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Proposition V.

( Argument ad absurdum).

Theorem. If two circles cut one another, they shall not have the same centre.

Suppose that E is the centre of both Os; and prove, on that supposition, that EF and Eg each = EC, and :: = each other; i.e. less = greater, which shows the supposition to be false.

PROPOSITION VI.

( Argument ad absurdum).

Theorem. If two circles touch each other internally, they shall not have the same centre.

Suppose that F is the centre of both Os; and prove, on that supposition, that and FB each = FC, and .. = each other; i. e. less = greater, which shows the supposition to be false.

PROPOSITION VII. Theorem. If a point be taken in the diameter of a circle, of all the straight lines which can be drawn from it to the circumference, the greatest is that in which the centre is, and the other part of that diameter is the least; and of any others, that which is the nearest to the line which passes through the centre, is always greater than the more remote: and from the same point only two equal straight lines can be drawn to the circumference, one on each side of the shortest line.

The proof of this proposition consists of two parts.

I. That AF is the greatest, and Fd the least, of all lines drawn from F to the O; and that FB > FC, and FC > FG.

II. That upon each side of the shortest line FD, there can be drawn from point F to the O only two right lines that = each other.

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Steps of the Demonstration to Part I. 1. Prove that AE + EF, i. e. AF > BF,

that (in AS BEF, CEF) base of > base cf,
that similarly cf > GF,
that GF + FE > ED,
that GF > FD.

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Steps of the Demonstration to Part II. (Argument

ad absurdum). 1. In the construction of the figure show that po

= fh. Then suppose that from point F to the another line Fk besides fh can be drawn

= Fg, and prove, on that supposition, 2. that FK = Fh; i. e. a line near to = one more

remote from that passing through the centre, which is impossible by the preceding part of the demonstration, and :. the supposition is false.

PROPOSITION VIII.

Theorem. If a point be taken without a circle, and straight lines be drawn to the circumference, of which one passes through the centre; of those which fall upon the concave circumference, the greatest is that which passes through the centre; and of the rest, that which is nearest to that which passes through the centre, is always greater than the more remote; but of those which fall on the convex circumference, the least is that between the point without the circle and the diameter; and of the rest, that nearer to the least is always less than the more remote; and only two equal straight lines can o ch be drawn from the point to the circumference, one on each side of the least.

The proof of this proposition consists of three parts.

I. That of those lines which fall on the concave part* of the O, the greatest is DA, and of the rest, DE > DF, and DF > DC.

II. That of those lines which fall on the convex O, is the least, and DK > DL, and DL > DH.

III. That from u there can only be drawn two equal right lines to the O, one on each side of the least line.

Steps of the Demonstration to Part 1. 1. Prove that AD = EM + MD,

that ad > ED, 3. that (in AS EMD, FMD) base DE > base di,

that similarly DF > DC.

Steps of the Demonstration to Part II. 1. That KD > GD, i. e. that GD < KD, 2. that MK + KD (drawn to point k within A MLD)

are < the side of the A, ML + LD, 3. that dk < DL, 4. that similarly dl< DH.

* If from a point outside a circle, as A, two lines AB, AC be drawn touching the circle, and the points of contact be joined by a right line BC, that part of the circumference lying within the A / thus formed is called the convex, and that part of the circumference D lying without this triangle is called the concave part of the circumference with respect to the point A.

Steps of the Demonstration to Part III. (Argument

ad absurdum). 1. In the construction of the figure, prove that

DK = DB. Then suppose that from point d to the o another line on besides DB can be drawn

= DK; and show, on that supposition, 2. that DB = Dx; i. e. a line near to the least =

one more remote, which, by the preceding part of the demonstration is impossible, and :: the supposition is false.

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Steps of the Demonstration. Suppose that d is not the centre of the O, but that some other point, as E, is the centre. Then prove, on that supposition, 1. That ng is the greatest of all lines drawn from

D to the O, 2. that pc (the one nearer to dG) > DB; and that

DB > DA, 3. that DA, DB, DC are given = each other, which

shows the supposition to be false, and :: that D is the centre.

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