D are equal to FM, MD; but the angle EMD is greater than the angle FMD; therefore the base ED is greater (24. 1.) than the base FD. In like manner it may be shewn that FD is greater than CD. Therefore DA is the greatest; and DE greater than DF, and DF than DC. And because MK, KD are greater (20. 1) than MD, and MK is equal to MG, the remainder KD is greater (5. Ax.) than the remainder GD, that is, GD is less than KD: And because MK, DK are drawn to the point K within the triangle MLD from M, D, the extremities of its side MD; MK, KD are less (21.1.) than ML, LD, whereof MK is equal to ML; therefore the remainder DK is less than the remainder DL: In like manner, it may be shewn that DL is less than DH: Therefore DG is the least, and DK less than DL, and DL than DH. LK GBN H C M F E A Also there can be drawn only two equal straight lines from the point D to the circumference, one upon each side of the least; at the point M, in the straight line MD, make the angle DMB equal to the angle DMK, and join DB; and because in the triangles KMD, BMD, the side KM is equal to the side BM, and MD common to both, and also the angle KMD equal to the angle BMD, the base DK is equal (4. 1.) to the base DB. But, besides DB, no straight line can be drawn from D to the circumference, equal to DK; for, if there can, let it be DN; then, because DN is equal to DK, and DK equal to DB, DB is equal to DN; that is, the line nearer to DG, the least, equal to the more remote, which has been shewn to be impossible. PROP. IX. THEOR. 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. Let the point D be taken within the circle ABC, from which there fall on the circumference 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, join DE, and produce it to the circumference in F, G; then FG is a diameter of the circle ABC: And because in FG, the diameter of the circle ABC, there is taken the point D which is not the centre, DG is the greatest line from it to the circumference, and DC greater (7. 3.) 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. F A DE G PROP. X. THEOR. One circle cannot cut another in more than two points. B If it be possible, let the circumference FAB cut the circumference DEF in more than two points, viz. in B, G, F; take the centre K of the circ'e ABC, and join KB, KG, KF; and because within the circle DEF there is taken the point K, from which more than two equal straight lines, viz. KB, KG, KF, fall on the circumference DEF, the point K is (9. 3.) the centre of the circle DEF; 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 impossible (5. 3.). Therefore one circumference of a circle cannot cut another in more than two points. E A D H K F 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. 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. A H D G F E C B For, if not, let it fall otherwise, if possible, as FGDH, and join AF, AG: And because AG, GF are greater (20. 1.) than FA, that is, than FH, for FA is equal to FH, being radii of the same circle; take away the common part FG, and the remainder AG is greater than the remainder GH. But AG is equal to GD, therefore GD is greater than GH; and 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. COR. 1. If two circles touch each other internally, the distance between their centre must be equal to the difference of their radii: for the circumferences pass through the same point in the line joining the centres. COR. 2. And, conversely, if the distance between the centres be equal o the difference of the radii, the two circles will touch each other internally. PROP. XII. THEOR. If two circles touch each other externally, the straight line which joins they 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 shall pass through the point of contact. B E For, if not, let it pass otherwise, if possible, FCDG, and join FA, AG: and because F is the centre of the circle ABC, AF is equal to FC: Also because G is the centre of the circle, ADE, AG is equal to GD. Therefore FA, AG are equal to FC, DG; wherefore the whole FG is greater than FA, AG; but it is also less (20. 1.), which is impossible: Therefore the straight line which joins the points F, G cannot pass otherwise than A CAD F G through the point of contact A; that is, it passes through A. COR. Hence, if two circles touch each other externally, the distance between their centres will be equal to the sum of their radii. And, conversely, if the distance between the centres be equal to the sum of the radii, the two circles will touch each other externally. PROP. XIII. THEOR. 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 points B, D; join BD, and draw (10. 11. 1.) GH, bisecting BD at right angles: Therefore because the points B, D are in the circumference of each of the circles, the straight line BD falls within each (2. 3.) of them: and therefore their centres aro (Cor. 1. 3.) in the straight line GH which bisects BD at right angles : therefore GH passes through the point of contact (11. 3), 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 in the inside in more points than one. K Nor can two circles touch one another on the outside 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, 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 the circle ACK is without the circle ABC and therefore the straight line AC is also without ABC; but, because the points A, C are in the circumference of the circle ABC, the straight line AC must be within (2. 3.) the same circle, which is absurd: therefore a circle cannot touch another on the outside in more than one point: and it has been shewn, that a circle cannot B touch another on the inside in more than one point. 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. A E G Take E the centre of the circle ABDC, and from it draw EF, EG, perpendiculars to AB, CD; join AE and EC. 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 (3. 3.) it: Wherefore 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; therefore AF is equal to CG: And because AE is equal to EC, the square of AE is equal to the square of EC: Now the squares of AF, FE are equal (47. 1.) to the square of AE, because the angle AFE is a right angle; and, for the like reasor, 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 FE is equal to the remaining square of EG, and the straight line EF is therefore 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 (3. Def. 3.): therefore AB, CD are equally distant from the centre. B D Next, if the straight lines AB, CD be equally distant from the centre, *hat is, if FE be equal to EG, AB is equal to CD. F'or, the same con struction 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: But AB is double cf AF, and CD double of CG; wherefore AB is equal to CD. The diameter is the greatest straight line in a circle; and of all others, that which is nearer to the centre is always greater than one more remote ; and the greater is nearer to the centre than the less. Let ABCD be a circle, of which the diameter is AD, and the centre E; and let BC be nearer to the centre than FG; AD is greater than any straight line BC which is not a diameter, and BC greater than FG. From the centre draw EH, EK perpendiculars to BC, FG, and join EB, EC, EF; and because AE is equal to EB, and ED to EC, AD is equal to EB, EC: But EB, EC are greater (20. 1.) than BC; wherefore, also, AD is greater than BC. F E B H And, because BC is nearer to the centre than FG, EH is less (4. Def. 3.) than EK; But, as was demonstrated in the preceding, BC is double of BH, and FG double of FK, and the squares of EH, HB are equal to the squares of EK, KF, of which the square of EH is less than the square of EK, because EH is less than EK; therefore the square of BH is greater than the square of FK, and the straight line BH greater than FK: and therefore BC is greater than FG. Next, let BC be greater than FG; BC is nearer to the centre than FG: that is, the same construction being made, EH is less than EK; because BC is greater than FG, BH likewise is greater than KF: but the squares of BH, HE are equal to the squares of FK, KE, of which the square of BH is greater than the square of FK, because BH is greater than FK; therefore the square of EH is less than the square of EK, and the straight line EH less than EK. COR. The shorter the chord is, the farther it is from the centre; and, conversely, the farther the chord is from the centre, the shorter it is. The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle; and no straight line can be drawn between that straight line and the circumference, from the extremity of the diameter, so as not to cut the circle. Let ABC be a circle, the centre of which is D, and the diameter AB and let AE be drawn from A perpendicular to AB, AE shall fall without the circle. |