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rallelepipedon made of them, is equal to the solid
viz. Let A be to B, as B is to C. I say, the Solid made of A, B, C, is equal to the equilateral Solid made of B, equiangular to that made on A, B, C.
Let E be a solid Angle contained under the three plane Angles DEG, GEF, FED; and make DE, GE, EF, each equal to B, and complete the solid
Parallelepipedon E K. Again, put LM equal to A, * 26 of this, and at the Point L, at the Right Line LM, make * a
folid Angle contained under the plane Angles NLX, XLM, MLN, equal to the solid Angle E; and make LN equal to B, and LX to C. Then because A is to B, as B is to C, and A is equal to LM, and B to LN, EF, EG, or ED, and C to LX; it shall be as LM is to EF, fo is GE to LX. And so the Sides about the equal Angles MLX, GEF, are recipro
cally proportional. Wherefore the Parallelogram + 14. 6. MX ist equal to the Parallelogram GF. And since
the two plane Angles GEF, XLM, are equal, and the Right Lines LN, ED, being equal, are erected at the angular Points containing equal Angles with the
Lines firit given, each to each; the Perpendiculars I Cor. 35. of drawn I from the Points ND, to the Planes drawn
thro' XLM, GEF, are equal one to another. Therefore the Solids LH, EK, have the fame Altitude; but
folid Parallelepipedons that have equal Bases, and the 31 of bis same Altitude, are equal to each other. Therefore
the Solid HL is equal to the Solid EK. But the Solid HL is that made of the three Right Lines A, B, C, and the Solid EK that made of the Right Line B. Therefore, if three Right Lines be proportional, the solid Parallelepipedon made of them, is equal to the solid Parallelepipedon made of the middle Line, if. it bé an
equilateral one, and equiangular to the aforesaid Parallelepipedon ; which was to be demonstrated.
THEOR E M. If four Right Lines be proportional, the solid Parallelepipedons fimilar, and in like manner described
from them, all be proportional. And if the Solid Parallelepipedons, being similar, and alike described, be proportional, then the Right Lines they are described from, hall be proportional. ET the four Right Lines AB, CD, EF, GH,
be proportional, viz, let AB be to CD, as EF is to GH, and let the similar and alike fituate Parallelepipedons KA, LC, ME, NG, be described from them. I fay, KA is to L C, as ME is to NG.
For because the solid Parallelepipedon KA is fimilar to LC, therefore K A to LC shall have * a Pro-* 33 of this. portion triplicate of that which AB has to CD. For the same Reason, the Solid ME to NG will have a triplicate Proportion of that which EF 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 ME is to the Solid N G. I say, as the Right Line AB is to the Right Line CD, so is the Right Line E F to the Right Line GH. For because AK to LC has a Proportion triplicate of t 33 of this. that which AB has to CD, and ME to N G has a Proportion triplicate of that which EF has to GH, and fince AK is to L C, as ME is to NG; it shall be as AB is to CD, so is E F to GH. Therefore, if four Right Lines be proportional, the solid Parallelepipedons fimilar, and in like Manner described from them, shall be proportional. And if the folid Parallelepipedons, being similar and alike described, be proportional, then the Right Lines they are described from fall be proportional; which was to be demonstrated.
Line bé drawn from a Point in one of the
ET the Plane CD be perpendicular to the Plane
, some Point E be taken in the Plane CD.' I say, a Perpendicular, drawn from the Point E to the Plane A B, falls on AD.
For if it does not, let it fall without the same, as 1 EF meeting the Plane A B in the Point F, and from
the Point F let F G be drawn in the Plane AB per
pendicular to AD; this shall be * perpendicular to of this.
the Plane CD; and join E G. Then because FG is perpendicular to the Plane CD, and the Right Line
EG in the Plane of CD touches it: The Angle Def. 3. of FGE shall be t a Right Angle. But EF is also bis.
at Right Angles to the Plane AB; therefore the Angle E F G is a Right Angle. And so two Angles
of the Triangle E F G are equal to two Right An17.
gles; which is I absurd. Wherefore if a Right Line, drawn from the Point E, perpendicular to the Plane AB, does not fall without the Right Line AD: And so it must necessarily 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 shall fall in the common Section of the Planes; which was to be demonstrated.
lelepipedon be divided into two equal Parts, and
the solid Parallelepipedon AF, be cut in half in
For join D Y, YE, BS, SG. Then because DX is parallel to O E, the alternate Angles DXY, YOE are * equal to one another. And because DX is equal * 29. 15 to OE, and Y X to YO, and they contain equal Angles, the Base DY shall be † equal to the Bass YE; + 4. I. and the Triangle DXY to the Triangle YOE, and the other Angles equal to the other Angles: Therefore the Angle XY D is equal to the Angle O YE; and fo DYĚ is I a Right Line. For the same Rea- † 14. 1. fon BSG is also a Right Line, and BS is equal to SG. Then because CA is equal and parallel to DB, as also to EG, DB shall 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, Y S are joined. Therefore DG, YS are t in one Plane. And since DE is + 7 of this, parallel to BG, the Angle EDT shall be * equal to * 29. 1. the Angle BGT, for they are alternate. But the Angle DTY, is I equal to the Angle GTS. Therefore [ 15. I. DTY, GTS are two Triangles, having two Angles of the one equal to two Angles of the other, as likewise one side of the one equal to one side of the other, viz. the Side D Y 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 of the other; and so DT is equal to TG, and YT to TS. Wherefore, if the Sides of the opposite Planes of a solid Parallelepipedon be divided into two equal Parts, and Planes be drawn thro their Sections; the common Sečtion of the Planes, and the Diameter of the solid Parallelepipedon shall divide each other into the equal Parts; which was to be demonstrated.
which is a Parallelogram, and the other on a
of equal Altitude. The Base 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 say, the Prism ABCDEF is equal to the Prism GHKLMN.
For complete the Solids AX, GO. Then because
the Parallelogram AF is double to the Triangle * 41. I. GHK, and since the Parallelogram HK is * dou
ble to the Triangle GHK, the Parallelogram AF shall be equal to the Parallelogram HK. But solid
Parallelepipedons, that stand, upon equal Bases, and † 31 ottis
. have the fame Altitude, are + equal to one another. Therefore the Solid AX is equal to the Solid GO.
But the Prilm ABCDEF is half the Solid AX, and 28 of this, the Prism GHKLMN is I half the Solid GO. There
fore the Prism ABCDEF is equal to the Prism GHKLMN. Wherefore, if there be two Prisms having equal Altitudes, the Base of one of which is a Parallelogram, and that of the other a Triangle, and if the Parallelogram be double to the Triangle, the said Prisms fhall be equal to each other,
The End of the EL EVENTH BOQK.