Book VI. tangle AH is equal to the given fquare 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 fhall 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 fquare upon the given straight line C be that to which the rectangle to be applied muft be equal. Bifect 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 defcribe the fquare BGHK, and complete the rectangle AGHL. and because AB is bifected in D, E KH FAD B G с 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 thefe equals take the fquare of DB, therefore the remaining rectangle AG, GB is equal to the fquare of BE, that is to the fquare 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 fquare upon C. wherefore the rectangle AH equal to the given square upon C, has been applied to the given ftraight line AB, exceeding by the fquare GK. Which was to be done. 3. To apply a rectangle to a given straight line which fhall be equal to a given rectangle, and be deficient by a fquare. but the given rectangle muft not be greater than the fquare upon the half of the given ftraight line. Let AB be the given ftraight line, and let the given rectangle be that which is contained by the flraight lines C, D, which is not greater than the fquare upon the half of AB. it is required to apply to AB a rectangle equal to the rectangle C, D, deficient by a fquare. 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 bifect it in G, and from the center G, at the distance GE defcribe 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 a, and a. 3. 3. the rectangle C, D, from the determination, is not greater than the fquare of AL the half of AB, wherefore the rectangle EA, AH is not greater than the fquare of AL, that is of KG. add to each the fquare of KE, therefore the fquare of AK is not greater than b. 6. 2. the fquares of EK, KG, that is than the fquare of EG; and confequently the ftraight line AK or GL is not greater than GE. Now, if GE be equal to GL, the circle EHF touches AB in L, and therefore the square of AL is equal to the rectangle LA, AH, that is to the given rectangle C, D; and that which was required is done. but if EG, GL be un E H K C D. G F c. 36. 3. MLN B Q PO A 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 fquare 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 e 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 fquare 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 fquare. 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 fquare. 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 defcribe D OP B N MA is equal to BN, and the rectangle AN, NB to MA, AN, 3. 35. 3 that is a 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 fquare 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 fame thing, Having given AB the fum of the fides of a rectangle, and the magnitude of it being likewife given, to find its fides. And the 4. Problem is the fame 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 fides of a rectangle, and the magnitude of it, to find the fides, PROP. XXXI. B. VI. In the Demonftration of this the inverfion of proportionals is twice neglected, and is now added, that the conclusion 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 fimilar 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 Demonftration of these cafes, to which this 32. Propofition was needful. and the 32. may be otherwise and something 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, and GF to HC; AF and FC are in a A ftraight line. Draw CK parallel a to FH, and let E 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, G K Book VI. a. 31. I, C b. 30. 1. FKC are equal. and AG is to GF, as (FH to HC, that is c) CK c. 34. I. to KF; wherefore the triangles AGF, CKF are equiangular a, 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 ". 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 fame ftraight line. First, Let the parallelograms ABCD, AEFG have the angle BAD common to both, and be fimilar, and fimilarly placed; ABCD, AEFG are about the fame diameter. e. 14. I. 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. Cor.19.5. wherefore the remainder DG is a to A G D the remainder EB, as GA to AE. but E F H fo is AG to GF; and FH, HC are pa- B K C rallel to AG, GF; and the triangles AGF, FHC are joined at one angle, in the point F; wherefore b. 32. 6. AF, FC are in the fame straight line ". Book XI. 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 fame ftraight 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 ». PROP. XXXIII. B. VI. The words "because they are at the center," are left out, as the addition of fome unskilful hand. • έτυχε, 1 In the Greek, as alfo in the Latin Tranflation, the words TUE," any whatever," are left out in the Demonftration of both parts of the Propofition, and are now added as quite neceffary. and in the Demonftration of the fecond 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 Propofition 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 Propofitions are added, because they are frequently made use of by Geometers. TH DEF. IX. and XI. B. XI. HE fimilitude of plane figures is defined from the equality of their angles, and the proportionality of the sides about the equal angles; for from the proportionality of the fides only, or only |