D • 4. 1. each; and the angle DBC is equal to the anglet ACB; + Hyp. therefore the base DC is equal to the base AB, and the triangle DBC is equal to the triangle * ACB, the less to the greater, which is absurd. Therefore AB is not unequal to AC, that is, it is equal to it. Wherefore, if two angles, &c. B Q. E. D. CoR.-Hence every equiangular triangle is also equilateral. PROP. VII. THEOR. CD Upon the same base, and on the same side of it, there cail- See N. not 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. If it be possible, upon the same base AB, and upon the same side of it, let there be two triangles ACB, ADB, which have their sides CA, DA terminated in the extremity A of the basc equal to one another, and likewise their sides, CB, DB, that are terminated in B. Join CD; then, in the case in which the vertex of each of the triangles is without the other triangle, because AC is equalt to AD, the angle ACD is + Hyp. . equal * to the angle ADC: but the angle) SACD is greater † than the angle BCD; therefore the) + 9 Ax. 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 + Hyp. equal * to the angle BCD; but it has been demonstrated * 5. 1. to be greater than it, which is impossible. 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 t to AD in the tri + Hyp.' angle ACD, the angles ECD, FDC ,E,F upon the other side of the base CD are equal * to one another: but the angle ECD is greater t than the angle BCD; wherefore the angle FDC is likewise greater than BCD; much more then is the angle BDC greater than the angle 5. 1. * 5. 1. + 9 Ax. 1 + Hyp. . * 5. 1. BCD. Again, because CB is equal t to DB, the angle BDC is equal * to the angle BCD; but BDC has 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 another, and likewise those which are terminated in the other extremity. Q. E. D. PROP. VIII. THEOR. А D G B CE 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 equal to them, 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. AB to DE, and AC to DF; and also the base BC equal to the base HF. The angle BAC shall be equal to the angle EDF. For, if the triangle ABC be applied to DEF, so that the point B may 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 the base BC coincides with the base EF, but the sides BA, CA do not coincide with the sides ED, FD, but have a different situation, as EG, FG; then upon the same base EF, and upon the same side of it, there can be two triangles that have their sides which are terminated in one extremity of the base, equal to one another, and likewise their sides terminated in the other extremity: but this is * impossible; therefore, if the base BC 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 EDF, and is equal * to it. Therefore, if two triangles, &c. Q. E. D. + Hyp. * 7.1. * 8 Ax. PROP. IX. PROB. A 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 3. 1. equal to AD; join DE, and upon it describe * an equila- * 1. 1. teral triangle DEF; then join AF: the straight line AF shall bisect the angle BAC. Because AD is equal + to AE, and AF is common to the two triangles DAF, EAF; the two sides DA, AF, are equal B F to the two sides EA, AF, each to each; and the base DF is equal + to the base EF; therefore + Constr. the angle DAF is equal* to the angle EAF: wherefore *8.1. the given rectilineal angle BAC is bisected by the straight line AF.' Which was to be done. + Constr. DA 小 A PROP. X. PROB. 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. Describe * upon it an equilateral triangle ABC, and * 1. 1. bisect * the angle ACB .by the straight line CD. AB *9.1. shall be cut into two equal parts in the point D. Because AC is equal † to CB, and CD common to the two triangles ACD, BCD; the two sides AC, CD, are equal to BC, CD, each to each; and the angle ACD is equal † to the angle BCD; therefore the + Constr. base AD is equal to the base * DB, and the straight line AB is divided into two equal parts in the point D. Which was to be done. + Constr. A * 4. 1. PROP. XI. PROB. line, from a given point in the same. Let AB be a given straight line, and C a point given See N. in it; it is required to draw a straight line from the point C at right angles to AB. * 3.1. * 1.1. + Constr. AD + Constr. * 8.1. * 10 Def. Take any point D in AC, and make * CE equal to CD, and upon DE describe * the equilateral triangle DFE, and join FC. The straight line FC drawn from the given point C, shall be at right angles to the given straight line AB. Because DC is equal t to 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 ; and the base DF is equal t 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 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 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 demonstrated, 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 drawt BE at right angles to AB; and because ABC is a straight line, the angle CBE is equal * to the angle EBA; in the same manner, because ABD is a straight line, the angle DBE is equal to the angle EBA; wherefore + the angle DBE is equal to the angle CBE, the less to the greater; which is impossible: therefore two straight lines cannot have a common segment. E * 10 Def. +1 Ax. A B 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 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 line perpendicular to AB from the point C. AF GB Take any point D upon the other side of AB, and from the centre C, at the distance CD, describe * the * 3 Post. circle EGF, meeting AB in F, G; bisect * FG in H, * 10. 1. and join CH. The straight line CH, drawn from the given point C, shall be perpendicular to the given straight line AB. Join CF, CG: and because FH is equal + to HG, and + Constr. HC common to the two triangles FHC, GHC, the two sides FH, HC are equal to the two GH, HC, each to each; and the base CF is equal * to the base * 15 Def. CG: therefore the angle CHF is equal * to the angle * 8. 1. CHG; and they are adjacent angles: but when a straight line, standing on another 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: there- +10 Def. fore, from the given point C, a perpendicular CH has been drawn to the given straight line AB. Which was to be done. PROP. XIII. THEOR. +-10 Def. A E A * * 11. 1. 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 shall either be two right angles, or shall together be equal to two right angles. For if the angle CBA be equal to ABD, each of them is a right tangle: but if not, from the point B draw BE at right angles to CD; therefore the angles CBE, EBD * are two right angles : and because the angle CBE is equal to the two angles CBA, ABE together, add the angle EBD to each of these equals; therefore the angles CBE, EBD are equal * to the three * 2 Ax. angles CBA, ABE, EBD. Again, because the angle DBA is equal to the two angles VBE, EBA, add to each of these equals the angle ABC; therefore the angles DBA, ABC are equal to the three angles DBE, EBA, +2 Ax. ABC: but the angles CBE, EBD have been demonstrated to be equal to the same three angles; and things D B -C D * 10 Def. |