PA ARALLELOGRAMS upon equal bafes and between A DE H Let ABCD, EFGH be parallelograms upon equal bases BC, FG, and between the fame parallels AH, BG; the parallelogram ABCD is equal to EFGH. Join BE, CH; and because BC is equal to FG, and FG to EH, BC is B equal to EH; and they В C F Book I. a. 34. I are parallels, and joined towards the fame parts by the straight lines BE, CH. but straight lines which join equal and parallel ftraight lines towards the fame parts, are themselves equal and parallel b; therefore EB, CH are both equal and parallel, and b. 33. I. EBCH is a parallelogram; and it is equal to ABCD, because it c. 35. I. is upon the fame base BC, and between the fame parallels BC, AD. for the like reason the parallelogram EFGH is equal to the fame EBCH. therefore also the parallelogram ABCD is equal to EFGH. Wherefore parallelograms, &c. Q. E. D. TRI PROP. XXXVII. THEOR. RIANGLES upon the same base, and between the Let the triangles ABC, DBC be upon the fame base 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 D F figures EBCA, DBCF is a parallelogram; and EBCA is equal b to b. 35. I. 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 paC C. 34. I. C Book I. rallelogram EBCA, because the diameter AB bifects it; and the triangle DBC is the half of the parallelogram DBCF, because the diameter DC bifects it. but the halves of equal things are d. 7. Ax. equal d; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D. a. 31. . TRI RIANGLES upon equal bases, and between the fame parallels, are equal to one another. Let the triangles ABC, DEF be upon equal bases 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 A D H C. 34. I. lels BF, GH; and the triangle ABC is the half of the parallelogram GBCA, because the diameter AB bifects it; and the triangle DEF is the half of the parallelogram DEFH, because the diameter DF bifects it. but the halve of equal things are d. 7. Ax. equal a; therefore the triangle ABC is equal to the triangle DEF. Wherefore triangles, &c. Q. E. D. 3. 31. I. EQUAL QUAL triangles upon the fame bafe, and upon the fame fide of it, are between the fame parallels. Let the equal triangles ABC, DBC be upon the fame base 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 A draw a AE parallel to BC, and join EC. the triangle ABC is A D' b. 37. I. equal to the triangle EBC, because it is upon the fame bafe BC, Book I. and between the fame parallels BC, AE. but the triangle ABC is equal to the triangle BDC; therefore alfo 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 B that no other line but AD is parallel to BC; AD is therefore parallel to it. Wherefore equal triangles upon, &c. Q. E. D. PROP. XL. THEOR. EQUAL triangles upon equal bases, 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 same parallels. Join AD; AD is parallel to BF. for if it is not, thro' A draw a AG parallel to BF, and join GF. the triangle ABC is equal to the triangle B b GEF, because they are upon equal bafes BC, EF, and between the b. 38. 1. same parallels BF, AG. but the triangle ABC is equal to the triangle DEF; therefore alfo 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 demonftrated 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. F a parallelogram and triangle be upon the fame bafe, and between the fame parallels; the paralJelogram shall be double of the triangle Book I. Let the parallelogram ABCD and the triangle EBC be upon the fame bafe BC, and between the fame parallels BC, AE; the parallelogram ABCD is double of the a. 10. I. b. 23. I. C. 31. I. A B.. DE C parts; wherefore ABCD is also double of the triangle EBC. therefore if a parallelogram, &c. Q. E. D. PROP. XLII. PROB. O defcribe a parallelogram that fhall be equal to Ta 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. Bifecta BC in E, join AE, and at the point E in the straight line EC make the angle CEF equal to D; and thro' A draw © AG parallel to EC, and thro' C draw c CG parallel to EF. therefore FECG is a parallelogram. and becaufe BE is equal to EC, the trid. 38. 1. angle ABE is likewise equal to the triangle AEC, fince they are upon equal bafes BE, EC and be D 'B E therefore the triangle ABC is dou tween the fame parallels BC, AG; ble of the triangle AEC. and the parallelogram FECG is likewife e. 41. 1. double of the triangle AEC, because it is upon the fame base, and between the fame parallels. therefore the parallelogram FECG is equal to the triangle ABC, and it has one of its angles CEF equal to the given angle Lwherefore there has been described a parallelogram FECG equal to a given triangle ABC, having one of Book I. its angles CEF equal to the given angle D. Which was to be done. THE HE 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, and EH, FG the parallelograms A H about AC, that is, thro' which D AC paffes, and BK, KD 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. a B G Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ADC, and because EKHA a. 34. 5. is a parallelogram, the diameter of which is AK, the triangle AEK is equal to the triangle AHK. by the fame reason, the triangle KGC is equal to the triangle KFC. then because the triangle A EK 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. T 10 a given straight line to apply a parallelogram, which fhall 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 ftraight line AB a parallelogram equal to the triangle C, and having an angle equal to D. |