233. If from a point without a circle a tangent and a secant be drawn, the tangent is a mean proportional between the whole secant and the external segment. Let PC and PB be a tangent and a secant drawn from the point P to the circle CAB. To prove that PB: PC PC : PA. Join CA and CB. = In the two triangles PCA and PCB the angle P is common, and PCA = PBC, being measured by one-half of the same arc CA; then (by 80), Z PAC = ≤ PCB. Therefore, the triangles are equiangular and are (by 223) similar, which gives P B C' PB : PC = PC : PA; or (by 199), PC2 = PB × PA. q.e.d. 234. COR. PC2 BPX PA; therefore (by 28), = PROPOSITION XX. PROBLEM. 235. To divide a given straight line into parts proportional to any number of given lines. Let AB, m, n, and o be given straight lines. It is required to divide AB into parts proportional to the given lines m, n, and o. It is known (from 213) that lines drawn parallel to the base of a triangle divide the other two sides proportionally. Therefore, form with AB a triangle by drawing an indefinite straight line from A; measure off on this line a part equal to m, say AC; then n, say CE, and o, say EF, and join BF, thus forming the triangle AFB. Through the points E and C draw lines parallel to FB, meeting AB in K and H; then AH: HK: KB = AC: CE: EF=m: n: 0. 236. COR. 1. By making AC = CE = EF, the line AB will be divided equally. 237. COR. 2. By making AC = m, AH=n, and CE=0, we would have (by 213), m:n::o:HK; that is, HK would be a fourth proportional to m, n, and o. EXERCISE. To divide a given straight line into three segments, A, B, and C, such that A and B shall be in the ratio of two given straight lines m and n, and B and C shall be in the ratio of two other straight lines o and p. PROPOSITION XXI. PROBLEM. 238. To find a mean proportional between two given straight lines. Let m and n be the two lines. To find a mean proportional to them. It is known (from 228) that the perpendicular from the circumference to the diameter is a mean proportional between the segments of the diameter. m n C D B Therefore, if a diameter be made of m A and n and a perpendicular erected at their point of union, the portion included between the diameter and the circumference will be the mean proportional required; that is, lay off AD = m and DB = n. Describe upon AB as a diameter a circle, erect (by 182) a perpendicular at D, and CD will be a mean proportional, or AD: CD CD: DB, or m: CD = CD: n. 239. The mean proportional between two lines is often called the geometric mean, while half their sum is called the arithmetic mean. PROPOSITION XXII. PROBLEM. 240. On a given straight line, to construct a polygon similar to a given polygon. similar when they are composed of the same number of similar triangles similarly placed; therefore, divide A-F into triangles by drawing the diagonals AC, AD; and AE. It is known (from 218) that triangles are similar when they are equiangular. Therefore, construct (by 187) on A'B' a triangle equiangular with ABC, say A'B'C'; then on A'C' a triangle equiangular with ACD, say A'C'D'. Likewise on A'D', A'D'E', and on A'E', A'E'F"; then will A'-F" be the polygon required. BOOK IV. AREAS OF POLYGONS. DEFINITIONS. 241. The Area of a surface is the numerical value of the ratio of this surface to another surface, called the Unit of Surface, or Superficial Unit. 242. The unit of surface is the square whose side is some Unit of Length, as an inch, a foot, a metre, etc., and the area is expressed as so many square inches, square feet, square metres, etc. 243. Two surfaces are equivalent when their areas are equal. 244. The projection of a point upon a straight line is the foot of the perpendicular let fall from the point upon the line. Thus A' is the projection of A. The projection of a limited straight line upon another straight line, is the portion of the latter included between the projection of the terminal points of the former. Thus A'B' is the projection of AB on XX'. |