This is the first Proposition in Euclid in which the technical term “duplicate ratio” occurs. My experience with pupils is, that they find it very difficult to understand either Euclid’s proof or his definition. On this account I submit the following alternative proof, which, however, makes use of a new definition of the duplicate ratio of two lines, viz. the ratio of the squares (see Annotations on V. Def. x.) described on these lines. On AB and DE describe squares, and through C and F draw lines parallel to AB and DE, and complete the rectangles AI, DN. Now, the triangles JAC, ODF are evidently equiangular. Hence JA : AC :: OD : DF [iv.]; but AC : AB :: DF : DE [iv.]. Hence JA : AB :: OD : DE (ex æquali); but AB = AG, and DE = DL; therefore JA : AG :: OD : DL. Again, JA : AG :: AI : square AH [i.], and OD : DL :: DN : square DM [i.]. Hence AI : AH :: DN : DM; therefore AI : DN :: AH : DM [V. xvi.]; hence △ABC : △DEF :: AB2 : DE2. Exercises. 1. If one of two similar triangles has its sides 50 per cent. longer than the homologous sides of the other; what is the ratio of their areas? 2. When the inscribed and circumscribed regular polygons of any common number of sides to a circle have more than four sides, the difference of their areas is less than the square of the side of the inscribed polygon. PROP. XX.—Theorem. Similar polygons may be divided (1) into the same number of similar triangles; (2) the corresponding triangles have the same ratio to one another which the polygons have; (3) the polygons are to each other in the duplicate ratio of their homologous sides. Dem.—Let ABHIJ, CDEFG be the polygons, and let the sides AB, CD be homologous. Join AH, AI, CE, CF. 1. The triangles into which the polygons are divided are similar. For, since the polygons are similar, they are equiangular, and have the sides about their equal angles proportional [Def. i.]; hence the angle B is equal to D, and AB : BH :: CD : DE; therefore [vi.] the triangle ABH is equiangular to CDE; hence the angle BHA is equal to DEC; but BHI is equal to DEF (hyp.); therefore the angle AHI is equal to CEF. Again, because the polygons are similar, IH : HB :: FE : ED; and since the triangles ABH, CDE are similar, HB : HA :: ED : EC; hence (ex aequali) IH : HA :: FE : EC, and the angle IHA has been proved to be equal to the angle FEC; therefore the triangles IHA, FEC are equiangular. In the same manner it can be proved that the remaining triangles are equiangular. 2. Since the triangle ABH is similar to CDE, we have [xix.]. In like manner, AHI : CEF in the duplicate ratio of AH : CE; hence ABH : CDE = AHI : CEF [V. xi.]. Similarly, AHI : CEF = AIJ : CFG. In these equal ratios, the triangles ABH, AHI, AIJ are the antecedents, and the