If D be not PROOF. Then FG is a diameter of the circle ABC (I.Def. 17). the centre. is taken the point D, not the centre; DGDC > DB > DA. But they are also equal. Therefore DG is the greatest straight line from D to the circumference, and DC is greater than DB, and DB greater than DA (III. 7); But these lines are likewise equal, by hypothesis; which is impossible. Therefore E is not the centre of the circle ABC. In like manner it may be demonstrated that any other point than D is not the centre; Therefore D is the centre of the circle ABC. Therefore, if a point, &c. Q.E.D. Proposition 10.-Theorem. One circumference of a circle cannot cut another in more than two points. CONSTRUCTION.-If it be possible, let the circumference If possible, B the two circles have the same centre. E Take the centre K of the circle ABC (III. 1), and join KB, KG, KF. PROOF.-Because K is the centre of the circle ABC, the radii KB, KG, KF are all equal. And because within the circle DEF there is taken the point K, from which to the circumference DEF fall more than two equal straight lines KB, KG, KF, therefore K is the centre of the circle DEF (III. 9). But K is also the centre of the circle ABC (Const.); Therefore the same point is the centre of two circles which cut one another; which is impossible (III. 5). Therefore, one circumference, &c. Q.E.D. Proposition 11.-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; and let F be the centre of the circle ABC, and G the centre of the circle ADE. The straight line which joins their centres, being produced, shall pass through the point of contact A. CONSTRUCTION.-For, if not, let it pass otherwise, if If not, possible, as FGDH. Join AF and AG. PROOF. Because AG, GF are greater than AF (I. 20), and AF is equal to HF (I. def. 15); Therefore AG, GF are greater than HF. Take away the common part GF, and the remainder AG is greater than the remainder HG. H AG> HG. E But AG But AG is equal to DG (I. Def. 15); Therefore DG is greater than HG, the less than the DG. 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 it. Therefore, if one circle, &c. Q.E.D. Proposition 12.-Theorem. If two circles touch each other externally in any point, the straight line which joins their centres shall pass through that point. Let the two circles ABC, ADE touch each other externally in the point A; and let F be the centre of the circle ABC, and G the centre of the circle ADE; The straight line which joins their centres shall pass through the point of contact A. CONSTRUCTION.-For, if not, let it pass otherwise, if possible, as FCDG. Join FA and AG. PROOF.-Because F is the centre of the circle ABC, FA is equal to FC (I. Def. 15). A And because G is the centre of the circle ADE, GA is equal to GD; ...DG> HG. If not, FG is> FA+AG, but it is Therefore FA, AG are equal to FC, DG (Ax. 2). But FG is also less than FA, AG (I. 20); which is imalso less. possible. Therefore the straight line which joins the centres of the circles shall not pass otherwise than through the point A, that is, it must pass through it. Therefore, if two circles, &c. Q.E.D. Proposition 13.-Theorem. One circle cannot touch another in more points than one, whether it touch it internally or externally. I. First, let the circle EBF touch the circle ABC internally in the point B. Then EBF cannot touch ABC in any other point. CONSTRUCTION.If it be possible, let EBF touch ABC in another point D; join BD, and draw GH bisecting BD at right angles (I. 10, 11). E E F G then PROOF.-Because the two points B, D are in the circumference of each of the circles, the straight line BD falls within each of them (III. 2). Therefore the centre of each circle is in the straight line GH, which bisects BD at right angles (III. 1 cor.) Therefore GH passes through the point of contact (III. 11). But GH does not pass through the point of contact, the point because the points B, D are out of the line of GH; which GH passes through ofcontact, is absurd. which it does not. Therefore one circle cannot touch another internally in more points than one. II. Next, let the circle ACK touch the circle ABC externally in the point A. Then ACK cannot touch ABC in any other point. CONSTRUCTION.-If it be possible, let ACK touch ABC in If possible another point C. Join AC. PROOF. Because the points A, C are in the circumference of the circle ACK, the straight line AC must fall within the circle ACK (III. 2). But the circle ACK is without the circle ABC (Hyp.); Therefore the straight line AC is without the circle ABC. But because the two points A, C are in the circumference of the circle ABC, the let it touch in C also; K then straight line AC falls within the circle ABC (III. 2); which absurd. is absurd. Therefore one circle cannot touch another externally in more points than one. And it has been shown that one circle cannot touch another internally in more points than one. Therefore, one circle, &c. Q.E.D. Proposition 14.-Theorem. 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. 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 (III. 1). From E draw EF, EG, perpendiculars to AB, CD (I. 12). PROOF. 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 (III. 3). Therefore AF is equal to FB, and AB is double of AF. For the like reason CD is double of CG. AF = CG, and But AB is equal to CD (Hyp.); therefore AF is equal to CG (Ax. 7). And because AE is equal to CE, the square on AE is equal to the square on CE. F = CG2 + EG2. B ..EF=EG. Here EF= EG, and A the But the squares on AF, FE are equal to square on AE, because the angle AFE is a right angle (I. 47). For the like reason the squares on CG, GE are equal to the square on CE; Therefore the squares on AF, FE are equal to the squares on CG, GE (Ax. 1). But the square on AF is equal to the square on CG, because AF is equal to CG; Therefore the remaining square on FE is equal to the remaining square on GE (Ax. 3); And therefore the straight line EF is equal to the straight line 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. PROOF. The same construction being made, it may be AF2+EF2 demonstrated, as before, that AB is double AF, and CD double of CG, and that the squares on EF, FA are equal to the squares on EG, GC. = CG2 + EG2. .'. AF = CG, &c. But the square on EF is equal to the square on EG, because EF is equal to EG (Hyp.); Therefore the remaining square on FA is equal to the remaining square on GC (Ax. 3), And therefore the straight line AF is equal to the straight line CG. But AB was shown to be double of AF, and CD double of CG; Therefore AB is equal to CD (Ax. 6); Therefore, equal straight lines, &c. Q.E.D. |