D B F K L • BI four circles touching them ; one inscribed, and the others each touching one side externally, and the other two produced. The centres of the external circles will be the intersections of the lines bisecting two exterior angles; and the line bisecting the remote interior angle will pass through the same point. The method of proof respecting the external circles is almost the same as that of the foregoing proposition. Cor. 1. Join AD. Then (I. 47. cor. 5.) the angles EAD, GAD are equal. Hence the three straight lines bisecting the three angles of a triangle meet in the same point, and that point is the centre of the inscribed circle. Cor. 2. AE is equal (III. 17. schol.) to AG, BE to BF, and CF to CG. Hence the sum of any side AB, and the remote segment, CF or CG, of either of the other sides, is equal to half the perimeter : and consequently either of these segments, CF or CG, is equal to what remains when the third side AB is taken from half the perimeter. Cor. 3. If the circle KHL touch BC, a side of the triangle ABC, in H, and the other sides produced in K and L, AK or AL is equal to half the perimeter. For (III. 17. schol.) BK is equal to BH, CL to CH, and AK to AL; therefore AK and AL are equal to the perimeter, and each of them to half the perimeter. Cor. 4. Since (cor. 2.) AB and FC are equal to half the perimeter, and AK equal (cor. 3.) to the same; if A B be taken away there remains BK or BH equal to CF. Cor. 5. If C be a right angle, CG or CF is evidently equal to DG or DF, the radius of the inscribed circle. Hence in a rightangled triangle the diameter of the inscribed circle is equal to the excess of the sum of the legs above the hypotenuse. It would appear, in like manner, that the diameter of the circle touching the hypotenuse externally, and the legs produced, is equal to the perimeter. Cor. 6. Since the triangles ADB, BDC, and CDA are respectively equal (1. 41.) to the rectangles under the radius of the inscribed circle, and the halves of AB,BC, and CA, it follows (11.1.) that the area of ABC is equal to the rectangle under the radius and half the perimeter. Hence, if the sides be given in numbers, the length of the radius may be computed by calculating (II. 13. note) the area, and dividing it by half the sum of the sides, or its double by their sum. Cor. 7. If straight lines be drawn from M, the centre of one of the external circles, to A, B, and C, the triangles AMB, AMC, and BMC, are respectively equal (I. 41.) to the rectangles under the radius of that circle and the halves of AB, AC, and BC. Hence, if the last of these be taken from the sum of the others, there remains the triangle ABC equal to the rectangle under the radius, and the excess of half the sum of AB and AC above the half of BC; or, which is the same, to the rectangle under the radius, and the excess* of half the perimeter above BC. The radius, therefore, of any of the external circles may be computed by dividing the area by the excess of half the perimeter above the side which the circle touches externally. PROP. V. PROB. To describe a circle about a given triangle. Let the given triangle be ABC; it is required to describe a circle about it. Find (III. 1. schol.) D the centre of the circle passing through A, B, C, the three angular points of the triangle, and the circle described from D as centre, at the distance of any of the angular points A, B, C, is evidently the circle required. Cor. 1. It is evident, (from this proposition, and from III. 1. schol.) that the three straight lines bisecting the three sides perpendicularly, meet in the same point, and that the point in which they meet is the centre of the circumscribed circle. Cor. 2. Through the centre D draw DG perpendicular to BC: the angle BDG or CDG is equal to BAC. For (I. 47. cor. 5.) the angles BDG, CDG are equal, as are also (III. 26.) the arcs BG, CG. Therefore (III. 27.) the straight line joining AG bisects the angle BAC: and (III. 20.) each of the angles BDG, CDG is double of either half of BAC. PROP. A. THEOR. + Any equilateral and equiangular rectilineal figure may have one circle described about it, and another in it; and the same point is the centre of both circles. Let ABCDE be a rectilineal figure, having all its sides equal, * This excess (cor. 2.) is AE or AG; and hence, if AM and EM were joined, the triangle AEM would be equal to half the triangle ABC. + This proposition and the next are introduced as lemmas, for the purpose of generalizing and abbreviating much of what follows in this book. G C and all its angles equal : one circle may be described about it, and another in it, and the same point is the centre of both. Bisect the angles BCD, CDE by the lines CF and DF: these (I. ax. 12.) will meet, if produced : let them be produced and meet in F, and join FB. Then, since the angles FCD, FDC are equal, being the halves of equal angles, FC is equal (I. 6.) to FD. Also, in the triangles BCF, DCF, the side BC is equal (hyp.) to DC, CF is common, and (const.) the angle BCF is equal to DCF: therefore (I. 4.) BF is equal to DF, and consequently to CF. In the same manner, the straight lines drawn from F to the other angles A and E may be proved to be equal to BF, CF, or DF: and therefore a circle described from F as centre, at a distance equal to any of these lines, will pass through all the angular points, and (IV. def. 2.) will