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Book IV.

Definitions.

Definition 1 figure

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.

Definition 3 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.

Definition 4 figure

IV.

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

Definition 5 figure

V.

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

Definition 6 figure

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.

Triangle is circumſcribed.

Definition 7 figure

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.

Proposition 1 figure

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

Draw Dotted red and red line , the diameter of Yellow circle;
and if Dotted red and red line = Blue line, then
the problem is ſolved.

But if Dotted red and red line be not equal to Blue line,
Dotted red and red line > Blue line (hyp.);
make Red dotted line = Blue line (B. 1. pr. 3.) with
Red dotted line as radius,
deſcribe Blue circle, cutting Yellow circle, and
draw Yellow line, which is the line required.
For Yellow line = Red dotted line = Blue line (B. 1. def. 15. conſt.)

Q. E. D.

Proposition II. Problem.

Proposition 2 figure

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

To any point of the given circle draw Red line, a tangent (B. 3. pr. 17.); and at the point of contact
make Left blue angle = Right blue angle (B. 1. pr. 23.)
and in like manner Left yellow angle = Right yellow angle , and draw Yellow line.

Becauſe Left blue angle = Right blue angle (conſt.)
and Left blue angle = Black angle (B. 3. pr. 32)
∴ Black angle = Right blue angle ; alſo
Outlined angle angle = Right yellow angle for the ſame reaſon.
∴ Left red angle = Right red angle (B. 1. pr. 32.),

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

Q. E. D.

Proposition III. Problem.

Proposition 3 figure

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

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

Make Right yellow angle = Left yellow angle (B. 1. pr. 23.) and Right blue angle = Left blue angle .
At the extremities of the three radii, draw Blue line, Yellow line and Red dotted line, tangents to the given circle. (B. 3. pr. 17.)

The four angles of Quadrilateral , taken together, are equal to four right angles. (B. 1. pr. 32.)
but Bottom right black angle and Top right black angle are right angles (conſt.)
∴ Right red angle + Right yellow angle = Two right angles , two right angles
but Left yellow and red angles = Two right angles (B. 1. pr. 13.)
and Right yellow angle = Left yellow angle (conſt.)
and ∴ Right red angle = Left red angle .

In the ſame manner it can be demonſtrated that
Right outlined angle = Left outlined angle ;
∴ Right red, black, and blue angles = Left red, black, and blue angles (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.

Proposition 4 figure

In a given triangle Outer triangle to inſcribe a circle.

Biſect Blue and yellow angles and Black and outlined angles (B. 1. pr. 9.) by Blue dotted line and Blue line; from the point where theſe lines meet draw Black dotted line, Yellow dotted line and Red dotted line reſpectively perpendicular to Black line, Yellow line and Red line.

In Bottom left triangle and Top left triangle
Yellow angle = Blue angle , Bottom red angle = Top red angle and Blue dotted line common,
∴ Yellow dotted line = Black dotted line (B. 1. pr. 4 and 26.)

In like manner, it may be ſhown alſo
that Red dotted line = Black dotted line,
∴ Black dotted line = Yellow dotted line = Red dotted line;
hence with any one of theſe lines as radius, deſcribe Black circle

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.

Proposition 5 figure
Proposition 5 figure
Proposition 5 figure

To deſcribe a circle about a given triangle.

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

In Triangle with black angle and Triangle with red angle
Blue dotted line = Blue line (conſt.),
Yellow line common,
Black angle = Red angle (conſt.),
∴ Black line = Black dotted line (B. 1. pr. 4.).

In like manner it may be shown that
Black thin line = Black dotted line.

∴ Black dotted line = Black line = Black thin line; 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.

Proposition 6 figure

In a given circle Red circle to inſcribe a ſquare.

Draw the two diameters of the circle ⊥ to each other, and draw Yellow line, Black line, Red line and Blue line

Square is a ſquare.

For ſince Yellow angle and Black angle are, each of them, in a ſemicircle, they are right angles (B. 3. pr. 31),
∴ Blue line ∥ Black line (B. 1. pr. 28):
and in like manner Red line ∥ Yellow line.

