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PROPOSITION XXII.

THEOREM XX.

If every two of three plane angles be greater than the third, and if the

straight lines which contain them be all equal ; a triangle may be made of the straight lines (D F, GI & AC) which subtend those angles. Hypothesis.

Thesis. 1. Any two of the three given V a, b, c, A A may be made of the straight are > the third, as b ta>c, or at

lines GÍ, DF & AC, which subc>b, or b + c> a.

tend those V.
II. The sides HG, HI, DE, EF, AB &
BC' which contain those V, are equal.

DeMoNSTRATION.
The three given V a, b,& c are either equal, or unequal.
CASE I If the v a, b,& c be equal.

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Because the sides which contain the V, are equal (Hyp. 2.)

1. The ADEF, GHI & A B C are equal.

P. 4. B. i. 2. Therefore D F=GI= A C. 3. Consequently, DF + AC >GI.

Ax.4. B. 1. 4. Wherefore a A may be made of those straight lines DF, AC & GI. P.22. B. 1.

A
CASE. II. If the given V a, b, & cbe unequal

Preparation
1. At the vertex of one of the V as B, make V ABL=ya, P.23. B. 1.
2. Make BL>DE.

P. 3. B. i. 3. Draw LC & L A.

Pof.i. B. 1. DEMONSTRATION. ECAUSE the two Va+care > V b (Hyp. 1.) & LB=HG

=BC=HI (Prep. 2. & Hyp. 2.) 1. The base L C will be > GI.

P.24. B. 1. But LC<LA +AC.

P.20. B. I. 2. Much more then Ġ I is <LA +AC.

But LA=DF (Prep. 1. & P. 4. B. 1). 3. Therefore GI is < DF + A C.

Ax. 1. B. 1. 4. Consequently, a Amay be made of the straight lines DF, AC&GI,

Which was to be demonstrated.

Because

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PROPOSITION XXIII. . PROBLEM III.

O make a solid angle-(P), which shall be contained by three given plane angles (A B C, DEF & GHI), any two of them being greater than the third, and all three together (V ABC+VDEF+YGHI) less than four right angles. Given.

Sought. 1. Three VABC, DEF & GHI, any two of A solid V P, contained by the

which are greater than the third, as VB three plane V B, E & H. E> VH, VB+H> VE, & VE+H

> VB. II. V B+E+H<4L

P.

B. 1.

{P.22.

Resolution. 1. Take A B at will, & make the sides B C, DE, EF, GH & HI equal to one another & to A B.

3. 2. Draw the bases AC, DF, & GI.

Pos.i.

1.B. 1. 3. With those three bases A C, D F & G I make a A LMN lo SP.27. B. 1. that N M be =GI, NL= AC, & LM=DF.

B.11. 4. Inscribe the AL M N in a O L M N.

P. 5. B. 4: Ś. From the center O, to the V L, M & N, draw the straight lines

LO, O N & O M. 6. At the point o, erect the LOP to the plane of the OLMN. P.12. B.II. 7. Cut O P so that the of LO+the o of P O be = to the of AB. 8. Draw the straight lines L P, PN & PM.

M m

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N

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A

M

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DEMONSTRATION.
BECAUS
ECAUSE PO is I to the plane of the OLMN (Ref. 6.)

I
1. The APOL will be right angled in 0 (Ref. 5. & 8.)

២ ) 2. Consequently, the of PO + the of O L is = to the of L P. P.47. B. 1.

But the of PO+ the of OL=O AB, (Ref. 7.) 3. Therefore the o of A B is = to the of LP, & AB=LP. P.40. B. 1. 4.

Likewise PN & PM are each to A B.

But N Mis = to GI, NL=AC, & LMDF, (Ref. 3). 5. Consequently, A NMP is = to the AGHI, A NPL =) A ABC, Á LPM=ADEF, VNPM=HH, VLPN (P. 8. B. i,

VLPNG EHB, & VLPM

But those three NPM, LPN & L P M form a solid V P. 6. Therefore a solid V P has been made, contained by the ihree given plane VB, E & H.

Which was to be done.

(Cor. 3.

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IN

BECAUSE

PROPOSITION XXIV. THEOREM XXT. N every parallelepiped (A H); the opposite planes (B D & C F; BE & FG; AF & B H) are similar & equal parallelograms. Hypothesis.

Thesis. In the given El B F, the plane B D is The opposite planes B D, C F; B E opposite to CFBE10 FG & AF 10 BH. & FG; AF & B H are = & CO

pgrs, each to each.

Preparation.
Draw the opposite diagonals EH & AG, also AC & D H.

DEMONSTRATION.
ECAUSE the plle. planes B D & C F are cut by the plane

A B CE.
1. The line B A is plle. to E C.

P.16. B.1. 2. Likewise C H is plle. to G B.

And the same pile. planes B D & C F being also cut by the plane

DGHF. 3. The line D G will be plle. to F H. 4. Likewise A E is plle. to B C & D F plle. to G H.

And because those plle. planes ( Arg. 1. 2. & 4.) are the opposite sides

of the quadrilateral figures A E C B & D FHG. 5. Those quadrilateral figures A ECB & D FHG, are pgrs. D.35. B. 1. 6. Likewise the other opposite planes B D & CF ; AF & BH are pgrs.

And since A B & B G are plle. to EC & CH, each to each (Arg. 1.&2). 7. V A B G is = to VECH.

P.10. B.u. But A B is = to E C & BG=CH.

:34: 8. Therefore the O ABG is: & v to the AECH.

4:

B. B D ABG.

4:

B. 6. . But those pgrs. have cach an V common with the equiangular A. 9. Consequently, the pgrs. BD & C F are = &.

D. 1. B. 6. io. It may be demonstrated after the fame manner that the pgr. B D is

:& to the pgr. CF, & pgr. A F is = &n3 to the pgr. B H. 11. Therefore the opposite planes of a El are = & pgrs.

Which was to be demonstrated.

P

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1.

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And the pgr. C F is double of the ALC}(P.41

. B.1.)

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PROPOSITION XXV. THEOREM XXII. F a parallelepiped (BED C) be cut by a plane (KIML) parallel to the opposite planes (A EFB & CGD H); it divides the whole into two parallelepipeds (viz. the BEMK & KMDC), which shall be to one another as their bases (BFL K & KLHC). Hypothefis.

Thesis. The EBEDC is divided into two The EBM: MC=bafe BL: BM & MC, by a plane KM, plle. to the base L C. oppofite planes B E & CD.

Preparation 1. Produce B C both ways, as also F H.

Pof.2. B. 1, 2. In B C produced take any number of lines = 10 BK & CK: as BO & TO each == to B K & CW= KC. P.

3.
3. Thro’ those points T, O & W, draw the straight lines TU,

OP & W X plle. to B F or CH, until they meet the other
plle. produced in the points U, P & X.

P.31. B. 1. 4. Thro the lines TU, OP & W X let the planes TR, OQ

& W Y pass, plle. to the planes B E & CD, which will meet
the plane AED G in SR, NQ & VY.

B. 1.

DEMONSTRATION. Because

ECAUSE the lines BO & TO, are each = to BK & CW =KC (Prep. 2.) & the lines OP, TU & W X plle. to B F or CH, meet FH produced, in the points, P, U & X (Prep. 3).

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