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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, etc.
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, etc. 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, etc. 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