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Let a plane pass through BA, AC, and let Dae be the common section of it with the given plane : :: AB, AC, DAE, are in one plane.

Then :: ca is I to the plane, and meets DAE, :: _CAE is. a right Z; for the same reason BAE is a right 2.

:: Z CAE = Z BAE, the less to the greater, which is impossible.

Also, from the same point above a plane there can be drawn but one perpendicular to that plane; for if there could be two they would be ll, which is absurd.

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PROPOSITION XV. Planes to which the same straight line is perpendicular are

parallel to one another. Let as be 1 to each of the planes CD, EF; these planes are Il to each other.

If not, they will meet if produced: let them meet; their common section is a straight line gh, in which take any point K, and join AK, BK.

Then : AB is 1 EF it is I to BK :: ZABK is a right 2 ; for the same reason KA B is a right Z; :: two _s of the A A B K are = two right < $, which is impossible ; :. the planes CD, EF, though produced, do not meet; therefore they are II.

PROPOSITION XVI. If two straight lines meeting one another be parallel to two other

straight lines which meet one another, but are not in the same plane with the first two; the plane which passes through these is parallel to the plane through the others. Let A B, BC, meeting in B, be || to DE, EF, meeting in E,

but which are not in the same plane with A B, BC: the planes through

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AB, BC, and DE, EF, are ||- From B, draw BG + to plane through ED, DF, through a, where BG meets the plane; draw gh || to ED, and GK || to EF.

And :: BG is 1 to plane DF, each of the S BGH, BGK, is a right Z; and :: BA is || to go, (for both are || to DE) the L GBA + Z BGH=two right / s.

But bgn: a right 2
GBA = right Z, :. GB is + to

For the same reason GB is

1 BC, and .. GB is 1 to the plane through AB, BC; and it is to plane through DE, EF.

.. BG is at once + to the plane through AB, BC, and the plane through DE, EF; hence these planes are parallel.

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BA.

FH or EG.

PROPOSITION XVII. If two parallel planes be cut by another plane, their common

sections with it are parallels. Let the parallel planes, AB, CD, be cut by EFHg, and let EF, HG, be their common sections: EF shall be parallel to GH.

If not, EF, GH, will meet if produced either on the side of

First let them be produced on the side of th, and meet in the point K.

Hence, :: EFK is in the plane A B, every point of EFk is in that

plane, :. K is a point in the plane For the same reason, K is a point in the plane CD; :. the planes AB, CD, meet if produced, which is contrary to the hypothesis.

:. EF and gu do not meet when produced on the side of FA. In the same manner it may be shown that they do not meet when produced on the side Ef: wherefore EF and ng are parallel.

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AB.

PROPOSITION XVIII.

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

in the same ratio.

Let A B, CD, be cut by the parallel planes GH, Kl, mn, in the points A, E, B; C, F, D; then AE

: EB :: CF : FD.

H

G

L

K

Join AC, BD, AD; let ad meet Kl in x; join Ex, XF.

Then :: the parallel planes KL, MN, are cut by EXDB, :: Ex is parallel to BD; and :: GH, KL, are cut by the plane A XFC, the common sections Ac, xf, are parallel. Then :: Ex is parallel to be in A ABD, and :: XF is parallel to Ac in A ADC,

... AE : EB :: AX : XD. And AX : XD :: CF : FD.

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B

M

.. AE : EB :: CF : FD.

PROPOSITION XIX. If a straight line be perpendicular lo a plane, every plane pass

ing through it shall be perpendicular to that plane. Let AB be 1 to plane ck; every plane through A B shall be I to CK.

Let plane de pass through AB; and let ce be the common section of D E and cK. Take any point F in CE, draw fg in DE, I to CE; then : AB is 1 to CK, < ABF is a right Z, but grb is a right Z, :. AB and Fg are ll, and AB is 1 to the plane ck, :. fg is 1 to the same plane. But one plane is to another plane, when the straight lines drawn in one of the planes, + to their

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common section, are also I to the other plane. And any line FG, in the plane de, drawn I to the line ce, is also to CK. .. the plane DE is = to ck. And the same may be proved of any other plane passing through AB.

PROPOSITION XX.

If two planes which intersect be each perpendicular to a third

plane, their common section shall be perpendicular to the same plane.

Let the two planes AB, BC, be each 1 to the third plane, and let Bd be the common section of AB, BC; BD shall be 1 to the third plane.

If it be not, from D in AB draw DE 1 AD, the common section of AB with the third plane; in bc draw DF + CD, the common section of Bc with the same plane.

Then :: AB is I to plane ADC, and dE is I to AD, :. DE is I to ADC; similarly, DF is 1 to ADC; .. from the same point d, two

1S, DE, DF, to the same plane ADC, have been

c drawn, which is impossible; .. the only line that can be drawn from D I to ADC is bd, the common section of AB and BC.

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PROPOSITION XXI.

If a solid angle be contained by three plane angles, any two of

them are greater than the third.
Let the solid 2 A, be contained by the three plane 25, BAC,

CAD, DAB; any two of them shall
be greater than the third.
If the three angles be all =,

the proposition is evident. If not, let BAC be not less than either of the

other two, and > DAB. At the В

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point A, in the plane through BA, AC, make BAE = { D AB. Make A E = AD; through E draw bec, cutting AB, ac, in B and ć, and join DB, DC.

And :: DA = AE, and AB is common, and DAB= ZEAB, :: DB=EB; but :: BD, DC, are > BC and BD=BE, :. Dc is > EC; and : DA = AE and Ac is common, also DC> EC, :: L DAC is > EAC; and DAB= _ BAE, :: DAC+ Z DAB is > Z BAC.

Also, _ BAC is not less than either of the other two; :: Z BAC + either of the other two, must be > than the remaining angle.

PROPOSITION XXII.

Every solid angle is contained by plane angles, which together

are less than four right angles.

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First, let the solid ¿ A be contained by the three plane 25, BAC, CAD, DAB; take in each of the lines AB, AC, AD, any points B, C, D; join BC, CD, DB. Then :: the <S CBA, ABD, DBC, make the solid 2 B, :: Z CBA + 2 ABD are > DBC; similarly 2 BCA + 2 ACD, are > L DCB; and / CDA + ZADB, are > BDC. i. ZS CBA, ABD, BCA, ACD, CDA, ADB, are > _S DBC, BCD, BDC; but _ DBC + _ DCB + BDC = two right <s.

i. Z CBA + 2 ABD + Z BCA + 2 ACD + 2 CDA + L ADB are > two right < 5; and : the three _s of each of the A $ ABC, ACD, ADB, are = two right <s, i. the nine _s of these As, viz., the < $ CBA, BAC, ACB, ACD, ADC, DAC, ADB, ABD, BAD, are = six right_$; and of these the < $ CBA, ABD, BCA, ACD, ADC, ADB, are > two right _s. .. the three remaining 2$, BAC, BAD, DAC, are four right 2S.

Next, let the solid < at A be contained by any number of

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