Book Ill. PROP. IX. THEOR. a If a point be taken within a circle, from which there fall more than two equal straight lines to the circumference, that point is the centre of the circle. Let the point D be taken within the circle ABC, from which to the circumference there fall more than two equal straight lines, viz. DA, DB, DC, the point D is the centre of the circle. For, if not, let E be the centre, DE shall be the greatest line from it to • 2. 3. the circumference, and DC greater a B than DB, and DB than DA: But they are likewise equal, which is impossible; Therefore E is not the centre of the circle ABC: In like manner it may be demonstrated, that no other point but D is the centre; D therefore is the centre. Wherefore, if a point be taken, &c. Q.E.D. PROP. X. THEOR, BD ONE circumference of a circle cannot cut another А H EK! a 9. 3. KG, KF, the point K isa the centre of the circle DEF: Boox III. But K is also the centre of the circle ABC; therefore the same point is the centre of two circles that cut one another, which is impossibleb. Therefore one circumference of a 15.3. circle cannot cut another in more than two points. Q.E.D. PROP. XI. THEOR. If two circles touch each other internally, the straight line. which joins their centres being produced shall pass through the point of contact. Let the two circles ABC, ADE touch each other 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 the centres F, G, being produced, passes through the point A. H For, if not, let it fall otherwise, if possible, as FGDH, and join AF, AG: And because AG, GF are greater a than FA, that is, than FH, for FA is equal to FH, both being from the E same centre; take away the common B part FG; therefore the remainder AG is greater than the remainder GH: But AG is equal to GD; therefore GD is greater than GH, the less than the greater, which is impossible. Therefore the straight line which joins the points F, G cannot fall otherwise than upon the point A, that is, it must pass through it. Therefore, if two circles, &c. Q.E.D. PROP. XII. THEOR. If two circles touch each other externally, the straight line which joins their centres, shall pass through the point of contact. 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 ADE: The straight line which joins the points F, G shall pass through the point of contact A. For, if not, let it pass otherwise, if possible, as FCDG, Book III . and join FA, AG: And because F is the centre of the cit. cle ABC, AF is equal E B A greater than FA, AG: * 20.1. But it is also lessa; which is impossible: Therefore the straight line which' joins the points F, G shall not pass otherwise than through the point of contact A, that is, it must pass through it. Therefore, if two circles, &c. Q.E.D. PROP. XIII. THEOR. See N. ONE circle cannot touch another in more points than one, whether it touches it on the inside or outside. For, if it be possibile, let the circle EBF touch the circle ABC in more points than one, and first on the inside, in • 10. 11. 1. the points B, D; join BD, and drawa GH bisecting BD at right angles : Therefore because the points B, D, are in H A D С с F G the circumference of each of the circles, the straight line b 2. 3. • Cor. 1. 3. BD falls within each b of them : and their centres are in the straight line GH which bisects BD at right angles : d 11. 3. Therefore GH passes through the point of contact d; but it does not pass through it, because the points B, D are without the straight line GH, which is absurd : Therefore one circle cannot touch another on the inside in more points than one. Nor can two circles touch one another on the outside Book III. in more than one point; For, if it be possible, let the circle ACK touch the circle ABC in the points A, C, and join AC: Therefore, K because the two points A, C are in the circumference of the circle ACK, the straight line AC which joins them shall fall within the circle ACK: And 12. 3. the circle ACK is without the circle ABC; and therefore the straight line AC is without this last circle; but because the points A, C are in the circumference of the circle ABC, the straight line AC must be within the B same circle, which is absurd : Therefore one circle cannot touch another on the outside in more than one point. And it has been shown, that they cannot touch on the inside in more points than one. Therefore, one circle, &c. Q.E.D. a PROP. XIV. THEOR. EQUAL straight lines in a circle are equally distant from the centre; and 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, they are equally distant from the centre. Take E the centre of the circle ABDC, and from it draw EF, EG, perpendiculars to AB, CD: 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 ait: Wherefore AF is A equal to FB, and AB double of AF. For the same reason CD is double of CG: And AB is equal to CD; therefore AF is equal to CG: And because AE is equal to EC, the square of AE is equal to the square of EC: But : BT the squares of AF, FE, are equalb to b 17. 1. the square of AE, because the angle AFE is a right angle; and for the like reason, the squares of EG, GC are equal to the square of EC: Therefore the squares of AF, FE, are a 3. 3. Book III. 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; a CD must be equally distant from the centre, Next, if the straight lines AB, CD be equally distant from the centre, that is, if FE be equal to EG; AB is equal to CD: For, the same construction being made, it may, as before, be demonstrated, 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; therefore the remaining square of AF is equal to the remaining square of CG; and the straight line AF is therefore equal to CG: And AB is double of AF, and CD double of CG; wherefore AB is equal to CD. Therefore equal straight lines, &c. Q.E.D. See n. The diameter is the greatest straight line in a cir cle; and, of all others, that which is nearer to the greater is nearer to the centre than the less. А В E H Kmeter, and BC greater than FG. E G С wherefore also AD is greater than BC. And, because BC is nearer to the centre than FG, EH |