each to each, to which the equal fides are oppofite. therefore the Book I. c PROP. XXXIV. THEOR. THE oppofite fides and angles of parallelograms are equal to one another, and the diameter bifects them, that is, divides them into two equal parts. N. B. A Parallelogram is a four fided figure of which the oppofite fides are parallel. and the diameter is the ftraight line joining two of its oppofite angles. Let ABCD be a parallelogram, of which BC is a diameter. the opposite fides and angles of the figure are equal to one another; and the diameter BC bifects it. a A B a. 29. 1. Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal to one another, and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal to one another. wherefore the two triangles ABC, C CBD have two angles ABC, BCA D in one, equal to two angles BCD, CBD in the other, each to each, there c Book I. therefore the triangle ABC is equal to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Q. E. D. PROP. XXXV. THEOR. See N. PARALLELOGRAMS upon the fame bafe and between the fame parallels, are equal to one another. See the ad Let the parallelograms ABCD, EBCF be upon the fame bafe BC and 3d Fi-and between the fame parallels AF, BC. the pa. allelogram ABCD fhall be equal to the parallelogram EBCF. gures. If the fides AD, DF of the parallelograms ABCD, DBCF oppofite to the bafe BC, be terminated in the fame point D; it is plain that each A D F 2. 34. I. a of the parallelograms is double of the But if the fides AD, EF oppofite to EBCF be not terminated in the fame C 3 point; then because ABCD is a parallelogram, AD is equal to BC; for the fame reafon, EF is equal to BC; wherefore AD is b. 1. Ax. equal b to EF; and DE is common; therefore the whole, or the c. 2. or 3. remainder, AE is equal to the whole, or the remainder DF; AB alfo is equal to DC; and the two EA, AB are therefore equal to Ax. c 4. 1. the two FD, DC, each to each; and the exterior angle FDC is ed. 29. 1. qual to the interior EAB; therefore the bafe EB is equal to the base FC, and the triangle EAB equal to the triangle FDC. take the triangle FDC from the trapezium ABCF, and from the fame trapezium take the triangle EAB; the remainders therefore are f. 3. Ax. equalf, that is, the parallelogram ABCD is equal to the parallelogram EBCF. therefore parallelograms upon the fame base &c. Q. E. D. PROP. PROP. XXXVI. THEOR. PARALLELOGRAMS upon equal bafes and between the fame parallels, are equal to one another. A DE H Let ABCD, EFGH be parallelograms upon equal bafes BC, FG, and between the fame parallels AH, BG; the parallelogram ABCD is equal to EFGH. Book I. are parallels, and joined towards the fame parts by the straight lines BE, CH. but straight lines which join equal and parallel straight lines towards the fame parts, are themselves equal and parallel; there b. 33. 17 fore EB, CH are both equal and parallel, and EBCH is a parallelogram; and it is equal to ABCD, because it is upon the fame bafe c. 3§. i. BC, and between the fame parallels BC, AD. for the like reafon the parallelogram EFGH is equal to the fame EBCH. therefore alfo the parallelogram ABCD is equal to EFGH. Wherefore parallelograms, &c. Q. E. D. PROP. XXXVII. THEOR. TRIANGLES upon the same base, and between the fame parallels, are equal to one another. E A D Let the triangles ABC, DBC be upon the fame bafe BC and between the fame parallels AD; BC. the triangle ABC is equal to the triangle DBC. Produce AD both ways to the points E, F, and thro' B a draw BE parallel to CA; and thro' C draw CF parallel to BD. therefore each of the b figures EBCA, DBCF is a parallelogram; and EBCA is equal to b. 35. 1: DBCF, because they are upon the fame bafe BC, and between the fame parallels BC, EF; and the triangle ABC is the half of the pa C rallelogram Book I. rallelogram EBCA, because the diameter AB bifects it; and the w triangle DBC is the half of the parallelogram DBCF, because the c. 34, 1. diameter DC bifects it. but the halves of equal things are equal &; d. 7. Ax. therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D. TRIA PROP. XXXVIII. THEOR. 'RIANGLES upon equal bafes, and between the fame parallels, are equal to one another. Let the triangles ABC, DEF be upon equal bafes BC, EF, and between the fame parallels BF, AD. the triangle ABC is equal to the triangle DEF. Produce AD both ways to the points G, H, and thro' B draw BG a. 31. i. parallel to CA, and thro' F draw FH parallel to ED. then each of the figures GBCA, DEFH is a parallelogram; and they are b. 36. 1. equal b to one another, because they are upon equal bafes BC, EF and between the fame parallels BF, G D H AN B é. 34. 1. GH; and the triangle ABC is the half CE F of the parallelogram GBCA, because the diameter AB bifects it; and the triangle DEF is the c half of the parallelogram DEFH, because the diameter DF bifects J. 7. Ax. it. but the halves of equal things are equal; therefore the triangle ABC is equal to the triangle DEF. Wherefore triangles, &c. Q. E. D. PROP. XXXIX. THEOR. EQUAL triangles upon the fame base, and upon the fame fide of it, are between the fame parallels. Let the equal triangles ABC, DBC be upon the fame bafe BC, and upon the fame fide of it; they are between the fame parallels. Join AD; AD is parallel to BC; for if it is not, thro' the point #. 51. 5. A draw * AE parallel to BC, and join EC. the triangle ABC is equal E D b; 37: 1 equal to the triangle EBC, because it is upon the fame bafe BC, Book I. and between the fame parallels BC, AE. A 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 lefs, which is impoffible. therefore AE is not parallel to BC. in the fame manner it can be demonftrated that no other line but AD is parallel to BC; AD is therefore parallel to it. Wherefore equal triangles upon, &c! Q. E. D. B C PROP. XL. THEOR. EQUAL triangles upon equal bafes, and towards the fame parts, are between the fame parallels. Let the equal triangles ABC, DEF be upon equal bafes BC, EF, and towards the fame parts; they are between the fame parallels. B G CE a. 3i. 1. F b. 38. 14 Join AD; AD is parallel to BC. for if it is not, thro' A draw AG parallel to BF, and join GF. the triangle ABC is equal to the triangle GEF, because they are upon equal bafes BC, FF, and between the fame 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 lefs, which is impoffible. therefore AG is not parallel to BF. and in the fame 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. a parallelogram and triangle be upon the fame bafe, and between the fame parallels; the parallelogram fhall be double of the triangle. |