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Book III.

Definitions.

I.

Equal circles are those whose diameters are equal.

Definition 2 figure

II.

A right line is said to touch a circle when it meets the circle, and being produced does not cut it.

Definition 3 figure

III.

Circles are said to touch one another which meet but do not cut one another.

Definition 4 figure

IV.

Right lines are said to be equally distant from the centre of a circle when the perpendiculars drawn to them from the centre are equal.

V.

And the straight line on which the greater perpendicular falls is said to be farther from the centre.

Definition 6 figure

VI.

A segment of a circle is the figure contained by a straight line and the part of the circumference it cuts off.

Definition 7 figure

VII.

An angle in a segment is the angle contained by two straight lines drawn from any point in the circumference of the segment to the extremities of the straight line which is the base of the segment.

Definition 8 figure

VIII.

An angle is said to stand on the part of the circumference, or the arch, intercepted between the right lines that contain the angle.

Definition 9 figure

IX.

A sector of a circle is the figure contained by two radii and the arch between them.

Definition 10 figure a

X.

Similar segments of circles are those which contain equal angles.

Definition 10 figure b

Circles which have the same centre are called concentric circles.

Proposition I. Problem.

Proposition 1 figure

To find the centre of a given circle Blue circle.

Draw within the circle any straight line Red and dotted red line , make Red line = Red dotted line, draw Black line ⊥ Red and dotted red line ; bisect Black line, and the point of bisection is the centre.

For, if it be possible, let any other point as the point of concourse of Blue line, Blue dotted line and Black dotted line be the centre.

Because in Top triangle and Bottom triangle

Blue line = Black dotted line (hyp. and B. 1, def. 15.) Red line = Red dotted line (const.) and Blue dotted line common, Blue and yellow angles = Black angle (B. 1, pr. 8.), and are therefore right angles; but Black and yellow angles = Right angle (const.) Black angle = Black and yellow angles (ax. 11.) which is absurd; therefore the assumed point is not the centre of the circle; and in the same manner it can be proved that no other point which is not on Black line is the centre, therefore the centre is in Black line, and therefore the point where Black line is bisected is the centre.

Q. E. D.

Proposition II. Theorem.

Proposition 2 figure

A straight line (Red line) joining two points in the circumference of a circle Red circle, lies wholly within the circle.

Find the centre of Red circle (B. 3. pr. 1.);
from the centre draw Black line to any point in Red line,
meeting the circumference from the centre;
draw Yellow line and Blue line.

Then Blue angle = Black angle (B. 1. pr. 5.)
but Yellow angle > Blue angle or > Black angle (B. 1. pr. 16.)
∴ Yellow line > Black line (B. 1. pr. 19.)
but Yellow line = Black and dotted black line ,
∴ Black and dotted black line > Black line;
∴ Black line < Black and dotted black line ;
∴ every point in Red line lies within the circle.

Q. E. D.

Proposition III. Theorem.

Proposition 3 figure

If a straight line (Black line) drawn through the centre of a circle Blue circle bisects a chord ( Red and dotted red line ) which does not pass through the centre, it is perpendicular to it; or, if perpendicular to it, it bisects it.

Draw Blue line and Yellow line to the centre of the circle.

In Left triangle and Right triangle
Yellow line = Blue line, Black line common, and
Red line = Red dotted line ∴ Black angle = Yellow angle (B. 1. pr. 8.)
and ∴ Black line ⊥ Red and dotted red line (B. 1. def. 10.)
Again let Black line ⊥ Red and dotted red line

Then in Left triangle and Right triangle
Blue angle = Red angle (B. 1. pr. 5.)
Black angle = Yellow angle (hyp.)
and Yellow line = Blue line
∴ Red line = Red dotted line (B. 1. pr. 26.)
and ∴ Black line bisects Red and dotted red line .

Q. E. D.

Proposition IV. Theorem.

Proposition 4 figure

If in a circle two straight lines cut one another, which do not both pass through the centre, they do not bisect one another.

If one of the lines pass through the centre, it is evident that it cannot be bisected by the other, which does not pass through the centre.

But if neither the lines Black line or Red line pass through the centre, draw Black dotted line from the centre to their intersection.

If Black line be bisected, Black dotted line ⊥ to it (B. 3. pr. 3.)
∴ Blue and yellow angles = Right angle and if Red line be
bisected, Black dotted line ⊥ Red line (B. 3. pr. 3.)
∴ Blue angle = Right angle
and ∴ Blue angle = Blue and yellow angles ; a part
equal to the whole, which is absurd:
∴ Black line and Red line
do not bisect one another.

Q. E. D.

Proposition V. Theorem.

Proposition 5 figure

If two circles Blue and red circles intersect, they have not the same centre.

Suppose it possible that two intersecting circles have a common centre; from such supposed centre draw Yellow line to the intersecting point, and Black and dotted black line meeting the circumferences of the circles.

