equal to it and the angle at A is half the angle BGC, and the angle at D half of the angle EHF; therefore the angle at A is equal to the angle at D. In equal circles, equal straight lines cut off equal arcs, the greater equal to the greater, and the less to the less. Let ABC, DEF be equal circles, and BC, EF equal straight lines in them, which cut off the two greater arcs BAC, EDF, and the two less BGC, EHF: the greater BAC is equal to the greater EDF, and the less BGC to the less EHF. Take K, L, the centres of the circles, and join BK, KC, EL, LF; and because the circles are equal, the straight lines from their centres are equal; therefore BK, KC are equal to EL, LF; but the base BC is also equal to the base EF; therefore the angle BKC is equal (5. 1.) to the angle ELF: and equal angles stand upon equal (25. 3.) arcs, when they are at the centres; therefore the arc BGC is equal to the arc EHF. But the whole circle ABC is equal to the whole EDF; the remaining part, therefore, of the circumference, viz. BAC, is equal to the remaining part EDF. PROP. XXVIII. THEOR. In equal circles, equal arcs are subtended by equal straight lines. Let ABC, DEF be equal circles, and let the arcs BGC, EHF also be equal; and join BC, EF: the straight line BC is equal to the straight line EF. Take K, L, the centres of the circles, and join BK, KC, EL, LF: and because the arc BGC is equal to the arc EHF, the angle BKC is equal (26. 3.) to the angle ELF: also because the circles ABC, DEF are equal, their radii are equal: therefore BK, KC are equal to EL, LF: and they contain equal angles; therefore the base BC is equal (1. 1.) to the base EF. In a circle, the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle. Let ABCD be a circle, of which the diameter is BC, and centre E; draw CA dividing the circle into the segments ABC, ADC, and join BA, AD, DC; the angle in the semicircle BAC is a right angle; and the angle in the segment ABC, which is greater than a semicircle, is less than a right angle; and the angle in the segment ADC, which is less than a semicircle, is greater than a right angle. Join AE, and produce BA to F ; and because BE is equal to EA, the angle EAB is equal (3. 1.) to EBA: also, because AE is equal to EC, the angle EAC is equal to ECA; wherefore the whole angle BAC is equal to the two angles ABC, ACB. But FAC, the exterior angle of the triangle ABC, is also equal (25. 1.) to the two angles ABC, ACB; therefore the angle BAC is equal to the angle FAC, and each of them is therefore a right (7. Def. 1.) angle; wherefore the angle BAC in a semicircle is a right angle. B F A D E с And because the two angles ABC, BAC of the triangle ABC, are together less (10. 1.) than two right angles, and BAC is a right angle, ABC must be less than a right angle; and therefore the angle in a segment ABC, greater than a semicircle, is less than a right angle. Also because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are equal (22. 3.) to two right angles; therefore the angles ABC, ADC are equal to two right angles; and ABC is less than a right angle; wherefore the other ADC is greater than a right angle. COR. From this it is manifest, that if one angle of a triangle be equal to the other two, it is a right angle, because the angle adjacent to it is equal to the same two; and when the adjacent angles are equal, they are right angles. If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line which touches the circle, shall be equal to the angles in the alternate segments of the circle. Let the straight line EF touch the circle ABCD in B, and from the point B let the straight line BD be drawn cutting the circle: The angles which BD makes with the touching line EF shall be equal to the angles in the alternate segments of the circle: that is, the angle FBD is equal to the angle which is in the segment DAB, and the angle DBE to the angle in the segment BCD. A D From the point B draw (Prob. 6. 1.) BA at right angles to EF, and take any point C in the arc BD, and join AD, DC, CB; and because the straight line EF touches the circle ABCD in the point B, and BA is drawn at right angles to the touching line, from the point of contact B. the centre of the circle is (19. 3.) in BA; therefore the angle ADB in a semicircle, is a right (29. 3.) angle, and consequently the other two angles, BAD, ABD, are equal (25. 1.) to a right angle; but ABF is likewise a right angle; therefore the angle ABF is equal to the angles BAD, ABD: take from these equals the common angle ABD, and there will remain the angle DBF equal to the angle BAD, which is in the alternate segment of the circle. And because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are equal (22. 3.) to two right angles; therefore the angles DBF, DBE, being likewise equal (6. 1.) to two right angles, are equal to the angles BAD, BCD; and DBF has been proved equal to BAD; therefore the remaining angle DBE is equal to the angle BCD in the alternate segment of the circle. E PROP. XXXI. THEOR. B C F 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. 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. 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 all equal, the rectangle AE.EC is likewise equal to the rectangle BE.ED. But let one of them BD pass through the B A E D D centre, and cut the other AC, which does not pass through the centre, at right angles in the point E; then, if BD be bisected in F, F is the which passes centre of the circle ABCD; join AF: and because BD, through the centre, cuts the straight line AC, which does not pass through the centre at right angles, in E, AE, EC are equal (3. 3.) to one another; and because the straight line BD is cut into two equal parts in the point F, and into two unequal in the point E, BE.ED (5. 2.) +EF2=FB2=AF2. But AF2 AE2 = +(37. 1.) EF2, therefore BE.ED + EF2 AE2+ EF2, and taking EF2 from each, BE. ED=AE2-AE.EC. F E B Next, Let BD, which passes through the centre, cut the other AC, D which does not pass through 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 (Prob. 7. 1.) FG perpendicular to AC; therefore AG is equal (3. 3.) to GC; wherefore AE.EC+(5. 2.) ÉG2 =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.EDGE.EH; therefore AE.EC-BE.ED. F E A G B PROP. XXXII. 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. 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 (18. 3.) angle: 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 (37.1.) EB2+BD2, because EBD is a right angle; therefore AD. DC+EB2=EB2+BD2, and taking EB2 from each, AD.DC=BD3. But, if DCA does not pass through the centre of the circle ABC, take the centre E, and draw EF perpendicular (Prob. 7. 1.) to AC, and join EB, EC, ED: and 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 likewise bisects (3. 3.) it; 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 (37. 1.) EC2 = FC2 + FE2, and ED2 = FD2+FE2, because DFE is a right angle; therefore AD.DC+ EC2 ED. 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. |