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rectangle which shall be equal to a given square; to a given Book VI. straight line, either deficient or exceeding by a square; as also to apply a rectangle which shall be equal to another given, to a given straight line, deficient or exceeding by a square; are very often made use of by geometers: And, on this account, it is thought proper, for the sake of beginners, to give their constructions as follow:

1. To apply a rectangle which shall be equal to a given square, lo a given straight line, deficient by a square ; but the given square must not be greater than that upon the half of the given line.

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, and this square, by the determination, is not greater than that upon half the straight line AB.

Bisect AB in D, and if the square upon AD be equal to the

square upon C, the thing required is done : But if it be not equal to it, AD must be greater than C, accord

L

H K ing to the determination :

F Draw DE at right angles to AB, and make it equal

А.

D to C; produce ED to F,

GB so that EF may be equal to AD or DB, and from

с the centre E, at the dis

E tance EF, describe a circle meeting AB in G, and upon GB describe the square GBKH, and complete the rectangle AGHL; also join EG: And because AB is bisected in D, the rectangle AG, GB, together with the square of DG is equal a to (the square of DB, that is, of EF or EG, that is a 5. 2. 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 A H 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.

х

Book VI. Bisect AB in D, and draw BE at right angles to it, so

that BE may 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,

E
and complete the rectangle
AGHL. And because AB is L

к\Н
bisected in D, and produced
to G, the rectangle AG, GB
together with the square of F A D B

G a 6. 2. DB is equala to (the square of DG, or DE, that is, to) the

C squares of EB, BD.

From each of these equals take the square of DB;

of DB; therefore the remaining rectangle AG, GB is equal to the square of BE, that is, to the square 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 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 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 of it, and make AE equal to C, and BF 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 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 an

other, and AE, LG, BF parallels: therefore AL and LB are a 3. 3. equal, also EK is equal to KH“, and the rectangle 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

b 6. 2.

than the square of AL, that is of KG: Add to each the square Book VI. of KE; therefore the square of AK is not greater than the squares of EK, KG; that is, than the square of

C EG; and consequently the

E straight line AK or GL

D is not greater than GE.

K

G Now, if GE be equal to GL, the circle EHF touches AB in L, and therefore the square of H

F AL is equal to the rect

M L N

c 36. 3. A

B angle EA, AH, that is, to the given rectangle C, D; and that which was Q Q

Ρ Ο required is done: But if EG, GL be unequal, EG must be the greater: And 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 to LN, and it has been proved that Al is equal to LB; d 3. 3. therefore AM is equal to NB, and the rectangle AN, NB equal to the rectangle NA, AM, that is, to the rectangle e e Cor.36.3. EA, AH, or the rectangle C, D: But the rectangle 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. 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.

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, ex

E ceeding by a square.

D D Draw AE, BF at right angles to AB, on the contrary sides of it, and make A E equal

G

OP to C, and BF equal to D: Join EF, and bisect it in G; and

L

B from the centre G, at the dis

M

N tance GE, describe a circle meeting AE again in H; join

H

Book VI. HF, and draw GL parallel to AE; let the circle meet AB

produced in M, N, and upon BN describe the square NBOP, and complete the rectangle ANPQ; because the angle EHF in a semicircle is equal to the right angle EAB, AB, and HF are parallels, and therefore A# and BF are equal, and the rectangle EA, AH equal to the rectangle 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 recta 35. 3. angle AN, NB to MA, AN, that is a 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 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 4th Problems in his “ Apollonius Batavus :" And afterwards the learned Dr Halley gave them in the Scholium of the 18th Prop. of the 8th Book of Apollonius's Conics restored by him.

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 rectangle, 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.

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 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 tlieir diameters in the same straight line: But there seems to have been another, and that a direct demonstration of these cases, to which this 32d Proposition Book VI. 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, HFC be two triangles which have two sides AG, GF, proportional to the two sides FH, HC, viz. AG to FG, as FH to HC; and let AG be parallel to FH, and GF to HC; AF and FC are in a straight line.

Draw CK parallel a to FH, and let it meet GF produced a 31. 1. in K. Because AG, KC are each of them parallel to FH, they are

G parallelo to one another, and there

A

D

b 30. 1, fore the alternate angles AGF,

F FKC are equal : And AG is to E

H GF, as (FH to HC, that is c) CK

c 34. 1. to KF; wherefore the triangles AGF, CKF are equiangular",

B

d 6. 6. and the angle AFG equal to the

c

K angle CFK: But GFK is a straight line; therefore AF and FC are in a straight line .

e 14. 1. The 26th Prop. is demonstrated from the 32d 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.

First, Let the parallelograms ABCD, AEFG have the angle BAD common to both, and be similar, and similarly placed, ABCD, A EFG 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: Wherefore the remainder DG is a to a Cor.19.5. the remainder EB, as GA to AE: But DG is equal to FH, EB to HC, and AE to GF: Therefore as FH to HC, so 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 same straight line 6.

b 32. 6. Next, Let the parallelograms KFHČ, GFEA, which are similar, and similarly placed, bave 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 AG is to GF, as FH to HC; therefore AF, FC are in the same straight Jine b.

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