of the angles ABC, BCD, CDA is a right angle; therefore the quadrilateral figure ABCD is rectangular, and it has been shown to be equilateral; therefore it is a square; and it is inseribed 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 eirele; 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 cire ele ABCD, and EA is drawn from the centre E to the point of contact A, 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 ÄEB is a right angle, as likewise is EBG, GH is parallel (28. 1.) to AC; for the same reason, AC is parallel to FK, and in like manner, GF, HK may each of them be demonstrated to be parallel to BED; therefore the figures GK,GC, AK, FB, BK are parallelograms; and GF is therefore equal (34. 1.) to HK, and GH to FK; and because AC is equal to BD, and also to each of the two GH, and BD to each of the two GF, HK: GH, FK are each of them equal to GF or HK; therefore the quadrilateral figure FGHK is equilate: ral. It is also rectangular; for, GBEA being a parallelogram, and AEB à right angle, AGB (34. 1.) is likewise a right angle: In the ame manner, it may be shown 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. B FK; K PROP. VIII. PROB. + To inscribe a circle in a given square. 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 paralle) to AD or BC; therefore each of the figures, AK, KB, AH, HD, AG, GC, BG, G]) is a parallelogram, and their oppasite sides are equal (34. 1.); and becanse that AD is equal to AB, N E and that AF is the half of AD, and AF the half of AB, AE is equal to AF; wherefore the sides opposite to these are equal. viz. FG to GE; in the same manner, it may be demonstrated, that GH, GK, are oach of them equal to FG or GE; therefore 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, will pass through the extremities of the other three; and will also touch the straight lines AB, BC, CD, DA, because the angles at the points E, F, H, K, are right (29. 1.) angles, and because the straight line which is drawn from the extremity of a diameter at right angles to it, touches the circle (16.3.); therefore each of the straight lines AB, BC, CI, DA touches the circle, which is therefore inscribed in the square ABCD. Which was to be done. B H с PROP. IX. PROB. To describe a circle about a given square. D E Let ABCD be the given square; it is required to describe a circle about it. Join AC, BD, cutting one another in E; and because DA is equal to AB, and AC common to the triangles DAC, BAC, the two sides DA, AC are equal to the two BA, AC, and the base DC is equal to the base BC; wherefore the angle DAC is equal (8. 1.) to the angle BAC, and the angle DAB is bisected by the straight line AC. In the same manner, it may be demonstrated, that the angles ABC, BCD, CDA are severally bisected by the straight lines BD, AC; therefore, because the angle DAB is equal to the angle ABC, and the angle EAB is the half of DAB, and EBA the half of ABC; the angle EAB is equal to the angle EBA: and the side EA (6. 1.) to the side EB. In the same manner, it may be demonstrated, that the straight lines EC, ED are each of them equal to EA, or EB; therefore the four straight lines EA, EB, EC, ED are equal to one another; and the circle described from the centre E, at the distance of one of them must pass through the extremities of the other three, and be described about the square ABCD. Which was to be done. B + PROP. X. PROB. To describe an isosceles triangle, having each of the an gles at the base double of the third angle. E Take any straight line AB, and divide (11. 2.) it in the point C, sa that the rectangle AB, BC may be equal to the square of AC; and from the centre A, at the distance AB, describe the circle BDE, in which place (1. 4.) the straight line BD equal to AC, which is not greater than the diameter of the circle BDE; join DA, DC, and about the triangle ADC describe (5. 4.) the circle ACD; the triangle ABD is such as is required, that is, each of the angles ABD, ADB is double of the angle BAD. Because the rectangle AB.BC is equal to the square of AC, and AC equal to BD, the rectangle AB.BC is equal to the square of BD; and because from the point B without the circle ACD two straight lines BCA, BD are drawn to the circumference, one of which cuts, and the other meets the circle, and the rectangle AB. BC contained by the whole of the cutting line, and the part of it without the circle, is equal to the square of BD which meets it; the straight line BD touches (37. 3.) the circle ACD. And because BD touches the circle, and DC is drawn from the point of cortict 1), the angle BDC is equal (32. 3.) to the angle DAC in the alternate segment of the circle ; to each of these add the angle CDA: therefore the whole angle BDA is equal to the two angles CDA, DAC; but the exterior angle BCD is equal (32. 1.) to the angles CDA, DAC; therefore also BDA is equal to BCD, but BDA is equal (5. 1.) to CDB, because the side AD is equal to the side AB; therefore CBD, or DBA is equal to BCD; and consequently the three angles BDA, DBA, BCD, are equal to one another. And because the angle DBC is equal to the angle BCD, the side BD is equal (6.1.) to the side DC; but BD was made equal to CA; therefore also CA is equal to CD, and the angle CDA equal (5. 1.) to the angle DAC; therefore the angles CDA, DAC together, are double of the angle DAC; but BCD is equal to the angles CDA, DAC (32. 