# Book II

## Definition I.

A rectangle or a right angled parallelogram is ſaid to be contained by any two of its adjacent or conterminous ſides.

Thus: the right angled parallelogram is ſaid to be contained by the ſides and ; or it may be briefly deſignated by · .

If the adjacent ſides are equal; i.e. = , then · which is the expreſſion for the rectangle under and is a ſquare, and

is equal to { · or 2 · or 2

## Definition II.

In a parallelogram, the figure compoſed of one of the parallelograms about the diagonal, together with the two complements, is called a Gnomon.

Thus and are called Gnomons.

## Proposition I. Problem.

The rectangle contained by two ſtraight lines, one of which is divided into any number of parts, · = { · + · + · is equal to the ſum of the rectangles contained by the undivided line, and the ſeveral parts of the divided line.

Draw and = (prs. 2. 3. B. 1.); complete parallelograms, that is to ſay,

Draw { } (pr. 31. B. 1.)

= + +
= ·
= · , = · ,
= ·

· = · + · + · .

Q. E. D.

## Proposition II. Theorem.

If a ſtraight line be divided into any two parts , the ſquare of the whole line is equal to the ſum of the rectangles contained by the whole line and each of its parts. 2 = { · + ·

Deſcribe (B. 1. pr. 46.)
Draw parallel to (B. 1. pr. 31.)
= 2
= · = ·
= · = ·
= +
2 = · + · .

Q. E. D.

## Proposition III. Theorem.

If a ſtraight line be divided into any two parts , the rectangle contained by the whole line and either of its parts, is equal to the ſquare of that part, together with the rectangle under the parts.

· = 2 + · , or, · = 2 + · .

Deſcribe (pr. 46, B. 1.)

Complete (pr. 31, B. 1.)

Then = + , but
= · and
= 2, = · ,
· = 2 + · :

In a ſimilar manner it may be readily ſhown that · = 2 + · .

Q. E. D.

## Proposition IV. Theorem.

If a ſtraight line be divided into any two parts , the ſquare of the whole line is equal to the squares of the parts, together with twice the rectangle contained by the parts.

2 = 2 + 2 + twice · .

Describe (pr. 46, B. 1.)
draw (poſt. 1.),
and { } (pr. 31, B. 1.)

= (pr. 5, B. 1.),

= (pr. 29, B. 1.)

=

by (prs. 6, 29, 34. B. 1.) is a ſquare = 2.

For the ſame reaſons is a ſquare = 2,
= = · (pr. 43. B. 1.)
but = + + + ,
2 = 2 + 2 + twice · .

Q. E. D.

## Proposition V. Theorem.

If a ſtraight line be divided into two equal parts and alſo into two unequal parts, the rectangle contained by the unequal parts, together with the ſquare of the line between the points of ſection, is equal to the ſquare of half that line.

· + 2 = 2 = 2,

Deſcribe (pr. 46, B. 1.), draw and { } (pr. 31, B. 1.)

= (p. 36, B. 1.)

= (p. 43, B. 1.)

(ax. 2.) = = ·
but = 2 (cor. pr. 4. B. 2.)
and = 2 (conſt.)

(ax. 2.) =
· + 2 =
2 = 2.

Q. E. D.

## Proposition VI. Theorem.

If a ſtraight line be biſected and produced to any point , the rectangle contained by the whole line ſo increaſed, and the part produced, together with the ſquare of half the line, is equal to the ſquare of the line made up of the half, and the produced part.

· + 2 = 2.

Deſcribe (pr. 46, B. 1.), draw
and { } (pr. 31, B. 1.)

= = (prs. 36, 43, B. 1.)

= = · ;
= 2 (cor. 4, B. 2.)

= 2 = (conſt. ax. 2.)

· + 2 = 2.

Q. E. D.

## Proposition VII. Theorem.

If a ſtraight line be divided into any two parts , the ſquares of the whole line and one of the parts are equal to twice the rectangle contained by the whole line and that part, together with the ſquare of the other parts.

2 + 2 = 2 · + 2

Deſcribe , (pr. 46, B. 1.).

Draw (poſt. 1.), and { } (pr. 31, B. 1.).

= (pr. 43, B. 1.),
add = 2 to both (cor. 4, B. 2.)

= = ·
= 2 (cor. 4, B. 2.)

+ + = 2 · + 2 = + ;

2 + 2 = 2 · + 2.

Q. E. D.

## Proposition VIII. Theorem.

If a ſtraight line be divided into any two parts , the ſquare of the ſum of the whole line and any one of its parts, is equal to four times the rectangle contained by the whole line, and that part together with the ſquare of the other part.

2 = 4 · · + 2,

Produce and make =

Conſtruct (pr. 46, B. 1.);
draw , } } } (pr. 31, B. 1.) 2 = 2 + 2 + 2 · · (pr. 4, B. II.)
but 2 + 2 = 2 · · + 2 (pr. 7, B. II.)
2 = 4 · · + 2.

