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.

Proposition I. Theorem.

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

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

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

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

∴ and its baſe are reſpectively
equimultiples of and the baſe .

In like manner and its baſe are reſpectively
equimultiples of and the baſe .

∴ If m or 6 times >= or <n or 5 times then m or 6 times >=<n or 5 times ,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.

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.

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

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

Let the ſame conſtruction remain,
becauſe ::::and ::::}
(B. 6. pr. 1)
but :::: (hyp.),
∴:::: (B. 5. pr. 11.)
∴= (B. 5. pr. 9);
but they are on the ſame baſe , and at the ſame ſide of it, and
∴∥ (B. 1. pr. 39).

Q. E. D.

Proposition III. Theorem.

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

And if a ſtraight line () drawn from any angle of a triangle divide the oppoſite ſide () into ſegments (, ) proportional to the conterminous ſides (, ), it biſects the angle.

In equiangular triangles ( and ) 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 , oppoſite to equal angles and 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,

In like manner it may be ſhown, that
::::; and by alternation, that
::::; but it has already been proved that
::::, and therefore, ex æquali,
:::: (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.

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

From the extremities of , draw and , making =,= (B. 1. pr. 23);
and conſequently = (B. 1. pr. 32),
and ſince the triangles are equiangular,
:::: (B. 6. pr. 4);
but :::: (hyp.);
∴::::, and conſequently = (B. 5. pr. 9).

In the like manner it may be ſhown that
=.

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

But = (conſt.)
and ∴=; for the ſame
reaſon =, and
conſequently = (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.

If two triangles ( and ) have one angle () of the one, equal to one angle () 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 extremeties of , one of the ſides
of , about , draw
and , making
=, and =; then = (B. 1. pr. 32), and two triangles being equiangular,
:::: (B. 6. pr. 4);
but :::: (hyp.);
∴:::: (B. 5. pr. 11),
and conſequently = (B. 5. pr. 9);
∴= in every reſpect.
(B. 1. pr. 4).

But = (conſt.),
and ∴=; and
ſince also =, = (B. 1. pr. 32);
and ∴ and are equiangular, with their equal angles oppoſite to homologous ſides.

Q. E. D.

Proposition VII. Theorem.

If two triangles ( and ) have one angle in each equal ( equal to ), the ſides about two other angles proportional ( :::: ), and each of the remaining angles ( and ) 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 and are each leſs than a right angle: then if it be ſuppoſed
that and contained by the proportional ſides
are not equal, let be the greater, and make
=.

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

∴= (B. 1. pr. 32), and therefore the triangles are equiangular.

But if and 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.

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

∴ and 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 is ſimilar to
; but has been ſhewn to be ſimilar
to ;∴ and are
ſimilar to the whole and to each other.

Q. E. D.

Proposition IX. Problem.

From a given ſtraight line () to cut off any required part.

From either extremity of the given line draw making any angle with ; and produce till the whole produced line contains as often as contains the required part.

Draw , and draw ∥. is the required part of .

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

Q. E. D.

Proposition X. Problem.

To divide a ſtraight line () ſimilarly to a given divided line ().

From either extremity of the given line
draw making any angle;
take , and equal to , and reſpectively (B. 1. pr. 2);
draw , and draw and ∥ to it.

Since
{}
are ∥ :::: (B. 6. pr. 2),
or :::: (conſt.),
and :::: (B. 6. pr. 2),
:::: (conſt.),
and ∴ the given line is divided ſimilarly to .

Q. E. D.

Proposition XI. Problem.

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

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

For ſince ∥, ∴:::: (B. 6. pr. 2);
but == (conſt.);
∴::: :.

Q. E. D.

Proposition XII. Problem.

To find a fourth proportional to three given lines
{}.

Draw and making any angle;
take =, and =, alſo =, draw , and ∥; (B. 1. pr. 31);
is the fourth proportional.

On account of the parallels,
:::: (B. 6. pr. 2);
but
{}={}
(conſt.);

To find a mean proportional between two given ſtraight lines {}.

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

Draw and .

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

Q. E. D.

Proposition XIV. Theorem.

I.

Equal parallelograms and , which have one angle in each equal, have the ſides about the equal angles reciprocally proportional ( :::: )

II.

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

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

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

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 and may be vertically oppoſite, that is to ſay, ſo that and may be in the ſame ſtraight line. Whence alſo and muſt be in the ſame ſtraight line (B. 1. pr. 14.)

If four ſtraight lines be proportional ( :::: ), the rectangle ( × ) contained by the extremes, is equal to the rectangle ( × ) 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 and draw and ⊥ to them and = and reſpectively: complete the parallelograms:

and .

And ſince,
:::: (hyp.)∴:::: (conſt.)∴= (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
=,= and =, ∴:::: (B. 6. pr. 14).

If three ſtraight lines be proportional ( :::: ) 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 =, andſince ::::,then ::::,∴×=× (B. 6. pr. 16).

But =, ∴×=×, or =^{2};
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 =, then
×=×. ∴:::: (B. 6. pr. 16), and
∴::::.

Q. E. D.

Proposition XVIII. Theorem.

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

Reſolve the given figure into triangles by drawing the lines and .

At the extremities of make
= and =: again at the extremities of make = and =: in like manner make
= and =.

Then is ſimilar to .

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

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

∴ by (B. 6. def. 1.)
is ſimilar and ſimilarly ſituated; and on the given line .

Q. E. D.

Proposition XIX. Theorem.

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

Let and be equal angles, and and let homologous ſides of the ſimilar triangles and and on the greater of theſe lines take a third proportional, ſo that

::::; draw .

