Describe (I. 54.) the triangle ABG, equal to the figure ABCDEF, and having the side AB, and angle A B C the same with it: divide (55.) the base BG in the given ratio in the point H: join AC and AD: through H draw HK parallel fo A C, to meet CD produced in K; through K draw KL parallel to AD to meet DE in L, and join AL: AL shall be the line required. For, by the construction, which is similar to that of I. 55., it is evident, that the figure ABCDL is equal to the triangle AB H:* but the whole figure is equal to the triangle AB G: therefore, the part A LEF is equal to the triangle AHG. Therefore, the parts of the figure are as the triangles A B H, AHG, that is (39.) as the bases BH, HG, that is, in the given ratio. Next, let P be the given point in the side A B, from which the line of division is to be drawn. Describe, as before, (I. 54.), the triangle A B G equal to the given figure, and, by drawing AG' parallel to PG, make the triangle PB G'equal to ABG (as in I. 56.): divide (55.) BG' in the given ratio in the point H: join PC and PD: through H draw HK parallel to P C, to meet C D produced in K: through K draw K L parallel to PD to meet DE in L, and join PL. Then, as before, the part PBCDL of the given figure, is equal to the triangle PB H, and the remaining part PAFEL to the triangle PH Gʻ. Therefore, the parts of the figure are as the bases B H, HG' (39.), that is, in the given ratio. Therefore, &c. PROP. 69. Prob. 19. Given a triangle A, a point B, and two straight lines CD, CE, forming an angle DCE; to describe a triangle which shall be equal to the triangle A, so that it may have the angle D CE for one of its angles, and the opposite side passing through the point B. In the solution of this problem we have three cases to consider; first, when the point B is in one of the given lines as CD; secondly, when the point B is without the given angle DCE; and thirdly, when B is within the angle DCE. Case 1. Let the point B be in the line CD. Take CD equal to a side of the triangle A, and upon CD (I.50.) describe a triangle CDF, which shall have its two remaining sides equal to the two remaining sides of the triangle A, each to each, and therefore (I. 7.) shall be equal to the triangle A in every respect: through F (I. 48.) draw FE parallel to DC: join DE, BE: through D (I. 48.) draw DG parallel to BE to meet CE in G, and join GB: the triangle CBG shall be the triangle required. For, because EB is parallel to GD, the triangle GEB (I. 27.) is equal to DEB: therefore, adding the triangle ECB to each, the whole triangle GCB is equal to E C D, that is (because E F is parallel to CD) to F C D, that is, to the given triangle A. Case 2. Let the point B be without the angle DCE. Through B (I. 48.) draw BD parallel to CE to meet CD in D, and by the construction pointed out in Case 1. describe a triangle DCF having the given side D C and the given angle DCE, and equal to the given triangle A in CF produced take the : point G such that CGX GF may be equal to BDXC F (56.): join BG, and let BG cut CD in H: CHG shall be the triangle required. Join HF. Then, because C G x GF is equal to B D×CF, CF: FG::CG : BD (38.), that is, since DHB and CHG are (I. 15.) equiangular, ::CH HD. Therefore (29.) H F is parallel to DG; and hence, as before, the triangles DHF and GHF are (I. 27) equal to one another, and CHG is equal to CDF, that is to A. Case 3. Let the point B be within the angle DCE. Through B (I. 48.) draw BD parallel to EC to meet CD in D, and, by the construction pointed out in Case 1, describe D F H a triangle D C F, having the given side CD and the given angle DCE, and equal to the given triangle A: in CF (if it be possible) take the point G such that CGXGF may be equal to BDX CF (56. Cor.): join B G, and let GB produced cut C D in H; CHG shall be the triangle required. Join HF. Then, by a demonstration which may be given in the same words as that of Case 2, the triangle C H G is equal to C D F, that is to the given triangle A. In this last case a solution will be impossible if BD exceed a fourth of CF; for then BD×CF will exceed a fourth of the square of CF, that is (I. 29. Cor. 2.) the square of half C F ; and no point G can be taken in CF such that CG x GF may exceed the square of half C F (56. N. B.). We may remark also that, whenever the solution is possible in the last case, two points G may be found such that CG × GF = BDxCF, (56.) and therefore two lines G H may be drawn satisfying the given conditions. When BD is exactly a fourth of CF, these two points coincide with one another and with the middle point of CF, and therefore the two solutions become identical. If B be in one of the lines DC, E C produced, or within the angle which is vertical to DCF, the solution will be manifestly impossible. G B D't which shall form with two given straight lines CD CE cutting one another in Ca triangle equal to a given triangle A;" we should have had four solutions, two of them corresponding to the angle in which B lies, and one for each of the adjacent angles; as is apparent from the foregoing constructions applied to the adjoined figure. When B D is equal to a fourth of C F, two of these become identical; when BD exceeds a fourth of C F, the same two become impossible, but the other two, viz. those which correspond to the adjacent angles, are always possible, except when B is in one of the lines as CD. This last-mentioned position is peculiar: it has likewise, however, two solutions, one for each of the adjoining angles. BOOK III. § 1. First Properties of the Circle 2. Of Angles in a Circle—§ 3. Rectangles under the segments of Chords -4. Regular Polygons, and Approximation to Circular Area—§ 5. Circle a Maximum of Area, and Minimum of Perimeter-§ 6. Simple and Plane Loci-§ 7. Problems. SECTION 1.