And becauſe Blue angle = Red angle (conſt.), and
Red dotted line = Black dotted line = Blue dotted line (B. 1. def. 15).
∴ Black line = Red line (B. 1. pr. 4);

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

Q. E. D.

Proposition VII. Problem.

Proposition 7 figure

About a given circle Blue circle to circumſcribe a ſquare.

Draw two diameters of the given circle perpendicular to each other, and through their extremities draw Blue line, Red line, Black line, and Yellow line tangents to the circle;

and Square is a ſquare.

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

∴ Blue line ∥ Blue dotted line; in the ſame manner it can be demonſtrated that Black line ∥ Blue dotted line, and alſo that Red line and Yellow line ∥ Red dotted line;

∴ Square is a parallelogram, and
becauſe Yellow angle = Top left red angle = Top right red angle = Bottom right red angle = Bottom left red angle
they are all right angles (B. 1. pr. 34):
it is alſo evident that Blue line, Red line, Black line and Yellow line are equal.

∴ Square is a ſquare.

Q. E. D.

Proposition VIII. Problem.

Proposition 8 figure

To inſcribe a circle in a given ſquare.

Make Blue line = Blue dotted line,
and Red line = Red dotted line,
draw Dotted yellow and yellow line ∥ Blue and dotted blue line ,
and Dotted black and black line ∥ Dotted red and red line
(B. 1. pr. 31.)

∴ Black square is a parallelogram;
and ſince Blue and dotted blue line = Dotted red and red line (hyp.)
Blue line = Red dotted line

∴ Black square is equilateral (B. 1. pr. 34.)

In like manner, it can be ſhown that
Blue square = Red square are equilateral parallelograms;
∴ Black dotted line = Yellow dotted line = Black line = Yellow line,

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.

Proposition 9 figure

To deſcribe a circle about a given ſquare Red and yellow square .

Draw the diagonals Blue and dotted blue line and Black and dotted black line interſecting each other; then,

becauſe Red triangle and Yellow triangle have
their ſides equal, and the baſe
Blue and dotted blue line common to both,
Yellow angle = Black angle (B. 1. pr. 8),
or Yellow and black angle is biſected: in like manner it can be ſhown
that Red and blue angle is biſected;
but Yellow and black angle = Red and blue angle ,
hence Black angle = Red angle their halves,
∴ Black line = Blue line; (B. 1. pr. 6.)
and in like manner it can be proved that
Blue line = Black line = Black dotted line = Blue dotted line.

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.

Proposition 10 figure

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 Black and dotted black line and divide it ſo that
Black and dotted black line × Black dotted line = Black line2 (B. 2. pr. 11.)

With Black and dotted black line as radius, deſcribe Red circle and place
in it from the extremity of the radius, Blue line = Black line, (B. 4. pr. 1.); draw Yellow line.

Then Large triangle is the required triangle.

For, draw Red line and deſcribe
Blue circle about Left triangle (B. 4. pr. 5.)

Since Black and dotted black line × Black dotted line = Black line2 = Blue line2,
∴ Blue line is a tangent to Blue circle (B. 3. pr. 37.)
∴ Yellow angle = Outlined angle (B. 3. pr. 32),
add Black angle to each,
∴ Yellow angle + Black angle = Outlined angle + Black angle ;
but Black angle + Yellow angle or Black and yellow angle = Red angle (B. 1. pr. 5):
ſince Yellow line = Black and dotted black line (B. 1. pr. 5.)
conſequently Red angle = Outlined angle + Black angle = Blue angle (B. 1. pr. 32.)
∴ Red line = Blue line (B. 1. pr. 6.)
∴ Blue line = Black line = Red line (conſt.)
∴ Outlined angle = Black angle (B. 1. pr. 5.)

∴ Red angle = Black and yellow angle = Blue angle = Outlined angle + Black angle = twice Outlined angle ; and conſequently each angle at the baſe is double of the vertical angle.

Q. E. D.

Proposition XI. Problem.

Proposition 11 figure

In a given circle Blue 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 Triangle equiangular to it; (B. 4. pr. 2.)

Biſect Yellow and blue angle and Outlined and red angle (B. 1. pr. 9.)
draw Red line, Blue line, Yellow line and Red dotted line.

