Wherefore, if the given triangle be acute angled, the centre Book IV. of the circle falls within it; if it be a right angled triangle, the centre is in the fide oppofite to the right angle, and, if it be an obtuse angled triangle, the centre falls without the triangle, beyond the fide oppofite to the obtufe angle. PROP. VI. PRO B. To infcribe a fquare in a given circle. Let ABCD be the given circle; it is required to inscribe a fquare in ABCD. Draw the diameters AC, BD at right angles to one ano- the fame reafon, BC, CD are B E D 2 4. I. femicircle; wherefore the angle BAD is a right b angle; for b 31. 3. the same reason each of the angles ABC, BCD, CDA is a right angle; therefore the quadrilateral figure ABCD is rectangular, and it has been fhown to be equilateral; therefore it is a fquare; and it is infcribed in the circle ABCD. Which was to be done. To defcribe a square about a given circle. Let ABCD be the given circle; it is required to defcribe a fquare about it. I 2 Draw Book IV. Draw two diameters AC, BD of the circle ABCD, at right angles to one another, and through the points A, B, C, D draw a 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 b 18. 3. bangles; for the fame reafon, the angles at the points B, C, D are right angles; and because the angle AEB is a right angle, as likewife is EBG, GH d 34. I. G B E F K € 28. 1. is parallel to AC; for the fame reason, AC is parallel to FK, and in like manner GF, HK may each of them be demonftrated to be parallel to BED; therefore the figures GK, GC, AK, FB, BK are parallelograms; and GF is therefore equal d to HK, and GH to FK; and becaufe AC is H equal to BD, and that AC 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 alfo rectangular; for GBEA being a parallelogram, and AEB a right angle, AGB d is likewife a right angle: In the fame manner, it may be fhown that the angles at H, K, F are right angles; therefore the quadrilateral figure FGHK is rectangular; and it was demonftrated to be equilateral; therefore it is a fquare; and it is defcribed about the circle ABCD. Which was to be done. 3 10. I. PROP. VIII. PROB. To infcribe a circle in a given square. Let ABCD be the given fquare; it is required to infcribe a circle in ABCD. Bifect a each of the fides AB, AD, in the points F, E, and 31. 1. through E draw b 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, 34.1. and their oppofite fides are equal; and because AD is equal I. A E D F K to AB, and that AE is the half of AD, and AF the half of Book IV. AB, AE is equal to AF; wherefore the fides oppofite to these are equal, viz. FG to GE; in the fame manner, it may be demonftrated, that GH, GK are each of them equal to FG or GE; therefore the four ftraight lines GE, GF, GH, GK, are equal to one another; and the circle defcribed from the centre G, at the distance of one of them, fhall pass through the extremities of the other three; and fhall also touch the ftraight lines AB, BC, CD, DA, because the angles at the points E, F, H, K are right angles, and because the ftraight line which is drawn from the extremity of a diameter, at right angles to it, touches the circlee; therefore each of the straight lines AB, BC, CD, DA e 16. 3. touches the circle, which therefore is infcribed in the fquare ABCD. Which was to be done. d B H C d 29. I. To defcribe a circle about a given fquare. Let ABCD be the given square; it is required to defcribe 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 fides 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 A a to the angle BAC, and the angle DAB is bifected by the ftraight line AC: In the fame manner, it may be demonftrated, that the angles ABC, BCD, CDA are feverally bifected by the ftraight lines BD, AC; therefore, because the angle DAB is equal to the angle ABC, and that the B E angle EAB is the half of DAB, and a 8. F. b 6. 1. Book IV. and EBA the half of ABC; the angle EAB is equal to the angle EBA; wherefore the fide EA is equal b to the fide EB: In the fame manner, it may be demonftrated, that the ftraight lines EC, ED are each of them equal to EA or EB; therefore the four ftraight 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, fhall pass through the extremities of the other three, and be described about the fquare ABCD. Which was to be done. a II. 2. b I. 4. C 5. 4. T O defcribe an ifofceles triangle, having each of the angles at the bafe double of the third angle. Take any ftraight line AB, and divide a it in the point C, fo that the rectangle AB, BC may be equal to the square of CA; and from the centre A, at the distance AB, describe the circle BDE, in which place b 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 the circle ACD; the triangle ABD is fuch as is required, that is, each of the angles ABD, ADB is double of the angle BAD. E Because the rectangle AB, BC is equal to the fquare of AC, and that AC is equal to BD, the rectangle AB, BC is equal to the fquare of BD; and because from the point B without the circle ACD two ftraight 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 fquare of BD which meets it; the straight line BD touches d the circle 37. 3. B ACD; and because BD touches the circle, and DC is drawn 32 3. drawn from the point of contact D, the angle BDC is equal e Book IV. to the angle DAC in the alternate fegment 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 to the angles CDA, DAC; · 32. 1. therefore alfo BDA is equal to BCD; but BDA is equal g g 5.1. to CBD, because the fide AD is equal to the fide AB; therefore CBD, or DBA is equal to BCD; and confequently the three angles BDA, DBA, BCD, are equal to one another: And because the angle DBC is equal to the angle BCD, the fide BD is equal h to the fide D; but BD was made equal to CA; therefore alfo CA is equal to CD, and the angle CDA equal g 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 afo 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 isofceles triangle ABD is described, having each of the angles at the bafe double of the third angle. Which was to be done. h 6. 1. T PRO P. XI. PROB. O infcribe an equilateral and equiangular pen- Let ABCDE be the given circle, it is required to infcribe an equilateral and equiangular pentagon in the circle ABCDE. Defcribe a an ifofceles triangle FGH, having each of the a 10.4. angles at G, H, double of the angle at F; and in the circle ABCDE infcribe b the triangle ACD equiangular to the b 2. 4. I 4 triangle |