Then Yellow line = Black line (B. 1. def. 15.)
and Yellow line = Black and dotted black line (B. 1. def. 15.)
∴ Black line = Black and dotted black line ; a part
equal to the whole, which is absurd:
∴ circles supposed to intersect in any point cannot have the same centre.

Q. E. D.

Proposition VI. Theorem.

Proposition 6 figure

If two circles Red and black circles touch one another internally they have not the same centre.

For, if it be possible, let both circles have the same centre; from such a supposed centre draw Blue and dotted blue line cutting both circles, and Yellow line to the point of contact.

Then Yellow line = Blue dotted line (B. 1. def. 15.)
and Yellow line = Blue and dotted blue line (B. 1. def. 15.)
∴ Blue dotted line = Blue and dotted blue line ; a part
equal to the whole, which is absurd;

therefore the assumed point is not the centre of both circles; and in the same manner it can be demonstrated that no other point is.

Q. E. D.

Proposition VII. Theorem.

Figure I.
Proposition 7 figure 1
Figure II.
Proposition 7 figure 2

If from any point within a circle Blue circle which is not the centre, lines { Black and dotted black line Red line Blue line are drawn to the circumference; the greatest of those lines is that ( Black and dotted black line ) which passes through the centre, and the least is the remaining part (Yellow line) of the diameter.

Of the others, that (Red line) which is nearer to the line passing through the centre, is greater than that (Blue line) which is more remote.

Fig. 2. The two lines ( Blue and dotted blue line and Red line) which make equal angles with that passing through the centre, on opposite sides of it, are equal to each other; and there cannot be drawn a third line equal to them, from the same point to the circumference.

Figure I.

To the centre of the circle draw Red dotted line and Blue dotted line; then Black dotted line = Red dotted line (B. 1. def. 15.) Black and dotted black line = Black line + Red dotted line > Red line (B. 1. pr. 20.) in like manner Black and dotted black line may be shewn to be greater than Blue line, or any other line drawn from the same point to the circumference. Again, by (B. 1. pr. 20.) Black line + Blue line > Blue dotted line = Yellow line + Black line, take Black line from both; ∴ Blue line > Yellow line (ax.), and in like manner it may be shewn that Yellow line is less than any other line drawn from the same point to the circumference. Again, in Red triangle and Blue triangle , Black line common, Black and yellow angles > Yellow angle , and Red dotted line = Blue dotted line

∴ Red line > Blue line (B. 1. pr. 24.) and Red line may in like manner be proved greater than any other line drawn from the same point to the circumference more remote from Black and dotted black line .

Figure II.

If Red angle = Yellow angle then Blue and dotted blue line = Red line, if not take Blue line = Red line draw Yellow and dotted yellow line , then in Left triangle and Right triangle , Black line common, Red angle = Yellow angle and Red line = Blue line ∴ Red dotted line = Yellow line (B. 1. pr. 4.) ∴ Red dotted line = Yellow and dotted yellow line = Yellow line a part equal to the whole, which is absurd:

∴ Red line = Blue and dotted blue line ; and no other line is equal to Red line drawn from the same point to the circumference; for if it were nearer to the one passing through the centre it would be greater, and if it were more remote it would be less.

Q. E. D.

Proposition VIII. Theorem.

The original text of this proposition is here divided into three parts.

Proposition 8 figure 1

I.

If from a point without a circle, straight lines { Black and dotted black line Red line Blue lineetc. } are drawn to the circumference; of those falling upon the concave circumference the greatest is that ( Black and dotted black line ) which passes through the centre, and the line (Red line) which is nearer the greatest is greater than that (Blue line) which is more remote.

Draw Blue dotted line and Red dotted line to the centre.

Then, Black and dotted black line which passes through the centre, is greatest; for since Black dotted line = Red dotted line, if Black line be added to both, Black and dotted black line = Black line + Red dotted line; but > Red line (B. 1. pr. 20.) ∴ Black and dotted black line is greater than any other line drawn from the same point to the concave circumference.

Again in Blue triangle and Red triangle , Blue dotted line = Red dotted line,
and Black line common, but Yellow and black angles > Yellow angle ,
∴ Red line > Blue line (B. 1. pr. 24.);
and in like manner Red line may be shewn > than any other line more remote from Black and dotted black line .

Proposition 8 figure 2

II.

Of those lines falling on the convex circumference the least is that (Black dotted line) which being produced would pass through the centre, and the line which is nearer to the least is less than that which is more remote.

For, since Red line + Red dotted line > Black and dotted black line (B. 1. pr. 20.)
and Red line = Black line,
∴ Red dotted line > Black dotted line (ax. 5.)
And again, since Blue line + Blue dotted line >
Red line + Red dotted line (B. 1. pr. 21.),
and Blue line = Red line,
∴ Red dotted line < Blue dotted line. And so of others.

