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PROPOSITION II. THEOREM.
Any regular polygon may be inscribed in a circle, and circumscribed about one.
Let ABCDE &c. be a regular polygon describe a circle through the three points A, B, C, the centre being O, and OP the perpendicular let fall from it, to the middle point of BC: draw AO and OD.
If the quadrilateral OPCD be placed upon the quadrilateral OPBA, they will coincide; for the side OP is common; the angle OPC=OPB, each being a right angle; hence the side PC will apply to its equal PB, and the point C will fall on B: besides, from the nature of the polygon, the angle PCD= PBA; hence CD will take the direction BA; and since CD= BA, the point D will fall on A, and the two quadrilaterals will entirely coincide. The distance OD is therefore equal to AO; and consequently the circle which passes through the three points A, B, C, will also pass through the point D. By the same mode of reasoning, it might be shown, that the circle which passes through the three points B, C, D, will also pass through the point E; and so of all the rest: hence the circle which passes through the points A, B, C, passes also through the vertices of all the angles in the polygon, which is therefore inscribed in this circle.
Again, in reference to this circle, all the sides AB, BC, CD, &c. are equal chords; they are therefore equally distant from the centre (Book III. Prop. VIII.): hence, if from the point O with the distance OP, a circle be described, it will touch the side BC, and all the other sides of the polygon, each in its middle point, and the circle will be inscribed in the polygon, or the polygon described about the circle.
Scholium 1. The point O, the common centre of the in scribed and circumscribed circles, may also be regarded as the centre of the polygon; and upon this principle the angle AOB is called the angle at the centre, being formed by two radii drawn to the extremities of the same side AB.
Since all the chords AB, BC, CD, &c. are equal, all the angles at the centre must evidently be equal likewise; and therefore the value of each will be found by dividing four right an gles by the number of sides of the polygon.
Scholium 2. To inscribe a regular polygon of a certain number of sides in a given circle, we have only to divide the circumference into as many equal parts as the polygon has sides: for the arcs being equal, the chords AB, BC, CD, &c. will also be equal; hence likewise the triangles AOB, BOC, COD, must be equal, because the sides are equal each to each; hence all the angles ABC, BCD, CDE, &c. will be equal; hence the figure ABCDEH, will be a regular polygon.
PROPOSITION III. PROBLEM.
To inscribe a square in a given circle.
Draw two diameters AC, BD, cutting each other at right angles; join their extremities A, B, C, D: the figure ABCD will be a square. For the angles AOB, BOC, &c. being equal, the A chords AB, BC, &c. are also equal: and the angles ABC, BCD, &c. being in semicircles, are right angles.
PROPOSITION IV. PROBLEM.
Scholium. Since the triangle BCO is right angled and isosceles, we have BC BO: √2: 1 (Book IV. Prop. XI. Cor. 4.); hence the side of the inscribed square is to the radius, as the square root of 2, is to unity.
In a given circle, to inscribe a regular hexagon and an equilateral triangle.
Suppose the problem solved, and that AB is a side of the inscribed hexagon; the radii AO, OB being drawn, the triangle AOB will be equilateral.
For, the angle AOB is the sixth part of four right angles; therefore, taking the right angle for unity, we shall have AOB==
and the two other angles ABO, BAO, of the same triangle, are together equal to 2-3 4 and being mutually equal,
each of them must be equal to ; hence the triangle ABO is equilateral; therefore the side of the inscribed hexagon is equal to the radi's.
Hence to inscribe a regular hexagon in a given circle, the radius must be applied six times to the circumference; which will bring us round to the point of beginning.
And the hexagon ABCDEF being inscribed, the equilateral triangle ACE may be formed by joining the vertices of the alternate angles.
Scholium. The figure ABCO is a parallelogram and even a rhombus, since AB=BC=CO=AO; hence the sum of the squares of the diagonals AC2+ BO2 is equivalent to the sum of the squares of the sides, that is, to 4AB2, or 4BO2 (Book IV. Prop XIV. Cor.): and taking away BO from both, there will remain AC2-3BO2; hence AC2 : BO2 :: 3 : 1, or AC : BO :: √3: 1; hence the side of the inscribed equilateral triangle is 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 a pentagon, and also a regular polygon of fifteen sides.
