1 8 If the polygon have six or more sides, from any angular point A draw lines A,A,, A,A, (fig. 8) to the third angular point on each side of A1, cutting off quadrilaterals.* Since the number of remaining sides between A, A, is even, a certain number of quadrilaterals may be formed by joining every second angular point from A, with A,. 4 8 All these quadrilaterals are inscribed in the circle; hence, by the proposition, sum of alternate angles A, A, A, &c. = twice as many right angles as there are quadrilaterals 2 (r−1) right angles, if 2r is the number of sides of the polygon; .. sum of alternate angles + 2 right angles = 2r right angles = as many right angles as there are sides. 1850. (A). In a circle, the angle in a semicircle is a right angle. (III. 31.) (B). The greatest rectangle that can be inscribed in a circle is a square. Here (B) is a direct application of the fact asserted in (A). Let ABCD (fig. 9) be any rectangle inscribed in a circle. Join AC. By the proposition, 4ABC is a right angle, ABC is a semicircle. Draw BE perpendicular to AC, and take F the centre. * If the polygon has six sides, the lines A4, A4, will coincide. Then rectangle ABCD = 2 triangle ABC = rectangle on base AC, and between same parallels as the triangle ABC AC.BE. Now AC, being the diameter of the circle, is constant; therefore the area of the rectangle is proportional to BE, and is greatest when BE is greatest, i. e. when it coincides with GF. Hence the greatest rectangle inscribed in the circle is AGCH, which is evidently a square. 1850. (4). Cut off a segment from a given circle which shall contain an angle equal to a given rectilineal angle. (B). Divide a circle into two segments, such that the angle in one of them shall be five times the angle in the other. We may consider (4) as a direct application of the proposition that, "if from a point in a circle two lines be drawn, one touching and the other cutting the circle, the angle between them is equal to the angle in the alternate segment of the circle." (B) may be regarded as a direct application of the same proposition to a slightly different problem. Draw AB touching the circle (fig. 10). Take any point B. in the tangent; and on AB describe an equilateral triangle ABC. Bisect the angle BAC by the line AD. Hence the circle is divided as required by the line AD. 1851. (4). Inscribe an equilateral and equiangular quindecagon in a given circle. (IV. 16.) (B). In a given circle inscribe a triangle, whose angles are as the numbers 2, 5, and 8. 3 89 8 19 2 15 Let A, A, ... A be an equilateral and equiangular quindecagon inscribed in the given circle. Draw the lines AA, Â ̧Ã Â ̧Ã ̧, cutting off arcs which are to one another as 2, 5, 8. The angles AAA, A‚Ã‚А, A.A,A,, which stand upon these arcs, will also be in this ratio; and therefore A,A,A, will be the triangle required. 3 1 89 8 1850. (A). Describe an isosceles triangle, having each of the angles at the base double of the third angle. (IV. 10.) (B). Shew that the base of the triangle is equal to the side of a regular pentagon inscribed in the smaller circle of the figure. In the investigation of (A) it is shewn that BD = DC (fig. 11). If O is the centre of the small circle, ¿COD = 2 ≤CAD. But by (4), the angles ABD, ADB are each double of BAD; therefore the sum of these 3 angles, or 2 right angles, Hence CD, to which BD is equal, is the side of a regular pentagon inscribed in the circle ACD. 1851. (A). If the angle of a triangle be divided into two equal angles by a straight line, which also cuts the base, the segments of the base have the same ratio which the other sides of the triangle have to one another. (VI. 3.) (B). If A, B, C, be three points in a straight line, and D a point at which AB and BC subtend equal angles, shew that the locus of the point D is a circle. Produce AB to a point O, such that OB = OD* (fig. 12). Also the angle AOD is common to the two triangles OCD, ODA. Hence these triangles are similar; .. OD: OC: AD: DC :: AB: BC by the proposition; or, since OD OB, OB: OC :: AB : BC; which shews that O is a fixed point. Hence the locus of D is a circle whose centre is O. 1849. (4). The sides about the equal angles of equiangular triangles are proportionals, and those sides which are opposite to the equal angles are homologous. (VI. 4.) (B). Apply this proposition to prove that the rectangle contained by the segments of any chord passing through a given point within a circle is constant. Let AB, CD (fig. 13) be any two chords of a circle intersecting in O. Join BC, AD. Since the angles in the same segment of a circle are equal, ZABC = <ADC, and BCD = 2 BAD. Hence the triangles BOC, DOA are equiangular. Therefore by (4), Therefore AO DO: OC: OB. : rect. of AO, OB = rect. of DO, OC. * From the middle point of BD draw a line at right angles to it, cutting AB produced in 0. 1850. (A). Find a third proportional to two given straight lines. (VI. 11.) (B). AB is a diameter, and P any point in the circumference of a circle; AP and BP are joined and produced if necessary: if from any point C of AB a perpendicular be drawn to AB, meeting AP and BP in points D and E respectively, and the circumference of the circle in a point F, shew that CD is a third proportional to CE and CF.* Draw PM perpendicular to AB, (fig. 14). From similar triangles CEB, MPB, CE: MP :: CB : MB; and from similar triangles CDA, MPA, 1851. CE: CF:: CF: CD. (A). If two straight lines be parallel, and one of them be at right angles to a plane, the other is at right angles to the same plane. (XI. 8.) (B). From a point E draw EC, ED perpendicular to two planes CAB, DAB, which intersect in AB, and from D draw DF perpendicular to the plane CAB, meeting it in F; shew that the line joining the points C and F, produced if necessary, is perpendicular to AB. Since EC, DF are perpendicular to the same plane, they are parallel (XI. 6), and therefore the points E, C, D, F, lie in one plane. * This can only be considered as one of the third class of riders. |