· 141. COROLLARY 1.-Hence every triangle is half a rectangle, having the same base and altitude. 142. COROLLARY 2.-Triangles which have equal bases and equal altitudes are equal. THEOREM 50.· F 143. Rectangles, of the same altitude, are to one another as their bases. Let ABCD, AEFD, be two rectangles, which have a com- D mon altitude AD; they are to one another as their bases AB, AE. E B For, suppose that the base AB contains seven equal parts, and that the base AE contains four similar parts; then, if A AB be divided into seven equal parts, AE will contain four of them. At each point of division draw a perpendicular to the base, and these will divide the figure ABCD into seven equal rectangles (138); and, as AB contains seven such parts as AE contains four, the rectangle ABCD will also contain seven such parts as the rectangle AEFD contains four; therefore the bases AB, AE, have the same ratio that the rectangles ABCD, AEFG, have. THEOREM 51. 144. Rectangles are to one another as the products of the numbers which express their bases and altitudes. Let ABCD, AEGF, be two rectangles, and let some line taken, as a unit, be contained m times in AB, the base of the one, and n times in AD, its altitude; also p times in AE, the base of the other, and q times in AF, its altitude; the rectangle ABCD shall be to 15 the rectangle AEGF, as the product mn is to the product pq. Let the rectangles be so placed that their bases AB, AE, may be in a straight line; then their altitudes AD, AF, shall also form a straight line (48). Complete the rectangle EADH; and, because this rectangle has the same alti tude as the rectangle ABCD, when EA and AB are taken as their bases; and the same altitude as the rectangle AEGF, when AD, AF, are taken as their bases; we have the rectangle ABCD: ADHE :: AB: AE::m: p.... ...(143) But.... • m : Ρ :: mn: pn therefore, ABCD: ADHE :: mn: pn. In like manner, ADHE: AEGF :: pn: pq But, placing the terms of these two sets of proportionals alternately, we have, Hence the rectangle EFGH will contain the superficial unit ABCD, as often as the numeral product pq contains unity. Consequently, the product pq will express the area of the rectangle, or will indicate how often it contains the unit of its surfaces. Thus, if EF contains the linear unit AB four times, and EH contains it three times, the area EFGH will be 3 x 4=12: that is, equal to twelve times a square whose side AB is=1. In consequence of the surface of the rectangle EFGH being expressed by the product of its sides, the rectangle, or its area, may be denoted by the symbol EF x FG, in conformity to the manner of expressing a product in arithmetic. However, instead of expressing the area of a square, made on a line AB, thus, AB > AB; it is thus expressed AB2. 146. NOTE. A rectangle is said to be contained by two of its sides, about any one of its angles. THEOREM 52. 147. The area of a parallelogram is equal to the product of its base and altitude. For the parallelogram ABCD is equal to the rectangle ABEF, which has the same base AB, and the same altitude (138), and this last is measured by AB × BE, or by AB x AF; that is the product of the base of the parallelogram and its altitude. D 148. COROLLARY.-Parallelograms of the same base are to one another as their altitudes; and parallelograms of the same altitudes are to one another as their bases. For, in the former case, put B for their common base, and A, a, for their altitudes; then we have B x A: Bxa :: A: a. And, in the latter case, put A for their common altitude, and B, b, for their bases; then B x A: b× A :: B: b. THEOREM 53. 149. The area of a triangle is equal to the product of the base by half its altitude. For the triangle ABC is half the parallelogram ABCE, which has the same base, BC, and the same altitude AD (140); but the area of the parallelogram is BC × AD (147), therefore, the area of the triangle is BC × AD, or BC × AD. A 150. COROLLARY.-Two triangles of the same base are to one another as their altitudes; and two triangles, of the same altitude, are to one another as their bases. THEOREM 54. 151. The area of every trapezoid, ABCD, is equal to the product of half the sum of its parallel sides, AB, DC, by its altitude, EF. H DE K Through I, the middle of the side BC, draw KL, parallel to the opposite side AD, and produce DC until it meet KL in K. In the triangles IBL, ICK, the side IB is equal to IC, and the angle B equal to C (74), the angle BIL equal to CIK; therefore the triangles are equal (54), and the side CK equal to BL. Now the parallelogram ALKD is the sum of the polygon ALICD; and the triangle CIK, and the trapezoid ABCD, is the sum of the same polygon and the triangle BIL; therefore, the trapezoid ABCD is equal to the parallelogram ALKD, and has, for its measure, AL x EF. Now AL is equal to DK, and BL equal to CK; and CD-DK - CK; but DK is equal to AL, and CK equal to BL; Therefore, CD = AL - BL. 152. A straight line, DE, drawn parallel to the side of a triangle ABC, divides the other sides AB, AC, proportionally, or so that AD: DB:: AE : EC. Join BE and DC: the two triangles BDE, CDE, have the same base, DE, and they have also the same altitude, because BC is parallel to DE; and, consequently, the triangles DBE and DCE are equal. Since the triangles BED and AED have the same altitude, they are to one another as their bases; and since the triangles CED and AED have the same altitude, they are to one another as their bases. Therefore, the triangle BDE: ADE :: BD: DA. But since the triangle BDE is equal to the triangle CED, therefore the triangle CED: ADE :: BD: DA. But the triangle CED: ADE :: CE : EA. THEOREM 56. 153. If the two sides, AB, AC, of a triangle be cut proportionally by the line DE, so that AD: DB:: AE: EC, the line DE shall be parallel to the remaining side of the triangle. For, if DE be not parallel to BC, some other line, DO, will be parallel to BC; then, by the preceding proposition, AD: DB:: AO: OC. And, by hypothesis, AD: DB:: AE: EC. Therefore, ... AO: OC :: AE : EC. D B And hence OC must be equal to EC; which is impossible, unless the point O coincide with E; therefore no line besides DE can be parallel to BC. THEOREM 57. 154. A line, BD, which bisects any angle, ABC, of a triangle, will divide the opposite side AC into two segments, AD, DC, which shall have the same ratio as the other two sides, AB, BC, of the triangle. From C, one extremity of the base, draw CE, parallel to BD, meeting AB produced in E. Then the angle ABD is B E side BC is equal to BE (63). Now, because ACE is a triangle, and BD is drawn parallel to one of its sides, AD: DC :: AB : BE (152) ; but, since BE is equal to BC; therefore AD: DC :: AB: BC. THEOREM 58. 155. Two equiangular triangles have their sides proportional, and are similar to each other. |