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Book II

Definition I.

Definition 1 figure

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 Rectangle is ſaid to be contained by the ſides Black line and Red line; or it may be briefly deſignated by Black line· Red line.

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

is equal to { Black line · Red line or Red line2 Black line · Red line or Black line2

Q. E. D.

Definition II.

Definition 2 figure

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

Thus Blue and yellow parallelograms and Red and yellow parallelograms are called Gnomons.

Q. E. D.

Proposition I. Problem.

Proposition 1 figure

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

Draw Black line Blue, red, and yellow lines and = Black line (prs. 2. 3. B. 1.); complete parallelograms, that is to ſay,

Draw { Blue, red, and yellow dotted lines Blue, red, and yellow lines Black dotted lines Black line } (pr. 31. B. 1.)

Yellow, blue, and red rectangles = Yellow rectangle + Blue rectangle + Red rectangle
Yellow, blue, and red rectangles = Blue, red, and yellow lines · Black line
Yellow rectangle = Blue line · Black line, Blue rectangle = Red line · Black line,
Red rectangle = Yellow line · Black line

Blue, red, and yellow lines · Black line = Blue line · Black line + Red line · Black line + Yellow line · Black line.

Q. E. D.

Proposition II. Theorem.

Proposition 2 figure

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

Deſcribe Yellow and red rectangles (B. 1. pr. 46.)
Draw Black line parallel to Black dotted line (B. 1. pr. 31.)
Yellow and red rectangles = Blue and red lines 2
Red rectangle = Black line · Blue line = Blue and red lines · Blue line
Yellow rectangle = Black line · Red line = Blue and red lines · Red line
Yellow and red rectangles = Red rectangle + Yellow rectangle
Blue and red lines 2 = Blue and red lines · Blue line + Blue and red lines · Red line.

Q. E. D.

Proposition III. Theorem.

Proposition 3 figure

If a ſtraight line be divided into any two parts Blue and red lines , 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.

Blue and red lines · Blue line = Blue line2 + Blue line · Red line, or, Blue and red lines · Red line = Red line2 + Blue line · Red line.

Deſcribe Red rectangle (pr. 46, B. 1.)

Complete Yellow rectangle (pr. 31, B. 1.)

Then Red and yellow rectangles = Red rectangle + Yellow rectangle , but
Red and yellow rectangles = Blue and red lines · Blue line and
Red rectangle = Blue line2, Yellow rectangle = Blue line · Red line,
Blue and red lines · Blue line = Blue line2 + Blue line · Red line:

In a ſimilar manner it may be readily ſhown that Blue and red lines · Red line = Red line2 + Blue line · Red line.

Q. E. D.

Proposition IV. Theorem.

Proposition 4 figure

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

Bottom blue and red lines 2 = Blue line2 + Red line2 + twice Blue line · Red line.

Describe Bottom and right sides of square from blue and red lines (pr. 46, B. 1.)
draw Diagonal black and dotted black lines (poſt. 1.),
and { Vertical blue and dotted black lines Bottom blue and red lines Horizontal red and dotted red lines Right blue and red lines } (pr. 31, B. 1.)

Blue angle = Yellow angle (pr. 5, B. 1.),

Blue angle = Red angle (pr. 29, B. 1.)

Yellow angle = Red angle

by (prs. 6, 29, 34. B. 1.) Blue square with triangle is a ſquare = Blue line2.

For the ſame reaſons Red square with triangle is a ſquare = Red line2,
Top yellow rectangle = Bottom yellow rectangle = Blue line · Red line (pr. 43. b. 1.)
but Blue square, red square, and yellow rectangles = Blue square with triangle + Top yellow rectangle + Bottom yellow rectangle + Red square with triangle ,
Bottom blue and red lines 2 = Blue line2 + Red line2 + twice Blue line · Red line.

Q. E. D.

Proposition V. Theorem.

Proposition 5 figure

If a ſtraight line be divided Bottom red, blue, and yellow lines into two equal parts and alſo Bottom red, blue, and yellow lines 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.

Red line · Bottom blue and yellow lines + Blue line2 = Yellow line2 = Red and blue lines ,

Deſcribe Blue, red, and yellow rectangles (pr. 46, B. 1.), draw Black line and { Red and dotted black line Red and dotted blue lines Dotted black and yellow lines Bottom red, blue, and yellow lines Red dotted line Red and dotted blue lines } (pr. 31, B. 1.)

Black rectangle = Blue and bottom yellow rectangles (p. 36, B. 1.)

Top yellow rectangle = Bottom yellow rectangle (p. 43, B. 1.)

(ax. 2.) Lower left gnomon rectangles = Black, red, and bottom yellow rectangles = Bottom blue and yellow lines · Red line
but Red rectangle = Blue line2 (cor. pr. 4. B. 2.)
and Blue, red, and yellow rectangles = Red and blue lines 2 (conſt.)

(ax. 2.) Blue, red, and yellow rectangles = Black and bottom yellow rectangles
Red line · Bottom blue and yellow lines + Blue line2 =
Yellow line2 = Red and blue lines 2.

Q. E. D.

Proposition VI. Theorem.

