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Book VI.

Definitions.

Definition 1 figure

I.

Rectilinear figures are said to be similar, when they have their several angles equal, each to each, and the sides about the equal angles proportional.

II.

Two sides of one figure are said to be reciprocally proportional to two sides of another figure when one of the sides of the first is to the second, as the remaining side of the second is to the remaining side of the first.

III.

A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment, as the greater segment is to the less.

IV.

The altitude of any figure is the straight line drawn from its vertex perpendicular to its base, or the base produced.

Definition 4 figure

Proposition I. Theorem.

Proposition 1 figure

Triangles and parallelograms having the same altitude are to one another as their bases.

Let the triangles Red triangle and Blue triangle have a common vertex, and their bases, Blue line and Red line in the same straight line.

Produce Blue and red line both ways, take successively on Red line produced lines equal to it; and on Blue line produced lines successively equal to it; and draw lines from the common vertex to their extremities.

The triangles Black and red triangles thus formed are all equal to one another, since their bases are equal. (B. 1. pr. 38.)

∴ Black and red triangles and its base are respectively
equimultiples of Red triangle and the base Blue line.

In like manner Blue and yellow triangles and its base are respectively
equimultiples of Blue triangle and the base Red line.

∴ If m or 6 times Red triangle > = or < n or 5 times Blue triangle then m or 6 times Blue line > = or < n or 5 times Red line, m and n stand for every multiple taken as in the fifth definition of the Fifth Book. Although we have only shown that this property exists when m equal 6, and n equal 5, yet it is evident that the property holds good for every multiple value that may be given to m, and to n.

∴ Red triangle : Blue triangle :: Blue line : Red line (B. 5. def 5.)

Parallelograms having the same altitude are the doubles of triangles, on their bases, and are proportional to them (Part 1), and hence their doubles, the parallelograms, are as their bases. (B. 5. pr. 15.)

Q. E. D.

Proposition II. Theorem.

Proposition 2 figure

If a straight line Black line be drawn parallel to any side Black dotted line of a triangle, it shall cut the other sides, or those sides produced, into proportional segments.

And if any straight line Black line divide the sides of a triangle or those sides produced, into proportional segments, it is parallel to the remaining side Black dotted line.

Part I.

Let Black line ∥ Black dotted line, then shall
Yellow line : Red dotted line :: Yellow dotted line : Blue dotted line.

Draw Red line and Blue line,
and Black, red, and yellow triangle = Black, dotted yellow, and blue triangle (B. 1. pr. 37);
∴ Black, red, and yellow triangle : Dotted blue, black, dotted red triangle :: Black, dotted yellow, and blue triangle : Dotted blue, black, dotted red triangle (B .5. pr. 7); but
Black, red, and yellow triangle : Dotted blue, black, dotted red triangle :: Yellow line : Red dotted line (B. 6. pr. 1),
∴ Yellow line : Red dotted line :: Yellow dotted line : Blue dotted line.
(B. 5. pr. 11).

Part II.

Let Yellow line : Red dotted line :: Yellow dotted line : Blue dotted line,
then Black line ∥ Black dotted line.

Let the same construction remain,
because Yellow line : Red dotted line :: Black, red, and yellow triangle : Dotted blue, black, dotted red triangle and Yellow dotted line : Blue dotted line :: Black, dotted yellow, and blue triangle : Dotted blue, black, dotted red triangle } (B. 6. pr. 1)
but Yellow line : Red dotted line :: Yellow dotted line : Blue dotted line (hyp.),
∴ Black, red, and yellow triangle : Dotted blue, black, dotted red triangle :: Black, dotted yellow, and blue triangle : Dotted blue, black, dotted red triangle (B. 5. pr. 11.)
∴ Black, red, and yellow triangle = Black, dotted yellow, and blue triangle (B. 5. pr. 9);
but they are on the same base Black dotted line, and at the same side of it, and
∴ Black line ∥ Black dotted line (B. 1. pr. 39).

Q. E. D.

Proposition III. Theorem.

Proposition 3 figure

A right line (Blue line) bisecting the angle of a triangle, divides the opposite side into segments (Black line, Black dotted line) proportional to the conterminous sides (Red line, Yellow line).

And if a straight line (Blue line) drawn from any angle of a triangle divide the opposite side ( Black and dotted black line ) into segments (Black line, Black dotted line) proportional to the conterminous sides (Red line, Yellow line), it bisects the angle.

Part I.

Draw Blue dotted line ∥ Blue line, to meet Red dotted line;
then, Yellow angle = Blue angle (B. 1. pr. 29),
∴ Black angle = Blue angle ; but Black angle = Red angle , ∴ Red angle = Blue angle ,
∴ Red dotted line = Yellow line (B. 1. pr. 6);
and because Blue line ∥ Blue dotted line,
Red dotted line : Red line :: Black dotted line : Black line (B. 6. pr. 2)
but Red dotted line = Yellow line;
∴ Yellow line : Red line :: Black dotted line : Black line (B. 5. pr. 7).

Part II.

Let the same construction remain,
and Red line : Red dotted line :: Black line : Black dotted line (B. 6. pr. 2);
but Black line : Black dotted line :: Red line : Yellow line (hyp.)
∴ Red line : Red dotted line :: Red line : Yellow line (B. 5. pr. 11).
and ∴ Red dotted line = Yellow line (B. 5. pr. 9),
and ∴ Blue angle = Red angle (B. 5. pr. 5); but since
Blue line ∥ Blue dotted line; Black angle = Red angle ,
and Yellow angle = Blue angle (B. 1. pr. 29);
∴ Blue angle = Red angle , and Yellow angle = Black angle ,
and ∴ Blue line bisects Yellow and black angle .

Q. E. D.

Proposition IV. Theorem.

Proposition 4 figure

In equiangular triangles ( Left triangle and Right triangle ) the sides about the equal angles are proportional, and the sides which are opposite to the equal angles are homologous.

