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Book I.

Definitions.

I.

A point is that which has no parts.

II.

A line is length without breadth.

III.

The extremities of a line are points.

IV.

A ſtraight or right line is that which lies evenly between its extremities.

V.

A ſurface is that which has length and breadth only.

VI.

The extremities of a ſurface are lines.

VII.

A plane ſurface is that which lies evenly between its extremeties.

VIII.

A plane angle is the inclination of two lines to one another, in a plane, which meet together, but are not in the ſame direction.

Definition 9 figure

IX.

A plane rectilinear angle is the inclination of two ſtraight lines to one another, which meet together, but are not in the ſame ſtraight line.

Definition 10 figure

X.

When one ſtraight line ſtanding on another ſtraight line makes the adjacent angles equal, each of theſe angles is called a right angle, and each of theſe lines is ſaid to be perpendicular to the other.

Definition 11 figure

XI.

An obtuſe angle is an angle greater than a right angle.

Definition 12 figure

XII.

An acute angle is leſs than a right angle.

XIII.

A term or boundary is the extremity of any thing.

XIV.

A figure is a ſurface encloſed on all ſides by a line or lines.

Definition 15 figure

XV.

A circle is a plane figure, bounded by one continued line, called its cicumference or periphery; and having a certain point within it, from which all ſtraight lines drawn to its circumference are equal.

XVI.

This point (from which the equal lines are drawn) is called the centre of the circle.

Definition 17 figure

XVII.

A diameter of a circle is a ſtraight line drawn through the centre, terminated both ways in the circumference.

Definition 18 figure

XVIII.

A ſemicircle is the figure contained by the diameter, and the part of the circle cut off by the diameter.

Definition 19 figure

XIX.

A ſegment of a circle is a figure contained by a ſtraight line, and the part of the circumference which it cuts off.

XX.

A figure contained by ſtraight lines only, is called a rectilinear figure.

XXI.

A triangle is a rectilinear figure included by three ſides.

Definition 22 figure

XXII.

A quadrilateral figure is one wich is bounded by four ſides. The ſtraight lines Blue line and Red line connecting the vertices of the oppoſite angles of a quadrilateral figure, are called its diagonal.

XXIII.

A polygon is a rectilinear figure bounded by more than four ſides.

Definition 24 figure

XXIV.

A triangle whoſe three ſides are equal, is ſaid to be equilateral.

Definition 25 figure

XXV.

A triangle which has only two ſides equal is called an iſoſceles triangle.

XXVI.

A ſcalene triangle is one which has no two ſides equal.

Definition 27 figure

XXVII.

A right angled triangle is that which has a right angle.

Definition 28 figure

XXVIII.

An obtuſe angled triangle is that which has an obtuſe angle.

Definition 29 figure

XXIX.

An acute angled triangle is that which has three acute angles.

Definition 30 figure

XXX.

Of four-ſided figures, a ſquare is that which has all its ſides equal, and all its angles right angles.

Definition 31 figure

XXXI.

A rhombus is that which has all its ſides equal, but its angles are not right angles.

Definition 32 figure

XXXII.

An oblong is that which has all its angles right angles, but has not all its ſides equal.

Definition 33 figure

XXXIII.

A rhomboid is that which has its oppoſite ſides equal to one another, but all its ſides are not equal, nor its angles right angles.

XXXIV.

All other quadrilateral figures are called trapeziums.

Definition 35 figure

XXXV.

Parallel ſtraight lines are ſuch as are in the ſame plane, and which being produced continually in both directions, would never meet.

Postulates.

I.

Let it be granted that a ſtraight line may be drawn from any one point to any other point.

II.

Let it be granted that a finite ſtraight line may be produced to any length in a ſtraight line.

III.

Let it be granted that a circle may be deſcribed with any centre at any diſtance from that centre.

Axioms.

I.

Magnitudes which are equal to the ſame are equal to each other.

II.

If equals be added to equals the ſums will be equal.

III.

If equals be taken away from equals the remainders will be equal.

IV.

If equals be added to unequals the ſums will be unequal.

V.

If equals be taken away from unequals the remainders will be unequal.

VI.

The doubles of the ſame or equal magnitudes are equal.

VII.

The halves of the ſame or equal magnitudes are equal.

VIII.

Magnitudes which coincide with one another, or exactly fill the ſame ſpace, are equal.

IX.

The whole is greater than its part.

X.

Two ſtraight lines cannot include a ſpace.

XI.

All right angles are equal.

Axiom 12 figure

XII.

If two ſtraight lines ( Red and blue lines ) meet a third ſtraight line (Black line) ſo as to make the two interior angles ( Yellow angle and Red angle ) on the ſame ſide leſs than two right angles, theſe two ſtraight lines will meet if they be produced on that ſide on which the angles are leſs than two right angles.

The twelfth axiom may be expreſſed in any of the following ways:

  1. Two diverging ſtraight lines cannot be both parallel to the ſame ſtraight line.
  2. If a ſtraight line interſect one of the two parallel ſtraight lines it muſt also interſect the other.
  3. Only one ſtraight line can be drawn through a given point, parallel to a given ſtraight line.

Elucidations.

Geometry has for its principal objects the expoſition and explanation of the properties of figure, and figure is defined to be the relation which ſubſiſts between the boundaries of ſpace. Space or magnitude is of three kinds, linear, ſuperficial, and ſolid.

