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the supposition is false; .. that AC does
not fall within the O,

2. that similarly AC does not fall on the O,
and that it falls without the O.

Part 2nd. That between AE and ○ no right line can be drawn which does not cut .

FE

H

B

D

Suppose that FA is between them without cutting the O; and prove, on that supposition,

1. that, in ▲ DAG, DA > DG,

2, that DH > DG, i. e., less > greater, which shows the supposition to be false.

PROPOSITION XVII.

Problem. To drawn a straight

line from a given point, either G without or in the circumference,

which shall touch a given circle.

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Part 1st. To draw a tangent to O BCD, from point

A without it.

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2. 3.

that

EBA is a right,

that AB touches the O BCD.

Part 2nd, in which given point D is on the of the O, is proved by Cor. to xvi. 3.

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Steps of the Demonstration.

Suppose that Fc is not

line drawn from F, as FG, is

that supposition,

DE.

DE, but that some other
Then prove, on

1. that GCF <rt. ▲ FGC,

2. that FC > FG,

3. that FB > FG, i. e., less > greater; which shows the supposition to be false, and . FG

is not DE,

4. that, similarly, no line but Fc, drawn from F,

is DE.

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but

Suppose that the centre of the O is not in AC, in some other direction, as F; and prove, on that supposition,

B

1. that FCE is a rt. ▲,

2. that FCE = ACE, i. e., less = greater, which shows that the supposition is false, and .. F

is not the centre,

3. that the centre is in AC.

PROPOSITION XX.

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

Steps of the Demonstration to Case 1st,

In which the centre E is within

BAC.

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that 2

EAB = ex. ▲ BEF of ▲ BAE,

3.

4.

that similarly 2 ≤ EAC = FEC,

that :. whole ▲ BEC = 2 whole ▲ BAC.

B

Steps of the Demonstration to Case 2nd,
In which centre E is without BDC.

E

B

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1. Prove that

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2.

that 2 EDC

GEC,

3.

4.

that similarly 2 ≤ gdb = ≤ geb,

that remaining ▲ BEC = 2 remaining

E

BDC.

PROPOSITION XXI.

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

Steps of the Demonstration to Case 1st,
In which the segment is > O.

1. Prove that BFD2 BAD,

2.

3.

that similarly

that. BAD

bfd = 2 ≤ bed,

BED.

E

Steps of the Demonstration to Case 2nd,

In which the segment is < 0.

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2.

that LS ADC + CBA

3.

the 3 s of the A CBA; which 2 rt. ≤8.

that, similarly, 48 BAD + DCB = 2 rt. ≤s.

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