For, fince the angle EAD is equal to the angle DAF (by Conft.), and the angle AED to the angle DFA, (being each of them right angles), the remaining angle EDA will also be equal to the remaining angle ADF (I. 28. Cor.)

The triangles ADE, DAF, therefore, being equiangular, and having the fide AD Common to both, the fide DE will alfo be equal to the fide DF (I. 21.)

And, in the fame manner, it may be proved, that the fide DG is equal to the fide DF.

The right lines DE, DG and DF are, therefore, all equal to each other; and the angles at the points F, E and G are right angles, by conftruction.

If, therefore, a circle be described from the centre D, with either of the distances DE, DG or DF, it will touch the fides in the points E, G, F (III. 10.) and be infcribed in the triangle ABC, as was to be done.

PROP. V. PROBLEM.

To circumfcribe a circle about a given triangle.

Let ABC be the given triangle; it is required to circumfcribe a circle about it.

Bifect the fides AC, CB with the perpendiculars DE, EF (1. 10 and 11.); and join DF.

Then, fince the angles EDF, DFE are less than two right angles (by Conft.), the lines DE, EF will meet each other (I. 25. Cor.)

Let E, therefore, be their point of interfection, and draw the lines EA, EC and EF.

Then, because AD is equal to DC (by Conft.), DE common, and the angle ADE equal to the angle EDC (being each of them right angles), the base AE will also be equal to the base EC (I. 4.)

And, in the fame manner, it may be proved, that EC is equal to EB; confequently EA, EC and EB are all equal to each other.

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

PROP. VI. PROBLEM.

To infcribe a fquare in a given circle.

Let ABCD be the given circle; it is required to inscribe a square in it.

Through E, the centre of the circle, draw any two diameters AC, BD at right angles to each other (I. 11, 12.), and join AB, BC, CD and DA; then will BCDA be the fquare required.

For fince the two fides BE, EA, are equal to the two fides ED, EA, and the angle BEA to the angle AED, (being each of them right angles), the base BA will be equal to the base AD (I. 4.)

And, in the same manner, it may be proved, that the fides BC, CD are each equal to the fides BA, AD; whence the figure BCDA is equilateral.

It is alfo rectangular: for fince BDA is a femi-circle, the angle BAD is a right angle (III. 12.)

And, for the fame reason, the angles ABC, BCD and CDA are each of them right angles.

The figure BCDA, therefore, being equilateral, and having all its angles right angles is a fquare, and it is infcribed in the circle ABCD, as was to be done.

PRO P. VII. PROBLEM.

To circumfcribe a fquare about a given

circle.

B

H

Let ABCD be the given circle; it is required to circumfcribe a fquare about it.

Draw any two diameters AC, BD at right angles to each other (I. 11, 12.); and through the points A, B, C, D, draw the tangents KF, FG, GH, HK (III. 10.); and join AB.

Then,

Then, fince the angles EAF, EBF are, each of them, right angles (III. 12.), the angles FAB, FBA will be, together, lefs than two right angles; whence the lines KF, FG will meet each other (I. 25. Cor.)

And, if the points A, D, D, C and C, B be joined, it may be proved, in like manner, that all the other lines FK, KH, HG and GF will meet each other.

And, fince the angles at the points A, B, C, D are right angles (III. 12.), as also the angles at the point E (by Conft.), the figure FH, and all the parts into which it is divided, will be parallelograms (I. 22, 23.)

But the opposite fides of parallelograms are equal to each other (I. 30.); whence the fides FG, GH, HK and KF, being each equal to the diameter AC, or BD, the figure FH will be equilateral.

It is, alfo, rectangular: for fince FE is a parallelogram, and BEA is a right angle (by Conft.), the angle F will, alfo, be a right angle (I. 28. Cor.)

And, in the fame manner, it may be proved, that the angles G, H and K are right angles.

The figure FH, therefore, being equilateral, and hav. ing all its angles right angles, is a fquare; and it is cir cumfcribed about the circle ABCD, as was to be done.

Let ABCD be the given fquare; it is required to infcribe a circle in it.

Bifect the fides AD, AB in the points F and G (I. 10.), and draw FH, GK parallel to AB and AD (I. 27.); then will the point of interfection E be the centre of the circle required.

For, fince AE is a parallelogram (by Conft.), the fide AF will be equal to the fide GE, and the fide AG to the fide EF (I. 30.)

But the fide AF is equal to the fide AG, (being each of them equal to half the fide of the fquare AD or AB), whence the fide GE will also be equal to the fide EF.

And, in the fame manner, it may be proved, that HE, EK are each equal to GE and EF.

The lines EF, EG, EH and EK are, therefore, all equal to each other; and the angles at the points F, G, H and K are right angles, by the nature of parallel lines.

If, therefore, a circle be defcribed from the point E, at the distance EF, EG, EH or EK, it will pass through the remaining points, and be infcribed in the fquare AC, as was to be done.

PROP.