a remaining points, and will be circumscribed about the triangle A B C. Let it be described as A B C. Therefore a circle is circumscribed about a giveni triangle. Which was to be done. Corollary. Hence it is manifeft, that if the centre of the circle falls within the triangle ABC, the angle BAC bėing in a segment greater than a semicircle, is less than a right angle ; if it falls in the right line BC, the angle in the semicircle will be a right angle; and if it falls without the triangle A B C, the angle in the segment less than a femicircle will be greater than a right angle. Wherefore if the given triangle ABC be an acute angled one, DE, EF will meet within the triangle; but if it has a right angle B A C, they will meet in the side bc: If B A C be an obtuse angle, they will meet without the triangle A B C. PROP. VI. PROBL. To inscribe a square in a given circle. Let the given circle be A B C D: it is required to inscribe a square in the circle A B C D. Draw [by 11. 1.] two diameters A C, B D at right an gles to one another; and join A By A B C, CD, DA. Then because be is equal to ED, for E is the centre, and E A is com mon and at right angles, the base AB B D [by 4. 1.) will be equal to the base AD; and by the same reason B C, CD are each equal to BA, BD; there fore the quadrilateral figure ABCD is equilateral. I say, it is also right angled. For because the right line BD is a diameter of the circle AB CD; BAD will be a semicircle: and so [by 31. 3] the angle B A D is a right angle: by the same reason, each of the angles A B C, BCD, CD A will be a right angle. Therefore the quadrilateral figure A B C D is right angled. And it has been proved to be equilateral too: Therefore it is a square, and inscribed in the circle A B C D. Therefore the square A B C D is inscribed in the given circle A B CD. Which was to be done. PROP. а a a PROP. VII. THEOR. To circumscribe a square about a given circle. K Let the given circle be A B CD. It is required to circumscribe a square about the given circle A B C D. Draw two diameters A C, BD of the circle ABCD at right angles to one another, and [by 17. 1.] thro' the points A, B, C, D draw the right lines F G, GH, H K, KF, to touch the circle ABCD in the points A, B, C, D. Then because F G touches the circle A B C D, and the right line ea is drawn from the centre E to the point of contact A ; the angles G ; A at a [by 18. 3.] will be right angles. By the same reason, the angles at the points B, C, D are right angles ; and E в. because the angle A E B is a right angle, and E B G also a right angle; the line G H [by 28. 1.] will be parallel to H AC: By the same reason A c is parallel to FK. In like manner we demonstrate that GF, HK are each parallel to BED. . Therefore GK, GC, AK, FB, B K are parallelograms; and so GF is [by 34. 1.] equal to HK, as also G H to FK; and because A c is equal to BD; but A c is equal to each of the lines GH, FK, and BD is equal to each of the lines GF, HK: Therefore each of the right lines HG, FK will be equal to each of the right lines GF, HK ; and so the quadrilateral figure F GHK is equilateral. I say it is also rectangular. For because GBEA is a parallelogram, and A E B is a right angle, [by 34. 1.] A G B will be a right angle. After the same manner we demonstrate that the angles at the points H, K, F are right angles, and so the quadrilateral figure F G H K is right angled. But it has been proved to be equilateral ; Therefore it is a square, a square, and is circumscribed about the circle ABCD. Therefore a square is circumscribed about a given circle. Which was to be done. PROP. VIII. PROBL. Let the given square be A B C D. It is required to infcribe a circle in the given Square ABCD. Divide [by 10. 1.) the sides A B, A D into equal parts, in the points F, E; and thro' £ draw [by 31. 1.] E parallel to A B or c D, and thro' F draw FK parallel to AD or BC. a a Then will AK, KB, AH, HD, AG, А. E D GC, BG, GD, be each of them parallelograms; and (by 34. 1.] their opposite sides are equal. And because E K A D is equal to A B, and A E is one half of a D, and A F one half of AB; A E will be equal to AF: and the B с H fides opposite to them are equal; therefore F G is equal to G E. In like manner we demonstrate that G H, GK are each equal to FG, G E. Therefore the four right lines G E, GF, GH, GK are equal to one another; and so a circle described with the centre G, and either of the distances GE, GF, GH, GK, will also pass thro’ the remaining points, and will touch the right lines A B, BC, CD, D A, because the angles at E, F, H, K are right angles : For if the circle cuts the right lines A B, B C, C D, D A, a right line drawn from the end of the diameter of a circle at right angles, will fall within the circle, which [by 16. 