BOOK II. and the figure FK is the rectangle

contained by CF, FA,

for AF is equal to FG:

and AD is the square of AB;
therefore FK is equal to AD:

Take away the common part AK,
and the remainder FH is equal to the
remainder HD:

And HD is the rectangle contained

by AB, BH,

for AB is equal to BD;

and FH is the square of AH.

AH:

Therefore the rectangle AB, BH is equal to the square of

Wherefore, the straight line AB is divided in H,

so that the rectangle AB, BH, is equal to the square of AH. Which was to be done.

[It is impossible to exhibit this Proposition in exact numbers. The reader may try with any given number, e. g. 12., and will find that it cannot be done.

Thus, if 7 and 5 be taken for the two parts,

12 multiplied by 5 equals 60, which is larger than the square of 7 or 49.

But if 8 and 4 be tried,

12 multiplied by 4 equals 48, which is less than 64, or the square of 8.]

PROP. XII. THEOR.

In obtuse angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square of the sides subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side 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 a

perpendicular 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

A

the squares of BC, CD,

and twice the rectangle BC, CD:

BOOK II.

a xii. 1.

b iv. 2.

equal to

the squares of BC, CD, DA, and twice the rectangle BC,

CD:

с

But the square of BA is equal to the squares of BD, DA,
because the angle at D is a right angle:
[Therefore the square of BA is equal to the squares of BC,
CD, DA, and twice the rectangle BC, CD:]

and the square of CA is equal to the squares of CD, DA:
Therefore the square of BA is equal to

e xlvii. 1.

BOOK II. 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 obtuse angled triangles, &c. Q. E. D.

[Before going on to the next Proposition, the reader should satisfy himself that he understands this by demonstrating it again in the case of AC being produced.

In this figure the square of AB is greater than the square of AC, CB, by twice the rectangle AC, CE.

COR. From this it appears, that the rectangle AC, CE is equal to the rectangle BC, CD.]

B

BOOK II.

PROP. XIII. THEOR.

In every triangle, the square of the side subtending any of the acute angles is less than the squares of the sides 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 opposite 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 sides containing it,

let fall the perpendicular a 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.

CASE I. First, let AD fall within the triangle ABC; and because the straight line CB

is divided into two parts in the
point D,

the squares of CB, BD are equal

to twice the rectangle contained

b

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: But the square of AB is equal to the squares of BD, DA,

с

because the angle BDA is a right angle;

[Therefore the squares of CB, BA are equal to

twice the rectangle CB, BD; and the squares of AD, DC:] 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 rectangle CB, BD.

a xii. 1.

b vii. 2.

c xlvii. 1.

BOOK II.

d xvi. 1.

xii. 2.

iii. 2.

8 xlvii. 1.

the squares of AC, CB, and twice the rectangle BC, CD: To these equals add the square of BC,

and the squares of AB, BC are equal to the square of AC,

and twice 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 the doubles of these are equal:

[that is, twice the rectangle DB, BC is equal to

twice the rectangle BC, CD and twice the square of BC:]
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, by twice the rectangle DB, BC.

CASE III. Lastly, let the side AC be

perpendicular to BC;

then is BC the straight line between the per-
pendicular and the acute angle at B;
and it is manifest, that the squares of AB,

BC are equal to the square of AC and
twice the square of BC:

Therefore, in every triangle, &c. Q. E. D.

B

[The last case, though necessary for the completion of the proof, is of no further use. It is only in reality an involved mode of stating, that the square on the side subtending the right angle is equal to the sum of the squares on the two other sides of a right angled triangle.]