Elements of geometry and trigonometry . iangle ABO isequilateral ; therefore the side of the inscribed hexagon is equalto the radius. Hence to inscribe a regular hexagon in a given circle, theradius must be applied six times to the circumierence ; whichwill bring us round to the point of beginning. And the hexagon ABCDEF being inscribed, the equilateraltriangle ACE may be formed by joining the vertices of thealternate angles. Scholium. The figure ABCO is a parallelogram and evena rhombus, since AB=BC=CO=AO ; hence the sum of thesquares of the diagonals AC~ + BO^ is equivalent to the sum ofthe
Elements of geometry and trigonometry . iangle ABO isequilateral ; therefore the side of the inscribed hexagon is equalto the radius. Hence to inscribe a regular hexagon in a given circle, theradius must be applied six times to the circumierence ; whichwill bring us round to the point of beginning. And the hexagon ABCDEF being inscribed, the equilateraltriangle ACE may be formed by joining the vertices of thealternate angles. Scholium. The figure ABCO is a parallelogram and evena rhombus, since AB=BC=CO=AO ; hence the sum of thesquares of the diagonals AC~ + BO^ is equivalent to the sum ofthe squares of the sides, that is, to 4AB-, or 4B0^ (Book XIV. Cor.) : and taking away BO- from both, there willremain AC2=3BO^ hence AC^ : BO- : : 3 : 1, or AC : BO: : \^3 : 1 ; hence the side of the inscribed equilateral triangleis to the radius as the square root of three is to unity. PROPOSITION V. PROBLEM. In a given circle, to inscribe a regular decagon; then apcntafi^n^and also a regular polygon of fifteen sides. LOOK V, 113. DiMile the raJius AO inextreme and mean ratio atthe point M (Book IV. Prob.ÎV.) ; take the eliord AB equalto OM the greater segment ;AB will be the side of theregular decagon, and will re-quire to be applied ten timesto the circumference. For, drawing MB, we haveby construction, AO : OM: : OM : AiM ; or, since AB= 0M, AO : AB : : AB :AM ; since the triangles ABO, AMB, have a common angle A,included between proportional sides, they are similar (BookIV. Prop. XX.). Now the viangle OAB being isosceles, AMBmust be isosceles also, and AB = BM ; but AB = OM ; hencealso MBnOM ; hence the triangle B^^iO is isosceles. Again, the angle A^IB being exterior to the isosceles trian-gle BMO, is double of the interior angle O (Book I. Cor. 6.) : but the angle AMB=MAB ; hence the trian-gle OAB is such, that each of the angles OAB or OBA, at itsBase, is double of O, tlie angle at its vertex ; hence the threeangles of the triangle arc together equal to
Size: 1567px × 1595px
Photo credit: © The Reading Room / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bo, bookcentury1800, booksubjectgeometry, booksubjecttrigonometry
