angles of equiangular triangles. Also, because the triangles K BA, EDC are similar, B K is to KA as DE to E C; and for the like reason, KA is to K L as EC to EF: therefore, ex æquali, BK: KL::DE: EF. And in the same manner it may be demonstrated, that the sides about the other equal angles of the two figures are proportionals. Therefore, the figure described upon A B is similar to the given figure CDEFG. (def. 14.) Method 2. Place A B parallel to CD, and join CA, D B. Then, if AB be not equal to CD, CA will not be parallel to DB (I. 21. and 14. Cor. 2.), and CA, DB will meet, if produced, in some point P: join P E, P F, PG: through B draw BK parallel to DE (I. 48.), and let it meet PE in K: through K draw KL parallel to EF, and let it meet PF in L: through L draw L M parallel to F G, and let it meet PG in M; and join AM: the figure ABKLM shall be similar to the given figure CDEFG. For, because PC:PA::PD: PB (29.), that is, :PE: PK, that is :: PF: PL, that is, PG: P M, AM is parallel to CG; and because the sides of the two figures are parallel, each to each, they are equiangular (I. 18.) with one another. Also, because the triangles CDP, A BP are similar, CD: DP::AB:B P, and for the like reason DP: DE: BP: BK; therefore, ex @quali, CD: DE::AB:BK. And in the same manner it may be demonstrated, that the sides about the other equal angles of the two figures are proportionals. Therefore the figure upon A B is similar to the given figure. When AB is equal to CD, CA is parallel to BD, and the lines EK, FL, GM, instead of being drawn to a point P, must be drawn parallel to CA or B D. The demonstration that the figure ABKLM, so constructed, will be both similar and PROP. 66. Prob. 16. To describe a rectilineal figure, which shall be similar to a given rectilineal figure CDEFG, and shall have its perimeter equal to a given straight line AQ. be Produce CD to R, (see the second figure of Prop. 65.) so that CR may equal to the perimeter of the given figure CDEFG; take A B (53.) a fourth proportional to CR, AQ, and CD, and upon AB (65.) describe a rectilineal figure AB KLM similar to the given figure. Then, because the figures AB KLM and CDEFG are similar, the perimeter of the first is to the perimeter of the other (43.), as AB to CD, that is (12.), as A Q to CR: but the perimeter of CDEFG is equal to CR: therefore (18. Cor.) the perimeter of AB KLM is equal to A Q, that is, to the given perimeter. Therefore, AB K L M is the figure required. Therefore, &c. PROP. 67. Prob. 17. (Euc. vi. 25.) D To describe a rectilineal figure which shall be similar to a given rectilineal figure ABC, and shall have its area equal to the area of another given rectilineal figure DEF. Find (1.58. Cor.) the straight lines G, H such that their squares may be equal to the figures ABC, DEF respectively: take KL a fourth proportional (53.) to G, H, and B C, and upon KL (65.) describe the figure K L M similar to the figure BČA. Then, because the figure A B C is to the figure MKL as the square of BC to the square of K L (43.), that is, (37. Cor. 4.) as the square of G to the square of H, and that the figure A B C is equal to the square of G, the figure K L M is (18.) equal to the square of H, that is, to the figure DEF: and it is similar to ABC; therefore it is the figure required, K M EE H PROP. 68. Prob. 18. To divide a given rectilineal figure ABCDEF in a given ratio, by a straight line drawn from one of its angles, or from a given point in one of its sides. In the first place, let A be the given angle from which the line of divi be drawn. is to H Describe (I. 54.) the triangle A B G; equal to the figure ABCDEF, and naving 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 to 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. B I G 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 H K parallel to P C, to meet CD produced in K: through K draw K L parallel to PD to meet DE in L, and join P L. Then, H • The line AH is wanting in this figure; and PH is in like manner wanting in the figure below, as also AG, AG', as before, the part PBCDL of the given figure, is equal to the_triangle PBH, and the remaining part PAFE L to the triangle PH G'. 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 DCE for one of its angles, and the opposite side passing through the point B. have three cases to consider; first, when In the solution of this problem we the point B is in one of the given lines without the given angle DCE; and as CD; secondly, when the point B is 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 remain. ing sides of the triangle A, each to each, and therefore (I. 7.) shall be equal to the A B 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.) 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 H F F G 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 BDxCF (56.): join BG, and let BG cut CD in H: CHG shall be the triangle required. Join HF. Then, because CGX GF is equal to BD× 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 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 BG, and let GB produced cut CD 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 CDF, that is to the given triangle A. E cl which shall form In this last case a solution will be impossible if BD exceed a fourth of CF; for then BDXCF will exceed a fourth of the square of CF, that is (I. 29. Cor. 2.) the square of half CF; 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 = BDXC F, (56.) and therefore two lines G H may be drawn satisfying the given conditions. When B D 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, EC produced, or within the angle which is vertical to DCF, the solution will be manifestly impossible. Scholium. Had the problem been proposed under the following form: " through a given point B to draw a straight line CHD D' BOOK III. $1. First Properties of the Circle- SECTION 1.-First Properties of the Def. 1. Any portion of the circum- When the chord passes 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 H 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 Isaid to touch one another, when they meet, but do not cut one another. centre C in the points A and B: it the part AB shall fall within the shall cut the circle in those points, and circle. join CD: take also the points E and F in the line A B, the former between A 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, C F 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. 9. A rectilineal figure the circle. Also, because every point, as E, of is said to be A B, is at a less distance from C than inscribed in a circle, when all its angular the radius, the part AB falls within 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. 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. 1. (EUC. iii. 2.) If a straight line meet the circumference of a circle in two points, it shall out the circle in those points; and the part of the straight line which is between them shall fall within the circle. Let the straight line AB meet the circumference of a circle having the 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 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. 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. 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 D E, bisects the base, that is, DF is equal to FE. (I. 6. Cor. 3.) PROP. 3. (EUC. iii. 3.) If a diameter cut any other chord at right angles, it shall bisect it; and Next let D E be bisected in F by the diameter A B; the angles D FA, É 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. |