B. 1. B. 1. B. 1. B. 1. For, if it be possible, let one of them AB be greater than the other, make AE equal to AX, and draw ZE and ZX. In the triangles ZAE,ZAX, the side ZA is common, (1) Constr, AE is equal to AX (1), and the angle ZAE is equal to (2) hypoth. ZAX (2), therefore the sides ZE and ZX are equal (3) Prop. 4 (3), but the line ZO is equal to ZX (4), therefore ZE (4) Def.15. is equal to Z0, a part equal to the whole, which is absurd. Therefore neither AB nor AX is greater than the other, and therefore they are equal. Part 2. Of those lines which are incident upon the concave circumference, that line AY which passes through the centre is greater than any other AX. (5) Def.15. Draw ZX, and ZY is equal to ZX (5), therefore if AZ be added to both, AY shall be equal to AZ and ZX taken together, but AZ and ZX together are (6) Prop.20 greater than AX (6), therefore AY is greater than AX. Fig. 12. Part 3. The line AB or AX which is nearer to the greatest, is greater than the more remote AD. If the given lines AX and AD be at the same side of AY, draw ZX and ZD; in the triangles AZX, AZD the sides AZ, ZX are equal to the sides AZ, ZD, and the angle AZX is greater than AZD, there(7) Prop.24 fore the side AX is greater than AD (7). But if the given lines AB and AD be at different sides of AY, make the angle ZAX equal to ZAB, and (8) Part AX shall be equal to AB (8), but AX is greater than AD, therefore AB is greater than AD. Fig. 13. Part 4. Of those lines incident on the convex cir cumference, that line AF which, if produced, would pass through the centre is less than any other AX. Draw ZF and ZX; ZX and XA are greater than (9) Prop.20 ZA (9), and therefore, if the equals ZX and ZF be taken away, AX is greater than AF. Part 5. That line AB or AX wbich is nearer to the least is less than the more remote AC. If the given lines AX and AC be at the same side of AZ, draw ZX and ZC; ZC and CA taken together (10) Prop. are greater than ZX and XA (10), take away the 21. B. 1.' equals ZC and ZX, and AC is greater than AX. But if the given lines AB and AC be at different B. 1. prec. B. I. First. and 4. sides of AZ, make the angle ZAX equal to ZAB, and AX shall be equal to AB(11), but AC is greater (11) Part than AX, and therefore greater than AB. Part 6. Only two equal lines can be drawn either to the concave or convex circumference. If any three lines be drawn, either one of them shall pass through the centre, and therefore be either greater or less than either of the others (12), or two (12) Part 2 must be at the same side of the line passing through (13) Part 5 the centre, and therefore unequal (13). Schol. As it is evident that any right line drawn to the convex circumference is less than any line drawn to the concave, it follows that if any three lines be drawn from a point without a circle to its circumference, only two of them can be equal. Cor. Hence, and from Part 5. Prop. 7. it is evident that there is no point except the centre, from which three equal right lines can be drawn to the circumference of a circle. and 5. PROP. IX. THEOR. If a point be taken within a circle, from which more than two equal right lines can be drawn to the circumference, that point is the centre of the circle. For if it were a point different from the centre, only (1) Prop. 7. two equal right lines could be drawn from it to the B. 3. circumference(1). PROP. X. THEOR. One circle (BDF) cannot cut another (BLF) in more Fig.14815. than two points. See N. For, if it be possible, let it cut the other in three points B, F, and C, let A be the centre of the circle BLF, and draw from it to the points of intersection the lines AB, AF, and AC, these lines are equal (1), (1) Def. 15. but as the circles intersect, they have not the same B. 1. (2) Prop.5. centre(2); therefore A is not the centre of the circle B. 3. BDF, and therefore as three right lines AB, AF, AC, are drawn from a point not the centre, these lines are (3) Cor. not equal (3), but it was shewn before that they were Prop.8.B-3. equal, which is absurd; the circles therefore do not intersect in three points. Schol. Hence it is evident that one circle cannot meet another in more than two points, Fig. 16. If two circles (ECF and DCL) touch one another internally, the right line joining their centres being For, if it be possible, let A be the centre of the cir- triangle BAC, the sides BA and AC (1)Prop.20 are greater than BC(1), and BC is equal to BD, as they are radii of the circle DCL, the lines BA and AC B. 1. 20 PROP. XII. THEOR. B Fig. 17. If two circles (AOC and BFC) touch one another externally, the right line joining their centres passes through a point of contact. For, if it be possible, let A and B be the centres, and let the right line Ą B joining them not pass through a point of contact, and from C a point of contact draw CA and CB to the centres. Because in the triangles ACB, the sides AC and CB are greater than AB (1), and the line AO is equal to (1) Prop.20 AC as they are radji of the circle AOC, and the line B.de BF is equal to BC as they are radii of the circle BFC, AO and BF taken together are greater than BA, a part greater than the whole, which is absurd. The centres are not therefore so placed that the line joining them can pass through any point but a point of contact. PROP. XIII. THEOR. and 20. See y. One circle cannot touch another, either externally or Fig 18.19, internally, in more points than one. For, if it be possible, let the circles ADÉ and BDF. touch one another internally in two points D and C, draw the line AB joining their centres, and produce it until it pass through one of the points of contact D, and draw AC, BC. Because BD and BC are radii of the same circle BDF, BD is equal to BC, and therefore if AB be added to both, AD shall be equal to AB and BC; but AD and AC are radii of the circle. ADE, therefore AD is equal to AC, and therefore AB and BC are equal to AC, but they are greater than it (1), which is absurd. (1)Prop.20 B. I. But if the points of contact are at the extremities Fig. '19. of the right line joining the centres, CD must be bisected in A and also in B, because it is a diameter of both circles, which is absurd. Next, if it be possible, let the two circles ADE and Fig. 20. BDF touch one another externally in two points D and C, draw the right line AB joining the centres of the circles and passing through one of the points of contact C, and draw AD and DB. Because AD and AC are radii of the circle ADE, (2) Prop.20 they are equal; and because BC and BD are radii B. l. of the circle BDF, they also are equal, therefore AD and BD together are equal to AB, but they are greater than it (2), which is absurd. to cir P. 16.B. 1. the CE F, pe. dra the tan (2) Cor. each of them is acute (2), but CBG is a right angle (3), P. 17. B.I. which is absurd. Therefore the right line drawn through (3) hypoth. B perpendicular to AB does not meet the circle again. Part 2. Let BF be perpendicular to AB, and let EB be a line drawn from a point between it and the circle, which, if it be possible, does not cut the circle. Because the angle CBF is a right angle, CBE is acute, draw CI perpendicular to BE, and it must (4) Cor. 1. fall at the side of the angle CBE (4). Then in the triangle BCI the angle CIB is greater (5) Prop.19 than CBI, therefore the side CB is greater than CI (5), but CO is equal to CB, and therefore CO is greater Schol. 1. Hence it is evident that the right line BF Schol. 2. It is also evident, that the right line, which makes at B an acute angle, however great, must meet the circle again. Fig. 23. Cor. 1. From this proposition is immediately deduced a method of drawing a tangent through any given point B in the circumference of a circle : draw through that point a diameter AB, and erect at the extremity of it a perpendicular BF. Fig. 25. Cor. 2. If the line CD be produced beyond its extremity C, and there be taken in the produced part any number of points, and from these points as centres circles be described through the point D, it is evident that each of these circles touch the line in the point D, and meet each other in no other point, whence the right line EF is cut by each of these circles in a different point, and is therefore infinitely divisible. CE COT EF for AE See N. |