be circumscribed about the rectilineal figure ABCDE. From F draw FG perpendicular to BC, and FH to CD. Then, since BC and CD are equal chords of the circumscribed circle, they are (III. 14.) equally distant from the centre : that is, FG is equal to FH : and a circle described from F as centre at the distance of either of these, will pass through G and H; and it will touch (III. 16.) BC and CD, because the angles at G and H are right angles : and in the same manner it might be shown, that the same circle would touch the other sides of the figure : it would therefore (IV. def. 3.) be inscribed in the rectilineal figure ABCDE: wherefore any equilateral and equiangular rectilineal figure, &c. Cor. BC is bisected (III. 3.) in G, and CD in H; and (const.) FG is perpendicular to BC, and FH to CD. Hence the centre of the inscribed or circumscribed circle may be found by assuming any three of the angular points, and following the method pointed out in the scholium to the first proposition of the third book : and the radius of the inscribed circle is the perpendicular drawn from F to any of the sides, bile that of the circumscribed circle, is the straight line drawn from the centre to any of the angles. PROP. B. THEOR. Ir any equilateral and equiangular rectilineal figure be inscribed in a circle, tangents to the circle, drawn through the angular points, will form an equilateral and equiangular figure of the same number of sides, described about the circle. Let ABCDE be an equilateral and equiangular rectilineal figure F А L K described in the circle ACE: if tangents be drawn to the circle through the points A, B, C, D, E, meeting in the points F, G, H, K, L, the figure FGHKL is also an equilateral and equiangular figure of the same number of sides as the figure ABCDE. For, since GH touches the circle in C, and CD cuts it, the angle DCH is equal (III. 32.) to any angle at the circumference standing on the arc CD; and because HK touches the circle in D, and CD cuts it, the angle CDH is also equal to any angle at the circumference standing on the same arc CD: therefore the angles DCH, CDH are equal to one another. For the same reason EDK, DEK are equal to any angle at the circumference standing on the arc DE: but (III. 28.) the arcs AB, BC, CD, &c., are all equal, because (hyp.) their chords are equal ; and (III. 27.) angles at the circumference standing on these arcs are equal ; and therefore the angles DCH, CDH, EDK, DEK, LEA, &c., are all equal. Hence (I. 6.) CH is equal to HD, DK to KE, &c. Also, in the triangles CHD, DKE, there are two angles of the one equal to two angles of the other, and (hyp.) the sides CD, DE, are equal : therefore (I. 26.) the sides CH, HD are equal to DK, KE, each to each, and the angle H to K. In the same manner, it would be shown that the angle K is equal to L, L to F, &c.; and therefore the figure FGHKL is equiangular. It is also equilateral; for its sides are the doubles of the equal lines FB, GC, &c. Therefore, if any equilateral and equiangular rectilineal figure, &c. 11 PROP. VI. PROB. To inscribe a square in a given circle. square in it. Draw the diameters AC, BD at right angles to one another ; and join AB, BC, CD, DA; ABCD is the square required. Because (I. def. 30.) BE is equal to ED, and EA is common, and at right angles to BD; BA is equal (I. 4.) to AD. Now, the angles CBA, BAD being angles in semicircles, are (III. 31.) right angles; and therefore (1. 28. part 2.) AD is parallel to BC. For the same reason AB is parallel to DC; and therefore (I. def. 24.) ABCD is a parallelogram. But it has been shown that BAD is a right angle, and that the sides BA, AD containing it are equal : ABCD is therefore (I. def. 27.) a square, and (IV. def. 2.) it is inscribed in the circle ABCD: which was to be done. 1 PROP. VII. PROB. To describe a square about a given circle. Let ABCD be the given circle; it is required to describe a square about it. Draw two diameters, AC, BD, of the circle ABCD, at right angles to one another, and (III. 17.) through the points A, B, C, D draw FG, GH, HK, KF, touching the circle: FGHK is the square required. For, by the preceding proposition, A, B, C, D are the angular points of the inscribed square ; and therefore (IV. B.) the figure FGHK formed by the tangents drawn through these points, is a square described abont the circle : which was to be done. D E H PROP. VIII. PROB. To inscribe a circle in a given square. Let ABCD be the given square ; it is required to inscribe a circle in it. Find (IV. A. cor.) the centre E of the inscribed circle, and draw (I. 12.) EF perpendicular to AD: from E as centre, with EF as radius, describe the circle FGHK: it is (IV. A. cor.) the circle required. * F E B C PROP. IX. PROB. To describe a circle about a given square. Let ABCD be the given square ; it is required to describe a circle about it. Find, as in the preceding proposition, the centre E; and from that centre, at the distance EA, describe the circle ABCD: it is (IV. A. cor.) the circle required. A D E PROP. X. PROB. To describe an isosceles triangle, having each of the angles at the base double of the third angle. Take any straight line AB, and divide it (II. 11.) in C, so that * In practice, the construction is effected most easily, both in this proposition and the following, by drawing the diagonals, as their point of intersection is the centre. |