Vertex A

Angles might properly be conſidered as a fourth ſpecies of magnitude. Angular magnitude evidently conſiſts of parts, and muſt therefore be admitted to be a ſpecies of quantity. The ſtudent muſt not ſuppoſe that the magnitude of an angle is affected by the length of the ſtraight lines which include it, and of whoſe mutual divergence it is the meaſure. The vertex of an angle is the point where the ſides or the legs of the angle meet, as A.

Angles diagram B C D E F G H

An angle is often deſignated by a ſingle letter when its legs are the only lines which meet together at its vertex. Thus the red and blue lines form the yellow angle, which in other ſyſtems would be called the angle A. But when more than two lines meet in the ſame point, it was neceſſary by former methods, in order to avoid confuſion, to employ three letters to deſignate an angle about that point, the letter which marked the vertex of the angle being always placed in the middle. Thus the black and red lines meeting together at C, form the blue angle, and has been uſually denominated the angle FCD or DCF. The lines FC and CD are the legs of the angle; the point C is its vertex. In like manner the black angle would be deſignated the angle DCB or BCD. The red and blue angles added together, or the angle HCF added to FCD, make the angle HCD; and ſo of the other angles.

When the legs of an angle are produced or prolonged beyond its vertex, the angles made by them on both ſides of the vertex are ſaid to be vertically oppoſite to each other: Thus the red and yellow angles are ſaid to be vertically oppoſite angles.

Superpoſition is the proceſs by which one magnitude may be conceived to be placed upon another, ſo as exactly to cover it, or ſo that every part of each ſhall exactly coincide.

A line is ſaid to be produced, when it is extended, prolonged, or has its length increaſed, and the increaſe of length which it receives is called its produced part, or its production.

The entire length of the line or lines which encloſe a figure, is called its perimeter. The firſt ſix books of Euclid treat of plane figures only. A line drawn from the centre of a circle to its circumference, is called a radius. The lines which include a figure are called its ſides. That ſide of a right angled triangle, which is oppoſite to the right angle, is called the hypotenuſe. An oblong is defined in the ſecond book, called a rectangle. All the lines which are conſidered in the firſt ſix books of the Elements are ſuppoſed to be in the ſame plane.

The ſtraight-edge and compaſſes are the only inſtruments, the uſe of which is permitted in Euclid, or plane Geometry. To declare this reſtriction is the object of the poſtulates.

The Axioms of geometry are certain general propoſitions, the truth of which is taken to be ſelf-evident and incapable of being eſtablished by demonſtration.

Propoſitions are thoſe reſults which are obtained in geometry by a proceſs of reaſoning. There are two ſpecies of propoſitions in geometry, problems and theorems.

A Problem is a propoſition in which ſomething is propoſed to be done; as a line to be drawn under ſome given conditions, a circle to be deſcribed, ſome figure to be conſtructed, &c.

The ſolution of the problem conſiſts in ſhowing how the thing required may be done by the aid of the rule or ſtraight-edge and compaſſes.

The demonſtration conſiſts in proving that the proceſs indicated in the ſolution really attains the required end.

A Theorem is a propoſition in which the truth of ſome principle is aſſerted. This principle muſt be deduced from the axioms and definitions, or other truths previously and independently eſtablished. To ſhow this is the object of demonſtration.

A Problem is analagous to a poſtulate.

A Theorem reſembles an axiom.

A Poſtulate is a problem, the ſolution of which is aſſumed.

An Axiom is a theorem, the truth of which is granted without demonſtration.

A Corollary is an inference deduced immediately from a propoſition.

A Scholium is a note or obſervation on a propoſition not containing an inference of ſufficient importance to entitle it to the name of a corollary.

A Lemma is a propoſition merely introduced for the purpoſe of eſtablishing ſome more important propoſition.

Proposition I. Problem.

Proposition 1 figure

On a given finite ſtraight line (Black line) to deſcribe an equilateral triangle.

Deſcribe Blue circle and black line and Red circle and black line (poſtulate 3.); draw Yellow line and Red line (poſt. 1.). then Triangle will be equilateral.

For Black line = Yellow line (def. 15.); and Black line = Red line (def. 15.), Yellow line = Red line (axiom. 1.);

and therefore Triangle is the equilateral triangle required.

Q. E. D.

Proposition II. Problem.

Proposition 2 figure

From a given point ( Blue and red lines ), to draw a ſtraight line equal to a given finite ſtraight line (Black line)

Draw Black dotted line (poſt. 1.), deſcribe Isoceles triangle (pr. 1.), produce Red line (poſt. 2.), deſcribe Blue circle and black line (poſt. 3.), and Red circle and red and yellow lines (poſt. 3.); produce Red line (poſt. 2.), then Blue line is the line required.

For Yellow and red lines = Blue and red lines (def. 15.), and Red line = Red line (conſt.), Yellow line = Blue line (ax. 3.), but (def. 15.) Black line = Yellow line = Blue line ; Blue line drawn from the given point ( Blue and red lines ), is equal the given line Black line.

Q. E. D.

Proposition III. Problem.

Proposition 3 figure

From the greater ( Black lines ) of two given ſtraight lines, to cut off a part equal to the leſs (Blue line).

Draw Red line = Blue line (pr. 2.); deſcribe Circle and red line (poſt. 3.), then Blue line = Black line.

For Red line = Black line (def. 15.), Blue line = Red line (conſt.); Blue line = Black line (ax. 1.).

