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In a given circle to inscribe a regular pertagon.

B

Let CDABE be the given circle; it is required to inscribe a regular pentagon in it.

Make the isosceles triangle ABC such, that each of the angles CAB, CBA may be double the angle ACB (IV. 10.)

Bisect the angles CAB, CBA with the lines AE, BD (I. 9.), and join the points AD, DC, CE, EB; then will ABECD be the pentagon required.

For, since the angles CAB, CBA are each double the angle ACB (by Conft.), and the lines AE, BD bisect them, the angles ACB, CAE, EAB, ABD and DBC are all equal to each other.

And fince equal angles stand upon equal circumferences (III. 21.), the arcs CD, DA, AB, BE and ec are also equal to each other.

But equal arcs are fubtended by equal chords (III.22.); consequently the sides CD, DA, AB, BE and Ec are, likewise, equal.

The figure ABCD is, therefore, equilateral : and it is also equiangular.

For, since the arc co is equal to the arc be, to each of them add DAB, and the are CDAB will be equal to the

are DABE.

But equal angles are subtended by equal arcs (III. 21.), whence the angle ces is equal to the angle dce.

And, in the same manner, it may be shewn, that each of the angles CDA, DAB, ABE are equal to the angle cel or DCE.

The pentagon ABECD, therefore, is both equilateral and equiangular; and it is inscribed in the given circle, as was to be done.

PROP. XII.

About a given circle to describe a regular pentagon.

H

E

А

Let ABCDE be the given circle; it is required to circumscribe it with a regular pentagon.

Inscribe the regular pentagon DeABC (IV. 11.), and through the points A, B, C, D, E, draw the tangents FG, GH, HK, KL and LF; also join the points 0, A, 0, 0, 0, C, 0,0 and 0, E.

Then, since the angles o EF, OAF are right angles (III. 12.), the angles oe A, CAE, taken together, are less than two right angles; whence the lines LF, FG will meet each other (I. 25. Cor.)

And, in the same manner, it may be proved, that all the other lines FG, GH, HK, KL and LF will meet each other.

And, fince oE, OA and oB are all equal to each other, and EA is equal to AB, the angles OEA, OAE, OAB and OBA will be all equal to each other (I.7.)

But the angles at the points E, A, B are also equal, being each of them right angles (III. 12.); consequently the angles AEF, EAF, BAG and ABG are likewise equal ; and the angle F equal to the angle G (I. 21.)

And, in the same manner, it may be shewn, that the angles G, H, K, L and F are all equal to each other.

Since, therefore, the triangles EFA, AGB, &c. are equiangular, and have their bases EA, AB, &c. equal to each other, the remaining fides EF, FA, AG, GB, &c. will also be equal (I. 21.):

And since LF, FG, &c. are the doubles of ET, FA, &c. the figure FGHKL is a regular pentagon; and it is circumscribed about the circle ABCDE, as was to be done.

PRO P. XIII. PROBLEM.

In a given regular pentagon to inscribe a circle.

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Let ABCDE be the given regular pentagon; it is required to inscribe a circle in it.

Bifect any two angles BCD, CDE with the right lines €0, OD (I. 9.), and the point of intersection o will be the centre of the circle required. к

For

For draw the lines os, on and oe, and let fall the perpendiculars oH, OK, OL, OF and oG (I. 12.):

Then, because cb is equal to cd (by Hyp.), co como mon, and the angle eco equal to the angle ocd (by Conft.), the angle ceo will also be equal to the angle odc (1.7.)

But the angle odc is equal to half the angle cde (by Conf.) and the angle cde is equal to the angle CBA (by Hyp.); consequently the angle cbo is also equal to half the angle CBA.

The angle CBA, therefore, is bifected by the line BO; and, in the same manner, it may be shewn, that the angles at the points A, E are bisected, by the lines AO, CE.

Again, because the triangles OGC, och are equiangular, and have oc common to each, the perpendicular og will be equal to the perpendicular oH (I. 21.)

And, in the same manner, it may be shewn, that'oHg OK, OL, of and og are all equal to each other.

If, therefore, a circle be described from the centre o, at either of these distances, it will pass through the remaining points, and be inscribed in the pentagon ABCDE, as was to be done.

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PRO P. XIV. PROBLEM.

To describe a circle about a given regular pentagon.

B

Let ABCDE be a given regular pentagon ; it is required to circumscribe it with a circle.

Bisect any two angles BCD, CDE, with the right lines CO, OD (I. 9.), and the point of intersection o will be the centre of the circle required.

For, draw the lines oB, on and oe:

Then, because cb is equal to cd (by Hyp.), co common, and the angle eco equal to the angle ocd (by Conft.), the angle cBo will also be equal to the angle odc (1. 4.)

But the angle odc is equal to half the angle cde, (by Conf.), and the angle cde is equal to the angle CBA (by Hyp.) ; whence the angle ceo is also equal to half the angle CBA.

The angle CBA, therefore, is bifected by the line BO; and, in the fame manner, it may be shewn, that the angles at the points A, E are bisected, by the lines AO, 0E.

Since, therefore, the angle ocd is equal to the angle ODC (by Hyp. and Ax. 7.), the side oc will also be equal to the side od (I. 5.)

And, in the same manner, it may be shewn, that oD, OE, OA, OB and oc are all equal to each other,

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