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the same parts, and .. AC is and || DF; and ·: AB, DE, EF, and the base AC to the base DF, .. the /DEF, (Prop. 9.)

BC are= the LABC

PROPOSITION LXXXVI.-PROBLEM.

To draw a straight line perpendicular to a plane, from a given point above it.

Let A be the given point above the plane BH; it is required to draw from the point A a

B

G

to the plane BH.

E

D

А

In the plane draw any | BC, and from the point A draw AD + to BC. If AD be also to the plane BH, the thing required is done; but if it be not, from the point D draw, in the plane BH, DE at nt Ls to BC; and from the point A draw AF through F draw HG || BC; and ED and DA, BC is at t Ls (Prop. 80) to the plane passing through AD, DE; and GH is || BC, .. (Prop. 83) GH is at Ls to the plane through AD, DE; and is .. to every meeting it in that plane. But AF, which is in the plane through AD, DE, meets it; wherefore GH is

to DE, and BC is at Ls to

to AF, and DF is also to AF; hence AF is to each of the Is GF, FD, and.. AF is to the plane in which GH and ED are, (Prop. 80), that is to the plane BH. Hence from the given point A, above the plane BH, the AF is drawn to that plane.

Cor. 1. If it be required from a point C in a plane to erect a perpendicular to that plane, take a point A above the plane, and draw AF perpendicular to the plane; then if from C a line be drawn parallel to AF, it will be the perpendicular required.

For, being parallel to AF, it will be perpendicular to the same plane to which AF is perpendicular.

Cor. 2. From the same point, whether without or in a plane, there can only be drawn one perpendicular to that plane.

PROPOSITION LXXXVII.-THEOREM.

Planes CD, EF, to which the same straight line AB is perpendicular, are parallel to one another.

If not, they shall meet one another when produced; let them meet their common section, will be a | GH, in which take any point K, and join AK, BK; then, ·.· AB is to the plane EF, it is (Def. 3), to the | BK, which

is in that plane; .. ABK is a r*L; for the same reason, BAK is a r; wherefore the two Ls ABK, BAK, of the AABK, are together equal to two Ls, which is impossible, (Prop. 10, cor.); .. the planes CD, EF, though produced, do not meet one another; that is, they are parallel, (Def. 8.).

PROPOSITION LXXXVIII.—THEOREM.

B

E

C

K

If two straight lines AB, BC, which meet one another, be respectively parallel to other two DE, EF, which meet one another, but are not in the same plane with the first two; the plane which passes through the first two is parallel to that which passes through the latter. Through the point B draw BG to the plane which passes through EF, ED, (Prop. 86), meeting it in G; and through G draw GK | EF, and GH ED. And BG is to the plane FD, it is (Def. 3) to GK and GH, which meet it in that plane; ..

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A

each of the Ls BGH, BGK is a L. And since AB and GH are each | ED, they are one another, (Prop. 84); and BGH is a L, .. GBA is also a L. Again GK is | BC, for each of them is || EF; .. since KGB is a L, GBC is also a L, (Prop. 17). Since. BG is at Ls to each of the lines BC and BA, it is to the plane AC, and it was drawn to the plane DF. then BG is to each of the planes AC, DF, these planes are parallel, (Prop. 87).

PROPOSITION LXXXIX.-THEOREM.

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Since

If two parallel planes AB, CD, be cut by a third EF, GH, their common sections EF, GH, with it are parallel.

For the Is EF and GH are in the same plane, namely EFGH, which cuts the planes AB and CD; and they do not meet though produced; for the planes in which they are do not meet; .. EF and GII are parallel.

A

B

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D

K

PROPOSITION XC.-THEOREM.

E

G

A

C

L

M

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If two parallel planes AB, CD be cut by a third plane EH, they have the same inclination to that plane. Let the Is EF and GH be the common sections of the plane EH, with the two planes AB and CD; and from K, any point in EF, draw in the plane EH the KM at Ls to

K

B

F

N

D

H

EF, and let it meet GH in L; draw also KN at rs to EF in the plane AB; and through the Is KM, KN, let a plane be made to pass, cutting the plane CD in the line LO.

