IN

Book H.

PROP. A. THEOR. See N. IN any triangle, the difference between the squares

of two of the sides is equal to twice the rectangle contained by the base, and the segment of it intercepted between the perpendicular drawn to it from the opposite angle, and the point in which it is bifected,

Let ABC be a triangle, of which the fide AB is greater than a 10. 1. AC; and bisect a the base BC in D, and draw 6 AE perpendib 12. 1. cular to BC; the square of AB is greater than the square of AC,

by twice the rectangle BC, DE. C 3. I. Produce BC to F, and make © EF equal to EB: and because

BF is double of BE, and BC double of BD; the remainder CF
is double of the remainder DE: And
because BF is divided into two equal
parts in the point E, and into two
unequal parts in the point C; the

rectangle BC, CF, together with the 3.5. 2. fquare of CE, is equal d to the square

of BE: add to these equals the
square of EA; and the rectangle BC, CF, together with the
squares of CE, EA, is equal to the squares of BE, EA: Buc

the rectangle BC, CF is double e of the rectangle BC, DE, bef 47. 1.

cause CF is double of DE; and the square of AC is equal to the squares of CE, EA, because CEA is a right angle; and the square of AB is equal to the squares of BE, EA; therefore twice the re&tarigle BC, DE together with the square of AC, is equal to the square of AB; that is, twice the rectangle BC, DE is equal to the excess of the square of AB above the

square of AC. Wherefore, &c. Q. E. D. g 4. 1.

Cor. And if AF be joined, the triangle ABF is isosceles 5; hence it is manifest, that if any straight line AC be drawn froin the vertex to the base of an isosceles triangle ABF; the square of the fide AB is equal to the square of the line AC, together with the rectangle BC, CF of the segments of the base.

D E

e 1. 2.

PROP. B. THEOR. "HÉ squares of two ficles of a triangle are toge

ther double of the square' of half the base, and of the square of the straight line drawn from the vertex to bifect the base.

Let

a 10. I.

b

Let ABC be a triangle; and let the base BC be bise&ed a in Book II.
D, and join AD: the squares of BA, AC together are double
of the squares of BD, DA.

If the angles at D be right angles, the square of BA is equal
to the squares of BD, DA, and the square of AC to the squares b 47. 1.
of CD, DA, or of BD, DA; therefore the squares of BA, AC
are double of the squares of BD, DA.

But, if the angles at D be not right angles, from A draw C 12. 1.
AE perpendicular to BC: and because BC is bifected in D, the
{quares of BE, EC are double d of

d 9. 2.
the squares of BD, DE: add to these
equals twice the square of EA; and
the squares of BE, EC, together with
twice the square of EA, are double of
the squares of BD, DE, EA: And

B
the square of DA is equal to the
squares of DE, EA, because DEA is a right angle; therefore
the squares of BE, EC, together with twice the square of EA,
are double of the squares of BD, DA: but the square of BA is
equal to the squares of BE, EA, and the square of AC to the
squares of CE, EA; therefore the squares of BA, AC are
equal to the squares of BE, EC, together with twice the square
of EA; and it has been demonstrated, that the squares of BE,
EC, together with twice the square of EA, are double of the
{quares of BD, DA ; therefore the squares of BA, AC are
double of the squares of BD, DA. Wherefore, &c. Q. E. D.

PROP. C. THEOR.

THE squares of the two diameters of a paral

lelogram are together equal to the squares of its four fides.

TH

Let ABCD be a parallelogram, of which the diameters are
AC, BD; the squares of AC, BD together are equal to the
squares of the four sides AB, BC, CD, DA.

Let AC, BD cut one another in E: and because AC meets the
parallels AD, BC, the alternate angles DAE, BCE are equal a : a 29. I.
and the angle AED is equal to its vertical angle BEC; there- b 15. 1.
fore two angles of the triangle AED are equal to two of the
triangle BEČ; and the fides AD, BC, opposite to equal angles,
are also equal ; therefore their other sides are equal“, each to c.6. 1.
each, viz. AE to EC, and BE to ED: and because AE is
drawn from the vertex A of the triangle BAD, to bifect the
H 2

bare

d B. 2.

Book II. base BD; the squares of BA, AD' are double d of the squares mof BE, EA. For the same reason, the squares of BC, CD

are double d of the squares of BE, EC, or of BE, EA, because EC is equal to EA: therefore the squares of BA, AD,

BC, CD are quadruple of the squares of BE, EA: but the e 3 Cor.4.2. square of BD is quadruple of

А

D the square of BE, because BD is double of BE; and the square AC is. quadruple of the square of AE ; therefore the squares of BD, AC are quadruple of the squares of BE, EA: and it has B been proved, that the squares of the four sides are quadruple of BE, EA; therefore the squares of BD, AC are equal to the squares of the four fides AB, BC, CD, DA. Wherefore, &c. 0. E. D.

Cor. Hence it is manifest, that the diameters of a parallelogram bisect one another.

THE

A.
N arch of a circle is
any part of the circumference.

BOOK. III.
B.
A chord is any straight line in a circle, terminated both ways by
the circumference.

I. Omitted.

II.
A straight line is said to touch

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

III.
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.

Book III.

VI.
V A segment of a circle is the figure con-

tained by a straight line, and the part
of the circumference it cuts off.

VII. Qmitted.

VIII.
An angle in a segment is the angle con-

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

IX.
And an angle is said to stand upon the

arch intercepted between the straight
lines that contain the angle.

X.
The sector of a circle is the figure con-

tained by two straight lines drawn from
the centre, and the arch between them.

XI.
Similar segments of a circle, are

those in which the angles are
equal, or which contain equal
angles.

PROP. I. PROB.

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 a it in D; b) 31. 1. from the point D drawb DC at right angles to AB, and pro

duce it to E, and bisect CE in F: The point F is the centre of the circle ABG.

For, if it be not, let, if possible, G be the centre, and join GA, GD, GB: Then, because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two fides AD, DG are equal to the two BD, DG, each to each ; and bale GA is equal to the base GB, because they are drawn from the

centre