Sidebilder
PDF
ePub

Book III. cause EBD is a right angle ; 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 re-
maining 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

А
circle, there be drawn two straight
lines cutting it, as AB, AC, the rect-
angles contained by the whole lines
and the parts of them without the

F
circle, are equal to one another, viz. D
the rectangle BA, AE to the rectan-
gle CA, AF: for each of them is
equal to the square of the straight
line AD which touches the circle.

o

С

B

PROP. XXXVII. THEOR.

See N.

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.

[ocr errors]

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. a 17. 3.

Draw a the straight line DE touching the circle ABC, find b 18. 3. its centre F, and join FE, FB, FD ; then FED is a right b an

gle : and because DE touches the circle ABC, and DCA cuts c 36. 3. 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; and

FE is equal to FB, wherefore DE, EF are equal to DB, BF; Book III. and the base FD is common to the

D two triangles DEF, DBF; therefore the angle DEF is equal d to the angle

d 8.1. DBF; but DEF is a right angle, therefore also DBF is a right angle: and FB, if produced, is a diameter,

E and the straight line which is drawn. B at right angles to a diameter, from the extremity of it, touches e the circle:

e 16. 3. therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E. D.

[merged small][ocr errors]

THE

ELEMENTS OF EUCLID.

BOOK IV.

DEFINITIONS.

I. Book IV. A RECTILINEAL figure is said to be inscribed in another

rectilineal figure, when all the angles of the inscribed figure See N.

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 an-
gular 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 in-
cribed 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 inscrib-

ed in a rectilineal figure, when the cir-
cumference of the circle touches each side
of the figure.

a

VI.

Book IV. 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.
A straight line is said to be placed in a circle, when the extremi-

ties 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 not greater than the diameter of the circle.

Lėt ABC be the given circle, and D the given straight line, not greater than the diameter of the circle.

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

А equal to D; but, if it is not, BC is greater than D; make CE equal a tu D, and from the cen.

a 3. 1.

E lre C, at the distance CE, de

с

B scribe the circle AEF, and join CA: therefore, because C is

F the centre of the circle AEF, CA is equal to CE; but D is

D equal to CE; therefore D is equal to CA: wherefore, in the circle ABC, a straight line 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.

Book IV. Let ABC be the given circle, and DEF the given triangle; it

is required to inscribe in the circle ABC a triangle equiangular

to the triangle DEF. a 17. 3. Draw a the straight line GAH touching the circle in the point b 23. 1. A, and at the point A, in the straight line AH, make b the angle

HAC equal to the angle DEF; and at the point A, in the straight line AG, make the an

G
gle GAB equal to the

А
angle DFE, and join D
BC: therefore because
HAG touches the cir-
cle ABC, and AC is
drawn from the point

E

F of contact, the angle

B

с c 32. 3.' HAC is equalc to the

angle ABC in the alter-
nate segment of the cir-
cle: but HAC is equal to the angle DEF; therefore also the an-
gle ABC is equal to DEF: for the same reason, the angle ACB

is equal to the angle DFE; therefore the remaining angle BAC d 32. 1. is equald to the remaining angle EDF: wherefore the triangle

ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC. Which was to be done.

PROP. III. PROB.

ABOUT a given circle to describe a triangle equi. angular to a given triangle.

Let ABC be the given circle, and DEF the given triangle ; it is required to describe a triangle about the circle ABC equiangular to the triangle DEF.

Produce EF both ways to the points G, H, and find the centre

K of the circle ABC, and from it draw any straight line KB; a 23. 1. at the point K, in the straight line KB, make a the angle

BKA equal to the angle DEG, and the angle BKC equal to the

angle DFH; and through the points A, B, C draw the straight ► 17. 3. lines LAM, MBN, NCL touching the circle ABC: there

fore because LM, MN, NL touch the circle ABC in the

points A, B, C, to which from the centre are drawn KA, KB, < 18.3. KC, the angles at the points A, B, C are rightc angles : and

because the four angles of the quadrilateral figure AMBK are

« ForrigeFortsett »