PROP. IX.

To circumfcribe a circle about a given fquare.

Let ABCD be the given fquare; 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 bafe BC to the bafe CD, the angle DAC will be equal to the angle CAB (I.7.): that is, the angle BAD will be bifected by the line AC.

And, in the fame manner, it may be proved, that all the other angles of the fquare are bifected 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 pass through the remaining points, and circumfcribe the square AC, as was to be done.

PRO P. X. PROBLEM.

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

Let ABC be the given circle; it is required to inscribe an ifofceles triangle in it, that fhall have each of the an gles at its base double the angle at the vertex.

Draw any diameter CE, and divide the radius DE in the point F, 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 deferibe the circle BDF (III. 18.), and draw the lines BD, BF..

Then, fince the rectangle of DE, EF is equal to the fquare 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.)

PROP. IX.

To circumfcribe a circle about a given fquare.

Let ABCD be the given fquare; 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 BAĎ will be bifected by the line AC.

And, in the fame manner, it may be proved, that all the other angles of the fquare are bifected 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 alfo 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.

A

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

PRO P. X. PROBLEM

To inscribe an isofceles 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 ifofceles triangle in it, that fhall have each of the an gles at its base double the angle at the vertex.

Draw any diameter CE, and divide the radius DE in the point F, 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 of DE, EF is equal to the fquare 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 if, to each of these 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 FEB, (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 construction, 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 FÉB, is alfo 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 ACB (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 isofceles, and has each of the angles at its base double the angle at the vertex; as was to be shewn.