regular circumscribed polygons, similar to each other and to the inscribed; Then since the triangles in all the polygons are similar we shall have and since is greater than i, C' is greater than I, so o: 0 ::c: C', and since C is less than O therefore C' is less than 0. Therefore C' lies between I and O. Now if C' differed from C by ever so small a quantity, we might have the difference between I and O still less, and then C would no longer lie between I and O, which is impossible. So C' does not differ from C, and we have r: Rc: C. COR. Let a, A be the areas of the circles; .. a: A :: rc : RC, (II. 14) :: r2 : R3, that is, circles have to each other the ratio of the squares of their radii. SECTION III. PROBLEMS. PROBLEM I. To divide a given straight line into two parts which shall be in a given ratio. Note. By a given ratio is meant the ratio of two given lines, or of two given numbers: and since two lines can always be found which have the ratio of two given numbers, it follows that a given ratio can always be represented by the ratio of two given lines. Let AB be the given line, C and D the lines which have the given ratio; then it is required to divide AB into two parts, which have to one another the ratio of C: D. Construction. From A draw a line AE making any angle with AB, and cut off parts AH, HK equal to Cand D respectively. Join KB, and draw HP parallel to KB. P will be the point of division required. Proof. For since HP is parallel to KB, that is, AB is divided into two parts which are to one another in the given ratio. COR. I. In the same manner a line may be divided into any number of parts which have to one another given ratios. COR. 2. Hence a line may be divided into any number of equal parts, the given ratios being all ratios of equality. Note. This construction divides the line internally into parts which have the given ratio. If it is required to divide it externally, HK must be measured in the opposite direction along AE, as in the figure. The proof will be the same as before. PROBLEM 2, To find a fourth proportional to three given straight lines. Let A, B, C be the given straight lines to which it is required to find a fourth proportional. Construction. Take any angle X, and on one of its arms take XD, DE equal to A, B respectively: and on the other arm take XF equal to C. Join DF, and draw EH parallel to DF, to meet XF produced in H. Proof. For since DF is parallel to EH, XD DE :: XF : FH, but XD, DE, and XF are equal to A, B, C respectively; therefore A : B :: C: FH; that is, FH is the fourth proportional required. COR. Hence a third proportional to two given straight lines can be found, by taking C= B. PROBLEM 3. To find a mean proportional between two given straight lines. Let A, B be the given straight lines: it is required to find a mean proportional between A and B. Construction. Take HK, KL in the same straight line, equal to A and B respectively. On HK describe a semi A B M K circle, and draw KM perpendicular to HL to meet the circumference in N. KM is the line required. Proof. Join HM, ML. Then since HML is a semicircle, HML is a right angle; therefore MK, the perpendicular from the right angle on the hypothenuse, is a mean proportional between the segments of the base; that is, MK is a mean proportional between HK and KL, or between A and B. PROBLEM 4. To divide a straight line in extreme and mean ratio. A line AB is said to be divided in extreme and mean ratio in C, when AB AC :: AC: CB. Let AB be the given straight line, which it is required to divide in extreme and mean ratio. |