:::: (B. 6. pr. 4);
∴:::: (B. 5. pr. 16, alt.),
but :::: (conſt.),
∴:::: conſequently = for they have the ſides about
the equal angles and reciprocally proportional (B. 6. pr. 15);
∴:::: but :::: (B. 6. pr. 1),
∴::::, that is to ſay, the triangles are to one another in the duplicate ratio of their homologous ſides
and (B. 5. def. 11).

Q. E. D.

Proposition XX. Theorem.

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 and , and and , reſolving the polygons into triangles. Then becauſe the polygons are ſimilar, =, and ::::

∴ and are ſimilar, and = (B. 6. pr. 6);
but = becauſe they are angles of ſimilar polygons;
therefore the remainders and are equal;
hence ::::, on account of the ſimilar triangles,
and ::::, on account of the ſimilar polygons,
∴::::, ex æquali (B. 5. pr. 22), and as theſe proportional ſides
contain equal angles, the triangles and are ſimilar (B. 6. pr. 6).

In like manner it may be ſhown that the
triangles and are ſimilar.

But is to in the duplicate ratio of
to (B. 6. pr. 19), and
is to in like manner, in the duplicate
ratio of to ;
∴::::, (B. 5. pr. 11);

Again is to in the duplicate ratio of
to , and is to in
the duplicate ratio of to ,

∴:::::::;

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 is to in the duplicate ratio of
to ;

∴ is to in the duplicate
ratio of to .

Q. E. D.

Proposition XXI. Theorem.

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

Since and are ſimilar, they are equiangular, and have the ſides about the equal angles proportional (B. 6. def. 1); and ſince the figures and are alſo ſimilar, they are equiangular, and have the ſides about the equal angles proportional; therefore and 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.

Part I.

If four ſtraight line be proportional ( :::: ), 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 a third proportional to and , and a third proportional
to and (B. 6. pr. 11);
ſince :::: (hyp.),:::: (conſt.)

:::: (hyp.),
∴:::: (conſt.)
and ∴::::. (B. 5. pr. 11).

Q. E. D.

Proposition XXIII. Theorem.

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

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

Since +=, and = (hyp.),
+=, and ∴ and form one ſtraight line (B. 1. pr. 14);
complete .

Since :::: (B. 6. pr. 1),
and :::: (B. 6. pr. 1),
has to a ratio compounded of the ratios of
to , and of to .

Q. E. D.

Proposition XXIV. Theorem.

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

As and have a
common angle they are equiangular;
but becauſe ∥ and are ſimilar (B. 6. pr. 4),
∴::::; and the remaining oppoſite ſides are equal to thoſe,
∴ and have the ſides about the equal
angles proportional, and are therefore ſimilar.

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

Since, therefore, each of the parallelograms
and is ſimilar to , they are ſimilar to each other.

Q. E. D.

Proposition XXV. Problem.

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

Upon deſcribe =, and upon deſcribe =, and having = (B. 1. pr. 45), and then
and will lie in the ſame ſtraight line (B. 1. prs. 29, 14),

Between and find a mean proportional
(B. 6. pr. 13), and upon deſcribe , ſimilar to , and ſimilarly ſituated.

Then =.

For ſince and are ſimilar, and
:::: (conſt.),
:::: (B. 6. pr. 20);
but :::: (B. 6. pr. 1);
∴:::: (B. 5. pr. 11);
but = (conſt.),
and ∴= (B. 5. pr. 14);
and = (conſt.); conſequently,
which is ſimilar to is alſo =.

Q. E. D.

Proposition XXVI. Theorem.

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

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

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

∴::::; but :::: (hyp.),
∴::::, and ∴= (B. 5. pr. 9.),
which is abſurd.

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

Q. E. D.

Proposition XXVII. Theorem.

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 be the given line, and unequal ſegments, and and equal ſegments;
then >.

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.

To divide a given ſtraight 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 =^{2}.

Biſect , or
make =; and if ^{2}=^{2}, the problem is ſolved.

But if ^{2}≠^{2}, then
muſt > (hyp.).

Draw ⊥=; make = or ; with as radius deſcribe a circle cutting the
given line; draw .

But ^{2}=^{2}+^{2} (B. 1. pr. 47);
∴×+^{2} =^{2}+^{2}, from both, take ^{2}, and ×=^{2}.

But = (conſt.),
and ∴ is ſo divided
that ×=^{2}.

Q. E. D.

Proposition XXIX. Problem.

To produce a given ſtraight 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 .

Make =, and
draw ⊥=; draw ; and
with the radius , deſcribe a circle
meeting produced.

Then =×, and is ∴=; and if from both theſe equals
be taken the common part , , which is the ſquare of , will be =, which is =×; that is ^{2}=×; ∴::::, and is divided in extreme and mean ratio (B. 6. def. 3).

Q. E. D.

Proposition XXXI. Theorem.

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

From the right angle draw perpendicular to ; then :::: (B. 6. pr. 8).

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

In equal circles (,), angles, whether at the centre or circumference, are in the ſame ratio to one another as the arcs on which they ſtand (::::); ſo alſo are ſectors.

Take in the circumference of any number of arcs ,, &c. each =, and alſo in the circumference of take any number of arcs ,, &c. each =, draw the radii to the extremities of the equal arcs.

Then ſince the arcs ,,, &c. are all equal, the angles ,,, &c. are alſo equal (B. 3. pr. 27);

∴ is the ſame multiple of which the arc is of ; and in the ſame manner is the ſame multiple of , which the arc is of the arc .

Then it is evident (B. 3. pr. 27),
if (or if m times ) >, =, < (or n times )
then (or m times ) >, =, < (or n times );

∴::::, (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.

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

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 be drawn, making =; then ſhall
×=×+^{2}.

About deſcribe (B. 4. pr. 5),
produce to meet the circle, and draw .

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 of draw ⊥; then
ſhall ×=× the
diameter of the deſcribed circle.