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Book VI.“ figure ABC to the parallelogram BE, so 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 1.4. Prop. of B. 5. as to conclude that the third is equal to tbe fourth, from the equality of the first and fecond, which is no where demonstrated in the Elements as we row have them. but tho' this Propofition, viz. the third of four proporcjonals is equal to the fourth, if the first be equal to the lecond, had been given in the Elements by Euclid, as very probably it was, yet he wouid not have made use of it in this place, because, as was faid, the conclusion could have been immodiutely deduced vithout this superfluous step by Permutation. this we have fheun 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. 23. B. 11. and twice in Prop. 2. B. 12. and in the 5. 11. 12. and 13. of that Book; in which places of B 12. 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 use of Permutation in the lihe cases, for the Moderns r.ot 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, preoccupied many so much, that they do not sufficiently understand the true nature of them.

Besides, tho' the rectilineal figure ABC, to which another is to be made similar, may be of any kind whaterer, yet in the Demonítration the Greek Text has “ triangle" instead of " rectilineal fi

gurc,” which error is corrected in the above named Oxford Edition.

PROP. XXVII. B. VI. The second Case of this has enaws, otherwise, prefixed to it, as if it was a different Demonstration, which probably has been done by fome unskilful Librarian. Dr. Gregory has rightly left it out the fcheme of this second Cafe ought to be marked with the fame letters of the Alphabet which are in the scheme of the first, as is now done.

PROP.

Book VI. PROP. 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 Cases 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

H K C, according to the determi- L nation. draw DE at right angles to AB, and make it e

GB qual to C; produce ED to F, A so that EF be equal to AD or

с DB, and from the center E, at the distance EF describe a

E 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 rect. angle AH, because GII is equal to GB. therefore the rectangle AH

a. 6. 2.

Book VI. 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 appicd must be equal.

Bitect 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,

I

KH and produced to G, the rectangle AG, GB together with the square of DB is equal a to (the square of

F' Α D BG DG, or DE, that is to) the squares of EB, BD. from each of these

С equals take the square 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 appiied to the given straight line AB, cxceeding 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 fquare 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 fame fide 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

K K

AE again in H; join HF and draw GK parallel to it, and GL pa- Book VI. rallel 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“, and the a. 3. 3. 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 of KE, therefore the square b of AK is not greater than b. 6. 2, the squares of EK, KG, that is than the square of EG; and consequently the straight line

C-
AK or GL is not greater than

D-
GE. Now, if GE be equal to
GL, the circle EHF touches

G
AB in L, and therefore the
square of AL is equal to the HY

F c. 36. 3. rectangle EA, AH, that is to

M L N

A the given rectangle C, D;' and

B В that which was required is · done. but if EG, GL be un

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 4 LN, and d. 3. 3. 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 o EA, AH or the rectangle C, D. but the e.Cor. 36.3. 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

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 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 GL parallel to AE; let the circle meet AB produced

E in M, N, and upon BN describe

C.
the square NBOP, and com-

D
plete the rectangle ANPQ. be-
cause the angle EHF in a semi-
circle is equal to the right angle

G

ОР
EAB, AB and HF are parallels,
and therefore AH and BF are M
equal, and the rectangle EA,
AH equal to the rectangle EA,

H
BF, that is to the rectangle C,

D. and because ML is equal to LN, and AL to LB, therefore MA a. 35. 3. is equal to BN, and the rectangle 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 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 likewise 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, 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.

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