SECOND LESSONS IN GEOMETRY. BOOK FIRST. Propositions. 1 P. To describe an equilateral triangle upon a given straight line (AB). Construction. From the extreme points, A and B, with the radius AB, describe the circles ACE, BCD (a), intersecting each other in the point C: join C to A and B (6). Demonstration. ABC is the required triangle (c): for AB one of its sides, is also a radius of each of the circles (d); and the sides AC, BC, are also radii, each equal to AB: therefore the three sides are equal to one another (e), and the triangle is equilateral; and it is described upon AB, the given straight line, which was to be done. Recite (a), postulate 3; (6), post. 1; (c), definition 23; (d), def. 15; (e), axiom 1. Corollary. An isosceles triangle may be constructed by joining the extreme points of two radii. 2 P. From a given point (A), to draw a straight line equal to a given straight line (BC). Constr. Join the points A, B, by the line AB (a), upon which describe the equilateral triangle ABD (b); produce the equal sides DA, • DB to E, F (c); upon the centre B, with the radius BC, describe the circle CGH; also, upon the centre D, with radius DG, describe the circle GKL (d): AL is the required line. K H D A C B L Dem. From the equal radii DG, DL, take the equals DB, DA; the remainders BG, AL, are equal (e); but BG, BC, are equal radii; therefore AL is equal to BC (f); and it is drawn from the given point A, which was to be done. G E F Recite (a), post. 1; (b), prop. 1; (c), post. 2; (d), post. 3; (e), ax. 3; (f), ax. 1. 3 P. From the greater (AB), of two given straight lines, to cut off a part equal to the less (C). Constr. Draw the line AD equal to C, (a); and from the centre A, with the distance AD, describe the circle DEF (6). Dem. AD and AE are equal radii (c); but AD is made equal to C; therefore AE is C equal to C (d); and it is a part cut off from AB, which was to be done. (b),post. 3; 4 Th. If two triangles (ABC, DEF), have two sides (AB, AC) of the one, equal to two sides (DE, DF), of the other, each to each; and have likewise the angles (A and D) contained by those sides, equal; their third sides (BC and EF) shall also be equal; and their areas shall be equal; also their other angles-namely (B to E and C to F) those to which the equal sides are opposite. D Because the three given parts are A adjacent in each triangle, and equal, each to each; therefore, the parts BA, A and AC, may be applied to the parts ED, D and DF, so that they shall fill the same space (a), and that the points B, A and C, shall severally coincide with the points E, D and F; hence the line BC shall fall on the line EF, and be B equal to it; otherwise, falling above or below the line EF, two straight lines would enclose a space, which is impossible (b). Therefore, also. the angles at B and C shall be equal to those at E and F, respectively; and the areas of the two triangles shall coincide and be equal (a). Wherefore, if two triangles, &c. Recite (a), ax. 8; (b), cor. to def. 4. CE Q. E. D. Note. This proposition is of very general use in the elements: it helps to demonstrate the 5th and 47th of the first book, and many others. It proves the triangles ABG, ACF, in the next diagram, to be equal; as also the triangles FBC, GCB. In this case, as in all others, two things are compared by means of a third; the third thing here taken is a portion of space, to which each of the triangles is applied. 5 Th. The angles (ABC, ACB), at the base of an isosceles triangle, are equal to one another; and if the equal sides (AB, AC,) be produced (to D and E), the angles (BCE, CBD,) below the base (BC) shall be equal. Constr. In BD take any point F; and from AE, the greater, cut off a part, AG, equal to AF (a); then from the equals AF, AG, take the equals AB, AC, the remainders, BF and CG, will be equal (6); join BG, CF (c). Dem. The two triangles ABG, ACF, are equal; having two sides, AB, AG, of the one, equal to two sides, AC, AF, of the other, each to each; and the angle A is common to both: therefore the bases, BG and CF, are equal, and likewise the angles ABG, ACF; as also the angles at F, G (d). The two triangles BCG, CBF, are also equal: for it is shown above, that FC, FB, and the angle F, in the one, are severally equal to GB, GC, and the angle G, in the other; and the base BC is common to both; therefore the remaining angles are equal, each to each, to which the equal sides are opposite; viz. CBG to BCF, and FBC to GCB, which are the angles below the base (d). Again. From the equal angles, ABG, ACF, take the equals CBG, BCF, the remainders, ABC, ACB, are equal, which are the angles at the base (e). Wherefore, the angles, &c. Q. E. D. Recite (a), p. 3; (b), def. 24, ax. 3; (c), post. 1; (d), p. 4; (e), ax. 3. Corollary. Hence every equilateral triangle is also equiangular. 