Proposition 8 figure 3

III.

Also the lines making equal angles with that which passes through the centre are equal, whether falling on the concave or convex circumference; and no third line can be drawn equal to them from the same point to the circumference.

For if Dotted red and yellow line > Blue dotted line, but making Yellow angle = Blue angle ;
make Red dotted line = Blue dotted line, and draw Dotted black and red line .
Then in Red triangle and Blue triangle we have Red dotted line = Blue dotted line,
and Black line common, and also Blue angle = Yellow angle ,
∴ Dotted black and red line = Blue line (B. 1. pr. 4.);
but Blue line = Black dotted line;
∴ Black dotted line = Dotted black and red line , which is absurd.
∴ Blue dotted line is not = Red dotted line, nor to any part
of Dotted red and yellow line , ∴ Dotted red and yellow line is not > Blue dotted line.
Neither is Blue dotted line > Dotted red and yellow line , they are
∴ = to each other.

And any other line drawn from the same point to the circumference must lie at the same side with one of these lines, and be more or less remote than it from the line passing through the centre, and cannot therefore be equal to it.

Q. E. D.

Proposition IX. Theorem.

Proposition 9 figure

If a point be taken within a circle Blue circle, from which more than two equal straight lines (Yellow dotted line, Yellow line, Blue line) can be drawn to the circumference, that point must be the centre of the circle.

For if it be supposed that the point Yellow and blue point in which more than two equal straight lines meet is not the centre, some other point Black and red dotted line must be; join these two points by Black line, and produce it both ways to the circumference.

Then since more than two equal straight lines are drawn from a point which is not the centre, to the circumference, two of them at least must lie at the same side of the diameter Red, black, and red dotted line ; and since from a point Red, yellow, and blue point , which is not the centre, straight lines are drawn to the circumference; the greatest is Black and red dotted line , which passes through the centre: and Blue line which is nearer to Black and red dotted line , > Yellow line which is more remote (B. 3. pr. 8.); but Blue line = Yellow line (hyp.) which is absurd.

The same may be demonstrated of any other point, different from Red, yellow, and blue point , which must be the centre of the circle.

Q. E. D.

Proposition X. Theorem.

Proposition 10 figure
Proposition 10 figure

One circle Blue circle cannot intersect another Red circle in more points than two.

For if it be possible, let it intersect in three points;
from the centre of Blue circle draw Black line, Yellow line
and Blue line to the points of intersection;
∴ Black line = Yellow line = Blue line
(B. 1. def. 15.),
but as the circles intersect, they have not the same centre (B. 3. pr. 5.):
∴ the assumed point is not the centre of Red circle, and

∴ as Black line, Yellow line and Blue line are drawn from a point and not the centre, they are not equal (B. 3. prs. 7, 8); but it was shewn before that they were equal, in which is absurd; the circles therefore do not intersect in three points.

Q. E. D.

Proposition XI. Theorem.

Proposition 11 figure

If two circles Blue circle and Black circle touch one another internally, the right line joining their centres, being produced, shall pass through a point of contact.

For if it be possible, let Black line join their centres, and produce it both ways; from a point of contact draw Red line to the centre of Blue circle, and from the same point of contact draw Blue dotted line to the centre of Black circle.

Because in Yellow triangle ; Black line + Red line > Blue dotted line
(B. 1. pr. 20.),
and Blue dotted line = Yellow and black line as they are radii of
Black circle,
but Black line + Red line > Yellow and black line ; take
away Black line which is common,
and Red line > Yellow and dotted yellow line ;
but Red line = Yellow dotted line,
because they are radii of Blue circle,
and ∴ Yellow dotted line > Yellow and dotted yellow line a part greater than the whole, which is absurd.

The centres are not therefore so placed, that a line joining them can pass through any point but a point of contact.

Q. E. D.

Proposition XII. Theorem.

Proposition 12 figure

If two circles Blue circle and Red circle touch one another externally, the straight line Red, black, and blue line joining their centres passes through the point of contact.

If it be possible, let Red, black, and blue line join the centres, and not pass through a point of contact; then from a point of contact draw Yellow dotted line and Yellow line to the centres.

Because Yellow dotted line + Yellow line > Red, black, and blue line (B. 1. pr. 20.),
and Red line = Yellow dotted line (B. 1. def. 15.),
and Blue line = Yellow line (B. 1. def. 15.),

∴ Red line + Blue line > Red, black, and blue line , a part greater than the whole, which is absurd.

The centres are not therefore so placed, that the line joining them can pass through any point but the point of contact.

Q. E. D.

Proposition XIII. Theorem.

