Scholium. The angle of a regular polygon is determined by the number of its sides. (29. 1.)

Any regular polygon may be inccribed in a circle, and circumscribed about one.

Let ABCDE &c., be a regular poylgon describe a circle through the three points A, B, C, (7. 2,) the centre being O, and OP the dicular let fall from it to the middle H

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point of BC join AO and OD.

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If the quadrilateral OPCD be pla

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ced upon the quadrilateral OPBA,

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they will coincide; for the side OP is common: the angle OPC=OPB, being right-angles; 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 through the vertices of all the angles in the polygon, which is therefore inscribed in this circle. (Def. 9. 2.)

Again, in reference to this circle, all the sides AB, BC, CD, &c., are equal chords; they are therefore equally distant from the centre; (8. 2:) 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. 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, (see the diagram of Prop. 4,) the chords AB, BC, CD, &c., will also be equal; hence likewise the triangles ABO, BOC, COD must be equal, because they are equiangular; therefore, all the angles ABC, BCD, CDE, &c., will be equal; consequently the figure ABCDE, &c., will be a regular polygon.

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 chords AB, BC, &c., are also equal; and the angles ABC, BCD, &c., being in semicircles, are right-angles.

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Scholium. Since the triangle BOC is right-angled and isosceles, we have BC: BO:: √2: 1; (11. 4. Cor. 4;) hence, The side of the inscribed square is to the radius, as the square root of 2 is to unity.

PROPOSITION IV. PROBLEM.

In a given circle, to inscribe a regular hexagon and an equilateral triangle.

Suppose the problem solved, (2. 5. Sch.,) 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 other two angles ABO, BAO of the same triangle, are together equal to 2-}=}; and being mutually equal, each of them must be equal to ; therefore the triangle ABO is equilateral; consequently, the side of the inscribed hexagon is equal to the radius. 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 we set out from.

And the hexagon ABCDEF being inscribed, the equilateral triangle ACE may be formed by joining the vertices of the alternate angles.

Cor. The regular inscribed polygon ACE of three sides is less than the regular inscribed polygon ABCDEF of six sides. For, if ACE be taken from ABCDEF, there will remain the three triangles ABC, CDE, EFA. In like manner it may be shown that any regular inscribed polygon of any number of sides, is less than one of twice that number of sides; and consequently, the greater the number of sides of a regular inscribed polygon, the nearer its area approaches to the area of the circle.

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 equal to the sum of the squares of the sides; that is, to 4AB2, or 4BO2; (14.4.

Cor.;) and taking away BO2 from both, there will remain AC-3BO; hence, (6.3. Sch.,) AC: BO:: 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.

In a given circle, to inscribe a regular decagon; then a pentagon, and a pentedecagon.

AB=OM, AO:AB::AB: AM; and since the triangles ABO, AMB have the common angle A, included between proportional sides, they are similar. (20. 4.) Now the triangle OAB being isosceles, AMB must be isosceles also, and AB=BM; besides AB=OM; hence also MB=OM; therefore the triangle BMO is isosceles.

Now, the angle AMB being exterior to the isosceles triangle BMO, is double of the interior angle O: (28. 1. Cor. 6:) but the angle AMB MAB; hence the triangle OAB is such, that each of its angles at its base, OAB or OBA, 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.

Again, by joining the alternate corners of the regular decagon, the pentagon ACEGI will be formed, also regular.

Moreover, AB being still the side of the decagon, let AL be the side of the hexagon; the arc BL will then, with reference to the whole circumference, be- or ; hence the chord BL will be the side of the pentedecagon, or regular polygon of fifteen sides. 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, the square may enable us successively to inscribe 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 pentadecagon, polygons of 30, 60,120, &c., sides.

A regular inscribed polygon ABCD, &c., being given, to circumscribe a similar polygon about the same circle.

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