EG perpendicular to AE, and take EF and FG each equal to AB; join AG, AF, DG, and DF. Set off on the line EN, from E, the distances EH and HK, each equal to AG; then set off, in the opposite direction, the distance KL equal to AF, and from L set off LM equal to DG; also set off MN equal to DF Then bisect EN at the point P; bisect EP at the point R, and, finally, trisect ER at the point T; then will CT be the circumference of the circle, nearly. For, by construction, we have, if we call the diameter a unit, CE=21; EL=2EH — KL=2√13 — √10; LM = √5; MN = √2. Therefore, and and, therefore, EN=2√13-√10+ √5+ √2; ET (2√13−√10+ √5+ √2); 1 12 CT = 21 + 1/2 (2 √13 − √10 + √5+ √2) = 3.1415922, etc., which is the ratio true to six decimals. For simplicity and accuracy, a better graphic method of finding this ratio can hardly be expected, or even desired. PROBLEMS, WHICH REFER TO THE THIRD AND FOURTH BOOKS. CONSTRUCTION OF PROPORTIONAL LINES. PROBLEM I. To divide a given line into any number of equal parts. To make a definite case, suppose we wish to divide AB into five equal parts. Through A draw any indefinite line AX, making an angle with AB. Lay off on this line any convenient length five times, as Ap, A X b Р Q R S B pq, qr, rs, sb. Join the last extremity b with B, and through the other points of division draw parallel to Bb, lines cutting AB in the points P, Q, R, and S (B. II., P. VI.). The straight line will thus be divided into five equal parts (B. III., T. I.). PROBLEM II. To divide a given line into parts proportional to given lines. As a definite case, suppose we wish to divide AB into three parts, which shall be to each other as the three lines P, Q, R. As in the last Problem, draw any indefinite line AX, making an angle with P. R Ф A b/x P/ Q' B AB. Make Ap, pq, qb respectively equal to P, Q, and R. Draw Bb, and parallel to it draw pP', qQ', and the line AB will be divided at P' and Q' as required (B. III., T. III.) Scholium. As a particular case, suppose we wish to divide a straight line AB into two parts proportional to the lines M and N. The straight line AB will thus be divided at D in the ratio required. For, the two triangles DAC, DBC', are evidently similar, and give AD: DB:: AC: BC':: M: N. If, instead of taking BC', equal to N, in the direction of BY, opposite to AX, we had taken BC", equal to N, in the same direction with AX, then the line CC" would have met AB produced at D', which is called the conjugate point of the point D (B. III., T. XII., S.). PROBLEM III. To find a fourth proportional to three given lines M, N, P. Form any angle, as XAY, and take on AX, AB=M, BC=N, and on AY, AD=P; then draw BD, and through the point C draw CE parallel to BD. The line DE will be the fourth proportional required. For, we have (B. III., T. III.), A M N ZA P Y E D AB: BC::AD: DE, or M:N::P:DE. X B C Cor. I. If P were equal to N, the above proportion would become M:N::N:DE. That is, DE would in this case be a third proportional to the two lines M and N. Cor. II. We deduce immediately the following problem: A point O being given within an angle YAX, to draw through O a straight line DOE, such that the segments DO, OE may be to each other in the ratio of M to N. Through the point O draw OB parallel to AY; find a fourth proportional to the three lines M, N, AB; make BE equal to this fourth proportional, and draw EOD, which will be the line required. For, since OB is parallel to AD, we have OD: OE::AB:BE::M: N. When M = N, it is sufficient to take on AX, BE=AB, and then to draw EOD. PROBLEM IV. To find a mean proportional between two given lines M and N. On the indefinite line AX, take AB=M, BC= N, and on AC, as a diameter, describe the semi-circumference ADC. Through B draw BD perpendicular to AC, and BD will be the mean proportional sought. For, we have (T. I., S. I.), M N D X B C AB:BD::BD: BC, or M: BD::BD:N. PROBLEM V. To divide a given line into mean and extreme ratio. We have already noticed this kind of division (T. III., S. I.). And we have also actually solved this problem algebraically (T. XI., S. I.). We will now proceed with its geometrical solution. Let AB be the line which we are required to divide. A D B Draw BO perpendicular to AB and equal to one half of AB; then draw AO, with O as a centre; with OB for a radius, describe the circumference meeting AO in C. Finally, make AD equal to AC, and the line AB will be divided at the point D as required. In effect, from the construction, AB is a tangent to the circle OB; and if AO is produced until it meets the circumference in C', we shall have (T. III.), Consequently, AC:AB::AB: AC. AC-AB: AB:: AB-AC: AC. Now, AC' AC+CC' = AC+AB; consequently, AC' AB=AC=AD, and AB-AC=AB-AD=BD· hence the proportion evidently becomes AD:AB::BD : AD, or interchanging means and extremes, AB: AD:: AD: BD. We will suppose the problem solved, and that MN, mn are the two tangents, the one exterior and the other interior, meeting the line of the centres OO' in two points C, C'. |