to one another, and likewise their sides CB, DB, which are terminated at B. Join CD. In the case in which the vertex of each triangle is without the other triangle; because AC is equal to AD, [Hypothesis. the angle ACD is equal to the angle ADC. [I. 5. But the angle ACD is greater than the angle BCD, [Ax. 9. therefore the angle ADC is also greater than the angle BCD ; much more then is the angle BDC greater than the angle BCD. Again, because BC is equal to BD, the angle BDC is equal to the angle BCD. [Hypothesis. [I. 5. But it has been shewn to be greater; which is impossible. But if one of the vertices as D, be within the other triangle Then because AC is equal to AD, in the triangle ACD, [Hyp. the angles ECD, FDC, on the other side of the base CD, are equal to one another. [I. 5. But the angle ECD is greater than the angle BCD, A E B [Axiom 9. therefore the angle FDC is also greater than the angle BCD ; much more then is the angle BDC greater than the angle BCD. Again, because BC is equal to BD, the angle BDC is equal to the angle BCD. [Hypothesis. [I. 5. But it has been shewn to be greater; which is impossible. The case in which the vertex of one triangle is on a side of the other needs no demonstration. Wherefore, on the same base &c. Q.E.D. 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 which is 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, namely AB to DE, and AC to DF, and also the base BC equal to the base EF: the angle BAC shall be equal to the angle EDF Ꭰ G For if the triangle ABC be applied to the triangle DEF, so that the point B may be on the point E, and the straight line BC on the straight line EF, the point C will also coincide with the point F, because BC is equal to EF. [Hyp. Therefore, BC coinciding with EF, BA and AC will 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 on the same base and on the same side of it there will be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise their sides which are terminated at the other extremity. But this is impossible. [I. 7. Therefore since the base BC coincides with the base EF, the sides BA, AC must coincide with the sides ED, DF. Therefore also the angle BAC coincides with the angle EDF, and is equal to it. [Axiom 8. Wherefore, if two triangles &c. Q.E.D. PROPOSITION 9. PROBLEM. To bisect a given rectilineal angle, that is to divide it into two equal angles. Join AF. The straight line AF shall bisect the angle BAC. Because AD is equal to AE, [Construction. and AF is common to the two triangles DAF, EAF, the two sides DA, AF are equal to the two sides EÁ, AF, each to each; and the base DF is equal to the base EF; [Definition 24. therefore the angle DAF is equal to the angle EAF. [1. 8. Wherefore the given rectilineal angle BAC is bisected by the straight line AF. Q.E.F. 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. triangle ABC, Describe on it an equilateral [I. 1. and bisect the angle ACB by the straight line CD, meeting AB at [I. 9. D. A AB shall be cut into two equal parts at the point D. [Definition 24. and CD is common to the two triangles ACD, BCD, the two sides AC, CD are equal to the two sides BC, CD, each to each; and the angle ACD is equal to the angle BCD ; therefore the base AD is equal to the base DB. [Constr. [I. 4. Wherefore the given straight line AB is divided into two equal parts at the point D. Q.E.F. PROPOSITION 11. PROBLEM. To draw a straight line at right angles to a given straight line, from a given point in the same. Let AB be the given straight line, and C the given point in it: it is required to draw from the point Ca straight line at right angles to AB. Take any point D in AC, and make CE equal to CD. [I. 3. On DE describe the equilateral triangle DFE, and join CF. [I. 1. The straight line CF drawn from the given point C shall be at right angles to the given straight line AB. Because DC is equal to CE, [Construction. and CF is common to the two triangles DCF, ECF; the two sides DC, CF are equal to the two sides EC, CF, each to each; and the base DF is equal to the base EF; [Definition 24. therefore the angle DCF is equal to the angle ECF; [I. 8. 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 the angles is called a right angle; [Definition 10. therefore each of the angles DCF, ECF is a right angle. Wherefore from the given point O in the given straight line AB, CF has been drawn at right angles to AB. Q.E.F. Corollary. By the help of this problem it may be shewn 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. Then, because ABC is a straight line, the angle CBE is equal to the angle EBA. E B [Hypothesis. [Definition 10. Also, because ABD is a straight line, the angle DBE is equal to the angle EBA. [Hypothesis. Therefore the angle DBE is equal to the angle CBE, [Ax. 1. the less to the greater; which is impossible. [Axiom 9. Wherefore two straight lines cannot have a common segment. санто PROPOSITION 12. PROBLEM. 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 the given point without it it is required to draw from the point Ca straight line perpendicular to AB. The straight line CH drawn from the given point C shall be perpendicular to the given straight line AB. Join CF, CG. Because FH is equal to HG, [Construction. and HC is common to the two triangles FHC, GHC; the two sides FH, HC are equal to the two sides GH, HC, each to each; [Definition 15. and the base CF is equal to the base CG; therefore the angle CHF is equal to the angle CHG ; [I. 8. 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 the angles is called a right angle, and the straight line which stands on the other is called a perpendicular to it. [Def. 10. Wherefore a perpendicular CH has been drawn to the given straight line AB from the given point C without it. Q.E.F. |