# Book IV.

## Definitions.

### I.

A rectilinear figure is ſaid to be inſcribed in another, when all the angular points of the inſcribed figure are on the ſides of the figure in which it is ſaid to be inſcribed.

### II.

A figure is ſaid to be deſcribed about another figure, when all the ſides of the circumſcribed figure paſs through the angular points of the other figure.

### III.

A rectilinear figure is ſaid to be inſcribed in a circle, when the vertex of each angle of the figure is in the circumference of the circle.

### IV.

A rectilinear figure is ſaid to be circumſcribed about a circle, when each of its ſides is a tangent to the circle.

### V.

A circle is ſaid to be inſcribed in a rectilinear figure, when each ſide of the figure is a tangent to the circle.

### VI.

A circle is ſaid to be circumſcribed about a rectilinear figure, when the circumference paſſes through the vertex of each angle of the figure.

is circumſcribed.

### VII.

A ſtraight line is ſaid to be inſcribed in a circle, when its extremities are in the circumference.

The Fourth Book of the Elements is devoted to the ſolution of problems, chiefly relating to the inſcription and circumſcription of regular polygons and circles.

A regular polygon is one whoſe angles and ſides are equal.

## Proposition I. Problem.

In a given circle to place a ſtraight line, equal to a given ſtraight line (), not greater than the diameter of the circle.

Draw , the diameter of ;
and if = , then
the problem is ſolved.

But if be not equal to ,
> (hyp.);
make = (B. 1. pr. 3.) with
deſcribe , cutting , and
draw , which is the line required.
For = = (B. 1. def. 15. conſt.)

Q. E. D.

## Proposition II. Problem.

On a given circle to inſcribe a triangle equiangular to a given triangle.

To any point of the given circle draw , a tangent (B. 3. pr. 17.); and at the point of contact
make = (B. 1. pr. 23.)
and in like manner = , and draw .

Becauſe = (conſt.)
and = (B. 3. pr. 32)
= ; alſo
= for the ſame reaſon.
= (B. 1. pr. 32.),

and therefore the triangle inſcribed in the circle is equiangular to the given one.

Q. E. D.

## Proposition III. Problem.

About a given circle to circumſcribe a triangle equiangular to a given triangle.

Produce any ſide , of the given triangle both ways; from the centre of the given circle draw , any radius.

Make = (B. 1. pr. 23.) and = .
At the extremities of the three radii, draw , and , tangents to the given circle. (B. 3. pr. 17.)

The four angles of , taken together, are equal to four right angles. (B. 1. pr. 32.)
but and are right angles (conſt.)
+ = , two right angles
but = (B. 1. pr. 13.)
and = (conſt.)
and = .

In the ſame manner it can be demonſtrated that
= ;
= (B. 1. pr. 32.)
and therefore the triangle circumſcribed about the given circle is equiangular to the given triangle.

Q. E. D.

## Proposition IV. Problem.

In a given triangle to inſcribe a circle.

Biſect and (B. 1. pr. 9.) by and ; from the point where theſe lines meet draw , and reſpectively perpendicular to , and .

In and
= , = and common,
= (B. 1. pr. 4 and 26.)

In like manner, it may be ſhown alſo
that = ,
= = ;
hence with any one of theſe lines as radius, deſcribe

and it will paſs through the extremities of the other two; and the ſides of the given triangle, being perpendicular to the three radii at their extremities, touch the circle (B. 3. pr. 16.), which is therefore inſcribed in the given triangle.

Q. E. D.

## Proposition V. Problem.

To deſcribe a circle about a given triangle.

Make = and = (B. 1. pr. 10.) From the points of biſection draw and and reſpectively (B. 1. pr. 11.), and from their point of concourſe draw , and and deſcribe a circle with any one of them, and it will be the circle required.

In and
= (conſt.),
common,
= (conſt.),
= (B. 1. pr. 4.).

In like manner it may be shown that
= .

= = ; and therefore a circle deſcribed from the concourſe of theſe three lines with any one of them as a radius will circumſcribe the given triangle.

