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XXIII. N equal circles (A I B C, EK F G), angles, wether at the centres or circumferences (A MC, ENG or AIC, EKG), as also the fedors (AMCm, EN G n) have the fame ratio with the arches (AmC, EnG) on which they stand, have to one another. Hypothesis.

Thesis. 1. The OAIBC,EKFG are=!0 one another. I. VAMC: VENG = AmC: Eng. II. The Vat the centers AMC,ENG&the II. VAIC: VEKG = AmC.: Eng.

Vat the OAIC, EKG stand upon the III. Sea, AMCm:Sed.ENGN=AmC:EnG. arches A m C, En G.

Preparation, 1. Join the chords A C, E G.

Pol.1. B.1. 2. In the O AIB C, draw the chords CD, DB &c, each

=to AC, & in the OEKFG a pareil number of cords
GH, HF &c, cach = to E G.

P. 1. B.4 3. Draw MD, M B &c, also N H, NF &c.

Pof 1. B.I. DEMONSTRATION. ECAUSE on one side the cords A C,CD, DB, & on the other the cords EG, GH, H F are == to one another (Prep. 2). 1. The arches Am C, CD, D B are all equal on the one side, as the

arches EnG, GH, HF are on the other, 2. Consequently, the V AMC, CMD, DMB &c, & ENG,G NH,

HNF &c, are also = to one another, on one side & the other. 3. Wherefore, the V AMB & the arch A C D B, are equimult. of the

YAMC & of the arch A m C.
4. Likewife, VEN F & the arch EGHF are equimult. of VENG,

& of the arch En G.
But becaufe the O AIB C, E KF G are equal (Hyp. 1).
According as the arch A CDB is , or the arch EGHF;
YAMB is also >,= or < VENF.

P.27. B.3. 5. Wherefore, VA MC: VENG= A mC:En G.

Which was to be demonstrated. 1. Moreover, V AMC being double of VAIC, & VENG double

of Y EKG (P. 20. B. 3). 6. It follows that VAMC: VENG = VAIC: V EKG. P.15. B.5. 7. Consequently, VAIC: VEKG=

AmC: En G.
Which was to be demonstrated. 11.


P.28. B.3.


D. 5. B.5.

P.11. B.5.

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À m,

PREP. 4. In the arches A C, CD, take the points m& o, & join
Cm; Co, D. &c.

Pof.1. B.1.
Since then the two sides A M, MC are = to the two sides CM, MD

(D. 15. B. ), & the VAMC, CMD are equal (Arg. 2). 8. The base AC is = to the base CD, & the AAMC = to the ACMD. P. B...

Moreover, because the arch Am C is = to the arch COD (Arg. 1). 9. The complement A IBDC of the first is = to the complement CAIBD of the second.

Ax.3. B 1. 10. Wherefore y Am C is = to y COD.

P.27. B.3. 1. Therefore the segment Am C is similar to the segment CoD.

Ax.2. B.3. Besides they are subtended by equal cords ( Arg. 8). 12. Consequently, the segment Á m C is = to the segment Co D. P.24. B.3.

But since the A AMC is also = to the ACMD (Arg. 8). 13. The sector A M C m is = to the sector CMD o.

Ax.2. B.1 Likewise, the sector D M B is equal to each of the two foregoing

AMCm, CM Do. 14. Therefore the sectors A MC,CMD, D MB are = to one another. 15. It is demonstrated after the fame manner, that the sectors ENG,

GNH, HN F are = to one another.
16. Wherefore, the sect. AMB DC, & the arch ACDB are equimult. of

the sect. A MC m, & of the arch A m C, the sec). ENFHG, & the
arch E GH F are equimult. of the sect. E NG n, & of the arch Eng.
But because the OAIB C, EKF G are equal (Hyp. 1).
If the arch ACD B be = to the arch EG H'F, the lect. A MBDC
is also = to the sect. EN FHG, as is proved by the reasoning em-
ployed in this third part of the demonstration to arg. 12 inclusively.
And, if the arch A CDB be > the arch EĞHF, the fect.
A MBDC is also > the sect. EN FHG, & if lefs, less.
Since then there are four magnitudes, the two arches A mC, EnG,
& the two sect. AMC m, ENG n. And of the arch A m C, &
se&t. A MC m, the arch ACDB & sect. AMBDC are any equi-
mult. whatever ; & of the arch En G, & sector ENG n, the arch
EGHF, & the sect. E NFHG are any equimult. whatever.
And it has been proved that, if arch A CD B be>, = or <the arch

EGH F, sect. Å MBDC is also >,=or the sect. ENFHG. 17.1t follows, that sect. A MC: fect. ENG=AmC:EnG,

Which was to be demonstrated, 111.

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D. 5. B. 3.

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HE angle at the center, is to four right angles, as the arch upon
which it stands, is to the circumference.
For (Fig. 1), VBCD:L=BD: to a quadrant of the .
Wherefore, quadrupling the consequents.
VB.CĎ : 4 L = BD: 0.

P.15. B.5.
HE arches E F, B D of unequal circles, are similar, if they fub-
tend equal angles C & C, either at their centers, or at their Ó (Fig. 2).
But yBCD or vect:4L3BD: 08hD. }(Cor.1.)
Confequently, EF:0EnF=BD: OB m D.
Therefore, the arches EF, B D are similar.





CB, C D cut off from concentric circumferences similar arches EF, BD (Fig. 2).

R E M A R K. I

T is in consequence of the proportionality establihed in Cor. 1. that an arch of a circle (B D) is called the MEASURE of its correspondent angle (B C D); that is of the angle at the center, subrended by this arch; the circumference of a circle being the only curve, wbose arches, increase or diminish in the ratio of ihe correspondent angles, about the same point. The whole circle is conceived to be divided into 360 equal parts, which are called DEGREES ; and each of bese degrees into 60 equal parts, called MINUTES and each minute into 60 equal parts, called seconds &c. in consequence of this hypothefis, & the correspondence established between the arches, & the angles, we are obliged to conceive all the angles abouí a point in the same plane (that is the sum of 4 L), as divided into 360 equal parts, in such a manner, that the angle of a degree is no more than the 360tb part of 4 L, or the goth of a L, & conSequently, of an amplitude to be subtended by the 360th part of the circumference.

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SOLID is that which hath length, breadth and thickness.

That which bounds a Solid is a superficies.

III. A straight line (A B) is perpendicular to a plane (PL) (Fig. 1), if it be perpendicular to all the lines (CD & F E), meeting it in this plane ; tbat is, The line (A B) will be perpendicular to the plane (PL), if it be perpendicular to the lines (CD & F E) which being drawn in the plane (PL.) pass through the point (B), so that the angles (ABC, A BD, ABE & ABF) are right angles.

IV. A plane (A B) (Fig. 2.) is perpendicular to a plane (P L), if the lines (DE & F G) drawn in one of the planes (as in A B) perpendicularly to the common section (A N) of the planes, are allo perpendicular to the other plane (PL). The common section of two planes is the line which is in the two planes : as the line (AN), wbicb is not only in the plane (A B), but also in the plane (PL); therefore if the lines D E & F G drawn perpendicular to A N in the plane AB are also perpendicular to the plane P L ; the plane A B will be perpendicular to tbe plane P L.

V. The inclination of a straight line (A B) to a plane, (Fig. 3.) is the acute angle (A'B E), contained by the straight line (AB), and another (BE) drawn from the point (B), in which A B meets the plane (PL), to the point (E) in which a perpendicular (AE) to the plane (PL) drawn from any point (A) of the line (AB) above the plane, meets the same plane.

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