32 respectively equal, the parallelograms are equal to one another. This will be shown by drawing diagonals subtending the equal angles; as (I. 4) the triangles thus formed are equal; and, by this proposition, the parallelograms are respectively double of those triangles. Cor. 3. If two straight lines make an angle, two others parallel to them contain an equal or a supplemental angle. the vertical angle made by producing BA, CA, through A, are Thus, A and each equal to D; while the angle made by BA and the continuation of CA, or by CA and the continuation of BA is equal to the supplement (I. def. 38) of D. Cor. 4. Hence we can divide a given straight line AB into any proposed number of equal parts. To do this, draw AC making any angle with AB, and draw (I. 31) BD parallel to AC. take any point E, and make (I.3) EF, FG, BH, HK, KL, each equal to AE; the number of segments on each of the parallels being less by one than the number of the parts into which AB is to be divided. Draw (I. post. 1) the straight lines EL, FK, GH, cutting AB in M, N, 0: AM, MN, NO, OB are the parts required. Draw MP parallel to A D In AC But AC; then (I. 33) EL, FK, GH are parallel ; and AE and MP are equal, each of them being equal (const. and I. 34) to EF. (I. 29) the angles EAM, AME are respectively equal to PMN, MNP; and therefore (I. 26) AM is equal to MN. manner, BO may be proved to be equal to ON: and by drawing In the same NQ parallel to AC, MN may be proved in a similar manner to be equal to NO.* upon the same PROP. XXXV. THEOR.-Parallelograms base, and between the same parallels, are equal to one another. Let the parallelograms ABCD, EBCF, (figs. 2 and 3) be upon the same base BC, and between the same parallels AF, BC; these parallelograms are equal to one another. C F If the sides AD, DF, of the parallelograms AC, BF (fig. 1) opposite to the base BC, be ter- B minated in the same point D, each of the parallelograms is * The construction might be effected as in the 9th of the sixth book without the parallel BD. The method here given, however, is easier in practice, from the facility with which the parallels EL, FK, &c. are drawn. This problem might be considered unnecessary here, as it is the same in substance as the 9th of the sixth book. Besides the preferable construction, however, which is here given, the problem is so useful, that the student should be early acquainted with it. double (1. 34) of the triangle BDC; and they are therefore equal (I. ax. 6) to one another. A DE F A E D F But if the sides AD, EF (figs. 2 and 3) opposite to the base BC, be not terminated in the same point; then, because AC is a parallelogram, DC is equal (I. 34) to AB. Also (I. 29, part 2) because AB is parallel to DC, B C B C and AF falls upon them, the angles FDC, EAB are equal; and because BE is parallel to CF, and AF falls upon them, the angles DFC, AEB are likewise equal: wherefore (I. 26, part 3) the triangles FDC, EAB are equal to one another. Take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB, and (I. ax. 3) the remainders are equal; that is, the parallelogram AC is equal to the parallelogram BF.* Therefore parallelograms, &c. PROP. XXXVI. THEOR.-Parallelograms upon equal bases, and between the same parallels, are equal to one another. Let AC, EG be parallelograms upon equal bases BC, FG, and between the same parallels AH, BG; the parallelogram AC is equal to EG. A DE H B C F G Join BE, CH; and because BC is equal (hyp.) to FG, and (I. 34) FG to EH; BC is equal (I. ax. 1) to EH; and (hyp.) they are parallel, and they are joined towards the same parts by the straight lines BE, CH: therefore (I. 33) BE, CH are both equal and parallel, and (I. def. 24) EBCH is a parallelogram; and it is equal (I. 35) to AC, because they are upon the same base BC, and between the same parallels BC, AH. For the like reason, the parallelogram EG is equal to the same parallelogram EBCH; because they are on the same base EH, and between the same parallels EH, BG. Therefore also the parallelogram AC is equal (I. ax. 1) to EG. Wherefore parallelograms, &c. PROP. XXXVII. THEOR.t-Triangles upon the same base," and between the same parallels, are equal to one another. * It will appear, from this proposition, that the perimeters of two equal parallelograms on the same base may differ in any degree whatever. It follows also, from the 19th proposition, that, of all equal parallelograms on the same base, the rectangle has the least perimeter. It may also be remarked, that Legendre and some other writers call only such figures equal as exactly coincide when applied to one another; while those which, as in this proposition, "without coinciding, contain the same space," they call equivalent. This seems to be an unnecessary refinement. + This proposition and the following afford instances in which triangles, though not as in propositions 4, 8, and 26, in every respect equal, are equal in magnitude. Let the triangles ABC, DBC be upon the same base BC, and between the same parallels AD, BC; ABC is equal to DBC. R A D Produce AD both ways to E and F, and through B draw (I. 31) BE parallel to CA, and through C draw CF parallel to BD.