Q. E. D.

## Proposition VI. Problem.

In a given circle to inſcribe a ſquare.

Draw the two diameters of the circle to each other, and draw , , and

is a ſquare.

For ſince and are, each of them, in a ſemicircle, they are right angles (B. 3. pr. 31),
(B. 1. pr. 28):
and in like manner .

And becauſe = (conſt.), and
= = (B. 1. def. 15).
= (B. 1. pr. 4);

and ſince the adjacent ſides and angles of the parallelogram are equal, they are all equal (B. 1. pr. 34); and , inſcribed in the given circle, is a ſquare.

Q. E. D.

## Proposition VII. Problem.

About a given circle to circumſcribe a ſquare.

Draw two diameters of the given circle perpendicular to each other, and through their extremities draw , , , and tangents to the circle;

and is a ſquare.

= a right angle, (B. 3. pr. 18.)
alſo = (conſt.),

; in the ſame manner it can be demonſtrated that , and alſo that and ;

is a parallelogram, and
becauſe = = = =
they are all right angles (B. 1. pr. 34):
it is alſo evident that , , and are equal.

is a ſquare.

Q. E. D.

## Proposition VIII. Problem.

To inſcribe a circle in a given ſquare.

Make = ,
and = ,
draw ,
and
(B. 1. pr. 31.)

is a parallelogram;
and ſince = (hyp.)
=

is equilateral (B. 1. pr. 34.)

In like manner, it can be ſhown that
= are equilateral parallelograms;
= = = ,

and therefore if a circle be deſcribed from the concourſe of theſe lines with any one of them as radius, it will be inſcribed in the given ſquare. (B. 3. pr. 16.)

Q. E. D.

## Proposition IX. Problem.

To deſcribe a circle about a given ſquare .

Draw the diagonals and interſecting each other; then,

becauſe and have
their ſides equal, and the baſe
common to both,
= (B. 1. pr. 8),
or is biſected: in like manner it can be ſhown
that is biſected;
but = ,
hence = their halves,
= ; (B. 1. pr. 6.)
and in like manner it can be proved that
= = = .

If from the confluence of theſe lines with any one of them as radius, a circle can be deſcribed, it will circumſcribe the given ſquare.

Q. E. D.

## Proposition X. Problem.

To conſtruct an iſoſceles triangle, in which each of the angles at the baſe ſhall be double of the vertical angle.

Take any ſtraight line and divide it ſo that
× = 2 (B. 2. pr. 11.)

With as radius, deſcribe and place
in it from the extremity of the radius, = , (B. 4. pr. 1.); draw .

Then is the required triangle.

For, draw and deſcribe

Since × = 2 = 2,
is a tangent to (B. 3. pr. 37.)
= (B. 3. pr. 32),
+ = + ;
but + or = (B. 1. pr. 5):
ſince = (B. 1. pr. 5.)
conſequently = + = (B. 1. pr. 32.)
= (B. 1. pr. 6.)
= = (conſt.)
= (B. 1. pr. 5.)

= = = + = twice ; and conſequently each angle at the baſe is double of the vertical angle.

Q. E. D.

## Proposition XI. Problem.

In a given circle to inſcribe an equilateral and equiangular pentagon.

Conſtruct an iſoſceles triangle, in which each of the angles at the baſe ſhall be double of the angle at the vertex, and inſcribe in the given circle a triangle equiangular to it; (B. 4. pr. 2.)

Biſect and (B. 1. pr. 9.)
draw , , and .

Becauſe each of the angles
, , , and are equal,

the arcs upon which they ſtand are equal (B. 3. pr. 26.) and , , , and which ſubtend theſe arcs are equal (B. 3. pr. 29.) and the pentagon is equilateral, it is alſo equiangular, as each of its angles ſtand upon equal arcs. (B. 3. pr. 27).

Q. E. D.

## Proposition XII. Problem.

To deſcribe an equilateral and equiangular pentagon about a given circle .

