Through the point A draw the diagonal AC cutting off the triangle ABC. Draw afterwards, through the point B, the line BI parallel to AC, meeting DC, produced, in I, then draw AI. The polygon AIDE, etc., is E B I C L equivalent to the given polygon ABCDE, etc., and it has one side less. For, since BI is parallel to AC, the two triangles AIC, ABC are equivalent (B. III., T. XIX., C.), and if to these two triangles we add the portion of the surface ACDE, etc., we shall have, in the first case, the polygon AIDE, etc., and in the second case, the polygon ABCDE, etc.; consequently these two polygons are equivalent. It is evident, moreover, since the side CI of the triangle AIC is the prolongation of the side DC, that the two sides AB, BC of the first polygon have been replaced by the single side AI; hence the second figure has one side less than the first. Actually operate upon the polygon AIDE, etc., as upon the first; that is to say, draw the diagonal AD cutting off the triangle AID; afterwards draw IL parallel to AD meeting ED produced in L; we shall thus obtain a new polygon ALE, etc., equivalent to the second and having one side less. In continuing, in this way, we shall necessarily reach a polygon of three sides, and the problem will be resolved. PROBLEM VI. To transform a polygon into a square. If the given polygon is a triangle, let b represent its base, and h its height, and a the side of the square sought. We have this condition, x2=b×th, from which we obtain the following proportion: b:x::x:h. Thus, the side of the square sought is a mean proportional between the base and half the height of the triangle. This line being constructed (P. IV.), we easily obtain the construction of the square. Whatever may be the polygon, we commence by transforming it into a triangle (P. VII.); afterwards we transform, as above, the triangle into a square. When the given polygon is a parallelogram, or a rectangle, or in short, any figure whose area, depends immediately upon the product of two lines, all that is necessary in order to obtain the side of an equivalent square is to construct a mean proportional between these two lines. Thus, in the case of a regular polygon, it is sufficient, after having developed on an indefinite straight line the perimeter of the polygon, to seek a mean proportional between the half perimeter and the radius of its inscribed circle. To transform a circle into a square, it is necessary first to rectify the circumference, and afterwards to determine a mean proportional between the radius and the rectified semi-circumference. Scholium. The quadrature of the circle is wholly dependent upon the rectification of a circumference; and thus, up to the present, we have not been able, by the assistance of the Ruler and the Compass, to construct rigorously a square equivalent to a circle, as we can do for rectilinear figures, since all known methods give only approximate values for the ratio of the circumference to the diameter. CONSTRUCTION OF SIMILAR POLYGONS UNDER CERTAIN CONDITIONS. PROBLEM IX. Upon a given line, to construct a polygon similar to a given polygon. Let ab be the given line, and ABCDEF the given polygon. After decomposing the given polygon into triangles by drawing diagonals from the corner A to the other corners, form at the points a and b two angles cab, abc, respectively equal to the angles CAB, ABC. We shall thus obtain a triangle abc similar B с d to the triangle ABC. Construct, in the same manner, on ac a triangle acd similar to the triangle ACD; and so on for the other triangles. The polygon abcdef thus obtained will be similar to the given polygon ABCDEF. PROBLEM X. Two similar polygons being given, it is required to construct a third polygon similar to the two first, and equivalent to their sum or to their difference. The solution of this problem is an immediate consequence of C. IV., T. XXIX., B. III. Let a, a' be two homologous sides of the given polygons A, A'. Upon these sides, regarded as the sides about the right angle, or as the hypotenuse and one of the sides of the right angle, construct a right-angled triangle; afterwards, on the third side a" of the triangle thus obtained, construct (P. IX.) a polygon A" similar to one of the given polygons. The polygon thus constructed will be the polygon required. For, by construction, a"2 = a2 + a'2, or a'12 = a2 — a'2; hence, (B. III., T. XXIX., C. IV.), To find a square which shall be to a given square in the ratio of two given lines. the chords AB, AC, and prolong them if necessary beyond B and C. Upon AB take AB'=a, the side of the given square, which we here suppose to be greater than AB, and draw B'C' parallel to BC. The line AC' will be the side of the square sought. In effect, the two similar triangles ABC, AB'C', give the proportion Consequently, ABAC : AB' : AC'. AB2: AC2:: AB'2: AC2; but we have (B. III., T. XIV., S. I.), hence, AB2: AC:: BD: DC, or::m:n; AB'2: AC2::m:n; or, putting a for AB', and changing the order of the terms of the proportion, we have m:n::a2: AC'2. PROBLEM XII. A polygon P being given, to construct a second polygon X similar to the first, and such that the first may be to the second as m to n. Let a denote the side of the given polygon, and x the homologous side of the sought polygon. We shall have The question is thus made to depend upon the preceding problem; being known, it is sufficient then to construct on this line a polygon similar to the given polygon. PROBLEM XIII. Having given two polygons, P and Q, to construct a third. polygon X similar to the first, and such that the second may be to the third in the ratio of m to n. Let a denote a side of the polygon P, and x the homologous side of the polygon X. Then we shall have, by the conditions of the problem, P:X:: a2x2, This being supposed, conceive the polygons P and Q transformed into equivalent squares (P. VIII.), having respectively for their sides Ρ and q. In the same manner let y represent the side of a square equivalent to X; the preceding proportions will be changed into the following: Now, the last of these proportions will make known y by the construction of Problem XI., and the second will then give a as a fourth proportional to the lines p, y, and a, which will be the side of the polygon sought, homologous to a. PROBLEM XIV. To construct a rectangle equivalent to a given square m2, and such that the sum or the difference of two adjacent sides may be equal to a given line a. First case. The given line a being the sum of the adjacent sides. Upon a straight line AB= a, as a diameter, describe a circumference; through A the extremity of the diameter AB, draw the perpendicular AC=m. Through C draw CL parallel to AB, meeting the circumference in the points E, E'. From the points E and E' draw EF, E'F' perpendicular to AB. I E E' L с K B F F' G |