PROP. XXXVII. PROBLEM. 297. To divide a given straight line in extreme and mean ratio (§ 296). Required to divide AB in extreme and mean ratio. Construction. Draw line BEL AB, and equal to AB. With E as a centre and EB as a radius, describe O BFG. Draw line AE cutting the circumference at F and G. On AB take AC = AF; on BA produced, take AD AG. Then, AB is divided at C internally, and at D externally, in extreme and mean ratio. Proof. Since AG is a secant, and AB a tangent, = (§ 283) (1) (?) (?) AB = = 2 BE = FG. But, AG + AB = AD + AB = BD. And by (2), AB+ AC = FG+ AF AG. = (?) .. BD: AG = AG : AB. .. AB: AG = AG : BD. (?) .. AB: AD = AD: BD. Therefore, AB is divided at D externally in extreme and mean ratio. 298. Cor. If AB be denoted by m, and AC by x, proportion (3) of § 297 becomes Multiplying by 4, and adding m2 to both members, 4x2 + 4 mx + m2 = 4 m2 + m2 = 5m2. Extracting the square root of both members, Since 2x + m = ±m√5. cannot be negative, we take the positive sign before the radical sign; then, 77. To inscribe in a given circle a triangle similar to a given triangle. (§ 261.) (Circumscribe a O about the given ▲, and draw radii to the vertices.) 78. To circumscribe about a given circle a triangle similar to a given triangle. (§ 262.) BOOK IV. AREAS OF POLYGONS PROP. I. THEOREM. 299. Two rectangles having equal altitudes are to each other as their bases. Note. The words "rectangle,” “parallelogram,” “triangle,” etc., in the propositions of Book IV., mean the amount of surface in the rectangle, parallelogram, triangle, etc. Case I. When the bases are commensurable. Given rectangles ABCD and EFGH, with equal altitudes AB and EF, and commensurable bases AD and EH. Proof. Let AK be a common measure of AD and EH, and let it be contained 5 times in AD, and 3 times in EH. Drawing Is to AD and EH through the several points of division, rect. ABCD will be divided into 5 parts, and rect. EFGH into 3 parts, all of which parts are equal. (§ 114) Given rectangles ABCD and EFGH, with equal altitudes AB and EF, and incommensurable bases AD and EH. Proof. Divide AD into any number of equal parts, and apply one of these parts to EH as a unit of measure. Since AD and EH are incommensurable, a certain number of the parts will extend from E to K, leaving a remainder KH < one of the equal parts. Draw line KL 1 EH, meeting FG at L. Now let the number of subdivisions of AD be indefinitely increased. Then the unit of measure will be indefinitely diminished, and the remainder KH will approach the limit 0. will approach the limit ABCD EFGH By the Theorem of Limits, these limits are equal. (?) 300. Cor. Since either side of a rectangle may be taken as the base, it follows that Two rectangles having equal bases are to each other as their altitudes. PROP. II. THEOREM. 301. Any two rectangles are to each other as the products of their bases by their altitudes. Given M and N rectangles, with altitudes a and a', and bases b and b', respectively. Proof. Let R be a rect. with altitude a and base b'. Then, since rectangles M and R have equal altitudes, they are to each other as their bases. (§ 299) (1) And since rectangles R and N have equal bases, they are to each other as their altitudes. (?) (2) 302. The area of a surface is its ratio to another surface, called the unit of surface, adopted arbitrarily as the unit of measure (§ 180). The usual unit of surface is a square whose side is some linear unit; for example, a square inch or a square foot. 303. Two surfaces are said to be equivalent (~≈), when their areas are equal. |
