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Construction. If it be possible, let the circumference ABC cut the circumference DEF in more than two points, viz., in B, G, F. Take the centre K of the circle ABC (III. 3), and join KB, KG, KF.

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Demonstration. Then because K is the centre of the circle ABC, therefore KB, KG, KF are all equal to each other (I. Def. 15); and because within the circle DEF there is taken the point K, from which to the circumference DEF fall more than two equal straight lines KB, KG, KF; therefore

1. The point K is the centre of the circle DEF (III. 9); but K is also the centre of the circle ABC (constr.); therefore 2. The point K is the centre of two circles that cut one another. which is impossible (III. 5).

Therefore, one circumference of a circle cannot cut another in more than two points. Q.E.D.

PROPOSITION 11.-Theorem.

If one circle touch another internally in any point, the straight line which joins their centres being produced, shall pass through that point of contact.

Let the circle ADE touch the circle ABC internally in the point A.

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Then the straight line which joins their centres, being produced, shall pass through the point of contact A.

Construction. For, if not, let it fall otherwise, if possible, as FGDB; let F be the centre of the circle ABC, and G the centre of the circle ADE. Join AF and AG.

Demonstration. Then, because two sides of a triangle are together greater than the third side (I. 20); therefore

1. FG, GA are greater than FA;

but FA is equal to FH (I. Def. 15); therefore 2. FG, GA are greater than FH;

take away from these unequals the common part FG; therefore 3. The remainder AG is greater than the remainder GH (Ax. 5);

but AG is equal to GD (I. Def. 15); therefore

4. GD is greater than GH;

the less than the greater, which is impossible; therefore the straight line which joins the points F, G, being produced, cannot fall otherwise than upon the point A; that is,

5. FG being produced must pass through the point A.

Therefore, if one circle, &c. Q.E.D.

PROPOSITION 12.-Theorem.

If two circles touch each other externally in any point, the straight line which joins their centres shall pass through that point of contact. Let the two circles ABC, ADE, touch each other externally in the point A.

Then the straight line which joins their centres shall pass through the point of contact A.

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Construction. For, if not, let, if possible, F and G be the centres,

and the line joining them pass otherwise, as FCDG; and join FA, AG.

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Demonstration. And because F is the centre of the circle ABC,

1. FA is equal to FC;

also, because G is the centre of the circle ADE,

2. GA is equal to GD;

therefore FA, AG are equal to FC, DG (Ax. 2); wherefore 3. The whole FG is greater than FA, AG;

but also

4. FG is less than FA, AG (I. 20);

which is impossible; therefore the straight line which joins the points F, G, cannot pass otherwise than through A the point of contact; that is,

5. FG must pass through the point A.

Therefore, if two circles, &c. Q.E.D.

PROPOSITION 13.-Theorem.

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

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First. Let the circle EBF touch the circle ABC internally in the point B.

Then EBF cannot touch ABC in any other point.

Construction. If it be possible, let EBF touch ABC in another point D; join BD, and draw GH bisecting BD at right angles (I. 11). Demonstration. Because the points B, D are in the circumferences of each of the circles; therefore

1. The straight line BD falls within each of the circles EBF, ABC (III. 2);

therefore their centres are in the straight line GH which bisects BD at right angles (III. 1. Cor.); therefore

but

2. GH passes through the point of contact (III. 11) ;

3. GA does not pass through the point of contact,

because the points B, D are without the straight line GH, which is absurd; therefore one circle cannot touch another on the inside in more points than one.

Secondly. Let the circle ACK touch the circle ABC externally in the point A.

Then ACK cannot touch ABC in any other point.

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Construction. If it be possible, let ACK touch ABC in another point C.

Join AC.

Demonstration. Because the two points A, C are in the circumference of the circle ACK; therefore

1. The straight line AC falls within the circle ACK (III. 2), but the circle ACK is without the circle ABC (hyp.); therefore 2. The straight line AC is without the circle ABC ; but, because the points A, C are in the circumference of the circle ABC,

3. The straight line AC must be within the circle ABC (III. 2), which is absurd; therefore one circle cannot touch another on the

outside in more than one point; and it has been shown, that they cannot touch on the inside in more points than one.

Therefore, one circle, &c. Q.E.D

PROPOSITION 14.-Theorem.

Equal straight lines in a circle are equally distant from the centre; and conversely, those which are equally distant from the centre, are equal

to one another.

Let the straight lines AB, CD, in the circle ABDC, be equal to one another.

Then AB and CD shall be equally distant from the centre.

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B

Construction. Take E the centre of the circle ABDC (III. 1), from E draw EF, EG perpendiculars to AB, CD (I. 12), and join EA, EC.

Demonstration. Then, because the straight line EF passing through the centre, cuts AB, which does not pass through the centre, at right angles; EF bisects AB in the point F (III. 3); therefore 1. AF is equal to FB, and AB double of AF.

For the same reason,

2. CD is double of CG;

but AB is equal to CD (hyp.); therefore

3. AF is equal to CG (Ax. 7).

And because AE is equal to EC (I. Def. 15),

'but

4. The square on AE is equal to the square on EC;

5. The squares on AF, FE are equal to the square on AE (I. 47),

because the angle AFE is a right angle; and for the same reason,

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