3. is in the fame plane CD; [by E is impoffible: Wherefore a perpendicular drawn from the point to the plane A B does not fall out of the right line A D: Therefore it falls in A D. Wherefore if one plane be at right angles to another, a right line drawn from any point in one plane perpendicular to the other, will fall in the common section of the planes. Which was to be demonftrated. PROP. XXXIX THEOR. If the fides of the oppofite planes of a folid parallelepipedon be bifected, and planes be drawn thro' the fections; the common fection of the planes, and the diameter of the folid parallelepidon, will mutually bifect each other.. D C K E For let the fides of the oppofite planes C F, A H of the parallelepipedon DF be bifected in the points K, L, M, N, X, P, 0, Q, and through the sections draw the planes K N, and let y s be the common fection of the planes, and DG the diameter of the folid parallelepipedon: I fay Y-s, DG do mutually cut one another into two equal parts, that is, Y T is equal to T-s, and D T equal to TG. For join D Y, YE, BS, S G. Then because D X is parallel to OE; the alternate angles DXY, YOE [by 29. 1.] are equal to one another; and because DX is equal to o E, and x Y to Yo, and they contain equal angles; the bafe DY B T M H P S N G [by Book XI. by 4. 1.] will be equal to the bafe YE and the triangle DXY is equal to the triangle y o E, and the reft of the angles equal to the reft of the angles: Therefore the an gle x Y D is equal to the angle o Y E; and fo [by 14. 1.] DY E is one right line. By the same reafon BSG is one right line, and BS is equal to`s G. And because by [34. 1.} CA is equal and parallel to D B, and C A is alfo equal and parallel to EG; DB will be equal to E G, and [by 30. 1.] parallel to it. But the right lines DE, BG join them: Therefore [by 33. 1] D E is parallel to BG, and in each of them fome points D, Y, G, s, are taken, and DG, YS are joined: Therefore [by 7. 11.] DG, Ys are both in one plane. Wherefore becaufe DE is parallel to B G, the angle EDT [by 29. 1.] will be equal to the angle B GT; for they are alternate angles. But the angle DTY [by 15. 1.] is equal to the angle GTS: Therefore there are two triangles D T Y, GT s, having two angles of the one equal to two angles of the other, each to each, and one fide of the one equal to one fide of the other, oppofite to one of the equal angles, viz. D Y equal to G 8; for they are the halves of DE, BG: Therefore [by 26.1.] the reft of the fides will he equal to the reft of the fides; and fo DT is equal to TG, and Y T equal to TS. If therefore in a folid parallelepipedon, &c. Which was to be demonftrated, PROP. XL. THEOR. If there be two prifms of the fame altitude, and the bafe of one of them be a parallelogram, and that of the other a triangle, and that parallelogram be double to the triangle; the prifms will be equal. Let A B C D E F G H K L M N be prisms of the fame altitude. Let the bafe of one of them be the paralleJogram AF, and that of the other the triangle G H K, and let the parallelogram AF be double to the triangle GHK: I fay the prifm A B C D E F is equal to the prism GH KL MN. For But gram AF will be equal to the parallelogram HK. [by 31. 11.] parallelepipedons ftanding upon equal bases, and having the fame altitude, are equal to one another : Therefore the folid Ax is equal to the folid GO. But the prifm A B C D E F is one half the solid Ax, and the prism 6 HK L M N one half the folid Go: Therefore the prifm A B C D E F is equal to the prifm G H K L M. If therefore two prifms, &c. Which was to be demonftrated. EUCLID's EUCLID's ELEMENTS. BOOK XII. LEM MA. If there be two unequal magnitudes propofed, and from the greater be taken away more than one half of it, and again from the remainder be taken away more than its half, and this be always done; there will at last remain fome magnitude that will be less than the leaft of the two given magnitudes. L ET there be two unequal magnitudes A B, C, whereof A B is the greater: Ifay if from A B be taken away more than one half, and from the remainder be again taken away more than its half, and this be always done, there will at last remain fome magnitude leffer than c. For if c be multiplied, it will fome time or other become greater than the magnitude A B. Let this be done, and let D E be a multiple of c greater than A B. Divide DE into the parts DF, FG, GE, each equal to c, and from A B take away BH more than its half, and again from AH take away HK more than one half of it, and do thus continually until there are the fame number of divifions in A B as in DE.. Let then these divifions A K, KH, HB, be equal in number to the divifions G F, FG, GE. And because D E is greater than A B, and there is taken away from DE lefs than one half of it, viz. EG, and from A B more than its half, viz. B H; the remainder G D will be greater than the remainder a H. Again, becaufe 1 H GD is greater than HA, and from GD After the fame manner we demonftrate the thing, when the halves of the given magnitudes are taken away. Otherwife. A D Let AB, C be two given magnitudes, and let c be the leaft of them. Then because c is the leaft, if it be multiplied, it will at length be greater than the magnitude A B. Let this be done, and let FM be the magnitude, which divide into the parts MH, HG, GF, each equal to c; and take away BE from AB, greater than one half of it; and from EA, take E D greater than one half of it; and doing thus continually until the number of divifions in AB are equal to the number of divifions in FM. Let these be BE, ED, DA; and KL, LN, NX, each equal to DA; and do this continually till the divifions of xx be equal in number to the divifions in FM. A Then because BE is greater than one half AB, BE will be greater Dthan EA; therefore BE is much greater than DA, but x N is equal E to DA: wherefore B E is greater than X N. Again, because ED is greater than one half EA; ED will be greater than DA. But NL is equal to DA: wherefore ED is greater than F BCM K But NL: therefore the whole DB is greater than XL: But LK is equal to DA. Therefore the whole A B will be greater than the whole x K. MF is greater than BA: Wherefore MF is much greater than x K And because X N, NL, L-K are equal to one another, |