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PROPOSITION XV. THEOREM XIII. F two straight lines (A B & A C) situated in the same plane (A X), and meeting one another (in A), be parallel, to two straight lines (D É & D F) meeting one another, and situated in another plane (D Z); those planes (A X & D Z) will be parallel. Hypothefis.

Thesis. A B&AC ftuated in the plane A X The plane A X in which are the lines & meeting each other in A, are plle. to AB@ AC is plle. to the plane DZ DE & EF meeting each other in D, & in which aretbe lines DEED F. fituated in the plane D Z.

Preparation.
1. From the point A let fall upon the plane D Z the I
AG.

P.1. BIT. 2. Draw G H plle, to DE, & GL plle. to DF.

P.31. B. 1.

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DEMONSTRATION.

BECA US

ECAUSE the lines GH & GL are plle. to D E & DF (Prep. 2). 1. They will be also plle. to A B & AC.

P. 9. B.11. And G L being plle. to A C. 2. The V CAG+ AGL are = 2 L.

P.29. B. 1. But YAGL is a L (Prep. 1). 3. Consequently, VCAG is also a L. 4. It may be demonstrated after the same manner that V BAG is a L 5. Therefore, G A is I to the plane A X.

4:

B.in. But G A is also I to the plane D Z (Prep. 1). 6. Wherefore, the plane A X is plle. to the plane D Z.

Which was to be demonstrated.

P.

P.14. B.u.

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PROPOSITION XVI. THEOREM XIV.
I

F two parallel planes (Z X & Y P) be cut by another plane (A B D C),
the common sections with it (CD & A B) are parallels.
Hypothefis.

Thesis. 1. The planes 2 X & P Y are plle.

The common Sections CD&AB II. Tbey are cut by the plane ABCD.

are plle.

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If not,

The lines A B & C D being produced will meet somewhere.

Preparation.
Produce them until they meet in F.

Pof.2. B. 1.
ECAUSE the straight lines B'A F & DCF meet in F.
1. The planes P Y & Z X in which those lines are, will also meet one

another : (B A F being entirely in the plane P Y, & DCF entirely
in the plane Z X).

P. 1. B.11. 2. Which is impoffible (Hyp. 1).

Wherefore, AB & C D do not meet one another. 4. Therefore, A B & C D are plle.

D.35. B. 1. Which was to be demonstrated.

3

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PROPOSITION XVII. THEOREM XV. If two Iraight lines (AC & BD) be cut by parallel planes (XZ, PY&QM): they shall be cut in the same ratio, (that is, A E:EF=BF:FH &c). Hypothesis.

Thesis. 1. AC&B D are two straight lines.

AE: EGBF;FH. II, Cut by the plle. planes X Z, PY&QM.

Preparation
1. Join the points A & B, also G&H.
2. Draw A H which will pass thro’ the plane P Y in the
point I.

Pof.1. B. 1. 3. Draw E I & IF.

DEMONSTRATION.
ECAUSE the plle. planes Z X & P Y are cut by the plane of

&
the A AB H.
1. A B is plle. to I F.

-P.16. B.11. 2. Likewise, E I is plle. to G H. 3. Consequently, AI:1H= BF: FH.

P. 2. B. 6. AI:IH = AE: EG. 5. Therefore, AE: EG=BF: F H.

P.IT. B. 5. Which was to be demonstrated.

4. And,

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IF

PROPOSITION XVIII. THEOREM XVI. F a straight line (A B) is perpendicular to a plane (Z X): every plane (as Q E) which passes thro' this line (A B) shall be perpendicular to this plane (Z X). Hypothesis.

Thesis. A B is I to the plane Z X.

Every plane (as Q E) which passes thro the L A B is I to the plane Ż X.

Preparation.
1. Let a plane Q E pass thro' A B, which will cut the plane
Z X in EF.

P. 3. B. I.
2. Take in this straight line E F, a point D at will.
3. From this point D, draw in the plane Q E, the line DC
plle. to A B.

P.31. B. I. DEMONSTRATION.

Beca

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D.

ECAUSE the straight line A B is I to the plane ZX, & DC is plle. to A B (Hyp. & Prep. 3). 1. The line D C is I to the plane Z X:

P. 8. B.u. 2. Consequently, CD is also I to the common section É F.

3.

B.I. 3. Therefore, the plane E Q in which the lines A B & C D are, is I to the plane Z X.

D. 4. B.11. And as the fame demonstration may be applied to any other plane

which passes thro’ the I AB, we may conclude, 4. That every plane which passes thro' this line is I to the plane Z X.

Which was to be demonstrated.

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PROPOSITION XIX. THEOREM XVII. F two planes (CD & E F) cutting one another be each of them perpendicular to a third plane (Z X); their common section (AB) shall be perpendicular to the same plane (Z X). Hypothesis.

Thesis. 1. The planes CD & E F are I to the plane Z X. The common seation A Bis I ļl. They cut one another in AB,

to the plane Z X.

DEMONSTRATION. Becau

ECAUSE CB, the common section of the plane CD with the
plane X Z is also in the plane X Z.

P.
3.

B.1. 1. There

may be erected at the point B in C B a I (P. 11. B. 11.) which will be in the plane CD (Hyp. 1.)

P.18. B.II. And because the line F B the common section of the planes FE & X Z is also in the plane X Z.

P. 3. B.In. 2. There may be erected at the faine point B & at the same side with the foregoing another I which will fall in the plane F E.

P.18. B.I. But from the point B only one I can be raised. 3. Consequently, those I must coincide, that is, those two lines must

form but one which is common to the two planes.
But those planes have only the line A B in common (Hyp. 2.)
Therefore A B is I to the plane X Z.

Which was to be demonstrated.

P.13. B.11.

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