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the angle CEB has been shown to be equal to ABE, the angle *Interc. 1. CEB is* greater than ABC ; still more is the angle CED (which Cor. 2.

is greater than CEB) greater than ABC. And because the angle BDG is greater than BDE, the angle EGD (which is equal to BDG) is greater than BDE; still more is the angle EGF (which is greater than EGD) greater than BDE. And the angle EGF is equal to DFG; therefore DFG is greater than BDE. And so on.

And by parity of reasoning, the like may be proved of every other series of equal straight lines that make equal angles with one another as supposed. Wherefore, universally, if two equal straight lines &c. Which was to be demonstrated.

NOMENCLATURE.—If the angle BAC be bisected by a straight line AZ of unlimited length, such straight line is called the axis of the series ; the point A is called the vertex; and the angle CAZ or BAZ, the angle with the axis.

Cor. 1. If the angular points of any one of the tesseras be joined diagonally as I and F, the angle GIF is greater than the angle ABC or ACB, which are the angles made by joining the extremities of the first two straight lines of the series.

For the angles ABC, CEB, BDE, EGD, DFG, GIF, &c. have been shown to be successively greater each than the preceding.

Cor. 2. If a figure be made by joining any two whatsoever of the points A, B, C, D, E, F, &c.; the angles at each of the two cusps of such figure shall be equal to one another.

For it may be shown by the same process as in Prop. XXVIII B above ; beginning from the middle angular point if the number of angular points is odd, and from the two middle ones if even.

Cor. 3. The chords BC, DE, FG, &c. are bisected at right

angles by the axis. + Hyp: For because AB ist equal to AC, and the portion of the axis

intercepted between A and the intersection of BC with the axis is | 1. 4.

common, and the angle BAZ equal to CAZ; BC ist bisected by

the axis, and the angles at the intersection are equal, and conse*I.Nom. 37. quently* right angles. And because the axis thus bisects the base of +1. 28 A. the tessera DBCE at right angles, itt bisects the opposite side DE

at right angles.

Cor. 4. If a straight line WX of unlimited length both ways, be made to pass through the vertex A at right angles to AZ, and afterwards be moved along AZ keeping ever at right angles to it; it will at all times (after passing through B and C) make with

Cor. l.

And so on.

such parts of the series as it cuts, interior angles on the side towards A each greater than ACB.

For when it passes through any of the chords, (which, by Cor. 3 above, are at right angles to the axis), as GF, it makes (by Prop. XXVIII B above) angles EGF and DFG, each greater than

ACB. And when it passes through any other points as M and * 1. 28 A. N, it makes* angles GMX and FNW, each greater than GIF, Cor. 2.

which is (by Cor. 1 above) greater than ACB.

SCHOLIUM.—Though it has been shown that the angle at the cusp goes on continually increasing, this is not sufficient to prove that it will ever arrive at a certain specified greatness; as, for instance, that it will ever be so great as a right angle, or as CAZ, or even as twice the angle ACB. Nevertheless it is allowable to establish what will be the results if the angle at the cusp is ever found to have arrived at a certain specified greatness. Which is what is done in the Proposition next following.

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PROPOSITION XXVIII C.
THEOREM.--In a series of equal straight lines as in the last

Proposition, if the angle at the cusp be ever greater than the
angle with the axis; the straight lines bisecting the several
angles of the series (being prolonged) shall have met in the axis,
in the part that is between the chord and the vertex.

Let AB, AC, BD, CE, &c. be a series of equal straight lines as in the last Proposition. If the angle at the cusp, as ILK, be ever greater than the angle with the w

0.
axis CAZ; the straight

U
lines bisecting the an-
gles ACE, CEG, &c.
I

H
of the series, being pro-
longed, shall have met

F
in the axis AZ, in the
part which is between

E

D the chord at U and the

B vertex A. 11.22.Cor.2. At the point L in the straight line LI, maket an angle ILV #Hyp. equal to CAZ; and because the angle ILK is# 'greater than *INTERC. 1. Cor. 2. CAZ, the angle ILV (which is equal to CAZ) will be* less than

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Cor. 12.

Cor. 2.

*). 28 B.

ILK. Also because the angle ILA is* equal to CAL, and CAL Cor. 2. +INTERC. 1. is less than CAZ; ILA ist less than CAZ, and than ILV which Cor. 2.

is equal to CAZ. Wherefore the straight line LV will lie

between the two straight lines LU, LA, and being prolonged will 1.13.Cor.5. cuts the perimeter of the triangle LUA; and because it cannot

cut it in LU or LA (for then two straight lines would inclose a space), it will cut it in UA, that is to say, it will cut AZ the axis in some point between U and A. Let it cut it in S; and join SC, SE, SG, &c. These shall be the straight lines bisecting

the 'angles ACE, CEG, EGI, &c. of the series. *Constr.

Because the angle CAS is* equal to ILS, and CAL to ILA, INTERC. l. the remaining angle SAL ist equal to SLA; wherefore in the Cor. 7.

triangle ASL, the side SL is equal to SA. If then SC be not also equal to SA, it must either be greater or less. And first, let

it be assumed that it is greater. But if SC be greater than SA, • I. 18. the angle SCA, which is opposite to SA the less, must be *less than

SAC, which is opposite to SC the greater; and the angle SAC is + Interc. 1. equal to half the angle BAC or tot half the angle ACE; therefore

the angle SCA must be less than half the angle ACE, and the #INTERC. 1.

angle SCE must be greater than half the angle ACE, and conse. quently greater than the angle SCA which is less than half. Therefore in the triangles SCA, SCE, because the sides SC, CA

are equal to the sides SC, CE respectively, but the angle SCE is "1. 24. greater than SCA, the third side SE must be* greater than the

third side SA. Again, join AE, and because SE is greater than + 1. 18.

