squares of BF and FC are equivalent to the squares of DG and GC. And since AB is greater than DE, its half BF is greater than DG, and consequently the square of BF is greater than the square of DG; the square of FC is, therefore, less than the square of GC, because, when joined to the squares of BF and DG, they produce the same amount, or the square of the radius of the circle. Hence the perpendicular FC itself is less than GC. Again, if the chord AB be nearer the centre than DE, it is also greater. For the same construction remaining: It is proved that the squares of BF and FC are together equal to the squares of DG and GC; but FC being less than GC, the square of FC is less than the square of GC, and consequently the square of BF is greater than the square of DG; whence the side BF is greater than DG, and its double or the chord AB greater than DE. PROP. XIV. THEOR. Circles are equal which have equal diameters. Let ABC and DEF be two circles of equal diameters, or described with the same distance GA or HD: they are equal. For if the circle ABC be applied to DEF, the centre G being laid on H, these circles must coincide; because, the radius or semidiameter GA being equal to HD, every point A of the circumference ABC must, after the superposition of the surfaces, find a corre sponding point D of the circumference DEF. F Cor. It is also manifest that, conversely, equal circles must have equal diameters. PROP. XV. 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 is equal to DGE. 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 D E 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. Cor. Hence, in the same or equal circles, equal arcs are subtended by equal chords, and terminate equal angles at the centre. PROP. XVI. THEOR. In the same or equal circles, equal chords subtend equal arcs of a like kind. If the chord AB be different from the diameter, it will evidently subtend at the same time two unequal portions of the circumference of a circle, the one terminating the angle ACB at the centre and less than the semicircumference, the other greater than this and terminating the reversed angle. For join CA, CB, and FD, FE. The two triangles CAB and FDE, having all the sides of the one equal to those of the other, are equal (I. 2.); and consequently the angle ACB is equal to DFE. Wherefore the arcs AGB and DIE, which terminate these equal angles, are (III. 15.) themselves equal; and hence the remaining portions AHB and DKE of the equal circumferences are likewise equal. This demonstration, it is evident, will likewise apply in the case of the same circle. PROP. XVII. PROB. To bisect an arc of a circle. Let it be required to divide the arc AEB into two equal portions. Draw the chord AB, and bisect it by the perpendicular EF (I. 7.), cutting AB in E: The arc AE is equal to EB. E D B 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. 16.), 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 EF. PROP. PROP. XVIII. PROB. Given an arc, to complete its circle. Let ADB be an arc; it is required to trace 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 AC is also equal to CD (I. 9.) because the angle DAC was made equal to ADC. Wherefore (III. 9. cor.) the three straight lines CA, CD, and CB being all equal, the point C is the centre of the circle. PROP. XIX. 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 terminates at the centre, is double of ADB the corresponding angle at the circumference. For join DC and produce it to the opposite circumference. 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. 8.); but the exterior angle ACB is equal to both of these (I. 34.) and therefore equal to double of either, or the angle ACB at the centre is double of the angle ADB at the circumference. Next, let the straight line DCE lie within the angle ADB. From what has been demonstrated, 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 D C D ADE, BDE, or the angle ADB at the circumference, B B |