alternate segment of the circle (III. 32); 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 wherefore the triangle ABC and it is inscribed in the the remaining angle EDF (I. 32); About a given circle to describe a triangle equiangular to a given triangle. Given the circle ABC, and the triangle DEF; it is required to describe a triangle about the circle ABC equiangular to the triangle DEF. (Const.) 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 the angle BKA equal to the angle DEG (I. 23), angle BKC equal to the angle DFH; and through the points A, B, and C, draw the straight lines LAM, MBN, and NCL, touching the circle ABC (III. 17). (Dem.) Therefore, because LM, MN, and NL touch the circle ABC in the points A, B, M I and the and C, to which from the centre are drawn KA, KB, and KC, the angles at the points A, B, and C are right angles (III. 18); 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 because two of them KAM and KBM are right angles, the other two AKB and AMB are together equal to two right angles. But the angles DEG and DEF are likewise equal to two right angles (I. 13); therefore the angles AKB and AMB are equal to the angles DEG and 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 (I. 32); wherefore the triangle LMN and it is described about is equiangular to the triangle DEF; the circle ABC. To inscribe a circle in a given triangle. in the triangle ABC. it is required to inscribe a circle (Const.) Bisect the angles ABC and BCA, by the straight lines BD and CD, meeting one another in the point D, from which draw DE, DF, and DG perpendiculars to AB, BC, and CA. (Dem.) And because the angle EBD is equal to the angle FBD, and because the the angle ABC being bisected by BD; right angle BED is equal to the right angle BFD, the two triangles EBD and FBD have two angles of the one equal to two angles of the other; and the side BD, which is opposite to the right angles in each, is common to both h; E F therefore their other sides are equal (I. 26); wherefore DE is equal to DF. For a like reason, DG is equal to DF; therefore the three straight lines DE, DF, and 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, and CA, because the angles at the points E, F, and G are right angles, and the straight line which is drawn from the extremity of a radius at right angles to it, touches the circle (III. 16, Cor.); therefore the straight lines AB, BC, and CA do each of them touch the circle, and the circle EFG is inscribed in the triangle ABC. PROPOSITION V. PROBLEM. To describe a circle about a given triangle. Given the triangle ABC; about ABC. it is required to describe a circle (Const.) Bisect (I. 10) AB and AC, in the points D and E, and from these points draw DF and EF at right angles to AB and AC (I. 11); DF and EF produced, will meet one the base another; for if DE be joined, the two angles FDE and FED are together less than two right angles (I. 29, Cor.). Let them meet in F, and join FA; also, if the point F be not in BC, join BF and CF. (Dem.) Then because AD is equal to DB, and DF common, and at right angles to AB, AF is equal to the base FB (I. 4). In like manner, it may be shewn that CF is equal to FA; and therefore BF is equal to FC; and FA, FB, and 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. COR.-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 acute-angled, the centre of the circle falls within it; if it be a right-angled triangle, the centre is in the hypotenuse; and if it be an obtuseangled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle. PROPOSITION VI. PROBLEM. To inscribe a square in a given circle. Given the circle ABCD; ABCD. it is required to inscribe a square in (Dem.) (Const.) Draw the diameters AC and BD at right angles to one another, and join AB, BC, CD, and DĂ. 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 and ADE; the base BA is equal to the base AD (I. 4); and, for the same reason, BC and CD are each of them equal to BA or AD; therefore 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 the angle BAD is a right angle (III. 31); B E for the same reason, each of and it has been the angles ABC, BCD, and CDA is a right angle; therefore the quadrilateral figure ABCD is rectangular, shewn to be equilateral; therefore it is a square; inscribed in the circle ABCD. H and it is PROPOSITION VII. PROBLEM. To describe a square about a given circle. Given the circle ABCD; it is required to describe a square about it. (Const.) Draw two diameters AC and BD, of the circle ABCD, at right angles to one another, and (III. 17) through the points A, B, C, and D, draw FG, GH, HK, and KF, touching the circle. (Dem.) And because FG touches the circle ABCD, EA is drawn from the centre E to the point of contact A, angles at A are right angles (III. 18); for the same reason, the angles at the points B, and the for the and in and H hence K like manner, GF, HK may each of them be 'demonstrated to be parallel to BED, therefore parallel to one another; the figures GK, GC, AK, FB, and BK are parallelograms; and GF is therefore equal to HK (I. 34), and GH to FK; and because AC is equal to BD, and that AC is equal to each of the two GH and FK; and BD to each of the two GF and HK; therefore GH, FK, GF, and HK are all equal; therefore the quadrilateral figure FGHK is equilateral. It is also rectangular; for GBEA being a parallelogram, and AEB a right angle, AGB is likewise a right angle. In the same manner it may be shewn that the angles at H, K, and F are right angles; FGHK is rectangular; lateral; therefore the quadrilateral figure and it was demonstrated to be equiand it is described about therefore it is a square; the circle ABCD. To inscribe a circle in a given square. Given the square ABCD; it is required to inscribe a circle in it. and (Const.) Bisect (I. 10) each of the sides AB and AD in the points F and E, through E draw EH parallel to AB or DC (I. 31), and through F draw FK parallel to AD or BC; therefore each of the figures AK, KB, AH, HD, AG, GC, BG, and GD is a parallelogram, opposite sides are equal (I. 34). and their K And because AD is equal to AB, and that AE is the half of AD, and AF the half of AB, AE is equal to AF; wherefore the sides opposite to these are equal-namely, FG to GE; in the same manner it may be demonstrated that GH and GK are each of them equal to FG or GE; therefore the four straight lines GE, GF, GH, and 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 shall also touch the straight lines AB, BC, CD, and DA, because the angles at the points E, F, H, and K are right angles (I. 29), and because the straight line which is drawn from the extremity of a radius, at right angles to it, touches the circle (III. 16, Cor.); therefore each of the straight lines AB, which therefore is BC, CD, and DA touches the circle, inscribed in the square ABCD. PROPOSITION IX. PROBLEM. To describe a circle about a given square. about it. it is required to describe a circle E (Dem.) (Const.) Join AC and BD, cutting one another in E. And because DA is equal to AB, and AC common to the triangles DAC and BAC, the two sides DA and AC are equal to the two BA and AC, and the base DC is equal to the base BC; wherefore the angle DAC is equal to the angle BAC (I. 8), and the angle DAB is bisected by the straight line AC. In the same manner it may be demonstrated that the angles ABC, BCD, and CDA are severally bisected by the straight lines BD and AC; therefore because the angle DAB is equal to the angle ABC, and that the angle EAB is the half B of DAB, and EBA the half of ABC, the angle EAB is equal to the angle EBA; wherefore the side EA is equal to the side EB (I. 6). In the same manner it may be demonstrated that the straight lines EC and ED are each of them equal to EA or EB; therefore the four straight lines EA, EB, EC, and ED are equal to one another; and the circle described from the centre E, at the distance of one of them, shall pass through the extremities of the other three, and be described about the square ABCD (Def. 6). PROPOSITION X. PROBLEM. To describe an isosceles triangle, having each of the angles at the base double of the third angle. (Const.) Take any straight line AB, and (II. 11) divide it |