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Book VI. " figure ABC to the parallelogram BE, fo is the rectilineal KGH

" to the parallelogram EF;" by which, it is plain, he thought it was not so evident to conclude that the second of four proportionals is equal to the fourth from the equality of the first and third, which is a thing demonstrated in the 14. Prop. of B. 5. as to conclude that the third is equal to the fourth, from the equality of the first and second, which is no where demonstrated in the Elements as we now have them. but tho' this Propofition, viz. the third of four proportionals is equal to the fourth, if the first be equal to the fe. cond, had been given in the Elements by Euclid, as very probably it was, yet he would not have made use of it in this place, because, as was said, the conclusion could have been immediately deduced without this fuperfluous step by Permutation. this we have shewn at the greater length, both because it affords a certain proof of the vitiation of the Text of Euclid, for the very fame blunder is found twice in the Greek Text of Prop. 2 3. B. 1 1. and twice in Prop. 2, B. 12. and in the 5. 11. 12. and 18. of that Book; in which places of B. 1 2. except the last of them, it is rightly left out in the Oxford Edition of Commandine's Translation. and also that Geometers may beware of making ufe of Permutation in the like cases, for the Mo lerns not unfrequently commit this mistake, and among others Commandine himself in his Commentary on Prop. 5. B. 3. p. 6. b. of Pappus Alexandrinus, and in other places. the vulgar notion of proportionals, has, it seems, pre-occupied many so much, that they do not sufficiency understand the true nature of them.

Besides, tho' the rectilineal figure ABC, to which anothes is to be made fimilar, may be of any kind whatever, yet in the De. monstration the Greek Text has “ triangle" instead of " rectilineal " figure," which error is corrected in the above-named Oxford Edition.

PRO P. XXVII. B. VI. The second Case of this has anaãs, otherwise, prefixed to it, as if it was a different Demonstration, which probably has been done by some unskilful Librarian. Dr. Gregory has rightly left it out. the scheme of this second Case ought to be marked with the same letters of the Alphabet which are in the scheme of the Grft, as is now done.

Book VI. PRO P. XXVIII. and XXIX. B. VI. These two Problems, to the first of which the 27. Prop. is necessary, are the most general and useful of all in the Elements, and are most frequently made use of by the antient Geometers in the solution of other Problems; and therefore are very ignorantly left out by Tacquet and Dechales in their Editions of the Elements, who pretend that they are scarce of any use. the Cafes of these Problems, wherein it is required to apply a rectangle which shall be equal to a given square, to a given 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 follows.

1. To apply a rectangle which shall be equal to a given square, to 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 of 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,

L

H K according to the determination. draw DE at right angles to AB,

F and make it equal to C; produce ED to F, fo that EF be A D

GB equal to AD or DB, and from the center E, at the distance EF describe a circle meeting

VE AB in G, and upon GB describe the square GBKH, and com. plete 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 to the square of DB, that is of EF or EG, that is 2. 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 En, that is, of C. but the rectangle AG, GB is the rectangle AH, because GH is equal to GB. therefore the reco

Book VI. tangle All is equal to the given square upon the straight line C.

wherefore the rectangle AH equal to the given fquare upon C, has heen applied to the given straight line AB, deficient by the square GK. Which was to be done.

2. To apply a rectangle which fall 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 center 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 bisected in D,

к and produced to G, the rectangle

AG, GB together with the square 8. 6. 2. of DB is equal to (the square of

F A D B G DG, or.DE, that is to) the squares of EB, BD. from each of these e

C quals take the square of DB, there. fore 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.

Iet 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 center G, at the distance GE describe a circle meeting

AE again in H; join HF and draw GK parallel to it, and GL Book VI. 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 BP, 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 ; alfo EK is equal to KH', and a. 3. 3. 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 than the square of AL, that is of KG. add to each the square of KE, therefore the square bof AK is not greater than b. 6. 2. the squares of EK, KG, that is than the square of EG; and

Cconsequently the straight line

E AK or GL is not greater than

1 GE. Now, if GE be equal

IK to GL, the circle EHF touches AB in L, and therefore the square of AL is equal to the H

F

C. 36. 3. rectangle EA, AH, that is to M L the given rectangle C,D; and

Б that which was required is done. but if LG, GL be un

Q

PO equal, 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 ML is equal to LN, and it has been proved that AL is d. 3. 3. 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, e.Cor. 36.3. 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

L N

А

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Book VI. 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 bifeat
it in G, and from the center 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 upon BN describe

E
the square NBOP, and complete
the rectangle ANPQ. because the
angle EHF in a semicircle is equal
to the right angle EAB, AB and

G
HF are parallels, and therefore AH

OP
and BF are equal, and the rectan-

А

B gle EA, AH equal to the rectan-M

N gle EA, BF, that is to the rectan

H
gle 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, 2. 35. 3. that is to the rectangle EA, AH or the rectangle C, D. there

fore 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 3. and 4. Problems in his Apollonius Batavus. and afterwards the learned Dr. Halley gave them in the Scholium of the 18. Prop. of the 8. B. of Apollonius's Conics restored by him.

The 3. Problem is otherwise enuntiated 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 likewife given, to find its fides.

And the 4. Problem is the same with this, To find a point N in the given straight line AB produced, fo 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.

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