D PROBLEM XIX. To find a mean proportional between two given lines. B Place AB, BC in a straight line; upon AC describe the semicircle ADC, and from the point B draw BD perpendicular to AC. Then will BD be the mean proportional required. For the perpendicular BD, let fall from a point in the circumference upon the diameter, is a mean proportional between the two segments of the diameter AB, BC (B. IV., Pr. 23, Cor.), and these segments are equal to the two given lines. PROBLEM xx. To divide a given line into two parts such that the greater part may be a mean proportional between the whole line and the other part. F B E Let AB be the given straight line; it is required to divide it into two parts at the point F, such that AB: AF:: AF: FB. At the extremity of the line AB erect the perpendicular BC, and make it equal to the half of AB. From C as a centre, with a radius equal to CB, describe a circle. Draw AC cutting the circumference in D, and make AF equal to AD. The line AB will be divided in the point F in the manner required. For, since AB is a perpendicular to the radius CB at its extremity, it is a tangent (B. III., Pr. 9); and, if we produce AC to E, we shall have AE: AB:: AB: AD (B. IV., Pr. 29). Therefore, by division (B. II.,Pr. 7), AE—AB:AB:: AB-AD: AD. But, by construction, AB is equal to DE, and therefore AE—AB is equal to AD or AF, and AB-AD is equal to FB. Hence AF: AB:: FB: AD or AF; and, consequently, by inversion (B. II., Pr. 5), AB: AF:: AF: FB. Schol. 1. The line AB is said to be divided in extreme and mean ratio. An example of its use may be seen in Book VI., Pr. 5. Schol. 2. Let AB=a; AF-AD=AC-CD. But a CD=2 15a2 α AC=√AB2+BC2=, a2 + = √5. Through a given point in a given angle, to draw a straight line so that the parts included between the point and the sides of the angle may be equal. Let A be the given point, and BCD the given C angle; it is required to draw through A a line BD, so that BA may be equal to AD. Through the point A draw AE parallel to BC, and take DE equal to CE. Through the points D and A draw the line BAD; it will be the B line required. A For, because AE is parallel to BC, we have (B. IV., Pr. 16) But DE is equal to EC; therefore DA is equal to AB. PROBLEM XXII. To construct a square that shall be equivalent to a given parallelogram or to a given triangle. First. Let ABCD be the given parallelogram, AB its base, and DE its altitude. Find a mean proportional between AB and DE (Prob. 19), and represent it by X; the square described on X will A be equivalent to the given parallelogram ABCD. D E For, by construction, AB: X:: X: DE; hence X2 is equal to AB×DE (B. II., Pr. 1, Cor.). But AB×DE is the measure of the parallelogram, and X2 is the measure of the square. Therefore the square described on X is equivalent to the given parallelogram ABCD. Secondly. Let ABC be the given triangle, BC its base, and AD its altitude. Find a mean proportional between BC and the half of AD, and represent it by Y. Then will the square described on Y be equivalent to the triangle ABC. D For, by construction, BC: Y:: Y: AD; hence Y2 is equivalent to BCX AD. But BC× AD is the measure of the triangle ABC; therefore the square described on Y is equivalent to the triangle ABC. PROBLEM XXIII. Upon a given straight line, to construct a rectangle equivalent to a given rectangle. G A HE Let AB be the given straight line, and CDFE the given rectangle. It is required to construct on the line AB a rectangle equivalent D to CDFE. Find a fourth proportional (Prob. 18) to the three lines AB, CD, CE, and let AG be that fourth proportional. The rectangle constructed on the lines AB, AG will be equivalent to CDFE. For, because AB: CD::CE: AG (B. II., Pr. 1), ABX AG=CD xCE. Therefore the rectangle ABHG is equivalent to the rectangle CDFE, and it is constructed upon the given line AB. PROBLEM XXIV. To construct a triangle which shall be equivalent to a given polygon. E Let ABCDE be the given polygon; it is required to construct a triangle equivalent to it. Draw the diagonal BD, cutting off the triangle BCD. Through the point C draw CF parallel to DB, meeting AB produced in F. B F Join DF, and the polygon AFDE will be equivalent to the polygon ABCDE. For the triangles BFD, BCD, being upon the same base BD, and between the same parallels BD, FC, are equivalent. To each of