* Therefore (I. def. 24) each of the figures EC, BF is a parallelogram; and (I. 35) they are equal to one another, because they are upon the same base BC, and between the same parallels BC, EF. But the triangle ABC is the half of the parallelogram EC, because (I. 34) the diagonal AB bisects it; and the triangle DBC is the half of the parallelogram BF, because the diagonal DC bisects it: but (I. ax. 7) the halves of equals are equal; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. PROP. XXXVIII. THEOR. Triangles upon equal bases, and between the same parallels, are equal to one another. Let the triangles ABC, DEF be upon equal bases, BC, EF, and between the same parallels BF, AD: ABC is equal to DEF. G H Produce AD both ways to G and H, and (I. 31) through B draw BG parallel to CA, and through F draw FH parallel to ED. Then (I. def. 24) each of the figures GC, EH is a parallelogram; and (I. 36) they are equal to one another, because they are upon equal bases BC, EF, and between the same parallels BF, GH; and (I. 34) the triangle ABC is half of the parallelogram GC; and the_triangle_DEF of the parallelogram EH. But (I. ax. 7) the halves of equals are equal: therefore the triangle ABC is equal to the triangle DEF. Wherefore triangles, &c. B C } Cor. 1. From this proposition and from the 36th, it is plain, that triangles and parallelograms between the same parallels, but upon unequal bases, are unequal; that which has the greater base being greater than the other. Cor. 2. Hence a straight line drawn from the vertex of a triangle to the point of bisection of the base, bisects the triangle; and if two triangles have two sides of the one respectively equal to two sides of the other, and the contained angles (I. def. 38) supplemental, the triangles are equal. * It follows from the first corollary to the 29th proposition, that AD, when produced, will meet BE and CF: and in the next proposition, AD produced will meet the corresponding lines on the same principle. Euclid often assumes without proof, that certain lines will meet. The omission of the proof, however, is of no consequence, as it can occasion no doubt or difficulty. PROP. XXXIX. THEOR.*-If equal triangles, not coinciding, stand on the same base, and on the same side of it, the straight line joining their vertices is parallel to the base. Let the equal triangles ABC, DBC be on the same_base BC and on the same side of it; join AD: AD is parallel to BC. For, if it be not, through A draw (I. 31) AE parallel to BC, and join EC. Then (I. 37) the triangles ABC, EBC are equal, because they are on the same base BC, and between the same parallels BC, AE. But the triangle ABC is equal (hyp.) to the triangle DBC: therefore also the triangle EBC is equal (I. ax. 1) to the triangle DBC, the less to the greater, which is impossible. Therefore AE is not parallel to BC; and in the same manner it can be demonstrated, that no other line drawn through A, except AD, is parallel to BC: AD is therefore parallel to it. Wherefore, if equal triangles, &c. B E PROP. XL. THEOR.-If equal triangles, not having a common vertex, stand upon equal bases in the same straight line, and on the same side of it, the straight line joining their vertices is parallel to the line on which they stand. Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the same straight line BF, and on the same side of it; join AD: AD is parallel to BF. A D For, if it be not, through A draw (I. 31) AG parallel to BF, and join GF. Then (I. 38) the triangles ABC, GEF are equal, because they are on equal bases BC, EF, and between the same parallels BF, AG. But the triangle ABC is equal (hyp.) to the triangle DEF: therefore also the triangle GEF is equal to the triangle DEF, B the less to the greater, which is impossible. Therefore AG is not parallel to BF; and in the same manner it can be demonstrated, that through A no other straight line can be drawn parallel to BF, but AD: AD is therefore parallel to BF. Wherefore, if equal triangles, &c. CE PROP. XLI. THEOR.-If a parallelogram and a triangle be upon the same base, and between the same parallels, the parallelogram is double of the triangle. Let the parallelogram ABCD and the triangle EBC be upon the same base BC, and between the same parallels BC, AE; the parallelogram is double of the triangle. * In this proposition and the following, we have two other principles by which lines are proved to be parallel, in addition to those contained in propositions 27, 28, 30, and 33. A D E Join AC. Then (1. 37) the triangles ABC, EBC are equal, because they are upon the same base BC, and between the same parallels BC, AE. But (I. 34) the parallelogram ABCD is double of the triangle ABC, because the diagonal AC bisects it: wherefore ABCD is also double of the triangle EBC. Therefore, if a parallelogram, &c. B PROP. XLII. PROB.-To describe a parallelogram equal to a given triangle, and having one of its angles equal to a given angle. Let ABC be the given triangle, and D the given angle: it is required to describe a parallelogram equal to ABC, and having one of its angles equal to D. Bisect (I. 10) BC in E, join AE, and (I. 23) at the point E, in the straight line BC, make the angle CEF equal to D; through A draw (I. 31) AG parallel to BC, and through C draw CG parallel to EF; FECG is the parallelogram required. F G B E C D For, since (const.) BE is equal to EC, the triangle ABE is equal (I. 38, cor. 2) to the triangle AEC: therefore the triangle ABC is double of the triangle AEC. The parallelogram EG is likewise double (I. 41) of the triangle AEC, because they are upon the same base, and between the same parallels. Therefore the parallelogram EG is equal (ax. 6) to the triangle ABC, and it has (const.) one of its angles CEF equal to D: wherefore there has been described a parallelogram EG equal to a given triangle ABC, and having one of its angles CEF equal to the given angle D: which was to be done. PROP. XLIII. THEOR.-If through any point in either diagonal of a parallelogram, straight lines be drawn parallel to the sides; of the four parallelograms thus formed, those through which the diagonal does not pass, and which are called the complements of the other two, are equal. Let BD be a parallelogram, of which AC is one of the diagonals: then, if through K a point in AC, the two straight lines EF and HG be drawn parallel to the sides, the parallelograms BK and KD are equal.* E A H F Because BD is a parallelogram, and AC its diagonal, the triangle ABC is equal (I. 34) to the triangle ADC: and, because EH is a parallelogram, the diagonal of which is AK, B G the triangles AEK, ÄHK are equal. For K EH and GF are generally called the parallelograms about the diagonal, while BK and KD are called the complements of EH and GF, because, with these, they complete the entire parallelogram BD. |