Let the equiangular triangles be so placed that two sides Black line , Black dotted line opposite to equal angles Red arc and Black arc may be conterminous and in the same straight line; and that the triangles lying at the same side of that straight line, may have the equal angles not conterminous,

i.e. Red angle opposite to Yellow angle , and Blue angle to Black angle .

Draw Yellow dotted line and Yellow line. Then, because
Blue angle = Black angle , Red line ∥ Yellow and dotted red line (B. 1. pr. 28);
and for a like reason, Blue dotted line ∥ Blue and dotted yellow line ,
∴ Parallelogram is a parallelogram.
But Black line : Black dotted line :: Yellow line : Red dotted line (B. 6. pr. 2);
and since Yellow line = Red line (B. 1. pr. 34),
Black line : Black dotted line :: Red line : Red dotted line; and by
alternation, Black line : Red line :: Black dotted line : Red dotted line (B. 5. pr. 16).

In like manner it may be shown, that
Blue line : Blue dotted line :: Black line : Black dotted line;
and by alternation, that
Blue line : Black line:: Blue dotted line : Black dotted line;
but it has already been proved that
Black line : Red line :: Black dotted line : Red dotted line,
and therefore, ex æquali,
Blue line : Red line :: Blue dotted line : Red dotted line
(B. 5. pr. 22),
therefore the sides about the equal angles are proportional, and those which are opposite to the equal angles are homologous.

Q. E. D.

Proposition V. Theorem.

Proposition 5 figure

If two triangles have their sides proportional (Blue dotted line : Black dotted line :: Blue line : Black line) and (Black dotted line : Red dotted line :: Black line : Red line) they are equiangular, and the equal angles are subtended by the homologous sides.

From the extremities of Black line, draw Yellow line and Yellow dotted line,
making Left blue angle = Right blue angle , Left red angle = Right red angle (B. 1. pr. 23);
and consequently Top black angle = Right yellow angle (B. 1. pr. 32),
and since the triangles are equiangular,
Red dotted line : Black dotted line :: Yellow line : Black line (B. 6. pr. 4);
but Red dotted line : Black dotted line :: Red line : Black line (hyp.);
∴ Red line : Black line :: Yellow line : Black line,
and consequently Red line = Yellow line (B. 5. pr. 9).

In the like manner it may be shown that
Blue line = Yellow dotted line.

Therefore, the two triangles having a common base Black line, and their sides equal, have also equal angles opposite to equal sides, i.e.

Left yellow angle = Left blue angle and Red outlined angle = Left red angle (B. 1. pr. 8).

But Left blue angle = Right blue angle (const.)
and ∴ Left yellow angle = Right blue angle ; for the same
reason Red outlined angle = Right red angle , and
consequently Bottom black angle = Right yellow angle (B. 1. 32);

and therefore the triangles are equiangular, and it is evident that the homologous sides subtend the equal angles.

Q. E. D.

Proposition VI. Theorem.

Proposition 6 figure

If two triangles ( Right triangle and Top triangle ) have one angle ( Right red angle ) of the one, equal to one angle ( Red arc ) of the other, and the sides about the equal angles proportional, the triangles shall be equiangular, and have those angles equal which the homologous sides subtend.

From the extremities of Black line, one of the sides
of Top triangle , about Red arc , draw
Yellow line and Yellow dotted line, making
Left red angle = Right red angle , and Left blue angle = Right blue angle ; then Black angle = Yellow angle
(B. 1. pr. 32), and two triangles being equiangular,
Blue dotted line : Black dotted line :: Yellow dotted line : Black line (B. 6. pr. 4);
but Blue dotted line : Black dotted line :: Blue line : Black line (hyp.);
∴ Yellow dotted line : Black line :: Blue line : Black line (B. 5. pr. 11),
and consequently Yellow dotted line = Blue line (B. 5. pr. 9);
∴ Top triangle = Bottom triangle in every respect.
(B. 1. pr. 4).

But Left blue angle = Right blue angle (const.),
and ∴ Blue arc = Right blue angle ; and
since also Red arc = Right red angle ,
Yellow arc = Yellow angle (B. 1. pr. 32);
and ∴ Right triangle and Top triangle are equiangular, with their equal angles opposite to homologous sides.

Q. E. D.

Proposition VII. Theorem.

Proposition 7 figure

If two triangles ( Left triangle and Right triangle ) have one angle in each equal ( Yellow arc equal to Blue angle ), the sides about two other angles proportional (Red line : Yellow line :: Red dotted line : Yellow dotted line), and each of the remaining angles ( Bottom red angle and Red arc ) either less or not less than a right angle, the triangles are equiangular, and those angles are equal about which the sides are proportional.

First let it be assumed that the angles Bottom red angle and Red arc are each less than a right angle: then if it be supposed
that Black angle and arc and Blue arc contained by the proportional sides
are not equal, let Black angle and arc be the greater, and make
Black arc = Blue arc .

Because Blue angle = Yellow arc (hyp.), and Black arc = Blue arc (const.)
∴ Yellow angle = Red arc (B. 1. pr. 32);
∴ Red line : Blue line :: Red dotted line : Yellow dotted line (B. 6. pr. 4),
but Red line : Yellow line :: Red dotted line : Yellow dotted line (hyp.)
∴ Red line : Blue line :: Red line : Yellow line;
∴ Blue line = Yellow line (B. 5. pr. 9),
and ∴ Bottom red angle = Top red angle (B. 1. pr. 5).

But Bottom red angle is less than a right angle (hyp.)
∴ Top red angle is less than a right angle; and ∴ Yellow angle must be greater than a right angle (B. 1. pr. 13), but it has been proved = Red arc and therefore less than a right angle, which is absurd. ∴ Black angle and arc and Blue arc are not unequal;
∴ they are equal, and since Blue angle = Yellow arc (hyp.)

