PA PROP. XXXVI. THEOR. ARALLELOGRAMS upon equal bases, and between Let ABCD,EFGH be A parallelograms upon equal bafes BC, FG, and between the fame parallels AH, BG; the parallelogram ABCD is equal to EFGH. DE H Join BE, CH; and B because BC is equal to Book I. FG, and FG to a EH, BC is equal to EH; and they are paral- a 34. 1. lels, and joined towards the fame parts by the ftraight lines BE, CH: But ftraight lines which join equal and parallel straight lines towards the fame parts, are themselves equal and parallel b; b 33. I, therefore EB, CH are both equal and parallel, and EBCHis a parallelogram; and it is equal c to ABCD, because it is upon c 35. I, 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 alfo the parallelogram ABCD is equal to EFGH. Wherefore parallelograms, &c. Q. E. D. PROP. XXXVII. THEOR. TRIANGLES upon the fame bafe, and between the E A D F Let the triangles ABC, DBC be upon the same base BC and between the fame parallels AD,BC: The triangle ABC is equal to the triangleDBC. Produce ADboth ways to the points E, F, and through B draw a BE parallel to CA; and thro' C draw CF parallel to BD: Therefore each of the figures EBCA, DBCF is a parallelogram; and EBCA is equal to DBCF, because b 35. I. they are upon the same base BC, and between the fame parallels BC, EF; and the triangle ABC is the half of the parallelo gram gram EBCA,because the diameter AB befects cit; and the triangle DBC is the half of the parallelogram DBCF, because the diameter DC bifects it: But the halves of equal things are equal d; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D. TR PROP. XXXVIII. THEOR. "RIANGLES upon equal bases, 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. G A D H Produce AD both ways to the points G, H, and through B draw BG parallel a to CA, and through F draw FH parallel to 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 be B CE F tween the fame pa € 34. 1. ₫ 7. Ax. 2 31. I rallels BF, GH; and the triangle ABC is the half of the parallelogram GBCA, because the diameter AB bìfects it; and the triangle DEF is the half of the parallelogram DEFH, because the diameter DF bifects it: But the halves of equal things are equal d; therefore the triangle ABC, is equal to the triangle DEF. Wherefore triangles, &c. Q. E. D. PROP. XXXIX. THE OR. EQUAL QUAL 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 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, through the point A draw a AE parallel to BC, and join EC: The triangle t ABC A E D b 37. I. ABC is equal b to the triangle EBC, because it is upon the fame Book I. base BC, 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 that no o B C ther line but AD is parallel to BC; AD is therefore parallel to it. Wherefore equal triangles upon, &c. Q. E. D. QUAL PROP. XL. THE OR, E triangles upon equal bases, in the fame ftraight line, and towards the fame parts, are between the fame parallels. Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the fame straight b 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 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. IF PROP. XLI. THEOR. F a parallelogram and triangle be upon the fame bafe, and between the fame parallels; the parallelogram fhall be double of the triangle. Let a 31. b 38. Ii Book 1. a 37. I. b 34. I. a IO. I. b 23. I. C 31. I. d 38. I. - € 41. I. Let the parallelogram ABCD and the triangle EBC be upon tlfe fame base BC, and between the fame parallels BC, AE; the parallelogram ABCD is double of the triangle EBC. A Join AC; then the triangle ABC B PROP. XLII. PRO B. DE defcribe a parallelogram that fhall be equal to a To 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 defcribe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D. AF G Bifect a BC in E, join AE, and at the point E in the straight line EC make b the angle CEF equal to D; and through A draw c AG parallel to EC, and thro' C draw CG ce parallel to EF: Therefore FECG is a parallelogram: And because BE is equal to EC, the triangle ABE is likewife equal to the triangle AEC, fince they are upon equal bafes BE, EC, and between the fame parallels BC, AG; therefore the triangle ABC is double of the B triangle AEC. And the parallelogram FECG is likewife double of the triangle AEC, because it is upon the same 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 D: wherefore there has been described a parallelogram D E C FECG FECG equal to a given triangle ABC, angles CEF equal to the given angle D. having one of its Book I. done. TH PROP. XLIII. THEOR. '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 parallelo grams about AC, that is, thro' A H D K F B G C ter, the triangle ABC is equal a to the triangle ADC; And, a 34. Fbecause 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. T PROP. XLIV. PRO B. O 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 ftraight line AB a parallelogram equal to the triangle C, and having an angle equal to D. Make |