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PRO P. XXXVIII. THEOR.
a point in one of the plains perpendicular to the other plain,
fall in the common section of the plain.
a def. 4.
b def. 3.
Let the plain CD be perpendicular to AB; and, in the plain
For, if not, let it fall without the common section, as EF, and draw FG in the plain EGF, perpendicular to ADą; then, because FG is perpendicular to the plain CD, and the right line EG in the plain CD touches it, the angle FGE is a right angleb; but EF is also at right angles to the plain AB; therefore the angle EFG is a right angle: Therefore two angles in the triangle FGE are equal to the two right angles; which cannot be ; therefore the right line drawn from the point E perpendicular to AB, does not fall without the line AD ; therefore must fall on it. Wherefore, &c.
PRO P. XXXIX. THEOR.
IE F the sides of the posite plains of a solid parallelopipedon
be divided into two equal parts, and plains be drawn thro' their common sections, the common seEtion of these plains, and the diameter of the folid parallelopipedon shall bisect each other.
2 29. I.
Let the sides of CF, AH, the opposite plains of the solid paral. lelopipedon AF, be each bisected in the points K, L, M, N, X, O,P, R; let the plains KN, XR, be drawn through the sections ; and let YS.be the common section of the plains, and DG the diameter of the solid ; then YS, DG, will bisect each other.
For, join DY, YE, BS, SG ; then, because DX is parallel to OE, the alternate angles DXY, YOE, are equal“; and, because DX is equal to OE, and YX to YO, and they contain e. qual angles, the base DY is equal to YE, and the triangle DXY to YOE, the angles in the one equal to the angles of the other, each to each", viz. the angle XYD equal to the angle QYE, and OEY to XDY; therefore DYE is a right line For the same reason, BSG is a right line, and BS equal to SG.
$ 4. I.
C 14. I.
Then, because CA is equal and parallel to DB, and to EG4, DB Book XI. is parallel to EG, and DE to GB d, but D, E, G, B, Y, S, are points taken in each of them. Join DG, YS; then DG, d 34. 1. YS, are in one plain f; and, since DE is parallel to GB, the angle EDT is equal to BGT"; but the angle DTY is equal to the a 29. r. angle GTS 8; therefore DTY, GTS, are two triangles, having 8 15. 1. the angles YDT, DTY, equal to the two angles SGT, GTS, each to each, and the side YD equal to G8; therefore DT is equal to TG, and YT to TSh. Wherefore, &c.
h 26. 1.
which is a parallalelogram, and the other a triangle, and, if the parallelogram be double the triangle, the prisms are equal to each other.
Let ABCDEF, GHKLMN, be two prisms of equal altitude ; let the parallelogram AF be the base of the one, and the triangle GHK the base of the other; and, if AF be double GHK, the prifms are equal.
For, compleat the solids AX, GO; then, because AF is double GHK, and the parallelogram HK double the triangle GHK"; the parallelograms AF, HK, are equal; therefore the a 41. I. folids AX, GO, are equal b; but the half of equal things are b. 31. equal; therefore the prism GHKLMN is equal to the prism ABCDEF; for each is half the folids GO, AX Wherefore, c 28. &c.
IMILAR polygons inscribed in circles, are to one another as the squares of the diameters of the circles.
Let ABCDE, FGHKL, be circles, in which are inscribed the similar polygons ABCDE, FGHKL ; let BM, GN, be the diameters of the circles ; then the polygon ABCDE is to FGHKL as BM square is to GN square.
For, Join BE, AM, GL, FN ; then, because the polygons are similar, the angles BAE, GFL, are equal; and BA is to
AE as GF to IL; wherefore the triangles ABE, FGL, are e a 6. 6. quiangular ~; that is, the angle AEB equal to FLG, and the
fides about them proportional; but the angle AMB is equal to b 21. 3. AEB b, and FLG to FNG b: therefore the angle A MB is equal € 31. 3. to FNG, the angle BAM to GFN, and the remaining angle
ABM to FGN: Therefore the triangles AMB, GFN, are equi. d 4. 6. angular, and BM is to GN as BA is to GFd; but the triangle
ABM is to the triangle FGN in the duplicate ratio of AB to
GF, and the polygon ABCDE is to the polygon FGHKL in f 20. 6.
the duplicate ratio of AB to GFf; but the triangle ABM is to the triangle FGN in the duplicate ratio of BM to GN; there
fore the polygon ABCDE is to the polygon FGHKL in the du& 22. 5. plicate ratio of BM to GNS. Wherefore, &c.