∴ Bottom red angle = Red arc (B. 1. pr. 32), and therefore the triangles are equiangular.

But if Bottom red angle and Red arc be assumed to be each not less than a right angle, it may be proved as before that the triangles are equiangular, and have the sides about the equal angles proportional. (B. 6. pr. 4).

Q. E. D.

Proposition VIII. Theorem.

Proposition 8 figure

In a right angled triangle ( Yellow and red triangle ), if a perpendicular (Black line) be drawn from the right angle to the opposite side, the triangles ( Yellow triangle , Red triangle ) on each side of it are similar to the whole triangle and to each other.

Because Red and yellow angle = Blue angle (B. 1. ax. II), and
Black angle common to Yellow and red triangle and Yellow triangle ;
Blue arc = Red angle (B. 1. pr. 32);

∴ Yellow and red triangle and Yellow triangle are equiangular; and consequently have their sides about the equal angles proportional (B. 6. pr. 4), and are therefore similar (B. 6. def. 1).

In like manner it may be proved that Red triangle is similar to
Yellow and red triangle ; but Yellow triangle has been shewn to be similar
to Yellow and red triangle ; ∴ Yellow triangle and Red triangle are
similar to the whole and to each other.

Q. E. D.

Proposition IX. Problem.

Proposition 9 figure

From a given straight line ( Yellow and dotted yellow line ) to cut off any required part.

From either extremity of the given line draw Blue and dotted blue line making any angle with Yellow and dotted yellow line ; and produce Blue and dotted blue line till the whole produced line Blue, dotted blue, and dotted black line contains Blue line as often as Yellow and dotted yellow line contains the required part.

Draw Red line, and draw Red dotted line ∥ Red line.
Yellow line is the required part of Yellow and dotted yellow line .

For since Red dotted line ∥ Red line
Yellow line : Yellow dotted line :: Blue line : Blue dotted line
(B. 6. pr. 2), and by composition (B. 5. pr. 18);
Yellow and dotted yellow line : Yellow line :: Blue and dotted blue line : Blue line;
but Blue and dotted blue line contains Blue line as often
as Yellow and dotted yellow line contains the required part (const.);
∴ Yellow line is the required part.

Q. E. D.

Proposition X. Problem.

Proposition 10 figure

To divide a straight line ( Thin blue, red, and yellow line ) similarly to a given divided line ( Blue, red, and yellow line ).

From either extremity of the given line
Thin blue, red, and yellow line draw Dotted blue, red, and yellow line making any angle;
take Blue dotted line, Red dotted line and Yellow dotted line
equal to Blue line, Red line and Yellow line respectively (B. 1. pr. 2);
draw Black thin line, and draw Black dotted line and Black line ∥ to it.

Since { Black thin line Black dotted line Black line } are ∥
Yellow thin line : Red thin line :: Yellow dotted line : Red dotted line (B. 6. pr. 2),
or Yellow thin line : Red thin line :: Yellow line : Red line (const.),
and Red thin line : Blue thin line :: Red dotted line : Blue dotted line (B. 6. pr. 2),
Red thin line : Blue thin line :: Red line : Blue line (const.),
and ∴ the given line Thin blue, red, and yellow line is divided similarly to Blue, red, and yellow line .

Q. E. D.

Proposition XI. Problem.

Proposition 11 figure

To find a third proportional to two given straight lines (Black line and Blue line).

At either extremity of the given line Black line
draw Dotted red and red line making an angle;
take Red dotted line = Blue line, and draw Yellow line;
make Blue dotted line = Blue line,
and draw Yellow dotted line ∥ Yellow line; (B. 1. pr. 31.)
Red line is the third proportional to Black line and Blue line.

For since Yellow line ∥ Yellow dotted line,
∴ Black line : Blue dotted line :: Red dotted line : Red line (B. 6. pr. 2);
but Blue dotted line = Red dotted line = Blue line (const.);
∴ Black line : Blue line :: Blue line : Red line.
(B. 5. pr. 7).

Q. E. D.

Proposition XII. Problem.

Proposition 12 figure

To find a fourth proportional to three given lines { Dotted blue, red, and yellow lines } .

Draw Blue and red line
and Yellow and black line making any angle;
take Blue line = Blue dotted line,
and Red line = Red dotted line,
also Yellow line = Yellow dotted line,
draw Black thin line,
and Black dotted line ∥ Black thin line; (B. 1. pr. 31);
Black line is the fourth proportional.

On account of the parallels,
Blue line : Red line :: Yellow line : Black line (B. 6. pr. 2);
but { Dotted blue, red, and yellow lines } = { Blue, red and yellow lines } (const.);

∴ Blue dotted line : Red dotted line :: Yellow dotted line : Black line. (B. 5. pr. 7).

Q. E. D.

Proposition XIII. Problem.

Proposition 13 figure

To find a mean proportional between two given straight lines { Dotted blue and red lines }.

Draw any straight line Blue and red line , make Blue line = Blue dotted line,
and Red line = Red dotted line; bisect Blue and red line ;
and from the point of bisection as a centre, and half the
line as a radius, describe a semicircle Semicircle ,
draw Black line ⊥ Blue line:
Black line is the mean proportional required.

Draw Yellow line and Yellow dotted line.

Since Angle is a right angle (B. 3. pr. 31),
and Black line is ⊥ from it upon the opposite side,
∴ Black line is a mean proportional between
Blue line and Red line (B. 6. pr. 8),
and ∴ between Blue dotted line and Red dotted line (const.).

Q. E. D.

Proposition XIV. Theorem.

Proposition 14 figure

I.

Equal parallelograms Blue parallelogram and Yellow parallelogram , which have one angle in each equal, have the sides about the equal angles reciprocally proportional (Red line : Black line :: Yellow line : Blue line)

II.

And parallelograms which have one angle in each equal, and the sides about them reciprocally proportional, are equal.