Becauſe each of the angles
Yellow angle , Blue angle , Black angle , Red angle and Outlined angle are equal,

the arcs upon which they ſtand are equal (B. 3. pr. 26.) and ∴ Black line, Red line, Blue line, Yellow line and Red dotted line 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.

Proposition 12 figure

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

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

Theſe five tangents will form the required pentagon.

Draw { Red dotted line Blue line Black dotted line Yellow dotted line } . In Left triangle and Right triangle
Black line = Red line (B. 1. pr. 47),
Black dotted line = Red dotted line, and Blue line common;
∴ Right outlined angle = Left black angle and Left red angle = Yellow angle (B. 1. pr. 8.)
∴ Yellow outlined and black angle = twice Left black angle , and Red and yellow angle = twice Yellow angle ;

In the ſame manner it can be demonſtrated that
Black outlined and black angle = twice Right black angle , and Blue and red angle = twice Blue angle ;
but Red and yellow angle = Blue and red angle (B. 3. pr. 27),
∴ their halves Yellow angle = Blue angle , alſo Left blue outlined angle = Right blue outlined angle , and
Black dotted line common;

∴ Left black angle = Right black angle and Red line = Yellow line,
∴ Red and yellow line = twice Red line;
In the ſame manner it can be demonſtrated
that Black and dotted blue line = twice Black line,
but Black line = Red line
∴ Black and dotted blue line = Red and yellow line ;

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

Black outlined and black angle = twice Right black angle and Yellow outlined and black angle = twice Left black angle ,
and therefore Left black angle = Right black angle ,
∴ Black outlined and black angle = Yellow outlined and black angle ; in the ſame manner it can be demonſtrated that the other angles of the deſcribed pentagon are equal.

Q. E. D.

Proposition XIII. Problem.

Proposition 13 figure

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

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

Make Left blue angle = Right blue angle , and Left red angle = Right red angle (B. 1. pr. 9.)

Draw Yellow dotted line, Black line, Red line, Red dotted line, &c.
Becauſe Black dotted and yellow line = Yellow and dotted black line , Left blue angle = Right blue angle ,
Black line common to the two triangles
Left big triangle and Right big triangle ;
∴ Red line = Yellow dotted line and Bottom yellow angle = Left red angle (B. 1. pr. 4.)

And becauſe Yellow angle = Red angle = twice Left red angle
∴ = twice Bottom yellow angle , hence Yellow angle is biſected by Red line.

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

Draw Blue line, Blue dotted line, &c. perpendicular to the ſides of the pentagon.

Then in the two triangles Left small triangle and Right small triangle
we have Left blue angle = Right blue angle , (conſt.), Black line common,
and Left black angle = Right black angle = a right angle;
∴ Blue line = Blue dotted line. (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 Yellow circle 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.

Proposition 14 figure

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

Biſect Black and yellow angle and Yellow and red angle by Red dotted line and Blue dotted line,
and from the point of ſection, draw Yellow line, Yellow dotted line, and Black line.

Black and yellow angle = Yellow and red angle ,
Left yellow angle = Right yellow angle , ∴ Blue dotted line = Red dotted line (B. 1. pr. 6);
and ſince in Left triangle and Right triangle ,
Blue line = Red line, and Red dotted line common,
alſo Black angle = Left yellow angle ;
∴ Black line = Blue dotted line (B. 1. pr. 4).

In like manner it may be proved that
Yellow dotted line = Yellow line = Black line, and
therefore Yellow dotted line = Black line = Red dotted line = Blue dotted line = Yellow line:

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.

Proposition 15 figure

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

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

Since Black line paſſes through the centres
of the circles, Left triangle and Right triangle are equilateral
triangles, hence Red angle = Blue angle = one-third of two right
angles; (B. 1. pr. 32) but Red, blue, and black angle = Two right angles (B. 1. pr. 13);

∴ Red angle = Blue angle = Black angle = one-third of Two right angles (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.

Proposition 16 figure

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

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

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

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

Their difference is = 1 / 15

∴ the arc ſubtended by Black dotted line = 1 / 15 difference of the whole circumference.

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

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