Then, if BD be bisected in F, Join AF. Because BD which passes through the center, cuts the straight line AC, which does not pass through the center, at right angles in E, therefore AE is equal to EC: (I. 3.) 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, therefore the rectangle BE, ED, together with the square on EF, is equal to the square on FB; (II. 5.) that is, to the square on FA: but the squares on AE, EF, are equal to the square on FA: (1. 47.) therefore the rectangle BE, ED, together with the square on EF, is equal to the squares on AE, EF: (1. ax. 1.) take away the common square on EF, and the remaining rectangle BE, ED is equal to the remaining square on AE; (1. ax. 3.) that is, to the rectangle AE, EC. Thirdly, let BD, which passes through the center, cut the other AC, which does not pass through the center, in E, but not at right angles. Then, as before, if BD be bisected in F, Fis the center of the circle. Join AF, and from F draw FG perpendicular to AC; therefore AG is equal to GC; (III. 3.) (1.12.) wherefore the rectangle AE, EC, together with the square on EG, is equal to the square on AG: (11. 5.) to each of these equals add the square on GF; therefore the rectangle AE, EC, together with the squares on EG, GF, is equal to the squares on AG, GF; (1. ax. 2.) but the squares on EG, GF, are equal to the square on EF; (1. 47.) and the squares on AG, GF are equal to the square on AF: therefore the rectangle AE, EC, together with the square on EF, is equal to the square on AF; that is, to the square on FB: but the square on FB is equal to the rectangle BE, ED, together with the square on EF; (11. 5.) therefore the rectangle AE, EC, together with the square on EF, is equal to the rectangle BE, ED, together with the square on EF; (I. ax. 1.) take away the common square on EF, and the remaining rectangle AE, EC, is therefore equal to the remaining rectangle BE, ED. (ax. 3.) Lastly, let neither of the straight lines AC, BD pass through the center. H E C B Take the center F, (III. 1.) and through E the intersection of the straight lines AC, DB, And because the rectangle AE, EC is equal, as has been shewn, to the rectangle GE, EH; and for the same reason, the rectangle BE, ED is equal to the same rectangle GE, EH; therefore the rectangle AE, EC is equal to the rectangle BE, ED. (1. ax. 1.) Wherefore, if two straight lines, &c. Q. E.D. PROPOSITION 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 on 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. Then the rectangle AD, DC shall be equal to the square on DB. Either DCA passes through the center, or it does not: first, let it pass through the center E. Join EB, therefore the angle EBD is a right angle. (III. 18.) And because the straight line AC is bisected in E, and produced to the point D, therefore the rectangle AD, DC, together with the square on EC, is equal to the square on ED: (11. 6.) but CE is equal to EB; therefore the rectangle AD, DC, together with the square on EB, is equal to the square on ED: but the square on ED is equal to the squares on EB, BD, (1. 47.) because EBD is a right angle: therefore the rectangle AD, DC, together with the square on EB, is equal to the squares on EB, BD: (ax. 1.) take away the common square on EB; therefore the remaining rectangle AD, DC is equal to the square on the tangent DB. (ax. 3.) Next, if DCA does not pass through the center of the circle ABC. D B F Take E the center of the circle, (III. 1.) draw EF perpendicular to AC, (1. 12.) and join EB, EC, ED. Because the straight line EF, which passes through the center, cuts the straight line AC, which does not pass through the center, at right angles; it also bisects AC, (III. 3.) therefore AF is equal to FC; and because the straight line AC is bisected in F, and produced to D, the rectangle AD, DC, together with the square on FC, is equal to the square on FD: (11. 6.) to each of these equals add the square on FE; therefore the rectangle AD, DC, together with the squares on CF, FE, is equal to the squares on DF, FE: (1. ax. 2.) but the square on ED is equal to the squares on DF, FE, (1. 47.) because EFD is a right angle; and for the same reason, the square on EC is equal to the squares on CF, FE; therefore the rectangle AD, DC, together with the square on EC, is equal to the square on ED: (ax. 1.) but CE is equal to EB; therefore the rectangle AD, DC, together with the square on EB, is equal to the square on ED: but the squares on EB, BD, are equal to the square on ED, (1. 