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PROPOSITION XXII. THEOREM XX.
IF every two of three plane angles be greater than the third, and if the
ftraight lines which contain them be all equal; a triangle may be made of the ftraight lines (D F, GI & A C) which fubtend those angles.
A▲ may be made of the ftraight lines GI, DF & AC, which fubtend thofe V.
The three given V a, b,
c are either equal, or unequal.
OECAUSE the fides which contain the V, are equal (Hyp. 2.)
I. The ADEF, GHI & ABC are equal.
2. Therefore DFGI=AC.
3. Confequently, DF+AC>G I.
P. 4. B. i.
Ax.4. B. 1.
4. Wherefore a A may be made of those straight lines DF, AC & GI. P.22. B. 1. CASE. II. If the given V a, b, c be unequal
1. At the vertex of one of the V as B, make VA BL=Va. P.23.
3. Draw LC & LA.
P. 3. B. 1.
Pof.i. B. 1.
BECAUSE the two Va+care > \ b (Hyp. 1.) & LB=HG
3. Therefore GI is <DF+AC.
4. Confequently, a A may be made of the ftraight lines D F, AC & G I.
Which was to be demonstrated.
P.24. B. 1.
P.20. B. 1.
Ax.1. B. 1.
O make a folid angle (P), which shall be contained by three given plane angles (A BC, DEF & GHI), any two of them being greater than the third, and all three together (ABC+VDEF+VGHI) less than four right angles.
1. Take A B at will, & make the fides BC, DE, EF, GH & HI equal to one another & to A B.
2. Draw the bafes AC, D F, & GI.
3. With those three bafes AC, DF & G I make a ▲ LMN fo that N M be=GI, NL AC, & L M=DF.
4. Infcribe the A L M N in a O L M N.
Pof.1. B. 1.
5. From the center O, to the VL, M & N, draw the straight lines LO, ON & O M.
6. At the point O, erect the OP to the plane of the
7. Cut OP fo that the of LO+the
8. Draw the straight lines L P, PN &
of P O be to the of AB.
BECAUSE PO is 1 to the plane of the LMN (Ref. 6.)
of LP. P.47.
1. The APOL will be right angled in O (Ref. 5. & 8.)
AC, & LM = DF, (Ref. 3).
P.40. B. 1. Cor, 3.
AABC, ÁLPM=ADEF, VNPM=VH, VLPNSP. 8. B.1.
VB, & VLPM V E.
But those three NPM, LPN & LPM form a folid
6. Therefore a folid VP has been made, contained by the three given plane B, E & H.
Which was to be done.
PROPOSITION XXIV. THEOREM XXI.
every parallelepiped (A H); the oppofite planes (BD & CF; BE & FG; AF & B H) are fimilar & equal parallelograms.
In the given BF, the plane BD is oppofite to CF, BE to FG & AF to BH.
The oppofite planes B D, C F; BE & FG; AF & BH are=& pgrs. each to each.
Draw the oppofite diagonals E H & AG, alfo AC & D H.
ECAUSE the plle. planes B D & CF are cut by the plane ABCE.
1. The line B A is plle. to E C.
2. Likewife C H is plle. to G B.
And the fame plle. planes B D & C F being alfo cut by the plane
3. The line D G will be plle. to F H.
4. Likewife A E is plle. to B C & D F plle. to G H.
And because thofe plle. planes (Arg. 1. 2. 4.) are the oppofite fides
5. Thofe quadrilateral figures A ECB & DFHG, are pgrs.
7. VABG is to VECH.
But A B is to EC & BG=CH.
8. Therefore the AABG is & to the AECH.
But the pgr. B D is double of the ▲ A B G.
And the Pgr. CF is double of the AECH (P.41. B.1.)
But thofe pgrs. have each an common with the equiangular A.
9. Confequently, the pgrs. B D & C F are & .
D.35. B. 1.
P. 4. B. 1.
4. B. 6.
D. 1. B. 6.
PROPOSITION XXV. THEOREM XXII.
F a parallelepiped (BEDC) be cut by a plane (KIML) parallel to the oppofite planes (AEFB & CGDH); it divides the whole into two parallelepipeds (viz. the BEMK & KMD C), which shall be to one another as their bafes (BFLK & KLHC).
1. Produce B C both ways, as alfo F H.
Pof.2. B. 1.
to BK &
2. In BC produced take any number of lines
Thro' the lines TU, OP & WX let the planes TR, OQ
ECAUSE the lines BO & TO, are each to BK & CW
KC (Prep. 2.) & the lines OP, TU & WX plle. to B F or CH, meet FH produced, in the points, P, U & X (Prep. 3).
P.31. B. 1.