Let BAC be an angle in the segment BAC of the circle ABEC: C B E then according as the segment BAC is less than, equal to, or greater than, a semicircle; so shall the angle BAC be greater than, equal to, or less than, a right angle. Let O be the centre. Join OB, OC, if necessary. Then, according as the segment BAC is less than, equal to, or greater than, a semicircle, the arc BEC is greater than, equal to, or less than, half the circumference of the circle, and therefore the angle BOC, which stands on the arc BEC, is greater than, equal to, or less than two right angles. But the angle BAC at the circumference is half the angle BOC at the centre on the same arc BEC; III. 15. therefore according as the segment BAC is less than, equal to, or greater than, a semicircle, the angle BAC is greater than, equal to, or less than, a right angle. Q.E.D. Ex. 46. The circles described on two of the sides of a triangle as diameters intersect on the base, or the base produced. Ex. 47. The four circles described on the sides of a rhombus as diameters have one point in common. THEOR. 18. A segment is less than, equal to, or greater than, a semicircle, according as the angle in it is greater than, equal to, or less than a right angle. Let BAC be a segment of the circle ABEC, BAC an angle in the segment: then, according as the angle BAC is greater than, equal to, or less than a right angle: so shall the segment BAC be less than, equal to, or greater than, a semicircle. For in the preceding Theorem it has been shewn that if the segment BAC is less than a semicircle, then the angle BAC is greater than a right angle; if the segment BAC is equal to a semicircle, then the angle BAC is equal to a right angle; and if the segment BAC is greater than a semicircle, then the angle BAC is less than a right angle. Now of these hypotheses one must be true, and of the conclusions no two can be true at the same time; hence, by the Rule of Conversion, the converse of each of the above Theorems is true. Q.E.D. DEF. 12. If the angular points of a rectilineal figure be on the circumference of a circle, the figure is said to be inscribed in the circle, and the circle to be circumscribed about the figure. THEOR. 19. The opposite angles of a quadrilateral inscribed in a circle are supplementary. Let ABCD be a quadrilateral inscribed in the circle ABCD : then shall the opposite angles BAD, BCD be supplementary. Let O be the centre of the circle. Join OB, OD. Then the angle BAD at the circumference standing on the arc BCD is half the angle BOD at the centre standing on the same arc BCD, III. 15. and the angle BCD at the circumference standing on the arc BAD is half the angle BOD at the centre standing on the same arc BAD ; therefore the sum of the angles BAD, BCD is half the sum of the two conjugate angles of which OB, OD are the arms; but the sum of two conjugate angles is four right angles, therefore the sum of the angles BAD, BCD is two right angles, that is, the angles BAD, BCD are supplementary. Q.E.D. COR. I. An exterior angle of a quadrilateral inscribed in a circle is equal to the interior opposite angle. COR. 2. If the opposite angles of a quadrilateral are supplementary a circle can be described about the quadrilateral. For if one of the angular points of the quadrilateral did not lie on the circumference of the circle (III. 12) passing through the other three, the angle at that point would not be supplementary to the opposite angle. III. 16, Cor. I. Ex. 48. Prove Theor. 19 by drawing the diagonals of the quadrilateral and shewing that the sum of the angles at the extremities of one diagonal is equal to the sum of the angles of either of the triangles into which the other diagonal divides the quadrilateral. Ex. 49. If a circle can be described about a parallelogram, the parallelogram must be a rectangle. Ex. 50. A circle can be described about every rectangle. Ex. 51. The quadrilateral ABCD is inscribed in a circle, AB, DC produced meet in E, and BC, AD produced meet in F ; prove that, if BEFD can also be inscribed in a circle, AC is a diameter of the first circle and EF of the other. EXERCISES. 52. From the foot of a perpendicular AD on the hypotenuse of a right-angled triangle ABC, perpendiculars DE, DF are drawn to the sides. Prove that B, E, F, C, are concyclic, i.e., lie on the circumference of a circle. 53. AB is a chord of a circle whose centre is C, and DE is the perpendicular let fall on AB from any point D in the circumference. Prove that the angles ADE, BDC are equal. 54. AB is a fixed chord in a circle APQB; PQ another chord of given length. Shew that if AP, BQ meet in R, R will lie in the circumference of a fixed circle for all positions of PQ. 55. AOB, COD are two diameters of the circle ACBD at right angles to each other. Equal lengths OE, OF are taken along OA, OD respectively. Shew that BF produced cuts DE at right angles, and that these two straight lines, when both produced, intercept one-fourth of the circumference. 56. Through P, one of the points of intersection of two circles APB, APC, a straight line BC is drawn at AP; BA, CA meet the circles again in Q and R. bisects the angle QPR. right angles to Shew that AP 57. If two opposite sides of a quadrilateral inscribed in a circle be equal, the other two will be parallel. 58. ABCD is a parallelogram: if a circle through A,B cut AD and BC in E and F respectively, prove that another circle can be drawn through E, F,C,D. Find the position of EF when the two circles are equal. 59. The circumscribing circles of the two triangles spoken of in Book I., Theor. 20, are in all cases equal to each other. 60. The middle points of all chords of a given circle which pass through a fixed point lie on the circumference of a certain circle. 61. Through one common point of two intersecting |