Figure I.
Proposition 13 Figure 1

One circle cannot touch another, either externally or internally in more points than one

Fig. 1. For if it be possible, let Yellow circle and Blue circle touch one another internally in two points; draw Blue line joining their centres, and produce it until it pass through one of the points of contact (B. 3. pr. 11.);

draw Red line and Black line,
But Blue dotted line = Black line (B. 1. def. 15.),
∴ if Blue line be added to both,
Blue and dotted blue line = Blue line + Black line;
but Blue and dotted blue line = Red line (B. 1. def. 15.),
and ∴ Blue line + Black line = Red line; but
Blue line + Black line > Red line (B. 1. pr. 20.),
which is absurd.

Figure II.
Proposition 13 Figure 2

Fig. 2. But if the points of contact be the extremities of the right line joining the centres, this straight line must be bisected in two different points for the two centres; because it is the diameter of both circle, which is absurd.

Figure III.
Proposition 13 Figure 3

Fig. 3. Next, if it be possible, let Yellow circle and Blue circle touch externally in two points; draw Red and dotted red line joining the centres of the circles, and passing through one of the points of contact, and draw Blue line and Black line.

Blue line = Red line (B. 1. def. 15.);
and Red dotted line = Black line (B. 1. def. 15.):
∴ Black line + Blue line = Red and dotted red line ; but
Black line + Blue line > Red and dotted red line (B. 1. pr. 20.),
which is absurd.

There is therefore no case in which two circles can touch one another in two points.

Q. E. D.

Proposition XIV. Theorem.

Proposition 14 figure

Equal straight lines ( Yellow and dotted yellow line Red and dotted red line ) inscribed in a circle are equally distant from the centre; and also, straight lines equally distant from the centre are equal.

From the centre of Blue circle draw
Black dotted line ⊥ to Red and dotted red line and Blue dotted line
⊥ Yellow and dotted yellow line , join Black line and Blue line.

Then Yellow line = half Yellow and dotted yellow line (B. 3. pr. 3.)
and Red line = 1 / 2 Red and dotted red line (B. 3. pr. 3.)
since Yellow and dotted yellow line = Red and dotted red line (hyp.)
∴ Yellow line = Red line,
and Black line = Blue line (B. 1. def. 15.)
∴ Black line2 = Blue line2;
but since Yellow angle is a right angle
Black line2 = Black dotted line2 + Red line2 (B. 1. pr. 47.)
and Blue line2 = Blue dotted line2 + Yellow line2 for the same reason,
∴ Black dotted line2 + Red line2 = Blue dotted line2 + Yellow line2
∴ Black dotted line2 = Blue dotted line2,
∴ Black dotted line = Blue dotted line.

Also, if the lines Red and dotted red line and Yellow and dotted yellow line be equally distant from the centre; that is to say, if the perpendiculars Black dotted line and Blue dotted line be given equal then Red and dotted red line = Yellow and dotted yellow line .

For, as in the preceding case,
Blue dotted line2 + Yellow line2 = Red line2 + Black dotted line2;
but Blue dotted line2 = Black dotted line2:

∴ Yellow line2 = Red line2, and the doubles of these
Yellow and dotted yellow line and Red and dotted red line are also equal.

Q. E. D.

Proposition XV. Theorem.

Figure I.
Proposition 15 figure 1

The diameter is the greatest straight line in a circle: and, of all others, that which is nearest to the centre is greater than the more remote.

Figure I.

The diameter Red and black line is > any line Blue line.
For draw Yellow line and Yellow dotted line.
Then Yellow dotted line = Black line
and Yellow line = Red line,
∴ Yellow line + Yellow dotted line = Red and black line
but Yellow line + Yellow dotted line > Blue line (B. 1. pr. 20.)
∴ Red and black line > Blue line.

Again, the line which is nearer the centre is greater than the one more remote.

First, let the given lines be Blue line and Red dotted line, which are at the same side of the centre and do not intersect; draw { Yellow dotted line, Yellow line, Black dotted line, Blue dotted line. } In Wide triangle and Narrow triangle ,
Yellow line and Yellow dotted line = Black dotted line and Blue dotted line;
Red and yellow angles > Yellow angle ,
∴ Blue line > Red dotted line (B. 1. pr. 24.)

Figure II.
Proposition 15 figure 2

Figure II.

Let the given lines be Red line and Yellow line which either are at different sides of the centre, or intersect; from the centre draw Dotted yellow and red line and Blue dotted line ⊥ Yellow line and Red line, make Blue dotted line = Yellow dotted line, and draw Blue line ⊥ Dotted yellow and red line .

Since Red line and Blue line are equally distant from
the centre, Red line = Blue line (B. 3. pr. 14.);
but Blue line > Yellow line (Pt. 1. B. 3. pr. 15.),
∴ Red line > Yellow line.

Q. E. D.

Proposition XVI. Theorem.

Proposition 16 figure

The straight line Yellow line drawn from the extremity of the diameter Black line of a circle perpendicular to it falls without the circle.

