Produce AD both ways to the points G, H, and through Book I. B draw BG parallela to CA, and through F draw FH paral lel to ED; then each of the figures GBCA, DEFH is a parallelogram; and they are equalb; and the triangle ABC is b 36.1. the half of the parallelogram GBCA, and the triangle DEF c 34. 1. is the half of the parallelogram DEFH; therefore the triangle ABC is equald to the triangle DEF. Wherefore, triangles, d 7. Ax &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. If it is not, through the point A drawa AE parallel to BC, and join EC. The trian- a 31. 1. gle ABC is equal to the triangle EBC: but the triangle ABC is equal to the triangle BDC; therefore also the triangle BDC is equal to the triangle EBC, the greater to the less, which is impossible. Therefore AE is not parallel to BC. In the same manner it can be demonstrated, that A B D C no other line but AD is parallel to BC; AD is therefore parallel to it. Wherefore, equal triangles upon, &c. Q. E. D. b 37. 1. Book I. PROP. XL. THEOR. 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 line BF, and to wards the same parts; D G they are between the same parallels. a 31. 1. Join AD; AD is parallel to BC. If it is not, through A drawa b 38. 1. AG parallel to BF, and join GF. The triangle ABC is equal to the triangle GEF; 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. In the same manner it can be demonstrated, that there is no other parallel to it but AD; AD is therefore parallel to BF. Wherefore, equal triangles, &c. Q. E. D. PROP. XLI. THEOR. IF a parallelogram and a triangle be upon the same base, and between the same parallels, the parallelogram is double of the triangle. Let the parallelogram ABCD and the triangle EBC be upon the same base BC, and between the same parallels BC, ! 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. Bisecta BC in E, join AE, and at the point E in the a 10. 1. straight line EC make the angle CEF equal to D; and b 23. 1. through A draws AG parallel to EC, and through C draw c 31.1. CG parallel to EF; then A FECG is a parallelogram. A AEC; therefore FECG is equal to the triangle ABC, F Book I. PROP. XLIII. THEOR. THE complements of the parallelograms which are about the diameter of any parallelogram are equal to one another. 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 paral a 34. 1, lelograms, which make up the E K B G C 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, Book I. Having the angle EBG equal to the angle D, and the side BE 1. in the same straight line with AB: produce FG to H, and through A draw AH parallel to BG or EF, and join HB. b 31. 1. Then because the straight line HF falls upon the parallels AH, EF, the angles AHF, HFE are together equal to two c 29.1. right angles; wherefore the angles BHF, HFE are less than two right angles; therefore HB, FE will meetd, if produced; d Cor. 29. 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 e 43. 1. equal to the triangle C; wherefore LB is equal to the triangle C; and because the angle GBE is equalf to the angle ABM, f 15. 1. and likewise to the angle D, the angle ABM is equal to the angle D. Therefore the parallelogram LB is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D. Which was to be done. PROP. XLV. PROв. TO describe a parallelogram equal to a given rectilineal 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. |