PROPOSITION XVII. THEOREM [1.-If three straight lines (A, B, and C) be proportionals, the rectangle under the extremes (A and C) is equal in area to the square on the mean (B). [2.] And if the rectangle under the extremes be equal in area to the square on the mean, the three straight lines are proportionals. DEMONSTRATION. Take D equal to B; and because A is to B, as B is to C, and that B is equal to D; A is to B, as D is to C(a). But if four straight lines be proportionals, the rectangle under the extremes is equal in area to that under the means (6). Therefore the rectangle under A, C is equal in area to that under B, D. But the rectangle under B, D is the square on B; because B is equal to D. Therefore the rectangle under A, C is equal in area to the square on B. [2.] The same construction being made, because the rectangle under A, C is equal in area to the square on B, and the square on B is equal to the rectangle under B, D, because B is equal to D; therefore the rectangle under A, C is equal in area to that under B, D: But if the rectangle under the extremes be equal in area to that under the means, the four straight lines are proportionals (b): Therefore A is to B, as D is to C; but B is equal to D: wherefore as A is to B, so is B to C. SCHOLIUM. The foregoing proposition is really only a particular case of the sixteenth proposition. PROPOSITION XVIII. PROBLEM.-On a given straight line (AB) to construct a rectilineal figure similar, and similarly situated to a given rectilineal figure. form the angle BAG equal to the angle at (a), and the angle ABG equal to the angle CDE (a); again, at the points G, B, in the straight line GB, form the angle BGH equal to the angle DFE (a); and the angle GBH equal to FDE (a); then the quadrilateral ABHG is similar and similarly situated to the quadrilateral CDEF. DEMONSTRATION. Because the angle A is equal to the angle C, and the angle ABG to CDF, therefore the remaining angle AGB is equal to the remaining angle CFD (6), wherefore the triangles AGB and CFD are equiangular; again, because the angle BGH is equal to the angle DFE, and the angle GBH to FDE, therefore the remaining angle GHB is equal to the remaining angle FED (6); wherefore the triangles BGH and DFE are equiangular. Then because the angle AGB is equal to the angle CFD, and BGH to DFE, the whole angle AGH is equal to the whole CFE: For the same reason, the angle ABH is equal to the angle CDE; also the angle at A is equal to the angle at C, and the angle GHB to FED. Therefore the rectilineal figure ABHG is equiangular to CDEF. But likewise these figures have their sides about the equal angles proportionals; because the triangles GAB, FCD, being equiangular, BA is to AG, as DC is to CF (c); and because AG is to GB, as CF is to FD; and as GB to GH, so, by reason of the equiangular triangles BGH, DFE, is FD to FE; therefore, ex æquali, AĞ is to GH, as CF is to FE (d). In the same manner it may be proved that AB is to BH, as CD is to DE; and GH is to HB, as FE is to ED (c). Wherefore, because the rectilineal figures ABHG, CDEF are equiangular, and have their sides about the equal angles proportionals, they are similar to one another (e). 2. Next let the given rectilineal figure be CDKEF. SOLUTION. Join DE, and upon the given straight line AB describe the rectilineal figure ABHG, similar, and similarly situated to the quadrilateral figure CDEF, by the former case; and at the points B, H, in the straight line BH, make the angle HBL equal to the angle EDK, and the angle BHL equal to the angle DEK (a); then the rectilineal figure ABLHG is similar, and similarly situated to the figure CDKEF. DEMONSTRATION. Because the angle HBL is equal to EDK, and the angle BHL to EDK, therefore the remaining angle L is equal to the remaining angle K (b): and because the figures ABHG, CDEF are similar, the angle GHB is equal to the angle FED (e); and the angle BHL is equal to DEK; therefore the whole angle GHL is equal to the whole angle FEK; and for the same reason the angle ABL is equal to the angle CDK: therefore the fivesided figures ABLHG, CDKEF are equiangular. And because the figures ABHG, CDEF are similar, GH is to HB, as FE is to ED (e); but as HB is to HL, so is ED to EK (c); therefore, ex aequali, GH is to HL, as FE is to EK (d); for the same reason, AB is to BL, as CD is to DK; and because the triangles BLÍ and DKE are equiangular, BL is to LH, as DK is to KE (c). Therefore, because the rectilineal figures ABLHG, CDKEF are equiangular, and have their sides about the equal angles proportionals, they are similar to one another (e). SCHOLIUM. Similar figures are said to be "similarly situated" when their homologous sides are parallel. PROPOSITION XIX. THEOREM.