these equals add the polygon ABDE; then will the polygon AFDE be equivalent to the polygon ABCDE; that is, we have found a polygon equivalent to the given polygon, and having the number of its sides diminished by one. In the same manner, a polygon may be found equivalent to AFDE, and having the number of its sides diminished by one; and, by continuing the process, the number of sides may be at last reduced to three, and a triangle be thus obtained equivalent to the given polygon. Scholium. By Prob. 22, any triangle may be changed into an equivalent square, and hence a square can always be found equivalent to any given polygon. This operation is called squar ing the polygon, or finding its quadrature. The problem of the quadrature of the circle consists in finding a square equivalent to a circle whose diameter is given. PROBLEM XXV. To construct a square equivalent to the sum or difference of two given squares. C First. To make a square equivalent to the sum of two given squares, draw two indefinite lines AB, BC at right angles to each other. Take AB equal to the side of one of the given squares, and BC equal to the side of the other. Join AC; it will be the side of the required square. A B For the triangle ABC, being right-angled at B, the square on AC will be equivalent to the sum of the squares upon AB and BC (B. IV., Pr. 11). Secondly. To make a square equivalent to the difference of two given squares, draw the lines AB, BC at right angles to each other, and take AB equal to the side of the less square. Then, from A as a centre, with a radius equal to the other side of the square, describe an arc intersecting BC in C; BC will be the side of the square required, because the square of BC is equivalent to the difference of the squares of AC and AB (B. IV., Pr. 11, Cor. 1). Scholium. In the same manner, a square may be made equivalent to the sum of three or more given squares; for the same construction which reduces two of them to one will reduce three of them to two, and these two to one. PROBLEM XXVI. Upon a given straight line, to construct a polygon similar to a given polygon. Let ABCDE be the given polygon, and FG be the given straight line; it is required, upon the line FG, to construct a polygon similar to ABCDE. A D . B I K F G H Draw the diagonals BD, BE. At the point F, in the straight line FG, make the angle GFK equal to the angle BAE, and at the point G make the angle FGK equal to the angle ABE. The lines FK, GK will inter sect in K, and FGK will be a triangle similar to ABE. In the same manner, on GK construct the triangle GKI similar to BED, and on GI construct the triangle GIH similar to BDC. The polygon FGHIK will be the polygon required. For these two polygons are composed of the same number of triangles, which are similar to each other, and similarly situated; therefore the polygons are similar (B. IV., Pr. 26, Cor.). PROBLEM XXVII. Given the area of a rectangle and the sum of two adjacent sides, to construct the rectangle. Let AB be a straight line equal to the sum of the sides of the required rectangle. Upon AB as a diameter, describe a semicircle. At the point A erect the perpendicular AC, and make it equal to the side of a square having the given area. Through C E B draw the line CD parallel to AB, and let it meet the circumference in D, and from D draw DE perpendicular to AB. Then will AE and EB be the sides of the rectangle required. A For (B. IV., Pr. 23, Cor.) the rectangle AE× EB is equivalent to the square of DE or CA, which is, by construction, equivalent to the given area. Also, the sum of the sides AE and EB is equal to the given line AB. Scholium. The side of the square having the given area must not be greater than the half of AB, for in that case the line CD would not meet the circumference ADB. PROBLEM XXVIII. Given the area of a rectangle and the difference of two adjacent sides, to construct the rectangle.. F angle required. B Let AB be a straight line equal to the dif ference of the sides of the required rectangle. Upon AB as a diameter describe a circle, and at the extremity of the diameter draw the tangent AC equal to the side of a square having the given area. Through the point C and the centre F draw the secant CE; then will CD, CE be the adjacent sides of the rect For (B. IV., Pr. 29) the rectangle CD × CE is equivalent to the |