Let Red line and Black line; and Yellow line and Blue line, be so placed that Red and black line and Yellow and blue line may be continued right lines. It is evident that they may assume this position. (B. 1. prs. 13, 14, 15.)

Complete Red parallelogram .

Since Yellow parallelogram = Blue parallelogram ;

∴ Yellow parallelogram : Red parallelogram :: Blue parallelogram : Red parallelogram (B. 5. pr. 7.)
∴ Red line : Black line :: Yellow line : Blue line (B. 6. pr. 1.)

The same construction remaining:
Red line : Black line :: { Yellow parallelogram : Red parallelogram (B. 6. pr. 1.) Yellow line : Blue line (hyp.) Blue parallelogram : Red parallelogram (B. 6. pr. 1.)
∴ Yellow parallelogram : Red parallelogram :: Blue parallelogram : Red parallelogram (B. 5. pr. 11.)
and ∴ Yellow parallelogram = Blue parallelogram (B. 5. pr. 9).

Q. E. D.

Proposition XV. Theorem.

Proposition 15 figure

I.

Equal triangles, which have one angle in each equal ( Blue angle = Red angle ), have the sides about the equal angles reciprocally proportional (Blue line : Black line :: Red line : Yellow line).

II.

And two triangles which have an angle of the one equal to an angle of the other, and the sides about the equal angles reciprocally proportional, are equal.

I.

Let the triangles be so placed that the equal angles Blue angle and Red angle may be vertically opposite, that is to say, so that Blue line and Black line may be in the same straight line. Whence also Red line and Yellow line must be in the same straight line (B. 1. pr. 14.)

Draw Black dotted line, then

Blue line : Black line :: Red triangle : Blue triangle (B. 6. pr. 1.) :: Yellow triangle : Blue triangle (B. 5. pr. 7.) :: Red line : Yellow line (B. 6. pr. 1.) ∴ Blue line : Black line :: Red line : Yellow line (B. 5. pr. 11.)

II.

Let the same construction remain, and
Red triangle : Blue triangle :: Blue line : Black line (B. 6. pr. 1.)
and Red line : Yellow line :: Yellow triangle : Blue triangle (B. 6. pr. 1.)

But Blue line : Black line :: Red line : Yellow line, (hyp.)
∴ Red triangle : Blue triangle :: Yellow triangle : Blue triangle (B. 5. pr. 11);
∴ Red triangle = Yellow triangle (B. 5. pr. 9.)

Q. E. D.

Proposition XVI. Theorem.

Proposition 16 figure

Part I.

If four straight lines be proportional (Yellow line : Blue line :: Red dotted line : Black dotted line), the rectangle (Yellow line × Black dotted line) contained by the extremes, is equal to the rectangle (Blue line × Red dotted line) contained by the means.

Part II.

And if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines are proportional.

Part I.

From the extremities of Yellow line and Blue line draw Black line and Red line ⊥ to them and = Black dotted line and Red dotted line respectively: complete the parallelograms:

Red rectangle and Yellow rectangle .

And since, Yellow line : Blue line :: Red dotted line : Black dotted line (hyp.) ∴ Yellow line : Blue line :: Red line : Black line (const.) ∴ Red rectangle = Yellow rectangle (B. 6. pr. 14),

that is, the rectangle contained by the extremes, equal to the rectangle contained by the means.

Part II.

Let the same construction remain; beause
Black dotted line = Black line, Red rectangle = Yellow rectangle
and Red line = Red dotted line,
∴ Yellow line : Blue line :: Red line : Black line (B. 6. pr. 14).

But Red line = Red dotted line,
and Black line = Black dotted line (const.)
∴ Yellow line : Blue line :: Red dotted line : Black dotted line (B. 5. pr. 7).

Q. E. D.

Proposition XVII. Theorem.

Proposition 17 figure

Part I.

If three straight lines be proportional (Red line : Blue line :: Blue line : Black line) the rectangle under the extremes is equal to the square of the mean.

Part II.

And if the rectangle under the extremes be equal to the square of the mean, the three straight lines are proportional.

Part I.

Assume Yellow line = Blue line, and since Red line : Blue line :: Blue line : Black line, then Red line : Blue line :: Yellow line : Black line, ∴ Red line × Black line = Blue line × Yellow line
(B. 6. pr. 16).

But Yellow line = Blue line,
∴ Blue line × Yellow line = Blue line × Blue line, or = Blue line2;
therefore, if the three straight lines are proportional, the rectangle contained by the extremes is equal to the square of the mean.

Part II.

Assume Yellow line = Blue line, then
Red line × Black line = Yellow line × Blue line.
∴ Red line : Blue line :: Yellow line : Black line (B. 6. pr. 16), and
∴ Red line : Blue line :: Blue line : Black line.

Q. E. D.

Proposition XVIII. Theorem.

Proposition 18 figure

On a given straight line (Black line) to construct a rectilinear figure similar to a given one ( Top rectilinear figure ) and similarly placed.

Resolve the given figure into triangles by drawing the lines Red dotted line and Yellow dotted line.

At the extremities of Black line make
Bottom blue angle = Top blue angle and Bottom red angle = Top red angle :
again at the extremities of Red line make Top black and red angle = Bottom red and black angle
and Bottom black angle = Top black angle : in like manner make
Bottom yellow angle = Top yellow angle and Bottom red and yellow angle = Top red and yellow angle .

Then Bottom rectilinear figure is similar to Top rectilinear figure .

It is evident from the construction and (B. 1. pr. 32) that the figures are equiangular; and since the triangles
Top yellow triangle and Bottom yellow triangle are equiangular; then by (B. 6. pr. 4),

Black line : Blue line :: Black dotted line : Blue thin line and Blue line : Red line :: Blue thin line : Red dotted line.

Again, because Top blue triangle and Bottom blue triangle are equiangular,
Red line : Blue dotted line :: Red dotted line : Yellow line;
∴ ex æquali,
Blue line : Blue dotted line :: Blue thin line : Yellow line (B. 6. pr. 22.)

