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8. An angle in a segment is one formed by two straight lines drawn from any point in the arc of the segment to the extremities of its chord.
NOTE. (i) It will be shewn in Proposition 21, that all angles in the same segment of a circle are equal.
NOTE. (ii) The angle of a segment (as distinct from the angle in a segment) is sometimes defined as that which is contained between the chord and the arc; but this definition is not required in any proposition of Euclid.
9. An angle at the circumference of a circle is one formed by straight lines drawn from a point on the circumference to the extremities of an arc: such an angle is said to stand upon the arc by which it is subtended.
10. Similar segments of circles are those which contain equal angles,
11. A sector of a circle is a figure bounded by two radii and the arc intercepted between them.
SYMBOLS AND ABBREVIATIONS.
In addition to the symbols and abbreviations given on page 11, we shall use the following:
o for circle, ce for circumference.
Let ABC be a given circle.
chord AB, and bisect AB at D.
1. 10. From D draw DC at right angles to AB; and produce DC to meet the o ce at E and C. Bisect EC at F.
1. 10. Then shall F be the centre of the O ABC. Proof. First, the centre of the circle must be in EC: for if not, suppose the centre to be at a point G outside EC.
Join AG, DG, BG.
and GD is common ; and GA=GB, for by supposition they are radii; :: the L GDA=the L GDB;
I. 8. .. these angles, being adjacent, are rt. angles. But the 2 CDB is a rt. angle;
Constr. .:. the L GDB=the 4 CDB,
Ax. 11. the part equal to the whole, which is impossible.
.:: G is not the centre. So it may be shewn that no point outside EC is the centre;
.:. the centre lies in EC. .: F, the middle point of the diameter Ec, must be the centre of the O ABC.
Q.E.F. COROLLARY. The straight line which bisects a chord of a circle at right angles passes through the centre.
PROPOSITION 2. THEOREM. If any two points are taken in the circumference of a circle, the chord which joins them falls within the circle.
Let ABC be a circle, and A and B any two points on its Oce
Then shall the chord AB fall within the circle. Construction. Find D, the centre of the O ABC; III. 1.
and in AB take any point E.
Join DA, DE, DB.
1. Def. 15. ... the DAB=the < DBA. But the ext. 2 DEB is greater than the int. opp. — DAE;
.. the 2 DEB is also greater than the - DBE. .. in the A DEB, the side DB, which is opposite the greater
angle, is greater than DE which is opposite the less : I. 19. that is to say, DE is less than DB, a radius of the circle ;
E falls within the circle. Similarly, any other point between A and B may be shewn to fall within the circle. .:. AB falls within the circle.
NOTE. A part of a curved line is said to be concave to a point, when for every chord (taken so as to lie between the point and the curve) all straight lines drawn from the given point to the intercepted arc are cut by the chord : if, when any chord whatever is taken, no straight line drawn from the given point to the intercepted arc is cut by the chord, the curve is said to be convex to that point.
Proposition 2 proves that the whole circumference of a circle is concave to its centre.
PROPOSITION 3. THEOREM. If a straight line drawn through the centre of a circle bisects a chord which does not pass through the centre, it shall cut the chord at right angles. Conversely, if it cuts the chord at right angles, it shall biscct it.
Let ABC be a circle ; and let CD be a st. line drawn through the centre, and AB a chord which does not pass through the centre. First. Let CD bisect the chord AB at F.
Then shall CD cut AB at rt. angles.
and join EA, EB. Proof.
Then in the A% AFE, BFE,
and FE is common ;
1. 8. .. these angles, being adjacent, are rt. angles ;
that is, DC cuts AB at rt. angles. Q.E.D. Conversely. Let CD cut the chord AB at rt. angles.
Then shall cd bisect AB at F. Construction. Find E the centre; and join EA, EB. Proof. In the EAB, because EA=EB,
1. Def. 15. .:. the EAB= the < EBA.
Then in the AS EFA, EFB, the LEAF = the LEBF,
Proved. Because and the 2 EFA= the 2 EFB, being rt. angles; Hyp.
and EF is common ;
1. 26. that is, CD bisects AB at F.
PROPOSITION 4. THEOREM.
If in a circle two chords cut one another, which do not both pass through the centre, they cannot both be bisected at their point of intersection.
Let ABCD be a circle, and AC, BD two chords which intersect at E, but do not both pass through the centre.
Then AC and BD shall not be both bisected at E.
CASE I. If one chord passes through the centre, it is a a diameter, and the centre is its middle point; .. it cannot be bisected by the other chord, which by hypothesis does not pass through the centre.
CASE II. If neither chord passes through the centre; then, if possible, suppose E to be the middle point of both;
that is, let AE EC; and BE = ED. Construction. Find F, the centre of the circle. III. 1.
Join EF. Proof. Because FE, which passes through the centre, bisects the chord AC,
Hyp. FEC is a rt. angle. And because Fe, which passes through the centre, bisects the chord BD,
Hyp. .:. the FED is a rt. angle.
III. 3. ... the L FEC= the į FED, the whole equal to its part, which is impossible. .. AC and BD are not both bisected at E.