PROP. IX.

To circumfcribe a circle about a given fquare.

Let ABCD be the given square; it is required to circumfcribe it with a circle.

Draw the diagonals AC, BD, and the point of interfection E will be the centre of the circle required.

For, fince the fides DA, AC are equal to the fides BA, AC, and the base BC to the base CD, the angle DAC will be equal to the angle CAB (I. 7.); that is, the angle BAD will be bisected by the line AC.

And, in the fame manner, it may be proved, that all the other angles of the fquare are bisected by the lines DB and CA.

But the angles CDA, DAB, being right angles, are equal to each other; whence the angles EDA, EAD are also equal to each other; and confequently the line ED is equal to the line EA (I. 7.)

And, in like manner, it may be fhewn, that the lines EB, EC are each equal to the lines ED, EA; whence the lines EA, EB, EC and ED are all equal to each other.

If, therefore, a circle be described from the point E, at either of the distances EA, EB, EC or ED, it will pafs through the remaining points, and circumfcribe the fquare AC, as was to be done.

PROP. X. PROBLEM.

To infcribe an ifofceles triangle in a given circle, that shall have each of the angles at its base double the angle at the vertex.

Let ABC be the given circle; it is required to infcribe an isofceles triangle in it, that shall have each of the angles at its base double the angle at the vertex.

Draw any diameter CE, and divide the radius DE in the point r'fo, that the rectangle of DE, EF may be equal to the fquare of FD (II. 22.)

From the point E apply the right lines EA, EB each equal to FD (IV. 1.), and join AB, AC, CB; then will ABC be the triangle required.

For, through the points D, F, B defcribe the circle BDF (III. 18.), and draw the lines BD, BF.

'Then, fince the rectangle DE, EF is equal to the square of FD, or its equal EB, the line EB. will touch the circle BDF, at the point в (III. 30.)

And, because EB is a tangent to the circle, and BF is a chord drawn from the point of contact, the angle EBF will be equal to the angle FDB in the alternate segment (III. 24.)

And

And if, to each of thefe angles, there be added the angle FBD, the whole angle DBE or FEB will be equal to the angles FDB, FBD, taken together.

But the angle DBE is equal to the angle DEB or FEE (I. 5.), and the angles FDB, FBD to the angle EFB (I. 28.); whence the angle FEB will be equal to the angle EFB, and the fide EB to the fide BF (I: 5.)

And fince EB is equal to FD, by conftruction, BF will also be equal to FD, and the angle FDB to the angle FBD (I. 5.)

These two angles, therefore, taken together, are double the angle FDB; whence the angle EFB, or its equal FEB, is also double the angle FDb.

But the angle FEB, or CEB, is equal to the angle CAB (III. 15.), and the angle FDB, or EDB, to the angle ACâ (III. 14. and I. Ax. 6.); confequently the angle CAB is alfo double the angle ACB.

And, fince EAC, EBC are right angled triangles (III. 16.), having EA equal to EB (by Conft.) and EC common, the remaining fide AC will be equal to the remaining fide CB (III. 8. Cor.)

The triangle ABC, therefore, is ifofceles, and has each of the angles at its bafe double the angle at the vertex; as was to be shewn.

PROP. XI. PROBLEM.

In a given circle to inscribe a regular pentagon.

B

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

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

Bifect 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, fince 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 ftand upon equal circumferences (III. 21.), the arcs CD, DA, AB, BE and EC are alfo equal to each other.

But equal arcs are fubtended by equal chords (III. 22.); confequently the fides CD, DA, AB, BE and EC are, likewife, equal.

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

For, fince the arc CD is equal to the arc BE, to each of them add DAB, and the arc CDAB will be equal to the arc DABE.

But equal angles are fubtended by equal arcs (III. 21.), whence the angle CEB is equal to the angle DCE.

And, in the same manner, it may be fhewn, that each of the angles CDA, DAB, ABE are equal to the angle

CEB or DCE.

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

PROP. XII.

About a given circle to describe a regular pentagon.

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

Infcribe 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 O, A, O, B, 0, C, O, D and o, E.

Then, fince the angles OEF, OAF are right angles (III. 12.), the angles OEA, OAE, taken together, are less than two right angles; whence the lines LF, FG will meet each other (I. 25. Cor.)

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