Proposition xxxvi. may be demonstrated by help of the preceding, in the following manner : B Let B be any point without the circle PEF, from which let BPF a line cutting, and BE a line touching the circle, be drawn. From the centre C draw CB, CP, CE; and with this same centre and CB as distance, describe a circle and produce BF to meet its circumference in A. Then (last prop.) the rectangle AP PB is equal to the difference of the squares of BC, PC; that is, to the difference of the squares of BC, CE; that is, to the square of BE.* But since a perpendicular CM from #18. 3. and 47. 1. +3 Ax. +2 Ax. C to AB will bisect both AB and FP it followst that AF is equal to BP, and therefore AP to BF,t consequently the rectangle BF BP is equal to the square of BE. Hence, if from any point, &c. PROP. XXXVI. THEOR. 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 purt 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 shall be equal to the square of DB. Either DCA passes through the centre, or it does not: first, let it pass through the centre E, and join EB; 18. 3. *6. 3. *47. 1. +1 Ax. 13 Ax. therefore the angle EBD is a right* B squares of EB, BD, because EBD is a right angle: therefore the rectangle AD DC, together with the square of EB, is equal to the squares of EB, BD: take away the common square of EB; therefore the remaining rectangle_AD·DC is equal to the square of the tangent, DB. But if DCA does not pass through the centre of the circle ABC, take* the centre E, and draw EF *12. 1. perpendicular to AC, and join #1. 3. #3. 3. *6. 2. +2 Ax. #47. 1. tl Ax. B F E EB, EC, ED. And because the square of ED: but the squares of EB, BD are #47. 1. equal to the square of ED, because EBD is a right angle; therefore the rectangle AD DC, together with the square of EB, is equal to the squares of EB, BD: take away the common square of EB; therefore the remaining rectangle AD-DC is equal to the square of DB. Wherefore, if from any point, &c. Q. E. D. †3 Ax. 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 of the straight line AD, which touches the circle. D As two lines touching the circle may be drawn from any point without it, it follows that they are equal to each other. PROP. XXXVII. THEOR. If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it; and 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 of 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, and DB meets it; if the rectangle AD. DC be equal to the square of DB; DB shall touch the circle. Draw the straight line DE, touching the circle *17. 3. ABC, findt its centre F, and join FE, FB, FD: 18. 3. then FED is a right angle: and because DE +1. 3. #36. 3. +1 Ax. touches the circle ABC, and DCA cuts it, the rectangle AD. DC is equal to the square of DE: but the rectangle AD. DC is by hypothesis equal to the square of DB; therefore the square of DE is equal to the square of DB; and the straight line DE equal to the straight line DB: and FE is +15 Def. 1. equal to FB; 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: but DEF was shown to be a right angle: therefore also DBF ist a right angle: and BF, if produced, is a diameter; and the straight line which is drawn +8. 1. +1 Ax. BA F at right angles to a diameter, from the extremity of #16. 3. it, touches the circle: therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E. D. THE ELEMENTS OF EUCLID. wwwww BOOK IV. www DEFINITIONS. I. A RECTILINEAL figure is said to be inscribed in another rectilineal figure, when all the angular points of the inscribed figure are upon the sides of the figure in which it is inscribed, each upon each. II. In like manner a rectilineal figure is said to be described about another when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each. III. A rectilineal figure is said to be inscribed in a circle, when all the angular points of the inscribed figure are upon the circumference of the circle. |