Q. E. D.

Proposition IV. Theorem.

Proposition 4 figure

If two triangles have two ſides of the one reſpectively equal to two ſides of the other, (Red line to Red line and Blue line to Blue line) and the angles ( Left yellow angle and Right yellow angle ) contained by thoſe equal ſides alſo equal; then their baſes or their ſides (Black line and Black line) are alſo equal: and the remaining and their remaining angles oppoſite to equal ſides are reſpectively equal ( Left blue angle = Right blue angle and Left red angle = Right red angle ): and the triangles are equal in every reſpect.

Let the two triangles be conceived, to be ſo placed, that the vertex of one of the equal angles, Left yellow angle or Right yellow angle ; ſhall fall upon that of the other, and Red line to coincide with Red line, then will Blue line coincide with Blue line if applied: conſequently Black line will coincide with Black line, or two ſtraight lines will encloſe a ſpace, which is impoſſible (ax. 10), therefore Black line = Black line, Left blue angle = Right blue angle and Left red angle = Right red angle , and as the triangles Left triangle and Right triangle coincide, when applied, they are equal in every reſpect.

Q. E. D.

Proposition V. Theorem.

Proposition 5 figure

In any iſoſceles triangle Isoceles triangle if the equal ſides be produced, the external angles at the baſe are equal, and the internal angles at the baſe are alſo equal.

Produce Red line, and Red line, (poſt. 2.), take Yellow line = Yellow line, (pr. 3.); draw Blue line = Blue line.

Then in Left triangle and Right triangle we have,
Left red and yellow lines = Right red and yellow lines (conſt.), Black angle common to
both, and Red line = Red line (hyp.) Left blue and yellow angles = Right blue and yellow angles ,
Blue line = Blue line and Left red angle = Right red angle (pr. 4.).

Again in Lower left triangle and Lower right triangle we have Yellow line = Yellow line,
Left red angle = Right red angle and Blue line = Blue line ,
Left yellow angle plus remaining angle = Right yellow angle plus remaining angle and Left yellow angle = Right yellow angle (pr. 4.) but
Left blue and yellow angles = Right blue and yellow angles , ∴ Left blue angle = Right blue angle (ax. 3.)

Q. E. D.

Proposition VI. Theorem.

Proposition 6 figure

In any triangle ( ) if two angles ( Yellow angle and Black angle ) are equal, the ſides ( Black lines and Blue line) oppoſite two them are alſo equal.

For if the ſides be not equal, let one of them Black lines be greater than the other Blue line, and from it cut off Black line = Blue line (pr. 3.), draw Yellow line.

Then Small triangle and , Black line = Blue line, (conſt.) Yellow angle = Black angle (hyp.) and Red line common, the triangles are equal (pr. 4.) a part equal to the whole, which is abſurd; neither of the ſides Black lines or Blue line is greater than the other, hence they are equal.

Q. E. D.

Proposition VII. Theorem.

Proposition 7 figure

On the ſame baſe (Black line) and on the ſame ſide of it there cannot be two triangles having their conterminous ſides (Red line and Red line, Blue line and Blue line) at both extremities of the baſe, equal to each other.

When two triangles ſtand on the ſame baſe, and on the ſame ſide of it, the vertex of one ſhall either fall outſide of the other triangle, or within it; or, laſtly, on one of its ſides.

If it be poſſible let the two triangles be conſtructed ſo that { Red line = Red line Blue line = Blue line } , then drawBlack dotted lineand,

Red and black angles = Blue angles (pr. 5.) Red angles < Blue angles and Red angles < Blue and yellow angles but (pr. 5.) Red angles = Blue and yellow angles } which is abſurd,

therefore the two triangles cannot have their conterminous ſides equal at both extremities of the baſe.

Q. E. D.

Proposition VIII. Theorem.

Proposition 8 figure

If two triangles have two ſides of the one reſpectively equal to two ſides of the other (Blue line = Blue line and Red line = Red line), and alſo their baſes (Black line = Black line), equal; then the angles ( Left angle and Right angle ) contained by their equal ſides are alſo equal.

If the equal baſes Black line and Black line be conceived to be placed upon the other, ſo that the triangles ſhall lie at the ſame ſide of them, and that the equal ſides Red line and Red line, Blue line and Blue line be conterminous, the vertex of the one muſt fall on the vertex of the other; for to ſuppoſe them not coincident would contradict the laſt propoſition.

Therefore the ſides Red line and Blue line, being coincident with Red line and Blue line,
Left angle = Right angle .

Q. E. D.

Proposition IX. Problem.

Proposition 9 figure

To biſect a given rectilinear angle ( Blue and yellow angles ).

Take Red line = Red line (pr. 3.) draw Yellow line, upon which deſcribe Bottom triangle (pr. 1.), draw Black line .

Becauſe Red line = Red line (conſt.) and Black line common to the two triangles and Blue line = Blue line (conſt.),
Blue angle = Yellow angle (pr. 8.)

Q. E. D.

Proposition X. Problem.

Proposition 10 figure

To biſect a given finite ſtraight line ( Black lines ).

Conſtruct Triangle (pr. 1.),
draw Red line, making Blue angle = Yellow angle (pr. 9.).

Then Black line = Black dotted line by (pr. 4.),
Yellow line = Blue line (conſt.) Blue angle = Yellow angle and
Red line common to the two triangles.

Therefore the given line is biſected.

Q. E. D.

Proposition XI. Problem.

