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PROPOSITION XXVI. THEOREM.

In equal circles, the arcs, which subtend equal angles, whether they be at the centres or at the circumferences, must be equal.

B

K

H

L

Let ABC, DEF be equal circles, and let ▲ s BGC, EHF at their centres, and ▲ 8 BAC, EDF at their Oces, be equal.

Then must arc BKC=arc ELF.

For, if ABC be applied to O DEF,

so that G coincides with H, and GB falls on HE,
then,

GB=HE, .. B will coincide with E.

And: 4 BGC= ▲ EHF, :. GC will fall on HF;

and

GC=HF, .. C will coincide with F.

Then B coincides with E and C with F,

... arc BKC will coincide with and be equal to arc ELF.

COR. Sector BGCK is equal to sector EHFL.

Q. E. D.

NOTE. This and the three following Propositions are, and will hereafter be assumed to be, true for the same circle as well as for equal circles.

PROPOSITION XXVII. THEOREM.

In equal circles, the angles, which are subtended by equal arcs, whether they are at the centres or at the circumferences, must be equal.

G

H

B

K

Let ABC, DEF be equal circles, and let s BGC, EHF at their centres, and 2 8 BAC, EDF at their Oces, be subtended by equal arcs BKC, ELF.

Then must BGC= ▲ EHF, and ▲ BAC=▲ EDF.

For, if ABC be applied to o DEF,

so that G coincides with H, and GB falls on HE,
then: GB=HE, .. B will coincide with E;
and arc BKC-arc ELF, .. C will coincide with F.
Hence, GC will coincide with HF.

Then BG coincides with EH, and GC with HF,

.. 4 BGC will coincide with and be equal to
Again, BAC-half of BGC,

and EDF-half of EHF,
.. L BAC LEDF.

EHF.

III. 20.

III. 20.

I. Ax. 7.

Q. E. D.

Ex. 1. If, in a circle, AB, CD be two arcs of given magnitude, and AC, BD be joined to meet in E, shew that the angle AEB is invariable.

Ex. 2. The straight lines joining the extremities of the chords of two equal arcs of the same circle, towards the same parts, are parallel to each other.

Ex. 3. If two equal chords, in a given circle, cut one another, the segments of the one shall be equal to the segments of the other, each to each.

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In equal circles, the arcs, which are subtended by equal chords, must be equal, the greater to the greater, and the less to the less.

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Let ABC, DEF be equal circles, and BC, EF equal chords, subtending the major arcs BAC, EDF,

and the minor arcs BGC, EHF.

=arc EHF. LF.

Then must arc BAC: = arc EDF, and arc BGC:
Take the centres K, L, and join KB, KC, LE,
KB=LE, and KC=LF, and BC=EF,
..L BKC = L ELF.

Then

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so that K coincides with L, and KB falls on LE,
then . ▲ BKC = ▲ ELF, .. KC will fall on LF ;

and
Then

KC:

=

LF, .. C will coincide with F.
B coincides with E, and C with F,

.. arc BAC will coincide with and be equal to arc EDF,

and arc BGC........

EHF.

I. c.

Q. E. D.

Ex. 1. If, in a circle ABCD, the arc AB be equal to the arc DC, AD must be parallel to BC.

Ex. 2. If a straight line, drawn from A the middle point of an arc BC, touch the circle, shew that it is parallel to the chord BC.

Ex. 3. If two chords of a circle intersect at right angles, the portions of the circumference taken alternately are together equal to half the circumference.

PROPOSITION XXIX. THEOREM.

In equal circles, the chords, which subtend equal arcs, must

be equal.

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Let ABC, DEF be equal circles, and let BC, EF be chords subtending the equal arcs BGC, EHF.

[blocks in formation]

Take the centres K, L.

Then, if ABC be applied to © DEF, so that K coincides with L, and B with E, and arc BGC falls on arc EHF,

arc BGC=arc EHF, .. C will coincide with F.

Then B coincides with E and C with F,

.. chord BC must coincide with and be equal to chord EF.

Q. E. D.

Ex. 1. The two straight lines in a circle, which join the extremities of two parallel chords, are equal to one another.

Ex. 2. If three equal chords of a circle, cut one another in the same point, within the circle, that point is the centre.

NOTE 4. On the Symmetrical properties of the Circle with regard to its diameter.

The brief remarks on Symmetry in pp. 107, 108 may now be extended in the following way:

A figure is said to be symmetrical with regard to a line, when every perpendicular to the line meets the figure at points which are equidistant from the line.

Hence a Circle is Symmetrical with regard to its Diameter, because the diameter bisects every chord, to which it is perpendicular.

B

Further, suppose AB to be a diameter of the circle ACBD, of which O is the centre, and CD to be a chord perpendicular to AB.

Then, if lines be drawn as in the diagram, we know that AB bisects

(1.) The chord CD,

III. 1.

(2.) The arcs CAD and CBD,

III. 26.

(3.) The angles CAD, COD, CBD, and the reflex

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These Symmetrical relations should be carefully observed, because they are often suggestive of methods for the solution of problems.

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