In like manner it may be shown that the remaining sides of the two figures are proportional.

∴ by (B. 6. def. 1.)
Bottom rectilinear figure is similar to Top rectilinear figure
and similarly situated; and on the given line Black line.

Q. E. D.

Proposition XIX. Theorem.

Proposition 19 figure

Similar triangles ( Yellow triangle and Red and blue triangle ) are to one another in the duplicate ratio of their homologous sides.

Let Black angle and Red angle be equal angles, and Dotted black and black line and let Blue line homologous sides of the similar triangles Yellow triangle and Red and blue triangle and on Dotted black and black line the greater of these lines take Black dotted line a third proportional, so that

Dotted black and black line : Blue line :: Blue line : Black dotted line;
draw Yellow dotted line.

Dotted black and black line : Yellow line :: Blue line : Red line (B. 6. pr. 4);
∴ Dotted black and black line : Blue line :: Yellow line : Red line (B. 5. pr. 16, alt.),
but Dotted black and black line : Blue line :: Blue line : Black dotted line (const.),
∴ Blue line : Black dotted line :: Yellow line : Red line
consequently Yellow triangle = Blue triangle for they have the sides about
the equal angles Black angle and Red angle reciprocally proportional (B. 6. pr. 15);
∴ Red and blue triangle : Yellow triangle :: Red and blue triangle : Blue triangle (B. 5. pr. 7),
but Red and blue triangle : Blue triangle :: Dotted black and black line : Black dotted line (B. 6. pr. 1),
∴ Red and blue triangle : Yellow triangle :: Dotted black and black line : Black dotted line,
that is to say, the triangles are to one another in the duplicate ratio of their homologous sides
Blue line and Dotted black and black line (B. 5. def. 11).

Q. E. D.

Proposition XX. Theorem.

Proposition 20 figure

Similar polygons may be divided into the same number of similar triangles, each similar pair of which are proportional to the polygons; and the polygons are to each other in the duplicate ratio of their homologous sides.

Draw Black line and Black dotted line, and Black thin line and Black thin dotted line, resolving the polygons into triangles. Then because the polygons are similar, Top black angle = Bottom black angle , and Blue line : Blue dotted line :: Red thin line : Red dotted line

∴ Top yellow triangle and Bottom yellow triangle are similar, and Top red angle = Bottom red angle (B. 6. pr. 6);
but Top blue and red angle = Bottom blue and red angle because they are angles of similar polygons;
therefore the remainders Top blue angle and Bottom blue angle are equal;
hence Black dotted line : Blue dotted line :: Black thin dotted line : Red dotted line,
on account of the similar triangles,
and Blue dotted line : Yellow line :: Red dotted line : Yellow line,
on account of the similar polygons,
∴ Black dotted line : Yellow line :: Black thin dotted line : Yellow line,
ex æquali (B. 5. pr. 22), and as these proportional sides
contain equal angles, the triangles Top red triangle and Bottom red triangle
are similar (B. 6. pr. 6).

In like manner it may be shown that the
triangles Bottom blue triangle and Top blue triangle are similar.

But Top yellow triangle is to Bottom yellow triangle in the duplicate ratio of
Black dotted line to Black thin dotted line (B. 6. pr. 19), and
Top red triangle is to Bottom red triangle in like manner, in the duplicate
ratio of Black dotted line to Black thin dotted line;
∴ Top yellow triangle : Bottom yellow triangle :: Top red triangle : Bottom red triangle , (B. 5. pr. 11);

Again Top red triangle is to Bottom red triangle in the duplicate ratio of
Black line to Black thin line, and Top blue triangle is to Bottom blue triangle in
the duplicate ratio of Black line to Black thin line,

∴ Top yellow triangle : Bottom yellow triangle :: Top red triangle : Bottom red triangle :: Top blue triangle : Bottom blue triangle ;

and as one of the antecedents is to one of the consequents, so is the sum of all the antecedents to the sum of all the consequents; that is to say, the similar triangles have to one another the same ratio as the polygons (B. 5. pr. 12).

But Top yellow triangle is to Bottom yellow triangle in the duplicate ratio of
Blue line to Red thin line;

∴ Top polygon is to Bottom polygon in the duplicate
ratio of Blue line to Red thin line.

Q. E. D.

Proposition XXI. Theorem.

Proposition 21 figure

Rectilinear figures ( Red triangle and Blue triangle ) which are similar to the same figure ( Yellow triangle ) are similar also to each other.

Since Red triangle and Yellow triangle are similar, they are equiangular, and have the sides about the equal angles proportional (B. 6. def. 1); and since the figures Blue triangle and Yellow triangle are also similar, they are equiangular, and have the sides about the equal angles proportional; therefore Red triangle and Blue triangle are also equiangular, and have the sides about the equal angles proportional (B. 5. pr. 11), and are therefore similar.

Q. E. D.

Proposition XXII. Theorem.

Proposition 22 figure

Part I.

If four straight lines be proportional (Black line : Blue line :: Red line : Yellow line), the similar rectilinear figures similarly described on them are also proportional.

Part II.

And if four similar rectilinear figures, similarly described on four straight lines, be proportional, the straight lines are also proportional.

Part I.

Take Blue dotted line a third proportional to Black line
and Blue line, and Red dotted line a third proportional
to Red line and Yellow line (B. 6. pr. 11); since Black line : Blue line :: Red line : Yellow line (hyp.), Blue line : Blue dotted line :: Yellow line : Red dotted line (const.)

∴ ex æquali,
Black line : Blue dotted line :: Red line : Red dotted line;
but Yellow triangle : Red triangle :: Black line : Blue dotted line (B. 6. pr. 20),
and Blue polygon : Outlined polygon :: Red line : Red dotted line;
∴ Yellow triangle : Red triangle :: Blue polygon : Outlined polygon (B. 5. pr. 11).

