PROP. VII. THEOR. 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 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 base, equal to one another, and likewise their sides, CB, DB, that are terminated in B, equal to one another. B Join C, D; then, in the case in which the vertex of each of the triangles is without the other triangle, because AC is +Hyp. equal to AD, the angle ACD is *5. 1. equal to the angle ADC: But the 19Ax. angle ACD is greater than the angle BCD; therefore the angle ADC is greater also than BCD; much more then is the angle BDC greater than the angle BCD. Hyp. Again, because CB is equal to DB, the angle BDC is equal to the angle BCD; but it has been demonstrated to be greater than BCD; which is impossible. But if one of the vertices, as D, be within the other triangle ACB; produce AC, AD to E, F. Then because AC is equalf to AD in the triangle #5. 1. Hyp. ACD, the angles ECD, FDC upon the other side of the base CD are *5. 1. equal to one another: but the angle ECD is greater than the angle T9 AX. BCD: wherefore the angle FDC is likewise greater than BCD; much more then A is the angle BDC greater than the angle F BCD. Again, because CB is equalt to DB, the angle +Hyp. BDC is equal to the angle BCD; but BDC has #5. 1. 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 one another, and likewise those which are terminated in the other extremity. Q. E. D. PROP. VIII. THEOR. 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. A For, if the triangle ABC be applied to DEF, so that the point B D G CE 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, +Hyp. BA and AC shall coincide with ED and DF: for, if the base BC coincide 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 coincide 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, *7. 1. *8 Ax. PROP. IX. PROB. 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. #3. 1. #1. 1. Take any point D in AB, and from AC cut* off AE equal to AD; draw DE, and upon it describe an equilateral triangle DEF, so that F and A may be on opposite sides of DE; then draw AF: the straight line AF shall bisect the angle BAC. Because AD is equalt to AE, and + Const. AF is common to the two triangles B A DA E E C DAF, EAF; the two sides DA, AF, are equal to the two sides EA, AF, each to each; and the base DF, is equal to the base EF; therefore the angle +Const. DAF is equal to the angle EAF: wherefore the given rectilineal angle BAC is bisected by the straight line AF. Which was to be done. #8. 1. 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. #1.1. #9.1. + Const. Describe* upon AB an equilateral triangle ABC, and bisect* the angle ACB by the straight line CD. AB 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 +Const. #4: 1. the base AD is equal to the base DB, A D two equal parts in the point D. Which was to be done. PROP. XI. PROB. To draw a straight line at right angles to a given straight line, from 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 Cat right angles to AB. Take any point D in AC, make* CE equal to CD, *3. 1. upon DE describe* the equilateral triangle DFE, and *1. 1. draw CF. The straight line CF, +Const. drawn from the given point C, shall be at right angles to the given straight line AB. Because DC is equalt 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: +Const. AD C E.B and the base DF is equal to the base EF; therefore the angle DCF is equal to the angle ECF: and they *8. 1. 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, a straight line CF has been drawn at right angles to AB. Which was to be done. *8 Def. 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 +11. 1. *8 Def. CBE is equal to the angle EBA; in like manner, because ABD is a straight line, the angle DBE is equal to the angle EBA: whereforet the MAX. angle DBE is equal to the angle A CBE, the less to the greater; which is E D impossible; therefore two straight lines cannot have a com mon segment. PROP. XII. PROB. To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point with out 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. Take any point D upon the other AF side of AB, and from the centre C, at #10. 1. H E the distance CD, describe the circle EGF meeting #3 Post. AB in F, G; bisect* FG in H, and join CH. The straight line CH, drawn from the given point C, shall be perpendicular to the given straight line AB. + Const. *12 Def. Draw CF, CG. Then because FH is equalt to HG, and 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 CG; therefore the angle CHF is equal to the angle CHG; #8. 1. 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 perpendiculart 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. 18 Def. PROP. XIII. THEOR. 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 the angle CBA is either equal to the angle ABD or it is not: if the angle CBA be equal to ABD, each of them |