BAC, EDF, in which the Angles BAC, EDF, firft given are; and right Lines AI, DK, be drawn to the Angles firft given, from the Points I, K, made by the Perpendiculars in the Planes; thefe right Lines will contain equal Angles GAM, HDK, with the elevated Lines AG, DH.

a

Make DH, AL equal, and GI, LM parallel; and MC perpendicular to AC; MB to AB; KF to DF; and KE to DE; and draw the right Lines BC, LB, LC, and EF, HF, HE; 28. 11. and let LM be perpendicular to the Plane BAC; and fo the Angles LMC, LMA, LMB b3 def.11. are; and by the fame Reason, the Angles HKF, HKD, HKE are right ones. Therefore AL LM2+ AM = LM+CM2+ AC2 =LC2+ AC; whence the Ang. ACL is

€47. I.

d 48. I.

47. I.

f 26. I.

4. I.

2

g

2

2

2

d

с

2

2

= ;

a right one. Again, AL = LM + MA = EM* + BM2 + BA* = BL + BA. Therefore the Ang. ABL is also a right one By the like Argument, the Angles DFH, DEH are right ones. Therefore AB f = DE; and BLEH ; and AC f = DF; and CL FH whence alfo BC EF, and the Ang. ABC DEF, and the Ang. ACB = DFE. Therefore the remainders of the right Angles CBM, h 3 ax. 1. BCM are h equal to the remainders FEK, 26.1. EFK. Therefore CM = FK, and fo* AM = DK. Whence if from LA2 = HD2 be taken away AM2 = DK', there will HK. Wherefore the Triangles LAM, HDK, are mutually equilateral to each other. Therefore the Ang. LAM" = HDK. Q. E. D. COROL.

47. I.

1

conft.

47. I.
3 ax. LM2

8.1.

i

1

remain

Hence if there be two plain Angles equal, and from the Vertices of thofe Angles, two right Lines be elevated, containing equal Angles

with the Lines firft given, each to each; then fhall Perpendiculars let fall from the extreme Points of the elevated Lines to the Planes of the Angles first given, be equal to one another, viz. LMHK.

PROP. XXXVI.

If three right Lines A (DE), B (DG) C (DF) be proportional, a folid Parallelepip. DH made of them, is equal to the folid Parallelepipedon IN made of the mean or middle Line DG (IL) if it be an equilateral one, and equiangular to the aforefaid one DH.

a

a hyp.

Because DE: IK :: IL: DF; therefore the a Pgr. LK FE, and on account of the Equa-14. 6. lity of the Plain Angles at E and I, and of the Lines GD, IM, the Altitudes of the Parallelepipedons are equal (from the preced. Corol.) therefore the Parallelepipedons themselves are 31. 11. equal to one another. Q. E. D.

PROP. XXXVII.

с

If four right Lines A, B, C, D, be proportional, the Solid Parallelepipedons A,B,C,D, fimilar, and in ike manner defcribed from them, will be proportional ;

and

33. II.

b Schol. 23.

5.

12. I.

C

and if folid Parallelepipedons being fimilar and alike defcribed, be proportional, (A: B:: C: D) then shall the right Lines A, B, C, D, they are described from, be proportional.

a

For the Ratio's of the Parallelepipedons are triplicate of the Ratio's of the Lines to one another. Therefore if A: B:: C: D, then shall the Ppp. A: Ppp. B:: Ppp. C: Ppp. D. and contrary-wife.

b

A

PROP.

IB

E

D

XXXVIII.

If the Plane AB be at right Angles to the Plane AC, and a Line EF be drawn from a Point E in one of the Planes (AB) perpendicular

to the other Plane AC; that Perpendicular EF, fball fall in AD the Common Section.

If poffible, let F fall without the Intersection

с

AD, draw FG in the Plane AC perpendicular

to AD, and join EG. Now the Angle FGE

a

is a right one; and EFG is fuppofed to be a 24 & 3. right Angle. Therefore in one Triangle EFG, def. 11. there are two right Angles; which is abfurd.

PROP. XXXIX.

b

17. I.

divide each other into two equal Parts.

d

с

с

f

are

a

с

е

4. I.

f Schol. 15.

1.

g

Draw the right Lines, SA, SC, TD, TB. Now because the Sides DO, OT, are equalto 34. I. the Sides BQ, QT, the alternate Angles TOD, d 29. 1. TQB are fo alfo ; therefore the Bafes DT, TB, as alfo the alternate Angles DTO, BTQ, equal to each other. Therefore VTB is right Line. In like manner, ASC is a right Line. 8 34. I. Again, AD is equal and parallel to FG, FG 8 to CB, and confequently AD to CB, and AC i to DB. Therefore AB, ST are in the fame Plane ABCD. And fo fince the Angles AVS, BVT, at the Vertex, and the alternate Angles ASV, BTV, are equal; and AS 'BT; there- 17 ax. I. fore shall AV BV, and SV "VT. Q. E.D.

COROL

h

Hence all Diameters of a Parallelepipedon do mutually bifect one another, in one Point, as V. PROP.

PROP. XL.

If there be two Prisms ABCEFD, GHMLIK, of the fame Altitude, one of which has a Parallelogram ABCF for its Bafe, and the other a Triangle GHM; and if the Parallelogram ABCF be the Double of the Triangle GHM; the Said Prisms ABCEFD, GHMLIK are equal.

a

equal on account

h

and Altitudes ;

For if the Parallelepipedons AN, GQ be compleated, thefe fhall be of the Equality of the Bafes

d

therefore likewife thefe Prifms, the Halves of them, fhall be equal.

SCHOL.

From what has been hitherto demonftrated, we have the Dimenfions of Triangular, or Quadrangular Prifms or Parallelepipedons; which is done by drawing or multiplying the Base into the Altitude.

As if the Altitude be 10 Feet, and the Bafe 100 fquare Feet, (which is measured by Schol. 35. 1. or by 41. 1.) multiply 100 by 10; and thence will arife 1000 Cubical Feet for the Solidity of the given Prism.

For as a Rectangle, fo likewife is a rectangular Parallelepipedon, produced by multiplying or drawing the Altitude into the Base. Therefore

any