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DEDUCTION (A).

If two straight lines meeting one another be parallel to two others that meet one another, and are not in the same plane with the first two, the first two and the other two shall contain equal angles, also the plane through the first two shall be parallel to the plane through the other two.

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Let AB, BC two straight lines meeting one another be to FG, GH, meeting one another, the latter two straight lines not being in the same plane with the former. Then shall ABC be =LFGH,

and the plane ABC be to plane FGH.

Cut off GF BA and GH = BC; join AC, FH, AF, CH,BG. Then. GF is = and || to AB, .. AFis and || to BG,

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=

and || to BG; and || to CH;

and to HF

(I. 23)

(1. 23)

(Prop. 12)

(I. 23)

Then. AB, BC, CA are respectively = FG, GH, HF;

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Also plane ABC shall be || to plane FGH; for if not, if possible, let them meet when produced, then their common section will be a straight line.

(Prop. 5)

Since AB, BC cannot both be to the common section one of them will meet it if produced.

Let AB produced meet it in X.

Thus X is in the plane containing the parallels AB, FG and also in the plane FGH; and is .. in their common section of which FG is a part.

Hence AB, FG meet if produced; which is impossible. .. plane ABC is to plane FGH.

C. G.

15

DEDUCTION (B).

If two straight lines be cut by parallel planes, they shall be cut proportionally.

X

Let the two straight lines AC, FH be cut by the parallel planes X, Y, Z in A, B, C and F, G, H respectively.

Then AB BC as FG : GH.

Join AH cutting the plane Y in the point, R.
Join AF, CH, BR, RG.

Then the parallel planes Y, Z are cut by the plane · ACH,

.. their common sections BR, CH are || to one another;

.. AB : BC as AR : RH.

Similarly it may be shewn that

FG: GH as AR : RH; .. AB : BC as FG: GH.

(Prop. 14)

(VI. I)

(v. 3)

DEFINITION.

One plane is said to be perpendicular to another plane when every straight line drawn in one of the planes perpendicular to the common section of the two planes is perpendicular to the other plane.

PROPOSITION XV.

If a straight line be perpendicular to a plane, every plane which passes through it shall be perpendicular to that plane.

ME

A Y

Let AB be a straight line 1 to the plane HKLM,
and CFG a plane passing through AB.

Then shall CFG be 1 to the plane HKL.

In the common section FG take any point X, and through X in the plane CFG draw XY1 to FG.

Then. AB is 1 to plane HKL, .. AB is 1 to FG;

.. ▲ ABX is a right ▲ ;

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(Def.)

.. the plane CFG is to the plane HKL.

PROPOSITION XVI.

If two planes which cut one another be each of them perpendicular to a third plane, their common section shall be perpendicular to the same plane.

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Let the two planes ANB, CNF which cut one another be each 1 to the plane PQ.

Then shall their common section SNV be 1 to the plane PQ.

For, if not, if possible, in plane ANB draw NX 1 to NB, and.. to the plane PQ, and in the plane CNF draw NY

1 to NF, and.. 1 to plane PQ,

... from the point N two straight lines have been drawn 1 to the plane PQ; which is impossible;

(Prop. 11)

.. SN is 1 to plane PQ.

DEFINITION.

A solid angle is that which is made by the meeting in one point of two or more plane angles, which are not in the same plane.

PROPOSITION XVII.

If a solid angle be contained by three plane angles any two of them are together greater than the third.

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Let the solid angle at A be contained by the three plane angles BAF, FAC, CAB.

Then shall any two of them BAF, FAC be together greater than the third CAB.

If either of the two angles BAF, FAC be = or> ▲ CAB, the proposition is evidently true.

But, if not, from 4 CAB cut off ▲ BAN=L BAF;

=

L

make AN AF; through N draw BNC meeting AB, AC in B and C; join BF and CF.

L

Then.. BA, AF and BAFare respectively = BA, AN and 2 BAN,

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Again. BF, FC are together > BC and BF is = BN;

.. FC is > NC.

Also AF, AC are respectively = AN, AC;

.. L FAC is > ↳ NAC;

(1. 19)

.. LS BAF, FAC are together > < s BAN, NAC,

i. e. > ▲ CAB.

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