ROP. IX. THEOR. Book III. IF a point be taken within a circle, from which there fall more than two equal straight lines upon the circumference, that point is the centre of the circle. G; Let the point D be taken within the circle ABC, from which there fall on the circumference more than two equal straight lines, DA, DB, DC; the point D is the centre of the circle. For, if not, let E be the centre; join DE, and produce it to the circumference in F, then FG F IG A c a 7.3. Book III. PROP. X. THEOR. ONE circle cannot cut another in more than two points. If it be possible, let the circumference FAB cut the circumference DEF in more than two points, B, В. D U K EI a 9. 3. b 5. 3. PROP. XI. THEOR. IF two circles touch each other internally, the straight line which joins their centres being produced will pass through the point of contact. era than FA, or FH. Take away the common part FG, and Book III. the remainder AG is greater than the remainder GH. But AG is equal to GD; therefore GD is greater than GH; and a 20. 1. it is also less, which is impossible. Therefore the straight line which joins the points F and G cannot fall otherwise than on the point A ; that is, it must pass through A. 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 will 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, will pass through the point of contact A. For, if not, let it pass otherwise, if possible, as FCDG, and join FA, AG. Because F is the centre of the circle ABC, AF is equal to FC; and because G is B В E the centre of the circle ADE, AG is equal to GD. Therefore FA, AG D are equal to FC, G wherefore the whole FG is greater than FA, AG; but it is also lessa, which is impossible. Therefore the straight a 20. 1. line which joins the points F, G, cannot pass otherwise than through the point of contact A ; that is, it passes through A. Th-refore, if two circles, &c. Q. E. D. DG; ONE circle cannot touch another in more points than one, whether it touches it on the inside or outside. For, if it be possible, let the circle EBF touch the circle ABC in more points than one, and first on the inside, in the a 10. 11. 1. points B, D; join BD, and drawa GH, bisecting BD at right angles. Because the points B, D are in the circumference of H Н F c bisects BD at right angles; therefore GH passes through the K A B the points A, C are in the circumference of the circle ABC, Book III. the straight line AC must be within the same circle, which is absurd. Therefore a circle cannot touch another on the outside in more than one point; and it has been shown that a circle cannot touch another on the inside in more than one point. Therefore, one circle, &c. Q. E. D. PROP. XIV. THEOR. EQUAL straight lines in a circle are equally dis. tant from the centre; and those which are equally distant from the centre are equal to each other. Let the straight lines AB, CD, in the circle ABDC, be equal to each other; they are equally distant from the centre. Take E the centre of the circle ABDC, and from it draw EF, EG perpendicular to AB, CD; join AE and EC. Then EF bisects AB, and EG bisects a 3. 3. CD. But AB is equal to CD, therefore AF is equal to CG. Because AE is equal to EC, the A А E square of AE is equal to the square of EC. · Now the squares of AF, FE are equal to the square of AE, and the squares of EG, b 47. 1. GC are equal to the square of EC; B D therefore the squares of AF, FE are equal to the squares of CG, GE. But the square of AF is equal to the square of CG, because AF is equal to CG; therefore the remaining square of FE is equal to the remaining square of EG; therefore the straight line EF is equal to EG; therefore AB, CD are equally distant from the centre. c 3. Def. 3. 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 it may be demonstrated, as before, that the squares of EF, FA are equal to the squares of EG, GC. But the square of FE is equal to the square of EG, because FE is equal to EG; therefore the square of AF is equal to the |