being equal, the third angle B, in the one, must also be equal to the third angle L, in the other; therefore, the two triangles ABC, and MKL, are equiangular, and the sides of the MKL, are all tangents to the circle; as was required.

Corol. If the triangle is equiangular, the triangle so described, touching the circle, will be a regular poligon, circumscribed to the circle.

§ 55. PROBLEM. To inscribe a circle in a given triangle.

APPLIC. In the triangle ABD, (fig. 61,) it is required to inscribe a circle, that shall touch its three sides internally.

Solution. Bisect the angle BAD, by the line AH, and likewise the angle ABD, by the line BI; from the point C, the intersection of these two lines, draw the perpendiculars CG, CF, and CE, upon the three sides of the triangle, these perpendiculars will be equal; and the circle described from C, as centre, with either one of them, will touch the three sides of the triangle ABD; as was required.

Proof. In the two triangles, CAG, and CAF, the angles at A, are equal, being the halves of the angle A, of the given triangle, the angles at F, and G, are right angles, and the side AC, is common to both triangles; therefore, these two triangles are equal, and the side CG, in the one triangle, opposite to the half of the angle of the triangle, is equal the CF, opposite in the other triangle, to the other half of the same angle. In exactly the same manner, it is proved, that the CE, is equal to CF, by the bisection of the angle B; therefore, the three perpendiculars CG, CF, and CE, are equal, and a circle can be drawn from the centre C, passing through the three points G, F, and E, which will touch the triangle ABD, internally, all the sides being at right angles to their corresponding radii respectively; as was to be proved.

Corol. The three lines bisecting the three angles of a triangle cut one another in one point, which is the centre of the inscribed circle.

$56. PROBLEM.

An equilateral triangle being giv en, inscribed in a circle: to find the sides and angles of the regular poligon of six sides, inscribed in the same circle.

APPLIC. The equilateral triangle ABC, inscribed in the circle AGDEBF, being given, (fig. 62,) to find the sides of the six sided regular poligon, (that is, the hexaedron,) inscribed in the same circle, and the angles formed by the radii and the sides of the hexagon, C, being the centre of the circumscribed circle.

Solution. From the centre C, of the circle, draw the perpendiculars upon the sides of the given triangle; protracted to the intersection of the circumference of the circle as CKF, CIE, and CHG; join the chords AG, GD, DE, EB, BF, and FA; these will be the sides of the regular figure of six equal sides inscribed in the same circle. These equal sides will be equal to the radius of the circle, and each angle formed by one of these sides and the radius drawn to its angular point, will be equal to one-third of two right angles.

Proof. The perpendiculars from the centre of the circle. upon the three sides of the equilateral triangle as chords, divide the arcs AGD, DEB, and BFA, equally in the points G, E, and F; therefore, also, all the chords AG, GD, DE, and so on, are equal to one another, and subtend equal angles at the centre C; and as there are six sides, each angle at C, is equal to the sixth part of four right angles, or the third part of two right angles. The figure AGDEBF, is a regular hexagon in the circle, and the sum of any equal number of these equal arcs will be equal, and correspond to an equal angle at the circumference; therefore, all the angles, as AGD, GDE, and so on, standing in equal segments, are equal to one another; and their sum being equal to twice as many right angles as the figure has sides wanting four: that is, eight right angles, each of these angles of the hexagon is equal eight sixths of one right angle, or two thirds of two right angles, and they are bisected

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by the radii, for these radii, as CA, and CG, being equal, the angles opposite to them are equal: that is, CAG=CGA, and the sum of these is also equal to two thirds of two right angles, because they make with the angle at C, together, two right angles, the angles at C, are each one third of two right angles; each of the angles of any of the triangles, as CAG, is, therefore, one third of two right angles, and the triangle is equilateral; therefore, also, the sides of the hexagon, as AG, GD, &c. are each equal to the radius of the circumscribed circle; all which was to be proved.

Corol. 1. The radius of the circle will, therefore, cut the circumference in six parts having equal chords and equal arcs.

Corol. 2. By the bisection of the chord of the hexagon, as here done for the triangle, the sides of the regular figure of twelve sides, (or the dodecagon,) inscribed in the circle, will be obtained; and so on, for any continued bisection.

