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DOX VI. EF or EG, that is, to the squares of ED, DG: Take away

the square of DG from each of these equals; therefore the remaining rectangle AG, GB, is equal to the square of ED, that is, of C: But the rectangle AG, GB, is the rectangle AH, because GH is equal to GB; therefore the rectangle AH is equal to the given square upon the straight line C. Wherefore the rectangle AH, equal to the given square upon C, has been applied to the given straight line AB, deficient by the square GK. Which was to be done.

2. To apply a rectangle which shall be equal to a given square, to a given straight line, exceeding by a square.

Let AB be the given straight line, and let the square upon the given straight line C be that to which the rectangle to be applied must be equal.

Bisect AB in D, and draw BE at right angles to it, so
that BE be equal to C; and having joined DE, from the
centre D at the distance DE describe a circle meeting AB
produced in G; upon BG describe the square BGHK, and
complete the rectangle AGHL. And because AB is bi-
sected in D, and produced to
G, the rectangle AG, GB,

together with the square of
6.2. DB, is equala to (the square І.

IKH
of DG, or DE, that is, to)
the squares of EB, BD.
From each of these equals, F A

B G take the square of DB; therefore the remaining rectangle

С AG, GB, is equal to the square of BE, that is, to the square upon C. But the rect angle AG, GB, is the rectangle AH, because GH is equal. to GB. Therefore the rectangle AH is equal to the square upon C. Wherefore the rectangle AH, equal to the given square upon C, has been applied to the given. straight line AB, exceeding by the square GK. Which was to be done;

3. To apply a rectangle to a given straight line which shall be equal to a given rectangle, and be deficient by a square.' But the given rectangle must not be greater than the square upon the half of the given straight line.

Let AB be the given straight line, and let the given rectangle be that which is contained by the straight lines C, D, which is not greater than the square upon the half of i AB; it is required to apply to AB a rectangle equal, to the rectangle C, D, deficient by a square.

Draw AE, BF, at right angles to AB, upon the same side Book VI. of it, and make AE equal to C, and BF to D; join EF, in and bisect it in G; and from the centre G, at the distance GE, describe a circle meeting AE again in H: Join HF, and draw GK parallel to it, and GL parallel to AE, meeting AB in L.

Because the angle EHF in a semicircle is equal to the right angle EAB, AB and HF are parallels, and AH and BF are parallels; wherefore AH is equal to BF, and the rectangle EA, AH, equal to the rectangle EA, BF, that is, to the rectangle C,

D: And because EG, GF are equal to one another, and AE, LG, BF parallels; therefore AL and LB are equal, also EK is equal to KH a and the rectangle • 3. S. C, D, from the determination, is not greater than the square of AL, the half of AB; wherefore the rectangle EA, AH, is not greater than the square of AL, that is, of KG: Add to each the square of KE ; therefore the square b of 6. 2. AK is not greater than the squares of EK, KG, that is, than the square of EG, and consequently the straight line

CAK or GL is not greater E

D than GE. Now, if GE be equal to GL, the circle EHF touches AB in L, and there

K

G fore the square of AL ise

€ 36. 3. equal to the rectangle EA, AH, that is, to the given rect- H

E angle C, D, and that which

A! M IL N

B was required is done : But if EG, GL, be unequal, EG must be the greater: and 0

Ρ Ο therefore the circle EHF cuts the straight line AB : let it cut it in the points M, N, and upon NB describe the square NBOP, and complete the rectangle ANPQ : Because LM is equal tod LN, and it has been proved that AL is equal to « 3. 3. LB; therefore AM is equal to NB, and the rectangle AN, NB, equal to the rectangle NA, AM, that is, to the rectanglee EA, AH; or the rectangle C, D: But the rectangle Cor. 36. 3. AN, NB, is the rectangle AP, because PN is equal to NB: Therefore the rectangle AP, is equal to the rectangle C, D; and the rectangle AP equal to the given rectangle C, D, has been applied to the given straight line AB, deficient by the square BP.

BP. Which was to be done. 4. To apply a rectangle to a given straight line that shall be equal to a given rectangle, exceeding by a square.

1

Bros VI: Let AB be the given straight line, and the rectangle C,

D, the given rectangle, it is required to apply a rectangle to AB equal to C, D, exceeding by a square.

