EBCA, because the diameter AB bisects it; and the tri- Book I. angle DBC is the half of the parallelogram DBCF, because the diameter DC bisects it; but the halves of equal things are c 34. 1. equal; therefore the triangle ABC is equal to the triangle d 7. Ax. DBC. Wherefore, "triangles," &c. Q. E. D. PROP. XXXVIII. THEOR. Triangles upon equal bases, and between the same parallels, are equal to one another. Let the triangles ABC, DEF be upon equal bases BC, EF, and between the same parallels BF, AD: The triangle ABC is equal to the triangle DEF. Produce AD both ways to the points G, H, and through B draw BG parallel to CA, and through F draw FH parallel to a 31. 1. a ED: Then each of the figures GBCA, DEFH is a parallelogram; and they are equal to b one another, because they are upon equal bases BC, EF, and between the same pa rallels BF, GH; and the triangle ABC is the half of the parallelogram GBCA, c 34. 1. because the diameter AB bisects it; and the triangle DEF is the half of the parallelogram DEFH, because the diameter DF bisects it; but the halves of equal things are equal; d d 7. Ax. therefore the triangle ABC is equal to the triangle DEF. Wherefore," triangles," &c. Q. E. D. PROP. XXXIX. THEOR. Equal triangles upon the same base, and upon the same side of it, are between the same parallels. Let the equal triangles ABC, DBC be upon the same base BC, and upon the same side of it; they are between the same parallels. Join AD; AD is parallel to BC; for, if it is not, through the point A draw AE parallel to BC, and join EC: The triangle a 31. 1. a C Book I. b 37. 1. b A D E ABC is equal to the triangle EBC, because it is upon the B PROP. XL. THEOR. C Equal triangles upon equal bases, in the same straight line, and towards the same parts, are between the same parallels. Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the same straight B A D G CE F line BF, and towards the same parts; they are between the same parallels. Join AD; AD is parallel to BC: For if it is not, a 31. 1. through 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: But the triangle ABC is equal to the triangle DEF: therefore also the triangle DEF is equal to the triangle GEF, the greater to the less, which is impossible: Therefore AG is not parallel to BF and in the same manner, it can be demonstrated, that there is no other parallel to it than AD; AD is therefore parallel to BF. Wherefore, " equal triangles," &c. Q. E. D. b 38. 1. PROP. XLI. THEOR. If a parallelogram and a triangle are upon the same base, and between the same parallels; the parallelogram is double the triangle. Let the parallelogram ABCD and the triangle EBC be up- Book I. on the same base BC, and between the same parallels BC, AE; the parallelogram ABCD is double the triangle EBC. a Join AC; then the triangle ABC is equal to the triangle EBC, because they are upon the same base BC, and between the same parallels BC, AE. But the parallelogram ABCD is double the triangle ABC, because the diameter AC divides it into two equal parts; wherefore A B DE a 37. 1. b 34. 1. C ABCD is also double the triangle EBC. Therefore, "if a parallelogram," &c. Q. E. D. PROP. XLII. PROB. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let ABC be the given triangle, and D the given rectilineal angle. It is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D. a b AF G c 31. 1. Bisect BC in E, join AE, and at the point E in the straight a 10. 1. line EC make the angle CEF equal to D; and through A b 23. 1. draw AG parallel to BC, and through C draw CG parallel to EF: Therefore FECG is a parallelogram And because BE is equal to EC, the triangle ABE is likewise equal to the triangle AEC, since they are upon equal bases BE, EC, and between the same parallels BC, AG: Therefore the triangle B E C ABC is double the triangle D d 38. 1. AEC. And the parallelogram FECG is likewise double e 41. 1. the triangle AEC, because it is upon the same base, and between the same parallels: Therefore the parallelogram FECG is equal to the triangle ABC, and it has one of its angles CEF equal to the given angle D: wherefore there has been described a parallelogram FECG equal to a given triangle ABC, Book I. having one of its angles CEF equal to the given angle D. Which was to be done. a 31.1 PROP. XLIII. THEOR. The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another. A H D K E F Let ABCD be a parallelogram, of which the diameter is AC, let EH, FG be the parallelograms about AC, that is, through which AC passes, and let BK, KD be the other parallelograms which make up the whole figure ABCD, which are therefore called the complements: the complement BK is equal to the complement KD. B G C Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal a to the triangle ADC: and because EKHA is a parallelogram, the diameter of which is AK, the triangle AEK is equal to the triangle AHK: By the same reason, the triangle KGC is equal to the triangle KFC: Then, because the triangle AEK is equal to the triangle AHK, and the triangle KGC to KFC; the triangle AEK, together with the triangle KGC is equal to the triangle AHK together with the triangle KFC: But the whole triangle ABC is equal to the whole ADC; therefore the remaining complement BK is equal to the remaining complement KD. Wherefore, "the complements," &c. Q. E. D. PROP. XLIV. PROB. To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle. It is required to apply to the straight line AB, a parallelogram equal to the triangle C, and having an angle equal to D. Make the parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D, so that BE be in the same straight line Book I. a 42. 1. с b with AB, and produce FG to H; and through A draw AH b 31. 1. parallel to BG or EF, and join HB. Then, because the straight line HF falls upon the parallels AH, EF, the angles AHF, HFE are together equal to two right angles; wherefore the c 29. 1. angles BHF, HFE are less than two right angles: But straight lines which with another straight line make the interior angles upon the same side less than two right angles, do meet if produced: Therefore HB, FE, will meet if pro-d 12. Ax. duced; let them meet in K, and through K draw KL parallel to EA or FH, and produce HA, GB to the points L, M: Then HLKF is a parallelogram, of which the diameter is HK, and AG, ME are the parallelograms about HK; and LB, BF are the complements; therefore LB is equal to BF: But BF is equal to the triangle C; where- e 43. 1. fore LB is equal to the triangle C; and because the angle GBE is equal to the angle ABM, and likewise to the angle f 15. 1. D; the angle ABM is equal to the angle D: Therefore the parallelogram LB, which is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D. e Which was to be done. PROP. XLV. PROB. To describe a parallelogram equal to a given rec- See N. tilineal figure, and having an angle equal to a given rectilineal angle. Let ABCD be the given rectilineal figure, and E the given rectilineal angle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E. Join DB, and describe the parallelogram FH equal to the a 42. 1. triangle ADB, and having the angle HKF equal to the angle E; and to the straight line GH apply the parallelogram GM equal b 44. 1. |