Book II.

t

PROP. XII. THEOR.

N obtuse angled triangles, if a perpendicular be drawn from any of the acute angles to the oppofite fide pro duced, the square of the fide fubtending the obtuse angle is greater than the squares of the fides containing the obtuse angle, by twice the rectangle contained by the fide upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse angle.

Let ABC be an obtuse angled triangle, having the obtuse angle ACB, and from the point A let AD be drawn perpen-a 12. dicular to BC produced: The square of AB is greater than the squares of AC, CB by twice the rectangle BC, CD.

:

Because the straight line BD is divided into two parts in the

point C, the square of BD is equal

to the squares of BC, CD, and twice the rectangle BC, CD: Το each of these equals add the square of DA; and the squares of BD, DA are equal to the squares of BC, CD, DA, and twice the rectangle BC, CD: But the square of BA is equal 'to the squares of BD, DA, be

A b 4.2

cause the angle at D is a right B

C 47.

CD

angle; and the square of CA is e

qual to the squares of CD, DA: Therefore the square of BA
is equal to the squares of BC, CA, and twice the rectangle BC,
CD; that is, the square of BA is greater than the squares of BC,
CA, by twice the rectangle BC, CD. Therefore, in obtufe
angled triangles, &c. Q. E. D.

PROP.

Book II.

See N.

a 12. 1.

PROP. XIII. THEOR.

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 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 square of AC, oppofite to the angle B, is less than the squares of CB, BA by twice the rectangle CB, BD.

Firít, Let AD fall within the triangle ABC; and because

b 7.2.

the straight line CB is divided in-
to two parts in the point D, the
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
But, the square of AB is equal

C 47.1.

to the fquares of BD, DA, because the angle BDA is a right angle; and the fquare 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.

A

d 16. 1.

C 12.2.

Secondly, Let AD fall with-
out the triangle ABC: Then, be-
cause the angle at D is a right
angle, the angle ACB is greater
dthan 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 these e-
quals add the square of BC, and the B

CD squares 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 becaufe 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 f. 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 Iquares of AB, BC by twice the rectangle DB, BC.

Laftly, Let the fide 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 equal to the square of AC, and twice the fquare of BC: Therefore, in every triangle, &c. Q.E. D.

C. 47. I.

TO

PROP. XIV. PROB.

O describe a square that shall be equal to a given See N rectilineal figure.

Let A be the given rectilineal figure; it is required to describe a square that shall be equal to A.

Describe the rectangular parallelogram BCDE equal to the a. 45. 1.

rectilineal figure A. If then the fides of it BE, ED are equal

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 at 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 rectange BE, EF,

E

to

C 47. I.

Book II. together with the square of EG, is equal to the square of GH: But the squares of HE, EG are equal to the square of GH: 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 described upon EH. Which was to be done.

:

THE

EQUAL circles are thofe

I.

equal,

of which the diameters are
or from the centers 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 centers coincide, the circles must likewise coincide, fince ' the straight lines from their centers are equal.'