36 Th. Parallelograms upon equal bases, and between the same parallels, are equal to one another. Let ABCD, EFGH be two parallelograms upon equal bases BC, FG, and between the A same parallels AH, BG; they are equal. Join BE, CH; and because the bases BC, FG are equal, and FG is equal to EH (a); therefore BC and EH are equal, and they are parallel; therefore also BE and CH, в which join their extremities, are equal and parallel (b). EBCH is therefore a parallelogram, and equal to ABCD; because they are upon the same base BC, and between the same parallels BC, AH (c). For the like reason, EFGH is equal to EBCH; hence ABCD equals EFGH (d). Wherefore, parallelograms, &c. Recite (a) p. 34; (b) p. 33; (c) p. 35; (d) ax. 1. Q. E. D. 37 Th. Triangles (ABC, DBC), upon the same base (BC), and between the same parallels (BC, AD) are equal to one another. Produce AD both ways to E, F; through B draw BE parallel to AC (a); and F through C draw CF parallel to BD. AD F A B C Each of the figures EBCA, DBCF. is a parallelogram; and they are equal (b), because they are upon the same base and between the same parallels, BC, EF: and the triangle ABC is half the parallelogram EBCA, because the diameter AB bisects it (c); and the triangle DBC is half the parallelogram DBCF, because the diameter DC bisects it; and the halves of equals are equal (d). Therefore, the triangle ABC is equal to the triangle DBC. Wherefore, triangles, &c. Recite (a) p. 31; (b) p. 35; (c) p. 34; (d) ax. 7. Q. E. D. 38 Th. Triangles (ABC, DEF), upon equal bases (BC, EF), and between the same parallels (BF, AD), are equal to one another. Produce AD both ways to G, H; and through B, draw BG parallel to AC (a); and C through F, draw FH parallel to ED. Then each of the figures BCAG, FEDH is a parallelogram; and they are equals (6), being upon equal bases and between the same parallels BF, GH. B AD H N CE But the triangle DEF is half the parallelogram DEFH; and the triangle ABC is half the parallelogram GBCA (c); because they are bisected by the diameters DF and AB: but the halves of equals (d) are equal; therefore the triangles ABC and Wherefore, triangles, &c. DEF are equal. Recite (a) p. 31; (b) p. 36; (c) p. 34; (d) ax. 7. Q. E. D. 39 Th. Equal triangles (ABC, DBC), upon the same base (BC), and upon the same side of it, are between the same parallels. Join AD;-AD is parallel to BC: for if not, A through the point A draw AE parallel to BC (a), and join CE. The triangles ABC, EBC, are equal (b), because they are on the same base BC, and between the same parallels BC, AE: but the triangles DBC and ABC are equal by hyp., therefore the triangles EBC and DBC are equal, the less to the greater, which is absurd. B E D C Therefore AE is not parallel to BC; neither is any other line but AD parallel to BC. Wherefore, equal triangles, &c. Recite (a) p. 31; (6), p. 37. Q. E. D. 40 Th. Equal triangles (ABC, DEF), upon equal bases (BC, EF), in the same straight line (BF), on one side of it, are between the same parallels. Join AD;-AD is parallel to BC; The for if not, through A (a) draw AG parallel to BF, and join GF. triangle ABC is equal to the triangle GEF, because they are upon equal bases BC, EF, and between the same parallels BF, AG (b). A D C But the triangle ABC is equal to B the triangle DEF; therefore GEF and DEF are equal, the less to the greater, which is impossible. Therefore AG is not parallel to BF; neither is any other line but AD. Wherefore, equal triangles, &c. Recite (a) p. 31; (b) p. 38. Q. E. D. 41 Th. If a parallelogram (ABCD), and a triangle (EBC), be upon the same base BC, and between the same parallels (BC, AE), the parallelogram is double of the triangle. Join AC; then the triangle ABC equals the triangle EBC, because they are upon the same base A BC, and between the same parallels BC, AE (a). But the parallelogram ABCD is double of the triangle ABC (b), because the diameter bisects it: wherefore ABCD is also double of the triangle EBC (c). Therefore, if a parallelogram, &c. Recite (a) p. 37; (b) p. 34; (c) ax. 7. 42 P. To describe a parallelogram that shall be equal to a given triangle (ABC,) and have one of its angles equal to a given rectilineal angle (D). Bisect BC in E (a); join AE, and at the point E, in the straight line EC, make the angle CEF equal to D (b); and through A, draw AG parallel to EC; and through C draw CG parallel to EF (c). AFG BE C EFGC is therefore a parallelogram; and because BE and EC are equal, the triangles AEB, AEC are also equal, since they are upon equal bases, and between the same parallels BC, AG (d); therefore, the triangle ABC is double of the triangle AEC: the parallelogram CEFG is likewise double of the same triangle, being upon the same base and between the same parallels (e); CEFG is therefore equal to ABC, and has an angle equal to the angle D. There fore the thing required has been done. Recite (a) p. 10; (b) p. 23; (d) p. 38; (e) p. 41. (c) p. 31; 43 Th. The complements of the parallelograms about the diameter (AC) of any parallelogram (ABCD), are equal to one another. The parallelograms about the diameter AC, or through which AC passes, are EH, GF; and the complements, which make up the whole figure, are E BK, KD, which are said to be equal. The diameter AC bisects the parallelogram ABCD; its parts AK, KC bisect also EH, GF (a); therefore, the triangles ABC, ADC are equal; also the triangles AEK, AHK, and KGC, KFC. Therefore, from ABC take AEK+KGC, the remainder is BK. AH BC D K F C Also, from ADC take AHK+KFC, the remainder is KD. therefore BK is equal to KD. Wherefore, the complements, &c. Q. E. D. Recite (a) p. 34; (b) ax. 3. 