D A 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 (3. 1.) off DP equal to AC, the less, and join DC; therefore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides DB, BC are equal to the two AC, CB, each to each; and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC is equal to the triangle (4. 1.) 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. Q. E. D. B COR. Hence every equiangular triangle is also equilateral. PROP. VII. THEOR. C 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, let there be two triangles ACB, ADB, upon the same base AB, and upon the same side of it, 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. Join C D; then, in the case 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 (5. 1.) to the angle ADC: but the 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 A C D Ꭰ B the angle BCD. Again, because CB is equal to DB, the angle BDC is equal (5. 1.) to the angle BCD; but it has been demon strated to be greater than it; which is impossible. * See note, G 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 side of the base CD are equal (5.1.) to one another, but the angle ECD is greater than the angle BCD; wherefore the angle FDC is likewise greater than BCD; much more than is the angle BDC greater than the angle BCD. Again, because CB is equal to DB, the angle BDC is equal (5. 1.) to the angle BDC; but BCD 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. A B 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. AB to DE, and AC to DF; A B be on E, and the D CE G F 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; (7. 1,) 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 (8. Ax.) to it. Therefore if two triangles, &c. Q. E. D. 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. A Take any point D in AB, and from AC cut (3. 1.) off AE equal to AD; join DE, and upon it describe (1. 1.) an equilateral triangle DEF; then join AF; the straight line AF bisects 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 to the two sides, EA, AF, each to each; and the base DF is equal to the base EF; therefore the angle DAF is equal (8. 1.) to the angle EAF; wherefore the given rectilineal angle BAC is bisected by the straight line B AF, which was to be done. PROP. X. PROB. D E C 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 (1. 1.) upon it an equilateral triangle ABC, and bisect (9. 1.) the angle ACB by the straight line CD. AB is 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 base AD is equal to the base (4. 1.) DB and the straight line AB is divided into two equal parts in the point De 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 A B 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.* F Take any point D in AC, and (3. 1.) make CE equal to CD, and upon DE describe (1. 1.) the 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. Because DC is equal to CE, and E B to each; and the base DF is equal to the base EF; therefore the angle DCF is equ (8. 1.) 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 (10. Def. 1.) 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 draw BE at right angles to AB; and because ABC is a straight * See note. 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. C H F G B D Take any point D upon the other side of AB, and from the centre C, at the distance CD, describe (3. Post.) the circle FDG meeting AB in F, G; and, bisect (10. 1.) FG A in H, and join CF, CH, CG; the straight line CH, drawn from the given point C, is perpendicular to the given straight line AB. Because FH is equal 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 (15. Def. 1.) to the base CG; therefore the angle CHF igequal (8. 1.) to the angle CHG; and they are adjacent angles; but when a straight line standing 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 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. THE angles which one straight line makes with another upon the 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 are either two right angles, or are together equal to two right angles. |