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Book II:

PRO P. XIII.

THEOR.

See N.

IN every triangle, the square 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 sides, and the straight line intercepted between the perpendicular let fall upon it from the oppofite angle, and the acute angle.

a 12. I.

b 7. 2.

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

First, Let AD fall within the triangle ABC ; and because
the straight line CB is divided in-
to two parts in the point D, the

А
squares of CB, BD are equal to
twice the rectangle contained by
CB, BD, and the square of DC:
To each of these equals add the
square of AD; therefore the
squares of CB, BD, DA are equal
to twice the rectangle CB, BD,
and the squares of AD, DC: B D

C
But the square of AB is equal
° to the squares of BD, DA, because the angle BDA is a right
angle; and the square of AC is equal to the squares of AD, DC:
Therefore the squares of CB, BA are equal to the square of
AC, and twice the rectangle CB, BD; that is, the square of
AC alone is less than the squares of CB, BA by twice the rect-
angle CB, BD.
Secondly, Let AD fall with-

A out the triangle ABC: Then, becaule the angle at D is a right angle, the angle ACB is greater d than a right angle; and there. fore the square of AB is equal to the squares of AC, CB, and twice the rectangle BC, CD: To thefe equals add the square of BC, and the

B

D squares

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squares of AB, BC are equal to the square of AC, and twice Book II. the square of BC, and twice the rectangle BC, CD: But because BD is divided into two parts in C, the rectangle DB, BC is equal to the rectangle BC, CD and the square of BC: And ? 3. 2. the doubles of these are equal : Therefore the squares of AB, BC are equal to the square of AC, and twice the rectangle DB, BC: Therefore the square of AC alone is less than the squares of AB,BC bytwice the rectangle DB, BC.

А Laftly, Let the lide AC be perpendicular to BC; then is BC the straight line between the perpendicular and the acute angle at B; and it is manifeft that the squares of AB, BC are e. qual to the square of AC, and twice the

C 47. I. {quare of BC : Therefore, in every triangle, &c. Q. E, D.

B с

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O describe a square that thall be equal to a given sce N.

Trectilineal figure.

Let A be the given rectilineal figure ; it is required to de• {cribe a square that shall be equal to A.

Describe a the rectangular parallelogram BCDE equal to the a 45. 1. rectilineal figure A. If then the sides of it BE, ED are equal to one another, it is a square, and

H what was required is now done : But if they are not e. A qual, produce one of them BE to F,

B

G E and make E Fequal to ED, and bifect

C

D BF in G, and from the centre G, at the distance GB, or GF, describe the semicircle 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 ac E, the rectangle BE, EF, together with the square of EG, is equal to the square of b 5. 2. GF: But GF is equal to GH; therefore the rectangle BE, EF, E

to.

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Book II. together with the square of EG, is equal to the square of GH: mo But the squares of HE, EG are equal to the square of GH: 0 47. 1.

Therefore the rectangle BE, EF, together with the square of EG, is equal to the squares of HE, EG: Take away the square of EG, which is common to both ; and the remaining rectangle BE, EF is equal to the square of EH : But the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED; therefore BD is equal to the square of EH; but BD is equal to the rectilineal figure A ; therefore the rectilineal figure A is equal to the square of EH: Wherefore a square has been made equal to the given rectilineal figure A, viz. the square deferibed upon EH. Which was to be done.

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ES

I.
QUAL circles are those of which the diameters are equal,

or from the centres of which the straight lines to the circumferences 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 centres coincide, the circles must likewise coincide, since
the straight lines from the centres are equal.

II.
A straight line is faid to touch

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

UI.
Circles are said to touch one

another, which meet, but
do not cut one another.

IV.
Straight lines are said to be equally di-

ftant from the centre of a circle,
when the perpendiculars drawn to
them from the centre are equal.

V.
And the straight line on which the

greater perpendicular falls, is said to
be farther from the centre.

VI.

C

E 2

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Book III.

VI.
W A segment 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 segment is the angle con-

tained by two straight lines drawn
from any point in the circumference
of the segment, to the extremities
of the straight line which is the base
of the segment.

IX.
And an angle is said to inäist or stand

upon the circumference intercepted
between the straight lines that con-
tain the angle.

X.
The sector of a circle is the figure contain:

ed by two straight lines drawn from the
centre, and the circumference between
them.

XI.
Similar fegments of a circle,

are those in which the an-
gles are equal, or which
contain equal angles.

PRO P. I. :P R, O B.

To find the centre of a given circle.

Let ABC be the given circle; it is required to find its centre.

Draw within it any straight line AB, and bisect it in D; from the point D drawb DC at right angles to AB, and pro: duce it to E, and bisca CE in F: The point F is the centre of the circle ABC.

a 10. 1.

11. 1.

Pors

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