Let C be the centre of a circle, and A a different point, from which two straight lines AD and AE are drawn to the circumference; of these lines, AD, which lies nearer to B the opposite extremity of the diameter, is greater than AE. For the triangles ADC and AEC have the side CD equal to CE, the side CA common to both, but the contained angle DCA greater than ECA ; wherefore (I. 18.) the base AD is likewise greater than the base AE. Cor. 1. Hence the straight line ACB, which passes through the centre, is the greatest of all those that can be drawn to the circum ference of the circle from the ec- centric point A. For it is evident from the Proposition, that the nearer the point D approaches to B, the greater is AD; consequently the point B forms the extreme limit of majority, or AB is the greatest line that can be drawn from A to the circumference. Cor. 2. Hence also, whether the eccentric point be within or without the circle, the straight line AH is the shortest that can be drawn from A to the circumference. For AE is less than AD, and AG less than AF; and the nearer the terminating point approaches to H, which is obviously the most remote from B, the shorter must be its distance from A. Wherefore the point H marks the limit of minority, and AH is the shortest line that can be drawn from A to the circumference of the circle, PROP. VIII. THEOR. From any eccentric point, not more than two equal straight lines can be drawn to the circumference, one on each side of the diameter. Let A be a point which is not the centre of the circle, and AD a straight line drawn from it to the circumference. Find the centre C (III. 5. cor.) join CA and CD, draw (I. 4.) CE making an angle ACE equal to ACD and cutting the circumference in E, and join AE: The straight lines AE, AD are equal. For the triangles ADC, AEC, having the side CD equal to CE, the side AC common, and the con- tained angle ACD equal to ACE, are equal (I. 3.), and consequently the base AD is equal to AE. But, except AE, no straight line can be drawn from A on the same side of the diameter HB, that shall be equal to AD: For if the line terminate in a point F between E and B, it will be greater than AE (III.7.); and if the line terminate in G between E and H, it will, for the same reason, be less than AE. Cor. 1. That point from which more than two equal straight lines can be drawn to the circumference, is the cen tre of the circle. - Cor. 2. Hence a circle will not cut another in more than two points. PROP. IX. THEOR. A circle may be described through three points which are not in the same straight line. Let A, B, C, be three points not lying in the same direction; the circumference of a circle may be made to pass through them. For (I. 7.) bisect AB by the perpendicular DF, and BC by the perpendicular EF. These straight lines DF, EF will meet; because, DE being joined, the angles EDF, DEF are less than BDF, BEF, and consequently are together less than two right angles, and DF, EF are not parallel (I. 22.), but concur to form a triangle whose vertex is F. Again, every circle that passes through the two points A and B, has its centre in the perpendicular DF (III.5.); and, for the same reason, every circle that passes through B and C has its centre in EF; consequently the circle which would pass through all the three points, must have its centre in F, the point common to both perpendiculars DF and EF. It is farther manifest, that there is only one circle which can be made to pass through the three points A, B, C ; for the intersection of the straight lines DF and EF, which marks the centre, is a single point. Cor. Hence the mode of describing a circle about a given triangle ABC. PROP. X. THEOR. Equal chords are equidistant from the centre of a circle; and chords which are equidistant from the centre, are likewise equal. Let AB, DE be equal chords inflected within the same circle; their distances from the centre, or the perpendiculars CF, CG, let fall-upon them, are equal. For the perpendiculars CF and CG bisect the chords AB and DE (III. 4.), and consequently BF, DG, the halves of these, are likewise equal. The right-angled triangles CBF and CDG, which are thus of the same character, having the two sides BC, BF equal respectively to DC, DG, and the corresponding angle BFC equal to DGC, are equal (I. 21.), and consequently the side FC is equal to GC. Again, if the chords AB, DE be equally distant from the centre, they are themselves equal. For the same construction remaining : The triangles CBF and CDG are still right-angled, or of the same character, and have now the two sides CB, CF equal to CD, CG, and the angle BFC equal to DGC; consequently they are equal, and the side BF equal to DG ; the doubles of these, therefore, or the whole chords AB, DE, are equal. 3 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. 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 G.C. 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 GCitself 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 |