Imágenes de páginas
PDF
EPUB
[ocr errors]

Then, since 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 ķF, FG will meet each other (1. 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, since the angles at the points M, B, , 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 sides of parallelograms are equal to each other (I. 30.); whence the sides FG, GH, Hk and KF, being each equal to the diameter ac, or BD, the figure Fh will be equilateral.

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

And, in the same 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 square ; and it is cirs cumscribed about the circle ABCD, as was to be done,

[ocr errors]
[ocr errors]
[blocks in formation]

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

Bisect the sides 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 intersection e be the centre of the circle required.

For, since he is a parallelogram (by Conft.), the fide AF will be equal to the side Ge, and the side AG to the fide EF (I. 30.)

But the side AF is equal to the side AG, (being each of them equal to half the side of the square AD or AB), whence the side Ge will also be equal to the side ef.

And, in the same 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 described from the point E, the distance EF, EG, EH Or Ek, it will pass through the remaining points, and be inscribed in the square. Ac, as was to be done.

PRO P.

PROP. IX.

To circumscribe a circle about a given square.

B

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

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

For, since the sides DA, AC are equal to the sides 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 same manner, it may be proved, that all the other angles of the square 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 consequently the line ed is equal to the line EA (I. 7.)

And, in like manner, it may be shewn, 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 circumscribe the square AC, as was to be done,

PRO P. X. PROBLEM.

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

D

E

Let ABC be the given circle; it is required to inscribe an isosceles 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 f'so, that the rectangle of DE, EF may be equal to the square 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 describe the circle BDF (III. 18.), and draw the lines BD, BF.

Then, since 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 B (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 these angles, there be added the angle red, the whole angle DBE or Fes will be equal to the angles FDB, EBD, taken together.

But the angle Dbe is equal to the angle des or fee (I. 5.), and the angleš FDB, FBD to the angle EFB (I. 28.); whence the angle Feb will be equal to the angle EFB, and the fide ÉB to the side BF (I. 5.).

And since EB is equal to , by construction, of will also be equal to $D, and the angle FDB to the angle FBD (I. 5.)

These two angles, therefore; taken together, are doua ble 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 ACB (III. 14. and I. Ax. 6.); consequently the angle CAB is also double the angle ACB.

And, since 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 base double the angle at the vertex ; as was to be shewn.

« AnteriorContinuar »