In the given draw two diameters PCQ, RCS at right 4s to one another.

Join PS, SQ, QR, RP.

Then PSQR shall be the square required.

thes at C are rights; .. they are one another; and the chords PS, SQ, QR, RP on which they stand are equal; (III. 18)

.. PSQR is equilateral.

Again, RPS is in a semicircle; .. it is a right 4. (III. 16) Similarly the others of PSQR are rights;

.. PSQR is rectangular, and it has been proved to be equilateral;

... PSQR is a square and it is inscribed in the given .

In the given draw two diameters PCQ, RCS at right 4s to one another:

at P, S, Q, R draw tangents intersecting in V, X, Y, Z.

VXYZ shall be the square required.

So VRZ, XSY are each || to PCQ and are ... I to one another.

Then VX and ZY are each

also VZ, XY are each =

=

RS; . VS, SZ are □s:

PQ; ·: VQ, PY are □s:

but RSPQ; .. VX, ZY, VZ, XY are one another;

.. VY is equilateral.

Also angle at X = L PCS; ·.· PXSC is a □. (1. 26) ..at X is a right; so s at Y, Z, V are rights; .. VY is rectangular.

Hence VY is a square and it is described about the .

.. LS BAC, BCA are each of them = half a right ▲ . (1. 24)

Similarly

ABD = half a right 4, and .. = BAC;

and is.. described about the square ABCD.

(I. 4)

Let ABCD be the given square:

it is required to inscribe a circle in ABCD.

Bisect AB, BC in P, Q.

Through P, Q draw PR, QS || to BC, AB, respectively. With O as centre and distance OP describe à : it shall be the one required.

...the opposite sides ofs are = one another; (1. 26) .. SO = AP, OQ = PB, PO = QB, and OR = QC ; .. SO, OQ, PO, OR are all = one another;

.. the will pass through Q, R, S:

also. SAP is a right .. OPA is a right 4. (1. 21) Similarly the s at Q, R, S are rights;

.. the sides of the square touch the .

.. the is inscribed in the square.

(III. 6)

We now proceed to solve the general problems of inscribed and circumscribed regular figures, the constructions in the case of squares having been given separately on account of their simplicity.

PROBLEM IO.

To inscribe a regular polygon having a given number of sides in a circle.

X

This problem can be solved if the Oce of a can be divided into the given number of equal arcs.

Let the Oce of the PQR be divided into a number of equal arcs PQ, QR, RS &c.

Draw the chords PQ, QR, RS, &c.

Then shall PQRS be a regular polygon.

For if the figure be turned about its centre C, so that P may coincide with the former position of Q, and.. Q with that of R, R with that of S and so on, ...the arcs PXQ, QYR, RZS, &c. are all equal ;

then PQ will coincide with that of QR, QR with that of RS, &c.;

.. PQ= QR = RS, &c.; .'. the figure is equilateral ; also the arms of the PQR will coincide with those of the QRS.

Hence PQR = 4 QRS = L RST, &c. ; .'. the figure is equiangular;

.. PQRS is a regular figure, and it is inscribed in the given .

COROLLARY. Hence we may inscribe in a circle regular figures of 3, 6, 12, &c.; 4, 8, 16, &c.; 5, 10, 20, &c.; ̈ 15, 30, &c. sides. (III. G. H. K.)