3.] is absurd. Therefore a circle defcribed with the centre G, and either of the distances G E, GF, GH, G K will not cut the right lines A B, BC, CD, DA. It will therefore touch them, and [by 5. def. 4.] will be inscribed in the square A B C D.... There'ore a circle is inscribed in a given square. Which was to be done. a PROP. IX. PROBL. Let the given square be A BCD. It is required to circumscribe a circle about the square A B C D. a For join Ą @, BD, cutting one another in the point E. Then because D A is equal to A B, and A c is common, the two sides DA, A c are equal to the two sides.B A, AC, and the base d c equal to the base Bc; [by 8. 3.) the an gle DAC will be equal to the angle A D BAC: Therefore the angle D A B is bifected by the right line A e. After E the same manner we demonstrate that each of the angles ABC, BCD, CDA, is bisected by the right lines AC, DB. Therefore because the angle D A B is B с equal to the angle ABC, and the angle E AB is one half of the angle DAB, and the angle EB A one half of the angle AB C; the angle É A B will be equal to the angle E B A: Wherefore [by 6. 1.] the fide E A is equal to the side E B. In like manner we demonAtrate that the right lines E C, E D are each equal to each of the right lines E A, E B. Therefore the four right lines E A, E B, E C, E D are equal to one another. Wherefore a circle described with the centre E, and either of the di{tances E A, E B, E C, E D will pass thro' the rest of the points, and will be cirumscribed about the square A B C D. Let it be described, as A B C D. Therefore a circle is circumscribed about a given square, Which was to be done. PRO P. X. PROBL. To make an isosceles triangle baving one of the angles at the base double to the remaining angle. . Let A B be any given right line, and [by 11. 2.] divide it in the point c, fo that the rectangle under A B, BC be equal to the square of cA; and about the centre A with the distance A B [by 3. poft.] describe a circle B D E, and [by 1. 4.] apply the right line B D in the circle B DE, which is not greater than the diameter, equal to c A, and joining DA, Dc circumscribe the circle A C D [by 5. 4.] about the triangle A C D. Then because the rectangle under A B, Bc is equal to the fquare of Ac; and A c is equal to BD; the rectangle N 2 B rectangle under A B, BC will be D cqual to the square of BD; and because the point B is taken without the circle ACD, and two right lines B C A, B D drawn from the ΑΣ point B fall upon the circle A CD, one of which cuts the circle, and the other meets it, and the rectangle under AB, BC is equal to the square of BD: the right line BD [by 37. 3.) will touch the circle A CD; and therefore because BD is a tangert, and Dc is drawn from the point of contact D; the angle B D c [by 32. 3.] will be equal to the angle D Ac in the alternate segment of the circle ; and since the angle B D c is equal to DAC; add CDA, which is common to both; then the whole angle B D A is equal ; to the two angles e DA, DAC. But the outward angle B C D [by 32. 1.] is equal to CDA, DAC; and therefore B D A is equal to BCD. But the angle B D A [by 5. 1.] is equal to the angle C B D, because the side A D is equal to the side AB: and therefore DBA will be equal to BCD. Wherefore the three angles B D A, D B A, BCD are equal to one another: And because the angle de c is equal to the angle B C D, the side B D [by 6. 1.] will be equal to the side Dc. But B D is made equal to cA; therefore AC is equal to CD; and so the angle CD a [by 5. 1.] is equal to the angle DAC: Therefore the angles CD A, DAC, taken together, are double to the angle DAC: But [by 32. 1.] the angle BCD also is equal to the angles C D A, DAC: Therefore B c is double to D A C. But BCD is equal to either of the angles B D A, D BA; and therefore either B D A or DBA is double to D A B. Therefore there is describe i an isosceles triangle A DB, having either of the angles at the base B D, double to the remaining angle. Which was to be done. PROP. XI. PROBL. P. To inscribe an equilateral and equiangular pentagon in a given circle Let the given circle be A B CDE. It is required to inscribe an equilateral and equiangular pentagon in the circle ABCDE. Let |