Q. E. D.

## Proposition IX. Theorem.

If a ſtraight line be divided into two equal parts , and alſo into two unequal parts , the ſquares of the unequal parts are together double the ſquares of half the line, and of the part between the points of ſection.

2 + 2 = 22 + 22.

Make and = or ,
Draw and ,
, , and draw .

= (pr. 5, B. 1.) = half a right angle. (cor. pr. 32, B. 1.)
= (pr. 5, B. 1.) = half a right angle. (cor. pr. 32, B. 1.)
= a right angle.

= = = (prs. 5, 29, B. 1.).
hence = , = = (prs. 6, 34, B. 1.)
2 = { 2 + 2, or + 2 = { 2 = 22 2 = 22 2 + 2 = 22 + 22.

Q. E. D.

## Proposition X. Theorem.

If a ſtraight line be biſected and produced to any point , the ſquares of the whole produced line, and of the produced part, are together double of the ſquares of the half line, and of the line made up of the half and produced part.

2 + 2 = 22 + 2 2.

Make and = to or ,
draw and ,
and { } (pr. 31, B. 1.); draw alſo.

= (pr. 5, B. 1.) = half a right angle (cor. pr. 32, B. 1.)
= (pr. 5, B. 1.) = half a right angle (cor. pr. 32, B. 1.)
= a right angle.
= = = = =
half a right angle (prs 5, 32, 29, 34, B. 1.),
and = , = = , (prs. 6, 34, B. 1.). Hence by (pr. 47, B. 1.) 2 = { 2 + 2 or 2 { + 2 = 22 + 2 = 2 2 2 + 2 = 22 + 2 2.

Q. E. D.

## Proposition XI. Problem.

To divide a given ſtraight line in ſuch a manner, that the rectangle contained by the whole line and one of its parts may be equal to the ſquare of the other.

· = 2.

Deſcribe (pr. 46, B. 1.),
make = (pr. 10, B. 1.),
draw ,
take = (pr. 3, B. 1.),
on deſcribe (pr. 46, B. 1.),

Produce (poſt. 2.).

Then, (pr. 6, B. 2.) · + 2 = 2 = 2 = 2 + 2 · = 2, or,
= =
· = 2.

Q. E. D.

## Proposition XII. Problem.

In any obtuſe angled triangle, the ſquare of the ſide ſubtending the obtuſe angle exceeds the ſum of the ſquares of the ſides containing the obtuſe angle, by twice the rectangle contained by either of theſe ſides and the produced parts of the ſame from the obtuſe angle to the perpendicular let fall on it from the oppoſite acute angle.

2 > 2 + 2 by 2 · .

By pr. 4, B. 2.
2 = 2 + 2 + 2 · :
2 + 2 = 2 (pr. 7, B. 1.) = 2 · · + 2 + { 2 2 } or + 2 (pr. 47, B. 1.). Therefore, 2 = 2 · · + 2 + 2: hence 2 > 2 + 2 by 2 · · .

Q. E. D.

## Proposition XIII. Problem.

In any triangle, the ſquare of the ſide ſubtending an acute angle, is leſs than the ſum of the ſquares of the ſides containing that angle, by twice the rectangle contained by either of theſe ſides, and the part of it intercepted between the foot of the perpendicular let fall on it from the oppoſite angle, and the angular point of the acute angle.

### First.

2 < 2 + 2 by 2 · · .

### Second.

2 < 2 + 2 by 2 · · .

Firſt, ſuppoſe the perpendicular to fall within the triangle, then (pr. 7, B. 2.)
2 + 2 = 2 · · + 2,
2 + 2 + 2 = 2 · · + 2 + 2
(pr. 47, B. 1.)
2 + 2 = 2 · · + 2, and 2 < 2 + 2 by 2 · · .

Next ſuppoſe the perpendicular to fall without the triangle, then (pr. 7, B. 2.)
2 + 2 = 2 · · + 2,
2 + 2 + 2 = 2 · · + 2 + 2 (pr. 47, B. 1.),
2 + 2 = 2 · · + 2, 2 < 2 + 2 by 2 · · .

Q. E. D.

## Proposition XIV. Problem.

To draw a right line of which the ſquare ſhall be equal to a given rectilinear figure.
To draw ſuch that 2 =

Make = (pr. 45, B. 1.),
produce until = ;
take = (pr. 10, B. 1.),

Deſcribe (poſt. 3.),
and produce to meet it: draw .
2 or 2 = · + 2 (pr. 5, B. 2.),
but 2 = 2 + 2 (pr. 47, B. 1.);
2 + 2 = · + 2
2 = · , and
2 = =

Q. E. D.