And EF and GH are the common sections of the plane EH, with the two parallel planes AB and CD, EF is || GH, (Prop. 89). But EF is at Ls to the plane that passes through KM and KN, (Prop. 80); it is at rs to the lines KM and KN; .. GH is also at rs to the same plane, (Prop. 83); and it is .. at rs to LM, LO, which it meets in that plane... since LM and LO are at Ls to LG, the common section of the two planes CD and EH, the LOLM is the inclination of the plane CD to the plane EH, (Def. 6). For the same reason the NKM is the inclination of the plane AB to the plane EH. But ...KN and LO are parallel, being the common sections of the parallel planes AB and CD, with a third plane; the interior LNKM is the exterior LOLM; that is, the inclination of the plane AB to the plane EH is = the inclination of the plane CD to the same plane EH.

PROPOSITION XCI.-THEorem.

If two straight lines AB, CD be cut by parallel planes GH, KL, MN, in the points A, E, B; C, F, D; they shall be cut in the same ratio: that is, AE: EB= CF: FD.

Join AC, BD, AD, and let AD meet the plane KL in the point X; and join EX, XF.

IE

X

the two parallel planes KL, MN, are cut by the plane EBDX, the common sections EX, BD are parallel, (Prop. 89). For the same reason, the two parallel planes GH, KL, are cut by the plane AXFC, the common sections AC, XF are parallel. And. EX is || BD, a side of the ABD, AE: EB=AX: XD,

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(Prop. 59). Again, XF is || AC, a side of the AADC, AX:XD CF: FD... AE: EB=CF : FD, (Alg. 102).

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PROPOSITION XCII.-THEOREM.

If a straight line AB be perpendicular to a plane CD, any plane CE touching it will be perpendicular to the same plane CD.

For let CBG be their line of common section, and from any point G, in CG, let EG be drawn to it in the plane CE. Also, let BF, GH be c to CG in the plane CD. Then AB is to the plane CD, ABF is a L; and since EG is || AB,. and GH is BF, the LEGH is also a

A

B

G

F

H

L, (Prop. 85). But the LEGH is the inclination of the plane CE to the plane CD, (Def. 6); .. the plane CE is at rs to the plane CD.

PROPOSITION XCIII.-THEOREM.

If two planes AB, BC, cutting one another, be each of them perpendicular to a third plane ADC, their common section BD will also be perpendicular to the same plane.

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From D in the plane ADC, draw
DE to AD, and DF to DC.
DE is to AD, the common section of
the planes AB and ADC, and the plane
AB is at Ls to ADC, DE is at *Ls

A

B

to the plane AB, and .. also to the | BD in that plane. For the same reason DF is at rs to DB. And since BD is at Ls to both the lines DE and DF, it is at rLs to the plane in which DE and DF are; that is, to the plane ADC, (Prop. 80).

SOLID GEOMETRY.

DEFINITIONS.

I. A solid angle is that which is formed by more than two plane angles at the same point, but not in the same plane.

II. Two solids bounded by planes are similar, when their solid angles are equal, and their plane figures similar, each to each.

III. A parallelopiped is a solid bounded by six planes, of which the opposite ones are parallel. If the adjacent sides be perpendicular to one another, it is a rectangular parallelopiped.

IV. A cube is a rectangular parallelopiped, of which the six sides are squares.

V. A prism is a solid of which the sides are parallelograms, and the ends are plane rectilineal figures.

VI. A pyramid is a solid of which the sides are triangles, having a common vertex, and the base any plane rectilineal figure. If the base be a triangle, it is a triangular pyramid. If a square, it is a square pyramid.

VII. A cylinder is a solid described by the revolution of a rectangle about one of its sides remaining fixed; which side is named the axis; and either of the circles described by its adjacent sides the base of the cylinder.

VIII. A cone is a solid described by the revolution of a right angled triangle about one of its sides remaining fixed, which is called the axis; and the circle described by the other side is the base of the cone.

IX. Cones or cylinders are similar when they are described by similar figures.

PROPOSITION XCIV.-THEorem.

If a solid angle A be contained by three plane angles BAC, CAD, DAB, any two of these are together greater than the third.

If all the Ls be, or if the two greater be, the proposition is evident. In any other case, let BAC be the greatest L, and let BAE be cut off from it, DAB. Through any B point E in AE, let the BEC be

drawn in the plane of the BAC to meet its sides in B and C. Make AD=EA, and join BD, DC. ..the As BAD, BAE have AD AE and AB common, and the included Ls BAE, BAD=, the base BE is = BD, (Prop. 5). But the two sides BD, DC are BE, EC, DC is EC; and since the side AE is = AD, and AC common to the two As ACD, ACE, but the base DC EC,.. the CAD is EAC, (Prop. 14). Consequently the sum of the Ls BAD, CAD is the [BAC.

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