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Similarly FA may be shewn to be greater than any other st. line drawn from F to the oce;

.. FA is the greatest of all such lines. (ii) In the A EFG, the two sides EF, FG are together greater than EG;

I. 20. and EG=ED, being radii of the circle ; .. EF, FG are together greater than ED. Take away the common part EF;

then FG is greater than FD. Similarly any other st. line drawn from F to the oce may be shewn to be greater than FD;

... FD is the least of all such lines.
(iii)

In the AS BEF, CEF,
BE=CE,

1. Def. 15. Because

and EF is common;
but the 2 BEF is greater than the 2 CEF;
.. FB is greater than FC.

I. 24. Similarly it may be shewn that FC is greater than FG.

(iv) Join EG, and at E in FE make the 2 FEH equal to the L FEG.

I. 23.
Join FH.
Then in the AS GEF, HEF,
GE=HE,

1. Def. 15. Because and EF is common ;

also the 2 GEF=the 2 HEF; Constr.

.. FG= FH. And besides FH no other straight line can be drawn from F to the o ce equal to FG.

For, if possible, let FK= FG.
Then, because FH = FG,

Proved. ... FK=FH, that is, a line nearer to FA, the greatest, is equal to a line which is more remote ; which is impossible. Proved.

Therefore two, and only two, equal st. lines can be drawn from F to the oce.

Q.E.D.

I. 4.

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PROPOSITION 8. THEOREM. If from any point without a circle straight lines are drawn to the circumference, of those which fall on the concave circumference, the greatest is that which passes through the centre; and of others, that which is nearer to the greatest is always greater than one more remote.

Of those which fall on the convex circumference, the least is that which, when produced, passes through the centre; and of others, that which is nearer to the least is always less than one more remote.

From the given point there can be drawn to the circumference two, and only two, equal straight lines, one on each side of the shortest line.

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Let BGD be a circle; and from A, any point outside the circle, let ABD, AEH, AFG, be drawn, of which AD passes through C, the centre, and AH is nearer than AG to AD.

Then of st. lines drawn from A to the concave Oce,
(i) AD shall be the greatest, and (ii) AH greater than AG.
And of st. lines drawn from A to the convex Oce,
(iii) AB shall be the least

, and (iv) AE less than AF. (v) Also two, and only two, equal st. lines can be drawn from A to the oce.

Construction. Join CH, CG, CF, CE.

Proof. (i) In the A ACH, the two sides AC, CH are together greater than AH:

1. 20. but CH=CD, being radii of the circle; .. AC, CD are together greater than AH :

that is, AD is greater than AH. Similarly AD may be shewn to be greater than any other st. line drawn from A to the concave

: . AD is the greatest of all such lines.

ce

I. 24.

(ii)

In the AS HCA, GCA,
HC=GC,

I. Def. 15.
Because

and CA is common;
but the 2 HCA is greater than the _ GCA;

.. AH is greater than AG. (iii) In the A AEC, the two sides AE, EC are together greater than AC;

I. 20. but EC=BC;

I. Def. 15. .. the remainder AE is greater than the remainder AB.

Similarly any other st. line drawn from A to the convex o ce may be shewn to be greater than AB;

.:. AB is the least of all such lines. (iv) In the A AFC, because AE, EC are drawn from the extremities of the base to a point E within the triangle, .: AF, FC are together greater than AE, EC.

I. 21. But FC= EC ;

1. Def. 15. .. the remainder AF is greater than the remainder AE. (v) At C, in AC, make the 2 ACM equal to the 2 ACE.

Join AM.
Then in the two AS ECA, MCA,
EC= MC,

1. Def. 15.
Because and CA is common;
also the L ECA = the 2 MCA;

Constr. .. AE= AM. And besides AM, no st. line can be drawn from A to the o co, equal to AE.

For, if possible, let AK= AE :
then because AM=AE,

Proved.

AM=AK; that is, a line nearer to AB, the shortest line, is equal to a line which is more remote ; which is impossible. Proved.

Therefore two, and only two, equal st. lines can be drawn from A to the O

Q.E.D.

I. 4.

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EXERCISE. Where are the limits of that part of the circumference which is concave to the point A?

H.S.E.

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PROPOSITION 9. THEOREM.[FIRST PROOF.] If from a point within a circle more than two equal straight lines can be drawn to the circumference, that point is the centre of the circle.

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Let ABC be a circle, and D a point within it, from which more than two equal st. lines are drawn to the Oce, namely DA, DB, DC.

Then D shall be the centre of the circle ABC.
Construction. Join AB, BC:
and bisect AB, BC at E and F respectively. I. 10.

Join DE, DF.
Proof.

In the As DEA, DEB,
EA= EB,

Constr.
Because and DE is common;
and DA= DB;

Нур. .:. the DEA= the 2 DEB; ..,' these angles, being adjacent, are rt. angles. Hence ED, which bisects the chord AB at rt. angles, must pass through the centre.

III. 1. Cor. Similarly it may be shewn that FD passes through the centre.

:: D, which is the only point common to ED and FD, must be the centre.

Q.E.D.

I. 8.

NOTE. Of the two proofs of this proposition given by Euclid the first has the advantage of being direct.

PROPOSITION 9. THEOREM. [SECOND PROOF.]

If from a point within a circle more than two equal straight lines can be drawn to the circumference, that point is the centre of the circle.

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Let ABC be a circle, and D a point within it, from which more than two equal st. lines are drawn to the oce, namely DA, DB, DC.

Then D shall be the centre of the circle ABC.

Construction. For if not, suppose, if possible, E to be the centre.

Join DE, and produce it to meet the o ce at F, G.

Proof. Because D is a point within the circle, not the centre, and because DF passes through the centre E;

.. DA, which is nearer to DF, is greater than DB, which is more remote :

III. 7. but this is impossible, since by hypothesis, DA, DB, are equal.

... E is not the centre of the circle. * And wherever we suppose the centre E to be, otherwise than at D, two at least of the st. lines DA, DB, DC may be shewn to be unequal, which is contrary to hypothesis.

... D is the centre of the O ABC. Q.E.D. * NOTE. For example, if the centre E were supposed to be within the angle BDC, then DB would be greater than DA; if within the angle ADB, then DB would be greater than DC; if on one of the three straight lines, as DB, then DB would be greater than both DA and DC.

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