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 about another figure, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each. III. A rectilineal figure is said to be inscribed in a circle, when all the angles of the inscribed figure are upon the circumference of the circle. IV. A rectilineal figure is said to be described about a circle, when each side of the circumscribed figure touches the circumference of the circle. V. In like manner, a circle is said to be inscribed in a rectilineal figure, when the circumference of the circle touches each side of the figure. 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. PROB. In a given circle to place a straight line, equal to a given straight line 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. Draw BC the diameter of the circle ABC; then, if A Therefore, because C is the centre of the circle AEF, CA is equal to CE: but D is equal to CE; therefore D is equal to CA. Wherefore, in the circle ABC, a straight line is placed equal to the given straight line D, which is not greater than the diameter of the circle. Which was to be done. 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. ⚫ 17. 3. + 23. 1. 32. 3. 32. 1. * 1. 3. + 23. 1. I 17. 3. 18. 3. 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 join BC: Therefore, because HAG touches the circle ABC, and AC is drawn from the point E of contact, the angle HAC is equal to the angle ABC in the alternate segment of the circle : Ꭰ F G B C H 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 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. 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 D Produce EF both ways to the points G, H, and 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, maket the angle BKA equal to the angle DEG, 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: A K Therefore, 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 four right angles, for it can be divided into two triangles, and that two of them KAM, KBM are right angles, the other two AKB, AMB are equal to two right angles : But the angles DEG, DEF are likewise equal || to two 13. 1. right angles; therefore the angles AKB, AMB are equal to the angles DEG, DEF, of which AKB is equal to DEG; wherefore 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: wherefore the tri- § 32. 1. angle LMN is equiangular to the triangle DEF; and it is described about the circle ABC. Which was to be done. PROPOSITION IV. PROB. To inscribe a circle in a given triangle. Let the given triangle be ABC; it is required to inscribe a circle in ABC. * Bisect the angles ABC, BCA by the straight lines BD, And because the angle EBD E * 9. 1. † 12. 1. G D F C Constr. BED is equal to the right angle BFD; the two triangles EBD, FBD have two angles of the one equal to two angles of the other; and the side BD, which is opposite to one of the equal angles in each, is common to both; therefore their other sides shall be equal; wherefore DE is equal to DF: || 26. 1.7 For the same reason, DG is equal to DF; Therefore the three straight lines DE, DF, DG are equal to one another, and the circle described from the centre D, at the distance of any of them, shall 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 right angles, and the straight line which is drawn from the extremity of a diameter at right angles to it, touches § the circle; there- § 16. 3. fore the straight lines AB, BC, CA do each of them touch 10. 1. † 11. 1. the circle, and the circle EFG is inscribed in the triangle ABC. Which was to be done. 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 4. 1. produced, meet one another: for, if they do not meet, they are parallel, wherefore AB, AC, which are at right angles to them, are parallel; which is absurd: let them meet in F, and join FA; also if the point F be not in BC, join BF, 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 shown 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, shall pass through the extremities of the other two, and be described about the triangle ABC. Which was to be done. COR. 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, 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: wherefore, if the given triangle be acuteangled, 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, the |