ore, because C is the centre of the equal to the given straight line D, circle AEF, CA is equal to CE, but which is not greater than the diaD is equal to CE ; therefore D is meter of the circle. Which was to equal to CA: Wherefore, in the be done. circle ABC, a straight line is placed PROP. II. PROB. In a given circle to inscribe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle, it is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF. Draw (17. 3.) the straight line GAH touching the circle in the point A, and at the point A, in the the alternate segment of the circle': straight line AH, make (23. 1.) the But HAC is equal to the angle DEF: angle HAC equal to the angle DEF; therefore also the angle ABC is equal and at the point A, in the straight to DEF: For the same reason, the line AG, make the angle GAB equal angle ACB is equal to the angle to the angle Dre, and join BC: DFE; therefore the remaining angle Therefore, because HAG touches the BAC is equal (32. 1.) to the remaincircle ABC, and AC is drawn from ing angle EDF: Wherefore the triangle the point of contact, the angle HAC ABC is equiangular to the triangle is equal (32. 3.) to the angle ABC in DEF, and it is inscribed in the circle ABC. Which was to be done. PROP. III. PROB. About a given circle to describe a triangle equiangular to a given triangle. Let ABC be the given circle, and (17. 3.) the circle ABC; Therefore, DEF the given triangle ; it is require because LM, MN, NL touch the ed to describe a triangle about the circle ABC in the points A, B, C, to circle ABC equiangular to the tri- which from the centre are drawn KA, angle DEF. KB, KC, the angles at the points A, Produce EF both ways to the B, C, are right (18. a) angles : And points G, H, and find the centre K because the four angles of the quaof the circle ABC, and from it draw drilateral figure AMBK are equal to any straight line KB; at the point K, four right angles, for it can be divid. in the straight line KB, make (23. 1.) ed into two triangles; and that two the angle BKA equal to the angle of them KAM, KBM are right angles, 1 DEG, and the angle BKC equal to the other two AKB, AMB are equal the angle DFH; and through the to two right angles : But the angles points A, B, C, draw the straight DEG, DEF are likewise equal (13.1.) lines LAM, MBN, NCL, touching to two right angles; therefore the maining 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 (32. 1.) to the remaining angle EDF: Wherefore the triangle LMN is equiangular to the triangle angles AKB, AMB are equal to the DEF: And it is described about the angles DEG, DEF of which AKB is circle ABC. Which was to be done. eqnal to DÊG; wherefore the re PROP. IV. PROB. To inscribe a circle in a given trianglc. A Let the given triangle be ABC, it equal angles in each, is common to is required to inscribe a circle in both; therefore their other sides shall ABC. be equal; (26. 1.) wherefore DE is Bisect (9. 1.) the angles ABC, BCA equal to DF: For the same reason by the straight lines BD, CD meet- DG is equal to DF; therefore the ing one another in the point D, from three straight lines DE, DF, DG are which draw (12. 1.) DE, DF, DG equal to one another, and the circle perpendiculars described from the centre D, at the to AB, BC, distance of any of them, shall pass CA: And bea through the extremities of the other cause the an two, and touch the straight lines AB, gle EBD is e BC, CA, because the angles at the qual to the an points E, F, G, are right angles, and gle FBD, for the straight line which is drawn froin the angle ABC the extremity of a diarneter at right is bisected by BD, and that the right angles to it, touches (16. 3.) the angle BED is equal to the right angle circle ; therefore the straight lines • BFD, the two triangles EBD, FBD AB, BC, CA do each of them touch have two angles of the one equal to the circle, and the circle EFG is intwo angles of the other, and the side scribed in the triangle ABC. Which BD, which is opposite to one of the was to be done. D E PROP. V. PROB. To describe a circle about a given triangle, Let the given triangle be ABC; it Bisect (10. 1.) AB, AC in the is required to describe a circle about points D, E, and' from these points ABC. draw DF, EF at right angles (il. 1.) B B 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, and 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, the base AF is equal (4. 1.) to the base FB: In like manner, it may be shewn that CF is equal to FA; and therefore BF is equal to FC; and FA, FB, 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 ABČ. Which was to be done. Cor. And it is manifest, that when which it is, being in a segment less the centre of the circle falls within than a semicircle, is greater than a the triangle, each of its angles is less right angle: Wherefore, if the given than a right angle, each of them be. triangle be acute angled, the centre ing in a segment greater than a se of the circle falls within it; if it be a micircle ; but, when the centre is in right angled triangle, the centre is in one of the sides of the triangle, the the side opposite to the right angle; angle opposite to this side, being in a and, if it be an obtuse angled triangle, semicircle, is a right angle ; and, if the centre falls without the triangle, the centre falls without the triangle, beyond the side opposite to the obo the angle opposite to the side beyond tuse angle. a PROP. VI. PROB. To inscribe a square in a given circle. a Let ABCD be the given circle ; it ing the diameter of the circle ABCD, required to inscribe a square in BAD is a semicircle ; wherefore the ABCD. angle BAD is a right (31. 3.) angle; Draw the diameters AC, BD at for the same reason each of the right angles to one another; and join angles ABC, BCD, CDA is a right AB, BC, CD, DA ; because BE is angle ; therefore the quadrilateral fiequal to ED, for E is the centre, and gure ABCD is rectangular, and it has that EA is common, and at right been shewn to be equilateral ; there. angles to BD; the base BA is equal fore it is a square; and it is inscrib(4. 