And if any straight line Red dotted line be drawn from a point within that perpendicular to the point of contact, it cuts the circle.

Part I

If it be possible, let Red line, which meets the circle again, be ⊥ Black line, and draw Blue line.

Then, because Blue line = Black line,
Yellow angle = Black angle (B. 1. pr. 5.),
and ∴ each of these angles is acute (B. 1. pr. 17.)
but Yellow angle = Right angle (hyp.), which is absurd, therefore
Red line drawn ⊥ Black line does not meet
the circle again.

Part II.

Let Yellow line be ⊥ Black line and let Red dotted line be drawn from a point Point between Yellow line and the circle, which if it be possible, does not cut the circle.

Because Blue and red angle = Right angle ,
∴ Blue angle is an acute angle; suppose
Dotted blue and black line ⊥ Red dotted line, drawn from the centre of the circle, it must fall at the side of Blue angle the acute angle.
∴ Outlined angle which is supposed to be a right angle, is > Blue angle ,
∴ Black line > Dotted blue and black line ;
but Blue dotted line = Black line,

and ∴ Blue dotted line > Dotted blue and black line , a part greater than the whole, which is absurd. Therefore the point does not fall outside the circle, and therefore the straight line Red dotted line cuts the circle.

Q. E. D.

Proposition XVII. Theorem.

Proposition 17 figure

To draw a tangent to a given circle Red circle from a given point, either in or outside of its circumference.

If the given point be in the circumference, as at Blue and black dotted point , it is plain that the straight line Blue line ⊥ Black dotted line the radius, will be the required tangent (B. 3. pr. 16.)

But if the given point Red and blue point be outside of the circumference, draw Red dotted and solid line
from it to the centre, cutting Red circle; and
draw Blue dotted line ⊥ Red dotted line, describe Yellow circle
concentric with Red circle radius = Red dotted and solid line ,
then Blue line will be the tangent required.

For in Bottom triangle and Top triangle
Red dotted and solid line = Black dotted and solid line , Blue and red angle common,
and Red dotted line = Black dotted line,
∴ (B. 1. pr. 4.) Bottom yellow angle = Top yellow angle = a right angle,
∴ Blue line is a tangent to Red circle.

Q. E. D.

Proposition XVIII. Theorem.

Proposition 18 figure

If a right line Blue dotted line be a tangent to a circle, the straight line Blue line drawn from the centre to the point of contact, is perpendicular to it.

For if it be possible, let Red solid and dotted line be ⊥ Blue dotted line,
then because Yellow angle = Right angle , Red yellow angle is acute (B. 1. pr. 17.)
∴ Blue line > Red solid and dotted line (B. 1. pr. 19.);
but Blue line = Red line,
and ∴ Red line > Red solid and dotted line ,
a part greater than the whole, which is absurd.

∴ Red solid and dotted line is not ⊥ Blue dotted line; and in the same manner it can be demonstrated, that no other line except Blue line is perpendicular to Blue dotted line.

Q. E. D.

Proposition XIX. Theorem.

Proposition 19 figure

If a straight line Blue line be a tangent to a circle, the straight line Yellow line, drawn perpendicular to it from a point of the contact, passes through the centre of the circle.

For, if it be possible, let the centre be without Yellow line, and draw Red dotted line from the supposed centre to the point of contact.

Because Red dotted line ⊥ Blue line (B. 3. pr. 18.)

∴ Yellow angle = Right angle , a right angle;
but Blue and yellow angles = Right angle (hyp.), and ∴ Yellow angle = Blue and yellow angles ,
a part equal to the whole, which is absurd.

Therefore the assumed point is not the centre; and in the same manner it can be demonstrated, that no other point without Yellow line is the centre.

Q. E. D.

Proposition XX. Theorem.

Figure I.
Proposition 20 figure 1

The angle at the centre of a circle, is double the angle at the circumference, when they have the same part of the circumference for their base.

Figure I.

Let the centre of the circle be on Red solid and dotted line
a side of Yellow angle .

Because Black line = Red line,
Yellow angle = Red angle (B. 1. pr. 5.).

But Blue angle = Yellow angle + Red angle ,
Blue angle = twice Yellow angle (B. 1. pr. 32).

Figure II.
Proposition 20 figure 2

Figure II.

Let the centre be within Top red and yellow angles , the angle at the circumference; draw Black line from the angular point through the centre of the circle;
then Top red angle = Bottom red angle , and Top yellow angle = Bottom yellow angle ,
because of the equality of the sides (B. 1. pr. 5).

Hence Bottom red angle + Top red angle + Top yellow angle + Bottom yellow angle = twice Top red and yellow angles .
But Black angle = Top red angle + Bottom red angle , and
Blue angle = Top yellow angle + Bottom yellow angle ,
∴ Black and blue angles = twice Top red and yellow angles .