Divide the radius AO in extreme and mean ratio at the point M (Book IV. Prob. IV.); take the chord AB equal to OM the greater segment; AB will be the side of the regular decagon, and will require to be applied ten times to the circumference.
For, drawing MB, we have by construction, AO: OM :: OM: AM; or, since AB =OM, AO: AB :: AB : AM; since the triangles ABO, AMB, have a common angle A, included between proportional sides, they are similar (Book IV. Prop. XX.). Now the 'riangle OAB being isosceles, AMB must be isosceles also, and AB=BM; but AB=OM; hence also MB=OM; hence the triangle BMO is isosceles.
Again, the angle AMB being exterior to the isosceles triangle BMO, is double of the interior angle O (Book I. Prop. XXV. Cor. 6.): but the angle AMB MAB; hence the triangle OAB is such, that each of the angles OAB or OBA, at its base, is double of O, the angle at its vertex; hence the three angles of the triangle are together equal to five times the angle O, which consequently is the fifth part of the two right angles, or the tenth part of four; hence the arc AB is the tenth part of the circumference, and the chord AB is the side of the regular decagon.
2d. By joining the alternate corners of the regular decagon, the pentagon ACEGI will be formed, also regular.
3d. AB being still the side of the decagon, let AL be the side of a hexagon; the arc BL will then, with reference to the whole circumference, be —, or 1; hence the chord BI. will be the side of the regular polygon of fifteen sides, or pentedecagon. It is evident also, that the arc CL is the third of CB.
Scholium. Any regular polygon being inscribed, if the arcs subtended by its sides be severally bisected, the chords of those semi-arcs will form a new regular polygon of double the number of sides: thus it is plain, that the square will enable us to inscribe successively regular polygons of 8, 16, 32, &c. sides. And in like manner, by means of the hexagon, regular polygons of 12, 24, 48, &c. sides may be inscribed; by means of the decagon, polygons of 20, 40, 80, &c. sides; by means of the pentedecagon, polygons of 30, 60, 120, &c. sides.
It is further evident, that any of the inscribed polygons will be less than the inscribed polygon of double the number of sides, since a part is less than the whole.
PROPOSITION VI. PROBLEM.
A regular inscribed polygon being given, to circumscribe a sım ilar polygon about the same circle.
Let CBAFED be a regular polygon. At T, the middle point of the arc AB, apply the tangent GH, which will be parallel to AB (Book III. Prop. X.); do the same at the middle point of each of the arcs BC, CD, &c.; these tangents, I by their intersections, will form the regular circumscribed polygon GHIK &c. similar to the one inscribed.
Since T is the middle point of the arc BTA, and N the middle point of the equal arc BNC, it follows, that BT-BN; or that the vertex B of the inscribed polygon, is at the middle point of the arc NBT. Draw OH. The line OH will pass through the point B.
For, the right angled triangles OTH, OHN, having the common hypothenuse OH, and the side OT-ON, must be equal (Book I. Prop. XVII.), and consequently the angle TOĤ= HON, wherefore the line OH passes through the middle point B of the arc TN. For a like reason, the point I is in the longation of OC; and so with the rest.
But, since GH is parallel to AB, and HI to BC, the angle GHI ABC (Book I. Prop. XXIV.); in like manner HIKBCD; and so with all the rest: hence the angles of the cir cumscribed polygon are equal to those of the inscribed one. And further, by reason of these same parallels, we have GH : AB :: OH: OB, and HI : BC :: OH: OB; therefore GH · AB: HI: BC. But AB=BC, therefore GH-HI. For the same reason, HI=IK, &c.; hence the sides of the circumscribed polygon are all equal; hence this polygon is regular and similar to the inscribed one.
Cor. 1. Reciprocally, if the circumscribed polygon GHIK &c. were given, and the inscribed one ABC &c. were required to be deduced from it, it would only be necessary to