35. All other quadrilateral figures are called trapezia. 36. Multilateral figures, or polygons, are contained by five or more sides; but the triangle and square may be included in the series. They are called regular when they are equilateral. NOTE. The series of regular polygons are equiangular: it may be said to begin with the triangle, and end with the circle. The Greeks named the regular polygons from their angles, viz: A trigon has three equal angles. A tetragon has four An undecagon has eleven " A dodecagon has twelve A trisdecagon has thirteen equal angles; a quindecagon, fifteen; and so upwards to infinity, which may be represented by the circle. 37. The perimeter is the sum of the lines which bound a figure: and it may be remarked, although it will be hereafter proved, that under equal perimeters, regular polygons contain greater areas than figures contain whose sides are unequal. Also, the perimeters of regular polygons being equal, that perimeter which has the greater number of sides, contains the greater area; and the circle contains the greatest area within equal bounds. 38. A theorem is a proposition which requires to be proved. POSTULATES, OR PETITIONS. The geometer requires permission to move freely in space, viz: 1. To draw a straight line from one point to another. 2. To produce a terminated straight line. 3. To describe a circle about any point with any radius. AXIOMS, OR MAXIMS. 1. Things which are equal to the same are equal to each other. 2. Add equals to equals, the sums will be equal. 3. Take equals from equals, the remainders will be equal. 4. Add equals to unequals, the sums will be unequal. 5. Take equals from unequals, the remainders will be unequal. 6. Doubles of the same are equal to each other. 7. Halves of the same are equal to each other. 8. Things which coincide, or fill the same space, are equal to each other. 9. The whole is equal to all its parts, and greater than its part. 10. All right angles are equal to one another; and so are all angles measured by equal arcs to equal radii. 11. Two intersecting straight lines cannot both be parallel to the same straight line, or to each other. Illustration of the Definitions. The angular point is marked by a letter, and when three or more lines meet in a point, their extremities are lettered; and in naming the angle, the letter at the angular point is read between the other two. Abbreviations used in the following work: P, after the number of the proposition, stands for problem; Th. for theorem; p., after recite, for proposition; ax., axiom; cor., corollary; def., definition; pos., postulate; hyp., hypothesis. The first figure refers to the proposition; the latter, to the book. 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 (b). 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 D 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; (b), 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 (3); 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. 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; B G K H therefore AL is equal to BC (f); and it is drawn from the given point A, which was to be done." Recite (a), post. 1; (b), prop. 1; (c), post. 2; (d), post. 3; (e), ax. 3; (ƒ), ax. ì. 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 (b). Dem. AD and AE are equal radii (c); but AD is made equal to C; therefore AE is equal to C (d); and it is a part cut off from AB, which was to be done. Recite (a), prop. 2; (c), def. 15; (b),post. 3; B E B 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. 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. 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 (b); 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 cach; 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). 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 DC, 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; (b), post. 1; (c), prop. 4. |