Figure III.
Proposition 20 figure 3

Figure III.

Let the centre be without Red angle and
draw Red line, the diameter.

Because Blue and yellow angles = twice Black and red angles ; and
Blue angle = twice Black angle (case 1.);
∴ Yellow angle = twice Red angle .

Q. E. D.

Proposition XXI. Theorem.

Figure I.
Proposition 21 figure 1

The angles ( Red angle , Blue angle ) in the same segment of a circle are equal.

Figure I.

Let the segment be greater than a semicircle, and draw Red line and Blue line to the centre.

Yellow angle = twice Red angle or twice = Blue angle (B. 3. pr. 20.);
∴ Red angle = Blue angle .

Figure II.
Proposition 21 figure 2

Figure II.

Let the segment be a semicircle, or less than a semicircle, draw Blue line the diameter, also draw Red line.

Yellow angle = Blue angle and Red angle = Black angle (case 1.)
∴ Yellow and red angles = Blue and black angles .

Q. E. D.

Proposition XXII. Theorem.

Proposition 22 figure

The opposite angles Red and blue angles and Yellow and black angles , Black and blue angles and Yellow and red angles of any quadrilateral figure inscribed in a circle, are together equal to two right angles.

Draw Red line and Black line the diagonals; and because angles in the same segment are equal Left blue angle = Top blue angle ,
and Right red angle = Top red angle ;
add Yellow and black angles to both.
∴ Red and blue angles + Yellow and black angles = Yellow and black angles + Left blue angle + Right red angle =
two right angles (B. 1. pr. 32.). In like manner it may be shown that,
Black and blue angles + Yellow and red angles = Two right angles .

Q. E. D.

Proposition XXIII. Theorem.

Proposition 23 figure

Upon the same straight line and upon the same side of it, two similar segments of circles cannot be constructed which do not coincide.

For if it be possible, let two similar segments
Blue segment and Red segment be constructed;
draw any right line Red line cutting both the segments,
draw Blue line and Yellow line.

Because the segments are similar,
Yellow angle = Blue angle (B. 3. def. 10.),
but Yellow angle > Blue angle (B. 1. pr. 16.)
which is absurd: therefore no point in either of the segments falls without the other, and therefore the segments coincide.

Q. E. D.

Proposition XXIV. Theorem.

Proposition 24 figure

Similar segments Red segment and Yellow segment , of circles upon equal straight lines (Black line and Blue line) are each equal to the other.

For, if Yellow segment be so applied to Red segment ,
that Blue line may fall on Black line, the extremities of
Blue line may be on the extremities Black line and
Red curve at the same side as Blue curve ;

because Blue line = Black line,
Blue line must wholly coincide with Black line;
and the similar segments being then upon the same straight line at the same side of it, must also coincide (B. 3. pr. 23.), and are therefore equal.

Q. E. D.

Proposition XXV. Problem.

Proposition 25 figure

A segment of a circle being given, to describe the circle of which it is the segment.

From any point in the segment draw Blue line and Black line bisect them, and from the points of bisection

draw Yellow line ⊥ Blue line
and Red line ⊥ Black line
where they meet is the centre of the circle.

Because Blue line terminated in the circle is bisected perpendicularly by Yellow line, it passes through the centre (B. 3. pr. 1.), likewise Red line passes through the centre, therefore the centre is in the intersection of these perpendiculars.

Q. E. D.

Proposition XXVI. Theorem.

Proposition 26 figure
Proposition 26 figure

In equal circles Blue circle and Red circle, the arcs Arc , Arc on which stand equal angles, whether at the centre or circumference, are equal.

First, let Yellow angle = Black angle at the centre,
draw Black line and Black dotted line.

Then since Blue circle = Red circle,
Triangle and Triangle dotted have
Blue line = Red line = Blue dotted line = Red dotted line,
and Yellow angle = Black angle ,
∴ Black line = Black dotted line (B. 1. pr. 4.).

But Red angle = Blue angle (B. 3. pr. 20.);
∴ Blue segment and Red segment are similar (B. 3. def. 10.);
they are also equal (B. 3. pr. 24.)

If therefore the equal segments be taken from the equal circles, the remaining segments will be equal;

hence Segment = Segment (ax. 3.);
∴ Arc = Arc .

But if the given equal angles be at the circumference, it is evident that the angles at the centre, being double of those at the circumference, are also equal, and therefore the arcs on which they stand are equal.

Q. E. D.

Proposition XXVII. Theorem.

Proposition 27 figure
Proposition 27 figure

In equal circles, Red circle an[d] Blue circle the angles Yellow angles and Red angles which stand upon equal arches are equal, whether they be at the centres or at the circumferences.

