D the centre, in E, but not at right angles; then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and from F draw (12. 1.) FG perpendicular to AC; therefore AG is equal (3. 3.) to GC; wherefore AE.EC+ (5. 2.) EG2 = AG2, and adding GF2 to both, AE.EC+EG2+GF2-AG2+GF2. Now EG2+GF2=EF2, and AG2+GF2=AF2; therefore AE.EC+ EF2=AF2=FB2. But FB2 =BE.ED+(5. 2.) EF2, therefore AE.EC+ EF2 =BE.ED+EF2, and taking EF2 from both, AE. EC-BE.ED. Lastly, let neither of the straight lines AC, BD pass through the centre: take the centre F, and through E, the intersection of the straight lines AC, DB, draw the diameter GEFH: and because, as has been shown, AE.EC=GE.EH, and BE.ED=GE.EH; therefore AE.EC-BE. ED. H G F E B T E A C B G 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 part of it without the circle, is equal to the square of the line which touches it. D Let D be any point without the circle ABC, and DCA, DB two straight lines drawn from it, of which DCA cuts the circle, and DB touches it: the rectangle AD.DC is 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; therefore the angle EBD is a right angle (18. 3.): and because the straight line AC is bisected in E, and produced to the point D, AD.DC+EC2=ED2 (6. 2.). But EC EB, therefore AD.DC + EB2 = ED2. Now ED2 (47. 1.) EB2+ BD2, because EBD = is a right angle; therefore AD.DC + EB2 = EB2 +BD2, and taking EB2 from each, AD.DC =BD2. But, if DCA does not pass through the centre of the circle ABC, take (1. 3.) the centre E, and draw EF perpendicular (12. 1.) to AC, and join EB, EC, ED; and because the straight line EF, which passes through the centre, cuts B E A the straight line AC, which does not pass through the centre, at right angles, it likewise bisects it (3. 3.); therefore AF is equal to FC; and because the straight line AC is bisected in F, and produced to D (6. 2.), AD.DC+FC2= - FD2; add FE2 to both, then AD.DC+FC2+ FE2=FD2+FE2. But (47. 1.) EC2=FC2+ FE2, and ED2-FD2+FE2, because DFE is a right angle; therefore AD.DC+EC2-ED2. Now, because EBD is a right angle, ED2= EB2+BD2 EC2+BD2, and therefore, AD. DC+EC2=EC2+ BD2, and AD.DC=BD2. COR. 1. 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. BA.AE-CA. AF; for each of these rectangles is equal to the square of the straight line AD, which touches the circle. COR. 2. It follows, moreover, that two tangents drawn from the same point are equal. COR. 3. And since a radius drawn to the point of contact is perpendicular to the tangent, it follows that the angle included by two tangents, drawn from the same point, is bisected by a line drawn from the centre of the circle to that point; for this line forms the hypotenuse common to two equal right angled triangles. B D F E A A D 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; 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 touches the circle. Draw (17. 3.) the straight line DE touching the circle ABC; find the centre F, and join FE, FB, FD; then FED is a right angle (18. 3.): and because DE touches the circle ABC, and DCA cuts it, the rectangle AD. DC is equal (36. 3.) to the square of DE; but the rectangle AD.DC is. by hypothesis, equal to the square of DB: therefore the square of DE 18 equal to the square of DB; and the straight line DE equal to the straight line DB: but FE is equal to FB, wherefore DE.EF are equal to DB, BF; and the base FD is common to the two triangles DEF, DBF; therefore the angle DEF is equal (8. 1.) to the angle DBF; and DEF is a right angle, therefore also DBF is a right angle: but FB, if produced, is a diameter, and the straight line which is drawn at right angles to a diameter, from the extremity of it, touches (16. 3.) the circle therefore DB touches the circle ABC. B D F A ADDITIONAL PROPOSITIONS. PROP. A. THEOR. A diameter divides a circle and its circumference into two equal parts; and, conversely, the line which divides the circle into two equal parts is a diameter. Let AB be a diameter of the circle AEBD, then AEB, ADB are equal in surface and boundary. Now, if the figure AEB be applied to the figure ADB, their common base AB retaining its position, the curve line AEB must fall on the curve line ADB; otherwise there would, in the one or the other, be points unequally distant from the centre, which is contrary to the definition of a circle. A E F B D Conversely. The line dividing the circle into two equal parts is a diameter. For, let AB divide the circle into two equal parts; then, if the centre is not in AB, let AF be drawn through it, which is therefore a diameter, and consequently divides the circle into two equal parts; hence the portion AEF is equal to the portion AEFB, which is absurd. COR. The arc of a circle whose chord is a diameter, is a semicircumference, and the included segment is a semicircle. PROP. B. THEOR. Through three given points which are not in the same straight line, one circumference of a circle may be made to pass, and but one. Let A, B, C, be three points not in the same straight line: they shall all lie in the same circumference of a circle. For, let the distances AB, BC be bisected by the perpendiculars DF, EF, which must meet in some point F; for if they were parallel, the lines DB, CB, perpendicular to them would also be parallel (Cor. 2. 