Proposition 6 figure

If a ſtraight line be biſected Blue and yellow lines and produced to any point Red, blue, and yellow lines , 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.

Red, blue, and yellow lines · Red line + Blue line2 = Red and blue lines 2.

Deſcribe Yellow rectangles and red and blue squares (pr. 46, B. 1.), draw Black line
and { Red and dotted blue lines Red and dotted black lines Dotted black line and yellow line Red, blue, and yellow lines Red dotted line Red and dotted black lines } (pr. 31, B. 1.)

Top yellow rectangle = Bottom yellow rectangle = Black rectangle (prs. 36, 43, B. 1.)

Lower left gnomon = Blue, yellow, and black rectangles = Red line · Red, blue, and yellow lines ;
Red rectangle = Blue line2 (cor. 4, B. 2.)

Yellow rectangles and red and blue squares = Red and blue lines 2 = Blue and red squares and yellow and black rectangles (conſt. ax. 2.)

Red, blue, and yellow lines · Red line + Blue line2 = Red and blue lines 2.

Q. E. D.

Proposition VII. Theorem.

Proposition 7 figure

If a ſtraight line be divided into any two parts Bottom blue and red lines , 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.

Bottom blue and red lines 2 + Red line2 = 2 Bottom blue and red lines · Red line + Blue line2

Deſcribe All rectangles and squares , (pr. 46, B. 1.).

Draw Black line (poſt. 1.), and { Right blue and black dotted lines Yellow line Top dotted black and red lines Bottom blue and red lines } (pr. 31, B. 1.).

Yellow rectangle = Black rectangle (pr. 43, B. 1.),
add Blue rectangle = Red line2 to both (cor. 4, B. 2.)

Yellow rectangle and blue square = Blue square and black rectangle = Bottom blue and red lines · Red line
Red rectangle = Blue line2 (cor. 4, B. 2.)

Yellow rectangle and blue square + Blue square and black rectangle + Red rectangle = 2 Bottom blue and red lines · Red line + Blue line2 = All rectangles and squares + Blue rectangle ;

Bottom blue and red lines 2 + Red line2 = 2 Bottom blue and red lines · Red line + Blue line2.

Q. E. D.

Proposition VIII. Theorem.

Proposition 8 figure

If a ſtraight line be divided into any two parts Blue and red line , 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.

Blue, red, and yellow line 2 = 4 · Blue and red line · Red line + Blue line2,

Produce Blue and red line and make Yellow line = Red line

Conſtruct All lines (pr. 46, B. 1.);
draw Black line, Blue and dotted blue line Blue and dotted black line } Blue, red, and yellow line Dotted black and red line Dotted and solid red line } Blue, red, and yellow line } (pr. 31, B. 1.) Blue, red, and yellow line 2 = Yellow line2 + Blue and red line 2 + 2 · Blue and red line · Yellow line (pr. 4, B. II.)
but Red line2 + Blue and red line 2 = 2 · Blue and red line · Red line + Blue line2 (pr. 7, B. II.)
Blue, red, and yellow line 2 = 4 · Blue and red line · Red line + Blue line2.

Q. E. D.

Proposition IX. Theorem.

Proposition 9 figure

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

Blue and yellow lines 2 + Red line2 = 2Blue line2 + 2Yellow line2.

Make Blue and dotted blue lines and = Blue line or Yellow and red lines ,
Draw Black dotted line and Dotted yellow and black lines ,
Red and dotted red lines Blue and dotted blue lines , Black and yellow lines Blue and yellow lines , and draw Black line.

Bottom yellow angle = Top yellow angle (pr. 5, B. 1.) = half a right angle. (cor. pr. 32, B. 1.)
Blue angle = Black angle (pr. 5, B. 1.) = half a right angle. (cor. pr. 32, B. 1.)
Yellow and red angles = a right angle.

Blue angle = Top red angle = Bottom red angle = Black angle (prs. 5, 29, B. 1.).
hence Red and dotted red lines = Red line, Blue dotted line = Black and yellow lines = Yellow line (prs. 6, 34, B. 1.)
Black line2 = { Blue and yellow lines 2 + Red and dotted red lines 2, or + Red line2 = { Black dotted line2 = 2Blue line2 (pr. 47, B. 1.) Black and dotted black lines = 2Yellow line2 Blue and yellow lines 2 + Red line2 = 2Blue line2 + 2Yellow line2.

Q. E. D.

Proposition X. Theorem.

Proposition 10 figure

If a ſtraight line Red and yellow line be biſected and produced to any point Red, yellow, and blue line , 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.

Red, yellow, and blue line 2 + Blue line2 = 2Yellow line2 + 2 Yellow and blue line 2.

Make Black and red lines and = to Red line or Yellow line,
draw Red and yellow line and Black and dotted black line ,
and { Red and dotted red line Black and red lines Yellow and dotted yellow line Yellow and blue line } (pr. 31, B. 1.); draw Black line alſo.