Proposition 11 figure

From a given point ( Black and red lines ), in a given ſtraight line ( Black and red lines ), to draw a perpendicular.

Take any point ( Red lines ) in the given line,
cut off Black line = Red line (pr. 3.),
conſtruct Triangle (pr. 1.),
draw Yellow line and it ſhall be perpendicular to the given line.

For Blue line = Blue line (conſt.)
Black line = Red line (conſt.)
and Yellow line common to the two triangles.

Therefore Red angle = Blue angle (pr. 8.)
Yellow line Black and red lines (def. 10.).

Q. E. D.

Proposition XII. Problem.

Proposition 12 figure

To draw a ſtraight line perpendicular to a given indefinite ſtraight line ( Black and yellow lines ) from a given (point Blue and red lines ) without.

With the given point Blue and red lines as centre, at one ſide of the lines, and any diſtance Black line capable of extending to the other ſide deſcribe Red curve ,

Make Black line = Yellow line (pr. 10.) draw Blue line, Blue line and Red line. then Red line Black and yellow lines .

For (pr. 8.) ſince Black line = Yellow line (conſt.) Red line common to both, and Blue line = Blue line (def. 15.)

Yellow angle = Blue angle , and
Red line Black and yellow lines (def. 10.).

Q. E. D.

Proposition XIII. Theorem.

Proposition 13 figure

When a ſtraight line (Yellow line) ſtanding upon another ſtraight line (Red line) makes angles with it; they are either two right angles or together equal to two right angles.

If Yellow line be to Red line then,
Yellow and red angles and Blue angle = Two right angles (def. 10.),

But if Yellow line be not to Red line,
draw Black line Red line; (pr. 11.)
Yellow angle + Blue and red angles = Two right angles (conſt.),
Yellow angle = Blue and red angles = Red angle + Blue angle
Yellow angle + Blue and red angles = Yellow angle + Red angle + Blue angle (ax. 2.)
= Yellow and red angles + Blue angle = Two right angles .

Q. E. D.

Proposition XIV. Theorem.

Proposition 14 figure

If two ſtraight lines (Blue line and Black line), meeting a third ſtraight line (Red line), at the ſame point, and at oppoſite ſides of it, make with it adjacent angles ( Yellow angle and Blue angle ) equal to two right angles; theſe ſtraight lines lie in one continuous ſtraight line.

For, if poſſible, let Yellow line, and not Black line,
be the continuation of Blue line,
then Yellow angle + Blue and red angles = Two right angles
but by the hypotheſis Yellow angle + Blue angle = Two right angles
Blue and red angles = Blue angle , (ax. 3.); which is abſurd (ax. 9.).
Yellow line , is not the continuation of Blue line, and the like may be demonſtrated of any other ſtraight line except Black line, Black line is the continuation of Blue line.

Q. E. D.

Proposition XV. Theorem.

Proposition 15 figure

If two ſtraight lines (Black line and Red line) interſect one another, the vertical angles Yellow angle and Black angle , Red angle and Blue angle are equal.

Yellow angle + Red angle = Two right angles
Black angle + Red angle = Two right angles
Yellow angle = Black angle .

In the ſame manner it may be ſhown that
Red angle = Blue angle

Q. E. D.

Proposition XVI. Theorem.

Proposition 16 figure

If a ſide of a triangle ( Large triangle ) is produced, the external angle ( Bottom black angle and arc ) is greater than either of the internal remote angles ( Top black angle or Blue angle ).

Make Blue line = Blue dotted line (pr. 10.).
Draw Red line and produce it until
Red dotted line = Red line; draw Yellow line.

In Left triangle and Right triangle ; Blue line = Blue dotted line
Left yellow angle = Right yellow angle and Red line = Red dotted line (conſt. pr. 15.),
Top black angle = Bottom black angle (pr. 4.),
Bottom black angle and arc > Top black angle .

In like manner it can be ſhown, that if Blue lines
be produced, Red angle > Blue angle , and therefore
Bottom black angle and arc which is = Red angle is > Blue angle .

Q. E. D.

Proposition XVII. Theorem.

Proposition 17 figure

Any of two angles of a triangle Triangle are together leſs than two right angles.

Produce Black line, then will
Red angle + = Two right angles

But Yellow angle > Blue angle (pr. 16.)
Red angle + Blue angle < Two right angles ,

and in the ſame manner it may be ſhown that any other two angles of the triangle taken together are leſs than two right angles.

Q. E. D.

Proposition XVIII. Theorem.

Proposition 18 figure

In any triangle Triangle if one ſide Red lines be greater than another Blue line, the angle oppoſite to the greater ſide is greater than the angle to the oppoſite to the leſs. i. e. Black and red angles > Yellow angle

Make Red line = Blue line (pr.3.), draw Yellow line.

Then will Blue angle = Black angle (pr. 5.);
but Blue angle > Yellow angle (pr. 16.);
Black angle > Yellow angle and much more
is Black and red angles > Yellow angle .

Q. E. D.

Proposition XIX. Theorem.

Proposition 19 figure

If in any triangle Triangle one angle Blue angle be greater than another Blue angle the ſide Blue line which is oppoſite to the greater angle, is greater than the ſide Red line oppoſite the leſs.

If Blue line be not greater than Red line then muſt
Blue line = or < Red line.

