PROPOSITION XXII. THEOREM XX.

C

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.

Thefis.

A▲ may be made of the ftraight lines GI, DF & AC, which fubtend thofe V.

DEMONSTRATION.

The three given V a, b,
CASE I If the V a, b,

BECAUSE

c are either equal, or unequal.
c be equal.

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

Preparation.

1. At the vertex of one of the V as B, make VA BL=Va. P.23.
2. Make B L-DE.

3. Draw LC & LA.

DEMONSTRATION.

B. 1.

P. 3. B. 1.

Pof.i. B. 1.

BECAUS

BECAUSE the two Va+care > \ b (Hyp. 1.) & LB=HG
BCHI (Prep. 2. & Hyp. 2.)

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.

To

PROPOSITION XXIII.

PROBLEM III.

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.
SP.27.
P.22. B.11.
P. 5. B. 4.

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 &

LMN.

of P O be to the of AB.

PM.

P.12. B.11.

M m

BECAUSE PO is 1 to the plane of the LMN (Ref. 6.)

to the

of LP. P.47.
SAx. 1.

1. The APOL will be right angled in O (Ref. 5. & 8.)
2. Confequently, the of PO+ the of O L is
But the of PO+ the of OLDA B, (Ref. 7.)
3. Therefore the of A B is to the of L P, & Á B = L P.
4. Likewife P N & PM are each to A B.
But N Misto GI, NL
5. Confequently, A NMP is

AC, & LM = DF, (Ref. 3).
to the AGHI, A NPL

B. 1.

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

LPN}

P.

6. Therefore a folid VP has been made, contained by the three given plane B, E & H.

Which was to be done.

IN

PROPOSITION XXIV. THEOREM XXI.

every parallelepiped (A H); the oppofite planes (BD & CF; BE & FG; AF & B H) are fimilar & equal parallelograms.

Hypothefis.

In the given BF, the plane BD is oppofite to CF, BE to FG & AF to BH.

Preparation.

Thefis.

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.

BECAUSE

DEMONSTRATION.

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

DGHF.

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
of the quadrilateral figures A ECB & D F HG.

5. Thofe quadrilateral figures A ECB & DFHG, are pgrs.
6. Likewife the other oppofite planes BD & CF; AF & BH are pgrs.
And fince A B & B G are plle. to EC & CH, each to each (Arg. 1.2).

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 & .
10.It may be demonftrated after the fame manner that the pgr. BD is
=&to the pgr. CF, & pgr. A F is & to the pgr. В Н.
11.Therefore the oppofite planes of a Dare & ∞ pgrs.
Which was to be demonftrated.

P.16. B.11.

D.35. B. 1.

P.10. B.11.
P.34
B. 1.

P. 4. B. 1.

P.

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 &

P.

2. In BC produced take any number of lines
CK: as BO & TO each to BK & CW = KC.
3. Thro' thofe points T, O & W, draw the ftraight lines TU,
OP & W X plle. to BF or CH, until they meet the other
plle. produced in the points U, P & X.

4.

Thro' the lines TU, OP & WX let the planes TR, OQ
& W Y pafs, plle. to the planes BE & CD, which will meet
the plane AEDG in SR, NQ & VY.

BECAUSE

DEMONSTRATION.

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.