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

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

First, let DCA pass through

the centre E, and join EB.

Then EBD is a right angle. [III. 18.
And because the straight line AC
is bisected at E, and produced to
D, the rectangle AD, DC together
with the square on EC is equal to
the square on ED.
[II. 6.

But EC is equal to EB;

therefore the rectangle AD, DC together with the square on EB is equal to the square on ED.

But the square on ED is equal to the

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squares on EB, BD, because EBD is a right angle. [I. 47. Therefore the rectangle AD, DC, together with the square on EB is equal to the squares on EB, BD.

Take away the common square on EB;

then the remaining rectangle AD, DC is equal to the square on DB.

[Axiom 3.

Next let DCA not pass through the centre of the circle ABC; take the centre E;

from E draw EF perpendicular to AC;

and join EB, EC, ED.

[III. 1.

[I. 12.

Then, 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 also bisects it; [III. 3. therefore AF is equal to FC.

And because the straight line AC is bisected at F, and produced to D, the rectangle AD, DC, together with the square on FC, is equal to the square on FD.

To each of these equals add the square on FE.
Therefore the rectangle AD, DC
together with the squares on
CF, FE, is equal to the squares
on DF, FE.
[Axiom 2.

But the squares on CF, FE are
equal to the square on CE, be-
cause CFE is a right angle; [I. 47.
and the squares on DF, FE are
equal to the square on DE.
Therefore the rectangle AD, DC,
together with the square on CE,
is equal to the square on DE.

But CE is equal to BE;

B

D

[II. 6.

therefore the rectangle AD, DC, together with the square on BE, is equal to the square on DE.

But the square on DE is equal to the squares on DB,

BE, because EBD is a right angle.

[I. 47. Therefore the rectangle AD, DC, together with the square on BE, is equal to the squares on DB, BE.

Take away the common square on BE;

then the remaining rectangle AD, DC is equal to the square on DB.

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

COROLLARY. If from any point without a circle, there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of them without the circles are equal to one another; namely, the rectangle BA, AE is equal to the rectangle CA, AF; for each of them is equal to the square on the straight line AD, which touches the circle.

E

A

[Axiom 3.

PROPOSITION 37. THEOREM.

If from any point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it, and 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 on the line which meets the circle, the line which meets the circle shall touch it.

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

Draw the straight line DE,

touching the circle ABC; [III. 17. find F the centre,

and join FB, FD, FE.

[III. 1.

Then the angle FED is a
[III. 18.

right angle.

And because DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal to the square on DE. [III. 36. But the rectangle AD, DC is equal to the square on DB. [Hyp.

Therefore the square on DEis equal to the square onDB;[Ax.1. therefore the straight line DE is equal to the straight line DB.

And EF is equal to BF;

[I. Definition 15. therefore the two sides DE, EF are equal to the two sides DB, BF each to each ;

and the base DF is common to the two triangles DEF, DBF; therefore the angle DEF is equal to the angle DBF. [I. 8. But DEF is a right angle ; [Construction.

therefore also DBF is a right angle.

And BF, 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. Corollary.

therefore DB touches the circle ABC.

Wherefore, if from a point &c.

Q.E.D.

BOOK IV.

DEFINITIONS.

1. A RECTILINEAL figure is said to be inscribed in another rectilineal figure, when all the angles of the inscribed figure are on the sides of the figure in which it is inscribed, each on each.

2. 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.

3. A rectilineal figure is said to be inscribed in a circle, when all the angles of the inscribed figure are on the circumference of the circle.

4. A rectilineal figure is said to be described about a circle, when each side of the circumscribed figure touches the circumference of the circle.

5. 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.

DO

6. A circle is said to be described about a rectilineal figure, when the circumference of the circle passes through all the angular points of the figure about which it is described.

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

PROPOSITION 1. PROBLEM.

In a given circle, to place a straight_line, equal to a given straight line, which is 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.

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and from the centre C, at the distance CE, describe the

circle AEF, and join CA.

Then, because C is the centre of the circle AEF,

CA is equal to CE;

but CE is equal to D;

therefore CA is equal to D.

[I. Definition 15.

[Construction. [Axiom 1.

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.

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