If Blue line = Red line then
Blue angle = Blue angle (pr. 5.);
which is contrary to the hypotheſis.
Blue line is not leſs than Red line; for if it were,
Blue angle < Blue angle (pr. 18.)
which is contrary to the hypotheſis:
Blue line > Red line.

Q. E. D.

Proposition XX. Theorem.

Proposition 20 figure

Any two ſides Blue line and Red line of a triangle Triangle taken together are greater than the third ſide (Black line).

Produce Blue line, and
make Blue dotted line = Red line (pr. 3.);
draw Yellow line.

Then becauſe Blue dotted line = Red line (conſt.),

Blue angle = Red angle (pr. 5.) Blue and yellow angles > Red angle (ax. 9.)

Blue line + Blue dotted line > Black line (pr. 19.) and Blue line + Red line > Black line.

Q. E. D.

Proposition XXI. Theorem.

Proposition 21 figure

If from any point ( Point ) within a triangle Triangle ſtraight lines be drawn to the extremities of one ſide (Blue dotted line), theſe lines muſt be together leſs than the other two ſides, but muſt contain a greater angle.

Produce Black line,
Blue line + Red line > Black lines (pr. 20.),
add Red lines to each,
Blue line + Red lines > Black lines + Red dotted line (ax. 4.)

In the ſame manner it may be ſhown that
Black lines + Red dotted line > Black line + Yellow line,
Blue line + Red lines > Black line + Yellow line,
which was to be proved.

Again Blue angle > Yellow angle (pr. 16.), and also Red angle > Blue angle (pr. 16.), Red angle > Yellow angle .

Q. E. D.

Proposition XXII. Theorem.

Proposition 22 figure

Given three right lines { Three dotted lines the ſum of any two greater than the third, to conſtruct a triangle whoſe ſides ſhall be reſpectively equal to the given lines.

Aſſume Black line = Black dotted line (pr. 3.). Draw Blue line = Blue dotted line and Red line = Red dotted line } (pr. 2.).

With Blue line and Red line as radii,
describe Blue circle and Red circle (poſt. 3.);
draw Yellow dotted line and Yellow line,
then will Triangle be the triangle required.

For Black line = Black dotted line, Yellow line = Red line = Red dotted line, and Yellow dotted line = Blue line = Blue dotted line. } (conſt.)

Q. E. D.

Proposition XXIII. Problem.

Proposition 23 figure

At a given point ( Point ) in a given ſtraight line ( Black lines ), to make an angle equal to a given rectilineal angle ( Red angle ).

Draw Red line between any two points in the legs of the given angle.

Conſtruct Triangle (pr. 22.). ſo that
Black line = Black thin line, Yellow line = Blue line
and Red line = Red thin line.

Then Blue angle = Red angle (pr. 8.).

Q. E. D.

Proposition XXIV. Theorem.

Proposition 24 figure

If two triangles have two ſides of one reſspectively equal to two ſides of the other (Blue line to Blue thin line and Red dotted line to Red thin line), and if one of the angles ( Left top angles ) contained by the equal ſides be greater than the other ( Right angle ), the ſide (Black line) which is oppoſite to the greater angle is greater than the ſide (Yellow line) which is oppoſite to the leſs angle.

Make Left angle = Right angle (pr. 23.),
and Red line = Red thin line (pr. 3.),
draw Blue dotted line and Black dotted line.
Becauſe Red line = Red dotted line (ax. 1. hyp. conſt.)
Blue and red angles = Yellow angle (pr. 5.)
but Red angle < Yellow angle
and Red angle < Black and yellow angles ,
Black line > Black dotted line (pr. 19.)
but Black dotted line = Yellow line (pr. 4.)
Black line > Yellow line.

Q. E. D.

Proposition XXV. Theorem.

Proposition 25 figure

If two triangles have two ſides (Red line and Blue line) of the one reſpectively equal to two ſides (Red thin line and Blue thin line) of the other, but their baſes unequal, the angle ſubtended by the greater baſe (Black line) of the one, muſt be greater than the angle ſubtended by the leſs baſe (Yellow line) of the other.

Yellow angle =, > or < Black angle Yellow angle is not equal to Black angle
for if Yellow angle = Black angle then Black line = Yellow line (pr. 4.)
which is contrary to the hypotheſis;

Yellow angle is not leſs than Black angle
for if Yellow angle < Black angle
then Black line < Yellow line (pr. 24.),
which is alſo contrary to the hypotheſis:

Yellow angle > Black angle .

Q. E. D.

Proposition XXVI. Theorem.

Proposition 26 figure Case II. Case I.

If two triangles have two angles of the one reſpectively equal to two angles of the other, ( Case 1: Left yellow angle = Case 1: Right yellow angle and Case 1: Left red angle = Case 1: Right black and blue angles ), and a ſide of the one equal to a ſide of the other ſimilarly placed with reſpect to the equal angles, the remaining ſides and angles are reſpectively equal to one another.

Case I.

Let Blue line and Blue line which lie between the equal angles be equal,
then Red line = Case 1: Right red lines .
For if it be poſſible, let one of them Case 1: Right red lines be greater than the other;
make Red line = Red line, draw Yellow line.

In Case 1: Left triangle and Case 1: Right triangle we have Red line
= Red line, Case 1: Left yellow angle = Case 1: Right yellow angle , Blue line = Blue line;
Case 1: Left red angle = Case 1: Right black angle (pr. 4.)
but Case 1: Left red angle = Case 1: Right black and blue angles (hyp.)
and therefore Case 1: Right black angle = Case 1: Right black and blue angles , which is abſurd; hence neither of the ſides Red line and Case 1: Right red lines is greater than the other; and they are equal;
Black line = Black line, and Case 1: Left arc = Case 1: Right arc , (pr. 4.).

