PROPOSITION XII. THEOREM. If two circles touch one another externally at any point, the straight line joining the centre of one with that point of contact must when produced pass through the centre of the other. Let ABC touch O ADE externally in the pt. A. Let O be the centre of O ABC. Join OA, and produce it to E. Then must the centre of ADE lie in AE. Join OP meeting the Os in B, D; and join AP. and PD=AP, :. OB and PD together =0A and AP together; :. OP is not less than OA and AP together. which is impossible; I. 20. ..the centre of O ADE cannot lie out of AE. Q. E. D. Ex. Three circles touch one another externally, whose centres are A, B, C. Shew that the difference between AB great as the difference between the and AC is half as diameters of the circles, whose centres are B and C. 8. E. II. 10 PROPOSITION XIII. THEOREM. One circle cannot touch another in more points than one, whether it touch it internally or externally. First let the ADE touch the © ABC internally in pt. A. Then there can be no other point of contact, Take any pt. E in the Oce of the O ADE, and join OE. Then from O, a pt. within or without the O ADE, two lines OA, OE are drawn to the Oce, of which OA' passes through the centre P, .. OA is greater than OE, III. 8, COR. and.. E is a point within the ABC, Similarly it may be shewn that every pt. of the Oce of ADE, except A, lies within the © ABC; .. A is the only point in which the Os meet. Next, let the Os ABC, ADE touch externally in the pt. A. Then there can be no other point of contact. во Take O the centre of the ABC. Then P, the centre of the ADE, lies in OA produced. III. 12. Take any pt. D in the Oce of the O ADE, and join OD. Then. from O, a pt. outside the O ADE, two lines OA, OD are drawn to the Oce, of which OA when produced passes through the centre P, .. OD is greater than OA; .D is a point without the ABC. III. 8. Similarly, it may be shewn that every pt. of the Oce of ADE, except A, lies without the © ABC; .. A is the only point in which the Os meet. Q. E. D. DEF. VIII. The DISTANCE of a chord from the centre is measured by the length of the perpendicular drawn from the centre to the chord. PROPOSITION XIV. THEOREM. Equal chords in a circle are equally distant from the centre; and conversely, those, which are equally distant from the centre, are equal to one another. B P Let the chords AB, CD in ○ ABDC be equal. Then must AB and CD be equally distant from the centre O. Draw OP and OQ 1 to AB and CD; and join AO, CO. Then P and Q are the middle pts. of AB and CD; 11. 3. and :: AB=CD, :. AP=CQ. Then · AP=CQ, and AO=CO, in the right-angled As AOP, COQ, and.. AB and CD are equally distant from O. Def. vш. Next, let AB and CD be equally distant from 0. Then must AB=CD. For ': OP=OQ, and AO=CO, in the right-angled as AOP, COQ, .. AP=CQ, and. AB= CD. 1. E. Cor. Q. E. D. Ex. In a circle, whose diameter is 10 inches, a chord is drawn, which is 8 inches long. If another chord be drawn, at a distance of 3 inches from the centre, shew whether it is equal or not to the former. PROPOSITION XV. THEOREM. The diameter is the greatest chord in a circle, and of all others that which is nearer to the centre is always greater than one more remote; and the greater is nearer to the centre than the less. Let AB be a diameter of the C ABDC, whose centre is O, and let CD be any other chord, not a diameter, in the O, nearer to the centre than the chord EF. Then must AB be greater than CD, and CD greater than EF. Draw OP, OQ 1 to CD and EF; and join OC, OD, OE. Then AO=CO, and OB=OD, .. AB=sum of CO and OD, and.. AB is greater than CD. I. Def. 13. I. 20. Def. VIII. Again, : CD is nearer to the centre than EF, ... sum of sqq. on OP, PC=sum of sqq. on OQ, QE. I. 47. But sq. on OP is less than sq. on OQ; .. sq. on PC is greater than sq. on QE; Next, let CD be greater than EF. Then must CD be nearer to the centre than EF. |