-First Properties of the Def. 1. Any portion of the circumference of a circle is called an arc; and the straight line which joins the extremities of an arc is called the chord of that arc. When the chord passes through the centre, it is a diameter, and the arcs upon either side of it, being equal to one another, are called, each of them, a semi-circumference.* 2. The figure which is contained by an arc and its chord is called a segment. * That every diameter divides a circle and its cir cumference into two equal parts, is evident from the symmetrical character of the circle. The same may be proved by doubling the figure upon the diameter; for, every point of the circumference being at the same distance from the centre, the parts so applied (whether of circumference or area) will coincide. 3. An angle in a segment is the angle contained by two straight lines drawn from any point of the are of the segment to the extremities of its base or chord. 4. A sector of a circle is the figure contained by any arc, and the radii drawn to its extremities. 5. Equal circles are those which have equal radii. By this is intended no more than that when the term "equal circles" may be hereafter used, those which have equal radii are to be understood. That such circles, however, have also equal areas, is at once evident by applying one to the other, so that their centres may coincide. 6. Concentric circles are those which have the same centre. 7. (Euc. iii. def. 2.) A straight line is said to touch a circle, when it meets the circumference in any point, but being produced does not cut it in that point. Such a line is frequently, for brevity's sake, called a tangent; and the point in which it meets the circumference is called the point of contact. 8. Circles are said to touch one another, when they meet, but do not cut one another. 9. A rectilineal figure is said to be inscribed in a circle, when all its angular points are in the circumference of the circle. Also, when this is the case, the circle is said to be circumscribed about the rectilineal figure. Join CA, CB: bisect A B in D, and join CD: take also the points E and F in the line A B, the former between A B, and join CE, CF. and B, the latter upon the other side of Then, because CA is equal to C B, CAB is an isosceles triangle; therefore (I. 6. Cor. 3.) C D, which is drawn from the vertex to the bisection of the base, is perpendicular to AB. And, because CE is nearer to the perpendicular than CB is, (I. 12. Cor. 2.) CE is less than CB; therefore the point E is within the circle: on the other hand, because C F is farther from the perpendicular than C B is, CF is greater than CB; and therefore the point F is without the circle. And the same may be said of the parts about A. Therefore the straight line in question cuts the circle in the points A and B. Also, because every point, as E, of A B, is at a less distance from C than the radius, the part AB falls within the circle. Therefore, &c. Cor. 1. A straight line cannot meet a circle in more than two points. Cor. 2. A straight line which touches a circle meets it in one point only. Cor. 3. A circle is concave towards 10. A rectilineal figure is said to be its centre. circumscribed about a circle, when all its sides touch the circle. Also, when this is the case, the circle is said to be inscribed in the rectilineal figure. PROP. 2. (EUC. iii. 16.) The straight line, which is drawn at right angles to the radius of a circle from its extremity, touches the circle; and no other straight line can touch it in the same point. For, C A being shorter (I. 12. Cor. 3.) than any other line which can be drawn from C to A T, every other point of AT lies without the circle: therefore AT meets the circle in the point A but does not cut it, that is, it touches the circle. In the next place, let DE be any other straight line passing through the same point A: DE shall cut the circle. For, CA not being perpendicular to DE, let CE be perpendicular to it. Then, because CE is less than CA (I. 12. Cor. 3.), the point E is within the circle. But if D be a point in DE upon the other side of A, C D will be farther from the perpendicular than CA; wherefore CD being (I. 12. Cor. 2.) greater than CA, the point D will be without the circle. Therefore the straight line DE cuts the circle; and the same may be demonstrated of every other straight line which passes through A, except the straight line AT only, which is at right angles to A C.. Therefore, &c. Cor. 1. (Euc. iii. 18.) If a straight line touches a circle, the straight line drawn from the centre to the point of contact, shall be perpendicular to the line touching the circle. Cor. 2. (Euc. iii. 19.) If a straight line touches a circle, and from the point of contact, a straight line be drawn at right angles to the touching line, the centre of the circle shall be in that line. Cor. 3. Tangents TA, TB which are drawn to a circle from the same point T, are equal to one another. For, CAT, CBT being right-angled triangles which have the common hypotenuse CT, and their sides CA, CB equal to one another, their remaining sides TA, TB are likewise equal. (I. 13.) Cor. 4. Tangents which are at the extremities of the same diameter are parallel to one another. PROP. 3. (EUC. iii. 3.) If a diameter cut any other chord at right angles, it shall bisect