Let ABC be a triangle having the angle ABC equal to the Book 1. angle ACB; the fide AB is alfo equal to the fide AC. For, if AB be not equal to AC, one of them is greater than A D the other: Let AB be the greater, and from it cut off DB e a 3. I. equal to it. Wherefore, if two angles, &c. Q E. D. UPON b 4. 5. 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 base equal to one another, and likewife thofe which are terminated in the other extremity. 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 the extremity A of the bafe, equal to one another, and like wife their fides CB, DB that are ter minated in B. 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 the angle BCD; therefore the angle A ADC is greater alfo than BCD; B much more then is the angle BDC greater than the angle BCD. B 2 a 5. I. Book I. 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, becaufe 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, becaufe CB is equal A to DB, the angle BDC is equal to the angle BCD; but BDC has been proved to be greater than the fame BCD; which is impoffible. The cafe in which the ver tex of one triangle is upon a fide of the other, needs no demonftration. a B Therefore upon the fame bafe, and on the fame fide 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 thofe which are terminated in the other extremity.. Q. E. D. I to F two triangles have two fides of the one equal to two fides of the other, each to each, and have like wife their bafes equal; the angle which is contained by the two fides of the one fhall 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, AC equal to the two fides DE, DF, each to each, viz. AB to DE, and AC to A DF; and alfo the bafe BC equal to the bafe EF. The angle BAC is equal to the angle EDF. For, if the tri DG angle ABC be ap- B plied to 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 BC is equal to EF; therefore BC coinciding with EF, BA and Book 1. AC fhall coincide with ED and DF; for, if the bafe BC caincides with the base EF, but the fides BA, CA do not coincide with the fides ED, FD, but have a different fituation, as EG, FG; then, upon the fame bafe 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 bafe equal to one another, and likewife their fides terminated in the other extremity: But this is impoflible a; therefore, if the bafe BC coin- a 7. I. 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. fore if two triangles, &c. Q. E. D. There. b. 8. Ax O bisect a given rectilineal angle, that is, to divide Tit Let BAC be the given rect lineal angle, it is required to bi fect it. A b 1. I. Take any point D in AB, and from AC cut a off AE e- a 3. 1. qual to AD; join DE, and upon it defcribe an equilateral triangle DEF; then join AF; the ftraight line AF bifects the angle BAC. Becaufe 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; and the bafe DF is e- B qual to the bafe EF; therefore the D E F C c 8. I. EAF; wherefore the given rectilineal angle BAC is bifected by the ftraight line AF. T Which was to be done. PROP. X. PRO B. O bifect a given finite ftraight line, that is, to divide Let AB be the given ftraight line; it is required to divide it into two equal parts. Defcribe a upon it an equilateral triangle ABC, and bifecta 1. the angle ACB by the ftraight line CD. AB is cut into two b 9. 1. equal parts in the point D. B 3 Becaufe Book I. € 4. I. See N. a 3. I. b I. I. € 8. I. O draw a ftraight line at right angles to a given ftraight line, from a given point in the fame. T& 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. Take any point D in AC, and a make CE equal to CD, and upon DE defcribeb the equila teral triangle DFE, and join Because DC is equal to CE, triangles DCF, ECF; the two AD F C E B two EC, CF, each to each; and the bafe DF is equal to the bafe 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 d 10. Def. equal to one another, each of them is called a right angle ; therefore each of the angles DCF, ECF, is a right angle. Wherefore, from the given point C, in the given ftraight line AB, FC has been drawn at right angles to AB. Which was I. to be done. COR. By help of this problem, it may be demonftrated, that two ftraight lines cannot have a common fegment. If it be poffible, let the two ftraight lines ABC, ABD have the fegment AB common to both of them. From the point B draw BE at right angles to AB; and becaufe ABC is a straight line, the angle CBE is equal a T PROP. XII. PROB. E D B O draw a ftraight line perpendicular to a given straight line of an unlimited length, from a given point without it. Let AB be the 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 ftraight line perpendicular to AB from the point C. Take any point D upon the other fide of AB, and from the centre C, at the distance CD, defcribeb the circle EGF C meeting AB in F, G; and bi. A F B D fect FG in H, and join CF, C IO. I. CH, CG; the ftraight line CH, drawn from the given point C, is perpendicular to the given straight line AB. с I. e 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; and the bafe CF is equal to the d 15. Def. bafe CG; therefore the angle CHF is equal to the angle CHG; and they are adjacent angles; but when a ftraight line standing on a ftraight line makes the adjacent angles equal to one another, each of them is a right angie, and the straight line which ftands 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. THI PROP. XIII. THEO R. 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. |