دو 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; there fore, because in the triangles DBC, ACB, A D Cb4.3. 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. U 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 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 wife their fides CB, DB that are terminated in B. C 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 to the angle ADC: But the angle ACD is greater than a 5. I. 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 the angle BCD; but it has been demonstrated to be gr than it; which is impoffible. B2 Bok L. a 5. I. But, if one of the vertices, as D, be within the other trig angle 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 base 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. Again, because CB is equal A to DB, the angle BDC is equal to the F C D B angle BCD; but BDC has been proved to be greater than the fame BCD; which is impossible. The cafe in which the ver. tex of one triangle is upon a side of the other, needs no demonstration. 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 likewise those which are terminated in the other extremity. Q.E. D. F two triangles have two fides of the one equal to two fides of the other, each to each, and have like wife their bases equal; the angle which is contained by the two fides 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 fides AB, bafe BC equal to DG angle BAC is e qual to the angle EDF. For, if the tri angle ABC be ap- B plied to DEF, fo that the point B be on E, and the straight line BC upon EF; the point C shall alfo coincide with the point F, Because BC : BC is equal to EF; therefore BC coinciding with EF, BA and Book J. AC shall coincide with ED and DF; for, if the base BC caincides with the base EF, but the fides BA, CA do not coincide with the fides ED, FD, but have a different situation, as EG, FG; then, upon the fame base EF, and upon the fame 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 sides terminated in the other extre- mity: But this is impoflible a; therefore, if the base BC coin- cides 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 to it. There. b. 8. Ax• fore if two triangles, &c. Q. E. D. T O bife ct a given rectilineal angle, that is, to divide Let BAC be the given rect lineal angle, it is required to bi a 7. I. fect it. Take any point D in AB, and from AC cut off AE e-a 3. I. qual to AD; join DE, and upon it de fcribe an equilateral triangle DEF; then join AF; the straight line AF bifects the angle BAC. Because AD is equal to AE, and E DAF, EAF; the two fides DA, AF, are equal to the two fides EA, AF, each to each; and the base DF is e- B qual to the base EF; therefore the angle DAF is equal to the angle EAF; wherefore the given rectilineal angle BAC is bisected PROP. X. PRов. O bifect a given finite straight line, that is, to divide 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 bisect a r. the angle ACB by the straight line CD. AB is cut into two b 9. r. equal parts in the point D. 1 See N. a 3. I. b 1. I. Totraight linerom draw a straight line at right angles to a given 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. 1 Take any point D in AC, and a make CE equal to CD, and upon DE describe the equila teral triangle DFE, and join F FC; the straight line FC drawn Because DC is equal to CE, C EB c 8. 1. Ι. fides DC, CF, are equal to the two EC, CF, each to each; and the base DF is 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 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 demonftrated, 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 BE at right angles to AB; and because ABC is a straight line, ag e. en in atr T E PROP. ΧΙΙ. PROB. O draw a straight line perpendicular to a given point without it. an 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 CD line perpendicular to AB from C 1 the point C. Take any point D upon the other fide of AB, and from E the centre C, at the distance CD, describe the circle EGF meeting AB in F, G; and bi. A F G B D fect FG in H, and join CF, CIO. I. CH, CG; the straight line CH, drawn from the given point C, 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; and the bafe CF is equal d to the d 15. Def. base CG; therefore the angle CHF is equal to the angle CHG; and they are adjacent angles; but when a straight line standing. I.. 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 trom the given point C a perpendicular CH has been drawn to the given straight line AB. Which was to be done. PROP. XIII. THEOR. HE angles which one straight line makes with an- TH angles, or are together equal to two right angles. B 4 Let |