angle A B C is the one half of the parallelogram e BCA; since the diameter A B cuts into halves, and the triangle DBC the one half of the parallelogram D BCF; because [by prop. 34.] the diameter Dc cuts it into halves. But the halves of equal things are themselves equal; therefore the triangle A C is equal to the triangle D BC. Therefore triangles constituted upon the same base, and between the same parallels, are the one equal to the other. Which was to be demonstrated, PROP. XXXVIII. THEOR. Triangles constituted upon equal bases, and between the same parallels, are equal to one another. Let the triangles A BC, DEF be constituted upon the equal bases B C, EF, and between the same parallels BF, AD: I say, the triangle ABC is equal to the triangle DE F. For continue out AD A D H both ways to the points G, H, and through B (by prop. 31.) draw BG parallel to c A, but through F draw FH parallel to DE. Then GBC A, D E F H are both parallelograms. B c E F But [by prop. 36.) GBCA is equal to DEFH, for they stand upon equal bases BC, E F, between the same parallels B F, G H. But the triangle a b c is one half of the parallelogram GBCA, because [by prop. 34.] the diameter A B cuts it in halves; and the triangle F E D is the one half of the parallelogram D E Fh; for the diameter Dr cuts it into halves. But the halves of equal things are themselves equal. Therefore the triangle A B C is equal to the triangle D E F. Wherefore triangles constituted upon equal bases, and between the fame parallels, are equal to one another. Which was to be demonstrated. PROP, PROP. XXXIX. THEOR. Equal triangles, constituted towards the same parts, upon the same base, are between the same parallels. E Let the equal triangles A B C, DBC be constituted upon the fane base b c towards the same parts: I say, they are between the fame parallels. For draw AD: I fay, this is parallel to Bc. For if not, through the point a [by prop. 31.) draw the right line A e parallel to the right line B c, and draw EC. Then [by prop. 37.] the triangle A B C is equal to the D triangle E BC: for they are both upon the same base B c, and between the same parallels b C, A E. But [by supposition] the triangle ABC is equal to the triangle D BC. Therefore also the triangle DBC is equal to the triangle E BC, the greater B C equal to the less; which is impossible : therefore a e is not parallel to Bc. After the same manner we demonstrate, that no other line except A D is parallel to BC. Therefore A D is parallel to BC. Wherefore equal triangles constituted towards the same parts, upon the same base, are between the same parallels. Which was to be demonstrated. PROP, XL. THEOR. Equal triangles constituted towards the same parts, upon equal bases, are between the same parallels. Let the equal triangles A B C, CdE be constituted upon equal bases BC, CE, towards the fame part: I say, they are also between the same parallels. For draw AD: I say, Ad is parallel to B E. For if it be not, through a draw ? A parallel to Be, and draw F E. Then [by prop. 38.] the triangle A B C is equal to the triangle FCE; for they are constituted upon equal bases BC, CE, and are between the same parallels B E, A F, But the triangle A B C is equal to the triangle DCE. And fo so the triangle Dc E shall be equal A B C rallel to BE. Wherefore ad will be parallel to B e. Therefore equal triangles constituted upon equal bases, towards the fame parts, are between the same parallels. Which was to be demonstrated. PROP. XLI. THEOR. If a parallelogram and a triangle have the same base, and are between the same parallels , the parcllelogram will be double to the triangle. For let the parallelogram ABC), and the triangle E B C have the same bale Bc, and be between the same parallels B C, AE: I say, the parallelogram ABCD will be double to the triangle B E C. For draw A C. Then the triangle A B C [hy prop. 37.] is equal to the triangle EBC; for it is constituted A E upon the same base BC, and between the same parallels B C, A E. But the parallelogram ABCD is double to the triangle A BC; for the diameter AC [by prop. 34.] cuts it into halves: wherefore the parallelogram ABCD в с will be double to the triangle E B C. If therefore a parallelogram and a triangle have the same base, and are between the fame parallels, the parallelogram will be double to the triangle, Which was to be demonstrated. PROP. XLII. PROBL. To constitute a parallelogram equal to a given triangle, and having one of its angles equal to a given right-lined angle. Let the given triangle be ABC, and the given right. lined angle d; it is required to constitute a parallelogram equal to the triangle A B C, having one of its angles equal to the given right-lined angle D. Biseet B C [by prop. 10.] in E, and draw A E; and with the right line E C, at the point E А. therein [by prop. 23-] make the angle ce F equal to the angle D. And [by prop. 31.] through Ag draw A G parallel to Ec; and D through c, draw c G parallel to FE: then is F ECG the paralB lelogram required. For because B e is equal to E C, the triangle AB E [by prop: 38.] will be equal to the triangle A EC; for they stand upon equal bases B E, E C, and are between the same parallels B C, AG. Therefore the triangle A B C is double to the triangle A E C. But [by prop. 41.] the parallelogram FECG is double to the triangle A EC; for it has the same base, and is between the same parallels: Therefore [by ax. 6.] the parallelogram FE C G is double to the triangle ABC, and has the angle C E F equal to the given angle D. • Therefore the parallelogram FECG is made equal to the triangle A B C, having the angle c E F equal to the given angle D. Which was to be done. W Ас; PROP. XLIII. THEOR. In every parallelogram, the complements of those parallelograms, which are about a diameter, are equal to one another ". Let there be a parallelogram AB CD, whose diameter is and let the parallelograms E H, FG be about the fame; then BK, KD are called the complements of them: I say, the complement B K is equal to the complement Ks. For becaufe A B CD is a parallelogram, and a c is its diameter, the triangle AB C [by prop. 34.] is equal to the triangle AD C. Again, because E K H A is a parallelogram, whose diameter is Ak, the triangle A E K will be equal to the triangle AHK. For the same reason the triangle KFC is equal to the triangle KGC, Therefore fince the triangle АЕК KFC. A EK is equal to the trian- B G с gle A HK, and the triangle KFC to the triangle KGC; the triangle A e K, together with the triangle kg c, is K R Ε E equal to the triangle AHK, together with the triangle A H D But the whole triangle ABC is also equal to the whole triangle ADC: Therefore the complement B K remaining, is [by ax. 3.) equal to the complement K D remaining, Therefore in every parallelogram the complements of those parallelograms which are about a diameter, are equal to one another. Which was to be demonstrated. It may, perhaps, be as well to express this proposition thus: If two right lines be drawn through a point in the diagonal of any parallelogram parallel to the fides, thereby making four lesle:r parallelograms within the greater, those two of the four lesser parallelograms, which fall without that diagonal, will be equal to one another, PROP. XLIV. PROBL. To apply * a parallelogram to a given right line equal to a given triangle, and having one angle equal to a given right-lined angle. Let the given right line be A B, the given triangle c, and the given right-lined angle D: It is required to apply a parallelograrn to the given right line A B, equal to the given triangle c, having one angle equal to D. Constitute [by prop. 42.] the parallelogram B EFG equal to the triangle c, F E K having one angle E BG equal to the angle D; and put B E in a right D line with a B. Produce M FG to H, and through a с B draw [by prop: 31.] AH parallel to BG, or E F. H Join H B. Then because d L That is, to make fuch a para"elegram, thit a given right line shall be one of its fid:s; one of its angles shall ie equal to a given right-lined an le, and the sarallelogram mall be equal to given triangle. * |