PROP. XII. THEOR. IF two circles touch each other externally in any point, the straight line which joins their centres passes through that point. Let the two circles ABC, ADE touch each other externally in the point A; and let F and G be their centres: the straight line which joins F and G passes through A. For, if not, let it pass otherwise, if possible, as FCDG, and join FA, AG. Then (I. def. 30.) because F and G are the centres of the circles, AF is equal to FC, and AG to GD; therefore FA, AG are equal to FC, DG; wherefore the whole FG is greater than FA, AG, which (I. 20.) is impossible: therefore the straight B line which joins the points F, G, cannot pass otherwise than through the point of contact A: it therefore passes through it; wherefore, if two circles, &c. PROP. XIII. THEOR. ONE circle can touch another in only one point, whether it touch it internally or externally. For, by the two preceding propositions, if two circles touch one another in any point, the straight line joining their centres passes through that point. If therefore the circles could touch one another in more points than one, there could be drawn through their centres more straight lines than one, not coinciding, which (I. def. 3.) is impossible. One circle, therefore, &c. PROP. XIV. THEOR. EQUAL chords in a circle are equally distant from the centre; and (2.) chords which are equally distant from the centre, are equal to one another. Let the chords AB, CD, in the circle ABDC, be equal to one another; they are equally distant from the centre. A C Take (III.1.) E the centre of the circle, and draw (I. 12.) EF, EG perpendiculars to AB, CD: join also EA, EC. Then, because the straight line EF, passing through the centre, cuts the chord AB, which does not pass through the centre, at right angles, it also (III. 3.) bisects 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 (I. ax. 7.) to CG. Then, in the right-angled triangles EFA, EGC, the sides EA, AF are equal to the sides EC, CG, each to B E each; therefore (I. 47. cor. 5.) the sides EF, EG are equal. But chords in a circle are said (III. def. 3.) to be equally distant from the centre, when the perpendiculars drawn to them from the centre are equal: therefore AB, CD are equally distant from the centre. Next, if the chords AB, CD be equally distant from the centre, that is, if FE be equal to EG, AB B is equal to CD. For, the same construction being made, it may, as before, be demonstrated, that AB F A E is double of AF, and CD of CG; and because the right-angled triangles EFA, EGC have the sides AE, EF equal to CE, EG, each to each, the sides AF, CG are also (I. 47. cor. 5.) equal to one another. But AB is double of AF, and CD of CG; wherefore AB is equal (I. ax. 6.) to CD. Therefore equal chords, &c. PROP. XV. THEOR. THE diameter of a circle is the greatest chord: (2.) of others, one nearer to the centre is greater than one more remote; and (3.) 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 BC, and BC than FG. 1. From the centre draw (I. 12.) EH, EK perpendiculars to BC, FG, and join EB, EC, EF. Then, because AE is equal to EB, and ED to EC, AD is equal to EB, EC: but EB, EC are greater (I. 20.) than BC; wherefore also AD is greater than BC. F G K E AB H 2. Again, because BC is nearer to the centre than FG, EH (III. def. 4.) is less than EK; and the hypotenuses EB, EF are equal. Therefore (I. 47. cor. 6.) BH is greater than FK. But, as was demonstrated in the preceding proposition, BC is double of BH, and FG of FK: therefore BC is greater than FG. 3. 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 FK: and therefore (I. 47. cor 6.) EH is less than EK. Therefore the diameter, &c. PROP. XVI. THEOR. THE straight line drawn at right angles to a diameter of a circle, through its extremity, falls without the circle; but any other straight line drawn through that point cuts the circle. Let ABC be a circle, of which D is the centre, and AB a diameter: if AE be drawn through A perpendicular to AB, it falls without the circle. E C F B In AE take any point F; and draw DF, meeting the circumference in C. Because DAF is a right angle, it is greater (I. 17.) than DFA; and therefore (I. 19.) DF is greater than DA. But (I. def. 30.) DA is equal to DC; therefore DF is greater than DC, and the point F is therefore without the circle: and in the same manner it may be shown that any other point in AE, except the point A, is without the circle. D E K H Again, any other straight line drawn through A, cuts the circle. Let AG be drawn in the angle DAE, and draw (I. 12.) DH perpendicular to AG, and meeting the circumference in K. Then, because DHA is a right angle, and DAH less than a right angle, the side DH is less (I. 19.) than the side DA. But (I. def. 30.) DK is equal to DA; therefore DH is less than DK; the point H is therefore within the circle; and AG cuts the circle, since its continuation through A, must fall on the opposite side of EAL, and must therefore be without the circle. Therefore the straight line, &c. Cor. From this it is manifest, that the straight line which is drawn at right angles to the diameter of a circle from its extremity, touches (III. def. 1.) the circle; and that it touches it only in one point, because at