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I. EQUAL circles are those of which the diameters are equal, or from the centres of which the straight lines to the circumferences are equal.
This is not a definition, but a theorem, the truth of which is evident; for, if the circles be applied to one another, so that their centres coincide, the circles must likewise coincide, since the straight lines from the centres are equal.
II. A straight line is said to touch a circle when it meets the circle, and being produced does not cut it.
III. Circles are said to touch one another, which meet but do not cut one another.
IV. Straight 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.
Definitions IV. and V.
VI. A segment of a circle is the figure contained by a straight line and the circumference it cuts off.
VII. The angle of a segment is that which is contained by the straight line and the circumference.
VIII. 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.
IX. And an angle is said to insist or stand upon the circumference intercepted between the straight lines that contain the angle.
Definitions VIII. and IX.
X. The sector of a circle is the figure contained by two straight lines drawn from the centre, and the circumference between them.
XI. Similar segments of a circle are those in which the angles are equal, or which contain equal angles.
PROPOSITION I.-- PROBLEM.
To find the centre of a given circle.
Draw within it any straight line a b, and bisect (i. 10) it in d; from the point d draw (i. 11) dc at right angles to a b, and produce it to e, and bisect ce in f: the point f is the centre of the circle a b c.
For, if it be not, let, if possible, g be the centre, and join ga, gde
gb. Then, because da is equal to db, and dg common to the two triangles a dg, bdg, the two sides a d, dg, are equal to the two bd, dg, each to each ; and the base ga is equal to the base gb, because they are drawn from the centre g*: therefore the angle adg is equal (i. 8) to the angle gdb. But when a straight line standing upon another straight line
f makes the adjacent angles equal to one another, each of the angles is a right angle (i. def. 10): therefore the angle gdb is a right angle : but a
d fdb is likewise a right angle : wherefore the angle fdb is equal to the angle g db, the greater to the less, which is impossible : therefore g is not the centre of the circle a b c. In the same manner it can be shewn, that no other point but f is the centre ; that is, f is the centre of the circle a b c. Which was to be found.
Cor. From this it is manifest, that if in a circle a straight line bisect another at right angles, the centre of the circle is in the line which bisects the other.
PROPOSITION II.--THEOREM. If any two points be taken in the circumference of a circle, the straight line
which joins them shall fall within the circle. LET a b c be a circle, and a, b any two points in the circumference; the straight line drawn from a to b shall fall within the circle.
For, if it do not, let it fall, if possible, without, as a eb; find (i. 3) d the centre of the circle a bc; and join a d, db, and produce df, any straight line meeting the circumference a b, to e: then because da is equal to db, the angle dab is equal (i. 5) to the angle dba; and because a e, a side of the triangle da e, is produced to b, the angle deb is greater (i. 16) than the angle d ae; but dae is equal to the angle dbe; therefore the angle d e b is greater than the angle dbe: but to the greater angle the greater side is opposite (i. 19); db is therefore greater
b than de : but db is equal to df; wherefore d f is greater than d e, the less than the greater, which is impossible : therefore the straight line drawn from a to b does not fall without the circle. In the same manner,
may be demonstrated that it does not fall upon the circumference ; it falls therefore within it. Wherefore, if any two points, &c. Q. E. D.
* Whenever the expression “straight lines from the centre,” or “ drawn from the centre," occurs, it is to be understood that they are drawn to the circumference.
PROPOSITION III.-THEOREM. If a straight line drawn through the centre of a circle bisect a straight line in
it which does not pass through the centre, it shall cut it at right angles ;
and if it cuts it at right angles, it shall bisect it. LET abc be a circle ; and let cd, a straight line drawn through the centre, bisect any straight line a b, which does not pass through the centre, in the point f; it cuts it also at right angles. Take (i. 3) e the centre of the circle, and join e a, e b. Then, because
a f is equal to fb, and fe common to the two triangles a fe, bfe, there are two sides in the one equal to two sides in the other, and the base e a is equal to the base e b; therefore the angle a fe is equal (i. 8) to the angle bfe: but when a straight line standing upon another makes the adjacent angles equal to one another, each of them is a right
angle (i. def. 10): therefore each of the angles a dl
afe, bfe, is a right angle; wherefore the straight
line cd, drawn through the centre, bisecting another a b that does not pass through the centre, cuts the same at right angles.
But let cd cut a b at right angles ; c d also bisects it; that is, a f is equal to fb.
The same construction being made, because e a, e b, from the centre are equal to one another, the angle ea f is equal (i. 5) to the angle ebf: and the right angle a fe is equal to the right angle bfe: therefore, in the two triangles e a f, e bf, there are two angles in one equal to two angles in the other, and the side ef, which is opposite to one of the equal angles in each, is common to both; therefore the other sides are equal
1; (i. 26); a f therefore is equal to fb. Wherefore, if a straight line, &c. Q. E. D.
If in a circle two straight lines cut one another which do not both pass
through the centre, they do not bisect each other. LET a b c d be a circle, and a c, b d two straight lines in it which cut one another in the point e, and do not both pass through the centre : a c, b d, do not bisect one another.
For, if it be possible, let a e be equal to e c, and be to ed : if one of the lines pass through the centre, it is plain that it cannot be bisected
by the other which does not pass through the d
centre : but if neither of them pass through the centre, take (i. 3) f the centre of the circle, and
join ef; and because fe, a straight line through bi
the centre, bisects another a c which does not pass through the centre, it shall cut it at right (iii. 3) angles : wherefore fea is a right angle. Again, because the straight line fe bisects the straight line bd, which does not pass through the centre, it shall cut it at right angles (iii. 3) : wherefore feb is a right angle : and fea was shewn to be a right angle ; therefore fea is equal to the angle fe b, the less to the greater, which is impossible : therefore à c, b d do not bisect one another. Wherefore, if in a circle, &c. Q. E. D.
PROPOSITION V.-THEOREM. If two circles cut one another, they shall not have the same centre. LET the two circles a b c, cd g cut one another in the points b, c ; they have not the same centre.
For, if it be possible, let e be their centre; join e c, and draw any straight line efg meeting them in f and g; and because e is the centre of d the circle a b c ce is equal to ef. Again, because
f e is the centre of the circle cdg, ce is equal to eg: but ce was shewn to be equal to ef: therefore ef is equal to eg, the less to the greater, which is impossible : therefore e is not the centre
e of the circles abc, cd g. Wherefore, if two circles, &c. Q. E. D.
PROPOSITION VÍ.—THEOREM. If twò circles touch one another internally, they shall not have the same
LET the two circles a bccde, touch one another internally in the point c; they have not the same centre.
For, if they have, let it be f; join fc and draw any straight line feb meeting them in e and b; and because f is the centre of the circle abc, cf is equal to fb; also, because f is the centre of the circle cde, cf is equal to fe: and cf was shewn to be equal to fb; therefore fe is
f equal to fb, the less to the greater, which is impossible : wherefore f is not the centre of the circles abc, cde. Therefore, if two cireles,
d &c. Q. E. D.
PROPOSITION VII.—THEOREM. If any point be taken in the diameter of a circle which is not the centre, of
all the straight lines which can be drawn from it to the circumference, the greatest is that in which the centre is, and the other part of that diameter is the least; and of any others, that which is nearer to the line which