-If triangles (ABC, DEF) are similar, they are to one another in the duplicate ratio of their homologous sides (BC, EF). CONSTRUCTION. Take BG a third proportional to BC, EF (a), so that BC is to EF, as EF is to BG: and join GA. DEMONSTRATION. Then, because AB is to BC, as DE is to EF; alternately, AB is to DE, as BC is to EF (6); but as BC. is to EF, so is EF to BG; therefore as AB is to DE, so is EF to BG (c); and the sides of the triangles ABG, DEF AA which are about the equal angles, are reciprocally proportional. But triangles which have the sides about two equal angles reciprocally proportional are equal to one another (d); therefore the triangle ABG is equal to the triangle DEF. And because as BC is to EF, so is EF to BG; and that if three straight lines be proportionals, the first is said to have to the third the duplicate ratio of that which it has to the second (e); therefore BC has to BG the duplicate ratio of that which BC has to EF. But as BC is to BG, so is the triangle ABC to the triangle ABG (ƒ); therefore the triangle ABC has to the triangle ABG the duplicate ratio of that which BC has to EF; but the triangle ABG is equal to the triangle DEF; wherefore also the triangle ABC has to the triangle DEF the duplicate ratio of that which BC has to EF. (d) VI. 15. COROLLARY. From this it is manifest, that if three straight lines be proportionals, as the first is to the third, so is any triangle upon the first to a similar, and similarly-described triangle upon the second. PROPOSITION XX. THEOREM. If polygons (ABCDE, FGHKL) are similar, they may be divided into the same number of similar triangles, having the same ratio to one another that the polygons have; and the polygons have to one another the duplicate ratio of that which their homologous sides (AB, FG) have. CONSTRUCTION. Join BE, EC, GL, LH. DEMONSTRATION. Because the polygon ABCDE is similar to the polygon FGHKL, the angle A is equal to the angle E D F B L G C K H (a) VI. Def. 1. (b) VI 6. (c) VI. 4. (d) V. 22. (e) VI. 19. (ƒ) V. 11. (g) V. 12. F, and BA is to AE, as GF is to FL (a); and because the triangles ABE, FGL have an angle in one equal to an angle in the other, and their sides about these equal angles proportionals, the triangles are equiangular (), and therefore similar (c); wherefore the angle ABE is equal to the angle FGL. And, because the polygons are similar, the whole angle ABC is equal to the whole angle FGH (a); therefore the remaining angle EBC is equal to the remaining angle LGH. And because the triangles ABE, FGL are similar, EB is to BA, as LG is to GF (a); and also, because the polygons are similar, AB is to BC, as FG is to GH (a); therefore, ex æquali, EB is to BC, as LG is to GH (d); that is, the sides about the equal angles EBC, LGH are proportionals; therefore the triangles EBC and LGH are equiangular (6), and similar (c). For the same reason, the triangle ECD is similar to the triangle LHK: therefore the similar polygons ABCDE, FGHKL are divided into the same number of similar triangles. Also these triangles have, each to each, the same ratio which the polygons have to one another, the antecedents being ABE, EBC, ECD, and the consequents FGL, LGH, LHK and the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which the side AB has to the homologous side FG. Because the triangles ABE, FGL, are similar, ABE has to FGL the duplicate ratio of that which the side BE has to the side GL (e). For the same reason, the triangle BEC has to GLH the duplicate ratio of that which BE has to GL. Therefore, as the triangle ABE is to the triangle FGL, so is the triangle BEC to the triangle GHL (ƒ). Again, because the triangles EBC, LGH are similar, EBC has to LGH the duplicate ratio of that which the side EC has to the side LH. For the same reason, the triangle ECD has to the triangle LHK the duplicate ratio of that which EC has to LH. As therefore the triangle EBC is to the triangle LGH, so is the triangle ECD to the triangle LHK (f); but it has been proved that the triangle EBC is likewise to the triangle LGH, as the triangle ABE to the triangle FGL. Therefore, as the triangle ABE is to the triangle FGL, so is the triangle EBC to the triangle LGH, and the triangle ECD to the triangle LHK. And because, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents (g), therefore as the triangle ABE is to the triangle |