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touch it; they are, therefore, tangents drawn through the given point A, as was required to be done.

PROP. XXXII. fig. 13. To inscribe a circle in a given triangle; ABC. Bisect any two angles, as those at A and B, by the lines AD, BD, meeting in the point D; from which point draw DE and DF perpendicular to the sides AB, BC; then, as the two triangles BDE and BDF have the angles at E and F right, therefore equal, and the angles at B also equal, the remaining angles at D will likewise be equal; and the side BD being cominon to both, those two triangles will be equal, and consequently the side DE will be equal to the side DF. In the same way, DG may be shown. to be equal to DE and DF; and a circle, described with any one of these lines for radius, would pass through the three points E, F, and G; that is, it would touch each side of the given triangle ABC, and therefore be inscribed within it, as required.

PROP. XXXIII. fig. 23, Plate 1. To find the centre of a given circle, or of an arch of a circle. Take any three points A, B, D in the given eircumference or arch, and drawing the right lines AB, BD, bisect them by the perpendiculars EC, FC, meeting together in the point C: this point will be the centre required.

In the same manner, a circle may be drawn to pass through three given points, provided they are not in a right line; or through the three angular points of a given triangle; that is, it may be thus described about a triangle.

PROP. XXXIV. fig. 14. Plate 2. To divide a right line AB into any number of equal parts, as 5. From either extremity, as B, draw the indefinite line BC, and taking any convenient distance in the compasses, set it off five times from B to the points numbered 1, 2, 3, 4 and 5, at

C; then join CA, and through these points draw lines parallel to CA to the line AB, and cutting it in the points also marked 1, 2, &c. then will this line be divided into five equal parts, as was required to be done,

In the same manner may a line be divided into parts proportional to any given lines or magnitudes.

PROP. XXXV. fig. 15. To find a mean proportional between two right lines, as C and D. Make AE equal to the given line C, and produce it to B, making EB equal to D; upon the middle of the whole line, AB, describe the semicircle AFB; on E draw the perpendicular EF, and join AF and FB.

The angle AFB, in a semicircle, is a right angle, and therefore equal to either of the angles at E: the angle at A is common to the two triangles FAE and FAB; and the angle at B is common to the same triangle FAB and FEB; these three triangles are therefore equiangular and similar; and their corresponding or homologous sides will be proportional; that is, AE will be to EF, as EF to EB; that is, the rectangle under the extremes, AE and EB, will be equal to that under the means: but the means in this case being the same quantity repeated, it follows that AE, multiplied by EB, will be equal to the square of EF; fore EF will be a mean proportional between AE and EF: but AE is equal to C, and EB to D; EF is therefore a mean proportional between the two given lines, C and D.

there

PROP. XXXVI. fig. 16. To find a fourth proportional to three given right lines, A, B, C. Draw the inde finite lines DE, DF: make DG equal to A, DH. equal to B, and DI equal to C: join GI, and draw HL parallel to it: then shall I be the fourth proportional required; for GI being parallel to HL, the triangles DGI and DHL will be similar, and consequently, their correspond

ing sides proportionals: that is DG: DH:: DI: DL; so that DL is a fourth proportional to DG, DH, and DI; that is, to the given lines A, B, and C: or A: B:: C; IL

PROP. XXXVII. fig. 17. To construct a square equivalent to a given parallelogram, ABCD. Draw BE perpendicular to AD, which will be the altitude of the parallelogram; find FG a mean proportional between BE and AD, (Prop. 35, fig. 15), and on FG construct the square FHIG: then, as BE: FG:: FG: AD, it follows that the square of FG will be equal to the rectangle under AD and BE: but this rectangle is equal to the parallelogram ABCD, consequently the square FHIG is made equal to the given parallelogram ABCD, which was required to be done.

If, instead of a parallelogram, it had been a triangle to which a square was to be made equal, the side of this square must have been found, a mean proportional between the altitude of the triangle and one half of the base, or between the whole base and one half of the altitude.

PROP. XXXVIII. fig. 18. To make a triangle equal to a given polygon, as, for example, to the irregular pentagon ABCDE. Join the angles C and A; through B draw BF parallel to CA, meeting AE produced in the point F, and draw FC: then the triangles CBA, CFA, standing on the same base, CA, and between the same parallels, or being of the same altitude, are equal to one another: again, joining CE, drawing DG parallel to it, and joining CG, we have the triangle CGE equal to the triangle CDE: but CFA and CGE are parts of the great triangle CFG; to these parts add the triangle ACE, which is common to the given pentagon and to the triangle CFG, and we shall have the whole triangle FCG, equal to the whole pentagon ABCDE. By the same process, may be constructed a triangle

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which shall be of equal area to a polygon of ber of sides.

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PROP. XXXIX. fig. 19. To inscribe a square in a given circle, ACDB. Draw two diameters; AB and CD, intersecting each other at right angles: join the extremities of these diameters, and these lines will form a square, ACBD, within the given circle. For the angles AOC, COB, BOD, and DOA, being all equal, the chords subtending these angles must also be equal: but the angles at A, C, B, and D, being angles in a semicircle, are all right; consequently the figure ACBD is a square inscribed in the given circle, as was required.

The square may be placed in any position, as GEHF, whose extremities are equally bounded by the circle.

PROP. XL. fig. 20. To describe a regular hexagon in a given circle, whose centre is C. Take the radius of the circle AC, and set it off on the circumference to B and D; draw the diameters AG, BE, and DF; join all the extremities of these lines, and the figure ABFGED will be the hexagon required: for AB and AD being each made equal to AC, which is also equal to BC and CD, these two two triangles, CBA, CAD, are equilateral and on account of the intersecting lines BE, AG, DF, the opposite triangles must also be equilateral; consequently, EG and GF each equal to the radius of the circle; that is, to AB and AD. In the same way, the remaining sides DE and BF may be proved to be each equal to the radius, or to AB and AD; consequently the six sides of the figure to be all equal; it is, therefore, a regular hexagon inscribed within the given circle, as was required.

If lines be drawn joining the alternate angles of the hexagon A, F, E, we obtain an equilateral triangle, inscribed within a given circle.

PROF.

PROF. XLI. fig. 21. To inscribe a regular octagon within a given circle. By the 39th Prop. construct a square whose angular points touch the circumference of the circle at ADBE; divide the sides AD and DB, into two equal parts; and through the sections and the centre C, draw diameters touching the circumference in the points F, G, H,I: then right lines joining these eight points will form the octagon AFDGBHEI, as was required to be done.

ON TRIGONOMETRY.

HAVING in the foregoing propositions explained the chief properties of certain plain geometrical figures, as the triangle, the square, the parallelogram, the circfe, &c. it is now time to furnish the student with some observations on the application of the properties of one of these figures, namely, the triangle, to sundry very important purposes.

The branch of geometry which regards the properties of triangles, is termed Trigonometry, a name borrowed from the Greek language, signifying the art of measuring triangles.

Trigonometry is divided into two parts, Plane and Spherical. Plane trigonometry is employed concerning such triangles as are formed by three sides, or lines, all lying in the same plane, such as those drawn on a sheet of paper, on a smooth even table, or the like. Spherical trigonometry relates to those triangles whose sides are portions of circles, such as may be described on the surface of a celestial or terrestial globe, where the sides are all curves, and situated in different planes. This latter branch of trigonometry being founded on a knowledge of parts of geometry that have not been explained in the preceding propositions, it is not intended to enter upon it in this work; the attention

of

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