6 Th. If two angles (B, C) of a triangle (ABC), be equal to one another, the subtending sides (AC, AB) of the equal angles shall be equal to one another. Constr. For if AB be not equal to AC, it must be less than it, or greater. Let AB be the greater, and from it cut off a part BD equal to AC, the less (a); and join CD (b). D A Dem. Because, in the triangles DBC, ACB, DB is equal to AC, and BC is common to both; therefore, the two sides DB, BC, are equal to the two AC, CB, each to each; and the angles DBC, ACB, are equal: therefore the bases DCy AB, are equal; and the triangle DBC is equal to the triangle ACB (c), the less to the greater, which is absurd. Therefore AB is not unequal to AC, but equal to it. Wherefore, if two angles, &c. B Q. E. D Recite (a), prop. 3; (6), post. 1; (c), prop. 4. 7 Th. Upon the same base (AB), and on one side of it, two triangles cannot have their sides (AC, AD) equal, which terminate in one extremity of it, and likewise their sides (BC, BD), which terminate in the other extremity. Join CD: then, in the case in which the vertex of each triangle is without the other; because AC and AD are equal, the angles ACD, ADC, are also equal (a). But the angle BCD is less than the angle ACD, therefore less than ADC, and still less than BDC. Again, because CB is equal to DB, the angles BCD, BDC, are equal: but BCD has been proved to be much less than BDC; both equal and less is impossible. And, in the case of one of the vertices, D, being within the other triangle ACB, produce AC, AD, to E, F; therefore, because ACD is an isosceles triangle, the angles ECD, FDC, beyond the base, are equal (b); but the angle BCD is less than the angle ECD, therefore less than FDC, and still less than BDC. Again, because BC and BD are equal, the angles BDC, BCD, are also equal: but BCD has been proved to be less than BDC, which is impossible. A A C D D E F B B The case in which the vertex of one triangle is upon the side of another, needs no demonstration. Therefore, upon the same base, &c. Q. E. D. 8 Th. If two triangles have two sides (AB, AC) of the one, equal to two sides (DE, DF) of the other, each to each, and have likewise their bases (BC, EF) equal; the angles (A, D) contained by the corresponding sides, shall be equal to one another. Demonstration. Apply the triangle ABC to DEF; so that the point B fall on E, and the base BC fall upon its equal EF; then the point C shall coincide with F, and the sides BA, AC, fall on ED, DF, each upon each; for, if the bases coincide, but the sides A fall in another direction, as on G, then upon one side of the base EF, there can be two triangles, with equal sides, which terminate in either extremity of the base; which is impossible (a). Wherefore, if the bases Q. E. D. coincide, the sides shall also coincide; and, consequently, the angles BAC, EDF, contained by those sides, shall be equal. Recite (a) prop. 7. 9 P. To bisect a given rectilineal angle (BAC); that is, to divide it into two equal angles. A Constr. In AB take any point D; make AE equal to AD (a); join DE (1), and upon it describe an equilateral triangle DEF (c); join AF: the straight line AF bisects the angle BAC. A Dem. Because AD is made equal to AE, and AF is common to the two triangles DAF, EAF, and the bases DF, EF, were made equal: therefore the angle DAF is equal to EAF (d); that is, the rectilineal angle BAC is bisected by the straight line AF; which was to be done. D B F Recite (a), prop. 3; (6), post. 1; (c), prop. 1; (d), prop. 8. E C 10. P. To bisect a given finite straight line (AB); that is, to divide it into two equal parts. Constr. Upon AB describe an equilateral triangle ABC (a), and bisect the angle ACB by the straight line CD (b). Argument. AC is made equal to BC, the angle ACD to BCD, and CD is common to the two triangles ACD, BCD: therefore, the bases AD and BD are equal, and AB is cut into two equal parts in the point D. Which was to be done. Recite (a), prop. 1; (6), prop. 9. D B 11 P. To draw a straight line (CF), at right angles to a given straight line (AB), from a given point (C), in the same. F A Construction. In AC take any point D, and make CE equal to CD (a); upon DE describe the equilateral triangle DFE (6); join FC (e). Argument. The sides CD, CF, are equal to the sides CE, CF; and the bases DF, EF, are also equal: therefore the included angles at C, being equal (d), and adjacent, the straight line FC is drawn from the given point C, at AD right angles to AB (e); which was to be done. Recite (a), prop. 3; (d), prop. 8; (b), prop 1; C EB (c), post. 1; |