describing the semi-circle on the other side of OA meeting the given circle in C, and then joining AC. 57. PROP. IX. If the radius of one circle be equal to the radius of another, the circles shall be equal in all respects. For, if one of the circles be 'applied to', or laid upon, the other so that their centres coincide, since the radii are equal, it is plain that the circumferences will coincide throughout; and, since the circumferences coincide in every part, it is evident that they enclose the same area, that is, the circles are equal to one another. 58. PROP. X. In the same circle, or in equal circles, equal arcs have equal chords; and conversely, equal chords have equal arcs. Let AB, CD be equal arcs of two equal circles; draw the chords AB, CD; then chord AB = chord CD. For, find the centres E, F of the circles, and suppose the one circle to be laid upon the other, so that the centre E shall be upon F; then since the radii are equal, the whole circumference of one will coincide with the whole circumference of the other; and, without altering this coincidence, if one of them be turned round the centre, in its own plane, until the point A coincides with the point C, the point B will coincide with D, because the arc AB arc CD. So then, since the point A falls upon C, and B upon D, the straight line joining A and B must coincide with that joining C and D; that is, chord AB= chord CD. = And what is proved of equal circles will obviously hold true for equal arcs of the same circle. And the converse also evidently follows, viz. that equal chords in the same circle, or in equal circles, subtend equal arcs. 59. PROP. XI. In the same circle, or in equal circles, equal arcs subtend equal angles at the centre. Let AB, CD be equal arcs of two equal circles, whose centres are E and F. Join AE, BE, CF, DF; then L AEB = = < CFD. D For, joining the chords AB, CD, by (58) chord AB= chord CD; also AE= CF, and BE=DF'; .. the two triangles AEB, CFD, have all the sides of the one equal to all the sides of the other, each to each, and .. the triangles are equal in all respects (23 Cor.). Consequently AEB = CDF, being the angles opposite to the equal sides AB, CD. 60. PROP. XII. To bisect a given arc of a circle. Let ABC be the given arc. It is required to divide it into two parts in the point B, so that the arc AB = arc BC. Join AC; bisect AC in D; from D draw DB at right A D angles to AC, intersecting the given arc in B. Then ABC is bisected in B. For, drawing the chords AB, BC, the two sides AD, DB, in the triangle ADB, are equal to the two sides CD, DB, in the triangle CDB; also 4 ADB = ▲ CDB, since each of them is a right angle, .. the third side AB = CB. But equal chords subtend equal arcs, (58), .. arc AB =arc BC. COR. If BD be produced, it will pass through the centre of the circle. And, conversely, the line drawn from the centre, bisecting the chord, will also bisect the arc. 61. PROP. XIII. Two parallel chords in any circle will intercept equal arcs. Let AB, CD be any two parallel chords in the same circle; the arc AC shall be equal to the arc BD. Find E the centre of the circle; and draw EF perpendi- A cular to AB, meeting the circumference of the circle in F. Then since CD is parallel to AB, EF is also perpendicular to CD (34 Cor. 3); .. both the chords AB, CD are bisected by C D B the straight line EF (49 Cor.): and .. both the arcs AFB, CFD, are bisected in F (60); that is, arc AF = arc BF, and arc CF = arc DF; but if equals be taken from equals the remainders will be equal, .. arc AC arc BD. = Conversely, if arc AC-arc BD, the chord AB is parallel to the chord CD. 62. PROP. XIV. If the distance between the centres of two circles, which are in the same plane, be equal to the sum or difference of their radii, the circles will touch each other at one point only; and the point of contact will be in the straight line which joins the centres, or in that line produced. D 1. Let A, and B be the centres of two circles so situated, in the same plane, that AB, the straight line joining the centres is equal to the sum of their radii. Let C be the point in which AB meets the circumference of the first circle, then AC= D B the radius of that circle; and since AC + BC= the sum of the radii, BC must be the radius of the other circle; and is a point in its circumference; that is, the two circumferences have the point C common to both. And they have no other point common: for, if CD be drawn from C at right angles to AB, since CD is a tangent to both circles at the point C, every point in it, except the point C, is without both, that is, no point but is common to the two, and .. they touch each other in that point. 2. Let AB, the straight line joining the centres of the two circles, be equal to the difference of the radii. Produce AB to meet the circumference of the greater circle, whose centre is A, in C; then since AC is the radius of the greater circle, and AB is the difference of the two radii, BC= the radius of the smaller, and .. C is a point in the circum AB @ ference of the latter; that is, C is a point common to both circumferences. That Cis the only point common to the two circumferences is shewn precisely as in the former case; and .. the circles touch each other at that point. In the former case the circles are said to touch each other externally, in the latter internally. 63. PROP. XV. If a straight line touch a circle, and from the point of contact a chord be drawn dividing the circle into two segments, the angle between the tangent and this chord shall be equal to the angle in the alternate segment** of the circle. E Let the straight line ABC touch the circle BDE in the point B; and let BD be a chord dividing the circle into two segments. From B draw BE at right angles to AB, to meet the circumference again in E, which will .. be a diameter of the circle (55 Cor. 3). Join DE; take any point F in the arc BD, and join BF, df. Then CBD=¿BED 'in the alternate segment'; and ▲ ABD = LBFD. A C By alternate segment is meant the segment on the other side of the chord. For, since BE is a diameter, BDE is a right angle (54), being the angle 'in a semicircle' ; .. ‹ BED + 2 EBD = a right angle (37) = ≤ CBD+ ≤ EBD, :. ¿ CBD=2 BED. Again, since BFDE is a 'quadrilateral inscribed in a circle', ‹ BFD + ▲ BED = two right angles (53) = 4 ABD +≤ CBD; and ≤ CBD has been shewn to be equal to < BED, :. ‹ ABD = ‹ BFD. [It might appear, at first sight, that by drawing BE at right angles to AB, we have proved only a particular case of the proposition; but it is not so, because BED=every other angle in the same segment' (52 Cor).] EXERCISES B. (1) Are all diameters of the same circle equal to one another? Shew that the diameter is greater than any other straight line drawn in the circle and terminated by the circumference. (2) Does the chord of an arc increase as the arc increases? State the limitations. (3) Can a circle be made up of segments? If so, of how many? (4) Can a circle be made up of sectors? If so, of how many? In what case will a sector become a segment? (5) Shew that the circumferences of circles which have the same centre cannot cut each other. (6) If the circumference of a circle be divided into four equal arcs, shew that the chords of any two of them, which are adjacent, are at right angles to each other. (7) If the circumference of a circle be divided into six equal parts, shew that the chord of each of them is equal to the radius. (8) If the radius of a given circle be equal to a given straight line, find the centre of the circle. (9) Make a circle of given radius, whose circumference shall pass through, 1st, one given point, 2ndly, two given points. (10) Can more than one circle be drawn whose circumference shall pass through three given points? (11) Shew that in particular cases a circle may be |