PROP. XI. THEOR. The greater chord is nearer the centre of the circle; and that chord which is nearer the centre is also the greater. Let the chord DE be greater than AB; its distance from the centre, or the perpendicular CG let fall upon it, is less than the distance CF. B D E For in the right-angled triangle BCF, the square of the hypotenuse BC is equivalent to the squares of BF and FC (II. 10.); and, for the same reason, the square of the hypotenuse DC of the right angled triangle DCG is equivalent to the squares of DG and GC. But the radii BC and DC are equal, and consequently their squares; wherefore the squares of DG and GC are equivalent to the squares of BF and FC. And since DE is greater than AB, its half DG, made by the perpendicular from the centre, is greater than BF, and consequently the square of DG is greater than the square of BF; the square of GC is, therefore, less than the square of FC, because, when conjoined with the squares of DG and BF, they produce the same amount, or the square of the radius of the circle. Hence the perpendicular GC itself is less than FC. Again, if the chord DE be nearer the centre than AB, it is also greater. For the same construction remaining: It has been proved that the squares of BF and FC are together equivalent to the squares of DG and GC; but GC being less than FC, the G square of GC is less than the square of FC, and consequently the square of DG is greater than the square of BF; whence the side DG is greater than BF, and its double, or the chord DE, greater than AB. PROP. XII. THEOR. In the same or equal circles, equal angles at the centre are subtended by equal chords, and terminated by equal arcs. If the angle ACB at the centre C be equal to DCE, the chord AB is equal to DE, and the arc AFB equal to DGE. A E For let the sector ACB be applied to DCE. The centre remaining in its place, the radius CA will lie on CD; and the angle ACB being equal to DCE, the radius CB will adapt itself to CE. And because all the radii are equal, their extreme points A and B must coincide with D and E; wherefore the straight lines which join those points, or the chords AB and DE, must coincide. But the arcs AFB and DGE that connect the same points, will also coincide; for any intermediate point F in the one, being at the same distance from the centre as every point of the other, must, on its application, find always a corresponding point G. The same mode of reasoning is applicable to the case of equal circles. B Cor. 1. Hence, in the same or equal circles, equal arcs are subtended by equal chords, and terminate equal angles at the centre. Cor. 2. Hence also, in the same or equal circles, equal chords must subtend equal arcs of a like kind, that is, arcs which are both greater or both less than a semicircumfe rence. Schol. The length of a chord in a circle is thus insufficient alone to determine the magnitude of the angle which it subtends at the centre. To remove the ambiguity, it is requisite to know, whether this angle be greater or less than two right angles. PROP. XIII. PROB. To bisect a given arc of a circle. Let it be required to divide the arc AEB into two equal portions. Draw the chord AB, and bisect it (I. 7.) by the perpendicular EF cutting the circumference AB in E: The arc AE is equal to EB. A E B D For the triangles ADE, BDE, have the side AD equal to BD, the side DE common, and the containing right angle ADE equal to BDE; they are (I. 3.) consequently equal, and the base AE equal to BE. But these equal chords AE, BE must subtend equal arcs of a like kind (III. 12. cor. 2.), and the arcs AE, BE are evidently each of them less than a semicircumference. F Cor. The correlative arc AFB is also bisected, by the perpendicular EDF at the opposite point F. PROP. XIV. PROB. An arc being given, to complete its circle. Let ADB be an arc; it is required to trace out the circle to which it belongs. Draw the chord AB, and bisect it by the perpendicular CD (I. 7.), cutting the arc in D, join AD, and from A draw AC making an angle DAC equal to ADC (I. 4.): The intersection C of this straight line with the perpendicular, is the centre of the circle required. For join CB. The triangles ACE and BCE, having the side EA equal to EB, the side EC common, and the contained angle AEC equal to BEC, are equal (I. 3.), and consequently AC is equal to BC. But (I. 11.) AC is also equal to CD, because the angle DAC D E C B was made equal to ADC. Wherefore (III. 8. cor. 1.) the three straight lines CA, CD, and CB being all equal, the point C is the centre of the circle. PROP. XV. THEOR. The angle at the centre of a circle is double of the angle which, standing on the same arc, has its vertex in the circumference. Let AB be an arc of a circle; the angle which it termi nates at the centre, is double of ADB the corresponding angle at the circumference. For join DC and produce it to the opposite circumfe rence. This diameter DCE, if it lie not on one of the sides of the angle ADB, must either fall within that angle or without it. First, let DC coincide with DB. And because AC is equal to DC, the angle ADC is equal to DAC (I. 10.); but the exterior angle ACB is equal to both of these (I. 30.), and therefore equal to double of either, or the angle ACB at the centre is double of the angle ADB at the circumference. D B Next, let the straight line DCE lie within the angle ADB. From what has been demon strated, it is apparent, that the angle ACE is double of ADE, and the angle BCE double of BDE; wherefore the angles ACE, BCE taken together, or the whole angle ACB, are double of the collected angles ADE, BDE, or the angle ADB at the circumference. E D B Lastly, let DCE fall without the angle ADB. Because the angle BCE is double of BDE, and the angle ACE is double of ADE; the excess of BCE above ACE, or the angle ACB at the centre, is double of the excess of BDE above ADE, that is, of the angle ADB at the circumference. E B D Cor. Hence if an equal circle be described from any point D in the circumference, its arc intercepted by the |