1. The pgrs. TP & BP are

to the pgr. BL; & pgr. CX= pgr. KH.P.35. B. 1. The planes AR or AQ & TF or OF being plle; & the plane NP plle. to the plane AF; moreover the lines SA & R E being plle. to the lines B T or F U. 2. The folid OQ E B will be a

3.

= & to the

BEM K. D.10. B.11. It may be demonstrated after the fame manner that the folid TRQO is & to EBEMK; alfo the folid CD YW is = & ∞ to

KMD C.

But there are as many equalOQEB, &c. as there are equal pgrs.
OF, TP, &c. & thofe together compofe the TE: more-
over there are as many equal pgrs. O F, &c. as there has been taken
ftraight lines, each to B K, which together are to T B.
Confequently, the TE is the fame multiple of the
that the parts (TO, OB) of the line TB taken together, are
multiples of the line B K.

=

5. Likewife the CDYW is the fame multiple of the that the line WC is of the line K C.

BEMK, the line TB will be >, or < the line
CDY W is >, or <

the line C W will be >, or<the line K C.

BEMK

KMDC

or < the
B K.

KMDC,

7. Confequently, the BEMK: KMDC=BK: KC. But BK: KC bafe B L: bafe K H.

8. Therefore BEMK: KMDC= bafe B L: bafe K H.

Which was to be demonftrated.

AT

Ta given point (A) in a given ftraight line (A B), to make a folid angle equal to a given folid angle (F).

Given.

I. A point A in a straight line A B. II. A falid angle F.

Refolution.

1. From any point I in one of the fections about the folid F, let fall a IL upon the oppofite plane G F H.

2. Draw LF, LG, LH, HI & GI in the planes which form the folid V.

3. In the given ftraight line A B, take A MFG.

4. At the point A, make a plane \ MAD the plane V GF H. 5. Cut off A D=FH.

6. In the fame plane M A D, make a plane Y MAE to the plane VGFL.

7. Cut off A E = FL.

8. At the point E, in the plane MAD erect the LEC.

9. Make EC

LI,

10.Draw A C.

Preparation.

Draw ME, ED, CD & C M in the planes, MA D, CAD & MAC.

P.11. B.11.

Pof.1. B. 1. P. 3. B. 1. P.23

B. 1.

P.

3. B. I

P.23. B. 1. P. 3. B. I. P.12. B.11.

Pofi. B. 1.

BECAUSE

DEMONSTRATION.

ECAUSE in the AG FH & MAD, the fides FG & FH

are to the fides A M & A D, each to each, (Ref. 3. & 5.) & ·
to V MAD, (Ref. 4).
to MD.

VGFH is

1. G H will be

2. Likewife in the AG FL & AME, GL is to ME.

3.

4.

Therefore if GL be taken from G H & ME from MD.
LH will be to ED.

=

And fince in the ALHI & EDC, ED is to LH, LI=
EC & the DEC & HLI, are L, (Arg. 3. Ref. 9. & D.3. B. 11).

1 H will be to C D.

Likewife in the AFLI & AEC, LI is to E C, & LF = AE, befides VFLI & VAEC, are L, (Ref. 7. 9. & D. 3. B.11). 5. Therefore FIAC.

6. It may be demonstrated after the fame manner that GI is MC. Since then the three fides HI, FI & FH of the AIFH are

to the three fides DC, AC & AD, of the CAD (Arg.4. &5).
VIFH will be to CA D.
Likewife AG FI is to the AMA C & VGFIVMAC.
Therefore the plane VGFH being to the plane V MAD,
(Ref. 4)

The plane IFH to the plane VCAD (Arg. 7).
And the plane VGFI to the plane V MAC, (Arg. 8).
Befides the plane VGFH, IF H & GFI, form a folid V F.
And the plane MAD, CAD & MAC, fimilarly fituated as these
already mentioned, form the folid

A.

9. It follows that the folid A is to the folid \ F.

P. 4. B. 1.

Ax.3. B. 1.

B. 1.

P. 4. B. 1.

P. 8. B. 1.

D. 9. B.11.

Which was to be done.

To

PROPOSITION XXVII. PROBLEM V.

O defcribe from a given ftraight line (A B), a parallelepiped fimilar, & fimilarly fituated to one given (HN).

1. At the point A in the line A B make a folid V CADB, =
to the folid VH, or LH MI.

2. Cut A C fo that HI:HL=AB: A C.
3. Alfo A D fo that HL: HMAC: A D. J
4. Complete the pgrs. A E, BD & BC.
5. Complete the AF.

THE

DEMONSTRATION.

HE three pgrs. A E, BD & BC being with the three pgrs. HG, MI & LI of the (Ref. 1. 2. 3. & 4. & D. 1. B. 6).

As alfo their oppofite ones.

P.26. B.11.

P.12. B. 6.

P.31. B. 1.

& fimilarly fituated H N, each to each

P.24. B.11.

AF, are

1. Confequently, the fix planes or pgrs. which form the

, & fimilarly fituated to the fix planes or pgrs. which form the

given

HŃ.

2. Therefore the

fituated to the given

AF defcribed from A B, is fimilar & fimilarly

IF

A

THEOREM XXIII.

F a parallelepiped (A B) be cut by a plane (FCDE) paffing thro' the diagonals (FC & ED) of the oppofite planes (BG & AH): it fhall be cut into two equal parts.

5.

From whence it follows that FCDE is a pgr.

P.24. B.11. P.33. B.11. P. 9. B.11. Ax.1. B. 1. P.33. B. 1.

D.35. B. 1.
P.24. B.11.

to the AHDEP.34. B. 1.

But the pgr. B C G F is & plle. to the pgr. H DAE. 6. Confequently, the ABCF & F G C are

& EDA.

&

Moreover, the pgrs. FE AG & GADC, are = & to the pgrs.
BHDC & BHEF, each to each.

7. Therefore all the planes which form the prifm B F D are & no
to all the planes which form the prifm DF G.

8. Therefore the prifm BFD or BHEDCF is & to the prifm D F G or DEFCG A.

9. Confequently, the plane F C D E, cuts the AB into two equal

parts.

Which was to be demonftrated.

P. 4. B. 1.

P.24. B.11.

D.10. B.11.

N n