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shall be equal to the remaining Circumference ELF. Therefore in equal Circles, equal Angles stand upon equal Circumferences, whether they be at their Centers, or at their Circumferences ; which was to be demonItrated.

PROPOSITION XXVII.

THEOREM,
Angles, that stand upon equal Circumferences in

equal Circles, are equal to each other, whether
they be at their Centers or Circumferences.
ET the Angles BGC, EHF, at the Centers

of the equal Circles ABC, DEF, and the Angles BAC, EDF, at their Circumferences, stand upon the equal Circumferences BC, EF. I say the Angle BGC is equal to the Angle EHF, and the Angle BAC to the Angle EDF,

For if the Angle BGC be equal to the Angle EHF, it is manifest that the Angle BAC is also equal to the

Angle EDF: But if not, let one of them be the * 23: 1. greater, as BGC, and make * the Angle B GK, at

the Point G, with the Line BG, equal to the Angle † 26 of rbis. EHF. But equal Angles stand + upon equal Circum

ferences, when they are at the Centers. Wherefore the Circumference BK, is equal to the Circumfęrence EF. But the Circumference E F is equal to the Circumference BC. Therefore BK is equal to BC, a lefs to a greater; which is absurd. Wherefore the Angle BGC is not unequal to the Angle EHF; and so it must be equal to it. But the Angle at A is one half of the Angle BGC; and the Angle at D one half of the Angle EHF. Therefore the Angle at A is equal to the Angle at D. Wherefore Angles, that stand upon equal Circumferences in equal Circles, are equal to each other, whether they be at their Centers or Circumferences; ; which was to be demonstrated,

PRO

PROPOSITION XXVIII.

THE ORE M.
In equal Circles, equal Right Lines cut off equal

Parts of the Circumferences; the greater equal
to the greater, and the lesser equal to the leser.
L
ET ABC, DEF, be equal Circles, in which

are the equal Right Lines BC, EF, which cut off the greater Circumferences BAC, EDF, and the lesser Circumferences BGC, EHF. I say the greater

Circumference B AC, is equal to the greater Circumference EDF, and the lefser Circumference BGC, to the lefser Circumference EHF.

For affume the Centers K and L of the Circles, and join BK, KC, EL, LF.

Because the Circles are equal, the Lines drawn from their Centers are * also equal. Therefore the * Def. I. two Şides BK, KC, are equal to the two Sides EL, LF; and the Base B C, is equal to the Base EF. Therefore the Angle BKC, is f equal to the Angle † 8. 1. ELF. But equal Angles stand I upon equal Circum- 26 of tbis. ferences, when they are at the Centers. Wherefore the Circumference BGC, is equal to the Circumference EHF, and the whole Circle ABC, equal to the whole Ćircle DEF; and so the remaining Circumference B AC, snall be equal to the remaining Circumference EDF. Therefore in equal Circles, equal Right Lines cut off equal Parts of the Circumferences; which was to be demonstrated.

PROPOSITION XXIX.

THEOREM.
In equal Circles, equal Right Lines subtend equal

Circumferences.

;

ET there be two equal Circles ABC, DEF;

and let the equal Circumferences BGC, EHF, be affumed in them, and BC, EF, joined. I say, the Right Line BC is equal to the Right Line EF.

# 1 of this. For find * the Centers of the Circles K, L, and

join BK, KC, EL, LF.

Then because the Circumference BGC is equal to

the Circumference EHF; the Angle BKC shall be +27 of tbis. t equal to the Angle ELF. And because the Circles

ABC, DEF, are equal, the Lines drawn from their I Defi 1.

Centers shall be equal. Therefore the two Sides
BK, KC, are equal to the two Sides EL, LF; and

they contain equal Angles: Wherefore the Base BC 44. I.

is fequal to the Base EF. And so in equal Circles equal Circumferences subtend equal Right Lines ; which was to be demonstrated.

PROPOSITION XXX.

PROBLEM.
To cut a given Circumference into two equal Parts.

L

ET the given Circumference be ADB. It is

IO. 1.

Join AB, which bisect * in C; and let the Right Line CD be drawn from the Point C at Right Angles to AB; and join AD, DB.

