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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 line&c. } 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