consequently FG: GE:: BG: CD; therefore (V. 6.) FG.CD= BG.GE; and since (III. 26.) BG.GE=CG.GD, it follows that CG.GD=FG.CD, and FG: CG:: GD: CD, and hence (V. 10.) CF: CG:: CG: CD.

PROP. V. THEOR.

If, from the vertex of a triangle, two straight lines be drawn, making equal angles with the sides and cutting the base; the squares of the sides are proportional to the rectangles under the adjacent segments of the base.

In the triangle ABC, let the straight lines BD and BE make the angle ABD equal to CBE; then AB: BC:: DA.AE; EC.CD.

For (III. 9. cor.) through the points B, D, and E describe a circle, meeting the sides AB and BC of the triangle in F and G, and join FG.

Because the angles DBF and EBG are equal, they stand

If the triangle ABC be right-angled at C, and the vertical

A

HDE

B

lines BD and BE cut the base internally; then BC2+AC.CE: BC2:: AE: CD. For make AH equal to EC. Because AB2: BC2:: DA.AE: EC.CD, and (II. 10.) AB2=AC2+BC2, therefore AC2+ BC2: BC2: DA.AE: EC.CD, and, by division, AC: BC2:: DA.DE-EC.CD: EC.CD. But, by successive decomposition, DA.AE-EC.CD= DA.AC-DA.EC-EC.CD-DA.AC-EC.AC-AC.HD; whence AC BC2:: AC. HD: EC.CD, and (V. 13. and cor.) AC.EC: BC2;: EC.HD: EC.CD, or (V. 3.) : : HD: CD; consequently (V. 9.) BC2+AC·EC: BC2:; HC: CD; but, AH being equal to EC, HC is equal to AE; wherefore BC2+ AC.EC: BC2:: AE: CD.

:

B

If the vertical lines BD, BE cut the base AC of a rightangled triangle ACB externally; then will BC2-AC.EC: BC2 :: AE CD. For make AH EC. It is demonstrated as before, that AC BC2:: :

D

H A

CE

DA.AE-EC.CD: EC.CD; but DA.AE-EC.CD=DA.AC+ DA.EC-EC.CD=DA.AC-EC.AC=AC.HD: wherefore AC2: BC2:: AC.HD: EC.CD, and AC.EC: BC2 : ; EC.HD: EC.CD :: HD : CD, and consequently BC2-AC.EC: BC? :: HC or AE: CD.

PROP. VI. THEOR.

The perpendicular within a circle, is a mean proportional to the segments formed on it by straight lines, 'drawn from the extremities of the diameter, through any point in the circumference.

Let the straight lines AEC and BCG, drawn from the ex

tremities of the diameter of a circle through a point C in the circumference, cut the perpendicular to AB; the part DF within the circle is a mean proportional between the segments DE and DG.

For the angle ACB, being in a semicircle, is a right angle (III. 19.), and the angle ABG is common to the two triangles ABC and GBD, which are, therefore, similar (VI. 11.). Hence the remaining angle BAC is equal to BGD, and consequently the triangles ADE and GDB are similar; wherefore AD: DE : : DG: DB, and (V. 6.) AD.DB= DE.DG. But (III. 26. cor.), the rectangle under AD and DB is equivalent to the square of DF; whence DE.DG=DF, and (V. 6.) DE: DF: DF: DG.

The Appendix to the books of Geometry cannot fail, by its novelty and singular beauty, to prove highly interesting. The first part is taken from a scarce tract of Schooten, who was Professor of Mathematics at Leyden, early in the seventeenth century. But the second and most important part is chiefly selected from a most ingenious work of Mascheroni, a celebrated Italian mathematician; which in 1798 was translated into French, under the title of Geometrie du Compas. It will be perceived, however, that I have adapted the arrangement to my own views, and have demonstrated the propositions more strictly in the spirit of the ancient geometry.

NOTES TO TRIGONOMETRY.

1. THE French philosophers have, at the instance of Borda, lately proposed and adopted the centesimal division of the quadrant, as easier, more consistent, and better adapted to our scale of arithmetic. On that basis, they have also constructed their ingenious system of measures. The distance of the Pole from the Equator was determined with the most scrupulous accuracy, by a chain of triangles extending from Calais to Barcelona, and since prolonged to the Balearic Isles. Of this quadrantal arc, the ten millionth part, or the tenth part of a second, and equal to 39.371 English inches, constitutes the metre, or unit of linear extension. From the metre again, are derived the several measures of surface and of capacity; and water, at its greatest degree of contraction, furnishes the standard of weights.

It would be most desirable, if this elegant and universal system were adopted, at least in books of science. Whether, with all its advantages, it be ever destined to obtain a general currency in the ordinary affairs of life, seems extremely questionable. At all events, its reception must necessarily be very slow and gradual; and, in the meantime, this innovation is productive of much inconvenience, since it not only deranges our habits, but lessens the utility of our delicate instruments and elaborate tables. The fate of the centesimal division may finally depend on the continued merit of the works framed after that model.

2. The remarks contained in the preliminary scholium, will obviate an objection which may be made against the succeeding demonstrations, that they are not strictly applicable, except when the arcs themselves are each less than a quadrant. But this in fact is the only case absolutely wanted, all the derivative arcs being at once comprehended under the definition of the sine or tangent. To follow out the various combinations, would require a fatiguing multiplicity of diagrams and

expression becomes sinB2+sin(A—B)sin (A+B)=sinA3, or sin(A+B)sin(AB)=sin A2-sinB2.

23. Instead of A in the last article, take its complement, and sin (A+B)sin (A—B) = cosA3—sinB3, or cos(A — B) cos(A+B)=cosA2-sinB2.

2

24. Compare art. 21. with 22, and (cos2B-cos2A)=sinA’— sinB2.

25. Comparing likewise art. 20. with 23, and (cos2A+ cos2B)=cos A-sinB3.

26. Resolve the difference of the squares in art. 22. into its factors, and sin(A+B)sin(A—B)=(sinA+sinB)(sin A—sinB). 27. Make a similar decomposition in art. 23, and cos(A+B) cos(A-B)=(cosA+sinB) (cos A—sinB).

24. In art. 18, instead of A and B take their halves, and sinA+sinB 2sin(A+B)cos‡(A—B).

25. Make the same change in art. 19, and sin A—sinB= 2sin(A-B)cos(A+B).

26. Change likewise art. 20, and cosB+cosA≈2cos (A+B) cos (A—B).

27. Do the same thing in art. 21, and cosB (A-B)sin(A+B).

cos A = 2sin

From the third additional proposition to Book III., a very. simple expression may be derived for the sum of the sines of progressive arcs. Suppose the diameter AO were drawn; then BE+CF+ DG=HG=HO+DO, or 2sinAB+ 2sin AC+ 2sinAD HO + sinAD, and sinAB+ sinAC + sinAD = HO+sinAD = AQ.tanBAO+ sinAD. Wherefore, in sin na vers nq.cota+ general, sin a + sin2a + sin3a. sin na. Hence the sum of the sines in the whole semicircle is cota. Thus, if the sines for each degree up to 180°, the radius being unit, were added together, the amount would be 114,58866.

...

4. On examining the formation of the successive terms of the first and second tables, it will appear that the coefficients are certain multiples of the powers of 2, whose exponents likewise at every step decrease by two. It is farther manifest, that if 1,