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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.
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.
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.
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
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.
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.
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.
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.
A B CE.
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.
B. B D ABG.
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.
And the pgr. C F is double of the ALC}(P.41
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.
OP & W X plle. to B F or CH, until they meet the other
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
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).