88. Cor. 2. In a given triangle (ABC) to inscribe a parallelogram similar to a given parallelogram (LMNP). Outline. Transfer LMNP so that NP may be parallel to BC. Through L and M draw parallels to AB and AC, thus forming a triangle similar to ABC, and having LMNP as an inscribed parallelogram. Then, by means of 86, inscribe in ABC a parallelogram similar to LMNP. Ex. 1. In a given triangle to inscribe a square. Ex. 2. In a triangle to inscribe a rectangle similar to a given one. RATIO OF SURFACES OF POLYGONS 89. All the ratios hitherto considered have been ratios of segments of lines. It will now be shown how to compare the surfaces of polygons. The ratio of the surfaces of two polygons will be called the ratio of the polygons. We begin with the polygons that are most easily compared, namely, two rectangles of equal altitudes, and thence advance, step by step, to the comparison of polygons in general. 90. THEOREM 16. If two rectangles have equal altitudes, then the ratio of the rectangles is equal to the ratio of their bases. Let the rectangles OABC, O'A'B'C' have the bases OA, O'A', and the equal altitudes AB, A'B'. Let the rectangles be denoted by R, R', and their bases by b, b'. C' B' B2 B3 of CB. Prolong O'A', and make a similar construction. The segment 0Am is equal to mb; and the rectangle standing on it is equal to mR. The segment O'A', is equal to nb'; and the rectangle standing on it is equal to nR'. Since the altitudes are equal, the pairs of magnitudes and rect. OBm, rect. OB'n base OA, base OA'n are in the same order of size (II. 22, III. 50), i.e., the pairs of multiples are in the same order of size, whatever m and n are; therefore the scale of relation of R and R' is everywhere similar to the scale of relation of b and b'; hence, by definition of equal ratios, R: R' = b: b'. Parallelograms of equal altitudes. 91. Cor. I. Two parallelograms of equal altitudes have a ratio equal to the ratio of their bases. Triangles of equal altitudes. 92. Cor. 2. Two triangles of equal altitudes have a ratio equal to the ratio of their bases. Cor. 3. Two parallelograms or triangles of equal bases have a ratio equal to the ratio of their altitudes. Ex. 1. Perpendiculars are drawn from any point within an equilateral triangle on the three sides: show that their sum is equal to the altitude of the triangle (IV. 59, 33). Ex. 2. A quadrangle is divided by its diagonals into four triangles that form a proportion. Ex. 3. If two triangles have their bases in the same straight line, and their vertices on the same line parallel to the bases, then any other parallel, cutting the sides, cuts off two triangles that form a proportion with the given triangles. Relation among four proportional lines. 93. THEOREM 17. If four lines form a proportion, then the rectangle of the extremes is equivalent to the rectangle of the means. Let a, b, c, d be four lines such that a a: b = c: d. To prove that the rectangle of a and d is equivalent to the rectangle of b and c. On one of the sides of any right angle lay off OA, OB equal to a, b; and on the other side lay off Oc, OD equal to c, d. Complete the rectangles AD and BC. Compare each of these rectangles with the rectangle BD, which is their common part. Since rectangles of equal altitudes have a ratio equal to the ratio of their bases, therefore therefore, by equality of ratios, rect. AD rect. BD rect. BC: rect. BD; since these equal ratios have a common consequent, hence the rectangles AD and BC are equivalent; that is, the rectangle of the extremes is equivalent to the rectangle of the means. Special case. 94. Cor. If three lines are proportional, the rectangle of the extremes is equivalent to the square on the mean. Ex. 1. Apply the theorem to prove III. 94. Converse of 93. 95. THEOREM 18. If two rectangles are equivalent, the sides of one will form the extremes, and the sides of the other the means, of a proportion. In figure of theorem 17, if rectangles AD and BC are equivalent, they have equal ratios to rectangle BD (IV. 25). Therefore, etc. Converse of 94. 96. Cor. If there are three lines such that the rectangle of the extremes is equivalent to the square on the mean, then the three lines form a proportion. EXTREME AND MEAN RATIO 97. Definition. If a given line is divided into two parts such that one of the parts is a mean proportional between the whole line and the other part, the line is said to be divided in extreme and mean ratio or in medial section. Application of 94. 98. PROBLEM 12. To divide a given line in extreme and mean ratio. By means of the construction in II. 89, divide the given line so that the rectangle of the whole line and one part is equivalent to the square on the other part. The line is then divided in extreme and mean ratio; because the latter part is, by 96, a mean proportional between the whole line and the first part. NOTE. This mode of division is the ancient sectio aurea (II. 89). Ex. If the radius of a circle is divided in extreme and mean ratio, the greater segment is equal to the side of an inscribed regular decagon (III. 122). MUTUALLY EQUIANGULAR PARALLELOGRAMS 99. From the comparison of two parallelograms of equal altitudes, or of equal bases (91, 92), we can advance to the comparison of any two mutually equiangular parallelograms. This is done by introducing an intermediate parallelogram having a side in common with each, and then compounding the two successive ratios. The ratio of the two given surfaces is thus expressed as a ratio compounded of two line-ratios by means of the following theorem. 100. THEOREM 19. If two parallelograms are mutually equiangular, then their ratio is equal to the ratio compounded of the ratios of two adjacent sides of the first to the respective adjacent sides of the second. Let ABCD and BEFG be two parallelograms that have the angles ABC and EBG equal. Let these parallelograms be noted by P and R. To prove that the ratio P: R equals the ratio compounded of the two ratios AB: BG and CB: BE. D C H P ? B Place the two parallelograms so that the sides AB and BG are in one line, and so that the equal angles ABC and GBE are vertically opposite. Then the sides CB and BE are in one line (I. 52). Complete the parallelogram BGHC, and denote it by Q. In the set of three magnitudes P, Q, R, the ratio P: R is, by definition (22), compounded of the successive ratios P : Q and Q: R. Now and P: Q=AB: BG, Q: RCB: BE. [91 Therefore the ratio of the two parallelograms P and R is equal to the ratio compounded of the ratios of their sides. |