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For if not, if possible let F be the Cent. of,

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.. F is not Cr. of ABC.

In like manner it may be shown that no other pt. which is not in CA, is the Cr. of O ABC;

i. e. the Cr. of O, is in CA.

Therefore, if a str. line, &c.

PROP. LV. THEOR.

20. 3 Eu.

The angle at the centre of a circle is double of the angle at the circumference upon the same arc, that is, upon the same part of the circumference.

ce

Let ABC be a ; BEC an at the Cr. E, and BAC an▲ at the Oe, having the same arc BC for their base: then shall BEC = 2/ BAC.

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Prop. 31.

:: ZEAB+ / EBA = 2 / EA3; EAB+≤ EBA = ext. ▲ BEF,

but

BEF=2/EAB:

Similarly
.. whole

FEC = 2 EAC,

BEC = 2. whole

BAC.

2ndly. Let Cr. E be without the BAC.

It may be demonstrated as before, that

FEC: = 2/ FAC,

FEB

= = 2/ FAB,

.. rem.

BEC

=

= 2. rem.

BAC.

Therefore, the angle at the centre, &c.

PROP. LVI. THEOR. 21. 3 Eu.

The angles in the same segment of a circle are equal to one another.

Let ABCD be a O, and s BAD, BED in the same segment BAED: then shall

=/ BED.

BAD

Fig. 2.

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1st Case. Let seg. BAED> semi O

Join BF, FD:

Fig. 1.

Then ::

{

BFD is at Cr.,

Prop. 55.

7 Ax. Fig. 2

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And both on the same part BCD of the ",

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< BFD 2/ BAD,

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BFD2 BED;

▲ BAD = BED.

се

2nd Case. Let seg. BAED be not > semi O. Draw AF to Cr. and prod. it to at C, Join CE and BD:

.. seg. BADC > semi O.

Ands in it are equal by the 1st Case:

Similarly

.. whole

BAC

BEC.

CAD = ▲ CED,

BAD = whole BED.

Wherefore, the angles, &c.

PROP. LVII.

THEOR. 22. 3 Eu.

The opposite angles of any quadrilateral figure inscribed in a circle, are together equal to two right angles.

Let ABCD be a quadrilat. fig. in

ABCD:

then any two of its opp. s shall be together

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BCA

=

ADB,

whole ADC;
ABC,

= ABC+2 ADC,

= 2 rt. Ls,

ZABC+ADC= 2 rt. S.

In like manner it may be shown that < BAD+/DCB= 2 rt. ▲ S.

Therefore, the opp. /s, &c.

PROP. LVIII. THEOR. 26. 3 Eu.

In the same or equal circles, equal angles at the centre or the circumference stand upon equal arcs.

Let ABG, CDH be equal Os, and the equal angles AEB, CFD at their centres, and

Prop. 56.

Prop. 31.

AGB, CHD at their circumferences; then shall arc AB = arc CD.

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For, let str. line AE be applied to str. line CF, so that pt. A may coincide with pt. C, and AE fall upon CF.

Also

Then AE = CF

.. pt. E coincides with pt. F.
JAE coincides with CF,
LAEB CFD,
EB falls upon FD.

And ...

And

=

EB = FD

... pt. B coincides with pt. D;

every point in arc AB is the same dist. from E, as every point in arc CD, is from F; and the points A, B, coincide with points C, D, arc AB coincides with arc CD, arc AB arc CD;

..

i. e.

.. the equals at the Cr. AEB, CFD, as also thes at the Oce AGB, CHD which are also equal (Ax. 7) stand upon equal arcs AB and CD.

Wherefore, in the same, &c.

PROP. LIX.

THEOR.

31. 3 Eu.

The angle in a semicircle is a right angle.

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