PROPOSITION IV. PROBLEM. To circumscribe a circle about a given triangle. Let ABC be the given triangle. Bisect the sides CA, CB with the perpendiculars DG, FG, which must meet in some point G; for, if they did not meet, they would be parallel, A and, if parallel, then also would the lines to which they D F B H are perpendicular be parallel; that is, CA, CB would be parallel, which is not the case: consequently the perpendiculars DG, FG must meet. Join GA, GB, GC; and comparing the two right-angled triangles DGC, DGA, we have DG common, and the side DC equal to DA, each being half the side AC; therefore GA is equal to GC (B. II, Prop. vш, Cor. 2). By comparing the two triangles HGA, HGB, we may, for similar reasons, show that GA is also equal to GB. Therefore the point G is equidistant from A, B and C. Hence, if, with G as a centre, a circumference be described passing through A, it will also pass through B and C, and consequently circumscribe the triangle ABC. Cor. 1. The three perpendiculars which bisect the three sides of a triangle, meet in the same point. Cor. 2. This problem is equivalent to describing acircumference of a circle through any three points not situated in the same straight line. (87.) A circle described so as to touch one of the sides of a triangle exteriorly, and the other two sides produced, is called an escribed circle; thus, D, F and G are centres of escribed circles, in reference to the triangle ABC. F E From this we see that escribed circles touch the three sides of a triangle, and are situated wholly without the triangle; while the inscribed circle also touches all the sides of a triangle, but is situated wholly within the triangle. (88.) Hence, four circumferences may always be described, which shall touch any three given lines, provided no two of the lines are parallel or coincident; for these lines, being sufficiently produced, will form, by their intersections, a triangle; which triangle we have just shown has three escribed circles and one inscribed circle, making in all four circles, each of which touches the three given lines. (89.) We have shown that the three lines which bisect the three angles of a triangle, meet at the centre of the inscribed circle (B. V, Prop. ш, Cor.). It also follows (B. V, Prop. 11, Cor.), that these lines, when produced, pass through the centres of the escribed circles. It is moreover obvious, that if the exterior angles of a triangle be bisected, these bisecting lines will meet at the centres of the escribed circles. Hence, if lines be drawn bisecting the angles, and the exterior angles of a triangle, they will intersect each other by threes at the centres of the inscribed and escribed circles; thus the three lines which bisect the angles will meet at the same point, giving the centre of the inscribed circle. Any one of the lines bisecting an angle of the triangle will intersect, at the same point, two of the lines which bisect the exterior angles, giving the centre of an escribed circle. PROPOSITION V. THEOREM. A circle may be circumscribed about any given regular polygon; so also may a circle be inscribed in any given regular polygon. Let ABCDFG be a regular polygon. Through any three consecutive corners of the polygon, as A, B, C, describe the cir- G cumference of a circle (B. V, Prop. Iv), and it will also pass through all the remaining corners of the polygon. A B F K H C D For, from the centre H, draw HA, HB, HC, HD; also draw HK perpendicular to the side BC. Now, comparing the two quadrilaterals HKBA, HKCD, we know that the angle HKB is equal to HKC, since each is a right angle; also the angle KBA is equal to KCD, since the polygon is regular moreover KB is equal to KC (B. III, Prop. 1). Therefore, if the quadrilateral HKBA be applied to the quadrilateral HKCD, so that HK may retain its present position, the side KB will coincide with KC, and BA will take the direction of CD; and since BA is equal to CD, the point A will coincide with D: consequently the point H is equidistant from A, B, C and D. In the same way all the corners of the polygon may be shown to be at the same distance from H; therefore this circle circumscribes the polygon. Again, in reference to this circle, the sides of the given polygon are chords; they are therefore equally distant from the centre (B. III, Prop. iv). Hence, if, with H as a centre, and with HK as a radius, a circle be described, it will touch all the sides of this polygon at their middle points, and will therefore be inscribed in it. Schol. 1. The point H, which is the common centre of the circumscribed and inscribed circles of the polygon, may be regarded as the centre of the polygon itself; and on this principle, the angle AHB is called the angle at the centre, being formed by the two radii drawn to the extremities of the same side. And since all the chords AB, BC, CD, &c. are equal, all the angles at the centre must be equal; and therefore each may be found by dividing four right angles by the number of sides of the polygon. Schol. 2. In order to inscribe a regular polygon of a certain number of sides, in a given circle, we must be able to divide the circumference into as many equal parts as the polygon is to have sides; and then, by joining these successive points of division, we shall form the polygon desired. For the arcs being equal, their chords must be equal; also the angles of the polygon will be equal, since they will be inscribed in equal segments of the same circle (B. III, Prop. VIII, Coг. 1). PROPOSITION VI. PROBLEM. To inscribe a square in a given circle. Cor. 3); hence the figure is equiangular. Again, since the arcs AC, CB, BD, DA are quadrants, they are equal, and therefore their corresponding chords are equal; so that the figure is also equilateral, and consequently it is a square. Scholium. Since the triangle AFC is right-angled and isosceles, we have (B. II, Prop. VII) AC-2 AF*. Hence AC: AF:: 2 : 1, and AC : AF :: √2: 1; that is, the side of an inscribed square is to the radius, as the square root of 2 is to 1. |