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Ex. 1. If one side of a quadrilateral figure inscribed in a circle be produced, the exterior angle is equal to the opposite angle of the quadrilateral.

Ex. 2. If the sides AB, DC of a quadrilateral inscribed in a circle be produced to meet in E, then the triangles EBC, EAD will be equiangular.

Ex. 3. Shew that a circle cannot be described about a rhombus.

Ex. 4. The lines, bisecting any angle of a quadrilateral figure inscribed in a circle and the opposite exterior angle, meet in the circumference of the circle.

Ex. 5. AB, a chord of a circle, is the base of an isosceles triangle, whose vertex C is without the circle, and whose equal sides meet the circle in D, E: shew that CD is equal to CE.

Ex. 6. If in any quadrilateral the opposite angles be together equal to two right angles, a circle may be described about that quadrilateral.

Propositions XXIII. and XXIV., not being required in the method adopted for proving the subsequent Propositions in this book, are removed to the Appendix. Proposition xxv. has been already proved.

NOTE 3. On the Method of Superposition, as applied
to Circles.

In Props. XXVI. XXVII. XXVIII. XXIX. we prove certain relations existing between chords, arcs, and angles in equal circles. As we shall employ the Method of Superposition, we must state the principles which render this method applicable, as a test of equality, in the case of figures with circular boundaries.

DEF. XIII.. Equal circles are those, of which the radii are

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For suppose ABC, A'B'C' to be circles, of which the radii are equal.

Then if

A'B'C' be applied to

ABC, so that O', the centre of A'B'C', coincides with O, the centre of ABC, it is evident that any particular point A' in the Oce of the former must coincide with some point A in Oce of the latter, because of the equality of the radii O'A' and OA.

Hence Oce A'B'C' must coincide with Oce ABC,

that is, A'B'C' = 0 ABC.

Further, when we have applied the circle A'B'C' to the circle ABC, so that the centres coincide, we may imagine ABC to remain fixed, while A'B'C' revolves round the common centre. Hence we may suppose any particular point B' in the circumference of A'B'C' to be made to coincide with any particular point B in the circumference of ABC.

Again, any radius O'A' of the circle A'B'C' may be made to coincide with any radius OA of the circle ABC.

Also, if A'B' and AB be equal arcs, they may be made to coincide.

Again, every diameter of a circle divides the circle into equal segments.

For let AOB be a diameter of the circle ACBD, of which O is the centre. Suppose the segment ACB to be applied to the segment ADB, so as to keep AB a common boundary: then the arc ACB must coincide with the arc ADB, because every point in each is equally distant from O.

B

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 4 s BAC, EDF at their ○ces, be equal.

Then must arc BKC-arc ELF.

For, if ABC be applied to © DEF,

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

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.. 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.

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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

F

F

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

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

L

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 ▲ EHF.

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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.

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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

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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.

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

Take the centres K, L, and join KB, KC, LE,
LF.
Then KB LE, and KC=LF, and BC=EF,
=ELF.

.. 4 BKC

Hence, if ✪ ABC be applied to © DEF,

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

and ·. KC = LF, .. C will coincide with F.
Then 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 chord AB be equal to the chord 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 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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