Draw five tangents through the vertices of the angles of any regular pentagon inſcribed in the given circle (B. 3. pr. 17).

Theſe five tangents will form the required pentagon.

Draw { } . In and
= (B. 1. pr. 47),
= , and common;
= and = (B. 1. pr. 8.)
= twice , and = twice ;

In the ſame manner it can be demonſtrated that
= twice , and = twice ;
but = (B. 3. pr. 27),
their halves = , alſo = , and
common;

= and = ,
= twice ;
In the ſame manner it can be demonſtrated
that = twice ,
but =
= ;

In the ſame manner it can be demonſtrated that the other ſides are equal, and therefore the pentagon is equilateral, it is alſo equiangular, for

= twice and = twice ,
and therefore = ,
= ; in the ſame manner it can be demonſtrated that the other angles of the deſcribed pentagon are equal.

Q. E. D.

## Proposition XIII. Problem.

To inſcribe a circle in a given equiangular and equilateral pentagon.

Let be a given equiangular and equilateral pentagon; it is required to inſcribe a circle in it.

Make = , and = (B. 1. pr. 9.)

Draw , , , , &c.
Becauſe = , = ,
common to the two triangles
and ;
= and = (B. 1. pr. 4.)

And becauſe = = twice
= twice , hence is biſected by .

In like manner it may be demonſtrated that is biſected by , and that the remaining angle of the polygon is biſected in a ſimilar manner.

Draw , , &c. perpendicular to the ſides of the pentagon.

Then in the two triangles and
we have = , (conſt.), common,
and = = a right angle;
= . (B. 1. pr. 26.)

In the ſame way it may be ſhown that the five perpendiculars on the ſides of the pentagon are equal to one another.

Deſcribe with any one of the perpendiculars as radius, and it will be the inſcribed circle required. For if it does not touch the ſides of the pentagon, but cut them, then a line drawn from the extremity at right angles to the diameter of a circle will fall within the circle, which has been ſhown to be abſurd. (B. 3. pr. 16.)

Q. E. D.

## Proposition XIV. Problem.

To deſcribe a circle about a given equilateral and equiangular pentagon.

Biſect and by and ,
and from the point of ſection, draw , , and .

= ,
= , = (B. 1. pr. 6);
and ſince in and ,
= , and common,
alſo = ;
= (B. 1. pr. 4).

In like manner it may be proved that
= = , and
therefore = = = = :

Therefore if a circle be deſcribed from the point where theſe five lines meet, with any one of them as a radius, it will circumſcribe the given pentagon.

Q. E. D.

## Proposition XV. Problem.

To inſcribe an equilateral and equiangular hexagon in a given circle .

From any point in the circumference of the given circle deſcribe paſſing through its centre, and draw the diameters , and ; draw , , , &c. and the required hexagon is inſcribed in the given circle.

Since paſſes through the centres
of the circles, and are equilateral
triangles, hence = = one-third of two right
angles; (B. 1. pr. 32) but = (B. 1. pr. 13);

= = = one-third of (B. 1. pr. 32), and the angles vertically oppoſite to theſe are all equal to one another (B. 1. pr. 15), and ſtand on equal arches (B. 3. pr. 26), which are ſubtended by equal chords (B. 3. pr. 29); and ſince each of the angles of the hexagon is double the angle of an equilateral triangle, it is alſo equiangular.

Q. E. D.

## Proposition XVI. Problem.

To inſcribe an equilateral and equiangular quindecagon in a given circle.

Let and be the ſides of an equilateral pentagon inſcribed in the given circle, and the ſide of an inscribed equilateral triangle.

The arc ſubtended by and } = 2 / 5 = 6 / 15 { of the whole circumference.

The arc ſubtended by } = 1 / 3 = 5 / 15 { of the whole circumference.

Their difference is = 1 / 15

the arc ſubtended by = 1 / 15 difference of the whole circumference.

Hence if ſtraight lines equal to be placed in the circle (B. 4. pr. 1), an equilateral and equiangular quindecagon will be thus inſcribed in the circle.

Q. E. D.