Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Either DCA passes through the centre, or it does not : first, let it pass through the centre E.

D

E

A

Join EB; therefore (III. 18.)

1. T'he angle EBD is a right angle: and because the straight line AC is bisected in E, and produced to the point D, (II. 6.)

2. The rectangle AD, DC, together with the square of EC, is equal to the square of ED; and CE is equal to EB; therefore

3. The rectangle AD, DC, together with the square of EB, is equal to the square of ED: but the square of ED is equal (I. 47.) to the squares of EB, BD, because EBD is a right angle; therefore (Ax. 1.)

4. The rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD: take away the common square of EB; therefore (Ax. 3.)

5. The remaining rectangle AD, DC, is equal to the square of the tangent DB. But if DCA does not pass through the centre of the circle ABC, take (III. 1.) the centre E, and draw EF perpendicular (I. 12.) to AC, and join EB, EC, ED.

D

в.

And because the straight line EF, which passes through the centre, cuts the straight line AC which does not pass through the centre, at right angles, it shall likewise bisect it (III. 3.); therefore

1. AF is equal to FC: and because the straight line AC is bisected in F, and produced to D, (II. 6.)

2: The rectangle AD, DC, together with the square of FC, is equal to the square of FD:

to each of these equals add the square of FE; therefore (Ax. 2.)

3. The rectangle AD, DC, together with the squares of CF, FE, is equal to the squares of DF, FE: but the square of ED is equal (I. 47.) to the squares of DF; FE, because EFD is a right angle; and the square of EC is likewise equal to the squares of CF, FE; therefore (Ax. 1.)

4. The rectangle AD, DC, together with the square of EC, is equal to the square of ED: and CE is equal to EB; therefore

5. The rectangle AD, DC, together with the square of EB, is equal to the square of ED: but the squares of EB, BD, are equal (I. 47.) to the square of ED, because EBD is a right angle; therefore

6. The rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD: take away the common square of EB; therefore (Ax. 3.)

7. The remaining rectangle AD, DC, is equal to the square

of DB. .

Wherefore, if from any point, &c. Q.E.D.

COR.-If from any point without a circle there be drawn two straight lines cutting it, as AB, CAC, the rectangles contained by the whole lines and the parts of them without the circle are equal to one another.

[blocks in formation]

Thus :

The rectangle BA, AE, is equal to the rectangle CA, AF: for each of them is equal to the square of the straight line AD which touches the circle.

PROP. XXXVII.-THEOREM.

If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square of the line which meets it, the line which meets shall touch the circle.

Let any point D be taken without the circle ABC, and from it let two straight lines DCA and DB be drawn, of which DCA cuts the circle, and DB meets it; if the rectangle AD, DC, be equal to the square of DB, DB touches the circle.

D

[blocks in formation]

Draw (IlI. 17.) the straight line DE, touching the circle ABC; find its centre F, and join FE, , FD. Then (III. Î8.)

1. FED is a right angle: and because DE touches the circle ABC, and DCA cuts it, (III. 36.)

2. The rectangle AD, DC, is equal to the square of DE: but the rectangle AD, DC, is, by hypothesis, equal to the square of DB: therefore (Ax. 1.)

3. The square of DE is equal to the square of DB, and the straight line equal to the straight line DB: and FE is equal (I. Def. 15.) to FB; wherefore DE, EF, are equal to DB, BF, each to each ; and the base FD is common to the two triangles DEF, DBF; therefore (I. 8.)

4. The angle DEF is equal to the angle DBF: but DEF was shown to be a right angle, therefore also

5. DBF is a right angle: and FB, if produced, is a diameter, and the straight line which is drawn at right angles to a diameter, from the extremity of it, touches the circle (III. 16.); therefore

6. DB touches the circle ABC. Wherefore, if from a point, &c.

Q.E.D.

EUCLID'S ELEMENTS OF GEOMETRY.

Book IV.

DEFINITIONS.

1.

A rectilineal figure is said to be inscribed in another rectilineal figure,

when all the angular points of the inscribed figure are upon the sides of the figure upon which it is inscribed, each upon each.

II.

In like manner, a figure is said to be described about another figure,

when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each.

III.

A rectilineal figure is said to be inscribed in a circle, when all the angular

points of the inscribed figure are upon the circumference of the circle.

IV.

A rectilineal figure is said to be described about a circle, when each side

of the circumscribed figure touches the circumference of the circle.

[ocr errors]

v.

In like manner, a circle is said to be inscribed in a rectilineal figure,

when the circumference of the circle touches each side of the figure.

VI.

A circle is said to be described about a rectilineal figure, when the cir

cumference of the circle passes through all the angular points of the figure about which it is described.

VII.

A straight line is said to be placed in a circle, when the extremities of it

are in the circumference of the circle.

PROP. I.-PROBLEM.

In a given circle to place a straight line, equal to a given straight line not greater than the diameter of the circle.

Let ABC be the given circle, and D the given straight line not greater than the diameter of the circle; it is required to place in the circle ABC a straight line equal to D.

Draw BC the diameter of the circle ABC: then, if BC is equal to D, the thing required is done; for in the circle ABC a straight line BC is placed equal to D. But, if it is not, BC is greater (Hyp.) than D; make CE equal (I. 3.) to D, and from the centre Č, at the distance CE, describe the circle AEF, and join CA.

[ocr errors][merged small]

Because C is the centre of the circle AEF, (I. Def. 15.)

1. CA is equal to CE; but D is equal (Constr.) to CE; therefore (Ax. 1.)

2. D is equal to CA. Wherefore, in the circle ABC, a straight line CA is placed equal to the given straight line D, which is not greater than the diameter of the circle.

Q.E.F.

« ΠροηγούμενηΣυνέχεια »