Part II.

Let the same construction remain:

Yellow triangle : Red triangle :: Blue polygon : Outlined polygon (hyp.),
∴ Black line : Blue dotted line :: Red line : Red dotted line (const.)
and ∴ Black line : Blue line :: Red line : Yellow line. (B. 5. pr. 11).

Q. E. D.

Proposition XXIII. Theorem.

Proposition 23 figure

Equiangular parallelograms ( Yellow parallelogram and Blue parallelogram ) are to one another in a ratio compounded of the ratios of their sides.

Let two of the sides Blue line and Yellow dotted line about the equal angles be placed so that they may form one straight line.

Since Red angle + Yellow angle = Two right angles ,
and Black angle = Red angle (hyp.),
Black angle + Yellow angle = Two right angles ,
and ∴ Red line and Black line form one straight line (B. 1. pr. 14);
complete Red parallelogram .

Since Yellow parallelogram : Red parallelogram :: Blue line : Yellow dotted line (B. 6. pr. 1),
and Red parallelogram : Blue parallelogram :: Black line : Red line (B. 6. pr. 1),
Yellow parallelogram has to Blue parallelogram a ratio compounded of the ratios of
Blue line to Yellow dotted line, and of Black line to Red line.

Q. E. D.

Proposition XXIV. Theorem.

Proposition 24 figure

In any parallelogram ( Parallelogram ) the parallelograms ( Red parallelogram and Blue parallelogram ) which are about the diagonal are similar to the whole, and to each other.

As Parallelogram and Blue parallelogram have a
common angle they are equiangular;
but because Red line ∥ Blue and dotted blue line
Blue triangle and Blue, red, and yellow triangle are similar (B. 6. pr. 4),
∴ Yellow line : Red line :: Yellow and dotted red line : Blue and dotted blue line ;
and the remaining opposite sides are equal to those,
∴ Blue parallelogram and Parallelogram have the sides about the equal
angles proportional, and are therefore similar.

In the same manner it can be demonstrated that the
parallelograms Parallelogram and Red parallelogram are similar.

Since, therefore, each of the parallelograms
Blue parallelogram and Red parallelogram is similar to Parallelogram ,
they are similar to each other.

Q. E. D.

Proposition XXV. Problem.

Proposition 25 figure

To describe a rectilinear figure which shall be similar to a given rectilinear figure ( Red triangle ), and equal to another ( Polygon ).

Upon Yellow line describe Blue rectangle = Red triangle ,
and upon Red line describe Outlined rectangle and yellow angle = Polygon ,
and having Yellow angle = Red angle (B. 1. pr. 45), and then
Yellow line and Black dotted line will lie in the same straight line (B. 1. prs. 29, 14),

Between Yellow line and Black dotted line find a mean proportional
Blue line (B. 6. pr. 13), and upon Blue line
describe Yellow triangle , similar to Red triangle , and similarly situated.

Then Yellow triangle = Polygon .

For since Red triangle and Yellow triangle are similar, and
Yellow line : Blue line :: Blue line : Black dotted line (const.),
Red triangle : Yellow triangle :: Yellow line : Black dotted line (B. 6. pr. 20);
but Blue rectangle : Outlined rectangle and yellow angle :: Yellow line : Black dotted line (B. 6. pr. 1);
∴ Red triangle : Yellow triangle :: Blue rectangle : Outlined rectangle and yellow angle (B. 5. pr. 11);
but Red triangle = Blue rectangle (const.),
and ∴ Yellow triangle = Outlined rectangle and yellow angle (B. 5. pr. 14);
and Outlined rectangle and yellow angle = Polygon (const.); consequently,
Yellow triangle which is similar to Red triangle is also = Polygon .

Q. E. D.

Proposition XXVI. Theorem.

Proposition 26 figure

If similar and similarly posited parallelograms ( Red paralellogram and Gnomon ) have a common angle, they are about the same diagonal.

For, if it be possible, let Arc
be the diagonal of Gnomon and
draw Yellow line ∥ Red line (B. 1. pr. 31).

Since Outlined paralellogram and Gnomon are about the same
diagonal Arc , and have Angle common,
they are similar (B. 6. pr. 24);

∴ Red line : Blue line :: Red and dotted red line : Blue, dotted blue, and dotted black line ;
but Red line : Blue and dotted blue line :: Red and dotted red line : Blue, dotted blue, and dotted black line (hyp.),
∴ Red line : Blue line :: Red line : Blue and dotted blue line ,
and ∴ Blue line = Blue and dotted blue line (B. 5. pr. 9.),
which is absurd.

∴ Arc is not the diagonal of Gnomon
in the same manner it can be demonstrated that no other
line is except Double line .

Q. E. D.

Proposition XXVII. Theorem.

Proposition 27 figure

Of all the rectangles contained by the segments of a given straight line, the greatest is the square which is described on half the line.

Let Yellow, red, and blue line be the given line, Yellow line and Red, and blue line unequal segments, and Yellow and red line and Blue line equal segments;
then Red and blue square > Blue and yellow rectangle .

For it has been demonstrated already (B. 2. pr. 5), that the square of half the line is equal to the rectangle contained by any unequal segments together with the square of the part intermediate between the middle point and the point of unequal section. The square described on half the line exceeds therefore the rectangle contained by any unequal segments of the line.

Q. E. D.

Proposition XXVIII. Problem.

Proposition 28 figure

To divide a given straight line ( Dotted red, blue, and dotted blue line ) so that the rectangle contained by its segments may be equal to a given area, not exceeding the square of half the line.

Let the given area be = Yellow dotted line2.

Bisect Dotted red, blue, and dotted blue line , or
make Dotted red and blue line = Blue dotted line;
and if Dotted red and blue line 2 = Yellow dotted line2,
the problem is solved.

But if Dotted red and blue line 2 ≠ Yellow dotted line2, then
must Dotted red and blue line > Yellow dotted line (hyp.).

