* 12. 1. * 5. 3. 5.2.

+2 Ax.

* 47.1.

*5.2.

† 1 Ax.

† 3 Ax.

† 1. 3.

† 1 Ax.

*

*

*

D

A

E
G C
B

the other AC, which does not pass through the centre,
in E, but not at right angles: then, as before, if BD be
bisected in F, F is the centre of the circle. Join AF,
and from F draw* FG perpendicular to AC; therefore
AG is equal to GC; wherefore the rectangle AE, EC,
together with the square of EG, is equal to the square
of AG: to each of these equals add the square of GF;
therefore the rectangle AE, EC, together with the
squares of EG, GF, is equal to the
squares of AG, GF: but the squares of
EG, GF, are equal to the square of
EF; and the squares of AG, GF are
equal to the square of AF: therefore
the rectangle AE, EC, together with the
square of EF, is equal to the square of AF; that is, to
the square of FB: but the square of FB is equal * to
the rectangle BE, ED, together with the square of EF;
therefore the rectangle AE, EC, together with the
square of EF, is equal † to the rectangle BE, ED, to-
gether with the square of EF: take away the common
square of EF, and the remaining rectangle AE, EC, is
therefore equal+ to the remaining rectangle BE, ED.
Lastly, let neither of the straight lines
AC, BD pass through the centre: take+
the centre F, and through E, the inter-
section of the straight lines AC, DB,
draw the diameter GEFH: and be-
cause the rectangle AE, EC, is equal, as
has been shewn, to the rectangle GE, EH; and, for the
same reason, the rectangle BE, ED is equal to the same
rectangle GE, EH; therefore the rectangle AE, EC is
equal to the rectangle, BE, ED. Wherefore, if two
straight lines, &c. Q. E. D.

H

F

C

G

B

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 of 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 the same: the rectangle AD, DC shall be equal to the square of DB.

*

B

Either DCA passes through the centre, or it does not: first, let it pass through the centre E, and join EB; therefore the angle EBD is a right angle: and because the straight line AC is bisected in E, and produced to the point D, the rectangle AD, DC, together with the square of EC, is equal * to the square of ED: but CE is equal to EB; therefore the rectangle AD, DC, together with the square of EB, is equal to

18. 3.

E

6.2.

the square of ED: but the square of ED is equal to 47.1. the squares of EB, BD, because EBD is a right angle: therefore the rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD: †1 Ax. take away the common square of EB; therefore the remaining rectangle AD, DC is equal † to the square of † 3 Ax. the tangent, DB.

B

A

D

E

1. 3.

3. 3.

* 6.2.

But if DCA does not pass through the centre of the circle ABC, take* the centre E, and draw EF perpendicular to AC, and join EB, EC, ED. And because 12. 1. 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; therefore AF is equal to FC: and because the straight line AC is bisected in F, and produced to D, 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 the rectangle AD, DC, together with the squares of CF, FE, is equal to the squares of DF, +2 Ax. FE: but the square of ED is equal to the squares of 47.1. DF, FE, because EFD is a right angle; and for the same reason the square of EC is equal to the squares of CF, FE; therefore the rectangle AD, DC, together with the square of EC is equal + to the square of ED: +1 Ax. but CE is equal to EB; therefore 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 to the square of ED, because EBD is a right angle; * 47. 1. therefore 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 the remaining rectangle AD, DC is equal † to the square of DB. †3 Ax. Wherefore, if from any point, &c. Q. E. D.

*

*

"

See N.

17.3.

† 1.3.

18. 3.

*36. 3.

+ 1 Ax.

COR. If from any point without a cir cle, 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 circle, are equal to one another, viz. the rectangle BA, AE, 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. THEOR.

D

B

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 shall touch the circle.

*

*

Draw the straight line DE, touching the circle ABC, find+ its centre F, and join FE, FB, FD: then FED is a right angle: and because DE touches the circle ABC, and DCA cuts it, 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 the square of DE is equal + to the square of DB; and the straight line DE equal to the straight line DB: +15 Def.1. and FE is equal 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 the angle DEF is equal* to the angle DBF: but DEF was shewn to be a right angle; 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,

* 8. 1.

† 1 Ax.

*

D

E

*Cor.16.3. from the extremity of it, touches the circle: therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E, D.

THE

ELEMENTS OF EUCLID.

BOOK IV.

DEFINITIONS.

I.

A RECTILINEAL figure is said to be inscribed in another See N. rectilineal figure, when all the angles of the inscribed figure are upon the sides of the figure in 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 angles 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.

V.

In like manner a circle is said to be inscribed in a rec

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

VI.

A circle is said to be described about a rec-
tilineal figure, when the circumference of
the circle passes through all the angular
points of the figure about which it is de-
scribed.

VII.

+ Hyp.

3.1.

+15 Def. 1.

+ Constr. † 1 Ax.

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. PROB.

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.

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 than D: make CE equal* to D, and from the centre C, at the distance CE, describe the circle AEF, and join CA: CA shall be equal to D

D

E

B

Because C is the centre of the circle AEF, CA is equal to CE: but D is equal + to CE; therefore 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. Which was to be done.

PROP. II. PROB.

In a given circle to inscribe a triangle equiangular to a given triangle.