From point A draw the straight line AE = CD (I. 2). AE, describe circle EFH cutting. AB in F. AF shall be equal to CD. For A is the centre of the circle EFH: AF = AE. II Thus from AB a part AF has been cut off equal to CD.— Q.E.F. PROPOSITION IV., THEOREM 1. If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another: then they shall also have their bases or third sides equal, and the two triangles shall be equal, and their other angles shall be equal, each to each, namely, those to which the equal sides are opposite. In the As ABC, DEF, let AB = DE, and AC = DF and BAC LEDF. Then shall BC EF and A ABC =ADEF and ABC = DEF and ACB = DFE. For if ▲ ABC be applied to A D Δ Δ E F ▲ DEF so that the point A may be upon D and the straight line AB on the straight line DE, then AB DE, the point B will coincide with point E. And AB coinciding with DE, AC will fall on DF, ... / BAC= LEDF. And AC= DF, the point C will coincide with point F. But point B was shown to coincide with point E, .. the base BC will coincide with the base EF; if not, let it fall otherwise as EOF; then the two straight lines BC, EF will enclose a space, which is impossible. (Post. 5.) I .. BC will coincide with EF and is equal to it (ax. 8), and A ABC will coincide with A DEF and is equal to it, The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced the angles on the other side of the base shall be equal to one another. Let ABC be an isosceles ▲, having side AB = side AC, and let AB and AC be produced to D and E: ▲ ABC shall be equal to ▲ ACB; F B G E GA, Then in As AFC, AGB, FA = and AC = AB, and ▲ FAC = ≤ GAB; .. base FC base GB, and ▲ AFC = A AGB (I. 4), and ▲ ACF = ▲ ABG and AFC = AGB. Also whole AF whole AG, of which the parts AB, AC are equal, .. remainder BF = remainder CG (ax. 3). Then in As BFC, GCB, BF = CG, and FC = GB, and BFC = ▲ CGB; .. A BFC = A CGB (I. 4), and ▲ FBC = ▲ GCB, and ▲ BCF = ▲ CBG (I. 4). And it has been shown that ▲ ABG = ▲ ACF, and that parts of these, ▲ CBG and ▲ BCF are equal. = .. remaining. ABC remaining ACB, which are the angles at the base of ▲ ABC. Also it has been shown that FBC= the angles on the other side of the base. GCB, which are Wherefore the angles, &c.-Q.E.D. Corollary.-Hence every equilateral triangle is also equi angular. PROPOSITION VI., THEOREM 3. If two angles of a triangle be equal to one another, the sides also which subtend, or are opposite to the equal angles, shall be equal to one another. Let ABC be a ▲, having ▲ ABC = ▲ ACB; the side AC shall be equal to the side AB. For if AC be not equal to AB one of them must be greater than the other. DB Let AB be the greater, and from it cut off AC the less (I. 3). Join DC. Then in As ABC, DBC, AC=DB (cons.), and BC is common, and ACB DBC (hyp.). B4 A .. ΔΑΒΟ =^ DBC, the less to the greater, which is impossible (ax. 9). .. AB is not unequal to AC, that is, it is equal to it. Wherefore if two angles, &c.-Q.E.D. Corollary.--Hence every equiangular triangle is also equi lateral. PROPOSITION VII., THEOREM 4. On the same base, and on the same side of it, there cannot be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise those which are terminated at the other extremity equal to one another. If it be possible on the same base AB, and on the same side of it, let there be two As ACB, ADB, having AC and AD terminated at A equal, and also CB, DB terminated at B equal to one another. Join CD. Case I. When the vertex of each triangle is without the other. .. AC AD, LACD = 2 ADC (I. 5). Again But ACD is greater than ▲ BCD (ax. 9), .. ADC is also greater than / BCD; much more then is BDC greater than B ▲ BCD. BC= BD, ▲ BDC = BCD. But it has been shown to be greater, which is impossible. = Then AC AD, in A ACD (hyp.), s ECD, FDC, on the other side of the base are equal (I. 5). B much more then BDC Again BC= BD in ▲ BDC, ▲ BDC = BCD. But it has been shown to be greater, which is impossible. The case in which the vertex of one ▲ is on a side of the other needs no demonstration. Wherefore on the same base, &c.— A B PROPOSITION VIII., THEOREM 5. 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 the sides of the As ABC, DEF be equal, each to each : AB= DE, AC = DF, and BC= EF; the ▲ BAC shall be equal to EDF. For if the ▲ ABC be applied to ▲ DEF, so that point B may be upon point E, and the straight line BC on EF, the point C will coincide with point F, ... BC= EF. .. BC coinciding with EF, BA and AC will coincide with ED and DF. D For if the base BC coincides with the base EF, but the sides BA, AC do not coincide, but have a different situation, as EG, GF, then on the same base and on the same side of it there will be two triangles having their sides which are B terminated at one extremity of the base equal, and likewise those which are terminated at the other extremity. But this is impossible (I. 7). ..since base BC coincides with base EF, the sides BA, AC must coincide with the sides ED, DF. .. BAC coincides with EDF, and is equal to it. Wherefore if two triangles &c.-Q.E.D. PROPOSITION IX., PROBLEM 4. To bisect a given rectilineal angle. Let BAC be the given rectilineal angle : It is required to bisect it. Take any point D in AB. From AC cut off AE = AD (I. 3); Join DE, and on the side of DE remote from A, describe the equilat. A DEF (I. 1). Join AF. The straight line AF shall bisect the given angle BAC. Then in the AS DAF, EAF, B A E D Wherefore the given angle BAC is bisected by the straight line AF.-Q.E.F. |