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PROP. XV. PROB.

In and about a given circle, to inscribe and circumscribe a square.

Let EADB be a circle in which it is required to inscribe a square. Draw the diameter AB, the perpendicular ED through the centre, andjoin AD, DB, BE, and EA: The inscribed figure ADBE is a square. The angles about the centre C, being right angles, are equal to each other, and are, therefore, subtended by equal chords AD, DB, BE, and AE, but - - - G. T) T-I one of the angles ADB, being in a se- . /2 N micircle, is (III. 19.) a right angle, s N and consequently ADBE is a square. Ak-e N • Next, let it be required to circum- N 2% scribe a square about the circle. F J. I Apply tangents FG, GH, HI, and FI at the extremities of the perpendicular diameters: These will form a square. For all the angles of the quadrilateral figure CG, being together equal to four right angles, and those at C, A, and D being each a right angle, the remaining angle at G is also a right angle, CG is a rectangle; and AC being equal to CD, it is likewise a square. In the same manner, CH, CI, and CF are proved to be squares; the sides FG, GH, HI, and IF of the exterior figure, being therefore the doubles of equal lines, are mutually equal, and the angle at G being a right angle, FH is consequently a square. Cor. Hence the circumscribing square is double of the inscribed square, and this again is double of the square described on the radius of the circle.

PROP. xvi. PROB.

In and about a given circle, to inscribe and circumscribe a regular pentagon.

Let ABCDE be a circle in which it is required to inscribe a regular pentagon. Construct an isosceles triangle having each of its angles at the base double of its vertical angle (IV. 4.), and equiangular to this, inscribe the triangle ACE within the circle (IV. 11.), draw AD, EB bisecting the angles CAE, CEA (I. 5.), and join AB, BC, CD, and DE: The figure ABCDE is a regular pentagon. For the angles AEB, BEC are each the half of CEA, and therefore equal to ACE; but the angles EAD, DAC are likewise equal to ACE. Hence these angles, being all equal, must stand on equal arcs (III. 16. cor.); and the chords of these arcs, or the sides AB, BC, CD, DE, and AE are equal (III. 12, cor.). And because the segments EAB, ABC, BCD, CDE, and DEA are evidently equal, (III. 16.), the interior angles of the figure are all equal, and it is, therefore, a regular pentagon. Next, let it be required to circumscribe a regular pentagon about the circle. At the points A, B, C, D, and E apply tangents; these will form a regular pentagon. For FAK being a tangent, the angle KAE is equal to ACE (III. 21.); and in like manner it is shown that the angles AEK, DEL, EDI, CDH, DCH, BCG, CBG,

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ABF, BAF are all equal to ACE. The isosceles triangles AKE, BFA, having, therefore, the angles at the base equal and the bases themselves AE, AB,-are equal - (I. 20.); for the same reason, the triangles BGC, CHD, DIE, EKA, are equal. Whence the internal angles of the figure are equal, and its sides, being double of those of the annexed triangles, are likewise equal: The figure is, therefore, a regular pentagon.

PROP. XVII. PROB. To inscribe a regular hexagon in a given circle.

Letit be required, in the circle FBD, to inscribe a hexagon.

Draw the radius OA, on which construct the equilateral triangle ABO (I. 1. cor.), and repeat the equal triangles about the vertex O: These triangles will compose a hexagon.

For the triangle ABO, being equilateral, each of its angles AOB is the third part of two right angles; and consequently six of such angles may be placed about the centre O. But the bases of the triangles AOB, BOC, COD, DOE, EOF, and FOA form the sides of the figure, and the angles at those bases its internal angles ; wherefore it is a regular hexagon.

Cor. 1. Tangents applied at the points A, B, C, D, E, F, would evidently form a regular circumscribing hexagon.—An equilateral triangle might be inscribed by joining the alternate points; and, by applying tangents at

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those points, an equilateral triangle would be made to circumscribe the circle. Cor. 2. The side AB of the inscribed hexagon is equal to the radius; and since ABD is a right-angled triangle, and the squares of AB and BD are equal to the square of AD or to four times the square of AO, the square of BD the side of an inscribed equilateral triangle is triple the square of the radius. Cor. 3. The perimeter of the inscribed hexagon is equal to six times the radius, or three times the diameter, of the circle. Hence the circumference of a circle being, from its perpetual curvature, greater than any intermediate system of straight lines, is more than triple its diameter.

PROP. XVIII. PROB. To inscribe a regular decagon in a given circle.

Let ADH be a circle, in which it is required to inscribe a regular decagon.

Draw the radius OA, and with OA as its side describe the isosceles triangle AOB, having each of its angles at the base double of its vertical angle (IV. 4.), repeat the equal triangles about the centre O: These triangles will composé a decagon.

For the vertical angle AOB of the component isosceles triangle, is the fifth part of two right angles (IV. 4. cor.), and consequently ten such angles can be placed about the point O. But the sides and angles of the resulting figure are all evidently equal; it is, therefore, a regular decagon.

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Cor. Hence a regular pentagon will be formed, by joining the alternate points A, C, E, G, I, and A. It is also manifest, that a decagon and a pentagon may be circumscribed about the circle, by applying tangents at their several angular points,

PROP. XIX. THEOR.

The square of the side of a pentagon inscribed in a circle, is equivalent to the squares of the sides of the inscribed hexagon and decagon.

Let ABCDEF be half of a decagon inscribed in a circle whose diameter is AF; the square of AC, the side of an inscribed pentagon, is equivalent to the square of AB the side of the inscribed decagon, and of the square of the radius AO, or the side of an inscribed hexagon.

For join AD, and draw OB, OC, and OD. Since the arc DEF is double of AB, the angle AOB at the centre is (III. 15.) evidently equal to OAD or OAG at the circumference; and because the arc BCDEF again is double of DEF, the angle OAB at the circumference is likewise equal to AOG at the centre. Whence the triangles AOB and AGO, having the an- gles OAB and AOB equal to AOG and OAG, and the interjacent side AO common, are equal (I.20.), and therefore the base AB is equal to OG. Conse- quently, (IV. 18.) GAO is an isosceles triangle having each of the angles at its base double the vertical angle;

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