Corol. 3. As the square is a regular figure of four sides, and can, therefore, be inscribed in the circle, the same process of bisection will give the side of the regular poligon of eight sides, and any subsequent bisection.

§ 57. PROBLEM. Two sides and the included angle in a quadrilateral figure that cannot be inscribed in a circle, being given, together with the angle subtended by these lines at the fourth point of the same, to describe the quadrilateral figure.

APPLIC. Let AB, and BC, and the angle ABC, which they include, be given, (fig. 63, a, b,) let the angle E, be the angle subtended at the fourth point by the line AB, and the angle F, the angle subtended at the same fourth point by the line BC; it is required to describe the quadrilateral figure ABCD, resulting from it, (provided the same cannot be inscribed in a circle.)

Solution. Upon the side AB, describe a segment of a circle BDA, which will contain at the circumference an angle equal to its corresponding angle E. Upon BC, and towards

the same side, as the arc BDA, describe a segment CDB, that will contain at the circumference an angle equal to its corresponding angle F; join CD, and DA, the figure ABCD, will be the quadrilateral figure required.

Proof. The arc BDA, described upon AB, as chord of the circle ABD, will contain at any of its points in the circumference an angle equal to the given angle E, by the construction; in the same manner, the arc BDC, described upon BC, as chord, will contain in any point of its circumference an angle equal to the given angle F, by the similar construction; these two arcs having one point B, common, must intersect one another in a second point, if the two circles have not the same centre, as is supposed; this intersection taking place at the point D, the two given angles will be subtended simultaneously in that point alone; therefore, the point D, where the two arcs intersect one another, is the fourth point of the quadrilateral figure required, which is completed by the lines DA, and DC; or ABCD.

Corol. 1. If the sum of the two given angles is less than two right angles, the fourth point sought lies without the triangle formed by the given sides and the included angle, if they are together equal to two right angles, the fourth point lies in the straight line between the two extreme points; and if the sum of the two given angles is more than two right angles, the fourth point lies within the triangle formed by the three other points, (as in fig. 63, b.)

Corol. 2. If the two given lines lie in one straight line, the fourth point forms the third point of a triangle opposite to the sum of the two other sides, as side of the triangle (figure 63, c.)

Corol. 3. If the sum of the two angles at the fourth point is less than two right angles, as well as when the two given lines lie in one and the same straight line, there are two fourth points possible, one on each side of the two given straight lines.

Corol. 4. If the centres of the two circles fall together, their circumferences also lie upon one another, and no intersection, therefore, no determination of the fourth point, is obtainable by these data.

(Note.) This is generally called the Problem of three points.

$58. THEOREM. Parallel chords in a circle intersect equal arcs of the circumference.

APPLIC. The chords AB, and DE, in the circle DAFBEG, (figure 64,) being parallel, the arcs AD, and BE, which they intersect in the circumference of the circle, will be equal.

Demonstr. Draw the CF, perpendicular to the chords, it will bisect both these chords, and the arcs AFB, and DFE, equally; join CD, CA, CB, and CE; the angle FCA, will be equal to the angle FCB, and the FCD, to the FCE; subtracting equals from equals, there remain the equal angles ACD, and BCE, at the centre; and as equal angles at the centre subtend equal arcs, the arcs AD, and BE, are equal; as was to be demonstrated.

$59. THEOREM. Two chords intersecting one another in a circle, subtend by their vertical angles arcs of the circumference, which are together equal to the two arcs subtended by the diameters parallel to these chords.

APPLIC. The two chords AB, and DE, in the circle AHDBGEIF, (figure. 65,) intersecting one another in K, the sum of the arcs DB, and AFIE, which the vertical angles DKB, and AKE, subtend, will be equal to the sum of the arcs FI, and HDBG, subtended by the vertical angles FCI, and HCG, at the centre C, of the circle.

Demonstr. The diameter HI, being parallel to the chord DE, the arcs HD, and EI, which they intersect at the circumference, are equal; and for the same reason, also, the arcs AF, and BG, are equal; if to the arc BD, be added the two arcs HD, and BG, the arc HDBG, will be the arc subtended by the diameters, under an angle C, at the centre equal to the angle K, of the intersecting chords; if on the other side from the arc AFIE, be taken away the two arcs AF, and EI, which are equal to the BG, and HD, the arc FI, will be sub