THEOREM XXVIII. The perimeters of two similar polygons are proportional to their homologous sides. And their areas are proportional to the squares of these sides. First. By reason of the similarity of these polygons, we have (T. IX.), AB: A'B':: BC: B'C':: CD: C'D': :, etc. Now the sum of these antecedents, AB+BC+CD+, etc., which makes the perimeter of the first polygon, is to the sum of their consequents. A'B' + B'C' + C'D' +, etc., which makes the perimeter E D E' D' F C F/ B A' B' C' of the second polygon, as any one antecedent is to its corresponding consequent, and therefore as AB is to A'B'. Secondly. Since these polygons are similar, they are each composed of the same number of similar triangles, which, compared together, are each to each, in the same ratio, that of the squares of the homologous sides. It follows, then, that the sum of the triangles which compose the first polygon, is to the sum of the triangles which compose the second polygon, as the squares of the homologous sides of the two polygons. That is, we shall have ABCDEF: A'B'C'D'E'F':: AB2: A'B'. Cor. The perimeters of similar figures are proportional to their homologous lines, and their areas are proportional to the squares of those lines. For, by the definition of similar figures, the ratio of any two homologous lines, which is called the ratio of similitude, is constantly the same. COMPARISON OF SQUARES CONSTRUCTED ON CERTAIN LINES. THEOREM XXIX. The square constructed on the hypotenuse of a right-angled triangle, is equivalent to the sum of the squares constructed respectively on the other two sides. This Theorem is not a fundamental one, like Theorem XIV., but its importance and its fecundity have given rise to several demonstrations founded solely upon the equality and equivalence of figures. We shall confine ourselves to the one generally given in works of geometry. K E of a right angle and of the common angle ABC; the lines AB and BF are equal, being sides of the BC and BD equal, for a like reason. BAD, BFC are equal (B. I., T. XX.). same square, so also are Hence the two triangles But the triangle ABD have the same base BD is half the rectangle BLMD, since they and the same altitude BL (T. XIX.). For the same reason the triangle BFC is half the square ABFG, for they have the same base BF and the same altitude AB; consequently, BLMD=ABFG. We can prove in the same manner, by drawing AE and BI, that LMEC ACIK; = and as the square BDEC is equivalent to BLMD+LMEC, there will result M N square BDEC = square ABFG+ square ACIK. Cor. I. The square MNPQ, constructed on the diagonal BD or AC of a square ABCD, is double the square itself. B -K D P This is obvious from the simple inspection of the figure. In effect, the four squares AKBM, AKDN, BKCQ, DKCP, are equal, and respectively the doubles of the triangles AKB, AKD, BKC, DKC, of which the sum is equal to the square ABCD. Cor. II. We have already shown that the squares ABFG, ACIK (see first figure) are respectively equivalent to the rectangles BLMD, LMEC. And since these rectangles and the square BCED have the common altitude BD, they are to each other as their bases BL, LC, and BC (T. XX.). Hence we obtain this relation: AB2: AC2: BC2:: BL: LC: BC. Cor. III. Since the two squares ABFG, ACIK are respectively equivalent to the rectangles BLMD, LMEC, we have AB2 = BC × BL, AC2 = BC × LC; from which we deduce these two proportions: BC:AB::AB: BL, BC: AC:: AC: LC. These two proportions, together with the relations of the preceding Corollary, correspond in all respects with the third portion of Theorem XIII., and Scholium I. of Theorem XIV. We have thus, by another method, established properties already demonstrated. Cor. IV. In short, if on the three sides of a right-angled triangle ABC, we suppose three similar polygons to be constructed, as these polygons will be proportional to the squares of their homologous sides (T. XXVIII.), and we have this relation be tween these sides: it follows, that, of the three similar polygons, the polygon constructed on the hypotenuse is equivalent to the sum of the polygons constructed on the sides containing the right angle. Scholium. The algebraic formulas, (p + q)2 = p2 + q2+2pq, ( p − q)2 = p2+q2 — 2pq, which have been made the bases of Theorems XV. and XVI., and the other well-known formula, being translated into geometrical language, will give rise to some new Theorems in regard to areas, with which we will close this Book. THEOREM XXX. The square constructed on the sum or on the difference of two lines, is equivalent to the sum of the squares constructed respectively on these lines, plus or minus twice their rectangle. The simple inspection of the figure is nearly sufficient to satisfy one of the truth of this double proposition. First. Let AE be the greater of the two lines, EB the less. Construct the squares AEIG, ABCD, and produce EI until it meets CD at F. L D F с M K G I A E B The square constructed on AB, the sum of the two lines, is evidently composed of the squares AEIG, IKCF, constructed on the line AE and on the line IK = EB, increased by the two rectangles BEIK, GIFD, which have for their respective bases GI= AE, EI = AE, and for altitudes GD = EB, IK = EB. Hence, we have square AB = square AE + square EB + twice rectangle AE×EB. AE+ Secondly. Let AB be the greater line, BE the less, that which gives AE for the difference of these two lines. Construct the same figure as in the first case, and, in addition, construct the square GDLM equal to IKCF. The square AEIG is equal to the square ABCD plus the square GDLM, minus the two rectangles EBCF, MIFL. Now, these two rectangles have respectively for bases FE=AB, IM = GK = AB, and for altitudes BE=IK = IF. Hence we have square AE= square AB+ square EB- twice rectangle ABxBE. THEOREM XXXI. The rectangle constructed on the sum and the difference of two lines, is equivalent to the difference of the squares constructed on these two lines. Let AB be the greater line, BE = BE' D the less, so that AE will represent the sum of these two lines, and AE' their difference. Construct on AE as a base, and AG= AE' as an altitude, the rectangle AENG, also the square ABCD, and draw E'IF perpendicular to AB, forming the square AE'IG. = F C I K E' B E The two rectangles BENK, GIFD are equal, having equal bases and equal altitudes, namely, BK = AG=GI=AE', EB =EB IK = IF=DG; hence it follows that the rectangle AENG is equivalent to the figure DFIKBA. But this figure is the difference of the squares constructed on AB and on IK=BE' = BE. |