Let the two straight lines, AC, BD, cut one another in the point E, within the circle ABCD. Then the rectangle contained by AE, EC shall be equal to the rectangle contained by BE, ED. First. Let AC, BD, pass each of them through the centre, so that E is the centre. Demonstration. Then, since AE, EC, BE, ED, are all equal (I. Def. 15); therefore 1. The rectangle AE, EC is equal to the rectangle BE, ED. Secondly. 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. D E B Construction. Then, if BD be bisected in F, F is the centre of the circle ABCD. Join AF. Demonstration. Because BD, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles in E, therefore 1. AE is equal to EC (III. 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 2. 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 (I. 47); therefore 3. The rectangle BE, ED, together with the square on EF, is equal to the squares on AE, EF (I. Ax. Î); take away the common square on EF, and 4. The remaining rectangle BE, ED is equal to the remaining square on AE (I. Ax. 3); that is, to the rectangle AE, EC. Thirdly. 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. Construction. Then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and from F draw FG perpendicular to AC (I. 12). 2. The rectangle AE, EC, together with the square on EG, is equal to the square on AG (II. 5) ; to each of these equals add the square on GF; therefore 3. The rectangle AE, EC, together with the squares on EG, GF, is equal to the squares on AG, GF (I. Ax. 2) ; but the squares on EG, GF, are equal to the square on EF (I. 47) ; and the squares on AG, GF are equal to the square on AF; therefore 4. 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 therefore 5. The square on FB is equal to the rectangle BE, ED, together with the square on EF (II. 5); 6. 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 7. 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 centre. Construction. Take the centre F (III. 1), and through E, the intersection of the straight lines AC, DB, draw the diameter GEFH. Demonstration. 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 1. The rectangle AE, EC is equal to the rectangle BE, ED (I. Ax. 1). Wherefore, if two straight lines, &c. Q.E.D. PROPOSITION 36.-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. First. Let it pass through the centre E. Construction. Join EB. Demonstration. Therefore 1. The angle EBD is a right angle (III. 18). D C E A And because the straight line AC is bisected in E, and produced to the point D, therefore 2. The rectangle AD, DC, together with the square on EC, is equal to the square on ED (II. 6); but CE is equal to EB; therefore 3. 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 (I. 47), because EBD is a right angle; therefore 4. 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 5. The remaining rectangle AD, DC is equal to the square on the tangent DB (Ax. 3). Secondly. Let DCA not pass through the centre of the circle ABC. Construction. Take E the centre of the circle (III. 1); draw EF perpendicular to AC (I. 12), and join EB, EC, ED. Demonstration. Because the straight line EF, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles; it also bisects AC (III. 3); therefore 1. AF is equal to FC; and because the straight line AC is bisected in F, and produced to D, 2. The rectangle AD, DC, together with the square on FC, is equal to the square on FD (II. 6) ; to each of these equals add the square on FE; therefore 3. The rectangle AD, DC, together with the squares on CF, FE, is equal to the squares on DF, FE (I. Ax. 2) ; but the square on ED is equal to the squares on DF, FE (I. 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 4. 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 5. 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 (I. 47), because EBD is a right angle ; therefore 6. The rectangle AD, DC, together with the square on EB, is equal to the squares on EB, BD; take away the common square on EB, and 7. The remaining rectangle AD, DC is equal to the square on DB (I. Ax. 3). Wherefore, if from any point, &c. Q.E.D. |
