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BECAU 5

ECAUSE FG is plle. to D E which is I to the plane BAC. (Hyp.111). 1. The line G F is I to the same plane B A C.

P. 8. B.II. And the V FGB, FGA & F G C are L.

D. 3. B.I. 2. Confequently, the o of A F is = to o of FG+ 0 of GA. P.47. B. 1.

But the [] of AG is = to D of AB + of B G. (Prep.3). Eg P.47. B. 1. 3. Therefore, the of A F is = to D'FG+ OAB + O BG. Ax.1. B. 1. But

the O GB + OF G are = to the OB F ( Prep.3). P.47. B. I. 4. Consequently, the AF is also = to the O BF + O AB. 5. Therefore, V A BF, is a L

P.48. B. 1. 6. It may

be demonstrated after the same manner that VFCA, is a L. 7. That also the VKIH & KLH, are Le

In the AFCA & KLH; the line H K is = to AF (Prep. 1.) the VACF & KL H, are L ( Arg. 6. & 7.), & the V FACE KHL, (Hyp. 1).

P-26. B. i. 8. Therefore the sides A C & C F are = to the sides HL & L K, each

to cach. 9. Likewise A B is = to HI & BF=IK. 10.Confequently, in the ABAC & IHL; the bases B C & I L are

equal and the V ACB & A B C = to the FHLI & HIL,
each to each.

P.
4.

B. 1. Therefore if those equal , be taken from the four LACG,

ABG, HLM & HIM. 11. The remaining will be equal, viz. VBCG=VILM & V C B G = V LIM.

Ax 5.B. 1. Since then the AGBC & IML have their bases B C & IL equal (Arg. 10).

And the V at those bases are equal, each to each, (Arg. 11). 12. The sides B G & C G will be = to the Gides IM & ML. P.26. B. 1.

In the ABAG & HIM, A B is = to HI (Arg. 9.) BG=IM, (Arg. 12.) & the V ABG & HIM are Lo. (Prep. 3.

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13. Consequently, AG=HM

P. 4. B. 1. But the of AF(=OAGO GF) (Arg. 2.) is = to the

of HK=OHM + KM) (Hyp. 1. & P. 47. B. 1.) because AF is = HK. (Prep. 1). If therefore from the OAF be taken the O GÀ, & from the HK, the DHM=OGA, (Arg. 13. & P. 46. B. 1. Cor 3). 1. The remainder, viz. the o of Gf will be = to the of K M. Ax.3. B. 1. 15. Consequently, GF=KM (Cor. 3. of P. 46. B. 1).

Infine, because in the two A AĞÉ & HKM, the sides AF,
AG & F G are = to the sides HK, HM & KM, each to each,

(Prep. 1. & Arg. 13. & 15).
16.The y FAG or Ď AE is = to the Y KH M.

P. 8. B. 1. Which was to be demonstrated.

14.

CORO L A R r.

F elevated two equal Araight lines A F & H K; containing with the respedive fides, ibe V BAF & FÁC equal to the VIHK & KHL; each to each, & there be let fall from those points F & K (of those elevated straight lines) the perpendiculars F G & KM on the planes B AC & IHL: tbose IFG & KM will be equal. (Arg. 15).

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PROPOSITION XXXVI. THEOREM XXXI. IF F three straight lines (A, B, C) be proportionals, the parallelepiped (D N),

A, D described from these three lines as its sides, is equal to the equiangular parallelepiped (EI), described from the mean proportional (B). Hypothesis.

Thesis. i. The raight lines A, B, C are proportionals, that The Elis=!o the EDN.

is, A: B = B : C. 11. The EPDN, is described from those three lines that

is, DK — A, MK=B, E KL=C. Ill. The equiangular EET, is described from the

mean f10; ortional B, that is, EF=FG=FH=B.

DEMONSTRATION.

Becau

ECAUSE DK:EF=E F or FH:KL (Hyp. 2).

And the plane VEFH is = to the plane VDKL (Hyp. 3). 1. The pgr. D L, base of E DN is = to the pgr. EH, base of ĆEI P.14. B. 6.

Moreover, the plane VGFE & GFH contained by the elevated
line F G, & the sides E F & FH, being = to the plane VMKD,
& MKL, contained by the elevated line KM, & D K, & KL,
each to each, (Hyp. 3.),, & F G = KM, (Hyp. 2. & 3). «
The I let fall from the point G, on the base É H, will
be = 10 the I let fall from the point M on the base D L.

(Cor. of P. 35. B. 11). 3. Consequently, EE I has the same altitude with ihe ET DN. D. 4. B. 6.

But the base E H of E I is = to the base D L of DN,

(Arg. 1). 4. Therefore, El is to the SDN.

P.31. B.11. Which was to be demonstrated.

2.

A

B

С

D

PROPOSITION XXXVII. THEOREM XXXII.

F four straight lines (A, B, C, & D) be proportionals, (that is, if, A : B = C: D: the similar and similarly described parallelepipeds, from the two first (A & B), will be proportional to the similar and similarly described parallelepipeds, from the two laft (C & D); and if the two similar and similarly described parallelepipeds, from the two lines (A & B); be proporcional to the two other similar and similarly described parallelepipeds, from the two other straight lines (C & D); the homologous sides of the first · (A & B), will be proportional to the homologous sides (C & D) of the last. Hypothesis.

Thesis. 1. A: B = C: D.

BA:B = EC: BD. II. From A & B there bas been described III. Also from C & D.

DemonSTRATION.

BECAU

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1. The

ECAUSE the 6 A is sy to the EB (Hyp. 2).
EA:B = A3 : B3.

P.33. B.11. 2. Likewise, the EC:ED. C3 : D:.

Cor. But the ratio of A to B being to the ratio of C to D (Hyp. 1). 3. It follows, that three times the ratio of A to B is = to three times the ratio of C to D, that is, A3: B3 = CS : D3.

Ax.6. B. 1. 4. Consequently, the A: B= C:D

P.11. B. 5.

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BECAUS

P.33. B.15. P.33. B.15.

ECAUSE the BA is to the BB (Hyp. 1). 1. The BA:6B= A*: Bs.

Likewife the EC is to the ED (Hyp. 2). 2. The 60:8D=(3:D2.

But the A:B=OC:ED (Hyp. 3). 3

Therefore, Aa : B3 = C3: D3. 4. Consequently, A:B=C:D.

Which was to be demonstrated.

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R E M A R K. BECAUSE I. ECAUSE the triangular prism is the half of its parallelepiped

(P. 28. B. 11.), it follows (Ax. 7. B. 1), that the same truth is applicable to

fimilur triangular prisms. 11. It may be also applied to similar polygon prisms; because they may be divided

by planes into triangular prisms. ( Remark 2. of P. 34. B. 11).

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