32 Th. If any side (BC) of a triangle be produced, the exterior angle (ACD) is equal to the two interior angles (A and B); and the three interior angles of every triangle are equal to two right angles. Argument. Through C draw ČE parallel to AB (a). Then because AC meets the parallels AB, CE, the alternate angles A and ACE are equal (6). Also, because BD meets the parallels AB, CE, the exterior DCE equals the interior B (c). Therefore, the interior angles A and B equal the angles ACE and DCE, that is, the exterior angle ACD. Again, to these equals add the angle ACB (d); therefore, the three interior angles A, B and ACB are equal to the two ACD, ACB, that is, io two right angles (e). Wherefore, if any side, &c. Q. E. D. Recite (a) p. 31; (b) p. 27; (c) p. 29; (d) ax. 2; (e) p. 13. Cor. 1. All the interior angles of any rectilineal figure and four right angles, are equal to twice as many right angles as the figure has sides. For, about a point within the figure, as many triangles may be formed as the figure has sides, each of whose angles shall equal two right angles; and the A angles about the point are equal to four right angles. Cor. 2. All the exterior angles of any rectilineal figure are equal to four right angles; for such figure may be reduced to a mere point; about which there cannot be more than four D right angles. B B 33 Th. The straight lines (AC, BD) which join the same ends of two equal and parallel straight lines (AB, CD, are themselves equal and parallel. Join the alternate ends BC (a); then since the equal and parallel lines AB, CD are met by BC, the alternate angles ABC, BCD are equal (6); and because AB equals CD, and BC is common; therefore the bases AC, BD are equal (c); also the alternate angles ACB, CBD (6); therefore also AC is parallel to BD. Wherefore, the straight lines which join, &c. Q. E. D. Recite (a) pos 1; (6) p. 27 ; (c) p. 4. с Definition. A parallelogram is a quadrilateral figure whose opposite sides are parallel. B 34 Th. The opposite sides and angles of a parallelogram are equal, and the diameter bisects it (a). Because ABCD is a parallelogram, its opposite sides are parallel (b); and because BC joins opposite angles (c) it meets the parallels AB, CD, and also the parallels AĆ, BD, and makes the two angles at B alternately equal to the two at C (d); so that the whole angles ABD, ACD are equal. Again, in the triangles ABC, DCB, the side BC is common; and their angles at B and C are alternately equal ; therefore the remaining angles at A and D are equal (e); also the sides AB to ČD aud AC to BD (f); therefore each of the triangles ABC, BCD is half of the parallelogram, which is bisected by BC. Wherefore, the opposite sides and angles, &c., Q. E. D. Recite (a), p. 9, 10; (6), def. above; (c), def. 16; (d), p. 27; (e), p. 32; (), p. 26. 35 Th. Parallelograms upon the same base (BC), and between the same parallels (AF, BC), are equal to one another. To make ABCD, BCFD, distinct parallelograms on the same base, AF must be greater than BC; and the figure ABCF will be a trapezoid (a). 1. If'AF be the double of BC, the straight lines DB, DC will bisect AF, and also each of the parallelograms; and the triangle DBC will be the half of each (6); hence the parallelograms ABCD, BCFD will be equal (c). 2. But if AF be greater than twice BC, DE B will be the excess; and if less, DE will be a common segment of AD, EF. To each, therefore, add the excess, or from each subtract the segment; the sums, or differences, AE, DF, will be equal (d). In either case, therefore, the sides CD, DF equal the sides BA, AE; and the exterior angle CDF equals the interior and opposite angle BAE (e), and the triangles CDF, BAE are equal (f ). From the trapezoid ABCF take each of these triangles; the remaining parallelograms ABCD, BCFE are equal (d). Wherefore, parallelograms, &c. Q. E. D. Recite (a), def. 34; (b), p. 34; (c), ax. 6; (d), ax. 2, 3; (e), p. 32; (f), p. 4. 8 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, B which join their extremities, are equal and parallel (6). 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. Q. E. D. Recite (a) p. 34; (b) p. 33; (c) p. 35; (d) ax. 1. C 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 through C draw CF parallel to BD. Each of the figures EBCA, DBCF is a parallelogram; and they are equal (6), because they are upon the same base and between the same parallels, BC, EF: and the triangle ABC is half the parallelogram EBČA, 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. Q. E. D. Recite (a) p. 31; (6) p. 33; (c) p. 34; (d) ax. 7. A 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 through F, draw FH parallel to ED. Then each of the figures BCAG, FEDH is a parallelogram; and they are equals (b), being upon equal bases and between the same parallels BF, GH. But the triangle DEF is half the parallelogram DEFH; and the triangle ABC is half the parallelogram GBCA B с Е (c); because they are bisected by the diameters DF and AB: but the halves of equals (d) are equal; therefore the triangles ABC and DEF are equal. Wherefore, triangles, &c. Q. E. D. Recite (a) p. 31; (b) p. 36; (c) p. 34; (d) ax. 7. 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;-AV 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 (6), 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. Therefore AE is not parallel to BC; neither is any other line but AD parallel to BC. Wherefore, equal triangles, &c. Q. E. D. Recite (a) p. 31; (b), p. 37. B В 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; for if not, through A (a) draw AG parallel to BF, and join GF. The triangle ABC is equal to the triangle GEF, because they are upon equal bases BC, EF, and between the same parallels BF, AG (6). But the triangle ABC is equal to CE 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. Q. E. D. Recite (a) p. 31; (b) p. 38. 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 tri. angle. Join AC ;-then the triangle ABC equals the triangle EBC, because they are upon the same base 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. Q. E. D 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). EFGC is therefore a parallelogram ; and because BE and EC are equal, the triangles AEB, AEC are also eqnal, 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. Therefore the thing required has been done. Recite (a) p. 10; (b) p. 23; (c) p. 31; (d) p. 38; (e) p. 41. B E с 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 АН through which AC passes, are EH, GF; and the complements , which make up the whole figure , are ENK BK, KD, which are said to be equal. The diameter AC bisects the parallelogram ABCD; its parts AK, KC bisect also ÉH, 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. Also, from ADC take AHK+KFC, the remainder is KD. But taking equals from equals the remainders are equal (b); therefore BK is equal to KD. Wherefore, the complements, &c. Q. E. D. Recite (a) p. 34; (b) ax. 3. в с |