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Steps of the Demonstration to Part 1st,
That the in○ is a right ≤.

1. Prove that

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EAB = ABE; and EAC = ▲ ACE, that .. whole ≤ BAC = ≤S ABC + ACB, that BAC

2.

3.

Part 2nd. That

/ FAC, and .. each is a rt. L.

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in segment > <, is proved by showing that since is a right,.. the ABC < right

.

BAC of ▲ ABC

Steps of the Demonstration to Part 3rd,
That in a segment < is > right .
1. Prove that S ABC + ADC = 2 right ≤s,
ADC > right L.

2.

that

PROPOSITION XXXII.

Theorem. If a straight line touches a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle.

E

Steps of the Demonstration.

B

1. Prove that centre of O is in AB, and .. seg. ACB

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PROPOSITION XXXIII.

Problem. Upon a given straight line to describe a segment of a circle containing an angle equal to a given rectilineal angle.

The proof of case 1st, in which is merely the showing that

because AHB is 0.

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c is a right,

AHB is also a right 2,

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Steps of the Demonstration to Case 2nd,
In which c is not right ≤.

1. Prove that (in AS AFG, BFG) base AG = base BG,

that a described from G with distance

2.

3.

4.

that

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GA will pass through B,

that AD touches the O,

DAB and . c▲ in alternate

segment AEB (or AHB in fig. 2.)

D

PROPOSITION XXXIV.

E

B

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Problem. From a given circle to cut off a segment, which shall contain an angle equal to a given rectilineal angle.

Steps of the Demonstration.

FBC

in alternate segment BAC,

in the alternate segment BAC = D.

PROPOSITION XXXV.

Theorem. If two straight lines within a circle cut each other, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other.

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B

A

D

E

centre of the O, and .. it is BE X ED.

D

E

B

Steps of the Demonstration to Case 2nd,

(In which BD passes through the centre, and cuts AC, which does not pass through the centre, at rt. s.)

1. Prove that AE = EC,

2.

3.

that BEX ED + EF2 = AE2 + EF2,
that BE XEDAE2, i. e. AE × EC.

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Steps of the Demonstration to Case 3rd,

(In which BD, passing through the centre, cuts AC, which does not pass through the centre, but not at rights).

1. Prove that AG = GC,

2.

that AE X EC + EG2 = AG2,

3.

that AEX EC + EG2 + GF2 = AG2 + GF2,

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The last case (in which neither AC nor BD pass through the centre) is proved by showing that

AEX EC and BE X ED are each GE X EH, and .. each other.

PROPOSITION XXXVI.

Theorem. If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it, the rectangle B contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.

E

Steps of the Demonstration to Case 1st, (In which DA passes through the centre).

1. Prove that EBD is a right 4,

2.

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3.

4.

that AD X DC + Ec2 = ED2,

that AD X DC + EB2 = EB2 + BD3,
that AD X DC

= BD2.

F

C

Steps of the Demonstration to Case 2nd, (In which DA does not pass through the centre).

1. Prove that AF FC,

2.

that AD X DC + FC2 = FD2,

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