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1. Equal circles are those of which the diameters are equal, or from the centres of which the straight lines to the circumference are equal.
2. A straight line is said to touch a circle, or to be a tangent to it, when it meets the circle, and being produced does not cut it.
3. Circles are said to touch one another, which meet, but do not cut one another.
4. 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:
5. And the straight line on which the greater perpendicular falls, is said to be farther from the centre.
6. An arc of a circle is any portion of the circumference, and the chord of an arc is a straight line joining its extremities.
7. A segment of a circle is the figure contained by an arc and its chord.
8. An angle in a segment is the angle contained by two straight lines drawn from any point in its arc to the extremities of its chord.
9. An angle is said to insist or stand upon the arc which subtends it.
10. A sector of a circle is the figure contained by two radii and the intercepted arc.
11. Similar segments of circles are those which contain equal angles.
12. Concentric circles are those which have the same centre.
To find the centre of a given circle.
(References-Prop. I. 8, 10, 11.) Given.-Let ABC be the given circle. Sought.—It is required to find the centre of the circle ABC.
Construction.—1. Draw within the circle any chord AB, and bisect it in D. (1. 10.)
2. From the point D draw DC at right angles to AB. (I. 11.)
3. Produce CD to E, and bisect CE in F. (I. 10.)
Then the point Fshall be the centre of the circle ABC.
Proof.-1. For if F be not the centre, if possible, let G be the centre; and join GA, GD, GB.
2. Then, because DA is equal to DB (const.), and DG common to the two triangles ADG, BDG;
3. The two sides AD, DG, are equal to the two sides BD, DG, each to each;
4. And the base GA is assumed to be equal to the base GB, being radii of the same circle;
5. Therefore the angle ADG is equal to the angle BDG. (I. 8.)
6. But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle; (I. def. 10.)
7. Therefore the angle GDB is a right angle.
9. Therefore the angle GDB is equal to the angle FDB (ax. 11), the less to the greater; which is impossible.
10. Therefore G is not the centre of the circle ABC. 11. In the same manner it may be shown that no point which is not in CE can be the centre.
12. And since the centre is in CE, it must be in F, its point of bisection.
13. Therefore F is the centre of the circle ABC: which was to be found.
Corollary. 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 2.-THEOREM. If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle.
(References-Prop. I. 5, 16, 19; III. 1.) Hypothesis.—Let ABC be a circle, and A and B any two points in the circumference.
Sequence.—The straight line drawn from A to B shall fall within the circle.
Construction.-1. Find D the centre of the circle ABC (III. 1), and join DA, DB.
2. In AB take any point E; join DE, and produce it to the circumference in F.
Demonstration.-1. Because DA is equal to DB, the angle DAB is equal to the angle DBA. (I. 5.)
2. And because AE, a side of the triangle DAE, is produced to B, the exterior angle DEB is greater than the interior and opposite angle DAE. (I. 16.)
3. But the angle DAE was proved to be equal to the anglo DBE;
4. Therefore the angle DEB is also greater than DBE. 5. But the greater angle is subtended by the greater side; (I. 19.)
6. Therefore DB is greater than DE; 7. But DB is equal to DF; therefore DF is greater than DE, and the point E is therefore within the circle.
8. In the same manner it may be proved that every point in AB lies within the circle. ,
9. Therefore the straight line AB lies within the circle. Conclusion.—Therefore, if any two points, &c. Q. E. D.
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 conversely, if it cut it at right angles, it shall bisect it.
(References—Prop. I. 5, 8, 26; III. 1.) Hypothesis I.—Let ABC be a circle, and let CD, a straight line drawn through the centre, bisect any straight line AB, which does not pass through the centre.
Sequence.-CD shall cut AB at right angles.
Construction.—Take E, the centre of the circle (III. 1), and join EA, EB.
Demonstration.-1. Because AF is equal to FB (hyp.), and FE common to the two triangles AFE, BFE, and the base EA equal to the base EB; (I. def. 15.) 2. Therefore the
angle AFE is equal to the angle BFE; (I. 8.)
3. Therefore each of the angles AFE, BFE, is a right angle; (I. def. 10.)
4. Therefore the straight line CD, drawn through the centre, bisecting another AB that does not pass through the centre, also cuts it at right angles. Hypothesis II.
Next, let CD cut AB at right angles. Sequence.-CD shall bisect AB; that is, AF shall be equal to FB. The same construction being made;
Demonstration.-l. Because the radii EA, EB, are equal, the angle EAF is equal to the angle EBF; (I.
5.) 2. And the angle AFE is equal to the angle BFE; (hyp.)
3. Therefore in the two triangles EAF, EBF, there are two angles in the one equal to two angles in the other, each to each,
and the side EF, which is opposite to one of the equal angles in each, is common to both;
4. Therefore their other sides are equal; (I. 26.) 5. Therefore AF is equal to FB. Conclusion.—Therefore, 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.
(References-Prop. III. 1, 3.)
Hypothesis I.-Let ABCD be a circle, and AC, BD, two straight lines in it, which cut one another at the point E, and do not both pass through the centre.
Sequence.—AC, BD, shall not bisect one another. 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 centre.
Hypothesis II.—But if neither of them pass through the centre, if possible, let AE be equal to EC, and BE to EĎ.
Construction.—Take F, the centre of the circle (III. 1), and join EF.
Demonstration.-1. Because FE, a straight line drawn through the centre, is assumed to bisect another line AC, which does not pass through the centre (hyp.), therefore it cuts it at right angles; (III. 3.)
2. Therefore the angle FEA is a right angle.
3. Again, because the straight line FE is assumed to bisect the straight line BD, which does not pass through the centre (hyp.), therefore it cuts it at right angles; (III. 3.) 4. Therefore the angle FEB is a right angle. 5. But the angle FEA was shown to be a right angle;
6. Therefore the angle FEA is equal to the angle FEB, the less to the greater, which is impossible;
7. Therefore AC, BD, do not bisect each other. Conclusion. Therefore, if in a circle, &c. Q. E. D.