has rightly left it out: The scheme of this second case Boo: ought to be marked with the same letters of the alphabet which are in the scheme of the first, as is now done. PROP. XXVIII. and XXIX. B. VI. THESE two problems, to the first of which the 27th prop. is necessary, are the most general and useful of all in the Elements, and are most frequently made use of by the ancient 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; are very often made use of by geometers : And, on this account, it is thought pro-, per, 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 : Bat 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 rectan: gle 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, accord. I. ing to the determination : Draw DE at right angles to AB, and make it equal to C; produce ED to F, so that EF be equal to AD or DB, and from the centre E, at the distance 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 to the square of DB, that is, of • 5. 2. DOK VI. EF or EG, that is, 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 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. 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 centre 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 of of DG, or DE, that is, to) 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.... :1. ? KH Draw AE, BF, at right angles to AB, upon the same side Book VI. of it, and make AE equal to C, and BF to D; join EF, i 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, I teisi: 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 a and the rectangle . 3. S. 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 b of 0 6. 2. AK is not greater than the squares of EK, KG, that is, than the square of EG, and consequently the straight line AK or GL is not greater than GE. Now, if ĞE be equal to GL, the circle EHF touches AB in L, and therefore the square of AL isc equal to the rectangle EA, AH, that is, to the given rect- H angle C, D, and that which A M L NZ was required is done : But if EG, GL, be unequal, EG must be the greater: and 0 I. PO 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 LMis equal tod LN, and it has been proved that AL is equal to d 3. 3. LB; therefore AM is equal to NB, and the rectangle AN, NB, equal to the rectangle NA, AM, that is, to the rectanglee EA, AH; or the rectangle C, D: But the rectangle Cor. 36. 3. 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. € 36, 3. L Brok - VI 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. equal to L., N, and AL to LB, therefore MA is equal to BN, .. 35. 3. and the rectangle AN, NB, to MA, AN, that is, a to the rect angle 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. A Willebrordus Snellius was the first, as far as I know, who gave these constructions of the 3d and ith Problems in his Apollonius Batavus: And afterwards the learned Dr. Hale ley gave them in the Scholium of the 18th Prop. of the Sth Book of Apollonius's Conics restored by him. fedit 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 rectangte, 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 : Ot, 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, Poox VI. PROP. XXXI. B. VI. In the demonstration of this, the inversion of propor. tionals is twice neglected, and is now added, that the conclusion 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 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 their diameters in the same straight lines : But there seems to have been another, and that a direct, demonstration of these cases, to which this 32d Proposition was needful : And the 32d may be otherwise, and something more briefly, demonstrated, as follows: PROP. XXXII. B. VI. Let GAF, HFC, be two triangles which have two sides Draw CK parallel a to FH, and let it meet GF produced in E 31.1. K: Because AG, KC, are each of them parallel to FH, they are parallel b to one another, and PL Berk 30. 1. therefore the alternate angles AGF, FKC, are equal: And AG is to GF, as (FH to HC that isch CK to KF ; wherefore the triangles AGF, CKF . 34. 1. are equiangular, and the angle AFG equal to the angle a 6. 6. CFK: But GFK is a straight line, therefore AF and FC are in a straight linee. e 14. 1. • The 26th Prop. is demonstrated from the 22d, 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. herefore to OntO FH, are each |