Book II.

See N.

2. 12. I.

b. 7. 200

PROP. XIII. THEOR.

IN every triangle the fquare of the fide fubtending

any of the acute angles, is less than the squares of the fides containing that angle, by twice the rectangle contained by either of these fides, and the straight line intercepted between the perpendicular let fall upon it from the oppofite angle, and the acute angle.

Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC one of the fides containing it let fall the perpendicular AD from the oppofite angle. the fquare of AC opposite to the angle B, is less than the fquares of CB, BA by twice the rectangle CB, BD.

A

First, Let AD fall within the triangle ABC; and because the ftraight line CB is divided into two parts in the point D, the fquares of CB, BD are equal b to twice the rectangle contained by CB, BD, and the fquare of DC. to each of thefe equals add the fquare of AD, therefore the fquares of CB, BD, DA are equal to twice the rectangle

CB, BD, and the fquares of AD, B

47. 1. DC. but the fquare of AB is equal

d. 16. I.

C. 12. 2.

Ꭰ

C

to the fquares of BD, DA, because the angle BDA is a right angle; and the fquare of AC is equal to the fquares of AD, DC. therefore the fquares of CB, BA are equal to the square of AC, and twice the rectangle CB, BD; that is, the fquare of AC alone is less than the fquares of CB, BA by twice the rectangle CB, BD.

d

Secondly, Let AD fall without the triangle ABC. then because the angle at D is a right angle, the angle ACB is greater than a right angle; and therefore the square of AB is equal to the fquares of AC, CB and twice the rectangle BC, CD. to thefe equals add the fquare of BC, and the fquares of AB, BC

e

D

are equal to the fquare of AC, and twice the fquare of BC, and Book II. twice the rectangle BC, CD. but because BD is divided into two parts in C, the rectangle DB, BC is equal f to the rectangle BC, f. 3. 2. CD and the fquare of BC. and the doubles of thefe are equal. therefore the fquares of AB, BC are equal to the fquare of AC, and twice the rectangle DB, BC. therefore the fquare of AC alone, is less than the squares of AB, BC, by twice the rectangle DB, BC.

Laftly, let the fide AC be perpendicular to BC; then is BC the ftraight line between the perpendicular and the acute angle at B. and it

C

is manifeft that the fquares of AB, BC are equal to the square of AC, and twice the square of BC. Therefore in every triangle, &c. Q. E. D.

A

·47. I.

B

PROP. XIV. PROB.

'O defcribe a fquare that shall be equal to a given See N. rectilineal figure.

Let A be the given rectilineal figure; it is required to defcribe

a fquare that shall be equal to A.

Defcribe the rectangular parallelogram BCDE equal to the rec-a. 45. 1. tilineal figure A. If then the fides of it BE, ED are equal to one ano

to ED, and bifect

BF in G; and from the center G, at the distance GB or GF defcribe the femicircle BHF, and produce DE to H, and join GH, therefore because the straight line BF is divided into two equal parts in the point G, and into two unequal at E, the rectangle BE, EF, together with the square of EG, is equal to the square of GF. but GF is equal to GH; therefore the rectangle BE, EF, together with the fquare of EG, is equal to the fquare of GH. but the fquares

b

.b. 5. 2

Book II. of HE, EG are equal to the fquare of GH. therefore the rectangle BE, EF together with the fquare of EG is equal to the

fquares of HE, EG.
take away the fquare
of EG, which is com-
mon to both; and
the remaining rect-
angle BE, EF is e-

qual to the fquare of

EH. but the rectan

gle contained by BE, EF is the parallelogram BD, because EF is equal to ED; therefore BD is equal to the fquare of EH. but BD is equal to the rectilineal figure A; therefore the rectilineal figure A is equal to the fquare of EH. wherefore a fquare has been made equal to the given rectilineal figure A, viz. the fquare defcribed upon EH. Which was to be done.

EQUAL circles

I.

are thofe of which the diameters are equal, or from the centers of which the straight lines to the cir cumferences are equal.

'This is not a Definition but a Theorem, the truth of which is 'evident; for if the circles be applied to one another, so that

their centers coincide, the circles muft likewife coincide, fince the straight lines from their centers are equal.'

II.

A straight line is faid to touch

a circle, when it meets the circle and being produced does not cut it.

III.

Circles are faid to touch one another, which meet but do

cut one another.

IV.

Straight lines are faid to be equally distant from the center of a circle, when the perpendiculars drawn to them from the center are equal.

V.

And the straight line on which the greater perpendicular falls, is faid to be farther from the center.

Book III.

2. 10. x.

b. 11. I.

VI.

A fegment of a circle is the figure con-
tained by a straight line and the cir-

cumference it cuts off.

VII.

"The angle of a segment is that which is contained by the straight "line and the circumference."

VIII.

An angle in a fegment is the angle con-
tained by two straight lines drawn
from any point in the circumference
of the fegment, to the extremities of
the straight line which is the base of
the fegment.

IX.

And an angle is faid to infift or stand
upon the circumference intercepted
between the straight lines that con-
tain the angle.

X.

The fector of a circle is the figure con-
tained by two straight lines drawn
from the center, and the circumfe-
rence between them.

XI.

Similar fegments of a circle,
are those in which the an-
gles are equal, or which
contain equal angles.

PROP. I. PROB.

To find the center of a given circle.

Let ABC be the given circle; it is required to find its center. Draw within it any straight line AB, and bisect it in D; from the point D draw b DC at right angles to AB, and produce it to E, and bifect CE in F. the point F is the center of the circle ABC.