UPON PROP. VII. THEOR. Book I. PON the fame bafe, 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, equal to one another. If it be poffible, 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 likewife their fides CB, DB, terminated in B, equal to one another. 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 angle ADC is greater also than B BCD; much more then is the angle BDC greater than the angle BCD. Again, becaufe CB is equal to DB, the angle BDC is equal a to the angle BCD; but it has been demonftrated to be greater than it; which is impoffible. a E/ F 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 ACD, the angles ECD, FDC upon the other fide of the bafe CD are equal to one another, but the angle ECD is greater than the angle BCD; wherefore the angle FDC is likewife greater than BCD; much more then is the angle BDC greater than the angle BCD. A. gain, becaufe CB is equal to DB, A B. the angle BDC is equal to the angle BCD; but BDC has a 5. 1. been Book I. 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 side 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 those which are terminated in the other extremity equal to one another. Q. E. D. PROP. VIII. THE OR. two triangles have two fides of the one equal to two fides of the other, each to each, and have Likewife their bafes equal; the angle which is contained by the two fides of the one fhall be equal to the angle contained by the two fides of the other. Let ABC, DEF be two triangles having the two fides AB, B C E F bafe EF. The angle BAC is equal to the angle EDF. For, if the triangle ABC be applied to the triangle DEF, fo that the point B be on E, and the ftraight 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, 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 fame fide of it, there can be two triangles EDF, EGF, that have their fides which are are terminated in one extremity of the bafe equal to one Book ́I. another, and likewife their fides terminated in the other extremity But this is impoffible a; therefore, if the base BC a 7. 1. coincides with the base EF, the fides BA, AC cannot but coincide with the fides ED, DF; wherefore likewife the angle BAC coincides with the angle EDF, and is equal b to s. Ax, it. Therefore if two triangles, &c. Q. E. D. T PROP. IX. PROB. O bisect a given rectilineal angle, that is, to di- Let BAC be the given rectilineal angle, it is required to bifect it. A Take any point D in AB, and from AC cut a off AE e- a 3. £. qual to AD; join DE, and upon it defcribe b an equilateral triangle DEF; then join AF; the ftraight 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; but the base DF is alfo equal to the base EF; therefore the angle DAF is equal to the angle EAF; wherefore the given rectilineal angle BAC is bifected by B D the straight line AF. Which was to be done. PROP. X. PROB. E c 8. F O bifect a given finite ftraight line, that is, to Ta Let AB be the given straight line; it is required to divide it into two equal parts. Defcribe a upon it an equilateral triangle ABC, and bifect a I. I. Þ the angle ACB by the ftraight line CD. AB is cut into b 9. 1. two equal parts in the point D. Because 3. I. bi. 1. é 3. I. O draw a ftraight line at right angles to a given Tftraight line, from a given point in the fame. Let AB be a given ftraight line, and C a point given in it; it is required to draw a ftraight line from the point C at right angles to AB. the F Take any point D in AC, and a make CE equal to CD, and upon DE defcribe equilateral triangle DFE, and join FC; the straight line FC, drawn from the given point C, is at right angles to the given straight line AB. A D C E B c Because DC is equal to CE, and FC common to the two triangles DCF, ECF, the two fides DC, CF are equal to the two EC, CF, each to each, but the base DF is alfo 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 ftraight line makes with another ftraight line are equal to one another, each of them is called a 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. PROP. T PROP. XII. PRO B. O draw a ftraight line perpendicular to a given ftraight line of an unlimited length, from a given point without it. Let AB be a given ftraight line, which may be produced to any length both ways, and let C be a point without it. It is required to draw a D B Book I. a 3. Poft. the distance CD, defcribe a the circle EGF meeting AB in F, G; and bifect b 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. I. 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 alfo equal c c 11. Def. to the base CG; therefore the angle CHF is equal to the angle CHG; and they are adjacent angles; but when a ftraight line ftanding on a ftraight 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. TH PRO B. XIII. THEOR. HE angles which one ftraight line makes with another upon the one fide of it, are either two right angles, or are together equal to two right angles. Let the ftraight 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 |