Case II.

Again, let Red line = Red line, which lie oppoſite the equal angles Case 2: Left red angle and Case 2: Right red angle . If it be poſſible, let Case 2: Right blue lines > Blue line, then take Blue line = Blue line, draw Yellow line.

Then in Case 2: Left triangle and Case 2: Right triangle we have Red line = Red line,
Blue line = Blue line and Case 2: Left yellow angle = Case 2: Right yellow angle ,
Case 2: Left red angle = Case 2: Right black angle (pr. 4.)
but Case 2: Left red angle = Case 2: Right red angle (hyp.)
Case 2: Right black angle = Case 2: Left red angle which is abſurd (pr. 16.).

Conſequently, neither of the ſides Blue line or Case 2: Right blue lines is greater than the other, hence they muſt be equal. It follows (by pr. 4.) that the triangles are equal in all reſpects.

Q. E. D.

Proposition XXVII. Theorem.

Proposition 27 figure

If a ſtraight line (Black line) meeting two other ſtraight line, (Red line and Blue line) makes with them the alternate angles Blue angle and Red angle ; Bottom yellow angle and Top yellow angle ) equal, theſe two ſtraight lines are parallel.

If Blue line be not parallel to Red line they ſhall meet when produced.

If it be poſſible, let thoſe lines be not parallel, but meet when produce; then the external angle Red angle is greater than Blue angle (pr. 16), but they are alſo equal (hyp.), which is abſurd: in the ſame manner it may be ſhown that they cannot meet on the other ſide; they are parallel.

Q. E. D.

Proposition XXVIII. Theorem.

Proposition 28 figure

If a ſtraight line (Black line), cutting two other ſtraight lines (Red line and Yellow line), makes the external equal to the internal and oppoſite angle, at the ſame ſide of the cutting line (namely, Black angle = Bottom blue angle or Yellow angle = Bottom red angle ), or if it makes the two internal angles at the ſame ſide ( Bottom red angle and Top blue angle , or Bottom blue angle and Top red angle ) together equal to two right angles, thoſe two ſtraight lines are parallel.

Firſt, if Black angle = Bottom blue angle , then Black angle = Top blue angle (pr. 15.),
Bottom blue angle = Top blue angle Red line Yellow line (pr. 27.).

Secondly, if Bottom blue angle + Top red angle = Two right angles ,
then Top red angle + Top blue angle = Two right angles (pr. 13.),
Bottom blue angle + Top red angle = Top red angle + Top blue angle (ax. 3.)
Bottom blue angle = Top blue angle
Red line Yellow line (pr. 27.)

Q. E. D.

Proposition XXIX. Theorem.

Proposition 29 figure

A ſtraight line (Blue line) falling on two parallel ſtraight lines (Yellow line and Red line), makes the alternate angles equal to one another; and alſo the external equal to the internal and oppoſite angle on the ſame ſide; and the two internal angles on the ſame ſide together equal to two right angles.

For if the alternate angles Blue and yellow angles and Bottom black angle be not equal, draw Black line, making Yellow angle = Bottom black angle (pr. 23).

Therefore Black lines Red line (pr. 27.) and therefore two ſtraight lines which interſect are parallel to the ſame ſtraight line, which is impoſſible (ax. 12).

Hence the alternate angles Blue and yellow angles and Bottom black angle are not unequal, that is, they are equal: Blue and yellow angles = Red angle (pr. 15); Red angle = Bottom black angle , the external angle equal to the internal and oppoſite on the ſame ſide: if Top black angle be added to both, then Bottom black angle + Top black angle = Top black and red angles = Two right angles (pr. 13). That is to ſay, the two internal angles at the ſame ſide of the cutting line are equal to two right angles.

Q. E. D.

Proposition XXX. Theorem.

Proposition 30 figure

Straight lines ( Red and blue lines ) which are parallel to the ſame ſtraight line (Yellow line), are parallel to one another.

Let Black line interſect { Red line Yellow line Blue line } ;
Then, Yellow angle = Blue angle = Red angle (pr. 29.),
Yellow angle = Red angle
Red line Blue line (pr. 27.)

Q. E. D.

Proposition XXXI. Problem.

Proposition 31 figure

From a given point Black and red point to draw a ſtraight line prallel to a given ſtraight line (Black line).

Draw Black line from the point Black and red point to any point Black and blue point in Blue line,
make Yellow angle = Red angle (pr. 23.),
then Red lines Blue line (pr. 27.).

Q. E. D.

Proposition XXXII. Problem.

Proposition 32 figure

If any ſide (Black line) of a triangle be produced, the external angle ( Black and red angles ) is equal to the ſum of the two internal and oppoſite angles ( Yellow angle and Bottom black angle ), and the three internal angles of every triangle taken together are equal to two right angles.

Through the point Black and yellow point draw
Blue line Red line (pr. 31.).

Then { Red angle = Yellow angle Bottom black angle = Top black angle } (pr. 29.),

Yellow angle + Bottom black angle = Black and red angles (ax. 2.),
and therefore
Yellow angle + Blue angle + Bottom black angle = Black, blue, and red angles = Two right angles (pr. 13.).

