o Nom. 5. A parallelogram of which only the opposite sides are equal, and the angles are not right angles, is called a rhomboid. NOM. 6. All other quadrilateral rectilinear figures beside these, are called trapeziums. Nom. 7. The perpendicular drawn from one of the angular points of a triangle to the side opposite or to its prolongation, is called the altitude of the triangle in that direction; and the side to which or to its prolongation such perpendicular is drawn, is (with relation to that altitude) called the base. Nom. 8. The perpendicular drawn from any point in one side of a parallelogram to the side opposite or to its prolongation, *1.33.Cor.2. [which perpendicular has been shown to be* of the same length from whatever point in the side it may be drawn], is called the altitude of the parallelogram in that direction; and the side to which or to its prolongation such perpendicular is drawn, is (with relation to that altitude) called the base. Nom. 9. When two or more triangles or parallelograms are placed between the same parallels, (which is only another way +1.33.Cor.2. of declaring thatt their altitudes are equal]; the sides which are placed in the same straight line in one of the parallels for this purpose, are in like manner called their bases. SCHOLIUM.—Though the terms altitude and base are borrowed from the relation between a building and its foundation, they are not confined to objects in any particular position with respect to the horizon ; any more than the term perpendicular, which originally meant the line determined by a plummet.' PROPOSITION XXXV. THEOREM.-Parallelograms upon the same base, and between the same parallels, are equal to one another. Let the parallelograms AC, BF, be upon the same base BC, and between the same parallels AF, BC. The parallelograms shall be equal to one another. First Case; if the parallelograms have two of A the sides opposite to the base BC, terminated in 11.33. the same point D ; each of the parallelograms is: *Interc. 1. double of the triangle BDC; and they are* there B C Cor. 10. fore equal to one another. R * Second Case ; if the side EF of the one A E D F parallelogram be terminated in some point within the side AD of the other. Because ABCD is a * I. 33. parallelogram, AD is* equal to BC; for the same B + Interc.1. • reason EF is equal to BC; wherefore AD ist equal to EF. From each of the equals AD, EF, take away Interc. I. ED; and the remainder AE ist equal to the remainder DF. Cor. 7. *). 33. AB also is* equal to DC, and BE to CF. Wherefore in the triangles BAE, CDF, the sides BA, AE are equal to the sides CD, DF respectively; and the third side BE is equal to the +1. 8. third side CF; therefore the triangles BAE, CDF aref equal to one another. Add to each the trapezium EBCD, and the sums INTERC. 1. will be f equal; that is, the parallelogram ABCD will be equal Cor. 4. to the parallelogram EBCF. Third Case ; if the side EF of the one А D E с Interc.1. AD and EF, add DE; and the sum AE is* equal to the sum DF. Cor. 4. Wherefore the triangles BAE, CDF have their sides respectively + 1.8. equal as before, and the triangles aret equal. From the trapezium ABCF take the triangle CDF, and from the same trapezium take INTERC. 1. the triangle BAE, and the remainders will bef equal; that is, the Cor. 7. parallelogram ABCD will be equal to the parallelogram EBCF. And by parity of reasoning, the like may be proved of all other parallelograms under the same conditions. Wherefore, universally, parallelograms upon the same base, &c. Which was to be de monstrated. See Note. COR. Every parallelogram is equal to the rectangle made by any one of its sides, and its altitude as measured to such side or to its prolongation. For such parallelogram and rectangle are parallelograms upon the same base and between the same parallels. B A B F * PROPOSITION XXXVI. same parallels, are equal to one another. DE H upon equal bases BC, FG, and between the same parallels AH, BG. The parallelogram ABCD is equal to EFGH. Join BE, CH. * Hyp. Because BC is* equal to FG, and FG tot EH, BC ist equal + J. 33. to EH. And they are* parallel, and joined towards the same IINTERC. 1. Cor. 1. parts by the straight lines BE, CH; therefore BE, CH aref equal Hyp. and parallel, and EBCH is af parallelogram. And because +1. 34. 1. 32 bis. EBCH, ABCD are parallelograms upon the same base BC, and Nom. between the same parallels AH, BG, they are* equal to one • I. 35. another. And because EBCH, EFGH are parallelograms upon the same base EH and between the same parallels, they are equal to one another. Therefore the parallelograms ABCD, + INTERC. 