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PROPOSITION X.

(Argument ad absurdum).

B

D

H

Theorem. One circumference

of a circle cannot cut another in E more than two points.

Suppose the FAB can cut O DEF in more than two points, as B, G, F, and prove, on that supposition, that K would be the centre of both Os, which (by v. 3.) is impossible, and .. the supposition is false.

PROPOSITION XI.

(Argument ad absurdum).

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Theorem. If two circles touch each other internally, the straight line which joins their centres, being produced, shall pass through the point of contact.

Steps of the Demonstration.

Suppose that the line which joins the centres does not pass through the point of contact A, but has some other direction, as BC; then prove, on that supposition, 1. That FG + GA > FB,

2. that GA > GB,

3. that GD > GB, i. e. less greater, which shows the supposition to be false, and .. that the line joining the centres must pass through 4.

PROPOSITION XII.

(Argument ad absurdum).

Theorem. If two circles touch each other exter

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Prove that the supposition, that the line joining the centres passes otherwise than through a is false, by showing that, on such supposition, the line FG would be at the same time both > and < FA + AG.

PROPOSITION XIII.

(Argument ad absurdum).

Theorem. One circle cannot touch another in more points than one, whether it touches on the inside or the outside.

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Steps of the Demonstration to Case 1st.

Having supposed it possible that ○ EBF can touch O ABC internally in more than one point, as в and », show, on that supposition,

1. That right line BD falls within each O,

2. that GH passes through the point of contact, which shows the supposition to be false.

K

B

Steps of the Demonstration to Case 2nd. Having supposed it possible that the

touch

ACK can

ABC externally in more than one point, as in

A and c; prove, on that supposition,

1. That right line AC falls within ○ ACK,

2. that ac is without O ABC,

3. that AC is also within O ABC, which shows the supposition to be false.

PROPOSITION XIV.

Theorem. Equal straight lines in a circle are equally distant from the centre; and those which are equally distant from the centre, are equal to each other.

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Steps of the Demonstration to Part 1st.

1. Prove that AF FB, and.. AB

that similarly CD = 2 CG,

2.

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7.

AF CG,

AE2- EC2,

2 AF,

that AF + FE2 = EG2 + GC2,

that

FE EG,

that. AB and CD are equally distant from the centre.

Steps of the Demonstration to Part 2nd. 1. Prove that AF2 = CG2, and .. af = CG, that AB CD.

2.

Theorem.

PROPOSITION XV.

The diameter is the

greatest straight line in a circle; and F
of all others, that which is nearer to
the centre is always greater than the K
more remote; and conversely, the
greater is nearer to the centre than
the less.

B

H

Part 1st. That the diameter is the greatest line in a○; and that a line nearer the centre is always >

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4. Prove that EH2 + HB2 = EK2 + KF2,
that HB2 > FK2 and .. HB > FK,
that BCFG.

5.

6.

Part 2nd. That the greater line BC is nearer to the centre than the less FG, i. e., that EH <EK.

Steps of the Demonstration.

1. Prove that BH > FK,

2.

that EH2 <EK2 and :. EH < ek.

B

PROPOSITION XVI.

(Argument ad absurdum).

Theorem. The straight line drawn at right angles

to the diameter of a circle from the extremity of it, falls without the circle; and no straight line can be drawn between that straight line and the circumference from the extremity, so as not to cut the circle.

Steps of the Demonstration.

Part 1st. That the right line from the extremity A, at rights to AB, shall fall without the .

Suppose it to fall within the O as AC; and prove, on that supposition,

1. That each of the S DAC, ACD (of the ▲ ADC) is a right, which is impossible, and ..

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