BOOK IV. DEFINITIONS, (406) I. A rectilinear figure is said to be inscribed in another, when all the angular points of the inscribed figure are on the sides of the figure in which it is said to be inscribed. (407) II. A figure is said to be described about another figure, when all the sides of the circumscribed figure pass through the angular points of the other figure. (408) III. A rectilinear figure is said to be in scribed in a circle, when the vertex (409) IV. A rectilinear figure is said to be circumscribed about a circle, when each of its sides is a tangent to the circle. (410) V. A circle is said to be inscribed in a rectilinear figure, when each side of the figure is a tangent to the circle, (411) VI. A circle is said to be circumscribed about a recti linear figure, when the circumference passes through the vertex of each angle of the figure. (412) VII. A right line is said to be inscribed in a circle, when its extremities are in the circumference of the circle. (413) The fourth book of the Elements is devoted to the solution of problems, chiefly relating to the inscription and circumscription of regular polygons and circles. A regular polygon is one whose angles and sides are equal. (414) In a given circle (BCA) to inscribe a right line equal to a given right line, which is not greater than the diameter of the circle. Draw a diameter A B of the circle, and if this is equal to the given line, the problem is solved. If not, take in it the segment A E equal to the given line (III, Book I.); from the centre A with the radius A E describe a circle E C, and draw to either intersection of it with the given circle the line A C; this line is equal to A E, and therefore to the given line, B C B PROPOSITION II. PROBLEM. (415) In a given circle (B A C) to inscribe a triangle equiangular to a given triangle (ED F.) Draw the line G H a tangent to the given circle in any point A; at the point A with the line A H make the angle HAC equal to the angle E, and at the same point with the line A G make the angle G A B equal to the angle D, and draw B C. Because the angle E is equal to HAC (const.), and HAC is equal to the angle B in the alternate segment (XXXII, Book III.), the angles E and B are equal ; also the angles D and C are equal, therefore the remaining angle F is equal to B AC (XXXII, Book I.), and therefore the triangle B AC inscribed in the given circle is equiangular to the given triangle EDF. G A PROPOSITION III. PROBLEM. (416) About a given circle (A BC) to circumscribe a triangle equiangular to a given triangle (EDF). Produce any side D F of the given triangle both ways to G and H; from the centre K of the given circle draw any radius K A. With this line at the point K make the angle BK A equal to the angle ED G, and at the other side of KA make the angle AKC equal to EFH, and draw the lines L M, L N, and M N, tangents to the circle in the points B, A and C. M с I K L A N G D H Because the four angles of the quadrilateral figure LBKA taken together are equal to four right angles (134), and the angles KBL and KA L are right angles (const.), the remaining angles AKB and A LB are together equal to two right angles; but the angles EDG and EDF are together equal to two right angles (XÍII, Book I.), therefore the angles A R B and A LB are together equal to E D G and E DF; but A KB and EDG are equal (const.), and therefore A L B and EDF are equal. In the same manner it can be demonstrated that the angles ANC and EFD are equal; therefore the remaining angle M is equal to the angle E (XXXII, Book I.), and therefore the triangle L M N circumscribed about the given circle is equiangular to the given triangle. А B (417) In a given triangle (B A C) to inscribe a circle. Bisect any two angles B and C by the right lines B D and C D, and from their point of concourse D draw D F perpendicular to any side B C; the circle described from the centre D with the radius D F is inscribed in the given triangle. Draw D E and D G perpendicular to B A and AC. In the triangles DEB, DFB the angles DEB and DBE are equal to the angles DFB and DBF (const.), and the side D B is common to both, therefore the sides D E and D F are equal (XXVI, Book I.): in the same manner it can be demonstrated that the lines D G and D F are equal; therefore the three lines DE, DF, and D G are equal, and therefore the circle described from the centre D with the radius DF passes through the points E and G; and because the angles at F, E, and G are right, the lines B C, BA, and AC are tangents to the circle (XVI, Book III.), therefore the circle F E G is inscribed in the given triangle. (418) It is assumed in the demonstration of the proposition, that the two bisectors of the angles B C of the triangle will meet at the same point. This, however, may be proved by showing that they make angles with B C which are together less than two right angles. In this demonstration, and in various other places, Euclid assumes, that any point whose distance from the centre of a