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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 extremities.

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 circumference 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 which 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 diagram B C D E F G H

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.

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, etc.

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 analogous 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 and 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 to 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 draw Black dotted line and,

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 line, 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 + Yellow 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 dotted line 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 thin 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 respectively 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 lines, (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 parallel to a given straight line (Blue 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. Theorem.

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 = Right triangle (pr. 8.)
∴ All shapes minus Right 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 rectilinear 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. 30.)

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 ( Red 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.

Books.

  1. Proposition 47 figure
    Book I. Basic plane geometry
  2. Proposition 7 figure
    Book II. Geometric algebra
  3. Proposition 21 figure 1
    Book III. Circles and angles
  4. Proposition 4 figure
    Book IV. Regular polygons
  5. Proposition 24 figure
    Book V. Ratios and proportions
  6. Proposition 3 figure
    Book VI. Geometric proportions