Draw Red line ⊥ Dotted red and blue line = Yellow dotted line;
make Red and dotted black line = Dotted red and blue line or Blue dotted line;
with Red and dotted black line as radius describe a circle cutting the
given line; draw Yellow line.

Then Red dotted line × Blue and dotted blue line + Blue line2 = Dotted red and blue line 2
(B. 2. pr. 5.) = Yellow line2.

But Yellow line2 = Red line2 + Blue line2 (B. 1. pr. 47);
∴ Red dotted line × Blue and dotted blue line + Blue line2
= Red line2 + Blue line2,
from both, take Blue line2,
and Red dotted line × Blue and dotted blue line = Red line2.

But Red line = Yellow dotted line (const.),
and ∴ Dotted red, blue, and dotted blue line is so divided
that Red dotted line × Blue and dotted blue line = Yellow dotted line2.

Q. E. D.

Proposition XXIX. Problem.

Proposition 29 figure

To produce a given straight line ( Blue and dotted blue line ), so that the rectangle contained by the segments between the extremities of the given line and the point to which it is produced, may be equal to a given area, i.e. equal to the square on Black line.

Make Blue line = Blue dotted line, and
draw Red dotted line ⊥ Blue dotted line = Black line;
draw Red line; and
with the radius Red line, describe a circle
meeting Blue and dotted blue line produced.

Then Blue, dotted blue, and yellow line × Yellow line + Blue dotted line2 =
Dotted blue, and yellow line 2 (B. 2. pr. 6.) = Red line2.

But Red line2 = Red dotted line2 + Blue dotted line2 (B. 1. pr. 47.)

∴ Blue, dotted blue, and yellow line × Yellow line + Blue dotted line2 =
Red dotted line2 + Blue dotted line2,
from both take Blue dotted line2,
and Blue, dotted blue, and yellow line × Yellow line = Red dotted line2;
but Red dotted line = Black line,
∴ Red dotted line2 = the given area.

Q. E. D.

Proposition XXX. Problem.

Proposition 30 figure

To cut a given finite straight line ( Red and dotted red line ) in extreme and mean ratio.

On Red and dotted red line describe the square Yellow and blue square (B. 1. pr. 46);
and produce Blue line, so that
Blue and dotted blue line × Blue dotted line = Red and dotted red line 2 (B. 6. pr. 29);
take Red line = Blue dotted line, and draw Black line ∥ Blue and dotted blue line , meeting Yellow line ∥ Red and dotted red line (B. 1. pr. 31).

Then Yellow and outlined rectangles = Blue and dotted blue line × Blue dotted line,
and is ∴ = Yellow and blue square ; and if from both these equals
be taken the common part Yellow rectangle ,
Outlined square , which is the square of Red line,
will be = Blue rectangle , which is = Red and dotted red line × Red dotted line;
that is Red line2 = Red and dotted red line × Red dotted line;
∴ Red and dotted red line : Red line :: Red line : Red dotted line,
and Red and dotted red line is divided in extreme and mean ratio (B. 6. def. 3).

Q. E. D.

Proposition XXXI. Theorem.

Proposition 31 figure

If any similar rectilinear figures be similarly described on the sides of a right angled triangle ( Triangle ), the figure described on the side ( Dotted blue and blue line ) subtending the right angle is equal to the sum of the figures on the other sides.

From the right angle draw Black line perpendicular to Dotted blue and blue line ;
then Dotted blue and blue line : Red line :: Red line : Blue line (B. 6. pr. 8).

∴ Red rectangle : Yellow rectangle :: Dotted blue and blue line : Blue line (B. 6. pr. 20).

but Red rectangle : Blue rectangle :: Dotted blue and blue line : Blue dotted line (B. 6. pr. 20).

Hence Blue dotted line + Blue line : Dotted blue and blue line
:: Blue rectangle + Yellow rectangle : Red rectangle ;
but Blue dotted line + Blue line = Dotted blue and blue line ;
and ∴ Blue rectangle + Yellow rectangle = Red rectangle .

Q. E. D.

Proposition XXXII. Theorem.

Proposition 32 figure

If two triangles ( Left triangle and Right triangle ), have two sides proportional (Blue line : Red line :: Blue dotted line : Red dotted line), and be so placed at an angle that the homologous sides are parallel, the remaining sides (Yellow line and Yellow dotted line) form one right line.

Since Blue line ∥ Blue dotted line,
Top yellow angle = Bottom yellow angle (B. 1. pr. 29);
and also since Red line ∥ Red dotted line,
Bottom yellow angle = Black angle (B. 1. pr. 29);
∴ Top yellow angle = Black angle ; and since
Blue line : Red line :: Blue dotted line : Red dotted line (hyp.),
the triangles are equiangular (B. 6. pr. 6);

∴ Red angle = Outlined angle :
but Top yellow angle = Bottom yellow angle ;

∴ Blue angle + Bottom yellow angle + Outlined angle = Blue angle + Top yellow angle + Red angle =
Two right angles (B. 1. pr. 32), and ∴ Yellow line and Yellow dotted line
lie in the same straight line (B. 1. pr. 14).

Q. E. D.

Proposition XXXIII. Theorem.

Proposition 33 figure
Proposition 33 figure

In equal circles (Red circle, Blue circle ), angles, whether at the centre or circumference, are in the same ratio to one another as the arcs on which they stand ( Black angle : Black outlined angle :: Thin black arc : Yellow dotted arc ); so also are sectors.

Take in the circumference of Red circle any number of arcs Thin red arc , Thin blue arc , etc. each = Thin black arc , and also in the circumference of Blue circle take any number of arcs Red dotted arc , Blue dotted arc , etc. each = Yellow dotted arc , draw the radii to the extremities of the equal arcs.