Black angle = Left yellow angle (pr. 5, B. 1.) = half a right angle (cor. pr. 32, B. 1.)
Left red angle = Right yellow angle (pr. 5, B. 1.) = half a right angle (cor. pr. 32, B. 1.)
Yellow angles = a right angle.
Right red angle = Left red angle = Right yellow angle = Left blue angle = Right blue angle =
half a right angle (prs 5, 32, 29, 34, B. 1.),
and Blue line = Red dotted line, Yellow and blue line = Yellow and dotted yellow line = Red and dotted red line , (prs. 6, 34, B. 1.). Hence by (pr. 47, B. 1.) Black line2 = { Red, yellow, and blue line 2 + Red dotted line2 or Blue line2 { + Red and yellow line 2 = 2Red line2 + Black and dotted black line 2 = 2 Yellow and dotted yellow line 2 Red, yellow, and blue line 2 + Blue line2 = 2Yellow line2 + 2 Yellow and blue line 2.

Q. E. D.

Proposition XI. Problem.

Proposition 11 figure

To divide a given ſtraight line Red and dotted red 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.

Red and dotted red line · Red dotted line = Red line2.

Deſcribe Yellow and blue square (pr. 46, B. 1.),
make Blue line = Blue dotted line (pr. 10, B. 1.),
draw Black line,
take Blue and yellow line = Black line (pr. 3, B. 1.),
on Yellow line deſcribe Blue square (pr. 46, B. 1.),

Produce Black dotted line (poſt. 2.).

Then, (pr. 6, B. 2.) Blue dotted, blue, and yellow line · Yellow line + Blue line2 = Blue and yellow line 2 = Black line2 = Red and dotted red line 2 + Blue line2 Blue dotted, blue, and yellow line · Yellow line = Red and dotted red line 2, or,
Yellow rectangle and blue square = Yellow and blue square Blue square = Blue rectangle
Red and dotted red line · Red dotted line = Red line2.

Q. E. D.

Proposition XII. Problem.

Proposition 12 figure

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.

Blue line2 > Black line2 + Red line2 by 2Black line · Black dotted line.

By pr. 4, B. 2.
Black and dotted black line 2 = Black line2 + Black dotted line2 + 2Black line · Black dotted line:
add Yellow line2 to both
Black and dotted black line 2 + Yellow line2 = Blue line2 (pr. 7, B. 1.) = 2 · Black line · Black dotted line + Black line2 + { Black dotted line2 Yellow line2 } or + Red line2 (pr. 47, B. 1.). Therefore, Blue line2 = 2 · Black line · Black dotted line + Black line2 + Red line2: hence Blue line2 > Red line2 + Blue line2 by 2 · Black line · Black dotted line.

Q. E. D.

Proposition XIII. Problem.

Proposition 13 figure First Second

In any triangle, the ſquare of the ſide ſutendeding 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.

Blue line2 < Black and dotted black line 2 + Red line2 by 2 · Black and dotted black line · Black line.

Second.

Blue line2 < Red line2 + Black line2 by 2 · Black line · Black and dotted black line .

Firſt, ſuppoſe the perpendicular to fall within the triangle, then (pr. 7, B. 2.)
Black and dotted black line 2 + Black line2 = 2 · Black and dotted black line · Black line + Black dotted line2,
add to each Yellow line2 then,
2 + Black line2 + Yellow line2 = 2 · Black and dotted black line · Black line + Black dotted line2 + Yellow line2
(pr. 47, B. 1.)
Black and dotted black line 2 + Red line2 = 2 · Black and dotted black line · Black line + Blue line2, and Blue line2 < Black and dotted black line 2 + Red line2 by 2 · Black and dotted black line · Black line.

Next ſuppoſe the perpendicular to fall without the triangle, then (pr. 7, B. 2.)
Black and dotted black line 2 + Black line2 = 2 · Black and dotted black line · Black line + Black dotted line2,
add to each Yellow line2 then
Black and dotted black line 2 + Yellow line + Black line2 = 2 · Black line · Black line + Black dotted line2 + Yellow line2 (pr. 47, B. 1.),
Red line2 + Black line2 = 2 · Black and dotted black line · Black line + Blue line2, Blue line2 < Red line2 + Black line2 by 2 · Black and dotted black line · Black line.

Q. E. D.

Proposition XIV. Problem.

Proposition 14 figure

To draw a right line of which the ſquare ſhall be equal to a given rectilinear figure.
To draw Blue line ſuch that Blue line2 = Rectilinear figure

Make Rectangle = Rectilinear figure (pr. 45, B. 1.),
produce Dotted blue and black solid line until Black line = Yellow line;
take Black and blue dotted line = Black line (pr. 10, B. 1.),

Deſcribe Semicircle (poſt. 3.),
and produce Yellow line to meet it: draw Red line.
Black line2 or Red line2 = Black dotted line · Dotted blue and black solid line + Blue dotted line2 (pr. 5, B. 2.),
but Red line2 = Blue line2 + Blue dotted line2 (pr. 47, B. 1.);
Blue line2 + Blue dotted line2 = Black dotted line · Dotted blue and black solid line + Blue dotted line2
Blue line2 = Black dotted line · Dotted blue and black solid line , and
Blue line2 = Rectangle = Rectilinear figure

Q. E. D.