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equilateral one, and equiangular to the aforesaid Parallelepipedon; which was to be demonstrated.

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If four Right Lines be proportional, the folid Parallelepipedons fimilar, and in like manner defcribed from them, fhall be proportional. And if the folid Parallelepipedons, being fimilar, and alike defcribed, be proportional, then the Right Lines they are defcribed from, hall be proportional.

LET the four Right Lines AB, CD, EF, GH,

be proportional, viz. let AB be to CD, as EF is to GH, and let the fimilar and alike fituate Parallelepipedons KA, LC, ME, NG, be defcribed from them. I fay, KÁ is to LC, as ME is to NG.

For because the folid Parallelepipedon K A is fimilar to LC, therefore KA to LC fhall have a Pro- *33 of this portion triplicate of that which AB has to CD. For the fame Reason, the Solid ME to NG will have a triplicate Proportion of that which E F has to GH. But AB is to CD, as EF is to GH. Therefore AK is to LC, as ME is to NG. And if the Solid AK be to the Solid LC, as the Solid M E is to the Solid NG. I fay, as the Right Line AB is to the Right Line CD, fo is the Right Line EF to the Right Line GH. For because AK to LC has a Proportion triplicate of +33 of this. that which AB has to CD, and ME to NG has a Proportion triplicate of that which EF has to GH, and fince AK is to LC, as ME is to NG; it shall be as AB is to CD, fo is EF to GH. Therefore, if four Right Lines be proportional, the folid Parallelepi pedons fimilar, and in like manner defcribed from them, hall be proportional. And if the folid Parallelepipedons, being fimilar and alike defcribed, be proportional, then the Right Lines they are defcribed from, fhall be proportional; which was to be demonstrated.

PRO

this.

this.

PROPOSITION XXXVIII.

THEOREM.

If a Plane be perpendicular to a Plane, and a Line be drawn from a Point in one of the Planes perpendicular to the other Plane, that Perpendicular Jhall fall in the common Section of the Planes.

LET the Plane CD be perpendicular to the Plane AB, let their common Section be AD, and let fome Point Ebe taken in the Plane CD. I fay, a Perpendicular, drawn from the Point E to the Plane AB, falls on AD.

For if it does not, let it fall without the fame, as EF meeting the Plane AB in the Point F, and from the Point Flet FG be drawn in the Plane A B perpendiDef. 4 of cular to AD; this fhall be * perpendicular to the Plane CD; and join EG. Then because F G is perpendicular to the Plane CD, and the Right Line EG in the Plane of CD touches it: The Angle FGE fhall + Def. 3.of beta Right Angle. But EF is alfo at Right Angles to the Plane Angle AB; therefore the Angle EFG is a Right Angle. And fo two Angles of the Triangle EFG, are equal to two Right Angles; which is abfurd. Wherefore if a Right Line, drawn from the Point E perpendicular to the Plane AB, does not fall without the Right Line AD: And fo it muftineceffarily fall on it. Therefore, if a Plane be perpendicular to a Plane, and a Line be drawn from a Point in one of the Planes perpendicular to the other Plane, that Perpendicular fhall fall in the common Section of the Planes; which was to be demonftrated.

17. 1.

PRO

PROPOSITION XXXIX.

THEOREM.

If the Sides of the oppofite Planes of a folid Parallelepi pedon be divided into two equal Parts, and Planes be drawn thro' their Sections; the common Section of the Planes, and the Diameter, the folid Parallelepipedon, fball divide eachother into two equal Parts.

LET the Sides of CF, AH, the oppofite Planes of the folid Parallelepipedon A F, be cut in half in the Points K, L, M, N, X, O, P, R, and let the Planes KN, XR, be drawn thro' the the Sections: Alfo let YS be the common Section of the Planes, and DG the Diameter of the folid Parallelepipedon. I fay, YS, DG, bifect each other,that is, YT is equal to TS, and DT to TG.

*

* 29. I.

For join DY, YE, BS, SG. Then because DX is parallel to OE, the Alternate Angles DXY, YOE are equal to one another. And because DX is equal OE, and YX to YO, and they contain equal Angles, the Bafe DY fhall be equal to the Bafe YE; † 4. 1. and the Triangle DXY to the Triangle YOE, and the other Angles equal to the other Angles: Therefore the Angle XÝD is equal to the Angle OYE; and fo DYE is a Right Line. For the fame Rea- ‡ 14. 1. fon BSG is alfo a Right Line, and BS is equal to SG. Then becaufe CA is equal and parallel to DB, as also to EG, DB fhall be equal and parallel to EG; and the Right Lines DE, GB, join them: Therefore DE is parallel to BG, and D, Y, G, S, are Points * 33. 1. taken in each of them, and DG, YS, are joined. Therefore DG, YS, are in one Plane. And fince DE is parallel to BG; the Angle EDT fhall be equal to * the Angle BGT, for they are Alternate. But the Angle DTY, is equal to the Angle GTS. Therefore 15.1. DTY, GTS are two Triangles, having two Angles of the one equal to two Angles of the other, as likewife one Side of the one equal to one Side of the other, viz. the Side DY equal to the Side GS: For they are Halves of DE, BG: Therefore they shall have the other Sides of one, equal to the other Sides

+ 7 of this.

29.1.

of

14. I.

of the other; and fo DT is equal to TG, and YT to TS. Wherefore, if the Sides of the oppofite Planes of a folid Parallelepepidon be divided into two equal Parts, and Planes be drawn thro' their Sections; the common Section of the Planes, and the Diameter of the folid Parallelepipedon, fhall divide each other into two equal Parts; which was to be demonstrated.

PROPOSITION XL.

THEOREM.

If of two triangular Prifms, one standing on a Bafe, which a Parallelogram, and the other on a Triangle, if their Altitudes from thefe Bafes are equal, and the Parallelogram double to the Triangle'; then those Prifms are equal to each other. ·

LE

ET ABCDEF, GHKLMN be two Prisms of equal Altitude. The Bafe of one of which is the Parallelogram AF, and that of the other, the Triangle GHK, and let the Parallelogram AF be double to the Triangle GHK. I fay the Prism AB CDEF is equal to the Prifm GHKLMN.

For compleat the Solids AX, GO. Then because the Parallelogram A F is double to the Triangle GHK, and fince the Parallelogram HK is * double to the Triangle G H K, the, Parallelogram AF fhall be equal to the Parallelogram HK. But folid Parallelepipedons. that ftand upon equal Bases, and have the fame Altitude, are † equal to one another. Therefore the folid A X is equal to the Solid G O. #28 of this. But the Prifm ABCDEF is half the Solid AX, and the

t31 of this.

Prifm GHKLMN is half the Solid GO. There-
fore the Prifm ABCDEF is equal to the Prism GH
KLMN. Wherefore, if there be two Prifms having
equal Altitudes, the Bafe of one of which is a Parallelo-
gram, and that of the other a Triangle, and if the Pa-
rallelogram be double to the Triangle, the faid Prifms
fhall be equal to each other.

The END of the ELEVENTH BOOK.
EUCLID's

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