PROP. XI.- THEOREM. If one circle touch another internally in any point, the straight line which joins their centres, being produced, shall pass through that point. Let the circle ADE touch the circle ABC internally in the point A. Then the straight line which joins their centres, being produced, shall pass through the point of contact A. B G A C CONSTRUCTION For, if not, let it fall otherwise, if possible, as FGDB; let F be the centre of the circle ABC, and G the centre of ADE. Join AF and AG. DEMONSTRATION Because two sides of a triangle are together greater than the third side, therefore AG, GF, are greater than FA; (1. 20) but FA is assumed to be equal to FB, (1. def. 15) both being from the same centre; therefore AG, GF, must be greater than FB take away the common part FG; then the remainder AG must be greater than the remainder GB; but AG is assumed to be equal to GD; (I. def. 15) therefore GD must be greater than GB, the less than the greater, which is impossible; therefore the straight line which joins the centres, being produced, cannot fall otherwise than upon the point A. that is, it must pass through the point of contact of the two circles. Therefore, if two circles, &c. Q. E. D. PROP. XII.- THEOREM. If two circles touch each other externally, the straight line which joins their centres shall pass through the point of contact. (References Prop. 1. 20.) Let the two circles ABC, ADE, touch each other externally in the point A. f Then the straight line which joins their centres, shall pass through the point of contact A, For, if not, let, if possible, F and G be the centres, and the line joining them pass otherwise as FCDG; and join FA, AG. DEMONSTRATION Then, because F is assumed to be the centre of the circle ABC, therefore AF must be equal to FC; also, because G is assumed to be the centre of the circle ADE therefore AG must be equal to GD; and therefore FA, AG, must be equal to FC, DG; wherefore the whole FG must be greater than FA, AG; but FG is also less than FA, AG; (1. 20) which is impossible. therefore the straight line which joins the centres of the circles, shall not pass otherwise than through A ; that is, it must pass through the point of contact of the two circles. Therefore, if two circles, &c. Q. E. D. PROP. XIII.- THEOREM. One circle cannot touch another in more points than one, whether it touches it in the inside or the outside. (References Prop. I. 10, 11; III. 1; Cor. 2, 11.) First. Let the circle EBF touch the circle ABC interually in the point B. Then EBF cannot touch ABC in any other point. If it be possible, let EBF touch ABC in another point D; join BD, and draw GH bisecting BD at right angles. (1. 10.) DEMONSTRATION Because the points B, D, are assumed to be in the circumference of each of the circles, the straight line BD must fall within each of the circles EBF, ABC; (III. 2) and their centres are in the straight line GH, which bisects BD at right angles; (III. 1, Cor.) therefore GH must pass through the point of contact; (III. 11) but GH does not pass through the point of contact, because the points B, D, are without the straight line GH; which is absurd; therefore the circle EBF cannot touch the circle ABC on the inside in more points than one. Secondly. Let the circle ACK touch the circle ABC externally in the point A. Then ACK cannot touch ABC in any other point. If it be possible, let ACK touch ABC in another point C. Join AC. DEMONSTRATION. Then, because the points A, C, are assumed to be in the circumference of the circle ACK, the straight line AC must fall within the circle ACK; (111. 2) but the circle ACK is without the circle ABC; (hyp.) therefore the straight line AC must be without the circle ABC; but, because the points A, C, are assumed to be in the circumference of the circle ABC, the straight line AC must be within the circle; (III. 2) which is absurd; therefore the circle ACK cannot touch the circle ABC on the outside in more than one point. And it has been shown that one circle cannot touch another on the inside in more points than one. Wherefore one circle, &c. Q. E. D. Equal straight lines in a circle are equally distant from the centre; and conversely, those which are equally distant from the centre, are equal to one another. (References - Prop. 1. 12, 47; III. 1, 3, def. 4.) Let the straight lines AB, CD, in the circle ABDC, be equal to one another. Then they shall be equally distant from the centre. CONSTRUCTION Take E the centre of the circle ABDC, (1. 1) and from it draw EF, EG, perpendiculars to AB, CD, (1. 12) and join EA, EC. DEMONSTRATION Then, because the straight line EF passing through the centre, cuts the straight line AB, which does not pass through the centre, at right angles, it also bisects it; (II. 3) therefore AF is equal to FB, and AB double of AF; for the same reason CD is double of CG; but AB is equal to CD; (hyp.) therefore AF is equal to CG. (ax. 7.) F And because AE is equal to EC, (1. def. 15) the square of AE is equal to the square of EC; but the squares of AF, FE, are equal to the square of AE, because the angle AFE is a right angle; (1. 47) and for the same reason, the squares of EG, GC, are equal to the square of EC; therefore the squares of AF, FE, are equal to the squares of CG, GE; of which the square of AF is equal to the square of CG, because AF is equal to CG; therefore the remaining square of EF is equal to the remaining square of EG, (ax. 3) and therefore the straight line EF is equal to EG; but 'straight lines in a circle are said to be equally distant from the centre, when the perpendiculars drawn to them from the centre are equal;' (III. def. 4) therefore AB, CD, are equally distant from the centre E. Next, let the straight lines AB, CD, be equally distant from the centre, that is, let FE be equal to EG. Then AB shall be equal to CD. The same construction being made, it may be demonstrated as before, that AB is double of AF, and CD double of CG, and that the squares of EF, FA, are equal to the squares of EG, GC; of which the square of FE is equal to the square of EG, because FE is equal to EG; (hyp.) therefore the remaining square of AF is equal to the remaining square of CG; (ax. 3) and therefore the straight line AF is equal to CG; but AB was shown to be double of AF, and CB double of CG; wherefore AB is equal to CD. Therefore, equal straight lines, &c. Q. E. D. |