Sidebilder
PDF
ePub
[blocks in formation]

PROPOSITION XVIII.

THEOREM XVI.

IF a ftraight line (A B) is perpendicular to a plane (Z X): every plane

(as QE) which paffes thro' this line (A B) fhall be perpendicular to this plane (Z X).

AB is

Hypothefis.
to the plane Z X.

Thefis.

Every plane (as QE) which passes thro' the LAB is to the plane Z X.

Preparation.

1. Let a plane QE pafs thro' A B, which will cut the plane

Z X in EF.

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

BECA

DEMONSTRATION.

ECAUSE the ftraight line A B is to the plane ZX, & DC is plle. to A B (Hyp. & Prep. 3).

1. The line DC is to the plane Z X.

2. Confequently, CD is alfo 1 to the common section E F.

P. 3. B. 1,

P.31. B. 1.

P. 8. B.11.

D. 3. B.11.

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

D. 4. B.11.

And as the fame demonftration may be applied to any other plane which paffes thro' the LA B, we may conclude,

4. That every plane which paffes thro' this line is to the plane Z X. Which was to be demonstrated.

[merged small][merged small][merged small][merged small][merged small][merged small][ocr errors]

IF

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 fection (A B) fhall be perpendicular to the fame plane (Z X).

Hypothefis.

I. The planes CD & E F are 1 to the plane Z X.
II. They cut one another in A B.

DEMONSTRATION.

Thefis.

The common fection A B is

to the plane Z X.

BECAUSE CB, the common fection of the plane CD with the

plane X Z is alfo in the plane X Z..

1. There may be erected at the point B in CB a ▲ (P. 11. B. 11.) which will be in the plane C D (Hyp. 1.)

And because the line F B the common fection of the planes F E
& X Z is alfo in the plane X Z.

2. There may be erected at the fame point B & at the fame fide with
the foregoing another which will fall in the plane F E.
But from the point B only one can be raised.

P. 3. B.11.

P.18. B.11.

P. 3. B.11.

P.18. B.11.

3. Confequently, thofe muft coincide, that is, those two lines must form but one which is common to the two planes.

P.13. B.11.

But thofe planes have only the line A B in common (Hyp. 2.)
Therefore A B is to the plane X Z.

Which was to be demonstrated.

C

A b

B

E

D

PROPOSITION XX. THEOREM XVIII.

IF three plane angles (CA B, BAD & DA C) form a folid angle A: any

two of those angles (as BAD & CA B) are greater than the third (CAD).

[blocks in formation]

BECAU

ECAUSE the VCA B, d&c + b are equal.

1. It follows that VCAB+d will be

c+b.

CASE II.

When of the three angles C A B, d&c + b two as CAB &

d are equal, & the third c + b is less than either of them.

BECAUSE

ECAUSE VCAB is >vc+b.

i. V CAB+Vd will be much >c+b.

Ax.4. B. 1.

Ax.4. B. X.

Which was to be demonftrated.

[blocks in formation]

THE

CASE III.

When the three angles are unequal, & b + c is > CAB or d.

Preparation.

1. At the point A in AC make VbVCAB in the plane
CAD.

2. Make A E=AB.

"P23. B.1. P. 3. B.1.

D. Pof.1. B.A.

3. From the point C draw thro' E the ftraight line C E D.
4: From the points C & D draw C B & B D.

HE ABCA & CA E have the fides A B & AE equal (Prep.2).

The fide C A common &

b=VCAB (Prep. 1).

1. Confequently, the fide B C is to the fide CE.

But in the ACBD the fides CB+ B D are > C D.

Therefore, if from CB+BD be taken away the part C B, &

from CD a part

to C E.

2. The remainder B D

will be > ED.

In the ABA D & E AD, the fides A B & A E are = (Prep. 2).

[blocks in formation]

& A D common.

But the bafe B D is > the bafe ED (Arg. 2).

3. Therefore, Vd is >

c.

If therefore, CA B be added one fide, & its equal ▷ b on the

P.25. B.1.

other.

4. VCAB+d will be >Vb+cor CAD.

Ax.4. B.1.

Which was to be demonftrated.

A

B

PROPOSITION XXI. THEOREM XIX.

ALL L the plane angles (B A C, CAD & DA B) which form a folid

angle (A); are lefs than four right angles.

The

Hypothefis.

BAC, CAD & DAB

form a folid, VA

The plane
are < 4 L.

Thefis.
BAC+CAD+DAB

Preparation.

1. In the fides BA, AC, & AD take the three points B, C, D.
2. Draw B C, B D & C D.

3.

three folid V ; CBA, ABD plane VBCA,

Let a plane BCD pafs thro' thofe lines, which will form
with the planes BAC, CAD & BAD,
viz. the folid B, formed by the plane
& CBD; the folid VC, formed by the
ACD & BCD, & infine, the folid V D, formed by the
plane CDA, A DB & B D C.

DEMONSTRATION,

ECAUSE the folid V D, is formed by the plane VC DA,

BECAU

[blocks in formation]

VACBA CD are > VBCD.

1. The

2. Likewife, VABD+ABC are > VDBC.

3.

And

4. Hence, the fix plane VCDA+ADB+ABD+ABC+ACB +ACD are the three plane VBDC+DBC+BCD.

Pof.1. B. 1.

D.11. B.11.

P.20. B.11.

But thofe three plane VBDC+DBC+BCD are 2 L. P.32. B.1. 5. Therefore, the fix plane VCDA+ADB+ABD+ &c. are > 2 L (Arg. 4.)

But the nine of the ABCA, CAD & D A B viz. the fix already mentioned (Arg. 5.) & the three remaining BAC, CAD & DAB are together to 6 L.

If therefore the fix VCDA+ADB+ABD+ABC+ACB
+ACD which are together > 2 L be taken away.

6. The remaining plane VBA C+CAD+DA B will be < 4 L.
But thofe plane BAC, CAD & D A B form a folid V A.
7. Confequently, the plane which form a folid V A are < 4 L.
Which was to be demonftrated.

P.32. B:1.

« ForrigeFortsett »