1.) to the base AD; and, for the ed in the circle ABCD. Which was same reason, BC, CD are each of to be done. them equal to BA, or AD; therefore the quadrilateral fi. gure ABCD is equilateral. It is also rectangular ; for the straight line BD, be D PROP. VII. PROB. K с To describe a square about a given circle. Let ABCD be the given circle; it HK, and GH to FK; and because is required to describe a square about AC is equal to BD, it. and that AC is equal o Draw two diameters AC, BD of to each of the two the circle ABCD, at right angles to GH, FK : and BD one another, and through the points to each of the two A, B, C, D, draw (17. 3.) FG, GH, GF, AK; GH, FK HK, KF touching the circle; and are each of them ebecause FG touches the circle ABCD, qual to GF or HK; therefore the and EA is drawn from the centre E quadrilateral figure FGHK is equilato the point of contact A, the angles teral. It is also rectangular; for at A are right (18. 3.) angles ; for GBEA being a parallelogram, and the same reason, the, angles at the AEB a right angle, AGB (34. 1.) is points B, C, D are right angles; and likewise a right angle: In the same because the angle AEB is a right manner, it may be shewn that the angle, as likewise is EBG, GH is pa- angles at H, K, F are right angles ; rallel (28. 1.) to AC; for the same therefore the quadrilateral figure reason, AC is parallel to FK, and in FGHK is rectangular, and it was delike manner GF, HK may each of monstrated to be equilateral; therethem be demonstrated to be parallel fore it is a square; and it is described to BED; therefore the figures GK, about the circle ABCD. Which was GC, AK, FB, BK are parallelograms; done. and GF is therefore equal (34. 1.) to PROP. VIII. PROB. To inscribe a circle in a given square. B H Let ABCD be the given square; GF, GH, GK are equal to one anoit is required to inscribe a circle in ther; and the circle described from the ABCD. centre G, at the disBisect (10. 1.) each of the sides tance of one of them, AB, AD, in the points F, E, and shall pass through the through E draw (31. 1.) EH parallel extremities of the oto AB or DC, and through F draw ther three, and touch FK parallel to AD or BC; therefore the straight lines AB, each of the figures AK, KB, AH, HI, BC, CD, DA; beAG, GC, BG, GD, is a parallelo- cause the angles at the points E, F, gram, and their opposite sides are H, K, are right (29. 1.) angles, and equal; (34. 1.) and because AD is that the straight line which is drawn equal to AB, and that AE is the half from the extremity of a diameter, at of AD, and AF the half of AB, AE is right angles to it, touches the.circle ; equal to AF; wherefore the sides (16. 3.) therefore each of the straigbt opposite to these are equal, viz. FG lines AB, BC, CD, DA touches the to GE ; in the same manner, it may circle, which therefore is inscribed in be demonstrated that GH, GK are the square ABCD. Which was to each of them egual to FG or GE; be done. therefore the four straight lines GE, PROP. IX. PROB. To describe a circle about a given square. Let ABCD be the given square; it verally bisected by the straight lines is required to describe a circle about BD, AC; therefore, because the angle it. DAB is equal to the angle ABC, and Join AC, BD cutting one another that the angle EAB is the half of in E; and because DA is equal to DAB, and EBA the half of ABC; the AB, and AC common to the triangles angle EAB is equal to the angle DAC, BAC, the two sides DA, AC EBA; wherefore the side EA is equal are equal to the two BA, AC, and (6. 1.) to the side EB: In the same the base DC is manner, it may be demonstrated, that equal to the base A the straight lines EC, ED are each of BC; wherefore them equal to EA or EB; therefore the angle DAC the four straight lines EA, EB, EC, is equal (8. 1.) to ED are equal to one another : and the angle BAC, the circle described from the centre E, and the angle B at the distance of one of them, shall DAB is bisected pass through the extremities of the by the straight line AC : In the same other three, and be described about marmer, it may be denionstrated that the square ABCD. Which was to be the ar gles ABC, BCD, CDA are se- done. PROP. X. PROB. To describe an isoceles triangle, having each of the angles at the buse double of the third angle. E B Take any straight line AB, and di- angle AB, BC, vide (11. 2.) it in the point C, so contained by the that the rectangle AB, BC, be equal whole of the cutto the squ’re of CA; and from the ting line, and the centre A, at the distance AB, describe part of it without the circle BDE, in which place (1. 4.) the circle, is equal the straight line BD equal to AC, to the square of which is not greater than the diame- BD which meets ter of the circle BDE; join DA, DC, it; the straight and about the triangle ADC describe line BD touches (37. 3.) the circle (5. 4.) the circle ACD; the triangle ACD; and because BD touches the ABD is such as is required, that is, circle, and DC is drawn from the each of the angler ABD, ADB is point of contact D, the angle BDC is double of the angle BAD. equal (32. 3.) to the angle DAC in Because the rectangle AB, BC is the alternate segment of the circle ; equal io the square of AC, and that to each of these add the angle CDA; AC is equal to BD, the rectangle AB, therefore the whole angle BDA is BC is equal to the square of BD; and equal to the two angles CDA, DAC; because from the point B without the but the exterior angle BCD is equal circle ACD, two straight lines BCA, (32. 1.) to the angles CDA, DAC; BD are drawn to the circumference, therefore also BDA is equal to BCD; one of which cuts, and the other but BDA is equal (5. 1.) to the angle meets the circle, and that the rect- CBD, because the side AD is equal |