but it has been demonstrated to be greater than it; which is impoffible. 4 But if one of the vertices, as D, be within the other tri angle ACB; produce AC, AD to E E, F; therefore, because AC is e F qual to AD in the triangle ACD, C D B the angle BDC is equal to the angle BCD; but BDC has been a 5. 1. Eook I. been proved to be greater than the fame BCD; which is impoflible. The cafe in which the vertex of one triangle is on a fide of the other, needs no demonstration. up Therefore, upon the fame base, and on the same side 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 likewife those which are terminated in the other extremity equal to one another. Q. E. D. PROP. VIII. THEOR. F two triangles have two fides of the one equal to two fides of the other, each to each, and have Likewife their bases equal; the angle which is contained by the two fides of the one shall be equal to the angle contained by the two fides of the other. 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 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 fituation as EG and FG; then, upon the fame bafe EF, and upon the same side of it, there can be two triangles EDF, EGF, that have their fides which are Book I. are terminated in one extremity of the base equal to one Tobi PROP, IX. PROE, O bisect a given rectilineal angle, that is, to di- Let BAC be the given rectilineal angle, it is required to bifect it. Take any point D in AB, and from AC cut a off AE e-23. t. qual to AD; join DE, and upon it describe b an equilateral triangle DEF; then join AF; the straight line AF bifects the angle BAC. D 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; 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. T O bifect r PROP. X. PROB. 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. br. 1. c 8. Describe a upon it an equilateral triangle ABC, and bisect a 1. 1. the angle ACB by the straight line CD. AB is cut into b 9. 1. two equal parts in the point D. Because 1 O draw a straight line at right angles to a given 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, 3. I. b1. 1. and upon DE describe the equilateral triangle DFE, F 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. Because DC is equal to CE, and FC common to the A two triangles DCF, ECF, the two fides 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. But, when the adjacent angles which one straight line makes with another ftraight line are equal to one another, each of them is called a d 7. def., 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. D C EB PROP. 1 PROP. XII. PRO B. draw a straight line perpendicular to given Titraight line of an unlimited length, from ven point without it. a gi 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 Book I. is required to draw a C straight line perpendicu lar to AB from the point a 3. Poft. the circle EGF meeting AB in F, G; and bisectb FG in bio. 1. H, and join CF, CH, CG; the straight line CH, drawn from the given point C, is perpendicular to the given straight line AB. I. d 8. 1. 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, each to each; now the bafe CF is also equal c c 11. Def. to the base CG; therefore the angle CHF is equald to the 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 it; therefore from the given point C a perpendicular CH has been drawn to the given straight line AB. Which was to be done. PROB. XIII. THEOR. THEO Let the straight line AB make with CD, upon one fide of it, the angles CBA, ABD; these are either two right angles, orare together equal to two right angles. C For |