is available for finding the true central angle of an imperfectly constructed theodolite, or of any similar instrument, in which the revolving limbs AF, FB, in the last figure but one, are not at the centre of the circle; thus, if the arc AC, as shown by one limb FA, is 50° 20′ and the arc BD, as shown by the other limb 50° 20′ 48° 12' 98° 32' FB, is 48° 12′ the true central angle =

2

49° 16'.

2. We may also apply Cor. 9 to determine the DFB, when the arcs AC and BD are given; for if arc AC 48°, and arc BD = 100°, then Z DFB 100° 48° 52°.

PROP. 27.-THEOR.

In equal circles the angles which stand upon equal arcs are equal to one another, whether they be at the centres or the circumferences.

E.1 Hyp. 1.

2 Hyp. 2.

Let ABC, DEF, be eq. Os, G and H being the centres;

and let s BGC, EHF, at G and H, and s BAC, EDF, at the Oces, stand upon the eq. arcs BC, EF; 3 Conc. then BGC =/EHF and Z BAC = / EDF.

SUP.-If

BGC = ▲ EHF, then (20, III., and Ax. 7) ▲ BAC = ▲ EDF; but if not, one must be the greater.

Let BGC > < EHF.

C.1 | Sup.

223, I.

At G in B G make

BGK EHF:

D.1 C.26, III. 2 H.2.Ax. 1 3 Ax. 9.

Then BGK = EHF, .. arc BK arc FE;
but arc EF arc BC, .. arc BK = arc BC;
i. e. the less the gr., which is impossible;

=

BGC not / EHF, i. e. BGC = = / EHF; at ABGC, and at D = } ≤ EHF;

at A

at A = 2 at D,

. In equal circles, the angles, &c.

Q. E. D.

COR. I.-In the same or in equal circles, ABC, DEF, the sectors BGC, EHF, which stand upon equal arcs, BC, EF, are equal, and conversely.

C. Pst. 1.

Join BC, EF; and from s K, L, draw KB,
KC, LE, LF.

D.1 Def.15,I.27,III. lines BG, GC EH, HF, and BGC

2 4, I.

3 Sub.

4 Ax. 3.

5 Def. 11, III.

= EHF.

=

.. base BC base EF, and ▲ BGC

= AEHF.

From eq. Os ABC, DEF, take equal ares
BC, EF;

.. rem. arc BAC rem. arc EDF;

. BKC: =

ELF, and seg. BKC

is sim. to seg. ELF;

6 D. 2 & 24, III. but base BC base EF,

7 Add.

8 Ax. 2.

=

... seg. BKC seg. ELF.

To the eq. As BGC, EHF, add the equal segs. BKC, ELF;

.. the sector BGCK the sector EHFL.

N.B. The Converse may be left for the student to demonstrato.

COR. II.-As in Cor. 7, P. 26. III., if the chords AE, CD, of a circle are parallel, they intercept equal arcs, and vice versâ.

SCH.-1. Propositions 26 and 27 are converse propositions; aud what is true of equal circles is true of equal arcs in the same circle.

2. As in Cor. 2, Prop. 15, I., all the angles formed by any number of lines diverging from a common centre are together equal to four rt. angles, so the sum of the angles at the centre of a circie subtended by arcs, which together make up the whole circumference, is equal to four rt. angles. Also the sum of the angles at the circumference subtended by those same arcs is equal to two rt. angles.

3. And, since eq. arcs of cq. circles subtend eq. angles, such cq. arcs contain similar segments.

USE AND APP.-1. By Cor. II. of this Prop., a parallel through a given point E to a given st. line CD may readily be drawn; for join EC, and make Z CEA=DCE, and AE is parallel to CD.

2. The principle on which the area of a sector is ascertained may be developed from Cor. I. of this proposition, for the area of the triangle BGC, added to the area of the segment BKC gives the area of the sector BGEK.

Or, when the rad. and BGC are given, by principles hercafter to be proved, the Area of the Sector =

Area of

X 360°.

BGC,

PROP. 28.-THEOR.

In equal circles, equal straight lines cut off equal arcs, the greater equal to the greater, and the less to the less.

CON. 1, III. Pst. 1.

DEM. Def. 1, III. 8, I. 26, III. Ax. 3

Let ABC and DEF be eq. Os, and BC, EF eq. st. lines in them;

and let BC, EF cut off two gr. arcs BAC, EDF, and two less BGC, EHF;

then the gr. arc BAC= the gr. EDF; and

the less BGC the less EHF.

C. 1, III. Pst. 1. Take K, L, centres of the Os, and join KB, KC,

D.1 H. Def. 1, III. ABCDEF,

2 H. 8, I. 3 26, III.

4 H. Ax. 3.

5 Rec.

O

.. KB, KC LE, LF, each to each; and BC = EF, .. / BKC = / ELF; .. the arc BG C the arc EHF;

but ABC = O DEF;

.. rem. arc BAC rem. arc EDF. Therefore, in equal circles, equal st. lines, &c.

Q. E. D.

SCH. As in other instances, the principle of the proposition extends to equal st. lines in the same circle, in which also such equal st. lines cut off equal

arcs.

PROP. 29.-THEOR.

In equal circles equal arcs are subtended by equal straight lines.

E.1 Hyp. 2 Conc.

and arc BGC = arc EHF,

then on joining BC, and EF, chord BC
= chord E F,

C. 1, III. Pst. 1. Find K, L centres of the Os, and join KB, KC,

LE, LF.

4 Rec.

EL, LF each to each;
ZELF,

chord E F.

Therefore, in equal circles equal arcs &c. Q.E. D.

COR. I.---By the same kind of demonstration it may be shown that, in the same or in equal circles, equal sectors stand upon equal arcs; and conversely.

COR. II.-From Prop. 26, 27 and 29, straight lines which intercept equal arcs are parallel; and parallel st. lines intercept equal arcs; for the alternate angles are equal.

USE AND APP.-1. We may declare generally that whatever has been proved with respect to equal circles is also true when applied to the same circle.

2. In Spherical Trigonometry, Props. 26, 27, 28, and 29 are of continual use. By means of Props. 27 and 28, THEODOSIUS demonstrated that the arcs of the circles of the Italian and Babylonian hours, comprehended between two parallels, are equal; and, in the same way, it may be proved that the arcs of circles of the astronomical hours, comprehended between the two parallels to the equator, are equal.