For if it be possible, let one of them
Red angles be greater than the other Yellow angles
and make
Yellow and blue angles = Red angles

∴ Black and red dotted arc = Black dotted arc (B. 3. pr. 26.)
but Black arc = Black dotted arc (hyp.)
∴ Black arc = Black and red dotted arc a part equal
to the whole, which is absurd; ∴ neither angle is greater than the other, and ∴ they are equal.

Q. E. D.

Proposition XXVIII. Theorem.

Proposition 28 figure
Proposition 28 figure

In equal circles Yellow circle and Black circle, equal chords Red line, Red dotted line cut off equal arches.

From the centres of the equal circles,
draw Black line, Blue line and Black dotted line, Blue dotted line;
and because Yellow circle = Black circle
Black line, Blue line = Black dotted line, Blue dotted line
also Red line = Red dotted line (hyp.)
∴ Red angle = Yellow angle
∴ Blue arc = Red arc (B. 3. pr. 26.)
∴ Yellow arc = Black arc (ax. 3.)

Q. E. D.

Proposition XXIX. Theorem.

Proposition 29 figure
Proposition 29 figure

In equal circles Yellow circle and Black circle the chords Red line and Red dotted line which subtend equal arcs are equal.

If the equal arcs be semicircles the proposition is evident. But if not,
let Black line, Blue line, and Black dotted line, Blue dotted line
be drawn to the centres;
because Blue arc = Red arc (hyp.)
and Red angle = Yellow angle (B. 3. pr. 27.);
but Black line and Blue line = Black dotted line and Blue dotted line
∴ Red line = Red dotted line (B. 1. pr. 4.);
but these are the chords subtending the equal arcs.

Q. E. D.

Proposition XXX. Problem.

Proposition 30 figure

To bisect a given arc Red and dotted red arc .

Draw Black and dotted black line ;
make Black line = Black dotted line,
draw Yellow line ⊥ Black and dotted black line , and it bisects the arc.
Draw Blue dotted line and Blue line.
Black line = Black dotted line (const.),
Yellow line is common,
and Blue angle = Red angle (const.)
∴ Blue dotted line = Blue line (B. 1. pr. 4.)
Red arc = Red dotted arc (B. 3. pr. 28.),
and therefore the given arc is bisected.

Q. E. D.

Proposition XXXI. Theorem.

Figure I.
Proposition 31 figure 1

In a circle the angle in a semicircle is a right angle, the angle in a segment greater than a semicircle is acute, and the angle in a segment less than a semicircle is obtuse.

Figure I.

The angle Yellow and black angles in a semicircle is a right angle.

Draw Red line and Blue and black line
Red angle = Yellow angle and Blue angle = Black angle (B. 1. pr. 5.)
Blue angle + Red angle = Yellow and black angles = the half of two
right angles = a right angle. (B. 1. pr. 32.)

Figure II.
Proposition 31 figure 2

Figure II.

The angle Blue angle in a segment greater than a semicircle is acute.

Draw Red line the diameter, and Blue line
∴ Blue and red angles = a right angle
∴ Blue angle is acute.

Figure III.
Proposition 31 figure 3

Figure III.

The angle Red angle in a segment less than semicircle is obtuse.

Take in the opposite circumference any point, to which draw Blue line and Red line.

Because Yellow angle + Red angle = Two right angles (B. 3. pr. 22.)
but Yellow angle < Right angle (part 2.),
∴ Red angle is obtuse.

Q. E. D.

Proposition XXXII. Theorem.

Proposition 32 figure

If a right line Blue line be a tangent to a circle and from the point of contact a right line Red line be drawn cutting the circle, the angle Bottom yellow angle made by this line with the tangent is equal to the angle Top yellow angle in the alternate segment of the circle.

If the chord should pass through the centre, it is evident the angles are equal, for each of them is a right angle. (B. 3. prs. 16, 31.)

But if not, draw Black line ⊥ Blue line from the point of contact, it must pass through the centre of the circle, (B. 3. pr. 19.)

∴ Black angle = Right angle (B. 3. pr. 31.)
Top yellow angle + Blue angle = Right angle = Blue and yellow angles (B. 1. pr. 32.)
∴ Top yellow angle = Bottom yellow angle (ax.).
Again Outlined, blue and yellow angles = Two right angles = Top yellow angle + Red angle (B. 3. pr. 22.)

∴ Outlined and blue angles = Red angle , (ax.), which is the angle in the alternate segment.

Q. E. D.

Proposition XXXIII. Problem.

Proposition 33 figure

On a given straight line Black line to describe a segment of a circle that shall contain an angle equal to a given angle Right angle , Obtuse angle , Black angle .

If the given angle be a right angle, bisect the given line, and describe a semicircle on it, this will evidently contain a right angle. (B. 3. pr. 31.)

If the given angle be acute or obtuse, make with the given line, at its extremity,

Yellow angle = Black angle , draw Blue line ⊥ Red line and
make Top red angle = Bottom red angle , describe Blue circle
with Blue line or Yellow line as radius, for they are equal.

