the points A, B, C, draw the straight lines LAM, MBN, NCL, touching (III, 17.) the circle ABC. D K GE FH M B Because LM, MN, NL, touch the circle ABC in the points A, B, C, to which from the centre are drawn KA, KB, KC, (III. 18.) 1. 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, KĂM, KBM, are right angles, the other two (Ax. 3.) 2. AKB, AMB, are equal to two right angles: but the angles DEG, DEF, are likewise equal (I. 13.) to two right angles; therefore (Ax. 1.) DEF, 3. The angles AKB, AMB, are equal to the angles DEG, of which AKB is equal (Constr.) to DEG; wherefore (Ax. 3.) In like manner, it may be demonstrated that 5. The angle LNM is equal to DFE; and therefore (I. 32. and Ax. 3.) 6. The remaining angle MLN is equal to the remaining angle EDF. Wherefore the triangle LMN is equiangular to the triangle DEF; and it is described about the circle ABC. Q.E.F. PROP. IV.-PROBLEM. To inscribe a circle in a given triangle. Let the given triangle be ABC; it is required to inscribe a circle in ABC. Bisect (I. 9.) the angles ABC, BCA, by the straight lines BD, CD, meeting one another in the point D, from which draw (I. 12.) 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 (Ax. 11.) 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 (I. 26.); wherefore DE is equal to DF: 1. 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 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 (III. 16.); therefore 4. The straight lines AB, BC, CA, do each of them touch the circle, and therefore the circle EFG is inscribed in the triangle ABC. Q.E.F. PROP. V.PROBLEM. To'describe a circle about a given triangle. Let the given triangle be ABC; it is required to describe a circle about ABC. Bisect (I. 10.) AB, AC, in the points D, E, and from these points draw DF, EF, at right angles (I. 11.) to AB, AC. [DF, EF, 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. 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, (I. 4.) 1. The base AF is equal to the base FB. In like manner it may be shown that 2. CF is equal to FA; and therefore (Ax. 1.) wherefore FA, FB, FC, are equal to one another, and 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. Q.E.F. 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, (III. 31.) 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 is a right angle, being in a semicircle; and if the centre falls without the triangle, the angle opposite to the side beyond which it is, is greater than a right angle, being in a segment less than a semicircle. Wherefore, 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, the centre falls without the triangle, beyond the side opposite to the obtuse angle. PROP. VI.-PROBLEM. 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 (III. 1. and I. 11.) at right angles to one another, and join AB, BC, CD, DA. Because BE is equal to ED, for E is the centre, and that EA is common, and at right angles to BD; (I. 4.) 1. The base BA is equal to the base AD: and for the same reason, therefore 2. BC, CD, are each of them equal to BA or AD; 3. The quadrilateral figure ABCD is equilateral. It is also rectangular; for the straight line BD, being the diameter of the circle ABCD, BAD is a semicircle; wherefore (III. 31.) 4. The angle BAD is a right angle; for the same reason, therefore 5. Each of the angles ABC, BCD, CDA, is a right angle; 6. The quadrilateral figure ABCD is rectangular, and it has been shown to be equilateral; therefore (I. Def. 30.) 7. The quadrilateral figure ABCD is a square; and it is inscribed in the circle ABCD. Q.E.F. PROP. VII.-PROBLEM. 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 (III. Ï7.) FG, GH, HK, KF, touching the circle. Because FG touches the circle ABCD, and EA is drawn from the centre E to the point of contact A, (III. 18.) 2. The angles at the points B, C, D, are right angles; and because the angle AEB is a right angle, as likewise is EBG, (I. 28.) 6. The figures GK, GC, AK, FB, BK, are parallelograms; and therefore (I. 34.) and because AC is equal to BD, and that AC is equal to each of the two GH, FK; and BD to each of the two GF, HK; therefore 8. GH, FK, are each of them equal to GF, or HK; 9. The quadrilateral figure FGHK is equilateral. It is also rectangular; for GBEA being a parallelogram, and AEB a right angle, likewise (I. 34.) and it was demonstrated to be equilateral; therefore (I. Def. 30.) Let ABCD be the given square; it is required to inscribe a circle in ABCD. Bisect (I. 10.) each of the sides AB, AD, in the points F, E; and through E draw (I. 31.) EH parallel to AB or DC; and through F draw FK parallel to AD or BC. 1. Each of the figures AK, KB, AH, HD, AG, GC, BG, GD, is a parallelogram, and (I. 34.) their opposite sides are equal; and because AD is equal (I. Def. 30.) to AB, and that AE is the half of AD, and AF the half of AB, (Ax. 7.) 2. AE is equal to AF; wherefore the sides opposite to these are equal, viz. 3. FG is equal to GE; in the same manner it may be demonstrated that 4, GH, GK, are each of them equal to FG or GE; therefore 5. The four straight lines GE, GF, GH, GK, are equal to one another; and the circle described from the centre G, at the distance of one of them, shall pass through the extremities of the other three, and touch the straight lines AB, BC, CD, DA, because the angles at the points E, F, H, K, are (I. 29.) right angles, and that the straight line which |