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 alfo 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 fame manner, it may be fhewn, 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 bafes EA, AB, &c. equal to each other, the remaining fides EF, FA, AG, GB, &c. will also be equal (I. 21.):

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

PRO P. XIII. PROBLEM.

In a given regular pentagon to inscribe a circle.

B

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

Bifect 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. K

For

For draw the lines OB, OA 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 common, and the angle вCO equal to the angle OCD (by Const.), the angle CBO will also be equal to the angle ODC (I.7.)

But the angle ODC is equal to half the angle CDE (by Conft.) and the angle CDE is equal to the angle CBA (by Hyp.); confequently the angle CBO is alfo equal to half the angle CBA.

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

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 fame manner, it may be fhewn, that OH, OK, OL, OF and OG are all equal to each other.

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

PROP. XIV. PROBLEM.

To defcribe a circle about a given regular pentagon.

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

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

For, draw the lines OB, OA and OE:

Then, because CB is equal to CD (by Hyp.), co common, and the angle BCO equal to the angle OCD (by Conft.), the angle CBO will also be equal to the angle ODC (I. 4.)

But the angle ODC is equal to half the angle CDE, (by Conft.), and the angle CDE is equal to the angle CBA (by Hyp.); whence the angle CBO is also equal to half the angle CBA.

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

Since, therefore, the angle OCD is equal to the angle ODC (by Hyp. and Ax. 7.), the fide oc will also be equal to the fide OD (I. 5.)

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

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

PROP. XV. PROBLEM.

In a given circle to infcribe a regular hexagon.

B

Let ACEF be a given circle; it is required to inscribe a regular hexagon in it.

Through the centre o draw the diameter AD, and make DC equal to DO (IV. 1.), and it will be the fide of the hexagon required.

For, draw the diameter CF, and make BE parallel to CD (I.27.); and join the points DE, EF, FA, AB and BC:

Then, fince Doc is an equilateral triangle, the angles ODC, OCD and DOC will be all equal to each other (I. 5. Cor.)

And, becaufe OE is parallel to CD, the angle EOD will be equal to the angle ODC (I. 24.), and the angle FOE to the angle OCD (I. 25.)

But the angles ODC, OCD are each equal to the angle DOC; therefore, the angles DOC, EOD and FOE are all equal to each other; as are alfo the oppofite angles FOA, AOB and BOC.

Since, therefore, the triangles COD, DOE, &c. have two fides, and the included angle of the one equal to two fides and the included angle of the other, they will be equal in all respects (I. 4.)

The fides CD, DE, EF, &c. are therefore all equal to each other, as are alfo the angles BCD, CDE, &c. whence ABCDEF is a regular hexagon; and it is infcribed in the circle ACEF, as was to be done.

SCHOLIUM. Befides the figures here conftructed, and those arising from thence by continual bisections, or taking the differences, no other regular polygon can be described, by any known method, purely geometrical.

It may also be observed that fome of these figures, as well as feveral others, in the former part of the work, may often be described in a much eafier way, for practical purposes; but the principles upon which they depend can only be obtained from the following books of the Elements.