Draw AE, BF, at right angles to AB, on the contrary sides of it, and make AE equal to C, and BF equal to D: Join EF, and bisect it in G; and from the centre G, at the distance GE, describe a circle meeting AE again in H; join HF, and draw GL parallel to AE; let the circle meet AB produced in M, N, and

E upon BN describe the

C
square BNOP, and com-

D
plete the rectangle ANPW;
because the angle EHF in
a semicircle is equal to the

G
Q

0 P
right angle EAB, AB and
HF are parallels, aud there- ΑΙ

IB
fore AH and BF are equal, M
and the rectangle EA, AH,
equal to the rectangle EA, H

F
BF, that is, to the rectangle C, D: And because ML is

equal to la, N, and AL to LB, therefore MA is equal to BN, • 35. 3. and the rectangle AN, NB, to MA, AN, that is, a to the rect

angle EA, AH, or the rectangle C, D: Therefore the rectangle AN, NB, that is, AP, is equal to the rectangle C,D; and to the given straight line AB the rectangle AP has been applied equal to the given rectangle C, D, exceeding by the square BP. Which was to be done,

Willebrordus Snellius was the first, as far as I know, who gave these constructions of the 3d and ith Problems in bis Apollonius Batavus : And afterwards the learned Dr. Halley gave them in the Scholium of the 18th Prop. of the Sth Book of Apollonius's Conics restored by him. f.

The 3d Problem is otherwise enunciated thus : To cut a given straight line AB in the point N, so as to make the rectangle AN, NB, equal to a given space: Or, which is the same thing, having given AB the sum of the sides of a rectangte, and the magnitude of it being likewise given, to find its sides.

And the 4th Problem is the same with this: To find a point N in the given straight line AB produced, so as to make the rectangle AN, NB, equal to a given space : Or, which is the same thing, having given AB the difference of the sides of a rectangle, and the magnitude of it, to find the

sides.

1. )

Poox VI.

PROP. XXXI. B. VI.

In the demonstration of this, the inversion of proportionals is twice neglected, and is now added, that the conclusion may be legitimately made by help of the 24th Prop. of B. 5, as Clavius had done.

PROP. XXXII. B. VI.

The enunciation of the preceding 26th Prop. is not general enough ; because not only two similar parallelograms that have an angle common to both, are about the same diameter; but likewise two similar parallelograms that have vertically opposite angles, have their diameters in the same straight lines : But there seems to have been another, and that a direct, demonstration of these cases, to which this 32d Proposition was needful : And the 32d may be otherwise, and something more briefly, demonstrated, as follows:

PROP. XXXII. B. VI.

If two triangles which have two sides of the one, &c.

Let GAF, ÉFC, be two triangles which have two sides AG, GF, proportional to the two sides FH, HC, viz. AG to GF, as FH to HC; and let AG be parallel to FH, and GF to HC; AF and FC are in

G

D a straight line. Draw CK parallela to FH,

* 31. 1. and let it meet GF produced in E

H K: Because AG, KC, are each of them parallel to FH, they are parallel to one another, and

BY

• 30. 1. therefore the alternate angles

С AGF, FKC, are equal : And AG is to GF, as (FH to HC, that isc) CK to KF; wherefore the triangles AGF, CKF < 34. I. are equiangulard, and the angle AFG equal to the angle a 6. 6. CFK: But GFK is a straight line, therefore AF and FC are in a straight linee

The 26th Prop. is demonstrated from the 22d, as follows: · If two similar and similarly placed parallelograms have an angle common to both, or vertically opposite angles ; their diameters are in the same straight line.

K

e

e 14. 1.

Book V. First, let the parallelograms ABCD, AEFG, have the

angle BAD common to both, and be similar, and similarly placed : ABCD, AEFG are about the same diameter.

Produce EF, GF, to H, K, and join FA, FC; then because the parallelograms ABCD, AEFG are similar, 'DA is to AB, as GA to AE: Where- A. G

D * Cor. 19. 5. fore the remainder DG isa to the

remainder EB, as GA to AE:
But DG is equal to FH, EB to

E

H HC, and AE to GF: Therefore as FH to HC, so is AG to GF; and FH, HC are parallel to AG, B! GF; and the triangles AGF,

K FHC are joined at one angle in the point F; wherefore " 32. 6. AF, FC are in the same straight lineb.

Next, Let the parallelograms KFHC, GFEA, which are similar and similarly placed, have their angles KFH, GFE vertically opposite; their diameters AF, FC are in the same straight line.

Because AG, GF are parallel to FH, HC; and that AG is to GF, as FH to HC; therefore AF, FC are in the same straight line b.

PROP. XXXIII. B. VI.

The words “because they are at the centre,” are left out as the addition of some unskilful hand.

In the Greek, as also in the Latin translation, the words O, ETUXŞ, “ any whatever," are left out in the demonstration of both parts of the proposition, and are now added as quite necessary; and, in the demonstration of the second part, where the triangle BGC is proved to be equal to CGK, the illative particle apa, in the Greek text, ought to be omitted.

The second part of the proposition is an addition of Theon's, as he tells us in his commentary on Ptolemy's Μεγαλη Συνταξις, p. 50.

PROP. B. C. D. B. VI.

These three propositions are added, because they are free quently made use of by geometers,

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