PROPOSITION XXXVIII. THEOREM. 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. Then the triangle ABC shall be equal to the triangle DEF. Produce AD both ways to the points G, H; and between the same parallels BF, GH. And because the diameter AB bisects the parallelogram GBCA, therefore the triangle ABC is the half of the parallelogram GBCA; (I. 34.) also, because the diameter DF bisects the parallelogram DEFH, therefore the triangle DEF is the half of the parallelogram DEFH; but the halves of equal things are equal; (ax. 7.) therefore the triangle ABC is equal to the triangle DEF. PROPOSITION XXXIX. THEOREM. 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. Then the triangles ABC, DBC shall be between the same parallels. Join AD; then AD shall be parallel to BC. if possible, through the point A, draw AE parallel to BC, (1. 31.) meeting BD, or BD produced, in E, and join EC. Then the triangle ABC is equal to the triangle EBC, (1. 37.) and between the same parallels BC, AE: but the triangle ABC is equal to the triangle DBC; (hyp.) the greater triangle equal to the less, which is impossible: In the same manner it can be demonstrated, that no other line drawn from A but AD is parallel to BC; Wherefore, equal triangles upon, &c. Q. E. D. PROPOSITION XL. THEOREM. 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 towards the same parts. Then they shall be between the same parallels. Join AD; then AD shall be parallel to BF. if possible, through A draw AG parallel to BF, (1. 31.) meeting ED, or ED produced in G, and join G.F. Then the triangle ABC is equal to the triangle GEF, (1. 38.) 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; (hyp.) therefore the triangle DEF is equal to the triangle GEF, (ax. 1.) the greater triangle equal 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 line drawn from A parallel to it but AD; AD is therefore parallel to BF. Wherefore, equal triangles upon, &c. Q. E.D. PROPOSITION XLI. THEOREM. If a parallelogram and a triangle be upon the same base, and between the same parallels; the parallelogram shall be double of the triangle. Let the parallelogram ABCD, and the triangle EBC be upon the same base BC, and between the same parallels BC, AE. Then the parallelogram ABCD shall be double of the triangle EBC. Then the triangle ABC is equal to the triangle EBC, (1. 37.) because they are upon the same base BC, and between the same parallels BC, AE. But the parallelogram ABCD is double of the triangle ABC, because the diameter AC bisects it; (1. 34.) wherefore ABCD is also double of the triangle EBC. Therefore, if a parallelogram and a triangle, &c. Q.E.D. 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. Bisect BC i E, (1. 10.) and join AE; at the point E in the straight line EC, make the angle CEF equal to the angle D; (1. 23.) through C draw CG parallel to EF, and through a draw AFG parallel to BC, (1. 31.) meeting EF in F, and CG in G. Then the figure CEFG is a parallelogram. (def. A.) And because the triangles ABE, AEC are on the equal bases BE, EC, and between the same parallels BC, AG; they are therefore equal to one another; (1. 38.) and the triangle ABC is double of the triangle AEC; but the parallelogram FECG is double of the triangle AEC, (1. 41.) because they are upon the same base EC, and between the same parallels EC, AG; therefore the parallelogram FECG is equal to the triangle ABC, (ax.6.) and it has one of its angles CEF equal to the given angle D. Wherefore, a parallelogram FECG has been described equal to the given triangle ABC, and having one of its angles CEF equal to the given angle D. Q.E.F. PROPOSITION XLIII. THEOREM. 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: and EH,GF the parallelograms about AC, that is, through which AC passes: also BK, KD the other parallelograms which make up the whole figure ABCD, which are therefore called the complements. Then the complement BK shall be equal to the complement KD. Because ABCD is a parallelogram, and AC its diameter, therefore the triangle ABC is equal to the triangle ADC. (1. 34.) Again, because EKHA is a parallelogram, and AK its diameter, therefore the triangle AEK is equal to the triangle AHK; (1. 34.) and for the same reason, the triangle KGCis equal to the triangle KFC. Wherefore the two triangles AEK, KGC are equal to the two triangles AHK, KFC, (ax. 2.) but the whole triangle ABC is equal to the whole triangle ADC; therefore the remaining complement BK is equal to the remaining complement KD. (ax. 3.) Wherefore the complements, &c. Q. E.D. 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 the angle Ď. Make the parallelogram BEFG equal to the triangle C, through A draw AH parallel to BG or EF, (1. 31.) and join HB. Then because the straight line HF falls upon the parallels AH, EF, therefore the angles AHF, HFE are together equal to two right angles; (1. 29.) wherefore the angles BHF, HFE are less than two right angles: but straight lines which with another straight line, make the two interior angles upon the same side less than two right angles, do meet if produced far enough: (ax. 12.) therefore HB, FE shall meet if produced; let them be produced and meet in K, through K draw KL parallel to EA or FH, and produce HA, GB to meet KL in the points L, M. Then HİKF is a parallelogram, of which the diameter is HK; and AG, ME, are the parallelograms about HK; therefore the complement LB is equal to the complement BF; (1.43.) but the complement BF is equal to the triangle C; (constr.) wherefore LB is equal to the triangle C. And because the angle GBE is equal to the angle ABM, (I. 15.) and likewise to the angle D; (constr.) therefore the angle ABM is equal to the angle D. (ax. 1.) Therefore to the given straight line AB, the parallelogram LB has been applied, equal to the triangle C, and having the angle ABM equal to the given angle D. Q.E.F. PROPOSITION XLV. PROBLEM. 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 that shall be equal to the figure ABCD, and having an angle equal to the given angle E. Describe the parallelogram FH equal to the triangle ADB, and having the angle FKH equal to the angle E; (1. 42.) to the straight line GH, apply the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E. (1. 44.) Then the figure FKML shall be the parallelogram required. Because each of the angles FKH, GHM, is equal to the angle E, therefore the angle FKH is equal to the angle GHM; add to each of these equals the angle KHG; therefore the angles FKH, KHG are equal to the angles KHG, GHM; but FKH, KHG are equal to two right angles; (1. 29.) therefore also KHG, GĦM are equal to two right angles; and because at the point H, in the straight line GH, the two straight lines KH, HM, upon the opposite sides of it, make the adjacent angles KHG, GHM equal to two right angles, therefore HK is in the same straight line with HM. (1.14.) therefore the angle MHG is equal to the alternate angle HGF; (1. 29.) add to each of these equals the angle HGL; therefore the angles MHG, HGL are equal to the angles HGF, HGL; but the angles MHG, HGL are equal to two right angles; (1. 29.) therefore also the angles HGF, HGL are equal to two right angles, and therefore FG is in the same straight line with GL. (I. 14.) |