Q. E. D.

Proposition XXXIII. Theorem.

Proposition 33 figure

Straight lines (Blue line and Yellow line) which join the adjacent extremities of two equal and parallel ſtraight lines (Red line and Red dotted line), are themſelves equal and parallel.

Draw Black line the diagonal.
Red line = Red dotted line (hyp.)
Yellow angle = Black angle (pr. 29.)
and Black line common to the two triangles;
Blue line = Yellow line, and Blue angle = Red angle (pr. 4.);
and Blue line Yellow line (pr. 27.).

Q. E. D.

Proposition XXXIV. Theorem.

Proposition 34 figure

The oppoſite ſides and angles of any parallelogram are equal, and the diagonal (Black line) divides it into two equal parts.

Since { Blue angle = Yellow angle Top red angle = Bottom red angle } (pr. 29.) and Black line common to the two triangles.

{ Red line = Red dotted line Yellow line = Blue line Top black angle = Bottom black angle } (pr. 26.)
and Blue and red angles = Red and yellow angles (ax.):

Therefore the oppoſite ſides and angles of the parallelogram are equal: and as the triangles Left triangle and Right triangle are equal in every reſpect (pr. 4,), the diagonal divides the parallelogram into two equal parts.

Q. E. D.

Proposition XXXV. Theorem.

Proposition 35 figure

Parallelograms on the ſame baſe, and between the ſame parallels, are (in area) equal.

On account of the parallels, Red angle = Blue angle ; Black angle = Black outlined angle ; Blue line = Red line } (pr. 29.) (pr. 29.) (pr. 34.)

But, Left triangle = Left triangle (pr. 8.)
All shapes minus Left triangle = Left parallelogram ,
and All shapes minus Left triangle = Right parallelogram ;
Left parallelogram = Right parallelogram .

Q. E. D.

Proposition XXXVI. Theorem.

Proposition 36 figure

Parallelograms ( Red parallelogram and Yellow parallelogram ) on equal baſes, and between the ſame parallels, are equal.

Draw Yellow line and Black dotted line,
Black line = Blue line = Red line, by (pr. 34, and hyp.);
Black line= and Red line;
Yellow line = and Black dotted line (pr. 33.)

And therefore Middle parallelogram is a parallelogram:
but Red parallelogram = Middle parallelogram = Yellow parallelogram (pr. 35.)
Red parallelogram = Yellow parallelogram (ax. 1.).

Q. E. D.

Proposition XXXVII. Theorem.

Proposition 37 figure

Triangles Large yellow triangle and Small yellow and black triangles on the ſame baſe (Black line) and between the ſame parallels are equal.

Draw Red dotted line Red line Blue dotted line Blue line } (pr. 31.)

Produce Black dotted line.

Large blue and yellow triangles and Small black, red, and yellow triangles are parallelograms on the ſame baſe, and between the ſame parallels, and therefore equal. (pr. 35.)

{ Large blue and yellow triangles = twice Large yellow triangle with red and black borders Small black, red, and yellow triangles = twice Small yellow and black triangles } (pr. 34.)

Large yellow triangle = Small yellow and black triangles .

Q. E. D.

Proposition XXXVIII. Theorem.

Proposition 38 figure

Triangles ( Red triangle and Blue triangle ) on equal baſes and between the ſame parallels are equal.

Draw Blue dotted line Blue line Red dotted line Red line } (pr. 31.)

Left parallelogram = Right parallelogram (pr. 36.);
Left parallelogram = twice Red triangle (pr. 34.),
and Right parallelogram = twice Blue triangle (pr. 34.),
Red triangle = Blue triangle (ax. 7.).

Q. E. D.

Proposition XXXIX. Theorem.

Proposition 39 figure

Equal triangles Yellow triangles and Black and yellow triangles on the ſame baſe (Black line) and on the ſame ſide of it, are between the ſame parallels.

If Blue line, which joins the vertices of the triangles, be not Black line, draw Red line Black line (pr. 31.), meeting Black dotted line.

Draw Yellow line.

Becauſe Red line Black line (conſt.)
Yellow triangles = Black, blue, and yellow triangles (pr. 37.):
but Yellow triangles = Black and yellow triangles (hyp.);

Black and yellow triangles = Black, blue, and yellow triangles , a part equal to the whole, which is abſurd.

Red line Black line; and in the ſame manner it can be demonſtrated, that no other line except
Blue line is Black line; Blue line Black line.

Q. E. D.

Proposition XL. Theorem.

Proposition 40 figure

Equal triangles ( Yellow triangle and Red triangle ) on equal baſes, and on the ſame ſide, are between the ſame parallels.

If Blue line which joins the vertices of triangles
be not Black and blue baseline ,
draw Red line Black and blue baseline (pr. 31.),
meeting Black dotted line.

Draw Yellow line.
Becauſe Red line Black and blue baseline (conſt.)
Yellow triangle = Red and blue triangles triangle but Yellow triangle = Red triangle
Red triangle = Red and blue triangles triangle , a part equal to the whole, which is abſurd.
Red line Black line: and in the ſame manner it can be demonſtrated, that no other line except
Blue line is Black line: Blue line Black line.

Q. E. D.

Proposition XLI. Theorem.

Proposition 41 figure

If a parallelogram Parallelogram and a triangle Blue, red, and yellow triangles are upon the ſame baſe Black line and between the ſame parallels Black dotted line and Black line, the parallelogram is double the triangle.