every point except A, it falls without the circle. It is also evident that there can be but one tangent at the same point of a circle. PROP. XVII. PROB. FROM a given point, either without a given circle, or in its circumference, to draw a straight line touching the circle. First, let A be a given point without the given circle BCD; it is required to draw from A a straight line touching the circle. Find (III. 1.) E the centre of the circle, and draw AE cutting the circumference in D; from the centre E, at the distance EA, describe (I. post. 3.) the circle AFG; from D draw (I. 11.) DF at right angles to EA, and draw EBF, and AB. AB touches the circle BCD. Because E is the centre of the circles, EA is equal to EF, and ED to EB; therefore the two sides AE, EB are equal to the two FE, ED, and they contain the angle AEF common F B' E P B H to the two triangles AEB, FED: therefore the angle EBA, being equal (I. 4. part 3.) to EDF, is a right angle. Secondly, if the given point be in the circumference of the circle, as the point D, draw DE to the centre E, and DF at right angles to DE; DF touches (III. 16. cor.) the circle. Cor. If AB be produced to H, AH is bisected (III. 3.) in B. Hence a chord in a circle touching a concentric one, is bisected at the point of contact. Schol. It is evident that from any point A without the circle, two tangents, AB and AB', may be drawn to the circle, and that these are equal to one another, being equal respectively to the equal lines DF and DF'.* PROP. XVIII. THEOR. IF a straight line touch a circle, the straight line drawn from the centre to the point of contact, is perpendicular to the line touching the circle. Let the straight line DE touch the circle ABC in C; take the centre F, and join FC: FC is perpendicular to DE. A B For, if it be not, from F draw FG perpendicular to DE, meeting the circle in B. Then (III. 16. cor.) FG is greater than FB; and (I. 17.) because FGC is a right angle, GCF is an acute angle: therefore (I. 19.) FC is greater than FG; but FC is equal to FB; therefore FB is greater than FG; which (I. ax. 9.) is impossible: wherefore FG is not perpendicular to DE. In the same manner it may be shown, that no other straight line drawn from F, except FC, is perpendicular to DE; that is, FC is perpendicular to DE. Therefore, if a straight line, &c. PROP. XIX. THEOR. D GE If a straight line touch a circle, a straight line drawn from the point of contact, at right angles to the tangent, passes through the centre. Let the straight line DE touch the circle ABC in C, and let CA be perpendicular to DE; the centre of the circle is in CA. The construction of the first case of this problem is as easily effected in practice, by describing a circle on AE as diameter, as its circumference will cut that of the given circle in the points B and B'. The reason of this is evident from the 31st proposition of this book. For, if not, let F be the centre, if possible, and join CF. Because DE touches the circle ABC, and FC is drawn from the centre to the point of contact, FC is perpendicular (III. 18.) to DE; therefore FCE is a right angle: but ACE is also (hyp.) a right angle; therefore FCE is equal to ACE, which (I. ax. 9.) is impossible: wherefore F is not the centre of the circle ABC. In the same manner it may be shown, that no other point which is not in CA, is the centre; that is, the centre is in CA. Therefore, if a straight line, &c. B Ꭰ C E PROP. XX. THEOR. THE angle at the centre of a circle is double of the angle at the circumference, upon the same base, that is, upon the same part of the circumference. In the circle ABC, let BEC be an angle at the centre, and BAC an angle at the circumference, which have the same arc BC for their base; BEC is double of BAC. Join A E, and produce it to F; and first, let E, the centre of the circle, be within the angle BAC. Because EA is equal to EB, the angle EAB is equal (1. 5.) to EBA; therefore the angles EAB, EBA are together double of EAB: but (I. 32.) the angle BEF is equal to EAB, EBA; therefore also BEF is double of EAB. For the same reason, the angle FEC is double of the angle EAC; therefore the whole angle BEC is double of the whole BAC. * Again, let E the centre of the circle be without the angle BAC: it may be demonstrated, as in the first case, that the angle FEC is double of FAC, and that FEB, a part of the first, is double of FAB a part of the other; therefore the remaining angle BEC is double of the remaining angle BAC. The angle at the centre, therefore, &c.t R F E For That, if two magnitudes be double of two others, each of each, the sum and difference of the first two are respectively double of the sum and difference of the other two, is thus proved by Playfair: "Let A and B, C and D be four magnitudes, such that A= 2C, and B=2D; then A+B=2(C + D). since AC+ C, and BD+ D, adding equals to equals, A+ B= (C + D) + (C + D) = 2(C + D). So also, if A be greater than B, and therefore C greater than D, since AC+ C, and B D + D, taking equals from equals, A-B=(C—D) + (C–D), that is A-B=2(C—D).” There is evidently a third case, viz., when AB or AC passes through the centre: but though this case is not given in a separate form, its proof is contained in that of either of the others. |