it; and conversely, if a diameter bisect any other chord, it shall cut it at right angles. Let AB be a diameter of the circle ADE, the centre of which is C; and, first, let it cut the chord DE at right angles in the point F: DE shall be bisected in F. Join CD, CE. Then, because CDE is an isosceles triangle, the straight line CF which is drawn from the vertex C at right angles to the base DE, bisects the base, that is, DF is equal to FE. (I. 6. Cor. 3.) Next let D E be bisected in F by the diameter A B; the angles DFA, E FA shall be right angles. For in the isosceles triangle CDE, the straight line which is drawn from the vertex C to the middle point of the base DE is at right angles to the base. (I. 6. Cor. 3.) Therefore, &c. Cor. 1. A diameter bisects all chords which are parallel to the tangent at either extremity of the diameter. Cor. 2. The straight line which bisects any chord at right angles passes through the centre of the circle. Cor. 3. If two circles have a common chord, the straight line which bisects it at right angles shall pass through the centres of both the circles. Cor. 4. (Euc. iii. 4.) It appears from the proposition, that two chords of a circle cannot bisect one another except they both of them pass through the centre. For, if only one of the chords pass through the centre, it cannot be bisected by the other which does not pass through the centre. And if, when neither of them passes through the centre, it were possible that they should bisect one another, the diameter passing through the supposed point of mutual bisection would be at right angles to each of them, which is absurd. PROP. 4. (Euc. iii. 15.) The diameter is the greatest straight line in a circle; and, of others, that which is nearer to the centre is greater than the more remote: also the greater is nearer to the centre than the less. distance CH be less than CK: the diameter A B shall be greater than the chord D E, and the chord D E shall be greater than the chord F G. Join CD, CE, CF. Then, because CA is equal to CD, and CB to CE, the whole A B is equal to CD and CE together: but C D and D E together (I. 10.) are greater than DE: therefore A B is greater than D E. Again, because CHD, CKF are right-angled triangles, and that CD-square is equal to C F-square, the squares of CH, Ĥ D together (I. 36.) are equal to the squares of CK, KF together: but the square of C H is less than the square of CK; therefore the square of HD is greater than the square of K F, that is, HD is greater than KF. And DE, FG are double of H D, K F respectively, because the perpendiculars C H, C K pass through the centre (3.): therefore DE is greater than F G. Next, let the chord DE be greater than FG; it shall also be nearer to the centre. For, C H and C K being drawn as before, the squares of CH, HD together are equal to the squares of CK, KF together; but the square of HD, which is half of D E, is greater than the square of KF, which is half of FG: therefore the square of CH is less than the square of C K, and CH is less than CK, that is, DE is nearer to the centre than F G is. Therefore, &c. Cor. (Euc. iii. 14.) 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. PROP. 5. (EUc. iii. 9.) If a point be taken, from which to the circumference of a circle there fall more than two equal straight lines, that point shall be the centre of the circle. Let ABC be a circle, and let D be a point taken, such that the three straight lines DA, DB, DC drawn from the point D to the circumference, are equal to one another: the point D shall be the centre of the circle ABC. Join A B and B C, and bisect them in the points E and Frespectively; and join DE, DF. Then, because D A B is an isosceles triangle, the straight line DE, which is drawn from the vertex D to the bisection of the base A B, is at right angles to AB (I. 6. Cor. 3.); and, because D E bisects the chord A B at right angles (3. Cor. 2.), it passes through the centre of the circle. In the same manner it may be shown that the straight line DF passes through the centre of the circle. But the only point through which each of the straight lines D E, D F passes, is their point of intersection D. Therefore D is the centre of the circle. Therefore, &c. Cor. 1. From any other point than the centre there cannot be drawn to the circumference of a circle more than two straight lines that are equal to one another, whether the point be within or without the circle. (Euc. iii. 7 and 8 parts of.) Cor. 2. It appears from the demonstration that if three points A, B and C be given which are not in the same straight line, a circle may be found, the circumference of which shall pass through the three points A, B and C; the circle, namely, which has for its centre the intersection of the two lines which bisect A B and B C at right angles. PROP. 6. If two circles have the same centre, either they shall coincide, or one of them shall fall wholly within the other. For if the radii of two concentric circles be equal to one another, it is manifest that every point in the circumference of the one must be at the same distance from their common centre, with every point in the circumference of the other; and therefore the two circumferences cannot but coincide, But if the radii be unequal, every point in the circumference of that which has the lesser radius is at a less distance from the common centre, and therefore must fall within the circumference of the greater circle. Therefore, &c. Cor. (Euc. iii. 5 and 6.) If two circles cut or touch one another, they cannot have the same centre. G |