1.); therefore also BCD is double of DAC. But BCD is equal to each of the angles BDA, DBA, and therefore each of the angles BDA, DBA, is double of the angle DAB; where. fore an isosceles trianglé, ABD is described, having each of the an. gles at the base double of the third angle. Which was to be done. e “ Cor. 4. The angle BAD is the fifth part of two right augles.. For since each of the angles ABD and ADB is equal to twice the angle BAD, they are together equal to four times BAD, and therefore all the three angles ABD, ADB, BAD, taken together, are equal to five times the angle BAD. But the three angles ABD, ADB, BAD are equal to two right angles, therefore five times the angle BAD is equal to two right angles; or BAD is the fifth part of two right angles.” 6 Cor. 2. Because BAD is the fifth part of two, or the tenth part of four right angles, all the angles about the centre A are together equal to ten times the angle BAD, and may therefore be divided into ten parts each equal to BAD. And as these ten equal angles at the centre, must stand on ten equal arches, therefore the arch BD is one-tenth of the circumference; and the straight line BD), that is AC, is therefore equal to the side of an equilateral decagon inscribed in the circle BDE.” . A F E B PROP. XI. PROB. To inscribe an equilateral and equiangular pentagon in a given circle. Let ABCDE be the given circle, it is required to inscribe an equilateral and equiangular pentagon in the circle ABCDE. Describe (10. 4.) an isosceles triangle FGH, having each of the an. gles at G, H, double of the angle at F; and in the circle ABCDE inscribe (2. 4.) the triangle ACD equiangular to the triangle FGH, so that the angle CAD be equal to the angle at F, and each of the angles ACD, CDA equal to the angle at G or H; wherefore each of the angles ACD, CDA is double of the angle CAD. Bisect (9. 1.) the angles ACD, CDA by the straight lines CE, DB; and join AB, BC, DE, EA. ABCDE is the pentagon required. Because the angles ACD, CDA are each of them double of CAD, and are bisected by the straight lines CE, DB, the five angles DAC, ACE, ECD,CDB, BDA are equal to one another; but equal angles stand upon equal (26. 3.) arches; therefore the five arches AB, BC, CD, DE, EA are equal to one another: and equal arches are subtended by equal (29. 3.) straight lines; therefore the five straight lines AB, BC, CD, DE, EA are equal to one another. Wherefore the pentagon ABCDE is equilateral. It is also equiangular; because the areh AB is equal to the areh DE; if to each be added BCD, the whole ABCD is equal to the whole EDCB: and the angle AED stands on the arch ABCD, and the angle BAE on the are H EDCB: therefore the angle BAE is equal (27.3.) to the angle AED: for the same reason, each of the angles ABC, BCD, CDE is equal to the angle BAE or AED: therefore the pentagon ABCDE is equiangular: and it has been shown that it is equilateral. Wherefore, in the given circle, an equilateral and equiangular pentagon has been inscribed. Which was to be done. Otherwise : 6 Divide the radius of the given circle, so that the rectangle confuined by the whole and one of the parts may be equal to the square of the other (11. 2.). Apply in the circle, on each side of a given point, a line equal to the greater of these parts; then (2. Cor, 10, 4.) each of the arches eut off will be one-tenth of the circumference, and therefore the arch made up of both will be one-fifth of the circumference; and if the straight line subtending this arch be drawn it will be the side ofan equilateral pentagon inscribed in the circle." pro PROP. XII. PROB. To describe an equilateral and equiangular pentagon about a given circle. Let ABCDE le the given circle, it is required to describe an eqnilateral and equiangular pentagon about the circle ABEDE. Let the angles of a pentagon, ins ribed in the circle, by the last position, be in the points, A, B, C, D, E, so that the arches, AB, BC, CD, DE, FA are equal (1:1. 4.); and through the points A, B, C, D, E draw GH, HK, KL, LM, MG, touching (17.3.) the circle; take the centre F, and join,FB, FK,FC,FL, FD. And because the straight line KL touches the circle ABCDE in the point C, to which FC is drawn from the centre F, FC is perpendicular (18.3.) to KL; therefore each of the angles at C is a right angle: for the same reason, the angles at the points B, D are right angles; and because FCK is a right angle the square of FK is equal (47. 1.) to the squares of FC, CK. For the same reason, the square of FK is equal to the squares of FB, BK: therefore the squares of FC, CK are equal to the squares of FB, BK, of which the square of FC is equal to the square of FB; the remaining square of CK is therefore equal to the remaining square of BK, and the straight line CK equal to BK: and because FB is equal to FC and FK common to the triangles BFK, CFK, the two BF, FK are equal to the two CF, FK; and the base BK is equal to the base KC; therefore the angle BFK is equal (8. 1.) to the angle KFC, and the angle BKF to FKC; wherefore the angle BFC is double of the angle KFC, and BKC double of FKC: for the same reason, the angle CFD is double of the angle CFL, and CLD double of CLF: and because the arch BC is equal to the arch CD the angle BFC is equal (27. 3.) to the angle CFD; and BFC is double of the angle KFC, and CFD double of CFL; therefore the angle KFC is |