Red line is a tangent to Blue circle (B. 3. pr. 16.)
∴ Black line divides the circle into two segments
capable of containing angles equal to
Outlined and red angles and Yellow angle which were made respectively equal
to Obtuse angle and Black angle (B. 3. pr. 32.)

Q. E. D.

Proposition XXXIV. Problem.

Proposition 34 figure

To cut off from a given circle Blue circle a segment which shall contain an angle equal to a given angle Red angle .

Draw Red line (B. 3. pr. 17.), a tangent to the circle at any point; at the point of contact make

Blue angle = Red angle the given angle;
and Segment contains an angle = the given angle.

Because Red line is a tangent,
and Black line cuts it, the
angle in Segment = Blue angle (B. 3. pr. 32.),
but Blue angle = Red angle (const.)

Q. E. D.

Proposition XXXV. Theorem.

Figure I.
Proposition 35 figure 1

If two chords { Blue and dotted blue line Black and dotted black line } in a circle intersect each other, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other.

Figure I.

If the given right lines pass through the centre, they are bisected in the point of intersection, hence the rectangles under their segments are the squares of their halves and are therefore equal.

Figure II.
Proposition 35 figure 2

Figure II.

Let Black dotted, red dotted, and black line pass through the centre, and
Blue and dotted blue line not; draw Yellow line and Red line.
Then Blue line × Blue dotted line = Yellow line2 − Red dotted line2 (B. 2. pr. 6.),
or Blue line × Blue dotted line = Black and red dotted line 2 − Red dotted line2.
∴ Blue line × Blue dotted line = Red dotted and black line × Black dotted line (B. 2. pr. 5.).

Figure III.
Proposition 35 figure 3

Figure III.

Let neither of the given lines pass through the centre, draw through their intersection a diameter Red dotted and red line ,

and Red dotted line × Red line = Blue line × Blue dotted line (Part. 2.),
also Red dotted line × Red line = Black line × Black dotted line (Part. 2.);
∴ Blue line × Blue dotted line = Black line × Black dotted line.

Q. E. D.

Proposition XXXVI. Theorem.

Figure I.
Proposition 36 figure 1

If from a point without a circle two straight lines be drawn to it, one of which Blue line is a tangent to the circle, and the other Black, dotted red, and red line cuts it; the rectangle under the whole cutting line Black, dotted red, and red line and the external segment Red line is equal to the square of the tangent Blue line.

Figure I.

Let Black, dotted red, and red line pass through the centre;
draw Yellow line from the centre to the point of contact;
Blue line2 = Dotted red, and red line 2 minus Yellow line2 (B. 1. pr. 47),
or Blue line2 = Dotted red, and red line 2 minus Red dotted line2,
∴ Blue line2 = Black, dotted red, and red line × Red line (B. 2. pr. 6).

Figure II.
Proposition 36 figure 2

Figure II.

If Dotted red, and red line do not pass through the centre,
draw Yellow dotted line and Blue dotted line.

Then Dotted red, and red line × Red line = Black line2 minus Blue dotted line2
(B. 2. pr. 6), that is,
Dotted red, and red line × Red line = Black line2 minus Yellow line2,
∴ Dotted red, and red line × Red line = Blue line2 (B. 3. pr. 18).

Q. E. D.

Proposition XXXVII. Problem.

Proposition 37 figure

If from a point outside a circle two straight lines be drawn, the one Black and dotted black line cutting the circle, the other Red line meeting it, and if the rectangle contained by the whole cutting line Black and dotted black line and its external segment Black dotted line be equal to the square of the line meeting the circle, the latter Red line is a tangent to the circle.

Draw from the given point
Blue line, a tangent to the circle, and draw from the
centre Yellow line, Red dotted line, and Blue dotted line,
Blue line2 = Black and dotted black line × Black dotted line (B. 3. pr. 36.)
but Red line2 = Black and dotted black line × Black dotted line (hyp.),
∴ Red line2 = Blue line2,
and ∴ Red line = Blue line;

Then in Red triangle and Blue triangle
Red dotted line and Red line = Blue dotted line and Blue line,
and Yellow line is common,
∴ Blue angle = Red angle (B. 1. pr. 8.);
but Red angle = Right angle a right angle (B. 3. pr. 18.),
∴ Blue angle = Right angle a right angle,
and ∴ Red line is a tangent to the circle (B. 3. pr. 16.).

Q. E. D.

Books.

  1. Proposition 47 figure
    Book I. Basic plane geometry
  2. Proposition 7 figure
    Book II. Geometric algebra
  3. Proposition 21 figure 1
    Book III. Circles and angles
  4. Proposition 4 figure
    Book IV. Regular polygons
  5. Proposition 24 figure
    Book V. Ratios and proportions
  6. Proposition 3 figure
    Book VI. Geometric proportions