PROPOSITION 15. PROBLEM. Let ABDF be the given circle. and draw a diameter AGD. Join CG, EG, and produce them to cut the Oce of the given circle at F and B. Join AB, BC, CD, DE, EF, FA. and DG=DE, being radii of the O EHC: .. GE, ED, DG are all equal, and the A EGD is equilateral. Hence the LEGD = one-third of two rt. angles. I. 32. Similarly the < DGC = one-third of two rt. angles. But the 4 EGD, DGC, CGB together=two rt. angles; 1. 13. .. the remaining - CGB=one-third of two rt. angles. ::. the three 4* EGD, DGC, CGB are equal to one another. And to these angles the vert. opp. — BGA, AGF, FGE are respectively equal : .. the <* EGD, DGC, CGB, BGA, AGF, FGE are all equal; .. the arcs ED, DC, CB, BA, AF, FE are all equal: III. 26. ... the chords ED, DC, CB, BA, AF, FE are all equal : III. 29. .. the hexagon is equilateral. Again the arc FA= the arc DE: Proved. to each of these equals add the arc ABCD ; then the arc FABCD = the arc ABCDE: hence the angles at the Oce which stand on these equal arcs are equal. that is, the 2 FED= the 2 AFE. III, 27. In like manner the remaining angles of the hexagon may be shewn to be equal. ... the hexagon is equiangular ; .. the hexagon ABCDEF is regular, and it is inscribed in the ABDF. Q.E.F. COROLLARY. The side of a regular hexagon inscribed in a circle is equal to the radius of the circle. SUMMARY OF THE PROPOSITIONS OF BOOK IV. The following summary will assist the student in remembering the sequence of the Propositions of Book IV. (i) Of the sixteen Propositions of this Book, Props. 1, 10, 15, 16 deal with isolated constructions. (ii) The remaining twelve Propositions may be divided into three groups of four each, as follows: (a) Group 1. Props. 2, 3, 4, 5 deal with triangles and circles. circles. (iii) In each group the problem of inscription precedes the corresponding problem of circumscription. Further, each group deals with the inscription and circumscription of rectilineal figures first and of circles afterwards. To inscribe a regular quindecagon in a given circle. BA E Let ABCD be the given circle. It is required to inscribe a regular quindecagon in the O ABCD. IV Construction. and let AC be one of its sides. and let AB be one of its sides. Proof. the arc AC, which is one-third of the Oce, contains five, and the arc AB, which is one-fifth of the Oce, contains three; .. their difference, the arc BC, contains two. Bisect the arc BC at E: III. 30. then each of the arcs BE, EC is one-fifteenth of the Oce. .:: if BE, EC be joined, and st. lines equal to them be placed sucessively round the circle, a regular quindecagon will be inscribed in it. Q.E.F. EXERCISES ON PROPOSITIONS 11-16.. 1. Express in terms of a right angle the magnitude of an angle of the following regular polygons : (i) a pentagon, (ii) a hexagon, (iii) an octagon, (iv) a decagon, (v) a quindecagon. 2. Any angle of a regular pentagon is trisected by the straight lines which join it to the opposite vertices. 3. In a polygon of ņ sides the straight lines which join any angular point to the vertices not adjacent to it, divide the angle into n - 2 equal parts. 4. Shew how to construct on a given straight line (i) a regular pentagon, (ii) a regular hexagon, (iii) a regular octagon. 5. An equilateral triangle and a regular hexagon are inscribed in a given circle ; shew that (i) the area of the triangle is half that of the hexagon ; (ii) the square on the side of the triangle is three times the square on the side of the hexagon. 6. ABCDE is a regular pentagon, and AC, BE intersect at H: shew that (i) AB=CH=EH. (ii) AB is a tangent to the circle circumscribed about the triangle BHC. (iii) AC and BE cut one another in medial section. 7. The straight lines which join alternate vertices of a regular pentagon intersect so as to form another regular pentagon. 8. The straight lines which join alternate vertices of a regular polygon of n sides, intersect so as to form another regular polygon of n sides. If n=6, shew that the area of the resulting hexagon is one-third of the given hexagon. 9. By means of iv. 16, inscribe in a circle a triangle whose angles are as the numbers 2, 5, 8. 10. Shew that the area of a regular hexagon inscribed in a circle is three-fourths of that of the corresponding circumscribed hexagon. NOTE ON REGULAR POLYGONS. B The following propositions, proved by Euclid for a regular pentagon, hold good for all regular polygons. 1. The bisectors of the angles of any regular polygon are concurrent. Let D, E, A, B, C be consecutive angular D points of a regular polygon of any number of sides. Bisect the _s EAB, ABC by AO, BO, which intersect at O. A For in the As EAO, BAO, and AO is common; Constr. : the L OEA=the L OBA. I. 4. But the L OBA is half the L ABC; Constr. also the 2 ABC=the L DEA, since the polygon is regular ; : the L OEA is half the L DEA: that is, EO bisects the L DEA. Similarly if O be joined to the remaining angular points of the polygon, it may be proved that each joining line bisects the angle to whose vertex it is drawn. That is to say, the bisectors of the angles of the polygon meet at the point O. Q.E.D. COROLLARIES. Since the L EAB=the L ABC; Нур. and since the 28 OAB, OBA are respectively half of the 28 EAB, ABC; :: the L OAB=the L OBA; .. OA=OB. Similarly OE=OA. Hence the bisectors of the angles of a regular polygon are all equal. Therefore a circle described with centre O, and radius OA, will be circumscribed about the polygon. Also it may be shewn, as in Proposition 13, that perpendiculars drawn from O'to the sides of the polygon are all equal. Therefore a circle described with centre O, and any one of these perpendiculars as radius, will be inscribed in the polygon. I. 6. |