Now because AC is equal to CB, and CD is common, the two Sides AC, CD, are equal to the two Sides BC, CD; but the Angle ACD is equal

to the Angle BCD; for each of them is a Right 4. I. Angle: Therefore the Base AD is f equal to the 28 of this. Base BD. But equal Right Lines cut I off equal

Circumferences. Wherefore the Circumference AD shall be equal to the Circumference BD. Therefore a given Circumference is cut into two equal Parts; which was to be done.

PRO

PROPOSITION XXXI.

TH EQRE M. In a Circle, the Angle that is in a Semicircle, is a Right Angle ; but the Angle in a greater Segment, is less than a Right Angle ; and the Angle in a leffer Segment, greater than a Right Angle : Moreover, the Angle of a greater Segment, is greater than a Right Angle; and the Angle of a lesser Segment, is less than a Right Line. LETE ET there be a Circle ABCD, whose Dia

meter is BC, and Center E; and join BA, AC, AD, DC. I say, the Angle which is in the Semicircle BAC is a Right Angle, that which is in the Segment ABC being greater than a Semicircle, viz. the Angle ABC, is less than a Right Angle ; and that which is in the Segment ADC, being less than a Semicircle, that is, the Angle ADC is greater than a Right Angle. For join AE, and produce B A to F. Then because BE is equal to E A, the Angle EAB fhall be * equal to the Angle E BA. And because * AE is equal to EC, the Angle ACE will be * equal to the Angle CAE. Therefore the whole Angle BAC is equal to the two Angles ABC, ACB; but the Angle FAC being without the Triangle ABC, is + equal to the two Angles ABC, ACỔ: † 32. I. Therefore the Angle BAC is equal to the Angle FAC; and so each of them is I a Right Angle. I Def. 10. I. Wherefore the Angle BAC in a Semicircle is a Right Angle

. And because the two Angles ABC, BAC, of the Triangle ABC", are less than two Right * 17. I. Angles, and BAC is a Right Angle; then ABC is less than a Right Angle, and is in the Segment ABC greater than a Semicircle.

And since ABCD is a quadrilateral Figure in a
Circle, and the opposite Angles of any quadrilateral
Figure described in a Circle, are equal to two Right + 22 of this.
Angles; the Angle ABC, ADC are equal to two
Right Angles, and the Angle ABC is less than a
Right Angle: Therefore the remaining Angle ADC

5. 1.

will be greater than a Right Angle, and is in the Segment ADC, which is less than a Semicircle.

I say, moreover, the Angle of the greater Segment contained under the Circumference ABC, and the Right Line AC, is greater than a Right Angle; and the Angle of the lesser Segment, contained under the Circumference ADC, and the Right Line AC is less than a Right Angle. This manifestly appears; for because the Angle contained under the Right Lines BA, AC, is a Right Angle, the Angle contained under the Circumference ABC, and the Right Line AC, will be greater than a Right Angle. Again, because the Angle contained under the Right Line CA, AF, is a Right Angle, therefore the Angle which is contained under the Right Line AC, and the Circumference ADC, is less than a Right Angle. Therefore, in a Circle, the Angle that is in a Semicircle, is a Right Angle; but the Angle in a greater Segment, is less than a Right Angle; and the Angle in a leffer Segment, greater than a Right Angle: Moreover, the Angle of a greater Segment, is greater than a Right Angle; and the Angle of a lefser Segment, is less than a Right Angle; which was to be demonstrated.

PROPOSITION XXXII.

THEOREM.
If any Right Line touches a Circle, and a Right

Line be drawn from the Point of Conta Et cut-
ting the Circle ; the Angles it makes with the
Tangent Line, will be equal to those which are
made in the alternate Segments of the Circle.
ET any Right Line E F touch the Circle ABCD

in the Point B, and let the Right Line BD be any how drawn from the Point B cutting the Circle. I say, the Angles which BD makes with the Tangent Line EF, are equal to thofe in the alternate Segments of the Circle'; that is, the Angle FBD is equal to an Angle made in the Segment DAB, viz. to the Angle DAB; and the Angle DBE equal to the Angle DCB, made in the Segment DCB. For:

Draw

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