circle is equal to the radius, must be on the circle. See (22). (419) If D A be drawn it will bisect the angle A. For E D and G D are equal, and A D the common side is opposite to right angles E and G, and therefore the triangles D A E and D A G are in every respect equal. Therefore the angles D A E and D A G are equal. Hence the lines bisecting the three angles of a triangle intersect at the same point, and that point is the centre of the inscribed circle. DI D H E H (420) The areas of the three triangles B D C, Ć D ), and A D B, are respectively equal to half the rectangles under the radius of the inscribed circle and the sides B C, CA, and B A of the given triangle. Hence the area of the given triangle is equal to the rectangle under the radius of the inscribed circle and the semiperimeter of the triangle. (421) Hence if the sides be given in numbers, the radius of the inscribed circle may be found by dividing the area (found by (186)). by the seniperimeter. *** (422) The problem to inscribe a circle in a triangle is a particular case of a more general problem, ‘To describe a circle touching three given right lines. 1°. If the three given lines be parallel to one another, the problem is obviously impossible, since no circle touching two of them could touch the third. 2°. If two of the lines A G, B H be parallel and the third AB intersect them. Draw the lines A D and B D bisecting the angles A and B. These will intersect, since they make angles with A B which are together less ? than two right_angles. Let them meet at D. Perpendiculars D F, DG, and DH to the three given lines from D are equal. This may be C proved as in the preceding proposition. Hence D is the centre and DF the radius of the circle. It appears from the diagram that there are two circles which touch the given right lines. 3o. Let the three given right lines intersect so as to form a triangle. In this case the circle is determined as in the proposition. But this is not the only circle which may be drawn touching the given right lines. Draw the lines C D and A D bisecting the external angles at A and C. These, as before, will meet at D, and perpendiculars DE, DF, DG on the given lines from this point are equal. Hence D is the centre and DF the radius of a circle touching the three given lines. The demonstration of this is the same exactly as that of the proposition. In the same manner two other circles may be described touching the given right lines as in the diagram. Thus if three right lines intersect so as to form a triangle, four different circles may be described each touching them all. By this case it appears that the bisector of any internal angle of a triangle, and those of the remaining external angles, intersect at the same point. 4o. If the three given lines intersect at the same point, no circle can be described touching them all. *** (423) It is plain the problem to describe a circle touching two B given right lines is indeterminate. We can however in this case determine the locus of its centre. 1°. If the two right lines be parallel. Draw the line A B intersecting them perpendicularly, and bisect it at C, and through C draw D E parallel to the given lines. This will be the locus of the centres. For if any other perpendi- D cular F G be drawn, a circle described on it as the diameter will touch the given lines. 2°. If the given lines intersect. Draw the lines A B and C D bịsecting the angles under the given lines. These lines will be the locus of the centres. The demonstrations will easily appear from that of Prop. IV. and from the annexed diagram. of B D PROPOSITION V. PROBLEM. (424) About a given triangle (B A C) to circumscribe a circle. r Bisect any two sides B A and A C of the given triangle, and through the points of bisection D and E draw D F and E F perpendicular to A B and A C, and from their point of concourse F draw to any angle A of the triangle BAC the line FA, the circle described from the centre F with the radius F A is circumscribed about the given triangle. Draw FB and FČ; in the triangles FDA, F D B the sides D A and D B are equal (const.), FD is common to both, and the angles at D are right, therefore the sides FA and F B are equal (IV, Book I.): in the same manner it can be demonstrated that the lines FA and FC are equal, therefore the three lines FA, FB, and FC are equal, and therefore the circle described from the centre F with the radius F A passes through B and C, and therefore is circumscribed about the given triangle BAC. (425) Cor.— If the centre F fall within the triangle, it is evident all the angles are acute, for each of them is in a segment greater than a semicircle. If the centre F be in any side of the triangle the angle opposite to that side is right, because it is an angle in a semicircle (XXXI, Book III.); and if the centre fall without the triangle the angle opposite to the side which is nearest the centre is obtuse, because it is an angle in a segment greater than a semicircle (XXXI, Book III.). |