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 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. V. PROB. To inscribe a square in a given circle. Let ABC be the given circle; it is requred 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, for E is the centre, and that EA is common, and at A right angles to BD ; the base BA is equal ~ZNS, (4.1.) to the base AD; and for the same / reason, BC, CD are each of them equal to gure ABCD is equilateral. It is also rect- N // angular; for the straight line BD, being N 2^ / the diameter of the circle ABCD, BAD is a N / / semicircle; wherefore the angle BAD is a N}^ right (31. 3.) angle; for the same reason C each 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 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 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 AEB is G A F a right angle, as likewise is EBG, GH - --T-s— is parallel (28. 1.) to AC ; for the same | 2- N reason, AC is parallel to FK, and in like / E. manner GF, HK may each of them be de- b monstrated to be parallel to BED ; there cause AC is equal to BD, and that AC C K is equal to each of the two GH, FK; 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 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 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 was described about the circle ABCD. Which was to be done. PROP. VII. PROB. 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 AD is equal to AB, and that AE is the half of AD, and AF the half of AB, EA is equal to AF; wherefore the sides opposite to these are equal, A E D viz. FG to GE : in the same manner, it -—-F---— may be demonstrated that GH, GK are each of them equal to FG or GE; therefore the four straight lines GE, GF, GH, G GK, are equal to one another; and the F - K 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; be- B H C cause the angles at the points E, F, H, K are right (29. 1.) angles, and that 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, CD, DA touches the circle, which therefore is inscribed in the square ABCD. Which was to be done PROP. IX. PROB. 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 that the angle EAB is the half of DAB, and EBA the half of ABC; the angle EAB is equal to the angle EBA ; wherefore the side EA is equal (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, shall pass through the extremities of the other three, and be described about the square ABCD. Which was to be done. PROP. X. PROB. To describe an isosceles triangle, having each of the angles at the base double of the third angle. Take any straight line AB, and divide (11. 2.) it in the point C, so that the rectangle AB, BC be equal to the square of CA; and from the centre A, at the distance AB, describe the circle BDE, in which place (1. 4.) the straight line BD is 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. s Because the rectangle AB, BC is equal to the square of AC, and that AC is equal to BD, the rectangle AB, BC is equal to the square of BD; and because from the point E 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 that 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 contact D, the angle BDC is equal (32. 3.) to the angle DAC B D 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 the angle CBD, 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 BDC is equal to the angle BOD, 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; therefore also BCD is double of DAC, and BCD is equal to each of the angles BDA, DBA.; each therefore of the angles BDA, DBA, is double of the angle DAB; wherefore an isosceles triangle ABD is described, having each of the angles at the base double of the third angle. Which was to be done. PROP. XI. PROP. 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 angles 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 A 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 each of the angles ACD, CDA is double of CAD, 9 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.) circumferences; therefore the five circumferences AB, BC, CD, DE, EA are equal to one another: and equal circumferences 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 circumference AB is equal to the circumference DE: if to each be added BCD, the whole ABCD is equal to the whole EDCB: and the angle AED stands on the circumference ABCD, and the angle BAE on the circumference 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. To describe an equilateral and equiangular pentagon about a given circle. Let ABCDE be the given circle; it is required to describe an equilateral and equiangular pentagon about the circle ABCDE. Let the angles of a pentagon, inscribed in the circle, by the last proposition, be in the points A, B, C, D, E, so that the circumferences AB, BC, CD, DE, EA are equal (11. 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 hine 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 circumference BC is equal to the circumference CD, the angle BFC is equal (27. 3.) to the angle CFD; and BFC is double of the G angle KFC, and CFD double of CFL; therefore the angle KFC is equal to the angle CFL; and the right angle FCK is equal to the right angle FCL: therefore, in the two triangles FKC, FLC, there are two angles of one equal to two angles of the other, each to each, and the side FC, which is adjacent to the equal angles in each, is common to both; therefore the other sides shall be equal (26. K C L 1.) to the other sides, and the third angle to the third angle: therefore the straight line KC is equal to CL, and the angle FKC to the |