Demonstration. Then, in the triangles DBC, ABC, because DB is equal to AC, and BC is common to both triangles, A B 1. The two sides DB, BC are equal to the two sides AC, CB, each to each; and the angle DBC is equal to the angle ACB (hyp.), therefore 2. The base DC is equal to the base AB (I. 4), and the triangle DBC is equal to the triangle ABC, the less equal to the greater, which is absurd (Ax. 9). Therefore 3. AB is not unequal to AC; that is, AB is equal to AC. Wherefore, if two angles, &c. Q.E.D. Cor. Hence an equiangular triangle is also equilateral. PROPOSITION 7.-Theorem. 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. B If it be possible, on 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. First. Let the vertex of each of the triangles be without the other. Construction. Join CD. Demonstration. Because AC is equal to AD in the triangle ACD; therefore, 1. The angle ACD is equal to the angle ADC (I. 5). But the angle ACD is greater than the angle BCD (Ax. 9); therefore also 2. The angle ADC is greater than BCD; much more therefore 3. The angle BDC is greater than the angle BCD. Again, because the side BC is equal to BD in the triangle BCD (hyp.), therefore 4. The angle BDC is equal to the angle BCD (I. 5). But the angle BDC was proved greater than the angle BCD, hence, the angle BDC is both equal to, and greater than, the angle BCD; which is impossible. Secondly. Let the vertex D of the triangle ADB fall within the triangle ACB. Construction. Produce AC to E, and AD to F, and join CD. Demonstration. Then, because AC is equal to AD in the triangle ACD, the angles upon the other side of the base CD, namely, 1. The angles ECD, FDC are equal to one another (I. 5); but the angle ECD is greater than the angle BCD (Ax. 9); therefore also 2. The angle FDC is greater than the angle BCD; much more then 3. The angle BDC is greater than the angle BCD. Again, because BC is equal to BD in the triangle BCD, therefore 4. The angle BDC is equal to the angle BCD (I. 5), but the angle BDC has been proved greater than BCD, wherefore the angle BDC is both equal to, and greater than the angle BCD, which is impossible. Thirdly. 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, &c. Q.E.D. PROPOSITION 8.-Theorem. 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 EF. B C E Then the angle BAC shall be equal to the angle EDF. Demonstration. For, if the triangle ABC be applied to DEF, so that the point B be on E, and the straight line BC on EF; then, because BC is equal to EF (hyp.), therefore B 1. The point C shall coincide with the point F, wherefore BC coinciding with EF, N E B 2. BA and AC shall coincide with ED, DF; for, if the base BC coincide with the base EF, but the sides BA, AC do not coincide with the sides ED, DF, but have a different situation, as EG, GF: Then, upon the same base, and upon the same side of it, there can be two triangles which have their sides which are terminated in one extremity of the base, equal to one another, and likewise those sides which are terminated in the other extremity; but this is impossible (I. 7). Therefore If the base BC coincide with the base EF, the sides BA, AC cannot but coincide with the sides ED, DF; wherefore, likewise 3. The angle BAC coincides with the angle EDF, and is equal to it. (Ax. 8.) Therefore 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. F E Let BAC be the given rectilineal angle. It is required to bisect it. Construction. In AB take any point ; from AC cut off AED equal to ADE(I. 3), and join DE; on the side of DE remote from A, describe the equilateral triangle DEF (I. 1), and join AF. Then the straight line AF shall bisect the angle BAC, Proof. Because AD is equal to AE (constr.), and AF is common to the two triangles DAF, EAF, and (constr.) therefore 1. The two sides DA, AF are equal to the two sides EA, AF, each to each; Wherefore 2. The base DF is equal to the base EF; 3. The angle DAF is equal to the angle EAF (I. 8). 4. The angle BAC is bisected by the straight line AF. Q.E.F. PROPOSITION 10.-Problem. 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 AB into two equal parts. D Construction. Upon AB describe the equilateral triangle ABC (I. 1), and bisect the angle ACB by the straight line CD meeting AB in the point D (I. 9). Then AB shall be cut into two equal parts in the point D. Proof. Because AC is equal to CB (constr.), and CD is common to the two triangles ACD, BCD; 1. The two sides AC, CD are equal to BC, CD, each to each; and (constr.) |