B с B с ms Ax. 1. в с 1. If the sides AD, DF, of the zms ABCD, DBCF, opp. to BC the base, be terminated in the same point D [as in fig. 1], :: each double the ABDC, Prop. 33. the ABCD= Om DBCF. Ax. 6. 2. If the sides AD, EF, be not terminated in the same point (figs. 2 and 3), then : ABCD, EBCF are ŞAD = BC, Prop. 33. = whole or remain. DF. remain. AE And ::: EA, AB = FD, DC, ea. to ea. Prop. 33. ext. LFDC = int. _ EAB, Prop. 28. base EB = base FC, A EAB=AFDC. Prop. 4. Take the A EAB from the trapezium ABCF; and from the same trapezium take the a FDC; then the remainders are =; that is, O" ABCD=OM EBCF. Wherefore parallelograms, &c. CoR.— Triangles upon the same base, and between the same parallels, are equal to one another. For if the diameters AC, FB, be drawn (figs. 2 and 3), the As ABC, FBC, • Ax. 2, 3. Ax. 3. are the halves of the equal Om ABCD, EBCF, .. the As are equal to one another. Ax. 7. PROP. XXXV. THEOR. 36. 1 Eu. Parallelograms upon equal bases, and between the same parallels, are equal to one another. Let ABCD, EFGH, be BC, FG, and between the same ||s, AH, BG; then ABCD = OM EFGH. upon = bases Ax. 1. Join BE, CH, BC FG, Нур. EH = FG, Prop. 33. BC= EH; str. lines EB, HC, join the extremities of = and || str. lines, they are themselves = and || : that is, BE is = and || to CH, Prop. 32. EBCH is a EBCH= ABCD, , since they are on the same base BC and beProp. 34. tween the same is. EBCH, Def. m Ax.l. zms ... ABCD = OMEFGH. Wherefore parallelograms, &c. Cor. 1.-Triangles upon the equal bases and between the same parallels are equal to each other. For draw the diams. AC, EG; the As ABC, EFG, are the halves of the equal ABCD, EFGH, and are therefore equal Ax. 7. to cach other. Cor. 2.-If a parallelogram and a triangle be upon the same or equal bases, and between the same parallels, the parallelogram is double the triangle. For the Om ABCD is double the A ABC, or double the A EFG, which is its equal, by last Cor. PROP. XXXVI. THEOR. 39. 1 Eu. Equal triangles upon the same base, and upon the same side of it, are between the same parallels. Let the = As ABC, DBC, be on the same base BC, and on the same side of it; then the As will be between the same ||s. Cor. Ax. 1. i. e. For if not, Prop. 30. draw AE || BC, and join EC. I being on the same base BC, and beProp.34, Then A ABC = AEBC tween the same is BC, AE; but ABC=A DBC, less =greater, .. AD || BC. CoR.--In the same way it may be shown, PROP. XXXVII. THEOR. 43. 1 Eu. m The complements of the parallelograms which are about the diameter of any parallelo- Let ABCD be a ", AC its diam.; and comp. BEKG = comp. KĀDF. с B G A ABC = A ADC; AAEK Prop.33. A AHK + AAx. 2. .A AEK + AKGC KFC: but the whole 4ABC = whole A ADC; the remains comp. (remains comp. BEKG KHDF. { Ax. 3. Wherefore the complements, &c. PROP. XXXVIII. PROB. 46, 1 Eu. To describe a square upon a given straight line. Let AB be the given str. line; it is required to descr. a square upon AB. с D E А B |