THE ELEMENTS OF EUCLID. BOOK IV. DEFINITIONS. See N. I. A RECTILINEAL figure is said to be inscribed in another rectilineal figure, when all the angles of the inscribed figure are upon the sides of the figure in which it is inscribed, each upon each. II. In like manner, a figure is said to be described III. A rectilineal figure is said to be inscribed in IV. A rectilineal figure is said to be described about a circle, V. In like manner, a circle is said to be inscribed VI. A circle is said to be described about a rectilineal figure, when the circumference of the circle passes through all the angular points of the figure about which it is described. VII. A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle. PROPOSITION I. PROBLEM. In a given circle, to place a straight line, equal to a given straight line, which is not greater than the diameter of the circle. Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle; it is required to place in the circle ABC, a straight line equal to D. + Hyp. E B3. 1. Draw BC the diameter of the circle ABC: then, if BC is equal to D, the thing required is done; for in the circle ABC, a straight line BC is placed equal to D: but, if it is not, BC is greater † than D: make CE equal✶ to D; and from the centre C, at the distance CE, describe the circle AEF, and join CA: CA shall be equal to D. Because C is the centre of the circle AEF, CA is equal † to † 15 Def. 1. CE: but D is equal to CE; therefore D is equal + to CA. † Constr. Wherefore, in the circle ABC, a straight line CA is placed † 1 Ax. equal to the given straight line D, which is not greater than the diameter of the circle. Which was to be done. D PROPOSITION II. PROB.—In a given circle, to inscribe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle ; it is required to inscribe in the circle ABC, a triangle equiangular to the triangle DEF. Find the centre, and through it draw any straight line BC, terminated both ways by the circumference; this line is a diameter. * 17. 3. * 23. 1. 32. 3. † Constr. +1 Ax. Draw the straight line GAH touching the circle in the point A; and at the point A, in the straight line AH, make * the angle HAC equal to the angle DEF; and at the point A, in the straight line AG, make the angle GAB equal to the angle DFE; and E F B * H Because HAG touches the circle ABC, and AC is drawn from the point of contact, the angle HAC is equal to the angle ABC in the alternate segment of the circle: but HAC is equal to the angle DEF; therefore also the angle ABC is equal to DEF: for the same reason, the angle ACB is equal to the angle DFE; therefore the remaining angle BAC is 32. 1. & equal to the remaining angle EDF: wherefore the triangle ABC is equiangular to the triangle DEF; and it is inscribed in the circle ABC. Which was to be done. 1 Ax. † 1. 3. 23. 1. 17. 3. 18. 3. PROPOSITION III. PROB. About a given circle, to describe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle; it is required to describe a triangle about the circle ABC, equiangular to the triangle DEF. L K GE H N Produce EF both ways to the points G, H; find the centre K of the circle ABC, and from it draw any straight line KB; at the point K, in the straight line KB, make the angle BKA equal to the angle DEG, M B and the angle BKC equal to the angle DFH; and through the points A, B, C, draw the straight lines LAM, MBN, NCL, touching the circle ABC: LMN shall be the triangle required. * Because LM, MN, NL touch the circle ABC in the points A, B, C, to which from the centre are drawn KA, KB, KC, the angles at the points A, B, C are right angles: and because the four angles of the quadrilateral figure AMBK are equal to * 13. 1. + Ax. + Constr. four right angles ‡, for it can be divided into two triangles, and that two of them KAM, KBM are right angles, therefore the other two AKB, AMB are equal † to two right angles: † 3 Ax. but the angles DEG, DEF are likewise equal to two right angles; therefore the angles AKB, AMB are equal † to the angles DEG, DEF; of which AKB is equal † to DEG; where fore the remaining angle AMB is equal † to the remaining angle DEF. In like manner, the angle LNM may be demonstrated to be equal to DFE; and therefore the remaining angle MLN is equal to the remaining angle EDF: there- 32. 1. & fore the triangle LMN is equiangular to the triangle DEF; and it is described about the circle ABC. Which was to be done. +3 Ax. 8 Ax. PROPOSITION IV. B.-To inscribe a circle in a given triangle. Let the given triangle be ABC; it is required to inscribe a circle in ABC. E See N. 9. 1. . 12. 1. † 11 Ax. Bisect * the angles ABC, BCA by the straight lines BD, CD meeting one another in the point D, from which draw DE, DF, DG perpendiculars to AB, BC, CA. And because the angle EBD is equal to the angle FBD, for the angle ABC is bisected by BD, and that the right angle BED is equal to the right angle BFD; therefore the two triangles EBD, FBD have two angles of the one, equal to two angles of the other, each to each; and the side BD, which is opposite to one of the equal angles in each, is common to both; therefore their other sides are equal *; wherefore DE is equal to 26. 1. DF for the same reason, DG is equal to DF; therefore DE B F is equal to DG: therefore the three straight lines DE, DF, † 1 Ax. DG are equal to one another, and the circle described from the centre D, at the distance of any of them, will pass through the extremities of the other two, and touch the straight lines AB, BC, CA, because the angles at the points E, F, G, are Suppose the line MK drawn: then, because the three angles of every triangle are together equal to two right angles, therefore all the angles of the 32. 1. two triangles AMK, BMK, are together equal to four right angles: but all the angles of these triangles are together equal to the angles of the figure ABMK; therefore all the angles of the figure AMBK are together equal to four right angles. 16. 3. * right angles, and the straight line which is drawn from the extremity of a diameter at right angles to it, touches the circle; therefore the straight lines AB, BC, CA do each of them touch the circle, and therefore the circle EFG is inscribed in the triangle ABC. Which was to be done. See N. * 10. 1. * 11. 1. PROPOSITION V. PROB.—To describe a circle about a given triangle. Let the given triangle be ABC; it is required to describe a circle about ABC. * Bisect * AB, AC, in the points D, E, and from these points draw DF, EF at right angles to AB, AC, DF, EF produced, one another : meet * 4. 1. are parallel; which is absurd let them meet in F, and join *See fig. 3. FA; also if the point F be not in BC, join BC, CF *. Then, because AD is equal to DB, and DF common, and at right angles to AB, the base AF is equal * to the base FB. In like manner it may be shewn, that CF is equal to FA; and therefore BF is equal to FC; and FA, FB, FC, are equal to one another wherefore the circle described from the centre F, at the distance of one of them, will pass through the extremities of the other two, and be described about the triangle ABC. Which was to be done. † 1 Ax. † 31. 3. † 31. 3. +31. 3. COROLLARY. And it is manifest, that when the centre of the circle falls within the triangle, each of its angles is less than a right angle †, each of them being in a segment greater than a semicircle; but when the centre is in one of the sides of the triangle, the angle opposite to this side, being in a semicircle t, is a right angle; and, if the centre falls without the triangle, the angle opposite to the side beyond which it is, being in a segment less than a semicircle +, is greater than a right angle: therefore, conversely, if the given triangle be acute-angled, the centre of the circle falls within it; if it be a right-angled triangle, the centre is in the side opposite to the right angle; and if it be an obtuse-angled triangle, |