Then since the arcs Thin black arc , Thin red arc , Thin blue arc , etc. are all equal, the angles Black angle , Red angle , Blue angle , etc. are also equal (B. 3. pr. 27);

∴ Black, red, and blue angle is the same multiple of Black angle which the arc Thin black, red, and blue arc is of Thin black arc ; and in the same manner Black, red, and blue outlined angle is the same multiple of Black outlined angle , which the arc Yellow, red, and blue dotted arc is of the arc Yellow dotted arc .

Then it is evident (B. 3. pr. 27),
if Black, red, and blue angle (or if m times Black angle ) >, =, < Black, red, and blue outlined angle
(or n times Black outlined angle )
then Thin black, red, and blue arc (or m times Thin black arc ) >, =, <
Yellow, red, and blue dotted arc (or n times Yellow dotted arc );

∴ Black angle : Black outlined angle :: Thin black arc : Yellow dotted arc , (B. 5. def. 5), or the angles at the centre are as the arcs on which they stand; but the angles at the circumference being halves of the angles at the centre (B. 3. pr. 20) are in the same ratio (B. 5. pr. 15), and therefore are as the arcs on which they stand.

It is evident that sectors in equal circles, and on equal arcs are equal (B. 1. pr. 4; B. 3. prs. 24, 27, and def. 9). Hence, if the sectors be substituted for the angles in the above demonstration, the second part of the proposition will be established, that is, in equal circles the sectors have the same ratio to one another as the arcs on which they stand.

Q. E. D.

Proposition A. Theorem.

Proposition A figure

If the right line (Red dotted line) bisecting an external angle Black and blue angle of the triangle Triangle meet the opposite side (Blue line) produced, that whole produced side ( Blue and dotted blue line ), and its external segment (Blue dotted line) will be proportional to the sides ( Black and dotted black line and Yellow line), which contain the angle adjacent to the external bisected angle.

For if Red line be drawn ∥ Red dotted line, then Blue angle = Outlined blue angle , (B. 1. pr. 29); = Black angle , (hyp.), = Yellow angle , (B. 1. pr. 29); and ∴ Black dotted line = Yellow line, (B. 1. pr. 6),
and Black and dotted black line : Yellow line :: Black and dotted black line : Black dotted line,
(B. 5. pr. 7);
But also,
Blue and dotted blue line : Blue dotted line :: Black and dotted black line : Black dotted line.
(B. 6. pr. 2);
and therefore
Blue and dotted blue line : Blue dotted line :: Black and dotted black line : Yellow line,
(B. 5. pr. 11).

Q. E. D.

Proposition B. Theorem.

Proposition B figure

If an angle of a triangle be bisected by a straight line, which likewise cuts the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square of the straight line which bisects the angle.

Let Yellow line be drawn, making Red angle = Blue angle ; then shall
Blue line × Black line = Red dotted line × Red line + Yellow line2.

About Blue, black, and red triangle describe Yellow circle (B. 4. pr. 5),
produce Yellow line to meet the circle, and draw Blue dotted line.

Since Red angle = Blue angle (hyp.),
and Yellow angle = Black angle (B. 3. pr. 21),
∴ Blue, yellow, and red triangle and Yellow, black, and blue triangle are equiangular (B. 1. pr. 32);
∴ Blue line : Yellow line :: Dotted yellow and yellow line : Black line (B. 6. pr. 4);
∴ Blue line × Black line = Yellow line × Yellow and dotted yellow line (B. 6. pr. 16)
= Yellow dotted line × Yellow line + Yellow line2 (B. 2. pr. 3);
but Yellow dotted line × Yellow line = Red dotted line × Red line (B. 3. pr. 35);
∴ Blue line × Black line = Red dotted line × Red line + Yellow line2.

Q. E. D.

Proposition C. Theorem.

Proposition C figure

If from any angle of a triangle a straight line be drawn perpendicular to the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle.

From Red angle of Blue, yellow, red, dotted red triangle
draw Yellow dotted line ⊥ Dotted red and red line ; then
shall Blue dotted line × Yellow line = Yellow dotted line × the
diameter of the described circle.

Describe Red circle (B. 4. pr. 5), draw its diameter
Blue line, and draw Black line; then because
Blue angle = Blue outlined and yellow angle (const. and B. 3. pr. 31);
and Yellow outlined angle = Red outlined angle (B. 3. pr. 21);
∴ Dotted blue, dotted yellow, and dotted red triangle is equiangular to Blue, yellow, and black triangle (B. 6. pr. 4);
∴ Blue dotted line : Yellow dotted line :: Blue line : Yellow line;
and ∴ Blue dotted line × Yellow line = Yellow dotted line × Blue line (B. 6. pr. 16).

Q. E. D.

Proposition D. Theorem.

Proposition D figure

The rectangle contained by the diagonals of a quadrilateral figure inscribed in a circle, is equal to both the rectangles contained by opposite sides.

Let Quadrilateral be any quadrilateral figure inscribed in Red circle;
and draw Dotted red and red line and Blue line; then
Dotted red and red line × Blue line =
Black dotted line × Black line + Yellow line × Blue dotted line.

Make Blue angle = Red angle (B. 1. pr. 23),
∴ Blue and black angle = Red and black angle ; and Yellow angle = Blue outlined angle (B. 3. pr. 21);

∴ Yellow line : Red line :: Blue line : Blue dotted line (B. 6. pr. 4);
and ∴ Red line × Blue line = Yellow line × Blue dotted line (B. 6. pr. 16); again,
because Blue angle = Red angle (const.),
and Yellow outlined angle = Red outlined angle (B. 3. pr. 21);
∴ Black dotted line : Red dotted line :: Blue line : Black line (B. 6. pr. 4);
and ∴ Red dotted line × Blue line = Black dotted line × Black line (B. 6. pr. 16);
but, from above,
Red line × Blue line = Yellow line × Blue dotted line;
∴ Dotted red and red line × Blue line = Black dotted line × Black line + Yellow line × Blue dotted line (B. 2. pr. 1.)

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