PROPOSITION XX. THEOREM. (Eucl. vI. 21.) Rectilinear figures, which are similar to the same rectilinear figure, are also similar to each other. Let each of the rectilinear figures A and B be similar to the rectilinear figure C. Then must the figure A be similar to the figure B. For. A is similar to C, .. A is equiangular to C, and A and C have their sides about the equals pro portionals. Again, B is similar to C, VI. Def. 1. .. B is equiangular to C, and B and C have their sides about the equals pro portionals. VI. Def. 1. Hence A and B are each equiangular to C, and have the sides about the equals of each of them and of C proportionals. .. A is equiangular to B, Ax. 1. and A and B have their sides about the equals pro portionals. V. 5. .. the figure A is similar to the figure B. VI. Def. 1. Q. E. D. PROPOSITION XXI. THEOREM. (Eucl. VI. 20.) Similar polygons may be divided into the same number of similar triangles, having the same ratio to one another, which the polygons have; and the polygons are to one another in the duplicate ratio of their homologous sides. Let ABCDE, FGHKL be similar polygons, and let AB be the side homologous to FG. I. The polygons may be divided into the same number of similar As. II. These As have each to each the same ratio which the polygons have. III. The polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which the side AB has to the side FG. Join BE, EC, GL, LH: then I... the polygon ABCDE is similar to the polygon FGHKL, For the same reason the ▲ ECD is similar to ▲ LHK. Thus the polygons are divided into the same number of similar As. II. A ABE is similar to ▲ FGL, .. ▲ ABE has to ▲ FGL the duplicate ratio of BE to GL. Δ So also, ▲ EBC has to ▲ LGH the duplicate ratio of BE to GL. .. A ABE is to ▲ FGL as A EBC is to ▲ LGH. Δ Again, ▲ EBC is similar to ▲ LGH, VI. 19. VI. 19. V. 5. .. A EBC has to ▲ LGH the duplicate ratio of VI. 19. VI. 19. V. 5. EC to LH. .. A EBC is to ▲ LGH as A ECD is to ▲ LHK. Now as one of the antecedents is to one of the consequents so are all the antecedents together to all the consequents together, V. 10. and ... ▲ ABE is to ▲ FGL as polygon ABCDE is to polygon FGHKL. III. Since ▲ ABE has to ▲ FGL the duplicate ratio of AB to FG, VI. 19. .. polygon ABCDE has to polygon FGHKL the duplicate ratio of AB to FG. V. 5. Q. E. D. COR. I. In like manner it may be proved, that similar figures of four or any number of sides, are to one another in the duplicate ratio of their homologous sides: and it has been already proved for triangles, VI. 19. Therefore, universally, similar rectilinear figures are to one another in the duplicate ratio of their homologous sides. COR. II. If MN be a third proportional to AB and FG, AB has to MN the duplicate ratio of AB to FG, VI. Def. 2. and .. AB is to MN as the figure on AB to the similar and similarly described figure on FG; that being true in the case of quadrilaterals and polygons, which has been already proved for triangles. VI. 19 Cor. PROPOSITION XXII. THEOREM. (Eucl. VI. 31.) In right-angled triangles, the rectilinear figure, described upon the side opposite to the right angle, is equal to the similar and similarly described figures upon the sides containing the right angle. D Let ABC be a right-angled ▲, having the right ▲ BAC. Then must the rectilinear figure, described on BC, be equal to the similar and similarly described figures on BA, AC. Draw AD1 to BC. Then ▲ ABC is similar to ▲ DBA, and.. BC is to BA as BA is to BD, VI. 12. and .. as BC is to BD so is the figure described on BC to the similar and similarly described figure on BA, VI. 21, Cor. 2. and.. as BD is to BC so is figure on BA to figure on BC. V. 12. For the same reason as DC is to BC so is figure on AC to figure on BC. Hence as BD, DC together are to BC so are figures on BA, AC together to figure on BC. But BD, DC together are equal to BC, and .. figures on BA, AC together = V. 22. figure on BC. V. 18. Q. E. D. NOTE.-The Proposition which follows is not given by Euclid, but is necessary to the proof of Prop. XXIV. If two rectilinear figures be equal and also similar, their homologous sides must be equal, each to each. E G D K Let the rectil. figs. ABCDE, FGHKL be equal and similar, and let DC and KH be homologous sides of the figures. Then must DC=KH. For, if not, let DC be greater than KH. Then DC is to DE as KH is to KL, .. DE is greater than KL. V. 14. Hence if ▲ KLH be applied to ▲ DEC, so that KH falls on DC and KL on DE (for HKL = 4 CDE), HL will fall entirely within ▲ DEC, ..A KLH is less than ▲ DEC. DEC is to ▲ KLH as figure ABCDE is to But figure FGHKL, and figure ABCDE=figure FGHKL ..▲ DEC = ▲ KLH, = or the greater the less, which is impossible. .. DC is not greater than KH. VI. 21. V. 18. Similarly it may, be shown that DC is not less than KH. .. DC=KH. Q. E. D. |