Book VI. tangle 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.

L

E

Bisect AB in D, and draw BE at right angles to it, fo 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, and produced to G, the rectangle AG, GB together with the square 6. 2. of DB is equal a to (the square of DG, or DE, that is to) the squares of EB, BD. from each of these equals take the square of DB, there

KH

FADBG

C

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 muft 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 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 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; alfo EK is equal to KH 2, and 2. 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 of AK is not greater than b. 6. 2. the squares of EK, KG, that

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 defcribe the square NBOP, and complete the rectangle ANPQ. because ML is equal to d 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 given rectangle, it is required to apply a rectangle to AB equal to C, D, exceeding by a square.

Book VI.

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 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 in M, N, and upon BN describe the square NBOP, and complete the rectangle ANPQ. because the angle EHF in a femicircle is equal

E

C

D

F

gle EA, BF, that is to the rectan

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,

35. 3 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 refstored 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 fides 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, 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 fides 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 conclufion may be legitimately made by help of the 24. Prop. of B. 5. as Clavius had done.

PROP. XXXII. B. VI.

The Enuntiation of the preceding 26. Prop. is not general enough; because not only two similar parallelograms that have an angle common to both, are about the fame diameter; but likewise two fimilar parallelograms that have vertically opposite angles, have their diameters in the same straight line. but there seems to have been another, and that a direct Demonstration of these cafes, to which this 32. Proposition was needful. and the 32. may be otherwife and fomething 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 parallel to FH,

Book VI.

and GF to HC; AF and FC are in a A

G

straight line.

F

Draw CK parallel to FH, and let E

D

it meet GF produced in K. because AG, KC are each of them parallel to FH, they are parallel to one another, and B

therefore the alternate angles AGF,

K

Cb. 30. 1.

FKC are equal. and AG is to GF, as (FH to HC, that is c) CK c. 34. 1. to KF; wherefore the triangles AGF, CKF are equiangular d, and d. 6. 6.

the angle AFG equal to the angle CFK. but GFK is a straight

line, therefore AF and FC are in a straight line e.

e. 14. I.

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

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

Book VI.

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; A

G

D

b. 32. 6.

Book XI.

AGF, FHC are joined at one angle, in the point F; wherefore
AF, FC are in the same straight line b.

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 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 center," are left out, as the addition of some unskilful hand.

In the Greek, as also in the Latin Translation, the words ὁ ἔτυχε, " any whatever," are left out in the Demonstration of both parts of the Propofition, 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 aga in the Greek Text ought to be omitted.

The fecond part of the Proposition is an addition of Theon's, as he tell us in his Commentary on Ptolomy's Μεγάλη Συντάξις, P. 50.

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

These three Propositions are added, because they are frequently made use of by Geometers.

DEF. IX. and XI. B. XI.

HE fimilitude of plane figures is defined from the equality of T their angles, and the proportionality of the fides about the equal angles; for from the proportionality of the sides only, or only