47.) because EBD is a right angle: therefore the rectangle A1), DC, together with the square on EB, is equal to the squares on EB, BD; take away the common square on EB; and the remaining rectangle AD, DC is equal to the square on DB. (1. ax. 3.) Wherefore, if from any point, &c. Q. E. D. COR. If from any point without a circle, there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of them without the circle, are equal to one another, viz. the rectangle BA, AE, to the rectangle CA, AF: for each of them is equal to the square on the straight line AD, which touches the circle. If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square on the line which meets it, the line which meets, shall touch the circle. Let any point D be taken without the circle ABC, and from it let two straight lines DCA and DB be drawn, of which DCA cuts the circle in the points C, A, and DB meets it in the point B. If the rectangle AD, DC be equal to the square on DB; Draw the straight line DE, touching the circle ABC, in the point E; (u. 17.) find F, the center of the circle, (1. 1.) and join FE, FB, FD. Then FED is a right angle: (III. 18.) and because DE touches the circle ABC, and DCA cuts it, therefore the rectangle AD, DC is equal to the square on DE: (111. 36.) but the rectangle AD, DC, is, by hypothesis, equal to the square on DB: therefore the square on DE is equal to the square on DB; (1. ax. 1.) and the straight line DE equal to the straight line DB: and FE is equal to FB; (1. def. 15.) wherefore DE, EF are equal to DB, BF, each to each; and the base FD is common to the two triangles DEF, DBF; therefore the angle DEF is equal to the angle DBF: (1. 8.) but DEF was shewn to be a right angle; therefore also DBF is a right angle: (1. ax. 1.) and BF, if produced, is a diameter; and the straight line which is drawn at right angles to a diameter, from the extremity of it, touches the circle; (III. 16. Cor.) therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E. D IN the Third Book of the Elements are demonstrated the most elementary properties of the circle, assuming all the properties of figures demonstrated in the First and Second Books. It may be worthy of remark, that the word circle will be found sometimes taken to mean the surface included within the circumference, and sometimes the circumference itself. Euclid has employed the word (Tεpipepsia) periphery, both for the whole, and for a part of the circumference of a circle. If the word circumference were restricted to mean the whole circumference, and the word are to mean a part of it, ambiguity might be avoided when speaking of the circumference of a circle, where only a part of it is the subject under consideration. A circle is said to be given in position, when the position of its center is known, and in magnitude, when its radius is known. Def. I. And it may be added, or of which the circumferences are equal. And conversely: if two circles be equal, their diameters and radii are equal; as also their circumferences. Def. 1. states the criterion of equal circles. Simson calls it a theorem ; and Euclid seems to have considered it as one of those theorems, or axioms, which might be admitted as a basis for reasoning on the equality of circles. Def. II. There seems to be tacitly assumed in this definition, that a straight line, when it meets a circle and does not touch it, must necessarily, when produced, cut the circle. A straight line which touches a circle, is called a tangent to the circle; and a straight line which cuts a circle is called a secant. Def. IV. The distance of a straight line from the center of a circle is the distance of a point from a straight line, which has been already explained in note to Prop. XI. page 53. Def. vI. x. An arc of a circle is any portion of the circumference; and a chord is the straight line joining the extremities of an arc. Every chord except a diameter divides a circle into two unequal segments, one greater than, and the other less than a semicircle. And in the same manner, two radii drawn from the center to the circumference, divide the circle into two unequal sectors which become equal when the two radii are in the same straight line. As Euclid, however, does not notice re-entering angles, a sector of the circle seems necessarily restricted to the figure which is less than a semicircle. A quadrant is a sector whose radii are perpendicular to one another, and which contains a fourth part of the circle. Def. VII. No use is made of this definition in the Elements. Def. XI. The definition of similar segments of circles as employed in the Third Book is restricted to such segments as are also equal. Props. XXIII, and xxiv. are the only two instances, in which reference is made to similar segments of circles. Prop. I. "Lines drawn in a circle," always mean in Euclid, such lines only as are terminated at their extremities by the circumference. If the point G be in the diameter CE, but not coinciding with the point F, the demonstration given in the text does not hold good. At the same time, it is obvious that G cannot be the centre of the circle, because GC is not equal to GE. |