Construction.—1. Bisect AB, AC, in the points D, E. (I. 10.) . 2. From the points D, E, draw DF, EF, at right angles to AB, AC. (I. 11.) 3. DF, EF, produced, will meet one another; for if they do not, they are parallel, and AB, AC, which are at right angles to them, are parallel; which is absurd: 4. Let them meet in F, and join FA; also if the point F be not in BC, join BF, CF. 5. And from the centre F, at the distance BF, CF, or AF, describe the circle ABC. Then the circle ABC shall be described about the triangle ABC. Proof.-1. Because AD is equal to BD (const.), and DF common, and at right angles to AB, 2. Therefore the base FA is equal to the base FB. (I. 4.) 3. In the same manner it may be shown that FC is equal to FA. 4. Therefore FB is equal to FC; (ax. 1.) 5. Therefore FA, FB, FC, are equal to one another, and the circle described from the centre F, at the distance of any one of them, will pass through the extremities of the other two. Conclusion.—Therefore, the circle ABC is described about the triangle ABC. Q. E. F. Corollary. 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 in 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. (ITI. 31.) Wherefore, conversely, 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 hypotenuse; and if it be an obtuse-angled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle. PROPOSITION 6.-PROBLEM. (References, Prop. I. 4, 11; III. 1, 31.) Construction.-1. Draw two diameters, AC, BD, at right angles to one another. (III. 1, I. 11.) 2. Join AB, BC, CD, DA. Then the figure ABCD shall be the square required. Proof.—1. Because BE is equal to DE, for E is the centre; 2. And that EA is common, and at right angles to BD; 3. Therefore the base BA is equal to the base DA. (I. 4.) 4. For the same reason, BC, DC, are each of them equal to BA, or DA; 5. Therefore the quadrilateral figure ABCD is equilateral. 6. It is also equiangular; 7. For the straight line BD being a diameter of the circle ABCD, BAD is a semicircle; (const.) 8. Therefore the angle BAD is a right angle. (III. 31.) 9. For the same reason, each of the angles ABC, BCD, CDA, is a right angle; 10. Therefore the quadrilateral figure ABCD is rectangular; and it has been shown to be equilateral; therefore it is a square (I. def. 30); and it touches the circumference of the circle at each of its angles. Conclusion.--Therefore, the square ABCD is inscribed in the circle ABCD. Q. E. F. PROPOSITION 7.-PROBLEM. To describe a square about a given circie. (References, Prop. I. 28, 34; III. 17, 18.) Given. Let ABCD be the given circle. Sought. It is required to describe a square about ABCD. Construction.—1. Draw two diameters AC, BD, at right angles to one another. (III. 1, I. 11.) 2. And through the points A, B, C, D, draw TG, GH, EK, KF, touching the circle. (III. 17.) Construction.-1. Bisect AB, AC, in the points D, E. (I. 10.) 2. From the points D, E, draw DF, EF, at right angles to AB, AC. (I. 11.) 3. DF, EF, produced, will meet one another; for if they do not, they are parallel, and AB, AC, which are at right angles to them, are parallel; which is absurd: 4. Let them meet in F, and join FA; also if the point F be not in BC, join BF, CF. 5. And from the centre F, at the distance BF, CF, or AF, describe the circle ABC. Then the circle ABC shall be described about the triangle ABC. Proof.-1. Because AD is equal to BD (const.), and DF common, and at right angles to AB, 2. Therefore the base FA is equal to the base FB. (I. 4.) 3. In the same manner it may be shown that FC is equal to FA. 4. Therefore FB is equal to FC; (ax. 1.) 5. Therefore FA, FB, FC, are equal to one another, and the circle described from the centre F, at the distance of any one of them, will pass through the extremities of the other two. Conclusion. Therefore, the circle ABC is described about the triangle ABC. Q. E. F. Corollary.--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 in 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. (III. 31.) Wherefore, conversely, 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 hypotenuse; and if it be an obtuse-angled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle. PROPOSITION 6.—PROBLEM. (References, Prop. I. 4, 11; III. 1, 31.) Construction.-1. Draw two diameters, AC, BD, at right angles to one another. (III. 1, 1. 11.) 2. Join AB, BC, CD, DA. Then the figure ABCD shall be the square required. Proof.-1. Because BE is equal to DE, for E is the centre; 2. And that EA is common, and at right angles to BD; 3. Therefore the base BA is equal to the base DA. (I. 4.) 4. For the same reason, BC, DC, are each of them equal to BA, or DA; 5. Therefore the quadrilateral figure ABCD is equilateral. 6. It is also equiangular; 7. For the straight line BD being a diameter of the circle ABCD, BAD is a semicircle; (const.) 8. Therefore the angle BAD is a right angle. (III. 31.) 9. For the same reason, each of the angles ABC, BCD, CDA, is a right angle; 10. Therefore the quadrilateral figure ABCD is rectangular; and it has been shown to be equilateral; therefore it is a square (I. def. 30); and it touches the circumference of the circle at each of its angles. Conclusion.--Therefore, the square ABCD is inscribed in the circle ABCD. Q. E. F. PROPOSITION 7.-PROBLEM. To describe a square about a given circle. (References—Prop. I. 28, 34; III. 17, 18.) Given.—Let ABCD he the given circle. Sought. It is required to describe a square about ABCD. Construction.-1. Draw two diameters AC, BD, at right angles to one another. (III. 1, 1. 11.) 2. And through the points A, B, C, D, draw TG, GII, HK, KF, touching the circle. (III. 17.) The figure GUKF shall be the square required. Proof.-1. Because FG touches the circle ABCD, and EA is drawn from the centre E to the point of contact A; (const.) 2. Therefore the angles at A are right angles. (III. 18.) 3. For the same reason, the angles at the points B, C, D, are right angles. 4. And because the angle AEB is a G A right angle, as likewise is EBG; 5. Therefore GH is parallel to AC. (I. 28.) 6. For the same reason, AC is parallel to FK. 7. In the same manner it may be shown that each of the lines GF, HK, is parallel T C K to BD; 8. Therefore the figures GK, GC, CF, FB, BK, are parallelograms; 9. Therefore GF is equal to HK, and GH to FK. (I. 34.) 10. And because AC is equal to BD, and that AC is equal to each of the two, GH, FK; and that BD is equal to each of the two, GF, HK; 11. Therefore GH, FK, are each of them equal to GF, or HK; * 12. Therefore the quadrilateral figure FGHK is equilateral. 13. It is also rectangular; 14. For AEBG being a parallelogram, and AEB a right angle, therefore AGB is also a right angle. (I. 34.) 15. In the same manner it may be shown that the angles at H, K, F, are right angles; 16. Therefore the quadrilateral figure FGHK is rectangular; and it has been shown to be equilateral; therefore it is a square (I. def. 30); and all its sides touch the circumference of the circle. Conclusion. Therefore, the square FGHK is described about the circle ABCD. 0. E. F.' PROPOSITION 8.—PROBLEM. To inscribe a circle in a given square. Construction.-1. Bisect each of the sides AB, AD, in the points F, E. (1. 10.) |