PROP. XXXVI. THEOR. PARALLELOGRAMS upon equal bases, and between the same parallels, are equal to one another. Boor I. because BC is equal to FG, and FG toa EH, BC is equal. 34. 1. to EH; and they are parallels, and joined towards the same parts by the straight lines BE, CH. But straight lines which join equal and parallel straight lines towards the same parts, are themselves equal and parallelb; therefore EB, 33. 1. CH, are both equal and parallel, and EBCH is a parallelogram; and it is equal to ABCD, because it is upon the same base BC, and between the same parallels BC, AD: For the like reason, the parallelogram EFGH is equal to the same EBCH: Therefore also the parallelogram ABCD is equal to EFGH. Wherefore parallelograms, &c. Q. E. D. 35. 1. PROP. XXXVII. THEOR. TRIANGLES upon the same base, and between the same parallels, are equal to one another. Let the triangles ABC, DBC, be upon the same base BC, and between the same parallels AD, BC: The triangle ABC is equal to the triangle DBC. Prodúce AD both ways to the points E, F, and through B drawa BE parallel to CA; and through C draw CF parallel to BD; therefore each of the figures EBCA, DBCF is a parallelogram; and EBCA is equalb to DBCF, because they are upon the same base BC, and between the same parallels BC, EF; and the triangle ABC is the half of the parallelogram, EBCA, because the D a 31.1. 6 35.1. с 34. 1. Book I. diameter AB bisects it; and the triangle DBC is the half of the parallelogram DBCF, because the diameter DC bi sects it: but the halves of equal things are equald; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q.E.D. 7 Ax. 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 31.1. B draw BG parallela to CA, and through F draw FH pa rallel to ED: Theng GBCA, DEFH, is cause they are upon B H E F equal bases BC, 34. 1. triangle ABC is the half of the parallelogram GBСА, 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 *7 Ax. equald; 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 31.1. the point A drawa AE parallel to BC, and join EC: The triangle ABC is equal to the trian- sible: Therefore AE is not parallel to BC. In the same manner, it can be demonstrated, 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, 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 rallel to BC: For, if it is not, through A drawa AG parallel to BF, and join GF: The triangle ABC is equal to the triangle GEF, because they are upon equal 33.1. 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 but AD: AD is therefore parallel to BF. Wherefore equal triangles, &c. Q.E.D. PROP. XLI. THEOR. Ir a parallelogram and triangle be upon the same base, and between the same parallels; the parallelogram shall be double of the triangle. Воок 1. Let the parallelogram ABCD and the triangle EBC be upon the same base BC, and between the same parallels BC, AE; the parallelogram ABCD A is double of the triangle EBC. DE Join AC; then the triangle ABC 37. 1. is equal to the triangle EBC, because they are upon the same base BC, and between the same parallels RC, AE. But the parallelogram 34. 1. ABCD is doubleb of the triangle ABC, because the diameter AC divides it into two equal parts; wherefore ABCD is also double of the triangle EBC. Therefore, if a parallelogram, &c. Q. E. D. 10 1. B PROP. XLII. PROB. 1 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 the angle CEF equal to D; and G 23. 1. straight line EC make $31. 1. through A draw AG parallel to EC, and through C draw CG parallel to EF: Therefore FECG is a parallelogram: And because BE is equal to EC, the triangle 493.1. ABE is likewise equald to the triangle AEC, since they are upon equal bases BE, EC, and between the same parallels BC, AG; therefore the triangle ABC is double of the triangle AEC. And the parallelogram FECG is likewise • 41. 1. double of 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; D B E C : wherefore there has been described a parallelogram FECG Book I. equal to a given triangle ABC, having one of its angles, CEF equal to the given angle D. Which was to be done. 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 E AC, and EH, FG, the paral- ment KD, Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equala to the triangle ADC: And, 34. 1. 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. PROв. 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. |