Draw Red line the diagonal;
Then Blue triangle = Blue, red, and yellow triangles (pr. 37.)
Parallelogram = twice Blue triangle (pr. 34.)
Parallelogram = twice Blue, red, and yellow triangles .

Q. E. D.

Proposition XLII. Theorem.

Proposition 42 figure

To conſtruct a parallelogram equal to a given triangle Black, blue, and yellow triangles and having an angle equal to a given rectilinear angle Yellow angle .

Make Black line = Black dotted line (pr. 10.)
Draw Yellow line.
Make Blue angle = Yellow angle (pr. 23.)

Draw { Red dotted line Red line Blue line Black line } (pr. 31.)

Parallelogram = twice Blue, and yellow triangles (pr. 41.)
but Blue, and yellow triangles = Black triangle (pr. 38.)
Parallelogram = Black, blue, and yellow triangles .

Q. E. D.

Proposition XLIII. Theorem.

Proposition 43 figure

The complements Blue parallelogram and Black parallelogram of the parallelograms which are about the diagonal of a parallelogram are equal.

Black parallelogram and red and yellow triangles = Blue parallelogram and red and yellow triangles (pr. 34.)
and Bottom red and yellow triangles = Top red and yellow triangles (pr. 34.)
Black parallelogram = Blue parallelogram (ax. 3.)

Q. E. D.

Proposition XLIV. Problem.

Proposition 44 figure

To a given ſtraight line (Black line) to apply a parallelogram equal to a given triangle ( Red triangle ), and having an angle equal to a given recilinear angle ( Yellow angle ).

Make Yellow parallelogram = Red triangle with Blue angle = Yellow angle (pr. 42.)
and having one of its ſides Black dotted line conterminous with and in continuation of Black line. Produce Blue line till it meets Yellow line Blue dotted line draw Red line produce it till it meets Red dotted line continued; draw Yellow dotted line Black dotted and solid lines meeting Yellow line produced, and produce Blue dotted line.

Yellow parallelogram = Blue parallelogram (pr. 43.)
but Yellow parallelogram = Red triangle (conſt.)
Blue parallelogram = Red triangle ; and
Blue angle = Red angle = Black angle = Yellow angle (pr. 29. and conſt.)

Q. E. D.

Proposition XLV. Problem.

Proposition 45 figure

To conſtruct a parallelogram equal to a given rectilinear figure ( Triangles ) and having an angle equal to a given rectilinear angle ( Red angle ).

Draw Red line and Blue line dividing the rectilinear figure into triangles.

Conſtruct Blue parallelogram = Blue triangle
having Yellow angle = Red angle (pr. 42.)
to Red line apply Yellow parallelogram = Yellow triangle
having Black angle = Red angle (pr. 44.)
to Blue line apply Red parallelogram = Red triangle
having Blue angle = Red angle (pr. 44.)
Parallelograms = Triangles
and Parallelograms is a parallelogram (prs. 29, 14, 30.)
having Blue angle = Red angle .

Q. E. D.

Proposition XLVI. Problem.

Proposition 46 figure

Upon a given ſtraight line (Black line) to conſtruct a ſquare.

Draw Blue line and = Black line (pr. 11. and 3.)

Draw Red line Black line, and meeting Yellow line drawn Blue line.

In Square Blue line = Black line (conſt.)
Yellow angle = a right angle (conſt.)
Red angle = Yellow angle = a right angle (pr. 29.),
and the remaining ſides and angles muſt be equal. (pr. 34.)
and Square is a ſquare. (def. 27.)

Q. E. D.

Proposition XLVII. Theorem.

Proposition 47 figure

In a right angled triangle Center triangle the ſquare on the hypotenuſe Red line is equal to the ſum of the ſquares of the ſides, (Blue line and Yellow line).

On Red line, Blue line and Yellow line deſcribe ſquares, (pr. 46.)

Draw Black dotted line Red dotted line (pr. 31.) alſo draw Black line and Black line.

Bottom yellow angle = Top yellow angle ,

To each add Black angle Black and bottom yellow angle = Black and top yellow angle ,
Red line = Red dotted line and Blue line = Blue dotted line;

Blue triangle and black angle = Red triangle and black angle .

Again, becauſe Yellow line Blue dotted line
Red square = twice Red triangle and black angle ,
and Blue rectangle = twice Blue triangle and black angle ;
Red square = Blue rectangle .

In the ſame manner it may be ſhown
that Black square = Yellow rectangle ;
hence Black and red squares = Yellow and blue rectangles .

Q. E. D.

Proposition XLVIII. Theorem.

Proposition 48 figure

If the ſquare of one ſide (Red line) of a triangle is equal to the ſquares of the other two ſides (Blue line and Black line), the angle ( Yellow angle ) ſubtended by that ſide is a right angle.

Draw Black dotted line Blue line and = Black line (prs. 11. 3.)
and draw Red dotted line alſo.

Since Black dotted line = Black line (conſt.)
Black dotted line2 = Black line2;
Black dotted line2 + Blue line2 = Black line2 + Blue line2,
but Black dotted line2 + Blue line2 = Red dotted line2 (pr. 47.),
and Black line2 + Blue line2 = Red line2 (hyp.)
Red dotted line2 = Red line2,
Red dotted line = Red line;
and Yellow angle = Red angle (pr. 8.),
conſequently Red angle is a right angle.

Q. E. D.