29. 1.), or else form but one straight line: but they meet in B, and ABC is not a straight line by hypothesis. Let then, FA, FB, and FC be drawn; then, because FA, FB meet AB at equal distances from the perpendicular, they are equal. For similar reasons FB, FC, are equal; hence the points A, B, C, are all equally distant from the point F, and consequently lie in the circumference of the circle, whose centre is F, and radius FA. It is obvious, that besides this, no other circumference can pass through the same points; for the centre, lying in the perpen D B C dicular DF bisecting the chord AB, and at the same time in the perpendicular EF bisecting the chord BC (Cor. 1. 3. 3.), must be at the intersection of these perpendiculars; so that, as there is but one centre, be but one circumference. PROP. C. THEOR. there can If two circles cut each other, the line which passes through their centres will be perpendicular to the chord which joins the points of intersection, and will divide it into two equal parts. Let CD be the line which passes through the centres of two circles cutting each other, it will be perpendicular to the chord AB, and will divide it into two equal parts. For the line AB, which joins the points of intersection, is a chord com mon to the two circles. And if a perpendicular be erected from the middle of this chord, it will pass (Cor. 1. 3. 3.) through each of the two centres C and D. But no more than one straight line can be drawn through two points; hence, the straight line which passes through the centres will bisect the chord at right angles. COR. Hence, the line joining the intersections of the circumferences of two circles, will be perpendicular to the line which joins their centres. SCHOLIUM. 1. If two circles cut each other, the distance between their centres will be less than the sum of their radii, and the greater radius will be also less than the sum of the smaller and the distance between the centres. For, CD is less (20. 1.) than CA+AD, and for the same reason, AD/AC+ CD. 2. And, conversely, if the distance between the centres of two circles be less than the sum of their radii, the greater radius being at the same time less than the sum of the smaller and the distance between the centres, the two circles will cut each other. For, to make an intersection possible, the triangle CAD must be possible. Hence, not only must we have CD<AC+AD, but also the greater radius AD<AC+CD; And whenever the triangle CAD can be constructed, it is plain that the circles described from the centres C and D, will cut each other in A and B. COR. 1. Hence, if the distance between the centres of two circles be greater than the sum of their radii, the two circles will not intersect each other. COR. 2. Hence, also, if the distance between the centres be less than the difference of the radii, the two circles will not cut each other. For, AC+CD>AD; therefore, CD>AD-AC; that is, any side of a triangle exceeds the difference between the other two. Hence, the triangle is impossible when the distance between the centres is less than the difference of the radii; and consequently the two circles cannot cut each other. PROP. D. THEOR. In the same circle, equal angles at the centre are subtended by equal arcs ; and, conversely, equal arcs subtend equal angles at the centre. Let C be the centre of a circle, and let the angle ACD be equal to the angle BCD; then the arcs AFD, DGB, subtending these angles, are equal. Join AD, DB; then the triangles ACD, BCD, having two sides and the included angle in the one, equal to two sides and the included angle in the other, are equal: so that, if ACD be applied to BCD, there shall be an entire coincidence, the point A coinciding with B, and D common to both arcs; the two extremities, therefore, of the arc AFD, thus coinciding with those of the arc BGD, all the intermediate parts must coincide, inasmuch as they are all equally distant from the centre. Conversely. Let the arc AFD be equal to the arc BGD; then the angle ACD is equal to the angle BCD. For, if the arc AFD be applied to the arc BGD, they would coincide; so that the extremities AD of the chord AD, would coincide with those of the chord BD; these chords are therefore equal: hence, the angle ACD is equal to the angle BCD (8. 1.). COR. 1. It follows, moreover, that equal angles at the centre are sub |