44 P. To a given straight line (AB) to apply a parallelogram equal to a given triangle (C), having an angle equal to a given rectilineal angle. Upon half the base of the given triangle C, make a parallelogram GBEF, equal to C (a); and let its angle GBE equal the given angle D (b); let AB, the given straight line, be the produced part of EB; through B produce GB to M; through A draw HL parallel to GM; produce FG to H; through B draw HK; produce FE to K; make KL parallel to EA. FE K C C M B HA LD The complements BL and BF are equal (c); but BF was made equal to the given triangle C; therefore, BL is equal to C (d); and it is applied to the given straight line AB, having one of its angles ABM, equal to the opposite vertical angle GBE, which was made equal to the given angle D. Therefore the thing required has been done. Recite (a) p. 42; (b) p. 23; (c) p. 43; (d) ax. 1. 45 P. To describe a parallelogram equal to a given rectilineal figure (ABCD), having an angle equal to a given rectilineal angle (P); and to apply the parallelogram to a given straight line (CM). Produce the side CB of the given rectilineal figure indefinitely to E (a); join DB (6), and parallel to it, through A, draw AE (c); join DE (6). A DG K Again, bisect CE in F (d); upon CF describe E BF the parallelogram CFGH, equal to the triangle DEC, and having an angle FCH equal to the given angle P (e). 0 C M NP Also, let CM, the line to which the parallelogram is to be applied, be in the same straight line with HC; produce FC to O; through M draw LN parallel to FO (c); produce GF to meet NL; through C draw LK; produce GH to meet LK; parallel to HCM draw KON (c.) 1. The triangles DAB, DEB, on the same base DB, and between the same parallels DB, AE, are equal to one another (f); to each add the triangle DBC; therefore, the triangle DEC is equal to the given rectilineal figure ABCD (g). 2. But the parallelogram CFGH was made equal to the triangle DEC; therefore CFGH is equal to the given rectilineal figure ABCD (b), and its angle FCH is equal equal to the given angle P. 3. The complements FH and MO are equal (i); and the angle MCO is equal to its vertical and opposite angle FCH, and therefore to the given angle P (k). Wherefore to CM is applied a parallelogram equal to a given rectilineal figure, having an angle equal to a given rectilineal angle. Recite (a) pos. 2; (d) p. 10; (e) p. 42; (f) p. 37; (g) ax. 4; (h) ax. 1; (i) p. 43; (k) p. 15; NOTE. This demonstration includes the corollary, as given in Simson's and Playfair's edition. 46 P. To describe a square upon a given straight line (AB.) Constr. From the point A draw AC at right an- C gles to AB (a); make AD equal to AB (6); draw DE parallel to AB, and BE parallel to AD (c). 1. Of the equal angles. The angle A is a right angle; and, because of the parallels, the two interior angles, A and D, are equal to two right angles (d); also, for the same reason, A equals B, and B equals E. Therefore, the figure has four right angles. D A E B 2. Of the equal sides. The sides AB, AD are made equal; and because the opposite sides of a parallelogram are equal (e), DE is equal to AB, and BE to AD. Hence also, the figure has four equal sides; and is therefore a square (f). Wherefore, the required square has been described. Recite (a) p. 11; (b) p. 3; (c) p. 31; (d) p. 29; (e) p. 34; (f) def. 30. 47 Th. In any right-angled triangle (ABC), the square upon the side (BC), subtending the right angle, is equal to the sum of the squares of the sides (AB, AC), containing the right angle. Construction. On BC describe the square BE, on AB the square BG, on AC the square AK (a); κ draw AL parallel to BD, or CE (b); join AD, CF; also AE, BK. Dem. Because of the adjacent right angles at C A, the lines BA, AH, and CA, AG, are straight lines (c), the one parallel to BF, the other to CK. The triangle BCF is upon the side BF, and between the same parallels as the square BG, of which it is equal to the half: Also, the triangle ABD is upon the side BA, and between the same parallels as the parallelogram BL, of which it is equal to the half (d). But the triangles BCF, ABD are equal; having two sides BF, BC, in the one, equal to two sides, BA, BD, in the other; and each of the two containing the angle ABC and a right angle (e). Therefore, the half of the square BG, is equal to the half of the parallelogram BL; hence the square is equal to the parallelogram (f). |