SA, the angle SEA must bet less than SAE. Add to each 11.

28 B. the equalt angles CEA, CAE, and the whole angle SEC must be* *INTERC. 1. Cor. 6. less than the whole angle SAC. But SAC is equal to half the INTERC. 1. angle BAC, or tot half the angle CEG ; therefore SEC must bef IINTERC.1.

less than half the angle CEG; and SEG must be greater than Cor. 2. half the angle CEG, and consequently greater than the angle

SEC which is less than half. Therefore in the triangles SEC,
SEG, because the sides SE, EC are equal to the sides SE, EG

respectively, but the angle SEG is greater than SEC, the third * I. 24.

side SG must be* greater than the third side SC. But SC was greater than SA; still more therefore must SG be greater than SA. In like manner, by joining AG, may be shown that because SG is greater than SA, the angle SGA must be less than SAG, and the angle SGE less than SAC, and consequently less than half EGI; whence the angle SGI must be greater than half EGI, and conse

Cor. 12.

Cor. 12.

quently greater than SGE which is less than half; wherefore the third side SI must be greater than the third side SE, and consequently greater than SA which has been shown to be less than SE; and so on, with each of the other straight lines in suc

cession. Wherefore it will follow that SL must be greater than *I.Nom.26. SA; which is impossible, for it is equal to it. The assumption*,

therefore, which involves this impossible consequence, cannot be true; or SC is not greater than SA.

And in the same way may be shown, that it cannot be less. But because SC is

neither greater than SA nor less, it is equal to it. And because + I. 5. SA is equal to SC, the angle SCA ist equal to SAC. And the | Hyp. Interc. 1. angle SAC iss half the angle BAC, therefore SCA is* equal to Cor. 1. half the angle BAC, and consequently equalt to half the angle PINTERC. 1.

ACE; whence the remainder SCE is equal to half the angle
ACE, and the angle ACE is bisected by the straight line CS.

And because SC, CE are equal to SC, CA respectively, and the 11, 4.

angle SCE is equal to the angle SCA, the third side SE ist equal

to the third side SA, and the angle SEC to the angle SAC; *Interc. 1. wherefore the angle SEC is equal to half the angle BAC, or to* half

the angle CEG, and the angle CEG is bisected by the straight line ES. And in the same way may be shown in succession, that all the other angles EGI, &c. of the series are severally bisected by

the straight lines drawn from the angular points G, &c. to S. +1. 14. Cor. But becauset only one straight line can bisect a given angle ;

straight lines which should have been drawn bisecting the several angles of the series, if sufficiently prolonged, would have necessarily met in S.

And by parity of reasoning, the like may be proved of every other series of straight lines under the same conditions. Wherefore, universally, in a series of equal straight lines &c. Which was to be demonstrated.

Cor. 12.

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CoR. 1. If the angle at the cusp be ever equal to the angle with the axis, the straight lines bisecting the angles of the series shall in like manner have met in the axis, at its intersection with the chord.

For the angle ILV will be equal to ILK, and the point S coincide with the point U.

Cor. 2. If a straight line of unlimited length WX moving along the axis as before, shall at any time when it cuts the series

but not in any of its angular points, make with it on the side remote from the vertex an angle MNX equal to or less than the angle with the axis CAZ, the straight lines bisecting the angles of the series shall in like manner have met in the axis, at a point between O which is the intersection of the axis with WX, and the vertex A.

For if the angle MNX is equal to CAZ, it is equal to NLV * Constr.

which is* equal to CAZ, and LV though prolonged ever +1.27.Cor.2. so far willt never meet NX ; still lessť, if the angle MNX is #1.27.Cor.3. less than NLV. Whereon may in either case be shown as

has been done before, that LV being prolonged will cut the axis in some point between 0 and A, as S; and that the straight lines bisecting the angles of the series will have met in S.

CoR. 3. If the series ever re-meets the axis, the straight lines bisecting the angles of the series shall have met in the axis, at a point nearer to the vertex.

For if P be the last of the angular points before the series re-meets the axis, it may be shown as was done with the angular point L, that if P and the vertex A be joined, the straight line bisecting the angle of the series at P will lie on the side of such joining straight line which is towards the axis; and consequently the line bisecting the angle of the series at P will cut the axis in some point as S, between the vertex and the point in which the series re-meets the axis. Whereon may be shown as before, that the straight lines bisecting the angles of the series will have met in S.

SCHOLIUM.—It would be easy to collect, that under the conditions stated in the Proposition above and its Corollaries, all the angular points of the series lie in the circumference of a circle whose centre is S; and con. sequently the series being continued will meet the axis, if it has not done so already. But as this is no where distinctly appealed to in anything that follows, it will be no further noticed.

PROPOSITION XXVIII D.
THEOREM.If the angles at the base of a tessera be less than right

angles, the angles opposite to the base cannot be right angles.

Let ABNM be a tessera, in which the angles BAM, ABN at the base AB are less than right angles. AMN, BNM which are opposite to the base, cannot be right angles.

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