Book VI. to) the fquares of ED, DG: Take away the fquare of DG from each of thefe equals; therefore the remaining rectangle AG, GB is equal to the fquare of ED, that is, of C: But the rectangle AG, GB is the rectangle AH, becaufe GH is equal to GB; therefore the rectangle AH is equal to the given fquare upon the ftraight line C. Wherefore the rectangle AH, equal to the given fquare upon C, has been applied to the given ftraight line AB, deficient by the fquare GK. Which was to be done.

a 6. 2.

2. To apply a rectangle which fhall be equal to a given fquare, to a given ftraight line, exceeding by a fquare.

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

Bifect AB in D, and draw BE at right angles to it, fo that BE be equal to C; and having joined DE, from the centre D at the distance DE defcribe a circle meeting AB produced in G; upon BG defcribe the square BGHK, and complete the rectangle AGHL. And becaufe AB is bifected in D, and produced to G, the rectangle AG, GB together with the fquare of DB

L

E

KH

B G

C

is equal to (the fquare of DG, FA D
or DE, that is, to) the fquares
of EB, BD. From each of thefe
equals take the fquare of DB;

therefore the remaining rectangle AG, GB is equal to the
fquare of BE, that is, to the fquare upon C. But the rectangle
AG, GB is the rectangle AH, because GH is equal to GB:
Therefore the rectangle AH is equal to the fquare upon C.
Wherefore the rectangle AH, equal to the given square upon
C, has been applied to the given ftraight line AB, exceeding
by the fquare 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 fquare. But the given rectangle muft not be greater than the square upon the half of the given ftraight line.

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

Draw

Draw AE, BF at right angles to AB, upon the fame fide of it, Book VI. and make AE equal to C, and BF to D: Join EF and bifect it in G; and from the centre G, at the distance GE, defcribe 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 femicircle 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 becaufe EG, GF are equal to one another, and AE, LG, BF parallels; therefore AL and LB are equal; alfo EK is equal to KH, and the rectangle C, D from the a 3. 3. determination, is not greater than the fquare of AL the half of AB; wherefore the rectangle EA, AH is not greater than the fquare of AL, that is of KG: Add to each the square of KE; therefore the fquare of AK is not greater than the b 6. 2. fquares of EK, KG, that is, than the fquare of EG; and confequently the kraight line AK or GL is not greater than GE. Now, if GE be equal to GL, the circle EHF touches AB in L, and therefore the fquare of AL is equal to the rectangle EA, AH, that is, to the given rectangle C, D; and that which was required is done : But if EG, GL be unequal, EG must be the greater; and

E

C

D

G

H

c 36. 3.

A

M

T

N

3

Q

PO

therefore the circle EHF cuts the ftraight line AB; let it cut it in the points M, N, and upon NB defcribe the fquare NBOP, and complete the rectangle ANPQ: Because ML is equal to LN, d 3. 3. and it has been proved that AL is equal to LB; therefore AM is equal to NB, and the rectangle AN, NB equal to the rectangle NA, AM, that is, to the rectangle EA, AH or the e Cor. 36. rectangle C, D: But the rectangle AN, NB is the rectangle 3. AP, becaufe 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 ftraight line AB, deficient by the fquare BP. Which was to be done. Y

4. Ta

Book VI.

4. To apply a rectangle to a given straight line that shall be equal to a given rectangle, exceeding by a fquare.

Let AB be the given ftraight 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 fquare.

E

Draw AE, BF at right angles to AB, on the contrary fides of it, and make AE equal to C, and BF equal to D: Join EF, and bifect it in G; and from the centre G. at the diftance GE, defcribe 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 upon BN defcribe the fquare NBOP, and complete the rectangle ANPQ; because the angle EHF in a femicircle is equal to the right angle EAB.

C

D

G

OP

L

B

AB and HF are parallels, and M
therefore AH and BF are e.
qual, and the rectangle EA,

N

H

AH equal to the rectangle

a 35. 3.

EA, BF, that is, to the rectangle C, D: And because ML is equal to LN, and AL to LB, therefore MA is equal to BN, and the rectangle AN, NB to MA, AN, that is, to the rectangle 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 ftraight line AB the rectangle AP has been applied equal to the given rectangle C, D, exceeding by the fquare BP. Which was to be done.

Willebrordus Snellius was the firft, as far as I know, who gave thefe conftructions of the 3d and 4th problems in his Appollonius Batavus: And afterwards the learned Dr Halley gave them in the Scholium of the 18th prop. of the 8th book of Apollonius's conics reftored by him.

The 3d problem is otherwife enunciated thus: To cut a given ftraight line AB in the point N, so as to make the rectangle AN, NB equal to a given fpace: Or, which is the fame thing, having given AB the fum of the fides of a rectangle, and the magnitude of it being likewife given, to find its fides.

And the 4th problem is the fame with this, To find a point
N in

N in the given ftraight line AB produced, fo as to make the Book VI. rectangle AN, NB equal to a given fpace: Or, which is the in fame thing, having given AB the difference of the fides of a rectangle, and the magnitude of it, to find the fides.

PROP. XXXI. B. VI.

In the demonftration of this, the inverfion of proportionals is twice neglected, and is now added, that the conclufion 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 fimilar parallelograms that have an angle common to both, are about the fame diameter; but likewife two fimilar parallelograms that have vertically oppofite angles, have their diameters in the fame ftraight line: But there feems to have been another, and that a direct demonstration of thefe cafes, to which this 32d propofition was needful: And the 32d may be otherwife and something more briefly demonstrated as follows.

PROP. XXXII. B. VI.

If two triangles which have two fides of the one, &c.
Let GAF, HFC be two triangles which have two fides AG,

GF, proportional to the two fides FH, HC, viz. AG to GF, as

FH to HC; and let AG be paral

lel to FH, and GF to HC; AF A

alternate angles AGF, FKC are e

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

The 26th prop. is demonftrated from the 32d, as follows. If two fimilar and fimilarly placed parallelograms have an angle common to both, or vertically oppofite angles; their diameters are in the fame ftraight line.

Y 2

Firfa

e 14. I.

Book VI.

First, Let the parallelograms ABCD, AEFG have the angle BAD common to both, and be fimilar, and fimilarly placed; ABCD, AEFG are about the fame diameter.

Produce EF, GF, to H, K, and join FA, FC: Then because the parallelograms ABCD, AEFG are fimilar, DA is to AB, as GA to AE; where

a Cor. 19, fore the remainder DG is to the A

5.

b 32.6.

E

G

remainder EB, as GA to AE: But
DG is equal to FH, EB to HC,
and AE to GF: Therefore as FH
to HC, fo is AG to GF; and
FH, HC are parallel to AG, GF;
and the triangles AGF, FHC are
joined at one angle, in the point
F; wherefore AF, FC are in the fame straight line .

B

K

D

H

C

Next, Let the parallelograms KFHC, GFEA, which are fimilar and fimilarly placed, have their angles KFH, GFE vertically oppofite; their diameters AF, FC are in the fame ftraight 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 fame ftraight line b.

PROP. XXXIII. B. VI.

The words "because they are at the centre," are left out, as the addition of fome unskilful hand.

ά ετυχε,

In the Greek, as alfo in the Latin tranflation, the words TUX, "any whatever," are left out in the demonftration of both parts of the propofition, and are now added as quite necef fary; and, in the demonftration of the fecond 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 fecond part of the propofition is an addition of Theon's, as he tells us in his commentary on Ptolomy's Μεγάλη Συντάξις, p. 50.

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

Thefe three propofitions are added, because they are frequently made ufe of by geometers.

DEF.