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BOOK XI.

a def. 4.

b def. 3.

C 17. I.

IF

PROP. XXXVIII. THEOR.

F a plain be perpendicular to a plain, and a line be drawn from a point in one of the plains perpendicular to the other plain, that perpendicular shall fall in the common fection of the plain.

Let the plain CD be perpendicular to AB; and, in the plain CD, any point E be taken, from which let fall the perpendicu lar EG; then EG fhall be perpendicular to the common fection AD.

For, if not, let it fall without the common fection, as EF, and draw FG in the plain EGF, perpendicular to AD; then, becaufe 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 alfo 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 muft fall on it. Wherefore, &c.

a 20. I.

I.

C 14. I.

PRO P. XXXIX. THEOR.

IF the fides of the pofite plains of a folid parallelopipedon be divided into two equal parts, and plains be drawn thro' their common fections, the common fection of these plains, and the diameter of the folid parallelopipedon fball bifect each other.

Let the fides of CF, AH, the oppofite plains of the folid parallelopipedon AF, be each bifected in the points K, L, M, N, X, O,P,R; let the plains KN, XR, be drawn through the fections; and let YS be the common fection of the plains, and DG the diameter of the folid; then YS, DG, will bifect each other.

For, join DY, YE, BS, SG; then, because DX is parallel to OE, the alternate angles DXY, YOE, are equal 2; and, becaufe DX is equal to OE, and YX to YO, and they contain equal angles, the bafe 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 OYE, and OEY to XDY; therefore DYE is a right line. For the fame reason, BSG is a right line, and BS equal to SG.

e 9. £ 2.

Then, becaufe CA is equal and parallel to DB, and to EG, 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 ; and, fince DE is parallel to GB, the angle EDT is equal to BGT; but the angle DTY is equal to the a 29. 1. angle GTS; therefore DTY, GTS, are two triangles, having g 15. 1. the angles YDT, DTY, equal to the two angles SGT, GTS, each to each, and the fide YD equal to G8; therefore DT is equal to TG, and YT to TS". Wherefore, &c.

h 26. I.

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IF F two triangular prisms of equal altitudes, the base of one of which is a parallalelogram, and the other a triangle, and, if the parallelogram be double the triangle, the prifms are equal to each other.

Let ABCDEF, GHKLMN, be two prifms of equal altitude; let the parallelogram AF be the bafe 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 folids 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 ; but the half of equal things are b 31. equal; therefore the prifm GHKLMN is equal to the prifm ABCDEF; for each is half the folids GO, AX. Wherefore, c 28. &c.

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a 6. 6.

C 31. 3.

S

PROP. I. THE OR.

IMILAR polygons infcribed in circles, are to one another as the fquares of the diameters of the circles.

Let ABCDE, FGHKL, be circles, in which are infcribed the fimilar polygons ABCDE, FGHKL; let BM, GN, be the diameters of the circles; then the polygon ABCDE is to FGHKL as BM fquare is to GN fquare.

For, Join BE, AM, GL, FN; then, because the polygons are fimilar, the angles BAE, GFL, are equal; and BA is to AE as GF to FL; wherefore the triangles ABE, FGL, are equiangular; 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, and FLG to FNG: therefore the angle AMB is equal to FNG, the angle BAM to GFN, and the remaining angle ABM to FGN: Therefore the triangles AMB, GFN, are equi 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 the duplicate ratio of AB to GF f; 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 plicate ratio of BM to GN. Wherefore, &c.

d 4. 6.

c 19. 6.

f

20. 6.

8 22 5.

D

2.

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