VI. lineal figure, when the circumference of VII. A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle. PROP. 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. E 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 cir А cle 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 B not, BC is greater than D; make CE equal (3. 1.) to D, and from the centre C, at the distance CE, describe the circle AEF, and join CA: Therefore, because C is the D 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. PROP. 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. Draw (17. 3.) the straight line GAH touching the circle in the point A, and at the point A, in the straight line AH, make (23. 1.) the angle HAC equal to the angle DEF; and at the point A, in the straight line AG, make the angle GAB equalto the angleDFE,and join G BC. Therefore, because А H HAG touches the circle ABC and AC is drawn from the point of contact, the angle D HAC is equal (32. 3.) to the angle ABC in the alternate segment of the circle : But HAC is equal to the angle B В 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 (32. 1.) 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. PROP. III. PROB. About a given circle to describe a triangle equiangular to a given tri angle. 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. 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, make (23. 1.) 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 (17. 3.) the circle ABC: 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 (18. 3.) angles. And because the four angles of the quadrilateral figure ÁMBK are equal to four right angles, for it can be divided into two triangles; and because two of them, KAM, KBM are right angles, L D the other two AKB, AMB are equal to two right angles : But the angles DEG, DEF are likewise equal (13. 1.) to two 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 (32. 1.) to the remaining angle EDF: Wherefore the triangle LMN is equiangular to the triangle DEF : And it is described about the circle ABC. Which was to be done. G D Let the given triangle be ABC ; it is required to inscribe a circle in ABC. Bisect (9. 1.) the angles ABC, BCA by the straight lines BD, CD meeting one another in the point D, from which draw (12. 1.) DE, DF, DG perpendiculars, to AB, BC, CA. A Then because the angle EBD is equal to the angle FBD, the angle ABC being bisected by BD; and because the right angle BĚD, is equal to the right angle BFD, the two triangles EBD, E E 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 are B В F equal (26.1.); wherefore DE is equal to DF. 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, will pass through the extremities of the other two, and will 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 (Cor. 16. 3.) the cir-.. cle. Therefore the straight lines AB, BC, CA, do each of them touch the circle, and the circle EFG isinscribed in the triangle ABC. Which was to be done. 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 ABC. Bisect (10. 1.) AB, AC in the points D, E, and from these points draw DF, EF at right angles (11. 1.) to AB, AC ; DF, EF produced will 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 (4. 1.) 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. Cor. 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 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, the centre falls without the triangle, beyond the side opposite to the obtuse angle. PROP. VI. PROB. To inscribe a square in a given circle, Let ABCD be the given circle ; it is required to inscribe a square in ABCD. Draw the diameters AC, BD at right angles to one another, and join AB, BC, CD, DA ; because BE is equal to ED, E being the centre, and because EA is at right angles to BD, and common to the triangles ABE, ADE ; the base BA is equal (4. 1.) to the base AD ; and, for the same reason, BC, CD are each of them equal to BA or AD ; E therefore the quadrilateral figure ABCD B D is equilateral. It is also rectangular ; for the straight line BD being a diameter of the circle ABCD, BAD is a semicircle ; wherefore the angle BAD is a right (31. 3.) angle ; for the same reason each N of the angles ABC, BCD, CDA is a right angle ; therefore the quadrilateral figure ABCD is rectangular, and it has been shewn to be equilateral ; therefore it is a square ; and it is inscribed in the circle ABCD. Which was to be done. 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 through the points A, B, C, D draw (17. 3.) FG, GH, HK, KF touching the circle ; and because FG touches the circle ABCD, and EA is drawn from the centre E to the point of contact Av, the angles at A are right (18. 3.) angles ; for the same reason, the angles at the points, B, C, D are right angles ; and because the angle AEB is a right angle, as likewise is EBG, GH is parallel (28. 1.) to AC ; for the same reason, G A А F Η с K to GF or HK ; therefore the quadrilateral figure FGHK is equilateral. It is also rectangular; for GBEA being a parallelogram, and AEB a right angle, AGB (34. 1.) is likewise a right angle: In the same manner, it may be shewn that the angles at H, K, F are right angles ; therefore the quadrilateral figure FGHK is rectangular ; and it was demonstrated to be equilateral; therefore it is a square ; and it is described about the circle ABCD. Which was to be done. D Let ABCD be the given square ; it is required to inscribe a circle in ABCD. Bisect (10. 1.) each of the sides AB, AD, in the points F, E, and through E draw (31. 1.) EH parallel to AB or DC, and through F draw FK parallel to AD or BC ; therefore each of the figures AK, KB, AH, HD, AG, GC, BG, GD is a parallelogram, and their opposite sides are equal (34. 1.); and because that AD is equal to AB, |