and it has been shown to be equilateral; therefore (I. Def. 30.) 7. The quadrilateral figure ABCD is a square; and it is inscribed in the circle ABCD. Q.E.F. PROP. VII.—PROBLEM. 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 (III. 17.) FG, GH, HK, KF, touching the circle. G F D therefore H с K Because FG touches the circle ABCD, and EA is drawn from the centre E to the point of contact 4, (III. 18.) 1. The angles at A are right angles; for the same reason, 2. The angles at the points B, C, D, are right angles; and because the angle AEB is a right angle, as likewise is EBG, (I. 28.) GH is parallel to AC; 3. for the same reason, 4. AC is parallel to FK; and in like manner it may be demonstrated that 5. therefore GF, HK, are each of them parallel to BED; 6. The figures GK, GC, AK, FB, BK, are parallelograms; and therefore (I. 34.) 7. GF is equal to HK, and GH to FK; and because AC is equal to BD, and that AC is equal to each of the two GH, FK; and BD to each of the two GF, HK; 8. GH, FK, are each of them equal to GF, or HK; 9. The quadrilateral figure FGHK is equilateral.. It is also rectangular; for GBEA being a parallelogram, and AEB a right angle, likewise (I. 34.) 10. AGB is a right angle: Let AB be the given straight line, and the angle C the given rectilineal angle; it is required to describe upon the given straight line AB a segment of a circle, containing an angle equal to the angle C. First, let the angle C be a right angle. Η A F B Bisect (I. 10.) AB in F, and from the centre F, at the distance FB, describe the semicircle AHB; therefore (III. 31.) The angle AHB in the semicircle is equal to the right angle at C. But if the angle C be not a right angle: H H At the point in the straight line AB, make (I. 23.) the angle BAD equal to the angle C, and from the point A draw (I. 11.) AE at right angles to AD; bisect (I. 10.) AB in F, and from F draw (I. 11.) FG at right angles to AB, and join GB. And because AF is equal to FB, and FG common to the triangles AFG, BFG, the two sides AF, FG, are equal to the two BF, FG, each to each; and the angle AFG is equal (I. Def. 10.) to the angle BFG; therefore (I. 4.) 1. The base AG is equal to the base GB; and the circle described from the centre G, at the distance GA, shall pass through the point B; let this be the circle AHB: and because from the point A the extremity of the diameter AE, AD is drawn at right angles to AE, therefore (III. 16. Cor.) 2. AD touches the circle AHB; and because AB drawn from the point of contact A cuts the circle, (III. 32.) 3. The angle DAB is equal to the angle in the alternate segment AHB: but the angle DAB is equal (Constr.) to the angle C; therefore also 4. The angle C is equal to the angle in the segment AHB. Wherefore, upon the given straight line AB, the segment AHB of a circle is described which contains an angle equal to the given angle C. Which was to be done. PROP. XXXIV.-PROBLEM. To cut off a segment from a given circle which shall contain an angle equal to a given rectilineal angle. Let ABC be the given circle, and D the given rectilineal angle; it is required to cut off a segment from the circle ABC that shall contain an angle equal to the angle D. Draw (III. 17.) the straight line EF touching the circle ABC in the point B; and at the point B in the straight line BF make (I. 23.) the angle FBC equal to the angle D: the segment BAC shall contain an angle equal to the given angle D. Because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, (III. 32.) 1. The angle FBC is equal to the angle in the alternate segment BAC of the circle: but the angle FBC is equal (Constr.) to the angle D; therefore (Ax. 1.) 2. The angle in the segment BAC is equal to the angle D. Wherefore the segment BAC is cut off from the given circle ABC containing an angle equal to the given angle D. Which was to be done. PROP. XXXV. THEOREM. If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other. B Let the two straight lines AC, BD, within the circle ABCD, cut one another in the point E: the rectangle contained by AE, EC, is equal to the rectangle contained by BE, ED. A E If AC, BD, pass each of them through the centre, so that E is the centre; it is evident that AE, EC, BE, ED, being (I. Def. 15.) all equal, likewise 1. The rectangle AE, EC, is equal to the rectangle BE, ED. But let one of them BD pass through the centre, and cut the other AC, which does not pass through the centre, at right angles, in the point E. F E Then if BD be bisected in F, Fis the centre of the circle ABCD. Join AF. And because BD, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles in E, (III. 3.) 1. AE, EC, are equal to one another: and because the straight line BD is cut into two equal parts in the point F, and into two unequal parts in the point E, (II. 5.) B 2. The rectangle BE, ED, together with the square of EF, is equal to the square of FB; that is, to the square of FA: but the squares of AE, EF, are equal (I. 47.) to the square of FA; therefore (Ax. 1.) 3. The rectangle BE, ED, together with the square of EF, is equal to the squares of AE, EF: take away the common square of EF, and (Ax. 3.) D 4. The remaining rectangle BE, ED, is equal to the remaining square of AE; that is, to the rectangle AE, EC. Next, let BD, which passes through the centre, cut the other AC, which does not pass through the centre, in E, but not at right angles. 1. AG is equal to GC; P E C B Then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and from F draw (I. 12.) FG perpendicular to AC; therefore (III. 3.) wherefore (II. 5.) 2. The rectangle AE, EC, together with the square of EG, is equal to the square of AG: to each of these equals add the square of GF; therefore (Ax. 2.) but the squares of EG, GF, are equal (I. 47.) to the square of EF; and the squares of AG, GF, are equal to the square of AF; therefore 4. The rectangle AE, EC, together with the square of EF, is equal to the square of AF; that is, to the square of FB: but (II. 5.) 5. The square of FB is equal to the rectangle BE, ED, together with the square of EF; therefore (Ax. 1.) 6. The rectangle AE, EC, together with the square of EF, is equal to the rectangle BE, ED, together with the square of EF: away the common square of EF, and therefore (Ax. 3.) take 7. The remaining rectangle AE, EC, is equal to the remaining rectangle BE, ED. Lastly, let neither of the straight lines AC, BD, pass through the centre. H T ♡ Take (III. 1.) the centre F, and through E, the intersection of the straight lines AC, DB, draw the diameter GEFH. And because the rectangle AE, EC, is equal, as has been shown, to the rectangle GE, EH; and for the same reason, the rectangle BE, ED, is equal to the same rectangle GE, EH; therefore (Ax. 1.) The rectangle AE, EC, is equal to the rectangle BE, ED. Wherefore, if two straight lines, &c. Q.E.D. PROP. XXXVI.-THEOREM. If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it. Let D be any point without the circle ABC, and let DCA, DB, be two straight lines drawn from it, of which DCA cuts the circle, and DB touches the same: the rectangle AD, DC, is equal to the square of DB. |