PROPOSITION I. THEOREM. The line, which bisects a chord of a circle at right angles, must contain the centre. Let the st. line CE bisect the chord AB at rt. angles in D. Then the centre of the must lie in CE. For if not, let O, a pt. out of CE, be the centre; and join OA, OD, OB. Then, in as ODA, ODB, · AD=BD, and DO is common, and OA = OB ; and .. ▲ ODB is a right ▲ . But 4 CDB is a right, by construction; .. LODB = L CDB, which is impossible; .. O is not the centre. I. c. I. Def. 9 Thus it may be shewn that no point, out of CE, can be the centre, and.. the centre must lie in CE. COR. If the chord CE be bisected in F, then F is the centre of the circle. PROPOSITION II. THEOREM. If any two points be taken in the circumference of a circle, the straight line, which joins them, must fall within the circle. Let A and B be any two pts. in the Oce of the O ABC. Then must the st. line AB fall within the ©. Take any pt. D in the line AB. Find O the centre of the O. Join OA, OD, OB. III. 1, Cor. I. A. I. 16. I. 19. ..the distance of D from O is less than the radius of the O, and .. D lies within the O. And the same may be shewn of any other pt. in AB. .. AB lies entirely within the O. Post. Q. E. D. PROPOSITION III. THEOREM. If a straight line, drawn through the centre of a circle, bisect a chord of the circle, which does not pass through the centre, it must cut it at right angles: and conversely, if it cut it at right angles, it must bisect it. E In the ABC, let the chord AB, which does not pass through the centre O, be bisected in E by the diameter CD. Then must CD be 1 to AB. Join OA, OB. Then in As AEO, BEO, · AE=BE, and EO is common, and OA=OB, Next let CD be 1 to AB. Then must CD bisect AB. For OA=0B, and OE is common, in the right-angled As AEO, BEO, .. AE=BE, that is, CD bisects AB. I. E. Cor. p. 43. Q. E. D. Ex. 1. Shew that, if CD does not cut AB at right angles, it cannot bisect it. Ex. 2. A line, which bisects two parallel chords in a circle, is also perpendicular to them. Ex. 3. Through a given point within a circle, which is not the centre, draw a chord which shall be bisected in that point. PROPOSITION IV. THEOREM. If in a circle two chords, which do not both pass through the centre, cut one another, they do not bisect each other. E Let the chords AB, CD, which do not both pass through the centre, cut one another, in the pt. E, in the O ACBD. Then AB, CD do not bisect each other. If one of them pass through the centre, it is plainly not bisected by the other, which does not pass through the centre. But if neither pass through the centre, let, if it be possible, AE=EB and CE=ED; find the centre O, and join OE. Then OE, passing through the centre, bisects AB, And OE, passing through the centre, bisects CD, LOEA: = ▲ OEC, which is impossible; .. AB, CD do not bisect each other. III. 3. III. 3. Q. E. D. Ex. 1. Shew that the locus of the points of bisection of all parallel chords of a circle is a straight line. Ex. 2. Shew that no parallelogram, except those which are rectangular, can be inscribed in a circle. PROPOSITION V. THEOREM. If two circles cut one another, they cannot have the same centre. If it be possible, let O be the common centre of the s ABC, ADC, which cut one another in the pts. A and C. Join OA, and draw OEF meeting the Os in E and F. Ex. Two circles, whose centres are A and B, intersect in C; through C two chords DCE, FCG are drawn equally inclined to AB and terminated by the circles: prove that DE and FG are equal. NOTE. Circles which have the same centre are called Concentric. |