1. EFGH (being each equal to EBCH) aret equal to one another. [In the same way might the parallelograms have been shown to be equal, if the points E, F, in which are terminated the sides of the parallelogram EFGH, had been situate within the points D and C, in which are terminated the sides of the other parallelogramı; or if one had been situate within, and one without; or if one, or both, had coincided with the point in which is terminated the side of the other parallelogram.] And by parity of reasoning, the like may be proved of all other parallelograms under the same conditions. Wherefore, universally, parallelograms upon equal bases, &c. Which was to be demonstrated. PROPOSITION XXXVII. THEOREM.-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 paral E AD F lels AD, BC. The triangle ABC is equal Interc.12. to the triangle DBC. Cor. 6. Prolong# AD both ways, and take* AE, B *1.3. 11. 34. DF, each equal to BC; and join BE, CF. Because AE, CB are Constr. equal* and parallelt, BE and CA are equal and parallel ; and + Hyp. BCAE is* a parallelogram. For the same reason BCFD is a *1. 32 bis. parallelogram. And because BCAE, BCFD are parallelograms Nom. upon the same base BC and between the same parallels BC, EF, + 1. 35. they aret equal to one another. But the triangle ABC is the 11. 33. half of the parallelogram BCAE, because the diagonal AB I bisects it; and for the same reason the triangle DBC is the half of the parallelogram BCFD; and the halves of equal magnitudes INTERC, 1. are* equal ; therefore the triangle ABC is equal to the triangle Cor. 12. DBC. And by parity of reasoning, the like may be proved of all other triangles under the same conditions. Wherefore, universally, triangles upon the same base, &c. Which was to be demonstrated. PROPOSITION XXXVIII. B THEOREM.-Triangles upon equal bases, and between the same parallels, are equal to one another. Let the triangles ABC, DEF be upon H equal bases BC, EF, and between the same parallels AD, BF. The triangle ABC is equal to the triangle DEF. СЕ +INTERC.12. Prolongt AD both ways, and take Cor. 6. AG equal to CB, and DH to EF ; and join BG, FH. | 1. 3. Constr. Because AG and CB are equal* and parallelt, BG and CAF + Hyp. are equal and parallel, and BCAG is* a parallelogram. For the 11. 34. I. 32 bis, same reason EFHD is a parallelogram. And because the paralNom. lelograms BCAG, EFHD are upon equalt bases BC, EF, and + Hyp. between the same parallels BF, GH, they aref equal to one another. But the triangle ABC is the half of the parallelogram I. 33. BCAG, because the diagonal AB* bisects it; and for the same reason the triangle DEF is the half of the parallelogram EFHD; Interc. l. and the halves of equal magnitudes aret equal ; therefore the Cor. 12. triangle ABC is equal to the triangle DEF. [In the same way might the triangles have been shown to be equal, if the point E in which is terminated a side of the triangle DEF, had been situate within the base BC of the other triangle, cr had coincided with the point C; or if, in addition to this, the #1.36. B point D had been on the other side of the point A, or had coincided with it.] And by parity of reasoning, the like may be proved of all other triangles under the same conditions. Wherefore, universally, triangles upon equal bases, &c. Which was to be demonstrated. PROPOSITION XXXIX. side of it, are between the same parallels. E Join AD; the straight lines AD and BC For if the angles DAB, ABC are not together equal to two right angles, let it be assumed that they are greater; and 1.28. from the point A draw* AE parallel to BC. Because +1.29.Cor.1. AE, BC are parallel, the angles EAB, ABC must bet together equal to two right angles; and because the angles DAB, ABC are supposed to be together greater than two right angles, INTERC. 1. they must be together greater than the angles EAB, ABC; and taking away the common angle ABC, the remaining angle "Interc. 1. DAB must be* greater than the remaining angle EAB. Cor. 8. Wherefore the straight line AE must lie within the straight +1.13.Cor.5. line AD, and being prolonged mustt meet BD. Let it meet it in F; and join FC. Because the triangles ABC, BCF are upon the same base BC and between the same parallels BC 11. 37. and AF, the triangle BCF must be f equal to the triangle *Hyp: ABC. But the triangle BCD is* equal to the triangle ABC; +INTERC. 1. therefore the triangle BCD must be equal to the triangle BCF, the 1.Nom.26. greater to the less, which is impossible. The assumption, there fore, which involves this impossible consequence, cannot be true; or DAB, ABC are not together greater than two right angles. And in the same way may be shown that ADC, DCB cannot be together greater than two right angles. Also they cannot be less ; for because the interior angles of the quadri*1.32.Cor.1. lateral figure ABCD are* together equal to four right angles, is ADC, DCB are together less than two right angles, the remaining angles DAB, ABC must be together greater than two right Cor, 2. * I |