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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 , &c. each = Thin black arc , and alſo in the circumference of Blue circle take any number of arcs Red dotted arc , Blue dotted arc , &c. 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 , &c. are all equal, the angles Black angle , Red angle , Blue angle , &c. 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. Book I. Basic plane geometry
  2. Book II. Geometric algebra
  3. Book III. Circles and angles
  4. Book IV. Regular polygons
  5. Book V. Ratios and proportions
  6. Book VI. Geometric proportions